Syzygies of curves via K3 surfaces Michael Kemeny - PowerPoint PPT Presentation

Syzygies of curves via K3 surfaces Michael Kemeny Humboldt-Universit¨ at zu Berlin August 28, 2015. Haeundae, Busan, Korea

Notation ◮ L denotes a globally generated line bundle on a smooth, projective curve C . ◮ For M any line bundle K i , j ( C , M ; L ) denotes the middle cohomology in the sequence i +1 i � � H 0 ( C , L ) ⊗ H 0 ( C , ( j − 1) L + M ) → H 0 ( C , L ) ⊗ H 0 ( C , jL + M ) i − 1 � H 0 ( C , L ) ⊗ H 0 ( C , ( j − 1) L + M ) → ◮ K i , j ( C , L ) is short for K i , j ( C , O C ; L ) ◮ The line bundle L is said to satisfy property ( N p ) if K i , j ( C , L ) = 0 for i ≤ p , j ≥ 2.

p -very ampleness A line bundle L is called p -very ample iff for any effective divisor D of degree p + 1 ev : H 0 ( C , L ) → H 0 ( D , L | D ) is surjective. Equivalently, L is not p v. a. iff φ L : C → P r admits a ( p + 1) secant p − 1 plane. Thus L is 0-very ample ⇔ L is globally generated. Thus L is 1-very ample ⇔ L is very ample.

Secant Conjecture Conjecture (Secant Conjecture, Green–Lazarsfeld, 86) Let L be a globally generated line bundle of degree d on a curve C of genus g such that d ≥ 2 g + p + 1 − 2 h 1 ( C , L ) − Cliff ( C ) . Then L fails property N p if and only if L is not p + 1 -very ample. The line bundle L is not p + 1-very ample iff φ L admits a ( p + 2)-secant p plane.

Special case: p = 0 Firstly the direction “ L not ( p + 1) v.a. = ⇒ L fails N p ” is straightforward and was shown by GL when they formulated the conjecture. So the conjecture concerns the other direction. For L very ample, condition N 0 is equivalent to φ L : C → P r being projectively normal . In this case the Secant Conjecture becomes: Theorem (Green–Lazarsfeld ’86) Let L be a very ample line bundle on C with deg( L ) ≥ 2 g + 1 − 2 h 1 ( C , L ) − Cliff ( C ) . Then φ L : C → P r is projectively normal.

Special case L = ω C We next consider the special case L = ω C . Assume p < Cliff( C ). Then ω C is p + 1-very ample. Indeed, otherwise ∃ D ∈ C p +2 such that H 0 ( ω C ) → H 0 ( D , ω C | D ) is non-surjective. Equivalently H 0 ( D , ω C | D ) → H 1 ( C , ω C ( − D )) is nonzero, so the surjection H 1 ( C , ω C ( − D )) ։ H 1 ( C , ω C ) is non-injective. Thus h 0 ( C , O ( D )) = h 1 ( C , ω C ( − D )) ≥ 2 so Cliff( C ) ≤ p which is a contradiction.

Green’s conjecture Thus the Secant Conjecture becomes, in case L = ω C : Conjecture (Green’s Conjecture, 1984) The canonical bundle ω C satisfies ( N p ) for p < Cliff ( C ) .

K3 Surfaces A K3 surface is a smooth, projective surface with h 1 ( X , O X ) = 0 and ω X ≃ O X . By the adjunction formula, if C ⊆ X is smooth curve then O ( C ) | C ≃ ω C . K3 surfaces enter the picture via the following proposition: Theorem (Lefschetz Thm. Green 1984, Farkas, K.- 2015) Let X be a K3 surface and let L , H be line bundles on X. Assume H is effective and base point free and either: 1. L the trivial line bundle O X , or 2. ( L · H ) > 0 . In addition assume H 1 ( X , qH − L ) = 0 for all q ≥ 0 . Then, for any smooth D ∈ | H | , restriction gives an isomorphism K p , q ( X , − L ; H ) ≃ K p , q ( D , − L ; ω D ) .

Proof of Lefschetz Thm Proof. By the assumptions we have, for each q ∈ Z , a short exact sequence s 0 → H 0 ( X , ( q − 1) H − L ) → H 0 ( X , qH − L ) → H 0 ( D , q ω D − L | D ) → 0 , where the first map is multiplication by the section s ∈ H 0 ( X , H ) defining D . Thus we have a s.e.s of graded Sym H 0 ( X , H ) modules � � � H 0 ( X , ( q − 1) H − L ) → H 0 ( X , qH − L ) → H 0 ( D , q ω D − L | D ) → 0 . 0 → q q q Taking Koszul cohomology leads to the l.e.s s → K p , q ( X , − L , H ) → K p , q ( B , H 0 ( X , H )) → K p , q − 1 ( X , − L , H ) s → K p − 1 , q ( X , − L , H ) → where B is the graded Sym H 0 ( X , H ) module � q H 0 ( D , q ω D − L | D ).

Proof Cont. Proof (Cont.) But multiplication by s induces the zero map on cohomology so we have K p , q ( B , H 0 ( X , H )) ≃ K p , q ( X , − L , H ) ⊕ K p − 1 , q ( X , − L , H ) . Next we have s 0 → H 0 ( X , O X ) → H 0 ( X , H ) → H 0 ( D , K D ) → 0 so choose a splitting H 0 ( X , H ) ≃ H 0 ( D , K D ) ⊕ C { s } . This gives p p p − 1 � H 0 ( X , H ) ≃ � H 0 ( D , K D ) ⊕ � H 0 ( D , K D ) . A diagram chase then gives K p , q ( B , H 0 ( X , H )) ≃ K p , q ( D , − L , ω D ) ⊕ K p − 1 , q ( D , − L , ω D ). Induction on p now gives the claim.

Consequences The Lefschetz Theorem has some interesting consequences for sections of K3 surfaces. For example, if H is very ample then it implies K p , q ( D 1 , ω D 1 ) ≃ K p , q ( X , H ) ≃ K p , q ( D 2 , ω D 1 ) for any two smooth sections D 1 , D 2 ∈ | H | . Thus, Green’s conjecture would imply that Cliff( D ) is constant for all D ∈ | H | . This was verified by Green and Lazarfseld in 87. In fact, one can always compute the Clifford index of sections of a very ample line bundle in terms of the Picard lattice of a K3 surface.

Lazarsfeld’s conjecture For our purposes, we will only need one special case: Theorem (Lazarsfeld, 86) Let H be a line bundle on a K3 surface, and assume that H is base point free and ( H ) 2 > 0 . Assume there is no decomposition H = L 1 ⊗ L 2 for line bundles L i with h 0 ( L i ) ≥ 2 for i = 1 , 2 . Then a general smooth curve C ∈ | H | is Brill–Noether–Petri general. In particular, Cliff( C ) = ⌊ g ( C ) − 1 ⌋ in the conclusion of the theorem above. 2

Green’s conjecture for curves on K3 The results on the previous slides suggest that it might be a good idea to try and prove Green’s conjecture for a curve C on a K3 surface. This was done, by C. Voisin in 2002 and 2005, in the case Pic( X ) ≃ Z [ C ] and by Aprodu–Farkas (2011) in general: Theorem (Voisin, Aprodu–Farkas) Green’s Conjecture holds for a smooth curve C ⊆ X lying on a K3 surface X. In particular we deduce by semicontinuity the fact that Green’s conjecture holds for general curves. But we also get even more, e.g. by applying the Thm to K3s with Picard rank two generated by classes C and E with ( E ) 2 = 0, ( C · E ) = k one sees that Green’s conjecture holds for general curves of each gonality (warning: this is not how this fact was first proven!).

Kernel Bundles We will need the kernel bundle description of Koszul cohomology. Let L be a globally generated line bundle on a variety. Set M L := Ker( H 0 ( X , L ) ⊗ O X ։ X ) , where the morphism is evaluation of sections. For any second line bundle F (or coherent sheaf), we have K p , q ( X , F ; L ) p +1 p � H 0 ( L ) ⊗ H 0 ( X , F ⊗ L q − 1 ) → H 0 ( � M L ⊗ F ⊗ L q )) ≃ coker( p +1 p +1 ≃ ker( H 1 ( � M L ⊗ F ⊗ L q − 1 )) → � H 0 ( L ) ⊗ H 1 ( F ⊗ L q − 1 ))

Kernel Bundles: Hyperelliptic curves In particular, if either H 0 ( � p M L ⊗ F ⊗ L q )) = 0 or H 1 ( � p +1 M L ⊗ F ⊗ L q − 1 )) = 0 then K p , q ( X , F ; L ) = 0. One more fact will be useful. If C is a hyperelliptic curve and E ∈ Pic 2 ( C ) denotes the g 1 2 then ω C ≃ ( g − 1) E . Further, the canonical morphism φ ω C : C → P g − 1 is the composite of φ E : C → P 1 with the Veronese v : P 1 ֒ → P g − 1 .

Kernel Bundles: Hyperelliptic curves The Euler sequence can be written 0 → Ω P g − 1 (1) → H 0 ( O (1)) ⊗ O P g − 1 → O (1) → 0 . Thus M ω C ≃ φ ∗ E (Ω P g − 1 (1)) . Further, v ∗ (Ω P g − 1 (1)) ≃ O P 1 ( − 1) ⊕ g − 1 . So we have M ω C ≃ O C ( − E ) ⊕ g − 1 .

Generic Secant Conjecture We can now prove our first main result. Theorem (Farkas, K.-) The Secant Conjecture holds for general curve C of genus g with a general line bundle L of degree d on C. The word general means that the property holds on a dense open subset of the appropriate moduli space.

Proof of Generic Secant Conj. Proof. We first state some simple reductions. In the case h 1 ( C , L ) ≥ 1, the claim reduces to generic Green’s Conj, by an straightforward argument of Koh–Stillman from 1989. So we may assume h 1 ( C , L ) = 0. Assume L is p + 1 very ample with h 1 ( L ) = 0 and d ≥ 2 g + p + 1 − Cliff( C ). We need to show L satisfies ( N p ). In fact, L is projectively normal under this bound on d (and h 1 ( L ) = 0), so by a result of GL we only need to prove K p , 2 ( C , L ) = 0. As h 1 ( C , L ) = 0, then L fails to be p + 1 very ample iff there exists D ∈ C p +2 such that h 0 ( C , L ( − D )) ≥ d − g − p . By Riemann–Roch this is equivalent to h 1 ( C , L ( − D )) ≥ 1 or − ( L − D − K C ) effective.

Proof of Secant Conj. Proof cont. In other words, if L is p + 1 very ample then L − K C / ∈ C p +2 − C 2 g − d + p . It is an easy exercise to prove that if a + b ≥ g then the difference map C a × C b → Pic a − b ( C ) is surjective. Thus we must have ( p + 2) + (2 g − d + p ) ≤ g − 1 or d ≥ g + 2 p + 3 (1) in addition to d ≥ 2 g + p + 1 − Cliff( C ) . (2)

Proof of Secant Conj. Proof cont. For simplicity, assume g = 2 i + 1 is odd, the even case is similar. It is easy to see h 1 ( C , L ) = 0 and K p , 2 ( C , L ) = 0 implies K p − 1 , 2 ( C , L ( − x )) = 0 for any x ∈ C with L ( − x ) globally generated. Using this we reduce to the case p ≥ i − 1. In this case inequalities (1),(2) reduce to d ≥ 2 p + 2 i + 4. Using the projective normality result (from two slides back) we see that it suffices to assume d = 2 p + 2 i + 4 .