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Topics in Classical Algebraic Geometry (ang)


Topics in Classical Algebraic Geometry. Part I
IGOR V. DOLGACHEV September 22, 2009

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Contents
1 Polarity 1.1 Polar hypersurfaces . . . . . . . . . . . . . . . . 1.1.1 The polar pairing . . . . . . . . . . . . . 1.1.2 The first polars . . . . . . . . . . . . . . 1.1.3 The second polars . . . . . . . . . . . . 1.1.4 The Hessian hypersurface . . . . . . . . 1.1.5 Parabolic points . . . . . . . . . . . . . . 1.1.6 The Steinerian hypersurface . . . . . . . 1.2 The dual hypersurface . . . . . . . . . . . . . . . 1.2.1 The polar map . . . . . . . . . . . . . . 1.2.2 Dual varieties . . . . . . . . . . . . . . . 1.2.3 The Pl¨ cker equations . . . . . . . . . . u 1.3 Polar polyhedra . . . . . . . . . . . . . . . . . . 1.3.1 Apolar schemes . . . . . . . . . . . . . . 1.3.2 Sums of powers . . . . . . . . . . . . . . 1.3.3 Generalized polar polyhedra . . . . . . . 1.3.4 Secant varieties . . . . . . . . . . . . . . 1.3.5 The Waring problems . . . . . . . . . . . 1.4 Dual homogeneous forms . . . . . . . . . . . . . 1.4.1 Catalecticant matrices . . . . . . . . . . 1.4.2 Dual homogeneous forms . . . . . . . . 1.4.3 The Waring rank of a homogeneous form 1.4.4 Mukai’s skew-symmetric form . . . . . . 1.5 First examples . . . . . . . . . . . . . . . . . . . 1.5.1 Binary forms . . . . . . . . . . . . . . . 1.5.2 Quadrics . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . Conics 2.1 Self-polar triangles . . . . . . . . . . . . 2.1.1 The Veronese quartic surface . . . 2.1.2 Polar lines . . . . . . . . . . . . 2.1.3 The variety of self-polar triangles iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 1 4 5 6 9 12 14 14 15 18 20 20 22 23 24 26 27 27 30 32 33 35 35 36 38 40 41 41 41 42 43

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iv 2.1.4 Conjugate triangles . . . . . Poncelet relation . . . . . . . . . . 2.2.1 Darboux’s theorem . . . . . 2.2.2 Poncelet curves . . . . . . . 2.2.3 Invariants of pairs of conics 2.2.4 The Salmon conic . . . . . Exercises . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

CONTENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 47 47 52 54 58 60 62 65 65 65 66 68 69 72 72 73 76 77 82 85 87 87 87 88 92 96 98 101 102 102 104 106 107 110 111 113 113 113 114 117 117 118

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3

Plane cubics 3.1 Equations . . . . . . . . . . . . . . . . . . 3.1.1 Weierstrass equation . . . . . . . . 3.1.2 The Hesse equation . . . . . . . . . 3.1.3 The Hesse pencil . . . . . . . . . . 3.1.4 The Hesse group . . . . . . . . . . 3.2 Polars of a plane cubic . . . . . . . . . . . 3.2.1 The Hessian of a cubic hypersurface 3.2.2 The Hessian of a plane cubic . . . . 3.2.3 The dual curve . . . . . . . . . . . 3.2.4 Polar polygons . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . Determinantal equations 4.1 Plane curves . . . . . . . . . . . . . . . . . 4.1.1 The problem . . . . . . . . . . . . 4.1.2 Plane curves . . . . . . . . . . . . 4.1.3 The symmetric case . . . . . . . . . 4.1.4 Examples . . . . . . . . . . . . . . 4.1.5 Quadratic Cremona transformations 4.1.6 A moduli space . . . . . . . . . . . 4.2 Determinantal equations for hypersurfaces . 4.2.1 Cohen-Macauley sheaves . . . . . . 4.2.2 Determinants with linear entries . . 4.2.3 The case of curves . . . . . . . . . 4.2.4 The case of surfaces . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . .

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5

Theta characteristics 5.1 Odd and even theta characteristics . . . . . . . . . . . 5.1.1 First definitions and examples . . . . . . . . . 5.1.2 Quadratic forms over a field of characteristic 2 5.2 Hyperelliptic curves . . . . . . . . . . . . . . . . . . . 5.2.1 Equations of hyperelliptic curves . . . . . . . . 5.2.2 2-torsion points on a hyperelliptic curve . . . .

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CONTENTS 5.2.3 Theta characteristics on a hyperelliptic curve . . . . . 5.2.4 Families of curves with odd or even theta characteristic 5.3 Theta functions . . . . . . . . . . . . . . . . . . . . . . . . . 5.3.1 Jacobian variety . . . . . . . . . . . . . . . . . . . . 5.3.2 Theta functions . . . . . . . . . . . . . . . . . . . . . 5.3.3 Hyperelliptic curves again . . . . . . . . . . . . . . . 5.4 Odd theta characteristics . . . . . . . . . . . . . . . . . . . . 5.4.1 Syzygetic triads . . . . . . . . . . . . . . . . . . . . . 5.4.2 Steiner complexes . . . . . . . . . . . . . . . . . . . 5.4.3 Fundamental sets . . . . . . . . . . . . . . . . . . . . 5.5 Scorza correspondence . . . . . . . . . . . . . . . . . . . . . 5.5.1 Correspondences on an algebraic curve . . . . . . . . 5.5.2 Scorza correspondence . . . . . . . . . . . . . . . . . 5.5.3 Scorza quartic hypersurfaces . . . . . . . . . . . . . . 5.5.4 Theta functions and bitangents . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Plane Quartics 6.1 Bitangents . . . . . . . . . . . . . . . . . . 6.1.1 28 bitangents . . . . . . . . . . . . 6.1.2 Aronhold sets . . . . . . . . . . . . 6.1.3 Riemann’s equations for bitangents 6.2 Quadratic determinant equations . . . . . . 6.2.1 Hesse-Coble-Roth construction . . 6.2.2 Symmetric quadratic determinants . 6.3 Even theta characteristics . . . . . . . . . . 6.3.1 Contact cubics . . . . . . . . . . . 6.3.2 Cayley octads . . . . . . . . . . . . 6.3.3 Seven points in the plane . . . . . . 6.4 Polar polygons . . . . . . . . . . . . . . . 6.4.1 Clebsch and L¨ roth quartics . . . . u 6.4.2 The Scorza quartic . . . . . . . . . 6.4.3 Polar hexagons . . . . . . . . . . . 6.4.4 A Fano model of VSP(f ; 6) . . . . 6.5 Automorphisms of plane quartic curves . . 6.5.1 Automorphisms of finite order . . . 6.5.2 Automorphism groups . . . . . . . 6.5.3 The Klein quartic . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

v 119 120 122 122 124 125 128 128 130 133 136 136 138 140 142 145 146 147 147 147 149 151 155 155 158 162 162 163 166 168 168 172 174 175 177 177 180 183 186 188

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vi 7 Planar Cremona transformations 7.1 Homaloidal linear systems . . . . . . . . . . . . . . . . . 7.1.1 Linear systems and their base schemes . . . . . . . 7.1.2 Exceptional configurations . . . . . . . . . . . . . 7.1.3 The bubble space of a surface . . . . . . . . . . . 7.1.4 Cremona transformations . . . . . . . . . . . . . . 7.1.5 Nets of isologues and fixed points . . . . . . . . . 7.2 First examples . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 Involutorial quadratic transformations . . . . . . . 7.2.2 Symmetric Cremona transformations . . . . . . . 7.2.3 De Jonqui` res transformations . . . . . . . . . . . e 7.2.4 De Jonqui` res involutions and hyperelliptic curves e 7.3 Elementary transformations . . . . . . . . . . . . . . . . . 7.3.1 Segre-Hirzebruch minimal ruled surfaces . . . . . 7.3.2 Elementary transformations . . . . . . . . . . . . 7.3.3 Birational automorphisms of P1 × P1 . . . . . . . 7.3.4 De Jonqui` res transformations again . . . . . . . . e 7.4 Characteristic matrices . . . . . . . . . . . . . . . . . . . 7.4.1 Composition of characteristic matrices . . . . . . . 7.4.2 The Weyl groups . . . . . . . . . . . . . . . . . . 7.4.3 Noether-Fano inequality . . . . . . . . . . . . . . 7.4.4 Noether’s Reduction Theorem . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . Del Pezzo surfaces 8.1 First properties . . . . . . . . . . . 8.1.1 Varieties of minimal degree 8.1.2 A blow-up model . . . . . . 8.2 The EN -lattice . . . . . . . . . . . 8.2.1 Lattices . . . . . . . . . . . 8.2.2 The EN -lattice . . . . . . . 8.2.3 Roots . . . . . . . . . . . . 8.2.4 Fundamental weights . . . . 8.2.5 Effective roots . . . . . . . 8.2.6 Cremona isometries . . . . 8.2.7 Lines on Del Pezzo surfaces 8.3 Anticanonical models . . . . . . . . 8.3.1 Anticanonical linear systems 8.3.2 Anticanonical model . . . . 8.4 Surfaces of degree d in Pd . . . . . 8.4.1 Projective normality . . . . 8.4.2 Surfaces of degree ≥ 7 . . . 8.4.3 Surfaces of degree 6 in P6 . 8.4.4 Surfaces of degree 5 . . . . 8.5 Quartic Del Pezzo surfaces . . . . .

CONTENTS 191 191 191 194 197 199 201 204 204 207 209 211 213 213 217 218 221 221 227 229 233 234 237 239 241 241 241 242 245 245 248 249 254 256 258 262 263 263 267 268 268 271 272 275 280

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CONTENTS 8.5.1 Equations . . . . . . . . . . . . . . . . . . . . . . 8.5.2 Cyclid quartics . . . . . . . . . . . . . . . . . . . 8.5.3 Singularities and lines . . . . . . . . . . . . . . . 8.5.4 Automorphisms . . . . . . . . . . . . . . . . . . . 8.6 Del Pezzo surfaces of degree 2 . . . . . . . . . . . . . . . 8.6.1 Lines and singularities . . . . . . . . . . . . . . . 8.6.2 The Geiser involution . . . . . . . . . . . . . . . . 8.6.3 Automorphisms of Del Pezzo surfaces of degree 2 8.7 Del Pezzo surfaces of degree 1 . . . . . . . . . . . . . . . 8.7.1 Lines and singularities . . . . . . . . . . . . . . . 8.7.2 Bertini involution . . . . . . . . . . . . . . . . . . 8.7.3 Rational elliptic surfaces . . . . . . . . . . . . . . 8.7.4 Automorphisms of Del Pezzo surfaces of degree 1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Historical Notes . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Cubic surfaces 9.1 More about the E6 -lattice . . . . . . . . . . . . . 9.1.1 Double-sixers . . . . . . . . . . . . . . . 9.1.2 Tritangent trios . . . . . . . . . . . . . . 9.2 Lines on a nonsingular cubic surface . . . . . . . 9.2.1 Schur’s quadrics . . . . . . . . . . . . . 9.3 Singularities . . . . . . . . . . . . . . . . . . . . 9.3.1 Non-normal cubic surfaces . . . . . . . . 9.3.2 Normal cubic surfaces . . . . . . . . . . 9.3.3 Canonical singularities . . . . . . . . . . 9.4 Determinantal equations . . . . . . . . . . . . . 9.4.1 Cayley-Salmon equation . . . . . . . . . 9.4.2 Hilbert-Burch Theorem . . . . . . . . . . 9.4.3 The cubo-cubic Cremona transformation 9.4.4 Cubic symmetroids . . . . . . . . . . . . 9.5 Representations as sums of cubes . . . . . . . . . 9.5.1 Sylvester’s pentahedron . . . . . . . . . 9.5.2 The Hessian surface . . . . . . . . . . . 9.5.3 Cremona’s hexahedral equations . . . . . 9.5.4 The Segre cubic primal . . . . . . . . . . 9.6 Automorphisms of cubic surfaces . . . . . . . . 9.6.1 Elements of finite order in Weyl groups . 9.6.2 Eckardt points . . . . . . . . . . . . . . 9.6.3 Subgroups of W (E6 ). . . . . . . . . . . 9.6.4 Automorphisms of finite order . . . . . . 9.6.5 Automorphisms groups . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

vii 280 283 285 286 288 288 290 293 294 294 294 296 297 303 304 307 307 307 311 313 314 318 318 319 320 324 324 327 331 332 336 336 339 340 343 352 352 353 355 358 365 371

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viii 10 Geometry of Lines 10.1 Grassmanians of lines . . . . . . . . . . . . 10.1.1 The tangent and the secant varieties 10.1.2 The incidence variety . . . . . . . . 10.1.3 Schubert varieties . . . . . . . . . . 10.2 Linear complexes of lines . . . . . . . . . . 10.2.1 Linear complexes and apolarity . . 10.2.2 6 lines . . . . . . . . . . . . . . . . 10.2.3 Linear systems of linear complexes 10.3 Quadratic complexes . . . . . . . . . . . . 10.3.1 Generalities . . . . . . . . . . . . . 10.3.2 Intersection of 2 quadrics . . . . . . 10.3.3 Kummer surface . . . . . . . . . . 10.4 Scrolls of lines . . . . . . . . . . . . . . . 10.4.1 Generalities . . . . . . . . . . . . . 10.5 Ruled surfaces . . . . . . . . . . . . . . . .

CONTENTS 375 375 377 378 383 387 389 391 395 396 396 398 400 404 404 406

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Chapter 1

Polarity
1.1
1.1.1

Polar hypersurfaces
The polar pairing

Let E be a finite-dimensional vector space over a field K of characteristic 0. We denote by S k E its symmetric kth power and let E ∗ denote its dual space of linear functions. We have a canonical bilinear pairing , : E ⊗ E∗ → K (1.1)

that can be extended, using the universal properties of symmetric products, to a bilinear pairing S k E ⊗ S d E ∗ → S d−k E ∗ , d ≥ k. (1.2) In coordinates, it can be described as follows. Pick up a basis (ξ0 , . . . , ξn ) of E and let (t0 , . . . , tn ) be the dual basis in E ∗ . We can identify an element of S d E ∗ with a homogeneous polynomial f of degree d in the variables t0 , . . . , tn and an element of S k E with a homogeneous polynomial ψ of degree k in variables ξi . Since ξi , tj = ∂ δij , we view each ξi as the partial derivative operator ∂i = ∂ti . Hence any element k ψ ∈ S E can be viewed as a differential operator Dψ = ψ(∂0 , . . . , ∂n ). The pairing (1.2) becomes ψ(ξ0 , . . . , ξn ), f (t0 , . . . , tn ) = Dψ (f ).
i i For any monomial ∂ i = ∂00 · · · ∂nn and any monomial tj = tj0 · · · tjn , we have n 0 j! j−i (j−i!) t

(1.3)

∂ i (tj ) =

if j − i ≥ 0 otherwise.

0 1

(1.4)

2 Here and later we use the vector notation: i! = i0 ! · · · in !,

CHAPTER 1. POLARITY

i = (i0 , . . . , in ) ≥ 0 ⇔ i0 , . . . , in ≥ 0,

, |i| = i0 + · · · + in .

This gives an explicit expression for the pairing (1.2). Consider a special case when ψ = (a0 ∂0 + · · · + an ∂n )k = k!
|i|=k

(i!)−1 ai ∂ i .

Then Dψ (f ) = k!
|i|=k

(i!)−1 ai ∂ i (f ).

(1.5)

Let P(E) = Pn . We view any f ∈ S d E ∗ as a section of OPn (d), and denote by V (f ) the hypersurface of degree d in Pn defined by f . It is considered as an effective divisor in Pn , not necessary reduced. We understand that V (0) = Pn (the zero divisor). We can also view ψ ∈ S k E as a polynomial defining a hypersurface of degree k in ˇ the dual projective space Pn = P(E ∗ ) (an envelope of class k). We view a0 ∂0 + · · · + an ∂n = 0 as a point a ∈ P(E) with projective coordinates [a0 , . . . , an ]. Definition 1.1. Let X = V (f ) be a hypersurface of degree d in Pn . The hypersurface Pak (X)) = V (Dak (f )) of degree d − k is called th k-th polar hypersurface of the point a with respect to the hypersurface V (f ). We will also call it the k-th polar hypersurface of V (f ) with respect to the point a. Example 1.1.1. Let d = 2, i.e.
n

f (t0 , . . . , tn ) =
i=0

αii t2 + i
0≤i<j≤n

αij ti tj

is a quadratic form. Then Pa (V (f )) = V (g), where
n

Da (f ) =
i=0

ai

∂f =2 ∂ti

ai αij tj ,
0≤i,j≤n

aji = aij .

The linear map a → 1 Da (f ) is a map from E to E ∗ which can be identified with an 2 element of E ∗ ⊗ E ∗ = (E ⊗ E)∗ which is the polar bilinear form associated to f with matrix (aij ). Example 1.1.2. Let Mn (K) be the vector space of square matrices of size n with coordinates tij . We view the determinant function ∆ : Mn (K) → K as an element of ∂∆ S n (Mn (K)∗ ), i.e. a polynomial of degree n in the variables tij . Let Cij = ∂tij . For any point A = (aij ) in Mn (K) the value of Cij at A is equal to the ij-th cofactor of A. Then
n

Dan−1 (∆) =
i,j=1

Mij (a)tij

1.1. POLAR HYPERSURFACES

3

is a linear function on Mn identified with the cofactor matrix adj(A) of A (called in the classical literature the adjugate matrix, not the adjoint matrix as is customary to call it now). Let us give another definition of the polar hypersurfaces Pak (X). Choose two different points a = [a0 , . . . , an ] and b = [b0 , . . . , bn ] in Pn and consider the line = a, b spanned by the two points as the image of the map ϕ : P1 → Pn , [t0 , t1 ] → t0 a + t1 b := [a0 t0 + b0 t1 , . . . , an t0 + bn t1 ]

(a parametric equation of ). The intersection ∩X is isomorphic to the positive divisor on P1 defined by the degree d homogeneous form ϕ∗ (f ) = f (t0 a + t1 b) = f (a0 t0 + b0 t1 , . . . , an t0 + bn t1 ). Using the Taylor formula at (0, 0), we can write ϕ∗ (f ) =
k+m=d

d! m k t t Akm (a, b), m!k! 0 1 ∂ d ϕ∗ (f ) (0, 0). ∂tk ∂tm 0 1

(1.6)

where Akm (p, q) = Using the Chain Rule, we get Akm (a, b) = k!
|i|=k

(i!)−1 ai ∂ i (f )(b) = Pak (f )(b) = Pbm ak (f ) (j)−1 bj ∂ j (f )(a) = Pbm (f )(a) = Pak bm (f ).
|j|=m

(1.7)

= m! Observe the symmetry

Akm (a, b) = Amk (b, a).
n

(1.8)

When we fix a and let b vary in P we obtain a hypersurface of degree d − k which is the k-th polar hypersurface of X = V (f ) with respect to the point a. When we fix b and vary a, we obtain the m-th polar hypersurface of X with respect to the point b. Since we are in characteristic 0, Dam (f ) = 0 for m ≤ d. To see this we use the Euler formula: n ∂f df = ti . ∂ti i=0 Applying this formula to the partial derivatives we obtain d(d − 1) . . . (d − k + 1)f = k!
|i|=k

(i!)−1 ti ∂ i (f )

(1.9)

(also called the Euler formula). It follows from this formula that a ∈ Pak (X) ⇔ a ∈ X. In view of (1.7) and (1.8), we have b ∈ Pak (X) ⇔ a ∈ Pbd−k (X). (1.11) (1.10)

4

CHAPTER 1. POLARITY

1.1.2

The first polars

Let us consider some special cases. Let X = V (f ) be a hypersurface of degree d. Obviously, any 0-th polar of X is equal to X, and, by (1.11), the d-th polar Pad (X) is empty if a ∈ X and Pn if a ∈ X. Now take k = 1, d − 1. Using (1.7), we obtain
n

Da (f ) =
i=0 n

ai

∂f , ∂ti

Dad−1 (f ) =
i=0

∂f (a)ti . ∂ti

Together with (1.11) this implies the following. Theorem 1.1.1. For any b ∈ X let Tb (X) denote the (embedded) tangent hyperplane of X at b if b is a nonsingular point or Pn otherwise. Then Tb (X) = Pbd−1 (X). Moreover, X ∩ Pa (X) = {b ∈ X : a ∈ Tb (X)} The following picture makes an attempt to show what happens in the case when X is a conic. Pa (X) ………… ………… ………… ………… ………… a iiii ……… gfed `abc iii X iiii ………… iiii iii iiii iiii i Figure 1.1: Polar line of a conic The set of first polars Pa (X) defines a linear system contained in the complete linear system OPn (d − 1) . The dimension of this system is ≤ n. Proposition 1.1.2. The dimension of the linear system of first polars ≤ r if and only if, after a linear change of variables, the polynomial f becomes a polynomial in r + 1 variables. Proof. Induction on n and n − r. The assertion is obvious if r = n. Assume r = n − 1. Let ci ∂i f = 0 be a nontrivial linear relation between the first partial derivatives. Consider an invertible linear change of variables
n

ti =
j=0

aij uj ,

i = 0, . . . , n,

1.1. POLAR HYPERSURFACES where ai0 = ci , i = 0, . . . , n. By the Chain Rule, ∂f = ∂u0
n

5

ci
i=0

∂f = 0. ∂ti

This proves the assertion in this case. Assume r < n − 1. By induction on n − r, we may assume that, after a linear change of variables, f depends only on the variables u0 , . . . , ur+2 . By induction on n, after a further change of variables, we may assume that f depends only on the variables v0 , . . . , vr+1 .

1.1.3

The second polars

The (d − 2)- polar of X = V (f ) is a quadric, called the polar quadric of X with respect to a. It is defined by the quadratic form q = Dad−2 (f ) = (d − 2)!
|i|=d−2

(i!)−1 ai ∂ i (f ).

By symmetry, b ∈ Pad−2 (X) ⇔ a ∈ Pb2 (X). Thus q=2
|i|=2

(i!)−1 ti ∂ i (f )(a).

By (1.10), each a ∈ X belongs to the polar quadric Pad−2 (X). Also, by Theorem 1.1.1, (1.12) Ta (Pad−2 (X)) = Pa (Pad−2 (X)) = Pad−1 (X) = Ta (X). This shows that the polar quadric is tangent to the hypersurface at the point a. Let us see where Pa2 (X) intersects X. Consider the line = a, b through two points a, b. Let ϕ : P1 → Pn be its parametric equation. The index of intersection i(X, )a of with X at the point a is equal to the multiplicity µ of the polynomial ϕ∗ (f ) at the point (1, 0) (if the line is contained in X, by definition, the multiplicity is equal to ∞). It follows from (1.6) that µ ≥ s + 1 ⇐⇒ b ∈ Pad−i (X), i = 0, . . . , s. (1.13)

For s = 0, by (1.10), this condition says that a ∈ X, i.e. i(X, )a ≥ 1. For s = 1, by Theorem 1.1.1, this condition implies that b, and hence , belongs to the tangent plane Ta (X). For s = 2, this condition says that belongs to the second polar Pad−2 (X). Definition 1.2. A line is called a flex tangent to X at a point a if i(X, )a > 2. Proposition 1.1.3. Let be a line through a point a. Then is a flex tangent to X at a if and only if it is contained in the intersection of Ta (X) with the polar quadric Pad−2 (X).

6

CHAPTER 1. POLARITY

Note that the intersection of a quadric hypersurface Q = V (q) with its tangent ¯ ¯ hyperplane H at a point a ∈ Q is a cone in H over the quadric Q in the image H of H in P(E/Ka). Corollary 1.1.4. Assume n ≥ 3. For each a ∈ X there exists a flex tangent line. The union of the flex tangent lines containing the point a is the cone Ta (X) ∩ Pad−2 (X) in Ta (X). Example 1.1.3. Assume a is a singular point of X. By Theorem 1.1.1, this is equivalent to Pad−1 (X) = Pn . By (1.12), the polar quadric Q is also singular at a and thus it is a cone over its image under the projection from a. The union of flex tangents is equal to Q. Example 1.1.4. Assume a is a nonsingular point of a surface X ⊂ P3 . A hyperplane which is tangent to X at a cuts out in X a curve C with a singular point a. If a is an ordinary double point of C, there are two flex tangents corresponding to the two branches of C at a. The polar quadric Q is nonsingular at a. It is a cone over a quadric ¯ ¯ Q in P1 . If Q consists of 2 points we have two flex tangents corresponding to the ¯ two branches of C at a. If Q consists of one point (corresponding to non-reduced hypersurface in P1 ), then we have one branch. The latter case happens only if Q is singular at some point b = a.

1.1.4

The Hessian hypersurface

Let Q(a) be a polar quadric of X = V (f ) at some point a ∈ Pn . The symmetric matrix defining the corresponding quadratic form is equal to the Hessian matrix of second partial derivatives of f He(f ) = ∂2f ∂ti ∂tj .
i,j=0,n

(1.14)

evaluated at the point a. The quadric Q(a) is singular if and only if the determinant of the matrix is equal to zero (the singular points correspond to the null-space of the matrix). The hypersurface He(X) = V (det He(f )) (1.15)

describes the set of points a ∈ Pn such that the polar quadric Pad−2 (X) is singular. It is called the Hessian hypersurface of X. Its degree is equal to (d − 2)(n + 1) unless it coincides with Pn . Proposition 1.1.5. The following is equivalent: (i) He(X) = Pn ; (ii) there exists a nonzero polynomial g(z0 , . . . , zn ) such that g(∂0 f, . . . , ∂n f ) ≡ 0.

1.1. POLAR HYPERSURFACES

7

Proof. This is a special case of a more general result about the jacobian of n + 1 polynomial functions f0 , . . . , fn defined by J(f0 , . . . , fn ) = det ( ∂fi ) . ∂tj

Suppose J(f0 , . . . , fn ) ≡ 0. Then the map f : Cn+1 → Cn+1 defined by the functions f0 , . . . , fn is degenerate at each point (i.e. dfx is of rank < n+ 1 at each point x). Thus the closure of the image is a proper closed subset of Cn+1 . Hence there is an irreducible polynomial which vanishes identically on the image. Conversely, assume that g(f0 , . . . , fn ) ≡ 0 for some polynomial g which we may assume to be irreducible. Then ∂g = ∂ti
n

i=0

∂fj ∂g (f0 , . . . , fn ) = 0, i = 0, . . . , n. ∂zj ∂ti

Since g is irreducible its set of zeroes is nonsingular on a Zariski open set U . Thus the vector ∂g ∂g (f0 (x), . . . , fn (x)), . . . , (f0 (x), . . . , fn (x) ∂z0 ∂zn
i is a nontrivial solution of the system of linear equations with matrix ( ∂fj (x)), where ∂t x ∈ U . Thus the determinant of this matrix must be equal to zero. This implies that J(f0 , . . . , fn ) = 0 on U hence it is identically zero.

Remark 1.1.1. It was claimed by O. Hesse that the vanishing of the Hessian implies that the partial derivatives are linearly dependent. Unfortunately, his attempted proof is wrong. The first counterexample was given by P. Gordan and E. Noether in [125]. Consider the polynomial f = t2 t2 + t3 t2 + t4 t0 t1 = 0. 0 1 Note that the partial derivatives ∂f = t2 , 0 ∂t2 ∂f = t2 , 1 ∂t3 ∂f = t0 t1 ∂t4

are algebraically dependent. This implies that the Hessian is identically equal to zero. We have ∂f ∂f = 2t0 t2 + t4 t1 , = 2t1 t3 + t4 t0 . ∂t0 ∂t1 Suppose that a linear combination of the partials is equal to zero. Then c0 t2 + c1 t2 + c2 t0 t1 + c3 (2t0 t2 + t4 t1 ) + c4 (2t1 t3 + t4 t0 ) = 0. 0 1 Collecting the terms in which t2 , t3 , t4 enters we get 2c3 t0 + c4 t4 = 0, 2c3 t1 = 0, c3 t1 + c4 t0 .

This gives c3 = c4 = 0. Since the polynomials t2 , t2 , t0 t1 are linear independent we 0 1 also get c0 = c1 = c2 = 0. The known cases when the assertion of Hesse is true are d = 2 (any n) and n ≤ 3 (any d) (see [125], [168], [42]).

8

CHAPTER 1. POLARITY

Recall that the set of singular quadrics in Pn is the discriminant hypersurface D2 (n) (n+1)(n+2) −1 2 in P defined by the equation t00  t01  det  .  . . t0n  t01 t11 . . . t1n ... ... . . . ...  t0n t1n   .  = 0. .  . tnn

(1.16)

By differentiating, we easily find that its singular points are defined by the determinants of n×n minors of the matrix. This shows that the singular locus of D2 (n) parametrizes quadrics defined by quadratic forms of rank ≤ n − 1 (or corank ≥ 2). Abusing the terminology we say that a quadric is of rank k if the corresponding quadratic form is of this rank. Note that dim Sing(Q) = corank Q − 1. Assume that He(f ) = 0. Consider the rational map p : Pn = P(E) → P(S 2 (E ∗ )) = P( )−1 defined by a → Pad−2 (X). Note that Pad−2 (f ) = 0 implies Pad−1 (f ) = 0 n and hence i=0 bi ∂i f (a) = 0 for all b. This shows that a is a singular point of X. Thus p is defined everywhere except maybe at singular points of X. So the map p is regular if X is nonsingular, and the pre-image of the discriminant hypersurface is equal to the Hessian of X. The pre-image of the singular locus Sing(D2 (n)) consists of the subset of points a ∈ He(f ) such that dim Sing(Pad−2 (X)) > 0. One expects that, in general case, this will be equal to the set of singular points of the Hessian hypersurface.
n+2 2

Here is another description of the Hessian hypersurface. Proposition 1.1.6. The Hessian hypersurface He(X) is the locus of singular points of the first polars of X. Proof. Let a ∈ He(X) and let b ∈ Sing(Pad−2 (X)). Then Db (Dad−2 (f )) = Dad−2 (Db (f )) = 0. Since Db (f ) is of degree d − 1, this means that Ta (Pb (X)) = Pn , i.e., a is a singular point of Pb (X). Conversely, if a ∈ Sing(Pb (X)) for b ∈ Pn , then Dad−2 (Db (f )) = 0, hence Db (Dad−2 (f )) = 0. This means that b is a singular point of the polar quadric with respect to a. Hence a ∈ He(X). Let us find the affine equation of the Hessian hypersurface. Applying the Euler formula (1.9), we can write t0 f0i = (d − 1)∂i f − t1 f1i − . . . − tn fni , t0 ∂0 f = df − t1 ∂1 f − . . . − tn ∂n f,

1.1. POLAR HYPERSURFACES

9

where fij denote the second partial derivative. Multiplying the first row of the Hessian determinant by t0 and adding to it the linear combination of the remaining rows with the coefficients ti , we get the following equality.   ∂0 f ∂1 f . . . ∂n f  f10 f11 . . . f1n  d−1   det  . det(He(f )) = . . . . . t0  .  . . . fn0 fn1 . . . fnn Repeating the same procedure but this time with the columns, we finally get  det(He(f )) =
d d−1 f

 ∂1 f (d − 1)2  det  . t2  . 0 . ∂n f

∂1 f f11 . . . fn1

 . . . ∂n f . . . f1n   . . .  . ... fnn

(1.17)

Let φ(z1 , . . . , zn ) be the dehomogenization of f with respect to t0 , i.e., f (t0 , . . . , td ) = td φ( 0 We have ∂f ∂2φ d−2 = td−1 φi (z1 , . . . , zn ), = t0 φij (z1 , . . . , zn ), 0 ∂ti ∂ti ∂tj where φi = ∂f , ∂zi φij = ∂2f . ∂zi ∂zj i, j = 1, . . . , n, tn t1 , . . . , ). t0 t0

Plugging in these expressions in (1.17), we obtain, that up to a nonzero constant factor,  t0
−(n+1)(d−2) d d−1 φ(z)

 φ1 (z)  det(He(φ)) = det  . .  . φn (z)

φ1 (z) φ11 (z) . . .

... ... . . .

φn1 (z) . . .

 φn (z) φ1n (z)   ,  φnn (z)

(1.18)

where zi = ti /t0 , i = 1, . . . , n. Remark 1.1.2. If f (x, y) is a real polynomial in three variables, the value of the determinant (1.18) at a point a of the hypersurface f (x) = 0 multiplied by f1 (a)2 +f2−1 2 +f3 (a)2 (a) is equal to the Gauss curvature of X(R) at the point a (see [108]).

1.1.5

Parabolic points

Let us see where He(X) intersects X. A glance at the expression (1.18) reveals the following fact.

10

CHAPTER 1. POLARITY

Proposition 1.1.7. Each singular point of X belongs to He(X). Let us see now when a nonsingular point a ∈ X lies in its Hessian hypersurface He(X). By Corollary 1.1.4, the flex tangent lines in Ta (X) sweep the intersection of Tp (X) with the polar quadric Pad−2 (X). If a ∈ He(X), then the polar quadric is singular at some point b. If n = 2, a singular quadric is the union of two lines, so this means that one of the lines is a flex tangent line. A nonsingular point a of a plane curve X such that there exists a flex tangent at a is called an inflection point or a flex of X. If n > 2, the flex tangents lines at a point a ∈ X ∩ He(X) sweep a cone over a singular quadric in Pn−2 . Such a point is called a parabolic point of X. The closure of the set of parabolic points is the parabolic hypersurface in X. Theorem 1.1.8. Let X be a hypersurface of degree d in Pn . If n = 2, then He(X) ∩ X consists of singular and inflection points of X. In particular, each nonsingular curve of degree ≥ 3 has an inflection point, and their number ≤ 3d(d − 2) or infinite. If n > 2, then the set X ∩He(X) consists of singular points and parabolic points. The parabolic hypersurface in X is either the whole X or a subvariety of degree (n + 1)d(d − 2) in Pn . Example 1.1.5. Let X be a surface of degree d in P3 . If a is a parabolic point of X, then Tp (X) ∩ X is a singular curve whose singularity at a is unibranched. In fact, otherwise X has at least two distinct flex lines which cannot sweep a cone over a singular quadric in P1 . The converse is also true. For example, a nonsingular quadric has no parabolic points, and all nonsingular points of a singular quadric are parabolic. A generalization of a quadratic cone is a developable surface. It is a special kind of a ruled surface (see [108] and later Chapters) which are characterized by the condition that the tangent plane does not change along a ruling. The Hessian surface of a developable surface contains this surface. The residual surface of degree 2d − 8 is called Pro-Hessian surface. A concrete example of a developable surface is the quartic surface (x0 x3 − x1 x2 )2 − 4(x2 − x0 x2 )(x2 − x1 x3 ) = 0. 1 2 It is the surface swept out by the tangent lines of a normal rational curve of degree 3. It is also the determinantal surface of a binary cubic, i.e. the surface parametrizing binary cubics a0 x3 + 4a1 x2 y + 6a2 xy 2 + a3 y 3 which have a multiple root. The Pro-Hessian of any quartic developable surface is the surface itself ([36]. Let us see when X has infinitely many inflection points. Certainly, this happens when X contains a line component; each its point is an inflection point. It must be also an irreducible component of He(X). The set of inflection points is a closed subset of X. So, if X has infinitely many inflection points, it must have an irreducible component consisting of inflection points. Each such component is contained in He(X). Conversely, each common irreducible component of X and He(X) consists of inflection points. We will prove the converse in a little more general form taking care of not necessary reduced curves.

1.1. POLAR HYPERSURFACES

11

Proposition 1.1.9. A polynomial f (x0 , x1 , x2 ) is a factor of its Hessian polynomial He(f ) if and only if each factor of f entering with multiplicity 1 is a linear polynomial. Proof. Since each point on a non-reduced component of X = V (f ) is a singular point (i.e. all the first partials vanish), and each point on a line component is an inflection point, we see that the condition is sufficient for X | He(f ). Suppose this happens and let R be a reduced irreducible component of the curve X which is contained in the Hessian. Take a nonsingular point of R and consider an affine equation of R with ˆ coordinates (x, y). We may assume that OR,x is included in OR,x ∼ K[[t]] such that = x = t, y = tr , where (0) = 1. Thus the equation of R looks like f (x, y) = y − xr + g(x, y), (1.19)

where g(x, y) does not contain terms ay, a ∈ K. It is easy to see that (0, 0) is an inflection point if and only if r > 2 with the flex tangent y = 0. We use the affine equation of the Hessian (1.18), and obtain that the image of   d f1 f2 d−1 f h(x, y) = det  f1 f11 f12  f2 f21 f22 in K[[t]] is equal to 

0

det −rtr−1 + g1 1 + g2

−rtr−1 + g1 −r(r − 1)tr−2 + g11 g12

 1 + g2 g12  . g22

Since every monomial entering in g is divisible by y 2 , xy or xi , i > r, we see that gy is divisible by t and gx is divisible by tr . Also g11 is divisible by tr−1 . this shows that   0 atr−1 + . . . 1 + ... g12  , h(x, y) = det atr−1 + . . . −r(r − 1)tr−2 + . . . 1 + ... g12 g22 where . . . denotes terms of higher degree in t. We compute the determinant and see that it is equal to r(r − 1)tr−2 + . . .. This means that its image in K[[t]] is not equal to zero, unless the equation of the curve is equal to y = 0, i.e. the curve is a line. In fact, we have proved more. We say that a nonsingular point of X is an inflection point of order r − 2 and write it as ordfl(X)x if one can choose an equation of the curve as in (1.19) with r ≥ 3. It follows from the previous proof that r − 2 is equal to the multiplicity i(X, He)a of the intersection of the curve and its Hessian at the point a. Let ordfl(X)x = i( , x)x − 2, where is the flex tangent line of X at x, be the order of the inflection point. For any nonsingular curve of degree d > 1, we have ordfl(X)x = 3d(d − 2).
x∈X

(1.20)

12

CHAPTER 1. POLARITY

1.1.6

The Steinerian hypersurface

Recall that Hessian hypersurface of a hypersurface X = V (f ) is the locus of points a such that the polar quadric Pa (X) is singular. The Steinerian hypersurface St(X) of X is the locus of singular points of the polar quadrics. Thus St(X) =
a∈He(X)

Sing(Pad−2 (X)).

(1.21)

The proof of Theorem 1.1.6 shows that it can be equivalently defined as St(X) = {a ∈ Pn : Pa (X) is singular.} We also have He(X) =
a∈St(X)

(1.22) (1.23)

Sing(Pa (X)).

A point b = [b0 , . . . , bn ] ∈ St(X) satisfies the equation   b0 . He(f )(a) ·  .  = 0, . bn where a ∈ He(X). This equation defines a subvariety HS(X) of Pn × Pn given by n + 1 equations of bi-degree (d − 2, 1). When the Steinerian map is defined, it is just its graph. The projection to the second factor is a closed subscheme of Pn with support at St(X). This gives a scheme-theoretical definition of the Steinerian hypersurface which we will accept from now on. It also makes clear why St(X) is a hypersurface, not obvious from the definition. The expected dimension of the image of the second projection is n − 1. The following argument confirms our expectation. It is known that the locus of singular hypersurfaces of degree d in P(V ) is a hypersurface Dn (d) ⊂ P(S d E ∗ ) of degree (n + 1)(d − 1)n defined by the discriminant of a general degree d homogeneous polynomial in n + 1 variables (the discriminant hypersurface). Let L be the projective subspace of P(S d−1 E ∗ )) which consists of first polars of X. Assume that no polar Pa (f ) is equal to zero. Then St(X) ∼ L ∩ Dn (d − 1). = So, unless L is contained in Dn (d − 1) we get a hypersurface. Moreover we obtain deg(St(X)) = (n + 1)(d − 2)n (1.24)

Assume that the quadric Pad−2 (X)(a) is of coank 1 (i.e. the matrix He(f )(a) is of rank n). Then the quadric Pad−2 (X) has a unique singular point b = [b0 , . . . , bn ], whose coordinates can be chosen to be any column or a row of the adjugate matrix

1.1. POLAR HYPERSURFACES

13

adj(He(f )) evaluated at the point a. Thus St(X) is the image of the Hessian hypersurface under the rational map st : He(X)− → St(X), a → Sing(Pad−2 (X),

given by polynomials of degree n(d − 2). We call it the Steinerian map. Of course, it is not defined when all polar quadrics are of corank > 1. Also, if first polar Pa (f ) has an isolated singular point for a general point a, we get a rational map st−1 : St(X)− → He(X), a → Sing(Pa (X)).

These maps are obvioulsy inverse to each other. It is a difficult question to determine the sets of indeterminacy points for both maps. Proposition 1.1.10. The Steinerian hypersurface coincides with the whole Pn if and only if X has a point of multiplicity ≥ 3. Proof. The first polars of X form a linear system of hypersurfaces of degree d − 1. By Bertini’s Theorem, a singular point of a general member of the linear system is one of the base points. Thus St(X) = Pn implies that X has a singular point. Without loss of generality we may assume that the points is [1, 0, . . . , 0]. Write the equation of X in the form f = tk gd−k (t1 , . . . , tn ) + tk−1 gd+1−k (t1 , . . . , tn ) + · · · + gd (t1 , . . . , tn ) = 0, 0 0 where the subscript indicates the degree of the polynomial. Then the first polar Pa (X) has the equation
k n k

a0
i=0

ktk−1−i gd−k+i + 0
s=1

as
i=0

tk−i 0

∂gd−k+i = 0. ∂ts

The largest power of t0 in this expression is at most k. The degree of the equation is d − 1. Thus a is a singular point of Pa (X) if and only if k ≤ d − 3, or, equivalently, when a is at least triple point of X. One can assign one more variety to a hypersurface X = V (f ). This is the Cayleyan variety. It is defined as the image Cay(X) of the rational map HS(X)− → G(1, Pn ), (a, b) → a, b,

where G(1, Pn ) denotes the Grassmannian of lines in Pn . The map is not defined at the intersection of the diagonal with HS(X). We know that HS(a, a) = 0 means that Pad−1 (X) = 0, and the latter means that a is a singular point of X. Thus the map is a regular map for a nonsingular hypersurface X. Note that in the case n = 2, the Cayleyan variety is a plane curve in the dual plane, the Cayleyan curve of X.

14

CHAPTER 1. POLARITY

Proposition 1.1.11. Let X be a hypersurface of degree d ≥ 3 with no singular points of multiplicity ≥ 3. Then deg Cay(X) =
n+1 n−1 ) 2 (d − 2)(1 + (d − 2) 1 n+1 (d − 2)(1 + (d − 2)n−1 ) 2 2

if d > 3 if d = 3,

where the degree is considered with respect to the Pl¨ cker embedding of the Grassu mannian G(1, Pn ). Proof. By Proposition 1.1.10, St(X) = P2 , hence HS(X) is a complete intersection of 3 hypersurfaces in Pn × Pn of bi-degree (d − 2, 1). It is known that the set of lines intersecting a codimension 2 linear subspace L is a hyperplane section of the Grassmannian G(1, Pn ) in its Pl¨ cker embedding. Write Pn = P(V ) and L = P(W ). u Let ω = w1 ∧ . . . ∧ wn−1 for some basis (w1 , . . . , wn−1 ) of W . The locus of pairs of points a = Cv1 , b = Cv2 lying on a line intersecting L is given by the equation v1 ∧ v2 ∧ ω = 0. This is a hypersurface L of bi-degree (1, 1) in Pn × Pn . Let h1 , h2 be the natural generators of H ∗ (Pn × Pn , Z). We have #HS(X) ∩ L = ((d − 2)h1 + h2 )n+1 (h1 + h2 ) = =
n+1 2 n+1 2

(d − 2)n +

n+1 2

(d − 2)

(d − 2)((d − 2)n−1 + 1).

If d = 3, we will see later that He(X) = St(X) and the Steinerian map is an involution. Thus to get the degree we have to divide the above number by 2. Remark 1.1.3. From the point of view of the classic invariant theory, the homogeneneous forms defining the Hessian and Steinerian hypersurfaces of V (f ) are examples of covariants of f . The form defining the Cayleyan of a plane curve is an example of a contravariant.

1.2
1.2.1

The dual hypersurface
The polar map

The linear space of first polars Pa (X) defines a linear subsystem of the complete linear system OPn (d − 1) of hypersurfaces of degree d − 1 in Pn . Its dimension is equal to n if the first partial derivatives of f are linearly independent. By Proposition 1.1.2 this happens if and only if X is not a cone. We assume that this is the case. Let us identify the linear system of first polars with P(E) = Pn by assigning to each a ∈ Pn the polar ˇ hypersurface Pa (X). Let pX : Pn − → Pn be the rational map defined by the linear system of polars. It is called the polar map. In coordinates, the polar map is given by [a0 , . . . , an ] → ∂f ∂f (a), . . . , (a) . ∂t0 ∂tn

ˇ Recall that a hyperplane Ha = V ( ai ξi ) in the dual projective space Pn is the point a = [a0 , . . . , an ] ∈ Pn . The pre-image of the hyperplane H under pX is the polar ∂f Pa (f ) = V ( ai ∂ti (x)).

1.2. THE DUAL HYPERSURFACE

15

If X is nonsingular, the polar map is a regular map given by polynomials of degree d − 1. One can view the polar map as the rational map that sends a point x to the polar hyperplane Pxd−1 (X) = H. A point in the pre-image of a hyperplane H is called a pole of H with respect to X. Proposition 1.2.1. Assume X is nonsingular. The ramification divisor Ram(pX ) of the polar map is equal to He(X). Proof. Note for any finite map φ : X → Y of nonsingular varieties, the ramification divisor Ram(φ) is defined locally by the determinant of the linear map of locally free sheaves φ∗ (Ω1 ) → Ω1 . The image of Ram(φ) in Y is called the branch divisor. Both Y X of the divisors may be nonreduced. We have the Hurwitz formula KX = φ∗ (KY ) + Ram(φ). (1.25)

´ The map φ is etale outside Ram(φ), i.e., for any point x ∈ X the homomorphism of local ring OY,φ(y) → OX,x defines an isomorphism of their formal completions. In particular, the pre-image φ−1 (Z) of a nonsingular subvariety Z ⊂ Y is nonsingular outside the support of Ram(φ). Applying this to the polar map we see that the singular points of Pa (X) = p−1 (Ha ) are contained in the ramification locus Ram(pX ) X of the polar map. On the other hand, we know that the set of singular points of first polars is the Hessian He(X). This shows that He(X) ⊂ Ram(pX ). Applying the Hurwitz formula, we have KPn = OPn (−n − 1), KPn = OPn (−n − 1), p−1 (KPn ) = ˇ ˇ ˇ X OPn ((−n − 1)(d − 1)). This gives deg(Ram)pX )) = (n + 1)(d − 2) = deg(He(X)). This shows that He(X) = Ram(pX ). What is the branch divisor? One can show that the pre-image of a hyperplane Ha is singular if and only if it is tangent to the branch locus of the map. The pre-image of Ha is the polar hypersurface Pa (X). Thus the set of hyperplanes tangent to the branch divisor is equal to the Steinerian St(X). This shows that the branch locus contains the dual variety of St(X). Another implication of this is the following. Corollary 1.2.2. Assume X is nonsingular. For any point a ∈ He(X) the polar hyperplane of X with the pole at a is tangent to the Steinerian St(X) at a.

1.2.2

Dual varieties

Recall that the dual variety X ∗ of a subvariety X in Pn = P(E) is the closure in the ˇ dual projective space Pn = P(E ∗ ) of the locus of hyperplanes in Pn which are tangent to X at some nonsingular point of X. When X = V (f ) is a hypersurface, we see that the dual variety is the image of X ˇ under the rational map given by the first polars. In fact, (∂0 f (x), . . . , ∂n f (x)) in Pn is n n the hyperplane V ( i=0 ∂i f (x)ti ) in P which is tangent to X at the point x. The following result is called the projective duality. Many modern text-books contain a proof (see [120], [132], [250]). Theorem 1.2.3. Assume that K is of characteristic 0 (as we always do). Then (X ∗ )∗ = X.

16

CHAPTER 1. POLARITY

It follows from any proof that for any nonsingular point y ∈ X ∗ , the dual of the ˇ embedded (or projective) tangent space Ty (X ∗ ) ⊂ Pn is equal to the closure of the set of nonsingular points x ∈ X such that the hyperplane y in Pn is tangent to X at x. In particular, the fibres of the duality map d : X ns → X ∗ , x → Tx (X), (1.26)

are open subsets of a projective subspace in Pn . Here and later X ns denotes the set of nonsingular points of a variety X. In particular, if X ∗ is a hypersurface, the dual space of P T (X ∗ )y must be a point, and hence the map d is birational. Let us apply this to our case when X is a nonsingular hypersurface. Then the map ˇ given by first polars is a regular map Pn → Pn defined by homogeneous polynomials of degree d − 1. It is a finite map (after applying the Veronese map it becomes a linear projection map). Therefore its fibres are finite sets. This shows that the dual of a nonsingular hypersurface is a hypersurface. Thus, the duality map, equal to the restriction of the polar map, is a birational isomorphism d : X ∼bir X ∗ . = The degree of the dual hypersurface X ∗ (if it is a hypersurace) is called the class of X. For example, the class of any plane curve of degree > 1 is well-defined. Example 1.2.1. Let Dd (n) be the discriminant hypersurface in P(S d E ∗ ). We would like to describe explicitly the tangent hyperplane of Dd (n) at its nonsingular point. Let ˜ Dd (n) = {(X, x) ∈ |OPn (d)| × Pn : x ∈ Sing(X)}. ˜ Let us see that Dd (n) is nonsingular and the projection to the first factor ˜ π : Dd (n) → Dd (n) is a resolution of singularities. In particular, π is an isomorphism over the open set Dd (n)ns of nonsingular points of Dd (n). ˜ The fact that Dd (n) is nonsingular follows easily from considering the projection n to P . For any point b ∈ Pn the fibre of the projection is the projective space of hypersurfaces which have a singular point at b (this amounts to n + 1 linear conditions on ˜ the coefficients). Thus Dd (n) is a projective bundle over Pn and hence is nonsingular. Let us see where π is an isomorphism. Let us choose projective coordinates Ai , |i| = d, in OPn (d) = P(S d E ∗ ) corresponding to the coefficients of a hypersurface of de˜ gree d and coordinates x0 , . . . , xn in Pn . Then Dd (n) is given by n+1 bihomogeneous equations of bi-degree (1, d − 1): ∂g (x) = ∂ts is Ai xi−es = 0,
|i|=d

s = 0, . . . , n,

(1.27)

where g = |i|=d ai ti and es is the s-th unit vector in Zn+1 . We identify the tangent ˜ space of P(S d E ∗ ) × P(E) with the space S d E ∗ /KF ⊕ E/Kb. Then T (Dd (n))(X,b)

1.2. THE DUAL HYPERSURFACE is defined by the (n + 1) ×  . . . i0 bi−e0 . . .  . . . . .  . . . . i−en . . . in b ...
n+d d

17

matrix M of partial derivatives of equations (1.27) i−e0 −e0 i−e0 −en  ... |i=d| i0 i0 ai b |i=d| i0 in ai b  . . . . , . . i−en −e0 i−en −en ... |i=d| in i0 ai b |i=d| in in ai b

i i−es where f = = 0 if i − es is not a non|i=d| ai t and it is underestood that b negative vector. The projection π is an isomorphism if its differential is injective and the map is one-to one. To make it one-to-one we must have dim Sing(X) = 0. Now look at the map π restricted to the set of pairs (V, x) with the property that x is a single singular point of V . Suppose dπ(X,b) has nontrivial kernel. A vector in the kernel is a vector c = (. . . , ci , . . . , c0 , . . . , cn ) satisfying M · c = 0 and (. . . , ci , . . .) = λ(. . . , ai , . . .). The last n + 1 columns of the matrix M form the Hessian matrix of f at b. Multiplying the matrices, we obtain

M · c = λ (f )(b) + He(f ) · c = 0, where (f )(b) is the vector of partial derivatives of f computed at the point b and c = (c0 , . . . , cn ). Since Sing(V (f )) = {b}, we have (f )(b) = 0, Thus the differential has nontrivial kernel if and only if the equation He(f )(b)x = 0 has a solution not proportional to b = (b0 , . . . , bn ). We may assume that b = (1, 0, . . . , 0). Write f = td−2 A(t1 , . . . , tn ) + . . .. Computing the Hessian matrix at the point b we see that it is 0 equal to   0 ... ... 0 0 a11 . . . a1n    . . . . , . . .  . . . . . 0 an1 ... ann where g = 0≤i,j≤n aij ti tj (see (1.18)). A solution not proportional to (1, 0, . . . , 0) exists if and only if det(aij ) = 0. By definition, this means that the singular point of X at b is not an ordinary double point. The image of the tangent space under the differential dπ is equal to the set of g ∈ T P(S d E ∗ )V (f ) = S d E ∗ /CF such that the linear system He(f ) · x = − (g)(b) has a solution. Since the corank of the Hessian matrix is equal to 1, any solution must be orthogonal to the kernel of the matrix, i.e. b · (g)(b) = 0. By Euler’s formula implies that g(b) = 0. Now we are ready to compute the dual variety of Dd (n). The condition g(b) = 0, where Sing(X) = {b} is equivalent to Dbd (f ) = 0. Thus the tangent hyperplane, considered, as a point in the dual space P(S d E) corresponds to the envelope bd = n ( s=0 bs ∂i )d . The set of such envelopes is the Veronese variety νd (Pn (E ∗ )), the ˇ image of Pn = Pn (E ∗ ) → P(S d E) = P(S d E ∗ ) under the map given by the complete linear system OPn (d) of hypersurfaces of degree d in the dual projective space. ˇ Thus (1.28) Dd (n)∗ ∼ νd (Pn ), =

18

CHAPTER 1. POLARITY

Of course, it is predictable. Recall that the Veronese variety is embedded naturally in |OPn (d)|∗ . Its hyperplane section can be naturally identified with a hypersurface of degree d in Pn . A tangent hyperplane is a hypersurface with a singular point, i.e. a point in Dd (n). Thus the dual of νd (Pn ) is Dd (n), and hence, by duality, the dual of Dd (n) is νd (Pn ). Example 1.2.2. Let Q = V (q) be a nonsingular quadric in Pn . Let A = (aij ) be a symmetric matrix defining Q, i.e. q(t) = t · A · t. A tangent hyperplane of Q at a point x = [x0 , . . . , xn ] ∈ Pn is the hyperplane
n n

t0
j=0

a0j xj + · · · + tn
j=0

anj xj = 0.

Thus the vector of coordinates y = (y0 , . . . , yn ) of the tangent hyperplane is equal to the vector A · x. Since A is invertible, we can write x = A−1 · y. We have 0 = x · A · x = (y · A−1 )A(A−1 · y) = y · A−1 · y = 0. Here we treat x or y as a row-matrix or as a column-matrix in order the matrix multiplication makes sense. Since A−1 = det(A)−1 adj(A), where adj(A) is the adjugate matrix, we obtain that the dual variety of Q is also a quadric given by the adjugate matrix of the matrix defining Q. The description of the tangent space of the discriminant hypersurface from Example 1.2.1 has the following nice application. Proposition 1.2.4. Let X be a hypersurface of degree n in Pn . Suppose x is a nonsingular point of the Steinerian hypersurface St(X). Then Sing(Px (X)) consists of an ordinary singular point y and Tx (St(X)) = {a ∈ Pn : y ∈ Pa (X)}. Proof. The linear system L of the first polars of X cuts the discriminant hypersurface Dd−1 (n) at a point x. Since St(X) = L ∩ Dd−1 (n) is nonsingular at x, the point x is a nonsingular point of Dd−1 (n), and hence defines the hypersurface Px (X) with ordinary double point y. This follows from the computations from Example 1.2.1. The description of the tangent space of Dd−1 (n) at its nonsingular point proves the assertion.

1.2.3

¨ The Plucker equations

Let C = V (f ) be an irreducible plane curve of degree d. If C is nonsingular, its first polar Pa (C) with respect to a general point in P2 intersects C at d(d − 1) points b such that a ∈ Tb (C). This shows that the pencil of lines through a contains d(d − 1) tangent lines to C. A pencil of lines in P2 is the same as a line in the dual plane. Thus we see that the dual curve C ∗ has d(d − 1) intersection points with a general line. In other words deg(C ∗ ) = d(d − 1). (1.29)

1.2. THE DUAL HYPERSURFACE

19

If C is singular, the degree of C ∗ must be smaller. In fact, all polars Pa (C) pass through singular points of C and hence the number of nonsingular points b such that a ∈ Tb (C) is smaller than d(d − 1). The difference is equal to the sum of intersection numbers of a general polar and the curve at singular points d(d − 1) − deg(C ∗ ) =
x∈Sing(C)

i(C, Pa (C))x .

(1.30)

Let us compute the intersection numbers assuming that C has only ordinary nodes and cusps. Assume x is an ordinary node. Choose a coordinate system such that x = [1, 0, 0] and write the equation in the form f = td−2 f2 (t1 , t2 ) + . . .. We may assume 0 that f2 (t1 , t2 ) = t1 t2 . Computing the partials and dehomogenizing the equations, we find that Pa (f ) = a1 φx + a2 φy , where φ = xy + . . . is the affine equation of the curve, and φx , φy its partials in x and y. Thus, we need to compute the dimension of the vector space C[x, y]/(φ, a1 φx + a2 φy ) = C[x, y]/(xy + . . . , a1 x + a2 y + . . .), where . . . denotes the terms of higher degree. It is easy to see that this number is equal to the intersection number at a node with a general line through the node. The number is equal to 2. If x is an ordinary cusp, the affine equation of C is y 2 + x3 + . . . and we have to compute the dimension of the vector space C[x, y]/(f, a1 fx + a2 fy ) = C[x, y]/(y 2 + x3 + . . . , a1 x2 + a2 y + . . .). It is easy to see that this number is equal to the intersection number at a cusp with a parabola whose tangent is equal to the line y = 0. The number is equal to 3. Thus we obtain Theorem 1.2.5. Let C be an irreducible plane curve of degree d. Assume that C has only ordinary double points and ordinary cusps as singularities. Then deg(C ∗ ) = d(d − 1) − 2δ − 3κ, where δ is the number of nodes and κ is the number of cusps. Note that that the dual curve C ∗ of a nonsingular curve of degree d > 2 is always singular. This follows from the formula for the genus of a nonsingular plane curve and the fact that C and C ∗ are birationally isomorphic. The polar map C → C ∗ is equal to the normalization map. A singular point of C ∗ corresponds to a line which is either tangent to C at several points, or is a flex tangent. We skip a local computation which shows that a line which is a flex tangent at one point with ordfl = 1 (an honest flex tangent) gives an ordinary cusp of C ∗ and a line which is tangent at two points which are not inflection points (honest bitangent) gives a node. Thus we obtain that ˇ the number δ of nodes of C ∗ is equal to the number of honest bitangents of C and the number κ of ordinary cusps of C ∗ is equal to the number of honest flex tangents to C ∗ . ˇ

20

CHAPTER 1. POLARITY

Assume that C is nonsingular and C ∗ has no other singular points except ordinary nodes and cusps. We know that the number of inflection points is equal to 3d(d − 2). Applying Theorem 1.2.5 to C ∗ , we get that d(d − 2)(d2 − 9) ˇ 1 δ = d(d − 1)(d(d − 1) − 1) − d − 9d(d − 2) = . 2 2 (1.31)

This is the (expected) number of bitangents of a nonsingular plane curve. For example, we expect that a nonsingular plane quartic has 28 bitangents. We refer for discussions of Pl¨ cker formulas to many modern text-books (e.g. u [107], [116], [129], [120]).

1.3
1.3.1

Polar polyhedra
Apolar schemes

Let E be a complex vector space of dimension n + 1. Recall from section 1.1 that we have a natural pairing S k E × S d E ∗ → S d−k E ∗ , (ψ, f ) → Dψ (f ), d ≥ k,

which extends the canonical pairing E × E ∗ → C. By choosing a basis in E and the dual basis in E ∗ , we view the ring Sym• E ∗ as the polynomial algebra C[t0 , . . . , tn ] and Sym• E as the ring of differential operators C[∂0 , . . . , ∂n ]. The polarity pairing is induced by the natural action of operators on polynomials. Definition 1.3. A homogeneous form ψ ∈ S k E is called apolar to a homogeneous form f ∈ S d E ∗ if Dψ (f ) = 0. We extend this definition to hypersurfaces in the obvious way. Lemma 1.3.1. For any ψ ∈ S k E, ψ ∈ S m E and f ∈ S d E ∗ , Dψ (Dψ (f )) = Dψψ (f ). Proof. By linearity and induction on the degree, it suffices to verify the assertions in the case when ψ = ∂i and ψ = ∂j . In this case they are obvious. Corollary 1.3.2. Let f ∈ S d E ∗ . Let APk (f ) be the subspace in S k E spanned by apolar forms of degree k to f . Then


AP(f ) =
k=0

APk (f )

is a homogeneous ideal in the symmetric algebra Sym• E. Definition 1.4. The quotient ring Af = Sym• E/AP(f ) is called the apolar ring of f .

1.3. POLAR POLYHEDRA

21

The ring Af inherits the grading of Sym• E. Since any polynomial ψ ∈ S r E with r > d is apolar to f , we see that Af is killed by the ideal md+1 = (∂0 , . . . , ∂n )d+1 . + Thus Af is an Artinian graded local algebra over C. Since the pairing between S d E and S d E ∗ has values in S 0 E ∗ = C, we see that AP(f )d is of codimension 1 in S d E. Thus (Af )d is a vector space of dimension 1 over C and coincides with the socle of Af , i.e. the ideal of elements of Af annulated by its maximal ideal. Note that the latter property characterizes Gorenstein graded local Artinian rings, see [103], [148]. Proposition 1.3.3. (Macaulay) The correspondence f → Af is a bijection between P(S d E) and graded Artinian quotient algebras Sym• E/I with one-dimensional socle. Proof. We have only to show how to reconstruct Cf from S(V )/I. The multiplication of d vectors in V composed with the projection to S d E/Id defines a linear map S d E → S d E/Id . Since (Sym• E/I)d is one-dimensional. Choosing a basis (Sym• E/I)d , we obtain a linear function on S d E. It corresponds to an element of S d E ∗ . This is our Cf . Recall that for any closed subscheme Z ⊂ Pn is defined by a unique saturated homogeneous ideal IZ in C[t0 , . . . , tn ]. Its locus of zeros in the affine space An+1 is the affine cone CZ over Z isomorphic to Spec(C[t0 , . . . , tn ]/IZ ). Definition 1.5. Let f ∈ S d E ∗ . A subscheme Z ⊂ P(E) is called apolar to f if its homogeneous ideal IZ is contained in AP(f ), or, equivalently, Spec(Af ) is a closed subscheme of the affine cone CZ of Z. This definition agrees with the definition of an apolar homogeneous form ψ. A homogeneous form ψ ∈ S k E is apolar to f if and only if the hypersurface V (ψ) is apolar to V (f ). Consider the natural pairing (Af )k × (Af )d−k → (Af )d ∼ C = (1.32)

defined by multiplication of polynomials. It is well defined because of Lemma 1.3.1. The left kernel of this pairing consists of ψ ∈ S k E mod AP(f ) such that Dψψ (f ) = 0 for all ψ ∈ S d−k E. By Lemma 1.3.1, Dψψ (f ) = Dψ (Dψ (f )) = 0 for all ψ ∈ S d−k E. This implies Dψ (f ) = 0. Thus ψ ∈ AP(f ) is zero in Af . This shows that the pairing (6.13) is a perfect pairing. This is one of the nice features of a Gorenstein artinian algebra. It follows that the Hilbert poynomial
d

HAf (t) =
i=0

dim(Af ))i ti = ad td + · · · + a0

is a reciprocal monic polynomial, i.e. ai = ad−i , ad = 1. It is an important invariant of a homogeneous form f .

22

CHAPTER 1. POLARITY

Example 1.3.1. Let f = ld be the dth power of a linear form l ∈ E ∗ . For any ψ ∈ S k E = (S k E ∗ )∗ we have Dψ (ld ) = d(d − 1) . . . (d − k + 1)ld−k ψ(l) = d!l[d−k] ψ(l),
1 where we set l[i] = i! li . Here we view ψ ∈ S d E as a homogeneous function on E ∗ . In n coordinates, L = i=0 ai ti , ψ = ψ(∂0 , . . . , ∂n ) and ψ(l) = ψ(a0 , . . . , an ). Thus we see that AP (f )k , k ≤ d, consists of polynomials of degree k vanishing at l. Assume d+1 for simplicity that l = t0 . The ideal AP (f ) is generated by ∂1 , ∂n , ∂0 . The Hilbert d polynomial is equal to 1 + t + · · · + t .

1.3.2

Sums of powers

From now on, to avoid the confusion, we denote by [l] the point in P(E ∗ ) defined by by the linear form l ∈ E. We will often identify it with the hyperplane V (li ). For any point a ∈ P(E ∗ ) we continue to denote by Ha the corresponding hyperplane in P(E). Suppose f ∈ S d E is equal to a sum of powers of nonzero linear forms
d d f = l1 + · · · + ls .

(1.33)

This implies that for any ψ ∈ S k E,
s s d li ) = d! i=1 i=1 d−k ψ(li )d(d − 1) · · · (d − k + 1)li

Dψ (f ) = Dψ (

(1.34)

In particular, taking d = k, we obtain that
d d l1 , . . . , l s ⊥ SdE

= {ψ ∈ S d E : ψ(li ) = 0, i = 1, . . . , s} = (IZ )d ,

where Z is the closed subscheme of points {[l1 ], . . . , [ls ]} ⊂ P(E ∗ ) corresponding to the linear forms li . d d This implies that the codimension of the linear span l1 , . . . , ls in S d E ∗ is equal to d d the dimension of (IZ )d , hence the forms l1 , . . . , ls are linearly independent if and only if the points [l1 ], . . . , [ls ] impose independent conditions on hypersurfaces of degree d in P(E). d d Suppose f ∈ l1 , . . . , ls , then (IZ )d ⊂ APd (f ). Conversely, if this is true, we have d d f ∈ APd (f )⊥ ⊂ (IZ )⊥ = l1 , . . . , ls . d If we additionally assume that (IZ )d ⊂ APd (f ) for any proper subset Z of Z, we obtain, after replacing the forms li s by proportional ones, that
d d f = l1 + · · · + ls .

Definition 1.6. A polar s-polyhedron of f is a set of hyperplanes Hi = V (li ), i = 1, . . . , s, in P(E) such that d d f = l1 + · · · + ls , and, considered as points [li ] in P(E ∗ ), the hyperplanes Hi impose independent conditions in the linear system |OP(E) (d)|.

1.3. POLAR POLYHEDRA

23

Note that this definition does not depend on the choice of linear forms defining the hyperplanes. Nor does it depend on the choice of the equation defining the hypersurface V (f ). The following propositions follow from the above discussion. Proposition 1.3.4. Let f ∈ S d E ∗ . Then Z = {[l1 ], . . . , [ls ]} form a polar s-polyhedron of f if and only if the following properties are satisfied (i) IZ (d) ⊂ APd (f );
d d (ii) l1 , . . . , ls are linearly independent in S d E ∗ or

(ii’) IZ (d) ⊂ APd (f ) for any proper subset Z of Z. Proposition 1.3.5. A set Z = {[l1 ], . . . , [ls ]} is a polar s-polyhedron of f ∈ S d E ∗ if and only if Z, considered as a closed subscheme of P(E), is apolar to f but no proper subscheme of Z is apolar to f .

1.3.3

Generalized polar polyhedra

Proposition 1.3.5 allows one to generalize the definition of a polar polyhedron. A polar polyhedron can be viewed as a reduced closed subscheme Z of P(E ∗ ) consisting of s points. Obviously, h0 (OZ ) = dim H 0 (P(E), OZ ) = s. More generally, we may consider non-reduced closed subschemes Z of P(E) of dimension 0 satisfying h0 (OZ ) = s. The set of such subschemes is parameterized by a projective algebraic variety Hilbs (P(E ∗ )) called the punctual Hilbert scheme of P(E ∗ ) of length s. Any Z ∈ Hilbs (P(E ∗ )) defines the subspace IZ (d) = P(H 0 (P(E), IZ (d)) ⊂ H 0 (P(E), OP(E) (d)) = S d E ∗ . The exact sequence 0 → H 0 (P(E), IZ (d)) → H 0 (P(E), OP(E) (d)) → H 0 (P(E), OZ ) → H 1 (P(E), IZ (d)) → 0 shows that the dimension of the subspace Z
d

(1.35)

= P(H 0 (P(E), IZ (d))⊥ ) ⊂ P(S d E)

(1.36)

is equal to h0 (OZ )−h1 (IZ (d))−1 = s−1−h1 (IZ (d)). If Z = Zred = {p1 , . . . , ps }, then Z d = vd (p1 ), . . . , vd (ps ) , where vd : P(E) → P(S d E) is the Veronese map. Hence dim Z = s − 1 if the points vd (p1 ), . . . , vd (ps ) are linearly independent. We say that Z is linearly d-independent if dim Z d = s − 1. Definition 1.7. A generalized s-polyhedron of f is a linearly d-independent subscheme Z ∈ Hilbs (P(E)) which is apolar to f .

24 Recall that Z is apolar to f if, for each k ≥ 0,

CHAPTER 1. POLARITY

IZ (k) = H 0 (P(E ∗ ), IZ (k)) ⊂ APk (f ).

(1.37)

In view of this definition a polar polyhedron is a reduced generalized polyhedron. The following is a generalization of Proposition 1.3.4. Proposition 1.3.6. A linear independent subscheme Z ∈ Hilbs (P(E)) is a generalized polar s-polyhedron of f ∈ S d E ∗ if and only if IZ (d) ⊂ APd (f ). Proof. We have to show that the inclusion in the assertion implies IZ (d) ⊂ APk (f ) for any k ≤ d. For any ψ ∈ S d−k E and any ψ ∈ I(Z)k , the product ψψ belongs to I(Z)d . Thus Dψψ (f ) = 0. By the duality, Dψ (f ) = 0, i.e. ψ ∈ APk (f ). Example 1.3.2. Let Z = m1 p1 + · · · + mk pk ∈ Hilbs (P(E)) be the union of fat points pk , i.e. at each pi ∈ Z the ideal IZ,pi is equal to the mi th power of the maximal ideal. Obviously,
k

s=
i=1

n+mi −1 mi −1

.

Then the linear system |OP(E) (d)−Z)| consists of hypersurfaces of degree which have singularity at pi of multiplicity ≥ mi for each i = 1, . . . , k. One can show (see [148], Theorem 5.3 ) that Z is apolar to f if and only if
d−m d−m f = l1 i +1 g1 + . . . + lk k +1 gk ,

where pi = V (li ) and gi is a homogeneous polynomial of degree mi − 1. Remark 1.3.1. It is not known whether the set of generalized s-polyhedra of f is a closed subset of Hilbs (P(E ∗ )). It is known to be true for s ≤ d + 1 since in this case dim I(Z)d = t := dim S d E − s for all Z ∈ Hilbs (P(E ∗ ) (see [148], p.48). This defines a regular map of Hilbs (P(E ∗ )) to the Grassmannian G(t, S d E) and the set of generalized s-polyhedra is equal to the preimage of a closed subset consisting of subspaces contained in APd (f ). Also we see that h1 (IZ (d)) = 0, hence Z is always linearly d-independent.

1.3.4

Secant varieties
vd : P(E ∗ ) → P(S d E ∗ ), l → ld ,

The notion of a polar polyhedron has a simple geometric interpretation. Let

be the Veronese map. Denote by Vern its image. Then f ∈ S d E ∗ \ {0} represents a d point [f ] in P(S d E ∗ ). A set of hyperplanes Hi = V (li ), i = 1, . . . , s, represents a set d of points [li ] in the Veronese variety Vern . It is a polar s-polyhedron of f if and only if d d d [f ] belongs to the linear span [l1 ], . . . , [ls ] , a (s − 1)-secant of the Veronese variety.

1.3. POLAR POLYHEDRA

25

Recall that for any irreducible nondegenerate projective variety X ⊂ Pr of dimension n its t-secant variety Sect (X) is defined to be the Zariski closure of the set of points in Pr which lie in the linear span of dimension t of some set of t + 1 linear independent points in X. Counting constants easily gives dim Sect (X) ≤ min((n + 1)(t + 1) − 1, r). The subvariety X ⊂ Pr is called t-defective if the inequality is strict. An example of a 1-defective variety is a Veronese surface in P5 . A fundamental result about secant varieties is the following lemma whose modern proof can be found, for example in [250], Proposition 1.10. Lemma 1.3.7. (A. Terracini) Let p1 , . . . , pt+1 be general t + 1 points in X and p be a general point in their span. Then Tp (Sect (X)) = Tp1 (X), . . . , Tpt+1 (X). The inclusion part Tp1 (X), . . . , Tpt+1 (X) ⊂ Tp (Sect (X)) is easy to prove. We assume for simplicity that t = 1. Then Sec1 (X) contains the cone C(p1 , X) which is swept out by the lines p1 , q, q ∈ X. Therefore Tp (C(p1 , X)) ⊂ Tp (Sec1 (X)). However, it is easy to see that Tp (C(p1 , X)) contains Tp1 (X). Corollary 1.3.8. Sect (X) = Pr if and only if for any t + 1 general points of X there exists a hyperplane section of X singular at these points. In particular, if r ≤ (n + 1)(t + 1) − 1, the variety X is t-defective if and only if for any t + 1 general points of X there exists a hyperplane section of X singular at these points. Example 1.3.3. Let X = Vern ⊂ P n −1 be the image of Pn under a Veronese map d defined by homogeneous polynomials of degree d. Assume (n+1)(t+1) ≥ d+n −1. n A hyperplane section of X is isomorphic to a hypersurface of degree d in Pn . Thus Sect (Vern ) = P(S d E ∗ ) if and only if for any t + 1 general points in Pn there exists a d hypersurface of degree d singular at these points. Take n = 1. Then r = d and r ≤ (n + 1)(t + 1) − 1 = 2t + 1 for t ≥ (d − 1)/2. Since t + 1 > d/2 there are no homogeneous forms of degree d which have t + 1 multiple roots. Thus the Veronese curve Rd = vd (P1 ) ⊂ Pd is not t-degenerate for t ≥ (d − 1)/2. Take n = 2 and d = 2. For any two points in P2 there exists a conic singular at these points, namely the double line through the points. This explains why a Veronese surface V22 is 1-defective. Another example is Ver2 ⊂ P14 and t = 4. The expected dimension of Sec4 (X) 4 is equal to 14. For any 5 points in P2 there exists a conic passing through these points. Taking it with multiplicity 2 we obtain a quartic which is singular at these points. This shows that Ver2 is 4-defective. 4
d+n

26

CHAPTER 1. POLARITY

The following corollary of Terracini’s Lemma is called the First main theorem on apolarity in [99]. The authors gave an algebraic proof of this theorem without using (or probably without knowing) Terracini’s Lemma. Corollary 1.3.9. A general form f ∈ S d E ∗ admits a polar s-polyhedron if and only if there exists linear forms l1 , . . . , ls ∈ V ∗ such that for any nonzero ψ ∈ S d E the ideal d−1 d−1 AP (ψ) ⊂ SymE ∗ does not contain {l1 , . . . , ls }. Proof. A general form f ∈ S d E ∗ admits a polar s-polyhedron if and only if the secant variety Secs (Vern ) is equal to the whole space. This means that the span of the tangent d d spaces at some points qi = V (li ], i = 1, . . . , s is equal to the whole space. By Terracini’s Lemma, this is equivalent to that the tangent spaces of the Veronese variety at the points qi are not contained in a hyperplane defined by some ψ ∈ S d E = (S d E ∗ )∗ . It remains to use that the tangent space of the Veronese variety at qi is equal to the d−1 projective space of all homogeneous forms of the form V (li l), l ∈ E ∗ {0} (see d Exercises). Thus, for any nonzero ψ ∈ S E, it is impossible that Pld−1 l (ψ) = 0 for i all l and i. But Pld−1 l (ψ) = 0 for all l if and only if Pld−1 (ψ) = 0. This proves the i i assertion. The following fundamental result is due to J. Alexander and A. Hirschowitz [4]. Theorem 1.3.10. Vern is t-defective if and only if d (n, d, t) = (2, 2, 1), (2, 4, 4), (3, 4, 8), (4, 3, 6), (4, 4, 13). In all these cases the secant variety Sect (Vern ) is a hypersurface. d For the sufficiency of the condition, only the case (4, 3, 6) is not trivial. It asserts that for 7 general points in P3 there exists a cubic hypersurface which is singular at these points. Other cases are easy. We have seen already the first two cases. The third case follows from the existence of a quadric through 9 general points in P3 . The square of its equation defines a quartic with 9 points. The last case is similar. For any 14 general points there exists a quadric in P4 containing these points. Corollary 1.3.11. Assume s(n+1) ≥ d+n . Then a general homogeneous polynomial n f ∈ C[t0 , . . . , tn ]d can be written as a sum of dth powers of s linear forms unless (n, d, s) = (2, 2, 2), (2, 4, 5), (3, 4, 9), (4, 3, 7), (4, 4, 14).

1.3.5

The Waring problems

The well-known Waring problem in number theory asks about the smallest number s(d) such that each natural number can be written as a sum of s(d) d-th powers of natural numbers. It also asks in how many ways it can be done. Its polynomial analog asks about the smallest number s(d, n) such that a general homogeneous polynomial of degree d in n + 1 variables can be written as a sum of s d-th powers of linear forms. The Alexander-Hirschowitz Theorem completely solves this problem. We have s(d, n) is equal to the smallest natural number s0 such that s0 (n + 1) ≥ n+d unless n (n, d) = (2, 2), (2, 4), (3, 4), (4, 3), (4, 4), where s(d, n) = s0 + 1. Other versions of the Waring problem ask the following questions:

1.4. DUAL HOMOGENEOUS FORMS

27

• (W1) Given a homogeneous form f ∈ S d E ∗ , study the subvariety VSP(f ; s)o of (P(E ∗ )(s) (the variety of power sums which consists of polar s-polyhedra of f or more general the subvariety VSP(f ; s) of Hilbs (P(E ∗ ) parametrizing generalized s-polyhedra. • (W2) For given s find the equations of the closure PS(s, d; n) in S d E ∗ of the locus of homogeneous forms of degree d which can be written as a sum of s powers of linear forms. Note that PS(s, d; n) is the affine cone over the secant variety Secs−1 (Vern ). d In the language of secant varieties, the variety VSP(f ; s)o is the set of linear independent sets of s points p1 , . . . , ps in Vern such that [f ] ∈ p1 , . . . , ps . The variety d VSP(f ; s) is the set of linearly independent Z ∈ Hilbs (P(E)) such that [f ] ∈ Z . Note that we have a natural map VSP(f ; s) → G(s, S d E),
d

Z → Z d,

where G(s, S E) is the Grassmannian of s-dimensional subspaces of S d E. This map is not injective in general. Also note that for a general form f the variety VSP(f ; s) is equal to the closure of VSP(f ; s)o in the Hilbert scheme Hilbs (P(E ∗ )) (see [148], 7.2). It is not true for an arbitrary form f . One can also embed VSP(f ; s)o in P(S d E ∗ ) by assigning to {l1 , . . . , ls } the product l1 · · · ls . Thus we can compactify VSP(f ; s)o by taking its closure in P(S d E ∗ ). In general, this closure is not isomorphic to VSP(f ; s). Proposition 1.3.12. Assume n = 2. For general f ∈ S d E ∗ the variety VSP(f ; s) is either empty or a smooth irreducible variety of dimension 3s − 2+d . d Proof. We consider VSP(f ; s) as the closure of VSP(f ; s)o in the Hilbert scheme Hilbs (P(E ∗ ). Recall that Z ∈ Hilbs (P(E ∗ ) is a generalized polar polyhedron of f if and only if f ∈ IZ (d)⊥ but this is not true for any proper closed subscheme Z of Z. Consider the incidence variety X = {(Z, f ) ∈ Hilbs (P(E ∗ ) × S d V ∗ : Z ∈ VSP(f ; s)}. It is known that for any nonsingular surface the punctual Hilbert scheme is nonsingular (see [110]). Let U be the open subset of the first factor such that for any point Z ∈ U , dim IZ (d) = dim S d E − s. The fibre of the first projection over Z ∈ U is an open Zariski subset of the linear space IZ (d)⊥ . This shows that X is irreducible and nonsingular. The fibres of the second projection are the varieties VSP(f ; s). Thus for an open Zariski subset of S d E ∗ the varieties VSP(f ; s) are empty or irreducible and nonsingular.

1.4
1.4.1

Dual homogeneous forms
Catalecticant matrices
apk : S k E → S d−k E ∗ , f ψ → Dψ (f ). (1.38)

Let f ∈ S d E ∗ . Consider the linear map (the apolarity map)

28

CHAPTER 1. POLARITY

Its kernel is the space APk (f ) of forms of degree k which are apolar to f . By the polarity duality, the dual space of S d−k E ∗ can be identified with S d−k E. Applying Lemma 1.3.1, we obtain
t

(apk ) = apd−k . f f

(1.39)

Assume that f =

s d i=1 li

for some li ∈ E ∗ . It follows from (1.34) that

d−k d−k apk (S k E) ⊂ l1 , . . . , ls , f

and hence rank(apk ) ≤ s. f


(1.40) apk f

If we choose a basis in E and a basis in E , then is given by a matrix of size k+n n+d−k × d−k whose entries are linear forms in coefficients of f . k Choose a basis ξ0 , . . . , ξn in E and the dual basis t0 , . . . , tn in E ∗ . Consider a monomial basis in S k E (resp. in S d−k E ∗ ) which is lexigraphically ordered. The matrix of apk with respect to these bases is called the kth catalecticant matrix of f and f is denoted by Catk (f ). Its entries cuv are parameterized by pairs (u, v) ∈ Nn+1 × Nn+1 with |u| = d − k and |v| = k. If we write f = d!
|i|=d

1 i ai t , i!

then cuv = au+v . This follows easily from the formula
i i ∂00 · · · ∂nn (tj0 · · · tjn ) = n 0 j! (j−i)!

if j − i ≥ 0 otherwise.

0

Considering ai as independent variables ti , we obtain the definition of a general catalecticant matrix Catk (d, n). Example 1.4.1. Let n = 1. Write f =  a0  a1  Catk (f ) =  .  . . ad−k
d d i=0 i

a1 a2 . . .

ai td−i ti . Then 1 0  ... ak . . . ak+1   .  .  ... . ... ad

ad−k+1

A matrix of this type is called a Hankel matrix or persymmetric matrix. It follows from (1.40) that f ∈ PS(s, d; 1) implies that all s + 1 × s + 1 minors of Catk (f ) are equal to zero. Thus we obtain that Secs−1 (Ver1 ) is contained in the subvariety of Pd defined d by (s + 1) × (s + 1)-minors of the matrices   t0 t1 ... tk  t1 t2 . . . tk+1    Catk (d, 1) =  . . .  , k = 1, . . . , min d − s, s . .   . . . ... . td−k td−k+1 . . . td

1.4. DUAL HOMOGENEOUS FORMS

29

For example, if s = 1, we obtain that the Veronese curve Ver1 ⊂ Pd satisfies the d equations ti tj − tk tl = 0, where i + j = k + l. It is well known that these equations generate the homogeneous ideal of the Veronese curve. Assume d = 2k. Then the Hankel matrix is a square matrix of size k + 1. Its determinant vanishes if and only if f admits a nonzero apolar form of degree k. The set of such f ’s is a hypersurface in C[t0 , t1 ]2k . It contains the Zariski open subset of forms which can be written as a sum of k powers of linear forms (see section 1.5.1). For example, take k = 2. Then the equation   a0 a1 a2 det a1 a2 a3  = 0 (1.41) a2 a3 a4 describes binary quartics f = a0 t4 + 4a1 t3 t1 + 6a2 t2 t2 + 4a3 t0 t3 + a4 t4 0 0 0 1 1 1 which lie in the Zariski closure of the locus of quartics represented in the form (α0 t0 + β0 t1 )4 + (α1 t0 + β1 t1 )4 . Note that a quartic of this form has simple roots unless it has a root of multiplicity 4. Thus any binary quartic with simple roots satisfying equation (1.41) can be represented as a sum of two powers of linear forms. The cubic hypersurface in P4 defined by equation (1.41) is equal to the 1-secant variety of a Veronese curve in P4 . Note that dim APi (f ) = dim Ker(api ) = f Therefore, dim(Af )i = rank Cati (f ), and
d n+k i

− rank Cati (f ).

HAf (t) =
i=0

rank Cati (f )ti .

(1.42)

It follows from (1.39) that rank Cati (f ) = rank Catd−i (f ) confirming that HAf (t) is a reciprocal monic polynomial. Suppose d = 2k is even. Then the coefficient at tk in HAf (t) is equal to rank Catk (f ). The matrix Catk (f ) is a square matrix of size n+k . One can show that for a general k f , this matrix is nonsingular. A polynomial f is called degenerate if det Catk (f ) = 0. Thus, the set of degenerate polynomials is a hypersurface (catalecticant hypersurface) given by the equation det Catk (2k, n) = 0. (1.43) The polynomial in variables ti , |i| = d, is called the catalecticant determinant.

30

CHAPTER 1. POLARITY

Example 1.4.2. Let d = 2. It is easy to see that the catalecticant polynomial is the discriminant polynomial. Thus a quadratic form is degenerate if and only if it is degenerate in the usual sense. The Hilbert polynomial of a quadratic form f is HAf (t) = 1 + rt + t2 , where r is the rank of the quadratic form. Example 1.4.3. Suppose f = td + · · · + td , s ≤ n. Then ti , . . . , ti are linearly s s 0 0 independent for any i, and hence rank Cati (f ) = s for 0 < i < d. This shows that HAf (t) = 1 + s(t + · · · + td−1 ) + td . Let P be the set of reciprocal monic polynomials of degree d. One can stratify the space S d E ∗ by setting, for any p ∈ P,
∗ S d Ep = {F ∈ S d E ∗ : HAf = p}.

If f ∈ PS(s, d; n) we know that rank Catk (f ) ≤ h(s, d, n)k = min(s,
n+k n

,

n+d−k n

).

One can show that for a general enough f , we have the equality (see [148], Lemma 1.7]). Thus there is a Zariski open subset of PS(s, d; n) which belongs to the strata d ∗ S d Ep , where p = i=0 h(s, d, n)i ti .

1.4.2

Dual homogeneous forms

In Chapter 1 we introduced the notion of a dual quadric. If Q = V (q), where q is ˇ a nondegenerate quadratic form, then the dual variety Q is a quadric defined by the quadratic form q whose matrix is the adjugate matrix of q. For any homogeneous form ˇ ˇ of even degree f ∈ S 2k E ∗ one can define the dual homogeneous form f ∈ S 2k E in a similar fashion using the notion of the catalecticant matrix. Let apk : S k E → S k E ∗ (1.44) f be the apolarity map (1.38). We can view this map as a symmetric bilinear form Ωf : S k E × S k E → C, Ωf (ψ1 , ψ2 ) = apk (ψ1 )(ψ2 ) = ψ2 , apk (ψ1 ) . f f (1.45)

Its matrix with respect to a monomial basis in S k E and its dual monomial basis in S k E ∗ is the catalecticant matrix Catk (f ). Let us identify Ωf with the associated quadratic form on S k E (the restriction of Ωf to the diagonal). This defines a linear map Ω : S 2k E ∗ → S 2 S k E ∗ , There is also a natural left inverse map of Ω P : S 2 S k E ∗ → S 2k E ∗ defined by multiplication S k E ∗ × S k E ∗ → S 2k E ∗ . All these maps are GL(E)equivariant and realize the linear representation S 2k E ∗ as a direct summand in the representation S 2 S k E ∗ . f → Ωf .

1.4. DUAL HOMOGENEOUS FORMS

31

Theorem 1.4.1. Assume that f ∈ S 2k E ∗ is nondegenerate. There exists a unique ˇ homogeneous form f ∈ S 2k E (the dual homogeneous form) such that ˇ Ω f = Ωf . ˇ ˇ Proof. We know that Ω(f ) is defined by the adjugate matrix adj(Catk (f )) = (c∗ ) so uv that ˇ Ωf = c∗ ξ u ξ v . uv Let ˇ f=
|u+v|=2k ∗ u+v d! . (u+v)! cuv ξ

Recall that the entries cuv of the catalecticant matrix depend only on the sum of the indices. Thus the entries of the cofactor matrix adj(Catk (f )) = (c∗ ) depend only on uv the sum of the indices. For any ti ∈ S k V ∗ , we have ˇ Pti (f ) =
u,v,u+v≥i (u+v)! u+v−i ∗ d! (u+v)! cuv (u+v−i)! ξ

=
j,|j|=k

d! ∗ j j! cij ξ

This checks that the matrix of the linear map S k E ∗ → S k E defined by Ωf is equal to ˇ the matrix adj(Catk (f )). Thus the quadratic form Ωf is the dual of the quadratic form ˇ Ωf . Recall that the locus of zeros of the quadric q in E ∗ is equal to the set of linear ˇ functions of the form l = bq (v) such that v, l = 0. The same is true for the dual form ˇ f . Its locus of zeros consists of linear forms l such that Ω−1 (lk ) ∈ S k E vanishes on f l. The degree k homogeneous form Ω−1 (lk ) is classically known as the anti-polar of l f (with respect to f ). Definition 1.8. Two linear forms l, m ∈ E ∗ are called conjugate with respect to a nondegenerate form f ∈ S 2k E ∗ if ˇ Ωf (lk , mk ) = f (lk mk ) = 0. ˇ
k Proposition 1.4.2. Suppose f is given by (1.33), where the powers li are linearly k ∗ independent in S E . Then each pair li , lj is conjugate with respect to f .

Proof. It follows from computation of Ωf in the proof of Proposition 1.4.3 that it suffices to check the assertion for quadratic forms. Choose a coordinate system such that 2 2 ˇ li = t0 , lj = t1 and f = t2 + t2 + · · · + t2 . Then f = ξ0 + · · · + ξn , where ξ0 , . . . , ξn n 0 2 are dual coordinates. Now the assertion is easily checked.

1.4.3

The Waring rank of a homogeneous form

Since any quadratic form q can be reduced to a sum of squares, one can characterize its rank as the smallest number r such that
2 2 q = l1 + · · · + lr

for some linear forms l1 , . . . , lr .

32

CHAPTER 1. POLARITY

Definition 1.9. Let f ∈ S d E ∗ . Its Waring rank wrk(f ) is the smallest number r such that d d f = l1 + · · · + lr (1.46) for some linear forms l1 , . . . , lr . Proposition 1.4.3. Let Ωf be the quadratic form on S k V associated to f ∈ S 2k E ∗ . Then the Waring rank of f is greater or equal than the rank of Ωf . Proof. Suppose (1.33) holds with d = 2k. Since Ωf is linear with respect to f , we have Ωf = Ωli . If we choose coordinates such that li is a coordinate function t0 , 2k 2k we easily compute the catalecticant matrix of li . It is equal to the matrix with 1 at the upper left corner and zero elsewhere. The corresponding quadratic form is equal to k (tk )2 . Thus Ωli = (li )2 and we obtain 2k 0
r r

Ωf =
i=1

Ωl i = 2k
i=1

k (li )2 .

Thus the rank of f is greater or equal than the rank of Ωf . Corollary 1.4.4. Suppose f is a nondegenerate form of even degree 2k, then wrk(f ) ≥
k+n n

.

A naive way to compute the Waring rank is by counting constants. Consider the map s : (E ∗ )r → C
d+n n

,

(l1 , . . . , lr ) →

d li .

(1.47)

If r(n + 1) ≥ d+n one expects that this map is surjective and hence wrk(f ) ≤ n r for general f . Here “general” means that the coefficients of f belong to an open
d+n

Zariski subset of the affine space C n . It follows from Theorem 1.3.10 that the only exceptional cases when it is false and the map s fails to be surjective are the following cases: • n = 2, d = 2, r = 2, wrk(f ) = 3; • n = 2, d = 4, r = 5, wrk(f ) = 6; • n = 3, d = 4, r = 9, wrk(f ) = 10; • n = 4, d = 3, r = 7, wrk(f ) = 8; • n = 4, d = 4, r = 14, wrk(f ) = 15; Proposition 1.4.5. Let f be a general homogeneous form of even degree 2k. Then wrk(f ) > rank Ωf except in the following cases, where the equality takes place,:

1.4. DUAL HOMOGENEOUS FORMS • k = 1; • n = 1; • n = 2, k ≤ 4; • n = 3, k = 2.

33

Proof. The first case is obvious. It follows from considering the map (1.47) that wrk(f ) ≥ n+2k /(n + 1). On the other hand the rank of Ωf for general f is equal to n dim S k E = n+k . n We know that the case n = 1 is not exceptional so that we can compute the Waring rank of f by counting constants and get wrk(f ) = k + 1 = rank Ωf . If n = 2, we get wrk(f ) ≥ (2k + 2)(2k + 1)/6 = (k + 1)(2k + 1)/3 and rank Ωf = k+2 = (k + 2)(k + 1)/2. We have (k + 1)(2k + 1)/3 > (k + 2)(k + 1)/2 2 if k > 4. By Theorem 1.3.10,  6 if k = 2,  wrk(f ) = 10 if k = 3,   15 if k = 3. This shows that wrk(f ) = rank Ωf in all these cases. If n = 3, we get wrk(f ) ≥ (2k + 3)(2k + 2)(2k + 1)/24 > unless k = 2. Finally, it is easy to see that for n > 3 wrk(f ) ≥ for k > 1.
2k+n 1 n+1 n k+3 3

= (k + 3)(k + 2)(k + 1)/6

>

k+n n

1.4.4

Mukai’s skew-symmetric form

Let ω ∈ Λ2 E be a skew-symmetric bilinear form on E ∗ . It admits a unique extension to a Poisson bracket {, }ω on S • E ∗ which restricts to a skew-symmetric bilinear form {, }ω : S k+1 E ∗ × S k+1 E ∗ → S 2k E ∗ . (1.48)

Recall that a Poisson bracket on a commutative algebra A is a skew-symmetric bilinear map A × A → A, (f, g) → {f, g} such that its left and right partial maps A → A are derivations. ˇ Let f ∈ S 2k E ∗ be a nondegenerate form and f ∈ S 2k E = (S 2k E ∗ )∗ be its dual 2 k+1 form. For each ω as above define σω,f ∈ Λ (S E)∗ by ˇ σω,f (f, g) = f ({f, g}ω ).

34

CHAPTER 1. POLARITY

Theorem 1.4.6. Let f be a nondegenerate form in S 2k E ∗ of Waring rank N . Assume that N = rank Ωf = n+2k . For any Z = { 1 , . . . , s } ∈ VSP(f ; s)o let Z k+1 be n k+1 the linear span of the powers li in S k+1 E ∗ . Then (i) P is isotropic with respect to each form σω,f ; (ii) apk−1 (S k−1 E) ⊂ Z f
k+1 ;

(iii) apk−1 (S k−1 E) is contained in the radical of each σω,f . f
k+1 k+1 Proof. To prove the first assertion it is enough to check that σω,f (li , lj ) = 0 for all i, j. We have k+1 k+1 ˇ k k ˇ kk σω,f (li , lj ) = f ({li , lj }ω ) = f (li lj )ω(li , lj ). k k ˇ kk By Proposition 1.4.2, f (li lj ) = Ωf (li , lj ) = 0. This checks the first assertion. ˇ For any ψ ∈ S k−1 V , N N 2k li ) = i=1 2k Dψ (li ) = (2k)! (k+1)! i=1 k+1 Dψ (lk−1 )li .

Dψ (f ) = Dψ (

This shows that apk−1 (S k−1 E) is contained in L(P). It remains to check that for any f ψ ∈ S k−1 V, G ∈ S k+1 E ∗ and any ω ∈ Λ2 E, one has σω,f (Dψ (f ), G) = 0. Choose coordinates t0 , . . . , tn in E and the dual coordinates ξ0 , . . . , ξn in E ∗ . The space Λ2 E is spanned by the forms ωij = ξi ∧ ξj . We have {Dψ (f ), g}ωij = Dξi (Dψ (f ))Dξj (g) − Pξj (Dψ (f ))Dξi (g) = Dξi ψ (f )Dξj (g) − Dξj ψ (f )Dξi (g) = Dψξi (f )Dξj (g) − Dψξj (f )Dξi (g). For any a, b ∈ S k E ∗ , ˇ f (ab) = Ωf (a, b) = Ω−1 (a), b . ˇ f Thus ˇ σωij ,f (Dψ (f ), g) = f (Dψξi (f )Dξj (g) − Dψξj (f )Dξi (g)) = ψξi , Dξj (g) − ψξj , Dξi (g) = Dψ (Dξi ξj (g) − Dξj ξi (g)) = Dψ (0) = 0.

Let Z = {[l1 ], . . . , [ls ]} ∈ VSP(f ; s)o be a polar s-polyhedron of a nondegenerate form f ∈ S 2k E ∗ and, as before, let Z k+1 be the linear span of k + 1th powers of the linear forms li . Let L(Z) = Z k+1 /apk−1 (S k−1 E). (1.49) f It is a subspace of W = S k+1 E ∗ /apk−1 (S k−1 E). By (1.39), f W ∗ = apk−1 (S k−1 E)⊥ = APk+1 (f ), f

1.5. FIRST EXAMPLES

35

where we identify the dual space of S k+1 E ∗ with S k+1 E. Now observe that P ⊥ k+1 is equal to IP (k + 1), where we identify P with the reduced closed subscheme of the dual projective space P(E ∗ ). This allows one to extend the definition of L(P) to any generalized polar s-polyhedron Z ∈ VSP(f ; s): L(Z) = IZ (k + 1)⊥ /apk−1 (S k−1 E) ⊂ S k E ∗ /apk−1 (S k−1 E). f f Proposition 1.4.7. Let f be a nondegenerate homogeneous form of degree 2k of Waring rank equal to Nk = n+k . Let Z, Z ∈ VSP(f ; s). Then k L(Z) = L(Z ) ⇐⇒ Z = Z . Proof. It is enough to show that IZ (k + 1) = IZ (k + 1) =⇒ Z = Z . Suppose Z = Z . Choose a subscheme Z0 of Z of length Nk − 1 which is not a subscheme of Z . Since dim IZ0 (k) ≥ dim S k E ∗ − h0 (OZ ) = n+k − Nk + 1 = 1, k we can find a nonzero ψ ∈ IZ0 (k). The sheaf IZ /IZ0 is concentrated at one point x and is annihilated by the maximal ideal mx . Thus mx IZ0 ⊂ IZ . Let ξ ∈ E be a linear form on E ∗ vanishing at x but not vanishing at any subscheme of Z . This implies that ξψ ∈ IZ (k + 1) = IZ (k + 1) and hence ψ ∈ IZ (k) ⊂ APk (f ) contradicting the nondegeneracy of f . It follows from Theorem 1.4.6 that each ω ∈ Λ2 E defines a skew-symmetric 2form σω,f on S k+1 E which factors through a skew-symmetric 2-form σω,f on W = ¯ S k+1 E/apk−1 (S k−1 E). We call this form a Mukai 2-form. For each P ∈ VSP(f ; Nk )o f the subspace L(Z) ⊂ W is isotropic with respect to σω,f . ¯

1.5
1.5.1

First examples
Binary forms

This is the case n = 1. The zero subscheme of a homogeneous form of degree d in 2 variables f (t0 , t1 ) is a positive divisor D = mi pi of degree d. Each such divisor is obtained in this way. Thus we can identify P(S d E ∗ ) with |OP(E) (d)| ∼ Pd = (d) d and also with the symmetric product P(E) = P(E) /Sd and the Hilbert scheme k Hilbd (P(E)). A generalized s-polyhedron of f is a positive divisor Z = i=1 mi [li ] ∗ 0 ⊥ of degree s in P(E ) such that [f ] ∈ Z = P(H (P(E), IZ (d)) ). Note that in our case Z is automatically linearly independent (because H 1 (IZ (d)) = 0). Obviously, H 0 (P(E), IZ (d)) consists of polynomials of degree d which are divisible by ψ = m m ξ1 1 · · · ξk k , where ξi ∈ AP(li )1 . In coordinates, if li = ai t0 + bi t1 , then ξi = bi ∂0 − ai ∂1 . Thus f is orthogonal to this space if and only if Pψψ (f ) = 0 for all ψ ∈ S d−s (E). By the apolarity duality this implies that Dψ (f ) = 0, hence ψ ∈ AP(f )s . Thus we obtain
m m Theorem 1.5.1. A positive divisor Z = V (l1 1 · · · lk k ) of degree is a generalized mk m1 s-polyhedron of f if and only if ξ1 · · · ξk ∈ AP(f )s .

36 Corollary 1.5.2. Assume n = 1. Then VSP(f ; s) = P(APs (f )). Note that the kernel of the map S s E → S d−s E ∗ ,
s d−s ∗

CHAPTER 1. POLARITY

ψ → Dψ (f )

is of dimension ≥ dim S E − dim S E = s + 1 − (d − s + 1) = 2s − d. Thus Dψ (f ) = 0 for some nonzero ψ ∈ S s E, whenever 2s > d. This shows that a f has always generilized polar s-polyhedron for s > d/2. If d is even, a binary form has an apolar d/2-form if and only det Catd/2 (f ) = 0. This is a divisor in the space of all binary d-forms. Example 1.5.1. Take d = 3. Assume that f admits a polar 2-polyhedron. Then f = (a1 t0 + b1 t1 )3 + (a2 t0 + b2 t1 )3 . It is clear that f has 3 distinct roots. Thus, if f = (a1 t0 + b1 t1 )2 (a2 t0 + b2 t1 )2 has a double root, it does not admit a polar 2-polyhedron. However, it admits a generalized 2-polyhedron defined by the divisor 2p, where p = (b1 , −a1 ). In the secant variety interpretation, we know that any point in P(S 3 E ∗ ) either lies on a unique secant or on a unique tangent line of the Veronese cubic curve. The space AP(f )2 is always onedimensional. It is generated either by a binary quadric (−b1 ξ0 + a1 ξ1 )(−b2 ξ0 + a2 ξ1 ) or by (−b1 ξ0 + a1 ξ1 )2 . Thus VSP(f ; 2)o consists of one point or empty but VSP(f ; 2) always consists of o one point. This example shows that VSP(f ; 2) = VSP(f ; 2) in general.

1.5.2

Quadrics

It follows from Example 1.3.1 that Sect (Vern ) = P(S 2 E ∗ ) if only if there exists a 2 quadric with t + 1 singular points in general position. Since the singular locus of a quadric V (q) is a linear subspace of dimension equal to corank(q) − 1, we obtain that Secn+1 (Vern ) = P(S 2 E ∗ ), hence any general quadratic form can be written as 2 a sum of n + 1 squares of linear forms l0 , . . . , ln . Of course, linear algebra gives more. Any quadratic form of rank n + 1 can be reduced to sum of squares of the coordinate functions. Assume that q = t2 + · · · + t2 . Suppose we also have q = n 0 2 2 l0 + · · · + ln . Then the linear transformation ti → li preserves q and hence is an orthogonal transformation. Since polar polyhedra of q and λq are the same, we see that the projective orthogonal group PO(n+1) acts transitively on the set VSP(f ; n+1)o of polar (n+1)-polyhedra of q. The stabilizer group G of the coordinate polar polyhedron is generated by permutations of coordinates and diagonal orthogonal matrices. It is isomorphic to the semi-direct product 2n Sn+1 (the Weyl group of root systems of types Bn , Dn ), where we use the notation 2k for the 2-elementary abelian group (Z/2Z)k . Thus we obtain Theorem 1.5.3. Let q be a quadratic form in n + 1 variables of rank n + 1. Then VSP(q; n + 1)o ∼ PO(n + 1)/2n · Sn+1 . = The dimension of VSP(q; n + 1)o is equal to 1 n(n + 1). 2

1.5. FIRST EXAMPLES

37

Example 1.5.2. Take n = 1. Using the Veronese map ν2 : P1 → P2 , we consider a nonsingular quadric Q = V (q) as a point p in P2 not lying on the conic C = V (t0 t2 − t2 ). A polar 2-gon of q is a pair of distinct points p1 , p2 on C such that p ∈ p1 , p2 . 1 It can be identified with the pencil of lines through p with the two tangent lines to C deleted. Thus W (q, 2)o = P1 \ {0, ∞} = C∗ . There are two generalized 2-gons 2p0 and 2p∞ defined by the tangent lines. Each of them gives the representation of q as o l1 l2 , where V (li ) are the tangents. We have VSP(f ; 2) = VSP(f ; 2) ∼ P1 . = In the next chapter we will discuss a good compactification of this space in the case n = 2. Let q ∈ S 2 E ∗ be a non-degenerate quadratic form. For each Z ∈ VSP(q; n + 1) the linear space L(Z) = Z 2 /Cq ⊂ S 2 E ∗ /Cq is of dimension n. It is an isotropic subspace of W = S 2 E ∗ /Cq with respect to any Mukai’s 2-form σω,q . This defines a ¯ map µ : VSP(q; n + 1) → G(n, W ), Z → L(Z). (1.50) By Proposition 1.4.7, the map is injective. The image of VSP(q, n + 1)o is contained in the locus G(n, W )µ of subspaces which are isotropic with respect to any Mukai’s 2form σq,ω . Since for general f the variety VSP(q, n+1) is the closure of VSP(q, n+1)o ¯ in the Hilbert scheme, the image of VSP(q; n + 1) is contained G(n, W )µ . Since all nonsingular quadrics are isomorphic, the assertion is true for any nondegenerate quadratic for f . Recall that the Grassmann variety G(n, W ) carries the natural rank n vector bundle S, the tautological bundle. Its fibre over a point L ∈ G(n, W ) is equal to L. It is a subbundle of the trivial bundle WG(n,W ) associated to the vector space W . We have a natural exact sequence 0 → S → WG(n,V ) → Q → 0, where Q is the universal quotient bundle, whose fibre over L is equal to W/L. We ∗ can consider each element σ of Λ2 W ∗ as a section of the trivial bundle Λ2 WG(n,W ) . Restricting σ to the subbundle S, we get a section of the vector bundle Λ2 S ∗ . Thus we can view a Mukai’s 2-form σq,ω as section sq,ω of Λ2 S ∗ . ¯ It follows from above that the image of the map (1.50) is contained in the set of common zeros of the sections sq,ω of Λ2 S ∗ . The next result has been proven already, by different method, in Chapter 2, Part I. Corollary 1.5.4. Let q be a nondegenerate quadratic form on a three-dimensional vector space E. Then the image of VSP(q; 3) in G(2, W ), embedded in the Pl¨ cker u space P(Λ2 W ), is a smooth irreducible 3-fold equal to the intersection X of G(2, W ) with a linear space of codimension 3. Proof. We have dim W = 5, so G(2, W ) ∼ G(2, 5) is of dimension 6. Hyperplanes in = the Pl¨ cker space are elements of the space P(Λ2 W ∗ ). Note that the functions sq,ω are u linearly independent. In fact, a basis ξ0 , ξ1 , ξ2 in E gives a basis ω01 = ξ0 ∧ ξ1 , ω02 = ξ0 ∧ ξ2 , ω12 = ξ1 ∧ ξ2 in Λ2 E. Thus the space of sections sq,ω is spanned by 3 sections s01 , s02 , s12 corresponding to the forms ωij . Without loss of generality we may assume that q = t2 + t2 + t2 . If we take a = t0 t1 + t2 , b = −t2 + t2 + t2 , we 0 1 2 2 0 1 2

38

CHAPTER 1. POLARITY

see that s01 (a, b) = 0, s12 (a, b) = 0, s02 (a, b) = 0. Thus a linear dependence between the functions sij implies the linear dependence between two of the functions. It is easy to see that no two functions are proportional. So our 3 functions sij , 0 ≤ i < j ≤ 2 span a 3-dimensional subspace of Λ2 W ∗ and hence define a codimension 3 projective subspace L in the Pl¨ cker space P(Λ2 W ). The image of VSP(q; 3) under the map u (1.50) is contained in the intersection G(2, E) ∩ L. This is a 3-dimensional subvariety of G(2, W ), and hence contains µ(VSP(q; 3)) as an irreducible component. We skip an argument, based on counting constants, which proves that the subspace L belongs to an open Zariski subspace of codimension 3 subspaces of Λ2 W for which the intersection L ∩ G(2, W ) is smooth and irreducible (see [79]. If n > 2, the vector bundle Λ2 S ∗ is of rank r = n > 1. The zero locus of 2 its nonzero section is of expected codimension equal to r. We have n+1 sections 2 sij of Λ2 S and dim G(n, E) = n( n+2 − n − 1). For example, when n = 3, we 2 have 6 sections sij each vanishing on a codimension 3 subvariety of 18-dimensional Grassmannian G(3, 9). So there must be some dependence between the functions sij . Remark 1.5.1. One can also consider the varieties VSP(q; s) for s > n + 1. For example, we have t2 0 t2 0 = =
1 2 (t0 1 2 (t0 1 + 2t1 )2 + 2 (t0 − 2t1 )2 − 4t2 , 1 1 + t1 )2 + 2 (t0 − t1 )2 − 1 (t2 + t1 )2 + 1 (t2 − t1 )2 . 2 2

This shows that VSP(q; n + 3), VSP(q; n + 4) are not empty for any nondegenerate quadric Q.

Exercises
1.1 Show that the first polar Pa (X) contains singular points of X. Suppose X is a plane curve and x ∈ X is its ordinary double point. Show that the pair consisting of the tangent line of Pa (X) at x and the line a, x is harmonic conjugate (see section 2.1.2) to the pair of tangents to the branches of X at x in the pencil of lines through x. If x is an ordinary cusp, then tangent line of Pa (X) at x is equal to the cuspidal tangent of X at x. 1.2 Show that a line contained in a hypersurface X belongs to all polars of X with respect to any point on this line. 1.3 Find the multiplicity of the intersection of a plane curve X with its Hessian at an ordinary double point and at an ordinary cusp of C. Show that the Hessian has a triple point at the cusp. 1.4 Suppose a hypersurface X in Pn has a singular point x of multiplicity m > 1. Prove that He(X) has this point as a point of multiplicity ≥ (n + 1)m − 2n. 1.5 Suppose a hyperplane is tangent to a hypersurface X along a closed subvariety Y of codimension 1. Show that Y is contained in He(X). 1.6 Suppose f is the product of d distinct linear forms li (t0 , . . . , tn ). Let A be the matrix of size (n + 1) × d whose ith column is formed by the coefficients of li (defined, of course up to proportionality). Let ∆I be the maximal minor of A corresponding to a subset I of [1, . . . , d] and fI be the product of linear forms li , i ∈ I. Show that X 2 2 He(f ) = (−1)n (d − 1)f n−1 ∆I fI .
I

EXERCISES
([176], p. 660). 1.7 Let n = 2. Assume He(V (f )) = P2 . Show that f is the union of concurrent lines.

39

1.8 Show that the locus of the points on the plane where the first polars of a plane curve X are tangent to each other is the Hessian of X and the set of common tangents is the Cayleyan curve . 1.9 Show that each flex tangent of a plane curve X, considered as a point in the dual plane, lies on the Cayleayan of X. 1.10 Show that the class of the Steinerian st(X) of a plane curve Xof degree d is equal to 3(d − 1)(d − 2) but its dual is not equal to Cay(X). 1.11 Let Dm,n ⊂ Pmn−1 be the image in the projective space of the variety of m × n matrices of rank ≤ min{m, n} − 1. Show that the variety ˜ Dm,n = {(A, x) ∈ Pmn−1 × Pn : A · x = 0} is a resolution of singularities of Dm,n . Find the dual variety of Dm,n . 1.12 Find the dual variety of the Segre variety sn (Pn ) ⊂ Pn
2

+2n

.

1.13 Prove that the degree of the dual variety of a nonsingular hypersurface of degree d in Pn is equal to d(d − 1)n−1 . 1.14 Let X be the union of k nonsingular conics in general position. Show that X ∗ is also the union of k nonsingular conics in general position. Check the Pl¨ cker formulas in this case. u 1.15 Let X has only δ ordinary nodes and κ ordinary cusps as singularities. Assume that the ˇ ˇ dual curve X ∗ has also only δ ordinary nodes and κ ordinary cusps as singularities. Find δ and ˇ κ in terms of d, δ, κ. ˇ 1.16 Give an example of a self-dual (i.e. X ∗ ∼ X) plane curve of degree > 2. = 1.17 Let f ∈ S 2 E ∗ . Show that the map V → V ∗ defined by ψ → Dψ (f ) corresponds to the symmetric bilinear form V × V → C associated to Q. 1.18 Show that the embedded tangent space of the Veronese variety Vern at a point represented d by the form ld is equal to the projectivization of the linear space of homogeneous polynomials of degree d of the form ld−1 m. 1.19 Show using the following steps that Ver4 is 6-defective by proving that for 7 general points 3 pi in P4 there is a cubic hypersurface with singular points at the pi ’s. (i) Show that there exists a Veronese curve R4 of degree 4 through the seven points, (ii) Show that the secant variety of R4 is a cubic hypersurface which is singular along R4 . 1.20 Let q be a nondgenerate quadratic form in n + 1 variables. Show that VSP(q; n + 1)o embedded in G(n, E) is contained in the linear subspace of codimension n. 1.21 Compute the catalecticant matrix Cat2 (f ), where f is a homogeneous form of degree 4 in 3 variables. 1.22 Let f ∈ S 2k E ∗ and Ωf be the corresponding quadratic form on S k E. Show that the quadric V (Ωf ) in P(S k E) is characterisezed by the following two properties: • Its pre-image under the Veronese map νk : P(E) → P(S k E) is equal to V (f ); • Ωf is apolar to any quadric in P(S k E ∗ ) which contains the image of the Veronese map P(E ∗ ) → P(S k E ∗ ).

40

CHAPTER 1. POLARITY

1.23 Let Ck be the locus in P(S 2k E) of hypersurfaces V (f ) such that det Catk (f ) = 0. Show that Ck is a rational variety. [Hint: Consider the rational map Ck − → P(S k E) which assigns to V (f ) the point defined by the subspace APk (f ) and study its fibres]. 1.24 Give an example of a polar 4-gon of the cubic t0 t1 t2 . 1.25 Find all binary forms of degree d for which VSP(f ; 2)o = ∅. 1.26 Let f be a form of degree d in n + 1 variables. Show that the variety VSP(f ; ` ´ irreducible variety of dimension n n+d . d `n+d´
d

)o is an

1.27 Describe the variety VSP(f ; 4), where f is a nondegenerate quadric in 3 variables.

Historical Notes
Although the polar lines of conics were known to mathematicians of Ancient Greece, the first systematic study of polars of curves of higher degree started in the works of E. Bobilier [21] and J. Pl¨ cker [192]. However, some of theory were known before u to G. Monge and J. Poncelet. According to the historical account in [101], vol. II, the name “polare” was introduced by J. Gergonne. Other historical information can be found in [18] and [188], p.279. The Hessian curve was first introduced by J. Steiner [231] who called it the Kerncurve. The name was coined by J. Sylvester in honor of O. Hesse whose paper [135] provided many fundamental properties of the curve. The Steinerian curve originates in the works of J. Steiner in more general setting of nets of plane curves (not necessary the net of polars). The name was given by G. Salmon and L. Cremona. The Cayleyan curve was introduced by A. Cayley in [33] who called it the pippiana. The name was proposed by L. Cremona. There are many beautiful results in the hessians in the classical literature, many of them can be found in standard text-books of that time (e.g. [46], [101], [210]). Excellent surveys of these results can be found in [18] and [188]. The theory of dual varieties, generalization of Pl¨ cker formulae to arbitrary dimenu sion is still a popular subject of modern algebraic geometry. It is well-documented in modern literature and for this reason this topic is barely touched here. The theory of apolarity is one of the forgotten topics of classical algebraic geometry. It originates from the works of Rosanes [203] and Reye [199]. We refer for survey of classical results to [188] and to a modern exposition of some of these results to [79] which we followed in these notes.

Chapter 2

Conics
2.1
2.1.1

Self-polar triangles
The Veronese quartic surface
P(E ∗ ) → P(S d E ∗ ), V (L) → V (Ld ),

Recall that the Veronese variety is defined to be the image of the map

where L is a nonzero linear form on E. Replacing E with the dual space E ∗ , we obtain a map νd : P(E) → P(S d E), Cv → Cv d , (2.1) where v is a nonzero vector in E. This map is called the Veronese map. Using the polarity pairing (1.2), one can identify the space S d E ∗ with the dual space (S d E)∗ . This allows one to identify (2.1) with the map given by the complete linear system |OP(E) (d)| of hypersurfaces of degree d. The Veronese variety is of dimension n and degree dn . More generally, one defines a Veronese variety as the image of Pn in PN (d,n)−1 , N (d, n) = n+d , under the map given by the complete linear system d |OPn (d)| and a choice of a basis in H 0 (Pn , OPn (d)). Let d = n = 2, this is the case of the Veronese quartic surface Ver2 . The pre-image 2 of a hyperplane H in P(S 2 E) ∼ P5 is a conic C. There are three sorts of hyperplanes = corresponding to the cases: C is nonsingular, C is a line-pair, C is a double line. In the first case H intersects the Veronese surface Ver2 = v2 (P2 ) transversally, in the second 2 case H is tangent to Ver2 at a single point, and in the third case H is tangent to Ver2 2 2 along a conic. Choosing a basis in E we can identify the space S 2 E with the space of symmetric 3×3 matrices. The Veronese surface is identified with matrices of rank 1. Its equations are given by 2×2 minors. The variety of matrices of rank ≤ 2 is the cubic hypersurface D2 (2) given by the determinant. It singular along the Veronese surface. Since any nonzero matrix of rank ≤ 2 can be written as a sum of matrices of rank 1, we see that the discriminant cubic hypersurface D2 (2) is equal to the first secant variety of Ver2 . 2 41

42

CHAPTER 2. CONICS

A linear projection of Ver2 from a point not lying in D2 (2) is an isomorphism onto 2 a quartic surface V4 in P4 , called the projected Veronese surface. The image of Ver2 under a linear projection from a point Q lying in D2 (2) but not 2 lying on the surface is a non-normal quartic surface V4 in P4 . To see this we may assume that Q = V (t2 + t2 ). The plane of conics V (at2 + bt0 t1 + ct2 ) contains Q and 0 1 0 1 intersects Ver2 along the conic of double lines V ((αt0 + βt2 )2 ). The projection maps 2 1 this conic two-to-one to a double line of the image of Ver2 . 2 The image of Ver2 under a linear projection from its point is a cubic scroll, the 2 image of P2 under a map given by the linear system of conics with one base point.

2.1.2

Polar lines

Let C be a nonsingular conic. For any point a ∈ P2 the first polar Pa (C) is a line, the polar line of a. For any line there exists a unique point a such that Pa (C) = l. The point a is called the pole of . The point a considered as a line in the dual plane is the ˇ polar line of the point with respect to the dual conic C. A set of three non-colinear lines 1 , 2 , 3 is called a self-polar triangle with respect to C if each i is the polar line of C with respect to the point of intersection of the other two lines. Recall that two unordered pairs {a, b}, {c, d} of points in P1 are called harmonic conjugate if −2ββ + αγ + α γ = 0, (2.2) where V (αt2 + 2βt0 t1 + γt2 ) = {a, b} and V (α t2 + 2β t0 t1 + γ t2 ) = {c, d}. It 0 1 0 1 follows that this definition does not depend on the order of points in each pair. It is easy to check that (2.2) is equivalent to the polarity condition Pcd (q) = Pab (q ) = 0, where V (q) = {a, b}, V (q ) = {c, d}. Proposition 2.1.1. Let 1 , 2 , 3 be a self-polar triangle of C and a = 1 ∩ 2 . Assume a ∈ C. Then the pairs of points 3 ∩ C and (b, c) = ( 1 ∩ 3 , 2 ∩ 3 ) on the line 3 are harmonic conjugate. Conversely, if {c, d} is a pair of points on 3 which is harmonic conjugate to the pair C ∩ 3 , then the lines a, b , a, c , 3 form a self-polar triangle of C. Proof. Consider the pair C∩ 3 as a quadric q in 3 . We have c ∈ Pb (C), thus Pbc (C) = 0. Restricting to 3 and using (2.3), we see that b, c form a harmonic pair with respect to q. Conversely, if Pbc (q) = 0, the polar line Pb (C) contains a and intersects 3 at c, hence coincides with a, c. Similarly, Pc (C) = a, b. The polar line = Pa (C) intersects the conic C at two points x, y such that a ∈ Tx (C) ∩ Ty (C). Borrowing terminology from the Euclidean geometry, we call three non-collinear lines in P2 a triangle. The lines themselves will be called the sides of the triangle. The three intersection points of pairs of sides are called the vertices of the triangle. (2.3)

2.1. SELF-POLAR TRIANGLES Let f = a00 t2 + a11 t2 + a22 t2 + 2a01 t0 t1 + 2a02 t0 t2 + 2a12 t1 t2 = 0 0 1 2

43

be the equation of a nonsingular conic C. Choose projective coordinates in P2 such that i = V (ti ). Then P[1,0,0] (X) = P[0,1,0] (X) = P[0,0,1] (X) = implies that 1 2 (t + t2 + t2 ). (2.6) 1 2 2 0 2 2 2 Conversely, any conic V (l1 + l2 + l3 ) where li are three linear independent linear forms, defines a self-polar triangle with the sides V (li ). ˇ Any triangle in P2 defines the dual triangle in the dual plane P2 . Its sides are the pencils of lines with the base point of one of the vertices. f= Proposition 2.1.2. The dual of a self-polar triangle of a conic C is a self-polar triangle ˇ of the dual conic C Proof. Choose the coordinate system such that the self-polar triangle is the coordinate triangle. Then C = V (t2 + t2 + t2 ) and the assertion is easily verified. 0 1 2 All of this is immediately generalized to nonsingular quadrics Q in Pn for arbitrary n. We leave the generalization to the reader. For example, the problem of reducing a quadratic form to the sum of squares (or to principal axes) is nothing more as the problem of finding a self-conjugate n + 1-polyhedron of Q.
1

2

2

∂f ) = V (a00 t0 + a01 t1 + a02 t2 ), ∂t0 ∂f = V( ) = V (a11 t1 + a01 t0 + a12 t2 ), ∂t1 ∂f = V( ) = V (a22 t2 + a02 t0 + a12 t1 ) ∂t2 = V(

(2.4) (2.5)

2.1.3

The variety of self-polar triangles

Let C be a nonsingular conic. The group of projective transformations of P2 leaving C invariant is isomorphic to the projective complex orthogonal group PO3 = O3 /(±I3 ) ∼ SO3 . = It is also isomorphic to the group PSL2 via the Veronese map ν2 : P1 → P2 , [t0 , t1 ] → [t2 , t0 t1 , t2 ]. 0 1

Obviously PO3 acts transitively on the set of self-polar triangles of C. We may assume that C is given by (2.6). The stabilizer subgroup of the self-polar triangle defined by the coordinate lines is equal to the subgroup generated by permutation matrices and orthogonal diagonal matrices. It is easy to see that it is isomorphic to the semi-direct product (Z/2Z)2 S3 . An easy exercise in group theory gives that this group is isomorphic to the permutation group S4 . Thus we obtain the following.

44

CHAPTER 2. CONICS

Theorem 2.1.3. The set of self-polar triangles of a nonsingular conic has a structure of a homogeneous space SO3 /Γ, where Γ is a finite subgroup isomorphic to S4 . Let us describe a natural compactification of the homogeneous space SO3 /Γ. Let V be a Veronese surface in P5 . We view P5 as the projective space of conics in P2 and V as its subvariety of double lines. A trisecant plane of V is spanned by three linear independent double lines. A conic C ∈ P5 belongs to this trisecant if and only if the corresponding three lines form a self-polar triangle of C. Thus the set of selfpolar triangles of C can be identified with the set of trisecant planes of the Veronese surface which contain C. The latter will also include degenerate self-polar triangles corresponding to the case when the trisecant plane is tangent to the Verones surface at some of its points of intersections. Projecting from C to P4 we will identify the set of self-polar triangles (maybe degenerate) with the set of trisecant lines of the projected ¯ Veronese surface X. This is a closed subvariety of the Grassmann variety G(1, P4 ) of 4 lines in P . Let E be a linear space of odd dimension 2k + 1 and let G(2, E) := G(1, P(E)) be the Grassmannian of lines in P(E). Consider its Pl¨ cker embedding G(2, E) → u P(Λ2 E). Any nonzero ω ∈ Λ2 E ∗ = Λ2 E ∗ defines a hyperplane Hω in P(Λ2 E). Consider ω as a linear map αω : E → E ∗ defined by αω (v)(w) = ω(v, w). The map αω is skew-symmetric in the sense that its transpose map coincides with −αω . Thus its determinant is equal to zero, and Ker(αω ) = {0}. Let v0 be a nonzero element of the kernel. Then for any v ∈ E we have ω(v0 , v) = αω (v)(v0 ) = 0. This shows that ω vanishes on all decomposable 2-vectors v0 ∧ v. This implies that the intersection of the hyperplane Hω with G(2, E) contains all lines which intersect the linear subspace Cω = P(Ker(αω )) ⊂ P(E) which we call the pole of the hyperplane Hω . Now recall the following result from linear algebra (see Exercise 2.1). Let A be a skew-symmetric matrix of odd size 2k + 1. Its principal submatrices Ai of size 2k (obtained by deleting the i-th row and the i-th column) are skew-symmetric matrices of even size. Let Pfi be the pfaffian s of Ai (i.e. det(Ai ) = Pf2 ). Assume that i rank(A) = 2k, or, equivalently, not all Pfi vanish. Then the system of linear equations A · x = 0 has one-dimensional null-space generated by the vector (a1 , . . . , a2k+1 ), where ai = (−1)i+1 Pfi . Let us go back to Grassmannians. Suppose we have an s + 1-dimensional subspace W in Λ2 E ∗ spanned by ω0 , . . . , ωs . Suppose for any ω ∈ W we have rank αω = 2k, or equivalently, the pole Cω of Hω is a point. It follows from the theory of determinant varieties that the subvariety {Cω ∈ P(Λ2 E ∗ ) : corank αω ≥ i}
i is of codimension 2 in P(Λ2 E ∗ ) [133], [157]. Thus, if s < 4, a general W will satisfy the assumption. Consider a regular map Φ : P(W ) → P(E) defined by ω → Cω . If we take ω = t0 ω0 + · · · + ts ωs so that t = (t0 , . . . , ts ) are projective coordinates in P(W ), we obtain that Φ is given by 2k + 1 principal pfaffian s of the matrix At defining ω. We shall apply the preceeding to the case dim E = 5. Take a general 3-dimensional subspace W of Λ2 E ∗ . The map Φ : P(W ) → P(E) ∼ P4 is defined by homogeneous = polynomials of degree 2. Its image is a projected Veronese surface S. Any trisecant

2.1. SELF-POLAR TRIANGLES

45

line of S passes through 3 points on S which are the poles of elements w1 , w2 , w3 from W . These elements are linearly independent since otherwise their poles lie on the conic image of a line under Φ. But no trisecant line can be contained in a conic plane section of S. We consider ω ∈ W as a hyperplane in the Pl¨ cker space P(Λ2 E). Thus any u trisecant line is contained in all hyperplanes defined by W . Now we are ready to prove the following. Theorem 2.1.4. Let X be the closure in G(1, P4 ) of the locus of trisecant lines of a projected Veronese surface. Then X is equal to the intersection of G(1, P4 ) with three linear independent hyperplanes. In particular, X is a Fano 3-fold of degree 5 with canonical sheaf ωX ∼ OX (−2). = Proof. We have already shown that the locus of poles of a general 3-dimensional linear W space of hyperplanes in the Pl¨ cker space is a projected Veronese surface S and its u tri-secant variety is contained in Y = ∩w∈W Hw ∩ G(1, P4 ). So, its closure X is also contained in Y . On the other hand, we know that X is irreducible and 3-dimensional (it contains an open subset isomorphic to a homogeneous space SO(3)/S4 ). By Bertini’s Theorem the intersection of G(1, P4 ) with a general linear space of codimension 3 is an irreducible 3-dimensional variety. This proves that Y = X. By another Bertini’s theorem, Y is smooth. The rest is the standard computation of the canonical class of the Grassmann variety and the adjunction formula. It is known that the canonical class of the Grassmannian G = G(m, Pn ) of m-dimensional subspaces of Pn is equal to KG = OG (−n − 1) (2.7)

(see Exercise 3.2). By the adjunction formula, the canonical class of X = G(1, P4 ) ∩ H1 ∩ H2 ∩ H3 is equal to OX (−2). Corollary 2.1.5. The homogeneous space X = SO(3)/S4 admits a smooth compact¯ ification X isomorphic to the intersection of G(1, P4 ), embedded via Pl¨ cker in P9 , u ¯ with a linear subspace of codimension 3. The boundary X \ X is an anticanonical divisor cut out by a hypersurface of degree 2. Proof. The only unproven assertion is one about the boundary. We use that the 3dimensional group G = SL2 acts transitively on a 3-dimensional variety X minus the boundary. For any point x ∈ X consider the map µx : G → X, g → g · x. Its fibre over the point x is the isotropy subgroup Gx of x. The differential of this map defines a linear map g = T (G)e → T (X)x . When we let x vary in X, we get a map of vector bundles φ : gX = g × X → T (X). Now take the determinant of this map
∗ Λ3 φ = Λ3 g × X → Λ3 T (X) = KX ,

where KX is the canonical line bundle of X. The left-hand side is the trivial line bundle over X. The map Λ3 (φ) defines a section of the anticanonical line bundle. The zeroes of this section are the points where the differential of the map µx is not injective, i.e., where dim Gx > 0. But this is exactly the boundary of X. In fact, the

46

CHAPTER 2. CONICS

boundary consists of orbits of dimension smaller than 3, hence the isotropy of each such orbit is of positive dimension. This shows that the boundary is contained in our anticanonical divisor. Obviously, the latter is contained in the boundary. Thus we see that the boundary is equal to the intersection of G(1, P4 ) with a quadric hypersurface.

¯ Remark 2.1.1. There is another construction of the variety X of self-polar triangles due to S. Mukai and H. Umemura [177]. Let V6 be the space of homogeneous binary forms f (t0 , t1 ) of degree 6. The group SL2 has a natural linear representation in V6 via linear change of variables. Let f = t0 t1 (t4 − t4 ). The zeroes of this polynomials are 0 1 the vertices of a regular octahedron inscribed in S 2 = P1 . The stabilizer subgroup of f in SL2 is isomorphic to the binary octahedron group Γ ∼ S4 . Consider the projective = linear representation of SL2 in P(V6 ) ∼ P5 . In the loc. cit. it is proven that the closure = ¯ ¯ X of this orbit in P(V6 ) is smooth and B = X \ X is the union of two orbits Kt5 t1 0 6 and Kt0 . The first orbit is of dimension 2. The isotropy subgroup of the first orbit is isomorphic to the multiplicative group K ∗ . The second orbit is one-dimensional and is contained in the closure of the first one. The isotropy subgroup is isomorphic to the subgroup of upper triangular matrices. They also show that B is equal to the image of P1 × P1 under a SL2 -equivariant map given by a linear system of curves of bi-degree (5, 1). Thus B is of degree 10, hence is cut out by a quadric. The image of the second orbit is a smooth rational rational curve in B and is equal to the singular locus of B. The fact that the two varieties are isomorphic follows from the theory of Fano 3-folds. 1 It can be shown that there is a unique Fano threefold V with Pic(V ) = Z 2 KV and 3 KV = 40. We will discuss this variety in a later chapter.

2.1.4

Conjugate triangles

Let C = V (f ) be a nonsingular conic, be a line in P2 , and p be its pole with respect to C. From the point view of linear algebra, the one-dimensional subspace defining p is orthogonal to the two-dimensional subspace defining with respect to the symmetric bilinear form defined by f . Given a triangle with sides 1 , 2 , 3 , the poles of the sides are the vertices of the triangle which is called the conjugate triangle. Its sides are the polar lines of the vertices of the original triangle. It is clear that this defines a duality in the set of triangles. Clearly, a triangle is self-conjugate if and only if it is a self-polar triangle. Let 1 , 2 , 3 be three tangents to C at the points p1 , p2 , p3 , respectively. They form a triangle which can be viewed as a circumscribed triangle. It follows from Theorem 1.1.1 that the conjugate triangle has vertices p1 , p2 , p3 . It can be viewed as an inscribed triangle. The lines 1 = p2 , p3 , 1 = p2 , p3 , 1 = p2 , p3 are polar lines with respect to the points q1 , q2 , q3 , respectively. Two lines in P2 are called conjugate with respect to C if the pole of one of the lines belongs to the other line. It is a reflexive relation on the set of lines. Obvioulsy, two triangles are conjugate if and only if each of the sides of the first triangle is conjugate to a side of the second triangle.

2.2. PONCELET RELATION

47

Now let us consider the following problem. Given two triangles without common sides, find a conic C such that the triangles are conjugate to each other with respect to the conic C. Assume that the first triangle is formed by the coordinate lines ti = 0. Using equations (2.4) it is easy to get a necessary and sufficient condition for this to be true. Let A be the 3 × 3 matrix whose rows are the coefficients of the linear equations defining the sides of the second triangle. Then the two triangles are conjugate to each other if and only if there exists an invertible diagonal matrix D such that the matrix D · A is symmetric. Or, equivalently, D · A · D−1 = At . (2.8)

It is easy to see that this gives one condition on the set of triangles. Remark 2.1.2. Consider a triangle (with an order on the set of sides) as a point in (P2 )3 . Then the ordered pairs of conjugate triangles with respect to some conic is a hypersurface in an open subset of (P2 )6 . Is it possible to find the equation of the closure X of this set? By symmetry, X is cut out by a hypersurface in the Segre embedding of the product. Remark 2.1.3. Let X = (P2 )[3] be the Hilbert scheme of P2 of 0-cycles of degree 3. It is a minimal resolution of singularities of the 3d symmetric product of P2 . Consider the open subset of X formed by unordered sets of 3 non-collinear points. We may view a point of U as a triangle. Thus any nonsingular conic C defines an automorphism gC of U of order 2. Its set of fixed points is equal to the variety of self-polar triangles of C. The automorphism of U can be viewed as a birational automorphism of X. The complement of U is a divisor. It is given by the determinant of the matrix formed by the coordinates of three general points. Since X is nonsingular, one can extend gC to an open subset of codimension ≥ 2. What is this set and how gC acts on it? The variety X is rational, i.e. birationally isomorphic to P6 (see Exercise 3.1). One can also ask to describe the group of birational automorphisms of P6 generated by the involutions gC .

2.2
2.2.1

Poncelet relation
Darboux’s theorem

Let C be a conic, and let T = { 1 , 2 , 3 } be a circumscribed triangle. A conic C which has T as an inscribed triangle is called the Poncelet related conic. Since passing through a point impose one condition, we have ∞2 Poncelet related conics corresponding to a fixed triangle T . Varying T we expect to get ∞5 conics, so that any conic is Poncelet related to C with respect to some triangle. But surprisingly this is wrong! A theorem of Darboux asserts that there is a pencil of divisors p1 + p2 + p3 such that the triangles T with sides tangent to C at the points p1 , p2 , p3 define the same Poncelet related conic. We shall prove it here. In fact, for the future use we shall prove a more general result. Instead of circumscribed triangles we shall consider circumscribed n-polygons. An n-polygon P in P2 is an ordered set of n ≥ 3 points (p1 , . . . , pn ) in P2 such that no

48

CHAPTER 2. CONICS

three points pi , pi+1 , pi+2 are colinear. The points pi are the vertices of P , the lines pi , pi+1 are called the sides of P (here pn+1 = p1 ). We say that two polygons are equal if the sets of their sides are equal. The number of n-polygons with the same set of vertices is equal to n!/2n = (n − 1)!/2. We say that P circumscribes a nonsingular conic C if each side is tangent to C. Given any ordered set (q1 , . . . , qn ) of n points on C, let i be the tangent lines to C at the points qi . Then they are the sides of the n-polygon P with vertices pi = i ∩ i+1 , i = 1, . . . , n ( n+1 = 1 ). This polygon circumscribes C. This gives a one-to-one correspondence between n-polygons circumscribing C and ordered sets of n points on C. Let P = (p1 , . . . , pn ) be an n-polygon that circumscribes a nonsingular conic C. A conic S is called Poncelet n-related to C with respect to P if all points pi lie on C. Let us start with any two conics C and S. We choose a point p1 on S and a tangent 1 to C passing through p1 . It intersects S at another point p2 . We repeat this construction. If the process stops after n steps (i.e. we are not getting new points pi ), we get an inscribed n-polygon in S which circumscribes C. In this case S is Poncelet related to C. The Darboux Theorem which will prove later says that if the process stops, then we can construct infinitely many n-polygons with this property starting from an arbitrary point on S. Consider the following correspondence on C × S: R = {(x, y) ∈ C × S : x, y is tangent to C at x}. Since, for any x ∈ C the tangent to C at x intersects S at two points, and, for any y ∈ S there are two tangents to C passing through y, we get that E is of bi-degree (2, 2). This means if we identify C, S with P1 , then R is a curve of bi-degree (2, 2). As is well-known R is a curve of arithmetic genus 1. Lemma 2.2.1. The curve R is nonsingular if and only if the conics C and S intersect at four distinct points. In this case, R is isomorphic to the double cover of C (or S) ramified over the four intersection points. Proof. Consider the projection map πS : R → S. This is a map of degree 2. A branch point y ∈ S is a point such that there only one tangent to C passing through y. Obvioulsy, this is possible only if y ∈ C. It is easy to see that R is nonsingular if and only if the double cover πS : R → S ∼ P1 has four branch points. This proves the = assertion. Note that the second projection map πC : R → C must also have 4 branch points, if R is nonsingular. A point x ∈ C is a branch point if and only if the tangent of C at x is tangent to S. So we obtain that two conics intersect transversally if and only if there are four different common tangents. Take a point (x[0], y[0]) ∈ R and let (x[1], y[1]) ∈ R be defined as follows: y[1] is the second point on S on the tangent to x[0], x[1] = x[0] is the point where the tangent of C at [x[1] contains y[1]. This defines a self-map τC,S : R → R. This map has no fixed points on R and hence, if we fix a group law on R, is a translation map ta with respect to a point a. Obviously, we get an n-polygon if and only if ta is of order

2.2. PONCELET RELATION

49

n, i.e. the order of a in the group law is n. As soon as this happens we can use the automorphism for constructing n-polygons starting from an arbitrary point (x[0], y[0]). This is the Darboux Theorem which we have mentioned in above. Theorem 2.2.2. (G. Darboux) Let C and S be two nondegenerate conics intersecting transversally. Then C and S are Poncelet n-related if and only if the automorphism τC,S of the associated elliptic curve R is of order n. If C and S are Poncelet n related, then starting from any point x ∈ C and any point y ∈ S there exists an n-polygon with a vertex at y and one side tangent to C at y which circumscribes C and inscribed in S. In order to give a more explicit answer when two conics are Poncelte related one needs to recognize when the automorphism τC,S is of finite order. Let us choose projective coordinates such that C is the Veronese conic t0 t2 − t2 = 0, the image of P1 1 under the map [t0 , t1 ] → [t2 , t0 t1 , t2 ]. By using a projective transformation leaving 0 2 C invariant we may assume that the four intersection points p1 , . . . , p4 of C and S correspond to the points [1, 0], [1, 1], [0, 1], [1, a] ∈ P1 , where a = 0, 1. Then R is isomorphic the elliptic curve given by the affine equation y 2 = x(x − 1)(x − a). The conic S belongs to the pencil of conics with base points p1 , . . . , p4 : (t0 t2 − t2 ) + λt1 (at0 − (1 + a)t1 + t2 ) = 0. 1 We choose the zero point in the group law on R to be the point (x[0], y[0]) = (p4 , p4 ) ∈ C × S. Then the automorphism τC,S sends this point to (x[1], y[1]), where y[1] = (λa, λ(1 + a) + 1, 0), x[1] = ((a + 1)2 λ2 , 2a(1 + a)λ, 4a2 ).

2a Thus x[1] is the image of the point (1, (a+1)λ ) ∈ P1 under the Veronese map. The point y[1] corresponds to one of the two roots of the equation

y2 =

2a 2a 2a ( − 1)( − a). (a + 1)λ (a + 1)λ (a + 1)λ

So we need a criterion characterizing points (x, ± x(x − 1)(x − a) of finite order. Note that different choice of the sign corresponds to the inversion involution on the elliptic curve. So, the order of the points corresponding to two different choices of the sign are the same. We have the following: Theorem 2.2.3. (A. Cayley). Let R be an elliptic curve with affine equation y 2 = g(x), where g(x) is a cubic polynomial with three distinct nonzero roots. Write


y=
k=0

ak xk .

50 Then a point (0, g(0)) is of order n if and only if a2 a3 . . . am+1 a3 a4 . . . am+1 a3 a4 . . . am+2 a4 a5 . . . am+2 ... ... . . . ... ... ... . . . ... am+1 am+2 a2m am+1 am+2 a2m

CHAPTER 2. CONICS

=

0,

n = 2m + 1,

=

0,

n = 2m.

Proof. We fix a square root c0 of g(0) and consider the point p = (0, c0 ). A necessary and sufficient condition for p to be a n-torsion point is that there exists a rational function f on R with a zero of order n at p and a pole of order n at the infinity point (∞, 0). We shall assume that n = 2k − 1 is odd. The other case is considered similarly. Since f is regular on the affine part, it must be a restriction of a polynomial f (x, y) of some degree d. Since the infinity is an inflection point, the degree of f must be equal to k − 1 and f (x, y) must have a zero of order 2k − 1 at (0, c0 ) and a pole of order k − 2 at ∞ infinity. Now we expand y = k=0 ak xk and put ym = a0 + a1 x + · · · + ak−1 xk−1 . We have y − yk x(y − yk−1 ) ... k−2 x (y − y2 ) = = = = ak xk + · · · + a2k−2 x2k−2 + . . . ak−1 xk + · · · + a2k−3 x2k−2 + . . . ... a2 xk + · · · + ak x2k−2 + . . . .

We can find n − 1 coefficients c0 , c1 , . . . , ck−2 such that the polynomial f (x, y) = c0 (y − yk ) + c1 x(y − yk−1 ) + · · · + ck−2 xk−2 (y − y2 ) vanishes at x = 0 of order 2k − 1 if and only if ak ak+1 ... a2k−2 ak−1 ak ... a2k−3 . . . a2 . . . a3 = 0. ... ... . . . ak

It is easy to see that this determinant is equal to one of the determinants from the assertion of the theorem. To apply the proposition we have to take α= 2a , (a + 1)λ β =1+ 2a , (a + 1)λ γ =a+ 2a . (a + 1)λ

2.2. PONCELET RELATION

51

Let us consider the variety Pn of pairs of conics (C, S) such that S is Poncelet n-related to C. We assume that C and S intersect transversally. We already know that Pn is a hypersurface in P5 × P5 . Obviously Pn is invariant with respect to the diagonal action of the group SL3 (acting on the space of conics). Thus the equation of Pn is an invariant of a pair of conics. This invariant was computed by F. Gerbradi [122]. It is of 1 bi-degree ( 4 T (n), 1 T (n)), where T (n) is equal to the number of elements of order n 2 in the abelian group (Z/nZ)2 . Let us look at the quotient of Pn by PSL3 . Consider the rational map β : P5 ×P5 → 2 (4) (P ) which assigns to (C, S) the point set C ∩ S. The fibre of β over a subset B of 4 points in general linear position is isomorphic to an open subset of P1 × P1 , where P1 is the pencil of conics with base point B. Since we can always transform such B to the set of points {[1, 0, 0], [0, 1, 0], [0, 0, 1], [1, 1, 1]}, the group PSL3 acts transitively on the open subset of such 4-point sets. Its stabilizer is isomorphic to the permutation group S4 generated by the following matrices:  0 −1 1 0 0 0  0 0 , 1  1 0 0  0 0 0 −1 , 1 0   1 0 −1 0 −1 −1 . 0 0 −1

The orbit space Pn /PSL3 is isomorphic to a curve in an open subset of P1 × P1 /S4 , where S4 acts diagonally. By considering one of the projection maps, we obtain that Pn /PSL3 is an open subset of a cover of P1 of degree N equal to the number of Poncelet n-related conics in a given pencil of conics with 4 distinct base points with respect to a fixed conic from the pencil. This number was computed by F. Gerbardi 1 [122] and is equal to 2 T (n). A modern account of Gerbardi’s result is given in [11]. A smooth compactification of Pn /PSL3 is the modular curve X 0 (n) which parametrizes the isomorphism classes of the pairs (R, e), where R is an elliptic curve and e is a point of order n in R. Proposition 2.2.4. Let C and S be two nonsingular conics. Consider each n-polygon inscribed in C as a subset of its vertices, and also as a positive divisor of degree n on C. The closure of the set of n-polygons inscribed in C and circumscribing S is either 1 empty, or a gn , i.e. a linear system of divisors of degree n. Proof. First observe that two polygons inscribed in C and circumscribing S which share a common vertex must coincide. In fact, the two sides passing through the vertex in each polygon must be the two tangents of S passing through the vertex. They intersect C at another two common vertices. Continuing in this way we see that the two polygons have the same set of vertices. Now consider the Veronese embedding vn of C ∼ P1 in Pn . An effective divisor of degree n is a plane section of the Veronese curve = Ver1 = vn (P1 ). Thus the set of effective divisors of degree n on C can be identified n ˇ ˇ with the dual projective space Pn . A hyperplane in Pn is the set of hyperplanes in Pn n ˇ which pass through a fixed point in P . The degree of an irreducible curve X ⊂ Pn of divisors is equal to the cardinality of the set of divisors containing a fixed general point of Ver1 . In our case it is equal to 1. n

52

CHAPTER 2. CONICS

2.2.2

Poncelet curves

Let C and S be two Poncelet n-related conics in the plane P2 = P(E). Recall that this means that there exist n points p1 , . . . , pn on C such that the tangent lines i = Tpi (C) meet on S. One can drop the condition that S is a conic. We say that a plane curve S of degree n − 1 is Poncelet related to the conic C if there exist n points as above such that the tangents to C at these points meet on S. Before we prove an analog of Darboux’s Theorem for Poncelet related curves of higher degree we have to relate these curves to curves of jumping lines of some special rank 2 vector bundles on the projective plane, so called the Schwarzenberger bundles. Let us write P1 = P(U ) for some vector space of dimension 2. Recall that there are natural bijections between the following sets • closed embeddings P1 → P2 ; • nonsingular conics in P2 ; • nonsingular conics in the dual P2 ; • isomorphisms P2 ∼ P(S 2 U ); = • isomorphism between P2 and P(U )(2) ; Fix an isomorphism P2 ∼ P(S 2 U ) defined by a choice of a conic C in P2 . Consider = the multiplication map S 2 U ⊗ S n−2 (U ) → S n U . It defines a rank 2 vector bundle Sn,C on P2 whose fibre at the point x = [q] ∈ P(S 2 U ) is equal to the quotient space S n (U )/qS n−2 (U ). One easily see that it admits a resolution of the form 0 → S n−2 (U )(−1) → S n U → Sn,C → 0, (2.9)

where we identify a vector space E with the vector bundle π ∗ E, where π is the structure map to the point. The vector bundle Sn,C is called the Schwarzenberger bundle associated to the conic C. Its dual bundle has the fibre over a point x = [q] equal to the dual space (S n U/qS n−2 U )∗ = {f ∈ S n U ∗ : Dq (f ) = 0} = Ker(apq ). (2.10)

If we embed the dual projective line P(U ∗ ) in P(S n U ∗ ) by means of the Veronese map, then the divisor of zeroes of q can be considered as a divisor V (q) of degree 2 on the Veronese curve Rn ⊂ P(S n U ∗ ), or, equivalently, as a 1-secant of Rn . A hyperplane containing this divisor is equal to V (qg) for some g ∈ S n−2 U . Thus the space P(Ker(apq )) can be identified with the ∗ projective span of V (q). In other words, the fibres of the dual projective bundle Sn,C are equal to the secants of the Veronese curve Rn . It follows from (2.9) that the vector bundle Sn,C has the first Chern class of degree n − 1 and the second Chern class is equal to n(n − 1)/2. Thus we expect that a general section of Sn,C has n(n − 1)/2 zeroes. We identify the space of sections of Sn,C with the vector space S n U . A point [s] ∈ P(S n U ) can be viewed as a hyperplane Hs in

2.2. PONCELET RELATION

53

P(S n U ∗ ). Its zeroes are the secants of Rn contained in Hs . Since Hs intersects Rn at n-points p1 , . . . , pn , any secant pi , pj is a secant contained in Hs . The number of such secants is equal to n(n − 1)/2. Recall that we can identify the conic with P(U ) by means of the Veronese map ν2 : P(U ) → P(S 2 U ). Similarly, the dual conic C ∗ is identified with P(U ∗ ). By using the Veronese map νn : P(U ∗ ) → P(S n U ∗ ) we can identify C ∗ with Rn . Now a point on Rn is a tangent line on the original conic C, hence n points p1 , . . . , pn from above are the sides i of an n-polygon circumscribing C. A secant pi , pj from above is a point in P2 equal to the intersection point qij = i ∩ j . And the n(n − 1)/2 points qij represent the zeroes of a section s of the Schwarzenberger bundle Sn,C . For any two linearly independent sections s1 , s2 , their determinant s1 ∧ s2 is a section of Λ2 Sn,C and hence its divisor of zeroes belongs to |OP2 (n − 1)|. When we consider the pencil s1 , s2 spanned by the two sections, the determinant of each member s = λs1 + µs2 has the zeroes on the same curve V (s1 ∧ s2 ) of degree m − 1. Let us summarize this discussion by stating and the proving the following generalization of Darboux’s Theorem. Theorem 2.2.5. Let C be a nonsingular conic in P2 and Sn,C be the associated Scwarzenberger rank 2 vector bundle over P2 . Then n-polygons circumscribing C are parameterized by P(Γ(Sn,C )). The vertices of the polygon Πs defined by a section s correspond to the subscheme Z(s) of zeroes of the section s. A curve of degree n − 1 passing through the vertices corresponds to a pencil of a sections of Sn,C containing s and is equal to the determinant of a basis of the pemcil. Proof. A section s with the subscheme of zeroes Z(s) with ideal sheaf IZ(s) defines the exact sequence s 0 → OP2 → Sn,C → IZ (n − 1) → 0. A section of IZ (n − 1) is a plane curve of degree n − 1 passing through Z(s). The image of a section t of Sn,C in Γ(IZ (n − 1)) is the discriminant curve s ∧ t. Any curve defined by an element from Γ(IZ (n − 1)) passes through the vertices of the n-polygon Πs and is uniquely determined by a pencil of sections containing s. One can explicitly write the equation of a Poncelet curve as follows. First we d d−1 d choose a basis ξ0 , ξ1 of the space U and the corresponding basis (ξ0 , ξ0 ξ1 , . . . , ξ1 ) d d−i i d n ∗ of the space S U . The dual basis in S U is ( i t0 t1 )0≤i≤d . Now the coordinates in the plane P(S 2 U ) are t2 , 2t0 t1 , t2 , so a point in the plane is a binary conic Q = 0 2 2 2 aξ0 + 2bξ0 ξ1 + cξ1 . For a fixed x = [Q] ∈ P(S 2 U ), the matrix of the multiplication map S n−2 U → S n U, G → QG is   a 2b a      ..  c 2b  .     .. ..   . . c K(x) =     .. ..  . . a     ..  . 2b c

54
n

CHAPTER 2. CONICS

n−i i A section of Sn,C is given by f = i=0 ci ξ0 ξ1 ∈ S n U . Its zeroes is the set of points x such that the vector c of the coefficients belongs to the column subspace of the matrix K(x). Now we vary f in a pencil of binary forms whose coefficient vector c belongs to the nullspace of some matrix A of size (n − 1) × (n + 1) and rank n − 1. The determinant of this pencil of sections is the curve in the plane defined by the degree n − 1 polynomial equation in x = [a, b, c]

det K(x) · A = 0. Note that the conic C corresponding to our choice of coordinates is V (t2 − t0 t2 ). 1 Remark 2.2.1. Recall that a section of Sn,C defines a n-polygon in the plane P(S 2 U ) corresponding to the hyperplane section Hs ∩ Rn . Its vertices is the scheme of zeroes Z(s) of the section s. Let π : X(s) → P2 be the blow-up of Z(s). For a general s, the linear system of Poncelet curves through Z(s) embeds the surface X(s) in P(S n U ∗ ) with the image equal to Hs ∩ Sec1 (Rn ). The exceptional curves of the blow-up are mapped onto the secants of Rn which are contained in Hs . These are the secants pi , pj , where Hs ∩ Rn = {p1 , . . . , pn }. The linear system defining the embedding is the proper transform of the linear system of curves of degree n − 1 passing through 1 2 n(n − 1) points of Z(s). This implies that the embedded surface X(s) has the degree equal to (n − 1)2 − 1 n(n − 1) = 1 (n − 1)(n − 2). This is also the degree of the secant 2 2 variety Sec1 (Rn ) . For example, take n = 4 to get that the secant variety of R4 is a cubic hypersurface in P4 whose hyperplane sections are cubic surfaces isomorphic to the blow-up of the six vertices of a complete quadrilateral.

2.2.3

Invariants of pairs of conics

The Poncelet theorem is an example of a porism which can be loosely stated as follows. If one can find one object satisfying a certain special property then there are infinitely many such objects. In case of Darboux’s theorem this is the property of the existence of a polygon inscribed in one conic and circumscribing the other conic. Here we consider another example of a porism between two conics. This time the relation is the following. Given two nonsingular conics C and S there exists a self-conjugate triangle with respect to C which is inscribed in S. We say that the two conics are conjugate or apolar. Proposition 2.2.6. Let S and C be two nonsingular conics defined by symmetric matrices A and B respectively. Then C admits a self-conjugate triangle which is inscribed in S if and only if Tr(AB −1 ) = 0. Moreover, if this condition is satisfied, for any point x ∈ S \ (S ∩ C) there exists a self-conjugate triangle inscribed in S with vertex at x. Proof. Let Q be an invertible 3 × 3 matrix. Replacing A with A = QT AQ and B with B = QT BQ we check that Tr(A B
−1

) = Tr(QT AB −1 (QT )−1 ) = Tr(AB −1 ).

2.2. PONCELET RELATION

55

This shows that the trace condition is invariant with respect to a linear change of variables. Thus we may assume that C = V (t2 +t2 +t2 ). Suppose there is a self-conjugate 0 1 2 triangle with respect to C which is inscribed in S. Since the orthogonal group of C acts transitively on the set of self-conjugate triangles, we may assume that the triangle is the coordinate triangle. Then the points [1, 0, 0], [0, 1, 0], and [0, 0, 1] must be on S. Hence S = V (at0 t1 + bt0 t2 + ct1 t2 ), and the condition Tr(AB −1 ) is verified. Let us show the sufficiency of the trace condition. Choose coordinates as above. Let S = V (at2 + bt2 + ct2 + 2dt0 t1 + 2et0 t2 + 2f t1 t2 ). (2.11) 0 1 2 The trace condition is a + b + c = 0. Let x = [x0 , x1 , x2 ] be any point on S and = V (x0 t0 + x1 t1 + x2 t2 ) be the polar line Px (C). Without loss of generality, we may assume that x2 = −1 so that we can write t2 = x0 t0 + x1 t1 and take t0 , t1 as coordinates on . The line intersects S at two points [c0 , c1 ] and [d0 , d1 ] which are the zeroes of the binary form q = at2 + bt2 + c(x0 t0 + x1 t1 )2 + 2dt0 t1 + 2(et0 + f t1 )(x0 t0 + x1 t1 ) 0 1 = (a + 2ex0 + cx2 )t2 + (b + 2f x1 + cx2 )t2 + 2(d + ex1 + f x0 + cx0 x1 )t0 t1 . 0 0 1 1 The line intersects C at the points y = (a0 , a1 , x0 a0 + x1 a1 ), z = (b0 , b1 , b0 x0 + b1 x1 ). Their coordinates on the line are the zeroes of the binary form q = t2 + t2 + (x0 t0 + x1 t1 )2 = (1 + x2 )t2 + 2x0 x1 t0 t1 + (1 + x2 )t2 . 0 1 0 0 1 1 It follows from Proposition 2.1.1 that the points x, y, z are the vertices of a self-polar triangle if and only if (2.2) holds. To check this condition we will use that a + b + c = 0 and ax2 + bx2 + c + dx0 x1 − ex0 − f x1 = 0. We have 0 1 (a + 2ex0 + cx2 )(1 + x2 ) + (b + 2f x1 + cx2 )(1 + x2 ) − 2(d + ex1 + f x0 + cx0 x1 )x0 x1 0 1 1 0 = a + b + 2ex0 + cx2 + 2f x1 + cx2 + (a + 2ex0 + cx2 )x2 + (b + 2f x1 + cx2 )x2 0 1 0 1 1 0 −2(d + ex1 + f x0 + cx0 x1 )x0 x1 Replacing a + b with −c = −ax2 − bx2 − 2dx0 x1 + 2ex0 + 2f x1 we check that the 0 1 sum is equal to zero. Thus starting from any point x on S we find that the triangle with vertices x, y, z is self-conjugate with respect to C. Remark 2.2.2. Let P2 = P(E) and C = V (q), S = V (f ), where q, f ∈ S 2 E ∗ . Let ˇ C = V (ψ), where ψ ∈ S 2 (E). Then the trace condition from Proposition 2.2.6 is ψ, f = 0, (2.12)

where the pairing is the polarity pairing (1.2). In other words, C is conjugate to S if and only if the dual conic of S is apolar to C.

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Consider the set of self-polar triangles with respect to C inscribed in S. We know that this set is either empty or of dimension ≥ 1. We consider each triangle as a set of its 3 vertices, i.e. as an effective divisor of degree 3 on S. Proposition 2.2.7. The closure X of the set of self-polar triangles with respect to C 1 which are inscribed in S, if not empty, is a g3 , i.e. a linear system of divisors of degree 3. Proof. First we use that two self-polar triangles with respect to C and inscribed in S which share a common vertex must coincide. In fact, the polar line of the vertex must intersect S at the vertices of the triangle. Then the assertion is proved using the argument from the proof of Proposition 2.2.4.
1 Note that a general g3 contains 4 singular divisors corresponding to ramification points of the corresponding map P1 → P1 . In our case these divisors correspond to 4 intersection points of C and S. Another example of a poristic statement is the following.

Theorem 2.2.8. Let T and T be two different triangles. The following assertions are equivalent: (i) there exists a conic S containing the vertices of the two triangles; (ii) there exists a conic Σ touching the sides of the two triangles; (iii) there exists a conic C with respect to which each of the triangles is self-polar. Moreover, when one of the conditions is satisfied, there is an infinite number of triangles inscribed in S, circumscribed around Σ, and all of these triangles are self-polar with respect to C. Proof. (iii)⇔ (ii) Let [l1 ], [l2 ], [l3 ] and [m1 ], [m2 ], [m3 ] be the sides of the two triangles ˇ considered as points in the dual plane P2 . Consider the linear systems V = |OP2 (2) − ˇ [l1 ] − [l2 [−[l3 ]| and W = |OP2 (2) − [m1 ] − [m2 ] − [m3 ]| of conics passing through ˇ the corresponding points. Let C = V (f ). We can write
2 2 2 f = a1 l1 + a1 l2 + a3 l3 = b1 m2 + b2 m2 + b3 m2 1 2 3

for some scalars ai , bi . For any V (ψ) ∈ V ∪ W we have ψ, f = 0. This shows that the span of V and W in |OP2 (2)| is contained in a hyperplane orthogonal to f . ˇ Thus V ∩ W = ∅ and a common conic vanishes at all li ’s and mi ’s. Hence the dual conic Σ is touching the sides of the two triangles. Reversing the arguments, we find that condition (ii) implies that there exists a conic V (f ) such that ψ, f = 0 for any 2 2 V (ψ) ∈ V ∪ W . Since, for any V (ψ) ∈ V ∪ W , ψ, li = ψ, li = 0, we obtain 2 2 2 that f belongs to the linear span of l1 , l2 , l3 , and also to the linear span of m2 , m2 , m2 . 1 2 3 This proves the equivalence of (ii) and (iii). More details for this argument can be seen in the later chapter about the apolarity theory. (iii)⇔ (i) This follows from Proposition 2.1.2. Let us prove the last assertion. Suppose one of the conditions of the theorem is satisfied. Then we have the conics C, S, Σ with the asserted properties with respect to

2.2. PONCELET RELATION

57

the two triangles T, T . By Proposition 2.2.7, the set of self-polar triangles with respect 1 to C inscribed in S is a g3 . By Proposition 2.2.4, the set of triangles inscribed in S and 1 1 circumscribing Σ is also a g3 . Two g3 ’s with 2 common divisors coincide. Let C = V (f ) and S = V (g) be two conics (not necessary nonsingular). Consider the pencil V (t0 f + t1 g) of conics spanned by C and S. The zeroes of the discriminant equation D = discr(t0 f + t1 g) = 0 correspond to singular conics in the pencil. In coordinates, if f, g are defined by symmetric matrices A = (aij ), B = (bij ), respectively, then D = det(t0 A + t1 B) is a homogeneous polynomial of degree ≤ 3. Choosing different system of coordinates replaces A, B by QT AQ, QT BQ, where Q is an invertible matrix. This replaces D with det(Q)2 D. Thus the coefficients of D are invariants on the space of pairs of quadratic forms on C3 with respect to the action of the group SL3 . To compute D explicitly, we use the following formula for the determinant of the sum of two n × n matrices X + Y :
n

det(X + Y ) =
k=1 1≤i1 <...<ik ≤n

∆i1 ,...,ik ,

(2.13)

where ∆i1 ,...,ik is the determinant of the matrix obtained from X by replacing the columns Xi1 , . . . , Xik with the columns Yi1 , . . . , Yik . Applying this formula to our case, we get D = ∆t3 + Θt2 t1 + Θ t0 t2 + ∆ t3 , (2.14) 0 0 1 1 where (2.15) ∆ = det A Θ = det(A1 A2 B3 ) + det(A1 B2 A3 ) + det(B1 A2 A3 ) = Tr(B · adj(A)) Θ = det(B1 B2 A3 ) + det(B1 A2 B3 ) + det(A1 B2 B3 ) = Tr(A · adj(B)) ∆ = det(B) where adj means the adjugate matrix of complementary minors. We immediately recognize the geometric meanings of vanishing of the coefficients of D. The coefficient ∆ (resp. ∆ ) vanishes if and only if C (resp. S) is a singular conic. If ∆, ∆ are nonzero, then the coefficient Θ (resp. Θ ) vanishes if and only if there exists a self-polar triangle of C inscribed in S (resp. a self-polar triangle of S inscribed in C). This follows from Proposition 2.2.6. We can also express the condition that the two conics are Poncelet related. Theorem 2.2.9. Let C and S be two nonsingular conics. A triangle inscribed in C and circumscribing S exists if and only if Θ 2 − 4Θ∆ = 0. Proof. Choose a coordinate system such that C = V (t0 t1 + t1 t2 + t0 t2 ). Suppose there is a triangle inscribed in C and circumscribing S. Applying an orthogonal transformation, we may assume that the vertices of the triangle are the references points [1, 0, 0], [0, 1, 0] and [0, 0, 1]. Let S = V (g), where g = at2 + bt2 + ct2 + 2dt0 t1 + 2et0 t2 + 2f t1 t2 . 0 1 2 (2.16)

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The condition that the triangle circumscribes S is that the points [1, 0, 0], [0, 1, 0], and ˇ [0, 0, 1] lie on the dual conic S. This implies that the diagonal entries bc − f 2 , ac − 2 2 e , ab − d of the matrix adj(B) are equal to zero. Therefore we may assume that g = α2 t2 + β 2 t2 + γ 2 t2 − 2αβt0 t1 − 2αγt0 t2 − 2βγt1 t2 . 0 1 2 We get 0 1 Θ = Tr 1 0 1 1  2 α Θ = Tr −αβ −αγ    1 0 1 · 2αβγ 2 0 2αγβ 2 −αβ β2 −βγ  2αβγ 2 2αγβ 2 0 2βγα2  = 4αβγ(α + β + γ), 2 2βγα 0   −αγ −1 1 1 −βγ   1 −1 1  = −(α + β + γ)2 , γ2 1 1 −1 ∆ = −4(αβγ)2 . This checks that Θ 2 − 4Θ∆ = 0. Let is prove the sufficiency of the condition. Take a tangent line 1 to S intersecting C at two points x, y and consider tangent lines 2 , 3 to S passing through x and y, respectively. The triangle with sides 1 , 2 , 3 circumscribes S and has two vertices on C. Choose the coordinates such that this triangle is the coordinate triangle. Then, we may assume that C = V (at2 + 2t0 t1 + 2t1 t2 + 2t0 t2 ) and S = V (g), where g is as 0 in (7.31). Computing Θ 2 − 4Θ∆ we find that it is equal to zero if and only if a = 0. Thus the coordinate triangle is inscribed in C. Remark 2.2.3. Choose a coordinate system such that C = V (t2 + t2 + t2 ). Then the 0 1 2 condition that S is Poncelet related to C with respect to triangles is easily seen to be equal to c2 − c1 c3 = 0, 2 where det(A − tI3 ) = (−t)3 + c1 (−t)2 + c2 (−t) + c3 is the characteristic polynomial of a symmetric matrix A defining S. This is a quartic hypersurface in the space of conics. The polynomials c1 , c2 , c3 generate the algebra of invariants of S 2 (C3 )∗ with respect to the group SL3 . (2.17)

2.2.4

The Salmon conic

One call also look for covariants or contravariants of a pair conics, that is, rational maps |OP2 (2)| × |OP2 (2)| → |OP2 (d)| or |OP2 (2)| × |OP2 (2)| → |OP2 (d)|∗ which are defined geometrically, i.e. not depending on bases of the projective spaces involved. Recall the definition of the cross-ratio of four distinct ordered points pi = [ai , bi ] on P1 [p1 p2 ; p3 , p4 ] = (p1 − p2 )(p3 − p4 )/(p1 − p3 )(p2 − p4 ), (2.18) where pi − pj = ai bj − aj bi .

2.2. PONCELET RELATION

59

It is immediately checked that the cross-ratio does not take the values 0, 1, ∞. It does not depend on the choice of projective coordinates. It is also invariant under a permutation of the four points equal to the product of two commuting transpositions. The permutation (12) changes r to −r/(1 − r) and the permutation (23) changes r to 1/r. Two pairs of points (p1 , p2 ) and (q1 , q2 ) are called harmonic conjugate if the crossratio [p1 q1 ; p2 q2 ] is equal to −1. This definition does not depend on the order of points in each pair, and hence extends to the notion of two harmonic unordered pairs of points, or, equivalently, two binary forms in two variables. In the latter terms it agrees with the earlier definition (2.2). If we identify the projective space of binary forms of f degree 2 with the projective plane, the relation (2.2) can be viewed as a symmetric hypersurface H of bi-degree (1, 1) in P2 × P2 . In particular, it makes sense to speak about harmonic conjugate pairs of maybe coinciding points. We immediately check that a double point is harmonically conjugate to a pair of points if and only if it coincides with one of the roots of this form. The expression in the left-hand side of this formula is the invariant of a pair (f, g) of binary quadratic forms defined by taking the coefficient at t for the discriminant invariant of f + tg. It is analogous to the invariants Θ and Θ for a pair of conics. Now let C = V (f ) and S = V (g) be a pair of conics. Consider the pencil of conics C(λ, µ) = V (λf + µg). Write the equation of the dual conic C(λ, µ)∗ in the form Aλ2 + ψλµ + Bµ2 = 0. It is easy to see that V (A) = C ∗ and V (B) = S ∗ and V (ψ) is the conic in the dual space defined by the symmetric matrix whose ij-entry is equal to the coefficient at λµ in det(λaij + µbij ), where (aij ), (bij ) are the matrices defining the dual conics. Considering a pencil of lines as a P1 one can define a cross-ratio of 4 ordered lines in a pencil. Four lines in a pencil define a harmonic pencil if the first two lines are harmonic conjugate to the last two lines. An example of a harmonic pencil is the set of lines t1 = 0, t2 = 0, t1 − t2 = 0, t1 + t2 = 0.. A Salmon conic associated to a pair of conics C and S is defined to be the locus of points x in OP2 such that the pairs of the tangents through x to C and to S are harmonic conjugate. Note that it makes sense even x lies on one of the conics, we consider the corresponding tangent as the double tangent. Let us see that this locus is indeed a conic. The dual statement is that the locus of lines which intersect two conics at two pairs of harmonic conjugate pairs of points is a conic in the dual plane. We use the computations from the proof of Proposition 2.2.6. Without loss of generality we assume that C is given by the equation t2 +t2 +t2 = 0 and 0 1 2 another one is given by a full equation (2.11). We work in the open subset α2 = 0 to use t0 and t1 as the coordinates on . The condition that the line = V (α0 t0 +α1 t1 +α2 t2 ) intersects C and S at harmonic conjugate pairs of points is
2 2 2 2 2 2 2 2 (aα2 + 2eα2 α0 + cα0 )(α2 + α1 ) + (bα2 + 2f α2 α1 + cα1 )(α2 + α0 )− 2 2(dα2 + eα1 α2 + f α0 α2 + cα0 α1 )α0 α1 = 0. 2 It is easy to see that α2 factors out from the left-hand-side leaving us with the equation of a conic 2 2 2 (b + c)α0 + (a + c)α1 + (a + b)α2 + 2eα0 α1 + 2f α1 α2 − 2dα0 α1 = 0.

(2.19)

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This is the equation of the dual of the Salmon conic. The following is a remarkable property of the Salmon conic. Theorem 2.2.10. Let C and S be two conics such that the dual conics intersect at four distinct points representing the four common tangents of C and S. Then the eight tangency points lie on the Salmon conic associated with C and S. Proof. Let x be a point where the Salmon conic meets C. Then the tangent line through x to C represents a double line in the harmonic pencil formed by the four tangents through x to C and S. As we remarked before the conjugate pair of lines must contain . Thus is a common tangent to C and S and hence x is one of the eight tangency points. Conversely, the argument is reversible and shows that every tangency point lies on the Salmon conic. Remark 2.2.4. The Salmon conic F is obviously a covariant of a pair of conics C, S. Its dual conic F ∗ is a contravariant of C, S. The jacobian J(C, C , F ) defined by the determinant of the jacobian matrix of the three quadratic polynomials is an example of a covariant of degree 3. Similarly we get the contravariant of degree 3 equal to the jacobian of the dual conics. It is proven in [126] that any covariant of C, S is given by a polynomial in homogeneous forms defining C, S, F, J(C, S, F ). Similarly any contravariant of C, S is given by a polynomial in homogeneous forms defining C ∗ , S ∗ , F ∗ , J(C ∗ , S ∗ , F ∗ ).

Exercises
2.1 Let E be a vector space of even dimension n = 2k over a field K of characteristic 0 and P 2 ∗ (e1 , . . . , en ) be a basis in E. Let ω = i<j aij ei ∧ ej ∈ Λ E and A = (aij )1≤i≤j≤n be the skew-symmetric matrix defined by the coefficients aij . Let Λk (ω) = ω ∧ · · · ∧ ω = ak!e1 ∧ · · · ∧ en for some a ∈ F . The element a is called the pfaffian of A and is denoted by Pf(A). (i) Show that Pf(A) = X
S∈S

(S)

Y
(i,j)∈S

aij ,

where S is a set of pairs (i1 , j1 ), . . . , (ik , jk ) such that 1 ≤ is < js ≤ 2k, s = 1, . . . , k, {i1 , . . . , ik , j1 , . . . , jk } = {1, . . . , n}, S is the set of such sets S, (S) = 1 if the permutation (i1 , j1 , . . . , ik , jk ) is even and −1 otherwise. (ii) Compute Pf(A) when n = 2, 4, 6. (iii) Show that, for any invertible matrix C, Pf(t C · A · C) = det(C)Pf(A). (iv) Using (iii) prove that det(A) = Pf(A)2 . (iv) Show that Pf(A) =
n X (−1)i+j−1 Pf(Aij )aij , i=1

EXERCISES

61

where Aij is the matrix of order n − 2 obtained by deleting the ith and jth rows and columns of A. (v) Let B be a skew-symmetric matrix of odd order 2k − 1 and Bi be the matrix of order 2k − 2 obtained from B by deleting the ith row and ith column. Show that the vector (Pf(B1 ), . . . , (−1)i+1 Pf(Bi ), . . . , Pf(B2k−1 )) is a solution of the equation B · x = 0. (vi) Show that the rank of a skew-symmetric matrix A of any order n is equal to the largest m such that there exists i1 . . . < im such that the matrix Ai1 ...im obtained from A by deleting ij th rows and columns, j = 1, . . . , m, has nonzero pfaffian . 2.2 Let P be a trisecant plane in the space of conics to the Veronese variety of double lines. Consider it as a point in the Grassmannian G(1, P(S 2 E ∗ )) ∼ G(1, P4 ). Show that the plane = of hyperplanes through P , considered as a point in the dual Grassmannian G(1, P(S 2 E)), is a 2-dimensional linear system of conics in the dual plane P(E ∗ ) with 3 base points corresponding to the double lines in P . 2.3 Let V = ν2 (P2 ) be a Veronese surface in P5 . (i) Show that a general 3-dimensional subspace L intersects V at 4 points.
∗ (ii) Let P be a plane in P5 and LP be the 2-dimensional linear system (a net) of conics ν2 (H) in P2 , where H is a hyperplane in P5 containing P . Show that P is a trisecant plane if and only the set of base points of LP consists of 3 points (counting with multiplicities). Conversely, the linear system of conics through 3 points defines a unique trisecant plane.

(iii) Show that the set of nets of conics with three base points (a subvariety of the Grassmannian of 2-planes in the space of conics) contains an irreducible divisor parametrizing nets with 3 distinct collinear points and an irreducible divisor parametrizing nets with 2 base points, one of them is infinitely near. (iv) Using (iii) show that the anticanonical divisor of degenerate triangles is irreducible. (v) Show that the trisecant planes intersecting the Veronese plane at one point (corresponding to net of conics with one base point of multiplicity 3) define a smooth rational curve in the boundary of the variety of self-polar triangles. Show that this curve is equal to the set of singular points of the boundary. 2.4 Let U ⊂ (P2 )(3) be the subset of the symmetric product of P2 parametrizing the sets of three distinct points. For each set Z ∈ U let LZ be the linear system of conics containing Z. Consider the map f : U → G(1, P4 ), Z → LZ ∈ |OP2 (2)|. (i) Consider the divisor D in U parametrizing sets of 3 distinct collinear points. Show that f (D) is a closed subvariety of G(1, P4 ) isomorphic to P2 . (ii) Show that the map f extends to the Hilbert scheme (P2 )[3] of 0-cycles Z with h0 (OZ ) = 3 (which admits a natural map π : (P2 )[3] → (P2 )(3) which is a resolution of singularities). ¯ (iii) Show that the closure D of π −1 (D) in the Hilbert scheme is isomorphic to a P3 -bundle ¯ over P2 and the restriction of f to D is the projection map to its base. (iv) Let P(S) be the projectivization of the tautological rank 3 vector bundle over the Grassmannian G(1, P4 ) and p : P → (P2 )[3] be its pull-back under the map f . Show that its fibre over a point Z is the linear system of conics with base scheme equal to Z. ˜ (v) Define the map f : P → |OP2 (2)| which assigns to a point in the fibre p−1 (Z) the corre˜ sponding conic in the net of conics though Z. Show that the fibre of f over a nonsingular conic C is isomorphic to the Fano variety of self-polar triangles of the dual conic C ∗ .

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˜ (vi) Let P s = f −1 (D2 (2)) be the pre-image of the hypersurface of singular conics. Describe ˜ the fibres of the projections p : P s → (P2 )[3] and f : P s → D2 (2). 2.5 Prove that the nth symmetric product of Pn is a rational variety. 2.6 Two points x, y are called conjugate with respect to a nonsingular conic C if the line x, y intersects C at two points which are harmonic conjugate to x, y. Prove that x and y are conjugate if and only if y ∈ Px (C) and x ∈ Py (C). 2.7 Prove that two unordered pairs {a, b}, {c, d} of points in P1 are harmonic conjugate if and only if there is an involution of P1 with fixed points a, b that switches c and d. 2.8 Prove the following Hesse theorem. If two pairs of opposite vertices of a quadrilateral are each conjugate for a conic, then the third pair is also conjugate. Such a quadrilateral is called a Hesse quadrilateral. Show that four lines form a polar quadrilateral for a conic if and only if it is a Hesse quadrilateral. 2.9 Show that any polar triangle of a conic can be extended to a polar quadrilateral. 2.9 Extend Darboux’s Theorem to the case of two conics which do not intersect transversally. 2.10 Show that the secant lines of a Veronese curve Rm in Pm are parameterized by the surface in the Grassmannian G(1, Pm ) isomorphic to P2 . Show that the embedding of P2 into the Grassmannian is given by the Schwarzenberger bundle. 2.11 Let U be a 2-dimensional vector space. Use the construction of curves of degree n − 1 Poncelet related to a conic to exhibit an isomorphism of linear representations Λ2 (S n U ) and S n−1 (S 2 U ) of SL(U ). 2.12 Assume that the pencil of sections of the Schwarzenberger bundle Sn,C has no base points. Show that the Poncelet curve associated to the pencil is nonsingular at a point x defined by a section s from the pencil if and only if the scheme of zeroes Z(s) is reduced. 2.13 Find the geometric interpretation of vanishing of the invariants Θ, Θ from (2.14) in the case when C or S is a singular conic. 2.14 Express the condition that two conics are tangent in terms of the invariants ∆, ∆ , Θ, Θ . 2.15 Let p1 , p2 , p3 , p4 be four distinct points on a nonsingular conic C. Let pi , pj denote the line through the points pi , pj . Show that the triangle with the vertices A = p1 , p3 ∩ p2 , p4 , B = p1 , p2 ∩ p3 , p4 and C = p1 , p4 ∩ p2 , p3 is a self-conjugate triangle with respect to C. 2.16 Show that two pairs {a, b}, {c, d} of points in P1 are harmonic conjugate if and only the cross-ratio [a, c; b, d] is equal to −1. 2.17 Let abcd be a quadrangle in P2 , and p, q be the intersection points of two pairs of opposite sides a, b, c, d and b, c, a, d. Let p , q be the intersection points of the line p, q with the diagonals a, c and b, d. Show that the pairs (p, q) and (p , q ) are harmonic conjugate . 2.18 Show that the pair of points on a diagonal of a complete quadrilateral defined by its sides is harmonic conjugate to the pair of points defined by intersection with other two diagonals.

2.19 Find the condition on a pair of conics that the associate Salmon conic is degenerate.

Historical Notes
The Poncelet ’s closure theorem which is the second part of Darboux’s Theorem 2.2.2 was first discoverd by Poncelet himself [194]. We refer to the excellent account of the history of the Poncelet related conics to [23]. Other elementary and non-elementary

Historical Notes

63

treatments of the Poncelet properties and their generalizations can be found in [11],[12], [127],[128]. The relationship between Poncelet curves and vector bundles is discussed in [240], [183], [241], [243]. Among many equivalent definitions of the Schwarzenberger bundles we chose one discussed in [83]. The papers of [177] and [140],[141] discuss the compactification of the variety of conjugate triangles. The latter two papers of N. Hitchin also discuss an interesting connection with Painleve equations. The notion of the conjugacy of conics is due to Rosanes [204]. Reye called two conjugate conics apolar [200]. The condition (2.12) for conjugate conics was first discovered by O. Hesse in [137]. He also proved that this property is poristic. The condition for Poncelet relation given in terms of invariants of a pair of conics (Theorem 2.2.9) was first discovered by A. Cayley [34], [35]. The invariants of a pair of quadrics in any dimension were studied by C. Segre [221]. A good modern discussion of Poncelet’s theorem and its applications can be found in [109]. The proof of Theorem 2.2.10 is due to J. Coolidge [52], Chaper VI, §3. The theorem was known to G. von Staudt [230] ((see [52], p. 66) and can be also found in Salmon’s book on conics [209], p. 345. Although Salmon writes in the footnote on p. 345 that “I believe that I was the first to direct the attention to the importance of this conic in the theory of two conics” , this conic was already known to Ph. La Hire [162]. p. 44 (see [52], p. 44 ).

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Chapter 3

Plane cubics
3.1
3.1.1

Equations
Weierstrass equation

Let X be a nonsingular projective curve of genus 1. By Riemann-Roch, for any divisor D of degree d > 0, we have dim H 0 (X, OX (D)) = d. The complete linear system D = P(H 0 (X, OX (D))) defines an isomorphism X ∼ C, where C is a curve of = degree d in Pd−1 (see [134], Chapter IV, Corollary 3.2). We consider here the case d = 3, i.e., a plane cubic model C = V (f ) of X. By Theorem 1.1.6, C has an inflection point p0 . Without loss of generality we may assume that p0 = [0, 0, 1] and the tangent line at this point has the equation t0 = 0. This implies that f = t0 q(t0 , t1 , t2 ) + at3 , where q is a quadratic polynomial. We may assume that q = 1 bt2 + t2 L(t0 , t1 ) + q (t0 , t1 ) for some quadratic polynomial q and a linear polynomial 2 l. Notice that b = 0, otherwise, we can express t2 as a rational function in t0 , t1 and obtain that E is a rational curve. So, we may assume that b = 1. Replacing t2 with 1 t2 + 2 l(t0 , t1 ) we may assume that l = 0. Now the equation looks as f = t0 t2 + at3 + bt2 t0 + ct1 t2 + dt3 = 0. 2 1 1 0 0
b By scaling, we may assume that a = 1. Replacing t1 with t1 + 3 t0 = 0, we may assume that b = 0. This gives us the Weierstrass equation of a nonsingular cubic:

t0 t2 + t3 + αt1 t2 + βt3 = 0 2 1 0 0

(3.1)

It is easy to see that C is nonsingular if and only if the polynomial x3 + αx + β has no multiple roots, or, equivalently, its discriminant ∆ = 4α3 + 27β 2 is not equal to zero. Two Weierstrass equations define isomorphic elliptic curves if and only if there exists a projective transformation transforming one equation to another. It is easy to see that it happens if and only if (α , β ) = (λ3 α, λ2 β) for some nonzero constant λ. This can be expressed in terms of the absolute invariant j= 4α3 4α3 . + 27β 2 (3.2)

65

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CHAPTER 3. PLANE CUBICS

Two elliptic curves are isomorphic if and only if their absolute invariants are equal. The projection [t0 , t1 , t2 ] → [t0 , t1 ] exhibits C as a double cover of P1 with the branch points [1, x], [0, 1], where x3 + αx + β = 0. The corresponding points [1, x, 0], and [0, 1, 0] on C are the ramification points. If we choose p0 = [0, 1, 0] to be the zero point in the group law on C, then 2p ∼ 2p0 for any ramification point p implies that p is a 2-torsion point. Any 2-torsion point is obtained in this way. Here we use that the group law on a cubic curve with the distinguished point p0 chosen as the zero point is given by the formula p ⊕ q ∈ p + q − p0 . (3.3) Note that, by Riemann-Roch, the complete linear system p + q − p0 consists of one point. It follows from the above computation that any nonsingular plane cubic V (f ) is projectively isomorphic to the plane cubic V (t2 t0 +t3 +αt1 t2 +βt3 ). The functions S : 2 1 0 0 F → α, T : F → β can be extended to invariants on the space of homogeneous cubic forms with respect to the groups of unimodular linear transformations. The explicit expresions of S and T in terms of the coefficients of f are rather long and can be found in many places (e.g. [80]). Definition 3.1. A nonsingular plane cubic V (f ) is called harmonic cubic (resp. equianharmonic cubic) if S(f ) = 0 (resp. T (f ) = 0). Theorem 3.1.1. Let C = V (f ) be a nonsingular plane cubic and c be any point on C. The following conditions are equivalent. (i) C is a harmonic (resp. equianharmonic) cubic. (ii) The cross-ratio of four roots of the polynomial t0 (t3 + αt1 t2 + βt3 ), taken in any 1 0 0 order, is equal to −1 (resp. a third root of −1); (iii) The group of automorphisms of C leaving the point c invariant is a cyclic group of order 4 (resp. 6). Proof. (i) ⇔ (ii) Direct computation. (ii) ⇔ (iii) Let G be the group of automorphisms of C leaving c fixed. By choosing a projective embedding of C given by the linear system |3c|, we obtain that C is isomorphic to a plane cubic V (f ) given by a Weierstrass equation f = 0 and G is isomorphic to the group of projective transformations of P2 leaving the point [0, 0, 1] invariant. By direct computation, it is easy to see that G consists of transformations T : [t0 , t1 , t2 ] → [λ2 t0 , λ2 t1 , λ3 t2 ] leaving f invariant. Now the assertion is easily verified.

3.1.2

The Hesse equation

Since any flex tangent line intersects the curve with multiplicity 3, applying formula(1.21), we obtain that the curve has exactly 9 inflection points. Using the group law on an elliptic curve with an inflection point as the zero, we can interpret any inflection point p as a 3-torsion point. This follows from (3.3) since the divisor of the rational function

3.1. EQUATIONS

67

l/l0 mod (f ), where l = 0 is the equation of the inflection tangent at p and l0 = 0 is the equation of the inflection tangent at p0 , is equal to 3p − 3p0 . This of course agrees with the fact the group X[3] of 3-torsion points on an elliptic curve X is isomorphic to (Z/3Z)2 . Let H be a subgroup of order 3 of X. Since the sum of elements of this group add up to 0, we see that the corresponding 3 inflection points p, q, r satisfy p + q + r ∼ 3p0 . It is easy to see that the rational function on C with the divisor p + q + r − 3p0 can be obtained as the restriction of the rational function m(t0 , t1 , t2 )/l0 (t0 , t1 , t2 ), where m = 0 defines the line containing the points p, q, r. There are 3 cosets with respect to each subgroup H. Since the sum of elements in each coset is again equal to zero, we get 12 lines, each containing three inflection points. Conversely, if a line contains three inflection points, the sum of these points is zero, and it is easy to see that the three points forms a conjugacy class with respect to some subgroup H. Each element of (Z/3Z)3 is contained in 4 cosets (it is enough to check this for the zero element). Thus we obtain a configuration of 12 lines and 9 points, each line contains 3 points, and each point is contained in 4 lines. This is the Hesse line arrangement (123 , 94 ). Let 1 , 2 be two inflection lines. Choose projective coordinates such that the equations of these lines are t0 = 0 and t1 = 0. Then it is easy to see that the equation of C can be wriiten in the form f (x0 , x1 , x2 ) = t0 t1 (at0 + bt1 + ct2 ) + dt3 , 2 (3.4)

where at0 + bt1 + ct2 = 0 is a third inflection line. Suppose the three lines are concurrent. Then the equation can be further transformed to the form t0 t1 (t0 + t1 ) + t3 = 0. Since the sets of three distinct points in P1 are projectively equivalent we can 2 change the coordinates to assume that the equation is t3 + t3 + t3 = 0. Obviously, it 0 1 2 is in the Hesse form. So we may assume that three lines are non-concurrent. Consider the equation (3.4). By scaling the coordinate t2 we may assume that c = 3. Let 3 be a primitive 3d root of 1. Define new coordinates u, v by the formula at0 + t2 = Then abF (x0 , x1 , x2 ) = ( 3 u +
2 3v 3u

+

2 3 v,

bt0 + t2 =

2 3u

+

3 v.

− t2 )( 2 u + 3

3v

− t2 )(−u − v + t2 ) + dt3 2

= −u3 − v 3 + (d + 1)t3 − 3uvt2 = 0. 2 Since C is nonsingular, we have d = 1. After scaling the coordinate t2 we arrive at the Hesse canonical form or second canonical form or Hesse equation of a plane cubic curve t3 + t3 + t3 + 6mt0 t1 t2 = 0. (3.5) 0 1 2 Here the expression for the last coefficient is given to simplify future computations. The condition that the curve is nonsingular is 1 + 8m3 = 0. The curve given given by this equation is singular if and if 8m3 + 1 = 0. (3.6)

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By reducing the Hesse equation to a Weiestrass forms one can express the absolute invariant (3.2) in terms of the parameter m: j= 64(m − m4 )3 . (1 + 8m3 )3 (3.7)

3.1.3

The Hesse pencil
λ(t3 + t3 + t3 ) + µt0 t1 t2 = 0. 0 1 2 (3.8)

Consider a pencil of plane cubics defined by the equation

It is called the Hesse pencil. Its base points are [0, 1, −1], [1, 0, −1], [1, −1, 0], [0, 1, − ], [1, 0, − 2 ], [1, − , 0], [0, 1, − 2 ], [1, 0, − ], [1, − 2 , 0], (3.9)

where = e2πi/3 . As is easy to see they are the nine inflection points of any nonsingular member of the pencil. The singular members of the pencil correspond to the values of the parameters (λ, µ) = (0, 1), (1, −3), (1, −3 ), (1, −3 2 ). The last three values correspond to the three values of m for which the Hesse equation defines a singular curve. Any triple of lines containing the nine base points belong to the pencil and define a singular member. Here they are: V (t0 ), V (t1 ),
2 2

V (t2 ), t2 ), V (t0 +
2 2

V (t0 + t1 + t2 ), V (t0 + t1 + V (t0 + t1 + t2 ), V (t0 + V (t0 +
2

t1 + t2 )
2

(3.10)

t1 +

t2 ), V (t0 + t1 + t2 ) t2 )

t1 + t2 ), V (t0 + t1 + t2 ), V (t0 + t1 +

We leave to a suspicious reader to check that (t0 + t1 + t2 )(t0 + t1 + (t0 + t1 + t2 )(t0 + (t0 +
2 2 2

t2 )(t0 +
2

2

t1 + t2 ) = t3 + t3 + t3 − 3t0 t1 t2 , 0 1 2
2

t1 +

t2 )(t0 + t1 + t2 ) = t3 + t3 + t3 − 3 t0 t1 t2 , 0 1 2 t 2 ) = t3 + t 3 + t3 − 3 2 t0 t1 t2 . 0 1 2

t1 + t2 )(t0 + t1 + t2 )(t0 + t1 +

The 12 lines (3.10) and 9 inflection points (3.9) form the Hesse configuration corresponding to any nonsingular member of the pencil. Choose [0, 1, −1] to be the zero point in the group law on C. Then we can define an isomorphism of groups φ : (Z/3Z)2 → X[3] by sending (1, 0) to [0, 1, − ], (0, 1) to [1, 0, −1]. The points of the first row is the subgroup H generated by φ((1, 0)). The points of the second row is the coset of H containing φ((0, 1)).

3.1. EQUATIONS
2

69

Remark 3.1.1. Note that varying m in P1 \ {− 1 , − 2 , − 2 , ∞} we obtain a family of 2 elliptic curves Xm with a fixed isomorphism φm : (Z/3Z)2 → Xm [3]. By blowing up the 9 base points we obtain a rational surface S(3) together with a morphism f : S(3) → P1 obtained from the rational map P2 − → P1 , [t0 , t1 , t2 ] → [t0 t1 t2 , t3 + t3 + t3 ] by re0 1 2 solving (minimally) the indeterminacy points. The fibre of f over a point (a, b) ∈ P2 is isomorphic to the member of the Hesse pencil corresponding to (λ, µ) = (−b, a). One can show that this is a modular family of elliptic curves with 3-level, i.e. the universal object for the fine moduli space of pairs (X, φ), where X is an elliptic curve and φ : (Z/3Z)2 → X[3] is an isomorphism of groups. There is a canonical isomorphism P1 ∼ Y , where Y is the modular curve of level 3, i.e. a nonsingular compactification = of the quotient of the upper half-plane H = {a + bi ∈ C : b > 0} by the group Γ(3) = {A = a b c d ∈ SL(2, Z) : A ≡ I3 mod 3}

which acts on H by Moebius transformations z → az+b . The boundary of H/Γ(3) cz+d in Y consists of 4 points (the cusps). They correspond to the singular members of the Hesse pencil.

3.1.4

The Hesse group

The Hesse group G216 is the group of projective transformations which preserve the Hesse pencil of cubic curves. First we see the obvious symmetries generated by the transformations τ : [t0 , , t1 , t2 ] → [t0 , 3 t1 , 2 t2 ]. (3.11) 3 σ : [t0 , t1 , t2 ] → [t2 , t0 , t1 ].
2

(3.12)

They define a projective representation of the group (Z/3Z) , called the Schr¨ dinger o representation. If we fix the group law by taking the origin to be [0, 1, −1], then the transformation (3.11) induces on each nonsingular fibre the translation automorphism by the point [0, 1, − ]. The transformation (3.11) is the translation by the point [1, 0, −1] and the transformation (3.12) is the translation by the point [1, 0, −1]. Theorem 3.1.2. The Hesse group G216 is a group of order 216 isomorphic to the semi-direct product (Z/3Z)2 SL(2, F3 ), where the action of SL(2, F3 ) on (Z/3Z)2 is the natural linear representation. Proof. Let σ ∈ G216 . It transforms a member of the Hesse pencil to another member. This defines a homomorphism G216 → Aut(P1 ). An element of the kernel K leaves each member of the pencil invariant. In particular, it leaves invariant the curve V (t0 t1 t2 ). The group of automorphisms of this curve is generated by homotheties

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[t0 , t1 , t2 ] → [t0 , at1 , bt2 ] and permutation of coordinates. Suppose σ induces a homothety. Since it also leaves invariant the curve V (t3 +t3 +t3 ), we must have 1 = a3 = b3 . 0 1 2 To leave invariant a general member we also need that a3 = b3 = bc. This implies that σ belongs to the subgroup generated by transformation (3.11). An even permutation of coordinates belongs to a subgroup generated by transformation (3.12). The odd permutation σ0 : [t0 , t1 , t2 ] → [t0 , t2 , t1 ] acts on the group of 3-torsion points of each nonsingular fibre as the inversion automorphism. Thus we see that K ∼ (Z/3Z)2 = σ0 .

Now let I be the image of Hes in Aut(P1 ). It acts by permuting the four singular members of the pencil and thus leaves the set of zeroes of the binary form ∆ = (8t3 + t3 )t0 1 0 invariant. It follows from the invariant theory that this implies that H is a subgroup of A4 . We claim that H = A4 . Consider the projective transformations given by the matrices     1 ε ε 1 1 1 σ1 = 1 ε ε2  , g2 = ε2 ε ε2  ε2 ε2 ε 1 ε2 ε The transformations σ0 , σ1 , σ2 generate a subgroup of PGL(3, C) isomorphic to the quaternion group Q8 with center (σ0 ). The transformation σ3 : [t0 , t1 , t2 ] → [εt0 , t2 , t1 ] 3 satisfies σ3 = σ0 . It acts by sending a curve Xm to Xεm . It is easy to see that the transformations σ1 , σ2 , σ3 , τ generate the group isomorphic to SL(2, F3 ). Its center is (σ0 ) and the quotient by the center is isomorphic to A4 . In other words, this group is the binary tetrahedral group. Note that the whole group can be generated by transformations σ, τ, σ0 , σ1 . Recall that a linear operator σ ∈ GL(V ) of a complex vector space of dimension n is called a complex reflection if it is of finite order and the rank of σ − idV is equal to 1. The kernel of σ − idV is a hyperplane Hv in V , called the reflection hyperplane of σ. It is an invariant with respect to σ and its stabilizer is a cyclic group. A complex reflection group is a finite subgroup G of GL(V ) generated by complex reflections. One can choose a unitary inner product on V such that any complex reflection σ from V can be written in the form sα,η : v → v + (η − 1)(v, α)v, where α is a vector of norm 1 perpendicular to the reflection hyperplane Hσ of σ, and η is a non-trivial root of unity of order equal to the order of σ. Recall the basic facts about complex reflection groups (see, for example, [229]): • The algebra of invariants (S • V )G ∼ C[t1 , . . . , tn ]G is freely generated by n = invariant polynomials f1 , . . . , fn (geometrically V /G ∼ Cn )). = • The product of degrees di of the polynomials f1 , . . . , fn is equal to the order of G.

3.1. EQUATIONS • The number of complex reflections in G is equal to (di − 1).

71

All complex reflections group were classified by G. Shephard and J. Todd [226]. There are 5 conjugacy classes of complex reflection subgroups of GL(3, C). Among them is the group isomorphic to a central extension of degree 3 of the Hesse group. It is generated by complex reflections sα,η , where V (α) is one of the 12 hyperplanes (3.10) in P(V ) and α is the unit normal vector (a, b, c) to the hyperplane V (at0 + bt1 + ct2 ) and η 3 = 1. Note that each reflection sα,η leaves invariant the hyperplanes with normal vector orthogonal to α. For example, s[1,0,0],ε leaves invariant the hyperplanes V (ti ), i = 0, 1, 2. This implies that each of the 12 complex reflections leave the Hesse pencil invariant. Thus the image of G in PGL(3, C) is contained in the Hesse group. It follows from the classification of complex reflection groups (or could be checked directly, see [229]) that it is equal to the Hesse group and the subgroup of scalar matrices from G is a cyclic group of order 3. Each of the 12 reflection hyperplanes defines 2 complex reflections. This gives 24 complex reflections in G. This number coincides with the number of elements of order 3 in Hes and so there are no more complex reflections ion G. Let d1 ≤ d2 ≤ d3 be the degrees of the invariants generating the algebra of invariants of G. We have d1 + d2 + d3 = 27, d1 d2 d3 = 648. This easily gives d1 = 6, d2 = 9, d3 = 12. There are obvious reducible curves of degree 9 and 12 in P2 invariant with respect to G. The curve of degree 9 is the union of the lines whose normal vectors are the coordinate vectors of the base points of the Hesse pencil. One can check that each such line intersects a nonsingular member of the pencil at nontrivial 2-torsion points with respect to the group law defined by the corresponding base point. For each nonsingular member of the Hesse pencil this line is classically called the harmonic line of the corresponding inflection point. The equation of the union of 9 harmonic lines is f9 = (t3 − t3 )(t3 − t3 )(t3 − t3 ) = 0. 0 1 0 2 1 2 The curve of degree 12 is the union of the 12 lines (3.10). Its equation is f12 = t0 t1 t2 [27t3 t3 t3 − (t3 + t3 + t3 )3 ] = 0 0 1 2 0 1 2 (3.14) (3.13)

A polynomial defining an invariant curve is a relative invariant of G (it is an invariant with respect to the group G = G ∩ SL(3, C)). One checks that the polynomials f9 is indeed an invariant, but the polynomial f12 is only a relative invariant. So, there exists another curve of degree 12 whose equation defines an invariant of degree 12. What is this curve? Recall that the Hesse group acts on the base of the Hesse pencil via the action of the tetrahedron group A4 . It has 3 special orbits with stabilizers of order 2,3 and 3. The first orbit consists of 6 points such that the fibres over these points are harmonic cubics. The second orbit consists of 4 points such that the fibres over these points are equiequianharmoniccubics. The third orbit consists of 4 points corresponding to singular members of the pencil. It is not difficult to check that the product of the equations of the equiequianharmoniccubics defines an invariant of degree 12. Its equation is f12 = (t3 + t3 + t3 )[(t3 + t3 + t3 )3 + 216t3 t3 t3 ] = 0 0 1 2 0 1 2 0 1 2 An invariant of degree 6 is f6 = 7(t6 + t6 + t6 ) − 6(t3 + t3 + t3 )2 . 0 1 2 0 2 3 (3.16) (3.15)

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CHAPTER 3. PLANE CUBICS

The product of the equations defining 6 harmonic cubics is an invariant of degree 18 f18 = (t3 + t3 + t3 )6 − 540t3 t3 t3 (t3 + t3 + t3 )3 − 5832t6 t6 t6 = 0 0 1 2 0 1 2 0 1 2 0 1 2 (3.17)

3.2
3.2.1

Polars of a plane cubic
The Hessian of a cubic hypersurface

Let X = V (f ) be a cubic hypersurface in Pn . We know that the Hessian He(X) is the locus of points a ∈ Pn such that the polar quadric Pa (V )) is singular. Also we know that, for any a ∈ He(X), Sing(Pa (X)) = {b ∈ P2 : Db (Da (f )) = 0}. Since Pb (Pa (X)) = Pa (Pb (X)) we obtain that b ∈ He(X). Theorem 3.2.1. The Hessian He(X) of a cubic hypersurface X contains the Steinerian St(X). If He(X) = Pn , then He(X) = St(X). For the last assertion one only needs to compare the degrees of the hypersurfaces. They are equal to n + 3. In particular, the rational map, if defined, st−1 : St(X) → He(X), a → Sing(Pa (X)) X (3.18)

is a birational automorphism of the Hessian hypersurface. We have noticed this already in Chapter 1. Proposition 3.2.2. Assume X has only isolated singularities. Then He(X) = Pn if and only if X is a cone over a cubic hypersurface in Pn−1 . Proof. Let W = {(a, b) ∈ Pn × Pn : Pa,b2 (X) = 0}. For each a ∈ Pn , the fibre of the first projection over the point a is equal to the first polar Pa (X). For any b ∈ Pn , the fibre of the second projection over the point b is equal to the second polar Pb2 (X) = V ( ∂i f (b)ti ). Let U = Pn \ Sing(X). For any b ∈ U , the fibre of the second projection is a hyperplane in Pn . This shows that p−1 (U ) is nonsingular. 2 The restriction of the first projection to U is a morphism of nonsingular varieties. The general fibre of this morphism is a regular scheme over the general point of Pn . Since we are in characteristic 0, it is a smooth scheme. Thus there exists an open subset W ⊂ Pn such that p−1 (W )∩U is nonsingular. If He(X) = 0, all polar quadrics Pa (X) 1 are singular, and a general polar must have singularities inside of p−1 (Sing(X)). This 2 means that p1 (p−1 (Sing(X))) = Pn . For any x ∈ Sing(X), all polar quadrics contain 2 x and either all of them are singular at x or there exists an open subset Ux ⊂ Pn such all quadrics Pa (X) are nonsingular at x for a ∈ Ux . Suppose that for any x ∈ Sing(X) there exists a polar quadric which is nonsingular at x. Since the number of isolated singular points is finite, there will be an open set of points a ∈ Pn such that the fibre p−1 (a) is nonsingular in p−1 (Sing(X)). This is a contradiction. Thus, there exists a 1 2

3.2. POLARS OF A PLANE CUBIC

73

point c ∈ Sing(X) such that all polar quadrics are singular at x. This implies that c is a common solution of the systems of linear equations He(f3 )(a) · X = 0, a ∈ Pn . Thus the first partials of f3 are linearly dependent. Now we apply Proposition 1.1.2 to obtain that X is a cone. Remark 3.2.1. The example of a cubic hypersurface in P4 which we considered in Remark 1.1.1 shows that the assumption of the theorem cannot be weakened. Its singular locus is the plane t0 = t1 = 0.

3.2.2

The Hessian of a plane cubic

Consider a plane cubic C = V (f ) with equation in the Hesse canonical form (3.5). 1 The partials of 3 f are t2 + 2mt1 t2 , 0 t2 + 2mt0 t2 , 1 t2 + 2mt0 t1 2 (3.19)

Thus the Hessian of C has the following equation: t0 He(C) = mt2 mt1 mt2 t1 mt0 mt1 mt0 = (1 + 2m3 )t0 t1 t2 − m2 (t3 + t3 + t3 ). 0 1 2 t2 (3.20)

In particular, the Hessian of the member of the Hesse pencil corresponding to the parameter (λ, µ) = (1, 6m), m = 0, is equal to t3 + t 3 + t3 − 0 1 2 1 + 2m3 t0 t1 t2 = 0, m2 m = 0, (3.21)

or, if (λ, µ) = (1, 0) or (0, 1), then the Hessian is equal to V (t0 t1 t2 ). Lemma 3.2.3. Let C be a nonsingular cubic. The following assertions are equivalent: (i) dim Sing(Pa (C)) > 0; (ii) a ∈ Sing(He(C)); (iii) He(C) is the union of three nonconcurrent lines; (iv) C is isomorphic to a Fermat cubic t3 + t3 + t3 = 0; 1 0 1 (v) He(C) is a singular cubic; (vi) C is an equianharmonic cubic. Proof. Use the Hesse equation for a cubic and for its Hessian. We see that He(C) is 3 singular if and only if either m = 0 or 1 + 8(− 1+2m )3 = 0. Obviously, m = 1 is a 6m2 solution of the second equation. Other solutions are , 2 . This corresponds to He(C), where C os of the form V (t3 + t3 + t3 ), or is given by the equation 1 1 0 t3 + t3 + t3 + 6 i t0 t1 t2 = ( i t0 + t1 + t2 )3 + (t0 + i t1 + t2 )3 0 1 1

74

CHAPTER 3. PLANE CUBICS +(t0 + t1 + i t2 )3 = 0,

where i = 1, 2, or t3 + t3 + t3 + 6t0 t1 t2 = (t0 + t1 + t2 )3 + (t0 + t1 + 0 1 1 +(t0 +
2 2

t2 )3

t1 + t2 )3 = 0.

This computation proves the equivalence of (iii), (iv), (v). Assume (i) holds. Then the rank of the Hessian matrix He is equal to 1. It is easy to see that the first two rows are proportional if and only if m(m3 − 1) = 0. It follows from the previous computation that this implies (iv). The corresponding point a is one of the three intersection points of the lines such that the cubic is equal to the sum of the cubes of linear forms defining these lines. Direct computation shows that (ii) holds. This shows the implication (i) ⇒ (ii). Assume (ii) holds. Again the previous computations show that m(m3 − 1) = 0 and the Hessian curve is the union of three lines. Again (i) is directly verified. The equivalence of (iv) and (vi) follows from Theorem 3.1.1 since the transformation [t0 , t1 , t2 ] → [t1 , t0 , e2πi/3 t2 ] generates a cyclic group of order 6 of automorphisms of C leaving the point [1, −1, 0] fixed. Corollary 3.2.4. Assume that C = V (f ) is not isomorphic to a Fermat cubic. Then the Hessian cubic is not singular, and the map a → Sing(Pa (C)) is an involution on C without fixed points. Proof. The only unproved assertion is that the involution does not have fixed points. A fixed point a has the property that Da (Da (f )) = Da2 (f ) = 0. It follows from Theorem 1.1.1 that this implies that a ∈ Sing(C). Remark 3.2.2. Consider the Hesse pencil of cubics with parameters (λ, µ) = (m0 , 6m) C(m0 , m) = V (m0 (t3 + t3 + t3 ) + 6m1 t0 t1 t2 ). 0 1 2 Taking the Hessian of each curve from the pencil we get the pencil H(λ) = V (λ0 t3 + t3 + t3 + 6λ1 t0 t1 t2 ). 0 1 2 The map C(m0 , m) → He(C(m0 , m)) defines a regular map H : P1 → P1 , [m0 , m1 ] → [t0 , t1 ] = [−m0 m2 , m3 + 2m3 ] 1 0 1 (3.22)

This map is of degree 3. For a general value of the inhomogeneous parameter λ = t1 /t0 , the pre-image consists of three points with inhomogeneous coordinate m = m1 /m0 satisyfing the cubic equation 6λm3 − 2m2 + 1 = 0. We know that the points
1 [λ0 , λ1 ] = [0, 1], [1, − 2 ], [1, − 2 ], [1, − 2 ]
2

(3.23)

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75

correspond to singular members of the λ-pencil. These are the branch points of the map H. Over each branch point we have two points in the pre-image. The points (m0 , m1 ) = (1, 0), (1, 1), (1, ), (1,
2

).

are the ramification points corresponding to cubics isomorphic to the Ferma cubic. A non-ramication point in the pre-image corresponds to a singular member. Let C(m) = C(1, m). If we fix a group law on a H(m) = He(C(m)), we can identify the involution described in Corollary 3.2.4 with the translation with respect to a non-trivial 2-torsion point η (see Exercises). Given a nonsingular cubic curve H(m) together with a fixed-point-free involution τ there exists a unique nonsingular cubic C(m) such that H(m) = He(C(m)) and the involution τ is the involution described in the corollary. Thus the 3 roots of the equation (3.23) can be identified with 3 nontrivial torsion points on H(m). We refer to Exercises for a reconstruction of C(m) from the pair (H(m), η). Recall that the Cayleyan curve of a plane cubic C is the locus of lines p, q in the dual plane such that a ∈ He(C) and b is the singular point of Pa (C). Each such line intersects He(C) at three points a, b, c. The following gives the geometric meaning of the third intersection point. Proposition 3.2.5. Let c be the third intersection point of a line ∈ Cay(C) and He(C). Then is a component of the polar Pd (C) whose singular point is c. The point d is the intersection point of the tangents of He(C) at the points a and b. Proof. Since b ∈ Sing(Pa (C)), we have Db (Da (f )) = 0. Similarly, we obtain that Pb (Pa (f )) = 0. This implies that Px (Pab (f )) = 0 for any x ∈ P2 . This means that the points a, b ∈ are conjugate with respect to all polar quadrics. Let U be the 2dimensional subspace of C3 defining the line . The restriction of the quadrics Px (C) to is defined by a quadratic form qx on U . Let bx be the corresponding polar bilinear form. Let a, b ∈ U be vectors spanning the lines a, b ∈ = P(U ). For all x ∈ P2 , we have bx (a, b) = 0. Consider the unique polar conic Qd = V (qd ) passing through the points a, b. We have 0 = 2bd (a + b) = qd (a + b) − qd (a) − qd (b) = qd (a + b). This means that the conic Qd intersects the line at three points corresponding to the vectors a, b, a + b. Thus is contained in Qd . Also this implies that Qd is a singular quadric, and hence d ∈ He(C) and its singular point c belongs to . Thus c is the third intersection point of with C. It remains to prove the last assertion. Chose a group law on the curve He(C) by fixing an inflection point as the zero point. We know that the Steiner involution is defined by the translation x → x ⊕ η, where η is a fixed 2-torsion point. Thus b = a ⊕ η. It follows from the definition of the group law on a nonsingular cubic that the tangents Ta (He(C)) and Tb (He(C)) intersect at a point d on He(C). In the group law d + 2a = 0, hence d = −2a. Since a, b, c lie on a line, we get c = −a − b in the group law. After subtracting, we get d − c = b − a = η. Thus the points x and c is an orbit of the Steiner involution. This shows that c is the singular point of Pd (C). By Proposition 1.2.4, Pd (C) contains the points a, b. Thus a, b is a component of Pd (C).

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It follows from the above proposition that the Cayleyan curve of a nonsingular cubic C parametrizes the line components of singular polar conics of C. It is also isomorphic to the quotient of He(C) by the Steinerian involution from Corollary 3.2.4 . Since this involution does not have fixed point the quotient map He(C) → Cay(C) is a unramified cover of degree 2. In particular, Cay(C) is a nonsingular curve of genus 1. Let us find the equation of the Cayleyan curve. A line belongs to Cay(X) of and only the restriction of the linear system of polar conics of X to is of dimension 1. This translates into the condition that the restriction of the partials of X to is a linear dependent set of three binary forms. So, write in the parametric form as the image of the map P1 → P2 given by [u, v] → [a0 u + b0 v, a1 u + b1 v, a2 u + b2 v]. The the pull-backs of the partials from (3.19) define 3 binary forms in u, v (a0 u + b0 v)2 + 2m(a1 u + b1 v)(a2 u + b2 v) and so on. The condition of linear dependence is given by the vanishing of the determinant  2  a0 + 2ma1 a2 2a0 b0 + 2m(a1 b2 + a2 b1 ) b2 + 2mb1 b2 0 det a2 + 2ma0 a2 2a1 b1 + 2m(a0 b2 + a2 b0 ) b2 + 2mb0 b2  1 1 a2 + 2ma0 a1 2a2 b2 + 2m(a0 b1 + a1 b0 ) b2 + 2mb0 b2 2 2 The coordinates of in the dual plane are [ξ0 , ξ1 , ξ2 ] = [a1 b2 − a2 b1 , a2 b0 − a0 b2 , a0 b1 − a1 b0 ]. Computing the determinant we find that the equation of Cay(X) in the coordinates ξ0 , ξ1 , ξ2 is 3 3 3 ξ0 + ξ1 + ξ2 + 6m ξ0 , ξ1 , ξ2 = 0, (3.24) where m = (1 − 4m3 )/6m. Using the formula (3.7) for the absolute invariant of the curve, this can be translated into an explicit relationship between the absolute invariant of an elliptic curve E and the isogeneous elliptic curve E/(te ), where te is the translation automorphism by a non-trivial 2-torsion point. Note that this agrees with the degree of the Cayleyan curve found in Proposition 1.1.11.

3.2.3

The dual curve

Write the equation of a general line in the form t2 = ξ0 t0 + ξ1 t1 and plug in the Hesse equation (3.21). The corresponding cubic equation has a multiple root if and only if the line is a tangent. We have (ξ0 t0 + ξ1 t1 )3 + t3 + t3 + 6mt0 t1 (ξ0 t0 + ξ1 t1 ) 0 1
3 3 2 2 = (ξ0 + 1)t3 + (ξ1 + 1)t3 + (3ξ0 ξ1 + 6mξ0 )t2 t1 + (3ξ0 ξ1 + 6mξ1 )t0 t2 = 0. 0 1 0 1

The condition that there is a multiple root is that the discriminant of the homogeneous cubic form in t0 , t1 is zero. The discriminant of the cubic form at3 +bt2 t1 +ct0 t2 +dt3 0 1 1 0 is equal to D = b2 c2 + 18abcd − 4ac3 − 4b3 d − 27a2 d2 .

3.2. POLARS OF A PLANE CUBIC After plugging in, we obtain

77

2 2 2 2 3 3 (3ξ0 ξ1 +6mξ0 )2 (3ξ0 ξ1 +6mξ1 )2 +18(3ξ0 ξ1 +6mξ0 )(3ξ0 ξ1 +6mξ1 )(ξ0 +1)(ξ1 +1) 3 2 3 2 3 3 −4(ξ0 + 1)(3ξ0 ξ1 + 6mξ1 ) − 4(ξ1 + 1)(3ξ1 ξ0 + 6mξ0 ) − [27(ξ0 + 1)2 (ξ1 + 1)2 3 3 2 2 4 4 = −27 + 864ξ0 ξ1 m3 + 648ξ0 ξ1 m − 648m2 ξ0 ξ1 − 648m2 ξ0 ξ1 + 648m2 ξ0 ξ1 2 2 6 6 3 3 3 3 3 3 +1296m4 ξ0 ξ1 − 27ξ1 − 27ξ0 + 54ξ0 ξ1 − 864ξ1 m3 − 864ξ0 m3 − 54ξ1 − 54ξ0 .

It remains to homogenize the equation and divide by (−27) to obtain the equation of the dual curve
6 6 6 3 3 3 3 3 3 ξ0 + ξ1 + ξ2 − (2 + 32m3 )(ξ0 ξ1 + ξ0 ξ2 + ξ2 ξ1 ) 3 3 3 2 2 2 −24m2 ξ0 ξ1 ξ2 (ξ0 + ξ1 + ξ2 ) − (24m + 48m4 )ξ0 ξ1 ξ2 = 0.

(3.25)

According to the Pl¨ cker formula (9.43) the dual curve of a nonsingular member, being u of geometric genus 1, must have 9 cusps. They correspond to the flex tangent of the original curve. The inflection points are given in (3.8). Computing the equations of the tangents we find the following singular points of the dual curve: [−2m, 1, 1], [1, −2m, 1], [1, 1, −2m], [−2mε, ε2 , 1], [−2mε, 1, ε2 ], [ε2 , −2mε, 1], [1, −2mε, 1, ε2 ], [1, ε2 , −2mε], [ε2 , 1, −2m]. One easily checks that the polar Pa (C) with pole at a base point of the Hesse pencil contains the flex tangent at a as a line component. This shows that the Caylean curve Cay(C) passes through the singular points of the dual cubic. The pencil spanned by the dual cubic and the Cayleyan cubic taken with multiplicity 2 is a pencil of sextic curves with 9 double points (an Halphen pencil of index 2). But this is another story.

3.2.4

Polar polygons

Since for any three general points in P2 there exists a plane cubic singular at these points (the union of three lines), a general ternary cubic form does not admit polar triangles. Of course this is easy to see by counting constants. A plane cubic curve projectively isomorphic to the cubic C = V (t3 + t3 + t3 ) will 1 2 0 be called a Fermat cubic. Obviously such a curve admits a polar 3-polyhedron (polar triangle). Proposition 3.2.6. A plane cubic admits a polar triangle if and only if either it is a Fermat cubic or it is equal to the union of three distinct concurrent lines.
3 3 3 Proof. Suppose C = V (l1 + l2 + l3 ). Without loss of generality we may assume that 3 3 l1 is not proportional to l2 . Thus, after coordinate change C = V (t3 + t3 + l3 ). If 1 0 l[t0 , t1 , t2 ] does not depend on t2 , the curve C is the union of three distinct concurrent lines. Otherwise we can change coordinates to assume that l = t2 and get a Fermat cubic.

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Remark 3.2.3. If C is a Fermat cubic, then its polar triangle is unique. Its sides are the three first polars of C which are double lines. By counting constants, we see that a general cubic admits a polar 4-polyhedron (polar quadrangle). We call a polar quadrangle {l1 , . . . , l4 } nondegenerate if it is defined by 4 points in P(E) no three of which are collinear. It is clear that a polar quadrangle is non-degenerate if and only if the linear system of conics in P(E) through the points 1 , . . . , 4 is an irreducible pencil (i.e. a linear system of dimension 1 whose general member is irreducible). This allows us to define a nondegenerate generalized polar quadrangle of C as a generalized polyhedron Z of C such that |IZ (2)| is an irreducible pencil. Lemma 3.2.7. C admits a degenerate polar quadrangle if and only if it is one of the following curves: (i) a Fermat cubic; (ii) a cuspidal cubic; (ii) the union of three concurrent lines (not necessary distinct); Proof. We have t 3 + t3 + t 3 = 0 1 2 where a = e2πi/3 . We have t0 t1 (9t0 + 15t1 ) = (t0 + t1 )3 + (t0 + 2t1 )3 − 2t3 − 5t3 . 0 1 Since the union of three distinct concurrent lines C is projectively equivalent to V (t0 t1 (9t0 + 15t1 )), we see that C admits a degenerate quadrangle. We also have t0 t2 = (2t0 + t1 )3 + (t0 − 4t1 )3 − 9t3 + 15t3 , 1 0 1 (2 + c3 )t3 = (t0 + at1 )3 + (t0 + bt1 )3 − (ct0 + dt1 )3 − (a3 + b3 − d3 )t3 , 0 1 where a3 + b3 = c2 d, a + b = cd2 , c3 + 2 = 0. This shows that case (iii) occurs. All cuspidal cubics are projectively equivalently. So it is enough to demonstrate a degenerate polar quadrangle for V (t3 + 6t2 t2 ). We have 0 1 t3 + 6t2 t2 = (t1 + t2 )3 + (t2 − t1 )3 − 2t3 + t3 . 0 1 2 0 Now let us prove the converse. Suppose
3 3 3 3 f = l1 + l 2 + l3 + l 4 ,

1 1 1 (t0 + t1 )3 + (t0 + at1 )3 + (t0 + a2 t1 )3 + t3 , 2 3 3 3

where l1 , l2 , l3 vanish at a common point a. Let ξ be a linear form on E corresponding to a. We have 1 2 2 2 2 2 Pξ (f ) = l1 ξ(l1 ) + l2 ξ(l2 ) + l3 ξ(l3 ) + l4 ξ(l4 ) = l4 ξ(l4 ). 3

3.2. POLARS OF A PLANE CUBIC

79

This shows that the first polar Pa (f ) = Pξ (f ) is either the whole P2 or a double line 2 V (l4 ). In the first case C is the union of three concurrent lines. Assume the second case occurs. We can choose coordinates such that a = [1, 0, 0] and l = V (t0 ). Write f = f0 t3 + f1 t2 + f2 t0 + f3 , 0 0 where fi are homogeneous forms of degree i in variables t1 , t2 . Then Pa (f ) = ∂0 (f ) = 3t2 f0 + 2t0 f1 + f2 . This can be proportional to t2 only if f1 = f2 = 0. 0 0 Thus V (f ) = V (t3 + f3 (t1 , t2 )). If f3 does not multiple linear factors, we can choose 0 coordinates such that f3 = t3 + t3 , and get the cubic. If f3 has a linear factor with 1 2 multiplicity 2, we reduce f3 to the form t2 t2 . This is the case of a cuspidal cubic. 1 Finally, if f3 is a cube of a linear form, we reduce the latter to the form t3 and get three 1 concurrent lines. Remark 3.2.4. The locus of Fermat cubics is isomorphic to the homogeneous space PSL(3)/32 .S3 . Its closure in P(S 3 E ∗ ) is a hypersurface F and consists of curves listed in the assertion of the previous lemma and also reducible cubics equal to the unions of irreducible conics with its tangent line. The explicit equation of the hypersurface F is given by the Aronhold invariant I4 of degree 4 in the coefficients of the cubic equation. Its formula can be found in mnay text-books in invariant theory (e.g. [80]). If the cubic is written in a Weierstrass form f = t0 t2 +t3 +at2 t1 +bt3 = 0, then 2 1 0 0 I4 (f ) = λa, for some nonzero constant λ independent of f . A nice expression forI4 in terms of a pfaffian of a skew-symmetric matrix was given by G. Ottaviani [187]. Lemma 3.2.8. The following properties equivalent (i) AP1 (f ) = {0}; (ii) dim AP2 (f ) > 2; (iii) V (f ) is equal to the union of three concurrent lines. Proof. By the apolarity duality (Af )1 × (Af )2 → (Af )3 ∼ C, = we have dim(Af )1 = 3 − dim AP1 (f ) = dim(Af )2 = 6 − dim AP2 (f ). Thus dim AP2 (f ) = 3 + dim AP1 (f ). This proves the equivalence of (i) and (ii). By definition, AP(f )1 = {0} if and only if Dψ (f ) = 0 for some nonzero linear operator ψ = ai ∂i . After a linear change of variables, we may assume that ψ = ∂0 , and then ∂0 (f ) = 0 if and only if C does not depend on t0 , i.e. C is the union of three concurrent lines. Lemma 3.2.9. Let Z be a nondegenerate generalized polar quadrangle of f . Then |IZ (2)| is a pencil of conics in P(E ∗ ) contained in the linear system |AP2 (f )|. Conversely, let Z be a 0-dimensional cycle of length 4 in P(E). Assume that |IZ (2)| is an irreducible pencil contained in |AP2 (f )|. Then Z is a nondegenerate generalized polar quadrangle of f .

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Proof. The first assertion follows from the definition of non-degeneracy and Proposition 1.3.6. Let us prove the converse. Let V (λq1 +µq2 ) be the pencil of conics |IZ (2)|. Since AP(f ) is an ideal, the linear system L of cubics of the form V (q1 l1 +q2 l2 ), where l1 , l2 are linear forms, is contained in P(AP3 (f )). Obviously it is contained in |IZ (3)|. Since |IZ (2)| has no fixed part we may choose Q1 and q2 with no common factors. Then the map E ∗ ⊕ E ∗ → IZ (3) defined by (l1 , l2 ) → q1 l1 + q2 l2 is injective hence dim L = 5. Assume dim |IZ (3)| ≥ 6. Choose 3 points in general position on an irreducible member C of |IZ (2)| and 3 non-collinear points outside C. Then find a cubic K from |IZ (3)| which passes through these points. Then K intersects C with total multiplicity 4 + 3 = 7, hence contains C. The other component of K must be a line passing through 3 non-collinear points which is absurd. So, dim |IZ (3)| = 5 and we have L = |IZ (3)|. Thus |IZ (3)| ⊂ P(AP3 (f )) and, by Proposition 1.3.6, Z is a generalized polar quadrangle of C.

Corollary 3.2.10. Suppose C = V (f ) is not the union of three concurrent lines. The subset of VSP(C; 4) consisting of nondegenerated generalized polar quadrangles is isomorphic to an open subset of the plane |AP2 (f )|∗ . Example 3.2.1. Let V (f ) be the union of an irreducible conic and its tangent line. After a linear change of variables we may assume that F = t0 (t0 t1 + t2 ). It is easy 2 2 2 to check that AP2 (f ) is spanned by ξ1 , ξ1 ξ2 , ξ2 − ξ0 ξ1 . It follows from Lemma 3.2.7 that f does not admit degenerate polar quadrangle. Thus any polar quadrangles of C is the base locus of an irreducible pencil in |AP2 (f )|. However, it is easy to see that all nonsingular conics in |AP2 (f )| are tangent at the point [0, 1, 0]. Thus no pencil has 4 distinct base points. This shows that VSP(f ; 4)o = ∅. Of course, VSP(f ; 4) = ∅. Any irreducible pencil in |AP2 (f )| defines a generalized 2 polar quadrangle. It is easy to see that the only reducible pencil is V (λ∂1 + µ∂1 ∂2 ). Thus VSP(f ; 4) contains a subvariety isomorphic to a complement of one point in P2 = |AP2 (f )|∗ . To compactify it by P2 we need to find one more generalized polar quadrangle. Consider the subscheme Z of degree 4 concentrated at the point [1, 0, 0] with ideal at this point generated by (x2 , xy, y 3 ), where we use inhomogeneous coordinates x = ξ1 /ξ0 , y = ξ2 /ξ0 . The linear system |IZ (3)| is of dimension 5 and consists of cubics of the form V (ξ0 ξ1 (aξ1 + bξ2 ) + g3 (ξ1 , ξ2 )). Thus Z is linearly 3-independent. One easily computes AP3 (f ). It is generated by all monomials except 2 2 2 2 ξ0 ξ1 and ξ0 ξ2 and also the polynomial ξ0 ξ2 − ξ0 ξ1 . We see that |IZ (3)| ⊂ P(AP3 (f ). Thus Z is a generalized polar quadrangle of C. It is non-degenerate since |IZ (2)| is 2 the pencil V (λξ1 + µξ1 ξ2 ). So, we see that VSP(C; 4) is isomorphic to the plane ∗ |AP2 (C)| . Example 3.2.2. Let V (f ) be an irreducible nodal cubic. Without loss of generality we may assume that f = t2 t0 + t3 + t2 t0 . The space of apolar quadratic forms is 2 1 1 2 2 2 spanned by ∂0 , ∂1 ∂2 , ∂2 − ∂1 . The net |AP2 (f )| is base point-free. It is easy to see that its discriminant curve is the union of three distinct non-concurrent lines. Each line

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81

defines a pencil with singular general member but without fixed part. So, VSP(f ; 4) = |AP2 (f )|∗ . Example 3.2.3. Let V (f ) be the union of an irreducible conic and a line which intersects the conic transversally. Without loss of generality we may assume that f = 2 2 2 t0 (t2 + t1 t2 ). The space of apolar quadratic forms is spanned by ξ1 , ξ2 , 6ξ1 ξ2 − ξ0 . 0 The net |AP2 (f )| is base point-free. It is easy to see that its discriminant curve is the union of a conic and a line intersecting the conic transversally. The line defines a pencil with singular general member but without fixed part. So, VSP(f ; 4) = |AP2 (f )|∗ . Example 3.2.4. Let V (f ) be a cuspidal cubic. Without loss of generality we may assume that f = t2 t0 + t3 . The space of apolar quadratic forms is spanned by 1 2 2 ξ0 , ξ0 ξ2 , ξ2 ξ1 . The net |AP2 (f )| has 2 base points [0, 1, 0] and [0, 0, 1]. The point [0, 0, 1] is a simple base point. The point [0, 1, 0] is of multiplicity 2 with the ideal locally defined by (x2 , y). Thus base point scheme of any irreducible pencil is not reduced. There are no polar 4-polyhedra defined by the base-locus of a pencil of conics in |AP2 (f )|. The discriminant curve is the inion of two lines, each defining a pencil with a fixed line. So |AP2 (f )|∗ minus 2 points parametrizes generalized polar 4-polyhedra. We know that V (f ) admits degenerate polar 4-polyhedra. Thus VSP(f ; 4)o is not empty and consists of degenerate polar 4-polyhedra. Example 3.2.5. Let V (f ) be a nonsingular cubic curve. We know that its equation can be reduced to a Hesse form V (t3 + t3 + t3 + 6at0 t1 t2 ), where 1 + 8a3 = 0. The 0 1 2 2 2 2 2 space of apolar quadratic forms is spanned by aξ0 ξ1 − ξ2 , aξ1 ξ2 − ξ0 , aξ0 ξ2 − ξ1 . The 3 curve V (f ) is Fermat if and only if a(a − 1) = 0. In this case the net has 3 ordinary base points and the discriminant curve is the union of 3 non-concurrent lines. The net has 3 pencils with fixed part defined by these lines. Thus the set of nondegenerate generalized polyhedrons is equal to the complement of 3 points in |AP2 (f )|∗ . We know that a Fermat cubic admits degenerate polar 4-polyhedra. Suppose V (f ) is not a Fermat cubic. Then the net |AP2 (f )| is base point-free. Its discriminant curve is a nonsingular cubic. All pencils are irreducible. There are no degenerate generalized polygons. So, VSP(f ; 4) = |AP2 (f )|∗ . Example 3.2.6. Assume that V (f ) = V (t0 t1 t2 ) is the union of 3 non-concurrent lines. 2 2 2 Then AP2 (f ) is spanned by ξ0 , ξ1 , ξ2 . The net |AP2 (f )| is base point-free. The discriminant curve is the union of three non-concurrent lines representing pencils without fixed point but with singular general member. Thus VSP(f ; 4) = |AP2 (f )|∗ . It follows from the previous examples that AP2 (f )| is base point-free net of conics if and only if C does not belong to the closure of the orbit of Fermat cubics. Theorem 3.2.11. Assume that C does not belong to the closure of the orbit of Fermat cubics. Then |AP2 (f )| is a base point-free net of conics and VSP(f ; 4) ∼ |AP2 (f )|∗ ∼ P2 . = = The variety VSP(f ; 4)o is isomorphic to the open subset of |AP2 (f )|∗ whose complement is the curve B of pencils with non-reduced base-locus. The curve B is a plane sextic with 9 cusps if V (f ) is a nonsingular curve, the union of three non-concurrent lines if V (f ) is an irreducible nodal curve or the union of three lines, and the union of a conic and its two tangent lines if V (f ) is the union of a conic and a line.

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Proof. The first assertion follows from the Examples 3.2.2-3.2.6. Since the linear system of conics |AP2 (f )| is base point-free, it defines a regular map φ : P(E) → |AP2 (f )|∗ . The pre-image of a line is a conic from N . The lines through a point q in |AP2 (f )|∗ define a pencil with base locus φ−1 (q). Thus pencils with non-reduced locus are parametrized by the branch curve B of the map φ. If C is a nonsingular cubic, we know from Example 3.2.5 that the discriminant curve ∆ is a nonsingular cubic. A line in |AP2 (f )| defines a pencil of conics. Its singular members are the intersection points of the line and ∆. It is easy to see that the pencil has exactly 3 singular members if and only its base point locus consists of 4 distinct points. Thus the curve B is the dual curve of ∆. By the duality, B is dual of ∆. We know from Part 1 that it is a sextic with 9 cusps. If C is an irreducible nodal curve, we know from Example 3.2.2 that ∆ is the union of three non-concurrent lines. The locus of lines intersecting ∆ not transversally is the union of three pencils of lines. As above B must be the union of three non-concurrent lines. If C is the union of a conic and and a line, we know from Example 3.2.3 that ∆ is the union of a conic and and a line intersecting the conic transversally. Obviously B must contain an irreducible component dual to the conic. Other irreducible components must be two tangent lines to the conic. Finally, if C is the union of three lines, the map φ is given by [t0 , t1 , t2 ] → [t2 , t2 , t2 ] and as is easy to see its branch locus is the union of the coordinate lines. 0 1 2 Let C ⊂ P(S 3 E ∗ ) ∼ P9 be the locus of three concurrent lines. For each V (f ) ∈ = P(S 3 E ∗ ) \ C, the space AP2 (f ) is 3-dimensional. This defines a regular map a : P(S 3 E ∗ )\C → G(3, S 2 E). Both the varieties are 9-dimensional. Fix a 3-dimensional subspace L of S 2 E and consider the linear map a : S 3 E ∗ → Hom(L, V ∗ ), ˜ a(f )(ψ) = Dψ (f ).

Its kernel consists of cubic forms C such that L ⊂ AP2 (f ). Note that the map a is a ˜ linear map from a 10-dimensional space to a 9-dimensional space. One expects that its kernel is 1-dimensional. This shows that, for a general point L ∈ G(3, S 2 E) the pre-image a−1 is a one-point. Thus the map a is birational.

Exercises
3.1 Find the Hessian form of a nonsingular cubic given by the Weierstrass equation. 3.2 Show that a cubic curve given by the Hessian equation (3.5) is a harmonic (resp. equianharmonic) cubic if and only if 1 − 20m3 − 8m6 = 0 (resp. m(m3 − 1) = 0). 3.3 Let H = He(C) be the Hessian cubic of a nonsingular plane cubic curve C not isomorphic to a Fermat cubic. Let τ : H → H be the Steinerian automorphism of H which assigns to a ∈ H the unique singular point of Pa (C).

EXERCISES

83

ˇ ˜ ˜ (i) Let H = {(a, ) ∈ H × P2 : ⊂ Pa (C)}. Show that the projection p1 : H → H is an unramified double cover. ˜ ∼ ˜ (ii) Show that H = H/ τ [Hint: for any (x, ) ∈ H, = a, τ (a) for a unique point a ∈ C]. 3.4 Let C = V (f ) ⊂ P2 be a nonsingular cubic. (i) Show that the set of second polars of C with respect to points on a fixed line is a conic in the dual plane. Its dual conic C( ) in P2 is called the polar conic of the line. (ii) Show that C( ) is equal to the set of poles of with respect to polar conics Px (C), where x ∈ . item[(iii)] What happens to the conic C( ) when the line is tangent to C? (iv) Show that the polar conic C( ) of a nonsingular cubic C coincides with the locus of points x such that Px (C) is tangent to . (v) Show that the set of lines such that C( ) is tangent to is the dual curve of C. (vi) Let = V (a0 t0 + a1 t1 + a2 t2 ). Show that C( ) can be given by the equation 1 0 0 a0 a1 a2 2 2 2 ∂ f ∂ f C ∂ f Ba0 B ∂t0 ∂t1 ∂t0 ∂t2 C ∂t2 0 B C 2 2 g(a, t) = det B ∂2f C = 0 ∂ f ∂ f ∂t1 ∂t2 A ∂t2 @a1 ∂t1 ∂t0 1 ∂2f ∂2f ∂2f a2 ∂t2 ∂t0 ∂t2 ∂t1 ∂t2
2

(vii) Show that the dual curve C of C can be given by the equation (the Schl¨ fli equation) a 1 0 0 ξ0 ξ1 ξ2 Bξ0 ∂ 2 g(ξ,T ) (ξ) ∂ 2 g(ξ,T ) (ξ) ∂ 2 g(ξ,T ) (ξ)C 2 C B ∂t0 ∂t1 ∂t0 ∂t2 ∂t0 C 2 2 2 det B Bξ1 ∂ g(ξ,T ) (ξ) ∂ g(ξ,T ) (ξ) ∂ g(ξ,T ) (ξ)C 2 ∂t1 ∂t0 ∂t1 ∂t2 ∂t1 A @ 2 2 2 g(ξ,T g(ξ,T ξ2 ∂∂t2 ∂t0 ) (ξ) ∂∂t2 ∂t1 ) (ξ) ∂ g(ξ,T ) (ξ) 2 ∂t
2



˛ ˛ 3.5 Let C ⊂ Pd−1 be an elliptic curve embedded by the linear system ˛OC (dp0 )˛, where p0 is a point in C. Assume d = p is prime. (i) Show that the image of any p-torsion point is an osculating point of C, i.e., a point such that there exists a hyperplane (an osculating hyperplane) which intersects the curve only at this point. (ii) Show that there is a bijective correspondence between the sets of cosets of (Z/pZ)2 with respect to subgroups of order p and hyperplanes in Pp−1 which cut out in C the set of p osculating points. (iii) Show that the set of p-torsion points and the set of osculating hyperplanes define a (p2 , p(p + 1)p )-configuration of p2 points and p(p + 1) hyperplanes (i.e. each point is p+1 contained in p + 1 hyperplanes and each hyperplane contains p points). (iv) Find a projective representaion of the group (Z/pZ)2 in Pp−1 such that each osculating hyperplane is invariant with respect to some cyclic subgroup of order p of (Z/pZ)2 . 3.6 A point on a nonsingular cubic is called a sextuple point if there exists an irreducible conic intersecting the cubic at this point with multiplicity 6. Show that there are 27 sextuple points. 3.7 The pencil of lines through a point on a nonsingular cubic curve C contains four tangent lines. Show that the twelve contact points of three pencils with collinear base points on C lie on 16 lines forming a configuration (124 , 163 ) (the Hesse-Salmon configuration).

84

CHAPTER 3. PLANE CUBICS

3.8 Show that the polar of the cubic with pole at its inflection point is the union of the tangent line at this point and the line which intersects the cubic at three points which are the nonzero 2-torsion points with respect to the group law with the pole equal to the zero point. Show that, in the Hesse pencil, these lines (called harmonic lines) depend only on the choice of the inflection point. Show that the nine harmonic lines and 12 singular points of singular members of the pencil form a configuration (94 , 123 ) (the dual Hesse configuration of lines). 3.9 Prove that the second polar of a nonsingular cubic C with respect to the point a on the Hessian He(C) is equal to the tangent line Ta (He(C)). 3.10 Let a, b be two points on the Hessian curve He(C) forming an orbit with the respect to the Steiner involution. Show that the line a, b is tangent to Cay(C) at some point d. Let c be the third intersection point of He(C) with the line a, b. Show that the pairs (a, b) and c, d are harmonic conjugate. 3.11 Show that from each point a on the He(C) one can pass three tangent lines to Cay(C). Let b be the singular point of Pa (C). Show that the set of the three tangent lines consists of the line a, b and the components of the reducible polar conic Pb (C). P 3.12 Let C = V ( 0≤i≤j≤k≤2 aijk ti tj tk ). Show that its Cayleyan curve Cay(C) can be given by the equation 0 1 a000 a001 a002 ξ0 0 0 B a112 a111 a112 0 ξ1 0 C B C B a022 a122 a222 0 0 ξ2 C B C = 0. det B C B2a012 2a112 2a112 0 ξ2 ξ1 C @2a002 2a012 2a022 ξ2 0 ξ1 A 2a001 2a011 2a012 ξ1 ξ0 ξ1 3.13 Show the group of projective transformations leaving a nonsingular plane cubic invariant is a finite group of order 18, 36 or 54. Determine these groups. 3.14 Let X be nonsingular projective curve X of genus 1. (i) Show that X is isomorphic to a curve in the weighted projective plane P(1, 1, 2) given by the equation t2 + p4 (t0 , t1 ) = 0, where p4 is a homogeneous polynomial of degree 4 (a 2 binary quartic). (ii) Show that a general binary quartic can be reduced by a linear change of variables to the form t4 + t4 + 6at2 t2 (Hint: write p4 as the product of two quadratic forms, and reduce 0 1 0 1 them simultaneously to sum of squares). (iii) Show that for X such reduction with a = ± 1 is always possible. 3 (iv) Show that the linear symmetries of the reduced quartic define a group of automorphisms of X which can be identifiied, after a choice of a group law on X, with the group of translations by 2-torsion points. (v) Show that the absolute invariant of X is related to the coefficient a from part (i) via the (1+3a2 )3 formula j = (9a2 −1)2 . (vi) Show that a harmonic (resp. √ equianharmonic) cubic corresponds to the binary quartics t4 ± 6t2 t2 + t4 (resp. t4 ± 2 −3t2 t2 + t4 ). 0 1 1 0 0 1 1 0 3.15 Find all ternary cubics C such that VSP(C; 4)o = ∅. 3.16 Show that a plane cubic curve belongs to the closure of the Fermat locus if and only if it admits a first polar equal to a double line or the whole space. 3.17 Fix 3 non-collinear points a, b, c in the plane and three non-concurrent lines 1 , 2 , 3 not containing any of the three points. Show that the locus of points x such that the lines a, x, b, x, c, x intersect the line

Historical Notes

85

Historical Notes
The discovery that any plane cubic can be written by a Weierstrass equation is due to Newton. It was Weierstrass who showed that the equation can be parametrized by elliptic functions, the Weierstrass functions ℘(z) and ℘(z) . The Hesse pencil was introduced and studied by O. Hesse [135],[136]. It has also known as the syzygetic pencil (see [46]). More facts about the Hesse pencils and and its connection to other constructions in modern algebraic geometry can be found in [7]. The equations of the Cayleyan curve of a plane cubic given in the Hesse form can be found in [210]. The equation from Exercised 3.12 is taken from [46]. The equation of the dual cubic curve given in the Hesse form can be found in [210]. The Schl¨ fli a equation from Exericise 3.4 was given by L. Schl¨ fli [212]. Its modern proof is given a in [120]. The polar polygons of a plane cubics were first studied by F. London [167] and G. Scorza [215]. A modern treatment of some of their results is given in [82] (see also [198] for related results). As always we refer for more historical information and survey of many results which were omitted in our our exposition to classical books [46], [188], [90], [213].

86

CHAPTER 3. PLANE CUBICS

Chapter 4

Determinantal equations
4.1
4.1.1

Plane curves
The problem

Here we will try to solve the following problem. Given a homogeneous polynomial f (t0 , . . . , tn ) find a d × d matrix A = (lij (t)) with linear forms as its entries such that f (t0 , . . . , tn ) = det(lij (t)). (4.1)

We will also try to find in how many essentially different ways one can do it. First let us reinterpret this problem geometrically and coordinate free. Let E be a vector space of dimension n + 1 and let V, W be vector spaces of dimension d. A square matrix corresponds to a linear map V → W , or an element of V ∗ ⊗ W . A matrix with linear forms corresponds to an element of E ∗ ⊗ V ∗ ⊗ W , or a linear map φ : E → V ∗ ⊗ W. We shall assume that the map φ is injective (otherwise the hypersurface V (f ) is a cone, so we can solve our problem by induction on the number of variables). Let φ : P(E) → P(V ∗ ⊗ W ) (4.2)

be the regular map of the associated projective spaces. Let Dd ⊂ P(V ∗ ⊗ W ) be the hypersurface parametrizing non-invertible linear maps. If we choose bases in V, W , then Dd is given by the determinant of a square matrix (whose entries will be coordinates in V ∗ ⊗ W ). The pre-image of Dd in P(E) is a hypersurface V (f ) of degree d. Our problem is to construct such a map φ in order that a given hypersurface is obtained in this way. sing Note that the singular locus Dd of the determinantal variety Dd corresponds to matrices of corank ≥ 2. It is easy to see that its codimension in P(V ∗ ⊗ W ) is equal to s 4. If the image of P(E) intersects Dd , then φ−1 (Dd ) will be a singular hypersurface. So, a nonsingular hypersurface of dimension ≥ 3 cannot be given by a determinantal equation. 87

88

CHAPTER 4. DETERMINANTAL EQUATIONS

4.1.2

Plane curves

Let us first consider the case of nonsingular plane curves C = V (f ) ⊂ P2 . Assume that C admits a determinantal form. As we have explained, the image of the map φ sing does not intersect Dd . Thus, for any x ∈ C, the corank of the matrix φ(x) is equal to 1 (here we consider a matrix up to proportionality since we are in the projective space). The kernel of this matrix is a one-dimensional subspace of V , i.e., a point in P(V ). This defines a regular map r : C → P(V ), x → Ker(φ(x)).

Now let t φ(x) : W ∗ → V be the transpose map. In coordinates, it corresponds to the transpose matrix. Its kernel is isomorphic to Im(φ(x))⊥ and is also one-dimensional. So we have another regular map l : C → P(W ∗ ), Let L = r∗ OP(V ) (1), M = l∗ OP(W ) (1). These are invertible sheaves on the curve C. We can identify V with a subspace of H 0 (C, L)∗ and W with a subspace of H 0 (C, M). Consider the composition of regular maps s2 r×l ψ : C −→ P(V ) × P(W ∗ ) −→ P(V ⊗ W ∗ ), (4.3) where s2 is the Segre map. It follows from the definition of the Segre map, that the tensor ψ(x) is equal to r(x) ⊗ l(x). It can be viewed as a linear map V ∗ → W ∗ . In coordinates, the matrix of this map is the product of the column vector defined by Ker(φ(x)) and the row vector defined by Ker(t φ(x)). It is a rank 1 matrix equal to the adjugate matrix of the matrix A = φ(x) (up to proportionality). Recall that a square matrix of rank 1 has a solution defined by any column of the adjugate matrix (since we have A · adj(A) = 0). Similarly, the kernel of the transpose of A is given by any row of the adjugate matrtix. Thus the projective coordinates of the matrix ψ(x) are cofactors of the matrix φ(x). Consider the rational map Adj : P(V ∗ ⊗ W )− → P(V ⊗ W ∗ ) (4.4) x → Ker(t φ(x)).

defined by taking the adjugate matrix. Recall that the adjugate matrix should be considered as a linear map Λd−1 V → Λd−1 W and we can identify P(Λd−1 V ∗ ⊗ Λd−1 W ) with P(V ⊗ W ∗ ). Although it is not well-defined on vector spaces, it is well-defined, as a rational map, on the projective spaces (see Example 1.1.2). Let Ψ = Adj ◦ φ, then ψ is equal to the restriction of Ψ to C. Since Adj is defined by polynomials of degree d − 1 (after we choose bases in V, W ), we have Ψ∗ OP(V ⊗W ∗ ) (1) = OP(E) (d − 1). This gives ψ ∗ OP(V ⊗W ∗ ) (1) = OP(E) (d − 1) ⊗ OC = OC (d − 1).

4.1. PLANE CURVES On the other hand, we get ψ ∗ OP(V ⊗W ∗ ) (1) = (s2 ◦ (r × l))∗ OP(V ⊗W ∗ ) (1) = (r × l)∗ s∗ OP(V ⊗W ∗ ) (1) = (r × l)∗ p∗ OP(V ) (1) ⊗ p∗ OP(W ∗ ) (1) 2 1 2 = r∗ OP(V ) (1) ⊗ l∗ OP(W ) (1) = L ⊗ M.

89

Here p1 : P (V ) × P(W ∗ ) → P(V ), p2 : P (V ) × P(W ∗ ) → P(W ∗ ) are the projection maps. Comparing the two isomorphisms, we obtain Lemma 4.1.1. L ⊗ M ∼ OC (d − 1) = (4.5)

Remark 4.1.1. It follows from Example 1.1.2 that the rational map (4.4) is given by the polars of the determinantal hypersurface. In fact, if A = (tij ) is a matrix with independent variables as entries, then ∂ det(A) = Mij , where Mij is the ijth cofactor ∂tij of the matrix A. The map Adj is a birational map since Adj(A) = A−1 det(A) and the map A → A−1 is obviously invertible. So, the determinantal equation is an example of a homogeneous polynomial such that the corresponding polar map is a birational map. Such a polynomial is called a homaloidal polynomial (see [81]).

1 Lemma 4.1.2. Let g = 2 (d − 1)(d − 2) be the genus of the curve C. Then

(i) deg(L) = deg(M) = 1 d(d − 1) = g − 1 + d; 2 (ii) H 0 (C, L) = V ∗ , H 0 (C, M) = W ; (iii) H i (C, L(−1)) = H i (C, M(−1)) = {0}, i = 0, 1. Proof. Let us first first prove (iii). A nonzero section of H 0 (C, L(−1)) is a section of L which defines a hyperplane in P(V ) which intersects the image r(C) of the curve C along a divisor r(D), where D is cut out in C by a line. Since all such divisors D are linear equivalent, we see that for any line the divisor r( ∩C) is cut out by a hyperplane in P(V ). Choose such that it intersects C at d distinct points x1 , . . . , xd . Choose bases in V and W . The image of φ( ) in P(V ∗ ⊗W ) = P(Matd ) is a pencil of matrices λA+µB. We know that there are d distinct values of (λ, µ) such that the corresponding matrix is of corank 1. Without loss of generality we may assume that A and B are nonsingular matrices. So we have d distinct λi such that the matrix A+λi B is singular. Let v1 , . . . , vd be the generators of Ker(A + λi B). The corresponding points in P(V ) are of course equal to the points r(xi ). We claim that the vectors v1 , . . . , vd are linearly independent vectors in V . The proof is by induction on d. Assume a1 v1 +· · ·+ad vd = 0. Then Avi + λi Bvi = 0 for each i = 1, . . . , d gives
d d d

0=A
i=1

ai vi =
i=1

ai Avi = −
i=1

ai λi Bvi .

90 We also have

CHAPTER 4. DETERMINANTAL EQUATIONS

d

d

0=B
i=1

ai vi =
i=1

ai Bvi .

Multiplying the second equality by λd and adding it to the first one, we obtain
d−1 d−1

ai (λd − λi )Bvi = B
i=1 i=1

ai (λd − λi )vi = 0.

Since B is invertible, this gives
d−1

ai (λi − λd )vi = 0.
i=1

By induction, the vectors v1 , . . . , vd−1 are linearly independent. Since λi = λd , we obtain a1 = . . . = ad−1 = 0. Since vd = 0, we also get ad = 0. Since v1 , . . . , vd are linearly independent, the points r(xi ) span P(W ). Hence no hyperplane contains these points. This proves H 0 (C, L(−1)) = 0. Similarly, we prove that H 0 (C, M(−1)) = 0. Applying Lemma 4.1.1 we get L(−1) ⊗ M(−1) ∼ OC (d − 3) = ωC , = where ωC is the canonical sheaf on C. By duality, H i (C, M(−1)) ∼ H 1−i (C, L(−1)), i = 0, 1. = This proves (iii). Let us prove (i) and (ii). Let H be a section of OC (1). The exact sequence 0 → L(−1) → L → L ⊗ OH → 0 gives, by passing to cohomology and applying (iii), H 1 (C, L) = 0. Replacing L with M and repeating the argument, we obtain that H 1 (C, M) = 0. We know that dim H 0 (C, L) ≥ dim V ∗ = d. Applying Riemann-Roch, we obtain deg(L) = dim H 0 (C, L) + g − 1 ≥ d + g − 1, Similarly, we get deg(M) ≥ d + g − 1. Adding up, and applying Lemma 4.1.1, we obtain d(d − 1) = deg OC (d − 1) = deg(L) + deg(M) ≥ 2d + 2g − 2 = d(d − 1). Thus all the inequalities in above are the equalities, and we get assertions (i) and (ii). (4.6)

4.1. PLANE CURVES

91

Now we would like to prove the converse. Let L and M be invertible sheaves on C satisfying (4.5) and properties from the previous lemma hold. Let r : C → P(V ), l : C → P(W ∗ ) be the maps given by the complete linear systems | L | and | M |. We define ψ : C → P(V ⊗ W ∗ ) to be the composition of r × l and the Segre map s2 . It follows from property (4.5) that the map ψ is the restriction of the map Ψ : P(E) → P(V ⊗ W ∗ ) given by a linear system of plane curves of degree d − 1. We can view this map as a tensor in S d−1 (E ∗ ) ⊗ V ⊗ W ∗ . In coordinates, it is a d × d matrix A(t) with entries from the space of homogeneous polynomials of degree d − 1. Since Ψ|C = ψ, for any point x ∈ C, we have rankA(x) = 1. Let M be a 2 × 2 submatrix of A(t). Since det M (x) = 0 for x ∈ C, we have f | det M . Consider a 3 × 3 submatrix N of A(t). We have det adj(N ) = det(N )2 . Since the entries of adj(N ) are determinants of 2 × 2 submatrices, we see that f 3 | det(N )2 . Since C is irreducible, this immediately implies that f 2 | det(N ). Continuing in this way we obtain that f d−2 divides all cofactors of the matrix A. Thus B = f 2−d adj(A) is a matrix with entries in E ∗ . Since rank B = rank adj(A), and rank A(x) = 1, we get that rank B(x) = d − 1 for any x ∈ C. So, if det B is not identically zero, we obtain that V (det(B)) is a hypersurface of degree d vanishing on C, hence det(B) = λf for some λ ∈ K ∗ . This shows that C = V (det(B)). To see that det(B) = 0, we have to use property (iii) of Lemma 4.1.2. Reversing the proof of this property, we see that for a general line in P(E) the images of the points xi ∈ ∩ C in P(V ) × P(W ∗ ) are the points (ai , bi ) such that the ai ’s span P(V ) and the bi ’s span P(W ∗ ). The images of the xi ’s in P(V ⊗W ∗ ) under the map Ψ span a subspace L of dimension d−1. If we choose coordinates so that the points ai and bi are defined by the unit vectors (0, . . . , 1, . . . , 0), then L corresponds to the space of diagonal matrices. The image of the line under Ψ is a Veronese curve of degree d − 1 in L. A general point Ψ(x), x ∈ , on this curve does not belong to any hyperplane in L spanned by d − 1 points xi ’s, thus it can be written as a linear combination of the points Ψ(xi ) with nonzero coefficients. This represents a matrix of rank d. This shows that det A(x) = 0 and hence det(B(x)) = 0. To sum up, we have proved the following theorem. Theorem 4.1.3. Let C ⊂ P2 be a nonsingular plane curve of degree d. Let Pic(C)g−1 be the Picard variety of isomorphism classes of invertible sheaves on C of degree g − 1 (or divisor classes of degree g − 1). Let Wg−1 ⊂ Picg−1 be the subset parametrizing invertible sheaves F with H 0 (C, F) = {0} (or effective divisors of degree g − 1). Let L0 ∈ Picg−1 \ Wg−1 , and M0 = ωC ⊗ L−1 . Then V = H 0 (C, L0 (1))∗ and 0 W = H 0 (C, M0 (1)) have dimension d and there is a unique regular map φ : P2 → P(V ∗ ⊗ W ) such that C is equal to the pre-image of the determinantal hypersurface Dd and the maps r : C → P(V ) and l : C → P(W ∗ ) given by the complete linear systems | L0 (1) | and | M0 (1) | coincide with the maps x → Ker(φ(x)) and x → Ker(t φ(x)), respectively. Conversely, given a map φ : P2 → P(V ∗ ⊗ W ) such that C = φ−1 (Dd ) there exists a unique L0 ∈ Picg−1 (C) such that V ∼ H 0 (C, L0 (1))∗ , = W ∼ H 0 (C, ωC (1) ⊗ L0 )−1 and the map φ is defined by L as above. = Remark 4.1.2. Let X be the set of d × d matrices A(t) with entries in E ∗ such that f = det A(t). The group G = GLd × GLd acts on the set by
−1 (σ1 , σ2 ) · A = σ1 · A · σ2 .

92

CHAPTER 4. DETERMINANTAL EQUATIONS

It follows from the Theorem that the orbit space X/G is equal to Picg−1 (C) \ Wg−1 . We map L0 → M0 = ωC ⊗ L−1 is an involution on Picg−1 \ Θ. It corresponds to 0 the involution on X defined by taking the transpose of the matrix.

4.1.3

The symmetric case

Let us assume that the determinant representation of a plane irreducible curve C of degree d is given by a pair of equal invertible sheaves L = M. It follows from Lemmas 4.1.1 and 4.1.2 that • L⊗2 ∼ OC (d − 1); = • deg(L) = 1 d(d − 1); 2 • H 0 (C, L(−1)) = {0}. Recall that the canonical sheaf ωC is isomorphic to OC (d − 3). Thus (L(−1))⊗2 ∼ ωC . = (4.7)

Definition 4.1. Let X be a curve with a canonical invertible sheaf ωX (e.g. a nonsingular curve, or a curve on a nonsingular surface). An invertible sheaf N whose tensor square is isomorphic to ωX is called a theta characteristic. A theta characteristic is called even (resp. odd) if dim H 0 (X, N ) is even (resp. odd). Using this definition we can express (4.7) by saying that L ∼ N (1), = where N is an even theta characteristic (because H 0 (C, N ) = {0}). Of course the latter condition is stronger. An even theta characteristic with no nonzero global sections (resp. with nonzero global sections) is called a non-effective theta characteristic (resp. effective theta characteristic). Rewriting the previous subsection in the special case L = M we obtain that V ∗ = 0 H (C, L) = H 0 (C, M) = W . The maps l = r given by the linear systems | L | and | M | and define a map r × r : C → P(V ) × P(V ). Its composition with the Segre map P(V ) × P(V ) → P(V ⊗ V ) and the projection to P(S 2 V ) defines a map φ : C → P(S 2 V ). In coordinates it is given by ψ(x) = r(x) · t r(x), ˜ ˜ where r(x) is the column of projective coordinates of the point r(x). Obvioulsy the ˜ image of the map ψ is contained in the variety of rank 1 quadrics in the dual space P(V ∗ ). It follows from the proof of Theorem 4.1.3 that there exists a linear map φ : P2 → P(S 2 V ∗ ) such that its composition with the rational map defined by taking the adjugate matrix equals, after restriction to C, the map ψ. The image of φ is a net N of

4.1. PLANE CURVES

93

quadrics in P(V ). The image of C is the locus of singular quadrics in N . For each point x ∈ C we denote the corresponding quadric by Qx . The rational map l (regular if C is nonsingular) is defined by assigning to a point x ∈ C the singular point of the quadric 1 Qx . The image X of C in P(V ) is a curve of degree equal to deg L = 2 d(d − 1). Proposition 4.1.4. The restriction map r : H 0 (P(V ), OP(V ) (2)) → H 0 (X, OX (2)) is surjective. Under the isomorphism H 0 (X, OX (2)) ∼ H 0 (C, L⊗2 ) = H 0 (C, OC (d − 1)) = the space of quadrics in P(V ) is identified with the space of plane curves of degree d − 1. The net of quadrics N is identified with the linear system of first polars of the curve C. Proof. Reversing the proof of property (iii) from Lemma 4.1.2 shows that the image of C under the map ψ : C → P(V ⊗ W ∗ ) spans the space. In our case, this implies that the image of C under the map C → P(S 2 V ) spans the space of quadrics in the dual space. If the image of C in P(V ) were contained in a quadric Q, then Q would be apolar to all quadrics in the dual space, a contradiction. Thus the restriction map r is injective. Since the spaces have the same dimension, it must be surjective. The composition of the map i : P2 → |OP(V ) (2)|, x → Qx , and the isomorphism |OP(V ) (2)| ∼ |OP2 (d − 1)| is a map s : P2 → |OP2 (d − 1)|. A similar map s is = given by the first polars x → Px (C). We have to show that the two maps coincide. Recall that Px (C) ∩ C = {c ∈ C : x ∈ Tc (C)}. In the next lemma we will show that the quadrics Qx , x ∈ Tc (C) form the line in N of quadrics passing through the singular point of Qc equal to r(c). This shows that the quadric Qr(x) cuts out in r(C) the divisor r(Px (C) ∩ C). Thus the curves s(x) and s (x) of degree d − 1 cut out the same divisor on C, hence they coincide. Lemma 4.1.5. Let W ⊂ S d V ∗ be a linear system of hypersurfaces of degree d, and D ⊂ P(W ) be the locus of singular hypersurfaces. Assume x ∈ D is a nonsingular point. Then the corresponding hypersurface has a unique ordinary double point y and the embedded tangent space Tx (D) is equal to the hyperplane of hypersurfaces containing y. Proof. Assume W = S d V ∗ . Then D coincides with the discriminant hypersurface D of all singular degree d hypersurfaces in P(V ). In this case the assertion follows from 1.2.1, where we described explicitly the tangent space of D at any point. Since D = P(W ) ∩ D and x ∈ D is a nonsingular point, the intersection is transversal and Tx (D) = Tx (D) ∩ P(W ). This is our assertion. We see that a pair (C, N ), where C is a plane irreducible curve and N is a noneffective even theta characteristic on C defines a net N of quadrics in P(V ), where V = H 0 (C, N (1)). Conversely, given a net N of quadrics in Pd−1 = P(V ). It is

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known that the singular locus of the discriminant hypersurface D2 (d−1) of quadrics in Pd−1 is of codimension 2. Thus a general net N intersects D2 (d−1) transversally along a nonsingular curve C of degree d. This gives a representation of C as a symmetric determinant and hence defines an invertible sheaf L and a non-effective even theta characteristic N . This gives a dominant rational map of varieties of dimension (d2 + 3d − 16)/2 (4.8) G(3, S 2 V ∗ )/PGL(V )− → |OP2 (d)|/PGL(3). The degree of this map is equal to the number of non-effective even theta characteristics on a general curve of degree d. We will see in the next chapter that the number of even theta characteristics is equal to 2g−1 (2g + 1), where g = (d − 1)(d − 2)/2 is the genus of the curve. A curve C of odd degree d = 2k + 3 has a unique vanishing even theta characteristic equal to N = OC (k) with h0 (N ) = (k + 1)(k + 2)/2. A general curve of even degree does not have vanishing even theta characteristics. Observe that under the isomorphism from Proposition 4.1.4, the variety of quadrics of rank 1 (i.e. double hyperplanes) is mapped isomorphically to the variety of plane curves of degree d − 1 which are everywhere tangent to the curve C. We call these curves contact curves of order d − 1. Thus any symmetric determinantal representation of C determines an algebraic system of dimension d − 1 of contact curves of degree d − 1. Proposition 4.1.6. Let C = V (f ), where C is equal to the determinant of a d × d symmetric matrix (lij ) of linear forms. Then the corresponding algebraic system of contact curves of degree d − 1, considered as a hypersurface in P(E) × P(V ) of bidegree (d − 1, 2), is given by the equation   l00 ... l0d u0  l11 ... l1d u1     . . .  = 0. . .  det  . (4.9) ... . .   . ld−11 . . . ld−1d−1 ud−1  u0 ... ud−1 0 Proof. Obviously the bordered determinant (4.9) can be written in the form
d−1

Aij ui uj ,
i,j=0

where Aij is the (ij)-cofactor of the matrix A. For any x ∈ C, the rank of the cofactor matrix adjA(x) is equal to 1. Thus the quadratic form on the dual space of P(V ) with coordinates u0 , . . . , ud−1 defined by the above equation is of rank 1. Hence it is equal to the double hyperplane H 2 = ( ai ui )2 , where (a0 , . . . , ad−1 ) ∈ P(W ) belongs to the null-space of the matrix A(x). Under the identification of Pd−1 = P(V ) with |OP2 (L)|∗ , the hyperplane H in the dual space corresponds to the point l(x), where l : C → Pd−1 is the map defined by the symmetric determinantal representation of C. This checks the assertion.

4.1. PLANE CURVES Consider the multiplication map H 0 (C, N (1)) ⊗ H 0 (C, N (1)) → H 0 (C, KC (2)).

95

(4.10)

Let use the notation V = H 0 (C, N (1)) ∼ Pd−1 as above. Passing to the projective = spaces we get a regular map P(V ) × P(V ) → P(KC (2)) = |OP2 (d − 1)|, (D1 , D2 ) → D1 + D2 It defines a hypersurface F ⊂ P(V ) × P(V ) × P(E) = Pd−1 × Pd−1 × P2 F = {(D1 , D2 , x) ∈ P(V ) × P(V ) × P(E) : x ∈ D1 + D2 } (4.12) (4.11)

defined by a 3-homogenous equation of 3-degree (1, 1, d − 1), symmetric in the variables of degree 1. In coordinates, it is equal to F :
0≤i,j≤d−1

aij (t0 , t1 , t2 )ui vj = 0,

where aij = aji are homogeneous forms of degree d − 1. The projection F → Pd−1 × Pd−1 is a family of curves of degree d−1 parametrized by Pd−1 ×Pd−1 . The projection to P2 is a family of divisors of type (1, 1) on Pd−1 × Pd−1 . For any x ∈ C, the set of divisors (D1 , D2 ) containing x is the union of two divisors Hx × Pd−1 and Pd−1 × Hx , where Hx is the hypersurface of divisors in |N (1)| containing the point x. Since this divisor is singular C is contained in the locus Σ of points parametrizing singular fibres of F → P2 . Note that Σ is given by the determinant of the matrix (aij ) from above, and its degree is equal to d(d − 1). We can find the equation of Σ using the following beautiful determinant identity due to O. Hesse [138]. Lemma 4.1.7. Let A = (aij ) be a square matrix of size k × k. For any x = (x1 , . . . , xk ) and y = (y1 , . . . , yk ) a11 a21 . . . ak1 x1 a11 a21 . − . . ak1 y1 a12 a22 . . . ak2 x2 a12 a22 . . . ak2 y2 ... ... . . . ... ... ... ... . . . a1k a2k . . . akk xk a1k a2k . . . x1 a11 x2 a21 . × . . . . . xk ak1 0 y1 x1 x2 . . . x2 0
2

a12 a22 . . . ak2 y2 a12 a22 . . . ak2

... ... . . . ... ... ... ... . . . ...

a1k a2k . . . akk yk

y1 y2 . . . yk 0

. . . akk . . . yk

a11 a21 = . . . ak1

a1k a2k . × U, . . akk

where U = U (a11 , . . . , akk ; x1 , . . . , xk ; y1 , . . . , yk ) is a polynomial of degree k − 2 in variables aij and of degree 2 in variables xi and yj .

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Replacing x, y by x = αx + βy, y = γ x + δy, the left-hand side changes by a constant multiple equal to the square of the determinant of matrix α β . This shows γ δ that the polynomial U depends only on the Pl¨ cker coordinates pij of the line x, y. u In the case when C = V (|A(t)|), we can interpret the determinantal equality as follows. We consider x and y as projective coordinates in Pd−1 . By Proposition 4.1.6, the left-hand side is equal to f g − h2 , where V (f ) ∩ C = 2D(x), V (g) ∩ C = 2D(y) and V (h) ∩ C = 2(D1 + D2 ). This shows that f g − h2 vanishes on the curve C = V (|A|). The residual curve is of degree 2(d − 1) − d = d − 2. Thus varying x, y, we get a family of curves of degree d − 2 parametrized by the Pl¨ cker coordinates of the u lines spanned by the points [x1 , . . . , xd ], [y1 , . . . , yd ]. We can view this as a family of quadric hypersurfaces in the Grassmannian of lines in Pd−1 parametrized by the plane P2 .

4.1.4

Examples

Take d = 2. Then Picg−1 (C) is one point represented by the divisor class of degree −1. It is obviously non-effective. Thus there is unique (up to the equivalence relation defined in Remark 4.1.2) representation of a conic as a determinant. For example, t0 t1 − t2 = det 2 t0 t2 t2 . t1

Take d = 3. Then Picg−1 (C) = Pic0 (C). If we fix a point x0 ∈ C, then x → [x − x0 ] defines an isomorphism from Pic0 (C) and the curve C. The divisor x − x0 is effective if and only if x = x0 . Thus we obtain that Pic0 (C) \ Wg−1 = C \ {x0 }. Let L0 = OC (D), where D is a divisor of degree 0. Then L = L0 (1) = OC (H + D), where H is a divisor of 3 collinear points. Similarly, M0 = OC (−D) and M = M0 (1) = OC (H −D). Note that any positive divisor of degree 3 is linearly equivalent to H + D for some degree 0 divisor D. Thus any line bundle L = L0 (1), where L0 ∈ Pic0 (C)\Wg−1 corresponds to a positive divisor of degree 3 not cut out by a line. The linear system |L| gives a reembedding C → C ⊂ P2 which is not projectively equivalent to the original embedding. The map r × l maps C to a curve X ⊂ P2 × P2 . Consider the restriction homomorphism α : V ∗ ⊗ W ∼ H 0 (P(V ), OP(V ) (1)) ⊗ H 0 (P(W ∗ ), OP(W ∗ ) (1)) ∼ = = H 0 (P(V ) × P(W ∗ ), p∗ OP(V ) (1) ⊗ p∗ OP(W ∗ ) (1)) → 1 2 → H 0 (X, p∗ OP (V ) (1) ⊗ p∗ OP(W ∗ ) (1) ⊗ OX ) ∼ = 1 2 ∼ H 0 (X, L ⊗ M) ∼ H 0 (X, OX (2)). = =

4.1. PLANE CURVES Lemma 4.1.8. The kernel of the restriction map α is of dimension 3. Let
2

97

aij xi yj = 0,
i,j=0

(k)

k = 1, 2, 3,

(4.13)

be the sections of bi-degree (1, 1) which span the kernel. Let X ⊂ P2 × P2 be the variety defined by these equations. Then X = (r × l)(C). Proof. The target space of α is of dimension 6 = dim H 0 (P2 , OP2 (2)). The domain of α is of dimension 9. In coordinates, an element of the kernel is a matrix A such that xAy = 0 for any (x, y) ∈ C. Since the image of C under the Segre map is equal to the image of an elliptic curve under a map defined by the complete linear system of degree 6, it must span P5 . Thus we have 6 linearly independent conditions on A. This shows that the kernel is of dimension 3. The projection of X to the first factor is equal to the locus of points [t0 , t1 , t2 ] such that the system
2 2 2

aij ti yj =
i,j=0

(k)

(
j=0 i=0

aij ti )yj = 0,

(k)

k = 1, 2, 3

has a nontrivial solution. The condition for this is  2  2 2 (1) (1) (1) i=0 ai0 ti i=0 ai1 ti i=0 ai2 ti    2 2  2 (2) (2) (2)   a ti det  ai1 ti ai2 ti  = 0 i0  i=0 i=0 i=0  2 2  2 (3)  (3) (3) ai0 ti ai1 ti ai2 ti
i=0 i=0 i=0

(4.14)

Thus, replacing [t0 , t1 , t2 ] with unknowns t0 , t1 , t2 , we obtain that the projection is either a cubic curve C or the whole plane. Assume that the second case occurs. Since the determinant of a matrix does not change after taking the transpose of the matrix, we see that the projection of X to the second factor is also the whole plane. This easily implies that X is a graph of a projective automorphism P2 → P2 . In appropriate coordinates X becomes the diagonal, and hence C embeds in P2 × P2 by means of the diagonal map P2 → P2 × P2 . But this means that L ∼ M ∼ OC (1). This contradicts = = our choice of L. Thus the projection of X and of C to the first factor is the cubic curve C equal to C reembedded by |L|. Similarly, the projection of X and of C to the second factor is the cubic curve C which is C reembedded by |M|. This implies that X = (r × l)(C). Since any matrix A(t) can be written in the form (4.14), we see that a determinantal equation of a plane cubic defines a model of the cubic as a complete intersection of three bilinear hypersurfaces in P2 × P2 .

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4.1.5

Quadratic Cremona transformations

Note that (4.14) gives a determinantal equation for the reembedded curve C = r(C). Let us see that different plane models of the same elliptic curve differ by a birational transformation of the plane. Let ϕ : X → Pn be a regular map from a variety X to a projective space. Recall that it is defined by an invertible sheaf F = f ∗ OPn (1) and a set of n + 1 sections (s0 , . . . , sn ). Two different maps differ by a projective automorphism of Pn if and only if they are defined by isomorphic sheaves and isomorphic sets of sections. Suppose we have an automorphism σ : X → X. Then the composition ϕ ◦ σ : X → Pn is defined by the invertible sheaf σ ∗ L and sections σ ∗ (s0 ), . . . , σ ∗ (sn ). Of course, the images of both maps ϕ and ϕ ◦ σ are the same, but there is no projective automorphism of Pn which induces the automorphism σ. However, in some cases one can find a birational automorphism T of Pn which does this job. Recall that, although T may be not defined on a closed subset Z ⊂ Pn , it could be defined on the whole X. This happens, for example, when X is a nonsingular curve and Z ∩ X is a set of points. In fact, we know that any rational map of nonsingular projective curve to a projective variety extends to a regular map. Assume T is given by a linear system |V | of hypersurfaces of degree m such that none of them vanish identically on the curve X. Let x1 , . . . , xk be the points on X ∩ Z. All polynomials f ∈ V intersect X with some multiplicity. Let mi be the minimal multiplicity (it is enough to compute it for a basis of V ). Then it is easy to see that the restriction of T to X is given by a linear system defined by the line bundle F = OX (m) ⊗ OX (−m1 x1 − . . . − mk xk ). This is the invertible sheaf which defines the regular map T ◦ ϕ : X → Pn . Sometimes this map defines a new embedding of X. Let us apply this to our situation. Fix a group law on an elliptic curve X with the zero point x0 . Let τx be the translation automorphism defined by a point x. Recall that τx (y) = x ⊕ y ∼ x + y − x0 . For any divisor D =
∗ τx (D) =

ni xi , we have
−1 ni τx (xi ) =

ni (xi

x) ∼

ni (xi + x0 − x)

=

ni xi + deg(D)(x0 − x).

In particular, we see that τx acts identically on divisors of degree 0 and in particular on divisors of functions. This allows one to define the action of τx on the divisor classes. Suppose we have two divisors D1 , D2 of the same degree m = 0. Then D1 − D2 is of degree 0. Thus we can find a degree 0 divisor G such that mG ∼ D1 − D2 (we use that the endomorphism of algebraic groups [m] : X → X, x → m · x is surjective). Let G ∼ xG − x0 for a unique point xG . Then
∗ τxG (D1 ) = D1 + m(x0 − xG ) = D1 − mG ∼ D2 .

(4.15)

This shows that translations act transitively on divisor classes of the same positive degree. Now suppose we have two embeddings of an elliptic curve φi : X → Pn , i = 1, 2, which are given by a complete linear systems defined by the corresponding invertible

4.1. PLANE CURVES

99

∗ sheaves L1 , L2 . By the above we can find a point x ∈ X such that τx L1 = L2 ∗ (recall that for any divisor D and any regular map ϕ : X → Y we have ϕ OY (D) = OX (ϕ∗ (D))). This shows that the embeddings φ2 : X → Pn and φ1 ◦tx : X → Pn are defined by the same invertible sheaf, and hence their images are projectively equivalent. But the image of φ1 ◦ τx is obviously equal to the image of φ1 . Thus there exists a projective transformation σ which sends φ1 (X) to φ2 (X) such that, for any y ∈ X,

σ(φ1 (τx (y)) = φ2 (y). Thus if we change φ1 by σ◦φ1 (by choosing different basis of the linear system defining φ1 ), we find that one can always choose bases in linear systems |L1 | and |L2 | such that the corresponding maps have the same image. In particular, any plane nonsingular cubic can be obtained as the image of an elliptic curve under a map defined by any complete linear system of degree 3. Now let us see how this implies that a translation automorphism of a nonsingular plane cubic can be realized by a certain Cremona transformation of the plane. Let T : P2 − → P 2 , [t0 , t1 , t2 ] → [f0 (t0 , t1 , t2 ), f1 (t0 , t1 , t2 ), f2 (t0 , t1 , t2 )]

be a rational map of P2 to itself given by polynomials of degree 2. The pre-image of a line V (a0 t0 + a1 t1 + a2 t2 ) is the conic V (a0 f0 + a1 f1 + a2 f2 ). The pre-image of a general point is equal to the intersection of the pre-images of two general lines, thus the intersection of two conics from the net L of conics spanned by f0 , f1 , f2 . If we want T to define a birational map we need the intersection of two general conics to be equal to 1. This can be achieved if all conics pass through the same set of three points p1 , p2 , p3 (base points). These points must be non-collinear, otherwise all polynomials have a common factor, after dividing, we get a projective transformation. Birational automorphisms of P2 (Cremona transformations) which are obtained by nets of conics through three non-collinear points are called quadratic transformations. If we choose a basis in P2 such that p1 = [1, 0, 0], p2 = [0, 1, 0], p3 = [0, 0, 1] and a basis in L given by the conics V (t1 t2 ), V (t0 t2 ), V (t0 t1 ), then the transformation is given by the formula T : [t0 , t1 , t2 ] → [t1 t2 , t0 t2 , t0 x1 ]. (4.16) This is called the standard Cremona transformation. In affine coordinates, it is given by T : (x, y) → (x−1 , y −1 ). Let C be a nonsingular cubic curve containing the base points p1 , p2 , p3 of a quadratic transformation T . Then the restriction of T to C is given by the complete linear system |2H − p1 − p2 − p3 |, where H is a line section of C. It is of degree 3, and hence defines an embedding of i : C → P2 such that i∗ OP2 (1) ∼ OC (2H − p1 − p2 − p3 ). Since = H = (2H − p1 − p2 − p3 ) − (H − p1 − p2 − p3 ), it follows from (4.15) that i∗ OC (1) ∼ OC (2H − p1 − p2 − p3 ), = x where 3(x − x0 ) ∼ p1 + p2 + p3 − H. As we have explained earlier, this implies that there exists a projective automorphism σ such that T = σ · T induces the translation automorphism τx on C.

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It follows from this that the group of translations acts transitively on the set of determinantal equations of C. One can change one discriminant equation to any other one by applying a quadratic transformation of P2 which leaves the curve invariant and induces a translation automorphism of the curve. Example 4.1.1. Let   t0 t2 t 2 f = t2 t1 + t2 t2 + t2 t0 = det −t1 t0 0  (4.17) 0 1 2 −t2 0 t1 Apply the Cremona transformation T : [t0 , t1 , t2 ] → [t0 t1 , t0 t2 , t1 t2 ]. We have (t0 t1 )2 t0 t2 + (t0 t2 )2 t1 t2 + (t1 t2 )2 t0 t1 = t0 t1 t2 (t2 t1 + t2 t2 + t2 t0 ). 0 1 2 Thus T transforms the curve to itself. Substituting (4.18) in A(t) from (4.17), we get    t0 t1 t1 t2 t1 t2 t0 0  = t0 t1 t2 det −t2 det −t0 t2 t0 t1 −t1 t2 0 t0 t2 −t1 Thus the new determinantal equation is  the entries of the matrix t2 t1 0  t2 0. t0 (4.18)

t0 f = det −t2 −t1

t2 t1 0

 t2 0. t0

(4.19)

However, it is projectively equivalent to the old one.     t0 t2 t0 t2 t2 1 0 0 −t2 t1 0  = 0 0 1 −t1 t0 0 1 0 −t2 0 −t1 0 t0

 t2 1 0  0 0 t1

0 0 1

 0 1 0

Let r be the right kernel map defined by the second matrix. We have r([0, 1, 0]) = [0, 0, 1], r([0, 0, 1]) = (0, 1, −1), r([1, 0, 0]) = [0, 1, 0]. Since the points (0, 1, −1), [0, 1, 0], [0, 0, 1] are on a line, we get L = r∗ OC (1) = OC (x1 + x2 + x3 ), where x1 = [0, 1, 0], x2 = [0, 0, 1], x3 = [1, 0, 0]. Thus the second determinantal equation corresponds to L0 = L(−1) = OC (x1 + x2 + x3 − H), where H is a line section. Doing the same for the first matrix we find the same invertible sheaf L0 . Note that 3H ∼ 3(x1 + x2 + x3 ) since V (t0 t1 t2 ) cuts out the divisor 3x1 + 3x2 + 3x3 . This shows that the Cremona transformation induces an automorphism of the curve C equal to translation tx , where x is a 3-torsion point. But we know from Lectrure 4 that such automorphism is induced by a projective transformation. This explains why we are not getting an essentially new determinantal equation.

4.1. PLANE CURVES

101

4.1.6

A moduli space

Let us consider the moduli space of pairs (C, A(t)), where C is a nonsingular plane curve of degree d, A(t) is a matrix of linear forms such that C = V (det(A(t))). We say that two pairs (C, A(t)) and (C, B(t)) are isomorphic if there exists invertible matrices C and D such that B(t) = CA(t)D. Equivalently we consider the space P(E ∗ ⊗ W ∗ ⊗ V ) modulo the natural action of the group G = GL(W ) × GL(V ) ∗ ((σ1 , σ2 )(x ⊗ y ⊗ z) = x ⊗ σ1 (y) ⊗ σ2 (z)). The determinant map A(t) → det(A(t)) is obvioulsy invariant and defines a map det : P(E ∗ ⊗ W ∗ ⊗ V )/G → |OP2 (d)|. We consider this map as a map of sets since there is an issue here whether the orbit space exists as an algebraic variety. But let us restrict this map over the subset |OP2 (d)|ns of nonsingular plane curves of degree d. When we know that the fibre of the map det over the curve C is bijective to Picg−1 (C) \ Wg−1 (C). There is an algebraic variety Picg−1 (the relative Picard scheme) and a divisor d calWd ⊂ Picd which admits a morphism p p : Picg−1 \ Wd → |OP2 (d)|ns d with fibres isomorphic to Picg−1 (C) \ Wg−1 (C). One can show that there exists a Zariski open subset P(E ∗ ⊗ W ∗ ⊗ V )ns of P(E ∗ ⊗ W ∗ ⊗ V ) such that its quotient by G is isomorphic to Picg−1 and the determinant map agrees with the projection p. d Since Picg−1 contains an open subset which is covered by an open subset of a d projective space, the variety Picg−1 is unirational. It is a very difficult question to d decide whether the variety Picg−1 is rational. It is known only for d = 3 and d = 4 d [111]. Let us sketch a beautiful proof of the rationality in the case d = 3 due to M. Van den Bergh [244]. Theorem 4.1.9. Assume d = 3. Then Pic0 is a rational variety. 3 Proof. A point of PiC 0 is a pair (C, L), where C is a nonsingular plane cubic and L is the isomorphism class of an invertible sheaf of degree 0. Let D be a divisor of degree 0 such that OC (D) ∼ L. Choose a line and let H = ∩ C = p1 + p2 + p3 . = Let pi + D ∼ qi , i = 1, 2, 3, where qi is a point. Since pi − qi ∼ pj − qj , we have pi + qj ∼ pj + qi . This shows that the lines pi , qj and pj , qi intersect at the same point rij on C. Thus we have 9 points: p1 , p2 , p3 , q1 , q2 , q3 , r12 , r23 , r13 . We have p1 + p2 + p3 + q1 + q2 + q3 + r12 + r23 + r13 ∼ ∼ (p1 +p2 +p3 )+(q1 +q2 +q3 )+(H −p1 −q2 )+(H −p1 −q3 )+(H −p2 −q3 ) ∼ 3H This easily implies that there is a cubic curve which intersects C at the nine points. Together with C we get a pencil of cubics with the nine points as the set of its base points. Let U = 3 × (P2 )3 /S3 , where S3 acts by σ : (p1 , p2 , p3 ), (q1 , q2 , q3 ) = (pσ(1) , pσ(2) , pσ(3) ), (qσ(1) , qσ(2) , qσ(3) ) .

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CHAPTER 4. DETERMINANTAL EQUATIONS

The variety U is easily seen to be rational. The projection to 3 /S3 ∼ P3 defines a = birational isomorphism between the product of P3 and (P2 )3 . For each u = (P, Q) ∈ U , let c(u) be the pencil of cubics through the points p1 , p2 , p3 , q1 , q2 , q3 and the points rij = pi , q , where (ij) = (12, (23), (13). Consider the set U of pairs (u, C), C ∈ c(u). The projection (u, C) → u has fibres isomorphic to P1 . Thus the field of rational functions on U is isomorphic to the field of rational functions on a conic over the field K(U ). But this conic has a rational point. It is defined by fixing a point in P2 and choosing a member of the pencil passing though this point. Thus the conic is isomorphic to P1 and K(U ) is a purely transendental extension of K(U ). Now we define a birational map from Pic0 to U . Each (C, L) defines a point of U by ordering 3 the set ∩ C, then defining q1 , q2 , q3 as above. The member of the corresponding pencil through pi ’s, qi ’s and rij ’s is the curve C. Conversely, a point (u, C) ∈ U defines a point (C, L) in Pic0 . We define L to be the invertible sheaf corresponding to 3 the divisor q1 + q2 + q3 . it is easy that these map are inverse to each other. Remark 4.1.3. If we choose a basis in each space E, V, W , then a map φ : E → Hom(W, V ) is determined by the matrices Ai = φ(ei ), where e1 = [1, 0, 0], e2 = [0, 1, 0], e3 = [0, 0, 1]. Our moduli space is the space of triples (A1 , A2 , A3 ) of d × d matrices up to the action of the group G = GLd × GLd simultaneously by left and right multiplication
−1 −1 −1 (σ1 , σ2 ) · (A1 , A2 , A3 ) = (σ1 A1 σ2 , σ1 A2 σ2 , σ1 A3 σ2 ).

Consider an open subset of maps φ such that A1 is an invertible matrix. Taking (σ1 , σ2 ) = (1, A−1 ), we may assume that A1 = Id is the identity matrix. The sta1 bilizer subgroup of (Id , A2 , A3 ) is the subgroup of (σ1 , σ2 ) such that σ1 σ2 = 1. Thus our orbit space is equal to the orbit space of pairs of matrices (A, B) up to simultaneous conjugation. The determinantal curve has the affine equation det(Id + XA + Y B) = 0. Compare this space with the space of matrices up to conjugation. As above this is reduced to the problem of description of the maps E → Hom(V, W ), where dim E = 2 instead of 3. The determinantal curve is replaced with a determinantal hypersurface in P1 given by the equation det(Id + XA) = 0. Its roots are (−λ−1 ), where λ are eigenvalues of the matrix A. If all roots are distinct (this corresponds to the case of a nonsingular curve!), a matrix is determined uniquely up to conjugacy by its eigenvalues, or equivalently by its characteristic polynomial. In the case of pairs of matrices, we need additional information expressed in terms of a point in Picg−1 \ Θ.

4.2
4.2.1

Determinantal equations for hypersurfaces
Cohen-Macauley sheaves

Recall that a finitely generated module M over a local Noetherian commutative ring A is called Cohen-Macaulay module if there exists a sequence a1 , . . . , an of elements in

4.2. DETERMINANTAL EQUATIONS FOR HYPERSURFACES

103

the maximal ideal of A such that n is equal to the dimension of the ring A/Ann(M ) and ai ∈ Ann(M/(a1 , . . . , ai−1 )M ), i = 1, . . . , n. If A is a Noetherian commutative ring, not necessary local, a finitely generated A-module is called Cohen-Macaulay if for any prime ideal p the localization Mp is a Cohen-Macaulay module over Ap . A Noetherian commutative ring is called a CohenMacaulay ring if, considered as a module over itself, it is a Cohen-Macaulay module. These definitions are globilized and give the notions of a Cohen-Macaulay scheme and a Cohen-Macaulay coherent sheaf. A coherent sheaf F on Pn is called an arithmetically Cohen-Macaulay sheaf (ACMsheaf) if the corresponding module Γ∗ (F) =
i∈Z

H 0 (Pn , F(j))

is a graded Cohen-Macaulay module over the ring of polynomials S = Γ∗ (OPn ). Using a local cohomology characterization of Cohen-Macaulay modules one shows that F is a ACM-sheaf if and only if the following conditions are satisfied (i) Fx is a Cohen-Macaulay module over OPn ,x for each x ∈ Pn ; (ii) H k (Pn , F(j)) = 0, for j ∈ Z, 1 ≤ k ≤ n − dim Supp(F), where Supp(F) denotes the support of F. It is known that for any Cohen-Macaulay module over a regular ring A depth(M ) + proj(M ) = dim A, where proj denotes the projective dimension, the minimal length of a free resolution of M . A global analog of this equality for ACM-sheaves is dim Supp(F) + proj(F) = n, where proj(F) denotes the projective dimension of F, the minimal length of a projective graded resolution for the module Γ∗ (F). Theorem 4.2.1. Let F be an ACM-sheaf over Pn such that dim Supp(F) = 1. Then there exists an exact sequence
r r

0→
i=0

OPn (fi ) −→
i=0

A

OPn (ei ) → F → 0.

(4.20)

Proof. Since proj(F) = 1, and hence we get a resolution of graded S-modules
r r

0→
i=0

S(fi ) →
i=0

S(ei ) → Γ∗ (F) → 0.

Passing to the corresponding sheaves in Pn we obtain the exact sequence from the assertion.

104

CHAPTER 4. DETERMINANTAL EQUATIONS

The map A is given by a r × r matrix whose ij entry is a homogeneous polynomial of degree ei − fj . We may assume that the resolution is minimal. To achieve this we must have aij = 0 whenever ei = fj . Clearly, the support F is given by the determininant of the matrix A. It is a hypersurface of some degree d. We must have d = (e1 + · · · + er ) − (f1 + · · · + fr ). (4.21)

Conversely if X = V (f ) is given as a determinant of a matrix A whose entries aij are homogeneous polynomials of degree ei − fj such that the equality (4.21) hold, then we get a resolution (4.20) defined by the matrix. The cokernel F will be an ACM-sheaf. Example 4.2.1. Take F = i∗ OV (k). Then, the minimal resolution is of course 0 → OPn (−d + k) → OPn (k) → F → 0. (4.22)

Here r = 1, f1 = −d + k, e1 = k. The equation is the tautological one X = det((f )), where (f ) is the 1 × 1 matrix with entry C. Note that according to the Lefschetz Theorem on Hyperplane Sections, Pic(V ) = ZOV (1) if n > 3. Thus (4.20) reduces to (4.22) and we cannot get any nontrivial determinantal equations for nonsingular hypersurfaces of dimension ≥ 3.

4.2.2

Determinants with linear entries

Let V be a hypersurface of degree d in Pn . Let M be an invertible sheaf on V . We will take F = i∗ (M), where i : V → Pn denotes the natural closed embedding. Then the condition (i) for a ACM-sheaf will be always satisfied (since Fx is isomorphic to OPn ,x /(tx ), where tx = 0 is a local equation of V ). Condition (ii) reads as H i (V, M(j)) = 0, 1 ≤ i ≤ n − 1, j ∈ Z. (4.23)

Assume the following additional conditions are satisfied. H 0 (V, M(−1)) = H n−1 (V, M(1 − n)) = 0. (4.24)

Consider the resolution (4.20), twist it by −1 and apply the exact sequence of cohomology. We must get
0→
r M i=0

H 0 (Pn , OPn (−fi − 1)) →
r M i=0

r M i=0

H 0 (Pn , OPn (−ei − 1)) → H 0 (Pn , F (−1)) → 0,
r M i=0

0 → H n−1 (Pn , F (1 − n)) →

H 0 (Pn , OPn (−fi − 2)) →

H 0 (Pn , OPn (−ei − 2)) → 0.

Here we used the standard facts (see [134]) that H k (Pn , OPn (j)) = 0, k = 0, n, j ∈ Z, H n (Pn , OPn (j)) ∼ H 0 (Pn , OPn (−n − 1 − j)). =

4.2. DETERMINANTAL EQUATIONS FOR HYPERSURFACES

105

Since fi < ei , (4.24) gives ei − 1 < 0 and −fi − 2 < 0, hence ei ≤ 0, fi ≥ −1. This implies ei = 0, fi = −1 for all i = 1, . . . , r. Applying (4.21), we get r = d. So, we obtain a resolution
d d

0→
i=0

OPn (−1) −→
i=0

A

OPn → F → 0.

(4.25)

This gives a determinantal expression of C as a d × d determinant with entries linear forms. It is convenient to rewrite the exact sequence in the form 0 → W1 ⊗ OPn (−1) −→ W2 ⊗ OPn → F → 0, where W1 , W2 are some linear spaces of dimension d, and T is a linear map T : V → Hom(W1 , W2 ), where Pn = P(V ). The determinantal hypersurface V (f ) is the pre-image in P(V ) of the variety of linear operators W1 → W2 of rank less than d. Applying the cohomology, we obtain a natural isomorphism H 0 (Pn , F) ∼ W2 . = (4.27)
T

(4.26)

Twisting (4.25) by OPn (−n) and applying the cohomology, we find a natural isomorphism (4.28) H n−1 (Pn , F(−n)) ∼ W1 ⊗ H n (Pn , OPn (−n − 1)) ∼ W1 . = = It follows from (4.26) and (4.27) that the invertible sheaf M is generated by global sections and defines a morphism
∗ lT : V → P(W2 ) = |M|∗ .

For any x ∈ V the point lT (x) is the projectivization of the dual space of the cokernel of the matrix T (x∗ ), where x = Kx∗ for some x∗ ∈ V . Twisting (4.26) by OPn (1) and applying the functor HomOPn (−, OPn )) to the exact sequence (4.25) we obtain an exact sequence
∗ ∗ 0 → W2 ⊗ OPn (−1) −→ W1 ⊗ OPn → Ext1 Pn (F(1), OPn )) → 0, O T∗

(4.29)

∗ ∗ where T ∗ : V → Hom(W2 , W1 ) is defined by T ∗ (v) = t T (v). Now we apply Grothendieck’s duality theorem (see [Conrad]) to obtain a natural isomorphism of sheaves

F = Ext1 Pn (F(1), OPn )) ∼ i∗ HomOV (F(1), OPn (d)) ∼ i∗ M∗ (d − 1). (4.30) = = O Let L = M∗ (d − 1). (4.31)

106 We can rewrite (4.29) in the form

CHAPTER 4. DETERMINANTAL EQUATIONS

∗ ∗ 0 → W2 ⊗ OPn (−1) −→ W1 ⊗ OPn → i∗ (L) → 0,

T∗

(4.32)

Applying cohomology we see that the sheaf L satisfies the same condition (4.24) as M. It follows from (4.26) and (4.27) that the invertible sheaf L is generated by global sections and defines a morphism rT : V → P(W1 ) = |L|∗ . For any x ∈ V the point rT (x) is the projectivization of the kernel of the matrix T (x), where x = Cx∗ for some x∗ ∈ V .

4.2.3

The case of curves

Assume n = 2, i.e. V is a plane curve C of degree d. Then the condition (ii) for a ACM-sheaf is vacuous. The condition (4.24) becomes H 0 (V, M(−1)) = H 1 (V, M(−1)) = 0. We will assume that C is irreducible and a reduced curve and M is an invertible sheaf on C satisfying the condition H 0 (C, M(−1)) = H 1 (C, M(−1)) = 0. Let ωC be the canonical sheaf of C and pa (C) = dim H 1 (C, OC ) = dim H 0 (C, ωC ) be the arithmetic genus of C. By Riemann-Roch (the reader unfamiliar with the Riemann-Roch on a singular curve may assume that C is nonsingular), deg M(−1)) = h0 (M(−1)) − h1 (M(−1)) + pa (C) − 1 = pa (C) − 1. Also we obtain L ⊗ M ∼ OOP2 (d − 1). =

If C is a nonsingular curve, everything agrees with the theory from the previous section. Example 4.2.2. Let C be a plane irreducible cubic curve. Then M(−1) must be an invertible sheaf of degree 0 with no nonzero sections. It is known that Pic0 (C) ∼ = C \ Sing(C) and has a structure of an algebraic group isomorphic to Gm if C is a nodal cubic and isomorphic to Ga if C is a cuspidal cubic. Any nonzero element of this group defines a determinantal representation of C. For any nonzero a ∈ C, we have 1  t1 t1 a2 t0 1 − a2 t0 t1 − at2  t0 t2 + 2t3 = det  t1 (4.33) 2 1 t1 t1 + at2 0

4.2. DETERMINANTAL EQUATIONS FOR HYPERSURFACES

107

Note that, for any t = [t0 , t1 , t2 ] ∈ C the rank of the matrix is equal to 2, as it should be because the sheaf L is invertible. We cannot get a symmetric determinannt representation in this way because L ∼ M would imply that L is a non-trivial 2= torsion point of Pic(C). However, the additive group does not have non-trivial torsion elements. On the other hand, we have   −t1 0 −t2 −t0 −t1  t0 t2 + t3 = det  0 (4.34) 2 1 −t2 −t1 0 The matrices have rank 1 at the singular point [1, 0, 0] of the curve. This shows that C admits symmetric determinantal representations not defined by an invertible sheaf on C. We refer to [16] for the theory of symmetric determinantal representaions of singular plane curves with certain type of singularities. One can show that any determinantal representation of a cuspidal cubic is equivalent either to one given in (4.33) or to one given in (4.34). The latter one corresponds to a non-invertible ACM sheaf E on C satisfying E ∼ HomOC (E, ωC ). = This sheaf ”compactifies” the Picard scheme of C.

4.2.4

The case of surfaces

Let V be a normal surface of degree d in P3 . We are looking for an invertible sheaf M on V such that F = i∗ (M) is a ACM-sheaf satisfying an additional assumption (4.24). It will give us a resolution (4.25). It follows from this resolution that M is generated by global sections. By Bertini’s theorem (in characteristic 0), a general section of M is a nonsingular curve C. Thus we can write M = OV (C) for some nonsingular curve C. Since V is a hypersurface in P3 , its local ring is a Cohen-Macaulay module over the corresponding local ring of P3 . Thus the first condition for an ACM sheaf is satisfied. Let us interpret the second ACM condition H 1 (V, M(j)) = 0, j ∈ Z. Recall that a subvariety X ⊂ Pn is called projectively normal if the restriction map r : H 0 (Pn , OPn (j)) → H 0 (X, OX (j)) is surjective for all j. If X is nonsingular in codimension 1, one can show that it is the same as requiring that the projective coordinate ring of X is normal. Suppose X ⊂ V for some hypersurface V . Then the restriction homomorphism r is the composition of the homomorphisms r1 : H 0 (Pn , OPn (j)) → H 0 (V, OV (j)) and r2 : H 0 (V, OV (j)) → H 0 (X, OX (j)). It is easy to see that V is projectively normal (the cokernel of r1 is H 1 (Pn , OPn (−d)) = 0). Thus X is projectively normal if r2 is surjective. The exact sequence 0 → JX (j) → OV (j) → OX (j) → 0, where JX is the sheaf of ideals of X in V , shows that r2 is surjective if and only if H 1 (X, JX (j)) = 0 for all j. Applying this to our case, where X = C ⊂ P3 , we get that C is projectively normal if and only if H 1 (V, OV (−C)(j)) = H 1 (V, ωV (−j) ⊗ OV (C))

108

CHAPTER 4. DETERMINANTAL EQUATIONS = H 1 (V, OV (C)(d − 4 − j)) = 0, j ∈ Z.

Here we used the adjunction formula for the canonical sheaf and the Serre Duality theorem. Thus we see that the ACM-condition is the condition for the projective normality of C. To get a resolution (4.25) we need the additional conditions H 0 (V, OV (C)(−1)) = H 2 (V, OV (C)(−2)) = 0. Together with the ACM condition this is equivalent to χ(OV (C)(−1)) = χ(OV (C)(−2)) = 0. Let OV (1) = OV (H). Consider the exact sequence 0 → OV (C)(−2) → OV (C)(−1) → OH (C − H) → 0. It gives χ(OH (C − H)) = χ(OV (C)(−1)) − χ(OV (C)(−2)). By Bertini’s Theorem we may assume that H is a nonsingular plane curve of degree d. By Riemann-Roch on H, we get deg(C) − d = deg(OH (C − H)) = d(d − 3)/2. This gives
1 χ(OV (C)(−1)) − χ(OV (C)(−2)) ⇐⇒ deg(C) = 2 d(d − 1).

(4.35)

(4.36)

(4.37)

The exact sequence 0 → OV (−1) → OV (C)(−1) → OC (C − H) → 0 gives χ(OC (C − H)) = χ(OV (C)(−1)) − χ(OV (−1)) = χ(OV (C)(−1)) − χ(OP3 )(−1) + χ(OP3 )(−d − 1) = χ(OV (C)(−1)) − Applying Riemann-Roch, on C, we get χ(OC (C −H)) = deg OC (C −H)+χ(OC ) = deg OC (C +KV −(d−3)H)+χ(OC )
1 = deg KC − (d − 3) deg(C) + χ(OC ) = − 2 (d − 3)d(d − 1) + g(C) − 1.

d . 3

Thus we see that χ(OV (C)(−1)) = 0 ⇐⇒ g(C) = 1 (d − 2)(d − 3)(2d + 1). 6

Together with (4.37) we see that condition (4.35) is equivalent to the conditions

4.2. DETERMINANTAL EQUATIONS FOR HYPERSURFACES (i) C is a projectively normal curve; (ii) deg(C) = 1 d(d − 1), 2 (iii) g(C) = 1 (d − 2)(d − 3)(2d + 1). 6

109

Example 4.2.3. Take d = 3. We get deg(C) = 3, g(C) = 0. We also have h0 (C) = dim H 0 (V2 , OV2 (C)) = 3. The linear system |C| maps S to P2 . This is a birational morphism which is inverse to the blow-up of 6 points in P2 . We will see later when we will be discussing cubic surfaces, that there are 72 such linear systems. Thus a cubic surface can be written in 72 essentially different ways as a 3 × 3 determinant. Example 4.2.4. Take d = 4. We get deg(C) = 6, g(C) = 3. The projective normality is equivalent to the condition that C is not hyperelliptic (Exercise 5.11). We also have h0 (C) = 4. According to Noether’s Theorem, the Picard group of a general surface of degree ≥ 4 is generated by a plane section. Since a plane section of a quartic surface is of degree 4, we see that a general quartic surface does not admit a determinantal equation. The condition that V4 contains a curve C as above imposes one algebraic condition on the coefficients of a quartic surface. Remark 4.2.1. Let V = V (det(A(t)) be a determinantal equation of a nonsingular surface of degree d in P3 . Let C ⊂ H be a nonsingular plane section of V . Then we obtain a determinantal equation of C. The left kernel sheaf for C is the restriction of the sheaf M to C, where M = OV (C)) is defined by the resolution (4.25). Since we know that deg(M ⊗ OC ) = d(d − 1)/2, we obtain another proof that deg(C) = d(d − 1)/2. Remark 4.2.2. For nonsingular hypersurfaces V in P3 the condition on M defining a symmetric determinant is that M is isomorphic to the sheaf L defined as the cokernel of the transpose of the matrix twisted by −1. Applying the functor HomOP3 (−, OV ) to the exact sequence
d 0 → OP3 (−1)d → OP3 → i∗ M → 0,

we obtain
d 0 → OP3 → OP3 (1)d → Ext1 P3 (i∗ M, OP3 ) → 0. O

Twisting by −1, we get i∗ L = Ext1 P3 (i∗ M, OP3 )(−1). O By the duality, we have ωV ∼ Ext1 P3 (i∗ OV , ωP3 ) ∼ Ext1 P3 (i∗ OV , OP3 (−4)). = = O O By standard properties of the sheaves Exti we have Ext1 P3 (i∗ M, OP3 )(−1) ∼ i∗ (M∗ ⊗ Ext1 P3 (i∗ OV , OP3 (−4))) ⊗ OP3 (3). = O O This gives L = M∗ ⊗ ωV (3).

110 Thus, if M ∼ L, we must have =

CHAPTER 4. DETERMINANTAL EQUATIONS

M⊗2 ∼ ωV (3) = OV (d − 1). = We also must have h0 (M(−1)) = 0. Note that Pic(V )[2] = {0} for a nonsingular surface in P . Thus there is at most one square root of OV (d − 1). When d = 2k + 1 is odd, the square root is isomorphic to M = OV (k) but it does not satisfy the condition h0 (M(−1)) = 0. So, there are no symmetric determinantal equations. When d = 2k we have no contradiction. However, in both cases the nonexistence of symmetric determinantal equations follows from the general fact that the determinantal variety of symmetric d × d matrices is singular in codimension 2. Thus any linear projective space of dimension 3 intersects it the singular locus and cuts out a singular surface. So, only singular surfaces admit a symmetric determinantal equation. We will return to this later.
3

Exercises
4.1 Show that any irreducible cubic curve admits a determinantal equation. 4.2 Let (t0 (t0 − t1 ), (t0 − t2 )(t0 − t1 ), t0 (t0 − t2 )) define a rational map from P2 to P2 . Show that it is a birational map and find its inverse. 4.3 Let C = V (f ) be a nonsingular plane cubic, p1 , p2 , p3 be three non-collinear points. Let (A0 , A1 , A2 ) define a quadratic Cremona transformation with fundamental points p1 , p2 , p3 . Let q1 , q2 , q3 be another set of three points such that the six points p1 , p2 , p3 , q1 , q2 , q3 are cut out by a conic. Let (B0 , B1 , B2 ) define a quadratic Cremona transformation with fundamental points q1 , q2 , q3 . Show that 0 1 A0 B0 A0 B1 A0 B2 −3 F det adj @A1 B0 A1 B1 A1 B2 A A2 B2 A2 B1 A2 B2 is a determinantal equation of C. 4.4 Find a determinantal equation of the cubic curve from Example 4.1.1 which is not equivalent to the equation from the example. 4.5 Find a determinantal equation for the Klein quartic V (t3 t1 + t3 t2 + t3 t0 ). 0 1 2 4.6 Find determinantal equations for a nonsingular quadric surface in P3 . 4.7 Let V ⊂ Matd be a linear subspace of dimension 3 of the space of d × d matrices. Show that`the locus of points x ∈ Pd−1 such that there exists A ∈ V for which x ∈ Ker(A) is defined ´ by d equations of degree d. In particular, for any determinantal equation of a curve C, the 3 images of C under the maps r : P2 → Pd−1 and l : P2 → Pd−1 are defined by such a system of equations. 4.8 Let V4 = V (det(A(t))) be a 4 × 4-determinantal equation of a nonsingular quartic surface and OV4 (C) be the corresponding invertible sheaf represented by a non-hyperelliptic curve C of genus 3 and degree 6. Show that L = OV4 (−C)(3) is isomorphic to OV4 (C ) for some other curve of genus 3 and degree 6. Find the interpretation of the sheaf L in terms of the determinantal equation. 4.9 Let C be a non-hyperelliptic curve of genus 3 and degree 6 in P3 .

Historical Notes

111

(i) Show that the homogeneous ideal of C in P3 is generated by four cubic polynomials f0 , f1 , f2 , f3 . (ii) P Show that the equation of any quartic surface containing C can be written in the form li Fi = 0, where li are linear forms. (iii) Show that (f0 , f1 , f2 , f3 ) define a birational map f from P3 to P3 . The image of any quartic containing C is another quartic surface. (iv) Show that the map f is the right kernel map for the determinantal representation of the quartic defined by the curve C. 4.10 Show that a curve of degre 6 and genus 3 in P3 is projectively normal if and only if it is not hyperelliptic. 4.11 Let C be a nonsingular plane curve of degree d and L0 ∈ Picg−1 (C). Assume that h0 (L0 ) = 0. Show that the image of C under the map given by the linear system L0 (1) is a singular curve.

Historical Notes
The fact that a general plane curve of degree d can be defined by the determinant of a symmetric d × d matrix with entries homogenous linear forms was first proved by A. Dixon [75]. However, for curves of degree 4 this was proved almost 50 years earlier by O. Hesse [139]. He also showed that it can be done in 36 ways. For cubic curves the representation follows from the fact that any cubic curve can be written in three ways as the Hessian curve. This fact was also proven by Hesse [135], p. 89. The first modern treatment of Dixon’s result was given in [14] and [237]. It was proved by L. Dickson [74] that that any plane curve can be written as the determinant of not necessarily symmetric matrix with linear homogeneous forms as its entries. The relationship between linear determinantal representations of an irreducible plane curve of degree d and line bundles of degree d(d − 1)/2 was first established in [49]. This was later elaborated by V. Vinnikov [246]. The theory of linear determinant representation for cubic surfaces was developed by L. Cremona [59]. Dickson proves in [74] that a general homogeneous form of degree d > 2 in r variables cannot be represented as a linear determinant unless r = 3 or r = 4, d ≤ 3. We refer to [16] for a survey of modern development of determinantal representations of hypersurfaces.

112

CHAPTER 4. DETERMINANTAL EQUATIONS

Chapter 5

Theta characteristics
5.1
5.1.1

Odd and even theta characteristics
First definitions and examples

We have already defined a theta characteristic, odd and even, on a nonsingular curve C (see section 4.1.3). In this chapter we will study them in more details . It follows from the definition that two theta characteristics, considered as divisor classes of degree g − 1 differ by a 2-torsion divisor class. Since the 2-torsion subgroup Jac(C)[2] is isomorphic to (Z/2Z)2g , there are 22g theta characteristics. However, in general, there is no canonical identification between the set TChar(C) of theta characteristics on C and the set Jac(C)[2]. One can say only that TChar(C) is an affine space over the vector space of Jac(C)[2] ∼ F2g . = 2 There is more structure on TChar(C). Recall that the group Jac(C)[2] is equipped with a natural symmetric bilinear form over F2 , called the Weil pairing. It is defined as follows (see [5], Appendix B). Let , be two 2-torsion divisor classes. Choose their representatives D, D with disjoint supports. Write div(f ) = 2D, div(f ) = 2D . Then f (D )/f (D) = ±1. Here f ( i xi ) = i f (xi ). Now we set , = 1 iff (D )/f (D) = −1 0 otherwise. , = 0. One can show

Note that the Weil paring is a symplectic form, i.e. satisfies that it is a non-degenerate symplectic form. For any ϑ ∈ TChar(C) define the function qϑ : Jac(C)[2] → F2 ,

→ h0 (ϑ + ) + h0 (ϑ).

Proposition 5.1.1. The function qϑ is a quadratic form on Jac(C)[2] whose associated symmetric bilinear form is equal to the Weil pairing. Later we shall see that there are two types of quadratic forms associated to a fixed non-degenerate symplectic form: even and odd. They agree with our definition of an 113

114

CHAPTER 5. THETA CHARACTERISTICS

even and odd theta characteristic. The number of even (odd) theta characteristics is equal to 2g−1 (2g + 1) (2g−1 (2g − 1)). An odd theta characteristic ϑ is obviously effective, i.e. h0 (ϑ) > 0. If C is a canonical curve, then divisor D ∈ |ϑ| satisfies the property that 2D is cut out by a hyperplane H in the space |KC |∗ , where C is embedded. Such a hyperplane is called a bitangent hyperplane. It follow from above that a canonical curve either has 2g−1 (2g − 1) bitangent hyperplanes or infinitely many. The latter case happens if and only if there exists a theta characteristic ϑ with h0 (ϑ) > 1. Such a theta characteristic is called vanishing theta characteristic. An example of a vanishing odd theta characteristic is the divisor class of a line section of a plane quintic curve. An example of a vanishing 1 even theta characteristic is the unique g3 on a canonical curve of genus 4 lying on a singular quadric. The geometric interpretation of an even theta characteristic is more subtle. If C is a plane curve we explained in the previous chapter how a non-vanishing (equivalently, non-effective) even theta characteristic determines a symmetric linear determinantal representation of C. The only known geometrical construction related to canonical curves is the Scorza construction of a quartic hypersurface associated to a canonical curve and a non-effective theta characteristic. We discuss this construction in section 5.5.

5.1.2

Quadratic forms over a field of characteristic 2

Recall that a quadratic form on a vector space V over a field F is a map q : V → F such that q(av) = a2 q(v) for any a ∈ F and any v ∈ V and the map bq : V × V → F, (v, w) → q(v + w) − q(v) − q(w)

is bilinear (it is called the polar bilinear form). We have bq (v, v) = 2q(v) for any v ∈ V . In particular, q can be reconstructed from bq if char(F ) = 2. In the case when char(F ) = 2, we get bq (v, v) ≡ 0, hence bq is a symplectic bilinear form. Two quadratic forms q, q have the same polar bilinear form if and only if q − q = l, where l(v + w) = l(v) + √ l(w), l(av) = a2 l(v) for any v, w ∈ V, a ∈ F . If F is a finite field of characteristic 2, l is a linear form on V , and we obtain bq = bq ⇐⇒ q = q +
2

,

(5.1)

for a unique linear form : V → F . Let e1 , . . . , en be a basis in V and A = (aij ) = (bq (ei , ej )) be the matrix of the bilinear form bq . It is a symmetric matrix with zeros on the diagonal if char(F ) = 2. It follows from the definition that
n n

q(
i=1

xi ei ) =
i=1

x2 q(ei ) + i
1≤i<j≤n

xi xj aij .

The rank of a quadratic form is the rank of the matrix A of the polar bilinear form. A quadratic form is called nondegenerate if the rank is equal to dim V . In coordinate-free way this is the rank of the linear map V → V ∗ defined by bq . The kernel of this map

5.1. ODD AND EVEN THETA CHARACTERISTICS

115

is called the radical of bq . The restriction of q to the radical is identically zero. The quadratic form q arises from a nondegenerate quadratic form on the quotient space. In the following we assume that q is nondegenerate. A subspace E of V is called singular if q|E ≡ 0. Each singular subspace is an isotropic subspace with respect to bq , i.e., bq (v, w) = 0 for any v, w ∈ E. The converse is true only if char(f ) = 2. Assume char(F ) = 2. Since bq is a nondegenerate symplectic form, n = 2k, and there exists a basis e1 , · · · , en such that the matrix of bq is equal to Jk = Thus
n n k

0k Ik

Ik . 0k

(5.2)

q(
i=1

xi ei ) =
i=1

x2 q(ei ) + i
i=1

xi xi+k .

Assume additionally that F ∗ = F ∗2 , i.e., each element in F is a square (i.e. F is a finite or algebraically closed field). Then, we can further reduce q to the form
2k n k

q(
i=1

xi ei ) = (
i=1

αi xi )2 +
i=1

xi xi+k ,

(5.3)

2 where q(ei ) = αi , i = 1, . . . , n. This makes (5.1) more explicit. Fix a nondegenerate symplectic form , : V ×V → F . Each linear function on V is given by (v) = v, η for a unique η ∈ V . By (5.1), two quadratic forms q, q with polar bilinear form equal to , satisfy q(v) = q (v) + v, η 2

for a unique η ∈ V . Choose a standard symplectic basis (i.e. the matrix of the bilinear form with respect to this basis is equal to (5.2)). The quadratic form defined by
2k k

q0 (
I=1

xi ei ) =
i=1

xi xi+k

has the polar bilinear form equal to the standard symplectic form. Any other form with the same polar bilinear form is defined by q(v) = q0 (v) + v, ηq 2 , where
2k

ηq =
i=1

q(ei )ei .

From now on we assume that F = F2 , the field of two elements. In this case a2 = a for any a ∈ F2 . The formula (5.1) shows that the set Q(V ) of quadratic forms

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associated to the standard symplectic form is an affine space over V with addition q + η, q ∈ Q(V ), η ∈ V , defined by (q + η)(v) = q(v) + v, η = q(v + η) + q(η). The number Arf(q) =
i=1 k

(5.4)

q(ei )q(ek+i )

(5.5)

is called the Arf invariant of q. One can show that it is independent of the choice of a standard symplectic basis. A quadratic form q ∈ Q(V ) is called even (resp. odd) if Arf(q) = 0 (resp. Arf(q) = 1). If we choose a standard symplectic basis for bq and write q in the form q0 + ηq , then we obtain
k

Arf(q) =
i=1

αi αi+k = q0 (ηq ) = q(ηq ).

(5.6)

In particular, if q = q + v = q0 + ηq + v, Arf(q + v) + Arf(q) = q0 (ηq + v) + q0 (ηq ) = q0 (v) + v, ηq = q(v) (5.7)

It follows from (5.6) that the number of even (resp. even) quadratic forms is equal to −1 −1 the cardinality of the set q0 (0) (resp. q0 (1)). We have
−1 |q0 (0)| = 2k−1 (2k + 1), −1 |q0 (1)| = 2k−1 (2k − 1).

(5.8)

This is easy to prove by using induction on k. Let Sp(V ) be the group of linear automorphisms of the symplectic space V . If we choose a standard symplectic basis then Sp(V ) ∼ Sp(2k, F2 ) = {X ∈ GL2k (F2 ) : t X · Jk · X = Jk .} = It is easy to see by induction on k that |Sp(2k, F2 )| = 2k (22k − 1)(22k−2 − 1) · · · (22 − 1).
2

(5.9)

The group Sp(V ) has 2 orbits in Q(V ), the set of even and the set of odd quadratic forms. An even quadratic form is equivalent to the form q0 and an odd quadratic form is equivalent to the form q1 = q0 + ek + e2k , where (e1 , . . . , e2k ) is the standard symplectic basis. Explicitly,
2k k

q1 (
i=1

xi ei ) =
i=1

xi xi+k + x2 + x2 . k 2k

The stabilizer subgroup Sp(V )+ (resp. Sp(V )− ) of an even quadratic form (resp. an odd quadratic form) is a subgroup of Sp(V ) of index 2k−1 (2k +1) (resp. 2k−1 (2k −1)). If V = F2k with the symplectic form defined by the matrix Jk , then Sp(V )+ (resp. 2 Sp(V )− ) is denoted by O(2k, F2 )+ (resp. O(2k, F2 )− ).

5.2. HYPERELLIPTIC CURVES

117

5.2
5.2.1

Hyperelliptic curves
Equations of hyperelliptic curves

Let us first describe explicitly theta characteristics on hyperelliptic curves. Recall that a hyperelliptic curve of genus g is a nonsingular projective curve X admitting a degree 2 map ϕ : C → P1 . By Hurwitz’s formula, there are 2g + 2 branch points p1 , . . . , p2g+2 in P1 . Let f2g+2 (t0 , t1 ) be a binary form of degree 2g + 2 whose zeroes are the branch points. The equation of C in P(1, 1, g + 1) is t2 + f2g+2 (t0 , t1 ) = 0. 2 (5.10)

Recall that a weighted projective space P(q) = P(q0 , . . . , qn ) is defined as the quotient of Cn+1 \ {0}/C∗ , where C∗ acts by t : [z0 , . . . , zn ] → [tq0 z0 , . . . , tqn zn ]. Here q = (q0 , . . . , qn ) are integers ≥ 1. In more scientific way, P(q0 , . . . , qn ) = Proj(C[t0 , . . . , tn ]), where C[t0 , . . . , tn ] is the polynomial algebra graded by the condition deg(ti ) = qi . We refer to [85] for the theory of weighted projective spaces and their subvarieties. Note that a hypersurface in P(q) is defined by a homogeneous polynomial where the unknowns are homogeneous of degree qi . Thus the equation (5.10) defines a hypersurface of degree 2g + 2. Although P(q) is a singular variety in general, it has a canonical sheaf ωP(q) = OP(q) (−|q|), where |q| = q0 + · · · + qn . Here the Serre sheaves are understood in the sense of theory of projective spectrums of graded algebras. There is also the adjunction formula for a hypersurface of degree d ωV = OV (d − |q|). In the case of a hyperelliptic curve, we have ωC = OC (g − 1). The morphism ϕ : C → P1 corresponds to the projection [t0 , t1 , t2 ] → [t0 , t1 ] and we obtain that ωC = ϕ∗ OP1 (g − 1). The weighted projective space P(1, 1, g + 1) is isomorphic to the projective cone in Pg+2 over the Veronese curve vg+1 (P1 ) ⊂ Pg+1 . The hyperelliptic curve is isomorphic to the intersection of this cone and a quadric hypersurface in Pg+1 not passing through the vertex of the cone. The projection from the vertex to the Veronese curve is the double cover ϕ : C → P1 . The canonical linear system |KC | maps C to Pg with the image equal to the Veronese curve vg−1 (P1 ). (5.11)

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5.2.2

2-torsion points on a hyperelliptic curve

Let c1 , . . . , c2g+2 be the ramification points of the map ϕ. We assume that ϕ(ci ) = pi . Obviously, 2ci − 2cj ∼ 0, hence the divisor class of ci − cj is of order 2 in Pic(C). Also, for any subset I of the set Bg = {1, . . . , 2g + 2}, we have αI =
i∈I

ci − #Ic2g+2 =
i∈I

(ci − c2g+2 ) ∈ Pic(C)[2].

Now observe that αBg =
i∈Bg

ci − (2g + 2)c2g+2 = div(φ) ∼ 0,

(5.12)

where φ = t2 /(bt0 − at1 )g+1 and p2g+2 = (a, b) (we consider the fraction modulo the equation (5.10) defining C). Thus ci − cj ∼ 2ci +
k∈Bg \{j}

ck − (2g + 2)c2g+2 ∼ αBg \{i,j} .

Adding to αI the zero divisor c2g+2 − c2g+2 we can always assume that #S is even. ¯ Also adding the principal divisor αBg , we obtain that αI = αI , where I denotes Bg \I. ¯ Bg ∼ 2g+2 Let F2 = F2 be the F2 -vector space of functions Bg → F2 , or, equivalently, subsets of Bg . The sum is defined by the symmetric sum of subsets I + J = I ∪ J \ (I ∩ J). The subsets of even cardinality form a hyperplane. It contains the subsets ∅ and Bg as a subspace of dimension 1. Let Eg denote the factor space. Elements of Eg are represented by subsets of even cardinality up to the complementary set (bifid maps in terminology of A. Cayley). We have Eg ∼ F2g , = 2 hence the correspondence S → αS defines an isomorphism I : Eg ∼ Pic(C)[2]. = Note that Eg carries a natural symmetric bilinear form e : E g × E g → F2 , e(I, J) = #I ∩ J mod 2. (5.14) (5.13)

This form is symplectic (i.e. e(I, I) = 0 for any I) and nondegenerate. If we choose a basis represented by the subsets Ai = {2i − 1, 2i}, Bi = {2i, 2i + 1}, i = 1, . . . , g, (5.15)

then the matrix of the bilinear form e will be equal to Jg from (5.2) Under the isomorphism (10.3.2), this bilinear form corresponds to the Weil pairing on 2-torsion points of the Jacobian variety of C.

5.2. HYPERELLIPTIC CURVES

119

Remark 5.2.1. The symmetric group S2g+2 acts on Eg via its action on Bg and preserves the symplectic form e. This defines a homomorphism sg : S2g+2 → Sp(2g, F2 ). If g = 1, Sp(2, F2 ) ∼ S3 , and the homomorphism s1 has the kernel isomorphic to the = Klein group (Z/2Z)2 . If g = 2, the homomorphism s2 is an isomorphism. If g > 2, the homomorphism sg is injective but not surjective.

5.2.3

Theta characteristics on a hyperelliptic curve

For any subset T of Bg set ϑT =
i∈T

ci + (g − 1 − #T c2g+2 = αT + (g − 1)c2g+2 .

We have 2ϑT ∼ 2αT + (2g − 2)c2g+2 ∼ (2g − 2)c2g+2 It follows from the proof of the Hurwitz formula that KC = ϕ∗ (KP1 ) +
i∈Bg

ci .

Choose a representative of KP1 equal to −2p2g+2 and use (5.12) to obtain KC ∼ (2g − 2)c2g+2 . Thus we obtain that ϑT is a theta characteristic. Again adding and subtracting c2g+2 ¯ we may assume that #T ≡ g + 1 mod 2. Since T and T define the same theta characteristic, we will consider the subsets up to taking the complementary set. We obtain a set Qg which has a natural structure of an affine space over Eg , the addition is defined by ϑT + αI = ϑT +I . Thus all theta characteristics are uniquely represented by the divisor classes ϑT , where T ∈ Qg . An example of an affine space over V = F2g is the space of quadratic forms q : 2 2g F2 → F2 whose associated symmetric bilinear form bq coincides with the standard symplectic form defined by the matrix (5.2). We identify V with its dual V ∗ by means of b0 and set q + l = q + l2 for any l ∈ V ∗ . For any T ∈ Qg we define the quadratic form qT on Eg by qT (I) = 1 (#(T + I) − #T ) = #T ∩ I + 1 #I = 1 #I + e(I, T ) 2 2 2 We have (all equalities are modulo 2)
qT (I + J) + qT (I) + qT (J) = 1 (#(I + J) + #| + #J) + e(I + J, T ) + e(I, T ) + e(J, T ) 2
1 = 2 (2#I + 2#J − 2#I ∩ J) = #I ∩ J.

mod 2.

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Thus each theta characteristic can be identified with an element of the space Qg = Q(Eg ) of quadratic forms on Eg with polar form e. Also notice that
1 (qT + αI )(J) = qT (J) + e(I, J) = 2 #J + e(T, J) + e(I, J) 1 = 2 #J + e(T + I, J) = qT +I (J).

Lemma 5.2.1. Let ϑT be a theta characteristic on a hyperelliptic curve C of genus g identified with a quadratic form on Eg . Then the following properties are equivalent (i) #T = g + 1 mod 4; (ii) h0 (ϑT ) = 0 mod 2; (iii) qT is even. Proof. Without loss of generality we may assume that p2g+2 is the infinity point (0, 1) in P1 . Then the field of rational functions on C is generated by the function y = t2 /t0 and x = t1 /t0 . We have ϑT =
i∈T

ci + (g − 1 − #T )c2g+2 ∼ (g − 1 + #T )c2g+2 −
i∈T

ci .

Any function φ from the space L(ϑT ) = {φ : div(φ) + ϑT ≥ 0} has a unique pole at c2g+2 of order < 2g + 1. Since the function y has a pole of order 2g + 1 at c2g+2 , 1 we see that φ = ϕ∗ (p(x)), where p(x) is a polynomial of degree ≤ 2 (g − 1 + #T ) in x. Thus L(ϑT ) is isomorphic to the linear space of polynomials p(x) of degree ≤ 1 (g − 1 + #T ) with zeroes at pi , i ∈ T . The dimension of this space is equal to 2 1 2 (g + 1 − #T ). This proves the equivalence of (i) and (ii). Let U = {1, 3, . . . , 2g + 1} ⊂ Bg (5.16) be the subset of odd numbers in Bg . If we take the standard symplectic basis in Eg defined in (5.15), then we obtain that qU = q0 is the standard quadratic form associated to the standard symplectic basis. It follows from (5.6) that qT is an even quadratic form if and only if T = U + I, where qU (I) = 0. Let I consists of k even numbers and s odd numbers. Then qU (I) = #U ∩ I + 1 #I = m + 1 (k + m) = 0 mod 2. Thus 2 2 #T = #(U + S) = #U + #I − 2#U ∩ S = (g + 1) + (k + m) − 2m = g + 1 + k − m. Then m + 1 (k + m) is even, hence 3m + k ≡ 0 mod 4. This implies that k − m ≡ 0 2 mod 4 and #T ≡ g + 1 mod 4. Conversely, if #T ≡ g + 1 mod 4, then k − m ≡ 0 mod 4 and qU (I) = 0. This proves the lemma.

5.2.4

Families of curves with odd or even theta characteristic

Let X → S be a smooth projective morphism whose fibre Xs over a point s ∈ S is a curve of genus g > 0 over the residue field κ(s) of s. Let Picn /S → S be the relative X Picard scheme of X /S. It represents the functor on the category of S-schemes defined by assigning to a S-scheme T the set of isomorphism classes of invertible sheaves on

5.2. HYPERELLIPTIC CURVES

121

X ×S T of relative degree n over T modulo tensor product with invertible sheaves coming from T . The S-scheme Picn /S → S is a smooth projective scheme over S. X Its fibre over a point s ∈ S is isomorphic to the Picard variety Picn s /κ(s) over the X field κ(s). The relative Picard scheme comes with a universal invertible sheaf U on ´ X ×S Picn /S (locally in etale topology). For any point y ∈ Picn /S over a point X X s ∈ S, the restriction of U to the fibre of the second projection over y is an invertible sheaf Uy on Xs ⊗κ(s) κ(y) representing a point in Picn (Xs ⊗ κ(y)) defined by y. For any integer m, raising a relative invertible sheaf into m-th power defines a morphism [m] : Picn /S → Picmn . X X /S Taking n = 2g − 2 and m = 2, the pre-image of the section defined by the relative canonical class ωX /S is a close subscheme of Picg−1 . It defines a finite cover X /S T C X /S → S of degree 22g . The pull-back of U to T C X /S defines an invertible sheaf T over P = X ×S T C X /S satisfying T ⊗2 ∼ ωP/T C X /S . By a theorem of Mumford [181], = the parity of a theta characteristic is preserved in an algebraic family, thus the function T C X /S → Z/2Z defined by y → dim H 0 (Uy , Ty ) mod 2 is constant on each connected component of T C X /S . Let T C ev /S (resp. T C odd ) be the closed subset of X /S X T C X /S , where this function takes the value 0 (resp. 1). The projection T C ev /S → S X (resp. T C odd → S) is a finite cover of degree 2g−1 (2g + 1) (resp. 2g−1 (2g − 1)). X /S It follows from above that T C X /S has at least two connected components. Now take S = |OP2 (d)|ns be the space of nonsingular plane curves C of degree d and X → |OP2 (d)|ns be the universal curve defined by {(x, C) : x ∈ C}. We set T C d = T C X /S , T C d
ev/odd

= T C X /S .

ev/odd

The proof of the following proposition can be found in [15]. Proposition 5.2.2. If d is even or d = 3, T C d consists of two irreducible components T C ev and T C odd . If d ≡ 1 mod 4, then T C ev is irreducible but T C odd has two d d d d irreducible components, one of which is the section of T C d → |OP2 (d)|ns defined by OP2 ((d − 3)/2). If d ≡ 3 mod 4, then T C odd is irreducible but T C ev has two d d irreducible components, one of which is the section of T C d → |OP2 (d)|ns defined by OP2 ((d − 3)/2). Let T C 0 be the open subset of T C ev corresponding to the pairs (C, ϑ) with h0 (ϑ) = d d 0. It follows from the theory of symmetric determinantal representations of plane curves that T C 0 /PGL(3) is an irreducible variety covered by an open subset of a Grassd mannian. Since the algebraic group PGL(3) is connected and acts freely on a Zariski open subset of T C 0 , we obtain that T C 0 is irreducible. It follows from the previous d d proposition that T C 0 = T C ev if d ≡ 3 mod 4. (5.17) d d Note that there exist coarse moduli space Mev and Modd of curves of genus tog g gether with an even (odd) theta characteristic. We refer to [56] for the proof of irreducibility of these varieties and for construction of certain compactifications of these spaces.

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5.3
5.3.1

Theta functions
Jacobian variety

Recall the classical definition of the Jacobian variety of a nonsingular projective curve C of genus g over C. We consider C as a compact oriented 2-dimensional manifold of genus g. We view the linear space H 0 (C, KC ) as the space of holomorphic 1-forms on C. By integration over 1-dimensional cycles we get a homomorphism of Z-modules ι : H1 (C, Z) → H 0 (C, KC )∗ , ι(γ)(ω) =
γ

ω.

The image of this map is a lattice Λ of rank 2g in the complex space H 0 (C, KC )∗ . The quotient by this lattice Jac(C) = H 0 (C, KC )∗ /Λ is a complex g-dimensional torus. It is called the Jacobian variety of C. Recall that the cap product ∩ : H1 (C, Z) × H1 (C, Z) → H2 (C, Z) ∼ Z equips the = group H1 (C, Z) ∼ Z2g with a nondegenerate symplectic form. Let α1 , . . . , αg , β1 , . . . , βg = be a standard symplectic basis, i.e., (αi , αj ) = (βi , βj ) = 0, (αi , βj ) = δij .

We choose a basis ω1 , . . . , ωg of holomorphic 1-differentials on C such that ωj = δij .
αi

(5.18)

Let τij =
βi

ωi .

The complex matrix τ = (τij ) is called the period matrix. The basis ω1 , . . . , ωg identifies H 0 (C, KC )∗ with Cg and the period matrix identifies the lattice Λ with the lattice Λτ = [τ Ig ]Z2g . The period matrix satisfies
t

τ = τ,

(τ ) > 0.

As is well-known (see [129]) this implies that Jac(C) is a projective algebraic group, i.e. an abelian variety. It is isomorphic to the Picard scheme Pic0 . C/C We consider any divisor D = nx x on C as a 0-cycle on C. The divisors of degree 0 are boundaries, i.e. D = ∂γ for some 1-chain β. By integrating over β we get a linear function on H 0 (C, KC ) whose coset modulo Λ = ι(H1 (C, Z)) does not depend on the choice of β. This defines a homomorphism of groups p : Div0 (C) → Jac(C). The Abel-Jacobi Theorem asserts that p is zero on principal divisors (Abel’s part), and surjective (Jacobi’s part). This defines an isomorphism of abelian groups aj : Pic0 (C) → Jac(C) (5.19)

which is called the Abel-Jacobi map. For any positive integer d let Picd (C) denote

5.3. THETA FUNCTIONS

123

the set of divisor classes of degree d. The group Pic0 (C) acts simply transitively on Picd (C) via addition of divisors. There is a canonical map ud : C (d) → Picd (C), D → [D], where we identify the symmetric product with the set of effective divisors of degree d. One can show that Picd (C) can be equipped with a structure of a projective algebraic variety (isomorphic to the Picard scheme Picd ) such that the map ud is a morphism C/C of algebraic varieties. Its fibres are projective spaces, the complete linear systems corresponding to the divisor classes of degree d. The action of Pic0 (C) = Jac(C) on Picd (C is an algebraic action equipping Picd (C with a structure of a torsor over the Jacobian variety. Let r Wg−1 = {[D] ∈ Picg−1 (C) : h0 (D) ≥ r + 1}.
0 In particular, Wg−1 was denoted by Wg−1 in Theorem 4.1.3, where we showed that the invertible sheaves L0 ∈ Picg−1 (C) defining a determinantal equation of a plane curve 0 of genus g belong to the set Picg−1 (C) \ Wg−1 . The fundamental property of the loci r Wg−1 is given by the following Riemann-Kempf Theorem

Theorem 5.3.1.
r Wg−1 = {x ∈ Wg−1 : multx Wg−1 ≥ r + 1}.

In particular, we get
1 Wg−1 = Sing(Wg−1 ).

From now on we will identify Pic0 (C) with the points on the Jacobian variety Jac(C) by means of the Abel-Jacobi map. For any theta characteristic ϑ the subset Θ = Wg−1 − ϑ ⊂ Jac(C) is a hypersurface in Jac(C). It has the property that h0 (Θ) = 1, [−1]∗ (Θ) = Θ (5.20)

where [m] is the multiplication by an integer m in the group variety Jac(C). Conversely, any divisor on Jac(C) satisfying these properties is a translate of Wg−1 by a theta characteristic. This follows from the fact that a divisor D on an abelian variety A satisfying h0 (D) = 1 defines a bijective map A → Pic0 (A) by sending a point x ∈ A to the divisor t∗ D − D, where tx is the is the translation map a → a + x in the group x variety, and Pic0 (A) is the group of divisor classes algebraically equivalent to zero. This fact implies that any two divisors satisfying properties (5.20) differ by translation by a 2-torsion point. We call a divisor satisfying (5.20) a symmetric theta divisor. An abelian variety that contains such a divisor is called a principally polarized abelian variety. Let Θ = Wg−1 − θ be a symmetric theta divisor on Jac(C). Applying Theorem 5.3.1 we obtain that, for any 2-torsion point ∈ Jac(C), we have mult Θ = h0 (ϑ + ). (5.21)

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CHAPTER 5. THETA CHARACTERISTICS

In particular, ∈ Θ if and only if θ + is an effective theta characteristic. According to ϑ, the symmetric theta divisors are divided into two groups: even and odd theta divisors.

5.3.2

Theta functions

The pre-image of Θ under the quotient map Jac(C) = H 0 (C, KC )∗ /Λ is a hypersurface in the complex linear space V = H 0 (C, KC )∗ equal to the zero set of some holomorphic function φ : V → C. This function φ is not invariant with respect to translations by Λ (only constants are because the quotient is compact). However, it has the property that, for any v ∈ V and any λ ∈ Λ, φ(v + λ) = eλ (v)(γ)φ(v), where eλ is an invertible holomorphic function on V . Such a function is called a theta function. The set of zeroes of φ does not change if we replace φ with φα, where α is an invertible holomorphic function on V . The functions eλ (v) will change into functions eλ (v) = eλ (v)φ(v + λ)φ−1 (v). One can show that, after choosing an appropriate α one may assume that eλ (v) = exp(2πi(aγ (v) + bγ )), where aγ is a linear function and bγ is constant. We will assume that that such choice has been made. It turns out the theta function corresponding to a symmetric theta divisor Θ can be given in coordinates defined by a choice of a normalized basis (5.18) by the following expression θ [ η ] (z; τ ) =
r∈Zg

exp πi (r +

1 2

) · τ · (r +

1 2

) + 2(z + 1 η) · (r + 2

1 2

)

(5.22)

where , η ∈ {0, 1}g considered as a column or a raw vector from Fg . The function 2 defined by this expression is called a theta function with characteristic. The invertible function eλ (z1 , . . . , zg ) for such a function is given by the expression eλ (z) = exp −πi(m · τ · m − 2z · m − · n + η · m), where we write λ = τ · m + n for some m, n ∈ Zg . One can check that θ [ η ] (−z; τ ) = exp(πi · η)θ [ η ] (z; τ ). (5.23)

This shows that θ [ η ] (−z; τ ) is an odd (resp. even) function if and only if · η = 1 (resp. 0). In particular, θ [ η ] (0; τ ) = 0 if the function is odd. It follows from (5.21) that this happens when the corresponding theta characteristic ϑ is odd unless it is an effective even theta characteristic. Taking η = 0, we obtain the Riemann theta function θ(z; τ ) =
r∈Zg

exp πi(r · τ · r + 2z · r).

5.3. THETA FUNCTIONS All other theta functions with characteristic are obtained from θ(z; τ ) by translate
1 θ [ η ] (z; τ ) = exp πi( · η + · τ · )θ(z + 2 τ · η + 1 2

125

; τ ).

In this way the set points on Cg (C) of the form 1 τ · + 1 η are identified with elements of 2 2 the 2-torsion group 1 Λ/Λ of Jac(C). The theta divisor corresponding to the Riemann 2 theta function is the translate of Wg−1 by a certain theta characteristic κ called the Riemann constant. Of course, there is no any distinguished theta characteristic, the definition of κ depends on the choice of a symplectic basis in H1 (C, Z). The multiplicity m of a point on a theta divisor Θ = Wg−1 − ϑ is equal to the multiplicity of the corresponding theta function defined by vanishing partial derivatives up to order m − 1. Thus the quadratic form defined by θ can be redefined in terms of the corresponding theta function as qϑ ( 1 τ · 2 + 1 η ) = mult0 θ 2
+ η+η

(z, τ ) + mult0 θ [ η ] (z, τ ).

It follows from (5.23) that this number is equal to ·η +η·η +η ·η . (5.24)

A choice of a symplectic basis in H1 (C, Z) defines a standard symplectic basis in H1 (C, F2 ) ∼ 1 Λ/Λ = Jac(C)[2]. Thus we can identify 2-torsion points 1 τ · + 1 η =2 2 2 with vectors ( , η ) ∈ F2g . The quadratic form corresponding to the Riemann theta 2 function is the standard one q0 (( , η )) = · η The quadratic form corresponding to θ [ η ] (z; τ ) is given by (5.24). The Arf invariant of this quadratic form is equal to Arf(qϑ ) = · η = q0 (( , η)).

5.3.3

Hyperelliptic curves again

In this case we can compute the Riemann constant explicitly. Recall that we identify 2-torsion points with subsets of even cardinality of the set Bg = {1, . . . , 2g + 2} which we can identify with the set of ramification or branch points. Let us define a standard symplectic basis in C by choosing the 1-cycle αi to be the path which goes from c2i−1 to c2i along one sheet of the Riemann surface C and returns to c2i−1 along the other sheet. Similarly we define the 1-cycle βi by choosing the points c2i and c2i+1 . Choose g holomorphic forms ωj normalized by the condition (5.18). Let τ be the corresponding period matrix. Notice that each holomorphic 1-form changes sign when we switch the sheets. This gives
c2i 1 2 δij c2g+2 c2g+2

=

1 2

ωj =
αi c2g+2 c2i−1

ωj =
c2i−1 c2g+2

ωj −
c2i c2g+2

ωj

=
c2i−1

ωj +
c2i

ωj − 2
c2i

ωj .

126 Since 2
c2i

CHAPTER 5. THETA CHARACTERISTICS

c2g+2

c2g+2

ω1 , . . . ,
c2i

ωg = aj(2c2i − 2c2g+2 ) = 0,

we obtain ι(c2i−1 + c2i − 2c2g+2 ) = 1 ei 2 mod Λτ , where, as usual, ei denotes the ith unit vector. Let Ai , Bi be defined as in (5.15). We obtain that aj(αAi ) = 1 ei mod Λτ . 2 Similarly, we find that ajc2g+2 (αBi ) = 1 τ · ei 2 mod Λτ .

Now we can match the set Qg with the set of theta functions with characteristics. Recall that the set U = {1, 3, . . . , 2g + 1} plays the role of the standard quadratic form. We have qU (Ai ) = qU (Bi ) = 0, i = 1, . . . , g. Comparing it with (5.24), we see that the theta function θ [ η ] (z; Ω) corresponding to ϑU must be the function θ(z; τ ). This shows that ιg−1 (ϑU ) = ιc2g+2 (ϑU − kc2g+2 ) = 0. c2g+2 Thus the Riemann constant κ corresponds to the theta characteristic ϑU . This allows one to match theta characteristics with theta functions with theta characteristics. Write any subset I of Eg in the form
g g i Ai i=1

I= where = ( 1, . . . ,
g ),

+
i=1

ηi Bi ,

η = (η1 , . . . , ηg ) are binary vectors. Then ϑU +I ←→ θ [ η ] (z; τ ).

In particular, ϑU +I ∈ TChar(C)ev ⇐⇒ · η = 0 mod 2. Example 5.3.1. We give the list of theta characteristics for small genus. We also list 2-torsion points at which the corresponding theta function vanishes. g=1 3 even “thetas”: ϑ12 = θ [ 1 ] 0 ϑ13 = θ [ 0 ] 0 ϑ14 = θ [ 0 ] 1 1 odd theta (α12 ), (α13 ), (α14 ).

5.3. THETA FUNCTIONS

127

ϑ∅ = θ [ 1 ] 1 g=2 10 even thetas: ϑ123 = θ [ 01 ] 10 ϑ124 = θ [ 00 ] 10 ϑ125 = θ [ 00 ] 11 ϑ126 = θ [ 11 ] 11 ϑ234 = θ [ 10 ] 01 ϑ235 = θ [ 10 ] 00 ϑ236 = θ [ 01 ] 00 ϑ245 = θ [ 11 ] 00 ϑ246 = θ [ 00 ] 00 ϑ256 = θ [ 00 ] 01

(α∅ ).

(α12 , α23 , α13 , α45 , α46 , α56 ), (α12 , α24 , α14 , α35 , α36 , α56 ), (α12 , α25 , α15 , α34 , α36 , α46 ), (α12 , α16 , α26 , α34 , α35 , α45 ), (α23 , α34 , α24 , α15 , α56 , α16 ), (α23 , α25 , α35 , α14 , α16 , α46 ), (α23 , α26 , α36 , α14 , α45 , α15 ), (α24 , α25 , α13 , α45 , α16 , α36 ), (α26 , α24 , α13 , α35 , α46 , α15 ), (α26 , α25 , α13 , α14 , α34 , α56 ).

6 odd thetas ϑ1 = θ [ 01 ] 01 ϑ2 = θ [ 11 ] 01 ϑ3 = θ [ 11 ] 01 ϑ4 = θ [ 10 ] 10 ϑ5 = θ [ 10 ] 11 ϑ6 = θ [ 01 ] 11 g=3 36 even thetas ϑ∅ , ϑijkl , 28 odd thetas ϑij . g=4 136 even thetas ϑi , ϑijklm 120 odd thetas ϑijk . (α∅ , α12 , α13 , α14 , α15 , α16 ), (α∅ , α12 , α23 , α24 , α25 , α26 ), (α∅ , α13 , α23 , α34 , α35 , α36 ), (α∅ , α14 , α24 , α34 , α45 , α46 ), (α∅ , α15 , α35 , α45 , α25 , α56 ), (α∅ , α16 , α26 , α36 , α46 , α56 ).

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CHAPTER 5. THETA CHARACTERISTICS

5.4
5.4.1

Odd theta characteristics
Syzygetic triads

We have already remarked that effective theta characteristics on a canonical curve C ⊂ Pg−1 correspond to hyperplanes everywhere tangent to C. We call them bitangent hyperplanes (not to be confused with hyperplanes tangent at ≥ 2 points). An odd theta characteristic is effective and determines a bitangent hyperplane, a unique one if it is non-vanishing. In this section we will study the configuration of bitangent hyperplanes to a canonical curve. Let us note here that a general canonical curve is determined uniquely by the configuration of its bitangent hyperplanes [28]. From now on we fix a non-degenerate symplectic space (V, ω) of dimension 2g over F2 . Let Q(V ) be the affine space of quadratic forms with associated symmetric bilinear form equal to ω. The Arf invariant divides Q(V ) into the union of two sets Q(V )+ and Q(V )− , of even or odd quadratic forms. Recall that Q(V )− is interpreted as the set of odd theta characteristics when V = Pic(C) and ω is the Weil pairing. For any q ∈ Q(V ) and v ∈ V , we have q(v) = Arf(q + v) + Arf(q). Thus the function Arf is the symplectic analog of the function h0 (ϑ) mod 2 for theta characteristics. ˜ The set V = V Q(V ) is equipped with a structure of a Z/2Z-graded vector space over F2 . It complements the addition on V (the 0-th graded piece) and the structure of an affine space on Q(V )(the 1-th graded piece) by setting q + q := v, where ˜ q = q + v. One can also extend the symplectic form on V to V by setting ω(q, q ) = q(q + q ), ω(q, v) = ω(v, q) = q(v).

Definition 5.1. A set of three elements q1 , q2 , q3 in Q(V ) is called a syzygetic triad (resp. azygetic triad) if Arf(q1 ) + Arf(q2 ) + Arf(q3 ) + Arf(q1 + q2 + q3 ) = 0 (resp. = 1). A subset of k ≥ 3 elements in Q(V ) is called an azygetic set if any subset of three is azygetic. Note that a syzygetic triad defines a set of four quadrics in Q(V ) that add up to zero. Such a set is called a syzygetic tetrad. Obviously, any subset of three in a syzygetic tetrad is a syzygetic triad. Another observation is that three elements in Q(V )− form an azygetic triad if their sum is an element in Q(V )+ . For any odd theta characteristic ϑ any divisor Dη ∈ |ϑ| is of degree g − 1. The condition is that four odd theta characteristics ϑi form a syzygetic tetrad means that the sum of divisors Dϑi are cut out by a quadric in Pg−1 . The converse is true if C does not have vanishing even theta characteristic. Let us now compute the number of syzygetic tetrads.

5.4. ODD THETA CHARACTERISTICS

129

Lemma 5.4.1. Let q1 , q2 , q3 be a set of three elemenst on Q(V ). The following properties are equivalent: (i) q1 , q2 , q3 is a syzygetic triad; (ii) q1 (q2 + q3 ) = Arf(q2 ) + Arf(q3 ); (iii) ω(q1 + q2 , q1 + q3 ) = 0. Proof. The equivalence of (i) and (ii) follows immediately from the identity q1 (q2 + q3 ) = Arf(q1 ) + Arf(q1 + q2 + q3 ). We have ω(q1 + q2 , q1 + q3 ) = q1 (q1 + q2 ) + q1 (q1 + q3 ) + q1 (q2 + q3 ) = Arf(q1 )+Arf(q2 )+Arf(q1 )+Arf(q3 )+q1 (q2 +q3 ) = Arf(q2 )+Arf(q3 )+q1 (q2 +q3 ). This shows the equivalence of (ii) and (iii). Proposition 5.4.2. Let q1 , q2 ∈ Q(V )− . The number of ways in which the pair can be extended to a syzygetic triad of odd theta characteristics is equal to 2(2g−1 +1)(2g−2 − 1). Proof. Assume that q1 , q2 , q3 is a syzygetic triad in Q(V )− . By the previous lemma, q1 (q2 + q3 ) = 0. Also, we have q2 (q2 + q3 ) = Arf(q3 ) + Arf(q2 ) = 0. Thus q1 and q2 vanish at v = q2 + q3 . Conversely, assume v ∈ V satisfies q1 (v) = q2 (v) = 0. Set q3 = q2 + v. We have Arf(q3 ) = Arf(q2 ) + q2 (v) = 1, hence q3 ∈ Q(V )− . Since q1 (v) = q1 (q2 + q3 ) = 0, by the previous lemma q1 , q2 , q3 is a syzygetic triad. Thus the number of the ways we can extend q1 , q2 to a syzygetic triad q1 , q2 , q3 is equal to the cardinality of the set
−1 −1 Z = q1 (0) ∩ q2 (0) \ {0, v0 },

where v0 = q2 + q3 . It follows from (5.6) that v ∈ Z satisfies ω(v, v0 ) = q2 (v) + q1 (v) = 0. By definition, q2 (v0 ) = Arf(q3 ) + Arf(q2 ) = 0, and hence, ω(v, v0 ) = 0 if and only if v ∈ Z. Thus any v ∈ Z is a representative of a nonzero element in ⊥ W = v0 /v0 ∼ F2g−2 on which q1 and q2 vanish. It is clear that q1 and q2 induce the = 2 same quadratic form q on W . It is an odd quadratic form. Indeed, we can choose a symplectic basis in V by taking as a first vector the vector v0 . Then computing the Arf invariant of q1 we see that it is equal to the Arf invariant of the quadratic form q. Thus we get #Z = 2(#Q(W )− − 1) = 2(2g−2 (2g−1 − 1) − 1) = 2(2g−1 + 1)(2g−2 − 1).

130

CHAPTER 5. THETA CHARACTERISTICS

Corollary 5.4.3. Let tg be the the number of syzygetic tetrads of odd theta characteristics on a nonsingular curve of genus g. Then tg = 1 g−3 2g 2 (2 − 1)(22g−2 − 1)(2g−2 − 1). 3

Proof. Let I be the set of triples (q1 , q2 , T ), where q1 , q2 ∈ Q(V )− and T is a syzygetic tetrad containing q1 , q2 . We count #I in two ways by projecting I to the set P of unordered pairs of distinct elements Q(V )− and to the set of syzygetic tetrads. Since each tetrad contains 6 pairs of from P, and each pair can be extended in (2g−1 + 1)(2g−2 − 1) ways to a syzygetic tetrad, we get #I = (2g−1 + 1)(2g−2 − 1) This gives tg = 1 g−3 2g 2 (2 − 1)(22g−2 − 1)(2g−2 − 1). 3
2g−1 (2g −1) 2

= 6tg .

Let V be a vector space with a symplectic or symmetric bilinear form. Recall that a linear subspace L is called isotropic if the restriction of the bilinear form to L is identically zero. Corollary 5.4.4. Let {q1 , q2 , q3 , q4 } be a syzygetic tetrad in Q(V )− . Then P = {q1 + qi , . . . , q4 +qi } is an isotropic 2-dimensional subspace in (V, ω) which does not depend on the choice of qi . Proof. It follows from Lemma 5.4.1 (iii) that P is an isotropic subspace. The equality q1 + · · · + q4 = 0 gives qk + ql = qi + qj , (5.25) where {i, j, k, l} = {1, 2, 3, 4}. This shows that the subspace P of V formed by the vectors qj + qi , j = 1, . . . , 4, is independent on the choice of i. One of its bases is the set (q1 + q4 , q2 + q4 ).

5.4.2

Steiner complexes

Let P be the set of unordered pairs of distinct elements in Q(V )− . The addition map in Q(V )− × Q(V )− → V defines a map s : P → V \ {0}. Definition 5.2. The union of pairs from the same fibre s−1 (v) of the map s is called a Steiner compex. It is denoted by Σ(v). It follows from (5.25) that any two pairs from a syzygetic tetrad belong to the same Steiner complex. Conversely, let {q1 , q1 }, {q2 , q2 } be two pairs from Σ(v). We have (q1 + q1 ) + (q2 + q2 ) = v + v = 0, showing that the tetrad (q1 , q1 , q2 , q2 ) is syzygetic.

5.4. ODD THETA CHARACTERISTICS

131

Proposition 5.4.5. There are 22g −1 Steiner complexes. Each Steiner complex consists of 2g−1 (2g−1 − 1) elements paired by translation q → q + v. An odd quadratic form q belongs to a Steiner complex Σ(v) if and only if q(v) = 0. Proof. Since 22g −1 = #(V \{0}), it suffices to show that the map s : P → V \{0} is surjective. The symplectic group Sp(V, ω) acts transitively on V \ {0} and on P , and the map s is obviously equivariant. Thus its image is a non-empty G-invariant subset of V \ {0}. It must coincide with the whole set. By (5.7), we have q(v) = Arf(q + v) + Arf(q). If q ∈ Σ(v), then q + v ∈ Q(V )− , hence Arf(q + v) = Arf(q) = 1 and we get q(v) = 0. Conversely, if q(v) = 0 and q ∈ Σ(v), we get q + v ∈ Q(V )− and hence q ∈ Σ(v). This proves the last assertion. Lemma 5.4.6. Let Σ(v), Σ(v ) be two Steiner complexes. Then #Σ(v) ∩ Σ(v ) = 2g−1 (2g−2 − 1) if ω(v, v ) = 0 2g−2 (2g−1 − 1) otherwise. = v for some

Proof. Let q ∈ Σ(v) ∩ Σ(v ). Then we have q + q = v, q + q q ∈ Σ(v), q ∈ Σ(v ). This implies that q(v) = q(v ) = 0.

(5.26)

Conversely, if these equalities hold, then q+v, q+v ∈ Q(V )− and q, q ∈ Σ(v), q, q ∈ Σ(v ). Thus we have reduced our problem to linear algebra. We want to show that the number of elements in Q(V )− which vanish at 2 nonzero vectors v, v ∈ V is equal to 2g−1 (2g−2 − 1) or 2g−2 (2g−1 − 1) depending on whether ω(v, v ) = 0 or 1. Let q be one such quadratic form. Suppose we have another q with this property. Write q = q + v0 for some α. We have q(v0 ) = 0 since q is odd and ω(v0 , v) = ω(v0 , v ) = 0. Let L be the plane spanned by v, v . Assume ω(v, v ) = 1, then we can include v, v in a standard symplectic basis. Computing the Arf invariant, we find that the restriction of q to L⊥ is an odd quadratic form. Thus it has 2g−2 (2g−1 − 1) zeroes. Each zero gives us a solution for v0 . Assume ω(v, v ) = 0. Then L is a singular plane for q since q(v) = q(v ) = q(v + v ) = 0. Consider W = L⊥ /L ∼ F2g−4 . The form q = 2 has 2g−3 (2g−2 − 1) zeroes in W . Any representative v0 of these zeroes defines the quadratic form q + v0 vanishing at v, v . Any quadratic form we are looking for is obtained in this way. The number of such representatives is equal to 2g−1 (2g−2 − 1). Definition 5.3. Two Steiner complexes Σ(v) and Σ(v ) are called syzygetic (resp. azygetic if ω(v, v ) = 0)(resp. ω(v, v ) = 1). Theorem 5.4.7. The union of three mutually syzygetic Steiner complexes Σ(v), Σ(v ) and Σ(v + v ) is equal to Q(V )− .

132 Proof. Since

CHAPTER 5. THETA CHARACTERISTICS

ω(v + v , v) = ω(v + v , v ) = 0, we obtain that the Steiner complex Σ(v + v ) is syzygetic to Σ(v) and Σ(v ). Suppose q ∈ Σ(v) ∩ Σ(v ). Then q(v + v ) = q(v) + q(v ) + ω(v, v ) = 0. This implies that Σ(v) ∩ Σ(v ) ⊂ Σ(v + v ) and hence Σ(v), Σ(v ), Σ(v + v ) share the same set of 2g−1 (2g−2 − 1) elements. This gives #Σ(v) ∪ Σ(v ) ∪ Σ(v + v ) = 6.2g−2 (2g−1 − 1) − 2.2g−1 (2g−2 − 1) = 2g−1 (2g − 1) = #Q(V )− .

Definition 5.4. A set of three mutually syzygetic Steiner complexes is called a syzygetic triad of Steiner complexes. A set of three Steiner complexes corresponding to vectors forming a non-isotropic plane is called azygetic triad of Steiner complexes. Let Σ(vi ), i = 1, 2, 3 be a azygetic triad of Steiner complexes. Then #Σ(v1 ) ∩ Σ(v2 ) = 2g−2 (2g−1 −1). Each set Σ(v1 )\(Σ(v1 )∩Σ(v1 )) and Σ(v1 )\(Σ(v1 )∩Σ(v1 )) consists of 2g−2 (2g−1 − 1) elements. the union of these sets forms the Steiner complex Σ(v3 ). The number of azygetic triads of Steiner complexes is equal to 1 22g−2 (22g −1) 3 (= the number of non-isortropic planes). We leave the proofs to the reader. Let S4 (V ) denote the set of syzygetic tetrads. By Corollary 5.4.4, each T ∈ S4 (V ) defines an isotropic plane PT in V . Let Isok (V ) denote the set of k-dimensional isotropic subspaces in V . Proposition 5.4.8. Let S4 (V ) be the set of syzygetic tetrads. For each tetrad T let PT denote the corresponding isotropic plane. The map S4 (V ) → Iso2 (V ), T → PT , is surjective. The fibre over a plane T consists of 2g−3 (2g−2 − 1) tetrads forming a partition of the intersection of the Steiner complexes Σ(v), where v ∈ P \ {0}. Proof. The surjectivity of this map is proved along the same lines as we proved Proposition 5.4.5. We use the fact the symplectic group Sp(V, ω) acts transitively on the set of isotropic subspaces of the same dimension. Let T = {q1 , . . . , q4 } ∈ S4 (V ). By definition, PT \ {0} = {q1 + q2 , q1 + q3 , q1 + q4 }. Suppose we have another tetrad T = {q1 , . . . , q4 } with PT = PT . Suppose T ∩ T = ∅. Without loss of generality we may assume that q1 = q1 . Then, after reindexing, we get q1 + qi = q1 + qi , hence qi = qi and T = T . Thus the tetrads T with PT = P are disjoint. Obviously any q ∈ T belongs to the intersection of the Steiner complexes Σ(v), v ∈ P \ {0}. It remains to apply Lemma 5.4.6. A closer look at the proof of Lemma 5.4.6 shows that the fibre over P can be identified with the set Q(P ⊥ /P )− . Combining Proposition 5.4.8 with the computation of the number tg of syzygetic tetrads, we obtain the known number of isotropic planes in V : #Iso2 (V ) = 1 2g (2 − 1)(22g−2 − 1) 3 (5.27)

5.4. ODD THETA CHARACTERISTICS

133

Let Iso2 (v) be the set of isotropic planes containing a nonzero vector v ∈ V . The set Iso2 (v) is naturally identified with nonzero elements in the symplectic space (v ⊥ /v, ω ), where ω is defined by the restriction of ω to v ⊥ . We can transfer the symplectic form ω to Iso(v). We obtain ω (P, Q) = 0 if and only if P + Q is an isotropic 3-subspace. Let us consider the set S4 (V, v) = α−1 (Iso2 (v)). This set consists of syzygetic tetrads that are invariant with respect to the translation by v. In particular, each tetrad from S4 (V, v) is contained in Σ(v). We can identify the set S4 (V, v) with the set of cardinality 2 subsets of Σ(v)/ v . There is a natural pairing on S4 (V, v) defined by T, T = 1 #T ∩ T 2 mod 2. (5.28)

Proposition 5.4.9. For any T, T ∈ S4 (V, v), ω (PT , PT ) = T, T . Proof. Let X = {{T, T } ⊂ S4 (V ) : αv (T ) = αv (T )}, Y = {{P, P } ⊂ Iso2 (v)}. We have a natural map αv : X → Y induced by αv . The pairing ω defines a function ˜ φ : Y → F2 . The corresponding partition of Y consists of two orbits of the stabilizer group G = Sp(V, ω)v on Y . Suppose {T1 , T2 } and {T1 , T2 } are mapped to the same subset {P, P }. Without loss of generality we may assume that T1 , T1 are mapped to P . Thus T 1 + T 2 , T 2 + T1 = T 1 , T 2 + T 1 , T 2 + T 1 , T 1 + T2 , T 2 = T 1 , T 2 + T 1 , T 2 . This shows that the function X → F2 defined by the pairing (5.28) is constant on fibres of αv . Thus it defines a map φ : Y → F2 . Both functions are invariant with respect ˜ to the group G. This immediately implies that their two level sets either coincide or are switched. However, #Iso2 (v) = 22g−2 − 1 and hence the cardinality of Y is equal to (22g−2 − 1)(22g−3 − 1). Since this number is odd, the two orbits are of different cardinalities. Since the map αv is G-equinvariant, the level sets must coincide. ˜

5.4.3

Fundamental sets

Recall that a standard symplectic basis in (V, ω) consists of vectors (v1 , . . . , v2g ) such that ω(vi , vj ) = 0 unless j = g + i. Suppose we have an ordered set S of 2g + 1 vectors (u1 , . . . , u2g+1 ) satisfying ω(ui , uj ) = 1 unless i = j. It defines a standard symplectic basis by setting vi = u1 + · · · + u2i−2 + u2i−1 , vi+g = u1 + · · · + u2i−2 + u2i , i = 1, . . . , g. Conversely, we can solve the ui ’s from the vi ’s uniquely to reconstruct the set S from a standard symplectic basis. Definition 5.5. A set of 2g + 1 vectors (u1 , . . . , u2g+1 ) satisfying ω(ui , uj ) = 1 unless i = j is called a normal system in (V, ω).

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CHAPTER 5. THETA CHARACTERISTICS

We have established a bijective correspondence between normal systems and standard symplectic bases. Recall that a symplectic form ω defines a non-degenerate null-system in V , i.e. a bijective linear map f : V → V ∗ such that f (v)(v) = 0 for all v ∈ V . Fix a basis (e1 , . . . , e2g ) in V and the dual basis (t1 , . . . , t2g ) in V ∗ and consider vectors ui = e1 + · · · + e2g − ei , i = 1, . . . , 2g and u2g+1 = e1 + · · · + e2g . Then there exists a unique null-system V → V ∗ that sends ui to ti and u2g+1 to tg+1 = t1 +· · ·+t2g . The vectors u1 , . . . , u2g+1 form a normal system in the corresponding symplectic space. Let (u1 , . . . , u2g+1 ) be a normal system. Denote the corresponding vectors by pi,2g+2 . We will identify nonzero vectors in V with points in the projective space P(V ). For any i, j = 2g + 2 consider the line spanned by pi,2g+2 and pi,2g+2 . Let pij be the third nonzero point in this line. Now do the same with points pij and pkl with the disjoint sets of indices. Denote this point by pijkl . Note that the residual point on the line spanned by pij and pjk is equal to pik . Continuing in this way we will be able to index all points in bb(V ) with subsets of even cardinality (up to complementary sets) of the set Bg = {1, . . . , 2g + 2}. This notation will agree with the notation of 2-torsion points for hyperelliptic curves of genus g. For example, we have ω(pI , pJ ) = #I ∩ J mod 2.

It is easy to compute the number of normal systems. It is equal to the number of standard symplectic bases in (V, ω). The group Sp(V, ω) acts simply transitively on such bases, so their number is equal to #Sp(2g, F2 ) = 2g (22g − 1)(22g−2 − 1) · · · (22 − 1).
2

(5.29)

Now we introduce the analog of a normal system for quadratic forms in Q(V ). Definition 5.6. A fundamental set in Q(V ) is an ordered azygetic set of 2g+2 elements in Q(V ). The number 2g + 2 is the largest possible cardinality of a set such that any three elements are azygetic. This follows from the following immediate corollary of Lemma 5.4.1. Lemma 5.4.10. Let B = (q1 , . . . , qk ) be an azygetic set. Then the set (q1 +q2 , . . . , q1 + qk ) is a normal system in the symplectic subspace of dimension k − 2 spanned by these vectors. The lemma shows that any fundamental set in Q(V ) defines a normal system in V , and hence a standard symplectic basis. Conversely, starting from a normal system (u1 , . . . , u2g+1 ) and any q ∈ Q(V ) we can define a fundamental set (q1 , . . . , q2g+2 ) by q1 = q, q2 = q + u1 , . . . , q2g+2 = q + u2g+1 . Since elements of in a fundamental system add up to zero, we get that the elements of a fundamental set also add up to zero.

5.4. ODD THETA CHARACTERISTICS

135

Proposition 5.4.11. There exists a fundamental set with all or all but one quadratic forms are even or odd. The number of odd quadratic forms in such a basis is congruent to g + 1 modulo 4. Proof. Let (u1 , . . . , u2g+1 ) be a normal system and (t1 , . . . , t2g+1 ) be its image under the map V → V ∗ defined by ω. Consider the quadratic form q=
1≤i<j≤2g+1

ti tj .

It is immediately checked that q(uk ) ≡
2g 2

= g(2g − 1) ≡ g

mod 4.

Passing to the associated symplectic basis we can compute the Arf invariant of q to get Arf(q) = This implies that Arf(q + t2 ) = Arf(q) + q(uk ) = k 0 if g ≡ 0, 3 mod 4 otherwise. 1 if g ≡ 1 mod 2 0 otherwise.

2 Consider the fundamental set formed by the quadrics q, 2q + yk , k = 1, . . . , 2g + 1. Thus if g ≡ 0 mod 4 the set consists of all even quadratic forms. If g ≡ 1 mod 4, the quadratic form q is odd, all other quadratic forms are even. If g ≡ 2 mod 4, all quadratic forms are odd. Finally, if g ≡ 3 mod 4, then q is even, all other quadratic forms are odd.

Definition 5.7. A fundamental set with all or all but one quadratic forms are even or odd is called a normal fundamental set One can show (see [50], p. 271) that any normal fundamental set is obtained as in the proof of the previous proposition. Choose a normal fundamental set (q1 , . . . , q2g+2 ) such that all the first 2g + 1 quadrics are of the same type. Any quadratic form q ∈ Q(V ) can be written in the form q2g+2 + t2 = q + t2 , i i
i∈I i∈I

where I is a subset of [1, 2g + 1]. We denote such a quadratic form by qS , where S = I ∪ {2g + 2} considered as a subset of 1, 2g + 2] modulo the complementary set. We can and will always assume that #S ≡ g + 1 mod 2.

The quadratic form qS can be characterized by the property that it vanishes on points pij , where i ∈ S and j ∈ {1, . . . , 2g + 2}. The following properties can be checked.

136 • qS + qT = pS+T ; • qS + pI = qS+I ;

CHAPTER 5. THETA CHARACTERISTICS

• qS (pT ) = 0 if and only if #S ∩ T + 1 #S ≡ 0 mod 2; 2 • qS ∈ Q(V )+ if and only if #S ≡ g + 1 mod 4. Again we see that a choice of a fundamental set defines the notation of quadratic forms which agrees with the notation of theta characteristics for hyperelliptic curves. Since fundamental sets are in a bijective correspondence with normal systems their number is given by (5.29)

5.5
5.5.1

Scorza correspondence
Correspondences on an algebraic curve

Let C be a nonsingular projective curve and R be a correspondence on C, i.e. a closed subvariety of C × C (not necessarily reduced or irreducible). Recall the standard terminology. We assume that R does not contain the diagonal and the projections of R to each factor is a finite map. The degrees (a, b) of these maps define the type of the correspondence. A correspondence is symmetric if it is invariant under the switch of the factors. One can define the inverse correspondence R−1 as the image of R under the switch of the factors so that R is symmetric if and only if R = R−1 . We assume that R is irreducible and does not coincide with the diagonal ∆ of C × C. A united point of a correspondence is a common point with the diagonal. It comes with the multiplicity. In the usual way a correspondence of type (a, b) defines a regular map from R to the symmetric products C (b) , the image of a point x under this map is the divisor R(x) of degree b equal to the intersection of R with {x} × C. It can be extended, by additivity, to an endomorphism of the Jacobian variety φR : Jac(C) → Jac(C). A correspondence R has valence ν if the divisor classes of R(x) + νx does not depend on x. Proposition 5.5.1. The following properties are equivalent: (i) the cohomology class [R] in H 2 (C × C, Z) is equal to [R] = (a + ν)[{x} × C] + (b + ν)[C × {x}] − ν∆, where x is any point on C; (ii) the divisor class of R(x) + νx does not depend on x; (iii) the homomorphism φR is equal to homomorphism [−ν] : Jac(C) → Jac(C) of the multiplication by −ν.

5.5. SCORZA CORRESPONDENCE

137

Proof. (i) ⇒ (ii) Let p1 , p2 : C × C → C be the projections. We use the well-known fact that the natural homomorphism of the Picard varieties p∗ (Pic0 (C)) ⊕ p∗ (Pic0 (C)) → Pic0 (C × C) 1 2 is an isomorphism. Fix a point x0 ∈ C and consider the divisor R + ν∆ − (a + ν)({x0 } × C) − (b + ν)(C × {x0 }). By assumption, it is algebraically equivalent to zero. Thus R + ν∆ ∼ p∗ (D1 ) + p∗ (D2 ) 1 2 for some divisors D1 , D2 on C. Thus the divisor class R(x)+νx is equal to the divisor class of the restriction of p∗ (D2 ) to {x} × C. Obviously, it is equal to the divisor class 2 of D2 , hence is independent of x. (ii) ⇔ (iii) This follows from the definition of the homomorphism φR . (ii) ⇒ (i). We know that there exists a divisor D on C such that the restriction R + ν∆ − p∗ (D) to any fibre of p1 is linearly equivalent to zero. By the seesaw 2 principle ([180] Chapter 2, Corollary 6), R + ν∆ − p∗ (D) ∼ p∗ (D ) for some divisor 2 1 D on C. This implies that [R] = deg D [{x} × C] + deg D[C × {x}] − ν[∆]. Taking the intersections with fibres of the projections, we find that a = deg D − ν and b = deg D − ν. Note that for a general curve C of genus g > 2 End(Jac(C)) ∼ Z = (see [158]), so any correspondence has valence. An example of a correspondence without valence is the graph of an automorphism of order > 2 of C. Observe that the proof of the proposition shows that for a correspondence R with valence ν R ∼ p∗ (D ) + p∗ (D) − ν∆, (5.30) 1 2 where D is the divisor class of R(x) + νx and D is the divisor class of R−1 (x) + νx. It follows from the proposition that the correspondenceR−1 has valence ν. The next corollary is known as the Cayley-Brill formula. Corollary 5.5.2. Let R be a correspondence of type (a, b) on a nonsingular projective curve C of genus g. Assume that R has valence equal to ν. Then the number of united points of R is equal to a + b + 2νg. This immediately follows from (5.30) and the formula ∆ · ∆ = 2 − 2g. Example 5.5.1. Let C be a nonsingular complete intersection of a nonsingular quadric Q and a cubic in P3 . In other words, C is a canonical curve of genus 4 curve without vanishing even theta characteristic. For any point x ∈ C, the tangent plane Tx (Q) cuts 1 out the divisor 2x + D1 + D2 , where |x + D1 | and |x + D2 | are the two g3 ’s on C defined by the two rulings of the quadrics. Consider the correspondence R on C × C defined by R(x) = D1 + D2 . This is a symmetric correspondence of type (4, 4) with 1 valence 2. Its 24 united points correspond to the ramification points of the two g3 ’s.

138

CHAPTER 5. THETA CHARACTERISTICS

5.5.2

Scorza correspondence

Let C be a nonsingular projective curve of genus g > 0 and ϑ is a non-effective thetacharacteristic on C. Let d1 : C × C → Jac(C), x, y) → [x − y] (5.31) be the difference map. Let Θ = Wg−1 − ϑ be symmetric theta divisor corresponding to ϑ. Define Rϑ = d−1 (Θ). 1 Set-theoretically, (Rϑ )red = {(x, y) ∈ C × C : h0 (x + ϑ − y) > 0}. Lemma 5.5.3. Rϑ is a symmetric correspondence of type (g, g), with valence equal to −1 and without united points. Proof. Since Θ is a symmetric theta divisor, the divisor d−1 (Θ) is invariant with re1 spect to the switch of the factors of X × X. This shows that Rϑ is symmetric. Fix a point x0 and consider the map i : C → Jac(C) defined by i(x) = [x − x0 ]. It is known (see [19], Chapter 11, Corollary (2.2)) that Θ · i∗ (C) = (C × {x0 }) · d∗ (Θ) = g. 1 This shows that Rϑ is of type (g, g). Also it shows that Rϑ (x0 ) − x0 + ϑ ∈ Wg−1 . For any point x ∈ C, we have h0 (ϑ + x) = 1 because ϑ is non-effective. Thus Rϑ (x) is the unique effective divisor linearly equivalent to x + ϑ. By definition, the valence of Rϑ is equal to −1. Applying the Cayley-Brill formula we obtain that Rϑ has no united points. Definition 5.8. The correspondence Rϑ is called the Scorza correspondence. Example 5.5.2. Assume g = 1 and fix a point on C equipping C with a structure of an elliptic curve. Then ϑ is a non-trivial 2-torsion point. The Scorza correspondence Rϑ is the graph of the translation automorphism defined by η. In general Rϑ could be neither reduced nor irreducible correspondence. However for general curve X of genus g everything is as expected. Proposition 5.5.4. Assume C is general in the sense that End(Jac(C)) ∼ Z. Then Rϑ = is reduced and irreducible. Proof. The assumption that End(Jac(C)) ∼ Z implies that any correspondence on = C × C has valence. This implies that the Scorza correspondence is irreducible curve and reduced. In fact, it is easy to see that the valence of the sum of two correspondences is equal to the sum of valences. Since Rϑ has no united points, it follows from the Cayley-Brill formula that the valence of each part must be negative. Since the valence of Rϑ is equal to −1, we get a contradiction.

5.5. SCORZA CORRESPONDENCE It follows from (5.30) that the divisor class of Rϑ is equal to Rϑ ∼ p∗ (ϑ) + p∗ (ϑ) + ∆. 1 2

139

(5.32)

Since KC×C = p∗ (KC )+p∗ (KC ), applying the adjunction formula and using that 1 2 ∆ ∩ R = ∅ and the fact that p∗ (ϑ) = p∗ (ϑ), we easily find 1 2 ωRϑ = 3p∗ (ωC ). 1 In particular, the arithmetic genus of Rϑ is given by pa (Rϑ ) = 3g(g − 1) + 1. (5.34) (5.33)

Note that the curve Rϑ is very special, for example, it admits a fixed-point free involution defined by the switching the factors of X × X. Proposition 5.5.5. Assume that C is not hyperelliptic. Let R be a symmetric correspondence on C × C of type (g, g), without united points and some valence. Then there exists a unique non-effective theta characteristic ϑ on C such that R = Rϑ . Proof. It follows from the Cayley-Brill formula that the valence ν of R is equal to −1. Thus the divisor class of R(x) − x does not depend on x. Since R has no united points, the divisor class D = R(x) − x is not effective, i.e., h0 (R(x) − x) = 0. Consider the difference map d1 : C × C → Jac(C). For any (x, y) ∈ R, the divisor 0 R(x) − y ∼ D + x − y is effective of degree g − 1. Thus d1 (R) + D ⊂ Wg−1 . Let σ : X × X → X × X be the switch of the factors. Then
0 0 φ(R) = d1 (σ(R)) = [−1](d1 (R)) ⊂ [−1](Wg−1 − D) ⊂ Wg−1 + D ,

where D = KC − D. Since R ∩ ∆ = ∅ and C is not hyperelliptic the equality d1 (x, y) = d1 (x , y ) implies (x, y) = (x , y ). Thus the difference map d1 is injective on R. This gives
0 0 R = d−1 (Wg−1 − D) = d−1 (Wg−1 − D ). 1 1

Restricting to {x} × C we see that the divisor classes D and D are equal. Hence D is a theta characteristic ϑ. By assumption, h0 (R(x) − x) = h0 (ϑ) = 0, hence ϑ is non-effective. The uniqueness of ϑ follows from formula (5.32). Let x, y ∈ Rϑ . Then the sum of two positive divisors (Rϑ (x) − y) + (Rϑ (y) − x) is linearly equivalent to x + ϑ − y + y + ϑ − x = 2ϑ = KC . This defines a map γ : Rϑ → |KC |, (x, y) → (Rϑ (x) − y) + (Rϑ (y) − x). Recall from [129], p. 360, that the theta divisor Θ defines the Gauss map G : Θ0 → |KC |, where Θ0 is the open subset of nonsingular points of Θ. It assigns to a point z the tangent space Tz (Θ) considered as a hyperplane in Tz (Jac(C)) ∼ H 1 (C, OC ) ∼ = = (5.35)

140

CHAPTER 5. THETA CHARACTERISTICS

H 0 (C, OC (KC ))∗ . More geometrically, G assigns to D − ϑ the linear span of the divisor D in the canonical space |KC |∗ (see [5], p. 246). Since the hyperplane section of the canonical C by the hyperplane γ(x, y) contains the divisors R(x) − y (and R(y) − x), and they do not move, we see that γ = G ◦ d1 . Lemma 5.5.6. γ ∗ (O|KC | (1)) ∼ ORϑ (Rϑ ) ∼ p∗ (KC ). = = 1

Proof. The Gauss map G is given by the normal line bundle OΘ (Θ). Thus the map γ is given by the line bundle d∗ (OΘ (Θ)) = ORϑ (d∗ (Θ) ∼ ORϑ (Rϑ ). = 1 1 It remains to apply formula (5.32). The Gauss map is a finite map of degree 2g−2 . It factors through the map Θ0 → g−1 Θ0 /(ι), where ι is the negation involution on Jac(C). The map γ also factors through the involution of X × X. Thus the degree of the map Rϑ → γ(Rϑ ) is equal to 2d(ϑ), where d(ϑ) is some numerical invariant of the theta characteristic ϑ. We call it the Scorza invariant. Let Γ(ϑ) := γ(Rϑ ). We considered it as a curve embedded in |KC |. Applying Lemma 5.5.6, we obtain Corollary 5.5.7. deg Γ(ϑ) = g(g − 1) . d(ϑ)

Remark 5.5.1. Let C be a canonical curve of genus g and Rϑ be a Scorza correspondence on C. For any x, y ∈ C consider the degree 2g divisor D(x, y) = Rϑ (x) + Rϑ (y) ∈ |KC + x + y|. Since |2KC − (KC + x + y)| = |KC − x − y| we obtain that the dimension of the linear system of quadrics through D(x, y) is of dimension 1 g−1 (2)| − 2g + 1. This shows that the set D(x, y) imposes 2 g(g + 1) − 2g = dim |OP one less condition on quadrics passing through this set. For example, when g = 3 we get that D(x, y) is on a conic. If g = 3 it is the base set of a net of quadrics. We refer to [84] and [105] for projective geometry of sets imposing one less condition on quadrics (called self-associated sets).

5.5.3

Scorza quartic hypersurfaces

The following construction due to G. Scorza needs some generality assumption on C. Definition 5.9. A pair (C, ϑ) is called Scorza general if the following properties are satisfied (i) Rϑ is a connected nonsingular curve;

5.5. SCORZA CORRESPONDENCE (ii) d(ϑ) = 1; (iii) Γ(ϑ) is not contained in a quadric.

141

We will see in the next chapter that a general canonical curve of genus 3 is Scorza general. For higher genus this was proven in [235]. We continue to assume that C is non-hyperelliptic. Consider the canonical embedding C → |KC |∗ ∼ Pg−1 and identify C with its image (the canonical model of C). = For any x ∈ C, the divisor Rϑ (x) consists of g points yi . If all of them distinct we have g hyperplanes γ(x, yi ) = Rϑ (x)−yi , or, g points on the curve Γ(ϑ). More generally, we have a map C → C (g) defined by the projection p1 : Rϑ → C. The composition of this map with the map γ (g) : C (g) → Γ(ϑ)(g) is a regular map φ : C → Γ(ϑ)(g) . Let H ∩ C = x1 + . . . + x2g−2 be a hyperplane section of C. Adding up the images of the points xi under the map φ we obtain g(2g − 2) points on Γ(ϑ). Proposition 5.5.8. Let D = x1 + . . . + x2g−2 be a canonical divisor on C. Assume (C, ϑ) is Scorza general. Then the divisors
2g−2

φ(D) =
i=1

φ(xi )

span the linear system of divisors on Γ(ϑ) cut out by quadrics. Proof. First note that the degree of the divisor is equal to 2 deg Γ(ϑ). Let (x, y) ∈ Rϑ and Dx,y = γ(x, y) = (Rϑ (x) − y) + (Rϑ (y) − x) ∈ |KC |. For any xi ∈ Rϑ (x) − y, the divisor γ(x, xi ) contains y. Similarly, for any xj ∈ Rϑ (y) − x, the divisor γ(y, xj ) contains x. This means that φ(Dx,y ) is cut out by the quadric Qx,y equal to the sum ˇ ˇ of two hyperplanes Hx , Hy corresponding to the points x, y ∈ C ⊂ |KC |∗ via the duality. The image of |KC | in Γ(ϑ)(g(2g−2)) spans a linear system L (since any map of a rational variety to Jac(Γ(ϑ)) is constant). Since Γ(ϑ) is not contained in a quadric, it generates |KC |. This shows that all divisors in L are cut out by quadrics. The quadrics Qx,y span the space of quadrics in |KC | since otherwise there exists a quadric in |KC |∗ apolar to all quadrics Qx,y . This would imply that for a fixed x ∈ C, the divisor Rϑ (x) lies in a hyperplane, the polar hyperplane of the quadric with respect to the point x. However, because ϑ is non-effective, Rϑ (x) spans Pg−1 . Thus dim L ≥ g(g + 1)/2, and, since no quadrics contains Γ(ϑ), L coincides with the linear system of quadrics sections of Γ(ϑ). Let E = H 0 (C, ωC )∗ so that |KC | = P(E ∗ ) and |KC |∗ = P(E). We can identify the space of quadrics in P(E) with P(S 2 E). Using the previous proposition we obtain a map P(E ∗ ) → P(S 2 E). The restriction of this map to the curve Γ(ϑ) is given by the linear system |OΓ(ϑ) (2)|. This shows that the map is given by quadratic polynomials, so defines a linear map α : S 2 (E ∗ ) → S 2 (E). The proof of the proposition implies that this map is bijective. Theorem 5.5.9. Assume (C, ϑ) is Scorza general. There exists a unique quartic hypersurface V (f ) in P(E) = Pg−1 such that the inverse linear map α−1 is equal to the polarization map ψ → Dψ (f ).

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CHAPTER 5. THETA CHARACTERISTICS

Proof. Consider α−1 : S 2 (E) → S 2 (E ∗ ) as a tensor U ∈ S 2 (E ∗ ) ⊗ S 2 (E ∗ ) ⊂ (E ∗ )⊗4 viewed as a 4-multilinear map E 4 → C. It is enough to show that U is totally symmetric. Then α−1 is defined by the apolarity map associated to a quartic hypersurface. Fix a reduced divisor Rϑ (x) = x1 +. . .+xg . Let Hi be the hyperplane in P(E) spanned by Rϑ (x) − xi . Choose a basis (t1 , . . . , tg ) in E ∗ such that Hi = V (ti ). It follows from the proof of Proposition 5.5.8 that the quadratic map P(E ∗ ) → P(S 2 E) assigns to the hyperplane Hi the quadric Qx,xi equal to the union of two hyperplanes associated to x and xi via the duality. The corresponding linear map α satisfies
g

α(t2 ) = ξj ( j
i=1

bi ξi ), j = 1, . . . , g,

where (ξ1 , . . . , ξg ) is the dual basis to (t1 , . . . , tg ), and (b1 , . . . , bg ) are the coordinates of the point x. This implies that
g

U (ξj ,
i=1

bi ξ i , ξ k , ξ m ) =

1 if j = k = m, = U (ξk , bi ξj , ξk , ξm ). 0 otherwise i=1

g

This shows that U is symmetric in the first and the thirst argument when the second argument belongs to the curve Γ(ϑ). Since the curve Γ(ϑ) spans P(E ∗ ), this is always true. It remains to use that U is symmetric in the first and the second argument, as well as in the third and the forth argument. Definition 5.10. Let (C, ϑ) be Scorza general pair consisting of a canonical curve of genus g and a non-effective theta characteristic ϑ. Then the quartic hypersurface V (f ) is called the Scorza quartic hypersurface associated to (C, ϑ). We will study the Scorza quartic plane curves in the case g = 3. Very little is known about Scorza hypersurfaces for general canonical curves of genus > 3. We do not even know whether they are nonsingular. However, it follows from the construction that it is always a nondegenerate in the sense of section 1.4.1.

5.5.4

Theta functions and bitangents

Let C be a nonsingular curve of genus g > 0. Fixing a point c0 on C allows one to define an isomorphism of algebraic varieties Picd (C) → Jac(C), [D] → [D − dc0 ].

The composition of this map with the map ud : C (d) → Picd (C) is and consider the map, called the Abel-Jacobi map ud (c0 ) : C (d) → Jac(C). If no confusion arises we drop c0 from this notation. For d = 1, this map defines an embedding u1 : C → Jac(C).

5.5. SCORZA CORRESPONDENCE

143

For the simplicty of the notation we will identify C with its image. For any c ∈ C the tangent space of C at a point c is a one-dimensional subspace of the tangent space of Jac(C) at c. By the unique translation automorphism we identify this space with T0 Jac(C) at the zero point. Under the Abel-Jacobi map the space of holomorphic 1form on Jac(C) is identified with the space of holomorphic forms on C. Thus we can identify T0 Jac(C) with the dual space H 0 (C, KC )∗ . As a result we obtain the canonical map of C ϕ : C → P(H 0 (C, KC )∗ ) = |KC |∗ ∼ Pg−1 . = If C is not hyperelliptc the canonical map is an embedding. We continue to identify H 0 (C, KC )∗ with T0 Jac(C). A symmetric odd theta divisor Θ = Wg−1 −ϑ contains the origin of Jac(C). If h0 (ϑ) = 1, this point is nonsingular and hence θ defines a hyperplane in T0 (Jac(C)), the tangent hyperplane T0 Θ. Passing to the projectivization we have a hyperplane in |KC |∗ . Proposition 5.5.10. The hyperplane in |KC |∗ defined by Θ is a bitangent hyperplane to the image ϕ(C) under the canonical map. Proof. Consider the difference map (5.31) d1 : C × C → Jac(C). In the case when Θ is an even divisor, we proved in (5.32) that d∗ (Θ) ∼ p∗ (θ) + p∗ (θ) + ∆. 1 1 2 (5.36)

Since two theta divisors are algebraically equivalent the same is true for an odd theta divisor. The only difference is that d∗ (Θ) contains the diagonal ∆ as the pre-image of 1 0. It follows from the definition of the Abel-Jacobi map u1 (c0 ) that u1 (c0 )(C) ∩ Θ = d−1 (Θ) ∩ p−1 (c0 ) = c0 + Dϑ , 1 1 where Dϑ is the unique effective divisor linearly equivalent to ϑ. Let G : Θ ns → P(T0 Jac(C)) be the Gauss map defined by translation of the tangent space at a nonsingular point of Θ to the origin. It follows from the proof of Torelli Theorem [5] that the Gauss map ramifies at any point where θ meets u1 (C). So, the image of the Gauss map intersects the canonical image with multiplicity ≥ 2 at each point. This proves the assertion. More explicitly, the equation of the bitangent hyperplane corresponding to Θ is given by the linear term of the Taylor expansion of the theta function θ [ η ] corresponding to Θ. Note that the linear term is a linear function on H 0 (C, KC )∗ , hence can be identified with a holomorphic differential
g

hΘ =
i=1

∂θ [ η ] (z, τ ) (0)ωi , ∂zi

where (z1 , . . . , zg ) are coordinates in H 0 (C, KC )∗ defined by a normalized basis ω1 , . . . , ωg of H 0 (C, KC ). The sections of H 0 (Jac(C), OJac(C) (Θ) ∼ C can be identified = with holomorphic half-order differentials. To make this more precise, i.e. describe how to get a square root of a holomorphic 1-differential, we use the following result (see [106], Proposition 2.2).

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CHAPTER 5. THETA CHARACTERISTICS

Proposition 5.5.11. Let Θ be a symmetric odd theta divisor defined by the theta function θ [ η ]. Then for all x, y ∈ C θ [ η ] (d1 (x − y))2 = hΘ (ϕ(x))hΘ (ϕ(y))E(x, y)2 , where E(x, y) is a certain section of OC×C (∆) (the prime-form). An attentive reader should notice that the equality is not well-defined in many ways. First, the vector ϕ(x) is defined only up to proportionality and the value of a section of a line bundle is also defined only up to proportionality. To make sense of this equality we pass to the universal cover of Jac(C) identified with H 0 (C, KC )∗ and to the universal cover U of C × C and extend the difference map and the map ϕ to the map of universal covers. Then the prime-form is defined by a certain holomorphic function on U and everything makes sense. As the equality of the corresponding line bundles, the assertion trivially follows from (5.36). Now let us fix a point y = c0 , so we can define the root function on C. It is a rational function on the universal cover of C defined by r [ η ] (x, c0 ) = It satisfies θ [ η ] (d1 (x − c0 )) E(x, c0 )
2

θ [ η ] (d1 (x − c0 )) . E(x, c0 )

= hΘ (ϕ(x))hΘ (ϕ(c0 )).

(5.37)

Note that E(x, y) satisfies E(x, y) = −E(y, x), since θ [ η ] is an odd function, we have r [ η ] (x, y) = r [ η ] (y, x) for any x, y ∈ C × C \ ∆. Thus every honest bitangent hyperplane of the canonical curve defines a rootfunction. Suppose we have two odd theta functions θ [ η ] , θ η . Then the ratio of the corresponding root functions is equal to
θ [ η ](d1 (x−c0 )) » – θ (d1 (x−c0 )) η

and its square is a rational function
1

on C, defined uniquely up to a constant factor depending on the choice of c0 . Its divisor
2 is equal to the difference 2ϑ − 2ϑ . Thus we can view the ratio as a section of KX with divisor θ−θ . This section is not defined on C but on the double cover of C corresponding to the 2-torsion point ϑ−ϑ . If we have two pairs ϑ1 , ϑ , ϑ2 , ϑ2 of odd theta characteristics satisfying ϑ1 − ϑ = ϑ2 − ϑ2 = , i.e. forming a syzygetic tetrad, the product of the two ratios is a rational function on C with divisor ϑ1 + ϑ2 − ϑ − ϑ2 . Following

Riemann [201] and Weber [247] we denote this function by

ϑ1 ϑ1 ϑ2 ϑ2 .

By Riemann-

Roch, h0 (ϑ1 + ϑ2 ) = h0 (KC + ) = g − 1, hence any g pairs (ϑ1 , ϑ ), . . . , (ϑg , ϑg ) of odd theta characteristics in a Steiner complex define g linear independent functions
ϑ1 ϑ1 ϑg ϑg , . . . , ϑg−1 ϑg−1 . ϑg ϑg

After scaling, and getting rid of squares by using (5.37) we

obtain a polynomial in hΘ1 (ϕ(x)), . . . , hΘg (ϕ(x)) vanishing on the canonical image of C.

EXERCISES

145

Example 5.5.3. Let g = 3. We get the equation of a hypersurface he canonical curve lies in a hypersurface the equations ϑ1 ϑ1 + ϑ2 ϑ2 + ϑ3 ϑ3 = 0. (5.38)

After getting rid of squares, we obtain the quartic equation of C (lm)2 + (pq)2 + (rs)2 − 2lmpq = 2lmrs − 2pqrs = (lm − pq − rs)2 − 4lmpq = 0, (5.39) where l, m, p, q, rs are the linear functions in z1 , z2 , z3 defining the linear terms of the Taylor expansion at 0 of the odd theta functions corresponding to three pairs in a Steiner complex. The number of possible ways to write the equation of a plane quartic in this form is equal to 63.20 = 1260. Remark 5.5.2. For any non-zero 2-totsion point, the linear system KC + | maps C to Pg−2 , the map is called the Prym canonical map. We have seen that the root functions
ϑ1 ϑ1 ϑ2 ϑ2

belong to H 0 (C, KC + ) and can be used to define the Prym canonical map.

For g = 3, the map is a degree 4 cover of P1 and we expressed the quartic equation of C as a degree 4 cover of P1 .

Exercises
5.1 Find 3 non-equivalent symmetric determinant expressions for the cubic curve given by a Weierstrass equation t0 t2 + t3 + at1 t2 + bt3 = 0. 0 0 2 1 5.2 Find a symmetric determinant expression for the Fermat quartic V (t4 + t4 + t4 ). 0 1 2 5.3 Let C be an irreducible plane curve of degree d with a (d − 2)-multiple point. Show that its normalization is a hyperelliptic curve of genus g = d−2. Conversely, show that any hyperelliptic curve of genus g admits such a plane model. 5.4 Show that a nonsingular curve of genus 2 has a vanishing theta characteristic but a nonsingular curve of genus 3 has a vanishing theta characteristic if and only if it is a hyperelliptic curve. 5.5 Show that a nonsingular nonhyperelliptic curve of genus 4 has a vanishing theta characteristic if and only if its canonical model lies on a quadric cone. 5.6 Show that a nonsingular plane curve of degree 5 does not have a vanishing theta characteristic. 5.7 Find the number of vanishing theta characteristics on a hyperelliptic curve of genus g. 5.8 Compute the number of syzygetic tetrads contained in a Steiner complex. 5.9 Show that the composition of two correspondences (defined as the composition of the multivalued maps defined by the correspondences) with valences ν and ν is a correspondence with valence −νν . 5.10 Let f : X → P1 be a non-constant rational function on a nonsingular projective curve X. Consider the fibred product X×P1 X as a correspondence on X×X. Show that it has valence and compute the valence. Show that the Cayley-Brill formula is equivalent to the Hurwitz formula. 5.11 Suppose that a nonsingular projective curve X admits a non-constant map to a curve of genus > 0. Show that there is a correspondence on X without valence.

146

CHAPTER 5. THETA CHARACTERISTICS

5.12 Show that any correspondence on a nonsingular plane cubic has valence unless the cubic is harmonic or equianharmonic. 5.13 Describe all symmetric correspondences of type (4, 4) with valence 1 on a canonical curve of genus 4. 5.14 Let Rϑ be the Scorza correspondence on a curve C. Prove that a point (x, y) ∈ Rϑ is singular if and only if x and y are ramifications points of the projections Rϑ → C.

Historical Notes
It is a too large task to discuss the history of theta functions. We only mention that the connection between odd theta functions with characteristics and bitangents to a quartic curves goes back to Riemann [201], [247]. There are numerous expositions of the theory of theta functions and jacobian varieties (e.g. [5], [47], [181]). The theory of fundamental sets of theta characteristics goes back to A. G¨ pel and J. Rosenhein. o Its good exposition can be found in Krazer’s book [160]. As an abstract symplectic geometry over the field of two elements it is presented in Coble’s book [51] which we followed. Some additional material can be found in [50] (see also a modern exposition in n [206]). The theory of correspondences on an algebraic curve originates from the Charles Principle of Correspondence [39] which is the special case of the Cayley-Brill formula in the case g = 0. We have already encountered with its application to Poncelet polygons in Chapter 2. This application was first found by A. Cayley [37]. He was also the first to extend Chasles’s Principle to higher genus [37] although with incomplete proof. The first proof of the Cayley-Brill formula was given by A. Brill [25]. The notion of valence (die Werthigeit) was introduced by Brill. The fact that only correspondences with valence exist on a general curve was first pointed out by A. Hurwitz [147]. He also showed the existence of correspondences without valence. We refer to [228] for a fuller history of the theory of correspondences. The study of correspondences of type (g, g) with valence −1 was initiated by G. Scorza [217], [218]. His construction of a quartic hypersurface associated to a noneffective theta characteristic on a canonical curve of genus g was given in [219]. A modern exposition of Scorza’ theory was first given in [82].

Chapter 6

Plane Quartics
6.1
6.1.1

Bitangents
28 bitangents

A nonsingular plane quartic C is a non-hyperelliptic genus 3 curve embedded by its canonical linear system |KC |. It has no vanishing theta characteristics, so the only effective theta characteristics are odd ones. The number of them is 28 = 22 (23 − 1). Thus C has exactly 28 bitangents . Each bitangent is tangent to C at two points that may coincide. In the latter case a bitangent is called a flex bitangent. We can specialize the results from section 5 of the previous chapter to the case g = 3 taking V = Pic(C)[2] with the symplectic form ω defined by the Weil pairing. Elements of Q(V )− will be identified with bitangents. The union of bitangents forming a syzygetic tetrad cuts out in C is a divisor of degree 8 equal to C ∩ Q for some conic Q. There are t3 = 315 syzygetic tetrads. There is a bijection between the set of syzygetic tetrad and the set of isotropic planes in Pic(C)[2] There are 63 Steiner complexes of bitangents. Each complex consists of 6 pairs of bitangents i , i such that the divisor class of i ∩C − i ∩C is a fixed nonzero 2-torsion divisor class. Two Steiner complexes have either 4 or 6 common bitangents, dependent on whether they are syzygetic or not. Each isotropic plane in Pic(C)[2] defines three Steiner complexes with common four bitangents. Two azygetic Steoner complexes have 6 common bitangents. The number of azygetic triads is equal to 336. In the following we will often identify an odd theta characteristic with the corresponding bitangent. Let i = V (li ), i = 1, . . . , 4, be four syzygetic bitangents and Q = V (q) be the corresponding conic. Since V (l1 l2 l3 l4 ) and V (q 2 ) cut out the same divisor on C we obtain that C can be given by an equation f = l1 l2 l3 l4 + q 2 = 0. 147 (6.1)

148

CHAPTER 6. PLANE QUARTICS

Conversely, if f can be written in the form (6.1), the linear forms li define four syzygetic bitangents. So we see that f can be written as in (6.1) in only finitely many ways. This is confirmed by “counting constants”. We have 12 constants for the linear forms and 6 constants for quadratic forms, they are defined up to scaling by λ1 , . . . , λ5 subject to the condition λ1 · · · λ4 = λ2 . Thus we have 14 parameters for quartic curves 5 represented in the form (6.1). This is the same as the number of parameters for plane quartics. Let l = 0, m = 0, p = 0, q = 0, r = 0, s = 0 be the equations of 6 bitangents such that (l, m, p, q) and (l, m, r, s) are two syzygetic tetrads. In other words, the three pairs (l, m), (p, q), (r, s) is a part of the set of 6 pairs in a Steiner complex of bitangents. By (6.1), we can write f = lmpq − q 2 = lmrs − q 2 for some quadratic forms q, q . Subtracting we have lm(pq − rs) = (q + q )(q − q ).
1 If l divides q +q and m divides q −q , then the quadric V (q) = V ( 2 [(q +q)+(q −q )]) passes through the point l ∩ m. But this is impossible since no two bitangents intersect at a point on the quartic. Thus we obtain that lm divides either q + q or q − q . Without loss of generality, we get lm = q + q , pq − rs = q − q , and hence 1 q = 2 (lm + pq − rs). Thus we define the quartic by the equation

−4f = −4lmpq+(lm+pq−rs)2 = (lm)2 −2lmpq−2lmrs−2pqrs+(pq)2 +(rs)2 = 0. (6.2) It is easy to see that this is equivalent to the equation √ √ √ lm + pq + rs = 0. (6.3) Thus we see that a nonsingular quartic can be written in 315 ways in the form of (6.1) and in 1260 = 6 ·63 ways in the form of (6.3). In the previous chapter we found 3 this equation by using theta functions. Remark 6.1.1. Consider the orbit space X = (C3 )6 /T , where T = {(z1 , z2 , z3 , z4 , z5 , z6 ) ∈ (C∗ )6 : z1 z2 = z3 z4 = z5 z6 } is a 14-dimensional algebraic torus. Any orbit T (l, m, p, q, r, s) ∈ X defines the quar√ √ √ tic curve V ( lm + pq + rs). We have shown that the map X → |OP2 (4)| ∼ P14 = is of degree 1260. The group PGL(3) acts naturally on both spaces. One can show that X/P GL(3) is a rational variety and we get a map X/PGL(3) → |OP2 (4)|/PGL(3) ∼ = M3 of degree 1260.
1 The projection from the intersection point of two bitangents defines a g4 with two 1 non-reduced members. The intersection point of three bitangents gives a g4 which is not expected on a general curve of genus 3. It is not known the maximal possible number of triple points of the arrangements of 28 lines formed by the bitangents. However, we can prove the following.

6.1. BITANGENTS

149

Proposition 6.1.1. No three bitangents forming an azygetic triad can intersect at one point. Proof. Let ϑ1 , ϑ2 , ϑ3 be the corresponding odd theta characteristics. The 2-torsion divisor classes ij = ϑi − ϑj form a non-isotropic plane. Let be a non-zero point in the orthogonal complement. Then qηi ( ) + qηj ( ) + ηij , = 0 implies that qηi takes the same same value at . We can always choose such that this value is equal to 0. Thus the three bitangents belong to the same Steiner complex Σ( ). Obviously, no two differ by , hence we can form 3 pairs from them. These pairs can be used to define the equation (6.3) of C. It follows from this equation that the intersection point of the three bitangents lies on C. But this is impossible because C is nonsingular. Remark 6.1.2. A natural question is whether the set of bitangents determines the quartic, i.e. whether two quartics with the same set of bitangents coincide. Surprizingly it has not been answered by the ancients. Only recently it was proven that the answer is yes [27] (for general curve), [164] (for any nonsingular curve).

6.1.2

Aronhold sets

We know that in the case g = 3 a normal fundamental set of 8 theta characteristics contains 7 odd theta characteristics. The corresponding unordered set of 7 bitangents is called an Aronhold set . It follows from (5.29) that the number of Aronhold sets is equal to #Sp(6, F2 )/7! = 288. A choice of an ordered Aronhold set defines a unique normal fundamental set that contains it. The eighth theta characteristic is equal to the sum of the characteristics from the Aronhold set. Thus an Aronhold set can be defined as an azygetic set of seven bitangents. A choice of an ordered Aronhold set allows one to index all 2-torsion divisor classes (resp. odd theta characteristics) by subsets of even cardinality (resp. of cardinality 2) of {1, . . . , 8}, up to complementary set. Thus we have 63 2-torsion classes ab , abcd and 28 bitangents ij corresponding to 28 odd theta characteristics ϑij . The bitangents from the Aronhold set correspond to the subsets (18, 28, . . . , 78). We also know that ϑA − ϑB = A+B . This implies, for example, that four bitangents A , B , C , D form a syzygetic tetrad if and only if A + B + C + D = 0. Following Cayley we denote a pair of numbers from the set {1, . . . , 8} by a vertical line |. If two pairs have a common number we make them intersect. For example, we have • Pairs of bitangents: 168 of type || and 210 of type ∨; • Triads of bitangents 1. (azygetic) 420 of type , 840 azygetic of type |||, 2. (syzygetic) 56 of type • Tetrads of bitangents: 1. (syzygetic) 105 azygetic of type ||||, 210 of type , W , 1680 of type ∨ |, and 280 of type WÕÕ;

150 2. (asygetic) 560 of type | ∨∨,

CHAPTER 6. PLANE QUARTICS W W , 280 of type WÕÕ , 1680 of type WÕÕ, 2520 of type

3. (non syzygetic but containing a syzygetic triad) 2520 ofWtype || ∨, 5040 of D W type | , 3360 of type , 840 of type  DÕÕ, 3360 of type WÕÕ W W There are two types of Aronhold sets: cc , WÕÕ . They are represented by the  sets of indices (12, 13, 14, 15, 16, 17, 18) and (12, 13, 23, 45, 46, 47, 48). The number of the former type is 8, the number of the latter nu type is 280. Note that the different types correspond to orbits of the subgroup of Sp(6, F2 ) isomorphic to the permutation group S8 . For example, we have two orbits of S8 on the set of Aronhold sets consisting of 8 and 280 elements. Lemma 6.1.2. Three odd theta characteristics ϑ1 , ϑ2 , ϑ3 in a Steiner complex Σ( ), no two of which differ by , are azygetic. Proof. Let ϑi = ϑi + , i = 1, 2, 3. Then the tetrads {ϑ1 , ϑ1 , ϑ2 , ϑ2 } and {ϑ1 , ϑ1 , ϑ3 , ϑ3 } are syzygetic and have two common theta characteristics. By Proposition 5.4.9, the corresponding isotropic planes do not span an isotropic 3-space. Thus ϑ1 −t2 , ϑ1 −ϑ2 = 1, hence ϑ1 , ϑ2 , ϑ3 is an azygetic triad. The previous lemma suggests a way to construct an Aronhold set from a Steiner set Σ( ). Choose another Steiner set Σ(η) azygetic to the first one. They intersect at 6 odd theta characteristics ϑ1 , . . . , ϑ6 , no two of which differ by . Consider the set {ϑ1 , . . . , ϑ5 , ϑ6 + , ϑ6 + η}. We claim that this is an Aronhold set. By the previous lemma all triads ϑi , ϑj , ϑk , i, j, k ≤ 5 are azygetic. Any triad ϑi , ϑ6 + , ϑ6 + η, i ≤ 5, is azygetic too. In fact qϑi ((ϑ6 + )−(ϑ6 +η)) = qϑi ( +η) = 0 since ϑi ∈ Σ( +η). So / the assertion follows from Lemma 5.4.1. We leave to the reader to check that remaining triads {ϑi , ϑj , ϑ6 + }, {ϑi , ϑj , ϑ6 + η}, i ≤ 5, are azygetic. Proposition 6.1.3. Any six lines in an Aronhold set are contained in a unique Steiner complex. We use that the symplectic group Sp(6, F2 ) acts transitively on the set of Aronhold sets. So it is enough to check the assertion for one Aronhold set. Let it correspond to the index set (12, 13, 14, 15, 16, 17, 18). It is enough to check that the first six are contained in a unique Steiner complex. For this it is enough to exhibit a 2-torsion divisor class such that qϑI ( ) = 0 for the first six subsets I and show its uniqueness. Equivalently we have to show that there exists a unique subset J of [1, 8] of cardinality 2 or 4 such that it icontains exactly one element from each I. Obviously, the only such subset is {1, 8}. Corollary 6.1.4. Any Steiner complex contains 26 azygetic hexads. Half of them are contained in another Steiner complex, necessarily azygetic to the first one. Any other hexad can be extended to a unique Aronhold set. Proof. Let Σ( ) be a Steiner complex consisting of 6 pairs of odd theta characteristics. Consider it as G-set, where G = (Z/2Z)6 whose elements, identified with subsets I of [1, 6], act by swithching elements in i-th pairs, i ∈ I. It is clear that G acts simply

6.1. BITANGENTS

151

transitively on the set of azygetic sextupes in Σ( ). For any azygetic complex Σ(η) the intersection Σ( ) ∩ Σ(η) is an azygetic hexad. Note that two syzygetic complexes have only 4 bitangents in common. The number of such hexads is equal to 26 − # ⊥ = 26 − 25 = 25 . Thus the set of azygetic hexads contained in a unique Steiner complex is equal to 25 .63. But this number is equal to the number 288.7 of subsets of cardinality 6 of Aronhold sets. By the previous proposition, all such sets are contained in a unique Steiner complex. Let (ϑ1 , . . . , ϑ7 ) be an Aronhold set. We use the corresponding normal fundamental set to index its elements by the subsets (18, 28, . . . , 78). By Proposition 6.1.3 the hexad ϑ2 , . . . , ϑ7 is contained in a unique Steiner complex Σ( ). Let η2 = ϑ2 + . The only 2-torsion point at which all quadrics q28 , . . . , q78 vanish is the point p18 corresponding to the subset {1, 8}. Thus qη2 = q28 + p18 = q12 . This shows that the bitangent defined by η2 corresponds to (12). similarly, we see that the bitangents corresponding to ηi + , i = 3, . . . , 7 corresponds to (1i).

6.1.3

Riemann’s equations for bitangents

Here we show how to write equations of all bitangents knowing the equations of an Aronhold set of bitangents. Let 1 , . . . , 7 be an Aronhold set of bitangents of C. By Proposition 6.1.1, any three lines are not concurrent. By a linear transformation, and reodering, we may assume 1 = V (t0 ), 2 = V (t1 ), 3 = V (t2 ), 4 = V (t0 + t1 + t2 ) and the remaining ones are V (a0i t0 + a1i t1 + a2i t2 ), i = 1, 2, 3. Theorem 6.1.5. The exist linear forms u0 , u1 , u2 such that C can be given by the equation √ √ √ t0 u0 + t1 u1 + t2 u2 = 0. The forms ui can be found from equations u0 + u1 + u2 + x0 + x1 + x2 u0 u1 u2 + + + k1 a01 x0 + k1 a11 x1 + k1 a21 x2 a01 a11 a21 u0 u1 u2 + + + k2 a02 x0 + k2 a12 x1 + k2 a22 x2 a02 a12 a22 u1 u2 u0 + + + k3 a03 x0 + k3 a13 x1 + k3 a23 x2 a03 a13 a23 =0 = = = 0 0 0,

where k1 , k2 , k3 can be found from solving the following linear equations:
1 a01 1  a11 1 a21



1 a02 1 a12 1 a22

1  a03 1  a13 1 a23

    λ1 −1 λ2  = −1 , · λ3 −1

152  λ0 a01 λ0 a02 λ0 a03 λ1 a11 λ1 a12 λ1 a13

CHAPTER 6. PLANE QUARTICS      λ2 a21 k1 −1 λ2 a22  · k2  = −1 , λ2 a23 k3 −1

The equations of the remaining 21 bitangents are • u0 = 0, u1 = 0, u2 = 0, • x0 + x1 + u2 = 0, x0 + x2 + u1 = 0, x1 + x2 + u0 = 0, • 9 of type • •
u0 a01

+ ki a1i x1 + ki a2i x2 = 0, i = 1, 2, 3,
u1 1−ki a0i a2i

u0 1−ki a1i a2i

+

+

u2 1−ki a01 a1i

= 0, i = 1, 2, 3, = 0, i = 1, 2, 3.

u0 a0i (1−ki a1i a2i )

+

u1 a1i (1−ki a0i a2i )

+

u2 a2i (1−ki a01 a1i )

Proof. Applying Proposition 6.1.3 we can find three Steiner complexes partitioned in pairs pairs ( 2 , ξ3 ), ( 3 , ξ2 ), ( 4 , ξ41 ), . . . , ( 7 , ξ71 ), ( 3 , ξ1 ), ( 1 , ξ3 ), ( 4 , ξ42 ), . . . , ( 7 , ξ72 ), ( 1 , ξ2 ), ( 2 , ξ1 ), ( 4 , ξ43 ), . . . , ( 7 , ξ73 ). (6.4)

We use here that the intersection of two Steiner complexes cannot cannot consist of five tangents. Now we have
2

− ξ3 =

3

− ξ2 ,

3

− ξ1 =

1

− ξ2 ,

1

− ξ2 =

2

− ξ1 ,

where we identify the bitangents with the corresponding odd theta characteristic. This implies that 1 − ξ1 = 2 − ξ2 = 3 − ξ3 , i.e. the three pairs ( 1 , ξ1 ), ( 2 , ξ2 ), ( 3 , ξ3 ) belong to the same Steiner complex Σ. One easily checks that
1

− ξ1 ,

1

− ξ2 =

2

− ξ2 ,

2

− ξ3 =

3

− ξ3 ,

3

− ξ1 = 0,

and hence Σ is syzygetic to the three complexes (6.4) and hence it does not contain i , i ≥ 4. Now we use the three pairs ( 1 , ξ1 ), ( 2 , ξ2 ), ( 3 , ξ3 ) to write C in the form (6.3) √ √ √ x0 u0 + x1 u1 + x2 u2 = 0, where η1 = V (u0 ), η2 = V (u1 ), η3 = V (u2 ). By (7.3.6), we can introduce the quadrics q1 q2 q3 such that
2 2 2 C = V (4x0 x1 u0 u1 − q3 ) = V (−4x0 x2 u0 u2 − q1 ) = V (−4x1 x2 u1 u2 − q3 ), (6.6)

= −x0 u0 + x1 u1 + x2 u3 , = x0 u0 − x1 u1 + x2 u3 , = x0 u0 + x1 u1 − x2 u3 ,

(6.5)

6.1. BITANGENTS

153

Now we use the first Steiner complex from (6.4) to do the same by using the first three pairs. We obtain C = V (4x1 u2 l4 l41 − q 2 ), where 4 = V (l4 ), l4 = x0 + x1 + x2 , ξ41 = V (l41 ), and q is a quadratic form. As in the proof of formula (6.3), we find that q1 − q = 2λ1 x1 u2 , q1 + q = From this we get q1 = λ1 x1 u2 + This gives l4 l41 l4 l42 l4 l43 = x2 u1 − λ1 (−x0 u0 + x1 u1 + x2 u3 ) + λ2 x1 u2 1 = x2 u1 − λ1 (−x0 u0 − x1 u1 + x2 u3 ) + λ2 x2 u0 1 = x2 u1 − λ1 (x0 u0 + x1 u1 − x2 u3 ) + λ2 x0 u1 . 1 (6.7) x2 u1 − l4 l41 = −x0 u0 + x1 u1 + x2 u3 . λ1 2(x2 u2 − l4 l41 . λ1

The last two equations give l4 ( l43 u2 x1 l42 + ) = x0 (−2u0 + λ3 u1 + ) + u0 (λ2 x2 + ). λ2 λ3 λ3 λ3 (6.8)

The lines V (l4 ), V (l0 ), and V (u0 ) belong to the third Steiner complex (6.4), and by lemma 6.1.2 form an azygetic triad. By Proposition 6.1.1, they cannot be concurrent. u This implies that the line V (α2 x2 + α1 ) passes through the intersection point of the 3 lines V (x0 ) and V (l4 ). This gives a linear dependence between the linear functions x l4 = a0 x0 + a1 x1 + a2 x2 , l0 = x0 and (α2 x2 + λ1 ) (we can assume that a0 = a1 = 3 a2 = 1 but will do it later). This can happen only if α2 = c1 a2 , 1 = c1 a1 , α3

1 for some constant c1 . Now α2 x2 + α3 x1 = c1 (a2 x2 + a1 x1 ) = c1 (l4 − a0 x0 ), and we can rewrite (6.8) in the form

c1 l4 ( This implies that

l42 l43 u1 u2 + − c1 u0 ) = x0 (−c1 (2 + a0 c1 )u0 + + ). λ2 λ3 a1 a2 l42 l43 k1 + = c1 u0 + x0 , α2 α3 c1 k1 l4 = −c1 (2 + c1 a0 )u0 + u1 u2 + , a1 a2 u0 u2 + , a0 a2 (6.9) (6.10)

for some constant k1 . Similarly, we get k1 l4 = −c2 (2 + c2 a1 )u1 +

154

CHAPTER 6. PLANE QUARTICS k1 l4 = −c3 (2 + c3 a2 )u2 + u1 u2 + . a0 a1

It is easy to see that this implies that k1 = k2 = k3 = k, c1 = −a0 , c2 = −a1 , c3 = −a2 . The equations (6.9) and (6.10) become l42 l43 k + = −a0 u0 − x0 , α2 α3 a0 kl4 = (6.11)

u1 u2 u0 + + . (6.12) a0 a1 a2 At this point, we can scale the coordinates to assume a1 = a2 = a2 = 1, k = −1, and obtain our first equation x0 + x1 + x2 + u0 + u1 + u2 = 0. Replacing l41 with l51 , l61 , l71 , and repeating the argument we obtain the remaining three equations relating u0 , u1 , u2 with x0 , x1 , x2 . It remains to find the constants k1 , k2 , k3 . We have found 4 linear equations relating 6 linear functions x0 , x1 , x2 , u0 , u1 , u2 . Since three of them form a basis in the space of linear functions , there must be one relation. We may assume that the first equation is a linear combination of the last three with some coefficients λ1 , λ2 , λ3 . This leads to the system of linear equations from the statement of the theorem. Finally, we have to find the equations of the bitangents. The equations (6.6) show that the lines V (u0 ), V (u1 ), V (u2 ) are bitangents. The equation (6.11) and similar equations l43 l41 k + = −a1 u1 − x1 , α3 α1 a1 l41 l42 k + = −a2 u2 − x2 , α1 α2 a2 after adding up give l41 l42 l43 + + = −k(a0 x0 + a1 x1 + a2 ), α1 α2 α3 l41 u0 = − k(a1 x1 + a2 x2 ), α1 a0 l42 u1 = − k(a0 x0 + a2 x2 ), α1 a1 l43 u2 = − k(a1 x0 + a2 x1 ). α1 a2 After our normalization k = −1, a0 = a1 = a2 = 1, we get three equations of the second type. Similarly, we get the expressions for l5i , l6i , l7i which are the nine equations of the third type. and then

6.2. QUADRATIC DETERMINANT EQUATIONS

155

Let us use the Aronhold set ( 1 , . . . , 7 ) to index bitangents by subsets of [1, 8] of cardinality 2. As we explained at the end of the previous section, the bitangents ξ1 , ξ2 , ξ3 correspond to the subsets (23), (13), (12). The bitangents ξ4k , ξ5k , ξ6k , ξ7k correspond to the subsets (k4), (k5), (k6), (k7), k = 1, 2, 3. What is left are the bitangents corresponding to the subsets (56), (57), (67), (45), (46), (47). The first three look like (23), (13), (12), both of type . The second three look like ξ5k , ξ6k , ξ7k , both W of type WÕÕ. To find the equations of bitangents of type , we interchange the roles of of the lines 1 , 2 , 3 with the lines 5 , 6 , 7 . Our lines will be the new lines analogous to the lines ξ1 = V (u0 ), ξ2 = V (u1 ), ξ3 = V (u2 ). Solving the system we find their equations. To find the equations of the triple of bitangents of type, we delete 4 from the original Aronhold set, and consider the Steiner complex containing the remaining lines as we did in (6.4). The lines making the pairs with 5 , 6 , 7 will be our lines. We find their equations following as we found the equations for ξ5k , ξ6k , ξ7k . Remark 6.1.3. We will see later in Chapter 10 that any seven lines in general linear position can be realized as an Aronhold set for a plane quartic curve. Another way to see it can be found in [248], p. 447).

6.2
6.2.1

Quadratic determinant equations
Hesse-Coble-Roth construction

Let C be a nonsingular plane quartic. Let a ∈ Pic0 (C) \ {0}. Consider the natural bilinear map µ : H 0 (C, OC (KC + a)) × H 0 (C, OC (KC − a)) → H 0 (C, OC (2KC )) defined by the tensor multiplication of the sections. The associated map of complete linear systems ϕ : |KC + a| × |KC − a| → |2KC | = |OP2 (2)| ∼ P5 . = (6.13)

assigns to a pair of divisors D ∈ |KC + a| and D ∈ |KC − a| the divisor D + D ∈ |2KC | cut out by a unique conic which we denote by D, D . If we choose a basis (s1 , s2 ) of H 0 (C, OC (KC + a)) and a basis (s1 , s2 ) of H 0 (C, OC (KC − a)), then the map m is given by (λs1 + µs2 , λ s1 + µ s2 ) → λλ a11 + λµ a12 + λ µa21 + µµ a22 , (6.14)

where a11 , a12 , a21 , a22 ∈ H 0 (C, OC (2KC )) ∼ H 0 (P2 , OP2 (2)) are identified with = homogeneous polynomials of degree 2 in variables t0 , t1 , t2 . Consider the variety W = {(D1 , D2 , x) ∈ |KC + a| × |KC − a| × P2 : x ∈ D1 , D2 }. If we identify |KC + a| and |KC − a| with P1 , we see that W ⊂ P 1 × P1 × P2 (6.15)

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is a hypersurface defined by the multi-homogeneous equation λλ a11 (t) + λµ a12 (t) + λ µa21 (t) + µµ a22 (t) = 0 of multi-degree (1, 1, 2). Consider the projections p1 : W → P1 × P1 , p 2 : W → P2 . (6.17) (6.16)

The fibre of p1 over a point ([λ, µ], [λ , µ ]) is isomorphic (under p2 ) to a conic. It is singular if and only if the discriminant of the conic (6.16) is equal to zero. It is easy to see that this is a bihomogeneous polynomial in the variables (λ, µ), (λ , µ ) of bidegree (3, 3). Thus the locus ∆1 of points [λ, µ], [λ , µ ] such that p−1 is a reducible 1 conic is a curve in P1 × P1 of bi-degree (3, 3). The fibre of p2 over a point x ∈ P2 is isomorphic (under the first projection) to a curve of bi-degree (1, 1) in P1 × P1 . Under the Segre isomorphism between P1 × P1 and a quadric in P3 , such a curve is isomorphic to a conic. This conic is reducible if and only if the equation λλ a11 (t) + λµ a12 (t) + λ µa21 (t) + µµ a22 (t) = 0 is a “cross” on P1 × P1 (i.e. the union of two lines belonging to different rulings). We can rewrite the equation in the form (λ, µ) · a11 (t) a12 (t) λ · a21 (t) a22 (t) µ .

It defines a reducible curve if and only if there exists (λ0 , µ0 ) such that, after plugging in λ = λ0 , µ = bµ0 , any (λ , µ ) will satisfy the equation. The condition for this is of course a (t) a12 (t) det 00 = 0. (6.18) a21 (t) a22 (t) This defines a homogeneous equation of degree 4 in variables t0 , t1 , t2 . It is not identically equal to zero otherwise the entries a11 , a21 must have a common linear factor. The corresponding conics cut out the divisors D1 + D1 , D2 + D1 , where Di = div(si ), Di = div(si ). Their common points form a divisor D1 ∈ |KC + a|. Since a = 0, D1 cannot be cut out by a line. Thus (6.18) defines a quartic curve. It must coincide with our curve C. To see this it is enough to show that each point of C satisfies the equation. Let x ∈ C, then we choose a unique D ∈ |KC + a| containing x and take any D ∈ |KC − a|. We obtain a subset of the conic p−1 (x) isomorphic to 2 P1 . This shows that p−1 (x) is a reducible conic. 2 Conversely, suppose C is given by a determinantal equation as above. For every x ∈ C we have the left and the right kernel of the corresponding matrix. These are one-dimensional vector spaces. The corresponding maps φi : C → P1 , i = 1, 2, are defined by quadratic polynomials (−a21 , a11 ) and (−a12 , a11 ), respectively. Note that the common zeroes of both coordinates belong to the curve C . Thus the linear system defined by the two conics has 4 base points on C and hence φi is given by a linear

6.2. QUADRATIC DETERMINANT EQUATIONS

157

system Vi of degree 4. We may assume that Vi is contained in |KC + di | for some divisor classes di of degree 0. It is easy to see that the base loci of the two linear systems add up to the zeroes of a11 . This immediately implies that d1 + d2 = 0. Thus we have proved: Theorem 6.2.1. An equation of a nonsingular plane quartic C can be written in the form a1 a2 = 0, a3 a4 where ai ’s are homogeneous forms of degree 2. Let X be the set of matrices A with quadratic forms as its entries such that C = V (det A) modulo the equivalence relation defined by A ∼ B if A = CBC for some constant invertible matrices. The set X is bijective to the set Pic0 (C) \ {0}. Remark 6.2.1. The previous theorem agrees with the general theory developed in Chapter 5. To define a quadratic determinant one considers the exact sequence
2 0 → OP2 (−2)2 → OP2 → i∗ (M) → 0.

We have h0 (M) = 2, h0 (M(−1)) = h1 (M) = 0, h1 (M(−1)) = 1.

By Riemann-Roch, deg(M) = 4, hence M = OC (KC + a), for some a ∈ Pic0 (C). Since h0 (M(−1)) = 0, we obtain a = 0. Remark 6.2.2. We assume that the reader is familiar with the theory of 3-folds. The variety W which was introduced in (6.15) is a Fano 3-fold. Its canonical sheaf is equal to ωW ∼ OP1 (−1) OP1 (−1) OP2 (−1). = If we use the Segre embedding P1 × P1 × P2 → P11 , then ωW can be identified with OW (−1). The variety W admits two structures of a conic bundle. They are induced by the projections p1 : W → P1 × P1 and p2 : W → P2 . The degeneration locus of the first map is a curve ∆1 of arithmetic genus 4, and the degeneration locus ∆2 of the second map is the curve C of genus 3. Note that each curve has a double cover defined by considering the irreducible components of the fibres. The double cover over ∆2 splits since each component corresponds to one of the two rulings of P1 × P1 . The ˜ double cover ∆1 → ∆1 does not split. For a general C, the curve ∆1 is nonsingular and the double cover over it is unramified. One shows that the intermediate Jacobian ˜ variety of W is isomorphic to the Prym variety of the cover ∆1 → ∆1 . It is aslo isomorphic to the Prym variety of the trivial cover over ∆2 which is the Jacobian of C. Thus we obtain that the intermediate Jacobian of W is isomorphic to Jac(C). Remark 6.2.3. Let V = H 0 (P1 × P1 × P2 , OP1 (1) OP1 (1) OP2 (2)).

It is a vector space of dimension 24. Let U be an open subset in P(V ) which consists of sections (6.16) such that the corresponding determinant (6.18) defines a nonsingular

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quartic curve. The group G = SL2 × SL2 acts naturally on U and the orbit space is isomorphic to the space Pic0 \ {zero section}. Let W be the 3-fold (6.17) defined 4 by a section from U . The projection W → P1 × P1 defines a curve ∆1 of bi-degree (3, 3) parametrizing singular fibres. It comes with a double cover defined by choosing a component of a reducible fibre. In this way we see that U/G is birationally isomorphic to the space of nonsingular curves of bi-degree (3, 3) on P1 × P1 together with an unramified double cover. If we further act by G = SL3 the orbit space is birationally isomorphic to the universal Jacobian space over M3 and, on the other hand, to the moduli space R4 of curves of genus 4 together with a nontrivial 2-torsion divisor class. It was proven by F. Catanese that the latter space is a rational variety.

6.2.2

Symmetric quadratic determinants

Assume now that a = is a 2-torsion divisor class. Then H 0 (C, OC (KC + )) = H 0 (C, OC (KC − )) and the bilinear map m is symmetric. The determinantal equation of C corresponding to must be given by a symmetric quadratic determinant q11 q12 2 = q11 q22 − q12 . (6.19) q12 q22 Thus we obtain the following. Theorem 6.2.2. An equation of a nonsingular plane quartic can be written in the form a1 a2 a2 = 0, a3

where a1 , a2 , a3 are homogeneous forms of degree 2. Let X be the set of symmetric 2×2 matrices A with quadratic forms as its entries such that C = V (det A) modulo the equivalence relation defined by A ∼ B if A = CBC for some constant nonsingular matrices. The set X is bijective to the set Pic(C)[2] \ {0}. Since ϕ(D1 , D2 ) = ϕ(D2 , D1 ), the map f factors through a linear map ϕ : P2 → |OP2 (2)|. ¯ Here we identify |KC + | with P1 and the symmetric square P1 × P1 /S2 with P2 (a set of k unordered points in P1 is a positive divisor of degree k, i.e. an element of |OP1 (k)| ∼ Pk ). Explicitly, we view P2 as P(S 2 H 0 (C, OC (KC + ))). The corre= sponding linear map S 2 H 0 (C, OC (KC + )) → S 2 H 0 (P2 , OP2 (3)) defines a regular map φ : P1 = |KC + | → |OP2 (2)| which is quadratic. Explicitly, φ((t0 , t1 )) = V (t2 a11 (t) + 2t0 t1 a12 (t) + t2 a22 (t)). 0 1

6.2. QUADRATIC DETERMINANT EQUATIONS

159

Let L( ) be a net of conics equal to the image of the map ϕ. By choosing a basis ¯ (s0 , s1 ) of H 0 (C, KC + ) we may assume that L( ) is spanned by the conics V (a11 ) = 2D1 , V (a12 ) = D1 , D2 , V (a22 ) = 2D2 ,

where D1 = div(s0 ), D2 = div(s1 ). In particular, we see that L( ) has no base points (since C has no pencils of divisors of degree 3). The set B( ) of singular conics in L( ) is a plane cubic isomorphic to a plane section of the discriminant hypersurface D2 (2). Its pre-image under the map P1 ×P1 → P2 is the degeneration curve ∆1 of the conic bundle W → P1 × P1 from (6.17). Remark 6.2.4. We can view φ as the composition of the Veronese map v2 : P1 → P2 and the map ϕ. Let v2 (P1 ) be the Veronese curve. The pre-image of D2 (2) under the ¯ map φ is the locus of zeroes of a binary sextic. It corresponds to the intersection scheme of the cubic B( ) and the conic v2 (P1 ). One can show that the Jacobian variety of the genus 2 curve corresponding to this binary sextic is isomorphic to the Prym variety of the pair (C, ). The curve C can be defined as the locus of point x ∈ P2 such that, viewed as hyperplanes in |OP2 (2)|, the pre-image φ−1 (x) is a degenerate binary quadratic form. We apply Hesse’s determinantal identity for bordered determinants from Lemma 4.1.7 and its interpretation to obtain the identity a11 a21 u0 a12 a22 u1 u0 a11 u1 × a21 0 v0 a12 a22 v1 v0 a11 v1 − a21 0 v0 a12 a22 v1 u0 u1 0
2

= |A|U (u0 , u1 , v0 , v1 ; t0 , t1 , t2 ).

Here [u0 , u1 ], [v0 , v1 ] are coordinates in each copy of |KC + | and U = 0 is the family of conics parametrized by |KC + | × |KC + |. Lemma 6.2.3. The cubic curve B( ) is nonsingular if and only if the linear system |KC + | does not contain a divisor of the form 2a + 2b. Proof. The plane section L( )∩D2 (2) is singular if and only if L( ) contains a singular point of D2 (2) represented by a double line, or if it is tangent to D2 (2) at a nonsingular point. We proved in Chapter 2, section 2.1.2 that the tangent hypersurface of D2 (2) at a nonsingular point represented by a reducible conic Q is equal to the space of conics passing through the singular point q of Q. If L is contained in the tangent hyperplane, then all conics from L( ) pass through q. But we have seen already that L is base point free. This shows that L( ) intersects transversally the nonsingular locus of D2 (2). In particular, B( ) is singular if and only if L( ) contains a double line. Assume that this happens. Then we get two divisors D1 , D2 ∈ |KC + | such that D1 + D2 = 2A, where A = a1 + a2 + a3 + a4 is cut out by a line . Let D1 = p1 + p2 + p3 + p4 , D2 = q1 + q2 + q3 + q4 . Then the equality of divisors (not the divisor classes) p1 + p2 + p3 + p4 + q1 + q2 + q3 + q4 = 2(a1 + a2 + a3 + a4 ) implies that either D1 and D2 share a point x, or D1 = 2p1 + 2p2 , D2 = 2q1 + 2q2 . The first case is impossible, since |KC + − x| is of dimension 0. The second case

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happens if and only if |KC + | contains a divisors D1 = 2a + 2b. The converse is also true. For each such divisor the line a, b defines a residual pair of points c, d such that D2 = 2c + 2d ∈ |KC + | and ϕ(D1 , D2 ) is a double line. Let ˇ I = {(x, ) ∈ B( ) × P2 : ⊂ ϕ(x)}. (6.20) The first projection p1 : I → B( ) is a double cover ramified at singular points of B( ). ˜ The image B( ) of the second projection is locus of lines in P2 which are irreducible components of reducible conics from L( ). It is a plane curve of some degree d in the dual plane or the whole plane. ˇ ˜ Lemma 6.2.4. The curve B( ) ⊂ P2 parametrizing irreducible components of reducible conics from the linear system L( ) is a plane cubic. If B( ) is nonsingular, ˜ then B( ) is also nonsingular and is isomorphic to an unramified double cover of B( ). ˜ Proof. Let us see that d = deg(B( )) = 3. A line in the dual plane is the pencil of lines in the original plane. Thus d is equal to the number of line components of reducible conics in L which pass through a general point q in P2 . Since q is a general point, we may assume that q is not a singular point of any reducible conic from L( ). Then there are d different reducible conics passing through q. We know that L( ) has no base points. Then q must be a base point of a pencil of conics in L( ). Note that a general pencil of conics in L( ) has 4 distinct base points. To see this we consider the regular map P2 → |L( )|∗ defined by the linear system |L( )|. Its degree is equal to 4, hence it general fibre consists of 4 distinct points. It is easy to check that a pencil of conics with 4 distinct base points contains 3 reducible conics. This shows that d = 3. If B( ) is nonsingular, its double cover p1 : I → B( ) ˇ ˜ is unramified, hence I is an elliptic curve. Its image B( ) = p2 (I) in P2 is a plane cubic. Note that two reducible conics f (D1 , D2 ) and f (D3 , D4 ) in B( ) share a common irreducible component if and only if D1 + D2 is cut out by two lines and and D3 + D4 is cut out by two lines and . Let A be the divisor on C cut out by . We know that no two divisors from |KC + | share a common point. Also no divisor is cut out by a line. This easily implies that Di ∩ consists of one point for each i = 1, . . . , 4. Since D1 +D2 ≥ A, D3 +D4 ≥ A, we see that contains at least 2 ramification points of the map C → P1 defined by the linear system |KC + |. Since we have only finitely many such points, we see that there are only finitely many such lines . In particular, ˜ the second projection p2 : I → B( ) is an isomorphism over a dense Zariski subset of ˜ B( ). ˜ If B( ) is nonsingular, then p2 : I → B( ) is a birational map of an elliptic curve to a cubic. Obvioulsy, this cubic must be nonsingular. Theorem 6.2.5. Let S = {( 1 , 1 ), . . . , ( 6 , 6 )} be a Steiner complex of 12 bitangents associated to a 2-torsion divisor class . Then the 12 bitangents, considered as points ˜ in the dual plane, lie on the cubic curve B( ). If we assume that |KC + | does not contain a divisor of the form 2p + 2q, then the cubic curve is nonsingular.

6.2. QUADRATIC DETERMINANT EQUATIONS

161

Proof. Let (ϑi , ϑi ) be a pair of odd theta characteristics corresponding to a pair ( i , i ) of bitangents from S. They define a divisor D = ϑi + ϑi ∈ |KC + | which is cut out ˜ by two lines. Thus f (D, D) ∈ B( ) and the bitangents i , i belong to B( ). The rest of the assertions follow from the previous lemmas. Remark 6.2.5. Let S1 , S2 , S3 be a syzygetic (azygetic) triad of Steiner complexes. ˜ ˜ ˜ They define three cubic curves B( ), B(η), B(η + ) which have 4 (resp. 6) points in common. ˜ Remark 6.2.6. The cubic B( ) has at most ordinary nodes as singularities. We know that the projection p1 : I → B( ) is a double cover unramified outside singular points of B( ) corresponding to double lines. If B( ) is an irreducible cuspidal cubic, the complement of the cusp is isomorphic to C and hence does not admit nontrivial unramified covers. If B( ) is the union of a conic and a line touching it at some point, then, again the complement of the singular point is the disjoint union of two copies of C and hence does not admit an unramified cover. Finally, if B( ) is the union of 3 concurrent lines, then the complement to the singular point is the disjoint union of three copies of C, no unramified covers again. Thus B( ) is nonsingular or a nodal cubic. It is easy to see that its cover I is again nonsingular or a nodal curve of arithmetic genus 1. The ˜ second projection p2 : I → B( ) is an isomorphism over the complement of finitely many points. It is easy to see that the image of a nodal curve is a nodal curve. Remark 6.2.7. Let ∆1 be a curve of bi-degree (3, 3) in P1 × P1 parametrizing singular fibres of the projection p1 : W → P1 ×P1 . Let π : P1 ×P1 → P1 ×P1 /S2 = P2 be the quotient map. The curve ∆1 is equal to π −1 (B( )). The cover π|∆1 : ∆1 → B( ) is a double cover ramified along the points where ∆1 intersects the diagonal. It consists of pairs (D, D), where D ∈ |KC + | such that 2D is cut out by a reducible conic. It is easy to see that this conic must be the union of two bitangents which form one of 6 pairs of bitangents from the Steiner complex associated to . The branch locus of π : P1 × P1 → P2 is a conic C. It can be identified with the Veronese curve v2 (P1 ) which we discussed in Remark 6.2.4. The cubic B( ) intersects it transversally at 6 points. If it is nonsingular, ∆1 is nonsingulat curve of genus 4. If B( ) is an irreducible cubic with a node (by the previous remark it cannot have a cusp), ∆1 is an irreducible curve of arithmetic genus 2 with two nodes. If B( ) is the union of a conic and a line intersecting each other transversally, then ∆1 is the union of a nonsingular elliptic curve of bi-degree (2, 2) and a nonsingular rational curve of bidgegree (1, 1) which intersect each other transversally. If B( ) is the union of three lines, then ∆1 is the union of three nonsingular rational curves of bi-degree (1, 1), each pair intersect transversally. Remark 6.2.8. Let V = H 0 (P1 × P2 , OP1 (2) OP2 (2)). It is a vector space of dimension 18. Let U be an open subset in P(V ) which consists of sections t2 a11 (t) + 2t0 t1 a12 (t) + t2 a22 (t) such that the corresponding determinant 0 1 (6.19) defines a nonsingular quartic curve. The group G = SL2 acts naturally on U via its action on P1 and the orbit space X is a cover of degree 63 of the space |OP2 (4)|ns of nonsingular plane quartics. The fibre over C4 is naturally identified with the set of nonzero 2-torsion divisor classes on C4 . Since X is obviously irreducible and of

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dimension 14, we obtain that X is an irreducible unramified finite cover of degree 63 of |OP2 (4)|ns . Let Z be the closed subset of U of sections such that the linear system of quadrics spanned by a11 (t), a12 (t), a22 (t) contains a double line. Its image in |OP2 (4)|ns is a closed set. Thus a general quartic satisfies the assumption of Lemma 6.2.3 for any . I do not know whether there exists a nonsingular quartic which does not satisfy these assumptions for any . If we further act on X by G = SL3 (C) via its natural action on P2 we obtain the orbit space birationally isomorphic to the space R3 of isomorphism classes of genus 3 curves together with a nontrivial divisor class of order 2. This space is known to be rational [87], [154]. This space is also birationally isomorphic to the space of bielliptic curves of genus 4 (see Exercise 6.11).
2 Let f = q1 q3 − q2 be an expression of f as a symmetric determinant. Consider a quadratic pencil of conics

q(λ, µ) := λ2 q1 + 2λµq2 + µ2 q3 = 0.

(6.21)

Then the condition that Q(λ, µ) is tangent to C is the vanishing of the discriminant 2 D = −q1 q3 + q2 on C. Since it is identically vanishes on C, we see that every conic from the pencil is tangent to our quartic C. Thus we obtain Corollary 6.2.6. A nonsingular plane quartic can be in 63 different ways represented as an evolute of a quadratic pencil of conics. Remark 6.2.9. A quadratic pencil of conics (6.21) can be thought as a subvariety X of P1 × P2 given by a bi-homogeneous equation of bi-degree (2, 2). The projection to P1 is a conic bundle with 6 degenerate fibres corresponding to six pairs of bitangents in the Steiner complex corresponding to the pencil. The projection to P2 is a double cover branched along the quartic C. Later on we will identify X with the Del Pezzo surface of degree 2 associated to a nonsingulart plane quartic.

6.3
6.3.1

Even theta characteristics
Contact cubics

We specialize the results from section 4.1.3. A nonsingular plane quartic has 36 = 22 (23 +1) even theta characteristic. None of them vanishes since C is not hyperelliptic. For any even theta characteristic ϑ the linear system |KC + ϑ| defines a symmetric determinant expression for C. Let P2 = P(E) and V ∗ = H 0 (C, OC (KC + ϑ)). Recall that the symmetric determinantal expression for C corresponding to ϑ defines a linear map E → S 2 V ∗ ⊂ Hom(V, V ∗ ) which, after projectivization, defines a linear map of projective spaces s : P2 → |OP3 (2)|,

6.3. EVEN THETA CHARACTERISTICS

163

where P3 = P(V ) = |KC + ϑ|∗ . The image of s is a net of quadrics N in P3 whose locus of singular quadrics is equal to C. The set of singular points of quadrics from N is a sextic model S of C, the image of C under a map given by the linear system |KC + ϑ|. The pre-image under s of a hyperplane cuts out a divisor D ∈ |KC +ϑ|. The divisor 2D ∈ |3KC | is cut out by a unique cubic. This cubic is called a contact cubic. When we vary D in |KC + ϑ| we get a 3-dimensional variety of contact cubics isomorphic to P3 . Thus we obtain 36 irreducible families of contact cubics. Explicitly, ϑ defines a symmetric determinantal representation C = V (|A|), where A = (lij ) is a 4 × 4 symmetric matrix of linear forms in coordinates [t0 , t1 , t2 ] in P2 . Contact cubics in the algebraic system corresponding to ϑ are parametrized by coordinates [u0 , u1 , u2 , u3 ] in P3 = |KC + ϑ|∗ and are given by the equation l11 l21 l31 l41 u0 l12 l22 l32 l42 u1
13 23 33 43

u2

l14 l24 l34 l44 u3

u0 u1 u2 = 0. u3 0

(6.22)

We can also interpret the determinantal identity from Lemma 4.1.7 which we write in the self-explanatory form A u A v A u − × v 0 v 0 u 0
2

= |A|U.

(6.23)

It shows that two contact cubics cut our on C a set of 12 points that lie on a cubic curve. Remark 6.3.1. Consider the set of nets of quadrics in P3 as the Grassmannian G(3, 10) of 3-dimensional subspaces in H 0 (P3 , OP3 (2)). Let U be an open subset defining nets N of conics such that the locus of singular conics defines a nonsingular plane quartic curve C ⊂ N together with an even theta characteristic ϑ. The group SL(4) acts G(3, 10) via its natural action in P3 . The orbit space is birationally isomorphic to the unramified cover of degree 36 of M3 parametrizing isomorphism classes of pairs (C, ϑ), where C is a nonsingular non-hyperelliptic curve of genus 3 and ϑ is an even theta characteristic. We will show later that this space is birationally isomorphic to M3 .

6.3.2

Cayley octads

The image of s is a net N (i.e. two-dimensional linear system) of quadrics. Take a basis Q1 , Q2 , Q3 of N . The base locus of N is the complete intersection of these quadrics. One expects to get 8 distinct points. Let us see that this is indeed true. Proposition 6.3.1. The set of base points of the net of quadrics N consists of 8 distinct points, no three of which are collinear, no four are coplanar. Proof. Suppose three points are on a line . This includes the case when two points coincide. This implies that is contained in all quadrics from N . Take a point x ∈ .

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For any quadric Q ∈ N , the tangent plane of Q at x contains the line . Thus the tangent planes form a pencil of planes through . Since N is a net, there must be a quadric which is singular at x. Thus each point of is a singular point of some quadric from N . However, the set of singular points of quadrics from N is equal to the nonsingular sextic S. This shows that no three points are collinear. Suppose that 4 points lie in a plane π. Restricting quadrics from N to π defines a linear system of conics through 4 points no three of which are collinear. It is of dimension 1. Thus, there exists a quadric in N which contains π. However, since C is nonsingular all quadrics in N are of corank ≤ 1. Definition 6.1. A set of 8 distinct points in P3 which is a complete intersection of 3 quadrics is called a Cayley octad. From now on we assume that a cayley octad satisfies the properties from Theorem 6.3.1. Let ϕ : C → S ⊂ P3 be the map defined by the linear system |KC + ϑ|. Its image is a sextic model of C given by the right kernel of the matrix defining the determinantal equation. Theorem 6.3.2. Let q1 , . . . , q8 be the Cayley octad defined by the net of quadrics N . Each line pi , qj intersects S at two point ϕ(pi ), ϕ(pj ). The line pi , pj is a bitangent of C. Proof. Fix a point q on ij = qi , qj different from qi , qj . Each quadric from N vanishing at q has 3 common points with . Hence it contains . Since vanishing at a point is one linear condition on the coefficients of a quadric, we obtain a pencil of quadrics in N such that is contained in its set of base points. Two quadrics intersect along a curve of degree 4. Thus the base locus of the pencil is a reducible curve of degree 4 which contains a line component. The residual curve is a twisted cubic. Take a nonsingular quadric in the pencil. Then the cubic is a curve of bi-degree (2, 1) and a line is a curve of bi-degree (0, 1). Thus they intersect at 2 points x, y (not necessary distinct). Any two nonsingular quadrics from the pencil do not intersect at these points transversally. Hence they have a common tangent plane. For each point, an appropriate linear combination of these quadrics will be singular at this point. The pencil does not have any other singular quadrics. Indeed, a singular point of such a quadric must lie on and hence define a singular point of the base locus. So it must be one of the two singular points x, y. No two quadrics from the same pencil share a singular point since the set C of singular quadrics does not contain a line. This shows that x, y ∈ S and the pencil of quadrics is equal to the image of a line in P2 under the map s. The line intersects C at two points pi , pj such that ϕ(pi ) = x, ϕ(pj ) = y. Thus is a bitangent. We can also see all even theta characteristics. Theorem 6.3.3. Let q1 , . . . , q8 be the Cayley octad associated to an even theta characteristic ϑ. Let ϑij be the odd theta characteristic corresponding to the lines qi , qj . Then any even theta characteristic different from ϑ can be represented by the divisor class ϑi,jkl = ϑij + ϑik + ϑil − KC .

6.3. EVEN THETA CHARACTERISTICS for some distinct i, j, k, l.

165

Proof. Suppose that ϑi,jkl is an odd theta characteristic ϑmn . Consider the plane π which contains the points qi , qj , qk . It intersects S at six points corresponding to the theta characteristics ϑij , ϑik , ϑjk . Since the planes cuts out divisors from |KC + ϑ|, we obtain ϑij + ϑik + ϑjk ∼ KC + ϑ This implies that ϑjk + ϑil + ϑmn ∼ KC + ϑ. Hence qj , qk and qi , ql lie in a plane π . The intersection point of the lines qj , qk and qi , ql is a base point of two pencils in N and hence is a base point of N . However, it does not belong to the Cayley octad. This contradiction proves the assertion. Remark 6.3.2. Note that ϑi,jkl = ϑj,ikl = ϑk,ijl = ϑl,ijk . Thus ϑi,jkl depends only on the choice of a subset of four in {1, . . . , 8}. Also it is easy to check that the complementary sets define the same theta characteristic. This shows that we get 35 = 8 /2 different even theta characteristics. Together with ϑ = ϑ∅ we 4 obtain 36 even theta characteristics. Observe now that the notations ϑij for odd thetas and ϑijkl , ϑ∅ agrees with the notation we used for odd even theta characteristics on curves of genus 3. For example, any set ϑ1,8 , . . . , ϑ78 defines an Aronhold set. Or, a syzygetic tetrad corresponds to four chords forming a spatial quadrangle, for example p 1 , p 3 , p 2 , p 4 , p2 , p 3 , p1 , p 4 . Here is another application of Cayley octads. Proposition 6.3.4. There are 1008 azygetic hexads of bitangents of C such that their 12 contact points lie on a cubic. Proof. Let 1 , 2 , 3 be an azygetic triad of bitangents. The corresponding odd theta characteristics add up to KC + ϑ, where ϑ is an even theta characteristic. Let O be the Cayley octad corresponding to the net of quadrics for which C is the Hessian curve and S ⊂ P3 = |KC + ϑ|∗ be the corresponding sextic model of C. We know that the restriction map |OP3 (2) → |OS (2)| = |OC (3KC )| = OP2 (3) is a bijection. We also know that the double planes in |OP3 (2) are mapped to contact cubics corresponding to ϑ. The cubic curve 1 + 2 + 3 is one of them. Using the interpretation of bitangents as chords of the Cayley octad, we see that this cubic corresponds to three chords spanning a plane in P3 . Obviously the chords must be of the form qi , qj , qi , qk , qj , qk , where 1 ≤ i < j < k ≤ 8. The number of such triples is 8 3 = 56. So, incidentally we see that the number of point [u0 , u1 , u2 , u3 ] such that the determinant (6.22) represents the union of three lines is equal to 56. Fixing such a triple of cords, we can find 5 = 10 triples disjoint from the fixed one. The sum of the six 2 corresponding odd theta characteristics is equal to 3K and hence the contact points are on a cubic. We can also see it by using the determinantal identity (6.23). The number of the hexads from the assertion of the proposition is now equal to 36·280 = 1008.

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In Salmon’s book [210] one can find possible types of such hexads. • 280 of type (12, 23, 31, 45, 56, 64); • 168 of type (12, 34, 35, 36, 37, 38); • 560 of type (12, 13, 14, 56, 57, 58). Recall that the three types correspond to three orbits of the permutation group S8 on the set of azygetic hexads whose contact points are on a cubic. Note that not any azygetic hexad has this property. For example, a subset of an Aronhold set does not have this property. For completeness sake let us give the number of not azygetic hexads whose contact points are on a cubic. Each such is the union of two syzygetic hexads with a unique common bitangent. using the classification of syzygetic tetrads octads one can find the number of such pairs. It is equal to 5040. Here is the list. • 140 of type (12, 23, 13, 14, 45, 15); • 1680 of type (12, 23, 34, 45, 56, 16); • 2520 of type (12, 34, 35, 36, 67, 68).

6.3.3

Seven points in the plane

Let p1 , . . . , p7 be seven points in the projective plane. We assume that the points satisfy the following conditions (*) no three of the points are collinear and no six lie on a conic. Consider the linear system L of cubic curves through these points. Proposition 6.3.5. The linear system of cubic curves through p1 , . . . , p7 is a net. The rational map P2 − → P2 = L∗ is of degree 2. It extends to a regular finite map of degree 2 φ : S → P1 , where S is the blow-up of the seven points. The branch curve of φ is a nonsingular plane quartic C. The ramification curve W is the proper transform of a curve of degree 6 with double points at each pi . Conversely, given a nonsingular plane quartic C, the double cover of P2 ramified over C is a nonsingular surface isomorphic to the blow-up of 7 points p1 , . . . , p7 in the plane satisfying (*). We postpone the proof of this proposition until Chapter 8. The surface S is a Del Pezzo surface of degree 2 . Let σ : S → P2 be the blowing-up map. The curves Ei = σ −1 (pi ) are exceptional curves of the first kind, (−1)-curves for short. We will often identify L with its proper transform in S equal to | − KS | = | − σ ∗ (KP2 ) − E1 − . . . − E7 |. The pre-image of a line in L∗ is a nonsingular member of L if and only if intersects transversally C. In this case it is a double cover of branched over ∩ C. The pre-image of a tangent line is a singular member, the singular points lie over the contact points. Thus, the pre-image of a general tangent line is an irreducible cubic curve with a singular point at σ(W ). The pre-image of a bitangent is a member of

6.3. EVEN THETA CHARACTERISTICS

167

| − KS | with singular points (they may coincide if the bitangent is a flex bitangent). It is easy to see that its image in the plane is either an irreducible cubic Fi with a double point at pi or the union of a line pi , pj and the conic Qij passing through the point pk , k = i, j. In this way we can account for all 28 = 7+21 bitangents. If denote the bitangents corresponding to Fi by i8 and the bitangents corresponding to pi , pj + Qij by ij we can even accommodate the notation of bitangents by subsets of cardinality 2 of [1, 8]. We will see below that this notation agrees with the previous notation. In particular, the bitangents corresponding to the curves Fi ’s form an Aronhold set. Note that quartic curve C does not determine the point p1 , . . . , p7 uniquely. There are many ways to define the blowing morphism σ : S → P2 . However, if we fix an Aronhold set of bitangents there is only one way, up to composition with a projective tranformation, to blow-down seven disjoint (−1)-curves such that the corresponding cubic curves Fi are mapped to the bitangents from the Aronhold set. We will see this later in Chapter 8. Thus a choice of an Aronhold set is equivalent to a choice of a set of seven points p1 , . . . , p7 defining C. Consider the universal family X = {(s, F ) ∈ S × L : s ∈ F }. The fibre of the first projection π1 : X → S over a point s ∈ S can be identified, via the second projection, with the pencil L(s) ⊂ L of curves passing through the point s. The second fibration π2 : X → L is an elliptic fibration, its fibres isomorphic to the corresponding members of L. It has 7 rational sections Ei defined by the (−1)-curves Ei ’s. The projection π2 : Ei → L is an isomorphism over L\{Fi }. The fibre over {Fi } is identified, via π1 , with Ei . Thus each section Ei is isomorphic to the ruled surface F1 . The restriction of the projection π1 to Ei is the map Ei → Ei defined by the ruling of F1 . There is another natural rational section E8 defined as follows. It is easy to see 1 that any g2 on a nonsingular cubic curve F is obtained by projection from a point p ∈ F to a line. The point p is the intersection point of F with the line spanned by 1 any divisor from the g2 . It was called by J. Sylvester the coresidual point of F (see 1 [210], p. 134). Take a general curve F ∈ L and restrict L to F . This defines a g2 on ∗ F parametrized by the image of F under the double cover φ : S → L . Let pF be the corresponding coresidual point. If F = Fi , then one can extend the definition of the coresidual point to the case of an irreducible cubic to obtain that pF = pi . If F is the union of line and a conic, then the definition extends too. The coresidual point is a point on the conic. Since two curves in L cannot intersect at three points, counting with multiplicities, given a point p ∈ P2 there exists a unique F ∈ L such that pF = p. Thus the correspondence F → pF allows one to identify the original plane P2 with the plane L. The linear system L defines a rational map of degree 2 from L to the dual plane L∗ . Let E8 ⊂ X be the closure of the set of points (σ −1 (pF ), F ). The projection π2 : E8 → L is the blow-up of the seven points corresponding to the curves Fi . The restriction of the projection π1 to E8 is an isomorphism onto S. Thus E8 defines a rational section of π2 and a regular section of π1 . The section E8 intersects the section −1 Ei along the exceptional curve of Ei identified with π2 ({Fi }) ∩ E8 . The first seven sections are disjoint.

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∗ Proposition 6.3.6. The linear system H = |π1 (σ ∗ OP2 (1)) ⊗ OX (E8 ))| has the base locus equal to E8 ∩ (E1 + · · · E7 ). Let τ : X → X be the blow-up of the base locus. The proper transform of H on X defines a birational morphism α : X → P3 . The proper transform of each divisor Ei , i = 1, . . . , 8 blows down to a point qi . The exceptional divisor of τ is blown down to the union of lines qi , q8 . The rational map π2 ◦ τ ◦ α−1 : P3 → L ∼ P2 is given by the linear system of quadrics through the = points q1 , . . . , q8 .

It follows from this proposition that the set of points q1 , . . . , q8 is a Cayley octad. Conversely, let q1 , . . . , q8 be a Cayley octad O. The net N of quadrics through O defines a rational map P3 − → P2 = N ∗ whose general fibre is a quartic elliptic curves. Projecting from q8 we obtain a net N of cubic curves through the points p1 , . . . , p7 , the projections of p1 , . . . , p7 . We can identify the linear systems L and N . The dual linear system N ∗ parametrizes the quartic curves of arithmetic genus 1 through the Cayley octad O. We can identify it with the plane L. The singular members of N and ˇ L are parametrized by the dual curve C of C. One can match the interpretation of bitangents via singular members of the net of cubic curves L and the net of quartic space curves N . The line qi , qj defines a pencil of quadrics that contains the line. Its base locus is a member of the net N and consists of the union of the line and a rational cubic curve intersecting the line at two points. If (i, j) = (i, 8), projecting from this point we get a member of L equal to the singular cubic curve Fi . Otherwise it is projected to the union of the line pi , pj and a conic Qij . This shows that the notation of bitangents via the Cayley octad, or the Aronhold set, or the set of 7 points agree. Remark 6.3.3. We know, via the determinantla representation of a quartic plane curve that a choice of an even theta characteristic defines the projective equivalence class of a Cayley octad from which C is reconstructed as the Hessian curve of the corresponding net of quadrics. On the other hand, a choice of an Aronhold set defines a choice of the projective equivalence class of a set of 7 points in the plane from which C is reconstructed as the branch curve of the double cover defined by the net of cubic curves through the seven points. Proposition 6.3.6 shows that an Aronhoild set defines a Cayley octad. However, this Cayley octad comes with a choice of one of its points. Conversely, a Cayley octad together with a choice of its points defines an Aronhold set via the projection from this point. This shows that the moduli space Mev admits 3 a rational map of degree 8 onto the moduli space of projective equivalence classes of sets of unordered 7 points in P2 .

6.4
6.4.1

Polar polygons
¨ Clebsch and Luroth quartics

Since 5 general points in P(E ∗ ) lie on a singular quartic (a double conic), a general quartic does not admit a polar 5-polyhedron (polar pentagon) although the count of constant suggests that this is possible. This remarkable fact was first discovered by J. L¨ roth in 1868. Suppose a quartic C admits a polar pentagon {[l1 ], . . . , [l5 ]}. Let u

6.4. POLAR POLYGONS

169

Q = V (q) be a conic in P(E ∗ ) passing through the points [l1 ], . . . , [l5 ]. Then q ∈ AP2 (f ). The space AP2 (f ) = {0} if and only det Cat2 (f ) = 0. Thus the set of quartics admitting a polar pentagon is the locus of the catalecticant invariant on the space P(S 4 E ∗ ). It is a polynomial of degree 6 in the coefficients of a homogeneous form of degree 4. Definition 6.2. A plane quartic admitting a polar pentagon is called a Clebsch quartic. A Clebsch quartic C = V (f ) is called non-degenerate if dim AP2 (f ) = 1. Thus the polar pentagon of a non-degenerate Clebsch quartic lies on a unique conic. We call it the apolar conic. The apolar conic is reducible if and only if the corresponding operator is the product of two linear operators. This means that the second polar Pab (f ) = 0 for some points a, b ∈ P(E). Proposition 6.4.1. Let f ∈ S 4 E ∗ be such that the second polar Dab (f ) = 0 for some a, b ∈ P(E). Then, in appropriate coordinate system f = f3 (t0 , t1 )t0 + f4 (t1 , t2 ), f = f3 (t1 , t2 )t0 + f4 (t1 , t2 ), a = b, a = b.

In particular, Daa (f ) = 0 if and only if V (f ) has a triple point. Proof. Suppose a = b. Choose coordinates such that a = [1, 0, 0], b = [0, 0, 1] and write
4

f=
i=0

fi (t1 , t2 )t4−i . 0
2

Then Daa (f ) =

∂ , ∂t2 0

2

Dab (f ) =

∂ ∂t2 ∂t0 (f )

= 0. Now the assertions easily follow.

We will assume that the apolar conic of a non-degenerate Clebsch quartic is irreducible. 4 4 Let {[l1 ], . . . , [l5 ]} be a polar pentagon of f such that f = l1 + · · · + l5 . For any 1 ≤ i < j ≤ 5, let aij = V (li ) ∩ V (lj ) ∈ P(E). We can identify aij with a linear operator ψij ∈ E (defined up to a constant factor). Obviously, Dψij (f ) coincides with the first polar Daij (f ). Applying ψij we obtain
4 4 Dψij (f ) = Dψij (l1 + · · · + l5 ) = 4 k=i,j 3 ψij (lk )lk .

Thus [lk ], k = i, j, form a polar triangle of Paij (f ). Since the associated conic is irreducible no three points among the [lij ]’s are linearly dependent. Thus Paij (V (f )) is a Fermat cubic. Lemma 6.4.2. Let f ∈ S 4 E ∗ . Assume that Dab (f ) = 0 for any a, b ∈ P(E). Let S be the locus of points a ∈ P(E) such that the first polar of V (f ) is isomorphic to a Fermat cubic or belongs to the closure of its orbit. Then S is a plane quartic.

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Proof. Let I4 : S 3 E ∗ → C be the Aronhold invariant vanishing on the locus of Fermat cubics (see Remark 3.2.4). It is a polynomial of degree 4 in coefficients of a cubic. Compose I4 with the polarization map E × S 4 E ∗ → S 3 E ∗ , (ψ, f ) → Dψ (f ). We get a bihomogeneous map of degree (4, 4) E × S 4 E ∗ → C. It defines a degree 4 homogeneous map S : S4E∗ → S4E∗ (6.24) This map is called the Clebsch quartic covariant. It assigns to a quartic form in 3 variables another quartic form in 3-variables. By construction, this map does not depend on the choice of coordinates. Thus it is a covariant of quartics, i.e. a GL(E)equivariant map from S 4 E ∗ to some S d E ∗ . By definition, the locus of a ∈ E such that S(f )(v) = 0 is the set of vectors v ∈ E such that I4 (Pa (f )) = 0, i.e., V (Pa (f )) belongs to the closure of the orbit of a Fermat cubic. Example 6.4.1. Assume that the equation of f is given in the form f = at4 + bt4 + ct4 + 6f t2 t2 + 6gt2 t2 + 6ht2 t2 . 0 1 2 1 2 0 1 0 1 Then the explicit formula for the Clebsch covariant gives S(f ) = a t4 + b t4 + c t4 + 6f t2 t2 + 6g t2 t2 + 6h t2 t2 , 0 1 2 1 2 0 1 0 1 where a b c f g h = = = = = = 6g 2 h2 6h2 f 2 6f 2 g 2 bcgh − f (bg 2 + ch2 ) − ghf 2 acf h − g(ch2 + af 2 ) − f hg 2 abf g − h(af 2 + bg 2 ) − f gh2

For a general f the formula for S is too long. Note that the Clebsch covariant S defines a rational map S : P(S 4 E ∗ )− → P(S 4 E ∗ ) (6.25)

Note that the map is not defined on the closed subset of quartics V (f ) such that V (Pa (f )) belongs to the closure of the orbit of a Fermat cubic for any a ∈ P(E). Proposition 6.4.3. The map S is not defined on V (f ) if and only if V (f ) is a Clebsch quartic admitting a reducible apolar conic. We refer for a proof to [82]. For any quartic curve C satisfying the assumption of the previous proposition, the curve S(C) := V (S(f )) will be called the Clebsch covariant quartic associated to C. We will show that for a general quartic C the Clebsch quartic S(C) comes with a certain non-effective theta characteristic ϑ such that the pair (S(C), ϑ) is Scorza general and the Scorza quartic associated to (C, ϑ) is equal to C.

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171

If C is a non-degenerate Clebsch quartic, then, as we explained in above, the vertices of its polar pentagon must belong to the Clebsch covariant quartic S(C). This gives Proposition 6.4.4. Let C = V (f ) be a nondegenerate Clebsch quartic. Then each polar pentagon of C is inscribed in the quartic curve V (S(f )). Lemma 6.4.5. A quartic curve C circumscribing a pentagon defined by 5 lines V (li ) can be written in the form C = V (g), where g = l1 · · · l5
i=1

ai . li

Proof. Consider the linear system of quartics passing through 10 vertices of a pentagon. The expected dimension of this linear system is equal to 4. Suppose it is larger than 4. Since each side of the pentagon contains 4 vertices, requiring that a quartic vanishes at some additional point on the side forces the quartic contain the side. Since we have 5 sides, we will be able to find a quartic containing the union of 5 lines, obviously a contradiction. Now consider the linear system of quartics whose equation can be wriitten as in the assertion of the lemma. The equations have 5 parameters and it is easy to see that the polynomials l1 · · · l5 /li , i = 1, . . . , 5, are linearly independent. Definition 6.3. A plane quartic circumscribing a pentagon is called a L¨ roth quartic. u Thus we see that for any Clebsch quartic C the quartic S(C) is a L¨ roth quartic. u One can prove that any L¨ roth quartic is obtained in this way from a unique Clebsch u quartic (see [82]). Since the locus of Clebsch quartics is a hypersurface (of degree 6) in the space of all quartics, the locus of L¨ roth quartics is also a hypersurface. Its u degree is equal to 54 ([174]) and the number is equal to one of the coefficients of the Donaldson polynomial for the projective plane (see [165]). Let C = V (f ) be a general Clebsch quartic. Consider the map c : VSP(f ; 5)o → P(S 2 E) (6.26)

defined by assigning to { 1 , . . . , 5 } ∈ V SP (f ; 5)o the unique conic passing through these points in the dual plane. This conic is apolar to C. The fibres of this map are polar pentagons of f inscribed in the apolar conic. We know that the closure of the set of Clebsch quartics is defined by one polynomial in coefficients of quartic, the catalecticant invariant. Thus the varierty of Clebsch quartics is of dimension 13. Consider 4 4 the map (E ∗ )5 → P(S 4 E ∗ ) defined by (l1 , . . . , l5 ) → V (l1 + · · · + l4 ). The image of this map is the variety of Clebsch quartics. A general fibre must be of dimension 15 − 13 = 2. However, scaling the li by the same factor, defines the same quartic. Thus the dimesnion of the space of all polar pentagons of a general Clebsch quartic is equal to 1. Over an open subset of the Clebsch locus, the fibres of c are irreducible one-dimensional varietes. Proposition 6.4.6. Let C = V (f ) be a nondegenerate Clebsch quartic and Q be its apolar conic. Consider any polar pentagon of C as a set of 5 points on Q (the dual of its sides). Then VSP(f ; 5)o is an open non-empty subset of a linear pencil on Q of degree 5.

172 Proof. Consider the correspondence X = {(x, { 1 , . . . ,
5 })

CHAPTER 6. PLANE QUARTICS

∈ Q × VSP(f ; 5)o : x = [li ] for some i = 1, . . . , 5}.

Let us look at the fibres of the projection to Q. Suppose we have two polar pentagons of f with the same side [l]. We can write
4 2 f − l 4 = l1 + · · · + l4 ,

f − λl4 = m4 + · · · + m4 . 1 4 For any ψ ∈ S 2 E such that ψ([li ]) = 0, i = 1, . . . , 4, we get Dψ (f ) = 12ψ(l)l2 . Similarly, for any ψ ∈ S 2 E such that ψ ([mi ]) = 0, i = 1, . . . , 4, we get Dψ (f ) = 12λψ (l)l2 . This implies that ψ(l)ψ − ψ (l)ψ is an apolar conic to C. Since C was a general Clebsch quartic, there is only one apolar conic. The set of V (ψ)’s is a pencil with base points V (li ), the set of V (ψ ) is a pencil with base points V (li ). This gives a contradiction unless the two pencils coincide. But then their base points coincide and the two pentagons are equal. This shows that the projection to Q is a one-to-one map. In particular, X is an irreducible curve. Now it is easy to finish the proof. The set of degree 5 positive divisors on Q ∼ P1 = is the projective space |OP1 (5)|. The closure P of our curve of polar pentagons lies in this space. All divisors containing one fixed point in their support form a hyperplane. Thus the polar pentagons containing one common side [l] correspond to a hyperplane section of P. Since we know that there is only one such pentagon and we take [l] in an open Zariski subset of Q, we see that the curve is of degree 1, i.e. a line. So our curve is contained in one-dimensional linear system of divisors of degree 5.

6.4.2

The Scorza quartic

Here we assume that C = V (f ) is not projectively equivalent to the quartics from Proposition 6.4.1. Let us study the map (6.25) in more detail. Let S = S(C). For any a ∈ S, the first polar Pa (f ) defines a Fermat curve (or its degeneration). As we saw in the proof of Lemma 3.2.7, these curves are characterized by the property that there exists a point b such that the first polar is a double line. This defines a correspondence RC = {(a, b) ∈ S × S : Pb (Pb (C)) is a double line}. Proposition 6.4.7. Let C = V (f ) be a general plane quartic. Then S = S(C) is a nonsingular curve and there exists a non-effective theta characteristic ϑ with d(ϑ) = 1 on S such that RC coincides with the Scorza correspondence Rϑ on S. Proof. Take linear forms li in general position and consider the Clebsch quartic C = 4 V ( li ). Then S(C) is the L¨ roth quartic given by the equation from Lemma 6.4.5. u It is checked directly that it is nonsingular. Thus the image of the Scorza map contains a nonsingular curve, and hence for general C the curve S(C) is nonsingular. The variety of nonsingular L¨ roth quartics is an open subset in a hypersurface in the space u of quartics. The image of this open subset in the moduli space M3 of curves of genus 3 is of codimension 1. Taking C general enough we may assume that S(C) does not

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173

admit a non-constant map to curves of genus 1 or 2. The moduli space of curves that admit such maps is of higher codimension in M3 . Thus we may assume that the image of the Scorza map contains an open subset U of nonsingular curves that do not admit non-constant maps to curves of genus 1 or 2. Assume that S(C) ∈ U . Applying Proposition 5.5.5, it suffices to check that RC is symmetric, of type (3, 3) and has valence −1. The symmetry of RC is obvious. We have a map from S to the closure F of the Fermat locus defined by a → V (Pa (f )). For any curve in F, except the union of three lines, the set of points such that the first polar is a double line is finite. It is equal to the set of double points of the Hessian curve and consists of 3 points for Fermat curves, one point for cuspidal cubics and 2 points for the unions of a conic and a line. If C is general enough the image of S in F does not intersect the locus of the unions of three lines (which is of codimension 2). Thus we see that each projection from RC to S is a finite map of degree 3. Thus RC is of type (3, 3). For any general point x ∈ S the first polar Px (C) is projectively equivalent to a Fermat cubic. The divisor RC (x) consists of the three vertices of its unique polar triangle. For any y ∈ RC (x) the side H = V (l) opposite to y is defined by Py (Px (C)) = Px (Py (C)) = V (l2 ). It is a common side of the polar triangles of Px (C) and Py (C). We have H ∩ S = y1 + y2 + x1 + x2 , where RC (x) = {y, y1 , y2 } and RC (y) = {x, x1 , x2 }. This gives y1 + y2 + x1 + x2 = (RC (x) − x) + (RC (y) − y) ∈ |KS |. Consider the map f : S → Pic2 (S) given by x → [R(x) − x]. Assume f is not constant. If we replace in the previous formula y with y1 or y2 , we obtain that f (y) = ¯ f (y1 ) = f (y2 ) = KS − f (x). Thus f is of degree ≥ 3 on its image S and factors to a finite map to the normalization W of the image. Since a rational curve does not admit non-constant maps to an abelian variety, we obtain that W is of positive genus. By Hurwitz formula, the genus of W is less or equal than 2. By assumption, S does not admit a non-constant map to W . Hence RC has valence v = −1. Using the CayleyBrill formula, we obtain that RC has no united points. In particular, Paa is never a double line. Thus RC satisfies all the assumption of Proposition 5.5.5. it remains to verify that d(ϑ) = 1. We may assume that C is non-degenerate in the sense of section 1.4.1. This means that the polarity map ψ → Pψ (C) is bijective. It follows from the definition of the correspondence RC = Rϑ that the curve Γ(ϑ) is the locus of lines l = V (L) such that Px,y (C) is the double line V (L2 ) for some (x, y) ∈ S(C). This implies that the map Rϑ → Γ(ϑ) is of degree 2, hence d(ϑ) = 1.

Example 6.4.2. Let C = V (f ) be a non-degenerate Clebsch quartic and S = S(C) be the L¨ roth quartic. It follows from Proposition 6.4.4 that each polar pentagon is u inscribed in S. If we take two vertices x, y of a pentagon, then Px,y (C) is a double line representing one of the sides of the pentagon. This means that the apolar conic of C is the curve Γ(ϑ), hence d(ϑ) = 3. The theta characteristic ϑ on a L¨ roth quartic u obtained in this way is called the pentagonal theta characteristic. We do not know any other example of a theta characteristic with d(ϑ) = 1.

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Recall from Proposition 6.4.6 that the polar pentagons of C are parametrized by P1 . Two pentagons cannot have a common vertex x since the three sides not containing x are equal to the irreducible components of the Hessian of Px (C) and other two sides are reconstructed from the vertices of the triangle formed by the Hessian. Assigning to x ∈ S the unique polar pentagon with vertex x we obtain a regular map ϕ : S → P1 of degree 10. Proposition 6.4.8. Let C be a general plane quartic such that the associated pair (S(C), ϑ) is Scorza general. Then the Scorza quartic associated to the pair (S(C), ϑ) is equal to C. Proof. Let C be the Scorza quartic associated to (S(C), ϑ). It follows immediately from (6.15) in the proof of Theorem 5.5.9 that for any point (x, y) ∈ Rϑ the second polar Px,y (C ) is a double line. This shows Px (C ) is a Fermat cubic, and hence S(C) = S(C ). Let Qev be the variety parametrizing Scorza general pairs (C, ϑ). Assigning to a pair the Scorza quartic curve, we define a map Qev → |OP2 (4)|. The Scorza map S defines a rational map S : |OP2 (4)| → T C ev . By Proposition 5.2.2 4 the variety T C ev is irreducible. As we have just showed this map admits a rational 4 section. Since both varieties are irreducible varieties of the same dimension, the section is dominant and injective, hence birational. This shows that the Scorza map is injective on an open Zariski subset and the assertion is proved. Passing to the rational quotients by PGL(3), we obtain Corollary 6.4.9. Let Mev be the moduli space of curves of genus g together with an 3 even theta characteristic. The birational map S : |OP2 (4)| → Qev has the inverse defined by assigning to a pair (C, ϑ) the Scorza quartic. It induces a birational isomorphism M3 ∼ Mev . = 3 The composition of this map with the forgetting map Mev → M3 is a rational self-map 3 of M3 of degree 36. Remark 6.4.1. The corollary generalizes to genus 3 the fact that the map from the space of plane cubics |OP2 (3)| to itself defined by the Hessian is a birational map to the cover |OP2 (3)|ev , formed by pairs (X, ), where is a non-trivial 2-torsion point (an even characteristic in this case). Note that the Hessian covariant is defined similarly to the Clebsch invariant. We compose the polarization map V × S 3 E ∗ → S 2 E ∗ with the discriminant invariant S 2 E ∗ → C.

6.4.3

Polar hexagons

A general quartic admits a polar 6-polyhedron (polar hexagon). It follows from Proposition 1.3.12 that the variety VSP(f ; 6) is a smooth irreducible 3-fold. Let us see how to construct polar hexagons of f explicitly. Let Z = {[l1 ], . . . , [l6 ]} ∈ VSP(f ; 6)o , and 4 4 f = l1 + · · · + l6 ,

6.4. POLAR POLYGONS where i.e.,
i

175

= V (li ). We know that each pair li , lj , i = j, is conjugate with respect to f ,
2 2 ˇ 22 Ωf (li , lj ) = f (li lj ) = 0. ˇ

Let ψi and ψj ∈ S 2 E be the anti-polars of li and lj with respect to f , i.e.
2 Dψi (f ) = li , 2 Dψj (f ) = lj .

It follows from (1.45) that ψi ([lj ]) = ψj ([li ]) = 0. Assume that ψi is irreducible. Then the map S 2 E × S 2 E → S 4 E, (α, β) → αψi + βψj

has one-dimensional kernel spanned by (β, −α). This easily implies that the dimension of the linear space L of quartic forms αψi + βψj is equal to 9. Thus L coincides with IZ (4). Note that any form from L vanishes on li , lj and common zeroes of A and B. This shows that Z = { i , j } ∪ V (A) ∩ V (B) is a polar hexagon of f . By Proposition 1.4.7 it must coincide with Z. This shows that the points k , k = i, j are reconstructed from the points i , j . It also suggests the following construction of polar hexagons of f. Start with any [l] ∈ P(E ∗ ) such that its anti-polar ψ is irreducible and does not vanish at l. This is an open condition on [l]. Note that the latter condition means that ˇ [l] does not belong to the quartic V (f ) ⊂ P(E ∗ ). Let [l ] ∈ V (ψ) and let ψ be the anti-polar of l . For any α, α ∈ S 2 E with α([l]) = α ([l ]) = 0, we have Dαψ+βψ (f ) = Dα (Dψ (f )) + Dα (Dψ (f )) = Dα (l2 ) + Dα (l 2 ) = 0. This shows that the linear space L of quartic forms αψ + βψ is contained in AP4 (f ). As before we compute its dimension to find that it is equal to 9. Thus L coincide with IZ (4), where Z = {[l], [l ]}∪(V (α)∩V (α )). By Proposition 1.3.4, Z is a generalized polar hexagon of f (an ordinary one if V (α) intersects V (α ) transversally). Note that this confirms the dimension of VSP(f ; 6). We can choose [l] in ∞2 ways, and then choose [l ] in ∞1 ways. Remark 6.4.2. Consider the variety VSP(F ; 6) = {([l], Z) ∈ P(E ∗ ) × VSP(f ; 6) : {[l]} ⊂ Z}. The projection to the second factor is a degree 6 map. The general fibre over a point [l] is isomorphic to the anti-polar conic V (ψ) of [l].

6.4.4

A Fano model of VSP(f ; 6)

Recall that each Z ∈ VSP(f ; 6) defines a subspace IZ (3) ⊂ AP3 (f ). Its dual space IZ (3)⊥ ⊂ W = S 3 E ∗ /ap1 (E) is an isotropic subspace with respect to Mukai’s 2f forms.

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Lemma 6.4.10. Let f be a non-degenerate quartic form and Z ∈ VSP(f ; 6). Then dim IZ (3) = 4. Proof. Counting constants based on the exact sequence (1.35) shows that dim IZ (3) ≥ 10 − 6 = 4. Assume dim IZ (3) > 4. Let Z1 be a closed subscheme of Z of length 5. Again counting constant we get IZ1 (2) = {0}. Let C be a conic from the linear system |IZ1 (2)|. Obviously C ∈ IZ (2) since otherwise AP2 (f ) = {0} contradicting the nondegeneracy assumption on f . Choose a 0-dimensional scheme Z0 of length 2 such that each irreducible component of C contains a subscheme of Z = Z0 ∪ Z1 of length ≥ 4. It is always possible. Now, counting constants, we see that dim IZ (3) > 4 − 2 = 2. By B´ zout’s Theorem, all cubics K from |IZ (3)| are of the form C + , where is e a line. Thus the residual lines form a linear system of lines of dimension ≥ 2, hence each line is realized as a component of some cubic K. However, |IZ (3)| ⊂ |IZ (3) and therefore all lines pass through the point in Z \ Z1 . This is obviously impossible. Applying the previous lemma we obtain a well-defined map µ : VSP(f ; 6) → G(3, W ) ∼ G(3, 7), = Z → IZ (3)⊥ .

By Proposition 1.4.7, this map is injective. Its image is contained in the locus of subspaces which are isotropic with respect to Mukai’s forms. Theorem 6.4.11. (S. Mukai) Let f ∈ S 4 E ∗ be a general quartic form in 3 variables. Then the map µ : VSP(f ; 6) → G(3, 7) is an isomorphism onto a smooth subvariety X equal to the locus of common zeroes of a 3-dimensional space of sections of the vector bundle Λ2 S, where S is the tautological vector bundle over the Grassmannian. The canonical class of X is equal to −H, where H is a hyperplane section of X in the Pl¨ cker embedding. u Proof. We refer to the proof to the original paper of Mukai [178], or to [79], where some details of Mukai’s proof are provided. Recall that a Fano variety of dimension n is a smooth projective variety X with ample −KX . If Pic(X) ∼ Z and −KX = mH, where H is an ample generator of the = Picard group, then X is said to be of index m. The degree of X is the self-intersection 1 number H n . The number g = 2 H n + 1 is called the genus. Remark 6.4.3. The variety X2 was omitted in the original classification of Fano varieties with the Picard number 1 due to Gino Fano. It was discovered by V. Iskovskikh. It has the same Betti numbers as the P3 . It was proven by Mukai that every such variety arises as a smooth projective model of W (f ; 6) for a unique quartic for V (f ). Remark 6.4.4. Another approach to Mukai’s description of VSP(f ; 6) for a general plane quartic V (f ) is due to K. Ranestad and F.-O. Schreyer[195]. It allows them also to extend Theorem 6.4.11 to other 2 cases where n = 2 and wrk(f ) = 2+k k (k = 3, 4). VSP(f ; 10) ⊂ G(4, 9) is a K3 surface of degree 38 in P20 , k = 3

6.5. AUTOMORPHISMS OF PLANE QUARTIC CURVES VSP(f ; 15)o ⊂ G(5, 11) is a set of 16 points, k = 4.

177

Although these descriptions were certainly known to Mukai, he did not have a chance to provide the details of his proofs. The approach of Ranestad and Schreyer is based on a well-known result of D. Bucksbaum and D. Eisenbud that any Gorenstein Artinian quotient R/I of the polynomial algebra in 3 variables admits a resolution
s s

0 → R(−d − 3) →
i=1

R(−ei ) →
i=1

R(−di ) → R → R/I → 0,

where the map φ is given by an alternating matrix whose entries aij are homogeneous polynomials of degree ei − dj and d is the degree of the socle of R/I. It follows easily that (s−1)(d+3) = 2(d1 +· · ·+ds ). In our case when R/I = Af , the ideal I = AP(f ) is generated by forms of degree k + 1. the dimension of this space is 2k + 3. This gives s = 2k + 3 and d1 = . . . = ds = k + 1. From this we obtain e1 = . . . = e2k+3 = k. This implies that the map φ is given by an alternating 2k + 3 × 2k + 3 matrix whose s entries are linear forms. The image of i=1 R(−k − 1)k+1 = C2k+3 in R is equal to APk+1 (f ) and can be identified with the dual of our space W . The map φ can be identified with a linear map V → Λ2 W ∗ . Its image corresponds to the subspace generated by our forms σij . Then the authors show that any apolar scheme of lenghth Nk defines a subspace of W of dimension k + 1 in which the image of V in Λ2 W ∗ vanishes. Remark 6.4.5. We refer to [172] for the beautiful geometry of the variety VSP(f ; 6), where V (f ) is the Klein quartic.

6.5
6.5.1

Automorphisms of plane quartic curves
Automorphisms of finite order

Since an automorphism of a nonsingular plane quartic curve C leaves KC invariant, it is defined by a projective transformation. Lemma 6.5.1. Let σ be an automorphism of order n > 1 of a nonsingular plane quartic C = V (f ). Then one can choose coordinates in such a way that a generator a b of the cyclic group (σ) is represented by a diagonal matrix diag[1, ζn , ζn ], where ζn is a primitive nth root of unity, and f is given in the following list. (i) (n = 2), (a, b) = (0, 1), t4 + t2 g2 (t0 , t1 ) + g4 (t0 , t1 ). 2 2 (ii) (n = 3), (a, b) = (0, 1), t3 g1 (t0 , t1 ) + g4 (t0 , t1 ). 2 (iii) (n = 3), (a, b) = (1, 2), f = t4 + αt2 t1 t2 + t0 t3 + t0 t3 + βt2 t2 . 0 0 1 2 1 2

178 (iv) (n = 4), (a, b) = (0, 1),

CHAPTER 6. PLANE QUARTICS

t4 + g4 (t0 , t1 ). 2 (v) (n = 4), (a, b) = (1, 2), t4 + t4 + t4 + αt2 t2 + βt0 t2 t2 . 0 1 2 0 2 1 (vi) (n = 6), (a, b) = (3, 2), t4 + t4 + αt2 t2 + t0 t3 . 0 1 0 1 2 (vii) (n = 7), (a, b) = (3, 1), t 3 t 1 + t3 t2 + t 0 t 3 . 0 1 2 (viii) (n = 8), (a, b) = (3, 7), t4 + t 3 t 2 + t1 t3 . 0 1 2 (ix) (n = 9), (a, b) = (3, 2), t4 + t 0 t 3 + t3 t1 . 0 1 2 (x) (n = 12), (a, b) = (3, 4), f = t 4 + t 4 + t0 t3 . 0 1 2 Here the subscripts in polynomials gi indicate their degree. Proof. Let us first choose coordinates such that σ acts by the formula σ : [x0 , x1 , x2 ] → b a [x0 , ζn x1 , ζn x2 ]. We will frequently use that f is of degree ≥ 3 in each variable. This follows from the assumption that f is nonsingular. Case 1: ab = 0, say a = 0. Write f as a polynomial in t2 . f = αt4 + t3 g1 (t0 , t1 ) + t2 g2 (t0 , t1 ) + t2 g3 (t0 , t1 ) + g4 (t0 , t1 ). 2 2 2 (6.27)

If α = 0, we must have 4b = 0 mod n. This implies that n = 2 or 4. In the first case g1 = g3 = 0, and we get case (i). If n = 4, we must have g1 = g2 = g3 = 0, and we get case (iv). If α = 0, then 3b = 0 mod n. This implies that n = 3 and g2 = g3 = 0. This gives case (ii). Case 2: ab = 0. Note the case when a = b = 0 is reduced to Case 1 by scaling the matrix of the transformation and permuting the variables. In particular, n > 2. Let p1 = [1, 0, 0], p2 = [0, 1, 0], p3 = [0, 0, 1] be the reference points. Case 2a: All reference points lie on C. This implies that the degree of f in each variable is equal to 3. We can write f in the form f = t3 a1 (t1 , t2 ) + t3 b1 (t0 , t2 ) + t3 c1 (t0 , t1 )+ 0 1 2

6.5. AUTOMORPHISMS OF PLANE QUARTIC CURVES t2 a2 (t1 , t2 ) + t2 b2 (t0 , t2 ) + t2 c2 (t0 , t1 ). 0 1 2

179

Since f is invariant, it is clear that any ti cannot enter in two different coefficients a1 , b1 , c1 . Without loss of generality, we may assume that f = t3 t1 + t3 t2 + t3 t0 + t2 a2 (t2 , t3 ) + t2 b2 (t0 , t2 ) + t2 c2 (t0 , t1 ). 0 1 2 0 1 2 Now we have a = 3a + b = 3b mod n. This easily implies n = 7 and we can take a generator of (g) such that (a, b) = (3, 1). By checking the eigenvalues of other monomials, we verify that no other monomials enters in f . This is case (vii). Case 2b: Only two reference points lie on the curve. By normalizing the matrix and permuting the coordinates we may assume that p1 = [1, 0, 0] does not lie on C. Then we can write f = t4 + t2 g2 (t1 , t2 ) + t0 g3 (t1 , t2 ) + g4 (t1 , t2 ), 0 0 where t4 , t4 do not enter in g4 . 1 2 Without loss of generality we may assume that t3 t2 enters in g4 . This gives 3a+b = 1 0 mod n. Suppose t1 t3 enters in g4 . Then a + 3b = 0 mod n. This gives n = 8, and 2 we may take a generator of σ corresponding to (a, b) = (3, 7). This is case (viii). If t1 t3 does not enter in g4 , then t3 enters in g3 . This gives 3b = 0 mod n. Together 2 2 with 3a + b = 0 mod n this gives n = 3 and we take g with (a, b) = (1, 2) or n = 9 and (a, b) = (3, 2). These are cases (iii) and (ix). Case 2c: Only one reference point lies on the curve. By normalizing the matrix and permuting the coordinates we may assume that p1 = [1, 0, 0], p2 = [0, 1, 0] do not lie on C. Then we can write f = t4 + t4 + t2 g2 (t1 , t2 ) + t0 g3 (t1 , t2 ) + g4 (t1 , t2 ), 0 1 0 where t4 , t4 do not enter in g4 . This immediately gives 4a = 0 mod n. Suppose t3 1 2 2 enters in g3 . Then 3b = 0 mod n, hence n = 6 or n = 12. It is easy to see that f = t3 + t3 + αt2 t2 + t0 t3 . 0 1 0 1 2 If n = 6, then (a, b) = (3, 2) and α may be different from 0. This is case (vi). If n = 12, then (a, b) = (3, 4) and α = 0. This is case (x). Case 2d: None of the reference point lies on the curve. In this case f = t4 + t4 + t4 + t2 g2 (t1 , t2 ) + t0 L3 (t1 , t2 ) + αt3 t2 + βt1 t3 . 0 1 2 0 1 2 Obviously, 4a = 4b = 0 mod n. This gives case (i) or n = 4 and (a, b) = (1, 2), (1, 3). These cases are isomorphic and give (v).

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6.5.2

Automorphism groups

Recall some standard terminology from the theory of linear groups. Let G be a subgroup of the general linear group GL(V ) of a complex vector space of dimension n. The group G is called intransitive if the representation of G in GL(V ) is reducible. Otherwise it is called transitive. The group G is called imprimitive if G contains an intransitive normal subgroup G . In this case V decomposes into a direct sum of G invariant proper subspaces, and elements from G permute them. We employ the notation from [55]:a cyclic group of order n is denoted by n, the semi-direct product A B is denoted by A : B, a central extension of a group A with kernel B is denoted by B.A. Theorem 6.5.2. The following is the list of all possible groups of automorphisms of a nonsingular plane quartic.
Type Order 168 96 48 24 16 9 8 6 6 4 3 2 Structure L2 (7) 42 : S3 4.A4 S4 4.22 9 D8 6 S3 22 3 2 Equation t3 t1 + t3 t2 + t3 t0 0 1 2 t4 + t4 + t4 1 2 √ 0 4 + t4 ± 2 −3t2 t2 + t4 t2 1 0 1 0 t4 + t4 + t4 + a(t2 t2 + t2 t2 + t2 t2 ) 0 1 2 0 1 0 2 1 2 t4 + t4 + at2 t2 + t4 2 1 0 1 0 t4 + t0 t3 + t1 t3 0 1 2 t4 + t4 + t4 + at2 t2 + bt2 t0 t1 2 0 1 0 1 2 t4 + at2 t2 + t4 + t1 t3 0 0 1 1 2 4 + t (t3 + t3 ) + at2 t t + bt2 t2 t0 0 1 2 0 1 2 1 2 t4 + t4 + t4 + at2 t2 + bt2 t2 + ct2 t2 2 0 1 2 0 0 1 0 1 t3 L1 (t0 , t1 ) + g4 (t0 , t1 ) 2 t4 + t2 g2 (t0 , t1 ) + t4 + at2 t2 + t4 2 2 0 0 1 1 Parameters

I II III IV V VI VII VIII IX X XI XII

√ a = 0, ±2 −3, ±6 a, b = 0 a=0

a=

√ −1± −7 2

b=0
(a − b)(b − c)(a − c) = 0

Table 6.1: Automorphisms of plane quartics Proof. Case 1. Let G be an intransitive group realized as a group of automorphisms of a nonsingular plane quartic. Since in our case n = 3, V must be the direct sum of one-dimensional subspaces Vi , or a one-dimensional subspace V1 and a 2-dimensional subspace V2 . Case 1a. V = V1 ⊕ V2 ⊕ V3 . Choose coordinates (t0 , t1 , t2 ) such that V1 is spanned by (1, 0, 0) and so on. Let σ ∈ G be an element of order n. Assume n > 4, i.e. n = 6, 7, 8, 9 or 12. It is clear that two elements of different orders > 4 cannot belong to G since otherwise G contains an element of order > 12. If n = 8, the equation of type (viii) from Lemma 6.5.1 can be transformed by a linear change of variables t1 , t2 to the equation of a surface of type of type II. If n = 12, then a linear change of variables t0 , t2 transforms the equation to one of type III. If n = 6, 7, 9, we get that G is a cyclic group for a general curve with equation (vi), (vii), (ix) from Lemma 6.5.1. This gives the rows of types VI, VIII, IX from the table. Assume G contains an element σ of order m ≤ 4. Again, if m does

6.5. AUTOMORPHISMS OF PLANE QUARTIC CURVES

181

not divide n, we get an element of order > 12 unless m = 4, n = 6. It is easy to check that in this case G is cyclic of order 12. If m divides n, we easily check that G is cyclic of order n. Assume n = 4. If σ has two equal eigenvalues, then the equation can be reduced to type (iv) from Lemma 6.5.1. One can show that the binary quartic g4 can be reduced by a linear transformation to the form t4 + t4 + at2 t2 (see Exercise 3.9). For general 0 1 0 1 a, the group of automorphisms of the curve is generated by the transformations σ = diag[i, i, 1], σ = diag[1, −1, 1], τ : [t0 , t1 , t2 ] → [t1 , t0 , t2 ]. The element σ generates the center and the quotient by the center is isomorphic to (Z/2Z)2 . This gives the group of Type V from the table. If σ has distinct eigenvalues, then the equation can be reduced to the form (v) from the Lemma. It has an additional automorphism of order 4 equal to τ from above. The group is isomorphic to the dihedral group D8 . Assume n = 3. If G contains an element of order = 3, we are in one of the previous cases. If σ has two equal eigenvalues, then the equation of C can be reduced to type (ii) from Lemma 6.5.1, where we may assume g1 = t0 . It is clear that G = σ . This gives Type XII. If g has distinct eigenvalues, we get equation of type (iii). It has additional symmetry defined by switching the variables t1 , t2 . This gives the permutation group S3 of Type X. Finally if n = 2 we get Types XIII if equation (i) from Lemma 6.5.1 has no additional symmetry of order 2. If we have additional symmetry of order 2, we may assume that it is given by diag[1, −1, −1]. This implies that g2 does not contain the monomial t0 t1 and the coefficient g4 does not contain the monomials t3 t1 , t0 t3 . But then we get 0 1 an additional symmetry defined by switching the variables t0 and t1 . After a linear transformation of variables, this case is reduced to Type XI. Case 1b. V = V1 ⊕ V2 , dim V2 = 2, where V2 is an irreducible representation of G. In particular, the image of G in GL(V2 ) is not abelian. Choose coordinates such that (1, 0, 0) ∈ V1 and V2 is spanned by (0, 1, 0) and (0, 0, 1). We have a natural homomorphism ρ : G → GL(V1 ) × GL(V2 ) ∼ C∗ × GL2 . = Since we are interested in projective representation, we may assume that the projection of ρ(G) to GL(V1 ) is trivial and identify G with a subgroup G of GL(V2 ). Write f = αt4 + t3 L1 (t1 , t2 ) + t2 g2 (t1 , t2 ) + t0 g3 (t1 , t2 ) + g4 (t1 , t2 ). 0 0 0 Since V2 is irreducible, g1 = 0. Since V (f ) is nonsingular, this implies that α = 0. Assume g2 = 0. Then g2 must be G -invariant. Since G is not abelian, this easily implies that, after a linear change of variables, we may assume that g2 = at1 t2 and G is generated by the transformations
−1 σ1 : [t1 , t2 ] → [ζn t1 , ζn t2 ], σ2 : [t0 , t1 ] → [t2 , t1 ],

ζn = e2πi/n

(6.28)

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They form the group isomorphic to the dihedral group D2n of order 2n. Since g3 = 0, then this group acts on the locus V (g3 ) of zeroes of g3 in P(V2 ). Since G is nonabelian, we obtain that n = 3 and G ∼ D6 ∼ S3 . It is easy to see that g3 = a(t3 +t3 ). = = 1 2 Finally, if g4 = 0, its set of zeroes must consist of two points taken with multiplicity 2. We must get g4 = bt2 t2 . This leads to the row Type VII in the Table 1. If g3 = 0, we 0 1 get g4 = at2 t2 which shows that the curve is singular in this case. 1 2 Assume g1 = g2 = 0 but g3 = 0. Since V (g3 ) is invariant and V2 is irreducible, V (g3 ) consists of three points permuted by G ∼ S3 . We can choose coordinates to = assume that g3 = t3 + t3 , where G acts as in (6.28) with n = 3. In order that V (g4 ) 1 2 be G -invariant we must have g4 = at2 t2 . So we get Type V again. 1 2 ¯ Finally we can consider the case g1 = g2 = g3 = 0. The group G must leave V (g4 ) invariant. Since the curve is nonsingular, V (g4 ) consists of 4 distinct points. After a linear change of variables we may assume that g4 = t4 + t4 + at2 t2 . We know from 0 1 1 2 Case 1a that G contains a subgroup G1 isomorphic to 4.22 . There are two special values of a when the group is bigger. These are the cases where the elliptic curve defined by √ the binary quartic is harmonic (a = ±6) or equiequianharmonica = ±2 −3 (see Exercise 3.9). In the first case the quartic acquires an additional symmetry of order 2 (resp. 3). To see the additional symmetries use the following identities:
x4 + y 4 √ x4 + y 4 + 2 −3x2 y 2 = = ´ 1` (x + y)4 + (x − y)4 + 6(x + y)2 (x − y)2 , (6.29) 8 √ ´ e−πi/3 ` (x + iy)4 + (x − iy)4 + 2 −3(x + iy)2 (x − iy)2 . 4

The first identity shows that the form g4 can be reduced to the form t4 + t4 . Hence 0 1 the curve is isomorphic to the Fermat quartic. The Hessian of this curve is the union of three lines which the group of automorphisms permutes. This easily implies that the group is the extension 42 : S3 of order 96. This is Type II. The second identity shows that V (g4 ) has an additional symmetry of order 3 defined by (t0 , t1 ) → 1 1 (t0 + it1 , t0 − it1 ) = √ (eπi/4 t0 + e3πi/4 t1 , eπi/4 t0 + e−πi/4 t1 ). 1−i 2

It√ multiplies g4 by e4πi/3 . The group G of automorphisms of the quartic V (t4 + t4 + 0 1 2 −3t2 t2 + t4 ) is generated by σ1 : [t0 , t1 , t2 ] → [t1 , t0 , t2 ], σ2 : (t0 , t1 , t2 ) → 0 1 2 [t0 , −t1 , t2 ], σ3 : [t0 , t1 , t2 ] → [t0 , t1 , it2 ], and σ4 : [t0 , t1 , t2 ] → [ t0 it1 t0 it1 + , − , eπi/3 t2 ]. 1−i 1−i 1−i 1−i

The element σ3 of order 4 generates the center of G. We have
−1 −1 2 2 2 3 σ1 = σ2 = 1, (σ1 σ2 )2 = σ3 , σ4 = 1, σ1 σ4 σ1 = σ4 σ3 /

Thus G ∼ 4.A4 . This is Type III. = Observe that the group 4.A4 contains 4 √ elements of order 12. One can verify (see Exercise 6.12) that the polynomial t4 +t4 +2 −3t2 t2 can be reduced to the polynomial 0 1 0 1

6.5. AUTOMORPHISMS OF PLANE QUARTIC CURVES

183

(t3 + t3 )t1 by a linear change of variables. This gives the equation of the curve from 0 1 case (x) of Lemma 6.5.1. Case 3. The group G contains a normal transitive imprimitive subgroup H. The group H contains a subgroup from Case 1 and the quotient by this subgroup permutes cyclically the coordinates. It follows from the list in Lemma 6.5.1 that it can happen only if f f f f = = = = t4 + αt2 t1 t2 + t0 (t3 + t3 ) + βt2 t2 0 0 1 2 1 2 t3 t1 + t3 t2 + t3 t0 0 1 2 t4 + t4 + t4 0 1 2 t4 + t4 + t4 + a(t2 t2 + t2 t2 + t2 t2 ). 0 1 2 0 1 0 2 1 2 (6.30) (6.31) (6.32) (6.33)

In the first curve we have the additional automorphism of order 2 interchanging t1 and t2 . This gives Type X. The second curve is the Klein quartic which will be discussed in the next section. The third curve is the Fermat quartic. We have seen this curve already in the previous case. In the forth case Aut(C) permutations and sign changes of coordinates. it is easy to see that this defines a subgroup G of Aut(C) of order 24. It acts by permutations on the set of 4 bitangents V (t0 ± t1 ± t2 ) of C. This easily shows that G is isomorphic to the permutation group S4 (or the octahedron group). One can show √ the full that 1 automorphism group of the curve coincides with S4 unless a = 2 (−1 ± 7). This is Type IV. In the latter case the curve is isomorphic to the Klein curve (see [112]). Case 4. G is a simple group. Here we use the classification of simple nonabelian finite subgroups of PGL(3) (see [20]). There are only two transitive simple groups. One is the group Gof order 168 isomorphic to the group of automorphisms of the Klein quartic. It contains an element σ of order 7 and element of order 3 from the normalizer of the group σ . Thus G contains a imprimitive subgroup of order divisible by 7. It follows from the previous classification that C must be as in case (x) with α = 0, so it is the Klein quartic. This is Type I. The other group is the Valentiner group of order 360 isomorphic to the alternating group A6 . It is known that latter group does not admit a 3-dimensional linear representation (a certain central extension of degree 3 does). Since any automorphism group of a plane quartic acts on the 3-dimensional linear space H 0 (C, ωC ) the Valentiner group cannot be realized as an automorphism group of a plane quartic.

6.5.3

The Klein quartic

Recall the following well-known result of A. Hurwitz (see [134] Chapter IV. ex. 2.5). Theorem 6.5.3. Let X be a nonsingular connected projective curve of genus g > 1. Then #Aut(X) ≤ 84(g − 1).

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For g = 3, the bound gives #Aut(X) ≤ 168 and it is achieved for the Klein quartic C = V (t3 t1 + t3 t2 + t3 t0 ). 0 1 2 (6.34)

Recall that we know that its group of automorphisms contains an element S of order 7 acting by the formula S : [t0 , t1 , t2 ] → [t0 ,
2

t1 ,

4

t2 ],

= e2πi/7 ,

where we scaled the action to represent the transformation by a matrix from SL3 . Another obvious symmetry is an automorphism G2 of order 3 given by cyclic permutation U of the coordinates. It is easy to check that U −1 SU = S 2 , (6.35)

so that the subgroup generated by S, U is a group of order 21 isomorphic to the semidirect product 7 : 3. By a direct computation one checks that the following unimodular matrix defines an automorphism T of C of order 2:   2 − 6 − 5 4− 3 i  2 √ − 5 4− 3 − 6 (6.36) 7 4 3 6 2 − − − 5 We have T U T = U 2, (6.37) so that the subgroup generated by U, T is the dihedral group of order 6. One checks that the 49 products S a T S b are all distinct. In particular, the cyclic subgroup (S) is not normal in the group G generated by S, T, U . Since the order of G is divisible by 2··3·7 = 42, we see that #G = 42, 84, 126, or 168. It follows from the Sylow theorem that the subgroup group (S) must be normal in the first three cases, so #G = 168, and by Hurwitz’s theorem Aut(C) = G = S, U, T . Lemma 6.5.4. The group G = Aut(C) is a simple group of order 168. Proof. Suppose H is a nontrivial normal subgroup of G. Assume that its order is divisible by 7. Since its Sylow 7-subgroup cannot be normal in H, we see that H contains all Sylow 7-subgroups of G. By Sylow’s Theorem, their number is equal to 8. This shows that #H = 56 or 84. In the first case H contains a Sylow 2-subgroup of order 8. Since H is normal, all its conjugates are in H, and, in particular, T ∈ H. The quotient group G/H is of order 3. It follows from (6.37) that the coset of U must be trivial. Since 3 does not divide 56, we get a contradiction. In the second case, H contains S, T, U and hence coincide with G. So, we have shown that H cannot contain an element of order 7. Suppose it contains an element of order 3. Since all such elements are conjugate, H contains U . It follows from (6.35), that the coset of S in G/H is trivial, hence S ∈ H contradicting the assumption. It remains to consider the case when H is a 2-group. Then #G/H = 2a · 3 · 7, with a ≤ 2. It follows from

6.5. AUTOMORPHISMS OF PLANE QUARTIC CURVES

185

Sylow’s Theorem that the image of the Sylow 7-subgroup in G/H is normal. Thus its pre-image in G is normal. This contradiction finishes the proof that G is simple. Remark 6.5.1. One can show that G ∼ PSL2 (F7 ) ∼ PSL3 (F2 ). = = The first isomorphism has a natural construction via the theory of automorphic functions. The Klein curve is isomorphic to a compactification of the modular curve X(7) corresponding to the principal congruence subgroup of level 7. The second isomorphism has a natural construction via considering a model of the Klein curve over a finite field of 2 elements (see [100]). The group Aut(C) has 3 orbits on C with non-trivial stabilizers of orders 2, 3, 7. They are of cardinality 84, 56 and 24, respectively. The orbit of cardinality 24 consists of inflection points of C. We know that a cyclic group of order 7 is normalized by an element of order 3. Thus the orbit is equal to the union of 8 sets each consisting of an orbit of a group of order 3. An example of such a group is the vertices of the triangle formed by the inflection tangent lines
2 2 t0 + t1 + t2 = 0, t0 + η3 t1 + η3 t2 = 0, t0 + η3 t1 + η3 t2 = 0.

This can be directly checked. From this it follows that the inflection points form the set of vertices of 8 triangles. We know that the inflection points are the intersection points of C and its Hessian given by the equation He(f ) = 5t2 t2 t2 − t0 t5 − t5 t2 − t1 t5 = 0. 0 1 2 1 0 2 The orbit of cardinality 56 consists of the tangency points of 28 bitangents of C. An example of an element of order 3 is a cyclic permutation of coordinates. It has 2 2 2 fixed points (1, η3 , η3 ) and (1, η3 , η3 ) on C. They lie on the bitangent with equation
2 4t0 + (3η2 + 1)t1 + (3η3 + 1)t2 = 0.

Define a polynomial of degree 14 by  ∂2f Ψ=
∂t2  ∂20  f 1 det  ∂t2 t0 ∂ f  ∂t2 t0 ∂ψ ∂t0

∂2f ∂t0 t1 ∂2f ∂t2 1 ∂2f ∂t2 t1 ∂ψ ∂t1

∂2f ∂t0 t2 ∂2f ∂t1 t2 ∂2f ∂t2 2 ∂ψ ∂t2

∂ψ ∂t0 ∂ψ   ∂t1  ∂ψ  ∂t2 



0

One checks that it is invariant with respect to G and does not contain f as a factor. Hence it cuts out in V (f ) a G-invariant positive divisor of degree 56. It must consists of a G-orbit of cardinality 56. One can compute it explicitly (classics did not need computers to do this) to find that Ψ = t14 + t14 + t14 − 34t0 t1 t2 (t10 t2 + . . .) − 250t0 t1 t2 (t2 t8 + . . .)+ 0 1 2 0 0 1

186

CHAPTER 6. PLANE QUARTICS 375t2 t2 t2 (t6 t2 + . . .) + 18(t7 t7 + . . .) − 126t3 t3 t3 (t0 t2 + . . .). 0 1 2 0 2 0 1 0 1 2 1

Here the dots mean monomials obtained from the first one by permutation of variables. The orbit of cardinality 84 is equal to the union of 21 sets, each consisting of 4 intersection points of C with the line of fixed points of a transformation of order 2. An example of such a point is ((
4



3

)( −

6

) 4, (

2



5

)( −

6

) ,(

4



3

)(

2



5

) 2 ).

Similarly to the above one considers the Jacobian determinant Ξ = J(f, g, h) of the polynomials f, g, h. It is a G-invariant polynomial of degree 21. Its zeroes on V (f ) give the orbit of 84 points. One computes to find that Ξ = t21 + t21 + t21 − 7t0 t1 t2 (t17 t2 + . . .) + 217t0 t1 t2 (t3 t15 . . .)− 0 1 2 0 0 1 308t2 t2 t2 (t13 t2 + . . .) − 57(t14 t7 + . . .) − 289(t7 t14 + . . .)+ 0 1 2 0 0 2 0 1 4018t3 t3 t3 (t2 t10 + . . .) + 637t3 t3 t3 (t9 t3 + . . .)+ 0 1 2 0 1 0 1 2 0 1 1638t0 t1 t2 (t10 t8 + . . .) − 6279t2 t2 t2 (t6 t9 + . . .)+ 0 1 0 1 2 0 1 7007t5 t5 t5 (t0 t5 + . . .) − 10010t4 t4 t4 (t5 t4 + . . .) + 3432t7 t7 t7 . 0 1 2 1 0 1 2 0 1 0 1 2

Exercises
6.1 Show that two syzygetic tetrads of bitangents cannot have two common bitangents. 6.2 Let Ct = V (tf + q 2 ) be a family of plane quartics over C depending on a parameter t. Assume that V (f ) is nonsingular and V (f ) and V (q) intersect transversally at 8 points p1 , . . . , p8 . Show that Ct is nonsingular for all t in some open neighborhood of 0 in usual topology and the limit of 28 bitangents when t → 0 is equal the set of 28 lines pi , pj . 6.3 Show that the locus of nonsingular quartics which admit a flex bitangent is a hypersurface in the space of all nonsingular quartics. 6.4 Consider the Fermat quartic V (t4 + t4 + t4 ). Find all bitangents and all Steiner complexes. 0 1 2 Show that it admits 12 flex bitangents. 6.5 An open problem: what is the maximal possible number of flex bitangents on a nonsingular quartic? 6.6 Show that a nonsingular plane quartic C admits 63 irreducible one-parameter families of conics which are tangent to C at 4 points. 6.7 Let S = {( 1 , 1 ), . . . , ( 6 , 6 )} be a Steiner complex of 12 bitangents. Prove that the six ` ´ intersection points i ∩ i lie on a conic and all 28 = 378 intersection points of bitangents lie 2 on 63 conics. [Hint: the conic is the Veronese curve from Remark 6.2.4]. 6.8 Find all possible types of azygetic hexads of bitangents. Which types are contained in a Steiner complex? 6.9 Show that the pencil of conic passing through the four contact points of two bitangents contains five members each passing through the contact points of a pair of bitangents. 6.10 Show that the linear system L( ) of conics associated to a nonzero 2-torsion divisor class is equal to the linear system of first polars of the cubic B( ).

EXERCISES

187

6.11 Show that a choice of ∈ Jac(C)[2] \ {0} defines a conic Q and a cubic B such that C is equal to the locus of points x such that the polar Px (B) is touching Q. 6.12 Let C = V (a11 a22 − a2 ) be a representation of a nonsingular quartic C as a symmetric 12 ˜ quadratic determinant corresponding to a choice of a 2-torsion divisor class . Let C be the unramified double cover of C corresponding to . Show that C is isomorphic to a canonical curve of genus 5 given by the equations a11 [t0 , t1 , t2 ] − t2 = a12 [t0 , t1 , t2 ] − t3 t4 = a22 [t0 , t1 , t2 ] − t2 = 0 3 4 in P4 . 6.13 A nonsingular curve is called bielliptic if it admits a double cover to an elliptic curve. Show that the moduli space of bielliptic curves of genus 4 is birationally isomorphic to the moduli space of isomorphism classes of genus 3 curves together with a nonzero 2-torsion divisor class. √ 6.14 Show that the curves V (t4 + t4 + t4 + 2 −3t2 t2 ) and V (t4 + t4 + t0 (t3 + t3 ) are 0 1 2 1 2 0 2 0 1 isomorphic. 6.15 A plane quartic C = V (f ) is called a Caporali quartic if VSP(f ; 4)o = ∅. (i) What is the dimension of the locus of the Caporali quartics? (ii) Show that the Clebsch covariant quartic S(C) is reducible. (iii) Find the intersection of the loci of Fermat quartics and Caporali quartics. (E. Caporali [26]). 6.16 Let q be a nondegenerate quadratic form in 3 variables. Show that W (q 2 ; 6)o is a homogeneous space for the group PSL(2, C). 6.17 Let C be a hyperelliptic curve of genus g. Show that the graph of the hyperelliptic involution has valence 2. 6.18 Let f = t3 t1 + t3 t2 + t3 t0 . Show that V (S(f )) = V (f ). 0 1 2 6.19 Show that the binary form f = t0 (t0 + 2t1 )2 does not admit nondegenerate polar 2th polyhedra. 6.20 Show that the locus of lines = V (l) such that the anti-polar of l2 with respect to a quartic curve V (f ) is a plane curve of degree 6 in the dual plane. 6.21 Show that the Clebsch covariant of the Fermat quartic C is equal to C. 6.22 Classifiy automorphism groups of irreducible singular plane quartics. 6.23 For each nonsingular plane quartic curve C with automorphism group G describe the ramification scheme of the cover C → C/G. 6.24 Let C be the Klein quartic. For any subgroup H of its automorphism group of C determine the genus of H and the ramification scheme of the cover C → C/H. 6.25 Analyze the action of the automorphism group of the Klein quartic C on the set of even theta characteristics. Show that there is only one which is invariant with respect to the whole group. Find the corresponding determinantal representation of C. 6.26 Let C be a general plane quartic. A triangle of lines is called a biscribed triangle of C if each side is a tangent line and each vertex is on C. (i) Show that for any biscribed triangle there is exists a unique contact cubic which is tangent to C at the vertices of the triangle and at the tangency points of its sides. (ii) Show that the contact cubic defined by a biscribed triangle corresponds to an even theta characteristic on C. Using this show that there are 288 = 8.36 biscribed triangles.

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CHAPTER 6. PLANE QUARTICS

6.27 Show that a smooth plane quartic admits an automorphism of order 2 if and only if among its 28 bitangents four form a syzygetic set of bitangents intersecting at one point. 6.28 Show that the dual curve of a general nonsingular plane quartic is a curve of degree 12 with 24 cusps and 28 nodes. 6.29 Consider the curve S (resp. T ) in the dual plane equal to the closure of the locus of lines that intersects a nonsingular quartic at 4 points with equianharmonic(resp. harmonic) cross-ratio. Show that C (resp. C ) is of degree 4 (resp. 6) and intersects the dual quartic at its 24 cusps.
4 4 4 4 6.30 Let C = l4 + l1 + l2 + l3 + l4 be a Clebsch quartic. Show that the pentagonal theta characteristic on the L¨ roth quartic S = S(C) defines the representation of S as the determinant u of the 4×4 matrix with diagonal entries l+li and off-diagonal entries equal to l (B. van Geemen).

6.31 Show that the seven points which have the same polar lines with respect to a conic and a cubic define Aronhold set for a L¨ roth quartic (H. Bateman [13]). u

Historical Notes
The fact that a general plane quartic curve has 28 bitangents was first proved in 1850 by C. Jacobi [150] athough the number was apparently known to J. Poncelet. The proof used Pl¨ cker formulas and so did not apply to any nonsingular curve. Using contact u cubics, Hesse extended this result to arbitrary nonsingular quartics [138]. The first systematic study of the configuration of bitangents began by O. Hesse [138],[139] and J. Steiner [233]. Although the Steiner’s paper does not contain proofs. They considered azygetic and syzygetic sets, Steiner complexes of bitangents although the terminology was introduced later by Frobenius [113]. Hesse’s approach was using the relationship between bitangents and Cayley octads. The notion of a Steiner group of bitangents was introduced by A. Cayley in [38]. Weber [247] changed it to a Steiner complex in order not to be confused with the terminology of group theory. The fact that the equation of a nonsingular quartic could be brought to the form (6.1) was first noticed by J. Pl¨ cker [193]. Equation (6.3) arising from a Steiner comu plex appears first in Hesse’s paper [139], §9. The determinantal identity for bordered determinants (6.23) appears in [138]. The number of hexads of bitangents with contact points on a cubic curve was first computed by Hesse [138] and Salmon [210]. The equation of a quartic as a quadratic determinant appeared first in Pl¨ cker [192], u p. 228 and in Hesse [139], §10. Both of them knew that it can be done in 63 different ways. Hesse also proves that the 12 lines of a Steiner complex, consider as points in the dual plane, lie on a cubic. More details appear in Roth’s paper [205] and Coble’s book [51]. Using his determinantal identity Hesse showed that a linear symmetric determinantal representation of a plane curve of degree d defines a d − 1-dimensional family of contact curves of degree d−1. However he acknowledges in [138] that he did not prove that general curve of degree d > 4 admits such a representation. This was proved much later by Dixon[75]. For quartic curves Hesse proves the existence of a determinantal representation in 36 differeent ways. The relationship between seven points in the projective plane and bitangents of a plane quartic was first given by S. Aronhold [6]. The fact that Hesse’s construction and

Historical Notes

189

Aronhold’ construction are equivalent via the projection from one point of a Cayley octad was first noticed by A. Dixon [76]. The relation of bitangents to theta functions with odd characteristics goes back to B. Riemann [201] and Weber [247] and was developed later by G. Frobenius [113], [115]. In particular, Frobenius found a relationship between the sets of seven points or Cayley octads with theta functions of genus 3. Coble’s book [51] has a nice exposition of Frobenius’s work. The equations of bitangents presented in Theorem 6.1.5 were first found by Riemann, with more details explained by Weber. The theory of covariants and contravariants of plane quartics was initiated by A. Clebsch in his fundamental paper in [43] about plane quartic curves. In this paper he introduces his covariant quartic S(C), the catalecticant invariant and shows that its vanishing is necessary for writing the equation of a quartic as a sum of five powers of linear forms. Much later G. Scorza [216] proved that the rational map S on the space of quartics is of degree 36 and related this number with the number of even theta characteristics. The interpretation of the apolar conic as the parameter space of inscribed pentagons was given by G. L¨ roth [169]. u The groups of automorphisms of nonsingular plane quartic curves were classified by S. Kantor [153] and A. Wiman [249]. The first two curves from our table were studied earlier by F. Klein [156] and W. Dyck [94]. Of course the Klein curve is the most famous of those and appears often in the modern literature (see, for example, [227]). The classical literature about plane quartics is enormous. We refer to [40] for a nice survey of classical results, as well as to his many original works on plane quartics which are assembled in [41]. Other surveys can be found in [188] and [102].

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CHAPTER 6. PLANE QUARTICS

Chapter 7

Planar Cremona transformations
7.1
7.1.1

Homaloidal linear systems
Linear systems and their base schemes

Here we recall some known definitions from the theory of linear systems and rational maps (see [163]). Let X be a nonsingular irreducible variety of dimension n and a be a sheaf of ideals on X. A resolution of a is a projective birational morphism π : Y → X of nonsingular varieties such that π −1 (a) := a · OY ∼ OY (−F ) for some effective = divisor F on Y . Using the universal property of blow-up of ideals, we see that a resolution of a is a resolution of singularities of the normalization of the blow-up of a. Let ν : B(a) → X be the normalization of the blow-up of the ideal a so that ¯ ν −1 (a) = OY (−F ) for some effective divisor. The ideal a = σ∗ OB(a) (−F ) is equal to the integral closure of the ideal a. In case when X is affine it consists of regular functions on X such that, considered as rational functions on B(a), they belong to the space H 0 (B(a), OB(a) (−F )). By the universal property of the blow-up, π factors π = π ◦ : Y → B(a) → X, hence σ∗ OB(a) (−F ) = π∗ OX (−F ) = a. We will be applying this to the case when a is equal to the base ideal of a linear system. To fix the notation, let us remind the definition. Let L be an invertible sheaf on X and V ⊂ H 0 (X, L) a linear subspace of positive dimension. We denote by |V | the projective space of lines P(V ) identified with the set of divisors Ds of zeroes of sections s from V \ {0}. In case V = H 0 (X, L) we employ the notation |L| or |D|, where L ∼ OX (D). The set of divisors Ds , s ∈ V \ {0}, is called the linear system = defined by the subspace V , in the case |V | = |L|, it is called a complete linear system. We assume that |V | has no fixed component (i.e. effective divisor F = 0 such that V is contained in the image of the natural map H 0 (X, L(−E)) → H 0 (X, L)). The evaluation map ev : V ⊗ OX → L (7.1) 191

192

CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

defines a map of sheaves V ⊗ L−1 → OX . Its image is an ideal sheaf b(|V |) in OX which is called the base ideal of |V |. The closed subscheme Bs(|V |) of X defined by this ideal is called the base scheme of |V | and its support is the base locus of |V |. We have Bs(|V |) = ∩D∈|V | D, where each D is identified with a closed subscheme of X. Let s0 , . . . , sm be a basis of V and Di = Dsi ∈ |V | be the corresponding divisors. Then
m

Bs(|V |) =
i=0

Di .

By definition, V ⊂ H 0 (X, L ⊗ b(|V |)). (7.2) Let ν : B → X be the normalization of the blow-up of the ideal sheaf b(|V |) and ν −1 (b(|V |) = OB (−F ). Applying the projection formula, we get σ∗ (σ ∗ L)(−F )) = L ⊗ b((|V |), hence H 0 (X, L ⊗ b((|V |)) = H 0 (B, σ ∗ L(−F )). Combining with (7.2), we obtain an injective linear map σ ∗ : V ⊂ H 0 (B, σ ∗ L(−F )). Let π : Y → X be a resolution of b(|V |) (also called a resolution of |V |). Since π factors through σ, and the direct image of OY (−F ) in B is equal to OB (−F ), we obtain an inclusion π ∗ : V → H 0 (Y, π ∗ L(−F )). (7.3) Let φ|V | : X− → P(V ∗ ) be the rational map defined by the linear system |V |. The rational map φ|V | is given by assigning to a point x ∈ X \ Bs(|V |) the hyperplane in |V | of sections vanishing at x. A choice of a basis in V defines projective coordinates in P(V ∗ ) and the explicit formula φ|V | (x) = [s0 (x), . . . , sm (x)]. (7.4) The rational map φ|V | is regular if and only if the base locus of |V | is empty, or equivalently, the evaluation map (7.1) is surjective. In this case |V | is called base point free. It is called very ample if its base locus is empty and φ|V | defines a closed embedding. Let π : Y → X be a resolution of |V |, then |π ∗ (V )| is a base-free linear system in

7.1. HOMALOIDAL LINEAR SYSTEMS

193

the complete linear system |σ ∗ (L)(−F ))|. Let σ : Y → P(V ∗ ) be the corresponding regular map. We obtain a commutative diagram Y  pppp σ  pp pp  p"   φ|V | • • • • • • •/ P(V ∗ ). X
π

(7.5)

It follows from the definition of a rational map defined by a base-free linear system that σ ∗ OP(V ∗ ) = (π ∗ L)(−F )) and σ ∗ (H 0 (P(V ∗ ), OP(V ∗ ) (1))) coincides with the image of V in H 0 (Y, (π ∗ L)(−F )). If we assume that the base ideal b(|V |) is integrally closed, then we can identify the complete linear system |(π ∗ L)(−F )| with |L⊗b(|V |)|, where we identify H 0 (X, L ⊗ b(|V |)) with a subspace of H 0 (X, L) via the inclusion of sheaves b(|V | → OX . Note that one can also define the proper inverse image f −1 (|V |) of a linear system |V | ⊂ L on X under a rational map f : X → X . We consider f as a regular map f : dom(f ) → X and use that any invertible sheaf of dom(f ) and its section can be uniquely extended to an invertible sheaf and its section on X. For any rational map φ : X → X , a commutative diagram Y  eee eeσ  ee  e   φ X • • • • • • •/ X ,
π

(7.6)

where π is a birational projective morphism and σ is a morphism, is called a resolution of indeterminacy points of f . We will always assume that Y is normal and π is an isomorphism over dom(f ). Thus a resolution of b(|V | defines a resolution of the rational map φ|V | . Consider a resolution of indeterminacy points of the rational map φ = φ|V | : X → P(V ∗ ). Then π −1 (|V |) defines a regular map σ, hence π −1 (b(|V |)) is an invertible sheaf. By the universal property of the blow-up, the map π factors through the blow-up of b(|V |). Note that the pair (π, σ) defines a regular map from Y to the normalization γφ of the graph Γφ of f , the Zariski closure of the graph of the map φ : dom(φ) → X ¯ in X ×X . It always defines a resolution of indeterminacy points of φ. When φ = φ|V | , the graph Γφ|V | is isomorphic to the blow-up of b(|V |). To define a rational map φ : X → X of projective varieties, we choose a very ample sheaf L on X which defines a closed embedding ι : X → Pn = |H 0 (X , L )∗ |. Let |V | = φ−1 (|L |) ⊂ |φ−1 (L )|. Then φ|V | = φ ◦ ι. Proposition 7.1.1. Assume that the image X of φ|V | is linearly normal in P(V ∗ ) and the map of φ|V | : X− → X is of degree 1. Then the map π ∗ : V → H 0 (Y, (π ∗ L)(−F )) is bijective. In particular, the base ideal b = b(|V |) is integrally closed and |V | = |H 0 (Y, L ⊗ b)|.

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

Proof. Recall that a projective subvariety Z ⊂ Pn is called linearly normal if it is a normal variety and the canonical restriction map H 0 (Pn , OPn (1)) → H 0 (Z, OZ (1)) is bijective. Let σ : Y → X ⊂ P(V ∗ ) be the map given by the linear system |π ∗ (V )| ⊂ |π ∗ L(−F )|. We know that π −1 (|V )|) = σ ∗ (|OP(V ∗ ) (1)|), so it suffices to prove that the linear system |σ ∗ (|OP(V ∗ ) (1)|)| is complete. By Zariski’s Main Theorem ([134], Chapter 3, §11), σ∗ OY = ι∗ OX , where ι : X → P(V ∗ ) is the closed embedding. We have H 0 (Y, σ ∗ OP(V ∗ ) (1)) = H 0 (P(V ∗ ), σ∗ (σ ∗ OP(V ∗ ) (1))) = H 0 (P(V ∗ ), OP(V ∗ ) (1) ⊗ σ∗ OY ) = H 0 (X , OX (1)). By definition of φ = φ|V | , we have |V | = φ−1 (|OP(V ∗ ) (1)|), hence π ∗ (|V |) = π ∗ (φ−1 (|OP(V ∗ ) (1)|)) = σ ∗ (|OP(V ∗ ) (1)|)). Since X is linearly normal, the restriction map H 0 (P(V ∗ ), OP(V ∗ ) (1)) → H 0 (X , OX (1)) is bijective. Hence σ ∗ (|OP(V ∗ ) (1)|) = |σ ∗ OP(V ∗ ) (1)|. This proves the assertion. Suppose φ = φ|V | is of finite degree deg φ = [C(X) : f ∗ (C(X ))], where X = φ(X) = σ(Y ) is the image of the rational map f . Consider a resolution of indeterminacy points of f . Let π −1 (b(|V |)) = OY (−F ) for some effective divisor F . It follows from the intersection theory on algebraic varieties (see [116], Chapter 4, §4) that deg f deg X = (π ∗ (D) − F )n ,
Y

where D ∈ |V | and deg X is the degree of X in the projective space P(V ∗ ). In particular, f is a birational map if and only if (π ∗ (D) − F )dim X = deg X .
Y

(7.7)

7.1.2

Exceptional configurations

From now on we assume that X is a nonsingular surface and |V | is a linear system without fixed components defining a rational map f : X → Pm . Let π : Y → X, σ : Y → Pm be its resolution of indeterminacy points. Resolving singularities of Y we assume that Y is a nonsingular surface. We will assume that Y is a minimal resolution of singularities. We know (see [134], Chapter V, §5) that any birational morphism π : Y → X of nonsingular projective surfaces can be factored into a composition of blow-ups with centers at closed points. Let
N 2 1 π : Y = YN −→ YN −1 −→ . . . −→ Y1 −→ Y0 = X

π

πN −1

π

π

(7.8)

be such a factorization. Here πi : Yi → Yi−1 is the blow-up of a point xi ∈ Yi−1 . Let
−1 Ei = πi (xi ),

Ei = (πi+1 ◦ . . . πN )∗ (Ei ).

(7.9)

7.1. HOMALOIDAL LINEAR SYSTEMS

195

The divisors Ei are called the exceptional configurations of the birational morphism π : Y → X. Note that Ei be not be a non-reduced deivisor. For any effective divisor D = 0 on X let multxi D be defined inductively in the following way. We set multx1 D to be the usual multiplicity of D at x1 . It is defined as the largest integer m such that the local equation of D at x1 belongs to the m-th power of the maximal ideal mX,x1 . Suppose multxi D is defined. We take the proper inverse −1 −1 transform πi (D) of D in Xi and define multxi+1 (D) = multxi+1 πi (D). It follows from the definition that
N

π −1 (D) = π ∗ (D) −
i=1

mi Ei ,

where mi = multxi D. Now suppose π : Y → X is a resolution of indeterminacy points of a rational map f defined by a linear system |V | ⊂ |L|. Let mi = min multxi D, i = 1, . . . , N.
D∈|V |

If D0 , . . . , Dt are divisors corresponding to a basis of V , then mi = min{multxi D0 , . . . , multxi Dt }, i = 1, . . . , N. It is clear that π −1 (|V |) = π ∗ (|V |) −
i=1 N N

mi Ei .

(7.10)

Let F = i=1 mi Ei , then π −1 (|V |) is contained in the complete linear system |π ∗ (L)(−F )|. Let b = b(|V |). The ideal sheaf π −1 (b) = b · OY is the base locus of π −1 (|V |) and hence coincides with OY (−F ). Applying (7.7), we obtain that deg φ|V | = (π ∗ (D) − F )2 , where D ∈ |L| and we consider φ|V | as a rational map from X onto its image X . Suppose that |L| is base point-free. Then we can choose D such that it does not contain base points of |V |. This gives D2 − F 2 = deg φ|V | deg X , (7.11)

Lemma 7.1.2. Let π : Y → X be a birational morphism of nonsingular surfaces and Ei , i = 1, . . . , N, be its exceptional configurations. Then Ei · Ej = −δij , Ei · KY = −1. Proof. This follows from the standard properties of the intersection theory on surfaces. For any morphism of nonsingular projective surfaces φ : X → X and two divisors D, D on X, we have φ∗ (D) · φ∗ (D ) = deg(φ)D · D . Also, if C is a curve such that φ(C) is a point, we have C · φ∗ (D) = 0. (7.13) (7.12)

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Applying (7.12), we have
2 −1 = Ei = (πi+1 ◦ . . . ◦ πN )∗ (Ei )2 = Ei2 .

Assume i < j. Applying (7.13) by taking C = Ej and D = (πi+1 ◦ . . . ◦ πj−1 )∗ (Ei ), we obtain
∗ 0 = Ej · πj (D) = (πj+1 ◦ . . . ◦ πN )∗ (Ej ) · (πj+1 ◦ . . . ◦ πN )∗ (D) = Ej · Ei .

This proves the first assertion. To prove the second assertion, we use that
∗ KYi+1 = πi (KYi ) + Ei .

By induction, this implies that
N

KY = π ∗ (KY0 ) +
i=1

Ei .

(7.14)

Intersecting with both sides and using (7.13), we get
N

KY · Ej =
i=1

2 Ei ·Ej = Ej = −1.

Assume now that φ|V | : X− → X is a birational rational map of nonsingular projective algebraic surfaces. By Bertini’s theorem ([134], Chapter II, Theorem 8.18), a general hyperplane section H of X is a nonsingular irreducible curve of some genus g. Since π −1 (|V |) has no base points, by another Bertini Theorem ([134], Chapter II, Corollary 10.9), its general member H is a nonsingular irreducible curve. Since H ∈ |σ ∗ (H )|, we obtain that H is of genus g and the map σ : H → σ(H) is an isomorphism. Using the adjunction formula, we obtain H · KY = 2g − 2 − H 2 = H
2

+ H · KX − H 2 .

Write H = π ∗ (D) − F and apply the projection formula, to obtain H · KY = D · KX − F · KY . Applying (7.11) and the lemma, we obtain Proposition 7.1.3. Suppose φ|V | : X− → X is a birational rational map of nonsingular projective algebraic surfaces. Let D ∈ |L|. Then (i) D2 −
N i=1

m2 = H i
N i=1

2

= deg X ;

(ii) D · KX −

mi = H · K X ;

7.1. HOMALOIDAL LINEAR SYSTEMS

197

7.1.3

The bubble space of a surface

Consider a factorization (7.8) of a birational morphism of nonsingular surfaces. Note that, if the morphism π1 ◦ · · · ◦ πi : Yi → X is an isomorphism on a Zariski open neighborhood of the point xi+1 , the points xi can be identified with its image in X. Other points are called infinitely near points in X. To make this notion more precise one introdices the notion of the bubble space of a surface X. Let B(X) be the category of birational morphisms π : X → X of nonsingular projective surfaces. Recall that a morphism from (X → X) to (X category is a regular map φ : X → X such that π ◦ φ = π .
π

→ X) in this

π

Definition 7.1. The bubble space X bb of a nonsingular surface X is the factor set X bb =
(X →X)∈B(X)
π

X /R,

where R is the following equivalence relation: x ∈ X is equivalent to x ∈ X if the rational map π −1 ◦ π : X − → X maps isomorphically an open neighborhood of x to an open neighborhood of x . It is clear that for any π : X → X from B(X) we have an injective map iX : X → X bb . We will identify points of X with their images. If φ : X → X is a morphism in B(X) which is isomorphic in B(X ) to the blow-up of a point x ∈ X , any point x ∈ φ−1 (x ) is called infinitely near point to x of the first order. This is denoted by x x . By induction, one defines an infinitely near point of order k, bb denoted by x k x . This defines a partial order on X . bb We say that a point x ∈ X is of height k, if x k x0 for some x0 ∈ X. This defines the height function on the bubble space ht : X bb → N. Clearly, X = ht−1 (0). Points of height zero are called proper points of the bubble space. They will be identified with points in X. Let ZX be the free abelian group generated by the set X bb . Its elements are integer valued functions on X bb with finite support. They added up as functions with bb values in Z. We write elements of ZX as finite linear combinations m(x)x, where bb x ∈ X and m(x) ∈ Z (similar to divisors on curves). Here m(x) is the value of the corresponding function at x. Definition 7.2. A bubble cycle is an element η = following additional properties (i) m(x) ≥ 0 for any x ∈ X bb ; (ii)
x x
bb

m(x)x of ZX

bb

satisfying the

mx ≤ mx .

We denote the subgroup of bubble cycles by Z+ (X bb ).

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Clearly, any bubble cycle η can be written in a unique way as a sum of bubble cycles Zk such that the support of ηk is contained in ht−1 (k). We can describe a bubble cycle by a weighted graph, called the Enriques diagram, by assigning to each point from its support a vertex, and joining two vertices by an ordered edge if one of the points is infinitely near to another point of the first order. The edge points to the vertex of lower height. We weight each vertex by the corresponding multiplicity. It is clear that the Enriques diagram is a tree. Let ξ = mx x be a bubble cycle. We order the points from the support of η such that xi xj implies j < i. We refer to such an order as an admissible order. We write N ξ = i=1 mi xi . Then we represent x1 by a point on X and define π1 : X1 → X to be the blow-up of X with center at x1 . Then x2 can be represented by a point on X1 as either infinitely near of order 1 to x1 or as a point equivalent to a point on X. We blow up x2 . Continuing in this way, we get a sequence of birational morphisms:
N 2 1 π : Yξ = YN −→ YN −1 −→ . . . −→ Y1 −→ Y0 = X,

π

πN −1

π

π

(7.15)

where πi+1 : Yi+1 → Yi is the blow-up of a point xi ∈ Yi−1 . Clearly, the bubble cycle N η is equal to the bubble cycle i=1 mi xi . Let L be an invertible sheaf on X and η be a bubble cycle with an admissible order and (7.15) be the corresponding sequence of blow-ups. Let Ei , i = 1, . . . , N, be the exceptional configurations. Set
N

|L − η| := {D ∈ |L| : π ∗ (D) −
i=1

mi Ei ≥ 0}.

This is a linear subsystem of |L|. Its elements D satisfy the following linear conditions. For any x ∈ η with ht(x) = 0 we must have multx D ≥ m(x). This condition depends only on the equivalence class of x. Let y ∈ η with ht(y) = 1 and y x for some x ∈ η. Then we must have multy (φ∗ (D) − m(x)E) ≥ my , where y is represented by a point on the exceptional curve E of the blow-up φ : S → X with center at x. Then we go to level 2 and so on. N Let F = i=1 mi Ei and aη = πη (OYη (−F )). It is an integrally closed ideal sheaf ¯ on X equal to the integral closure b of the base ideal b of the linear system |L − η|. We have |L − η| = |H 0 (X, L ⊗ aη )|. The following formula is known as the Hoskin-Deligne formula (see [68], [142]) Proposition 7.1.4.
N

length(aη ) := dim H 0 (X, OX /aη ) = The exact sequence

1 2 i=1

mi (mi + 1).

0 → L ⊗ aη → L → L ⊗ OX /aη → 0

7.1. HOMALOIDAL LINEAR SYSTEMS shows that dim |L − η| ≥ dim |L| − length(aη ).

199

(7.16)

Thus the Hoskin-Deligne formula justifies the count of constants, passing through a point with multiplicity m imposes m(m + 1)/2 conditions. The following proposition follows from Proposition 7.1.1. Proposition 7.1.5. Let |V | be a linear system in |L| without fixed components. Suppose it defines a birational rational map onto a projectively normal surface X in P(V ∗ ). There exists a unique bubble cycle η such that |V | = |L − η|.

7.1.4

Cremona transformations

A birational map f : Pn − → Pn is called a Cremona transformation. The group Bir(Pn ) of birational transformations of Pn is denoted by Cr(n) and is called the Cremona group. It is isomorphic to the group of automorphisms of the field of rational functions on Pn identical on constants. In other words Cr(n) ∼ AutC (C(z1 , . . . , zn )). = As any rational map defined on Pn , it is given by an n-dimensional linear system |V | ⊂ |OPn (d)| for some d ≥ 1. We assume that the linear system has no fixed component. The number d is called the degree of the Cremona transformation. A choice of a basis in V gives an explicit formula: φ : [x0 , . . . , xn ] → [f0 (x0 , . . . , xn ), . . . , fn (x0 , . . . , xn )], where fi (t0 , . . . , tn ) are homogeneous polynomials of degree d without common factor of positive degre. A linear system |V | defining a Cremona transformation is called a homaloidal linear system. Its base ideal b(|V |) is the sheaf of ideals associated to the homogeneous ideal generated by the polynomials f0 , . . . , fn . Its base scheme is the closed subscheme of Pn defined by the equations f0 = . . . = fn = 0. As we explained above there is a resolution of indeterminacy points of φ , Y } eee ee σ }} } ee }} e ~}} φ n • • • • • • •/ n P P
π

(7.17)

where Y is a nonsingular and π and σ are birational morphisms. Let E be the exceptional divisor of the resolution defined by the property π −1 (b(|V |)) · OY ∼ OY (−E). = Applying (7.7), we have σ ∗ (H)n = (π ∗ (dH) − E)n = 1. where H is a hyperplane in Pn . Applying Proposition 7.1.1, we obtain (7.18)

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Proposition 7.1.6. The base ideal b of a homaloidal linear system is integrally closed and its proper inverse transform on a resolution of indeterminacy points coincides with the complete linear system |π ∗ OPn (d)(−F )|, where π −1 (b) = OY (−F ). Let us specialize to the case of a Cremona transformation f : P2 − → P2 of degree d. In this case L ∼ OP2 (1) and L ∼ OP2 (d). Since ωP2 ∼ OP2 (−3), applying = = = Proposition 7.1.5 gives Proposition 7.1.7.
N

1 = d2 −
i=1 N

m2 , i mi .
i=1

3

= 3d −

It follows from Proposition 7.1.6 that a homaloidal linear system is equal to |OP2 (d)− η| for some bubble cycle. It is called the bubble cycle of the homaloidal net or of the Cremona transformation it defines. Theorem 7.1.8. A bubble cycle η = i=1 mi xi on P2 is equal to the bubble cycle of a homaloidal net of degree d if and only if |OP2 (d) − η| contains an irreducible divisor and equalities (7.19) and (7.19) hold. Proof. We have already proved the necessity of the conditions. Consider the linear system |V | = |OP2 (d) − η|. It follows from the conditions that
1 2 d(d N

+ 1) − 2 =

1 2

mi (mi + 1).

Applying the Hoskin-Deligne formula and (7.16), we obtain that dim |V | ≥ 2. By assumption, the linear system |V | has no fixed components. Let π : Y = Yη → P2 and π −1 (|V | be the proper transform of |V | on Yη . It has no base points and defines a regular map σ : Yη → Pn , n ≥ 2, which resolves φ. By (7.19) and (7.18) this map is birational on its image X and deg X = 1. This proves the assertion. The vector (d; m1 , . . . , mN ) is called the characteristic of the homaloidal net. It depends on an admissible order of the bubble cycle η. Of course, not any vector (d; m1 , . . . , mN ) satisfying equalities (7.19) and (7.19) is realized as the characteristic vector of a homaloidal net. There are other necessary conditions for a vector to be realized as the characteristic (d; m1 , . . . , mN ) for a homaloidal net. For example, if m1 , m2 correspond to points of height 0 of largest multiplicity, a line through the points should intersect a general member of the net non-negatively. This gives the inequality d ≥ m1 + m2 . Next we take a conic through 5 points with maximal multiplicities. We get 2d ≥ m1 + · · · + m5 .

7.1. HOMALOIDAL LINEAR SYSTEMS

201

Then we take cubics through 9 points, quartics through 14 points and so on. The first case which can be ruled out in this way is (5; 3, 3, 1, 1, 1, 1, 1). It satisfies the equalities from the theorem but does not satisfy the condition m ≥ m1 + m2 . We will discuss the description of characteristic vectors later in this chapter.

7.1.5

Nets of isologues and fixed points

Let φ : P2 − → P2 be a Cremona transformation. Let p be a point in the plane. Consider the locus of points Cφ (p) such that x, φ(x), p are collinear. This locus is called isologue of p, the point p is called its center. In terms of equations, if φ is given by polynomials (f0 (t), f1 (t), f2 (t)) of degree d and p = (a0 , a1 , a2 ), then   a0 a1 a2 t1 t2  = 0 (7.19) Cφ (p) : det  t0 f0 (t) f1 (t) f2 (t) It follows immediately that deg Cφ (p) = d + 1 unless Cφ (p) = P2 . As we will see later, this happens for special De Jonqui` res transformations. From now on we assume e that this is not the case for any point p. Then Cφ (p) is a curve of degree d + 1. It passes through the base points of φ (because the last row in the determinant is identical zero for such point) and it passes through the fixed points of φ, i.e. points x ∈ dom(φ) such that φ(x) = x (because the last two rows are proportional). Also Cφ (p) contains its center p (because the first two rows are proportional). One more observation is that Cφ (p) = Cφ−1 (p). When p varies in the plane we obtain a net of isologues. If F is the one-dimensional component of the set of fixed points, then F is a fixed component of the net of isologues. Remark 7.1.1. It follows from the definition that the isologue curve C(p) is projectively generated by the pencil of lines though p and the pencil of curves φ−1 ( ). Recall that given two pencils P and P of plane curve of degree d1 and d2 and a projective isomorphism α : P → P , the union of points Q ∩ α(Q), Q ∈ P, is a plane curve C. Assuming that the pencils have no common base points, C is a plane curve of degree d1 + d2 . To see this we take a general line and restrict P and P to it. We obtain two 1 1 linear series gd and gd on . The intersection C ∩ consists of points common to a 1 1 divisor from gd and gd . The number of such points is equal to the intersection of the 1 1 diagonal on P × P with a curve of bi-degree (d, d ), hence it is equal to d + d . It follows from the definition that C contains the base points of the both pencil. Proposition 7.1.9. Assume that φ has no infinitely near base points. Then the multiplicity of a general isologue curve at a base point x of multiplicity m is equal to m. Proof. Let u, v be local affine parameters at x. For each homogeneous polynomial p(t0 , t1 , t2 ) vanishing at x with multiplicity ≥ m let [p]k (u, v) be the degree k homogeneous term in the Taylor expansion at x. If V (f ) is a general member of the

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homoloidal net, then [f ]k = 0 for k < m and [fm ] = 0. Let Bm be the space of binary forms of degree m in variables u, v. Consider the linear map α : C3 → Bm , (a, b, c) → [(bt2 − ct1 )f0 (t) + (ct0 − at2 )f1 (t) + (at1 − bt0 )f2 (t)]m . The map is the composition of the linear map C3 → C3 defined by (a, b, c) → [bt2 − ct1 ]0 , [ct0 − at2 ]0 + [at1 − bt0 ]0 ) and the linear map C3 → Bm defined by (a, b, c) → [af0 + bf1 + cf2 ]m . The rank of the first map is equal to 2, the kernel is generated by (t0 ]0 , [t1 ]0 , [t2 ]0 ). Since no infinitely near point is a base point of the homaloidal net, the rank of the second map is greater or equal than 2. This implies that the map α is not the zero map. Hence there exists an isologue curve of multiplicity equal to m. Remark 7.1.2. Coolidge claims in [54], p. 460, that the assertion is true even in the case of infinitely near points. The following example shows that this is wrong. Consider the quadratic transformation defined by (f0 , f1 , f2 ) = (t0 t2 + t2 , t1 t2 + t2 , t2 ). It has 2 0 2 one base point (0, 1, 0) and two infinitely near points, all of multiplicity 1. In affine coordinates x = t0 /t1 , y = t1 /t1 the equations of the curves are (xy + y 2 , x2 + y, y 2 ). The affine equations of the isologue curves are linear combinations of the minors of the matrix x 1 y xy + y 2 x2 + y y 2 The minors are x3 − y 2 , −x2 , −y 3 . We see that the multiplicity of all isologue curves at the base point (0, 1, 0) are equal to 2. Corollary 7.1.10. Assume that the homaloidal net has no infinitely near base points and the net of isologues has no fixed component. Then the number of fixed points of φ is equal to d + 2, counting with appropriate multiplicities. Proof. Take two general points p, q in the plane. In particular, the line = p, q does not pass through the base points of the homoloidal net and the fixed points. Also p = Cφ (q) and q ∈ Cφ (p). Consider a point x in the intersection Cφ (p) ∩ Cφ (q) which is neither a base point nor a fixed point. Then p, q ∈ x, φ(x), hence x ∈ ∩ Cφ (p) ∩ Cφ (q). Conversely, if x ∈ ∩ Cφ (p) and x = p, then x, φ(x), p are collinear and, since q ∈ , we get that x, φ(x), q are collinear. This implies that x ∈ Cφ (q). This shows that the base points of the pencil of isologue curves Cφ (p), p ∈ , consists of base points of the homaloidal net, fixed points and d points on (counted with multiplicities). The N base points of the homoloidal net contribute i=1 m2 to the intersection. Applying i N Theorem 7.1.8, we obtain that fixed points contribute d + 2 = (d + 1)2 − d − i=1 m2 i to the intersection. The multiplicity of a fixed points is the index of intersection of two general isologue curves. Note that the Cremona transformation from Remark 7.1.2 has no fixed points. Remark 7.1.3. The assumption that φ has no infinitely near points implies that the graph Γ of φ is a nonsingular surface in P2 × P2 isomorphic to the blow-up of the base scheme of the homaloidal net. Let h1 , h2 be the pre-images of the cohomology classes of lines under the projections. They generate the cohomology ring H ∗ (P2 × P2 , Z). Let [Γ] be the cohomology class of Γ and [∆] be the cohomology class of the diagonal

7.1. HOMALOIDAL LINEAR SYSTEMS

203

∆. Write [Γ] = ah2 + bh1 h2 + ch2 . Since the pre-image of a general point under φ is a 1 2 point, we have [Γ] · h2 = 1. Replacing φ with φ−1 , we get [Γ] · h2 = 1. Since a general 2 1 line intersects the pre-image of a general line at d points we get [Γ] · h1 · h2 = d. This gives [Γ] = h2 + dh1 h2 + h2 . (7.20) 1 2 Similarly, we get [∆] = h2 + h1 h2 + h2 . 1 2 This implies that [Γ] · [∆] = d + 2. This confirms the assertion of the previous corollary. In fact, one can use the argument for another proof of the corollary if we assume (that follows from the corollary) that no point in the intersection Γ ∩ ∆ lies on the exceptional curves of the projections. The net of isologue curves without fixed components is a special case of a Laguerre net. It is an irreducible net of plane curves of degree d generated by the curves V (f0 ), V (f1 ), V (f2 ) such that t0 f1 (t) + t1 f2 (t) + t2 f( t) = 0. (7.22) (7.21)

Replacing the first row in the determinant defining an isologue curve with x0 , x1 , x2 we see that the net of isologue curves is a Laguerre net. Take two general curves Cλ = V (λ0 f0 + λ1 f1 + λ2 f2 ) and Cµ = V (µ0 f0 + µ1 f1 + µ2 f2 ) from the net. Let p = [a0 , a1 , a2 ] belong to Cλ ∩ Cµ . Assume that p is not a base point. Then (f0 (a), f1 (a), f2 (a)) is a nontrivial solution of the system of linear equations with the matrix of coefficients equal to   λ0 λ1 λ2 µ0 µ1 µ2  . a0 a1 a2 This implies that the points λ = [λ0 , λ1 , λ2 ], µ = [µ0 , µ1 , µ2 ], a = [a0 , a1 , a2 ] are collinear. Thus all intersection points of Cλ and Cµ besides base points lie on the line. Conversely, suppose a non-base point point a = λ, µ lies on a line λ, µ and belongs to the curve Cλ . Then (f0 (a), f1 (a), f2 (a)) is a non-trivial solution of λ0 t0 + λ1 t1 + λ2 t2 = 0, a0 t0 + a1 t1 + a2 t2 = 0, hence satisfies the third equation µ0 t0 +µ1 t1 +µ2 t2 = 0. This shows that a ∈ Cλ ∩Cµ . Thus we see that the intersection Cλ ∩ Cµ consists of d − 1 non-base points hence the number of base points counting with multiplicities is equal to d2 − d + 1. Now let N be an irreducible net of plane curves of degree d with the property that any two its general members intersect at d − 1 collinear points outside the base locus. Let us see that N is a Laguerre net. We follow the proof from [54], p. 423. Let V (f1 ), V (f2 ) be two general members intersecting at d − 1 points on a line l = 0 not passing through base points. Let pi be the residual point on V (fi ). Choose a general line l1 = 0 passing through p2 and a general line l2 = 0 passing through p1 . Then

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V (l1 f1 ) and V (l2 f2 ) contain the same set of d + 1 points on the line l = 0, hence we can write l1 f1 + cl2 f2 = lf3 (7.23) for some polynomial f3 of degree d and some constant c. For any base point q of the net, we have l1 (q)f1 (q) + cl2 (q)f2 (q) = l(q)f3 (q). Since l(q) = 0 and f1 (q) = f2 (q) = 0, we obtain that f3 (q) = 0. Thus the curve V (f3 ) passes through each base point and hence belongs to the net N . This shows that N has a basis V (f1 ), V (f2 ), V (f3 ) satisfying (7.23). Changing the basis of the net, we may assume that l1 = t0 , cl2 = t1 , −l = t2 and t0 f1 + t1 f2 + t2 f3 = 0 proving that N is a Laguerre net. Example 7.1.1. Take a net of curves of degree 3 with 7 base points. Then it is a Laguerre net since two residual intersection points of any two general members are on a line. One can prove this invoking the Hilbert-Burch Theorem 9.4.6. Applying this theorem we obtain that the homogeneous ideal is generated by the maximal minors of a matrix l1 l2 l3 φ1 φ2 φ3 where li are linear forms and φi are quadratic forms. Since the minors fi , in appropriate order, satisfy the equation l1 f1 − l2 f2 + l3 f3 = 0, we obtain that the net is a Laguerre net.

7.2
7.2.1

First examples
Involutorial quadratic transformations

Take d = 2. We find m2 = 1, i mi = 3. This easily implies m1 = m2 = i m3 = 1, N = 3. The birational transformation of this type is called a quadratic transformation. The homaloidal linear system consists of conics passing through a bubble cycle x1 + x2 + x3 . We have encountered these transformations in section 4.1.5 Assume x1 , x2 , x3 are proper points. They are not collinear, since otherwise all conics have a common line component. Let g be a projective transformation which sends the points x1 , x2 , x3 to the points p1 = [0, 0, 1], p2 = [0, 1, 0], p3 = [1, 0, 0]. Then T ◦ σ −1 is given by the linear system of conics through the points p1 , p2 , p3 . We can choose a basis formed by the conics V (t1 t2 ), V (t0 t2 ), V (t0 t1 ). The corresponding Cremona transformation is given by the formula τ1 : [t0 , t1 , t2 ] → [t1 t2 , t0 t2 , t0 t1 ]. In affine coordinates z1 = t1 /t0 , z2 = t2 /t0 , the transformation is given by 1 1 (x, y) → ( , ). x y (7.25) (7.24)

Thus any quadratic transformation with no infinitely near base points is equal to σ ◦ τ1 ◦ σ for some projective transformations σ, σ . Note that τ1 ◦ τ1 : [t0 , t1 , t2 ] → [t0 t2 t0 t1 , t1 t2 t0 t1 , t1 t2 t0 t2 ) = t0 t1 t2 [t0 , t1 , t2 ] = [t0 , t1 , t2 ].

7.2. FIRST EXAMPLES

205

Thus τ1 is an involution. However, in general, σ ◦ τ1 ◦ σ is not an involution. The transformation τ1 is called the standard quadratic transformation. Assume now that x1 and x2 are proper points and x3 1 x1 . Again, after a linear change of variables, we may assume that x1 = [0, 0, 1], x2 = [1, 0, 0] and x2 corresponds to the tangent direction t0 = 0. The homaloidal linear system consists of conics which pass through x1 , x3 and have a common tangent t0 = 0 at x1 . We can take a basis formed by the conics V (t0 t2 ), V (t0 t1 ), V (t2 ). The corresponding Cre1 mona transformation is given by the formula τ2 : [t0 , t1 , t2 ] → [t2 , t0 t1 , t0 t2 ]. 1 (7.26)

Any quadratic transformation with one infinitely near base point is equal to σ ◦ τ2 ◦ σ for some projective transformations σ, σ . In the affine coordinates as above, the transformation is given by 1 y (x, y) → ( , 2 ). x x (7.27)

Assume now that x3 x2 x1 . By a linear change of variables we may assume that x1 = [0, 0, 1], x2 corresponds to the tangent direction t0 = 0, and x3 lies on the proper transform of the line t2 = 0. The homaloidal linear system consists of conics which pass through x1 and have a common tangent t0 = 0, and after blowing up x1 still intersect at one point. We can take a basis formed by the conics V (t0 t2 − t2 ), V (t2 ), V (t0 t1 ). The corresponding Cremona transformation is given by 1 0 the formula τ3 : [t0 , t1 , t2 ] → [t2 , t0 t1 , t2 − t0 t2 ]. (7.28) 0 1 Any quadratic transformation with one infinitely near base point is equal to σ ◦ t2 ◦ σ for some projective transformations σ, σ . In affine coordinates, the transformation is given by (x, y) → (x, x2 − y). (7.29)

Note that the case x2 1 x1 , x3 1 x1 is not realized since a general member of the linear system is singular at x1 . It is easy to see that the quadratic transformations τi are involutorial. . Proposition 7.2.1. Let φ be an involutorial quadratic Cremona transformation. Then there exists a projective transformation g such that g ◦φ◦g −1 = τi for some i = 1, 2, 3. Proof. We assume that φ has no infinitely near base point and prove that i = 1 in this case. We leave other cases to the reader. Let p1 , p2 , p3 be the base points. Choose a projective transformation g such that g(p1 ) = [1, 0, 0], g(p2 ) = [0, 1, 0], g(p3 ) = [0, 0, 1]. Then φ = g ◦ φ ◦ g −1 is an involution and has the base points [1, 0, 0], [0, 1, 0], [0, 0, 1]. We can choose a basis of the homaloidal net of conics through these points in the form (t1 t2 , t0 t2 , t0 t1 ). This shows that the transformation φ is given by the formula φ (x) = [a1 x1 x2 +b1 x0 x2 +c1 x0 x1 , a2 x1 x2 +b2 x0 x2 +c2 x0 x1 , a3 x1 x2 +b3 x0 x2 +c3 x0 x1 ].

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The image of the line t0 is the point [a1 , a2 , a3 ]. Since φ is an involution, this point must be a base point of φ . Similarly, we obtain that the points [b1 , b2 , b3 ] and [c1 , c2 , c3 ] are base points. Thus we may assume that the transformation φ = σ ◦ τ1 , where σ is the projective transformation which permutes the coordinates. It is directly verified by iteration that σ must be the identity. Example 7.2.1. The first historical example of a Cremona transformation is the inversion map. Recall the inversion transformation from the plane geometry. Given a circle of radius R, a point x ∈ R2 with distance r from the center of the circle is mapped to the point on the same ray at the distance R/r. The following picture illustrates this in the case R = 1.   •  1  r       • r   1 

Figure 7.1: In the affine plane C2 the transformation is given by the formula (x, y) → ( x2 Ry Rx , ). 2 x2 + y 2 +y

In projective coordinates, the transformation is given by the formula (x0 , x1 , x2 ) → (x2 + x2 , Rx1 x0 , Rx2 x0 ). 1 2 Note that the transformation has three base points [1, 0, 0], [0, 1, i], [0, 1, −i]. It is an involution and transforms lines not passing through the base points to conics (circles in the real affine plane). The lines passing though one of the base points are transformed to lines. The lines passing through the origin (1, 0, 0) are invariant under the transformation. The conic x2 + x2 − Rx2 = 0 is the closure of the set of fixed points. 1 2 0 Example 7.2.2. Let C1 and C2 be two conics intersecting at 4 distinct points. For each general point x in the plane let φ(x) be the intersection of the polar lines Px (C1 ) and Px (C2 ). Let us see that this defines an involutorial quadratic transformation with base points equal to the singular points of three reducible conics in the pencil generated by C1 and C2 . It is clear that the transformation φ is given by three quadratic polynomials. Since Px (C1 ) ∩ Px (C2 ) is equal to Px (C) ∩ Px (C ) for any two different members of the pencil, taking C to be a reducible conic and x to be its singular point, we obtain that

7.2. FIRST EXAMPLES

207

φ is not defined at φ. Since the pencil contains three reducible members, we obtain that φ has three base points, hence φ is given by a homaloidal net and hence is a birational map. Obviously, x ∈ Pφ(x) (Ci ) ∩ Pφ(x) (C2 ), hence φ is an involution. Note that fixed point of the transformation are the base points of the pencil of conics.

7.2.2

Symmetric Cremona transformations

Assume that the bubble cycle defining the homoloidal net η consists of points taken with equal multiplicity m. The Cremona transformations defined by such a bubble cycle are called symmetric. Then the necessary conditions are d2 − N m2 = 1, 3d − N m = 3.

Multiplying the second equality by m and subtracting from the first one, we obtain d2 − 3dm = 1 − 3m. This gives (d − 1)(d + 1) = 3m(d − 1). The case d = 1 corresponds to a projective transformation. Assume d > 1. Then we get d = 3m − 1 and hence 3(3m − 1) − N m = 3. Finally, we obtain (9 − N )m = 6, This gives us 4 cases. (i) m = 1, N = 3, d = 2; (ii) m = 2, N = 6, d = 5; (ii) m = 3, N = 7, d = 8; (iii) m = 6, N = 8, d = 17. The first case is obviously realized by a quadratic transformation with 3 fundamental points. The second case is realized by the linear system of plane curves of degree 5 with 6 double points. Take a bubble cycle η = 2x1 + . . . + 2x6 , where the points xi in the bubble space do not lie on a proper transforms of a conic and no three lie on the proper transforms of a line. I claim that the linear system |V | = |OP2 (2) − η| is homaloidal. The space of plane quintics is of dimension 20. The number of conditions for passing through a point with multiplicity ≥ 2 is equal to 3. Thus dim |OP2 (2) − η| ≥ 2. It is easy to see that the linear system does not have fixed components. For example, if the fixed component is a line, it cannot pass through more than 2 points, hence the residual components are quartics with 4 double points, obviously reducible. If the fixed component is a conic, then it passes through at most 5 points, hence the residual components are cubics with at least one double points and passing through the remaining points. It is easy to see that the dimension of such linear system is at most 1. If the fixed component is a cubic, then by the previous analysis we may assume that it is irreducible. Since it has at most one singular point, the residual conics pass through at least 5 points and the dimension of the linear system is equal to zero (or it is empty). Finally, if the fixed component is a quartic, then the residual components are lines passing through 3 points, again a contradiction. d = 3m − 1.

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Applying Bezout’s Theorem, we see that two general members of our linear system intersect at 1 point outside of the base locus, also their genus is equal to 0. Thus the linear system is a homaloidal. Assume that all base points are proper points in the plane. Then the linear system blows down the six conics, each passing through 5 base points. Ahomaloidal cycles of type (iii) is realized by a Geiser involution. We consider an irreducible net N of cubic curves through 7 general points x1 , . . . , x7 in the plane. For any general point x the subpencil of the net which consists of cubics passing through x has the ninth base point. The Geiser transformation assigns this point to x. It is clear that this transformation is involutorial and its base points are x1 , . . . , x7 . To determine its degree, consider the rational map f : P2 − → N ∗ given by the net. We have used this map already in Proposition 6.3.5 from Chapter 6. A pencil in N is a point in N ∗ and its pre-image consists of the base points of the net outside of the base locus of N . Thus f is of degree 2. The restriction of f to a general line is given by the linear series of degree 3 and dimension 2, hence C = f ( ) is a rational cubic curve. The pre-image of a general line in N ∗ is a member of N , i.e. a cubic through the base points. The pre-image of C is a curve of degree 9 passing through the base points with multiplicity 3. It consists of the union of and the curve φ( ). Thus deg φ( ) is equal to 8. Since φ is an involution, φ( ) is a general member of the homaloidal net defining the Geiser involution. A homaloidal cycle of type (iv) is realized by a Bertini involution. We consider a pencil of cubic curves through a general set of 8 points x1 , . . . , x8 . Let q be its ninth base point. For any general point x in the plane let E(p) be the member of the pencil containing x. Let φ(x) be the residual point in the intersection of E(p) with the line x, q. The transformation x → φ(x) is the Bertini involution. If we take q as the origin in the group law on a nonsingular E(p), then φ(x) = −x. Consider the web N of of curves of degree 6 whose general member passes through each point pi with multiplicity 2. The restriction of N to any E(p) is a pencil with 1 1 fixed part 2p1 + . . . + 2p8 and the moving part g2 . One of the members of this g2 is the divisor 2q cut out by 2E(p ), p = p . As we have seen in section 6.3.3 of Chapter 6, the members of this pencil are cut out by lines through the coresidual point on E(p). This point must coincide with the base point of the pencil. Thus members of the pencil are divisors x + φ(x). Now we use that N defines a degree 2 rational map f : P2 − → Q ⊂ P3 , where Q is a singular irreducible quadric in P3 . The image of the point q is equal to the singular point of Q. The restriction of f to a general line is given by the linear system of dimension 3 and degree 3. Its image of a line in Q is a rational curve R of degree 6 intersecting each line on Q at 3 points and not passing through the singular point of Q. It is easy to see that it is a singular curve of arithmetic genus 4 cut out by a cubic hypersurface. Since the pre-image of a hyperplane section under f is a curve from N , the pre-image of R is a curve of degree 18 passing through the base points of N with multiplicities 6. As in the previous case, we see that the pre-image of R is equal to the union of and φ( ). Since φ( ) is a member of the homaloidal linear system defining φ we obtain that the characteristic of φ is equal to (17, 6, 6, 6, 6, 6, 6, 6, 6).

7.2. FIRST EXAMPLES

209

7.2.3

De Jonqui` res transformations e
N N

Assume that there exists a point q in the support of η with multiplicity d − 1. We have d2 − (d − 1)2 −
i=2

= 1,

3d − (d − 1) −
i=2

= 3.

This easily implies

N i=2

mi (mi − 1) = 0, hence N = 2d − 1.

m2 = . . . = mN = 1,

For simplicity of the exposition we assume that the base scheme is reduced, i.e. all base points are proper. The homaloidal system must consist of curves of degree d with singular point q of multiplicity d − 1 (monoidal curves) passing simply through 2d − 2 points x1 , . . . , x2d−2 . The corresponding Cremona transformation is called De Jonqui` res transformation. e Changing the projective coordinates in the source plane, we may assume that q = [0, 0, 1]. Then the equation of a curve from the homaloidal linear system must look like f (t0 , t1 , t2 ) = t2 fd−1 (t0 , t1 ) + fd (t0 , t1 ) = 0. (7.30)

Since a general curve from the homaloidal linear system intersects a line through the points q, xi = xj with degree d + 1 > d, no such line can exist. The base points satisfy the condition • no three points q, xi , xj , 1 < i < j are collinear. Let us see that this condition is sufficient for the existence of the homaloidal net. Counting constants, we see that the linear system curves of degree d − 1 passing through the points x1 , . . . , x2d−2 with point of multiplicity d − 2 at q is non-empty and its expected dimension is equal to zero. If it contains an irreducible curve, Bezout’s Theorem implies that the linear system consists of this curve. Suppose there is an reducible curve C in the linear system. A general line through the point q intersects C at one point p = q. This implies that C = C1 + 1 + . . . + k , where C1 is an irreducible curve of degree d − 1 − k and i are lines passing through q. It follows from the assumption that C1 passes through at least 2d − 2 − k points xi ’s. Let us assume that it does not happen, i.e. • no proper subset of 2d − 2 − k points xi ’s lie on an irreducible curve of degree d − 1 − k with singular point q of multiplicity d − 2 − k One can show that this condition is equivalent to the condition that the inverse of the transformation has no infinitely near base points. Let Γ be the unique irreducible curve of degree d − 1 with singular point of multiplicity d − 2 at q and passing through the points x1 , . . . , x2d−2 . Its equation must be of the form g(t0 , t1 , t2 ) = t2 gd−2 (t0 , t1 ) + gd−1 (t0 , t1 ) = 0, (7.31) where the subscript indicates the degree of the binary form. The union of this curve and the pencil of lines through q is a pencil contained in the homaloidal net. Let V (f ),

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where f as in (7.30), be a curve from the net which is not contained in the pencil. The Cremona transformation φ : P2 − → P2 defined by the homaloidal net is obtained by a choice of a basis in the homaloidal net. Consider the Cremona transformation φ : [t0 , t1 , t2 ] → [t0 g(t0 , t1 , t2 ), t1 g(t0 , t1 , t2 ), f (t0 , t1 , t2 )] (7.32)

defined by the choice of a special basis. Any other transformation with the same homaloidal net is equal to the composition s ◦ φ, where s is a projective transformation. It is easy to see that T = s ◦ φ transforms the pencil of lines through q to the pencil of lines through the point s(q). A De Jonqui` res transformation is called special if e s(q) = q. A special transformation is given by the formula [t0 , t1 , t2 ] → [(a1 t0 + a2 t1 )g, (a3 t0 + a4 t1 )g, (a5 t0 + a6 t1 )g + a7 f ] In affine coordinates x = t1 /t0 , y = t2 /t0 it is given by the formula T : (x, y) → ( ax + b r1 (x)y + r2 (x) , , cx + d r3 (x)y + r4 (x) (7.34) (7.33)

where ri (x) are certain rational functions in x. All such transformations form a subgroup of the Cr(2). It is called a De Jonquir` res subgroup. Conversely, if a Cremona e transformation leaves invariant a pencil of lines, then it can be considered as an automorphism of the field C(x, y) leaving invariant the subfield C(x). It is easy to see that it can be given by a formula (7.34). After homogenizing, we get a formula of type (7.33). It is easy to invert the transformation φ defined by formula (7.32). We find that T −1 is a De Jonqui` res transformation given by the formula e [t0 , t1 , t2 ] → [t0 g (t0 , t1 , t2 ), t1 g (t0 , t1 , t2 ), fd (t0 , t1 , t2 )], where g (t0 , t1 , t2 ) f (t0 , t1 , t2 ) = t2 gd−2 (t0 , t1 ) − fd−1 (t0 , t1 ) = −t2 gd−1 (t0 , t1 ) + fd (t0 , t1 ). (7.35)

Observe that φ−1 is also an De Jonqui´ res transformation. Note that, if fd−1 = −gd−1 , e the transformation is an involution. Note the following properties of a De Jonquir` res transformation. The lines p1 , pi e are blown down to 2d − 2 points q1 , . . . , q2d−2 . The curve Γ is blown down to a point q . If we resolve the map by π : X → P2 , then the exceptional curve π −1 (q) is mapped to the curve Γ of order d − 1 with (d − 2)-multiple point y1 . The exceptional curves π −1 (pi ) are mapped to lines q , yi . It is easy to see that the net of isologues of a De Jonqui`es transformation is defined r unless it is a special De Jonqui`es transformation. Thus it has d + 2 fixed points. Let r us find the locus of fixed points of the special transformation T given by (7.32). [t0 , t1 , t2 ] → [(a1 t0 + a2 t1 )g + b1 f, (a3 t0 + a4 t1 )g + b2 f, (a5 t0 + a6 t1 )g + +a7 f )] (7.36)

7.2. FIRST EXAMPLES They satisfy rank t0 t0 g t1 t1 g t2 f = 1.

211

(7.37)

Since we are excluding the point [0, 0, 1], this condition is equivalent to the equation t2 g − f = t2 gd−2 (t0 , t1 ) + t2 gd−1 (t0 , t1 ) = t2 fd−1 (t0 , t1 ) + fd (t0 , t1 ). 2 (7.38)

The closure of this set is a plane curve X of degree d with a (d − 2)-multiple point at q. It is birationally isomorphic to a hyperelliptic curve of genus g = d − 2. The corresponding double cover f : X → P1 is defined by the projection [t0 , t1 , t2 ] → [t0 , t1 ]. Its branch points are given by the discriminant of the quadratic equation (in the variable t2 ): D = (gd−1 − fd−1 )2 + 4fd gd−2 . We have 2d − 2 = 2g + 2 points as is expected. A space construction of a De Jonqui` res transformation due to Cremona [61]. Cone sider a rational curve R of bi-degree (1, d − 2) on a nonsingular quadric Q in P3 . Let L be a line on Q which intersects R at d − 2 distinct points. For each point x in the space there exists a unique line joining a point on L and on R. In fact, the plane spanned by x and L intersects R at a unique point r outside R ∩ L and the line x, r intersects L a unique point s. Take two general planes Π and Π and consider the following birational transformation φ : Π− → Π . Take a general point p ∈ Π, find the unique line joining a point r ∈ R and a point s ∈ L. It intersects Π at the point φ(p). For a general line in Π the union of lines r, s, r ∈ R, s ∈ L, which intersect is a ruled surface of degree d. Its intersection with Π is a curve of degree d. This shows that the transformation φ is of degree d. It has 2d − 2 simple base points. They are d − 1 points in Π ∩ R and d − 1 points which are common to the line Π ∩ Π and the d − 1 lines joining the point L ∩ Π with the points in the intersection Π ∩ R. Finally the point L ∩ Π is a base point of multiplicity n − 1. Identifying Π and Π” by means of an isomorphism, we obtain a De Jonqui` res transformation. e

7.2.4

De Jonqui` res involutions and hyperelliptic curves e

1 Let C be a hyperelliptic curve of genus g and g2 be its linear system defining a degree 2 1 map to P1 . Consider the linear system |D| = |g2 +a1 +· · ·+ag |, where a1 , . . . , ag ∈ C. We assume that the divisor D1 = a1 + · · · + ag is not contained in the linear system 1 |(g − 2)g2 | or, equivalently, |KC − D| = ∅. By Riemann-Roch, dim |D| = 2, hence the linear system |D| defines a map ϕ : C → P2 . The image of ϕ is a plane curve Hg+2 of degree g + 2 with a g-multiple point q, the image of the divisor D1 . By choosing projective coordinates such that q = [0, 0, 1], we can write Hg+2 by an equation t2 fg (t0 , t1 ) + 2t2 fg+1 (t0 , t1 ) + fg+2 (t0 , t1 ) = 0. (7.39) 2

Let be a general line through q. It intersects Hg+2 at two points a, b not equal to q. For any point x ∈ l let y be the fourth point such that the pairs (a, b) and (x, y) are harmonic conjugate.

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We would like to define a birational map T : P2 − → P2 whose restriction to a general line through q takes a point to its harmonic conjugate. Notice that such map is not defined at the points where a line through q is tangent to Hg+2 . it is also undefined at the point q. Let x1 , . . . , x2g+2 be the tangency points. They correspond to the ramification points of the double map C → P1 . It is a fair guess that the transformation T must be a De Jonqui` res transformation defined by the linear system e |(g + 2) − (g + 1)q − x1 − . . . − x2g+2 | and the curve Hg+2 must be the curve of fixed points. Let us see this. Consider the first polar of Hg+2 with respect to the point q. Its equation is t2 fg (t0 , t1 ) + fg+1 (t0 , t1 ) = 0. We know that it passes through the tangency points x1 , . . . , x2g+2 . Also it follows from the equation that it has a g-multiple point at q. It suggests that the first polar is the curve Γ which was used to define a De Jonqui` res transformation. Thus we take d = g+2 and e gi = fi , in equation (7.31). To show that we get an involutorial transformation, we need to check that the curve V (t2 fg+1 (t0 , t1 ) + fg+2 (t0 , t1 )) belongs to the linear system (7.40). The points xi = [1, ai , bi ] belong to the intersection of curves Γ and Hg+2 . In appropriate coordinate system, we may assume that bi = 0. Plugging fg (1, ai ) = −fg+1 (1, ai )/bi in the equation of Hg+2 , we obtain b2 ( i −fg+1 (1, ai ) ) + 2bi fg+1 (1, ai ) + fg+2 (1, ai ) bi = bi fg+1 (1, ai ) + fg+2 (1, ai ) = 0. Thus the curve given by the equation t2 fg+1 (t0 , t1 ) + fg+2 (t0 , t1 ) = 0 e belongs to the linear system (7.40). So, we can define the De Jonqui` res transformation by the formula t0 t1 t2 = t0 t2 fg (t0 , t1 ) + fg+1 (t0 , t1 ) = t1 t2 fg (t0 , t1 ) + fg+1 (t0 , t1 ) = −t2 fg+1 (t0 , t1 ) − fg+2 (t0 , t1 ). (7.41) (7.40)

This transformation is an involution. It follows from (7.38) that the curve of fixed points is the curve Hg+2 . Its restriction to a line l = V (t1 − tt0 ) is given by the formula t0 t1 t2 = t2 fg (1, t) + t0 fg+1 (1, t) = t t2 fg (1, t) + t0 fg+1 (1, t) = −t2 fg+1 (1, t) − t0 fg+2 (1, t).

7.3. ELEMENTARY TRANSFORMATIONS In affine coordinates t2 /t0 on the line t1 − tt0 = 0, the transformation is x→y= This gives xyfg (1, t) + (x + y)fg+1 (1, t) + fg+2 (1, t) = 0.
2

213

−xfg+1 (1, t) − fg+2 (1, t) . xfg (1, t) + fg+1 (1, t) (7.42)

The pair (x, y) satisfies the quadratic equation z − z(x + y) + xy = 0 and the pair (a, b), where a, b are the points of intersection of the line with Hg+2 satisfies the quadratic equation z 2 fg (1, t) + 2zfg+1 (1, t) + fg+2 (1, t) = 0. It follows from the definition (2.2) of harmonic conjugates that equation (7.42) expresses the condition that the pairs (x, y) and (a, b) are harmonic. Definition 7.3. The Cremona transformation defined by the formula (7.41) is called the De Jonqui` res involution defined by the hyperelliptic curve Hg+2 (7.39). It is denoted e by IHg+2 . Remark 7.2.1. By conjugating the De Jonqui` res involution with a Cremona transfore mation given by the formula (x0 , x1 , x2 ) = (xg+1 , xg x1 , fg (x0 , x1 )x2 + fg+1 (t0 , t1 ), 0 0 we may assume that the hyperelliptic curve (7.39) is given by the equation t2 t2g + fg+2 (t0 , t1 )fg (t0 , t1 ) − fg+1 (t0 , t1 )2 = 0. 2 0 The formula (7.41) simplifies. In affine coordinates it is given by y =− f (x) , y x = x, (7.43)

where f (x) is the dehomogenized polynomial fg+2 (t0 , t1 )fg (t0 , t1 ) − fg+1 (t0 , t1 )2 . For any polynomial f this defines an involutary Cremona transformation which is conjugate to IHg+2 for some g.

7.3
7.3.1

Elementary transformations
Segre-Hirzebruch minimal ruled surfaces

First let us recall the definition of a minimal ruled surface Fn . If n = 0 this is the surface P1 × P1 . If n = 1 it is isomorphic to the blow-up of one point in P2 with the ruling π : F1 → P1 defined by the pencil of lines through the point. If n > 1, we consider the cone in Pn+1 over a Veronese curve vn (P1 ) ⊂ Pn , i.e., we identify Pn−1 with a hyperplane in Pn and consider the union of lines joining a point not on the hyperplane with all points in vn (P1 ). The surface Fn is a minimal resolution of its vertex. The exceptional curve of the resolution is a smooth rational curve En with 2 En = −n. The projection from the vertex of the cone, extends to a morphism p :

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Fn → P1 which defines a ruling (a P1 -bundle). The curve En is its section, called the exceptional section. In the case n = 1, the exceptional curve E1 of the blow-up F1 → P2 is also a section of the corresponding ruling p : F1 → P1 . It is also called the exceptional section. The ruling p : Fn → P1 is a projective vector bundle isomorphic to the projectivization of the vector bundle V(En ), where En = OP1 ⊕ OP1 (n). Recall that for any locally free sheaf E of rank r + 1 over a scheme S one defines the vector bundle V(E) as the scheme Spec(Sym(E)). Locally, when we choose a trivialization E|U ∼ OU = r+1 over an open affine set U ⊂ S, we get
r Sym(E)|U ∼ Sym(OU ) ∼ O(U )[t0 , . . . , tr ] = =

and Spec(E)|U ∼ Ar+1 . A local section U → V(E) is defined by a homomorphism = U Sym(E) → O(U ) of O(U )-algebras, and hence by a linear map E|U → O(U ). Thus the sheaf of local sections of the vector bundle V(E) is isomorphic to the dual sheaf E ∗. The projectivization of V(E) is the scheme P(E) = Proj(Sym(E)). It comes with the natural morphism p : P(E) → S. In the same notation as above,
r+1 P(E)|U ∼ Proj(Sym(OU )) ∼ Proj(O(U )[t0 , . . . , tr ]) ∼ Pr . = = = U

By definition of the projective spectrum, we have an invertible sheaf OP(E) (1). Its r+1 sections over p−1 (U ) are homogeneous elements of degree 1 in Sym(OU ). This gives for any k ≥ 0, p∗ OP(E) (k) ∼ Symk (E). = Note that for any invertible sheaf L over S, we have P(E ⊗ L) ∼ P(E) as schemes, = however the sheaves O(1) are different. For any scheme π : X → S over S a morphism of S-schemes f : X → P(E) is defined by an invertible sheaf L over X and a surjection φ : π ∗ E → L. Then we trivialize P(E) over U , the surjection φ defines r + 1 sections of L|π −1 (U ). These define a local map x → [s0 (x), . . . , sr (x)] from π −1 (U ) to p−1 (U ) = Pr . These U maps are glued together to define a global map. We have L = f ∗ OP(E) (1). In particular, taking X = P(E) and f the identity morphism, we obtain a surjection p∗ E → OP(E) (1). When we push it down, we get the identity map p∗ p∗ E = E → p∗ OP(E) (1). Example 7.3.1. Let us take X = S. Then an S-morphism S → P(E) is a section s : S → P(E). It is defined by an invertible sheaf L on S and a surjection φ : E → L. We have L = s∗ OP(E) (1). Let N = Ker(φ). This is a locally free sheaf of rank r. Example 7.3.2. Take x = Spec(K) to be a point in S, and i : x → S be its inclusion in S. Then an invertible sheaf on a point is the constant sheaf Kx and i∗ E = Ex = E/mx E = Ex is the fibre of the sheaf . The inclusion of x in S is defined by a surjection ∗ Ex → Kx , i.e. by a point in the projective space P(Ex ) (if we prefer to define a projective space as the set of lines). Thus we see that the fibres of the projective bundle P(E) can be identified with the projective spaces P((Ex )∗ ).

7.3. ELEMENTARY TRANSFORMATIONS

215

Lemma 7.3.1. Let s : S → P(E) be a section and N = Ker(E → s∗ OP(E) (1). Let us identify S with s(S). Then N ⊗ L−1 is isomorphic to the conormal sheaf of s(S) in P(E). Proof. Recall (see [134], Proposition 8.12) that for any closed embedding i : Y → X of a S-scheme defined by the ideal sheaf I we have an exact sequence
1 I/I 2 → i∗ Ω1 X/S → ΩY /S → 0,

(7.44)

where the first homomorphism is injective if i is a regular embedding (e.g. X, Y are regular schemes). The sheaf I/I 2 is called the conormal sheaf of Y in X and is ∗ denoted by NY /X . Its dual sheaf is called the normal sheaf of Y in X and is denoted by NY /X . Also recall that the sheaf Ω1n of regular 1-forms on projective space can be defined P by the exact sequence 0 → Ω1n → OPn (−1) → OPn → 0. P It is generalized to any projective bundle
∗ 0 → Ω1 P(E)/S → p E ⊗ OP(E) (−1) → OP(E) → 0.

(7.45)

Here the homomorphism p∗ E ⊗ OP(E) (−1) → OP(E) is equal to the homomorphism p∗ E → OP(E) (1) after twisting by −1. Thus
∗ ∼ Ω1 P(E)/S (1) = Ker(p E → OP(E) (1)).

(7.46)

Applying s∗ to both sides we get ∼ s∗ Ω1 P(E)/S (1) = N . (7.47)

Consider the morphism s : S → P(E) as a morphism of S-schemes. It is equal to the composition of an isomorphism s : S → s(S) and the closed embedding s(S) → P(E) of S-schemes. Since Ω1 s(S)/S = {0}, we get from (7.44)
∗ Ns(S)/P(E) ∼ i∗ Ω1 = P(E)/S .

Applying to both sides s∗ , we obtain from (7.47) s∗ Ns(S)/P(E) ⊗ OP(E) (−1) ∼ N ⊗ L−1 ∼ Ns(S)/P(E) . = = ∗ This proves the assertion. Let us apply this to minimal ruled surfaces Fn . It is known that any locally free sheaf over P1 is isomorphic to the direct sum of invertible sheaves. Suppose E is of rank 2. Then E ∼ OP1 (a) ⊕ OP1 (b) for some integers a, b. Since the projective bundle = P(E) does not change if we tensor E with an invertible sheaf, we may assume that a = 0 and b = n ≥ 0.

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Proposition 7.3.2. Let p : S → P1 be a morphism of a nonsingular surface such that all fibres are isomorphic to P1 . Suppose S has a section E with E 2 = −n for some n ≥ 0, then S ∼ Fn . = Proof. Let f be the divisor class of a fibre of p and s be the divisor class of the section E. For any divisor class d on S such that d · f = a, we obtain (d − as) · f = 0. If d represents an irreducible curve C, this implies that p(C) is a point, and hence C is a fibre. Writing every divisor as a linear combination of irreducible curves, we obtain that any divisor class is equal to af + bs for some integers a, b. Let us write KP(E) = af + bs. By adjunction formula, applied to a fibre and the section s, we get −2 = (af + bs) · f, This gives KS = (−2 − n)f − 2s. Assume n = 0. Consider the linear system |nf + s|. We have (nf + s)2 = n, (nf + s) · ((−2 − n)f − 2s) = −2 − n. By Riemann-Roch, dim |nf + s| ≥ n + 1. The linear system |nf + s| has no base points because it contains the linear system |nf | with no base points. Thus it defines a regular map P(E) → Pn . Since (nf + s) · s = 0, it blows down the section s to a point p. Since(nf + s) · f = a, it maps fibres to lines passing through p. The degree of the image is (nf + s)2 = n. Thus the image of the map is a surface of degree n equal to the union of lines through a point. It must be a cone over the Veronese curve vn (P1 ) if n > 1 and P2 if n = 1. The map is its minimal resolution of singularities. This proves the assertion in this case. Assume n = 0. We leave to the reader to check that the linear system |f + s| maps S isomorphically to a quadric surface in P3 . Corollary 7.3.3. P(OP1 ⊕ OP1 (n)) ∼ Fn . = Proof. The assertion is obvious if E = OP1 ⊕ OP1 . Assume n > 0. Consider the section of P(E) defined by the surjection φ : E = OP1 ⊕ OP1 (n) → L = OP1 , (7.49) (7.48) −2 + n = (af + bs) · s = a − 2nb.

corresponding to the projection to the first factor. Obviously N = Ker(φ) ∼ OP1 (n). = Applying the lemma, we get
∗ Ns(P1 )/P(E) ∼ OP1 (n). =

Now, if C is any curve on a surface X, its ideal sheaf is isomorphic to OX (−C) and hence the conormal sheaf is isomorphic to OX (−C)/OX (−2C). Consider the exact sequence ∗ 0 → OX (−2C) → OX (−C) → NC/X → 0.

7.3. ELEMENTARY TRANSFORMATIONS
∗ Tensor it with OX (C) we obtain that NC/X ⊗ OX (C) ∼ OC . This implies = ∗ NC/X = OX (−C) ⊗ OC , NC/X = OX (C) ⊗ OC .

217

(7.50)

In particular, we see that the degree of the invertible sheaf NC/X on the curve C is equal to the self-intersection C 2 . Thus we obtain that the self-intersection of the section s defined by the surjection (7.49) is equal to −n. It remains to apply the previous proposition.

7.3.2

Elementary transformations

Let p : Fn → P1 be a ruling of Fn (the unique one if n = 0). Let x ∈ Fn and Fx be the fibre of the ruling containing x. If we blow up x, the proper inverse transform of Fx is an exceptional curve of the first kind. We can blow it down to obtain a nonsingular surface S . The projection p induces a morphism p : S → P1 with any fibre isomorphic to P1 . Let En be the exceptional section or any section with self-intersection 0 if n = 0 (such a section is of course equal to a fibre of the second ruling of F0 ). Assume that x ∈ En . The proper inverse transform of En on the blow-up has self-intersection equal to −n, and its image in S has the self-intersection equal to −n + 1. Applying Proposition 7.3.2, we obtain that S ∼ Fn−1 . This defines a birational map = elmx : Fn − → Fn−1 .

Figure 7.2:

Assume that x ∈ En . Then the proper inverse transform of En on the blow-up has self-intersection −n − 1 and its image in S has the self-intersection equal to −n − 1. Applying Proposition 7.3.2, we obtain that S ∼ Fn+1 . This defines a birational map = elmx : Fn − → Fn+1 . A birational map elmx is called an elementary transformation.

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

Remark 7.3.1. Let E be a locally free sheaf over a nonsingular curve B. As we explained in Example 7.3.2, a point x ∈ P(E) is defined by a surjection Ex → Kx , where Kx is the constant sheaf on x. Composing this surjection with the natural surjection E → Ex , we get a surjective morphism of sheaves φx : E → Kx . Its kernel Ker(φx ) is a subsheaf of E which has no torsion. Since the base is a regular one-dimensional scheme, the sheaf Ex is locally free. Thus we have defined an operation on locally free sheaves. It is also called an elementary transformation. Consider the special case when B = P1 and E = OP1 ⊕ OP1 (n). We have an exact sequence 0 → E → OP1 ⊕ OP1 (n) −→ Kx → 0. The point x belongs to the exceptional section En if and only if φx factors through OP1 → Kx . Then E ∼ OP1 (−1) ⊕ OP1 (n) and = P(E ) ∼ P(E (1)) ∼ P(OP1 ⊕ OP1 (n + 1)) ∼ Fn+1 . = = = This agrees with our definition of elmx . If x ∈ En , then φx factors through OP1 (n), and we obtain E ∼ OP1 ⊕ OP1 (n − 1). In this case P(E ) ∼ Fn−1 and again it agrees = = with our definition of elmx . Let x, y ∈ Fn . Assume that x ∈ En , y ∈ En and p(x) = p(y). Then the composition tx,y = elmy ◦ elmx : Fn − → Fn is a birational automorphism of Fn . Here we identify the point y with its image in elmx (Fn ). Similarly we get a birational automorphism ty,x = elmy ◦ elmx of Fn . We can also extend this definition to the case when y 1 x, where y does not correspond to the tangent direction defined by the fibre passing through x or the exceptional section (or any section with self-intersection 0). We blow up x, then y, and then blow down the proper transform of the fibre through x and the proper inverse transform of the exceptional curve blown up from x.
φx

7.3.3

Birational automorphisms of P1 × P1

We will often identify F0 = P1 × P1 with a nonsingular quadric Q in P3 . Let us fix a point x0 ∈ Q. The linear projection px0 : Q \ {x0 } → P2 defines a birational map. Let l1 , l2 be two lines on Q passing through x0 and q1 , q2 be their projections. The inverse map p−1 blows up the points q1 , q2 and blows down the proper transform of the line x0 q1 , q2 . For any birational automorphism T of P2 the composition p−1 ◦ T ◦ px0 is a x0 birational transformation of Q. This defines an isomorphism of groups Φx0 : Bir(P2 ) ∼ Bir(Q), T → p−1 ◦ T ◦ px0 . = x0 Explicitly, choose coordinates in P3 such that Q = V (z0 z3 −z1 z2 ) and x0 = [0, 0, 0, 1]. The inverse map p−1 can be given by the formulas x0 [t0 , t1 , t2 ] → [t2 , t0 t1 , t0 t2 , t1 t2 ]. 0

7.3. ELEMENTARY TRANSFORMATIONS If T is given by the polynomials f0 , f1 , f2 , then Φx0 (T ) is given by the formula [z0 , z1 , z2 , z3 ] → [f0 (z )2 , f0 (z )f1 (z ), f0 (z )f2 (z ), f1 (z )f2 (z )], where fi (z ) = fi (z0 , z1 , z2 ).

219

(7.51)

Remark 7.3.2. Let z1 , . . . , zn ∈ Q be F -points of T different from x0 . Let T −1 (x0 ) be a point if T −1 is defined at x0 or the principal curve of T corresponding to x0 with x0 deleted if it contains it. The Cremona transformation Φx0 (T ) is defined outside the set q1 , q2 , px0 (z1 ), . . . , px0 (zn ), px0 (T −1 (x0 )). Here, we also include the case of infinitely near fundamental points of T . If some of zi ’s lie on a line li or infinitely near to points on li , their image under px0 is considered to be an infinitely near point to qi . Let Aut(Q) ⊂ Bir(Q) be the subgroup of biregular automorphisms of Q. It acts naturally on Pic(Q) = Zf + Zg, where f = [l1 ], g = [l2 ]. The kernel Aut(Q)o of this action is isomorphic to Aut(P1 ) × Aut(P1 ) ∼ PGL2 × PGL2 . The quotient group = is of order 2, and its nontrivial coset can be represented by the automorphism τ of Aut(P1 × P1 ) defined by (a, b) → (b, a). Proposition 7.3.4. Let σ ∈ Aut(Q)o . If σ(x0 ) = x0 , then Φx0 (σ) is a quadratic transformation with fundamental points q1 , q2 , px0 (σ −1 (x0 )). If σ(x0 ) = x0 , then Φx0 (σ) is a projective transformation. Proof. It follows from Remark 7.3.2 that Φx0 (σ) has at most 3 fundamental points if σ(x0 ) = x0 and at most 2 fundamental points if σ(x0 ) = x0 . Since any birational map with less than 3 fundamental points (including infinitely near) is regular, we see that in the second case Φx0 (σ) is a projective automorphism. In the first case, the image of the line q1 , q2 is equal to the point px0 (σ(x0 )). Thus Φx0 (σ) is not projective. Since it has at most 3 fundamental points, it must be a quadratic transformation. Remark 7.3.3. In general, the product of two quadratic transformations is not a quadratic transformations. However in our case all quadratic transformations coming from Aut(Q) have a common pair of fundamental points and hence their product is a quadratic transformation. The subgroup Φx0 (Aut(Q)) of Cr(2) = Bir(P2 ) is an example of a subgroup of the Cremona group Cr(2) which is isomorphic to an algebraic linear group. According to a theorem of Enriques-Fano, any subgroup of Cr(2) which is isomorphic to a linear algebraic group, is contained in a subgroup isomorphic to Aut(Fn ) for some n. There is even a generalization of this result to the group Cr(n) = Bir(Pn ) (see [73]). Instead of minimal ruled surfaces one considers smooth toric varieties of dimension n. Take two points x, y no two on the same fibre of each projection p1 : F0 → P1 , p2 : F0 → P1 . Let x = F1 ∩ F2 , y = F1 ∩ F2 , where F1 , F1 are two fibres of p1 and F2 , F2 are two fibres of p2 . Then tx,y is a birational automorphism of F0 . Proposition 7.3.5. Φx0 (tx,y ) is a product of quadratic transformations. If x0 ∈ {x, y}, then Φx0 (tx,y ) is a quadratic transformation. Otherwise Φx0 (tx,y ) is the product of two quadratic transformation. Proof. Assume first that y is not infinitely near to x. Suppose x0 coincides with one of the points x, y, say x0 = x. It follows from Remark 7.3.2 that Φx0 (T ) is defined

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

outside q1 , q2 , px0 (y). On the other hand, the image of the line q1 , px0 (y) is a point. Here we assume that the projection F0 → P1 is chosen in such a way that its fibres are the proper transforms of lines through q1 under p−1 . Thus Φx0 (T ) is not regular with x0 at most three F -points, hence is a quadratic transformation. If x0 = x, y, we compose tx,y with an automorphism σ of Q such that σ(x0 ) = x. Then Φx0 (tx,y ◦ σ) = Φx0 (tx0 ,σ−1 (y) ) = Φx0 (tx,y ) ◦ Φx0 (σ). By the previous lemma, Φx0 (σ) is a quadratic transformation. By the previous argument, Φx0 (tx0 ,σ−1 (y) ) is a quadratic transformation. Also the inverse of a quadratic transformation is a quadratic transformation. Thus Φx0 (tx,y ) is a product of two quadratic transformations. Now assume that y x. Take any point z = x. Then one can easily checks that tx,y = tz,y ◦ tx,z . Here we view y as an ordinary point on tx,z (F0 ). Proposition 7.3.6. Let T : Fn − → Fm be a birational map. Assume that T commutes with the projections of the minimal ruled surfaces to P1 . Then T is a composition of biregular maps and elementary transformations. Proof. Let (X, π, σ) be a resolution of T . Let p1 : Fn → P1 and p2 : Fm → P1 be the projections. We have φ = p1 ◦ π = p2 ◦ σ : X → P1 . Let a1 , . . . , ak be points in P1 such that Ci = φ−1 (ai ) = π ∗ (p1 (ai )) is a reducible 1 curve. We have π∗ (Ci ) = p−1 (ai ) and σ∗ (Ci ) = p−1 (ai ). Let Ei be the unique com1 2 ponent of Ci which is mapped surjectively to p−1 (ai ) and Ei be the unique component 1 of Ci which is mapped surjectively to p−1 (ai ). Let π be a composition of blow-ups of 2 points x1 , . . . , xN and let f be a composition of blow-ups of points y1 , . . . , yN . The pre-images in X of the maximal points (with respect to the partial order defined by ) are irreducible curves with self-intersection −1. Let E be a component of Ci with E 2 = −1 which is different from Ei , Ei . We can reorder the order of the blow-ups to assume that π(E) = xN and f (E) = yN . Let πN : X → XN −1 be the blow-up xN and fN : X → YN −1 be the blow-up yN . Since πN and fN are given by the same linear system, there exists an isomorphism t : XN −1 ∼ YN −1 . Thus, we can replace = the resolution (X, π, f ) with (XN −1 , π1 ◦ . . . ◦ πN −1 , f1 ◦ . . . ◦ fN −1 ◦ t). Continuing in this way, we may assume that xN and yN are the only maximal points of π and σ such that p1 (xN ) = p2 (yN ) = ai . Let E = π −1 (xN ) and E = f −1 (yN ). Let R = E be a component of φ−1 (ai ) which intersects E. Let x = π(R). Since xN x, and no other points is infinitely near to x, we get R2 = −2. Blowing down E, we get that the image of R has self-intersection −1. Continuing in this way we get two possibilities Ci = Ei + Ei ,
2 Ei = Ei 2 = −1, Ei · Ei = 1, 2 Ei = Ei 2 = −1,

Ci = Ei + R1 + · · · + Rk + Ei ,

7.4. CHARACTERISTIC MATRICES
2 Ri = −1, Ei · R1 = . . . = Ri · Ri+1 = Rk · Ei = 1

221

and all other intersections are equal to zero. In the first case, T = elmxN . In the second case, let g : X → X be the blow-down Ei , let x = π(R1 ∩ Ei ). Then T = T ◦ elmx , where T satisfies the assumption of the proposition. Continuing in this way we write T as the composition of elementary transformations.

7.3.4

De Jonqui` res transformations again e

Let T be a De Jonqui` res transformation of degree d with fundamental points p1 , . . . , e p2d−1 . Consider the pencil of lines through p1 . The restriction of the linear system |d − η| to a general line from this pencil is of degree 1, and hence maps this line to a line. Since each such line intersects X at 2 points different from p1 , the image of is equal to . Thus T leaves any line from the pencil invariant or blows down it to a point. Let us blow up p1 to get a birational map π1 : S1 → P2 . The surface S1 is isomorphic −1 to F1 . Its exceptional section is E1 = π1 (p1 ). The proper transform of the curve Γ ¯ is a nonsingular curve Γ. It intersects E1 at d − 1 points z1 , . . . , z2d−2 corresponding to the branches of Γ at p1 . Let l1 , . . . , ld−1 be the fibres of the projection φ : S1 → P1 ¯ corresponding to the lines p1 , pi , where i = 2, . . . , 2d − 1. The curve C passes through −1 the points pi = π1 (pi ) ∈ li . Let π : X → P2 , f : X → P2 be the resolution of ¯ T obtained by blowing up the cycle η. The map factors through π : X → S which ¯ is the blow-up with center at the points pi . The proper transform of Γ = Γ in X ¯ is an exceptional curve of the first kind. The map f blows down the proper inverse transforms of the fibres li and the curve Γ . If we stop before blowing down Γ we get a surface isomorphic to S1 . Thus T can be viewed also as a birational automorphism of F1 which is the composition of 2d − 2 elementary transformations F1 − → F0 − → F1 − → . . . − → F0 − → F1 . If we take x0 to be the image of l1 under elmp2 , to define an isomorphism Φx0 : ¯ Bir(F0 ) → Bir(P2 ), then we obtain that T = Φx0 (T ), where T is the composition of transformations tpi ,pi+1 ∈ Bir(F0 ), where i = 3, 5, . . . , 2d − 3. Applying Proposition ¯ ¯ 7.3.5, we obtain the following. Theorem 7.3.7. Any De Jonqui` res transformation is a composition of quadratic transe formations.
elmp2 ¯

7.4

Characteristic matrices
. Xe ee }} π } σ ee }} ee ~}} φ P2 • • • • • • •/ P2

Consider a resolution (7.6) of a Cremona transformation φ

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

Obviously, it gives a resolution of the inverse transformation φ−1 . The roles of π and σ are interchanged. Let
1 2 M σ : X = XM −→ XM −1 −→ . . . −→ X1 −→ X0 = P2

σ

σM −1

σ

σ

(7.52)

be the factorization into a sequence of blow-ups similar to the one we had for π. It defines a bubble cycle ξ and the homaloidal net |d h − ξ| defining φ−1 . Let E1 , . . . , EM be the corresponding exceptional configurations. We will always take for X a minimal resolution. It must be isomorphic to the minimal resolution of the graph of φ. Lemma 7.4.1. Let E1 , . . . , EN be the exceptional configurations for π and E1 , . . . , EM be the exceptional configurations for σ. Then N = M. Proof. Let S be a nonsingular projective surface and π : S → S be a blow-up map. Then the Picard group Pic(S ) is generated by the pre-image π ∗ (Pic(S)) and the divisor class [E] of the exceptional curve. Also we know that [E] is orthogonal to any divisor class from π ∗ (Pic(S) and this implies that Pic(S ) = Z[E] ⊕ φ∗ (Pic(S)). In particular, taking S = P2 , we obtain, by induction that
N

Pic(X) = π ∗ (Pic(P2 ))
i=1

[Ei ].

This implies that Pic(X) is a free abelian group of rank N + 1. Replacing π with σ, we obtain that the rank is equal to 1 + M . Thus N = M . Remark 7.4.1. It could happen that all exceptional configurations of π are irreducible (i.e. no infinitely points are used to define π) but some of the exceptional configurations of σ are reducible. This happens in the case of the transformation given in Exercise 7.2. Definition 7.4. An ordered resolution of a Cremona transformation is the diagram (7.6) together with an order of a sequence of the exceptional curves for σ and π. Any ordered resolution of T defines two bases in Pic(X). The first basis is e : e0 = σ ∗ ( ), e1 = [E1 ], . . . , eN = [EN ] The second basis is e : e0 = π ∗ ( ), e1 = [E1 ], . . . , eN = [EN ]. Write e0 = de0 −
i=1 N N

mi ei ,

ej = dj e0 −
i=1

mij ei , j > 0.

7.4. CHARACTERISTIC MATRICES The matrix

223

d  −m1  A= .  . . −mN



d1 −m11 . . . −mN 1

... ... . . .

 dN −m1N   .  .  .

(7.53)

. . . −mN N

is called the characteristic matrix of T with respect to an ordered resolution. It is the matrix of change of basis from e to e . The first column of A is the vector (d, −m1 , . . . , −mN ), where (d; m1 , . . . , mN ) is the characteristic of φ. We write other columns in the form (dj , −m1j , . . . , −mN j ). They describe the exceptional configurations Ej of σ. If dj > 0, then images of Ej in P2 under the map π is called a total principal curves or total P-curves of φ. Its degree is equal to dj . It passes through the base points xk of φ with multiplicities ≥ mjk . The equality takes place if and only if no irreducible component of Ej is mapped to xk under the map X → Xk−1 . The total P -curve could be reducible, its irreducible components are principal curves or P -curves. Note that each Ei contains an irreducible component Ei with self-intersection −1. Under the map π it cannot be mapped to a point. In fact, assume that it is blown down to a point xi , a base point of height 1 of φ. Since the self-intersection increases under blowing-down, Ei = Ej for some j. Let α : X → X be the blowing-down of Ei . Then π ◦ α−1 : X → P2 is a regular map, and π ◦ α−1 : X → P2 is a regular map. Thus X is nonsingular and resolves the indeterminacy points of φ. This contradiction proves the claim and shows that the image of Ei is a principal curve. The characteristic matrix defines a homomorphism of free abelian groups φA : Z1+N → Z1+N . We equip Z1+N with the standard hyperbolic inner product where the norm of a vector v = (a0 , a1 , . . . , aN ) defined by v 2 = a2 − a2 − . . . − a2 . 0 1 N The group Z1+N equipped with this integral quadratic form is is customary denoted by I 1,N . It is an example of a quadratic lattice, a free abelian group equipped with an integral valued quadratic formt. We will discuss quadratic lattices in Chapter 9. Since both bases e and e are orthonormal with respect to the inner product, we obtain that the characteristic matrix is orthogonal, i.e. belongs to the group O(I 1,N ) ⊂ O(1, N ). Recall that the orthogonal group O(1, N ) consists of N + 1 × N + 1 matrices M such that M −1 = JN +1 · t M · JN +1 , (7.54) where JN +1 is the diagonal matrix diag[1, −1, . . . , −1]. In particular, the characteristic matrix A−1 of φ−1 satisfies   d m1 ... mN  −d1 −m11 . . . −mN 1    A−1 = J · At · J =  . . . .  . . .   . . . . . −dN −m1N . . . −mN N

(7.55)

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

Remark 7.4.2. A Cremona map given by polynomials (f0 , f1 , f2 ) can be considered as a regular map C3 → C3 of degree 1. Its divisor of critical points is equal to the jacobian determinant   ∂f ∂f ∂f
0 0 0

 ∂t0 1 jac(f0 , f1 , f2 ) := det  ∂f0 ∂t
∂f2 ∂t0

∂t1 ∂f1 ∂t1 ∂f2 ∂t1

∂t2 ∂f1  ∂t2  ∂f2 ∂t2

=0

Since the degree of the jacobian is equal to 3d−3 we expect that the degree of the union of principal curves is equal to 3d − 3. Using (7.55), we find that (d, d1 , . . . , dN ) is the N characteristic vector of the transformation φ−1 . Hence it satisfies 3d − i=1 di = 3. N So, we confirm that the sum i=1 di of the degrees of principal curves is equal to 3d − 3. Recall that for any rational map φ : X − → X of irreducible algebraic varieties, one can define the image φ(Z ) of an irreducible subvariety of X and the pre-image φ−1 (Z) of an irreducible subvariety of X. We choose an open subset U where φ is defined, and define φ(Z) to be the closure of φ(U ∩ Z ) in X. Similarly, we choose an open subset U of X, where φ−1 is defined and define φ−1 (Z) to be equal to the closure of φ−1 (U ∩ Z) in X. The image of a total principal curve π(Ej ) under the Cremona map is equal to the image of the base point yj of φ−1 in the plane. It is the unique base point yi of φ−1 of height 0 unique such that yj yi . Conversely, any irreducible curve blown down to a point under φ coincides with a P -curve. Proposition 7.4.2. Let φ : P2 − → P2 be a Cremona transformation with F -points x1 , . . . , xN and F -points y1 , . . . , yN of φ−1 . Let A be the characteristic matrix A. Let C be an irreducible curve on P2 of degree n which passes through the points yi with multiplicities ni . Let n be the degree of φ(C) and let ni be the multiplicity of φ(C) at xi . Then the vector v = (n , −n1 , . . . , −nN ) is equal to A−1 · v, where v = (n, −n1 , . . . , −nN ). Proof. Let (X, π, σ) be a minimal resolution of φ. The divisor class of the proper inverse transform π −1 (C) in X is equal to v = ne0 − ni ei . If we rewrite it in terms of the basis (e0 , e1 , . . . , eN ) we obtain that it is equal to v = n e0 − ni ei , where v = Av. Now the image of π −1 (C) under σ coincides with φ(C). By definition of the curves Ei , the curve φ−1 (C) is a curve of degree n passing through the fundamental points yi of φ−1 with multiplicities ni . Let C be a principal curve of φ and ce0 − i=1 ci be the class of π −1 (C). Let v = (c, −c1 , . . . , −cN ). Since φ(C) is a point, A · v = −ej for some j. Example 7.4.1. The following matrix is a characteristic matrix of the standard quadratic transformation τ1 or its degenerations τ2 , τ3 .   2 1 1 1 −1 0 −1 −1  A= (7.56) −1 −1 0 −1 . −1 −1 −1 0
N

7.4. CHARACTERISTIC MATRICES

225

Consider the case φ = τ1 . Since φ = φ−1 , the fundamental points p1 , p2 , p3 of φ and φ−1 are the same and we choose the same order on them. Let E1 , E2 , E3 be the exceptional curves of π and E1 , E2 , E3 be the exceptional curves of σ. We know that T blows down the line ij = pi , pj to the point pk , where {i, j, k} = {1, 2, 3}. The linear system |σ ∗ (e0 )| is |2e0 − e1 − e2 − e3 |, the proper inverse transform of ij in X has the divisor class e0 − ei − ej . Thus ek = e0 − ei − ej . This gives us the matrix (7.56). Now assume that φ = τ2 . The resolution π : X → P2 is the composition of the blow-up of the point p1 = [0, 0, 1], followed by blowing up an infinitely near point p2 corresponding to the tangent direction t0 = 0, and followed by the blowing up the point 2 2 2 p3 = [1, 0, 0]. Let E1 = E1 +E2 , E2 = E2 , E3 = E3 . Here E1 = −2, E2 = E3 = −1. −1 −1 Since τ2 = τ2 , we may assume that the fundamental points of φ are the same p1 , p2 , p3 . It is easy to see that under the map σ, the proper transform of the line t0 = 0 is blown down to the point p3 , the proper transform of the line x1 together with the curve E1 is blown down to the point p1 . Thus e1 = (e0 − e1 − e3 ) + (e1 − e2 ) = e0 −e2 −e3 , e2 = e0 −e1 −e3 , e3 = e0 −e1 −e2 . We get the same matrix. Note that the second column describes the P -curve as a curve from the linear system | − p2 − p3 |. Here p2 is infinitely near point to p1 . By definition, p2 + p3 is not a bubble cycle since p1 is absent. So, | − p2 − p3 | is not representing a curve on P2 . In fact, E1 is reducible and contains a component which is blown down to a point under π. Now assume φ = τ3 . The resolution π is the composition of the blow-up of p = [0, 0, 1], followed by the blow-up the infinitely near point corresponding to the direction t0 = 0, and then followed by the blow-up the intersection point of the proper transform of the line l = V (t0 ) with the exceptional curve of the first blow-up. We have E1 = 2 2 2 E1 +E2 +E3 , E2 = E2 +E3 , E3 = E3 . Here E1 = E2 = −2, E3 = −1. The blowing 2 down σ : X → P consists of blowing down the proper inverse transform of the line equal to e0 − e1 − e2 , followed by the blowing down the image of E2 and then blowing down the image of E1 . We have e1 = (e0 −e1 −e2 )+(e2 −e3 )+(e1 −e2 ) = e0 −e2 −e3 , e2 = (e0 − e1 − e2 ) + (e2 − e3 ) = e0 − e1 − e3 , e3 = e0 − e1 − e2 . Again we get the same matrix. Observe that the canonical class KX is an element of Pic(X) which can be written in both bases as
N n

KX = −3e0 +
i=1

ei = −3e0 +
i=1

ei .

This shows that the matrix A considered as an orthogonal transformation of I 1,N leaves the vector kN = −3e0 + e1 + · · · + eN = (−3, 1, . . . , 1) invariant. Here, ei denotes the unit vector in Z1+N with (i + 1)th coordinate equal to 1 and other coordinates equal to zero. The matrix A defines an orthogonal transformation of (ZkN )⊥ .

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

Lemma 7.4.3. The following vectors form a basis of (ZkN )⊥ . N ≥ 3 : α0 N = 2 : α0 N = 1 : α0 = e0 − e1 − e2 − e3 , αi = ei − ei+1 , i = 1, . . . , N, = e0 − 3e1 , α1 = e1 − e2 = e0 − 3e1 .

Proof. Obviously the vectors αi are orthogonal to the vector kN . Suppose a vector v = N N −1 (a0 , a1 , . . . , aN ) ∈ (ZkN )⊥ . Thus 3a0 + i=1 ai = 0, hence −aN = 3a0 + i=1 ai . Assume N ≥ 3. We can write v = a0 (e0 − e1 − e2 − e3 ) + (a0 + a1 )(e1 − e2 ) + (2a0 + a1 + a2 )(e2 − e3 )
N −1

+
i=3

(3a0 + a1 + · · · + ai )(ei − ei+1 ).

If N = 2, we write v = a0 (e0 − 3e1 ) + (3a0 + a1 )(e1 − e2 ). If N = 1, v = a0 (e0 − 3e1 ). It is easy to compute the matrix QN = (aij ) of the restriction of the inner product to (ZkN )⊥ with respect to the basis (α0 , αN −1 ). We have (−8), If N ≥ 3, we have  −2   1 (aij ) = 1    0 if i = j, if |i − j| = 1 and i, j ≥ 1 if i = 0, j = 3 otherwise. if N = 1, −8 3 , 3 −2 if N = 2.

For N ≥ 3, the matrix A + 2IN is the incidence matrix of the following graph (the Coxeter-Dynkin diagram of type T2,3,N −3 ).

For 3 ≤ N ≤ 8 this is the Coxeter-Dynkin diagram of the root system of the semisimple Lie algebra sl3 ⊕ sl2 of type A2 + A1 if N = 3, of sl5 of type A4 if N = 4, of so10 of type D5 if N = 5 and of the exceptional simple Lie algebra of type EN if N = 6, 7, 8. We have 2 kN = 9 − N.

7.4. CHARACTERISTIC MATRICES

227

This shows that the matrix QN is negative definite if N < 9, semi-negative definite with one-dimensional null-space for N = 9, and of signature (1, N − 1) for N ≥ 10. By a direct computation one checks that its determinant is equal to N − 9. Proposition 7.4.4. Assume N ≤ 8. There are only finitely many posssible characteristic matrices. In particular, there are only finitely many possible characteristics of a homaloidal net with ≤ 8 base points. Proof. Let G be the group of real matrices M ∈ GL(N ) such that t M QN M = QN . Since QN is negative definite for N ≤ 8, the group G is isomorphic to the orthogonal group O(N ). The latter group is a compact Lie group. A characteristic matrix belongs to the subgroup O(QN ) = G ∩ GL(N, Z). Since the latter is discrete, it must be finite. There are further properties of characteristic matrices for which we refer to [1] for the modern proofs. The most important of these is the following Clebsch Theorem. Theorem 7.4.5. Let A be the characteristic matrix. There exists a bijection β : N → N such that for any set I of columns with di = n, i ∈ I, there exists a set of rows J with #I = #J such that µj = β(a), j ∈ J. Note that subtracting two columns (or rows) with the same first entry, and taking the inner product square, we easily get that they differ only at two entries by ±1. This implies a certain symmetry of the matrix if reorder the columns and rows according to the Clebsch Theorem. We refer for the details to [1]

7.4.1

Composition of characteristic matrices

Suppose we have two birational maps φ : P2 − → P2 , φ : P2 − → P2 . We would like to compute the characteristic matrix of the composition φ ◦ φ. Let X X } eee | fff ff σ π }} σ π || | ee ff }} ee || f ~}} ~|| P2 • • • • • • •/ P2 P2 • • • • • • •/ P2 be resolutions of φ and φ . We want to construct a resolution of φ ◦ φ. Let
N 2 1 σ : X = XN −→ XN −1 −→ . . . −→ X1 −→ X0 = P2

(7.57)

σ

σN −1

σ

σ

be a composition of blow-ups of σ and
M 2 1 π : Y = YM −→ YM −1 −→ . . . −→ Y1 −→ Y0 = P2

π

πM −1

π

π

be a composition of blow-ups of π . Let x1 , . . . , xN be the fundamental points of T and y1 , . . . , yN be the fundamental points of φ−1 . Let x1 , . . . , xM be the fundamental points of T and y1 , . . . , yM be the fundamental points of φ −1 . For simplicity we will assume that no infinitely near points occur as fundamental points of φ, φ , φ−1 , φ −1 . We refer to the general case to [1].

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First some of the fundamental points of φ−1 may coincide with fundamental points of φ . This happens when a P-curve of φ contains a fundamental point of φ . Let us assume that yi = xi , i = 1, . . . , r. In this case the fibred product of X → P2 and Y → P2 contains Ei (1) × Ei , i = 1, . . . , r, as irreducible components. When we throw them away, we obtain an ordered resolution (Z, π ◦ g, σ ◦ h) of φ ◦ φ, where g : Z → X is a composition of blow-ups xr+1 , . . . , xM and h : Z → Y is the composition of the blow-ups of yr+1 , . . . , yN . Consider the following bases of Pic(Z). e1 = g ∗ (e0 ), g ∗ (e1 ), . . . , g ∗ (eN ), h∗ (er+1 ), . . . , h∗ (eM ) , e2 = g ∗ (e0 (1) ), g ∗ (e1 (1) ), . . . , g ∗ (eN (1) ), h∗ (er+1 ), . . . , h∗ (eM ) , e2 = h∗ (e0 (2) ), h∗ (e1 (2) ), . . . , h∗ (eM (2) ), g ∗ (er+1 (1) ), . . . , g ∗ (eN
(2) (2) (2) (1) (2) (2) (1) (1) (1) (2) (2) σ π (2)

) ,

e3 = h∗ (e0 ), h∗ (e1 ), . . . , h∗ (eM ), g ∗ (er+1 (1) ), . . . , g ∗ (eN (1) ) , Note that g ∗ (e0 ) = h∗ (e0 (2) ). The transition matrix from basis e1 to basis e2 is ˜ A1 = A1 0M −r,N 0N,M −r , IM −r
(1)

where A1 is the characteristic matrix of φ. The transition matrix from basis e2 to basis e2 is   Ir+1 0r+1,N −r 0r+1,N −r IN −r  . P =  0N −r,r+1 0N −r,N −r 0M −r,r+1 IM −r 0M −r,M −r The transition matrix from basis e2 to basis e3 is ˜ A2 = A2 0N −r,M 0M,N −r , IN −r

where A2 is the characteristic matrix of τ2 . The characteristic matrix of t2 ◦ t1 is equal to the product ˜ ˜ A = A1 ◦ P ◦ A2 . In the special case, when r = N , i.e., all fundamental points of φ−1 are fundamental points of φ , we obtain that the characteristic matrix of φ ◦ φ is equal to A1 0M −N,N 0N,M −N IM −N · A2 . (7.58)

7.4. CHARACTERISTIC MATRICES Example 7.4.2. Assume that r = 0, i.e. no F -point of φ−1 φ . Then the characteristic matrix of φ ◦ φ is equal to  ... ddM dd dd1 d1  −d m1 −d1 m1 . . . −dM m1 −m11   . . . . . . . . .  . . . . .  . −d mN −d1 mN . . . −d mN −m1N M   −m1 m11 ... m1M 0   . . . . . . . . . .  . . . . . −mM m1M ... mM M 0

229 coincide with a F -point of ... ... . . .  dN −mN 1   .  .  .  . . . −mN N   ... 0   . .  . .  . . ... 0

with the obvious meanings of d, mi , mij , d , mj , mij . In particular we see that the degree of the composition is equal to the product of the degrees of the factors. Example 7.4.3. Consider the standard quadratic transformation τ1 with base points −1 x1 , x2 , x3 . Let y1 , y2 , y3 be the base points of τ1 . Let τ be a Cremona transformation with base points x1 , . . . , xM and base points y1 , . . . , yM of τ −1 . Assume that yi = xi for i ≤ r. Let A be the characteristic matrix of the composition τ ◦ τ1 . If r = 3, we obtain from (6.30)   2 1 1 1 0 0 ... 0 −1 0 −1 −1 0 0 . . . 0     d d1 ... dM −1 −1 0 −1 0 0 . . . 0   −m1 −m11 . . . −m1M       A = −1 −1 −1 0 0 0 . . . 0 ·  . . . . . . . .  0 . . . . . . 0 1 0 . . . 0  .  . . . .    . . . . . . . . −mM −mM 1 . . . −mM M . . . . . . .  . . . . . . . . . 0 ... ... 0 0 0 ... 1 Here we choose some order on the points y1 , y2 , y3 which affects the matrix A1 . For example, we obtain that the characteristic of the composition map is equal to (2d−m1 −m2 −m3 ; d−m2 −m3 , d−m1 −m3 , d−m1 −m2 , m4 , . . . , mM ) (7.59) Assume r < 3. We leave to the reader to check that the characteristic of the composition map is equal to (2d − m1 − m2 ; d − m2 , d − m1 , d − m1 − m2 , m3 , . . . , mM ), (2d − m1 ; d, d − m1 , d − m1 , m2 , . . . , mM ), r=1 r = 2, (7.60) (7.61)

It is not difficult to see that the same formulae are true in the case when some of the points yi = xi are infinitely near.

7.4.2

The Weyl groups

Let EN = (ZkN )⊥ ∼ ZN equipped with the quadratic form obtained by the restriction = of the inner product in I 1,N . Assume N ≥ 3. For any vector α ∈ EN with α2 = −2,

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

we define the following element in O(EN ): rα : v → v + (v, α)α. It is called a reflection with respect to α. It leaves the orthogonal complement to α pointwisely fixed, and maps α to −α. Definition 7.5. The subgroup W (EN ) of O(EN ) generated by reflections rαi is called the Weyl group of EN . The following proposition is stated without proof. It follows from the theory of groups generated by reflections. Proposition 7.4.6. The Weyl group W (EN ) is of infinite index in O(EN ) for N > 10. For N ≤ 10, O(EN ) = W (EN ) (τ ), where τ 2 = 1 and τ = 1 if N = 7, 8, τ = −1 if N = 9, 10 and τ is induced by the symmetry of the Coxeter-Dynkin diagram for N = 4, 5, 6. Note that any reflection can be extended to an orthogonal transformation of the lattice I 1,N (use the same formula). The subgroup generated by reflections rαi , i = 0, acts as the permutation group SN of the vectors e1 , . . . , eN . Lemma 7.4.7. (Noether’s inequality) Let v = (d, m1 , . . . , mN ). Assume d > 0, m1 ≥ . . . ≥ mN ≥ 0, m3 = 0, and (i) (ii)
n i=1 N i=1

m2 = d2 + a,; i mi = 3d − 2 + a, m1 + m2 + m3 ≥ d + 1.

where a = ±1. Then Proof. We have m2 + · · · + m2 = d2 − 1, 1 N m1 + · · · + mN = 3d − 3.

Multiplying equality (ii) by m3 and subtracting it from the first one, we get m1 (m1 − m3 ) + m2 (m2 − m3 ) −
i≥4

mi (m3 − mi ) = d2 + a − 3m3 (d −

2−a 3 ).

We can rewrite the previous equality in the form (d− 2−a )(m1 +m2 +m3 −d− 2−a ) = (m1 −m3 )(d− 2−a −m1 )+(m2 −m3 )(d− 2−a −m2 )+ 3 3 3 3 +
i≥4

mi (m3 − mi ) + a + ( 2−a )2 . 3

Since a + b2 ≥ 0 and equality (i) implies mi < d, we obtain that the right-hand side positive. Since m3 = 0 we get d > 1 if a = −1 and d ≥ 1 if a = 1. In any case we have d − 2−a > 0. This implies that m1 + m2 + m3 > d + 2−a > d. 3 3

7.4. CHARACTERISTIC MATRICES Corollary 7.4.8. m1 > d/3.

231

We can apply Noether’s Lemma to the case when v = (d, m1 , . . . , mN ) is the characteristic vector of a homaloidal net or when de0 − mi ei is the class of an exceptional configuration. Definition 7.6. Let v = de0 − i=1 mi ei ∈ I 1,N . We say that v is of homaloidal type (resp. exceptional type) if it satisfies conditions (i) and (ii) from above with a = −1 (resp. a = 1). We say that v is of proper homaloidal (exceptional type) if there exists a Cremona transformation whose characteristic matrix has v as the first (resp. second column). Lemma 7.4.9. Let v = de0 − i=1 mi ei belong to the WN -orbit of e1 . Then d ≥ 0. N Let η = i=1 xi be a bubble cycle and αη : I 1,N → Pic(Yη ) be an isomorphism of lattices defined by choosing some admissible order of η. Then αη (v) is an effective divisor. Proof. The assertion is true for v = e1 . In fact, αη (v) is the divisor class of the first exceptional configuration E1 . Let w = s1 ◦ · · · ◦ s1 ∈ WN be written as the product of simple reflections and v = w(e1 ) = (d , m1 , . . . , mN ). Let us show by induction on the length of w as the minimal product of simple reflections that d ≥ 0. The assertion is obvious if k = 1 since v = e0 − ei − ej or differs from v by a permutation of the mi ’s. Suppose the assertion is true for t = k. Without loss of generality we may assume that sk+1 is the reflection with respect to some root e0 − e1 − e2 − e2 . Then d = 2d − m1 − m2 − m3 < 0 implies 4d2 < (m1 + m2 + m3 )2 ≤ 3(m2 + m2 + m2 ), 1 2 3 2 hence d2 − m2 − m2 − m2 < − d . If d ≥ 2, this contradicts condition (i) of the 1 2 3 3 exceptional type. If d = 1, we check the assertion directly by listing all exceptional types. To prove the second assertion, we use Riemann-Roch theorem applied to the divisor class D = αη (v). We have D2 = −1, D·KYη = −1, hence h0 (D)+h0 (KYη −D) ≥ 1. Assume h0 (KYη − D) > 0. Intersecting KY − D with e0 = αη (e0 ), we obtain a negative number. However the divisor class e0 is nef on Yη . This shows that h0 (D) > 0 and we are done. Lemma 7.4.10. Let v be a proper homaloidal type. Then it belongs to the WN -orbit of the vector e0 . Proof. Let v = de0 − i=1 mi ei be a proper homaloidal type and η be the correN sponding homaloidal bubble cycle. Let w ∈ WN and v = w(v) = d e0 − i=1 mi ei . We have mi = ei · v = w−1 (ei ) · v. Since w−1 (ei ) represents an effective divisor on Yη and v is the characteristic vector of the corresponding homaloidal net, we obtain w−1 (ei ) · v ≥ 0, hence mi ≥ 0. Obviously, mi ≥ 0. We may assume that v = e0 , i.e. the homaloidal net has at least 3 base points. Applying the Noether inequality, we find mi , mj , mk such that mi + mj + mk > d. We choose maximal possible such mi , mj , mk . After reordering, we may assume that m1 ≥ m2 ≥ m3 ≥ . . . ≥ mN . Note that this preserves the
N n N

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CHAPTER 7. PLANAR CREMONA TRANSFORMATIONS

properness of the homaloidal type since the new order on η is still admissible. Applying the reflection s with respect to the vector e0 −e1 −e2 −e3 , we obtain a new homaloidal N type v = d e0 − i=1 mi ei with d = 2d − m1 − m2 − m3 < d. By above, each mi ≥ 0. So, we can apply Noether’s inequality again until we get w ∈ WN such that the number of nonzero coefficients mi of v = w(v) = is at most 2 (i.e. we cannot apply Noether’s inequality anymore). A direct computation shows that such vector must be equal to e0 . Remark 7.4.3. It follows from Proposition 7.4.2 that the composition of a quadratic transformation with base points xi , xj , xk and a Cremona transformation with characteristic vector v has characteristic vector equal to v = s(w ) where s is the reflection with respect to the vector e0 − ei − ej − ek . It is important to understand that the proof does not show that v is obtained in this way, and, in particular, is a proper homaloidal type. If this were true we obtain a proof that any Cremona transformation is the product of quadratic transformations. This is the content of the Noether Theorem below whose proof is different. The original proof of Noether was along these lines, where he wrongly presumed that one can always perform a standard quadratic transformation with base points equal to the highest multiplicities, say m1 , m2 , m3 . The problem here is that the three points x1 , x2 , x3 may not represent the base points of a standard Cremona transformation when one of the following three cases happen for the three base points x1 , x2 , x3 of highest multiplicities (i) x2 x1 , x3 x1 );

(ii) the base ideal in an affine neighborhood of x1 is equal to (u2 , v 3 ). Theorem 7.4.11. Let A be a characteristic matrix of a homaloidal net. Then A belongs to the Weyl group W (EN ). Proof. Let A1 = (d, −m1 , . . . , −mN ) be the first column of A. Applying the previous lemma, we obtain w ∈ WN , identified with a (N + 1) × (N + 1)-matrix such that the w · A1 = e0 . Thus the matrix A = w · A has the first column equal to the vector (1, 0, . . . , 0). Since A is an orthogonal matrix (with respect to the hyperbolic inner product), it must be the block matrix of the unit matrix I1 of size 1 and an orthogonal matrix O of size n − 1. Since O has integer entries it is equal to the product of a permutation matrix P and the diagonal matrix with ±1 at the diagonal. Since A · kN = kN and w · kN = kN , this easily implies that O is the identity matrix IN . Thus w · A = IN +1 and A ∈ WN . Proposition 7.4.12. Every vector v in the WN -orbit of e0 is a proper homoloidal type. Proof. Let v = w(e0 ) for some w ∈ WN . Write w as the composition of simple reflections sk ◦ · · · s1 . Choose an open subset U of (P2 )N such that the point set (x1 , . . . , xN ) ∈ U satisfies the following conditions: (i) xi = xj for i = j; (ii) if s1 = se0 −ei −ej −ek , then xi , xj , xk are not collinear; (iii) let α1 be the involutive quadratic transformation with base points xi , xj , xk and (y1 , . . . , yN ) be the point set with yi = xi , yj = xj , yk = xk and yh = α1 (xh ) for

7.4. CHARACTERISTIC MATRICES

233

h = i, j, k. Then (y1 , . . . , yN ) satisfies (i) and (ii) for s2 . Next do it again by taking s3 and so on. It is easy to see that in this way U is a non-empty Zariski open subset of (P2 )N such that w(e0 ) represents the characteristic vector of a homaloidal net. Corollary 7.4.13. Every vector v in the WN -orbit of e1 can be realized as a proper exceptional type. Proof. Let v = w(e1 ) for some w ∈ WN . Then η be a bubble cycle realizing the homaloidal typew(e0 ) and φ be the corresponding Cremona transformation with characteristic matrix A. Then v is its second column, and hence corresponds to the first exceptional configuration E1 for φ−1 . Remark 7.4.4. It follows from Proposition 7.4.12 (resp. Corollary 7.4.13) that any N vector v = de0 − i=1 mi ei ∈ WN · e0 (resp. v ∈ WN · e1 ) \ {e1 , . . . , eN )) satisfies d > 0, mi ≥ 0. We do not know a purely group theoretical proof of this fact.

7.4.3

Noether-Fano inequality

Let T be a Cremona transformation of P2 defined by a linear system |d − mi xi |. We order the multiplicities m1 ≥ . . . ≥ mN . Obvioulsy, we may assume that x1 ∈ P2 . Assume that one of the points x2 and x3 is not infinitely near to x1 of the first order. Then replacing T with T ◦ Q, where Q is a quadratic transformation such that the fundamental points of Q−1 are equal to x1 , x2 , x3 , we obtain that T ◦ Q is given by a linear system of degree 2d − m1 − m2 − m3 < d (see (7.59)). Continuing in this way we obtain that Qk ◦ · · · Q1 ◦ T is given by a linear system of degree 1, i.e. a projective transformation. Unfortunately, this proof is wrong (as was the original proof of M. Noether). The reason is that at a certain step, maybe even at the first one, a quadratic transformation cannot be applied because of infinitely near points x2 x1 , x3 x1 . We will give a modified version of this proof due to V. Iskovskikh. First we generalize Corollary 7.4.8 to birational maps of any rational surfaces. The same idea works even for higher-dimensional varieties. Let T : S → S be a birational map of surfaces. Let π : X → S, σ : X → S be its resolution. Let |H | be a linear system on X without base points. Let σ ∗ (H ) ∼ π ∗ (H) −
i

mi Ei

for some divisor H on S and exceptional configurations Ei of the map π. Since |H | has no base points, |f ∗ (H )| has no base points. Thus σ ∗ (H ) intersects nonnegatively any curve on X. In particular, σ ∗ (H ) · Ei = −mi Ei2 = mi ≥ 0. (7.62)

This can be interpreted by saying that T −1 (H ) belongs to the linear system |H − η|, where η = mi xi is a bubble cycle on S.

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Theorem 7.4.14. (Noether-Fano inequality) Assume that there exists some integer m0 ≥ 0 such that |H + mKS | = ∅ for m ≥ m0 . For any m ≥ m0 such that |H + mKS | = ∅ there exists i such that mi > m. Moreover, we may assume that xi ∈ S, i.e. ht(xi ) = 0. Proof. We know that KX = π ∗ (KS ) +
i

Ei . Thus we have the equality in Pic(X) (m − mi )Ei .

σ ∗ (H ) + mKX = (π ∗ (H + mKS )) +

Applying σ∗ to the left-hand side we get the divisor class H + mKS which, by assumption cannot be effective. Since |π ∗ (H+mKS )| = ∅, applying σ∗ to the right-hand side, we get the sum of an effective divisor and the image of the divisor i (m−mi )Ei . If all m − mi are nonnegative, it is also an effective divisor, and we get a contradiction. Thus there exists i such that m − mi < 0. The last assertion follows from the fact that mi ≥ mj if xj xi . Note that Example 7.4.4. Assume S = S = P2 , H = d and H = . We have |H + KS | = | − 2 | = ∅. Thus we can take m0 = 1. If d ≥ 3, we have for any 1 ≤ a ≤ d/3, |H + aKS | = |(d − 3a) | = ∅. This gives mi > d/3 for some i. This is Corollary 7.4.8. Example 7.4.5. Let S = Fn and S = Fr be the minimal Segre-Hirzebruch ruled surfaces. Let |H | = |f | be the linear system defined by the ruling on S . It has no base points, so we can write [σ ∗ (H )] = π ∗ (af +bs)− mi ei , where f, s the divisors classes of a fibre and the exceptional section on S, and mi ≥ 0. Here (X, π, σ) is a resolution of T . Thus H = af + bs. Recall that KS = −2s − (2 + n)f, KS = −2s − (2 + r)f . Thus |H + KS | = |(−1 − n)f − 2s| = ∅. We take m0 = 1. We have |af + bs + mKS | = |(a − m(2 + n))f + (b − 2m)s|. Assume that 1<b≤ 2a . 2+n

If m = [b/2], then m ≥ m0 and both coefficients a − m(2 + n) and b − 2m are nonnegative. Thus we can apply Theorem 7.4.14 to find an index i such that mi > m ≥ b/2. In the special case, when n = 0, i.e. S = P1 × P1 , the inequality b ≤ a implies that there exists i such that mi > b/2. Similar argument can be also applied to the case S = P2 , S = Fr . In this case, |H| = |a | and |aH + mKS | = |(a − 3m) |. Thus, we can take m = [a/3], and find i such that mi > a/3.

7.4. CHARACTERISTIC MATRICES

235

7.4.4

Noether’s Reduction Theorem

We shall prove the following. Theorem 7.4.15. The group Bir(F0 ) is generated by biregular automorphisms and a birational automorphism tx,y for some pair of points x, y. Applying Proposition 7.3.5, we obtain the following Noether’s Reduction Theorem. Corollary 7.4.16. Bir(P2 ) is generated by projective automorphisms and quadratic transformations. Now let us prove Theorem 7.4.15. Let T : Fn − → Fm be a birational map. Let Pic(Fn ) = Zf + Zs, Pic(Fm ) = Zf + Zs ,

where f, f are the divisor classes of fibres, and s, s are the divisor classes of exceptional sections. Similar to the case of birational maps of projective plane, we can define an ordered resolution (X, π, σ) of T and its characteristic matrix A. We have two bases in Pic(X) e : π ∗ (f ), s = π ∗ (s), ei = [Ei ], i = 1, . . . , N, e : π ∗ (f ), π ∗ (s ), ei = [Ei ], i = 1, . . . , N. for simplicity of notation, let us identify f, s, f , s with their inverse transforms in Pic(X). As in the case of Cremona transformations, one can define the characteristic matrix of T . For example, its first column (a, b; m1 , . . . , mN ) expresses that the preimage of the linear system |f | on Fm is the linear system |af + bs − η|, where η = mi xi is a bubble cycle over Fn . The first column of the inverse matrix defines preimage of |f | under T −1 (the same as the image under T ). Example 7.4.6. Let T = elmx : Fn − → Fn±1 . Let f, s, e be the classes of a fibre, the exceptional section, and the exceptional curve E on the blow-up π : X → Fn of x. Suppose |s − x| = ∅, i.e., x does not lie on the exceptional divisor. Let f : X → Fn−1 ¯ be the blow-down the proper transform F of F . Then f = f, If |s − x| = ∅, we have f = f, s = s − e, e = f − e. s = s + f − e, e = f − e.

It is easy to see that these transformations are inverse to each other, as it should be. Thus we get f =f , f =f , s=s −e, e=f −e, if |s − x| = ∅, otherwise.

s=s +f −e,

e=f −e,

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Let T : Fn − → Fm . Composing T with elmx , we get a map elmx ◦ T : Fn − → Fm±1 . The image of |f | on Fm±1 is equal to |(a − mi )f + bs − (b − mx )x −
y=x

my y,

if |s − x| = ∅, if |s − x| = ∅,

(7.63)

|(a + b − mi )f + bs − (b − mx )x −
y=x

my y|,

where x is the image of the proper transform of the fibre passing through x. Lemma 7.4.17. Let T : F0 − → F0 be a birational automorphism equal to a composition of elementary transformations. Then T is equal to a composition of biregular automorphisms of F0 and a transformation tx,y for a fixed pair of points x, y, where y is not infinitely near to x. Proof. It follows from Proposition 7.3.5 that tx,y , where y 1 x can be written as a composition of two transformations of type tx ,y with no infinitely near points. Now notice that the transformations tx,y and tx ,y for different pairs of points differ by an automorphism of F0 which sends x to x and y to y . Suppose we have a composition T of elementary transformations. F0 −− → F1 −− → . . . −− → F1 −− → F0 If no F0 occurs among the surfaces Fn here, then T is a composition of even number k of elementary transformations preserving the projections to P1 . It is clear that not all points xi are images of points in F0 lying on the same exceptional section as x1 . Let xi be such a point (maybe infinitely near to x1 ). Then we compose T with txi ,x1 to obtain a birational map T : F0 − → F0 which is a composition of k − 2 elementary transformations. Continuing in this way we write T as a composition of transformations tx ,y . If F1 −− →F0 −− →F1 occurs, then elmxi may be defined with respect to another projection to P1 . Then we write as a composition of the switch automorphism τ and the elementary transformation with respect to the first projection. Then we repeat this if such (F0 , elmj ) occurs again. Let T : F0 − → F0 be a birational transformation. Assume the image of |f | is equal to |af + bs − mx x|. Applying the automorphism τ , if needed, we may assume that b ≤ a. Thus, using Example 7.4.5, we can find a point x with mx > b/2. Composing T with elmx , we obtain that the image of |f | in F1 is the linear system |a f + bs − mx x − y=x my y|, where mx = b − mx < mx . Continuing in this way using formula (7.63), we get a map T : F0 − → Fq such that the image of |f | is the linear system |a f + bs − mx x|, where all mx ≤ b/2. If b = 1, we get all mi = 0. Thus T is everywhere defined and hence q = 0. The assertion of the theorem is verified. Assume b ≥ 2. Since all mi ≤ b/2, we must have, by Example 7.4.5, b> 2a . 2+q
elmxi−1 elmxi elmx1 elmx2 elmxk−1 elmxk

EXERCISES Since the linear system |a f + bs | has no fixed components, we get (a f + bs ) · s = a − bq ≥ 0.

237

Thus q ≤ a /b < (2 + q)/2, and hence q ≤ 1. If q = 0, we get b > a . Applying τ , we will decrease b and will start our algorithm again until we either arrive at the case b = 1, and we are done, or arrive at the case q = 1, and b > 2a /3 and all mx ≤ b/2. Let π : F1 → P2 be the blowing down the exceptional section s to a point q. Then the image of a fibre |f | on F1 under π is equal to | − q|. Hence the image of our linear system in P2 is equal to |a − (a − b)q − p=q mp p|. Obviously, we may assume that a ≥ b, hence the coefficient at q is non-negative. Since b > 2a /3, we get a − b < a /3. By Example 7.4.5, there exists a point p = q such that mp > a /3. Let π(x) = p and E1 be the exceptional curve corresponding to x and S be the exceptional section in F1 . If x ∈ S, the divisor class s − e1 is effective and is represented by the proper inverse transform of S in the blow-up of x. Then (a f + bs − mx e1 −
i>1

mi ei ) · (s − e1 ) ≤ a − b − mx < 0.

This is impossible because the linear system |a f + bs − mx x − y=x y| on F1 has no fixed part. Thus x does not lie on S. If we apply elmx , we arrive at F0 and may assume that the new coefficient at f is equal to a − mx . Since mx > a /3 and a < 3b/2, we see that a − mx < b. Now we apply τ to decrease b. Continuing in this way we obtain that T is equal to a product of elementary transformations and automorphisms of F0 . We finish the proof of Theorem 7.4.15 by applying Lemma 7.4.17. Corollary 7.4.18. The group Cr(2) of Cremona transformations of P2 is generated by projective automorphisms and the standard Cremona transformation t0 . Proof. It is enough to show that the standard quadratic transformations τ2 and τ3 are generated by t0 and projective transformations. Let τ2 has fundamental points at p1 , p2 and an infinitely near point p3 1 p1 . Choose a point q different from p1 , p2 , p3 and not lying on the line p1 , p2 . Let T be a quadratic transformation with F -points at p1 , p2 , q. It is easy to check that T ◦τ2 is a quadratic transformation with F -points (p1 , p2 , τ2 (q)). Composing it with projective automorphisms we get the standard quadratic transformation t1 . Now let us consider the standard quadratic transformation τ3 with F -points p3 p2 p1 . Take a point q which is not on the line in the linear system | − p1 − p2 |. Consider a quadratic transformation T with F -points p1 , p2 , q. It is easy to see that T ◦ τ3 is a quadratic transformation with F -points p1 , p2 , τ3 (q). Composing it with projective transformations we get the standard quadratic transformation τ2 . Then we write τ2 as a composition of τ1 and projective transformations.

Exercises
7.1 Consider a minimal resolution X of the standard quadratic transformation τ1 . Show that τ1 lifts to an automorphism σ of X. Show that σ has 4 fixed points and the orbit space X/(σ) is

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isomorphic to the cubic surface with 4 nodes given by the equation t0 t1 t2 + t0 t1 t3 + t1 t2 t3 + t0 t2 t3 = 0. 7.2 Consider the rational map defined by [t0 , t1 , t2 ] → [t1 t2 (t0 − t2 )(t0 − 2t1 ), t0 t2 (t1 − t2 )(t0 − 2t1 ), t0 t1 (t1 − t2 )(t0 − t2 )]. Show that it is a Cremona transformation and find the Enriques diagram of the corresponding bubble cycle. 7.3 Let C be a plane curve of degree d with a singular point p. Let π : X → P be a sequence of P2 blow-ups which resolves the singularity. Define the bubble cycle η(C, p) = mi xi as follows: x1 = p and m1 = multp C, x2 , . . . , xk are infinitely near points to p of order 1 such that the proper transform C of C under the blow-up at p contains these points, mi = multxi C , i = 2, . . . , k, and so on. (i) Show that the arithmetic genus of the proper transform of C in X is equal to 1 (d−1)(d− 2 P 1 2) − 2 i mi (mi − 1). (ii) Describe the Enriques diagram of η(C, p), where C = V (tb−a ta + tb ), p = [1, 0, 0], and 1 2 0 a ≤ b are positive integers. 7.4 Show that two hyperelliptic plane curves Hg+2 and Hg+2 are birationally isomorphic if and only if there exists a De Jonqui` res transformation which transforms one curve to another. e 7.5 Let Hg+2 be a hyperelliptic curve given by the equation (7.39). Consider the linear system of hyperelliptic curves Hq+2 = V (t2 gq (t0 , t1 ) + 2t2 gq+1 (t0 , t1 ) + gq+2 (t0 , t1 )) such that 2 fg gq+2 − 2fg+1 gq+1 + fg+2 gq = 0. Show that (i) the curves Hq+2 exist if q ≥ (g − 2)/2; (ii) the branch points of Hg+2 belong to Hq+2 and vice versa; (iii) the curve Hq+2 is invariant with respect to the De Jonqui` res involution IHg+2 defined e by the curve Hg+2 and the curve Hg+2 is invariant with respect to the De Jonqui` res e involution IHq+2 defined by the curve Hq+2 ; (iv) the involutions IHg+2 and IHq+2 commute with each other; (v) the fixed locus of the composition Hg+2 ◦ Hq+2 is given by the equation V (fg+q+3 ), where 0 1 fg fg+1 fg+2 fg+q+3 = det @gq gq+1 gq+2 A 1 −t2 t2 . 2 (vi) the De Jonqui` res transformations which leave the curve Hg+2 invariant form a group. e It contains an abelian subgroup of index 2 which consist of transformations which leave Hg+2 pointwisely fixed. 7.6 Show that any De Jonqui` res transformation of finite order leaves a pencil of lines invariant. e

7.7 Consider the linear system La,b = |af + bs| on Fn , where s is the divisor class of the exceptional section, and f is the divisor class of a fibre. Assume a, b ≥ 0. Show that (i) La,b has no fixed part if and only if a ≥ nb; (ii) La,b has no base points if and only if a ≥ nb,

EXERCISES

239

(ii) Assume b = 1 and a ≥ n. Show that the linear system La,1 maps Fn in P2a−n+1 onto a surface Xa,n of degree 2a − n; (iii) Show that the surface Xa,n is isomorphic to the union of lines va (x), va−n (x), where va : P1 → Pa , v2a−n : P1 → Pa−n are the Veronese maps, and Pa and Pa−n are identified with two disjoint projective subspaces of P2a−n+1 . 7.8 Show that the surface Xa,n ⊂ P2a−n+1 contains a nonsingular curve C of genus g = 2a − n + 2 which is embedded in P2a−n+1 by the canonical linear system |KC |. 7.9 Find the automorphism group of the surface Fn . 7.10 Show that a projective automorphism T of P2 which fixes two points is equal to Φx0 (g) for some automorphism of F0 and a point x0 ∈ F0 . 7.11 Compute a characteristic matrix of a De Jonqui` res transformation. e 7.12 Compute a characteristic matrix of symmetric Cremona transformations. 7.13 Let C be an irreducible plane curve of degree d > 1 passing through the points x1 , . . . , xn with multiplicities m1 ≥ . . . ≥ mn . Assume that its proper inverse trans¯ form under the blowing up the points x1 , . . . , xn is a smooth rational curve C with ¯ C 2 = −1. Show that m1 + m2 + m3 > d. 7.14 Let (m, m1 , . . . , mn ) be the characteristic vector of a Cremona transformation. Show that the number of base points with mi > m/3 is less than 9. 7.15 Compute the characteristic matrix of the composition T ◦ T of a De Jonqui` res e transformation T with F -points p1 , p2 , . . . , p2d−1 and characteristic vector (d, d − 1, 1, . . . , 1) and a quadratic transformation T with F -points p1 , p2 , p3 . 7.16 Let σ : A2 → A2 be an automorphism of the affine plane given by a formula (x, y) → (x + P (y), y), where P is a polynomial of degree d in one variable. Consider σ as a Cremona transformation. Compute its characteristic matrix. In the case d = 3 write as a composition of projective transformations and quadratic transformations. 7.17 Show that every Cremona transformation is a composition of the following maps (“links”): (i) the switch involution τ : F0 → F0 ; (ii) the blowup σ : F1 → P2 ; (iii) the inverse σ 1 : P2 − → F1 ; (iv) an elementary transformation elmx : Fq − → Fq±1 . 7.18 Show that any Cremona transformation is a composition of De Jonqui` res transe formations and projective automorphisms. 7.19 Let x0 = (0, 1) × (1, 0), ×(1, 0) ∈ P1 × P1 , y0 = τ (x0 ). Show that ty0 ,x0 is given by the formula [u0 , u1 ] × [v0 , v1 ] → [u0 , u1 ] × [u0 v1 , u2 v0 ]. Check that the composition T = τ ◦ ty0 ,x0 satisfies T 3 = id. 7.20 Let C1 and C2 be two plane conics that span an irreducible pencil of conics. For any point x in the plane let T (x) be the intersection of the polar lines Px (C1 ) and Px (C2 ). Show that T is a quadratic Cremona transformation.

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Historical Notes
A comprehensive history of the theory of Cremona transformations can be found in several sources [145], [228], [54]. Here we give only a brief sketch. The general study of plane Cremona transformations was first initiated by L. Cremona in his two papers [60] and [61] published in 1863 and 1864. However examples of birational transformations were known since the antiquity, for example, the inversion transformation. The example of a quadratic transformation which we presented in Example goes back to Poncelet [194], although the first idea of a general quadratic transformation must be credited to C. MacLaurin [170]. It was generally believed that all birational transformations must be quadratic and much work was done in developing the general theory of quadratic transformations. The first transformation of arbitrary e degree was constructed in 1859 by E. De Jonqui` res in [70], the De Jonqui` res transe formations. His memoir remained unpublished until 1885 although an abstract was published in 1864 [69]. In his first memoir [60] Cremona gives a construction of a general De Jonqui` res transformation without reference to De Jonqui` res. We reproe e duced his construction in section 7.2.3. Cremona gives the credit to De Jonqui` res in e his second paper. Symmetric transformations of order 5 were first studied by M. Sturm [234], of order 8 by C. Geiser [121], and of order 17 much later by E. Bertini [17]. In his second paper Cremona lays foundation of the general theory of plane birational transformations. He introduces the notion of fundamental points and principal curves, establishes the equalities (7.1.7), proves that the numbers of base points of the transformation and its inverse coincide, proves that principal curves are rational and computes all possible characteristic vectors up to degree 10. The notion of a homaloidal linear system was introduced by Cremona later, first for space transformations in [62] and then for plane transformations in [63]. The word homaloid means flat and was used by J. Sylvester to mean a linear subspace of a projective space. More generally it was applied by A. Cayley to rational curves and surfaces. Cremona also introduces the net of isologues and proves that the number of fixed points of a general transformation of degree d is equal to d + 2. In the special case of De Jonqui` re transformations this e was also done by De Jonqui` re in [70]. The notion of isologue curves belongs to him e as well as the formula for the number of fixed points. The first major result in the theory of plane Cremona transformations after Cremona’s work was Noether’s Theorem. The statement of the theorem was guessed by W. Clifford in 1869 [48]. The original proof of M. Noether in [185] based on Noether’s inequality contained a gap which we explained in Remark 7.4.3. Independently, J. Rosanes found the same proof and made the same mistake [202] . In [186] Noether tried to correct his mistake, taking into account the presence of infinitely near base points of highest multiplicities where one cannot apply a quadratic transformation. He took into account the case of infinitely near points with different tangent direction but overlooked the cuspidal case. The theorem was accepted for thirty years until in 1901 C. Segre pointed out that the cuspidal case was overlooked [220]. In the same year, G. Castelnuovo [31] gave a complete proof along the same lines as used in this chapter. In 1916 J. Alexander [3] raised objections to Castelnuovo’s proof and gives a proof without using De Jonqui` res transformations [3]. This seems to be a still accepted proof. It e is reproduced, for example, in [2].

Historical Notes

241

The characteristic matrices of Cremona transformation were used by S. Kantor [153] and later by P. Du Val [91]. The latter clearly understood the connection to reflection groups. The description of proper homaliodal and exceptional types as obits of the Weyl groups were essentially known to H. Hudson. There are numerous modern treatment of this started from M. Nagata [182] and culminated in the monography of M. Alberich-Carramiana [1]. A modern account of Clebsch’s theorem and its history can be also found there. We intentionally omitted the discussion of finite subgroups of the Cremona group Cr(2), the modern account of this classification and the history can be found in [88].

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Chapter 8

Del Pezzo surfaces
8.1
8.1.1

First properties
Varieties of minimal degree

Recall that a subvariety X ⊂ Pn is called nondegenerate if it is not contained in a proper linear subspace. Let d = deg(X). We have the following well-known (i.e., can be found in modern text-books, e.g. [129], [132]) result. Theorem 8.1.1. Let X be an irreducible nondegenerate subvariety of Pn of dimension k and degree d. Then d ≥ n − k + 1, and the equality holds only in one of the following cases: (i) X is an irreducible quadric hypersurface; (ii) a Veronese surface v2 (P2 ) in P5 (iii) a cone over a Veronese surface v2 (P2 ) in P5 ; (iv) a rational normal scroll. Recall that a rational normal scroll is defined as follows. Choose k disjoint linear subspaces L1 , . . . , Lk in Pn which span the space. Let ai = dim Li . We have k 1 i=1 ai = n − k + 1. Consider Veronese maps vai : P → Li and define Xa1 ,...,ak ;n to be the union of linear subspaces spanned by the points va1 (x), . . . , vak (x), where x ∈ P1 . It is clear that dim Xa1 ,...,ak ;n = k and it is easy to see that deg Xa1 ,...,ak ;n = a1 + · · · + ak . Corollary 8.1.2. Let S be an irreducible nondegenerate surface in Pn of degree d. Then d ≥ n − 1 and the equality holds only in one of the following cases: (i) X is a nonsingular quadric in P3 ; (ii) X is an irreducible quadric cone in P3 ; (ii) a Veronese surface v2 (P2 ) in P5 ; 243

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(iii) a rational normal scroll Xa,n ⊂ P2a−n+1 . Del Pezzo surfaces come next. Definition 8.1. A Del Pezzo surface is a nonsingular surface with ample −KS . A weak Del Pezzo surface is a nonsingular surface with −KS nef and big. Recall that a divisor D is called nef if for any irreducible curve C the intersection number C · D is non-negative. It is called big if D2 > 0. Note that if we require C · D > 0 instead of C · D ≥ 0, then D is an ample divisor. This follows from the Moishezon-Nakai criterion of ampleness .

8.1.2

A blow-up model

Lemma 8.1.3. Let S be a weak Del Pezzo surface. Then, any irreducible curve C on S with negative self-intersection is a smooth rational curve with C 2 = −1 or −2. Proof. By adjunction C 2 + C · KS = deg ωC = 2 dim H 1 (C, OC ) − 2 By definition of a weak Del Pezzo surface, we have C · KS ≤ 0. Thus 0 > C 2 > −2 and H 1 (C, OC ) = 0. It is easy to show that the latter equality implies that C ∼ P1 = (the genus of the normalization of an irreducible curve is less or equal to the arithmetic genus defined as dim H 1 (C, OC ) and the difference is positive if the curve is singular). We will call a smooth rational curve with negative self-intersection −n a (−n)curve. Lemma 8.1.4. Let S be a weak Del Pezzo surface. Then H i (S, OS ) = H 2 (S, OS ) = 0. Proof. We write 0 = −KS + KS and apply the following Ramanujam’s Vanishing theorem([163], vol. I, Theorem 4.3.1): for any nef and big divisor D on a nonsingular projective variety X H i (X, OX (KX + D)) = 0, i > 0. Theorem 8.1.5. Let S be a weak Del Pezzo surface. Then, either X ∼ Fn , n = 0, 2, = or X is obtained from P2 by blowing up N ≤ 8 points in the bubble space. Proof. Since KS is not nef, a minimal model for S is either a minimal ruled surface V (over some base curve B) or P2 . Since H 1 (S, OS ) = 0, we must have B ∼ P1 = (use that the projection p : V → B satisfies p∗ OV ∼ OB and this defines a canonical = injective map H 1 (B, OB ) → H 1 (V, OV )). Thus V = Fn or P2 . Assume V = Fn . If n > 2, the exceptional section of V has self-intersection r < −2. Its proper inverse transform on S has self-intersection ≤ r. This contradicts Lemma 8.1.3. Thus n ≤ 2.

8.1. FIRST PROPERTIES

245

If n = 1, then composing the map S → F1 with p, we get a birational morphism S → P2 . Assume X ∼ Fn , where n = 0, 2. Then the birational morphism f : = X → Fn is equal to the composition of φ : S → V and a blow-up b : V → Fn of a point p ∈ Fn . Assume n = 0, and let 1 , 2 be two lines on F0 containing p. Let V → P2 be the blow-down of the proper transforms of the lines. Then the composition S → V → P2 is a birational morphism to P2 . Assume n = 2. The point p does not belong to the exceptional section since otherwise its proper inverse transform in S has self-intersection < −2. Let be the fibre of p : F2 → P1 which passes through p. Then elmp maps F2 to F1 and hence blowing down the proper inverse transform of defines a birational morphism S → V → F1 . Composing it with the birational morphism F1 → P2 , we get a birational morphism π : S → P2 . The last assertion follows from the known behavior of the canonical class of S under a blow-up. If π : S → P2 is a birational morphism which is a composition of N blow-ups, then 2 2 KS = KP2 − N = 9 − N. (8.1)
2 By definition, KS > 0, so N < 9.

2 Definition 8.2. The number d = KS is called the degree of a weak Del Pezzo surface.

Lemma 8.1.6. Let X be a nonsingular projective surface with H 1 (X, OX ) = 0. Let C be an irreducible curve on X such that | − KX − C| = ∅ and C ∈ | − KX |. Then C ∼ P1 . = Proof. We have −KX ∼ C + D for some nonzero effective divisor D, and hence KX + C ∼ −D ∼ 0. This shows that |KX + C| = ∅. By Riemann-Roch, 0 = dim H 0 (X, OX (KX + C)) = 1 ((KX + C)2 − (KX + C) · KX ) + 1 2
1 − dim H 1 (X, OX ) + dim H 2 (X, OX ) ≥ 1 + 2 (C 2 + KX · C) = dim H 1 (C, OC ).

This H 1 (C, OC ) = 0, and as we noted earlier, this implies that C ∼ P1 . = Proposition 8.1.7. Let S be a weak Del Pezzo surface. ¯ ¯ (i) Let f : S → S be a blowing down a (−1)-curve E. Then S is a weak Del Pezzo surface. (ii) Let π : S → S be the blowing-up with center at a point x not lying on any 2 (−2)-curve. Assume KS > 1. Then S is a weak Del Pezzo surface. ¯ Proof. (i) We have KS = f ∗ (KS ) + E, and hence, for any curve C on S, we have ¯ KS · C = f ∗ (KS ) · f ∗ (C) = (KS − E) · f ∗ (C) = KS · f ∗ (C) ≤ 0. ¯ ¯
2 2 ¯ Also KS = KS + 1 > 0. Thus S is a weak Del Pezzo surface. ¯ 2 2 2 (ii) Since KS > 2, we have KS = KS − 1 > 0. By Riemann-Roch, 2 1 dim | − KS | ≥ 2 ((−KS )2 − (−KS · KS )) = KS ≥ 0.

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CHAPTER 8. DEL PEZZO SURFACES

Thus | − KS | = ∅, and hence, any irreducible curve C with −KS · C < 0 must be a proper component of some divisor from | − KS | (it cannot be linearly equivalent to −KS because (−KS )2 > 0). Let E = π −1 (x). We have −KS · E = 1 > 0. So we ¯ may assume that C = E. Let C = f (C). We have ¯ ¯ −KS · C = π ∗ (−KS ) · C − E · C = −KS · C − multx (C). ¯ Since f∗ (KS ) = KS and C = E, the curve C is a proper irreducible component of ¯= ¯ some divisor from | − KS |. By Lemma 8.1.6, C ∼ P1 . Thus multx C ≤ 1 and hence ¯ − 1. This gives −KS · C = 0 and x ∈ C and hence C is a ¯ ¯ ¯ 0 > −KS · C ≥ −KS · C (−2)-curve. Since x does not lie on any (−2)-curve we get a contradiction. Definition 8.3. A blowing down structure on a weak Del Pezzo surface S is a composition of birational morphisms
N 2 1 π : S = SN −→ SN −1 −→ . . . −→ S1 −→ P2 ,

π

πN −1

π

π

where each π : Si → Si−1 is the blow-up a point xi in the bubble space of P2 Recall from section 7.4 that a blowing-down structure of a weak Del Pezzo surface defines a basis (e0 , e1 , . . . , eN ) in Pic(S), where e0 is the class of the full pre-image of a line and ei is the class of the exceptional configurations Ei defined by the point xi . We call it geometric basis. A blowing-down structure defines an isomorphism of free abelian groups φ : ZN +1 → Pic(S) such that φ(kN ) = KS , where kN = −3e0 +e1 +. . .+eN . We call such an isomorphism a geometric marking. Definition 8.4. A pair (S, φ), where S is a weak Del Pezzo surface and φ is a marking (resp. geometric marking) φ : ZN +1 → Pic(S) is called a marked weak Del Pezzo surface (resp. geometrically marked weak Del Pezzo surface). Corollary 8.1.8. Let η = i=1 xi be a bubble cycle on P2 and Sη be its blow-up. Then Sη is a weak Del Pezzo surface if and only if (i) r ≤ 8; (ii) the Enriques diagram of η is the disjoint union of chains; (iii) |OP2 (1) − η | = ∅ for any η ⊂ η consisting of four points; (iv) |OP2 (2) − η | = ∅ for any η ⊂ η consisting of 7 points. Proof. The necessity of condition (i) is clear. We know that S does not contain curves with self-intersection < −2. In particular, any exceptional cycle Ei of the birational morphism π : S → P2 contains only smooth rational curves E with E 2 = −1 or −2. This easily implies that the bubble points corresponding to each exceptional configuration Ei represent a totally ordered chain. This checks condition (ii). Suppose (iii) does not hold. Let D be an effective divisor from the linear system |OP2 (1) − η |. We can change the admissible order on η to assume that η = x1 +
r

8.2. THE EN -LATTICE

247

x2 + x3 + x4 . Then the divisor class of the proper transform of D in Yη is equal to e0 − e1 − e2 − e3 − e4 − i≥4 mi ei . Its self-intersection is obviously ≤ −3. Suppose (iv) does not hold. Let D ∈ |OP2 (2) − η |. Arguing as above we find that the divisor class of the proper transform of D is equal to If D is reducible, one of its 7 components proper inverse transform on Sη is equal to the proper to 2e0 − i=1 ei − i≥7 mi ei . Its self-intersection is again ≤ −3. −1 Let us prove the sufficiency. Let EN = πN (xN ) be the last exceptional configuration of the blow-down Yη → P2 . It is an irreducible (−1)-curve. Obviously η = η − xN satisfies conditions (i)-(iv). By induction, we may assume that S = Sη is a weak Del Pezzo surface. Applying Proposition 8.1.7, we have to show that xN does not lie on any (−2)-curve on S . Condition (ii) implies that it does not lie on any irreducible component of the exceptional configurations Ei , i = N . We will show in the next section that any (−2)-curve on a week Del Pezzo surface S of degree ≤ 7 is either blown down to a point under the canonical map Sη → P2 or equal to the proper inverse transform of a line through 3 points, or a conic through 5 points. If xN lies on the proper inverse transform of such a line (resp. a conic), then condition (iii) (resp. (iv)) is not satisfied. This proves the assertion. A set bubble points satisfying conditions (i)-(iv) is called a set of points in almost general position. We say that the points are in general position if the following holds. (i) all points are proper points; (ii) no three points are on a line; (iii) no 6 points on a conic; (iv) no cubic passes through the points with one of the point being a singular point. Proposition 8.1.9. The blow up of N ≤ 8 points in P2 is a Del Pezzo surface if and only if the points are in general position.

8.2
8.2.1

The EN -lattice
Lattices

A (quadratic) lattice is a free abelian group M ∼ Zr equipped with a symmetric bilin= ear form M ×M → Z. A relevant example of a lattice is the second cohomology group modulo torsion of a compact 4-manifold (e.g. a nonsingular projective surface) with respect to the cup-product. Another relevant example is the Picard group modulo numerical equivalence of a nonsingular projective surface equipped with the intersection pairing. The values of the symmetric bilinear form will be often denoted by (x, y) or x · y. We write x2 = (x, x). The map x → x2 is an integral valued quadratic form on M . Conversely, such a quadratic form q : M → Z defines a symmetric bilinear form by the formula (x, y) = q(x + y) − q(x) − q(y). Note that x2 = 2q(x).

248 Let M ∗ = HomZ (M, Z) and ιM : M → M ∗ ,

CHAPTER 8. DEL PEZZO SURFACES

ιM (x)(y) = x · y.

We say that M is nondegenerate if the homomorphism ιM is injective. In this case the group Disc(M ) = M ∗ /ιM (M ) is a finite abelian group. It is called the discriminant group of M . If we choose a basis to represent the symmetric bilinear form by a matrix A, then the order of Disc(M ) is equal to | det(A)|. The number disc(M ) = det(A) is called the discriminant of M . A different choice of a basis changes A to t CAC for some C ∈ GL(n, Z), so it does not change det(A). A lattice is called unimodular if |disc(M )| = 1 Tensoring M with reals, we get a real symmetric bilinear form on MR ∼ Rr . We = can identify M with an abelian subgroup of the inner product space Rr generated by a basis in Rr . The Sylvester signature (t+ , t− , t0 ) of the inner product space MR is called the signature of M . We write (t+ , t− ) if t0 = 0. For example, the signature of H 2 (X, Z)/Tors ∼ Zb2 for a nonsingular projective surface X is equal to (2pg + = 1, b2 − 2pg − 1), where pg = dim H 0 (X, OX (KX )). The signature on Num(X) = Pic(X)/num ∼ Zρ is equal to (1, ρ − 1) (this is called the Hodge Index Theorem, see = [134], Chap. V, Thm. 1.9). Let N ⊂ M be a subgroup of M . The restriction of the bilinear form to N defines a structure of a lattice on N . We say that N together with this form is a sublattice of M . We say that N is of finite index m if M/N is a finite group of order m. Let N ⊥ = {x ∈ M : x · y = 0, ∀y ∈ N }. Note that N ⊂ (N ⊥ )⊥ and the equality takes place if and only if N is a primitive sublattice (i.e. M/N is torsion-free). We will need the following lemmas. Lemma 8.2.1. Let M be a nondegenerate lattice and N be its nondegenerate sublattice of finite index m. Then |disc(N )| = m2 |disc(M )|. Proof. Since N is of finite index in M the restriction homomorphism M ∗ → N ∗ is injective. We will identify M ∗ with its image in N ∗ . We will also identify M with its image ιM (M ) in M ∗ . Consider the chain of subgroups N ⊂ M ⊂ M ∗ ⊂ N ∗. Choose a basis in M, a basis in N, and the dual bases in M ∗ and N ∗ . The inclusion homomorphism N → M is given by a matrix A and the inclusion N ∗ → M ∗ is given by its transpose t A. The order m of the quotient M/N is equal to | det(A)|. The order of N ∗ /M ∗ is equal to | det(t A)|. They are equal. Now the chain from above has the first and the last quotient of order equal to m and the middle quotient is of order |disc(M )|. The total quotient N ∗ /N is of order |disc(N )|. The assertion follows.

8.2. THE EN -LATTICE

249

Lemma 8.2.2. Let M be a unimodular lattice and N be its nondegenerate primitive sublattice. Then |disc(N ⊥ )| = |disc(N )|. Proof. Consider the restriction homomorphism r : M → N ∗ , where we identify M with M ∗ by means of ιM . Its kernel is equal to N ⊥ . Composing r with the projection N ∗ /ιN (N ) we obtain an injective homomorphism M/(N + N ⊥ ) → N ∗ /ιN (N ). Notice that N ⊥ ∩ N = {0} because N is a nondegenerate sublattice. Thus N ⊥ + N = N ⊥ ⊕ N is of finite index i in M . Also the sum is orthogonal, so that the matrix representing the symmetric bilinear form on N ⊕ N ⊥ can be chosen to be a block matrix. This shows that disc(N ⊕ N ⊥ ) = disc(N )disc(N ⊥ ). Applying Lemma 8.2.1, we get #(M/N + N ⊥ ) = |disc(N ⊥ )||disc(N )| ≤ #(N ∗ /N ) = |disc(N )|.

This gives |disc(N ⊥ )| ≤ |disc(N )|. Since N = (N ⊥ )⊥ , exchanging the roles of N and N ⊥ , we get the opposite inequality. Lemma 8.2.3. Let N be a nondegenerate sublattice of a unimodular lattice M . Then ιM (N ⊥ ) = Ann(N ) := Ker(r : M ∗ → N ∗ ) ∼ (M/N )∗ . = Proof. Under the isomorphism ιM : M → M ∗ the image of N ⊥ is equal to Ann(N ). Since the functor HomZ (?, Z) is left exact, applying it to the exact sequence 0 → N → M → M/N → 0, we obtain an isomorphism Ann(N ) ∼ (M/N )∗ . = A morphism of lattices σ : M → N is a homomorphism of abelian groups preserving the bilinear forms. If M is a nondegenerate lattice, then σ is necessary injective. We say in this case that σ is an embedding of lattices. An embedding is called primitive if its image is a primitive sublattice. An invertible morphism of lattices is called an isometry. The group of isometries of a lattice M to itself is denoted by O(M ) and is called the orthogonal group of M . Let MQ = M ⊗ Q ∼ Qn with the symmetric bilinear form of M extended to = a symmetric Q-valued bilinear form on MQ . The group M ∗ can be identified with the subgroup of MQ consisting of vectors v such that (v, m) ∈ Z for any m ∈ M . Suppose that M is nondegenerate lattice. The finite group Disc(M ) can be equipped with a quadratic form defined by q(¯) = (x, x) x mod Z,

where x denotes a coset x + ιM (M ). If M is an even lattice, i.e. m2 ∈ 2Z for all ¯ m ∈ M , then we take values modulo 2Z. The group of automorphism of Disc(M ) leaving the quadratic form invariant is denoted by O(Disc(M )). The proof of the next lemma can be found in [184].

250

CHAPTER 8. DEL PEZZO SURFACES

Lemma 8.2.4. Let M ⊂ N be a sublattice of finite index. Then the inclusion M ⊂ N ⊂ N ∗ ⊂ M ∗ defines the subgroup N/M in Disc(M ) = M ∗ /M such that the restriction of the quadratic form of Disc(M ) to it is equal to zero. Conversely, any such subgroup defines a lattice N containing M as a sublattice of finite index. The group O(M ) acts naturally on the dual group M ∗ preserving its bilinear form and leaving the subgroup ιM (M ) invariant. This defines a homomorphism of groups αM : O(M ) → O(Disc(M )). Lemma 8.2.5. Let N be a primitive sublattice in a nondegenerate lattice M . Then an isometry σ ∈ O(N ) extends to an isometry of M acting identically on N ⊥ if and only if σ ∈ Ker(αN ).

8.2.2

The EN -lattice

Let I 1,N = ZN +1 equipped with the symmetric bilinear form defined by the diagonal matrix diag(1, −1, . . . , −1) with respect to the standard unit basis e0 = (1, 0, . . . , 0), e1 = (0, 1, 0, . . . , 0), . . . , eN = (0, . . . , 0, 1) of ZN +1 . Any basis defining the same matrix will be called an orthonormal basis. The lattice I 1,N is a unimodular lattice of signature (1, N ). Consider the special vector in I 1,N defined by
N

kN = (−3, 1, . . . , 1) = −3e0 +
i=1

ei .

(8.2)

We define the EN -lattice as a sublattice of I 1,N given by EN = (ZkN )⊥ .
2 Since kN = 9 − N , it follows from Lemma 8.2.2, that EN is a negative definite lattice for N ≤ 8. Its discriminant group is a cyclic group of order 9 − N . Its quadratic 1 form is given by the value on its generator is equal to − 9−N mod Z (or 2Z if N is odd).

Lemma 8.2.6. Assume N ≥ 3. The following vectors form a basis of EN α1 = e0 − e1 − e2 − e3 , αi = ei−1 − ei , i = 2, . . . , N. The matrix of the symmetric bilinear form of EN with respect to this basis is equal to   −2 0 0 1 0 0 0 0 ... 0  0 −2 1 0 0 0 0 0 . . . 0   0 1 −2 1 0 0 0 0 . . . 0   1 0 1 −2 1 0 0 0 . . . 0   CN =  0 0 0 1 −2 1 0 0 . . . 0   0 0 0 0 1 −2 1 0 . . . 0    . . . . . . . . . . . . . . . . . . .  . . . . . . . . . . . 0 0 0 . . . . . . 0 0 0 −2 1

8.2. THE EN -LATTICE

251

Proof. By inspection, each αi is orthogonal to kN . Suppose we have a vector (a0 , a1 , . . . , aN ) orthogonal to kN . Then 3a0 + a1 + · · · + aN = 0. (8.3) We can write this vector as follows (a0 , a1 , . . . , aN ) = a0 α1 + (a0 + a1 )α2 + (2a0 + a1 + a2 )α3 +(3a0 + a1 + a2 + a3 )α4 + . . . + (3a0 + a1 + . . . + aN −1 )αN We use here that (8.3) implies that the last coefficient is equal to −aN . We leave the computation of the matrix to the reader. One can express the matrix CN by means of the incidence matrix AN of the following graph with N vertices We have CN = −2IN + AN . • • • • • • • N =3 • ··· • • N ≥4

Table 8.1: Coxeter-Dynkin diagram of type EN

8.2.3

Roots

A vector α ∈ EN is called a root if α2 = −2. If we write α as a vector (d, m1 , . . . , mN ) ∈ I 1,N , then the conditions are
N N

d2 −
i=1

m2 = −2, 3d − i
i=1 N i=1

mi = 0.

Using the inequality (

N i=1

mi )2 ≤ N

m2 , it is easy to find all solutions. i

Proposition 8.2.7. Let N ≤ 8 and αij αijk = ei − ej , 1 ≤ i < j ≤ N, = e0 − ei − ej − ek , 1 ≤ i < j < k ≤ N,

Any root in EN is equal to one of the following: N= 3 : ±αij , ±α1,2,3 . Their number is 8. N= 4 :±αij , ±αi,j,k . Their number is 20. N= 5 : ±αij , ±αi,j,k . Their number is 40.

252

CHAPTER 8. DEL PEZZO SURFACES

N= 6 :±αij , ±αi,j,k , ±(2e0 − e1 − . . . − e6 ). Their number is 72. N= 7 : ±αij , ±αi,j,k , ±(2e0 − e1 − . . . − e7 − ei ). Their number is 112. N= 8 :±αij , ±αi,j,k , ±(2e0 − e1 − . . . − e7 − ei − ej ), ±(3e0 − e1 − . . . − e8 − ei ). Their number is 240. For N ≥ 9, the number of roots is infinite. From now on we assume 3 ≤ N ≤ 8. An ordered set B of roots {β1 , . . . , βr } is called a root basis if they are linear independent over Q and βi · βj ≥ 0. A root basis is called irreducible if it is not equal to the union of non-empty subsets B1 and B2 such that βi · βj = 0 if βi ∈ B1 and βj ∈ B2 . The symmetric r × t-matrix C = (aij ), where aij = βi · βj is called the Cartan matrix of the root basis. Recall that a symmetric Cartan matrix is a symmetric negative (positive) definite matrix C = (aij ) of size n with aii = −2(2) and aij ≥ 0(≤ 0) for i = j. All such matrices can be classified. Each Cartan matrix is a block-sum of irreducible Cartan matrices. There are two infinite series of irreducible matrices of types An and Dn and three exceptional irreducible matrices of type En , where n = 6, 7, 8. The matrix C + 2In (C − 2In ), where C is an irreducible Cartan matrix, is the incidence matrix of the Coxeter-Dynkin diagram of type An , Dn , En .

An Dn

• •

• ··· • • ... •

• • • •

E6





• •





E7





• •







E8





• •









Table 8.2: Coxeter-Dynkin diagrams of types A,D, E For 3 ≤ n ≤ 5, we will use En to denote the Coxeter-Dynkin diagrams of types A2 + A1 (N = 3), A4 (N = 4) and D5 (N = 5).

8.2. THE EN -LATTICE

253

Since EN is a negative definite lattice for N ≤ 8, any root basis generates a negative definite sublattice. Hence the matrix of the symmetric form satisfies the conditions of a Cartan matrix. This gives the following. Proposition 8.2.8. The Cartan matrix C of an irreducible root basis in EN is equal to an irreducible Cartan matrix of type Ar , Dr , Er with r ≤ N . Definition 8.5. A canonical root basis in EN is a root basis with Cartan matrix of type EN . An example of a canonical root basis is the basis formed by the roots β1 = α123 , βi = αi−1,i , i = 2, . . . , N. (8.4)

Theorem 8.2.9. Any canonical root basis is obtained from a unique orthonormal basis (v0 , v1 , . . . , vn ) in I 1,N such that kN = −3v0 + v1 + . . . + vN by the formula β1 = v0 − v1 − v2 − v3 , βi = vi−1 − vi , i = 2, . . . , N. (8.5)

Proof. Given a canonical root basis (β1 , . . . , βN ) we solve for vi in the system of equations (8.5). We have
N

vi = vN +
i=2

βi , i = 1, . . . , N − 1,
N

v0 = β1 + v1 + v2 + v3 = β1 + 3vN + 3
i=4 N

βi + 2β3 + β2 ,

−kN = 3v0 − v1 − · · · − vN = 9vN + 9
i=4 N N

βi + 6β3 + 3β2

−(vN +
i=2

βi ) − (vN +
i=3

βi ) − . . . − (vN + βN ) − vN .

This gives vN = − 1 (kN + 3β1 + 2β2 + 4β3 + (9 − i)βi+1 ), 9−N i=3
N

Intersecting both sides with βi we find (vN , βi ) = 0, i = 1, . . . , N −1, and (vN , βN ) = 1. Thus all vi belong to (kN ⊥ EN )∗ . The discriminant group of this lattice is isomorphic to (Z/(9 − N )) ⊕ (Z/(9 − N )) and the only isotropic subgroup of order 9 − N is the diagonal subgroup. This shows that E∗ is the only sublattice of (kN ⊥ EN )∗ of N index 9 − N , hence vi ∈ E∗ for all i. It is immediately checked that (v0 , v1 , . . . , vN ) N is an orthonormal basis and kN = −3v0 + v1 + . . . + vN . Corollary 8.2.10. Let O(I 1,N )kN be the stabilizer of the vector kN in O(I 1,N ). Then O(I 1,N )kN acts simply transitively on the set of canonical root bases in EN .

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CHAPTER 8. DEL PEZZO SURFACES

Let β = (β1 , β2 , . . . , βN ) be a canonical root basis and α be a root. By applying a unique σ ∈ O(I 1,N )kN we may assume that σ(β1 ) = e0 − e1 − e2 − e3 , σ(βi ) = ei − ei+1 , i ≥ 2, and σ(α) is one of the vectors from Lemma 8.2.7. It is immediately checked that each such vector is equal to a linear combination of the roots σ(αi ) with either all non-negative or all non-positive integer coefficients. So each canonical root basis β = (β1 , . . . , βN ) defines a partition of the set of roots R R = R+ R− ,

where R+ is the set of non-negative linear combinations of βi . The roots from R+ (R− ) are called positive (negative) roots with respect to the root basis β. It is clear that R− = {−α : α ∈ R+ }. For any canonical root basis β, the subset
1,N Cβ = {x ∈ IR : (x, βi ) ≥ 0}

is called a Weyl chamber with respect to β. A subset of a Weyl chamber which consists of vectors such that (v, βi ) = 0 for some subset I ⊂ {1, . . . , N } is called a face. A face corresponding to the empty set is equal to the interior of the Weyl chamber. The face corresponding to the subset {1, . . . , N } is spanned by the vector kN . For any root α and any x ∈ I 1,N , let rα (v) = v + (v, α)α. It is immediately checked that rα ∈ O(I 1,N )kN , rα (α) = −α and rα (v) = v if (v, α) = 0. The isometry rα is called the reflection in the root α. By linearity rα acts 1,N as an orthogonal transformation of the real inner product space R1,N := IR . The following is a basis fact from the theory of finite reflection groups. We refer for the proof to numerous text-books on this subject (e.g. [24], [152]). Theorem 8.2.11. Let C be a Weyl chamber defined by a canonical root basis β. Let W (EN ) be the subgroup of O(EN ) generated by reflections rβi . For any x ∈ R1,N there exists w ∈ W (EN ) such that w(x) ∈ C. If x, w(x) ∈ C , then x = w(x) and x belongs to a face of C. The union of Weyl chambers is equal to R1,N . Two Weyl chambers intersect only along a common face. Corollary 8.2.12. The group W (EN ) acts simply transitively on canonical root bases, and Weyl chambers. It coincides with the group O(I 1,N )kN . The first assertion follows from the theorem. The second assertion follows from Corollary 8.2.10 since W (EN ) is a subgroup of O(I 1,N )kN . Corollary 8.2.13. O(EN ) = W (EN ) × τ , where τ is an isometry of EN which is realized by a permutation of roots in a canonical basis leaving invariant the Coxeter-Dynkin diagram. We have τ = 1 for N = 7, 8 and τ 2 = 1 for N = 7, 8.

8.2. THE EN -LATTICE

255

Proof. By Lemma 8.2.5, the image of the restriction homomorphism O(I 1,N )kN → O(EN ) is equal to the kernel of the homomorphism α : O(EN ) → O(Disc(EN )). It is easy to compute O(Disc(EN )) and find that it is isomorphic to Z/2Z. Also it can be checked that α is surjective and the image of the symmetry of the CoxeterDynkin diagram is the generator of O(Disc(EN )). It remains to apply the previous corollary. The definition of the group W (EN ) does not depend on the choice of a canonical basis and hence coincides with the definition of Weyl groups W (EN ) from Chapter 7. Note that Corollary 8.2.12 also implies that W (EN ) is generated by reflections rα for all roots α in EN . This is true for N ≤ 10 and is not true for N ≥ 11. Proposition 8.2.14. If N ≥ 4, the group W (EN ) acts transitively on the set of roots. Proof. Let (β1 , . . . , βN ) be a canonical basis from (8.4). Observe that the subgroup of W (EN ) generated by the reflections with respect to the roots β2 , . . . , βN is isomorphic to the permutation group SN . It acts on the set {e1 , . . . , eN } by permuting its elements and leaves e0 invariant. This implies that SN acts on the roots αij , αijk , via its action on the set of subsets of {1, . . . , N } of cardinality 2 and 3. Thus it acts transitively on the set of roots αij and on the set of roots αijk . Similarly, we see that it acts transitively on the set of roots 2e0 − ei1 − . . . − ei6 and −k8 − ei if N = 8. Also applying rα to α we get −α. Now the assertion follows from the following computation rβ1 (−k8 − e8 ) = 2e0 − e1 − e4 − . . . − e8 , rβ1 (2e0 − e1 − . . . − e6 ) = α456 , rβ1 (α124 ) = α34

Similarly, one defines the Weyl group associated to any Cartan matrix and CoxeterDynkin diagram of some type T . We consider the negative lattice with the symmetric bilinear form defined by the Cartan matrix. We call it a root lattice of the corresponding type. A basis in this lattice in which the matrix of the symmetric bilinear form is equal to the Cartan matrix is called a root basis. The subgroup of the orthogonal group of a root lattice generated by reflection in basis vectors is the Weyl group W (T ) of type T . It acts transitively on the set of root bases and coincides with the group generated by reflections in all roots. The types of roots bases in the lattice EN can be classified by the following procedure due to A. Borel and J. De Siebenthal [22] and, independently by E. Dynkin [95]. Let D be the Coxeter-Dynkin diagram. Consider the extended diagram by adding one more vertex which is connected to other edges as shown on the following extended Coxeter-Dynkin diagrams. Consider the following set of elementary operations over the diagrams D and their disconnected sums D1 + . . . + Dk . Extend one of the components Di to get the extended diagram. Consider its subdiagram obtained by deleting subset of vertices. Now all possible root bases are obtained by applying recursively the elementary operations to the initial Coxeter-Dynkin diagram of type EN and all its descendants.

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CHAPTER 8. DEL PEZZO SURFACES

˜ An

„ • „„„„„• · · · • hhhhh • „„„„ hhhhh h • • • • • ··· • • • • • • • • • •

˜ Dn

˜ E6



˜ E7







• •







˜ E8





• •











˜ ˜ ˜ Table 8.3: Extended Coxeter-Dynkin diagrams of types A, D, E

8.2.4

Fundamental weights

Let β = (β1 , β2 , . . . , βN ) be a canonical root basis in EN . Consider its dual basis (ω1 , . . . , ωN ) in E∗ ⊗ Q defined by ωi (βj ) = δij . Its elements are called fundamental N weights. We use the expressions for βi from Theorem 8.2.9. Let us identify E∗ with N ⊥ (kN )∗ = I 1,N /ZkN . Then we can take for representatives of ωj the following vectors from I 1,N ω1 ω2 ω3 ωi = = = = v0 , v0 − v1 , 2v0 − v1 − v2 , vi + . . . + vN , i = 4, . . . , N.

Definition 8.6. A vector in I 1,N is called an exceptional vector if it belongs to the W (EN )-orbit of ωN . Proposition 8.2.15. A vector v ∈ I 1,N is exceptional if and only if kN · v = −1 and

8.2. THE EN -LATTICE v 2 = −1. The set of exceptional vectors is the following N = 3, 4 N =5 N =6 N =7 N =8 : : : : :

257

ei , e0 − ei − ej ; ei , e0 − ei − ej , 2e0 − e1 − . . . − e5 ; ei , e0 − ei − ej , 2e0 − e1 − . . . − e6 + ei ; ei , e0 − ei − ej , 2e0 − e1 − . . . − e7 + ei + ej ; −k7 − ei ; ei , e0 − ei − ej , 2e0 − e1 − . . . − e7 + ei + ej ; −k8 + ei − ej ; −k8 + e0 − ei − ej − ek , −k8 + 2ei1 − . . . − ei6

The number of exceptional vectors is given by the following table. N # 3 6 4 10 5 16 6 27 7 56 8 240

Proof. Similarly to the case of roots we solve the equations
N N

d2 −
i=1

m2 = −1, 3d − i
i=1

mi = 1

First we immediately get the inequality (3d−1)2 ≤ N (d2 +1) which gives 0 ≤ d ≤ 4. If d = 0, the condition m2 = d2 + 1 and kN · v = −1 gives the vectors ei . If d = 1, i this gives the vectors e0 − ei − ej , and so on. Now we use the idea of Noether’s inequality from Chapter 7 to show that all these vectors (d, m1 , . . . , mN ) belong to the same orbit of W (EN ). We apply permutations from SN to assume m1 ≥ m2 ≥ m3 , then use the reflection rα123 to decrease d. Corollary 8.2.16. The orders of the Weyl groups W (EN ) is given by the following table

N #W (EN )

3 12

4 5!

5 24 .5!

6 23 .32 .6!

7 26 .32 .7!

8 27 .33 .5.8!

Proof. Observe that the orthogonal complement of eN in I 1,N is isomorphic to I N −1 . Since e2 = −1, by Lemma 8.2.5, the stabilizer subgroup of eN in O(I 1,N ) is equal N to O(I 1,N −1 ). This implies that the stabilizer subgroup of eN in W (EN ) is equal to W (EN −1 ). Obviously, W (E3 ) ∼ S3 × S2 and W (E3 ) ∼ S5 . Thus #W (E5 ) = = = 16#W (E4 ) = 24 .5!, #W (E6 ) = 27#W (E5 ) = 23 .32 .6!, #W (E7 ) = 56#W (E6 ) = 26 .32 .7!, #W (E8 ) = 240#W (E7 ) = 27 .33 .5.8!. The following proposition implies that the exceptional vectors are similar to one of lines.

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Proposition 8.2.17. For any two different exceptional vectors v, w 0 ≤ (v, w) ≤ 1. Proof. This can be seen directly from the list however we prefer to give a proof independent of the classification. Since (v, κN ) = (w, kN ), we have v − w ∈ EN . Since EN is a negative definite lattice we have (v − w)2 = −2 − 2(v, w) < 0. This gives (v, w) ≥ 0. Assume (v, w) > 1. Let h = −kN − v − w. We have (v + w)2 = −2 + 2(v, w) > 1 and h2 = 3 − 4 + (v + w)2 > 0, (h, −kN , h) = 1. This implies that the matrix h2 (h, −kN ) (h, −κN ) (−kN )2 is positive definite. Since the signature of I 1,N is (1, N ) this is a contradiction.

8.2.5

Effective roots

Let S be a weak Del Pezzo surface of degree d = 9 − N . Together with a geometric marking φ : I 1,N → Pic(S). The intersection form on Pic(S) equips it with a structure of a lattice. Since φ sends an orthonormal basis of I 1,N to an orthonormal basis of ⊥ Pic(S), the isomorphism φ is an isomorphism of lattices. The image KS of EN is isomorphic to the lattice EN . The image of a root α ∈ EN is a divisor class D such that D2 = −2 and D · KS = 0. We say that α is an effective root if φ(α) is an effective divisor class. Let i∈I ni Ri be its effective representative. Since −KS is nef, we obtain that Ri · KS = 0. Since 2 2 KS , we also get Ri < 0. Together with the adjunction formula this implies that each Ri is a (−2)-curve. Since a (−2)-curve does not move, we will identify it with its divisor class. Proposition 8.2.18. Let S be a weak Del Pezzo surface of degree d = 9 − N . The number r of (−2)-curves on S is less or equal than N . The sublattice NS of Pic(S) generated by (−2)-curves is a root lattice of rank r.
⊥ Proof. Since each nodal curve is contained in KS and Ri · Rj ≥ 0 for i = j, it suffices to prove that the set of (−2)-curves is linearly independent over Q. Suppose that this is not true. Then we can find two disjoint sets of curves Ri , i ∈ I, and Rj , j ∈ J, such that ni Ri ∼ mj R j , i∈I j∈J

where ni , mj are some non-negative rational numbers. Taking intersection of both sides with Ri we obtain that Ri · (
i∈I

ni Ri ) = R i · (
j∈J

mj Rj ) ≥ 0.

This implies that (
i∈I

ni Ri )2 =
i∈I

ni Ri · (
i∈I

ni Ri ) ≥ 0.

8.2. THE EN -LATTICE

259

Since (ZKS )⊥ is negative definite, this could happen only if i∈I ni Ri ∼ 0. Since all coefficients are non-negative, this happens only if all ni = 0. For the same reason each mi is equal to 0. It is clear that, if α is a nodal root, then −α is not a nodal root. Let η = x1 + . . . + xN be the bubble cycle defined by the blowing down structure S → P2 defining the geometric marking. It is clear that φ(αij ) = [Ei − Ej ] is effective if and only if xi i−j xj . It is also clear that αij is nodal if and only if i = j + 1. A root αijk is effective if and only if there exist points xi , xj , xk such that xi i−i xi , xj j−j xj , xk k−k xk , and there is a line in the plane whose proper transform in S belong to the class e0 − ei − ej − ek . The root αi ,j ,k is a nodal root. The root 2e0 − ei1 − . . . − ei6 is nodal if and only if its image in Pic(S) is the divisor class of the proper transform of an irreducible conic passing through the points xi1 , . . . , xi6 . The root 3e0 − e1 − . . . − e8 − ei is nodal if and only if its image in Pic(S) is the divisor class of the proper transform of an irreducible cubic with double points at xi and passing through the rest of the points. Definition 8.7. A Dynkin curve is a reduced connected curve R on a projective nonsingular surface X such that its irreducible components Ri are −2-curves and the matrix (Ri · Rj )ij is a Cartan matrix. The type of a Dynkin curve is the type of the corresponding root system. Under a geometric marking a Dynkin curve on a weak Del Pezzo surface S corresponds to an irreducible root base in the lattice EN . We use the Borel-De SiebenthalDynkin procedure to determine all possible root bases in EN . Theorem 8.2.19. Let R be a Dynkin curve on a projective nonsingular surface X. There is a birational morphism f : X → Y , where Y is a normal surface satisfying the following properties (i) f (R) is a point; (ii) the restriction of f to X \ R is an isomorphism; (iii) f ∗ ωY ∼ ωX . = Proof. Let H be a very ample divisor on X. Since the intersection matrix of compon nents of R = i=1 Ri has non-zero determinant, we can find rational numbers ri such that
n

(
i=1

ri Ri ) · Rj = −H · Rj ,

j = 1, . . . , n.

It is easy to see that the entries of the inverse of a Cartan matrix are nonpositive. Thus all ri ’s are nonnegative numbers. Replacing H by some multiple mH, we may assume that all ri are nonnegative integers. Let D = ri Ri . Since H + D is an effective

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divisor and (H +D)·Ri = 0 for each i, we have OX (H +D)⊗ORi = ORi . Consider the standard exact sequence 0 → OX (H) → OX (H + D) → OD → 0. Replacing H by mH, we may assume, by Serre’s Theorem, that H 1 (X, OX (H)) = 0 and OX (H) is generated by global sections. Let s0 , . . . , sN −1 be sections of OX (H) which define an embedding in PN −1 . Consider them as sections of OX (H + D). Let sN +1 be a section of OX (H + D)) which maps to 1 ∈ H 0 (X, OD ). Consider the map f : X → PN defined by the sections (s0 , . . . , sN ). Then f (D) = (0, . . . , 0, 1) and f |X ⊂ D is an embedding. So we obtain a map f : X → PN satisfying properties (i) and (ii). Since X is normal, f factors through a map f : X → Y , where Y is normal. Let ωY be the canonical sheaf of Y (it is defined to be equal to the sheaf j∗ ωY \f (R) , where j : Y \ f (R) → Y is the natural open embedding). We have ωX = f ∗ ωY ⊗ OX (A) for some divisor A. Since KX · Ri = 0 for each i, and f ∗ ωY ⊗ ORi = ORi we get A·Ri = 0. Since the intersection matrix of R is negative definite we obtain A = 0. Definition 8.8. A point y ∈ Y of a normal variety Y is called a canonical singularity if there exists a resolution π : X → Y such that π ∗ ωY ∼ ωX . In the case dim Y = 2, = a canonical singularity is called a RDP (rational double point). We state the next well-known theorem without proof. Theorem 8.2.20. Let y ∈ Y be a RDP and π : X → Y be a resolution such that π ∗ ωY ∼ ωX . Then π −1 (y) is a Dynkin curve. Moreover (Y, y) is analytically equiva= lent to one of the following singularities An Dn E6 E7 E8 : : : : : z 2 + x2 + y n+1 = 0, n ≥ 1 z 2 + y(x2 + y n−2 = 0, n ≥ 4 z 2 + x3 + y 4 = 0 z 2 + x3 + xy 3 = 0 z 2 + x3 + y 5 = 0 (8.6)

The corresponding Dynkin curve is of respective type An , Dn , En . Remark 8.2.1. The singularity of type A1 is called in classical and modern literature a node. For this reason a (−2)-curve is also called nodal although the same term is also used for an irreducible singular curve with ordinary double points (nodes) as a singularities.

8.2.6

Cremona isometries

Definition 8.9. An orthogonal transformation σ of Pic(S) is called a Cremona isometry if σ(KS ) = KS and σ sends any effective class to an effective class. The group of Cremona isometries will be denoted by Cris(S).

8.2. THE EN -LATTICE It follows from Corollary 8.2.12 that Cris(S) is a subgroup of W (S).

261

Proposition 8.2.21. An isometry σ of Pic(S) is a Cremona isometry if and only if it preserves the canonical class and sends a (−2)-curve to a (−2)-curve. Proof. Clearly, any Cremona isometry sends the class of an irreducible curve to the class of an irreducible curve. Since it also preserves the intersection form, it sends a (−2)-curve to a (−2)-curve. Let us prove the converse. Let D be an effective class in Pic(S) with D2 ≥ 0. Then −KS · D > 0 and (KS − D) · D < 0. This gives −KS · σ(D) > 0, σ(D)2 ≥ 0. Since 2 (KS − σ(D)) · (−KS ) = −KS + σ(D) · KS < 0, we have |KS − σ(D)| = ∅. By Riemann-Roch, |σ(D)| = ∅. So it remains to show that σ sends any (−1)-curve to an effective divisor class. This follows from the next lemma. Lemma 8.2.22. Let D be a divisor class with D2 = D · KS = −1. Then D = E + R, 2 where R is a nonnegative sum of (−2)-curves, and E is either a (−1)-curve or KS = 1 2 and E ∈ | − KS | and E · R = 0, R = −2. Moreover D is a (−1)-curve if and only if for each (−2)-curve Ri on S we have D · Ri ≥ 0. Proof. Let e0 = π ∗ ( ), where π : S → P2 is a birational morphism and is a line. We know that e2 = 1, e0 · KS = −3. Thus ((D · e0 )KS + 3D) · e0 = 0 and hence 0 (D · e0 )KS + 3D
2 2 = −6D · e0 − 9 + (D · e0 )2 KS < 0.

Thus −6D·e0 −9 < 0 and hence D·e0 > −9/6 > −2. This shows that (KS −D)·e0 = −3 − D · e0 < 0, and since e0 is nef, we obtain that |KS − D| = ∅. Applying Riemann-Roch we get dim |D| ≥ 0. Write an effective representative of D as a sum of irreducible components and use that D · (−KS ) = 1. Since −KS is nef, there is only one component E entering with coefficient 1 satisfying E · KS = −1, all other components are (−2)-curves. If D ∼ E, then D2 = E 2 = −1 and E is a (−1)-curve. Let π : S → S be a birational morphism of a weak Del Pezzo surface of degree 1 (obtained by blowing up 8 − k points on S in general position not lying on E). We identify E with its pre-image in S . Then (E + KS ) · KS = −1 + 1 = 0, hence, by Hodge Index Theorem, either S = S and E ∈ | − KS |, or
2 (E + KS )2 = E 2 + 2E · KS + KS = E 2 − 1 < 0.

Since E · KS = −1, E 2 is odd. Thus, the only possibility is E 2 = −1. If E ∈ | − KS |, we have E · Ri = 0 for any (−2)-curve Ri , hence E · R = 0, R2 = −2. Assume R = 0. Since −1 = E 2 + 2E · R + R2 and E 2 ≤ 1, R2 ≤ −2, we get E · R ≥ 0, where the equality take place only if E 2 = 1. In both cases, we get −1 = (E + R)2 = (E + R) · R + (E + R) · E ≥ (E + R) · R. Thus if D = E, we get D · Ri < 0 for some irreducible component of R. This proves the assertion.

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Corollary 8.2.23. Let R be the set of effective roots of a marked Del Pezzo surface (S, φ). Then the group of Cremona isometries Cris(S) is isomorphic to the subgroup of the Weyl group of EN which leaves the subset R invariant. Let W (S)n be the subgroup of W (S) generated by reflections with respect to (−2)curves. It acts on a marking ϕ : I 1,N → Pic(S) by composing on the left. Lemma 8.2.24. Let C n = {D ∈ Pic(S) : D · R ≥ 0 for any (−2)-curve R}. For any D ∈ Pic(S) there exists w ∈ W (S)n such that w(D) ∈ C n . If D ∈ C n and w(D) ∈ C n for some w ∈ W (S)n , then w(D) = D. In other words, C n is a fundamental domain for the action of W (S)n in Pic(S). Proof. The set of (−2)-curves form a root basis in the Picard lattice Pic(S) and W (S)n is its Weyl group. The set C n is a chamber defined by the root basis. Now the assertion follows from the theory of finite reflection groups which we have already employed for a similar assertion in the case of canonical root bases in EN . Theorem 8.2.25. For any marked weak Del Pezzo surface (S, ϕ), there exists w ∈ W (S)n such that (S, w ◦ ϕ) is geometrically marked weak Del Pezzo surface.
2 Proof. We use induction on N = 9 − KS . Let ei = φ(ei ), i = 0, . . . , k. It follows from the proof of Lemma 8.2.22, that each ei is an effective class. Assume e1 is the class of a (−1)-curve E1 . Let πN : S → SN −1 be the blowing down of E1 . Then e0 , e2 , . . . , eN are equal to the pre-images of the divisor classes e0 , e2 , . . . , eN which define a marking of Sk−1 . By induction, there exists an element w ∈ W (S1 )n such that w(e0 ), w(e2 ), . . . , w(eN ) define a geometric marking. Since πN (E1 ) does not lie on any (−2)-curve (otherwise S is not a weak Del Pezzo surface), we see that for ∗ any (−2)-curve R on SN −1 , πN (R) is a (−2) curve on S. Thus, under the canonical ∼ π ∗ (Pic(SN −1 ) ⊥ Ze1 , we can identify W (SN −1 )n with a isomorphism Pic(S) = N subgroup of W (S)n . Applying w to (e0 , . . . , eN ) we get a geometric marking of S. If e1 is not a (−1)-curve, then we apply an element w ∈ W (S)n such that w(e1 ) ∈ n C . By Lemma 8.2.22, w(e1 ) is a (−1)-curve. Now we have a basis w(e0 ), . . . , w(eN ) satisfying the previous assumption.

Let ϕ : I 1,N → Pic(S) and ϕ : I 1,N → Pic(S) be two geometric markings corresponding to two blowing-down structures π = π1 ◦. . .◦πN and π = π1 ◦. . .◦πN . Then T = π ◦ π −1 is a Cremona transformation of P2 and w = ϕ ◦ ϕ −1 ∈ W (EN ) is its characteristic matrix. Conversely, if T is a Cremona transformation with F -points x1 , . . . , xN such that their blow-up is a weak Del Pezzo surface S, a characteristic matrix of T defines a pair of geometric markings ϕ, ϕ of S and an element w ∈ W (EN ) such that ϕ = ϕ ◦ w. Corollary 8.2.26. Assume S is a Del Pezzo surface. Then any marking ϕ : I 1,N → Pic(S) is a geometric marking. The Weyl group of EN acts simply transitively on the set of markings by composing on the right.

8.2. THE EN -LATTICE

263

Corollary 8.2.27. There is a bijection from the set of geometric markings on S and the set of left cosets W (S)/W (S)n . Proof. The group W (S) acts simply transitively on the set of markings. By Theorem 8.2.25, each orbit of W (S)n contains a unique geometric marking. Corollary 8.2.28. The group Cris(S) acts on the set of geometric markings of S. Proof. Let (e0 , . . . , eN ) defines a geometric marking, and σ ∈ Cris(S). Then there exists w ∈ W (S)n such that ω(σ(e0 )), . . . , ω(σ(eN )) defines a geometric marking. Since σ(e1 ) is a (−1)-curve E1 , it belongs to C n . Hence, by Lemma 8.2.24, we get ¯ ¯ w(σ(e1 )) = σ(e1 ). This shows that w ∈ W n (S), where S → S is the blow-down 2 n σ(E1 ). Continuing in this way, we see that w ∈ W (P ) = {1}. Thus w = 1 and we obtain that σ sends a geometric marking to a geometric marking. Example 8.2.1. The action of Cris(S) on geometric markings is not transitive in general. For example, consider 6 distinct points x1 , . . . , x6 in P2 lying on an irreducible conic C. Let S be their blow-up and φ be the corresponding geometric marking. This is a weak Del Pezzo surface with a (−2)-curve R equal to the proper inverse transform of the conic. Let T be the quadratic transformation with F -points at x1 , x2 , x3 . Then C ∈ |2 − x1 − x2 − x3 | and hence is equal to T −1 (l), where is a line in P2 . This line contains the points qi = T (pi ), i = 4, 5, 6. Let q1 = T (p2 , p3 ), q2 = T (p1 , p3 ), q3 = T (p1 , p2 ). Then the blow-up of the points q1 , . . . , q6 is isomorphic to S and defines a geometric marking φ . Let w be the corresponding element of the Weyl group W (Q6 ) = W (E6 ). We have R = 2e0 − e1 − . . . − e6 = e0 − e4 − e5 − e6 . However, the element w ∈ W (S) defined by the two bases sends ei to ei . Thus w(R) = R and hence w ∈ Cris(S). Note that w is the reflection with respect to the root α = e0 − e1 − e2 − e3 and α · R = −1, so that rα (R) = R − α = (2e0 − e1 − . . . − e6 ) − (e0 − e1 − e2 − e3 ) = e0 − e4 − e5 − e6 . Since the points p4 , p5 , p6 are not collinear, this is not an effective class. The group Cris(S) in this case consists of permutations of the vectors e1 , . . . , e6 and is isomorphic to S6 . Its index in W (E6 ) is equal to 72. The group W (S)n is generated by the 1 reflection rα . Thus we get 2 #W (E6 ) = 36 · 6! geometric markings and the group Cris(S) has 36 orbits on this set. Let S be a weak Del Pezzo surface of degree d and Aut(S) be its group of biregular automorphims. By functoriality Aut(S) acts on Pic(S) leaving the canonical class KS invariant. Thus Aut(S) acts on the lattice QX = (ZKS )⊥ preserving the intersection form. Let ρ : Aut(S) → O(QX ), σ → σ ∗ , be the corresponding homomorphism. Proposition 8.2.29. The image of ρ is contained in the group Cris(S). If S is a Del Pezzo surface, the kernel of ρ is trivial if d ≤ 5. If d ≥ 6, then the kernel is a linear algebraic group of dimension 2d − 10.

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Proof. Clearly, any automorphism induces a Cremona isometry of Pic(S). We know that it is contained in the Weyl group. An element in the kernel does not change any geometric basis of Pic(S). Thus it descends to an automorphism of P2 which fixes an ordered set of k = 9 − d points in general linear position. If k ≥ 4 it must be the identity transformation. Assume k ≤ 3. The assertion is obvious when k = 0. If k = 1, the surface S is the blow-up of one point. Each automorphisms leaves the unique exceptional curve invariant and acts trivially on the Picard group. The group Aut(S) is the subgroup of Aut(P2 ) fixing a point. It is a connected linear algebraic group of dimension 6 isomorphic to the semi-direct product of C2 GL(2). If k = 2, the surface S is the blow-up of two distinct points p1 , p2 . Each automorphisms leaves the proper inverse transform of the line p1 , p2 . It either leaves the exceptuional curves E1 and E2 invariant, or switches them. The kernel of the Weyl reprsentation consists of elements which do not switch E1 and E2 . It is isomorphic to the subgroup of Aut(P2 ) which fixes two points in P2 and is isomorphic to the group G of invertible matrices of the form   1 0 ∗ 0 ∗ ∗ 0 0 ∗ Its dimension is equal to 4. The image of the Weyl representation is a group of order 2. So Aut(S) = G C2 . If k = 3, the surface S is the blow-up of 3 non-collinear points. The kernel of the Weyl representation is isomorphic to the group of invertible diagonal 3 × 3 matrices modulo scalar matrices. It is isomorphic to a 2-dimension torus (C∗ )2 . Corollary 8.2.30. Assume that d ≤ 5, then Aut(S) is isomorphic to a subgroup of the Weyl group W (E9−d ). We will see later examples of automorphisms of weak Del Pezzo surfaces of degree 1 or 2 which act trivially on Pic(S).

8.2.7

Lines on Del Pezzo surfaces

Let E be a (−1)-curve on a weak Del Pezzo surface S. Then −KS · E = 1. We will see in the next section that the linear system | − KS | defines an embedding of a Del Pezzo surface of degree d ≥ 3 in Pd . Thus the image of E is a line. For this reason, we call any (−1)-curve on S a line. Since a line does not move in a linear system we may identify it with its divisor class. Since P2 , F0 or F2 has no lines (in our definition!), we will assume that S is a blow-up of 0 < N ≤ 8 points x1 , . . . , xN in P2 . Also, if N = 1, S contains only one line, the exceptional curve of the blow-up. If N = 2, S contains either 3 lines if x2 is not infinitely near to x1 and contains 2 lines otherwise. So, we may assume that N ≥ 3. Let φ : I 1,N → Pic(S) be a geometric marking. The pre-image of a line on S is an exceptional vector. Since the number of exceptional vectors is finite, we obtain that a weak Del Pezzo surface has only finitely many lines.

8.3. ANTICANONICAL MODELS

265

If S is a Del Pezzo surface, the number of lines on it is given in Proposition 8.2.15. The situation with weak Del Pezzo but not Del Pezzo surfaces is more complicated. The number of lines depends on the structure of the set of (−2)-curves. Theorem 8.2.31. Fix a line E on a weak Del Pezzo surface. There is a natural bijection Lines on S ←→ W (S)n \W (S)/W (S)E . Proof. Let E be a line and w ∈ W (S). By Lemma 8.2.22 there exists g ∈ W (S)n such that g(w(E)) is a line. This line is the unique line l(E, w) in the orbit of w(E) with respect to the action of W (S)n . By Lemma 8.2.24, for any line E there exists w ∈ W (S) such that w(E) = E . This shows that the map ΦE : W (S) → set of lines, w → l(E, w),

is surjective. Suppose that l(E, w) = l(w , E). Then gw(E) = w (E) for some g ∈ W (S)n . Thus, w−1 gw (E) = E, and hence w −1 gw ∈ W (S)E and w ∈ W (S)n wW (S). Conversely, each w in the double coset W (S)n wW (S) defines the same line l(E, w).

8.3
8.3.1

Anticanonical models
Anticanonical linear systems
1 dim H 0 (S, OS (−rKS )) = 1 + 2 r(r + 1)d.

2 Lemma 8.3.1. Let S be a weak Del Pezzo surface with KS = d. Then

Proof. By Ramanujam’s Vanishing theorem, for any r ≥ 0 and i > 0, H i (S, OS (−rKS )) = H i (S, OS (KS + (−r − 1)KS )) = 0. The Riemann-Roch Theorem gives
1 1 dim H 0 (S, OS (−rKS )) = 2 (−rKS − KS ) · (−rKS )) + 1 = 1 + 2 r(r + 1)d.

(8.7)

Theorem 8.3.2. Let S be a weak Del Pezzo surface of degree d and N be the union of (−2)-curves on S. Then (i) | − KS | has no fixed part. (ii) If d > 1, then | − KS | has no base points. (iii) If d > 2, | − KS | defines a regular map φ to Pd which is an isomorphism outside ¯ N . The image surface S is a normal nondegenerate surface of degree d. The image of each connected component of N is a RDP of φ(S).

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(iv) If d = 2, | − KS | defines a regular map φ : S → P2 . It factors as a birational ¯ ¯ morphism f : S → S onto a normal surface and a finite map π : S → P2 of degree 2 branched along a curve of degree 4. The image of each connected ¯ component of N is a RDP of S. (v) If d = 1, | − 2KS | defines a regular map φ : S → P3 . It factors as a birational ¯ ¯ morphism f : S → S onto a normal surface and a finite map π : S → Q ⊂ P3 of degree 2, where Q is a quadric cone. The morphism π is branched along a curve of degree 6 cut out on Q by a cubic surface. The image of each connected ¯ component of N under f is a RDP of S. Proof. The assertions are easily verified if S = F0 or F2 . So we assume that S is obtained from P2 by blowing up k = 9 − d points xi . (i) Assume there is a fixed part F of | − KS |. Write | − KS | = F + |M |, where |M | is the mobile part. If F 2 > 0, by Riemann-Roch,
1 1 dim |F | ≥ 2 (F 2 − F · KS ) ≥ 2 (F 2 ) > 0,

and hence F moves. Thus F 2 ≤ 0. If F 2 = 0, we must also have F · KS = 0. Thus F = ni Ri , where Ri are (−2)-curves. Hence [f ] ∈ (ZKS )⊥ and hence F 2 ≤ −2 (the intersection form on (ZKS )⊥ is negative definite and even). Thus F 2 ≤ −2. Now M2 −KS · M
2 2 = (−KS − F )2 = KS + 2KS · F + F 2 ≤ KS + F 2 ≤ d − 2, 2 = KS + KS · F ≤ d.

Suppose |M | is irreducible. Since dim |M | = dim | − KS | = d, the linear system |M | defines a rational map to Pd whose image is a non-degenerate irreducible surface of degree ≤ d − 3 (strictly less if |M | has base points). This contradicts Theorem 8.1.1. Now assume that |M | is reducible, i.e. defines a rational map to a non-degenerate curve W ⊂ Pd of some degree t. By Theorem 8.1.1, we have t ≥ d. Since S is rational, W is a rational curve, and then the pre-image of a general hyperplane section is equal to the disjoint sum of t linearly equivalent curves. Thus M ∼ tM1 and d ≥ −KS · M = −tKS · M1 ≥ d(−KS · M1 ). Since −KS · M = 0 implies M 2 < 0 and a curve with negative self-intersection does 2 not move, this gives −KS · M1 = 1, d = t. But then M 2 = d2 M1 ≤ d − 2 gives a contradiction. (ii) Assume d > 1. We have proved that | − KS | is irreducible. A general member of | − KS | is an irreducible curve C with ωC = OC (C + KS ) = OC . If C is smooth, then it is an elliptic curve and the linear system |OC (C)| is of degree d > 1 and has no base points. The same is true for a singular irreducible curve of arithmetic genus 1. This is proved in the same way as in the case of a smooth curve. Consider the exact sequence 0 → OS → OS (C) → OC (C) → 0. Applying the exact sequence of cohomology, we see that the restriction of the linear system |C| = | − KS | to C is surjective. Thus we have an exact sequence of groups 0 → H 0 (S, OS ) → H 0 (S, OS (C)) → H 0 (S, OC (C)) → 0.

8.3. ANTICANONICAL MODELS Since |OC (C)| has no base points, we have a surjection H 0 (S, OC (C)) ⊗ OC → OC (C). This easily implies that the homomorphism H 0 (S, OS (C)) ⊗ OC → OS (C)

267

is surjective. Hence |C| = | − KS | has no base points. (iii) Assume d > 2. Let x, y ∈ S be two points outside E. Let f : S → S be the blowing up of x and y. By Proposition 8.1.7, blowing them up, we obtain a weak Del Pezzo surface S of degree d − 2. We know that the linear system | − KS | has no fixed components. Thus dim | − KS − x − y| = dim | − KS − Ex − Ey | ≥ 1. This shows that | − KS | separates points. Also, the same is true if y 1 x and x does not belong to any (−1)-curve E on S or x ∈ E and y does not correspond to the tangent direction defined by E. Since −KS · E = 1 and x ∈ E, the latter case does not happen. ¯ Since φ : S → S is a birational map given by a complete linear system | − KS |, its image is a nondegenerate surface of degree d = (−KS )2 . Since −KS · R = 0 ¯ for any (−2)-curve, we see that φ blows down R to a point p. If d = 3. S is a cubic surface with isolated singularities (the images of connected components of N ). It is well-known that a hypersurface with no singularities in codimension 1 is a normal ¯ variety. Thus S is a normal surface. If d = 4, then S is obtained by a blow-up one point on a weak Del Pezzo surface S of degree 3. This point does not lie on a (−2)-curve. ¯ ¯ ¯ Thus, S is obtained from S by a linear projection from a nonsingular point. Since S ¯ must be normal too (its local rings are integral extensions of local rings is normal, S ¯ ¯ of S , and their fields of fractions coincide). Continuing in this way we see that S is normal for any d > 2. ¯ The fact that singular points of S are RDP is proven in the same way as we have proved assertion (iii) of Theorem 8.2.19. (iv) Assume d = 2. By (ii), the linear system | − KS | defines a regular map 2 φ : S → P2 . Since KS = 2, the map is of degree 2. Using Stein’s factorization [134], ¯ it factors through a birational morphism onto a normal surface f : S → S and a finite ¯ → P2 . Also we know that f∗ OS = OS . A standard Hurwitz’s degree 2 map π : S ¯ formula gives ωS ∼ π ∗ (ωP2 ⊗ L), (8.8) ¯ = where s ∈ H 0 (P2 , L⊗2 ) vanishes along the branch curve W of π. We have OS (KS ) = ωS = (π ◦ f )∗ OP2 (−1) = f ∗ (π ∗ OP2 (−1)). ¯ It follows from the proof of Theorem 8.2.19 (iii) that singular points of S are RDP. ∗ Thus f ωS = ωS , and hence ¯ f ∗ ωS ∼ f ∗ (π ∗ OP2 (−1)). ¯ =

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Applying f∗ and using the projection formula and the fact that f∗ OX = OY , we get ωS ∼ π ∗ OP2 (−1). It follows from (8.8) that L ∼ OP2 (2) and hence deg W = 4. = ¯ = Proof of (v). Let π : X → P2 be the blow-up of 8 points x1 , . . . , x8 . Then | − KS | is the proper inverse transform of the pencil |3 − x1 − . . . − x8 | of plane cubics passing through the points x1 , . . . , x8 . Let x9 be the ninth intersection point of two cubics generating the pencil. The point x9 = π −1 (x9 ) is the base point of | − KS |. By Bertini’s Theorem, all fibres except finitely many, are nonsingular curves (the assumption that the characteristic is zero is important here). Let F be a nonsingular member from | − KS |. Consider the exact sequence 0 → OS (−KS ) → OS (−2KS ) → OF (−2KS ) → 0. (8.9)

The linear system |OF (−2KS )| on F is of degree 2. It has no base points. We know from (8.7) that H 1 (S, OS (−KS )) = 0. Thus the restriction map H 0 (S, OS (−2KS )) → H 0 (F, OF (−2KS )) is surjective. By the same argument as we used in the proof of (ii), we obtain that | − 2KS | has no base points. By Lemma 8.3.1, dim | − 2KS | = 3. Let φ : S → P3 be a regular map defined by | − 2KS |. Its restriction to any nonsingular member F of | − KS | is given by the linear system of degree 2 and hence is of degree 2. Therefore the map f is of degree t > 1. The image of φ is a surface of some degree k. Since (−2KS )2 = 4 = kt, we conclude that k = t = 2. Thus the image of φ is a quadric surface Q in P3 and the images of members F of −KS | are lines lF on Q. I claim that Q is a quadric cone. Indeed, all lines lF intersect at the base point φ(x9 ) of | − KS |. This is possible only of Q is a cone. Let S → S → Q be the Stein factorization. Note that a (−2)-curve R does not pass through the base point x9 of | − KS | (because −KS · R = 0). Thus π(x9 ) is a nonsingular point q of S . Its image in Q is the vertex q of Q. Since φ is a finite map, the local ring OS ,q is a finite algebra over OQ,q of degree 2. After completion, we may assume that OS ,q ∼ C[[u, v]]. If u ∈ OQ,q , then v satisfies a monic equation = 1 v 2 + av + b with coefficients in OQ,q , where after changing v to v + 2 a we may assume that a = 0. Then OQ,q is equal to the ring of invariants in C[[u, v]] under the automorphism u → u, v → −v which is easily to see isomorphic to C[[u, v 2 ]]. However, we know that q is a singular point so the ring OQ,q is not regular. Thus we may assume that u2 = a, v 2 = b and then OQ,q is the ring of invariants for the action (u, v) → (−u, −v). This action is free outside the maximal ideal (u, v). This shows that the finite map φ is unramified in a neighborhood of q with q deleted. In particular, the branch curve Q of φ does not pass through q. We leave to the reader to repeat the argument from the proof of (iv) to show that the branch curve W of φ belongs to the linear system |OQ (3)|.
π φ

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269

8.3.2

Anticanonical model

Let X be a normal projective algebraic variety and D be a Cartier divisor on X. It defines the graded algebra


R(X, D) =
r=0

H 0 (S, OS (rD))

which depends only (up to isomorphism) on the divisor class of D in Pic(X). Assume R(X, D) is finitely generated, then XD = ProjR(X, D) is a projective variety. If s0 , . . . , sn are homogeneous generators of R(X, D) of degrees q0 , . . . , qn there is a canonical closed embedding into the weighted projective space XD → P(q0 , . . . , qn ). Also the evaluation homomorphism of sheaves of graded algebras R(X, D) ⊗ OX → Sym• (L) defines a morphism ϕcan : X = Proj(Sym• (L)) → XD . For every r > 0 the inclusion of subalgebras Sym• (H 0 (X, OX (rD))) → R(X, D) defines a rational map τr : XD − → P(H 0 (X, OX (rD)∗ )). The rational map φ|rD| : X− → P(H 0 (X, OX (rD)∗ ) defined by the complete linear system |rD| factors through ϕ φ|rd| : X − → XD − → P(H 0 (X, OX (rD)∗ )). A proof of the following proposition can be found in [67], 7.1. Proposition 8.3.3. Suppose |rD| has no base points for some r > 0 and Ddim X > 0. Then (i) R(X, D) is a finitely generated algebra; (ii) XD is a normal variety; (iii) dim XD = maxr>0 dim φ|rD| (X); (iv) if dim XD = dim X, then ϕ is a birational morphism. We apply this to the case when X = S is a weak Del Pezzo surface and D = −KS . Applying the previous Proposition, we easily obtain that X−KS ∼ S, = ¯
ϕ τr

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¯ where we use the notation of Theorem 8.3.2. The variety S is called the anticanonical ¯ → Pd is a closed embedding, model of S. If S is of degree d > 2, the map τ1 : S hence R(S, −KS ) is generated by d + 1 elements of order 1. If d = 2, the map τ1 is the double cover of P2 . This shows that R(S, −KS ) is generated by 3 elements s0 , s1 , s2 of degree 1 and one element s3 of degree 2 with a relation s2 + f4 (s0 , s1 , s2 ) = 0 for 3 ¯ some homogeneous polynomial f4 of degree 2. This shows that S is isomorphic to a hypersurface of degree 4 in P(1, 1, 12) given by an equation t2 + f4 (t0 , t1 , t2 ) = 0. 3 In the case d = 1, by Lemma 8.3.1 we obtain that dim R(S, −KS )1 = 2, dim R(S, −KS )2 = 4, dim R(S, −3KS ) = 7. Let s0 , s1 be generators of degree 1, s2 be an element of degree 2 which is not in S 2 (R(S, −KS )1 ) and let s3 be an element of degree 3 which is not in the subspace generated by s3 , s0 s2 , s2 s1 , s3 , s2 s0 , s2 s1 . The subring R(S, −KS ) generated by 0 1 0 1 s0 , s1 , s2 , s3 is isomorphic to C[t0 , t1 , t2 , t3 ]/(F (t0 , t1 , t2 , t3 )), where F = t2 + t3 + f4 (t0 , t1 )t2 + f6 (t0 , t1 ) 3 2 where f4 (t0 , t1 ) and f6 (t0 , t1 ) are binary forms of degree 4 and 6. The projection [t0 , t1 , t2 , t3 ] → [t0 , t1 , t2 ] defines a double cover of the quadratic cone Q ⊂ P3 is isomorphic to the weighted projective plane P(1, 1, 2). Using Theorem 8.3.2 one can ¯ show that the rational map S− → Proj R(S, −KS ) is an isomorphism. This shows ¯ that the anticanonical model S of a weak Del Pezzo surface of degree 1 is isomorphic to a hypersurface V (F ) of degree 6 in P(1, 1, 2, 3). Remark 8.3.1. The singularities of the branch curves of the double cover S → P2 (d = 2) and S → Q ((d = 1) are simple singularities. This means that in appropriate analytic (or formal) coordinates they are given by one of the following equations: An Dn E6 E7 E8 : : : : : x2 + y n+1 = 0, n ≥ 1 y(x2 + y n−2 = 0, n ≥ 4 x3 + y 4 = 0 x3 + xy 3 = 0 x3 + y 5 = 0 (8.11) (8.10)

This easily follows from Theorem 8.2.20.

8.4
8.4.1

Surfaces of degree d in Pd
Projective normality

We saw that the image of a weak Del Pezzo surface of degree d > 2 under the map given by the linear system | − KS | is a nondegenerate surface of degree d in Pd . We call this surface an anticanonical model of S. It is a normal surface with canonical singularities. Let us prove the converse. First we need the following.

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Lemma 8.4.1. Let C be a nondegenerate nonsingular irreducible curve of degree d > 2 in Pd−1 . Then g ≤ 1 and the equality takes place if and only if the restriction map r : H 0 (Pd−1 , OPd−1 (1)) → H 0 (C, OC (1)) is surjective. Proof. Let H be a hyperplane section of C. We have dim H 0 (C, OC (H)) = d + 1 − g + dim H 1 (C, OC (H)) ≥ d, (8.12)

and the equality takes place if and only if C is projectively normal. Thus we obtain g ≤ 1 + dim H 1 (C, OC (H)) = 1 + dim H 0 (C, OC (KC − H)). (8.13)

If |KC − H| = ∅, then g ≤ 1 and the equality takes place if and only if r is surjective. Assume |KC − H| = ∅. By Clifford’s Theorem [134], dim H 0 (C, OC (KC − H)) ≤ 1 +
1 2

deg(KC − H) = 1 + (g − 1 − 1 d) = g − 1 d, 2 2

1 unless KC = H or C is a hyperellptic curve and KS − H = kg2 for some k > 0. 1 If we are not in one of the exceptional cases, we obtain g ≤ 1 + g − 2 d which is a contradiction. If KC = H, we get d = 2g −2 and (8.12) gives g = 2g −2+1−g +1 ≥ 1 2g − 2, hence g = 2 and d = 2g − 2 = 2, a contradiction. If KC − H = kg2 , then 1 1 1 H = (g − 1)g2 − kg2 = (g − 1 − k)g2 . Since deg H > 2, we get k < g − 2. Thus dim H 0 (C, OC (KC − H)) = k + 1, and (8.13) gives g ≤ 1 + k + 1 < 2 + g − 2 = g, a contradiction again. Thus |KC − H| = ∅ and we are done.

Theorem 8.4.2. Let X be a nondegenerate normal surface of degree d in Pd . Assume that X is not a projection of a surface of d in Pd+1 and has at most canonical singularities. Then X is an anticanonical model of a weak Del Pezzo surface S. Proof. First of all V is a rational surface. By projection from a general point on the surface we obtain that X is birationally isomorphic to a surface of degree d−1 in Pd−1 . Continuing in this way we obtain that X is birationally isomorphic to a cubic surface Y in P3 . Since X has canonical singularities, Y cannot be a cubic cone. We will see in the next Chapter that an irreducible cubic surface is rational except when it is a cone. Let π : S → X be a minimal resolution of singularities. Since X has only canonical singularities, π ∗ ωX ∼ ωS . Since X is normal, π∗ (OS ) = OX , and by the projection = formula, π∗ ω S ∼ ω X . = This implies that the canonical homomorphism H 1 (X, ωX ) → H 1 (S, ωS ) is injective. Since S is a nonsingular rational surface, we get H 1 (S, ωS ) ∼ H 1 (S, OS ) = 0. Thus = H 1 (X, ωX ) ∼ H 1 (X, OX ) = 0. = Let C be a general hyperplane section of X. Since X is normal, it is a smooth curve. By the previous lemma, its genus is 0 or 1. The exact sequence 0 → OX → OX (1) → OC (1) → 0. (8.14)

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shows that the restriction homomorphism H 0 (X, OX (1)) → H 0 (C, OC (1)) is surjective. If C is of genus 0, we have h0 (OC (1)) = deg OC (1) + 1 = d + 1. This implies that h0 (OX (1)) = d + 1, hence |OX (1)| is not complete and X is a projection of a surface in Pd+1 . Thus we may assume that C is an elliptic curve. Let us identify it with its pre-image under π. By the adjunction formula, OC = ωC = OS (KS + C) ⊗ OC . By Riemann-Roch,
1 dim H 0 (S, OS (KS + C)) = 2 (KS · C + C 2 ) + 1 = 1.

Thus |KS +C| consists of an isolated curve D. Since D·π ∗ (H) = 0 for any hyperplane section H of V not passing through the singularities, we obtain that each irreducible component R of D is contained in the exceptional curve of the resolution π. Since V has only canonical singularities, R is a (−2)-curve. Since (KS + C) · R = KS · R + C · R = 0 for any irreducible component of a resolution, we get D2 = 0. Since the sublattice of Pic(S) generated by the components of a Dynkin curve is negative definite, we get D = 0. Thus KS + C ∼ 0 and −KS = π ∗ OV (1) is nef and big. 2 So, S is a weak Del Pezzo surface of degree KS = d. Clearly, S is its anticanonical model. Corollary 8.4.3. Let X be a nondegenerate normal surface of degree d in Pd . Assume that X has at most canonical singularities and is not a projection of a surface of degree d in Pd+1 . Then d ≤ 9. Moreover, V is either surface of degree 8 in P8 isomorphic to the image of Fn (n = 0, 2) under the map defined by the linear system | − 2KFn |, or a projection of the Veronese surface v3 (P2 ) ⊂ P9 . Proof. Use that a weak Del Pezzo surface of degree ≥ 3 not isomorphic to F0 or F2 is the blow-up of N ≤ 8 bubble points x1 , . . . , xN in P2 and the linear system | − KS | = |3 − x1 − . . . − xN |. It is a subsystem of the complete linear system |3 | defining a Veronese map v3 : P2 → P9 . Along the same lines one proves that any normal surface of degree d in Pd is one of the following surfaces: (i) a projection of a normal surface of degree d in Pd+1 ; (ii) a cone over a normal elliptic curve lying in a hyperplane; (iii) an anticanonical model of a weak Del Pezzo surface. Recall that a nondegenerate subvariety X of a projective space Pn is called projectively normal if X is normal and the natural restriction map H 0 (Pn , OPn (m) → H 0 (X, OX (m)) is surjective for all m ≥ 0. This can be restated in terms of vanishing of cohomology H 1 (Pn , IX (m)) = 0, m > 0 (resp. m = 1),

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273

where IX is the ideal sheaf of X. It is known that X is projectively normal if and only if its projective coordinate ring k[X] is a normal domain. One shows that the integral closure is the ring k[X] = H 0 (X, OX (m)).
m≥0

Theorem 8.4.4. Let X be an anticanonical model of a weak Del Pezzo surface of degree d ≥ 4. Then X is projectively normal. Proof. The linear normality follows immediately from exact sequence (8.14) (use that h0 (OC (1)) = deg OC (1) = d). Let H be a general hyperplane. Tensoring the exact sequence 0 → OPn (m − 1) → OPn (m) → OH (m) → 0 with IX we get an exact sequence 0 → IX (m − 1) → IX (m) → IH∩X (m) → 0. (8.15)

We know from the proof of Lemma 8.4.1 that E = H ∩ X is an irreducible linearly normal curve of genus 1. Let H be a general hyperplane in H = Pd−1 . The exact sequence 0 → IE (m − 1) → IE (m) → IH ∩E (m) → 0 gives a surjection H 1 (E, IE (1)) → H 1 (E, IE (2)). By linear normality of E we obtain H 1 (E, IE (1)) = H 1 (E, IE (2)) = 0. Continuing in this way we prove the projective normality of E ⊂ H. The same cohomology game with exact sequence (8.15) gives a surjective map rm : H 1 (X, IX (m − 1)) → H 1 (X, IX (m)) for m ≥ 1. The exact sequence 0 → IX → OPd → OX → 0 together with vanishing of H 1 (X, OX ) and H 1 (Pd , OPd ) show that H 0 (X, IX ) = 0. The surjectivity of r1 implies H 1 (X, IX (1)) = 0. Now the surjectivity of r2 implies H 1 (X, IX (2)) = 0, and so on.

8.4.2

Surfaces of degree ≥ 7

A weak Del Pezzo surface of degree 9 is isomorphic to P2 and its anticanonical model is isomorphic to the Veronese surface v3 (P2 ) of degree 9 in P8 . It does not contain lines. A Del Pezzo surface of degree 8 is isomorphic to either to P1 × P1 or to F1 , the blow-up of one point in the plane. The anticanonical model of the first surface is a hyperplane section of the Veronese 3-fold v2 (P3 ) ⊂ P9 . It does not contain lines. The anticanonical model of F1 is a projection of the Veronese surface v3 (P2 ) from its point. It contains one line. A weak Del Pezzo surface of degree 8 is isomorphic to the ruled surface F2 . Its bicanonical model is a section of of the Veronese 3-fold by a tangent hyperplane. The anticanonical model of a weak Del Pezzo surface of degree 7 is a projection of the Veronese surface from a secant line of the surface. If the secant line is a tangent line, the projection acquires a singular point of type A1 .

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Let ρ : Aut(S) → O(Pic(S)) be the representation of the automorphism group in the Picard group of a Del Pezzo surface S. We described its kernel in section 8.2.6. If S = F0 , then the image is the group of order 2 which permutes the divisor classes of the two rulings on F0 . If S is of degree 2, then again the image is of order 2. It acts by permuting the classes of the two exceptional curves. In the remaining cases, the image is trivial.

8.4.3

Surfaces of degree 6 in P6

Let X be a nondegenerate surface of degree 6 in P6 with at most canonical singularities. By Theorem 8.4.2 its minimal resolution σ : S → X is a weak Del Pezzo surface of degree 6 and X is isomorphic to its anticanonical model. Let π : S → P2 be the blowing-down structure on S. It is the blow-up of three points x1 , x2 , x3 in an almost general position. In this case it means that the corresponding bubble cycle η = x1 + x2 + x3 , up to admissible order, is one of the following (i,i’) x1 , x2 , x3 are three proper non-collinear (collinear) proper points; (ii, ii’) x2 (iii, iii’) x3 x1 , x3 are non-collinear (collinear) proper points; x2 x1 are non-collinear (collinear) r points.

In the cases (i),(ii) and (iii) the bubble cycle η = x1 + x2 + x3 defined the homaloidal net |OP2 (2) − η| defining a quadratic Cremona transformation φ. Thus S defines a resolution (π, σ) of φ with the birational morphism σ : S → P2 defined by blowing down 3 exceptions configurations with the divisor classes e0 −e1 −e2 , e0 −e1 −e3 , e0 − e2 − e3 , where e0 , e1 , e2 , e3 is a geometric basis defined by π. Let Γφ be the graph of φ. The canonical map α : S → Γφ is a resolution of singularities. Let τ : S → P8 be the composition s Φ : S → Γφ → P2 × P2 → P8 , where the last map is the Segre map. This map is given by the linear system |e0 + e0 |, where e0 = 2e0 − e1 − e2 − e3 and |e0 | is the homaloidal net defining the quadratic map φ. Since e0 + e0 = 3e0 − e − 1 − e2 − e3 = −KS we obtain that Φ is defined by the anticanonical linear system on S and hence its image is a surface of degree 6 in P6 . This shows that X is isomorphic to the intersection of the Segre variety s(P2 × P2 ) with a linear subspace of codimension 2. It also implies that X ∼ Γφ . = Since there are three possible quadratic transformations up to composition with a projective automorphism, we obtain three non-isomorphic surfaces X; • X1 corresponding to three non-collinear proper points x1 , x2 , x3 ; • X2 corresponding to three non-collinear points x2 • X3 corresponding to three non-collinear points x3 x1 , x3 ; x2 x1 .

The surface Xi is isomorphic to the graph of the quadratic transformation τi from section 7.2.1. The surface X1 is nonsingular and has 6 lines on it. They form a hexagon. The surface X2 has one RDP of type A1 . It has 3 lines, two of them pass through the

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singular point. The surface X3 has one RDP of type A2 . It has 1 line passing through the singular point. Let X1 , X2 , X3 be the surfaces corresponding to cases (i)’, (ii)’ and (iii)’, respectively. They are not linear sections of the Segre variety. We leave to check the following properties of the surfaces to the reader. The surface X1 has a unique RDP of type A1 and has three lines passing through this point. The surface X2 has two RDPs of type A1 . It has 2 lines, one of them joins the two singular points. The surface X3 has 2 singular points of types A2 and A1 . It has one line passing through the point of type A2 . Remark 8.4.1. The surfaces X1 , X2 , X2 , X3 are examples of toric surfaces. They contain an open Zariski subset isomorphic to a complex torus (C∗ )2 which acts on X extending its action on itself by translations. If x1 = [1, 0, 0], x2 = [0, 1, 0], x3 = [0, 0, 1], then the torus in X1 is the pre-image in X1 of the complement of the coordinate axes ti = 0. Its complement in X1 is the hexagon of lines. The same description of the torus is true in the case X3 (resp. X2 ) if we choose x2 to be the tangent direction t2 = 0 (resp. and x3 = [0, 0, 1]). The corresponding fans defining the toric surfaces are the following.             • (−1, −1) X1       oo ooo ooo oo •o (−2, −1) X2 •   (1, 1) (−2, y 1)  • yyy  yyy  yyy  yc c cc cc cc cc c • (1, −1) X3

j jjjj jjjj j • (−3, −1)

X2

Let Sec(X) be the secant variety of X. Its expected dimension is equal to 5. In fact, we have Proposition 8.4.5. Let S be a weak Del Pezzzo surface whose anticanonical model X is contained in the Severi variety S2,2 = s(P2 × P2 ) ⊂ P8 . Then dim Sec(X) = 5.

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Proof. It is well know that the secant variety of S2,2 is the determinantal cubic hypersurface D of degenerate 3 × 3 matrices whose entries are 9 unknowns in P8 . Let z be a general point in D and consider the entry variety Yz = Tz D ∩ S2,2 representing the points on Y lying on the secants through z. The cone over Yz with vertex at z is the union of secants passing through z. Since the projection of a cubic from its point is a quadric and Tz X is a plane, it is easy to see that Yz is a conic. Since X is the intersection of S2,2 by a linear subspace of codimension 2, the end varieties of Sec(X) must be either conics or pairs of points. Since X contains only one-dimensional family of conics (represented by lines through points x1 , x2 , x3 ), we get that through any general point of Sec(X) passes only one conic. This easily gives that dim Sec(X) = 5, as is expected. Theorem 8.4.6. Assume X is a nonsingular surface of degree 6 in P6 . Then X is projectively equivalent to the subvariety given by equations   t0 t 1 t 2 rank t3 t0 t4  ≤ 2. t5 t 6 t 0 The secant variety Sec(X) is the cubic hypersurface defined by the determinant of this matrix. Proof. The surface X is isomorphic to the blow-up of three non-collinear proper points. We may assume that x1 = [1, 0, 0], x2 = [0, 1, 0], x2 = [0, 0, 1]. The linear system | − KX | is the proper transform of the linear system of cubics through the three points. It is generated by the cubics V (f ), where f is a monomial in coordinates z0 , z1 , z2 in 3 3 3 P2 different from z0 , z1 , z2 . Let us order them in the following way
2 2 2 2 2 2 z0 z 1 z2 , z 1 z2 , z 1 z2 , z 0 z2 , z 0 z2 , z 0 z1 , z 0 z1 .

The surface X is projectively equivalent to the image of P2 under the rational map given by these 7 homogeneous polynomials t0 , . . . , t6 . The relations between these polynomials are the minors of the matrix. This shows that X is contained in the intersection of 9 quadrics given by the minors. On the other hand, by Lemma 8.3.1, we have h0 (X, OX (−2KS )) = 19. Since X is projectively normal, the dimension of the linear system of quadrics containing X is equal to 8. This shows that the nine minors generate the linear system of quadrics containing X [146]. Let B be the base scheme of the linear system of quadrics. Suppose dim B > dim X. A general hyperplane section of S is a projectively normal elliptic curve of degree 6 in P5 . It is known that it is given by 9 linearly independent quadrics. This shows that dim B = dim X. It follows from the theory of determinant varieties [5], Chapter 2, Prop. 4.1, that B is a Cohen-Macaulay surface containing a nonsingular surface X. It must coincide with X. Since the rank of the sum of two matrices of rank ≤ 1 is at most 2, we see that Sec(X) is contained in the determinantal cubic hypersurface. By the previous proposition, it must be its irreducible component. However, by loc. cit., the determinantal cubic is irreducible.

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277

Let us describe the group of automorphisms of a Del Pezzo surface of degree 6. The surface is obtained by blowing up 3 non-collinear points p1 , p2 , p3 . We may assume that their coordinates are [1, 0, 0], [0, 1, 0], [0, 0, 1]. We know from section 8.2.6 that the kernel of the representation ρ : Aut(S) → O(Pic(S)) is a 2-dimensional torus. The root system is of type A2 + A1 , so the Weyl group is isomorphic to S3 × C2 . Let us show that the image of the Weyl representation is the whole group. The subgroup S3 is the image of the group of automorphisms of S induced by automorphisms of the projective plane which which permute the coordinates. The generator of the cyclic group of order 2 is induced by the standard Cremona transformation. It is easy to see this as follows. The surface S is isomorphic to the blow-up of 2 points on P1 × P1 not lying on the same fibre of any ruling. By coordinate change, we may assume that the points have coordinates x1 = ((1, 0), (1, 0)) and x2 = ((0, 1), (0, 1)). The torus is represented by homotheties on each factor of the product. The standard Cremona transformation is represented by the automorphism given in inhomogeneous coordinates on the factors by (x, y) → (1/x, 1/y). The subgroup S3 is generated by two elements of order 2 defined by a switch and the automorphism of S induced by the product of elementary transformations elmx1 elmx2 . We leave to the reader to verify the following. Theorem 8.4.7. Let S be a Del Pezzo surface of degree 6. Then Aut(S) ∼ (C∗ )2 = (S3 × S2 ).

If we represent the torus as the quotient group (C∗ )3 by the diagonal subgroup ∆(C∗ ), then the subgroup S3 acts permutations of factors, and the cyclic subgroup S2 acts by the negation. Note that the Weyl group S3 × S2 realized by automorphisms acts on the set of lines on S. Its incidence graph is a hexagon. The group S3 is isomorphic to the dihedral group D6 and acts on the graph in the same way as the dihedral group acts by symmetries of a regular hexagon. The generator of S2 is also a symmetry but it is not induced by a motion of the plane.

8.4.4

Surfaces of degree 5

Let X be a nondegenerate surface of degree 5 in P5 with at most canonical singularities. By Theorem 8.4.2 its minimal resolution σ : S → X is a weak Del Pezzo surface of degree 5 and X is isomorphic to its anticanonical model. Proposition 8.4.8. Let X be a nonsingular Del Pezzo surface of degree 5 in P5 . Then X is isomorphic to a linear section of the Grassmann variety G(2, 5) of lines in P4 . Proof. We use some elementary facts about Grassmannians which we recall in a later Chapter. It is known that the degree of G = G(2, 5) in the Pl¨ cker embedding is equal u to 5 and dim G = 6. Also is know that the canonical sheaf is equal to OG (−5). By the adjunction formula, the intersection of G with a general linear subspace of codimension 4 is a nonsingular surface X with ωX ∼ OX (−1). This must be a Del Pezzo surface = of degree 5. Since all Del Pezzo surfaces of degree 5 are isomorphic the assertion follows.

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Remark 8.4.2. Let E be the restriction of the tautological rank 2 quotient bundle on G to X. Then E is isomorphic to a rank 2 bundle on S generated by 5 sections with Chern classes c1 = −KS and c2 = 2. One can show that this vector bundle is given by an extension 0 → OS → E → IZ (3) → 0, where IZ is the ideal sheaf of the closed subscheme defined by any two points x, y such that x1 , . . . , x4 , x, y are in an almost general position. Corollary 8.4.9. Let X be a Del Pezzo surface of degree 5 in P5 . Then its homogeneous ideal is generated by 5 linearly independent quadrics. Proof. Since X is projectively normal, applying Lemma 8.3.1, we obtain that the linear system of quadrics containing X is of dimension 4. It is known that the homogeneous ideal of the Grassmannian G(2, 5) is generated by 5 quadrics. So, the restriction of this linear system to its linear section, we obtain the quadrics containing X define X scheme theoretically. Let X be an anticanonical model of a Del Pezzo surface of degree 5. The linear system of cubics in P5 containing X is of dimension 24. Let us see that any nonsingular cubic fourfold containing X is a rational ( the rationality of a general cubic fourfold is unknown at the moment). Lemma 8.4.10. Let X be an anticanonical model of weak Del Pezzo surface S of degree 5 in P5 . For any general point z ∈ P5 there exists a unique secant of X containing z. Proof. It is known that Sec(X) = P5 . This follows from Severi’s Theorem that any non-degenerate surface in P5 with secant variety of dimension 4 is a Veronese surface (see [250]). Let x, y ∈ X such that z ∈ = x, y. We may assume that they are distinct nonsingular points on X. Consider the projection p : X− → P3 be the projection with center . Its image is a cubic surface isomorphic to the anticanonical model of the blow up of S at the pre-images x , y of x, y on S. Here we use that the points x , y do not lie on (−2)-curves on S, hence the blow-up of x , y is a weak Del Pezzo surface of degree 3. The map p is an isomorphism outside x, y. Suppose z belongs to another secant = x , y . Then the plane generated by and defines a point on the cubic surface such that the pre-image under the projection map π contains x , y . This contradiction proves the assertion. Theorem 8.4.11. Let F be an irreducible cubic fourfold containing a non-degenerate nonsingular surface of degree 5 in P5 . Then F is a rational variety. Proof. Consider the linear system |IX (2) of quadrics containing X. It defines a morphism Y → P4 whose fibres are proper transforms of secants of X. This shows that the subvariety of G(2, 6) formed by secants of X is isomorphic to P4 . Take a general point in z ∈ F . By the previous lemma there exists a unique secant of X passing through z. By Bezour’s Theorem no other point outside X lies on this secant. This gives rational injective map F − → P4 defined outside X. Since a general secant intersects F at three points, with to of them on X, we see that the map is birational.

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279

Remark 8.4.3. According to a result of A. Beauville [16], Proposition 8.2, any smooth cubic fourfold containing X is a pfaffian cubic hypersurface, i.e. is given by the determinant of a skew-symmetric matrix with linear forms as its entries. Conversely, any pfaffian cubic fourfold contains a nondegenerate surface of degree 5, i.e. an anticanonical weak Del Pezzo or a scroll. Let us look at singularities and lines on a weak Del Pezzo surface of degree 5. Let π : S → P2 be the blowing down structure on S. It is the blow-up of four points non-collinear points x1 , x2 , x3 , x4 in an almost general position. In this case it means that the corresponding bubble cycle η = x1 + x2 + x3 + x4 , up to admissible order, is one of the following (i) x1 , x2 , x2 , x3 , x4 are proper points; (ii) x2 (iii) x3 (iv) x2 (v) x4 x2 , x3 , x4 ; x2 x1 , x4 ; x3 , x2 x1 . x1 ;

x1 , x4 x3 x2

As we have seen in section 8.2.5 the singularities of X correspond to root bases in the lattice E4 . The possibilities are A1 , A1 + A1 , A2 , A1 + A2 , A3 , A4 . All these cases are realized: A1 : x2 A1 + A1 : x2 A2 : x3 x2 x1 , x2 , x3 , no three points are collinear; x1 , x4 x3 , no three points are collinear;

x1 , x4 , no three points are collinear; x2 x2 x1 , x4 ; x1 , x2 , x3 are collinear; x1 , x4 , x1 , x2 , x3 are not collinear; x2 x1 , x1 , x2 , x3 are collinear.

A1 + A2 : x3 A3 : x4 x3

A4 : x4

x3

Since any set of four points in general position is projectively equivalent to the set x1 = [1, 0, 0], x1 = [1, 0, 0], x1 = [1, 0, 0], x1 = [1, 0, 0], we obtain that all Del Pezzo surfaces of degree 6 or 5 are isomorphic. A Del Pezzo surface of degree 5 has 10 lines. The union of them is a divisor in | − 2KS |. The incidence graph of the set of 10 lines is the famous Peterson graph. The number of lines on a weak Del Pezzo surface depends on the structure of its Dynkin curves, or, equivalently, singularities of its anticanonical model. It is easy to derive the following table for the number of lines on a singular anticanonical model of a weak Del Pezzo surface of degree 5.

280

CHAPTER 8. DEL PEZZO SURFACES T iiii! T iiii !! TTT ii ! TT ii 11 rrii rr TT !! 11 rr iiiiiiTTT TT i 11 TT 11 TT TT 11 11 T 11 11 ØØ ØØ 1 11 ØØ ØØ …… ØØ 11 vvvv ……………BØØØ Ø BB …… vv …… ØØ ………… BB ØØ ………… B Ø ……ØØ Table 8.4: Peterson graph A1 7 A1 + A1 4 A2 3 A1 + A2 1 A3 1 A4 1

Table 8.5: Lines a Del Pezzo surface of degree 5 ∼ The Weyl group W (E4 ) is isomorphic to the Weyl group W (A4 ) = S5 . If S has only one (−2)-node, the group of Cremona isometries is a isomorphic to S4 . It acts on the set of 10 elements with 7 orbits. There are 3 orbits with stabilizer of order 12 and 4 orbits with stabilizer order 24. Let us study automorphisms of a Del Pezzo surface of degree 5. Theorem 8.4.12. Let S be a Del Pezzo surface of degree 5. Then Aut(S) ∼ S5 . = Proof. The group S5 is generated by its subgroup isomorphic to S4 and an element of order 5. The subgroup S4 is realized by projective transformations pemuting the points p1 , . . . , p4 . The action is realized by the standard representattion of S4 in the hyperplane z1 + · · · + z4 = 0 of C4 identified with C3 by the projection to the first 3 coordinates. An element of order 5 is realized by a quadratic transformation with fundamental points x1 , x2 , x3 defined by the formula T : [t0 , t1 , t2 ] → [t0 (t2 − t1 ), t2 (t0 − t1 ), t0 t2 ]. (8.16)

It maps the line t0 = 0 to the point x2 , the line t1 = 0 to the point x4 , the line t2 = 0 to the point p1 , the point x4 to the point x3 . Note that the group of automorphisms acts on the graph of 10 lines and defines an isomorphism with its group of symmetries. Let S be a Del Pezzo surface of degree 5. The group Aut(S) ∼ S5 acts on linearly = on the space V = H 0 (S, OS (−KS )) ∼ C6 . Let us compute the character of this = representation. Choose the following basis in the space V : (t2 t1 − t0 t1 t2 , t2 t2 − t0 t1 t2 , t2 t0 − t0 t1 t2 , t2 t2 − t0 t1 t2 , t2 t0 − t0 t1 t2 , t2 t1 − t0 t1 t2 ). 0 0 1 1 2 2 (8.17)

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281

Let s1 = (12), s2 = (23), s3 = (34), s4 = (45) be the generators of S5 . It follows from the proof of Theorem 8.4.12 that s1 , s2 , s3 generate the subgroup of Aut(S) which is realized by projective transformations permuting the points x1 , x2 , x3 , x4 . The last generator is realized by a quadratic transformation T . Choose the following representatives of the conjugacy classes in S5 ; g1 = (12), g2 = (123) = s2 s1 , g3 = (1234) = s3 s2 s1 , g4 = (12345) = s4 s3 s2 s1 , g5 = (12)(34) = s1 s3 , g6 = (123)(45) = s3 s2 s1 s4 . The subgroup generated by s1 , s2 acts by permuting the coordinates t0 , t1 , t2 . The generator s3 acts as the projective transformation s3 : (y1 , . . . , y6 ) → (−y1 , y1 +y2 , −y3 , y3 +y4 , −y1 −y−2+y4 +y6 , y2 −y3 −y4 +y5 ), where (y1 , . . . , y6 ) is the basis from (8.17). Finally s4 acts by formula (8.16). The simple computation gives the character vector of the representation χ = (χ(1), χ(g1 ), χ(g2 ), χ(g3 ), χ(g4 ), χ(g5 ), χ(g6 )) = (6, 0, 0, 0, 1, −2, 0). Using the character table of S5 we find that χ is the character of an irreducible representation isomorphic to the second exterior power of the standard 4-dimensional representation of S5 (see [118], p. 28). Now let us consider the linear representation of S5 on the symmetric square S 2 (V ). Its character χS 2 V can be easily found using the standard facts about linear representation of finite groups. Using the formula
1 χS 2 V (g) = 2 (χ(g)2 + χ(g 2 )),

we get χS 2 V = (21, 3, 0, −1, 1, 5, 0). Taking the inner product with the character of the trivial representation we get 1. This shows that dim S 2 V S5 = 1. Similarly, we find that dim S 2 V contains one copy of the one-dimensional sign representation of S5 . The equation of the union of 10 lines is defined by the equation F = t0 t1 t2 t3 (t2 − t2 )(t2 − t2 )(t2 − t2 ) = 0. 0 1 0 2 1 2 It is immediately to check that F transforms under S5 as the sign representation. It is less trivial but straightforward to find a generator of the vector space S 2 V S5 . It is equal to G=2 t4 t2 − 2 t4 tj tk − t3 t2 tk + 6t2 t2 t2 . i j i i j 0 1 2 Its singular points are the reference points. In another coordinate system, the equation looks even prettier: t6 + t6 + t6 + (t2 + t2 + t2 )(t4 + t4 + t4 ) − 12t2 t2 t2 = 0. 0 1 2 0 1 2 0 1 2 0 1 2 (see [96]). The singular points are ([1, −1, −1], [−1, 1, −1], [−1, −1, 1], [1, 1, 1]. The S5 -invariant plane sextic W = V (G) is called the Wiman sextic. Its proper transform on S is a smooth curve of genus 6 in | − 2KS |. All curves in the pencil of sextics spanned by V (λF + µG) (the Wiman pencil) are A5 -invariant. It contains two S5 invariant members V (F ) and V (G).

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Remark 8.4.4. It is known that a Del Pezzo surface of degree 5 is isomorphic to the GIT-quotient of the space (P1 )5 by the group SL(2) (see [84]). The group S5 is realized naturally by the permutation of factors. The isomorphism is defined by assigning to any point x on the surface the five ordered points (p1 , . . . , p5 = x), where p1 , . . . , p4 are the tangent directions of the conic in the plane passing through the points x1 , x2 , x3 , x4 , x.

8.5
8.5.1

Quartic Del Pezzo surfaces
Equations

Here we study in more details weak Del Pezzo surfaces of degree 4. They are obtained by blowing up 5 points in P2 and hence vary in a family. Surfaces of degree 3 will be studied in the next Chapter. Theorem 8.5.1. Let X be an anticanonical model of a weak Del Pezzo surface S of degree 4. Then S is a complete intersection of two quadrics in P4 . Moreover, if X is nonsingular, then the equations of the quadrics can be reduced, after a linear change of variables, to the diagonal forms:
4 4

t2 = i
i=0 i=0

ai t2 = 0, i

where ai = aj for i = j. Proof. By Theorem 8.4.4, X is projectively normal in P4 . This gives the exact sequence 0 → H 0 (P4 , IX (2)) → H 0 (P4 , OP4 ) → H 0 (X, OX (2)) → 0. By Lemma 8.3.1, dim H 0 (X, OX (2)) = dim H 0 (S, OS (−2KS )) = 13. This implies that X is the base locus of a pencil t0 Q0 + t1 Q1 = 0 of quadrics. Assume X is nonsingular. The determinant equation shows that it contains 5 singular quadrics counting with multiplicities. Let us see that we have 5 distinct singular quadrics in the pencil. This follows from the general properties of linear systems of quadrics. We will give the details in the second volume of the book. Here we only sketch the facts without proof. A linear pencil of quadrics is defined by a line in the projective space |OPd (2)| of quadrics in Pd . Let ∆ be the discriminant hypersurface. Its singular locus consists of quadrics of corank > 1. None of these quadrics is contained in our pencil since otherwise the base locus is obviously singular. For any nonsingular point Q ∈ ∆, the tangent space of ∆ at Q can be identified with the linear space of quadrics passing through the singular point of Q. Again, since the base locus X of our pencil is a nonsingular surface, we obtain that X does not contain a singular point of a singular

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283

quadric from the pencil. This shows that the line intersects ∆ transversally, and hence contains exactly 5 = deg ∆ of singular quadrics. Now we are in business. Let p1 , . . . , p5 be the singular points of singular quadrics from the pencil. We claim that the points span P4 . In fact, otherwise we obtain a pencil P of quadrics in some hyperplane H with ≥ 5 singular members, hence all quadrics in this pencil must be singular. By Bertini Theorem, there is a point q ∈ H singular for all quadrics, hence the base locus of the pencil consists of the union of 4 lines through q taken with multiplicities. The base locus of P is the hyperplane section H ∩ X of X. Since X is nonsingular and its hyperplane section is a curve of arithmetic genus 1, it is easy to see that it does not contain the union of 4 lines with a common point. So, the points p1 , . . . , p5 span P4 . Choose coordinates such that p1 = (1, . . . , 0) and so on. Let Qt = V (ft ) be a quadric from the pencil. Its first polar with respect to pi t is given by the equation lt = ∂Fi = 0. Since the polar does not depend on t we may ∂x assume that lt = ai (t0 , t1 )li , where li is a linear function in the variables x0 , . . . , x4 and ai (t0 , t1 ) = ai t0 + bi t1 is a linear function in variables t0 , t1 . The linear forms li are obviously linear independent since otherwise all quadrics have a common singular point contradicting the nonsingularity of the base locus. Now again choose coordinates to assume that li = xi . By Euler formula,
4 4 4

ft =
i=0

ai (t0 , t1 )x2 = t0 ( i
i=0

ai x2 ) + t1 ( i
i=0

bi x2 ). i

Note that the singular quadrics from the pencil correspond to (t0 , t1 ) = (bi , −ai ). Their singular points are (1, 0, . . . , 0) and so on. After a linear change of variables in the coordinates of the pencil we may assume that bi = 1, i = 0, . . . , 4. This gives the equations from the assertions of the theorem. Since all points (bi , ai ) = (1, ai ) are distinct we see that ai = aj for i = j. Let X be an anticanonical model of a Del Pezzo surface S of degree 4. It is a nonsingular quartic surface given by the equations from Theorem 8.5.1. Following the classic terminology an anticanonical model of a weak Del Pezzo surface of degree 4 is called a Segre quartic surface. One can say more about equations of singular weak Del Pezzo quartics. Let Q be a pencil of quadrics in Pn . We view it as a line in the space of symmetric matrices of size n + 1 spanned by two matrices A, B. Assume that Q contains a nonsingular quadric, so that we can choose B to be a nonsingular matrix. Consider the λ-matrix A + λB and compute its elementary divisors. Let det(A + λB) = 0 has r different roots α1 , . . . , αr . For every root αi we have elementary divisors of the matrix A + λB (λ − αi )ei , . . . , (λ − αi )ei
(1) (si )

,

ei

(1)

≤ . . . ei

(si )

.

The Segre symbol of the pencil Q is the collection [(e1 . . . e1 1 )(e2 . . . , e2 2 ) . . . (e(1) . . . , e(sr ) )]. r r It is a standard result in linear algebra (see [119] or [144] that one can simultaneously reduce the pair of matrices (A, B) to the form (A , B ) (i.e. there exists an invertible
(1) (s ) (1) (s )

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matrix C such that CAC t = A , CBC t = B ) such that the corresponding quadratic forms Q1 , Q2 have the following form
r si

Q1 Q2 where

=
i=1 j=1 si r

p(αi , ei j ), q(ei j ),
i=1 j=1 s

s

=

e

e−1

p(α, e) q(e)

= α
i=1 e

ti te+1−i +
i=1

ti+1 te+1−i ,

=
i=1

ti te+1−i

It is understood here that each p(α, e) and q(e) are written in disjoint sets of variables. This implies the following. Theorem 8.5.2. Let X and X be two complete intersections of quadrics and P, P be the corresponding pencils of quadrics. Assume that P and P contains a nonsingular quadric. Let H and H be the set of singular quadrics in P and P considered as sets marked with the corresponding part of the Segre symbol. Then X is projectively equivalent to X if and only if the Segre symbols of P and P coincide and there exists a projective isomorphism φ : P → P such that φ(H) = H and the marking is preserved Applying this to our case n = 4, we obtain the following possible Segre symbols, where r is the number of singular quadrics in the pencil r = 5 [11111], r = 4 [(11)111], [2111], r = 3 [(11)(11)1], [(11)21], [311], [221], [(12)11]; r = 2 [14], [(13)1], [3(11)], [32], [(12)2], [(12)(11)]; r = 1 [5], [(14)]. Note that the case [(1, 1, 1, 1, 1)] leads to linearly dependent matrices A, B, so it is excluded for our purpose. Also in cases [(111)11], [(1111)1], [(112)1], [(22)1], there is a reducible quadric in the pencil, so the base locus is a reducible. Finally, the cases [(23)], [(23)], [(113)], [(122)], [(1112)] correspond to cones over a quartic elliptic curve.

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285

8.5.2

Cyclid quartics

Let X be a nonsingular quartic surface in P4 . Let us project X to P3 . First assume that the center of the projection p lies on X. Then the image of the projection is a cubic surface Y in P3 isomorphic to the blow-up of X at the point p. Let π : X ∼ S → P2 be = the blowing-down map. Since S is a Del Pezzo surface, the inverse of π is the blow-up of 5 distinct points p1 , . . . , p5 no three of which are collinear. Let p6 = π(p). The cubic surface Y is an anticanonical model of the blow-up of the bubble cycle p1 + · · · + p6 . If π(p) ∈ {p1 , . . . , p5 }, no line pi , pj , i < j ≤ 5, contains p6 and moreover p6 does not lie on the conic through p1 , . . . , p5 , we obtain a nonsingular cubic surface. The conditions are of course equivalent to that p does not lie on any of 16 lines on X. If it does we obtain a singular cubic surface with double rational points. The types depend on the set of lines containing p. Now let us assume that the center of the projection p does not lie on X. Let Qp be the unique quadric from the pencil which contains p. We assume that Q is a nonsingular quadric. We will see that the projection of X from p is a quartic surface singular along a nonsingular conic. In classical literature such a quartic surface is called a cyclide quartic surface. Theorem 8.5.3. Assume that the quadric Qp is nonsingular. Then the projection Y of X from p is a quartic surface in P3 which is singular along a nonsingular conic. Any irreducible quartic surface in P3 which is singular along a nonsingular conic arises in this way from a Segre quartic surface X in P4 . The surface X is nonsinguar if and only if Y is nonsingular outside the conic. Proof. First of all let us see that Y is indeed a quartic surface. If not, the projection is a finite map of degree 2 onto a quadric. In this case the restriction of the pencil of quadrics containing X to a line through p intersecting X has 2 base points. This implies that there is a quadric in the pencil containing this line and hence containing the point p. Since Qp is the unique quadric containing p, we see that it contains all lines connecting p with some point on X. Since the lines through a point on a nonsingular quadric are contained in the tangent hyperplane at this point, we wee that X is contained in a hyperplane which contradicts the non-degeneracy of the surface. Let H be the tangent hyperplane of Qp at p and C = H ∩ X. The intersection H ∩ Qp is an irreducible quadric in H with singular point at p. The curve C lies on this quadric and is cut out by a quadric Q ∩ H for some quadric Q = Q from the pencil. Thus the projection from p defines a degree 2 map from C to a nonsingular conic C equal to the projection of the cone H ∩ Qp . It span the plane in P3 equal to the projection of the hyperplane H. Since the projection defines a birational isomorphism from X to Y which is not an isomorphism over the conic K, we see that Y is singular along C. It is also nonsingular outside C (since we assume that X is nonsingular). Conversely, let C be a nonsingular conic in P3 . Consider the linear system |IC (2)| of quadrics through C. Choose coordinates to assume that C is given by equations t0 = t2 +t2 +t2 = 0. Then |IC (2)| is spanned by quadrics V (t0 ti ) and V (t2 +t2 +t2 ). 3 1 3 1 It defines a rational map f : P3 − → P4 given by the formula [y0 , . . . , y4 ] = [t2 , t0 t1 , t0 t2 , t0 t3 , t2 + t2 + t2 ]. 0 1 2 3

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Its image is the nonsingular quadric Q1 given by the equation
2 2 2 y1 + y2 + y3 − y0 y4 = 0.

(8.18)

The inverse rational map is of course the projection from the point [0, . . . , 0, 1] (the image of the plane t0 = 0). Let Y be an irreducible quartic surface in P3 singular along C. Its equation must be of the form (t2 + t2 + t2 )2 + 2t0 (t2 + t2 + t2 )g1 (t1 , t2 , t3 ) + t2 g2 (t0 , t1 , t2 , t3 ) = 0, 1 3 1 3 0 (8.19)

where g1 and g2 are homogeneous polynomials of degree 1 and 2, respectively. Its image lies on Q1 and is cut out by the quadric Q2 with equation
2 y4 + y4 g1 (y1 , y2 , y3 ) + g2 (y0 , y1 , y2 , y3 ) = 0.

(8.20)

Thus the image of Y is a Segre quartic X in P4 defined by equations (8.18) and (8.19). Obviously, X does not contain the point p = [0, 0, 0, 0, 1]. Then map f is an isomorphism outside the plane t0 = 0. Hence X is nonsingular if and only Y has no singular points outside C. Remark 8.5.1. After some obvious linear change of variables we can reduce the equation of a cyclide surface to the equation (t2 + t2 + t2 )2 + t2 g2 (t0 , t1 , t2 , t3 ) = 0, 1 1 3 0 (8.21)

The analog of a quartic cyclide surface in P2 is a quartic curve with two double points (a cyclide curve). Let be the line through the nodes. We may assume that its equation is x0 = 0 and the coordinates of the points are [0, 1, i], [0, 1, −i]. Then the equation of a cyclide curve can be reduced to the form (x2 + x2 )2 + x2 g2 (x0 , x1 , x2 ) = 0. 1 2 0 A conic passing through singular points of the cyclide is the projectivized circle (x1 − ax0 )2 + (x2 − bx0 )2 − cx2 = 0. 0 The linear system of circles maps P2 to P3 with the image a quadric in P3 . The coordinates in P2 corresponding to a choice of 4 linear independent circles are called in the classical literature the tetrahedral coordinates. An equation of a circle in this coordinates is a linear equation. A cyclide curve is given by a quadric equation in tetrahedral coordinates. Similarly quadric surfaces in P3 which contain the singular conic of the cyclide surface with equation (8.20) are projectivized balls (t1 − at0 )2 + (t2 − bt0 )2 + (t3 − t0 )2 − ct2 = 0 0 in P3 . The linear system of such balls is 4-dimensional and maps P3 to P4 with the image a quadric hypersurface. A choice of 5 linearly independent balls defines a pentaspherical coordinates in P3 . The equation of a ball in these coordinates is a linear

8.5. QUARTIC DEL PEZZO SURFACES

287

equation. By choosing a special basis formed by “orthogonal balls”, one my assume that the quadric hypersurface in P4 is given by the sum of squares of the coordinates. In these special pentahedral coordinates many geometric relationships between balls are expressed easier in terms of their linear equations (see [155]). In pentaspherical coordinates the equation of a cyclide quartic surface is the intersection of two quadrics. This explains the relation with quartic surfaces in P4 . It remains to consider the projection of a nonsingular Segre surface from a point p lying on a singular quadric Q from the pencil. First we may assume that p is not the singular point of Q. Then the tangent hyperplane H of Q at p intersects Q along the union of two planes. Thus H intersects X along the union of two conics intersecting at two points. This is a degeneration of the previous case. The projection is a degenerate cyclide surface. It is an irreducible quartic surface singular along the union of two lines intersecting at one point. Its equation can be reduced to the form t2 t2 + t2 g2 (t0 , t1 , t2 , t3 ) = 0. 1 2 0 Finally let us assume that the center of the projection is the singular point of a cone from the pencil. We have already observed that in this case we have a degre 2 map X → V , where V is a nonsingular quadric in P3 . The branch locus of this map is a nonsingular quartic elliptic curve of bi-degree (2, 2). If we choose the diagonal equations of X as in Theorem 8.5.1, and take point p = (1, 0, 0, 0, 0), then cone with vertex at p is given by the equation (a2 − a1 )t2 + (a3 − a1 )t2 + (a3 − a1 )t2 + (a4 − a1 )t2 = 0. 1 2 3 4 It is projected to the quadric with the same equations in coordinates [t1 , . . . , t4 ] in P3 . The branch curve is cut out by the quadric with the equation t2 + t2 + t2 + t2 = 0. 1 2 3 4 A more general cyclid quartic surfaces are obtained by projection from singular quartic surfaces in P3 . They have been all classified by C. Segre [222].

8.5.3

Singularities and lines

Applying the procedure of Borel-De Sibenthal-Dynkin, we obtain the following list of types of root bases in E5 . D5 , A3 + 2A1 , D4 , A4 , 4A1 , A2 + 2A1 , A3 + A1 , A3 , 3A1 , A2 + A1 , A2 , 2A1 , A1 . All of these types can be realized as the types of root bases defined by (−2)-curves. First we give the answer in terms of the blow-up model of X. D5 : x5 x4 x3 x2 x1 , x1 , x2 , x3 are collinear; A3 + 2A1 : x3 x2 x1 , x5 x4 , x1 , x4 , x5 are collinear; D4 : x4 x3 x2 x1 , x1 , x2 , x5 are collinear;

288

CHAPTER 8. DEL PEZZO SURFACES

A4 : x5 x4 x3 x2 x1 ; 4A1 : x2 x1 , x4 x3 , x1 , x2 , x5 andx3 , x4 , x5 are collinear; 2A1 + A2 : x2 x1 , x4 x3 , x1 , x2 , x5 andx3 , x4 , x5 are collinear; A1 + A3 : x3 x2 x1 , x5 4 , x1 , x4 , x5 are collinear; A3 : x4 x2 x1 ; or x3 x2 x1 , x1 , x4 , x5 are collinear; A1 + A2 : x3 x2 x1 , x5 x4 , x1 , x4 , x5 are collinear; 3A1 : x2 x1 , x4 , x3 , x1 , x3 , x5 are collinear; A2 : x3 x2 x1 ; 2A1 : x2 x1 , x3 x2 , or x1 , x2 , x3 , x1 , x4 , x5 are collinear; A1 : x1 , x2 , x3 are collinear. This can be also stated in terms of equations indicated in the next table. The number of lines is also easy to find by looking at the blow-up model. We have the following table (see [236]). D5 [(41)] 1 A3 [41] 5 A3 + 2A1 [(21)(11)] 2 A2 + A1 [32] 6 D4 [(31)1] 2 3A1 [(11)21] 6 A4 [5] 3 A2 [311] 8 4A1 [(11)(11)1] 4 2A1 [(11)111] 8 A2 + 2A1 [3(11)] 4 2A1 [221] 9 A3 + A1 [(21)2] 3 A1 [2111] 12 A3 [(21)11] 4

Table 8.6: Lines and singularities on a weak Del Pezzo surface of degree 4 Example 8.5.1. The quartic surfaces with singularities points of type 4A1 or 2A1 + A3 have a remarkable property that they admit a double cover ramified only at the singular points. We refer to [57] for more details about these quartics surfaces. The projections of these surfaces to P3 are cubic symmetroid surfaces discussed in the next Chapter. The cover is the quadric surface F0 in the first case and the quadric cone Q in the second case.

8.5.4

Automorphisms

By Theorem 8.5.1 a Del Pezzo surface of degree 4 is isomorphic to a nonsingular surface of degree 4 in P4 given by equations
4 4

f1 =
i=0

t2 = 0, i

f2 =
i=0

ai t2 = 0, i

where the coefficients ai are all distinct. We know that the representation of Aut(S) in W (S) ∼ W (D5 ) is injective. = Proposition 8.5.4. W (D5 ) ∼ 24 = S5 ,

where 2k denotes the elementary abelian group (Z/2Z)k .

8.5. QUARTIC DEL PEZZO SURFACES

289

Proof. Of course, this is a well-known fact from the theory of reflection groups. However, we give a geometric proof exhibiting the action of W (D5 ) on Pic(S). Fix a geometric basis e0 , . . . , e5 corresponding to a blow-up model of S and consider 5 pairs of pencils of conics defined by the linear systems Li = |e0 −ei |, Li = |−KS −(e0 −ei )| = |2e0 −e1 −e2 −e3 −e4 −e5 +ei |, i = 1, . . . , 5.

Let α1 , . . . , α5 be the canonical root basis defined by the geometric basis and ri be the corresponding reflections. Then r2 , . . . , r5 generate S5 and act by permuting the 5 pairs of pencils. Consider the product r1 r5 . It is immediately checked that it switches L4 with L4 and L5 with L5 leaving Li , Li invariant for i = 1, 2, 3. Similarly, a conjugate of r1 r5 in W (D5 ) does the same for some other pair pair of the indices. The subgroup generated by the conjugates is isomorphic to 24 . Its elements switch the Li with Li in an even number of pairs of pencils. This defines a surjective homomorphism W (D5 ) → S5 with kernel containing 24 . Comparing the orders of the groups we see that the kernel is 24 and we have an isomorphism of groups asserted in the proposition. The image of the addition map |Li | × |Li | → | − KS | defines a 3-dimensional linear system contained in | − KS |. It defines the projection ψi : S → P3 . Di · Di = 2 for Di ∈ Li , Di ∈ Li , the degree of the map is equal to 2. So the image of ψ is a quadric in P3 . This shows that the center of the projection is the vertex of one of the five singular quadric cones in the pencil of quadrics containing the anticanonical model X of S. The deck transformation gi of the cover is an automorphism and we obtain that the subgroup 24 is contained in Aut(S) contains a subgroup H isomorphic to 24 . One can come to the same conclusion by looking at the equations of X. The group of projective automorphims generated by the transformations which switch ti to −ti realizes the subgroup 24 . Let G be the subgroup of W (D5 ) realized by permutations of the set {e1 , . . . , e5 }. It is isomorphic to S5 and W (S) is equal to the semi-direct product H G. Now suppose that Aut(S) contains an element g ∈ H. Composing it with elements from H, we may assume that g ∈ G. Since elements of G leave e0 invariant, g is realized by a projective transformation of P2 leaving the set of points x1 , . . . , x5 invariant. Since there is a unique conic through these points, the group is isomorphic to a finite group of PSL(2) leaving invariant a binary quintic without multiple roots. All these groups can be easily found. It follows from the classification of finite subgroups of SL(2) and their algebra of invariants that the only possible groups are the cyclic groups C2 , C3 , C4 , C5 , the permutation group S3 , and the dihedral group D10 of order 10. The corresponding binary forms are projectively equivalent to the following binary forms: (i) C2 : u0 (u2 − u2 )(u2 + au2 ), a = −1, 1; 0 1 0 1 (ii) C4 : u0 (u2 − u2 )(u2 + u2 ); 0 1 0 1 (iii) C3 , C6 : u0 u1 (u0 − u1 )(u0 − ηu1 )(u0 − η 2 u2 ), η = e2πi/3 ; 1 (iv) C5 , D5 : (u0 − u1 )(u0 − u1 )(u0 −
2

u1 )(u0 −

3

u1 )(u0 −

4

u1 ),

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CHAPTER 8. DEL PEZZO SURFACES

where = e2πi/5 . The corresponding surfaces are projectively equivalent to the following surfaces (i) C2 : t2 + t2 + a(t2 + t2 ) + t2 = t2 − t2 + b(t2 − t2 ) = 0, a = −b, b; 0 2 1 3 4 0 2 1 3 (ii) C4 : t2 + t2 + t2 + t2 + t2 = t2 + it2 − t2 − it2 = 0; 0 1 2 3 4 0 1 2 3 (iii) S3 : t2 + ηt2 + η 2 t2 + t2 = t2 + η 2 t2 + ηt2 + t2 = 0; 0 1 2 3 0 1 2 4 (iv) D10 : t2 + t2 + 0 1
2 2 t2

+

3 2 t3

+

4 2 t4

=

4 2 t0

+

3 2 t1

+

2 2 t2

+ t2 + t2 = 0. 3 4

8.6
8.6.1

Del Pezzo surfaces of degree 2
Lines and singularities

Let S be a weak Del Pezzo surface of degree 2. Recall that the anticanonical linear system defines a birational morphism φ : S → X, where X is the anticanonical model of S isomorphic to the double cover of P2 branched along a plane quartic curve W with at most simple singularities. Let φ : S → P2 be the composition of φ and the double cover map σ : X → P2 . The restriction of φ to a line E is a map of degree −KS · E = 1. Its image in the plane is a line . The pre-image of is the union of E and a divisor D ∈ | − KS − E|. Since −KS · D = 1, the divisor D is equal to E + R, where E is a line and R is the union of (−2)-curves. Also we immediately find that E · D = 2, D2 = −1. There are three possible cases: (i) E = E , E · E = 2; (ii) E = E , E · E = 1; (iii) E = E , E = E . In the first case, the image of E is a line tangent to W at two nonsingular points. The image of D −E is a singular point of W . By Bezout’s Theorem, cannot pass through the singular point. Hence D = E and is a bitangent of W . In the second case, E · D − E = 1. The line passes through the singular point φ(D − E and is tangent to W at a nonsingular point. Finally, in the third case, is a component of W . Of course, when S is a Del Pezzo surface, the quartic W is nonsingular, and we have 56 lines paired into 28 pairs corresponding to 28 bitangents of W . Let π : S → P2 be the blow-up of seven points x1 , . . . , x7 in general position. Then 28 pairs of lines are the proper inverse transforms of the isolated pairs of curves: 21 pairs: a line through xi , xj and the conic through the complementary five points 7 pairs: a cubic with a double point at xi and passing through other points plus the exceptional curve π −1 (xi ). We use the procedure of Borel-De Siebenthal-Dynkin to compile the list of root bases in E7 . f It is convenient first to compile the list of maximal (by inclusions) root bases of type A, D, E (see [152] §12. Here D2 = A1 + A1 and D3 = A3 .

8.6. DEL PEZZO SURFACES OF DEGREE 2 Type An Dn E6 E7 E8 rank n − 1 Ak + An−k−1 An−1 , Dn−1 D5 E6 rank n Dk + Dn−k , k ≥ 2 A1 + A5 , A2 + A2 + A2 A1 + D6 , A7 , A2 + A5 D8 , A1 + E7 , A8 , A2 + E6 , A4 + A4

291

Table 8.7: Maximal root bases r 7 6 5 ≤4 Types E7 , A1 + D6 , A7 , 3A1 + D4 , A1 + 2A3 , A5 + A2 , 7A1 E6 , D5 + A1 , D6 , A6 , A1 + A5 , 3A2 , 2A1 + D4 , 2A3 , 3A1 + A3 , 6A1 , A1 + A2 + A3 , A2 + A4 D5 , A5 , A1 + D4 , A1 + A4 , A1 + 2A2 , 2A1 + A3 , 3A1 + A2 , A2 + A3 , 5A1 D4 , Ai1 + . . . + Aik , i1 + . . . + ik ≤ 4 Table 8.8: Root bases in the E7 -lattice From this easily find the following table of root bases in E7 : Note that there are two roots bases of types A1 + A5 , A2 + 2A1 , 3A1 , A1 + A3 and 4A1 which are not equivalent with respect to the Weyl group. The simple singularities of plane quartics were classified by P. Du Val [92], Part III. A1 : one node; 2A1 : two nodes; A2 : one cusp; 3A1 : irreducible quartic with three nodes; 3A1 : a cubic and a line; A1 + A2 : one node and one cusp; A3 : one tacnode (two infinitely near ordinary double points); 4A1 : a nodal cubic and a line; 4A1 : two conics intersecting at 4 points; 2A1 + A2 : two nodes and one cusp; A1 + A3 : a node and a tacnode; A1 + A3 : cubic and a tangent line; A4 : one rhamphoid cusp (two infinitely near cusps); 2A2 : two cusps; D4 : an ordinary triple point; 5A1 : a conic and two lines; 3A1 + A2 : a cuspidal cubic and a line; 2A1 + A3 : two conics tangent at one point;

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CHAPTER 8. DEL PEZZO SURFACES

2A1 + A3 : a(nodal cubic and its tangent line; A1 + A4 : a rhamphoid cusp and a node; A1 + 2A2 : a cusp and two nodes; A2 + A3 : a cusp and a tacnode; A5 : one oscnode (two infinitely near cusps); A5 : a cubic and its flex tangent; D5 : nodal cubic and a line tangent at one branch; A1 + D4 :a nodal cubic and line through the node; E6 : an irreducible quartic with one e6 -singularity; D6 : triple point with one cuspidal branch; A1 + A5 : two conics intersecting at two points with multiplicities 3 and 1. A1 + A5 : a nodal cubic and its flex tangent; 6A1 : four lines in general position; 3A2 : a three cuspidal quartic; 2A1 + D4 : two lines and conic through their intersection point; D5 + A1 : cuspidal cubic and a line through cusp’ 2A3 : two conics intersecting at two points with multiplicities 2; 3A1 + A3 : a conic plus its tangent line plus another line; A1 + A2 + A3 : cuspidal cubic and its tangent; A6 : one oscular rhamphoid cusp (three infinitely near x1 A2 + A4 :one rhamphoid cups and a cusp. E7 : cuspidal cubic and its cuspidal tangent; A1 + D6 : conic plus tangent line, and another line through point of contact; D4 + 3A1 : four lines with three concurrent; A7 :two irreducible conics intersecting at one point; A5 + A2 : cuspidal cubic and a flex tangent; 3A2 + A1 : conic and two tangent lines. Note that all possible root bases are realized except 7A1 (this can be realized in characteristic 2). One can compute the number of lines but this rather tedious. For example, in the case A1 we have 44 lines and the nodal Weyl group W (S)n acts on the set W (S)/W (S)E with 6 orbits of cardinality 2 and 44 orbits of cardinality 1. This gives that a one-nodal quartic has 21 bitangents (i.e. lines with two nonsingular points of tangency. x2 x1 cusps);

8.6.2

The Geiser involution

Let S be a weak Del Pezzo surface of degree 2. Consider the degree 2 regular map φ : S → P2 defined by the linear system | − KS |. In the blow-up model of S, the linear system | − KS | is represented by the net of cubic curves N with seven base bubble

8.6. DEL PEZZO SURFACES OF DEGREE 2

293

points x1 , . . . , x7 in P2 . It is an example of a Laguerre net considered in Chapter 7. If S is a Del Pezzo surface, then φ is a finite map of degree 2 and any subpencil of | − KS | has no fixed component. Any pencil contained in N has no fixed components and has 2 points outside the base points of the net. Assigning the line through these points, we will be able to identify the plane P2 with the net N , or with | − K|. The inverse map is defined by using the coresidual points of Cayley. For every nonsingular 1 member D ∈ N , the restriction of | − KS | to D defines a g2 realized by the projection from the coresidual point on D. This map extends to an isomorphism N → P2 .

Let X ⊂ P(1, 1, 1, 2) be the anticanonical model of S. The map φ factors through the birational map σ : S → X that blows down the Dynkin curves and a degree 2 ¯ finite map φ : X → P2 ramified along a plane quartic curve with simple singularities. ¯ The deck transformation γ of the cover φ is a birational automorphism of S called the Geiser involution. In fact, the Geiser involution is a biregular automorphism of S. Since σ is a minimal resolution of singularities of X, this follows from the existence of a equivariant minimal resolution of singularities of surfaces [166] and the uniqueness of a minimal resolution of surfaces.

Proposition 8.6.1. The Geiser involution γ has no isolated fixed points. Its locus of fixed points is the disjoint union of smooth curves W +R1 +. . .+Rk , where R1 , . . . , Rk are among irreducible components of Dynkin curves. The curve W is the normalization of the branch curve of the double cover φ : S → P2 . A Dynkin curve of type A2k has no fixed components, a Dynkin curve of type A2k+1 has one fixed component equal to the central component. A Dynkin curve of type D4 , D5 , D6 , E6 , E7 have fixed components marked by square on their Coxeter-Dynkin diagrams.

294 D4 • • D5 • • • D6 • • • E6 • • • • E7 • • • • •

CHAPTER 8. DEL PEZZO SURFACES

















• Assume that S is a Del Pezzo surface. Then the fixed locus of the Geiser involution is a smooth irreducible curve W isomorphic to the branch curve of the cover. It belongs to the linear system | − 2KS | and hence its image in the plane is a curve of degree 6 with double points at x1 , . . . , x7 . It is equal to the jacobian curve of the net of cubics, i.e. the locus of singular points of singular cubics from the set. It follows from the Lefschetz fixed-point-formula that the trace of γ in Pic(S) ∼ H 2 (S, Z) is equal to = e(W ) − 2 = −6. This implies that the trace of σ on QS = (KS )⊥ is equal to −7. Since rankQS = 7 this implies that γ acts as the minus identity on QS . It follows from the theory of finite reflection groups that the minus identity isogeny of the lattice E7 is represented by the element w0 in W (E7 ) of maximal length as a word in simple reflections. It generates the center of W (E7 ). We can also consider the Geiser involution as a Cremona involution of the plane. It coincides with the Geiser involution described in Chapter 7. The characteristic matrix of a Geiser involution with respect to the bases e0 , . . . , e7 and σ ∗ (e0 ), . . . , σ ∗ (e7 ) is the following.   8 3 3 3 3 3 3 3 −3 −2 −1 −1 −1 −1 −1 −1   −3 −1 −2 −1 −1 −1 −1 −1   −3 −1 −1 −2 −1 −1 −1 −1   (8.22) −3 −1 −1 −1 −2 −1 −1 −1   −3 −1 −1 −1 −1 −2 −1 −1   −3 −1 −1 −1 −1 −1 −2 −1 −3 −1 −1 −1 −1 −1 −1 −2

8.6. DEL PEZZO SURFACES OF DEGREE 2

295

We can consider this matrix as the matrix of the element w0 ∈ O(I 1,7 ) in the basis e0 , e1 , . . . , e7 .

8.6.3

Automorphisms of Del Pezzo surfaces of degree 2

Let S be a Del Pezzo surface of degree 2. The Geiser involution γ belongs to the center of W (S). The quotient group Aut(S)/ γ is the group of automorphisms of the branch cover φ : S → P2 . We use the classification of automorphisms of plane quartic curves from Chapter 6. Let G be a group of automorphisms of the branch curve V (f ) given by a quartic polynomial f . Let χ : G → C∗ be the character of G defined by σ ∗ (f ) = χ(σ)f . Let G = {(g , α) ∈ G × C∗ : χ(g ) = α2 }. This is a subgroup of the group G × C∗ . The projection to G defines an isomorphism G ∼ 2.G . The extension splits if and only if χ is equal to the square of some character = of G . In this case G ∼ G × 2. The group G acts on S given by equation (8.10) by = (σ , α) : [t0 , t1 , t2 , t3 ] → [σ ∗ (t0 ), σ ∗ (t1 ), σ ∗ (t2 ), αt3 ]. Any group of automorphisms of S is equal to a group G as above. This easily gives the classification of possible automorphism groups of Del Pezzo surfaces of degree 2.
Type Order Structure Equation t2 + t3 t1 + t3 t2 + t3 t0 3 0 1 2 t2 + t4 + t4 + t4 3 0 1 2 + t4 + at2 t2 + t4 0 0 1 1 2 + t4 + t4 + t4 + t3 2 1 0 +a(t2 t2 + t2 t2 + t2 t2 ) 0 1 0 2 1 2 t2 + t4 + t4 + at2 t2 + t4 3 2 0 0 1 1 t2 + t4 + t0 t3 + t1 t3 3 0 1 2 t2 + t4 + t4 + t4 + at2 t2 + bt2 t0 t1 3 2 0 1 0 1 2 t2 + t3 t0 + t4 + t4 + at2 t2 3 2 0 1 0 1 2 + t4 + at2 t t + t (t3 + t3 ) + bt2 t2 t3 2 0 2 2 0 1 1 0 1 t2 + t4 + t4 + t4 3 2 1 0 2 t2 + bt2 t2 + ct2 t2 +at2 0 1 2 0 1 t2 + t3 t0 + f4 (t0 , t1 ) 3 2 2 + t4 + t2 f (t , t ) + f (t , t ) t3 4 0 1 2 2 2 0 1 t2 + f4 (t0 , t1 , t2 ) 3 + t2 3 t4 2 a2 = −12 a=
√ −1± −7 2

Parameters

I II III IV V VI VII VIII IX X XI XII XIII

336 192 96 48 32 18 16 12 12 8 6 4 2

2 × L2 (7) 2 × (42 : S3 ) 2 × 4A4 2 × S4 2 × AS16 18 2 × D8

a2 = 0, −12, 4, 36 a, b = 0

2×6
2 × S3

23 6 22 2

distinct a, b, c = 0

Table 8.9: Groups of automorphisms of Del Pezzo surfaces of degree 2 We leave to a curious reader the task of classifying automorphism groups of weak Del Pezzo surfaces. Notice that in the action of Aut(S) in the Picard group they correspond to certain subgroups of the group Cris(S). Also the action is not necessary faithful, for example the Geiser involution acts act trivially on Pic(S) in the case of a weak Del Pezzo surface with singularity of type E7 .

296

CHAPTER 8. DEL PEZZO SURFACES

8.7
8.7.1

Del Pezzo surfaces of degree 1
Lines and singularities

Let S be a weak Del Pezzo surface of degree 1. Its anticanonical model X is a finite cover of degree 2 of a quadratic cone Q ramified over a curve B in the linear system |OQ (3)| with at most simple singularities. The list of types of possible Dynkin curves is easy to compile. First we observe that all diagrams listed for the case of the E7 lattice are included in the list. Also all the diagrams A1 + T , where T is from the previous list are included. We give only the new types r 8 7 6 Types E8 , A8 , D8 , 2A4 , A1 + A2 + A5 , A3 + D5 , 2D4 , A2 + E6 , A3 + D5 , 4A2 D7 , A2 + D5 , A3 + A4 , A3 + D4 A2 + D4 Table 8.10: Root bases in the E8 -lattice Note that there are two root bases of types A7 , 2A3 , A1 + A5 , 2A1 + A3 and 4A1 which are not equivalent with respect to the Weyl group. The following result of P. Du Val [92] will be left without proof. Note that Du Val uses the following notation: A1 = [], An = [3n−1 ]n ≥ 2, Dn = [3n−3,1,1 ], n ≥ 4, E6 = [33,2,1 ], E7 = [34,2,1 ], E8 = [35,2,1 ]. Theorem 8.7.1. All types of root bases in E8 can be realized by Dynkin curves except the cases 7A1 , 8A1 , D4 + 4A1 . In fact, describes explicitly the singularities of the branch sextic similarly to the case of weak Del Pezzo surfaces of degree 2. The number of lines on a Del Pezzo surface of degree 1 is equal to 240. Note the coincidence with the number of roots. The reason is simple, for any root α ∈ E8 , the sum −k8 + α is an exceptional vector. The image of a line under the cover φ : S → Q is a conic. The plane spanning the conic is a tritangent plane, a plane touching the branch sextic W at three points. There are 120 tritangent planes, each cut out a conic in Q which splits under the cover in the union of two lines intersecting at three points. Note the effective divisor D of degree 3 on W such that 2D is cut out by a tritangent plane is an odd theta characteristic on W . This gives another explanation of the number 120 = 23 (24 − 1).

8.7.2

Bertini involution

Let S be a weak Del Pezzo surface of degree 1. Consider the degree 2 regular map φ : S → Q defined by the linear system | − 2KS |. In the blow-up model of S, the linear system | − 2KS | is represented by the web W of sextic curves with eight base bubble points x1 , . . . , x8 in P2 . If S is a Del Pezzo surface, then φ is a finite map of degree 2.

8.7. DEL PEZZO SURFACES OF DEGREE 1

297

Let X ⊂ P(1, 1, 2, 3) be the anticanonical model of S. The map φ factors through the birational map σ : S → X that blows down the Dynkin curves and a degree 2 ¯ finite map φ : X → Q ramified along a curve of degree 6 cut out by a cubic surface. ¯ The deck transformation β of the cover φ is a birational automorphism of S called the Bertini involution. As in the case of the Geiser involution, we prove that the Bertini involution is a biregular automorphism of S.

Proposition 8.7.2. The Bertini involution β has one isolated fixed point, the base point of | − KS |. The one-dimensional part of the locus of fixed points is the disjoint union of smooth curves W + R1 + . . . + Rk , where R1 , . . . , Rk are among irreducible components of Dynkin curves. The curve W is the normalization of the branch curve of the double cover φ : S → Q. A Dynkin curve of type A2k has no fixed components, a Dynkin curve of type A2k+1 has one fixed component equal to the central component. A Dynkin curve of type D4 , D7 , D8 , E8 have fixed components marked by square on their Coxeter-Dynkin diagrams. The fixed components of Dynkin curves of other types given in the diagrams from Proposition 8.6.1

D7



• •









D8



• •











E8













• Assume that S is a Del Pezzo surface. Then the fixed locus of the Bertini involution is a smooth irreducible curve W of genus 4 isomorphic to the branch curve of the cover and the base point of | − KS |. It belongs to the linear system | − 3KS | and hence its image in the plane is a curve of degree 9 with triple points at x1 , . . . , x8 . It follows from the Lefschetz fixed-point-formula that the trace of β in Pic(S) ∼ H 2 (S, Z) is = equal to 1 + e(W ) − 2 = −7. This implies that the trace of σ on QS = (KS )⊥ is equal to −8. Since rankQS = 8 this implies that γ acts as the minus identity on QS . It follows from the theory of finite reflection groups that the minus identity isogeny of the lattice E7 is represented by the element w0 in W (E8 ) of maximal length as a word in simple reflections. It generates the center of W (E8 ). We can also consider the Bertini involution as a Cremona involution of the plane. It coincides with the Bertini involution described in Chapter 7. The characteristic matrix of a Geiser involution with respect to the bases e0 , . . . , e8 and σ ∗ (e0 ), . . . , σ ∗ (e8 ) is

298 the following.  17 −6  −6  −6  −6  −6  −6  −6 −6 6 −3 −2 −2 −2 −2 −2 −2 −2 6 −2 −3 −2 −2 −2 −2 −2 −2 6 −2 −2 −3 −2 −2 −2 −2 −2 6 −2 −2 −2 −3 −2 −2 −2 −2

CHAPTER 8. DEL PEZZO SURFACES

6 −2 −2 −2 −2 −3 −2 −2 −2

6 6 −2 −2 −2 −2 −2 −2 −2 −2 −2 −2 −3 −2 −2 −3 −2 −23

 6 −2  −2  −2  −2  −2  −2  −2 −3

We can consider this matrix as the matrix of the element w0 ∈ O(I 1,8 ) in the basis e0 , e1 , . . . , e 8 .

8.7.3

Rational elliptic surfaces

We know that the linear system | − KS | is an irreducible pencil with one base point x0 . Let τ : F → S be its blow-up. The proper inverse transform of | − KS | in F is a base-point-free pencil of curves of arithmetic genus 1. It defines an elliptic fibration ϕ : F → P1 . The exceptional curve E = τ −1 (x0 ) is a section of the fibration. Conversely, let ϕ : F → P1 be an elliptic fibration on a rational surface F which admits a section E and relative minimal in the sense that no fibre contains a (−1)curve. It follows from the theory of elliptic surfaces that −KF is the divisor class of a fibre and E is a (−1)-curve. Blowing down E, we obtain a rational surface S with 2 KS = 1. Since KF is obviously nef, we obtain that KS is nef, so S is a weak Del Pezzo surface of degree 1. Let ϕ : F → P1 be a rational elliptic surface with a section E. The section E defines a rational point e on a generic fibre Fη , considered as a curve over the functional field K of the base of the fibration. It is a smooth curve of genus 1, so it admits a group law with the zero equal to the point e. It follows from the theory of relative minimal models of surfaces that any automorphism of Fη over K extends to a biregular automorphism of F over P1 . In particular, the negation automorphism x → −x extends to an automorphism of F fixing the curve E. Its descent to the blowing down of E is the Bertini involution. Let D be a Dynkin curve on S. The point x0 cannot lie on D. In fact, otherwise the proper transform R of a component of D that contains x0 is a (−3)-curve on F . However, −KF is nef on F hence KF · R ≤ 0 contradicting the adjunction formula. This implies that the pre-image τ ∗ (D) of D on F is a Dynkin curve contained in a fibre. The whole fibre is equal to the union of τ ∗ (D) + R, where R is a (−2)-curve intersecting the zero section E. Kodaira’s classification of fibres of elliptic fibrations shows that the intersection graph of the irreducible components of each reducible fibre is equal to one of the extended Coxeter-Dynkin diagrams. The classification of Dynkin curves on a weak Del Pezzo surfaces of degree 1 gives the classification of all possible collections of reducible fibres on a rational elliptic surface with a section.

8.7. DEL PEZZO SURFACES OF DEGREE 1 The equation of the anticanonical model in P(1, 1, 2, 3) t2 + t3 + f4 (t0 , t1 )x + f6 (t0 , t1 ) = 0, 3 2

299

(8.23)

after dehomogenization t = t1 /t0 , x = t2 /t2 , y = t3 /t3 become the Weierstrass 1 1 equation of the elliptic surface y 2 + x3 + a(t)x + b(t) = 0. The classification of all possible singular fibres of rational elliptic surfaces (not necessary reducible) in terms of the Weierstrass equation was done by several people, e.g. [190].

8.7.4

Automorphisms of Del Pezzo surfaces of degree 1

Let S be a Del Pezzo surface of degree 1. We identify it with its anticanonical model (8.23) The vertex of Q has coordinates [0, 0, 1] and its pre-image in the cover consist of one point (0, 0, 1, a), where a2 +1 = 0 (note that (0, 0, 1, a) and (0, 0, 1, −a) represent the same point on P(1, 1, 2, 3). This is the base point of | − KS |. The members of | − KS | are isomorphic to genus 1 curves with equations y 2 + x3 + f4 (t0 , t1 )x + ¯ f6 (t0 , t1 ) = 0. Our group G acts on P1 via a linear action on (t0 , t1 ). The locus of 3 2 zeros of ∆ = f4 + 27f6 is the set of points in P1 such that the corresponding genus 1 curve is singular. It consists of a simple roots and b double roots. The zeros of f4 are either common zeros with f6 and ∆, or represent nonsingular elliptic curves with automorphism group isomorphic to Z/6Z. The zeros of f6 are either common zeros with f4 and ∆, or represent nonsingular elliptic curves with automorphism group ¯ isomorphic to Z/4Z. The group G leaves both sets invariant. ¯ is determined up to conjugacy by its set of points in P1 with nonRecall that G ¯ trivial stabilizers. If G is not cyclic, then there are three orbits in this set of cardinalities ¯ n/e1 , n/e2 , n/e3 , where n = #G and (e1 , e2 , e3 ) are the orders of the stabilizers. Let Γ be a finite noncyclic subgroup of PGL(2). We have the following possibilities. (i) Γ = D2k , n = 2k, (e1 , e2 , e3 ) = (2, 2, k); (ii) Γ = T , n = 12, (e1 , e2 , e3 ) = (2, 3, 3); (iii) Γ = O, n = 24, (e1 , e2 , e3 ) = (2, 3, 4); (iv) Γ = I, n = 60, (e1 , e2 , e3 ) = (2, 3, 5). ¯ If Γ is a cyclic group of order n, there are 2 orbits of cardinality 1. ¯ The polynomials f4 and f6 are relative invariants of G. Each orbit defines a binary form (the orbital form) with the set of zeros equal to the orbit. One can show that any projective invariant is a polynomial in orbital forms. This immediately implies that ¯= ¯= G ∼ A5 and if G ∼ S4 , then f4 = 0. ¯ ¯ We choose to represent G by elements of SL(2), i.e. we consider G as a quotient of a binary polyhedral subgroup G ⊂ SL(2) by its intersection with the center of SL(2). ¯ A projective invariant of G becomes a relative invariant of G. We use the description

300

CHAPTER 8. DEL PEZZO SURFACES

of relative invariants and the corresponding characters of G from [229]. This allows us to list all possible polynomials f4 and f6 . ¯ The following is the list of generators of the groups G, possible relative invariants f4 , f6 and the corresponding character. We use that a multiple root of f6 is not a root of f4 (otherwise the surface is singular). ¯ Case 1. G is cyclic of order n. Here n denote a primitive n-th root of 1. n 2 3 at4 0 f4 2 2 + bt0 t1 + ct4 1 t0 t1 (at2 + bt2 ) 0 1 t0 (at3 + bt3 ) 0 1 t1 (at3 + bt3 ) 0 1 t2 t2 0 1 at4 + bt4 0 1 t0 t3 1 t3 t1 0 t2 t2 0 1 t4 0 t4 0 t3 t1 0 t3 t1 0 t2 t2 0 1 t4 0 3 t0 t1 t2 t2 0 1 t4 0 χ(σ) 1 -1
2 3 3

at6 0

4

1 -1 i -i 1
2 5 2 5 5 5

5

1
3

6

1t6 + 0

6 t6 1 2 n

f6 2 2 2 2 + t0 t1 (bt0 + ct1 ) + dt6 1 t0 t1 (at4 + bt2 t2 + ct4 ) 0 0 1 1 ct6 + dt3 t3 + et6 0 0 1 1 t0 t2 (ct3 + dt3 ) 1 0 1 t2 t1 (ct3 + dt3 ) 0 0 1 t2 (ct4 + dt4 ) 0 0 1 t0 t1 (at4 + bt4 ) 0 1 t2 (ct4 + dt4 ) 1 0 1 t3 t3 0 1 t0 (at5 + t5 ) 0 1 t1 (at5 + t5 ) 0 1 t0 (t5 + t5 ) 0 1 t1 (t5 + t5 ) 0 1 t0 (t5 + t5 ) 0 1 t0 (t5 + t5 ) 0 1 t 6 + t6 ) 0 1 t6 + t 6 0 1 -1 t0 (t5 + t5 ) 0 1 t6 1 t0 t5 1

χ(σ) 1 1 1
2 3 3

-i -1 i 1
3 5 2 5 3 5 2 5 3 5 3 5

-1 -1
3 5 −3 n −2 n

>6

¯ Case 2. G = Dn is a dihedral group of order n = 2k. It is generated by two matrices 0 0 i σ1 = 2k −1 , σ2 = . i 0 0 2k (i) k = 2: f4 = a(t4 +t4 )+bt2 t2 , χ(σ1 ) = χ(σ2 ) = 1, 0 1 0 1 f4 = at0 t1 (t2 −t2 ), χ(σ1 ) = χ(σ2 ) = −1, 0 1

f4 = a(t4 −t4 ), χ(σ1 ) = 1, χ(σ2 ) = −1, f4 = at0 t1 (t2 +t2 ), χ(σ1 ) = −1, χ(σ2 ) = 1; 0 1 0 1 f6 = at0 t1 a(t4 + t4 ) + bt2 t2 , χ(σ1 ) = 1, χ(σ2 ) = −1, 0 1 0 1 f6 = a(t6 + t6 ) + bt2 t2 (t2 + t2 ), χ(σ1 ) = −1, χ(σ2 ) = −1, 0 1 0 1 0 1 f6 = a(t6 − t6 ) + bt2 t2 (t2 − t2 ), χ(σ1 ) = −1, χ(σ2 ) = 1, 0 1 0 1 0 1 f6 = t0 t1 (t4 − t4 ), χ(σ1 ) = χ(σ2 ) = 1, 0 1

8.7. DEL PEZZO SURFACES OF DEGREE 1

301

Note that the symmetric group S3 acts on the set of 3 exceptional orbits by projective transformations. This shows, that up to linear change of variables, we have the following essentially different cases. f4 = a(t4 + t4 ) + bt2 t2 , χ(σ1 ) = χ(σ2 ) = 1, 0 1 0 1 f4 = at0 t1 (t2 + t2 ), χ(σ1 ) = −1, χ(σ2 ) = 1; 0 1 f6 = at0 t1 a(t4 + t4 ) + bt2 t2 , χ(σ1 ) = 1, χ(σ2 ) = −1, 0 1 0 1 f6 = t0 t1 (t4 − t4 ), χ(σ1 ) = χ(σ2 ) = 1, 0 1 (ii) k = 3: f4 = t2 t2 , χ(σ1 ) = χ(σ2 ) = 1, 0 1 f6 = t6 + t6 + at3 t3 , χ(σ1 ) = χ(σ2 ) = −1; 0 1 0 1 f6 = t6 − t6 , χ(σ1 ) = χ(σ2 ) = 1; 0 1 (iii) k = 4: f4 = t4 ±t4 , χ(σ1 ) = χ(σ2 ) = ±1, 0 1 (iv) k = 6: f4 = at2 t2 , 0 1 f6 = t6 ± t6 , χ(σ1 ) = χ(σ2 ) = 0 1 1; ¯ Case 3. G = A4 . It is generated by matrices σ1 = i 0 , 0 −i σ2 = 0 i , i 0 1 σ3 = √ 2
−1 8 5 8 −1 8 8

f6 = at0 t1 (t4 ±t4 ), χ(σ1 ) = χ(σ2 ) = 0 1

1.

.

Up to the variable change t0 → it0 , t1 → t1 , we have only one case √ f4 = t4 + 2 −3t2 t2 + t4 , f6 = t0 t1 (t4 − t4 ). 0 0 1 1 0 1 ¯ Case 4. G = S4 . It is generated by matrices σ1 =
8

0
−1 8

0

,

σ2 =

0 i , i 0

1 σ3 = √ 2

−1 8 5 8

−1 8 8

.

There is only one, up to a change of variables, orbital polynomial of degree ≤ 6. It is f6 = t0 t1 (t4 − t4 ). 0 1 The corresponding characters are χ(σ1 ) = −1, χ(σ2 ) = 1, χ(σ3 ) = 1.

In this case f4 = 0. In the next theorem we list all possible groups G = Aut(S)/ T and their lifts ¯ G to subgroups of Aut(S). We extend the action of G on the coordinates t0 , t1 to an action on the coordinates t0 , t1 , t2 . Note that not all combinations of (f4 , f6 ) admit such an extension. In the following list, the vector a = (a0 , a1 , a2 , a3 ) will denote the transformation [t0 , t1 , t2 , t3 ] → [a0 t0 , a1 t1 , a2 t2 , a3 t3 ]. The Bertini transformation T corresponds to the vector (1, 1, 1, −1).

302 1. Cyclic groups G

CHAPTER 8. DEL PEZZO SURFACES

(i) G = 2, G = ((1, −1, 1, 1), T ) ∼ 22 = f4 = at4 + bt2 t2 + ct4 , 0 0 1 1 (ii) G = 2, G = ((1, −1, −1, i)), f4 = at4 + bt2 t2 + ct4 , 0 0 1 1 (iii) G = 3, G = ((1,
3 , 1, −1))

f6 = dt6 + et4 t2 + f t2 t4 + gt6 . 0 0 1 0 1 1

f6 = t0 t1 (dt4 + et2 t2 + f t4 ). 0 0 1 1

∼ 6, = f6 = at6 + bt3 t3 + ct6 ; 0 0 1 1

f4 = t0 (at3 + bt3 ), 0 1 (iv) G = 3, G = ((1,
3 , 3 , −1)),

f4 = t2 t2 , 0 1 (v) G = 3, G = 6, a = (1, 1,
3 , −1),

at6 + bt3 t3 + ct6 ; 0 0 1 1

f4 = 0, (vi) G = 4, G = ((i, 1, −1, i), T ) ∼ 4 × 2, = f4 = at4 + bt4 , 0 1 (vii) G = 4, G = ((i, 1, −i, − 8 )) ∼ 8, = f4 = at2 t2 , 0 1 (viii) G = 5, G = ((1,
5 , 1, −1))

f6 ;

f6 = t2 (ct4 + dt4 ), 0 0 1

f6 = t0 t1 (ct4 + dt4 ), 0 1

∼ 10, = f6 = t0 (bt5 + t5 ), 0 1

f4 = at4 , 0 (ix) G = 6, G = ((1,
6 , 1, 1), T )

∼ 2 × 6, = f6 = at6 + bt6 , 0 1

f4 = t4 , 0 (x) G = 6, G = (( 6 , 1,
2 3 , 1), T )

∼ 2 × 6, = f6 = at6 + bt6 , 0 1

f4 = t2 t2 , 0 1 (xi) G = 6, G = ((−1, 1, f4 = 0, (xii) G = 10, G = ((1,
3 , 1), T )

∼ 2 × 6, =

f6 = dt6 + et4 t2 + f t2 t4 + gt6 , 0 0 1 0 1 1 ∼ 20, = f6 = t0 t5 , 1

10 , −1, i)

f4 = at4 , 0

8.7. DEL PEZZO SURFACES OF DEGREE 1 (xiii) G = 12, G = ((
2 12 , 1, 3 , −1), T )

303

∼ 2 × 12, = f6 = t6 , 1

f4 = at4 , 0 (xiv) G = 12, G = ((i, 1,
12 , 8 ))

∼ 24, = f6 = t0 t1 (t4 + bt4 ), 0 1 ∼ 30, = f6 = t0 (t5 + t5 ). 0 1

f4 = 0, (xv) G = 15, G = ((1,
5 , 3 , 30 ))

f4 = 0, 2. Dihedral groups (i) G = 22 , G = D8 , f4 = a(t4 + t4 ) + bt2 t2 , 0 1 0 1

f6 = t0 t1 [c(t4 + t4 ) + dt2 t2 ], 0 1 0 1

σ1 : [t0 , t1 , t2 , t3 ] → [t1 , −t0 , t2 , it3 ], σ2 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ],
−1 −1 4 2 2 σ1 = σ2 = 1, σ1 = T, σ2 σ1 σ2 = σ2 .

(ii) G = 22 , G = 2.D4 , f4 = a(t4 + t4 ) + bt2 t2 , 0 1 0 1 f6 = t0 t1 (t4 − t4 ), 0 1

σ1 : [t0 , t1 , t2 , t3 ] → [t0 , −t1 , −t2 , it3 ], σ2 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , −t2 , it3 ],
2 σ1 = σ 2 = (σ1 σ2 )2 = T.

(iii) G = D6 , G = D12 , f4 = at2 t2 , 0 1 σ1 : [t0 , t1 , t2 , t3 ] → [t0 , f6 = t6 + t6 + bt3 t3 , 0 1 0 1 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ],

3 t1 , 3 t2 , −t3 ], σ2

−1 −1 3 2 σ1 = T, σ2 = 1, σ2 σ3 σ2 = σ1 .

(v) G = D8 , G = D16 , f4 = at2 t2 , 0 1 σ1 : [t0 , t1 , t2 , t3 ] → [ 8 t0 ,
4 σ1

f6 = t0 t1 (t4 + t4 ), 0 1 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ],
−1 = σ1 .

=

−1 8 t1 , −t2 , it3 ], σ2 −1 2 T, σ2 = 1, σ2 σ1 σ2

(vi) G = D12 , G = 2.D12 , f4 = at2 t2 , 0 1 σ1 : [t0 , t1 , t2 , t3 ] → [t0 , We have
−1 −1 6 2 3 σ1 = σ2 = σ3 = 1, σ2 σ1 σ2 = σ1 σ3 .

f6 = t6 + t6 , 0 1 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ], σ3 = T.

2 6 t1 , 3 t2 , t3 ], σ2

304 3. Other groups (i) G = A4 , G = 2.A4

CHAPTER 8. DEL PEZZO SURFACES

√ f4 = t4 + 2 −3t2 t2 + t4 , 0 0 1 2  σ1 =
i 0 0 0 0 −i 0 0 0 0 1 0 0 0 , 0 1

f6 = t0 t1 (t4 − t4 ), 0 1
0 0 1 0 0 0 , 0 1



 σ2 =

0 i 0 0

i 0 0 0



 1 σ3 = √ 2  

−1 8 5 8

−1 8 8

0 0

0 0

0 √0 2 0

3

0 0  . 0  √ 2



(ii) G = 3 × D4 , G = 3 × D8 f4 = 0, f6 = t0 t1 (t4 + at2 t2 + t4 ), 0 0 1 1

(iii) G = 3 × D6 , G = 6.D6 ∼ 2 × 3.D6 = f4 = 0, f6 = t6 + at3 t3 + t6 , 0 0 1 1

It is generated by σ1 : [t0 , t1 , t2 , t3 ] → [t0 , t1 , 3 t2 , t3 ], σ2 = [t0 , t1 , t2 , t3 ] → −1 [t0 , 3 t1 , t2 , t3 ], σ3 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ]. We have σ3 ·σ2 ·σ3 = −1 4 σ2 σ1 . (iv) G = 3 × D12 , G = 6.D12 f4 = 0, f6 = t6 + t6 , 0 1

It is generated by σ1 : [t0 , t1 , t2 , t3 ] → [t0 , t1 , 3 t2 , t3 ], σ2 = [t0 , t1 , t2 , t3 ] → −1 [t0 , 6 t1 , t2 , t3 ], σ3 : [t0 , t1 , t2 , t3 ] → [t1 , t0 , t2 , t3 ]. We have σ3 ·σ2 ·σ3 = −1 σ2 σ1 . (v) G = 3 × S4 , G = 3 × 2.S4 f4 = 0, 
8

f6 = t0 t1 (t4 − t4 ), 0 1
0 0 , 0 i

0
−1 8

0 −1 0 0 0 √ 2 0



 σ2 = 

σ1 =  0 0
0

0 0
−1 8 8

0 1 0 0

1 0 0 0

0 0 −1 0 0 1 0 0

0 0 , 0 i 0 0
3



 1  σ3 = √  2

−1 8 5 8

0 0

0 0

0 0  , 0  √ 2

 σ4 =

1 0 0 0

0

0 0 . 0 1



The following table gives a list of the full automorphism groups of Del Pezzo surfaces of degree 1.

EXERCISES
Type Order Structure f4 0 0 0 0 a(t4 + αt2 t2 + t4 ) 0 0 1 1 at2 t2 0 1 t4 0 t4 0 at2 t2 0 1 t2 t2 0 1 0 t4 0 t4 0 t4 + t4 + at2 t2 0 1 0 1 at4 + t4 0 1 t4 + t4 + at2 t2 0 1 0 1 0 t0 (at3 + bt3 ) 0 1 g2 (t2 , t2 0 1 g2 (t2 , t2 0 1 f4 (t0 , t1 ) t6 0 f6 t0 t1 (t4 − t4 ) 0 1 t6 + t6 0 1 3 t3 + t6 + at0 1 1 t0 (t5 + t5 ) 0 1 t0 t1 (t4 − t4 ) 0 1 t6 + t6 1 0 t6 1 t0 t5 1 t0 t1 (t4 + t4 ) 1 0 t6 + at3 t3 + t6 0 0 1 1 g3 (t2 , t2 ) 0 1 at6 + t6 0 1 t0 (at5 + t5 ) 0 1 bt0 t1 (t4 − t4 ) 0 1 t2 (bt4 + ct4 ) 0 0 1 t0 t1 (b(t4 + t4 ) + ct2 t2 ) 0 1 0 1 f6 (t0 , t1 ) ct6 + dt3 t3 + t6 0 0 1 1 t0 t1 f2 (t2 , t2 ) 0 1 g3 (t2 , t2 ) 0 1 f6 (t0 , t1 )

305
Parameters

I II III IV V VI VII VIII IX X XI XII XIII XIV XV XVI XVII XVIII XIX XX XXI

144 72 36 30 24 24 24 20 16 12 12 12 10 8 8 8 6 6 4 4 2

3 × (T : 2) 3 × 2D12 6 × D6 30 T 2D12 2 × 12 20 D16 D12 2×6 2×6 10 Q8 2×4 D8 6

a=0 √ α = 2 −3 a=0

a=0 a=0 a=0 √ a = 2 −3

a=0

b=0

6
4 22 2

Table 8.11: Groups of automorphisms of Del Pezzo surfaces of degree 1

Exercises
8.1 Prove that H 1 (S, OS ) = 0 for a weak Del Pezzo surface S without using the KodairaRamanujam’s Vanishing Theorem. 8.2 Let f : X → X be a resolution of a surface with canonical singularities. Show that R1 f∗ (OX ) = 0. 8.3 Describe all possible types of simple singularities which may occur on a plane curve of degree 4. 8.4 Let G(2, 5) be the Grassmannian of lines in P4 embedded in P9 by the Pl¨ cker embedding. u Show that the intersection of G(2, 5) with a general linear subspace of codimension 4 is an anticanonical model of a weak Del Pezzo surface of degree 5. 8.5 Let S be a weak Del Pezzo surface of degree 6. Show that its anticanonical model is isomorphic to a section of the Segre variety s(P2 × P2 ) in P8 by a linear subspace of codimension 2. 8.6 Let S be a weak Del Pezzo surface of degree 5. Show that its anticanonical model contains 5 pencils of conics and the group of automorphisms Aut(X) on this set of pencils defines an isomorphism Aut(S) → S5 .

306

CHAPTER 8. DEL PEZZO SURFACES

8.7 Prove that any non-degenerate surface of degree 5 in P5 is isomorphic to an anticanonical model of a Del Pezzo surface or a scroll. 8.8 Describe all weak Del Pezzo surfaces which are toric varieties (i.e. contain an open Zariski subset isomorphic to the torus (C∗ )2 such that each translation of the torus extends to an automorphism of the surface). 8.9 Describe all possible singularities on a weak Del Pezzo surface of degree d ≥ 5. 8.10 A Dupont cyclide surface is a quartic cyclide surface with 4 isolated singular points. Find an equation of such a surface. 8.11 Show that a weak quartic Del Pezzo surface is isomorphic to a minimal resolution of the double cover of the plane branched along the union of two conics. Show that the surface is a Del Pezzo surface if and only if the conics intersect transversally. 8.12 Let v, w be two exceptional vectors in I 1,N . Without using their explicit classification show that 0 ≤ v · w ≤ 1. 8.13 Let S be a Del Pezzo surface of degree 4 obtained by blowing up 5 points in the plane. Show that there exists a projective isomorphism from the conic containing the five points and the pencil of quadrics whose base locus is an anticanonical model of S such that the points are sent to singular quadrics. 8.14 Show that the anticanonical model of a Del Pezzo surface of degree 8 isomorphic to a quadric is given by the linear system of plane quartic curves with two fixed double points. 8.15 Show that the linear system of quadrics with 8 − d base points in general map P3 onto a 3-fold in Pd+1 of degree d. Show that an anticanonical model of a Del Pezzo surface of degree 8 > d ≥ 3 is projectively equivalent to a hyperplane section of this threefold. 8.16 Show the projection of an anticanonical Del Pezzo surfasce of degree d ≥ 3 for a general point is a surface of degree d in Pd−1 with the double curve of degree d(d − 3)/2. 8.17 Show that the Wiman pencil of 4-nodal plane sextics contains two 6-nodal rational curves and 2 unions of 5 conics [98].

Historical Notes
As the name suggests, P. Del Pezzo was the first who laid the foundation of the theory. In his paper of 1887 [71] he proves that a non-ruled nondegenerate surface of degree d in Pd can be birationally projected to a cubic surface in P3 from d − 3 general points on it. He showed that the images of the tangent planes at the points are skew lines on the cubic surface and deduced from this that d ≤ 9. He also gave a blow-up model of Del Pezzo surfaces of degree d ≥ 3, found the number of lines and studied some singular surfaces. J. Steiner was probably the first who related 7 points in the plane with curves of genus 3 by proving that the locus of singular points of the net of cubic curves is a plane sextic with nodes at the seven points [232]. A. Clebsch should be considered as a founder of the theory of Del Pezzo surfaces of degree 2. In his memoir [46] on rational double plane he considers a special case of double planes branched along a plane quartic curve. He shows that the pre-images of lines are cubic curves passing through a fixed set of 7 points. He identifies the branch curve with the Steiner sextic and relates the Aronhold set of 7 bitangents with the seven base points. Although C. Geiser was the first to discover the involution defined by the double cover, he failed to see the double plane construction.

Historical Notes

307

E. Bertini in [17], while describing his birational involution of the plane, proves that the linear system of curves of degree 6 with eight double base points has the property that any curve from the linear system passing through a general point x must also pass through a unique point x (which are in the Bertini involution). He mentions that the same result was proved independently by L. Cremona. This can be interpreted by saying that the linear system defines a rational map of degree 2 onto a quadric surface. Bertini also shows that the set of fixed points of the involution is a curve of degree 9 with triple points at the base points. The quartic cyclids in P3 with a nodal conic were first studied in 1964 by G. Darboux[64] and M. Moutard [175] and a year later by E. Kummer [161]. The detailed exposition of Darboux’ work can be found in [65], [66]. Some special types of these surfaces were considered much earlier by Ch. Dupin [89]. Kummer was the first to observe the existence of five quadratic cones whose tangent planes cut out two conics on the surface (the Kummer cones). They correspond to the five singular quadrics in the pencil defining the corresponding quartic surface in P4 . A. Clebsch finds a plane representation of a quartic cyclide by considering a web of cubics through five points in the plane [44]. He also finds in this way the configuration of 16 lines previously discovered by Darboux and proves that the Galois group of the equation for the 16 lines is isomorphic to 24 S5 . An ‘epoch-making memoir’ ([225], p. 141) of C. Segre [222] finishes the classification of quartic cyclids by considering them as projections of a quartic surface in P4 . At he same time he classifies the anticanonical models of weak Del Pezzo surfaces of degree 4 in terms of pencil of quadrics they are defined by. One easily deduces from his classification the classification of singular points on weak anticanonical models of weak Del Pezzo surfaces. The classification of lines was found by other method by G. Timms [236]. The classification of double singular points on algebraic surfaces in P3 started from the work of G. Salmon [208] who introduced the following notation C2 for an ordinary node, Bk for binode (the tangent cone is the union of two different planes) which depend on how the intersection of the planes intersect the surface, an unode Uk with tangent cone being a double plane. The indices here indicates the difference k between the degree of the dual surface and the dual of the nonsingular surface of the same degree. This nomenclature can be applied to surfaces in spaces of arbitrary dimension if the singularity is locally isomorphic to the above singularities. For anticanonical Del Pezzo surfaces the defect k cannot exceed 8 and all corresponding singularities must be rational double points of types A1 = C2 , Ak−1 = Bk , Dk−2 = Uk , k = 6, 7, E6 = U8 . Much later P. Du Val [92] have characterized these singularities as ones which do not affect the conditions on adjunctions, the conditions which can be applied to any normal surface. He showed that each RDP is locally isomorphic to either a ∗ node C2 , or binode Bk , or an unode Uk , or other unnodes U8 ∗ = E6 , U8 = E7 and ∗ ∗ E10 = E8 (he renamed U8 with E8 . A modern treatment of RDP singularities was given by M. Artin [8]. In the same series of papers Du Val classifies all possible singularities of anticanonical models of weak Del Pezzo surfaces of any degree and relates them to Coxeter’s classification of finite reflection groups. For Del Pezzo surfaces of degree 1 and 2 this classification have been rediscovered in terms of possible root bases in the corresponding root lattices by T. Urabe [242]. The relationship of this classification to the study of

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the singular fibres of a versal deformation of a simple elliptic singularities was found e by H. Pinkham [191] J. M´ rindol [173] and E. Looijenga (unpublished). We refer to modern texts on Del Pezzo surfaces [225], [171], [72], [159].

Chapter 9

Cubic surfaces
9.1
9.1.1

More about the E6 -lattice
Double-sixers

Let us study the E6 -lattice in more details. A sixer in E6 is a set of 6 mutually orthogonal exceptional vectors in I 1,6 . An example of a sixer is the set {e1 , . . . , e6 }. Lemma 9.1.1. Let {v1 , . . . , v6 } be a sixer. Then there exists a unique root α such that (vi , α) = 1, i = 1, . . . , 6.

Moreover, (w1 , . . . , w6 ) = (rα (v1 ), . . . , rα (v6 )) is a sixer satisfying (vi , wj ) = 1 − δij . The root associated to (w1 , . . . , w6 ) is equal to −α. Proof. The uniqueness is obvious since v1 , . . . , v6 are linear independent, so no vector is orthogonal to all of them. Let v0 = 1 (−k6 + v1 + · · · + v6 ) ∈ R1,6 . 3

First we show that v0 ∈ I 1,6 . Since M ∗ = M it is enough to show that, for any x ∈ I 1,6 , (v0 , x) ∈ Z. Consider the sublattice N of I 1,6 spanned by v1 , . . . , v6 , k6 . We have (v0 , vi ) = 0, i > 0, and (v0 , k6 ) = −3. Thus (v0 , M ) ⊂ 3Z. By computing the discriminant of N , we find that it is equal to 9. By Lemma 8.2.1 N is a sublattice of index 3 in I 1,6 . Hence for any x ∈ I 1,6 we have 3x ∈ N . This shows that (v0 , x) = Now let us set α = 2v0 − v1 − . . . − v6 . 309 1 (v0 , 3x) ∈ Z. 3

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We check that α is a root, and (α, vi ) = 1, i = 1, . . . , 6. It remains to check the second assertion. Since rα preserves the symmetric bilinear form, {w1 , . . . , w6 } is a sixer. We have (vi , wj ) = vi , rα (vj ) = vi , vj + (vj , α)α = (vi , vj ) + (vi , α)(vj , α) = (vi , vj ) + 1 = 1 − δij . Finally we check that
2 (rα (vi ), −α) = rα (vi ), −rα (α) = −(vi , α) = 1.

The two sixers with opposite associated roots form a double-six of exceptional vectors. We recall the list of exceptional vectors in E6 in terms of the standard orthonormal basis in I 1,6 .

ai bi cij

= ei , i = 1, . . . , 6; = 2e0 − e1 − . . . − e6 + ei , i = 1, . . . , 6; = e0 − ei − ej , 1 ≤ i < j ≤ 6.

(9.1) (9.2) (9.3)

Theorem 9.1.2. The following is a list of 36 double-sixes with corresponding associated roots: 1 of type D a1 a2 a3 a4 a5 a6 αmax b1 b2 b3 b4 b5 b6 −αmax 15 of type Dij ai aj 20 of type Dijk ai cjk aj cik ak cij clm bn cmn bl cln bm αijk . −αijk bi bj cjk cik cjl cil cjm cim cjn cin αij , −αij

Proof. We have constructed a map from the set of sixers (resp. double-sixes) to the set of roots (resp. pairs of opposite roots). Let us show that no two sixers {v1 , . . . , v6 } and {w1 , . . . , w6 } can define the same root. Since w1 , . . . , w6 , k6 span a sublattice of finite index in I 1,6 , we can write
6

vi =
j=1

aj wj + a0 k6

(9.4)

9.1. MORE ABOUT THE E6 -LATTICE

311

with some aj ∈ Q. Assume that vi = wj for all j. Intersecting both sides with α, we get 1 = a0 + · · · + a6 . (9.5) Intersecting both sides with −k6 , we get 1 = a1 + · · · + a6 − 3a0 , hence a0 = 0. Intersecting both sides with wj we obtain −aj = (vi , wj ). Applying Proposition 8.2.17, we get aj ≤ −1. This contradicts (9.5). Thus each vi is equal to some wj . Proposition 9.1.3. The group W (E6 ) acts transitively on sixes and double-sixes. The stabilizer subgroup of a sixer (resp. double-six) is of order 6!, 2.6!. Proof. We know that the Weyl group W (EN ) acts transitively on the set of roots and the number of sixers is equal to the number of roots. This shows that all sixers form one orbit. The stabilizer subgroup of the sixer (a1 , . . . , a6 ) (and hence of a root) is the group S6 . The stabilizer of the double-six D is the subgroup S6 , sα0 of order 2.6!. It is easy to see that two different double-sixes can share either 4 or 6 exceptional vectors. More precisely, we have #D ∩ Dij = 4, #D ∩ Dijk = 6, #Dij ∩ Dkl = #Dij ∩ Dklm = 4 if #{i, j} ∩ {k, l} = 0 6 otherwise 4 if #{i, j} ∩ {k, l, m} = 0, 2 6 otherwise 4 if #{i, j} ∩ {k, l} = 1 6 otherwise

#Dijk ∩ Dlmn =

A pair of double-sixes is called a syzygetic duad (resp. azygetic duad) if they have 4 (resp. 6) exceptional vectors in common. The next lemma is an easy computation. Lemma 9.1.4. Two double-sixes with associated roots α, β form a syzygetic duad if and only if (α, β) ∈ 2Z. This can be interpreted as follows. Consider the vector space ¯ Q = Q/2Q ∼ F6 = 2 equipped with the quadratic form
1 q(x + 2Q) = 2 (x, x)

(9.6)

mod 2.

Notice that the lattice E6 is an even lattice, i.e. its quadratic form x → x2 takes only even values. So the definition makes sense. The associated symmetric bilinear form is the symplectic form (x + 2Q, y + 2Q) = (x, y) mod 2.

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¯ Each pair of opposite roots ±α defines a vector v in Q with q(v) = 1. It is easy to see that the quadric q has Arf invariant (see Chapter 5, Part I) equal to 1 and hence vanishes on 28 vectors. The remaining 36 vectors correspond to 36 pairs of opposite roots or, equivalently, double-sixes. Note that we have a natural homomorphism of groups W (E6 ) ∼ O(6, F2 )− = (9.7)

obtaned from the action of W (E6 ) on Q/2Q. It is an isomorphism. This is checked by verifying that the automorphism v → −v of the lattice Q does not belong to the Weyl group W and then comparing the known orders of the groups. It follows from above that an syzygetic pair of double-sixes corresponds to orthogonal vectors v, w. Since q(v + w) = q(v) + q(w) + (v, w) = 0 = 1 + 1 + 0 = 0, we see that each nonzero vector in the isotropic plane spanned by v, w comes from a double-six. A triple of pairwise syzygetic double-sixes is called a syzygetic triad of doublesixes. They span an isotropic plane. Similarly we see that a pair of azygetic double¯ sixes span a non-isotropic plane in Q with three nonzero vectors corresponding to a triple of double-sixes which are pairwise azygetic. It is called an azygetic triad of double-sixes. We say that three azygetic triads form a Steiner compex of triads of double sixes if ¯ the corresponding planes in Q are mutually orthogonal. It is easy to see that an azygetic triad contains 18 exceptional vectors and thus defines a set of 9 exceptional (the omitted ones). The 27 exceptional vectors omitted from three triads in a Steiner complex is equal to the set of 27 exceptional vectors E∗ . There are 40 Steiner complexes of triads: 6 10 of type Γijk,lmn = (D, Dijk , Dlmn ), (Dij , Dik , Djk ), (Dlm , Dln , Dmn ), 30 of type Γij,kl,mn = (Dij , Dikl , Djkl ), (Dkl , Dkmn , Dlmn ), (Dmn , Dmij , Dnij ). Theorem 9.1.5. The Weyl group W (E6 ) acts transitively on the set of triads of azygetic double-sixes with stabilizer subgroup isomorphic to the group S3 × (S3 S2 ) of order 432. It also acts transitively on Steiner complexes of triads of double-sixes. A stabilizer subgroup is a maximal subgroup of W (E6 ) of order 1296 isomorphic to the wreath product S3 S3 . Proof. We know that a triad of azygetic double-sixes corresponds to a pair of roots (up to replacing the root with its negative) α, β with (α, β) = ±1. This pair spans a root sublattice Q of E6 of type A2 . Fix a root basis. Since the Weyl group acts transitively on the set of roots, we find w ∈ W such that w(α) = αmax . Since (w(β), αmax ) = (β, α) = 1, we see that w(β) = ±αijk for some i, j, k. Applying elements from S6 , we may assume that w(β) = −α123 . Obviously, the roots α12 , α23 , α45 , α56 are orthogonal to w(α) and w(β). These roots span a root sublattice of type 2A2 . Thus we obtain that the orthogonal complement of Q in E6 contains a sublattice of type

9.1. MORE ABOUT THE E6 -LATTICE

313

2A2 ⊥ A2 . Since |disc(A2 )| = 3, it follows easily from Lemma 8.2.1 that Q⊥ is a root lattice of type 2A2 . Obviously any automorphism which leaves the two roots α, β invariant leaves invariant the sublattice Q and its orthogonal complement Q⊥ . Thus the stabilizer contains a subgroup isomorphic to W (A2 ) × W (A2 ) × W (A2 ) and the permutation of order 2 which switches the two copies of A2 in Q⊥ . Since W (A2 ) ∼ = S3 we obtain that a stabilizer subgroup contains a subgroup of order 2.63 = 432. Since its index is equal to 120, it must coincide with the stabilizer group. It follows from above that a Steiner complex corresponds a root sublattice of type A2 ⊥ A2 ⊥ A2 contained in E6 . The group W (A2 ) S3 of order 3.432 is contained in the stabilizer. Since its index is equal to 40, it coincides with the stabilizer. Remark 9.1.1. The notions of syzygetic (azygetic) pairs, triads and a Steiner complex of triads of double-six is analogous to the notions of syzygetic (azygetic) pairs, triads, and a Steiner complex of bitangents of a plane quartic (see Chapter 6). In both cases we deal with a 6-dimensional quadratic space F6 . However, the difference is that the 2 quadratic forms are of different types.

9.1.2

Tritangent trios

A triple of v1 , v2 , v3 of exceptional vectors is called a tritangent trio if v1 + v2 + v3 = −k6 . If we view exceptional vectors as cosets in I 1,6 /Zk6 this is equivalent to saying that the cosets add up to zero. It is easy to list all tritangent trios. Lemma 9.1.6. There 45 tritangent trios: 30 of type ai , bj , cij , 15 of type cij , ckl , cmn , {i, j} ∪ {k, l} ∪ {m, n} = {1, 2, 3, 4, 5, 6}, i = j,

Theorem 9.1.7. The Weyl group acts transitively on the set of tritangent trios. Proof. We know that the permutation subgroup S6 of the Weyl group acts on tritangent trios by permuting the indices. Thus it acts transitively on the set of tritangent trios of the same type. Now consider the reflection with respect to the root α123 . We have rα123 (a1 ) = e1 + α123 = e0 − e3 − e4 = c34 , rα123 (b2 ) = (2e0 − e1 − e3 − e4 − e5 − e6 ) − α123 = e0 − e5 − e6 = c56 , rα123 (c12 ) = e0 − e1 − e2 = c12 . Thus w transforms the tritangent trio (a1 , b2 , c12 ) to the tritangent trio (c12 , c34 , c56 ). This proves the assertion.

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Remark 9.1.2. The stabilizer subgroup of a tritangent triplet is a maximal subgroup of W (E6 ) of index 45 isomorphic to the Weyl group of the root system of type F4 . Let Π1 = {v1 , v2 , v3 } and Π2 = {w1 , w2 , w3 } be two tritangent trios with no common elements. We have (vi , w1 + w2 + w3 ) = −(vi , k6 ) = 1. and by Proposition 8.2.17, (vi , wj ) ≥ 0. This implies that there exists a unique j such that (vi , wj ) = 1. After reordering, we may assume j = i. Let ui = −k6 − vi − wi . Since u2 = −1, (ui , k6 ) = −1, the vector ui is a exceptional vector. Since i
3 3 3

u1 + u2 + u3 =
i=1

(−k6 − vi − wi ) = −3k6 −
i=1

vi −
i=1

wi = −k6 ,

we get a new tritangent trio Π3 = (u1 , u2 , u3 ). The union Π1 ∪ Π2 ∪ Π3 contains 9 lines vi , wi , ui , i = 1, 2, 3. There is a unique triple of tritangents trios which consists of the same lines. It is formed by tritangents trios Πi = (vi , wi , ui ), i = 1, 2, 3. It is easy to see that any pair of triples of tritangents trios which consist of the same set of 9 lines is obtained in this way. Such a pair of triples of tritangent trios is called a pair of conjugate triads of tritangent trios. We can easily list all conjugate pairs of triads of tritangent trios: ai (I) bk cik bj cjk ak cij aj , bi cij (II) cln ckm ckl cim cjn cmn cjk , cil ai (III) bk cik bj al cjl cij ckl cmn

Here the conjugate triad can be read as the columns of the matrix. Altogether we have 20 + 10 + 90 = 120 different triads. There is a bijection from the set of pairs of conjugate triads to the set of azygetic triads of double-sixes. The 18 exceptional vectors contained in the union of the latter is the complementary set of the set of 9 exceptional vectors defined by a triad in the pair. Here is the explicit bijection. ai bk cik cij cln ckm ai bk cik bj cjk ak ckl cim cjn bj al cjl cij aj bi cmn cjk cil cij ckl cmn ↔ Dij , Dik , Djk

↔ D, Dikn , Djlm

↔ Dmn , Djkm , Djkn

Recall that the set of exceptional vectors omitted from each triad entering in a Steiner complex of triads of azygetic double-six is the set of 27 exceptional vectors. Thus a Steiner complex defines three pairs of conjugate triads of tritangent trios which contains all 27 exceptional vectors. We have 40 such triples of conjugate pairs.

9.2. LINES ON A NONSINGULAR CUBIC SURFACE

315

Theorem 9.1.8. The Weyl group acts transitively on the set of 120 conjugate pairs of triples of tritangent trios. A stabilizer subgroup H is contained in the maximal subgroup of W (E6 ) of index 40 realized as a stabilizer of a Steiner complex. The quotient group is a cyclic group of order 3. Proof. This follows from the established bijection between pairs of conjugate triads and triads of azygetic double-sixes and Theorem 9.1.5. In fact it is easy to see directly the transitivity of the action. It is clear that the permutation subgroup S6 acts transitively on the set of pairs of conjugate triads of the same type. Since the Weyl group acts transitively on the set of tritangent trios, we can send a tritangent trio (cij , ckl , cmn ) to a tritangent trio (ai , bj , cij ). As is easy to see from inspection that this sends a conjugate pair of type III to a pair of conjugate triads of type I. Also it sends a conjugate pair of type II to type I or III. Thus all pairs are W -equivalent. Remark 9.1.3. Note that each monomial entering into the expression of the determinant of the matrix expressing a conjugate pair of triads represents three orthogonal exceptional vectors. If we take only monomials corresponding to even permutations (resp. odd) we get a partition of the set of 9 exceptional vectors into the union of 3 triples of orthogonal exceptional vectors such that each exceptional vector from one triple intersects has non-zero intersection with two exceptional vectors from any other triple.

9.2

Lines on a nonsingular cubic surface

Let S be an nonsingular cubic surface in P3 . It is an anticanonical model of a Del Pezzo surface of degree 3. It has 27 lines (the images of exceptional vectors under any geometric marking), it has 45 triples of coplanar lines. These are the images of 45 tritangents trios of exceptional vectors. The corresponding plane is called a tritangent plane. Under isomorphism φ : I 1,6 ∼ Pic(S), we can transfer all the notions and the = statements from the previous sections to the Picard lattice Pic(S). Two different geometric markings π, π of S define a Cremona transformation of π ◦ π −1 : P2 − → P2 . If π : S → P2 and π : S → P2 be the blowing-down structures corresponding to an ordered double-six of type D, then the Cremona transformation is the symmetric involutorial transformation of degree 5 with characteristic matrix equal to  5 −2  −2  −2  −2  −2 −2 2 0 −1 −1 −1 −1 −1 2 −1 0 −1 −1 −1 −1 2 −1 −1 0 −1 −1 −1 2 −1 −1 −1 0 −1 −1 2 −1 −1 −1 −1 0 −1  2 −1  −1  −1  −1  −1 0

(9.8)

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9.2.1

Schur’s quadrics

Let Q ∈ S 2 E ∗ be a quadratic form on a finite-dimensional vector space V . Recall that the apolarity map defines a linear map apQ : E → E ∗ , v → Pv (Q) 1 which we identify with Q. For any linear subspace L ⊂ E, we have the polar subspace with respect to Q L⊥ = {x ∈ E : bQ (x, y) = 0, ∀y ∈ L} = Q(L)⊥ = ∩v∈L Pv (Q)⊥ . Q (9.9)

Here for any subspace W of E, we denote by W ⊥ the subspace of linear functions on E which are identical zero on W (the annulator of W , or the dual of W ). If Q is nondegenerate, then L⊥ = Q−1 (L⊥ ). Q If M is a linear subspace of E ∗ , we define its polar subspace with respect to Q by
⊥ MQ = Q(M ⊥ ). ⊥ If Q is nondegenerate, then MQ is the orthogonal complement of M ∗ with respect to ˇ on E ∗ defined by the linear map Q−1 : E ∗ → E. the dual quadratic form Q All of this can be extended to the projective space P(E) and a quadric hypersurface Q in P(E). For example, for any linear subspace L ⊂ P(E), the dual subspace L⊥ is a linear subspace of P(E ∗ ) spanned by the hyperplanes (considered as points in P(E ∗ )) containing L. Thus the dual H ⊥ of a hyperplane H ⊂ P(E) is the point a ∈ P(E ∗ ) corresponding to this hyperplane. Also, if Q is a quadric hypersurface in P(E) and L is a linear subspace of P(E), then the polar subspace of L with respect to Q is equal to

L⊥ = ∩a∈L Pa (Q). Q Also, for any linear subspace W of P(E ∗ ),
⊥ WQ = (∩a∈W ⊥ Pa (Q))⊥

(9.10)

(9.11)

Obviously, it is enough to do the intersection for a spanning set of the subspace. Let 1 , . . . , 6 be a set of skew lines on a nonsingular cubic surface S ⊂ P3 = P(E). A nonsingular quadric Q in P3 defines six skew lines 1 , . . . , 6 , where i is polar to i with respect to Q. The following beautiful result of Ferdinand Schur [214] shows that there exists a unique nonsingular quadric Q such that the set 1 , . . . , 6 is a set of skew lines on S (which together with 1 , . . . , 6 makes a double-six). Theorem 9.2.1. Let (l1 , . . . , l6 ), (l1 , . . . , l6 ) be a double-six of lines on a nonsingular cubic surface S. There exists a unique nonsingular quadric in P3 such that li is the polar line of li with respect to Q for each i = 1, . . . , 6.

9.2. LINES ON A NONSINGULAR CUBIC SURFACE

317

Proof. Fix an ordered double-six ( 1 , . . . , 6 ), ( 1 , . . . , 6 ) on a nonsingular cubic surface S. Choose a geometric marking φ : I 1,6 → Pic(S) such that φ(ei ) = ei = [ i ], i = 1, . . . , 6. Then the lineat system φ(e0 ) defines a birational map π : S → P2 which blows the lines i to the points pi . The image of the lines i is the conic Ci passing through all pj except pi . The pre-image of li with respect to φ is the exceptional vector bi . Let φ : I 1,6 → Pic(S) be the geometric marking such that φ (ei ) = i . It is obtained from φ by composing φ with the reflection sαmax ∈ O(I 1,6 ). We have φ (e0 ) = φ(sαmax (e0 )) = φ(5e0 − 2e1 − . . . − 2e6 ). Thus the linear system |e0 | = |5e0 − 2e1 − . . . − 2e6 | defines a birational map π : S → P2 which blows down the lines li to points qi . Note that there is no canonical identification of two P2 ’s. One views them as different planes 1 P2 and 2 P2 . For any line in 1 P2 , its full pre-image in S belongs to the linear system |e0 |. Since e0 · (−KS ) = 3, the curves in |e0 | are rational curves of degree 3 (maybe reducible). Similarly, the pre-images of lines in 2 P2 are rational curves of degree 3 on S. Now π ∗ ( ) + π ∗ ( ) ∈ |e0 + 5e0 − 2e1 − . . . − 2e6 | = | − 2KS |. Thus the union of two rational curves π ∗ ( ) and π ∗ ( ) is cut out by a quadric Ql,l in P3 = | − KS |∗ . Note that the intersection of a quadric and a cubic is a curve of arithmetic genus 4. Our curves are reducible curves of arithmetic genus 4. When we vary and , the corresponding quadrics span a hyperplane H in | − 2KS |. The map
1P

ˇ 2 × 2 P2 → H, ˇ

(l, l ) → Ql.l

is isomorphic to the Segre map. ˇ Recall that our surface S lies in P3 ∼ | − KS |∗ . Consider the dual space P3 = = ˇ be the quadric in this space which is apolar to all quadrics in the | − KS | and let Q hyperplane H, i.e. orthogonal to H with respect to the apolarity map S 2 (H 0 (S, OS (−KS )) × S 2 (H 0 (S, OS (−KS )∗ ) → C. In particular, if a quadric from H is a pair of planes Π1 ∪ Π2 corresponding to points ˇ ˇ a = Π⊥ and b = Π⊥ in P3 , then Pa,b (Q) = 0. 1 2 Now choose 3 special lines pi , pj , pi , pk , pj , pk in the first plane and similar lines qi , qj , qi , qk , qj , qk in the second plane. Then Rij = π ∗ (pi , pj ) = lij + li + lj , where ∗ ij = φ(cij ). Similarly, Rjk = π (qj , qk ) = lj + lk + ljk , where
6 6 6

ljk ∼ e0 − lj − lk = (5e0 − 2
i=1

ei ) − (2e0 −
i=1

ei + lj ) − (2e0 −
i=1

ei + ek )

= e0 − ej − ek ∼ lij . Thus the lines jk and ij coincide. Now notice that the curve Rij + Rjk is cut out by the reducible quadric Hij ∪ Hjk , where Hij is the tritangent plane containing the lines i , j , ij and Hjk is the tritangent plane containing the lines j , k , jk .

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Let a ∈ P(E ∗ ) and Ha = a⊥ be the corresponding hyperplane in P(E). If a, b Q = 0 for a, b ∈ P(E ∗ ), then ˇ ˇ Q(a) = (Ha )⊥ ∈ Hb . ˇ Q Let Pij = (Hij )⊥ . Since each pair of planes Hab , Hbc , considered as points in the dual ˇ Q ˇ space, are orthogonal with respect to Q, the point Pij belongs to Hki ∩ Hjk ∩ Hji . It is easy to see that this point is aj ∩ bi . Since ai ∩ bj ∈ Hij , the points Pij and Pji are ˇ polar to each other with respect to Q. Similarly, we find that the points (Pki , Pik ) are ˇ polar with respect to Q, hence the lines i and i are polar with respect to Q.

Figure 9.1: ˇ ˇ Let us show that Q is a nondegenerate quadric. Suppose Q is degenerate, then its ˇ is a linear space of positive dimension equal to the kernel set of singular points Sing(Q) ˇ of the symmetric bilinear form associated to Q. Thus, for any subspace L of P3 , the ⊥ ˇ lies in Sing(Q)⊥ . Therefore the points Pij lie in ˇ polar subspace LQ with respect to Q ˇ 3 a proper subspace of P . But this is obvioulsy impossible, since some of these points ˇ lie on a pair of skew lines and span P3 . Thus we can define the dual quadric Q of Q and obtain that the lines i and i are polar with respect to Q. Let us show the uniqueness of Q. Suppose we have two quadrics Q1 and Q2 such that i = ( i )⊥i , i = 1, . . . , 6. Let Q be a singular quadric in the pencil spanned by Q1 Q and Q2 . Let K be its space of singular points. Assume first that dim K = 0. Without loss of generality we may assume that K ∈ l1 ∪ l2 . Then dim(l1 )⊥ = dim(l2 )⊥ = 1. On the other hand, l1 ⊂ (l1 )⊥ , l2 ⊂ (l2 )⊥ . Q Q Q Q ⊥ Thus we have the equalities. But now l1 ⊂ KQ = P3 , hence K ⊂ (l1 )⊥ = l1 and Q similarly, K ⊂ l2 . Since l1 , l2 are skew, we get a contradiction. Assume now that dim K = 1. Since K cannot intersect all six lines li (otherwise it is contained in S and there are no such lines in S), we may assume that K does not

9.2. LINES ON A NONSINGULAR CUBIC SURFACE

319

intersect l1 . Then, as above, l1 = (l1 )⊥ and K = l1 . Now, K does not intersect l2 . Q Repeating the argument, we obtain that K = l2 . Thus l1 = l2 , which is a contradiction. Finally assume that dim K = 2. Then K intersects all lines. Then (li )⊥ are all of Q dimension ≥ 2 and contain K. Since K may contain at most two lines from the doublesix, we may assume that (l1 )⊥ = (l2 )⊥ = K. Since l1 ⊂ (l1 )⊥ = K, l2 ⊂ (l2 )⊥ = K, Q Q Q Q we see that the lines l1 , l2 are coplanar and hence intersect. This is a contradiction. Definition 9.1. The quadric Q is called the Schur quadric with respect to a given double-six. Consider the intersection curve C of the Schur quadric Q with the cubic surface S. Obviously it belongs to the linear system |−2KS |. Let π : S → P2 , π : S → P2 be the birational morphisms defined by the double-six (l1 , . . . , l6 ), (l1 , . . . , l6 ) corresponding to Q. The image of C under π (or π ) is a curve of degree 6 with double points at the points pi = π(li ). We call this curve the Schur sextic. It is defined as soon as we choose 6 points on P2 such that S is isomorphic to the blow-up of these points. Proposition 9.2.2. The six double points of the Schur sextic are bi-flexes, i.e. the tangent line to each branch is tangent to the branch with multiplicity ≥ 3. Proof. Let li ∩ Q = {a, b} and li ∩ Q = {a , b }. We know that Pa (Q) ∩ Q = {x ∈ Q : a ∈ P T (Q)x }. Since li = (li )⊥ , we have Q li ∩ Q = (Pa (Q) ∩ Pb (Q)) ∩ Q = {a , b }. This implies that a , b ∈ P T (Q)a and hence the lines a, a , a, b span the tangent space of Q at the point a. The tangent plane P T (Q)a contains the line li and hence intersects the cubic surface S along li and a conic Ka . We have P T (Ka ) = P T (S)a ∩ P T (Q)a = P T (Q ∩ S)a . Thus the conic Ka and the curve C = Q ∩ S are tangent at the point a. Since the line li is equal to the proper inverse transform of the conic Ci in P2 passing through the points pj , j = i, the conic Ka is the proper inverse transform of some line in the plane passing through pi . The point a corresponds to the tangent direction at pi defined by a branch of the Schur sextic at pi . The fact that Ka is tangent to C at a means that the line is tangent to the branch with multiplicity ≥ 3. Since similar is true, when we replace a with b, we obtain that pi is a bi-flex of the Schur sextic. Remark 9.2.1. A bi-flex is locally given by an equation whose Taylor expansion looks like xy + xy(ax + by) + f4 (x, y) + . . .. This shows that one has impose 5 conditions to get a bi-flex. To get 6 bi-flexes for a curve of degree 6 one has to satisfy 30 linear equations. The space of homogeneous polynomials of degree 6 in 3 variables is of dimension 28. So, the fact that such sextics exist is very surprising. ⊥ Also observe that the set of quadrics Q such that lQ = l for a fixed pair of skew lines (l, l ) is a linear (projective) subspace of codimension 4 of the 9-dimensional space of quadrics. So the existence of the Schur quadric is quite unexpected!

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I do not know whether for a given set of 6 points on P2 defining a nonsingular cubic surface, there exists a unique sextic with bi-flexes at these points. Example 9.2.1. Let S be the Clebsch diagonal surface given by 2 equations in P4 :
4 4

ti =
i=0 i=0

t3 = 0. i

(9.12)

It exhibits an obvious symmetry defined by permutations of the coordinates. Let a = √ √ 1 5), a = a = 1 (1 − 5) be two roots of the equation x2 − x − 1 = 0. One 2 (1 + 2 checks that the skew lines l : t1 + t3 + at2 = at3 + t2 + t4 = at2 + at3 − t5 = 0 and l : t1 + t 2 + a t 4 = t 3 + a t 1 + t4 = a t 1 + a t 4 − t 5 = 0 lie on S. Applying to each line even permutations we obtain a double-six. The Schur quadric is t2 = 0. i

9.3
9.3.1

Singularities
Non-normal cubic surfaces

Let X be an irreducible cubic surface in P3 . Assume that X is not normal. Then its singular locus contains a one-dimensional part C. Let C1 , . . . , Ck be irreducible components of C and mi be the multiplicity of a general point ηi of Ci as a point on X. A general section of X is a plane cubic curve H. Its intersection points with Ci are singular points of multiplicity mi . Their number is equal to di = deg(Ci ). By Bertini’s theorem, H is irreducible. Since an irreducible plane cubic curve has only one singular point of multiplicity 2, we obtain that C is irreducible and of degree 1. Let us choose coordinates in such a way that C is given by the equations t0 = t1 = 0. Then the equation of X must look like l0 t2 + 2l1 t0 t1 + l2 t2 = 0, 0 1 where li , i = 0, 1, 2, are linear forms in t0 , t1 , t2 . This shows that the left-hand side contains t2 and t3 only in degree 1. Thus we can rewrite the equation in the form t2 f + t3 g + h = 0. (9.13)

where f, g, h are binary forms in t0 , t1 , the first two of degree 2, and the third one of degree 3. Suppose f, g have no common zeros. Then the map P1 → P1 defined by (f, g) is of degree 2, and hence has two ramification points. This implies that f = al2 + bm2 , g = a l2 + b m2 for some linear polynomials l, m. After linear change of variables we may assume that l = t0 , m = t1 . Thus every monomial in the left-hand side of the equation (9.13) is divisible either by t2 or by t2 . Thus we can rewrite it in the form 0 1

9.3. SINGULARITIES

321

pt2 + qt2 , where p, q are linear forms in t0 , t1 , t2 , t3 . Without loss of generality, we 0 1 may assume that p has a non-zero coefficient at t3 . After a linear change of variables we may assume that p = t3 . If q has zero coefficient at t2 , our surface is a cone over a singular plane cubic. If the coefficient is non-zero, after a linear change of variables we may assume that q = t3 and the equation becomes t2 t2 + t3 t2 = 0. 0 1 Suppose f, g has one common non-multiple zero. After a linear change of variables t0 , t1 , we may assume that f = t0 t1 , g = t0 (t0 + t1 ) and the equation becomes t2 t0 t1 + t3 t0 t1 + t3 t2 + t0 t1 (at0 + bt1 ) + ct3 + dt3 = 0. 0 0 1 After the linear change of variables t2 → t2 + t3 + at0 + bt1 , we reduce the equation to the form t2 t0 t1 + t3 t2 + dt3 = 0. 0 1 Obviously, d = 0. Multiplying by d2 and changing t0 → dt0 , t1 → dt1 , t2 → d−1 t2 , we may assume that d = 1. Finally, if g is proportional to f , say g = λf , replacing t2 with t2 + λt3 , we reduce the equation 9.13 to the form t2 f + h = 0. In this case X is again a cone. Summarizing we get Theorem 9.3.1. Let X be an irreducible non-normal cubic surface. Then, either X is a cone over an irreducible singular plane cubic, or it is projectively equivalent to one of the following cubic surfaces singular along a line: (i) t2 t2 + t2 t3 = 0; 0 1 (ii) t2 t0 t1 + t3 t2 + t3 = 0. 0 1 The two surfaces are not projectively isomorphic. ¯ The last assertion follows from considering the normalization X of the surface X. In both cases it is a nonsingular surface, however in (i), the pre-image of the singular line is irreducible, but in the second case it is reducible. t3 → t3 + ct0 ,

9.3.2

Normal cubic surfaces

A normal cubic surface S has only isolated singularities. Let p be a singular point. Choose projective coordinates such that p = [1, 0, 0, 0]. Then the equation of the surface can be written in the form t0 f2 (t1 , t2 , t3 ) + f3 (t1 , t2 , t3 ) = 0, where f2 and f3 are homogeneous polynomials of degree given by the subscripts. If f2 = 0, the surface is a cone over a nonsingular plane cubic curve. If f2 = 0, then

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p is a singular point of multiplicity 2. Projecting from p we see that S is birationally isomorphic to P2 . Let π : S → S be a minimal resolution of singulartities of S. The sheaf R1 π∗ OS has support only at singular points of S. Since S is normal, π∗ OS = OS . Applying the Leray spectral sequence we obtain an exact sequence 0 → H 1 (S, OS ) → H 1 (S , OS ) → H 0 (S, R1 π∗ OS ) → H 2 (S, OS ). Since S is a nonsingular rational surface, we have H 1 (S , OS ) = 0. The canonical sheaf of S is OS (−1), hence, by Serre’s duality, H 2 (S, OS ) = H 0 (S, ωS ) = 0. Thus we obtain that H 0 (S, R1 π∗ OS ) = 0. This shows that, for any singular points s ∈ S, we have (R1 π∗ OS )s = 0. As is known (see [197]) this characterizes canonical singularities (or RDP) of a surface. This gives Theorem 9.3.2. Let S be a normal cubic surface in P3 . Then S is either a cone over a nonsingular plane cubic curve or an anticanonical model of a weak Del Pezzo surface of degree 3.

9.3.3

Canonical singularities

From now on we assume that X is a cubic surface with canonical singularities, i.e. X an anticanonical model of a weak Del Pezzo surface S of degree 3 All possible Dynkin curves on S can be easily found from the list of root bases in E7 . These are all root bases in E7 of rank ≤ 6 except of types D6 , D5 + A1 and 2A1 + D4 . These do not occur since the discriminants of the corresponding sublattices of E6 do not satisfy the assertion of Lemma 8.2.1. Let us list the remaining types of root bases
s

(r = 6) (r = 5) (r = 4) (r = 3) (r = 2) (r = 1)

E6 , A6 , D4 + A2 ,
k=1 s

Aik , i1 + · · · + is = 6 Aik , i1 + · · · + is = 5

D5 , D4 + A1 ,
k=1 s

D4 ,
k=1

Aik , i1 + · · · + is = 4

A3 , A2 + A1 , 3A1 A2 , A1 + A1 A1 .

Lemma 9.3.3. Let p0 = (1, 0, 0, 0) be a singular point of V (f3 ). Write f3 = t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 ), where g2 , g3 are homogeneous polynomials of degree 2,3. Let p = (t0 , t1 , t2 , t3 ) ∈ V (f3 ). If the line p0 , p is contained in V (f3 ), then the point q = (t1 , t2 , t3 ) is a common point of the conic V (g2 ) and the cubic V (g3 ). If moreover, p is a singular point of V (f3 ), then the conic and the cubic intersect at q with multiplicity > 1.

9.3. SINGULARITIES Proof. This is easy to verify and is left to the reader.

323

Corollary 9.3.4. V (f3 ) has at most 4 singular points. Moreover, if V (f3 ) has 4 singular points, then each point is of type A1 . Proof. Let p0 be a singular point. Choose coordinates such that p0 = (1, 0, 0, 0) and apply Lemma 9.3.3. Suppose we have more than 4 singular points. The conic and the cubic will intersect at at least 4 singular points with multiplicity > 1. Since they do not share an irreducible component (otherwise f3 is reducible), this contradicts B´ zout’s e Theorem. Suppose we have 4 singular points and p0 is not of type A1 . Since p0 is not an ordinary double point, the conic V (g2 ) is reducible. Then the cubic V (g3 ) intersects it at 3 points with multiplicity > 1 at each point. It is easy to see that this also contradicts B´ zout’s Theorem. e Lemma 9.3.5. The cases, Ai1 +· · ·+Aik , i1 +· · ·+ik = 6, except the cases 3A2 , A5 + A1 do not occur. Proof. Assume M = Ai1 +· · ·+Aik , i1 +· · ·+ik = 6. Then dM = (i1 +1) · · · (ik +1). Since 3|dM , one of the numbers, say i1 + 1, is equal either to 3 or 6. If i1 + 1 = 6, then M = A5 + A1 . If i1 + 1 = 3, then (i2 + 1) . . . (ik + 1) must be a square, and i2 + · · · + ik = 4. It is easy to see that the only possibility are i2 = i3 = 2 and i2 = i3 = i4 = i5 = 1. The last possibility is excluded by applying Corollary 9.3.4. Lemma 9.3.6. The cases D4 + A1 and D4 + A2 do not occur Proof. Let p0 be a singular point of S of type D4 . Again, we assume that p0 = (1, 0, 0, 0) and apply Lemma 9.3.3. As we have already noted, the singularity of type D4 is analytically (or formally) isomorphic to the singularity z 2 + xy(x + y) = 0. This shows that the conic V (g2 ) is a double line . The plane z = 0 cuts out a germ of a curve with 3 different branches. Thus there exists a plane section of S = V (f3 ) passing through p0 which is a plane cubic with 3 different braches at P . Obviously, it must be a union of 3 lines with a common point at p0 . Now the cubic V (g3 ) intersects the line at 3 points corresponding to the lines through p0 . Thus S cannot have more singular points. Let us show that all remaining cases are realized. We will exhibit the corresponding Del Pezzo surface as a the blow-up of 6 bubble points p1 , . . . , p6 in P2 . A1 : 6 points in P2 on an irreducible conic. A2 : p3 4A1 : p2 A3 : p4 A4 : p5 3A1 :p2 2A2 : p3
1 1

p1 .
1

p1 , p 4
1 1

1

p2 .
1

p3 p4
1

p2
1

p1 .
1 1 1 1

A2 + A1 :p3
1 1

p2
1

p1 , p 5
1

p4 .

p3
1 1

p2

p1 . p5 .
1

p1 , p 4 p2

p3 , p 6

p1 , p 6

p5

p4 .

324 A + 3 + A1 : p4 A5 :p6 D4 : p2
1 1

CHAPTER 9. CUBIC SURFACES p3
1 1 1

p2
1 1

1

p1 , p 6
1

1

p5 .

p5

1

p4
1

p3
1

p2
1

p1 .

1

p1 , p 4
1

p3 , p 6
1

p5 and p1 , p3 , p5 are collinear. p4 , and | − p1 − p2 − p3 | = ∅.
1

A2 + 2A1 : p3 A4 + A1 : p5 D5 : p5
1

p2 p3 p2 p3
1 1

p1 , p 5
1

p4

p3 p2

p2

p1 and |2 − p1 − . . . − p6 | = ∅.

p4

1

1

1

p1 and | − p1 − p2 − p6 | = ∅.
1

4A1 : p1 , . . . , p6 are the intersection points of 4 lines in a general linear position. 2A2 + A1 : p3 A3 + 2A1 : p4 A5 + A1 : p6 E6 :p6
1 1 1 1 1 1 1

p1 , p 6 p2
1 1 1

p5
1 1

1 1

p4 and | − p1 − p2 − p3 | = ∅. p5 and | − p1 − p2 − p3 | = ∅.
1

p1 , p 6 p2

p5 p4

p4 p3

p3 p2

p1 and |2 − p1 − . . . − p6 | = ∅.

p5

p1 and | − p1 − p2 − p3 | = ∅.

3A2 : p3 1 p2 1 p1 , p6 1 p5 1 p4 , | − p1 − p2 − p3 | = ∅, | − p4 − p5 − p6 | = ∅. Projecting from a singular point and applying Lemma 9.3.3 we see that each singular cubic surface can be given by the following equation. A1 : V (t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 )), where V (g2 ) is a nonsingular conic which intersects V (g3 ) transversally. A2 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )),where V (t1 t2 ) intersects V (g3 ) transversally. 2A1 : V (t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 )), where V (g2 ) is a nonsingular conic which is simply tangent to V (g3 ) at one point. A3 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (t1 t2 ) intersects V (g3 ) at the point [0, 0, 1] and at other 4 distinct points. A2 + A1 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (g3 ) is tangent to V (t2 ) at [1, 0, 0]. A4 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (g3 ) is tangent to V (t1 ) at [0, 0, 1]. 3A1 : V (t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 )), where V (g2 ) is nonsingular and is tangent to V (g3 ) at 2 points. 2A2 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) intersects V (g3 ) transverally and V (t2 ) is a flex tangent to V (g3 ) at [1, 0, 0]. A3 + A1 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (g3 ) passes through [0, 0, 1] and V (t1 ) is tangent to V (g3 ) at a point [1, 0, 0]. A5 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) is a flex tangent of V (g3 ) at the point [0, 0, 1]. D4 : V (t0 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) intersects transversally V (g3 ). 1 A2 + 2A1 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (g3 ) is tangent V (t1 t2 ) at two points not equal to [0, 0, 1]. A4 + A1 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (g3 ) is tangent to V (t1 ) at [0, 0, 1] and is tangent to V (t2 ) at [1, 0, 0]. D5 : V (t0 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) is tangent to V (g3 ) at [0, 0, 1]. 1 4A1 : V (t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 )), where V (g2 ) is nonsingular and is tangent to V (g3 ) at 3 points.

9.3. SINGULARITIES

325

2A2 + A1 : V (t0 g2 (t1 , t2 , t3 ) + g3 (t1 , t2 , t3 )), where V (g2 ) is tangent to V (g3 ) at 2 points [1, 0, 0] with multiplicity 3. A3 +2A1 : V (t0 t1 t2 +g3 (t1 , t2 , t3 )), where V (g3 ) passes through [0, 0, 1] and is tangent to V (g1 ) and to V (g2 ) at one point not equal to [0, 0, 1]. A5 + A1 : V (t0 t1 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) is a flex tangent of V (g3 ) at the point [0, 0, 1] and V (t2 ) is tangent to V (g3 ). E6 : V (t0 t2 + g3 (t1 , t2 , t3 )), where V (t1 ) is a flex tangent of V (g3 ). 1 3A2 : V (t0 t1 t2 +g3 (t1 , t2 , t3 )), where V (t1 ), V (t2 ) are flex tangents of V (g3 ) at points different from [0, 0, 1]. Remark 9.3.1. Applying a linear change of variables, one can simplify the equations. For example, in the case XXI, we may assume that the inflection points are [1, 0, 0] and [0, 1, 0]. Then g3 = t3 +t1 t2 L(t1 , t2 , t3 ). Replacing t0 with t0 = t0 +L(t1 , t2 , t3 ), 3 we reduce the equation to the form t 0 t 1 t 2 + t3 = 0 3 (9.14)

Another example is the E6 -singularity (case XX). We may assume that the flex point is [0, 0, 1]. Then g3 = t3 + t1 g2 (t1 , t2 , t3 ). The coefficient at t2 is not equal to zero, 2 3 otherwise the equation is reducible. After a linear change of variables we may assume that g2 = t2 + at2 + bt1 t2 + ct2 . Replacing t0 with t0 + at1 + bt2 , we may assume 3 1 2 that a = b = 0. After scaling the unknowns, we get t0 t2 + t1 t2 + t3 = 0. 1 2 2 (9.15)

The following table gives the classification of possible canonical singularities of a cubic surface, the number of lines and the class of the surface (i.e., the degree of the dual surface). Type I II III IV V VI VII VIII IX X XI Singularity ∅ A1 A2 2A1 A3 A2 + A1 A4 3A1 2A2 A3 + A1 A5 Lines 27 21 15 16 10 11 6 12 7 7 3 Class 12 10 9 8 8 7 8 6 6 6 6 Type XII XIII XIV XV XVI XVII XVIII XIX XX XXI Singularity D4 A2 + 2A1 A4 + A1 D5 4A1 2A2 + A1 A3 + 2A1 A5 + A1 E6 3A2 Lines 6 8 4 3 9 5 5 2 1 3 Class 6 5 5 5 4 4 4 4 4 3

Table 9.1: Singularities of cubic surfaces Note that the number of lines can be checked directly by using the equations. The map from P2 to S is given by the linear system of cubics generated by V (g3 ), V (t1 g2 ),

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CHAPTER 9. CUBIC SURFACES

V (t2 g2 ), V (t3 g2 ). The lines are images of lines or conics which has intersection 1 with a general member of the linear system. We omit the computation of the class of the surface.

9.4
9.4.1

Determinantal equations
Cayley-Salmon equation

Let l1 , l2 , l3 be three skew lines in P3 . Let Pi be the pencil of planes through the line li . Let us identify Pi with P1 and consider the rational map f : P1 × P1 × P1 − → P 3 which assigns to the triple of planes (Π1 , Π2 , Π3 ) the intersection point Π1 ∩ Π2 ∩ Π3 . This map is undefined at a triple (Π1 , Π2 , Π3 ) such that the line lij = Πi ∩ Πj is contained in Πk , where {i, j, k} = {1, 2, 3}. The line lij obviously intersects all three lines. The union of such lines is the nonsingular quadric Q containing l1 , l2 , l3 (count parameters to convince yourself that any 3 skew lines are contained in a unique nonsingular quadric). A plane from Pi intersects Q along li and a line mi on Q from another ruling. The triple belongs to the set fo undeterminacy locus I of f if and only if m1 = m2 = m3 . Consider the map φ : P1 × P1 × P1 → (P1 )3 , (Π1 , Π2 , Π3 ) → (m1 , m2 , m3 ).

We see that I = φ−1 (∆), where ∆ is the small diagonal. Obvioulsy φ is an isomorphism, so I is a smooth rational curve. Let ∆ij be one of the three diagonals (the locus of points with equal ith and jth coordinates). Its pre-image Di = φ−1 (∆ij ) is blown down under f to the line lk . In fact, if (Π1 , Π2 , Π3 ) ∈ D12 , then m1 = m2 and Π1 ∩ Π2 ∩ Π3 = m1 ∩ l3 . Clearly, D12 , D13 , D23 are divisors on P1 × P1 × P1 of degree (1, 1, 0), (1, 0, 1), (0, 1, 1), respectively. The map f can now be resolved by blowing up the curve I, followed by blowing down the proper inverse transforms of the divisors Dij to the lines lk . One should compare it with the standard birational map from the quadric P1 × P1 to P2 defined by the projection from a point. Note that in coordinates, f (Π1 , Π2 , Π3 ) is the line of solutions of a system of 3 linear equations, thus depends linearly in coefficients of each equations. This shows that the rational map is given by a linear system of divisors of degree (1, 1, 1). Let S be a cubic surface containing the lines l1 , l2 , l3 . Its full pre-image under f is a divisor of degree (3, 3, 3). It contains the divisors D12 , D13 , D23 whose sum is the divisor of degree (2, 2, 2). Let R be the residual divisor of degree (1, 1, 1). It is equal to the proper inverse transform of S. Let R=V
i,j,k=0,1

ai,j,k λi µj γk .

where (λ0 , λ1 ), (µ0 , µ1 ), (γ0 , γ1 ) are coordinates in the pencils P1 , P2 , P3 . Thus we obtain

9.4. DETERMINANTAL EQUATIONS

327

Theorem 9.4.1. (F. August) Any cubic surface containing 3 skew lines l1 , l2 , l3 can be generated by 3 pencils of planes in the following sense. There exists a correspondence R of degree (1, 1, 1) on P1 × P2 × P3 such that S = {x ∈ P3 : x ∈ Π1 ∩ Π2 ∩ Π3 for some (Π1 , Π2 , Π3 ) ∈ R} Let us rewrite R in the form R = V (λ0 A0 (µ0 , µ1 , γ0 , γ1 ) + λ1 A1 (µ0 , µ1 , γ0 , γ1 )), where A0 , A1 are bihomogeneous forms in (λ0 , λ1 ) and (γ0 , γ1 ). Suppose that S contains two distinct lines l, m which intersect l2 , l3 but do not intersect l1 . Let Π2 = l, l2 , Π3 = l, l3 . Since any plane Π in P1 intersects , the point (Π, Π2 , Π3 ) is mapped to S but not contained in any divisor D12 , D13 , D23 . Thus it belongs to R. Since Π is arbitrary, the point (Π2 , Π3 ) is the intersection point of the curves V (A0 ), V (A1 ) in P2 × P3 . This shows that the curves V (A0 ), V (A1 ) of bi-degree (1, 1) intersect at two distinct points. Change coordinates µ, γ to assume that these points are ((0, 1), (1, 0)) and ((1, 0), (0, 1)). Plugging in the equations of A0 = 0, A1 = 0, we see that V (A0 ), V (A1 ) belong to the pencil spanned by the curves V (µ0 γ0 ), V (µ1 γ1 ). Changing the coordinates (λ0 , λ1 ) we may assume that R = V (λ0 µ0 γ0 + λ1 µ1 γ1 ). The surface S is the set of solutions (t0 , t1 , t2 , t3 ) of the system of equations λ0 l1 (t0 , t1 , t2 , t3 ) = λ1 m1 (t0 , t1 , t2 , t3 ) µ0 l2 (t0 , t1 , t2 , t3 ) = µ1 m2 (t0 , t1 , t2 , t3 ) γ0 l3 (t0 , t1 , t2 , t3 ) = γ1 m3 i(t0 , t1 , t2 , t3 ). where li , mi are linear forms and λ0 µ0 γ0 + λ1 µ1 γ1 = 0. multiplying the left-hand sides and the right-hand-sides, we get S = V (l1 l2 l3 + m1 m2 m3 ). (9.17) (9.16)

Corollary 9.4.2. Assume additionally that S contains 2 distinct lines which intersect two of the lines l1 , l2 , l3 but not the third one. Then S can be given by the equation l1 l2 l3 − m1 m2 m3 = 0. (9.18)

An equation of cubic surface of this type is called a Cayley-Salmon equation. Observe that S contains the lines ij = V (li , mj ). Obviously, ii = li and 23 , 32 are the two lines which intersect 2 , 3 but not 1 . The lines 12 , 21 intersect 1 , 2 but not l3 (since otherwise V (A0 ), V (A1 ) in above have more than 2 intersection points). Similarly, we see that 13 , 31 intersect 1 , 3 but not 2 . Thus we have 9 different lines.

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CHAPTER 9. CUBIC SURFACES

As is easy to see they form a pair of two conjugate triads of tritangent planes (which can be defined as in the nonsingular case) l11 l21 l31 l12 l22 l32 l13 l23 . l33 (9.19)

Thus the condition on S imposed in Corollary 2.2.9 implies that S contains a pair of conjugate triples of tritangent planes. Conversely, such a set of 9 lines gives a Cayley-Salmon equation of S. In fact, each triple of the tritangent planes defines S along the same set of 9 lines. Thus S is contained in the pencil spanned by surfaces V (l1 l2 l3 ), V (m1 m2 m3 ). It is clear that two Cayley-Salmon equations defining the same set of 9 lines can be transformed to one another by a linear change of variables. Thus the number of essentially different Cayley-Salmon equations is equal to the number of pairs of conjugate triads of tritangent planes. Theorem 9.4.3. Let S be a normal cubic surface. The number of different CayleySalmon equations for S is equal to 120 (type I), 10 (type II), 1 (type III,IV, VIII), and zero otherwise. Proof. We know that the number of conjugate pairs of triads of tritangent trios of exceptional vectors is equal to 120. Thus the number of conjugate triads of triples of tritangent planes on a nonsingular cubic surface is equal to 120. It follows from the proof of Corollary 2.2.9 that a pair of conjugate triples of tritangent planes on a singular surface exists only if we can find 3 skew lines and 2 lines which intersect two of them but not the third. Also we know that the number of lines on S must be at least 9. So we have to check only types II − V I and V III. We leave to the reader to verify the assertion in these cases. Corollary 9.4.4. Let S be a nonsingular cubic surface. Then S is projectively equivalent to a surface V (t0 t1 t2 + t3 (t0 + t1 + t2 + t3 )l(t0 , . . . , t3 )). A general S can be written in this form in exactly 120 ways (up to projective equivalence). Proof. Consider a Cayley-Salmon equation l1 l2 l3 + m1 m2 m3 = 0 of S. let (9.19) be the corresponding 9 lines on S. If l1 , l2 , l3 , mj are linear independent, we choose a coordinate system such that l1 = t0 , l2 = t1 , l3 = t2 , mj = t3 . If not, the lines 1j , 2j , l3j intersect at one point pj = l1 ∩ l2 ∩ l3 . Assume that this is not the case for all j so that S is projectively equivalent to V (t0 t1 t2 + t3 m2 m3 ). let m2 = ai ti . If one of the ai ’s is equal to zero, say a3 = 0, the linear form m2 is a linear combination of coordinates t0 , t1 , t2 . We have assumed that this does not happen. Thus, after scaling the coordinates we may assume that m2 = ti . this gives the promised equation. Since we can start with any conjugate pair of triads of tritangent planes, the previous assumption is not satisfied only if any such pair consists of tritangent planes containing three concurrent lines. We will see later that the number of such tritangent planes on a nonsingular surface is at most 18. So we can always start with a conjugate triad of tritangent planes for which each plain does not contain concurrent lines.

9.4. DETERMINANTAL EQUATIONS

329

Corollary 9.4.5. let S be a normal surface of type I − IV or V III. Then there exists a 3 × 3 matrix A(t) with linear forms in t0 , . . . , t3 such that S = V det(A(t)) . Proof. Observe that l1 l1 l2 l3 + m1 m2 m3 = det  0 m3  m1 l2 0  0 m2  . l3

9.4.2

Hilbert-Burch Theorem

By other methods we will see that Corollary 9.4.5 can be generalized to any normal cubic surface of type different from XX. We will begin with the approach using the following well-known result from Commutative Algebra (see [103]). Theorem 9.4.6. (Hilbert-Burch) Let I be an ideal in polynomial ring R such that depth(I) = codimI = 2 (thus R/I is a Cohen-Macaulay ring). Then there exists a projective resolution 0 −→ Rn−1 −→ Rn −→ R −→ R/I −→ 0. The i-th entry of the vector (a1 , . . . , an ) defining φ1 is equal to (−1)i ci , where ci is the complementary minor obtained from the matrix A defining φ2 by deleting its ith row. We apply this theorem to the case when R = C[X0 , X1 , X2 ] and I is the homogeneous ideal of a closed 0-dimensional subscheme Z of P2 = Proj(R) generated by 4 linear independent homogeneous polynomials of degree 3. Let IZ be the ideal sheaf of Z. Then (IZ )m = H 0 (P2 , IZ (m)). By assumption H 0 (P2 , IZ (2)) = 0. (9.20)
φ2 φ1

Applying the Hilbert-Burch Theorem, we find a resolution of the graded ring R/I 0 −→ R(−4)3 −→ R(−3)4 −→ R −→ R/I → 0, where φ2 is given by a 3×4 matrix A(X) whose entries are linear forms in X0 , X1 , X2 . Passing to the projective spectrum, we get an exact sequence of sheaves 0 −→ W2 ⊗ OP2 (−4) −→ W1 ⊗ OP2 (−3) −→ IZ −→ 0, where W2 , W1 are vector spaces of dimension 3 and 4. Twisting by OP2 (3), we get the exact sequence 0 −→ W2 ⊗ OP2 (−1) −→ W1 ⊗ OP2 −→ IZ (3) −→ 0.
˜ φ2 ˜ φ1 φ2 φ1 φ2 φ1

(9.21)

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Taking global sections, we obtain W1 = H 0 (P2 , IZ (3)). Twisting by OP2 (−2), and using a canonical isomorphism H 2 (P2 , OP2 (−3)) ∼ C, we obtain that W2 = = H 1 (P2 , IZ (1)). The exact sequence 0 → IZ (1) → OP2 (1) → OZ → 0 shows that
0 W2 ∼ Coker(H 0 (P2 , OP2 (1)) → H 0 (OZ )) ∼ Coker C3 → Ch (OZ ) . = =

Since dim W2 = 3, we obtain that h0 (OZ ) = 6. Thus Z is a 0-cycle of length 6. ˜ Now we see that the homomorphism φ2 of vector bundles is defined by a linear map t : E → Hom(W2 , W1 ), (9.22)
∗ where P2 = P(E). We can identify the linear map t with the tensor E ∗ ⊗ W2 ⊗ W1 . Let us now view this tensor as a linear map ∗ ∗ u : W1 → Hom(E, W2 ).

(9.23)

In plain language, if t is viewed as a system of 3 linear equations with unknowns t0 , t1 , t2 , t3 whose coefficients are linear forms in variables x0 , x1 , x2 , then u is the same system rewritten as a system of 3 equations with unknowns x0 , x1 , x2 whose coefficients are linear forms in variables t0 , t1 , t2 , t3 . The linear map (9.22) defines a rational map
∗ f : P(E) → P(W1 ) = |IZ (3)|∗ ,

[v] → P(t(v)(W2 )⊥ ).

This is the map given by the linear system |IZ (3)|. In coordinates, it is given by maximal minors of the matrix A(X) defining φ2 . For any α ∈ t(v)(W2 )⊥ , we have u(α)(v) = 0. This shows that rank(u(α)) < 3. Thus S is contained in the locus of ∗ [α] such that α belongs to the pre-image of the determinantal locus in Hom(E, W2 ), i.e. the locus of linear maps of rank < 3. It is a cubic hypersurface in the space ∗ Hom(E, W2 ). Thus the image S of f is contained in a determinantal cubic surface S. Since the intersection scheme of two general members C1 , C2 of the linear system |IZ (3)| is equal to the 0-cycle Z of degree 6, the image of f is a cubic surface. This gives a determinantal representation of S. Theorem 9.4.7. Assume S is a normal cubic surface which does not have a singular point of type E6 . Then S admits a determinantal representation S = V (det(A)), where A is a matrix whose entries are linear forms. A surface with a singular point of type E6 does not admit such a representation. Proof. Assume S has no singular point of type E6 and let X be a minimal resolution of S. Then the set of (−2)-curves is a proper subset of the set of roots of the lattice ⊥ ⊥ KX and span a proper sublattice M of Pic(X). Let α be a root in KX which does not belong to M . Since the Weyl group W (X) acts transitively on the set of roots, we can choose a marking φ : I 1,6 → Pic(X) such that α = 2e0 − e1 − . . . − e6 . Let w ∈ W (X)n be a an element of the Weyl group generated by reflections with

9.4. DETERMINANTAL EQUATIONS

331

respect to (−2)-curves such that w ◦ φ is a geometric marking defining a geometric basis e0 , e1 , . . . , e6 . Since w preserves M , w(α) = 2e0 − e1 − . . . − e6 is not a linear combination of (−2)-curves. However any effective root x is a linear combination of (−2)-curves (use that x · KX ) = 0 and for any irreducible component E of x with E 2 = −2 we have E · KX < 0). Now X is obtained by blowing up a set of 6 points (maybe infinitely near) not lying on a conic. It is easy that this blow-up is isomorphic to the blow-up of a 0-dimensional cycle Z of length 6. Blowing up a sequence of k infinitely near points pk 1 . . . 1 p1 is the same as to blow up the ideal (x, y k ). The linear system |IZ (3)| is equal to the linear system of cubics through the points p1 , . . . , p6 . The ideal IZ is generated by a basis of the 3-dimensional linear system |IZ (3)| defining a rational map P2 − → S ⊂ P3 . Thus we can apply the Hilbert-Burch Theorem to obtain a determinant representation of S. Assume S has a singular point of type E6 and A(t) = (Aij )1≤i,j≤3 . Consider the system of linear equations
3

lij (t0 , . . . , t3 )xj = 0,
j=1

i = 1, 2, 3.

(9.24)

For any x = [x0 , x1 , x2 ] ∈ P2 the set of points p = (t0 , t1 , t2 , t3 ) such that A(t)x = 0 is a linear space. Consider the rational map π : S → P2 which assigns to t ∈ S the solution x of A(t)x = 0. Since π is not bijective, there exists a line on S which is blown down to a point (a1 , a2 , a3 ). This means that the equations (9.24) with [x0 , x1 , x2 ] substituted with (a1 , a2 , a3 ) define three planes intersecting along a line. Thus the three planes are linearly dependent, hence we can write
3 3 3

α(aj
j=1 3

l1j ) + β(aj
j=1

l2j ) + γ(
j=1

l3j )

=
j=1

aj (αl1j + βl2j + γl3j ) = 0.

for some α, β, γ not all zeros. Choose coordinates for x such that (a1 , a2 , a3 ) = [1, 0, 0]. Then we obtain that the entries in the first column of A(t) are linearly dependent. This allows us to assume that l11 = 0 in the matrix A(t). The equations l21 = l31 = 0 define the line . The equations l12 = l13 = 0 define a line m. Obviously, l = m since otherwise S has equation −l12 l21 l33 + l12 l31 l23 + l13 l21 l32 − l13 l31 l23 = 0, which shows that the line l = m is the double line of S. So, we see that S has at least two lines, but a surface of type XX has only one line. We have already seen that each time S is represented as the image of S under a rational map given by the linear system |IZ (3)|, where Z is a 0-cycle of length 6 satisfying condition (9.20), we can write S by a determinantal equation. A minimal resolution of indeterminacy points defines a blowing down morphism π : X → P2

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of the weak Del Pezzo surface X isomorphic to a minimal resolution of S. The inverse map is given by assigning to t ∈ S the nullspace N (A(t)). Changing Z to a projectively equivalent set replaces the matrix A(t) by a matrix A(t)C, where C is an invertible scalar matrix. This does not change the equation of S. Thus the number of essentially different determinantal representations is equal to the number of linear systems |e0 | on X which define a blowing down morphism X → P2 such that the corresponding geometric markings (e0 , e1 , . . . , e6 ) of X satisfy |2e0 − e1 − . . . − e6 | = ∅. This gives Theorem 9.4.8. The number of essentially different determinantal representations of S is equal to the number of unordered geometric markings (e0 , e1 , . . . , e6 ) of Pic(X) such that |2e0 − e1 − . . . − e6 | = ∅. It is equal to 72 if S is nonsingular.
∗ Consider again (9.22) as a tensor t ∈ W1 ⊗ V ∗ ⊗ W2 which defines a linear map ∗ ∗ t : W1 → Hom(W2 , V ). We have the corresponding rational map

g : S → P(W2 ) ∼ P2 , =

[w] → Ker(t∗ ).

∗ Let P3 = P(W1 ). Consider the projective resolution

0 −→ OP3 (−1) ⊗ W2 −→ OP3 ⊗ V ∗ −→ F → 0, where φ is defined by the linear map viewed as a 3 × 3 matrix with entries in W1 = ∗ (W1 )∗ . Since the determinant of the matrix is equal to the equation of S, the sheaf F is locally isomorphic to OS . We can write it as F = OS (D) for some divisor class D. Taking global sections, we obtain H 0 (S, OS (D)) ∼ V ∗ . Thus the linear system = |D| on S defines our rational map π : S → P(E). Twisting by OP3 (−3) and taking cohomology, we obtain an isomorphism W2 ∼ H 2 (S, OS (−3H + D)), where H is a = hyperplane section. Since OS (H) ∼ OS (−KS ), we obtain an isomorphism = W2 ∼ H 2 (S, OS (3KS + D)) ∼ H 0 (S, OS (−2KS − D))∗ . = = Let π : X → P(W2 ) be the rational map of a minimal resolution defined by g. The pre-image of D on X is equal to the class e0 and −KS = 3e0 − e1 − . . . − e6 , where (e0 , e1 , . . . , e6 ) is a geometric marking of X defined by the blowing down morphism π . Thus π is given by the linear system |5e0 − 2e1 − . . . − 2e6 |. If S is nonsingular, π blows down 6 skew lines l1 , . . . , l6 to 6 points p1 , . . . , p6 on P(E) and the map g = π blows down the proper inverse transforms m1 , . . . , m6 of conics Cj through the points pi , i = j to some points q1 , . . . , q6 in P(W2 ). The lines (l1 , . . . , l6 ; m1 , . . . , m6 ) form a double-six on S. It follows from above, that the lines (m1 , . . . , m6 ) define a determinantal representation of S corresponding to the transpose of the matrix A(t). Remark 9.4.1. We can also deduce Theorem 9.4.8 from the theory of determinantal representations from Chapter 4. Applying this theory we obtain that S admits a determinantal equation with entries linear forms if it contains a projectively normal curve C such that H 0 (S, OS (C)(−1)) = H 2 (S, OS (C)(−2)) = 0. (9.25)

φ

9.4. DETERMINANTAL EQUATIONS

333

Moreover, the set of non-equivalent determinantal representation is equal to the set of divisor classes of such curves. Let f : X → S be a minimal resolution and C = f ∗ (C). Since f ∗ OS (−1) = OX (KX ), the conditions (9.25) are equivalent to H 0 (X, OX (C + KX )) = H 2 (X, OX (C + 2KX )) = 0. (9.26)

Since C is nef, H 1 (X, OX (C + KX )) = 0. Also H 2 (X, OX (C + KX )) = H 0 (X, OX (−C )) = 0. By Riemann-Roch,
1 0 = χ(X, OX (C + KX )) = 2 ((C + KX )2 − (C + KX ) · KX ) + 1 1 = 2 (C 2

+ C · KX ) + 1.

Thus C is a smooth rational curve, hence C is a smooth rational curve. As we know from Chapter ??, a projectively normal rational curve in P3 must be of degree 3. Thus −KX · C = 3, hence C 2 = 1. The linear system |C | defines a birational map π : X → P2 . Let e0 = [C ], e1 , . . . , e6 be the corresponding geometric basis of Pic(X). We have KX = −3e0 + e1 + · · · + e6 and the condition 0 = H 2 (X, OX (C + 2KX )) = H 0 (X, OX (−C − KX )) = 0 is equivalent to |2e0 − e1 − . . . − e6 | = ∅. (9.27)

9.4.3

The cubo-cubic Cremona transformation

Consider a system of linear equations in variables t0 , t1 , t2 , t3
3

lij (z0 , z1 , z2 , z3 )tj = 0, i = 1, 2, 3, 4,
j=1

(9.28)

where lij are linear forms in variables z0 , z1 , z2 , z3 . It defines a rational map Φ : P3 − → P3 by assigning to (a0 , . . . , a3 ) ∈ P3 the space of solutions of the system (9.28) with ai substituted in zi . In coordinate-free way, we can consider the system as a linear map t : W → Hom(W1 , W2 ), where dim W = 4, dim W1 = 4, dim W2 = 3. The map Φ : P(W ) → P(W1 ) is defined by sending w ∈ W to the linear space Ker(t(v)) ⊂ W1 . The inverse map Φ−1 is defined by rewriting the system as a system with unknowns zi . Or, in corrdinate-free language, by viewing the tensor t ∈ W ∗ ⊗ ∗ W1 ⊗ W2 as a linear map W1 → Hom(W, W2 ). Let D be the set of linear maps (considered up to proportionality) W1 → W2 of rank ≤ 2. The map Φ is not defined at the pre-image of D in P(W ). It is given by the common zeros of the four maximal minors ∆i of the matrix (lij (Z)). The map Φ is given by [z0 , . . . , z3 ] → [∆1 , −∆2 , ∆3 , −∆4 ]. Lemma 9.4.9. The scheme-theoretical locus Z of common zeros of the cubic polynomials ∆i is a connected curve of degree 6 and arithmetic genus 3.

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CHAPTER 9. CUBIC SURFACES

Proof. We apply the Hilbert-Burch theorem to the ring R = C[z0 , z1 , z2 , z3 ] and the homogeneous ideal I of Z. We get a resolution 0 −→ R(−4)3 −→ R(−3)4 −→ I −→ 0.
φ2 φ1

(9.29)

Twisting by n and computing the Euler-Poincar´ characteristic we obtain the Hilbert e polynomial of the scheme Z P (Z; n) = χ(P3 , OZ (n)) = 6n − 2. This shows that Z is one-dimensional, and comparing with Riemann-Roch, we see that deg(Z) = 6 and χ(OZ ) = −2. The exact sequence 0 → IZ → OP3 → OZ → 0 gives dim H 0 (OZ ) = 1 if and only if H 1 (P3 , IZ ) = 0. The latter equality follows from considering the resolution of IZ . Thus Z is connected and pa (Z) = 1−χ(OZ ) = 3. The inverse map is also given by cubic polynomials. This explains the classical name for the transformation T , the cubo-cubic transformation. The pre-images of planes under Φ are cubic surfaces in P(W ) containing the curve Z. The images of planes under Φ are cubic surfaces in P(W1 ) containing the curve Z , defined similarly to Z for the inverse map Φ−1 . Now let V be a 3-dimensional subspace of W . Then restricting t : W → Hom(W1 , W2 ) to V we obtain a determinantal representation of the cubic surface S = Φ(P(E)) ⊂ P(W1 ). The map Φ : P(E) − → P(W1 ) is not defined at the set P(E) ∩ Z. Tensoring (9.29) with OP(E) we obtain a projective resolution for the ideal sheaf of Z ∩ P(E) in P(E) (use that Tor1 (R/J, I) = 0, where J is the ideal generated by the hyperplane in W defining V see [103], Exercise A3.16). If P(E) intersects Z transversally at 6 points, we see that S is a nonsingular cubic surface.

9.4.4

Cubic symmetroids

A cubic symmetroid is a hypersurface in Pn admitting a representation as a symmetric (3 × 3)-determinant whose entries are linear forms in n + 1-variables. Here we will be interested in cubic symmetroid surfaces. An example of a cubic symmetroid is a 4-nodal cubic surface   t0 0 t2 t1 −t3  t0 t1 t2 + t0 t1 t3 + t0 t2 t3 + t1 t2 t3 = det  0 −t3 t3 t2 + t3 It is called the Segre cubic surface. By choosing the singular points to be the reference points [1, 0, 0, 0], [0, 1, 0, 0], [0, 0, 1, 0], [0, 0, 0, 1], it is easy to see that cubic surfaces with 4 singularities of type A1 are projectively isomorphic. Note that the condition for a determinantal representation of a cubic surface with canonical singularities to be a symmetric determinantal representation is the existence of an isomorphism OS (C) ∼ OS (C)(2). =

9.4. DETERMINANTAL EQUATIONS This is obviously impossible for a nonsingular cubic surface.

335

Lemma 9.4.10. Let L ⊂ |OP2 (2)| be a pencil of conics. Then it is projectively isomorphic to one of the following pencils: (i) λ(t0 t1 − t0 t2 ) + µ(t1 t2 − t0 t2 ) = 0; (ii) λ(t0 t1 + t0 t2 ) + µt1 t2 = 0; (iii) λt2 + µ(t0 t1 + t0 t2 + t1 t2 ) = 0; 2 (iv) λt2 (t2 − t0 ) + µt1 (t0 + t2 ) = 0; (v) λt2 + µ(t0 t2 + t2 ) = 0; 0 1 (vi) λt2 + µt2 = 0; 0 1 (vii) λt0 t1 + µt0 t2 = 0; (viii) λt0 t1 + µt2 = 0. 0 Proof. Let C1 = V (f1 ), C2 = V (f2 ) be two generators of a pencil. If C1 and C2 have a common irreducible component we easily reduce it, by a projective transformation to cases (vii) or (viii). Assume now that C1 do not have a common component. Let k = #C1 ∩ C2 . Assume k = 4. Then, no three of the intersection points lie on a line since otherwise the line is contained in both conics. By a linear transformation we may assume that the intersection points are [1, 0, 0], [0, 1, 0], [0, 0, 1], (1, 1, 1). The linear system of conics passing through these points is given in (i). Assume k = 3. After a linear change of variables we may assume that C1 , C2 are tangent at [1, 0, 0] with tangency direction t1 + t2 = 0 and intersect transversally at [0, 1, 0] and [0, 0, 1]. The linear system of conics passing through the three points is λt0 t1 + µt0 t2 + γt1 t2 = 0. The tangency condition gives λ = µ. This gives case (ii). Assume k = 2. Let [1, 0, 0] and [0, 1, 0] be the base points. First we assume that C1 and C2 are tangent at both points. Obviously, one of the conics from the pencil is the double line t2 = 0. We can also fix the tangency directions to be t1 + t2 = 0 at 2 [1, 0, 0] and t0 + t2 = 0 at [0, 1, 0]. The other conic could be t0 t1 + t0 t2 + t1 t2 = 0. This gives case (iii). Now we assume that C1 and C2 intersect transversally at [1, 0, 0] and with multiplicity 3 at [0, 1, 0] with tangency direction t0 + t2 = 0. A conic passing through [1, 0, 0] and tangent to the line t0 + t2 = 0 at the point [0, 1, 0] has equation at2 + 2 b(t0 t1 + t1 t2 ) + dt0 t2 = 0. It is easy to check that the condition of triple tangency is a + d = 0. This gives case (iv). Finally assume that k = 1. Obviously the pencil is spanned by a conic and its tangent line taken with multiplicity 2. By a projective transformation it is reduced to form given in case (v) if the conic is irreducible and case (vi) if the conic is a double line . Theorem 9.4.11. Let S be an irreducible cubic symmetroid. Assume that S has only canonical singularities. Then S is projectively isomorphic to one of the following determinantal surfaces:

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(i) C3 = V (t0 t1 t2 + t0 t1 t3 + t0 t2 t3 + t1 t2 t3 ) with four RDP of type A1 ; (ii) C3 = V (t0 t1 t2 + t1 t2 − t2 t2 ) with two RDP of type A1 and one RDP of type A3 ; 3 3 (iii) C3 = V (t0 t1 t2 − t2 (t0 + t2 ) − t1 t2 ) with one RDP of type A1 and one RDP of 3 2 type A5 . Proof. Let A = (lij ) be a symmetric 3 × 3 matrix with linear entries lij (t0 , t1 , t2 , t3 ) defining the equation of S. It can be written in the form A(t) = t0 A0 + t1 A1 + t2 A2 + t3 A3 , where Ai , i = 1, 2, 3, 4, are symmetric 3 × 3 matrices. Let W be a linear system of conics spanned by the conics   t0 Ci = [t0 , t1 , t2 ] · A · t1  = 0. t2 The matrices Ai are linearly independent since otherwise S = V (det A(t)) is a cone with vertex (c0 , c1 , c2 , c3 ), where ci Ai = 0. Thus W is a web of conics. Let P2 = P(E) so that W = P(W ) for a 4-dimensional linear subspace W of S 2 E ∗ . Consider the polarity S 2 E ∼ (S 2 E ∗ )∗ . Then the projectivization of the dual of W = is a pencil L of apolar conics in dual projective space P(E ∗ ). Since the apolarity is equivariant with respect to the representation of SL(3) in S 2 E and in S 2 E ∗ , we see that we may assume that L is given in one of the cases from the previous lemma. Here we have to replace the unknowns ti with the differential operators ∂i . We list the corresponding dual 4-dimensional spaces of quadratic forms. (i) t0 t2 + t1 t2 + t2 t2 + 2t3 (t0 t2 + t1 t2 + t1 t2 ) = 0; 0 1 2 (ii) t0 t2 + t1 t2 + t2 t2 + 2t3 (t0 t1 − t0 t2 ) = 0; 0 1 2 (iii) t0 t2 + t1 t2 + 2t2 (t0 t1 − t1 t2 ) + 2t3 (t0 t2 − t1 t2 ) = 0; 0 1 (iv) t0 t2 + t1 t2 + 2t2 (t2 + t0 t2 ) + 2t3 (t1 t2 − t0 t1 ) = 0; 0 1 2 (v) t0 (2t0 t2 − t2 ) + t1 t2 + 2t2 t0 t1 + 2t3 t1 t2 = 0; 1 2 (vi) t0 t2 + 2t1 t0 t1 + 2t2 t1 t2 + 2t3 t0 t2 = 0; 2 (vii) t0 t2 + t1 t2 + t2 t2 + 2t3 t0 t1 = 0; 0 1 2 (viii) t0 t2 + t1 t2 + 2t2 t0 t2 + 2t3 t1 t2 = 0. 1 2 The corresponding determinantal varieties are the following. (i)  t0 det t3 t3 t3 t1 t3  t3 t3  = t0 t1 t2 + t2 (−t0 − t2 − t1 + t3 ) = 0. 3 t2

It has 4 singular points (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (1, 1, 1, 1) and hence is isomorphic to the 4-nodal cubic surface from case (i).

9.4. DETERMINANTAL EQUATIONS (ii) t0 det  t3 −t3  t3 t1 0  −t3 0  = t0 t1 t2 + t1 t2 − t2 t2 = 0. 3 3 t2

337

It has 2 ordinary nodes (0, 1, 0, 0), (0, 0, 1, 0) and a RDP (1, 0, 0, 0) of type A3 . (iii) t0 det t2 t3   t2 t3 t1 −2(t2 + t3 ) = 4(t2 +t3 )2 t0 +(t2 +t3 )t2 t3 +t1 t2 = 0. 3 −2(t2 + t3 ) 0

The surface has a double line given by t3 = t2 + t3 = 0. This case is excluded. (iv) t0 det −t3 t2  −t3 t1 t3  t2 t3  = 2t0 t1 t2 − t2 (t0 + 4t2 ) − t1 t2 = 0. 3 2 2t2

The surface has one ordinary double point (1, 0, 0, 0) and one RDP (0, 1, 0, 0) of type A5 . (v) 0 det t2 t0  t2 −t0 t3  t0 t3  = t1 t2 + 2t0 t2 t3 + t3 = 0. 2 0 t1

The surface has a double line t0 = t2 = 0. This case has to be excluded. (vi) 0 det t1 t3  t1 0 t2  t3 t2  = −t1 (t0 t1 + 2t2 t3 ) = 0. t0

This surface is reducible, the union of a plane and a nonsingular quadric. (vii)  t0 det  0 0 0 t1 t3  0 t3  = t0 (t1 t2 − t2 ) = 0. 3 t2

The surface is the union of a plane and a quadratic cone. (viii) 0 det  0 t2 The surface is a cone.  0 t0 t3  t2 t3  = t0 t2 = 0. 2 t1

338

CHAPTER 9. CUBIC SURFACES

Remark 9.4.2. If S is a cone over a plane cubic curve C. Then S admits a symmetric determinantal representation if and only if C admits such a representation. We refer to Chapter ? for determinantal representations of plane cubics. If S is irreducible non-normal surface, then S admits a symmetric determinantal representation. This corresponds to cases (iii) and (v) from the proof of the previous theorem. Case (iii) gives a surface isomorphic to the surface from case (i) of Theorem 9.1.2 and case (v) gives a surface isomorphic to the surface isomorphic to the surface from case (ii) of Theorem 9.1.2. We also see that a reducible cubic surface which is not a cone admits a symmetric determinantal representation only if it is the union of an irreducible quadric and a plane which intersects the quadric transversally. The 4-nodal cubic surface exhibits an obvious symmetry defined by the permutation group S4 . It also admits a double cover ramified only over its singular points. In fact, all three determinantal normal cubic surfaces with singularities of types 4A1 , 2A1 +A3 and A1 + A5 admit such a cover. It is defined by the family {( , Q) ∈ (P2 )∗ × W : ⊂ Q}, where W is the web of conics whose Hessian surface is the cubic surface. Note that the three cubic surfaces can be obtained as the projections of quartic surfaces with singularities of types 4A1 and 2A1 + A3 which have the similar covering property. This cover can be seen in many different ways. We give only one, the others can be found in Exercises. We consider only the Segre cubic surface S. First, note that S can be obtained as the blow-up of the vertices of a complete quadrangle. Let X be a weak Del Pezzo surface of degree 2 obtained as a minimal resolution of the double cover of P2 branched along the union of the four sides of the complete quadrangle. The double cover extends to a double cover f : X → X of a minimal resolution X of S branched over the exceptional curves of the singularities. The ramification locus of f consists of the union of 4 disjoint (−1)-curves. Blowing them down we get a weak Del Pezzo surface Y of degree 6. The cover descends to a double cover of S ramified over the nodes.

9.5
9.5.1

Representations as sums of cubes
Sylvester’s pentahedron

Counting constants we see that it is possible that a general homogeneous cubic form in 4 variables can be written as a sum of 5 cubes of linear forms in finitely many ways. Since there are no cubic surfaces singular at 5 general points, the theory of apolarity tells us that the count of constants gives a correct answer. The following result of J. Sylvester gives more: Theorem 9.5.1. A general homogeneous cubic form in 4 variables can be written uniquely as a sum 3 3 3 3 3 f = l 1 + l2 + l 3 + l4 + l 5 , where li are linear forms in 4 variables. Proof. Suppose
5 5 3 li i=1

f=

=
i=1

m3 . i

9.5. REPRESENTATIONS AS SUMS OF CUBES

339

ˇ Let xi , yi be the points in P3 corresponding to the hyperplanes V (li ), V (mi ). Consider ˇ the linear system of quadrics in P3 which pass through the points x5 , y1 , . . . , y5 . If x5 is not equal to any yj , this is a linear projective subspace of dimension 3. Applying the corresponding differential operators to f we find 4 linearly independent relations between the linear forms l1 , l2 , l3 , l4 . This shows that the points x1 , x2 , x3 , x4 are coplanar. It does not happen for general f . Thus we may assume that x5 = y5 , so that we can write m5 = λ5 l5 for some λ5 . After subtraction, we get
4 3 li + (1 − λ3 )l5 = 5 i=1 i=1 4

m3 . i

Now we consider quadrics through y1 , y2 , y3 , y4 . This is defined by 6-dimensional linear space. Its elements define linear relations between the forms l1 , . . . , l5 . Since the dimension of the linear span of these forms is equal to 4 (the genericity assumption), ˇ we obtain that there exists a 4-dimensional linear system of quadrics in P3 vanishing at x1 , . . . , x5 , y1 , . . . , y4 . Assume that all of these points are distinct. Two different quadrics from the linear system intersect along a curve B of degree 4. If this curve is irreducible, a third quadric intersects it at ≤ 8 points. So, the curve B must be reducible. Recall that the linear system of quadrics through an irreducible curve of degree 3 is of dimension 2. Thus, one of the quadrics in our linear system must contain a curve of degree ≤ 2 and so the base locus contains a curve of degree ≤ 2. The dimension of a linear system of quadrics containing an irreducibel conic is of dimension 4 and its base locus is equal to the conic. Since the points xi ’s are not coplanar, we see that this case does not occur. Assume that the base locus contains a line. Then 3 linearly independent quadrics intersect along a line and a cubic curve R. A forth quadric will intersect R at ≤ 4 points outside . Thus we have at most 4 points among x1 , . . . , x5 , y1 , . . . , y4 which do not lie on . This implies that 5 points lie on the line . Since no three points among the xi ’s and yj ’s can lie on a line, we obtain a contradiction. Thus one of the xj ’s coincide with some yi . We may now assume that m4 = λ4 l4 and get
3 3 3 3 li + (1 − λ3 )l4 + (1 − λ3 )l5 = 4 5 i=1 i=1 3

m3 . i

Take a plane through y1 , y2 , y3 and get a linear dependence
2 3 2 2 2 a1 l1 + a2 l2 + a3 l3 + a4 (1 − λ3 )l4 + a5 (1 − λ2 )l5 . 4 5

Here a4 , a5 = 0 since otherwise the points y1 , y2 , y3 and x4 = y4 or x4 = y5 are coplanar. A linear dependence between squares of linear forms means that the corresponding points in the dual space do not impose independent conditions on quadrics. The subvariety of (P3 )5 of such 5-tuples is a proper closed subset. By generality assumption we may assume that our set of 5 points x1 , . . . , x5 is not in this variety. Now we get λ3 = λ3 = 1 and 4
3 3 li i=1 3

=
i=1

m3 . i

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CHAPTER 9. CUBIC SURFACES

2 2 2 A plane through y1 , y2 , y3 gives a linear relation between l1 , l2 , l3 which as we saw before must be trivial. Thus the points x1 , x2 , x3 , y1 , y2 , y3 lie in the same plane Π. A linear system Q of quadrics in P3 through 3 non-collinear points does not have unassigned base points (i.e. its base locus consists of the three points). Since linear relations between l1 , l2 , l3 form a one-dimensional linear space, Q contains a hyperplane of quadrics vanishing at additional 3 points x1 , x2 , x3 . Thus the linear system of quadrics through a set of 4 points x1 , x2 , x3 , yi or y1 , y2 , y3 , xi has 2 unassigned base points lying in the same plane. By restricting the linear system to the plane we see that this is impossible unless all 6 points are collinear. This is excluded by the generality 3 3 assumption. This final contradiction shows that mi = λi li and i=1 (1 − λ3 )li = 0. i 3 3 Since a general form cannot be written as a sum of 4 cubes, we get λi = 1 and li = m3 i for all i = 1, . . . , 6.

3 Assume that F = i=1 li , where the linear forms li are linearly independent. Let x1 , . . . , x5 be the corresponding points in the dual P3 . We can apply a projective transformation to assume that

5

x1 = (1, 0, 0, 0), x2 = (0, 1, 0, 0), x3 = (0, 0, 0, 1), x4 = (0, 0, 0, 1), x5 = (1, 1, 1, 1). This implies that V (f ) is projectively isomorphic to the cubic surface V (λ0 t3 + λ1 t3 + λ2 t3 + λ3 t3 + λ4 (t0 + t1 + t2 + t3 )3 ). 0 1 2 3 By Sylvester’s Theorem the choice of this equation depends only on the order of the linear forms li . This shows that the moduli space of general cubic surfaces (defined as the orbit space of some open Zariski subset of cubic surfaces with respect to the action of the projective linear group of P3 ) is birationally isomorphic to C5 /S5 × C∗ . Consider the natural map C5 → C5 given by the elementary symmetric functions s1 , . . . , s5 . It shows that C5 /S5 ∼ C5 and = C5 /S5 × C∗ ∼ P(1, 2, 3, 4, 5). = This gives Corollary 9.5.2. The moduli space of cubic surfaces is a rational variety of dimension 4. Remark 9.5.1. One can construct the moduli space M of cubic surfaces using geometric invariant theory. It is defined as the projective spectrum of the graded ring of invariants


(9.30)

R=
n=0

S n (S 3 (C4 )∗ )∗ )SL(4) .

The explicit computation of this ring made by Clebsch and Salmon shows that this ring is generated by invariants of degree 8, 16, 24, 32, 40, 100, where the square of the last invariant is a polynomial in the first 5 invariants. This easily shows that the graded

9.5. REPRESENTATIONS AS SUMS OF CUBES

341

subalgebra R[2] of R generated by the elements of even degree is freely generated by the first 5 invariants. This gives an isomorphism M ∼ P(1, 2, 3, 4, 5). = One should compare it with isomorphism (9.30), which gives only a birational model of M. I have no explanation for this.

9.5.2

The Hessian surface

The Sylvester theorem gives the equation of the Hessian surface of a general cubic surface.
3 Definition 9.2. Let S = V (f ) be a general cubic surface and f = i=1 li be its equation. The set of 5 planes V (li ) is called the Sylvester pentahedron. The points V (li , lj , lk ), 1 ≤ i < j < k ≤ 5, are called the vertices, and the lines V (li , lj ), 1 ≤ i < j ≤ 5, are called the edges. 5

Theorem 9.5.3. Let S = V (f ) be a general cubic surface and He(S) be the Hessian 5 3 surface of S. Assume f = i=1 li . Then He(S) contains the edges of the Sylvester pentahedron, and the vertices are its ordinary double points. The equation of He(S) can be written in the form 5 a2 i = 0, l1 l2 l3 l4 l5 li i=1 where
5 i=1

ai li = 0.

Proof. Recall that He(S) = {x ∈ P3 : Px (S) is singular }.
2 For any point x ∈ V (li , lj ) we have Dx (f ) = k=i,j λk lk . This is a quadric of rank ≤ 3. Thus each edge is contained in He(S). Since each vertex lies in 3 noncoplanar edges it must be a singular point. Observe that any edge contains 3 vertices. Any quartic containing 10 vertices and two general points on each edge contains the 10 edges. Thus the linear system of quartics containing 10 edges is of dimension 5 34 − 30 = 4. Obviously any quartic with equation i=1 λii = 0 contains the edges. l Thus the equation of He(S) can be written in this form. We derive the same conclusion in another way which will also allow us to compute the coefficients λi . Consider the isomorphism from S to a surface in P4 given by 2 equations 5 3 zi = i=1 i=1 5

ai zi = 0.

This isomorphism is given by the map P3 → P4 defines by the linear forms li . The polar quadric V (Px (f )) is given by 2 equations in P4
5 2 li (a)zi

=
i=1

ai zi = 0.

342 It is singular if and only if the matrix l1 (x)z1 a1 l2 (x)z2 a2 l3 (x)z3 a3

CHAPTER 9. CUBIC SURFACES

zl4 (x)z4 a4

l5 (x)z5 a5

is of rank 1 at some point z = (z1 , . . . , z5 ) = (l1 (t), . . . , l5 (t)). This can be expressed 5 by the equalities li (x) = λai /li (t), i = 1, . . . , 5. Since i=1 ai li (x) = 0, we obtain
5 5

0=
i=1

ai li (x) =
i=1

a2 /li (t). i

This gives the asserted equation of He(S). Remark 9.5.2. Recall that the Hessian of any cubic hypersurface admits a birational automorphism σ which assigns to the polar quadric of corank 1 its singular point. Let X be a minimal nonsingular model of He(S). It is a K3 surface. The birational automorphism σ extends to a biregular automorphism of X. It exchanges the proper inverse transforms of the edges with the exceptional curves of the resolution. One can show that for a general S, the automorphism of X has no fixed points, and hence the quotient is an Enriques surface. Remark 9.5.3. Not every cubic surface, even a nonsingular one, can be written by a 3 Sylvester equation. For example, consider the Ferma surface V ( i=1 t3 ). Its Hessian i is V (t0 t1 t2 t3 ) and it does not contain the edges of a pentahedron.

9.5.3

Cremona’s hexahedral equations

It follows from Sylvester’s Theorem that a general cubic surface is isomorphic to a hyperplane section of the Fermat cubic hypersurface in P4 . This defines a dominant rational map from P4 to the moduli space Mnscub of nonsingular cubic surfaces (which exists a geometric quotient of the corresponding open subset of cubic surfaces). It follows from the proof of Corollary 9.5.2 that this map is birational. Cremona’s hexahedral equations which we discuss here allows one to define a regular map of degree 36 from an open Zariski subset U of P4 to Mnscub . Its fibres can be viewed as a choice of a double-six on the surface. Theorem 9.5.4. (L. Cremona) Assume that a cubic surface S is not a cone and admits a Cayley-Salmon equation (e.g. S is a nonsingular surface). Then S is isomorphic to a cubic surface in P5 given by the equations
6 6 6

t3 = i
i=1 i=1

ti =
i=1

ai ti = 0.

(9.31)

Proof. Let F = V (l1 l2 l3 +m1 m2 m3 ) be a Cayley-Salmon equation of S. Let us try to find some constants such that after scaling the linear forms they add up to zero. Write li = λi li , mi = µi mi , i = 1, 2, 3.

9.5. REPRESENTATIONS AS SUMS OF CUBES

343

Since S is not a cone, four of the linear forms are linearly independent. After reordering the linear forms, we may assume that the linear forms l1 , l2 , l3 , m1 are linearly independent. Let m2 = al1 + bl2 + cl3 + dm1 , m3 = a l1 + b l2 + c l3 + d l4 . The constants λi , µi must satisfy the following system of equations λ1 + aµ2 + a µ3 λ2 + bµ2 + b µ3 λ3 + cµ2 + c µ3 µ1 + dµ2 + d µ3 λ1 λ2 λ3 + µ1 µ2 µ3 = 0 = 0 = 0 = 0 = 0.

The first four linear equations allow us to express linearly all unknowns in terms of µ2 , µ3 . Plugging in the last equation, we get a cubic equation in µ2 /µ3 . Solving it, we get a solution. Now set z1 = l 2 + l 3 − l 1 , z4 = µ2 + µ3 − µ1 , z 2 = l3 + l 1 − l2 , z5 = µ3 + µ1 − µ2 , z3 = l1 + l2 − l3 , z6 = µ1 + µ2 − µ3 .

One checks that these six linear forms satisfy the equations from the assertion of the theorem. Corollary 9.5.5. (T. Reye) A general homogeneous cubic form f in 4 variables can be written as a sum of 6 cubes in ∞4 different ways. In other words, dim VSP(f ; 6)o = 4. Proof. This follows from the proof of the previous theorem. Consider the map (C4 )6 → C20 ,
3 3 (l1 , . . . , l6 ) → l1 + · · · + l6 .

It is enough to show that it is dominant. We show that the image contains the open subset of nonsingular cubic surfaces. In fact, we can use a Clebsch-Salmon equation l1 l2 l3 + m1 m2 m3 for S = V (f ) and apply the proof of the theorem to obtain that, up to a constant factor, 3 3 3 3 3 3 f = z1 + z2 + z3 + z4 + z5 + z6 .

Now let us see in how many ways one can write a surface by a Cremona hexahedral equation. Suppose a nonsingular S is given by equations (9.31) which one calls Cremona hexahedral equations. They allow us to locate 15 lines on S such that the remaining lines form a double-six. The equations of these lines in P5 are
6

zi + zj = 0, zk + zl = 0, zm + zn = 0,
i=1

ai zi = 0,

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CHAPTER 9. CUBIC SURFACES

where {i, j, k, l, m, n} = {1, 2, 3, 4, 5, 6}. Let us denote the line given by the above equations by lij,kl,mn . Let us identify a pair a, b of distinct elements in {1, 2, 3, 4, 5, 6} with a transposition (ab) in S6 . We have the product (ij)(kl)(mn) of three commuting transpositions corresponding to each line lij,kl,mn . The group S6 admits a unique (up to a composition with a conjugation) outer automorphism which sends each transposition to the product of three commuting transpositions. In this way we can match lines lij,kl,mn with exceptional vectors cab of the E6 -lattice. To do it explicitly, one groups together 5 products of three commuting transpositions in such a way that they do not contain a common transposition. Such a set is called a total and the triples (ij, kl, mn) are called synthemes. Here is the set of 6 totals T1 T2 T3 T4 T5 T6 = = = = = = (12)(36)(45), (12)(36)(45), (12)(35)(46), (12)(34)(56), (12)(34)(56), (12)(35)(46), (13)(24)(56), (13)(25)(46), (13)(24)(56), (13)(25)(46), (13)(26)(45), (13)(26)(45), (14)(26)(35), (14)(23)(56), (14)(25)(36), (14)(26)(35), (14)(25)(36), (14)(23)(56), (15)(23)(46), (15)(26)(34), (15)(26)(34), (15)(24)(36), (15)(23)(46), (15)(24)(36), (16)(25)(34) (16)(24)(35) (16)(23)(45) (16)(23)(45) (15)(24)(35) (16)(25)(34).

Two different totals Ta , Tb contain one common product (ij)(kl)(mn). The correspondence (a, b) → (ij)(kl)(mn) defines the outer automorphism α : S6 → S6 , (9.32)

For example, α((12)) = (12)(34)(56) and α((23) = (13)(45)(56). After we matched the lines lij,kl,mn with exceptional vectors cab , we check that this matching defines an isomorphism of the incidence subgraph of the lines with the subgraph of the incidence graph of 27 lines on a cubic surface whose vertices correspond to exceptional vectors cab . Theorem 9.5.6. Each Cremona hexahedral equations of a nonsingular cubic surface S defines an ordered double-six of lines. Conversely, a choice of an ordererd double-six defines uniquely Cremona hexahedral equations of S. Proof. We have seen already the first assertion of the theorem. If two surfaces given by hexahedral equations define the same double-six, then they have common 15 lines. Obviously this is impossible. Thus the number of different hexahedral equations of S is less or equal than 36. Now consider the identity (z1 + · · · + z6 ) (z1 + z2 + z3 )2 + (z4 + z5 + z6 )2 − (z1 + z2 + z3 )(z4 + z5 + z6 )
3 3 = (z1 + z2 + z3 )3 + (z4 + z5 + z6 )3 = z1 + · · · + z6

+3(z2 + z3 )(z1 + z3 )(z1 + z2 ) + 3(z4 + z5 )(z5 + z6 )(z4 + z6 ). It shows that Cremona hexahedral equations define a Cayley-Salmon equation (z2 + z3 )(z1 + z3 )(z1 + z2 ) + (z4 + z5 )(z5 + z6 )(z4 + z6 ) = 0,

9.5. REPRESENTATIONS AS SUMS OF CUBES

345

where we have to eliminate one unknown with help of the equation ai zi = 0. Applying permutations of z1 , . . . , z6 , we get 10 Cayley-Salmon equations of S. Each 9 lines formed by the corresponding conjugate pair of triads of tritangent planes are among the 15 lines determined by the hexahedral equation. It follows from the classification of the conjugate pairs that we have 10 such pairs composed of lines cij ’s (type II). Thus a choice of Cremona hexahedral equations defines exactly 10 Cayley-Salmon equations of S. Conversely, it follows from the proof of Theorem 9.5.4 that each Cayley-Salmon equation gives three Cremona hexahedral equations (unless the cubic equation has a multiple root). Since we have 120 Cayley-Salmon equations for S we get 36 = 360/10 hexahedral equations for S. They match with 36 double-sixes.

9.5.4

The Segre cubic primal

Let p1 , . . . , pm be a set of points in Pn , where m > n + 1. For any ordered subset (pi1 , . . . , pin+1 ) of n+1 points we denote by (i1 . . . in+1 ) the determinant of the matrix whose rows are projective coordinates of the points (pi1 , . . . , pin+1 ) in this ortder. We consider (i1 . . . in+1 ) as a section of the invertible sheaf ⊗n+1 p∗j OPn (1) on (Pn )m . j=1 i It is called a bracket-function. A monomial in bracket functions such that each index i ∈ {1, . . . , m} occurs exactly d times defines a section of the invertible sheaf
n

Ld =
i=1

p∗ OPn (d). i

According to the Fundamental Theorem of Invariant Theory (see [80]) the subspace m (Rn )(d) of H 0 ((Pn )m , Ld ) generated by such monomials is equal to the space of invariants H 0 ((Pn )m , Ld )SL(n+1 , where the group SL(n + 1) acts linearly on the space of sections via its diagonal action on (Pn )m . The graded ring
∞ m Rn = d=0 m (Rn )(d)

is a finitely generated algebra. Its projective spectrum is isomorphic to the GIT-quotient
m Pn := Pn )m //SL(n + 1) m of (Pn )m by SL(n + 1). Let r1 , . . . , rN be homogeneous generators of Rn . The ss ss m complement of it set of common zeroes U admits a regular map U → Pn . The set U ss does not depend on the choice of generators. Its points are called semi-stable. Let m U s be the largest open subset such that the fibres of the restriction map U s → Pn are orbits. Its points are called stable. It follows from the Hilbert-Mumford numerical stability criterion that a points set (p1 , . . . , pm ) in P1 is semi-stable (resp. stable) if and only if at most 1 m (resp. < 1 m) 2 2 points coincide. We have already seen the definition of the bracket-functions in the case m = 4. They define the cross-ratio of 4 points

[p1 , p2 , p3 , p4 ] =

(12)(34) . (13)(24)

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CHAPTER 9. CUBIC SURFACES

The cross-ratio defines the rational map (P1 )4 − → P1 . It is defined on the open set U s of points where no more that 2 coincide and it is an orbit space over the complement of three points 0, 1, ∞. In the case of points in P2 the condition of stability (semi-stability) is that at most 1 2 2 1 3 m (resp. < 3 m) coincide and at most 3 m (resp. < 3 m) points are on a line. Proposition 9.5.7. Let (q1 , · · · , q6 ) be a set of distinct points in P1 . The following conditions are equivalent. (i) There exists an involution of P1 such that the pairs (q1 , q2 ), (q3 , q4 ), (q5 , q6 ) are orbits of the involution. (ii) The binary forms gi , i = 1, 2, 3 with zeros (q1 , q2 ), (q3 , q4 ) and q5 , q6 ) are linearly dependent. (iii) Let pi be the image of qi under a Veronese map P1 → P2 . Then the lines p1 , p2 , p3 , p4 , p5 , p6 are concurrent. (iv) The bracket-function (14)(36)(25)−(16)(23)(54) vanishes on the set (q1 , . . . , q6 ). Proof. (i) ⇔ (ii) Let f : P1 → P1 be the degree 2 map defined by the involution. Let f be given by (t0 , t1 ) → (g1 (t0 , t1 ), g2 (t0 , t1 )), where g1 , g2 are binary forms of degree 2. By choosing coordinates in the target space we may assume that f (q1 ) = f (q2 ) = 0, f (q3 ) = f (q4 ) = 1, f (q5 ) = f (q6 ) = ∞, i.e. g1 (q1 ) = g1 (q2 ) = 0, g2 (q3 ) = g2 (q4 ) = 0, (g1 − g2 )(q5 ) = (g1 − g2 )(q6 ) = 0. Obviously, the binary forms g1 , g2 , g3 = g1 − g2 are linearly dependent. Conversely, suppose g1 , g2 , g3 are linearly dependent. By scaling we may assume that g3 = g1 − g2 . We define the involution by (t0 , t1 ) → g1 (t0 , t1 ), g2 (t0 , t1 )). (ii) ⇔ (iii) Without loss of generality we may assume that qi = (1, ai ) and g1 = t2 − (a1 + a2 )t0 t1 + a1 a2 t2 , g2 = t2 − (a3 + a4 )t0 t1 + a3 a4 t2 , g3 = t2 − (a5 + 1 0 1 0 1 a6 )t0 t1 + a5 a6 t2 . The condition that the binary forms are linearly dependent is 0   1 a1 + a2 a1 a2 (9.33) det 1 a3 + a4 a3 a4  = 0. 1 a5 + a6 a5 a6 The image of qi under the Veronese map (t0 , t1 ) → (t2 , t0 t1 , t2 ) is the point pi = 0 2 (1, ai , a2 ). The line pi , pj has the equation i   x0 x1 x2 det  1 ai a2  = (aj − ai )(ai aj x0 − (ai + aj )x1 + x2 ) = 0. i 1 aj a2 j Obviously the lines are concurrent if and only if (9.33) is satisfied. (iii) ⇔ (iv) We have    2  1 a1 + a2 a1 a2 a b2 c2 1 a3 + a4 a3 a4  · −a −b −c 1 a5 + a6 a5 a6 1 1 1

9.5. REPRESENTATIONS AS SUMS OF CUBES  (a − a1 )(a − a2 ) (b − a1 )(b − a2 ) (c − a1 )(c − a2 ) = (a − a3 )(a − a4 ) (b − a3 )(b − a4 ) (c − a3 )(c − a4 ) (a − a5 )(a − a6 ) (b − a5 )(b − a6 ) (c − a5 )(c − a6 ) 

347

Substituting a = a1 , b = a3 , c = a5 and computing the determinant we obtain that it is equal to   0 a2 − a3 a5 − a2 0 a4 − a5  (a3 − a5 )(a5 − a1 )(a1 − a3 ) det a1 − a4 a6 − a1 a3 − a6 0 = (a3 −a5 )(a5 −a1 )(a1 −a3 )[(a1 −a4 )(a3 −a6 )(a5 −a2 )+(a6 −a1 )(a2 −a3 )(a4 −a5 )]. Since the points are distinct the vanishing of determinant (9.33) is equivalent to vanishing of (a1 − a4 )(a3 − a6 )(a5 − a2 ) + (a6 − a1 )(a2 − a3 )(a4 − a5 ), i.e. the vanishing of the bracket-function (14)(36)(25) − (16)(23)(54) on our point set.

We let [ij, kl, mn] := (il)(kn)(jm) − (jk)(lm)(ni). (9.34)

For example, [12, 34, 56] = (14)(36)(25)−(16)(23)(45). The expressions [ij, kl, mn] 6 are elements of the linear space (R1 )(1). Note as multi-linear functions on V they belong to the 5-dimensional irreducible component of the representation of S6 in (V ∗ )⊗6 corresponding to the partition 6 = 3 + 3. In other words, the expression is invariant under permuting elements in the same pair, it is invariant under permuting the pairs by an even permutation, and changes the sign under permuting the pairs by an odd permutation. It is known (and is easy to check) that the linear representation of type 3 + 3 is isomorphic to the composition of α with the tensor product of the sign-representation with the standard irreducible representation of S6 in the space Vst = {(a1 , . . . , a6 ) ∈ C6 : a1 + . . . + a6 = 0}. Let us identify the ordered set (1, 2, 3, 4, 5, 6) with the ordered set of points (∞, 0, 1, 2, 3, 4, 5) of the projective line P1 (F5 ). The group PSL(2, F5 ) ∼ A5 identified with = the group of Moebius transformations z → az+b acts naturally on this set. Let cz+d u0 = [∞0, 14, 23] and let ui , i = 1, . . . , 4, be obtained from u0 via the action of the transformation z → z + i. Let U1 := u0 + u1 + u2 + u3 + u4 = [∞0, 14, 23] + [∞1, 20, 34] + [∞2, 31, 40] + [∞3, 42, 01] + [∞4, 03, 12] . Obviously U1 is invariant under the subgroup of order 5 generated by the transformation z → z + 1. It is also invariant under the transformation τ : z → −1/z. It is well-known that A5 is generated by these two transformations. The orbit of U∞ under the group A6 acting by permutations of ∞, 0, . . . , 4 consists of 6 functions

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U1 , U1 , U1 , U4 , U5 , U6 . We will rewrite them now returning to our old notation of indices by the set (1, 2, 3, 4, 5, 6) U 
1

0   = 

U2  U3  U4    U5 U6

[12, 36, 45] 0

[13, 24, 56] [15, 26, 34] 0

[14, 35, 26] [13, 46, 52] [16, 32, 45] 0

[15, 46, 23] [16, 35, 24] [14, 25, 36] [12, 34, 56] 0

[16, 25, 34] 1 [14, 56, 23] 1 [12, 35, 46] 1 · . [15, 36, 24] 1    [13, 45, 26] 1 0 1

  

(9.35) where the matrix is skew-symmetric. We immediately observe that U1 + U2 + U3 + U4 + U5 + U6 = 0. (9.36)

Next observe that the triples of pairs [ij, kl, mn] in each row of the matrix constitute 1 a total from (9.32). The permutation group S6 acts in the R6 as an irreducible 5dimensional linear representation defined by the Young tableau of type (2, 2, 2). The composition of this representation with the outer automorphism (9.32) is isomorphic to the standard representation Vst of S6 tensored with the sign representation. For example, (12) acts as U1 → −U2 , U2 → −U1 , Ui → Ui , i > 2. Under the outer automorphism (9.32) of S6 we have (12) → (12)(36)(45), (3456) → (1524) One immediately checks that (U1 , . . . , U6 ) are transformed under σ = (12)(3456) to (−U2 , −U1 , −U4 , −U5 , −U6 , −U3 ). This implies that the space of invariant functions is one-dimensional and is spanned by the function U1 − U2 . On the other hand, we check immediately that the function [12, 36, 54] is invariant under σ. This gives U1 − U2 = c[12, 36, 54] for some scalar c. Evaluating these functions on a point set (p1 , . . . , p6 ) with p1 = p2 , p3 = p6 , p4 = p5 we find that c = 6. Now applying permutations we obtain
U1 − U2 U1 − U5 U2 − U4 U3 − U4 U4 − U5 = = = = = 6[12, 36, 54], U1 − U3 = 6[13, 42, 65], U1 − U4 = 6[14, 53, 26], (9.37) 6[15, 64, 32], U1 − U6 = 6[16, 52, 34], U2 − U3 = 6[15, 26, 34], 6[13, 46, 52], U2 − U5 = 6[16, 35, 24], U2 − U6 = 6[14, 23, 56], 6[16, 45, 32], U3 − U5 = 6[14, 52, 63], U3 − U6 = 6[12, 46, 53], 6[12, 43, 56], U4 − U6 = 6[15, 36, 24], U5 − U6 = 6[13, 54, 62].

Similarly, we find that U1 + U2 is the only anti-invariant function under σ and hence must be to c(12)(36)(45). After evaluation the functions at a point set (p1 , . . . , p6 ) with p1 = p3 , p2 = p4 , p5 = p6 we find that c = 4. In this way we get the relations
U1 + U2 U1 + U5 U2 + U4 U3 + U4 U4 + U5 = = = = = 4(12)(36)(45), U1 + U3 = 4(13)(42)(56), U1 + U4 = 4(41)(53)(26), (9.38) 4(15)(46)(32), U1 + U6 = 4(16)(25)(34), U2 + U3 = 4(15)(26)(43), 4(13, )(46)(25), U2 + U5 = 4(16)(35)(24), U2 + U6 = 4(14)(23)(56), 4(16)(54)(32), U3 + U5 = 4(14)(25)(63), U3 + U6 = 4(12)(46)(53), 4(12)(34)(56), U4 + U6 = 4(15)(36)(24), U5 + U6 = 4(13)(45)(62).

9.5. REPRESENTATIONS AS SUMS OF CUBES Using (9.36), we obtain
U1 U2 U3 U4 U5 U6 = = = = = =

349

(12)(36)(45) + (13)(42)(56) + (14)(35)(26) + (15)(46)(32) + (16)(25)(34)(9.39) (12)(36)(45) + (13)(46)(25) + (14)(56)(23) + (15)(26)(43) + (16)(24)(53) (12)(53)(46) + (13)(42)(56) + (14)(52)(36) + (15)(26)(43) + (16)(23)(45) (12)(34)(56) + (13)(46)(25) + (14)(35)(26) + (15)(24)(36) + (16)(23)(45) (12)(34)(56) + (13)(54)(26) + (14)(52)(36) + (15)(46)(32) + (16)(24)(53) (12)(53)(46) + (13)(54)(26) + (14)(56)(23) + (15)(36)(24) + (16)(25)(34)

We see that our functions are in bijective correspondence with 6 totals from above. We call the functions U1 , . . . , U6 the Joubert functions. It is easy to see that the functions Ui do not vanish simultaneously on semi-stable point sets. Thus they define a morphism
6 J : P1 → P5 ,

Theorem 9.5.8. The morphism J defined by the Joubert functions is an isomorphism onto the subvariety S3 of P5 given by the equations
5 5

zi =
i=0 i=0

3 zi = 0.

(9.40)

6 Proof. It is known that the graded ring R1 is generated by the following bracketfunctions (standard tableaux)

(12)(34)(56), (12)(35)(46), (13)(24)(56), (13)(25)(46), (14)(25)(36)
6 (see [84]). The subspace of R1 (1) generated by the Joubert functions is invariant with 6 respect to S6 . Since R1 (1) is an irreducible representation, this implies that the relation Ui = 0 spans the linear relations between the Joubert functions. Consider the sum Σ= Ui3 . Obviously it is invariant with respect to A6 . One immediately checks that an odd permutation in S6 transforms each sum Σ to −Σ. This implies that Σ = 0 whenever two points pi and pj coincide. Hence Σ must be divisible by the product of 15 functions (ij). This product is of degree 5 in coordinates of each point but Σ is of degree 3. This implies that Σ = 0. Since the functions Ui generate the graded 6 6 6 ring R1 , by definition of the space P1 , we obtain an isomorphism from P1 to a closed subvariety of S3 . Since the latter is irreducible and of dimension equal to the dimension 6 of P1 , we obtain the assertion of the theorem.

The cubic threefold S3 is called the Segre cubic primal. We will often consider it as a hypersurface in P4 . It follows immediately by differentiating that the cubic hypersurface S3 has 10 double points. They are the points p = (1, 1, 1, −1, −1, −1) and others obtained by permuting the coordinates. We will see in a later chapter that this is maximal possible for a cubic hypersurface of dimension 3 with isolated singularities. A point p is given by the equations zi + zj = 0, 1 ≥ i ≤ 3, 4 ≥ j ≤ 6. Using (9.36) this implies that p is the image of a point set with p1 = p4 = p6 or p2 = p3 = p5 . Thus the singular points of the Segre cubic primal are the images of semi-stable but not stable point sets.

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Also S3 has 15 planes with equations zi + zj = zk + zl = zl + zm = 0. Let us see that they are the images of point sets with two points coincide. Without loss of generality we may assume that z1 + z2 = z3 + z4 = z5 + z6 = 0. Again from (9.36), we obtain that (12)(36)(45), (16)(23)(45) and (13)(26)(45) vanish. This happens if and only if p4 = p5 . We know that the locus of point sets (q1 , . . . , q6 ) such that the pairs (qi , qj ), (qk , ql ), and (qm , qn ) are orbits of an involution are defined by the equation [ij, kl, mn] = 0. By (9.37), we obtain that they are mapped to a hyperplane section of S3 defined by the equation za − zb = 0, where α((ab)) = (ij)(kl)(mn). It follows from Cremona hexahedral equations that a nonsingular cubic surface is isomorphic to a hyperplane section of the Segre cubic. In a theorem below we will make it more precise. But first we need some lemmas. Lemma 9.5.9. Let p1 , . . . , p6 be six points in P2 . Let {1, . . . , 6} = {i, j} ∪ {k, l} ∪ {m, n}. The condition that the lines pi , pj , pk , pl , pm , pn are concurrent is (ij, kl, mn) := (kli)(mnj) − (mni)(klj) = 0. (9.41)

Proof. Let P2 = P(V ), and let vi ∈ V represent a point pi ∈ P2 . The wedge product (ijx) = vi ∧ vj ∧ v can be identified with a linear function V → Λ3 V ∼ C considered = as a linear function on V . It vanishes on the line pi , pj . The expression (kli)(mnj) − (mni)(klj) is the value of (klx) ∧ (mnx) ∈ Λ2 V ∗ on vi ∧ vj ∈ Λ2 V . The condition that the lines are concurrent is that (ijx) ∧ (klx) ∧ (mnx) = 0 in Λ3 V ∗ . The function Λ2 V → Λ3 V ∗ , (vi , vj ) → (ijx) ∧ (klx) ∧ (mnx) coincides (up to a multiplicative factor) with the function (klx) ∧ (mnx) since they are both alternating and bilinear can be identified with (ijx) ∈ V ∗ . This proves the assertion.
6 We have already noted that [ij, kl, mn] ∈ R1 (1) are transformed by S6 in the same way as (ij)(kl)(mn) up to the sign representation. Similarly, the functions 6 (ij, kl, mn) ∈ R2 (1) are transformed by S6 in the same way as the functions (ijk)(lmn) up to the sign representation. However, the corresponding irreducible components in (V ∗ )⊗6 exchange the type. The functions (ij, kl, mn) belong to the component of type (2, 2, 2) and the functions (ijk)(mnl) belong to the component of type (3, 3). Let U  0 (12, 36, 45) (13, 42, 65) (14, 53, 26) (15, 46, 32) (16, 52, 34) 1 ¯

U2   ¯ U3   U4  =  ¯   ¯
U5 ¯ U6

¯

1

0

(15, 26, 43) 0

(13, 46, 25) (16, 32, 45) 0

(16, 24, 53) (14, 52, 63) (12, 43, 56) 0

(14, 56, 32) 1 (12, 53, 46) 1 · . (15, 36, 24) 1    (13, 54, 62) 1 0 1

Observe that the matrix used here is obtained from the matrix in (9.35) defining the functions Ui by replacing [ij, kl, mn] with (ij, kl, mn). Lemma 9.5.10. We have the relation ¯ ¯ ¯ U1 + U2 + U3 = −6(146)(253) (9.42)

and similar relations obtained from this one by permuting the numbers (1, . . . , 6).

9.5. REPRESENTATIONS AS SUMS OF CUBES Proof. Adding up, we get ¯ ¯ ¯ U1 + U2 + U3 = [(14, 53, 26) + (14, 52, 63) + (14, 56, 23)]

351

+[(16, 52, 34) + (16, 24, 53) + (16, 32, 45)] + [(46, 31, 52) + (46, 15, 32) + (46, 12, 53)]. Next we obtain
(14, 53, 26) + (14, 52, 63) + (14, 56, 23) = (142)(536) − (146)(532) + (146)(523) −(143)(526) + (142)(563) − (143)(562) = −2(146)(253), (16, 52, 34) + (16, 24, 53) + (16, 32, 45) = (163)(524) − (164)(523) + (165)(243) −(163)(245) + (164)(325) − (165)(324) = −2(146)(253), (46, 31, 52) + (46, 15, 32) + (46, 12, 53) = (465)(312) − (462)(315) + (463)(152) −(461)(153) + (465)(123) − (463)(125) = 2[(465)(312) − (462)(315) + (463)(152)].

Now we use the following Pl¨ cker relation u (ijk)(lmn) − (ijl)(kmn) + (ijm)(kln) − (ijn)(klm) = 0. It gives (465)(312) − (462)(315) + (463)(152) = −(146)(253). Collecting all of this together, we get the assertion. Let (p1 , . . . , p6 ) be a fixed ordered set of 6 points in P2 . Consider the following homogeneous cubic polynomials in coordinates x = (x0 , x1 , x2 ) of a point in P2 .
F1 F2 F3 F4 F5 F6 = = = = = = (12x)(36x)(54x) + (13x)(42x)(65x) + (14x)(53x)(26x) + (15x)(64x)(32x) + (16x)(25x)(43x) (12x)(36x)(54x) + (13x)(46x)(52x) + (14x)(65x)(23x) + (15x)(26x)(34x) + (16x)(24x)(35x) (12x)(35x)(46x) + (13x)(42x)(65x) + (14x)(52x)(63x) + (15x)(26x)(34x) + (16x)(32x)(45x) (12x)(43x)(56x) + (13x)(46x)(52x) + (14x)(53x)(26x) + (15x)(63x)(24x) + (16x)(32x)(45x) (12x)(43x)(56x) + (13x)(54x)(62x) + (14x)(52x)(63x) + (15x)(64x)(32x) + (15x)(24x)(35x) (12x)(35x)(46x) + (13x)(54x)(62x) + (14x)(65x)(23x) + (15x)(63x)(24x) + (16x)(25x)(43x)

(9.43)

Theorem 9.5.11. The rational map P2 → P5 given by the polynomials F1 , . . . , F6 has the image given by the equations
3 3 3 3 3 3 z1 + z2 + z3 + z4 + z5 + z6 = 0,

(9.44)

z1 + z2 + z3 + z4 + z5 + z6 = 0, a1 z1 + a2 z2 + a3 z3 + a4 z4 + a5 z5 + a6 z6 = 0, ¯ ¯ where (a1 , . . . , a6 ) are the values of (U1 , . . . , U6 ) at the point set (p1 , . . . , p6 ).

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Proof. Take x = (1, 0, 0), then each determinant (ijx) is equal to the determinant (ij) for the projection of p1 , . . . , p6 to P1 . Since all the bracket-functions are invariant with respect to SL(3) we see that any (ijx) is the bracket function for the projection of the points to P1 with center at x. This shows that the relations for the functions Ui imply the similar relations for the polynomials Fi . This is what classics called the Clebsch 5 transference principle. Let us find the additional relation of the form i=0 ai zi = 0. Consider the cubic curve C = a1 F1 (x) + . . . + a6 F6 (x) = 0, where a1 , . . . , a6 are as in the assertion of the theorem. We have already noted that (ij, kl, mn) are transformed by S6 in the same way as (ij)(kl)(mn) up to the sign representation. Thus the expression i ai Fi (x) is transformed to itself under an even permutation and transformed to − i ai Fi (x) under an odd permutation. Thus the equation of the cubic curve is invariant with respect to the order of the points p1 , . . . , p6 ). Obviously C vanishes at the points pi . Suppose we prove that C vanishes at the intersection point of the lines p1 , p2 and p3 , p4 , then by symmetry it vanishes at the intersection points of all possible pairs of lines, and hence contains 5 points on each line. Since C is of degree 3 this implies that C vanishes on 15 lines, hence C is identical zero and we are done. So, let us prove that the polynomial C vanishes at p = p1 , p2 ∩ p3 , p4 . Recall from analytic geometry (or multi-linear algebra) that p can be represented by the vector (v1 × v2 ) × (v3 × v4 ) = (v1 ∧ v2 ∧ v3 )v4 − (v1 ∧ v2 ∧ v4 )v3 = (123)v4 − (124)v3 . Thus the value of (ijx) at p is equal to (ijp) = (123)(ij4) − (124)(ij3) = (12)(ij)(34). Applying the transference principle to (9.38), we obtain F1 (x) + F2 (x) = 4(12x)(36x)(45x), F4 (x) + F5 (x) = 4(12x)(34x)(56x), F1 (x) + F6 (x) = 4(16x)(25x)(34x), F3 (x) + F6 (x) = 4(12x)(53x)(46x), F2 (x) + F3 (x) = (15x)(26x)(43x). This implies that F1 + F2 , F4 + F5 , F1 + F6 , F3 + F6 , F2 + F3 all vanish at p. Thus the value of C at p is equal to (a4 −a5 )F4 (p)+(a2 +a6 −a1 −a3 )F6 (p) = (a4 −a5 )(F4 (p)+F6 (p))+(a2 +a6 +a5 −a1 −a3 −a4 )F6 (p) = (a4 − a5 )(F4 (p) + F6 (p)) + (a2 + a5 + a6 )(F1 (p) + F3 (p). Here we used that a1 + . . . + a6 = 0 and F1 (p) + F3 (p) + 2F6 (p) = 0. Using Lemma 9.5.10, we find a4 −a5 = (a4 +a1 +a2 )−(a5 +a1 +a2 ) = 6(125)(436)−6(126)(435) = 6(12, 43, 56). a2 + a5 + a6 = 6(346)(125). (9.45)

9.5. REPRESENTATIONS AS SUMS OF CUBES Using (9.38) and (9.45), we get F4 (p) + F6 (p) = (51p)(42p)(36p) = (42p)(12, 34, 15)(12, 36, 34), F1 (p) + F3 (p) = (13p)(42p)(56p) = (42p)(12, 56, 34)(12, 13, 34).
1 Collecting this together we obtain that the value of 6 C at p is equal to

353

(12, 43, 56)(42p)[(12, 34, 15)(12, 36, 34) + (125)(436)(12, 13, 34)). It remains to check that (12, 34, 15)(12, 36, 34) + (125)(436)(12, 13, 34) = (125)(314)(123)(364) + (125)(463)(123)(134) = 0.

6 ¯ Remark 9.5.4. Since Ui ∈ R2 (1), we obtain a regular map 6 Φ : P2 → P4

¯ ¯ defined by (U1 , . . . , U6 ). It can be shown that this map is of degree 2 (see [84]). It factors through the association involution. Let (p1 , . . . , p6 ) be a general point set and S be a cubic surface isomorphic to its blow-up. The point set defines six skew lines 1 , . . . , 6 on S. Let 1 , . . . , 6 be the six skew lines such that ( 1 , . . . , 6 ) and ( 1 , . . . , 6 ) form the double-six. The blowing down of 1 , . . . , 6 define an ordered set of points (p1 , . . . , p6 ) whose orbit is the value of the association involution on the 6 orbit of (p1 , . . . , p6 ). This agrees with Theorem 9.5.6. Note that the graded ring R2 ¯i (or (ijk)(lmn)) and an element G from R6 (2). The is generated by the functions U 2 6 association involution is the identity on (R2 ) and sends G to −G. 6 The morphism Φ is S6 -equivariant, where S6 acts on P2 via permuting the factors 2 6 4 in (P ) and acts in P via the composition of the standard irreducible 5-dimensional representation and the outer automorphism α. This representation corresponds to the partition 6 = 2 + 2 + 2 and differs from the representation of type 3 + 3 by the signrepresentation. Remark 9.5.5. It is easy to see that the functions (ij)(kl)(mn) restricted to the point 6 sets lying on the Veronese conic define the function [ij][kl][mn](ij)(kl)(mn) in R1 (2). 6 ∼ Under the isomorphism P1 = S3 these functions are quadrics that pass through singular points of S3 . This can be interpreted by asserting that the composition of rational maps
6 6 P1 − → P2 −→ P4 ν2 Φ

is given by polar quadrics of the Segre cubic. One can show that its image is a hypersurface of degree 4 isomorphic to the dual hypersurface of S3 (see[84]).

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9.6
9.6.1

Automorphisms of cubic surfaces
Elements of finite order in Weyl groups

Let W be the Weyl group of a simple root system of type A, D, E. The conjugacy classes of elements of finite order can be classified. We will follow the classification due to R. Carter. We know that each w ∈ W is equal to the product of reflections with respect to roots α. Let l(w) be the smallest number k of roots such that w can be written as such a product. This number is equal to the number of eigenvalues of w in QC different from 1. If w = rα1 . . . rαl(w) , then the corresponding roots are linearly independent. 2 2 Each element w can be written as the product w = w1 w2 , where w1 = w2 = 1 and l(w) = l(w1 ) + l(w2 ). Moreover, if w1 = rα1 . . . rαl(w1 ) , where the roots αi . The same is true for w2 = rβ1 · · · rβl(w2 ) . Eacg such decomposition defines the Carter graph Γ(w) of w. Its vertices correspond to each root in the decomposition of w = rα1 . . . rαl(w1 ) rβ1 · · · rβl(w2 ) . Two vertices corresponding to αi and βj are joined by an edge if , where (αi , βj ) = 0. Example 9.6.1. Suppose Q is the root lattice of type An . Its Weyl group is the symmetric group Sn+1 . Let σ = (i1 , . . . , i2k ) be a 2k-cycle. It is the product of transpositions (i1 i2 ) . . . (i2k−1 i2k ). Write σ = (i1 i2 )(i3 i4 )(i3 i4 )(i2 i3 )(i3 i4 ) . . . (ik−1 ik ) = (i1 i2 )(i3 i4 )(i2 i4 ) . . . (ik−1 ik ). Continuing in this way we will be able to write σ as the product of two involutions σ = [(i1 i2 )(i3 i4 ) . . . (i2k−1 i2k )][(i2 i3 )(i4 i5 ) . . . (i2k−2 i2k−1 ). Each transposition (ij) is the reflection with respect to the root ei − ej , where we consider the lattice Q as the suballtice of Zn+1 formed by the vectors perpendicular to e1 + · · · + en+1 . Now it is easy to see that the Carter graph of σ is the Dynkin diagram of type A2k . The same conclusion can be derived in the case when σ is cycle of odd length. Since any permutation is the product of commuting cycles, we obtain the Coxeter graph of a permutation is the disconnected sum of graphs of typer Ak . Carter proves that two elements in the Weyl group W (Q) are conjugate if and only if there graphs coincide. Anytime we embed the root lattice Q in a root lattice Q , we get an embedding of the Weyl groups W (Q) → W (Q ). Since the conjugacy class in W (Q) is determind by the Carter graph, we see that two nonconjugate elements in the subgroup stay nonconjugate in the group. Each Carter graph Γ has the following properties: • Its vertices correspond to linearly independent roots. • Each subgraph of Γ which is a cycle contains even number of vertices. • Each graph without cycles is a Dynkin diagram of some root lattice Q . The corresponding sublattice of Q defines the embedding of W (Q ) in W (Q) and the conjugacy class corresponding to Γ can be represented by the Coxeter element in W (Q).

9.6. AUTOMORPHISMS OF CUBIC SURFACES

355

Recall that the Coxeter element is a product of reflections corresponding to simple roots (forming a root basis of Q). Its order is the Coxeter number of the Weyl group. The graphs corresponding to subroot lattices can be described by the Borel-De Siebenthal-Dynkin. Let w ∈ W (Q), its characteristic polynomial in its linear action on QC can be read off from the diagram. diagram. The following table contains the list of connected components of Carter’s graphs, the orders of the corresponding elements and the characteristic polynomial. Graph Ak Dk Dk (a1 ) Dk (a2 ) . . . Dk (a k −1 ) 2 E6 E6 (a1 ) E6 (a2 ) E7 E7 (a1 ) E7 (a2 ) E7 (a3 ) E7 (a4 ) E8 E8 (a1 ) E8 (a2 ) E8 (a3 ) E8 (a4 ) E8 (a5 ) E8 (a6 ) E8 (a7 ) E8 (a8 ) Order k+1 2k − 2 l.c.m(2k − 4, 4) l.c.m(2k − 6, 6) . . . even k 12 9 6 18 14 12 30 6 30 24 20 12 18 15 10 12 6 Characteristic polynomial tk + tk−1 + · · · + 1 (tk−1 + 1)(t + 1) (tk−2 + 1)(t2 + 1) (tk−3 + 1)(t3 + 1) . . . k 2 (t 2 + 1) (t4 − t2 + 1)(t2 + t + 1) t6 + t 3 + 1 2 2 2 (t − t + 1) (t + t + 1) (t6 − t3 + 1)(t + 1) t7 + 1 4 2 3 (t − t + 1)(t + 1) (t5 + 1)(t2 − t + 1) (t2 − t + 1)3 (t + 1) 8 7 t + t − t 5 − t4 − t 3 + t + 1 t8 − t 4 + 1 8 6 t − t + t4 − t 2 + 1 (t4 − t2 + 1)2 6 3 (t − t + 1)(t2 − t + 1) 8 t − t7 + t 5 − t4 + t 3 − t + 1 (t4 − t3 + t2 − t + 1)2 4 (t − t2 + 1)(t2 − t + 1)2 (t2 − t + 1)4

Table 9.2: Carter graphs and characteristic polynomials

9.6.2

Eckardt points

A point of intersection of three lines in a tritangent plane is called an Eckardt point. As we will see later the locus of nonsingular cubic surfaces with an Eckardt point is of codimension 1 in the moduli space of cubic surfaces. Proposition 9.6.1. There is a bijective correspondence between Eckardt points on a

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nonsingular cubic surface S and automorphisms of order 2 with one isolated fixed point. Proof. Let p ∈ S be an Eckardt point and let π : S → S be the blow-up of p. This is a Del Pezzo surface of degree 2. The pre-image of the linear system | − KS − p| is the linear system | − KS |. It defines a degree 2 regular map f : S → P2 whose restriction to S \ π −1 (p) ∼ S \ {p} is the linear projection of S with center at p. Let R1 , R2 , R3 = be the proper inverse transforms of the lines in S from the tritangent plane defined by p. These are (−2)-curves on S . Their image in P2 is a singular point of the branch curve B of degree 4. The image of E = π −1 (p) is a line passing through 3 singular points of a quartic curve. It must be an irreducible component of B. Thus B is the union of a line and a cubic curve C which intersect at three distinct points x1 , x2 , x3 . Let X be the double cover of the blow-up of P2 at the points x1 , x2 , x3 ramified along the proper transform of the curve B. We have a birational map of f : S − → X which is a regular map outside the union of curves Ri . It is easy to see that it extends to the whole S by mapping the curves Ri isomorphically to the pre-images of the points xi under the map X → P2 . Thus f : S → X is a finite map of degree 2, and hence is a Galois cover of degree 2. The corresponding automorphism of S leaves the curve E pointwisely invariant, and hence descends to an automorphism σ of the cubic surface S. Since it must leave | − KS | invariant, it is induced by a linear projective transformation g of P3 . Its set of ¯ fixed points in P3 is the point p and a plane which intersects S along a curve C . The linear projection from p maps C isomorphically to the plane cubic C. Thus σ has one isolated fixed point on S. Conversely, assume S admits an automorphism σ of order 2 with one isolated fixed point p. As above we see that σ is induced by a projective transformation σ . Diag¯ onalizing the corresponding linear map of C4 , we see that g has one eigenspace of ¯ dimension 1 and one eigensubspace of dimension 3. Thus in P3 it fixes a point and a plane Π. The fixed locus of σ is the point p and a plane section C not passing through p. Let P be the tangent plane of S at p. It is obviously invariant and its intersection with S is a cubic plane curve Z with a singular point at p. Its intersection with C gives 3 fixed nonsingular points on Z. If Z is irreducible, its normalization is isomorphic to P1 which has only 2 fixed point of any non-trivial automorphism of order 2. Thus Z is reducible. If it consists of a line and a conic, then one of the components has 3 fixed points including the point p. Again this is impossible. So we conclude that Z consists of three concurrent lines and hence a tritangent plane. It is clear that the automorphism associated to this tritangent plane coincides with σ. Example 9.6.2. Consider a cubic surface given by equation f3 [t0 , t1 , t2 ] + t3 = 0, 3 where C = V (f3 ) is a nonsingulat plane cubic. Let be a flex tangent of C. It is easy to see that the pre-image of C under the projection [x0 , x1 , x2 , x3 ] → [x0 , x1 , x2 ] splits in the union of three lines passing through a coomon point (the pre-image of the inflection point). Thus the surface contains 9 Eckardt points. Note that the corresponding 9 tritangent planes contain all 27 lines.

9.6. AUTOMORPHISMS OF CUBIC SURFACES Example 9.6.3. Consider a cubic surface given by an equation in P4 of the form
4 4

357

ai ti =
i=0 i=0

t3 = 0, i

where ai = 0. Assume a1 = a2 . Then the point p = [1, −1, 0, 0, 0] is an Eckardt point. In fact, the tangent plane at this point is t0 = t1 = 0. It cuts out the surface along the union of three lines intersecting at the point p. Similarly, we have an Eckardt point whenever ai = aj for some i = j. Thus we may have 1,3,6 or 10 Eckardt points dependent on whether we have two,three,four or five equal ai ’s. For the future need let us prove the following. Proposition 9.6.2. Let p1 and p2 be two Eckardt points on S such that the line p1 , p2 is not contained in S. Then intersects S in a third Eckardt point. =

Proof. Let σ be an automorphism of S defined by the projection from the point p1 . Then l intersects S at the point p3 = g(p2 ). Note the projection of any line passing through p2 must contain a singular point of the branch locus since otherwise intersects the line component of the branch locus at a nonsingular point and hence passes through p1 . Thus intersects one of the lines passing through p1 and the plane spanned by these two lines cuts out S in an additional line passing through p3 . In this way we find three lines through p3 . Proposition 9.6.3. No more than two Eckardt points lie on a line contained in the surface. Proof. Consider the linear projection from one of the Eckardt points p1 . Its branch curve is the union of a plane cubic C and a line intersecting at three points. The second Eckardt point p2 is projected to one of the intersection points, say q. The plane spanned by the lines p1 , p2 and one of the other 2 lines passing through p2 is a tritangent plane with Eckardt point p2 . Since it is invariant with respect to the involution σ defined by p1 , the point p2 is a fixed point and hence lies on the curve of fixed points of σ. The projection of the tritangent plane is a line which intersects C only at the point q. Hence q is an inflection point. Clearly this shows that if there is a third Eckardt point p3 , it must coincide with p2 .

9.6.3

Subgroups of W (E6 ).

We will need some known information about the structure of the Weyl group W (E6 ). Lemma 9.6.4. Let H be a maximal subgroup of W (E6 ). Then one of the following cases occurs: (i) H ∼ 24 : S5 of order 24 5! and index 27; = (ii) H ∼ S6 × 2 of order 2.6! and index 36; = (iii) H ∼ 31+2 : 2S4 of order 1296 and index 40; = +

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(iv) H ∼ 33 : (S4 × 2) of order 1296 and index 40; = (v) H ∼ 2.(A4 × A4 ).2 .2 of order is 1152 and index 45. = Here we use the ATLAS [55] notations for cyclic groups: Z/nZ = n and semidirect products: H G = H : G, 31+2 denotes the group of order 33 of exponent p, + A.B is a group with normal subgroup isomorphic to A and quotient isomorphic to B. We recognize a group from (i) as the stabilizer subgroup of a exceptional vector (or a line on a cubic surface). If we choose a simple root basis (α0 , . . . , α5 ) such that the ∗ exceptional vector is equal to α5 , then H is generated by the reflections si = sαi , i = 5. It is naturally isomorphic to the Weyl group W (D5 ). A group H of type (ii) is the stabilizer subgroup of a double-six. The subgroup S6 permutes the lines, the subgroup 2 switches the two sixers. In the geometric root basis α0 = e0 − e1 − e2 − e3 , αi = ei − ei+1 , the stabilizer subgroup of the doublesixdouble-six (e1 , . . . , e6 ; e1 , . . . , e6 ), where ei = 2e0 − e1 − . . . − e6 + ei , generated by permutations of ei ’s and the reflection with respect to the maximal root 2e0 − e1 − . . . − e6 . A group of type (iv) is the stabilizer subgroup of a Steiner triad of a double-sixes. A group of type (v) is the stabilizer subgroup of a tritangent plane (or a triple of exceptional vectors added up to 0). Proposition 9.6.5. W (E6 ) contains a unique normal subgroup W (E6 ) . It is a simple group and its index is equal to 2. Proof. Choose a root basis (α0 , . . . , α5 ) in the root lattice Q of type E6 . Let s0 , . . . , s5 be the corresponding simple reflections. Each element w ∈ W (E6 ) can be written as a product of the simple reflections. Let (w) is the minimal length of the word needed to write w as such a product. for example, (1) = 0, (si ) = 1. one shows that the function : W (E6 ) → Z/2Z, w → (w) mod 2 is a homomorphism of groups. Its kernel W (E6 ) is a subgroup of index 2. The restriction of the function to the subgroup H ∼ S6 generated by the reflections s1 , . . . , s5 is the sign function. = Suppose K is a normal subgroup of W (E6 ) . Then K ∩ H is either trivial or equal to the alternating subgroup A6 of index 2. It remains to use that H × (r) is a maximal subgroup of W (E6 ) and r is a reflection which does not belong to W (E6 ) . Remark 9.6.1. Recall that we have an isomorphism (9.7) of groups W (E6 ) ∼ O(6, F2 )− . = The subgroup W (E6 ) is isomorphic to the commutator subgroup of O(6, F2 )− . Let us mention other realizations of the Weyl group W (E6 ). Proposition 9.6.6. W (E6 ) ∼ SU4 (2), = where U4 (2) is the group of linear transformations with determinant 1 of F4 preserving 4 a non-degenerate Hermitian product with respect to the Frobenius automorphism of F4 .

9.6. AUTOMORPHISMS OF CUBIC SURFACES

359

Proof. Let F : x → x2 be the Frobenius automorphism of F4 . We view the expression
3 3

x3 = i
i=0 i=0

xi F(xi )

as a nondegenerate hermitian form in F4 . Thus SU4 (2) is isomorphic to the subgroup 4 of the automorphism group of the cubic surface S defined by the equation t3 + t 3 + t3 + t 3 = 0 0 1 2 3 ¯ over the field F2 . The Weyl representation (which is defined for nonsingular cubic surfaces over fields of arbitrary characteristic) of Aut(S) defines a homomorphism SU4 (2) → W (E6 ). The group SU4 (2) is known to be simple and of order equal to 1 ∼ 2 |W (E6 )|. This defines an isomorphism SU4 (2) = W (E6 ) . Proposition 9.6.7. W (E6 ) ∼ SO(5, F3 ), = W (E6 ) ∼ SO(5, F3 )+ , =

where SO(5, F3 )+ is the subgroup of elements of spinor norm 1. ¯ Proof. Let Q = Q/3Q. Since the determinant of the Cartan matrix of type E6 is equal to 3, the symmetric bilinear form defined by v + 3Q, w + 3Q = −(v, w) mod 3

is degenerate. It has one-dimensional radical spanned by the vector v0 = 2α1 + α1 + 2α4 + α5 mod 3Q

The quadratic form q(v) = (v, v) mod 3 defines a non-degenerate quadratic form on ¯ V = Q/F3 v0 ∼ F5 . We have a natural injective homomorphism W (E6 ) → O(5, F2 ). = 3 Comparing the orders, we find that the image is a subgroup of index 2. It must coincide with SO(5, F3 ). Its unique normal subgroup of index 2 is SO(5, F3 )+ . Remark 9.6.2. Let V be a vector space of odd dimension 2k + 1 over a finite field Fq equipped with a non-degenerate symmetric bilinear form. An element v ∈ V is called a plus vector (resp. minus vector) if (v, v) is a square in F∗ (resp. is not a square ∈ F∗ ). q q The orthogonal group O(V ) has three orbits in P(E): the set of isotropic lines, the set of lines spanned by a plus vector and the set of lines spanned by a minus vector. The isotropic subgroup of a non-isotropic vector v is isomorphic to the orthogonal group of the subspace v ⊥ . The restriction of the quadratic form to v ⊥ is of Witt index k if v is a plus vector and of Witt index k − 1 if v is a minus vector. Thus the stabilizer group is isomorphic to O(2k, Fq )± . In our case when k = 2, q = 3, we obtain that ¯ minus vectors correspond to cosets of roots in Q, hence the stabilizer of a minus vector is isomorphic to the stabilizer of a douible six, i.e. a maximal subgroup of W (E6 ) of index 36. The stabilizer subgroup of a plus vector is a group of index 45 and isomorphic to the stabilizer of a tritangent plane. The stabilizer of an isotropic plane is a maximal subgroup of type (iii), and the stabilizer subgroup of an isotropic line is a maximal subgroup of type (iv).

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9.6.4

Automorphisms of finite order

Since any automorphism of a nonsingular cubic surface S preserves | − KS |, it is induced by a projective transformation. After diagonalization we may assume that any automorphism is represented by a diagonal matrix with roots of unity as its entries. Lemma 9.6.8. Let S = V (f ) be a nonsingular cubic surface which is invariant with respect to a projective transformation σ of order n > 1. Then, after a linear change of variables, f is given in the following list. Also, a generator of the group σ can be a b c defined by (x0 , x1 , x2 , x3 ) → (x0 , ζn , ζn x2 , ζn x3 ), where ζn is a primitive nth root of unity. (i) (n = 2), (a, b, c) = [0, 0, 1], f = t2 l1 [t0 , t1 , t2 ] + t3 + t3 + t3 + at0 t1 t2 . 3 0 1 2 (ii) (n = 2), (a, b, c) = (0, 1, 1), f = t0 t2 (t2 + at3 ) + t1 t3 (t3 + bt3 ) + t3 + t3 . 0 1 (iii) (n = 3), (a, b, c) = [0, 0, 1], f = t3 + t3 + t3 + t3 + at0 t1 t2 . 3 0 1 2 (iv) (n = 3), (a, b, c) = (0, 1, 1), f = l3 (t0 , t1 ) + m3 (t2 , t3 ). (v) (n = 3), (a, b, c) = (0, 1, 2), f = l3 (t0 , t1 ) + t2 t3 l1 (t0 , t1 ) + t3 + t3 . 2 3 (vi) (n = 4), (a, b, c) = (0, 2, 1), f = t2 t2 + t3 + t3 + t2 (t0 + at1 ). 3 0 1 2 (vii) (n = 4), (a, b, c) = (2, 3, 1), f = t 3 + t0 t2 + t 1 t 2 + t1 t2 . 0 1 3 2 (viii) (n = 5), (a, b, c) = (4, 1, 2), f = t2 t1 + t 2 t 2 + t2 t3 + t 2 t 0 , 0 1 2 3 (ix) (n = 6), (a, b, c) = (0, 3, 2), f = l3 (t0 , t1 ) + t3 + t2 (t0 + at1 ). 3 2

9.6. AUTOMORPHISMS OF CUBIC SURFACES (x) (n = 6), (a, b, c) = (0, 2, 5), f = t 3 + t 3 + t2 t2 + t 3 . 0 1 3 2 (xi) (n = 6), (a, b, c) = (4, 2, 1), f = t2 t1 + t3 + t3 + t3 + λt0 t1 t2 . 3 0 1 2 (xii) (n = 6), (a, b, c) = (4, 1, 3), f = t3 + bt0 t2 + t2 t1 + t3 . 0 3 2 1 (xiii) (n = 8), (a, b, c) = (4, 3, 2), f = t 2 t 1 + t 2 t 3 + t0 t2 + t 3 . 3 2 1 0 (xiv) (n = 9), (a, b, c) = (4, 1, 7), f = t 2 t 1 + t 2 t 2 + t2 t3 + t 3 . 3 1 2 0 (xv) (n = 12), (a, b, c) = (4, 1, 10), f = t 2 t 1 + t2 t3 + t 3 + t3 . 3 2 0 1 Here the subscripts in li , mi indicates the degree of the polynomial.

361

Proof. Choose a coordinate system where σ diagonalizes as in the statement of the lemma. Let p1 = (1, 0, 0, 0), . . . , p4 = (0, 0, 0, 1) be the reference points. They are fixed under the action of σ in P3 . We will use frequently that f is of degree ≥ 2 in each variable. This follows from the assumption that the surface is nonsingular. We will also give a normal form of the equation with minimal number of parameters which is easy to get and is left to the reader. Case 1: Two of a, b, c, say a, b, are equal to zero. Write f as a polynomial in t3 . Assume p3 ∈ V (f ). Then f = t3 + t2 l1 [t0 , t1 , t2 ] + t3 l2 [t0 , t1 , t2 ] + l3 [t0 , t1 , t2 ]. 3 3
3c Since f is an eigenvector with the eigenvalue equal to ζn and l3 = 0, we must have n = 3 and l1 = l2 = 0. This is case (iii). Assume p3 ∈ V (f ). Then

f = t2 l1 [t0 , t1 , t2 ] + t3 l2 [t0 , t1 , t2 ] + l3 [t0 , t1 , t2 ]. 3 As above this gives n = 2, l2 = 0. This is case (i). Case 2: One of (a, b, c), say a, is equal to zero. Write f as a polynomial in the form f = l3 (t0 , t1 ) + t0 l2 (t2 , t3 ) + t1 m2 (t2 , t3 ) + m3 (t2 , t3 ). Assume that l2 = m2 = 0. If m3 is of degree 3 in t3 or t2 , say t2 , then 3b = 0 mod n. If l3 is of degree in t3 too, we get 3c = 0 mod n, hence n = 3. Without

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loss of generality we may assume that (b, c) = (1, 1) or (2, 1). In the first case m3 is any polynomial in t2 , t3 of degree ≥ 2 in t2 , t3 . This is case (iv). In the second case m3 = t3 + t3 . This is a special case of case (v). 3 2 If m3 is of degree 2 in t3 , then l3 contains t2 t2 , hence 2c + b = 0 mod n. This 3 gives n = 6, (a, b, c) = (0, 2, 5). This is case (x). Assume now that l2 or m2 is not equal to zero. If t2 , t2 do not enter in l2 and m2 , 2 3 then t0 t1 must enter in one of them. This gives b + c = 0 mod n. If t3 or t3 enters in 2 2 m3 , then 3b = 0 mod n or 3c = 0 mod n. This gives n = 3, (a, b, c) = (0, 1, 2) or (0, 2, 1). This is case (v). If t3 , t3 do not enter in m3 , then t2 t3 and t2 t2 both enter and 2 3 2 3 we get 2b + c = b + 2c = 0 mod n. This again implies n = 3 and we are in case (v). Now we may assume that t2 enters in l2 or m2 , then 2b = 0 mod n. If t2 also 2 3 enters in l2 or m2 , then 2c = 0 mod n. This implies n = 2 and m3 = 0. This is case (ii). If t2 does not enter in l2 and m2 , then m3 is of degree ≥ 2 in t3 . If t3 enters in m3 , 3 3 then 3c = 0 mod n, hence n = 6 and (a, b, c) = (0, 3, 2). Thus f = l3 (t0 , t1 ) + t2 m1 (t0 , t1 ) + t3 . 2 3 This gives case (ix). If t2 and t3 do not enter in l2 and in m2 but t2 t3 enters in one of these polynomials, 3 2 then we get b + c = 0 mod n. If t2 t2 enters in l3 , then b + 2c = 0 mod n, hence 3 4c = 0 and n = 4, (a, b, c) = (0, 2, 1) or (0, 2, 3). This is case (vi). Case 3: 0, a, b, c are all distinct. Note that If two of (a, b, c) are equal, then, by scaling and permuting coordinates we will be in the previous Cases. This obviously implies that n > 3. Also monomials t2 tj and ti t2 cannot both enter in f . i j Case 3a: All the reference points Pi belong to the surface. In this case f does not contain cubes of the variables ti and we can write f = t2 A1 (t1 , t2 , t3 ) + t2 B1 (t0 , t2 , t3 ) + t2 C1 (t0 , t1 , t3 ) + t2 D1 [t0 , t1 , t2 ], 0 1 2 3 where A1 , B1 , C1 , D1 are nonzero linear polynomials. Since all 0, a, b, c are distinct, each of these linear polynomials contains only one variable. If the coefficients at ti and tj contain the same variable tk , then the plane V (tk ) is tangent to the surface along a line. It is easy to see that this does not happen for a nonsingular surface. Thus without loss of generarily we may assume that f = t 2 t 1 + t2 t2 + t 2 t 3 + t 2 t 0 . 0 1 2 3 Then a+b = 2b+c−a = 2c−a = 0 mod n. This implies n = 5, (a, b, c) = (4, 1, 2). This is case (viii). Case 3b. Three reference point belong to the surface. By scaling and permuting variables we may assume that p1 does not belong to V (f ). The equation contains t3 but does not contain the cubes of other variables. 0 Since f is σ-invariant, t2 does not enter in f . We can write 0 f = t3 + t0 f2 (t1 , t2 , t3 ) + f3 (t1 , t2 , t3 ). 0 (9.46)

9.6. AUTOMORPHISMS OF CUBIC SURFACES

363

Each line i = p1 , pi does not belong to the surface and contains 2 fixed point of σ. Suppose each line i intersect V (f ) only at one point pi . Then f2 does not contain squares of the variables and f3 contains squares of each variable but not cubes. Without loss of generality we may assume that f3 contains t2 t3 . Then 2b+c = 0 mod n. Since 2 t2 t2 does not enter in f3 , the monomial t2 t1 must enter. This gives 2c+a = 0 mod n. 3 3 Solving for (a, b, c) we find that n = 9, (a, b, c) = (4, 1, 7). The polynomial f2 cannot contain ti tj and hence is equal to zero. This gives us case (xiv). Now we are in the situation when one of the lines i intersects V (f ) at a point p different from pi . If there is no other point in the intersection, then p is a third fixed point of σ on the line. This is impossible, and therefore i intersects the surface at three distinct points pi , p, q. Since σ permutes P and Q, we see that the restriction of σ 2 to i is the identity. Without loss of generality we may assume that i = 2 and, hence 2a = 0 mod n. Obviously, t2 t2 , t2 t3 do not enter in f3 and one of t2 or t2 does not 1 1 2 3 enter in f2 . Assume t2 does not enter in f2 . Then t2 t1 or t2 t3 enters in g3 . In the first 2 2 2 case 2b + a = 0 mod n. This gives n = 4, (a, b, c) = (2, 1, 3) or (2, 3, 1). This leads to case (vii). In the second case, we get 2b + c = 0 mod n. Since t2 t2 does not enter in f3 , t2 t1 3 3 must enter giving 2c + a = 0 mod n. This easily gives n = 8 and (a, b, c) = (4, 3, 2). This leads to case (xiii). Case 3c. Two reference points do not belong to the surface. We may assume that p1 , p2 are not in the surface. Thus t3 , t3 enter in f , hence 3a = 0 mod n. We may 0 1 assume that f is as in (9.46), where t1 enters in f3 . Clearly, t2 does not enter in g2 . If 1 t2 (or t2 ) enters in g2 , then 2c = 0 mod n, and we get n = 6, (a, b, c) = (2, b, 3) or 3 2 (4, b, 3). Since b = 3, t2 does not enter in g2 . Thus t2 t3 or t2 t1 enter in g3 . In the first 2 2 2 case 2b + c = 0 mod 6, hence 2b = 3 mod 6 which is impossible. Thus t2 t1 enters 2 giving 2b + a = 0 mod 6. This gives case (a, b, c) = (4, 1, 3) or (2, 5, 3). This is case (xii). Now we may assume that t2 and t2 do not enter in f2 . If t2 t1 enters, we are led to 3 2 2 the previous case (xii). So we may assume that t2 t3 enters giving 2b + c = 0. This 2 implies that t2 t1 enters, hence 2c+a = 0 mod n. This easily gives n = 12, (a, b, c) = 3 (4, 1, 10). This is case (xv). Case 3d. Three reference points do not belong to the surface. We may assume that p1 , p2 , p3 are not in the surface. Thus t3 , t3 , t3 enter in f , 0 1 2 hence 3a = 3b = 0 mod n. We may assume that f is as in (9.46), where t1 , t2 enter in f3 . Clearly, t2 , t2 do not enter in f2 . If t2 enters in f2 , then 2c = 0 mod n, and we 3 1 2 get n = 6, (a, b, c) = (2, 4, 3) or (4, 2, 3). This gives case (xi). Assume t2 does not enter in f2 . Without loss of generality we may assume that 3 2 t3 t1 enters in f3 . This gives 2c + a = 0 mod n. From this follows that n = 6 and (a, b, c) = (4, 2, 1). This case is isomorphic to case (xi). Case 3e. No reference point belongs to the surface. In this case each t3 enters in f , hence 3a = 3b = 3c = 0 mod n. This is i impossible for n > 3.

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In the natural representation of Aut(S) in W (E6 ) each nontrivial automorphism σ defines a conjugacy class in W (E6 ). The following table gives the list of the conjugacy classes. This can be found in [55], [29], [171]. It is given in the following table. Atlas 1A 2A 2B 2C 2D 3A 3C 3D 4A 4B 4C 4D 5A 6A 6C 6E 6F 6G 6H 6I 8A 9A 10A 12A 12C Carter ∅ 4A1 2A1 A1 3A1 3A2 A2 2A2 D4 (a1 ) A1 + A3 2A1 + A3 A3 A4 E6 (a2 ) D4 A1 + A5 2A1 + A2 A1 + A2 A1 + 2A2 A5 D5 E6 (a1 ) A1 + A4 E6 D5 (a1 ) Manin c25 c3 c2 c16 c17 c11 c6 c9 c4 c5 c19 c18 c15 c12 c21 c10 c8 c7 c10 c23 c20 c14 c25 c13 c24 Ord 6 2 2 2 2 3 3 3 4 4 4 4 5 6 6 6 6 6 6 6 8 9 10 12 12 |C(w)| 51840 1152 192 1440 96 648 216 108 96 16 96 32 10 72 36 36 24 36 36 12 8 9 36 12 12 Tr 7 -1 3 5 1 -2 4 1 3 1 -1 3 2 2 1 -1 0 2 -1 1 1 1 0 0 2 Char (t − 1)7 p4 (t − 1)3 1 p2 (t − 1)5 1 p1 (t − 1)6 p3 (t − 1)4 1 p3 (t − 1) 2 p2 (t − 1)5 p2 (t − 1)3 2 2 (t + 1)2 (t − 1)3 p1 p3 (t − 1)3 p2 p3 (t − 1)3 1 p3 (t − 1)4 p4 (t − 1)3 2 p2 (t − t + 1)2 (t − 1) p2 (t2 − t + 1)(t − 1)3 1 p1 p5 (t − 1) p2 p2 (t − 1)3 1 p1 p2 (t − 1)4 p1 p2 (t − 1)2 2 p5 (t − 1)2 p1 (t4 + 1)(t − 1)2 (t6 + t3 + 1)(t − 1) p1 p4 (t − 1)2 4 2 p2 (t − t + 1)(t − 1) (t3 + 1)(t2 + 1)(t − 1)2

x x x

x x x x x

x x x x x

x x x

Table 9.3: Conjugacy classes in W (E6 ) Here we mark with the cross the conjugacy classes realized by automorphisms of nonsingular cubic surfaces. Also |C(w)| denotes the cardinality of the centralizer of an element w from the conjugacy class, TrPic denotes the trace in the Picard lattice (equal to the trace in the root lattice plus 1), Char denotes the characteristic polynomial in Pic(S) and pe (t) = te + te−1 + · · · + 1. To determine to which conjugacy class our σ corresponds under the Weyl representation we use the topological Lefschetz fixed-point formula. The next theorem rewrites the list from Lemma 9.6.8 in the same order, renaming the cases with indication to which conjugacy class they correspond. Also, we simplify the formulae for f by scaling, and reducing a cubic ternary form to the Hesse form, and a cubic binary form to sum of cubes, and a quadratic binary forms to the product of the variables. Each time we use that the forms are non-degenerate because the surface is

9.6. AUTOMORPHISMS OF CUBIC SURFACES nonsingular.

365

Theorem 9.6.9. Let S be a nonsingular cubic surface admitting a non-trivial automorphism σ of order n. Then S is equivariantly isomorphic to one of the following surfaces V (f ) with σ = [x0 , a x1 , b x2 , c x3 ]. (9.47) n n n • 4A1 (n = 2), (a, b, c) = (0, 0, 1), f = t2 f1 (t0 , t1 , t2 ) + t3 + t3 + t3 + αt0 t1 t2 . 3 0 1 2 • 2A1 (n = 2), (a, b, c) = (0, 1, 1), f = t0 t2 (t2 + αt3 ) + t1 t3 (t2 + βt3 ) + t3 + t3 . 0 1 • 3A2 (n = 3), (a, b, c) = (0, 0, 1), f = t3 + t3 + t3 + t3 + αt0 t1 t2 . 0 1 2 3 • A2 (n = 3), (a, b, c) = (0, 1, 1), f = t 3 + t3 + t 3 + t3 . 0 1 2 3 • 2A2 (n = 3), (a, b, c) = (0, 1, 2), f = t3 + t3 + t2 t3 (t0 + at1 ) + t3 + t3 . 0 1 2 3 • D4 (a1 ) (n = 4), (a, b, c) = (0, 2, 1), f = t2 t2 + f3 (t0 , t1 ) + t2 (t0 + αt1 ). 3 2 • A3 + A1 (n = 4), (a, b, c) = (2, 1, 3), f = t 3 + t 0 t 2 + t1 t2 + t 1 t 2 . 0 1 3 2 • A4 (n = 5), (a, b, c) = (4, 1, 2), f = t 2 t 1 + t2 t2 + t 2 t 3 + t2 t0 . 0 1 2 3 • E6 (a2 ) (n = 6), (a, b, c) = (0, 3, 2), f = t3 + t3 + t3 + t2 (αt0 + t1 ). 0 1 3 2 • D4 (n = 6), (a, b, c) = (0, 2, 5), f = f3 (t0 , t1 ) + t2 t2 + t3 . 3 2

366 • A5 + A1 (n = 6), (a, b, c) = (4, 2, 1),

CHAPTER 9. CUBIC SURFACES

f = t2 t1 + t3 + t3 + t3 + λt0 t1 t2 . 3 0 1 2 • 2A1 + A2 (n = 6), (a, b, c) = (4, 1, 3), f = t3 + βt0 t2 + t2 t1 + t3 . 0 3 2 1 • D5 (n = 8), (a, b, c) = (4, 3, 2), f = t 2 t 1 + t2 t3 + t 0 t 2 + t3 . 3 2 1 0 • E6 (a1 ) (n = 9), (a, b, c) = (4, 1, 7), f = t 2 t 1 + t2 t2 + t 2 t 3 + t3 . 3 1 2 0 • E6 (n = 12), (a, b, c) = (4, 1, 10), f = t2 t1 + t 2 t 3 + t3 + t 3 . 3 2 0 1 Proof. We will be computing the trace of σ ∗ by using Lemma ??. We use the classification from Lemma 9.6.8. Order 2. In case (i), the fixed locus is the nonsingular elliptic curve given by equations t3 = f3 = 0 and isolated point (0, 0, 0, 1). The Euler-Poincar´ characteristic of the fixed e locus is equal to 1. Hence the trace in Pic(S) is equal to −1. This gives the conjugacy class 2A. In case (ii), the fixed locus is the line t0 = t1 = 0 and three isolated points lying on the line t2 = t3 (not contained in the surface). The Euler-Poincar´ e characteristic of the fixed locus is equal to 5. Hence the trace in Pic(S) is equal to 3. This gives the conjugacy class 2A1 . Order 3. In case (iii), the fixed locus is a nonsingular elliptic curve given by equations t3 = f3 = 0. The Euler-Poincar´ characteristic of the fixed locus is equal to 0. Hence the e trace in Pic(S) is equal to −2. This gives the conjugacy class 3A2 . In case (iv), the fixed locus is the set of 6 points lying on the lines t0 = t1 = 0 and t2 = t3 = 0. Here we use that the polynomials f3 , g3 do not have multiple roots since otherwise S is singular. The Euler-Poincar´ characteristic of the fixed locus is equal to e 6. Hence the trace in Pic(S) is equal to 4. This gives the conjugacy class A2 . In case (iv) , the fixed locus consists of 3 points lying on the line t2 = t3 = 0. Hence the trace in Pic(S) is equal to 1. This gives the conjugacy class 2A2 . In case (v), the fixed locus is the set of 3 points lying on the line t2 = t3 = 0. Again we use that f3 does not have multiple roots. The Euler-Poincar´ characteristic e of the fixed locus is equal to 3. Hence the trace in Pic(S) is equal to 1. This gives the conjugacy class 2A2 . Order 4.

9.6. AUTOMORPHISMS OF CUBIC SURFACES

367

In case (vi), the fixed locus is the set of 5 points lying on the lines t0 = t1 = 0 and two reference points P3 = (0, 0, 1, 0) and P4 = (0, 0, 0, 1). The Euler-Poincar´ e characteristic of the fixed locus is equal to 5. Hence the trace in Pic(S) is equal to 3. This gives the conjugacy class D4 (a1 ) or 4D. To distinguish the two classes, we notice that σ 2 acts as in case (i). This implies that σ 2 belongs to the conjugacy class 4A1 . On the other hand, the characteristic polynomial of 4D shows that 4D2 is the conjucacy class 2A1 . Thus we have the conjugacy class D4 (a1 ). In case (vii), we have three isolated fixed points (1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0). Thus the trace is equal to 1. This gives the conjugacy class A1 + A3 . Order 5. This is the unique conjugacy class of order 5. It is realized in case (viii). We have 4 isolated fixed points confirming that the trace is equal to 2. Order 6. In case (ix) we have 4 isolated fixed points so that the trace is equal to 2. This gives possible conjugacy classes E6 (a2 ), D4 , A1 + A2 . We know that our surface is σ 2 equivariantly isomorphic to a surface from case (iii). Thus σ 2 belongs to the conjugacy class 3A2 . Using the characteristic polynomials we check that only the square of the conjugacy class E6 (a2 ) is equal to 3A2 . In case (x) we have 4 isolated fixed points. This gives that the trace is equal to 2. It is clear that σ 3 acts as in case (i), thus σ 3 belongs to 4A1 . Also σ 2 acts as in case (iv). This shows that σ 2 belongs to A2 . Comparing the characteristic polynomials, this leaves only the possibility that σ belongs to D4 . In case (xi) we have only one isolated fixed point (0, 0, 0, 1). This gives that the trace is equal to −1 and hence σ belongs to A1 + A5 . In case (xii) we have 2 isolated fixed points so that the trace is equal to 2. The only conjugacy class with trace zero is 2A1 + A2 . Order 8. D5 is the unique conjugacy class of order 8. Its trace is 1. This agrees with case (xiii), where we have 3 fixed points. Order 9. E6 (a1 ) is the unique conjugacy class of order 9. Its trace is 1. This agrees with case (xiv), where we have 3 fixed points. Order 12. We have 2 fixed points giving the trace of σ equal to 0. This chooses the conjugacy class E6 . Remark 9.6.3. Some of the conjugacy classes (maybe all ?) are realized by automorphisms of minimal resolutions of singular surfaces. Also two non-conjugate elements from Aut(S) may define the same conjugacy class in W (E6 ). An example is an automorphism σ from the conjugacy class 3A2 and its square.

9.6.5

Automorphisms groups

In the following table we use the notation H3 (3) for the Heisenberg group of unipotent 3 × 3-matrices with entries in F3 .

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Theorem 9.6.10. The following is the list of all possible groups of automorphisms of nonsingular cubic surfaces.

Type

Order

Structure

I II III IV

648 120 108 54

3 : S4 S5 H3 (3) : 4 H3 (3) : 2

3

t3 0 t3 0

f (t0 , t1 , t2 , t3 ) 3 + t3 + t3 + t3 t0 1 2 3 t2 t1 + t0 t2 + t2 t2 + t3 t2 0 2 3 1 + t3 + t3 + t3 + 6at1 t2 t3 1 2 3 + t3 + t3 + t3 + 6at1 t2 t3 1 2 3

Parameters

20a3 + 8a6 = 1

V VI VII VIII IX X XI

24 12 8 6 4 4 2

S4 S3 × 2 8 S3 4 22 2

t3 + t0 (t2 + t2 + t2 ) 0 1 2 3 +at1 t2 t3 t3 + t3 + at2 t3 (t0 + t1 ) + t3 + t3 2 3 0 1 t3 2 + t2 t2 + t2 t1 + t3 + t0 t2 3 2 0 1 3 + at t (t + bt ) + t3 + t3 t3 2 3 0 1 0 1 t2 t2 + t2 t1 + t3 + t0 t2 + at3 3 2 0 1 1 t2 (t1 + t2 + at3 ) + t3 + t3 0 1 2 +t3 + 6bt1 t2 t3 3 t3 + t3 + t3 + 6at1 t2 t3 1 2 3 +t2 (t1 + bt2 + ct3 ) 0

a − a4 = 0, 8a3 = −1, 3 20a + 8a6 = 1 9a3 = 8a 8a3 = −1, a=0 a3 = −1 a=0 8b3 = −1 b3 , c 3 = 1 b3 = c3 8a3 = −1,

Table 9.4: Groups of automorphisms of cubic surfaces

Proof. Let S be a nonsingular cubic surface and G be a subgroup of Aut(S). Suppose G contains an element of order 3 from the conjugacy class A2 . Applying Theorem 9.6.9, we see that S is isomorphic to the Fermat surface V (t3 + t3 + t3 + t3 ). It 1 2 3 0 has 27 lines given by the equations t0 + t1 = 0, t2 + ηt3 = 0, 3 = η 3 = −1, or their transforms under permuting the variables. It is clear that any automorphism of S permutes the planes ti + tj = 0 and hence Aut(S) consists of permutations of the variables and multiplying the variables by cube roots of unity. This gives case I. It is easy to see that each plane ti + tj = 0 is a tritangent plane with an Eckardt point. Thus we have 18 Eckardt points, maximal possible. Assume that G contains an element of order 5. Applying Theorem 9.6.9, we see that S is isomorphic to the Clebsch diagonal surface t2 t1 + t0 t2 + t2 t2 + t3 t2 = 0. 0 2 3 1 (9.48)

9.6. AUTOMORPHISMS OF CUBIC SURFACES Consider the embedding of S in P4 given by the linear functions z0 z1 z2 z3 z4 = = = = = t0 + t 1 + t2 + t 3 3 4 2 ζ 5 t0 + ζ 5 t 1 + ζ 5 t 2 + ζ 5 t 3 3 4 2 ζ 5 t0 + ζ 5 t 1 + ζ 5 t2 + ζ 5 t 3 4 2 3 ζ 5 t0 + ζ 5 t 1 + ζ 5 t2 + ζ 5 t 3 2 3 4 ζ 5 t0 + ζ 5 t 1 + ζ 5 t2 + ζ 5 t 3

369

(9.49)

Then one easily checks that i=0 zi = 0 and (9.48) implies that also This shows that S is isomorphic the following surface in P4 :
4 3 zi = i=0 i=0 4

4

4 3 i=0 zi

= 0.

zi = 0

(9.50)

These equations exhibit an obvious symmetry which is the group S5 . The line z0 = z1 + z2 = z3 + z4 = 0 lies on S. Its S5 -orbit consists of 15 lines. The remaining 12 lines form a doublesix. Their equations are as follows. Let ω be a primitive 5th root of unity. Let σ = (a1 , . . . , a5 ) be a permutation of {0, 1, 2, 3, 4}. Each line σ spanned by the points [ω a1 , . . . , ω σ5 ] and [ω −a1 , . . . , ω −a5 ] belongs to the surface. This gives 12 = 5!/2.5 different lines. One checks immdiately that two lines σ and σ intersect if and only if σ = σ ◦ τ for some odd permutation τ . The group S5 (as well its subgroup S4 ) acts transitively on the double-six. The alternating subgroup stabilizes a sixer. Observe that S has 10 Eckardt points [1, −1, 0, 0, 0] and other ones obtained by permutations of coordinates. Also notice that any point, say [1, −1, 0, 0] is joined by a line in the surface to three other points [0, 0, 1, −1, 0], [0, 0, 0, 1, −1], [0, 0, 1, 0, −1]. The graph whose vertices are Eckardt points and edges the lines is a famous tri-valent Peterson graph whose group of symmetry is isomorphic to S5 . Assume G is larger than S5 . Consider the representation of G in the symmetry group of the graph of Eckardt points. Its image is equal to S5 , hence its kernel is non-trivial. Let H be a maximal subgroup of W (E6 ) which contains G. It follows from Lemma 9.6.4 that G must contain S6 or an involution. The restriction of the representation to S6 must be trivial, since the kernel is non-trivial and is not equal to A6 . This is impossible since S6 contains our S5 . If the kernel contains an involution, then the involution fixes 10 points. Since no involutions in W (E6 ) has trace equal to 8, we get a contradiction. Thus Aut(S) ∼ S5 . = Assume that G contains an element σ : [t0 , t1 , t2 , t3 ] → [ 3 t0 , t1 , t2 , t3 ] from the conjugacy class 3A2 . Then we are in case (iii) of Theorem 9.6.9. The plane cubic curve C = V (t3 +t3 +t3 +at1 t2 t3 ) has the projective group of automorphisms isomorphic to 1 2 3 32 : 2. Its normal subgroup 32 is generated by a cyclic permutation of coordinates and the transformation [t1 , t2 , t3 ] → [t1 , 3 t2 , 2 t3 ]. Together with σ this generates a group 3 G1 of order 54 isomorphic to 3.(32 : 2). Note that for a special value a, C may acquire

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an additional isomorphism of order 4 or 6). It happens when 1 − 20a3 − 8a6 = 0 or √ a(a3 − 1) = 0, respectively (see Exercises to Chapter 3). If ζ = −1+ 3 is a root of the 2 first equation, then the extra automorphism of order 4 is given by the formula [t1 , t2 , t3 ] → [t1 + t2 + t3 , t1 + ζt2 + ζ 2 t3 , t1 + ζ 2 t2 + ζt3 ]. It is easy to see that the group Aut(S) is isomorphic to 3.(32 : 4), the center is generated by the transformation which multiplies x0 by a third root of unity. If a4 = a, the curve C is projectively isomorphic to the Fermat cubic t3 + t3 + t3 , 1 2 3 hence we are in case I. According to example 9.6.2 S contains 9 tritangent planes with Eckardt points. Each plane is the preimage of a line under the projection to the plane Π containing the curve f of fixed points. Suppose there is a symmetry τ not belonging to G1 . Since G1 acts transitively on the set of Eckardt points, we may assume that τ fixes a tritangent plane containing an Eckardt point. Thus τ fixes the plane Π and hence is an automorphism of the plane cubic C. This proves that G = G1 if C has no extra automorphisms, G = G if C has an automorphism of order 4, and S is of type I if C has an automorphism of order 6. Assume that S contains an element σ of order 8. Then S is isomorphic to the surface from case (xiii) of Theorem 9.6.9. The only maximal subgroup of W (E6 ) which contains an element of order 8 is a subgroup H of order 1152. As we know it stabilizes a tritangent plane. In our case the tritangent plane is t2 = 0. It has the Eckardt point x = (0, 0, 0, 1). Thus G = Aut(S) is a subgroup of the linear tangent space T Sx . If any element of G acts identically on the set of lines in the tritangent plane, then it acts identically on the projectivized tangent space, and hence G is a cyclic group. Obvioulsy this implies that G is of order 8. Assume that there is an element τ which permutes cyclically the lines. Let G be the subgroup generated by σ and τ . Obviously, τ 3 = σ k . Since G does not contain elements of order 24, we may assume that k = 2 or 4. Obviously, τ normalizes σ since otherwise we have two distinct cyclic groups of order 8 acting on a line with a common fixed point. It is easy to see that this is impossible. Since Aut(Z/8Z)) ∼ (Z/2Z)2 this implies that σ and τ commute. Thus = στ is of order 24 which is impossible. This shows that Aut(S) ∼ Z/8Z. = It is easy to see that the square of the conjugacy classes D4 , 2A1 +A2 is equal to A2 , the square of E6 (a2 ) is equal to 3A2 , and the square of A1 + A5 is equal to 2A2 . Also the cube of the conjugacy class E6 (a1 ) and the fourth power of E6 is equal to 3A2 . Since surfaces with automorphism of order 3 from the conjugacy classes 3A2 , A2 , and also with an automorphisms of order 5 and 8 have been already classified we may assume that Aut(S) does not contain elements of order 5,8,9,12. By the previous analysis we may assume that any element of order 3 belongs to the conjugacy class 2A2 , and elements of order 6 to the conjugacy class E6 (a2 ) or A1 + A5 . Assume Aut(S) contains an element σ from conjugacy class 2A2 . Then the surface is σ-equivariantly isomorphic to the surface from case (v) of Theorem 9.6.9. t3 + t3 + t2 t3 t0 + f3 (t0 , t1 ) = 0. 2 3 We can reduce it to the form t3 + t3 + t2 t3 (t0 + at1 ) + t3 + t3 = 0. 2 3 0 1 (9.51)

9.6. AUTOMORPHISMS OF CUBIC SURFACES

371

The fixed points of σ are the points qi = (ai , bi , 0, 0), where f3 (ai , bi ) = 0. Observe that we have 3 involutions σi , i = 0, 1, 2, defined by
i 2i (t0 , t1 , t2 , t3 ) → (t0 , t1 , ζ3 t3 , ζ3 t2 ). i The set of fixed points of σi is the nonsingular plane section t3 = ζi t2 and an isolated i fixed point pi = (0, 0, 1, −ζ3 ). Thus each σi belongs to the conjugacy class 4A1 . The i 2i point pi is an Eckardt point in the tritangent plane t0 − ζ3 t2 − ζ3 t3 = 0. Notice that they lie on the line t0 = t1 = 0. This line is uniquely determined by σ, it is spanned by isolated fixed point of σ in P3 . It is immediately checked that σi σj = (σj σi )−1 = σ i+2j for i = j. This implies that the group G1 generated by σ, σ0 , σ1 , σ2 is isomorphic to S3 . The element σ belongs to a unique such group determined by the line . The elements of order 2 in G1 correspond to the Eckardt points on l. Suppose one of the fixed point of σ, say q1 , is an Eckardt point. A straightforward computation shows that this happens only if in equation (9.51) b3 = −1. Also it shows that only one of the fixed points could be an Eckardt point. Let P3 be the 3-Sylow subgroup of G = Aut(S). It is a cyclic group of order 3. Since |Aut(S)| = 2a 3b , the Sylow Theorems gives that the number of 3-Sylow subgroups divides 2a 3b and ≡ 1 mod 3. This shows that this number is equal to 22k . If k > 1, then G is contained in a maximal subgroups of W (E6 ) of order divisible by 24 . This is a group isomorphic to 24 : S5 or S6 × 2. In the first case G stabilizes a line in S, and then any element of order 3 has 2 fixed points on this line. But, as we saw in above the fixed points of an element of order 3 do not lie on a line contained in the surface. In the second case, we use that any subgroup of S6 containing 16 subgroups of order 3 must coincide with S6 . Certainly it is impossible. Thus k = 0 or 1. If k = 0, then G has a unique cyclic subgroup σ of order 3. So, either Aut(S) = G1 ∼ S3 , or G contains an involution τ ∈ G1 . If G = S3 , the natural homomorphism = G/ σ → Aut( σ ) ∼ Z/2Z has a non-trivial kernel of order 2a−1 . Let τ be an = element of order 2 from the kernel. Since it commutes with σ, it leaves invariant the set of 3 collinear fixed points of σ. Thus it fixes a = 1 or 3 fixed point of σ. If τ is of type 2A1 , then it has 5 isolated fixed points. The group σ leaves this set invariant and has 2 or 5 fixed points in this set. This shows that τ must be of type 4A1 and hence its isolated fixed point is one of the fixed points of σ which is an Eckardt point. As we have observed earlier, there could be only one such point. Hence there is only one additional element of order 2. The line joining this point with an Eckardt point pi must be contained in S, since otherwise, by Proposition 9.6.2 we have a third Eckardt point on this line. Thus τ commutes with any involution σi in G1 . Hence Aut(S) ∼ S3 × 2. The involution τ fixes the isolated fixed points of each σi . This = shows that in the one-dimensional subspace x0 = x1 = 0, it has 3 fixed points. This implies that it is the identity in this subspace. Thus τ acts nontrivially only on the variables t0 , t1 . This implies that b = 1 and the equation of S is of type VI. It is easy to see that automorphism of order 6 in G belongs to the conjugacy class A1 + A5 . If k = 1, G is isomorphic to a transitive subgroup of S4 which contains an S3 . It must be isomorphic to S4 . Each subgroup of Aut(S) isomorphic to S3 defines a line with 3 Eckardt points. Since any two such subgroups have a common element of order 2, each line intersects other 3 lines at one point. This shows that the four lines are

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coplanar and form a complete quadrangle in this plane. Also, since each of the three diagonals di has only two Eckardt points on it, we see that each diagonal is contained in the surface. Now choose coordinates such that the plane of the quadrangle has equation t0 = 0 and the diagonals have the equations t0 = ti = 0. The equation of the surface must now look as follows. at3 + t2 f1 (t1 , t2 , t3 ) + t0 f2 (t1 , t2 , t3 ) + ct1 t2 t3 = 0. 0 0 The group Aut(S) leaves the quadrangle invariant and hence acts by permuting the coordinates t1 , t2 , t3 and multiplying them by ±1. This easily implies that the equation can be reduced to the form of type V. The surface with automorphism group isomorphic to S3 has equation of type VIII. Assume that Aut(S) contains an element σ from conjugacy class 2A1 . Then the equation of the surface looks like at0 t2 t3 + t1 (t2 + t2 + bt2 t3 ) + t3 + t3 = 0. 2 3 0 1 It exhibits an obvious symmetry of order 3 defined by
2 [t0 , t1 , t2 , t3 ] → [t0 , t1 , ζ2 t2 , ζ2 t3 ].

Thus we are in one of the above cases. Suppose Aut(S) contains an element σ of order 4. If σ belongs to the conjugacy class 4B, then σ 2 belongs to 2A1 and hence this case has been already considered. If σ belongs to 4A1 then the equation of the surface looks like t2 t2 + t3 + at3 + t2 (t0 + t1 ) = 0. 3 0 1 2 Here we have to assume that the surface is not isomorphic to the surface of type VII. It follows from the proof of the next corollary that in all previous cases, except type VII, the automorphism group is generated by involutions of type 4A1 . Thus our surface cannot be reduced to one of the previous cases. Finally it remains to consider the case when only involutions of type 4A1 are present. Suppose we have 2 such involution. They define two Eckardt points p1 and p2 . In order the involution commute the line joining the two points must be contained in S. Suppose we have a third involution defining a third Eckardt point p3 . Then we have a tritangent plane formed by the lines pi , pj . Obviously it must coincide with each tritangent plane corresponding to the Eckardt points pi . This contradiction shows that we can have at most 2 commuting involutions. This gives the last two cases of our theorem. The condition that there is only one involution of type 4A1 is that the line l1 (t0 , t1 , t2 ) = 0 does not pass through a flex point of f3 (t0 , t1 , t2 ) = 0. Corollary 9.6.11. Let Aut(S)o be the subgroup of Aut(S) generated by involutions of type 4A1 . Then Aut(S)o is a normal subgroup of Aut(S) such that the quotient group is either trivial or a cyclic group of order 2 or 4. The order is 4 could occur only for the surface of type VII. The order 2 occurs only for surfaces of type X.

EXERCISES

373

Proof. We do it case by case. For surfaces of type I, the group Aut(S) is generated by transformations of type [t0 , t1 , t2 , t3 ] → [t0 , t1 , t3 , t2 ], where 3 = 1. It is easy to see that it is an involution of type 4A1 corresponding to the Eckardt point (0, 0, 1, − ). For surfaces of type II given by equation (9.50), the group Aut(S) is generated by transpositions of coordinates. They correspond to involutions of type 4A1 associated with Eckardt points of type [1, −1, 0, 0, 0]. In the case of surfaces of type III, we use that a line in P3 joining 2 Eckardt points contains the third Eckardt point. Thus any such line generate a subgroup isomorphic to S3 . We have 12 lines which contain 9 flex points. They are the projections of these line in P2 from the center of projection (1, 0, 0, 0). One can show that the group generated by these 12 subgroups must coincide with the whole group. The remaining cases follow from the proof of the theorem.

Exercises
10.1 Let Y ⊂ P4 be the image of P2 under a rational map given by the linear system of conics through a fixed point p. (i) Show that Y is a surface of degree 3 and its projection from a general point O in P4 is a non-normal cubic surface in P3 of type (i) from Theorem 9.3.1. (ii) Show that the projection of Y from a point O lying in the plane spanned by the image of the exceptional curve of the blow-up of P2 at p and the image of a line through p is a non-normal cubic surface of type (ii) from Theorem 9.3.1. (iii) Show that any non-normal cubic surface in P3 which is not a cone can be obtained in this way. 10.2 Show that the dual of the 4-nodal cubic surfrace is isomorphic to the quartic surface given by the equation √ √ √ √ t0 + t1 + t2 + t3 = 0. 10.3 Let T : (x, y) → (x−1 , y −1 ) be the standard Cremona transformation. Show that T extends to a biregular automorphism σ of a weak Del Pezzo surface X of degree 6 and the orbit space X/(σ) is isomorphic to a 4-nodal cubic surface. 10.4 Show that a cubic surface can be obtained as the blow-up of 5 points on P1 × P1 . Find the conditions on the 5 points such that the blow-up is isomorphic to a nonsingular cubic surface. 10.5 Compute the number of m-tuples of skew lines on a nonsingular surface for m = 2, 3, 4, 5. 10.6 Suppose a quadric intersects a cubic along the union of three conics. Show that the three planes defined by the conics pass through three lines in a tritangent plane 10.7 Let Γ and Γ be two twisted cibics in P3 containing a common point P . For a general plane Π through P let Π ∩ Γ = {P, p1 , p2 }, Π ∩ Γ = {p, p1 , p2 } and f (P ) = p1 , p1 ∩ p1 , p2 . ˇ Consider the set of planes through P as a hyperplane H in the dual space P3 . Show that the 3 image of the rational map H− → P , Π → f (Π) is a nonsingular cubic surface and every such cubic surface can be obtained in this way

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10.8 Show that the linear system of quadrics in P3 spanned by quadrics which contain a degree 3 rational curve on a nonsingular cubic surface S can be spanned by the quadrics defined by the minors of a matrix defining a determinantal reprfesentation of S. 10.9 Show that the linear system of cubic surfaces in P3 containing 3 skew lines defines a birational map from P3 to P1 × P1 × P1 . 10.10 Show that non-normal cubic surfaces are scrolls, i.e. contain a one-dimensional family of lines. 10.11 Show that all singular surfaces of type V II, X, XI, XIII − XXI are isomorphic and there are two non-isomorphic surfaces of type XII. 10.12 Prove that the linear system of cubic surfaces in P3 containing three skew lines defines a birational map from P3 to P7 whose image is equal to the Segre variety (P1 )3 . 10.13 Compute the number of determinantal representations of a singular cubic surface. 10.14 Find determinantal representations of an irreducible non-normal cubic surface. 10.15 Let be a line on a cubic surface with canonical singularities and E be its proper inverse transform on the corresponding weak Del Pezzo surface X. Let N be the sublattice of Pic(X) spanned by irreducible components of exceptional divisors of π : X → S. Define the multiplicity of by m(l) = #{σ ∈ O(Pic(X)) : σ(E) − E ∈ N } . #{σ ∈ O(Pic(X)) : σ(E) = E}

Show that the sum of the multiplicities is always equal to 27. 10.16 Show that every line on a nonsingular cubic surface S is tangent to the Hessian surface H(S) at two points. Also show that the 24 points obtained in this way from a double-six of lines are the intersection points of S and the Schur quadric associated to the double-six. 10.17 ([77]) Consider a Cayley-Salmon equation l1 l2 l3 − l1 l2 l3 = 0 of a nonsingular cubic surface. (i) Show that the six linear polynomials li , li satisfy the following linear equations
3 X j=1

aij lj =

3 X j=1

aij lj = 0, i = 1, 2, 3,

where

3 X i=1

aij = 0, j = 1, 2, 3,

ai1 ai2 ai3 = ai1 ai2 ai3 , i = 1, 2, 3.

(ii) Show that for each i = 1, 2, 3 the nine planes aij li − aij lj = 0, i, j = 1, 2, 3 intersect by three in six lines. The 18 lines obtained in this way form three double-sixes associated to the pair of conjugate triads defined by the Cayley-Salmon equation. (iii) Show that the Schur quadrics defined by the three double-sixes can be defined by the equations 3 3 X X 2 a2j a3j lj − a2j a3j lj 2 = 0,
j=1 3 X j=1 2 a1j a3j lj − j=1 3 X j=1

a1j a3j lj 2 = 0,

EXERCISES
3 X j=1 2 a1j a2j lj − 3 X j=1

375
a1j a2j lj 2 = 0.

10.18 ([78]) Prove the following theorem of Schl¨ fli: Given five skew lines in P3 and a line a intersecting them all, there exists a unique cubic surface that contains a double six including the seven lines. P 3 P 10.19 Consider the Cremona hexahedral equations xi = xi = 0 and xi − xj = 0. Show that these equations define a 4-nodal cubic surface. 10.20 Show that the pull-back of a bracket-function (ijk) under the Veronese map is equal to (ij)(jk)(ik). 10.21 Show that the condition that 6 points in P2 lie on a conic is (134)(156)(235)(246) − (135)(146)(234)(256 = 0. 10.22 Let S be the cubic surface obtained by blowing up a semi-stable ordered point set (p1 , . . . , p6 ) in P2 . Use the Clebsch transference principle to give the following interpretation of the mor6 6 phism Ψ : P2 → P1 ∼ S3 ⊂ P5 . = (i) Show that the projection of the set (p1 , . . . , p6 ) to P1 with center at a general point x ∈ 6 P2 is a set of distinct points in P1 whose orbit in P1 depends only on the image of 6 (p1 , . . . , p6 ) in P2 .
6 (ii) Show that the projection map extends to a morphism from S to Pa that coincides with the map Ψ.

(iii) Show that the image (q1 , . . . , q6 ) of a points set (p1 , . . . , p6 ) is a set such that the pairs (q1 , q2 ), (q3 , q4 ), (q5 , q6 ) are orbits of an involution of P1 if and only if the triples (p1 , p3 , p5 ), (p1 , p4 , p6 ), (p2 , p3 , p6 ) and (p2 , p4 , p5 ) are on lines which form a quadrilateral. 10.22 Show that the Segre cubic primal is isomorphic to a tangent hyperplane section of the cubic fourfold with 9 lines given by the equation xyz − uvw = 0 (Perazzo primal [189], [10].

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CHAPTER 9. CUBIC SURFACES

Chapter 10

Geometry of Lines
10.1 Grassmanians of lines
I = {v ⊗ w ∈ V ⊗ V, v, w ∈ V } of indecomposable vectors as well as I + = {v ⊗ w + v ⊗ w, v, w ∈ V } and I − = {v ⊗ w − w ⊗ v, v, w ∈ V }. The elements v ⊗ w + v ⊗ w of I + are denoted by vw, the elements v ⊗ w − v ⊗ w of I − are denoted by v ∧ w. The subsets I + and I − respectively generate the subspaces
2

Let V be a complex vector space of dimension n + 1, in V ⊗ V we have the subset

S 2 V and

V

of symmetric and antisymmetric tensors. Both I + and I − are affine cones with vertex at the origin in the affine space V ⊗ V andthe same is true for I. The corresponding projectivizations define some well known projective varieties: I ⊂ P(V ⊗ V ) is the Segre variety s2 (Pn × P2 ). |I + | ⊂ P(S 2 V ) is the quotient of Pn × Pn under the natural involution sending (x, y) to (y, x). It is equal to the image of the diagonal of Pn × Pn under the Segre map. Finally
2

|I − | ⊂ P( 377

V)

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CHAPTER 10. GEOMETRY OF LINES

is the Grassmann variety G(2, V ) of 2-dimensional linear subspaces of V , or, equivalently, lines in Pn = P(E). Recall that this description is obtained by assigning to any nonzero element v ∧ w ∈ I − the linear subspace spanned by the vectors v, w. Conversely any spanning set v, w of a 2-dimensional subspace of V defines a unique element v ∧ w in |I − |. Note that
2

V ⊗ V = S2V ⊕

V

and hence G(2, V ) is equal to the image of the projection of the Segre variety to 2 P( V ) with center at |I + |. Similarly, |I + | is the projected Segre variety with center at G(2, V ). 2 To obtain a system of equations defining G(2, V ) observe that V is canonically ∗ ∗ isomorphic to the space of alternating bilinear forms ω : V × V → C and that
2

I − \ {0} = {ω ∈

V : rank(ω) = 2}.
2

Therefore, after fixing a basis in V , we can identify V with the space of (n + 1) × (n + 1) skew-symmetric matricesA = (pij )and G(2, V ) with the locus of matrices of rank 2 (there are no nonzero skew-symmetric matrices of rank ≤ 1). The entries 2 pij , i < j, define projective coordinateson P( V ), the so called Pl¨ cker coordinates. u In particular, G(2, V ) is the zero set of the 4 × 4 pfaffian s of A. In fact, a stronger assertion is true. Proposition 10.1.1. The homogeneous ideal of G(2, V ) ⊂ P( V ) is generated by 4 × 4 pfaffians of a general skew-symmetric matrix of size n + 1. Remark 10.1.1. Another way to look at a point L of G(2, V ) as a 2 × (n + 1) matrix X whose rows are the coordinate vectors of a basis of L. A different choice of a basis changes X to CX, where C is an invertible matrix of size 2. Thus G(2, V ) can be viewed as the orbit space Mat2,n+1 /GL(2) of the set of rank 2 matrices of size 2×n+1 with respect to the action of GL(2) by left multiplications. The Pl¨ cker coordinates can u be identified with n+1 maximal minors of X. They are SL(2) invariant polynomials 2 on the space Mat2,n+1 . The First Fundamental Theorem of invariant theory asserts that they generate the algebra of SL(2)-invariant polynomials. The Second Fundamental Theorem asserts that the relations between the maximal minors are generated by the pfaffian s. Proposition 10.1.2. G(2, V ) is a smooth, irreducible variety of dimension 2(n − 1). Proof. This follows from the fact that G(2, V ) can be covered by open affine subsets isomorphic to affine space A2(n−1) . If we choose to represent points of G(2, V ) as skew-symmetric matrices, then the open subsets are matrices with one entry pij not equal to zero. All other 2(n − 1) entries above the diagonal could be arbitrary. If we choose to represent points of G(2, V ) as 2 × (n + 1) matrices, the open subset corresponds to matrices with one minor det Mij not equal to zero. Multiplying the −1 matrix by Mij on the left we may assume that Mij = I2 is the identity matrix. Then the remaining 2(n − 1) entries could be arbitrary.
2

10.1. GRASSMANIANS OF LINES

379

Remark 10.1.2. A reader familiar with the general notion of a Hilbert scheme should observe that G(2, V ) is the Hilbert scheme of subschemes of P(E) with Hilbert polynomial P (t) = t + 1. Since the normal bundle Nl of any line is isomorphic to OP1 (1)n−1 , we see that H 1 (l, Nl ) = 0, H 0 (l, Nl ) = 2(n − 1). The first equality implies that the Hilbert scheme is smooth at l, the second one gives the dimension of the Zariski tangent space.

10.1.1

The tangent and the secant varieties
n+1 4

By 10.1.1 the ideal of G(2, V ) is generated by  pij pkl − pik pjl + pil pjk

0  −pij = Pfaf  −pik −pil

quadratic forms of rank 6:  pij pik pil 0 pjk pjl   −pjk 0 pkl  −pjl −pkl 0

with 1 ≤ i < j < k < l ≤ n + 1. If n = 3 then G is the Klein quadric V (p12 p34 − p13 p24 + p14 p23 ) ⊂ P5 defining the Grassmannian of lines in P3 . A point p ∈ P( vector space generated by an alternating bilinear form wp : V ∗ × V ∗ → C. Since wp is alternating, the rank of wp is always even. Definition 10.1. The rank of p is the rank of wp . It is clear that wp has rank ≤ 2k iff there exist matrices s1 , . . . , sk of rank 2 such that wp = s1 + . . . + sk . In other words p has rank ≤ 2k if and only ifp ∈ S, where S is a space of dimension k − 1 which is at least k-secant to G. This gives the following. Proposition 10.1.3. The variety
2 2

V ) is a 1-dimensional

Gk =: {p ∈ P(

V ) : p has rank ≤ 2k + 2}

is the k-secant variety Seck (G) of G = G(2, V ). Let t=[ n−3 ], 2
2

then t is the maximal number k such that Seck G = P( V ). So the Pl¨ cker space is u stratified by the rank of its points and the strata are the following: Pn \ Sect (G), Sect (G) \ Sect−1 , . . . , Sec1 \ G, G. (10.1)

380

CHAPTER 10. GEOMETRY OF LINES

It follows from the previous remarks that Seck (G) \ Seck−1 (G) is the orbit of a matrix of rank 2k + 2 and size n + 1 under the action of GL(n + 1). Therefore dim Seck (G) = dim GL(n + 1)/Hk , where Hk is the stabilizer of a skew symmetric matrix of rank 2k + 2 (e.g. with the standard symplectic matrix J2k+2 in the left upper corner and zero elsewhere). An easy computation gives the following. Proposition 10.1.4. Let 0 ≤ k ≤ t, then dk = dim Seck (G) = (k + 1)(2n − 2k − 1) − 1. Let X ⊂ Pr be a reduced and non degenerate variety: the k-th defect of X can be defined as δk (X) = min (k + 1) dim X + k, r − dim Seck (X), which is the difference between the expected dimension of the k-secant variety of X and the effective one. We say that X is k-defective if Seck (X) is a proper subvariety and δk (X) > 0. Example 10.1.1. Let n = 2t + 3, then Sect (G) ⊂ P( V ) is the pfaffian hypersurface 2 of degree t + 2 in P( V ) parametrizing singular skew-symmetric matrices (pij ) of size 2t + 4. The expected dimension of Sect (G) is equal to 4t2 + 8t + 5 which is larger 2 than dim P( V ) = 2t+4 − 1. Thus dt (G) = dim Sect (G) + 1 and δt (G) = 1. 2 In the special case n = 5, we have t = 2 and dim G = 8. Recall that a nondegenerate subvariety X ⊂ Pr with dim X = [ 2r ] − 1 is called a Severi-Zak variety if 3 Sec1 (X) = Pr . There are four non-isomorphic Severi-Zak varieties and G(2, 6) is one of them. The other three are the Veronese surface in P5 , the Segre variety s2 (P2 × P2 ) in P8 and the E6 -variety of dimension 16 in P26 .
2

10.1.2

The incidence variety
Z = {(x, l) ∈ Pn × G : x ∈ l}

Consider the incidence correspondence:

and the corresponding projections: p : Z → P3 , q : Z → G.

The fibre of p over a point x is isomorphic to the set of lines containing the point x. If x = [v] for some v ∈ V , then a line containing p corresponds to a point in P(V /Cv). Thus the fibres of p are isomorphic to Pn−1 . The fibre of q over a line l ∈ G is isomorphic to l ∼ P1 . = Let us see that both projections are the structure projections of the corresponding projective bundles. Recall that we identify a vector space V over a field K with K-points of the affine space Spec(S • V ∗ ), where S • denotes the graded symmetric algebra of V . The projectivization P(E) is the set of points of the projective scheme

10.1. GRASSMANIANS OF LINES

381

Proj(S • V ∗ ). Similarly, any locally free sheaf E over a scheme X defines the vector bundle E = V(E) = Spec(S • E ∗ ) and the projective bundle P(E) = Proj(S • E ∗ ). The sheaf of local sections of E is isomorphic to E and the set of rational points of the fibre Ex over a point x ∈ X with residue field k(x) can be identified with the linear space E ⊗OX k(x). We will identify locally free sheaves with the corresponding vector bundles. Thus an exact sequence of vector bundles means an exact sequence of the corresponding sheaves. Note that our notation is dual to one used in [134], where P(E) = Proj(S • E). Let π : P(E) → X be the canonical structure morphism of a X-scheme. There