The generic Green-Lazarsfeld secant conjecture
aa r X i v : . [ m a t h . AG ] M a r THE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE
GAVRIL FARKAS AND MICHAEL KEMENY
Abstract.
Using lattice theory on special K Introduction
For a smooth curve C of genus g and a very ample line bundle L ∈ Pic d ( C ), the Koszulcohomology groups K p,q ( C, L ) of p -th syzygies of weight q of the embedded curve φ L : C → P r are obtained from a minimal free resolution of the graded S := Sym H ( C, L )-moduleΓ C ( L ) := M n ≥ H (cid:0) C, L ⊗ n (cid:1) . The graded Betti diagram of (
C, L ) is the table obtained by placing in the p th column and q throw the graded Betti number b p,q := dim K p,q ( C, L ). The resolution is said to be natural ifat most one Betti number along each diagonal is non-zero, which amounts to the statement b p, · b p +1 , = 0, for all p . In two landmark papers, Voisin has shown that the resolution of ageneral canonical curve of each genus is natural, see [V1], [V2].We now fix a Prym curve of genus g , that is, a pair [ C, η ], where C is a smooth curve of genus g and η = O C is a 2-torsion point, η ⊗ = O C . Prym curves of genus g form an irreducible modulispace R g whose birational geometry is discussed in [FL] and [FV]. It has been conjectured in[CEFS] that the resolution of a general Prym-canonical curve φ K C ⊗ η : C → P g − is natural. As explained in [CEFS], this statement, which came to be known as the Prym-GreenConjecture , reduces to one single vanishing statement in even genus, namely K g − , (cid:0) C, K C ⊗ η (cid:1) = 0 , and to the following two vanishing statements in odd genus:(1) K g − , (cid:0) C, K C ⊗ η (cid:1) = 0 and K g − , (cid:0) C, K C ⊗ η (cid:1) = 0 . The even genus case of the Prym-Green Conjecture is divisorial in moduli, that is, the locus Z g := n [ C, η ] ∈ R g : K g − , (cid:0) C, K C ⊗ η (cid:1) = 0 o is the degeneracy locus of a morphism betweenvector bundles of the same rank over R g . The conjecture has been verified computationally in[CEFS] for all g ≤
18, with the exceptions of g = 8 ,
16. The Prym-Green Conjecture is expectedto fail for genera g = 2 n , with n ≥
3, for reasons which are mysterious. We prove the following:
Theorem 1.1.
The Prym-Green Conjecture holds for a general Prym curve of odd genus.
In particular, via Theorem 1.1, we completely determine the shape of the resolution of a gen-eral Prym-canonical curve C ⊂ P g − of odd genus g = 2 i + 5. Precisely, if S := C [ x , . . . , x g − ],then the Prym-canonical ideal I C ⊂ S has the following resolution: ←− I C ←− S ( − b , ←− · · · ←− S ( − i ) b i − , ←− S ( − i − b i, ⊕ S ( − i − b i, ←−←− S ( − i − b i +1 , ←− · · · ←− S ( − i − b i +2 , ←− , where b p, = p (2 i − p + 1)2 i + 3 (cid:18) i + 4 p + 1 (cid:19) for p ≤ i, and b p, = ( p + 1)(2 p − i + 1)2 i + 3 (cid:18) i + 4 p + 2 (cid:19) for p ≥ i. In particular, the resolution is natural but fails to be pure , precisely in the middle. In thissense, the resolution of a general Prym-canonical curve of odd genus has the same shape as thatof a general canonical curve of even genus [V1].The proof of Theorem 1.1 is by specialization to
Nikulin surfaces . A Nikulin surface [FV],[vGS] is a K X equipped with a double cover f : e X → X branched along eight disjointrational curves N , . . . , N . In particular, the sum of the exceptional curves is even, that is,there exists a class e ∈ Pic( X ) such that e ⊗ = O X ( N + · · · + N ). If C ⊂ X is a smooth curveof genus g disjoint from the curves N , . . . , N , then the restriction e C ∈ Pic ( C ) is a non-trivialpoint of order two, that is, [ C, e C ] ∈ R g . Theorem 1.2.
Let X be a general Nikulin surface endowed with a curve C ⊂ X of odd genus g ≥ , such that C · N j = 0 , for j = 1 , . . . , . Then K g − , (cid:0) C, K C ⊗ e C (cid:1) = 0 and K g − , (cid:0) C, K C ⊗ e C (cid:1) = 0 . In particular, the Prym-Green Conjecture holds generically on R g . The techniques developed for the Prym-Green Conjecture, in particular the use of K Secant Conjecture , predicting that the shape of the resolution ofa line bundle of sufficiently high degree is determined by its higher order ampleness properties:Precisely, if L is a globally generated degree d line bundle on a curve C of genus g such that(2) d ≥ g + p + 1 − h ( C, L ) − Cliff( C ) , then L fails property ( N p ) if and only if L is not ( p +1)-very ample, that is, the map φ L : C → P r induced by the linear series | L | embeds C with a ( p + 2)-secant p -plane. The case h ( C, L ) = 0of the Secant Conjecture easily reduces to the ordinary Green’s Conjecture, that is, to the case L = K C , see [KS]. If L carries a ( p + 2)-secant p -plane, it is straightforward to see [GL1] that K p, ( C, L ) = 0, hence the Secant Conjecture concerns the converse implication. The SecantConjecture is a refinement of Green’s result [G], asserting that every line bundle L ∈ Pic d ( C )with d ≥ g + p + 1 satisfies property ( N p ). Thus to study the Secant Conjecture for generalcurves, it suffices to consider the case of non-special line bundles L ∈ Pic d ( C ), when d ≤ g + p .Our first result answers completely this question in the case when both C and L are general: Theorem 1.3.
The Green-Lazarsfeld Secant Conjecture holds for a general curve C of genus g and a general line bundle L of degree d on C . For an integer d ≤ g + p and a non-special line bundle L ∈ Pic d ( C ), we consider the variety V p +1 p +2 ( L ) := (cid:8) D ∈ C p +2 : h ( C, L ( − D )) ≥ d − g − p (cid:9) of ( p + 2)-secants p -planes to C in the embedding φ L : C → P d − g . The line bundles possessingsuch a secant are those lying in a translate difference variety in the Jacobian, precisely(3) V p +1 p +2 ( L ) = ∅ ⇔ L − K C ∈ C p +2 − C g − d + p . HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 3
For a general curve C , a general line bundle L ∈ Pic d ( C ) verifies V p +1 p +2 ( L ) = ∅ if and only if(4) d ≥ g + 2 p + 3 . In particular, the Secant Conjecture (and Theorems 1.3 and 1.7) are vacuous if inequality (4) isnot fulfilled. Of particular importance is the divisorial case of the Secant Conjecture, when thevariety n [ C, L ] ∈ Pic dg : K p, ( C, L ) = 0 o is the degeneration locus of a morphism between vectorbundles of the same rank over the universal degree d universal Jacobian Pic dg . This divisorialcase can be recognized by requiring that both inequalities (2) and (4) be equalities. We find g = 2 i + 1 , d = 2 g = 4 i + 2 , p = i − . In this case, we establish the Secant Conjecture in its strongest form, that is, for all C and L : Theorem 1.4.
The Secant Conjecture holds for every smooth curve C of odd genus g and everyline bundle L ∈ Pic g ( C ) , that is, one has the equivalence K g − , ( C, L ) = 0 ⇔ Cliff( C ) < g −
12 or L − K C ∈ C g +12 − C g − . In the extremal case of the Secant Conjecture in even genus, that is, when deg( L ) = 2 g + 1,we prove the expected statement for general curves and arbitrary line bundles on them: Theorem 1.5.
The Secant Conjecture holds for a Brill–Noether–Petri general curve C of evengenus and every line bundle L ∈ Pic g +1 ( C ) , that is, K g − , ( C, L ) = 0 ⇔ L − K C ∈ C g +1 − C g − . Theorem 1.5 is proved in Section 6, by regarding this case as a limit of the divisorial situation,occurring for odd genus. The strategy for proving Theorem 1.4 is to interpret the G-L SecantConjecture as an equality of divisors on the moduli space M g, g of 2 g -pointed smooth curvesof genus g via the global Abel-Jacobi map M g, g → Pic gg . For g = 2 i + 1, denoting by π : M g, g → M g the morphism forgetting the marked points, we consider the Hurwitz divisor M g,i +1 := { [ C ] ∈ M g : W i +1 ( C ) = ∅} of curves with non-maximal Clifford index and itspull-back Hur := π ∗ ( M g,i +1 ). We introduce two further divisors on M g, g . Firstly, the locusdescribed by the condition that the embedding φ O C ( x + ··· + x g ) : C ֒ → P g have extra syzygies Syz := n [ C, x , . . . , x g ] ∈ M g, g : K i − , (cid:0) C, O C ( x + · · · + x g ) (cid:1) = 0 o . Theorem 1.3 implies that K i − , ( C, L ) = 0 for a general line bundle L of degree 2 g on a generalcurve C , that is, Syz is indeed a divisor on the moduli space. Secondly, we consider the locus ofpointed curves such that the image of φ O C ( x + ··· + x g ) has an ( i + 1)-secant ( i − Sec := n [ C, x , . . . , x g ] ∈ M g, g : O C ( x + · · · + x g ) ∈ K C + C i +1 − C i − o . Obviously
Sec is a divisor on M g, g and using [GL1], we know that the difference Syz − Sec is effective. For any curve C with maximal Clifford index, the cycles K C + C i +1 − C i − and (cid:8) L ∈ Pic g ( C ) : K i − , ( C, L ) = 0 (cid:9) have the same class in H ∗ (Pic g ( C ) , Q ). It follows that Syz = Sec + π ∗ ( D ) , where D is an effective divisor on M g . Via calculations in the Picard group of the moduli space M g, g , we shall establish the equality of divisors D = M g,i +1 , and thus prove the G-L SecantConjecture in the case d = 2 g . An essential role is played by the following calculation: G. FARKAS AND M. KEMENY
Theorem 1.6.
The class of the divisor
Syz on M g, g is given by the formula: [ Syz ] = 12 i (cid:18) ii − (cid:19)(cid:16) − (6 i + 2) λ + (3 i + 1) g X j =1 ψ j (cid:17) ∈ CH ( M g, g ) . Here λ is the Hodge class on M g, g , whereas ψ , . . . , ψ g are the cotangent classes correspond-ing to marked points. The divisor class [ Sec ] has been computed in [Fa1] Theorem 4.2:[
Sec ] = 12 i (cid:18) ii − (cid:19)(cid:16) − i + 10 i − i − λ + (3 i + 1) g X j =1 ψ j (cid:17) ∈ CH ( M g, g ) . Comparing these expressions to the class of the Hurwitz divisor, famously computed in [HM],[
Hur ] = 12 i (cid:18) ii − (cid:19)(cid:16) i + 2)2 i − λ − i + 12 i − δ irr − · · · (cid:1) ∈ CH ( M g ) , we obtain the following equality of divisors at the level of M g, g :(5) [ Syz ] = [
Sec ] + i · [ Hur ] ∈ CH ( M g, g ) . For a curve C with Cliff( C ) < i , it is easy to show that dim K i − , ( C, L ) ≥ i (see Proposition2.6), which implies that the divisor π ∗ ( D ) − i · Hur is still effective on M g, g . On the other hand, π ∗ ([ D − i · M g,i +1 ]) = 0; since the map π ∗ : Pic( M g ) → Pic( M g, g ) is known to be injective,and on M g there can exist no non-trivial effective divisor whose class is zero (see for instance[Fa2, Lemma 1.1]), we conclude that D = i · M g,i +1 .When (2) becomes an equality and C is general, we give an answer which is more precise thanthe one provided in Theorem 1.3, as to which line bundles have unexpected syzygies. Theorem 1.7.
Let C be a curve of genus g which is Brill–Noether–Petri general. For p ≥ ,we set d = 2 g + p + 1 − Cliff( C ) . If L ∈ Pic d ( C ) is a non-special line bundle such that the secant variety V g − p − g − p − (2 K C − L ) hasthe expected dimension d − g − p − , then K p, ( C, L ) = 0 . The condition that dim V g − p − g − p − (2 K C − L ) be larger than expected can be translated into acondition on translates of divisorial difference varieties, and Theorem 1.7 can be restated:(6) If K p, ( C, L ) = 0 , then L − K C + C d − g − p − ⊂ C d − g − p − − C g − d + p . Clearly, if L ∈ Pic d ( C ) fails to be ( p + 1)-very ample, that is, L − K C ∈ C p +2 − C g − d + p , then L also satisfies condition (6). We show in Section 2, that for a general curve C , the locus V ( C ) := n L ∈ Pic d ( C ) : dim V g − p − g − p − (2 K C − L ) > d − g − p − o is a subvariety of Pic d ( C ) of dimension 2 g + 2 p − d + 2.We describe our proof of Theorem 1.3, starting with the case of odd genus g = 2 i + 1.Comparing the inequalities (2) and (4), we distinguish two cases. If p ≥ i −
1, Theorem 1.3can be easily reduced to the case d = 2 p + 2 i + 4. Observe that in this case, the secant locus K C + C p +2 − C g − d + p is a divisor inside Pic d ( C ). We use K Theorem 1.8.
We fix positive integers p and i ≥ with p ≥ i − , p ≥ and a general K surface X with Pic( X ) = Z · C ⊕ Z · H , where C = 4 i + 2 , H = 4 p + 4 , C · H = 2 p + 2 i + 4 . Then K j, ( C, H C ) = 0 for j ≤ p . In particular, the G-L Secant Conjecture holds for general linebundles of degree p + 2 i + 4 on C . HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 5
The proof of Theorem 1.8 relies on Green’s [G] exact sequence · · · → K p, ( X, H ) → K p, ( C, H C ) → K p − , (cid:0) X, O X ( − C ) , H (cid:1) → · · · , where the Koszul cohomology group whose vanishing has to be established is the one in themiddle. We look at the cohomology groups on the extremes. For a smooth curve D ∈ | H | , viathe Lefschetz hyperplane principle [G], we write the isomorphism K p, ( X, H ) ∼ = K p, ( D, K D ).Recall that Green’s Conjecture [G] predicts the equivalence K p, ( D, K D ) = 0 ⇔ p < Cliff( C ) . Green’s Conjecture holds for curves on arbitrary K D ) = p + 1, hence K p, ( D, K D ) = 0.Furthermore, we establish the isomorphism K p − , (cid:0) X, O X ( − C ) , H ) ∼ = K p − , ( D, O D ( − C ) , K D )(see Proposition 2.3). This latter group is zero if and only if(7) H (cid:16) D, p ^ M K D ⊗ K ⊗ D ( − C D ) (cid:17) = 0 , where M K D is the kernel bundle of the evaluation map H ( K D ) ⊗ O D → K D . We produce aparticular K three , on which both curves C and D specialize tohyperelliptic curves such that the condition (7) is satisfied.The other possibility, namely when p ≤ i −
1, can be reduced to the case when p = i −
1, thatis, d = 2 g . As already pointed out, this is the only divisorial case of the G-L Secant Conjecture.When the genus g is even, we show in Section 4 the following result: Theorem 1.9.
Let g = 2 i ≥ be even. Let C be a general curve of genus i and let L be ageneral line bundle on C of degree p + 2 i + 3 , where p + 1 ≥ i . Then K j, ( C, L ) = 0 for j ≤ p . This is achieved via proving Theorem 4.8. We specialize C to a curve lying on K X such that Pic( X ) = Z · C ⊕ Z · H , where this time C = 4 i − H = 4 i + 4 and C · H = 2 p + 2 i + 3.As in the odd genus case, the vanishing in question is established for the line bundle L = H C .The lattice theory required to show that K j, ( X, H ) = 0 for j ≤ p in the even genus case is moreinvolved that for odd genus, but apart from this, the two cases proceed along similar lines.We close the Introduction by discussing the connection between the Prym-Green and theSecant Conjecture respectively. First, observe that in odd genus g = 2 i + 5, Prym-canonicalline bundles fall within the range in which inequality (2) holds. In particular, the vanishing K i − , ( C, K C ⊗ η ) = 0 is predicted by Theorem 1.7. Overall however, the Prym-Green Con-jecture lies beyond the range covered by the Secant Conjecture. For instance, we have seenthat K i, ( C, K C ⊗ η ) = 0, despite the fact that a general Prym-canonical line bundle K C ⊗ η is( i + 1)-very ample (equivalently, η / ∈ C i +2 − C i +2 , see [CEFS, Theorem C]). Structure of the paper:
Section 2 contains generalities on syzygies on curves and K K M g, g needed to complete the proof of the divisorial case of the Secant Conjecture (Theorem 1.4). Acknowledgment:
We are grateful to M. Aprodu, D. Eisenbud, J. Harris, R. Lazarsfeld, F.-O.Schreyer, and especially to C. Voisin for many useful discussions related to this circle of ideas.
G. FARKAS AND M. KEMENY Generalities on Syzygies of Curves
In this section we gather some general results on syzygies of curves which will be of use. Wefix a globally generated line bundle L and a sheaf F on a projective variety X . We then formthe graded S := Sym H ( X, L )-moduleΓ X ( F , L ) := M q ∈ Z H ( X, F ⊗ L ⊗ q ) . Following [G], we denote by K p,q ( X, F , L ) the space of p -th syzygies of weight q of the moduleΓ X ( F , L ). Often F = O X , in which case we denote K p,q ( X, L ) := K p,q ( X, O X , L ). Geometri-cally, one studies Koszul cohomology groups via kernel bundles. Consider the vector bundle M L := Ker (cid:8) H ( X, L ) ⊗ O X ։ L (cid:9) , where the above map is evaluation. We quote the following description from [La2]: K p,q ( X, F , L ) ≃ coker n p +1 ^ H ( X, L ) ⊗ H ( X, F ⊗ L q − ) → H ( X, p ^ M L ⊗ F ⊗ L q ) o ≃ ker n H ( X, p +1 ^ M L ⊗ F ⊗ L q − ) → p +1 ^ H ( X, L ) ⊗ H ( X, F ⊗ L q − ) o In particular, for a non-special line bundle L , we have the equivalence, cf. [GL3, Lemma 1.10]: K p, ( X, L ) = 0 ⇔ H (cid:16) X, p +1 ^ M L ⊗ L (cid:17) = 0 . Using the above description of Koszul cohomology, it follows that the difference of Bettinumbers on any diagonal of the Betti diagram of a non-special line bundle L on a curve C is anEuler characteristic of a vector bundle on C , hence constant. Precisely,(8) dim K p +1 , ( C, L ) − dim K p, ( C, L ) = ( p + 1) · (cid:18) d − gp + 1 (cid:19)(cid:16) d + 1 − gp + 2 − dd − g (cid:17) . The following fact is well-known and essentially trivial:
Proposition 2.1.
Let C be a smooth curve and L a globally generated line bundle on C with h ( C, L ) = 0 and K p, ( C, L ) = 0 . If x ∈ C is a point such that L ( − x ) is globally generated, then K p − , ( C, L ( − x )) = 0 .Proof. Follows via the above description of Koszul cohomology, by using the exact sequence0 → p ^ M L ( − x ) ⊗ L → p ^ M L ⊗ L → p − ^ M L ( − x ) ⊗ L ( − x ) → . (cid:3) Syzygies of K surfaces. The following result, while simple, is essential in the proof ofTheorem 1.3, for it allows us to ultimately reduce the vanishing required in the G-L SecantConjecture to a vanishing of the type appearing in a slightly different context in the statementof the Minimal Resolution Conjecture of [FMP].
Lemma 2.2.
Let X be a K3 surface, and let L and H be line bundles with H effective and basepoint free. Assume ( H · L ) > and H ( X, qH − L ) = 0 for q ≥ . Then for each smooth curve D ∈ | H | , we have that K p,q ( X, − L, H ) ≃ K p,q ( D, − L D , K D ) for all p and q . HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 7
Proof.
The proof proceeds along the lines of [G, Theorem 3.b.7]. By the assumptions, we havea short exact sequence of Sym H ( X, H )-modules0 → M q ∈ Z H (cid:0) X, ( q − H − L (cid:1) → M q ∈ Z H (cid:0) X, qH − L (cid:1) → M q ∈ Z H (cid:0) D, qK D − L D (cid:1) → , where the first map is multiplication by the section s ∈ H ( X, H ) defining D . Let B denotedthe graded Sym H ( X, H )-module L q ∈ Z H ( D, qK D − L D ), and write its associated Koszulcohomology groups as K p,q ( B, H ( X, H )). The above short exact sequence induces a long exactsequence at the level of Koszul cohomology(9) · · · → K p,q − ( X, − L, H ) → K p,q ( X, − L, H ) → K p,q ( B, H ( X, H )) → K p − ,q ( X, − L, H ) → · · · The maps K p,q − ( X, − L, H ) → K p,q ( X, − L, H ) are induced by multiplication by s , and henceare zero, see also [G, 1.6.11]. Choose a splitting H ( X, H ) ≃ C { s } ⊕ H ( D, K D ) . This induces isomorphism f p : p ^ H ( X, H ) ≃ p − ^ H ( D, K D ) ⊕ p ^ H ( D, K D ) . The Koszul cohomology of the module B is computed by the cohomology of the complex A • → p ^ H ( D, K D ) ⊗ H (cid:0) D, ( q − K D − L D (cid:1) M p +1 ^ H ( D, K D ) ⊗ H (cid:0) D, ( q − K D − L D (cid:1) → p − ^ H ( D, K D ) ⊗ H ( D, qK D − L D ) M p ^ H ( D, K D ) ⊗ H ( D, qK D − L D ) → · · · where the maps being equal to ( − d p,q − , d p +1 ,q − ), with d p +1 ,q − : p +1 ^ H ( D, K D ) ⊗ H (cid:0) D, ( q − K D − L D (cid:1) → p ^ H ( D, K D ) ⊗ H ( D, qK D − L D )and its shift d p,q − being Koszul differentials (note that we have used that s vanishes along D ).Thus we have K p,q ( B, H ( X, H )) ≃ K p,q ( D, − L D , K D ) ⊕ K p − ,q ( D, − L D , K D )and hence via the sequence (9), we obtain K p,q ( X, − L, H ) ⊕ K p − ,q ( X, − L, H ) ≃ K p,q ( D, − L D , K D ) ⊕ K p − ,q ( D, − L D , K D ) . For p = 0, this becomes K ,q ( X, − L, H ) ≃ K ,q ( D, − L D , K D ). The claim follows by inductionon p . (cid:3) Now let, X be a K3 surface, and let L, H ∈ Pic( X ) and D ∈ | H | be as in the hypotheses ofthe lemma above. Assume further that H ( X, H − L ) = 0, and let C ∈ | L | be a smooth, integralcurve. Following [G, Theorem 3.b.1], we have a long exact sequence · · · → K p,q ( X, H ) → K p,q ( C, H C ) → K p − ,q +1 ( X, − L, H ) ≃ K p − ,q +1 ( D, − L D , K D ) → · · · Thus we have:
Proposition 2.3.
In the above situation, assume K p,q ( X, H ) = K p − ,q +1 ( D, − L D , K D ) = 0 .Then K p,q ( C, H C ) = 0 . We will also make use of the following result of Mayer’s [M, Proposition 8].
Proposition 2.4.
Let X be a K3 surface and let L ∈ Pic( X ) be a big and nef line bundle.Assume there is no smooth elliptic curve F with ( F · L ) = 1 . Then L is base point free. G. FARKAS AND M. KEMENY
We now turn our attention to the Green-Lazarsfeld Secant Conjecture [GL1]. We fix a generalcurve C of genus g and an integer d ≥ g + 1 − Cliff( C ). Using [ACGH, p. 222], we observe thata general line bundle L ∈ Pic d ( C ) is projectively normal, that is, the multiplication mapsSym n H ( C, L ) → H ( C, L ⊗ n )are surjective for all n . Since obviously also H ( C, L ) = 0, via [GL3, Lemma 1.10], we concludethat φ L : C → P d − g verifies property ( N p ) for some integer p ≥ d ≥ g + p + 1 − Cliff( C ),(that is, K j, ( C, L ) = 0 for all j ≤ p ), if and only if one single vanishing K p, ( C, L ) = 0 holds.Proposition 2.1 is used to reduce the proof of Theorem 1.3 to the following cases:
Proposition 2.5.
Let C be a general curve of genus g . In order to conclude that the G-L SecantConjecture holds for C and for general line bundles on C in each degree, it suffices to exhibit anon-special line bundle L ∈ Pic d ( C ) such that K p, ( C, L ) = 0 , in each of the following cases: (1) g = 2 i + 1 , d = 2 p + 2 i + 4 and p ≥ i − g = 2 i , d = 2 p + 2 i + 3 and p ≥ i − .Proof. As C is general, Cliff( C ) = ⌊ g − ⌋ . We explain the case g = 2 i + 1, the remaining evengenus case being similar. In the case p ≥ i −
1, the two inequalities (2) and (4), that is, d ≥ g + p + 1 − Cliff( C ) and d ≥ g + 2 p + 3respectively, reduce to the single inequality d ≥ p + 2 i + 4 . If d is even, we write d = 2 q + 2 i + 4, where q ≥ p . By assumption, we can find a line bundle L ∈ Pic d ( C ) with K q, ( C, L ) = 0. We may assume L to be projectively normal and then, asexplained, it follows that K p, ( C, L ) = 0. If d is odd, we write d = 2 q + 2 i + 5, where again q ≥ p . By assumption, there exists a line bundle L ′ ∈ Pic d +1 ( C ) with K q +1 ( C, L ′ ) = 0. We set L := L ′ ( − x ), where x ∈ C is a general point. By Proposition 2.1, we find that K q, ( C, L ) = 0,hence K p, ( C, L ) = 0 as well.In the range p ≤ i −
1, the inequalities (2) and (4) reduce to the inequality d ≥ i + p + 3 (cid:0) = 2 g + p + 1 − Cliff( C ) (cid:1) . If d ≤ i + 2, we apply the assumption in degree 4 i + 2 to find a line bundle L ′ ∈ Pic i +2 ( C )such that K i − , ( C, L ′ ) = 0. We then choose a general effective divisor D ∈ C i +2 − d and set L := L ′ ( − D ) ∈ Pic d ( C ). Via Proposition 2.1, we conclude that K d − i − , ( C, L ) = 0, hence K p, ( C, L ) = 0, as well. If, on the other hand d ≥ i + 2, then for even degree, we write d = 2 q + 2 i + 4, where q ≥ i − d is odd is analogous. (cid:3) Especially significant in our study is the divisorial case of the G-L Secant Conjecture: g = 2 i + 1 , d = 4 i + 2 , p = i − . Using (8), note that in this case dim K i − , ( C, L ) = dim K i, ( C, L ). For a pair [
C, L ] ∈ Pic gg ,we consider the embedding φ L : C ֒ → P g and denote by I ( L ) the graded ideal of φ L ( C ). Thenext observation, to be used in the proof of Theorem 1.7 provides a lower bound for the numberof syzygies of L , when the curve C has Clifford index less than maximal. Proposition 2.6.
Let C be a smooth curve of genus g = 2 i + 1 having gonality at most i + 1 and L a line bundle of degree g on C . Then dim K i, ( C, L ) ≥ i. HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 9
Proof.
It suffices to prove the statement for a general ( i + 1)-gonal curve [ C ] ∈ M i +1 ,i +1 . Wefix such a C , then denote by A ∈ W i +1 ( C ) a pencil of minimal degree and by φ A : C → P the associated covering. Let L ∈ Pic g ( C ) a suitably general line bundle such that φ L ( C ) isprojectively normal, h ( C, L ⊗ A ∨ ) = i + 1 and the multiplication map µ A,L ⊗ A ∨ : H ( C, A ) ⊗ H ( C, L ⊗ A ∨ ) → H ( C, L )is an isomorphism. The vector bundle E := ( φ A ) ∗ ( L ) induces an ( i + 1)-dimensional scroll X := P ( E ) ֒ → P H ( C, L ) ∨ of degree i + 1 containing φ L ( C ). The equations of X inside P i +1 are obtained by taking the (2 × µ A,L ⊗ A ∨ , see[Sch]. Precisely, if { σ , σ } is a basis of H ( C, A ) and { τ , . . . , τ i +1 } is a basis of H ( C, L ⊗ A ∨ )respectively, then X is cut out by the quadrics q ℓ := ( σ τ ℓ ) · ( σ τ i +1 ) − ( σ τ i +1 ) · ( σ τ ℓ ) ∈ K , ( C, L ) , where ℓ = 1 , . . . , i . Recall that there is an isomorphism K i, ( C, L ) ∼ = K i − , (cid:0) I ( L ) , H ( C, L ) (cid:1) := Ker n i − ^ H ( C, L ) ⊗ I ( L ) → i − ^ H ( C, L ) ⊗ I ( L ) o . By direct computation, for ℓ = 1 , . . . , i , we write down the following syzygies: γ ℓ := i X j =1 ( σ τ ) ∧ . . . ∧ ( d σ τ j ) ∧ . . . ∧ ( σ τ i ) ⊗ q ℓ ∈ i − ^ H ( C, L ) ⊗ I ( L ) . Since the quadrics q , . . . , q i are independent in K , , ( C, L ), we find that dim K i, ( C, L ) ≥ i. (cid:3) Syzygies and translates of difference varieties.
For a curve C and a, b ≥
0, let C a − C b ⊂ Pic a − b ( C )be the difference variety, consisting of line bundles of the form O C ( D a − E b ), where D a and E b are effective divisors on C having degrees a and b respectively. A result of [FMP] provides anidentification of the divisorial difference variety, valid for each smooth curve of genus g :(10) C g − j − − C j = Θ V j Q C . The right-hand-side here denotes the theta divisor of the vector bundle Q C := M ∨ K C , that is,Θ V j Q C := n ξ ∈ Pic g − j − ( C ) : h (cid:0) C, j ^ Q C ⊗ ξ ) ≥ o . We fix non-negative integers j ≤ g − a ≤ g − j − V ( C ) := n L ∈ Pic g − j − a − ( C ) : L + C a ⊂ C g − j − − C j o . Obviously C g − j − a − − C j ⊂ V ( C ). For hyperelliptic curves, one has set-theoretic equality: Proposition 2.7.
Let C be a smooth hyperelliptic curve of genus g and fix integers j ≤ g − and a ≤ g − j − . The following equivalence holds for a line bundle L ∈ Pic g − j − a − ( C ) : L + C a ⊂ C g − j − − C j ⇔ L ∈ C g − j − a − − C j . Proof.
Via (10), the hypothesis L + C a ⊂ C g − j − − C j can be reformulated cohomologically: h (cid:16) C, j ^ M K C ⊗ K C ( − L − D a ) (cid:17) ≥ , for every effective divisor D a ∈ C a . If C is hyperelliptic and A ∈ W ( C ) denotes the hyperelliptic pencil, then the kernel bundle splits M K C = (cid:0) A ∨ ) ⊕ ( g − , and the previous condition translates into h (cid:0) C, A ⊗ ( g − j − ⊗ ( − L − D a ) (cid:1) ≥ h ( C, ( g − − j ) A − L ) ≥ a + 1 . On C any complete linear series g rd with d ≤ g − A ⊗ r ( y + · · · + y d − r ),with y , . . . , y d − r ∈ C . It follows that there exists a divisor E = x + · · · + x g − a − such that L = A ⊗ ( g − j − a − ( − E ). Denoting by x ′ ℓ ∈ C the hyperelliptic conjugate of x ℓ , we obtain L = O C (cid:0) x ′ + · · · + x ′ g − − a − j − x g − a − j − · · · − x g − a − (cid:1) ∈ C g − j − a − − C j . (cid:3) A consequence of the above is that for a general curve C , we have an equality of cycles V ( C ) = ( C g − j − a − − C j ) + V ′ ( C ) , where V ′ ( C ) is a residual cycle of dimension at most g − a −
1. Contrary to our initial expectation,which was later tempered by Claire Voisin, the residual cycle V ′ ( C ) can in general be non-emptyand have dimension much smaller than g − a −
1, as the following example shows:
Proposition 2.8.
Let C be a general curve of genus g = 2 i and A ∈ W i +1 ( C ) a pencil ofminimal degree. We set L := K C − A ∈ Pic i − ( C ) . Then L + C ⊂ C i − − C , but however L / ∈ C i − − C .Proof. For a point x ∈ C , let p + · · · + p i ∈ C i be the divisor such that x + P ij =1 p j ∈ | A | . Since h ( C, K C ⊗ A ∨ ) = i , there exists an effective divisor D ∈ C i − with D + P i − j =1 p j ∈ | K C ⊗ A ∨ | .It follows, that L ( x ) = (cid:0) K C ⊗ A ∨ (cid:1) ⊗ A ∨ ( x ) = O C ( D + p + · · · + p i − ) ⊗ A ∨ ( x ) = O C ( D − p i ) ∈ C i − − C. On the other, we claim that
L / ∈ C i − − C . Else, there exists a point y ∈ C such that H ( C, L ( y )) = 0. Via the Base Point Free Pencil Trick, this is equivalent to saying that themultiplication map H ( C, A ) ⊗ H ( C, K C ⊗ A ∨ ( y )) → H ( C, K C ( y )) is not injective. Since h ( C, K C ⊗ A ∨ ( y )) = h ( C, K C ⊗ A ∨ ), this implies that the Petri map associated to the pencil A is not injective, a contradiction. (cid:3) Remark . The structure of the residual cycle V ′ ( C ) remains mysterious. One case that isunderstood via the exercises in [ACGH, p.276 ] is that when C is a general curve of genus 4 and a = j = 1. Then V ( C ) = ( C − C ) + (cid:8) ∓ ( K C − A ) (cid:9) , where A ∈ W ( C ), that is, V ′ ( C ) is a0-dimensional cycle.3. The generic Green–Lazarsfeld secant conjecture for curves of odd genus
In this section we prove Theorem 1.8. We begin by recalling a few basic facts. Let h be aneven lattice of rank ρ + 1 ≤
10 and signature (1 , ρ ). The moduli space of h -polarized K3 surfacesexists as a quasi-projective algebraic variety, is nonempty, and has at most two componentsboth of dimension 19 − ρ , which locally on the period domain are interchanged by complexconjugation, [Dol]. Complex conjugation here means that a complex surface X with complexstructure J is sent to ( X, − J ).We fix integers g = 2 i + 1 with i ≥ p ≥ i − p ≥
1. Let Θ g,p be the rank two latticewith ordered basis { H, η } and intersection form: (cid:18) p + 4 2 p − i p − i − (cid:19) HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 11
We let L denote the class H − η . Notice that ( H · L ) = 2 p + 2 i + 4 and ( L ) = 4 i . We denoteby ˆΘ g,p the rank three lattice with ordered basis { H, η, E } and intersection form p + 4 2 p − i p − i − . Obviously Θ g,p can be primitively embedded in ˆΘ g,p . By the surjectivity of the period mapping,there exist smooth K3 surfaces Z g and ˆ Z g respectively with Picard lattices isomorphic to Θ g,p and ˆΘ g,p respectively, and such that H is big and nef. Lemma 3.1.
Let α, β ∈ ˆΘ g,p . Then ( α · β ) is even and ( α ) is divisible by four.Proof. The first claim is clear, as all entries in the above rank three matrix are even. For thesecond claim, write α = aH + bη + cE for a, b, c ∈ Z and compute( α ) = 4 a ( p + 1) − b + 4 ab ( p − i ) + 4 ac. (cid:3) Corollary 3.2.
Let Z g respectively ˆ Z g be K3 surfaces with Picard lattices isomorphic to Θ g,p respectively ˆΘ g,p . Suppose a divisor α in Θ g,p respectively ˆΘ g,p is effective. Then ( α ) ≥ and α is base point free and nef.Proof. Suppose a divisor α in Θ g,p resp. ˆΘ g,p is effective. Since there are no ( −
2) classes in Θ g,p or in ˆΘ g,p , necessarily ( α ) ≥ α is nef. Since there do not exist classes F with ( F ) = 0,( F · α ) = 1, the class α is base point free by Proposition 2.4. (cid:3) Our next task is to study the Brill–Noether theory of curves in the linear system | H | . Lemma 3.3.
There exists a K3 surface Z g with Pic( Z g ) ≃ Θ g,p and H big and nef. For ageneral such K surface, a general curve D ∈ | H | is Brill–Noether–Petri general, in particular Cliff( D ) = p + 1 .Proof. We have an obvious primitive embedding Θ g,p ֒ → ˆΘ g,p . Note that on ˆ Z g , any class ofthe form aH + bη ∈ Pic( ˆ Z g ) with a < E .Let Z g be a general Θ g,p -polarized K3 surface which deforms to ˆ Z g ; i.e. a general element inan least one of the components of Θ g,p -polarized K3 surfaces. It suffices to establish that thehyperplane class admits no decomposition H = A + A for divisors A , A with h ( Z g , A i ) ≥ i = 1 ,
2. Indeed, this follows from the proof of [La1, Lemma 1.3]. For precise details, werefer to [K, Lemma 5.2]).Suppose we have such a decomposition on a general surface Z g . Then the A i would deformto effective divisors on ˆ Z g , so we could write A i = a i H + b i η for a i , b i ∈ Z , with a i ≥ a + a = 1. Without loss of generality, we may assume that a = 0, so ( A ) = − b . FromCorollary 3.2, this forces b = 0, so h ( ˆ Z g , A ) = 1, which is a contradiction. If Z cg is the complexconjugate of Z g with the induced Θ g,p -polarization, then the claim above clearly also holds forthe image of H in Pic( Z cg ). (cid:3) Corollary 3.4.
Let Z g be a general Θ g,p -polarized K3 surface. Then K j, ( Z g , H ) = 0 for j ≤ p .Proof. From the above lemma, Cliff( D ) = p + 1 for D ∈ | H | general. Thus the result followsfrom [AF1, Thm. 1.3]. (cid:3) Whereas for a general Θ g,p -polarized K D ∈ | H | and C ∈ | L | respectively, are Brill-Noether general, this is no longer the case for ˆΘ g,p -polarized K D and C become hyperelliptic. Lemma 3.5.
Let ˆ Z g be a general ˆΘ g,p -polarized K3 surface. Then H (cid:16) D, j ^ M K D ⊗ K D ⊗ η D (cid:17) = 0 for j ≤ p and for a general curve D ∈ | H | .Proof. As D is hyperelliptic, we have the following splitting j ^ M K D ≃ O D ( − jE ) ⊕ ( p +2 j )and K D ∼ = 2( p + 1) E D , see [FMP, Prop. 3.5]. We need to show H (cid:0) D, η D ((2 p + 2 − j ) E D ) (cid:1) = 0.Since ((2 p + 2 − j ) E + η ) = −
4, we have h ( ˆ Z g , (2 p + 2 − j ) E + η ) = h ( ˆ Z g , (2 p + 2 − j ) E + η ) = 0and thus h ( ˆ Z g , (2 p + 2 − j ) E + η ) = 0, using Corollary 3.2. Lastly, we compute((2 p + 2 − j ) E + η − H ) = 4( i + j ) − p − ≤ i − p ) − ≤ − h ( ˆ Z g , (2 p + 2 − j ) E + η − H ) = 0. Thus H ( D, η D ((2 p + 2 − j ) E D ) (cid:1) = 0 asrequired. (cid:3) As an immediate corollary we have:
Corollary 3.6.
There is a nonempty open subset of the moduli space of Θ g,p -polarized K3surfaces such that K j − , ( D, η D − K D , K D ) = 0 for all j ≤ p and D ∈ | H | . We may now conclude this section by establishing the G-L Secant Conjecture for general linebundle on general curves of odd genus.
Proof of Theorem 1.8.
We retain the notation. We have established that there exists aK3 surface Z g with Picard lattice Θ g,p such that H is big and nef, K j, ( Z g , H ) = 0 and K j − , ( D, − L D , K D ) = 0 for j ≤ p for each smooth curve D ∈ | H | . We have h ( Z g , η ) = h ( Z g , η ) = 0, h ( Z g , L ) = 0. For q ≥ qH − L ) = ( q − (4 p + 4) + 4( q − p − i ) − ≥ ( q − q − p + 4) − − ≥ p − . Since p ≥ H ( Z g , qH − L ) = 0 for q ≥
2, using Corollary 3.2 (in the case p = 1, use that qH − L is primitive). Thus Proposition 2.3 applies, and for each smooth curve C ∈ | L | we have K j, ( C, H C ) = 0 for j ≤ p . Note finally that h ( C, H C ) = h ( Z g , H ) = 2 p + 4, that is, H C isnon-special. (cid:3) The generic Green–Lazarsfeld Secant Conjecture for curves of even genus
Suppose g = 2 i for i ≥ p ≥ i −
1. Let Ξ g,p be the following rank two lattice with orderedbasis { H, η } and intersection form (cid:18) p + 4 2 p − i + 12 p − i + 1 − (cid:19) . We set L = H − η and note that ( H · L ) = 2 p + 2 i + 3 and ( L ) = 2 g −
2. We let ˆΞ g,p be therank three lattice of signature (1 ,
2) having ordered basis { H, η, E } and intersection form p + 4 2 p − i + 1 22 p − i + 1 − . Obviously Ξ g,p can be primitively embedded in ˆΞ g,p . By the Torelli theorem [Dol], there existsmooth K3 surfaces Z g respectively ˆ Z g with Picard lattices isomorphic to Ξ g,p respectively ˆΞ g,p ,and such that H is big and nef. HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 13
Lemma 4.1.
Let ˆ Z g be a K3 surface with Picard lattice given by ˆΞ g,p and such that H is bigand nef. Then E and H are effective and base point free.Proof. Both E and H are effective since they have positive intersection with H . We claim that E is base point free. As ( E ) = 0, it suffices to prove that E is nef, [H, Prop. 2.3.10]. Let R = aH + bE + cη be the class of a smooth rational curve, for a, b, c ∈ Z , and assume for acontradiction that ( R · E ) <
0, that is, a <
0. Then ( R − aH ) = ( bE + cη ) = − c ≤
0. Onthe other hand ( R − aH ) = − a (4 p + 4) − a ( R · H ) >
0, since a < H is nef. Wehave reached a contradiction, thus E is nef.To conclude that H is base point free, since H is big and nef, it suffices to show that thereis no smooth elliptic curve F with ( H · F ) = 1 (cf. Proposition 2.4). Suppose such an F existsand write F = aH + bE + cη . Since E is the class of an integral elliptic curve, E is nef and2 a = ( F · E ) ≥
0. We have − c = ( F − aH ) = a ( − a (4 p + 4)), which is only possible if a = c = 0 and so F = bE . Since ( bE · H ) = 2 b , this is a contradiction. (cid:3) Lemma 4.2.
Let ˆ Z g be a K3 surface with Pic( ˆ Z g ) = ˆΞ g,p , such that H is big and nef. Then noclass of the form aE + bη , for b = 0 , can be effective.Proof. Suppose aE + bη is effective, where b = 0. Since ( aE + bη ) = − b , there must be anintegral component R of aE + bη with ( R · aE + bη ) <
0. Since R is integral and not nef, wemust have ( R ) = −
2. Since ( aE + bη · E ) = 0 and E is nef, we have ( R · E ) = 0 and thus R isof the form xE + yη for x, y ∈ Z . But then ( R ) = − y = −
2, which is a contradiction. (cid:3)
Lemma 4.3.
Let ˆ Z g be as above. Then any class A := H + cη with c ∈ Z and satisfying ( A ) ≥ is nef.Proof. Suppose by contradiction that A is not nef. As ( A ) ≥ A · E ) = 2 >
0, the class A is effective. Thus there exists an integral base component R of A with ( R ) = − R · A ) < R = xH + yE + zη for x, y, z ∈ Z ; as ( R ) = −
2, we obtain x = 0. Since A − R is effective( R is a base component of A ), intersecting with E gives x = 1 and A − R = − yE + ( c − z ) η .From Lemma 4.2 we have c = z and then y ≤
0. If y = 0, then we would have R = A whichcontradicts that ( A ) ≥
0, so y <
0. But now R = H + cη + yE = A + yE , so we write − R ) = ( R · A ) + y ( E · R ) = ( R · A ) + 2 y ≤ − , for ( R · A ) < y <
0. This is a contradiction. (cid:3)
Corollary 4.4.
Let ˆ Z g be as above, and set L = H − η . Then H ( ˆ Z g , qH − L ) = 0 for q ≥ .Proof. For q = 0, note that ( L ) = 2 g − >
0, so L is big and nef by the previous lemma and H ( ˆ Z g , − L ) = 0. For q = 1, note that ( η ) = − η nor − η are effective byLemma 4.2. Thus H ( ˆ Z g , η ) = 0. For the remaining cases, it suffices to show 2 H − L = H + η is big and nef, since H is big and nef. We have ( H + η ) = 4 p + 2(2( p − i ) + 1) > (cid:3) We have seen that the line bundle L is big and nef. We now show that it is base point free. Lemma 4.5.
For ˆ Z g as above, the class L = H − η is base point free.Proof. As L is big and nef, it suffices to show that there is no smooth elliptic curve F with( L · F ) = 1. Suppose such an F were to exist, and write F = aH + bE + cη . As ( F ) = 0, weobtain that a = 0. As ( F · E ) ≥
0, we find a >
0. We calculate − c + a ) = ( F − a ( H − η )) = a (2 g − − a = a ( a (2 g − − . As a > g − >
0, this is a contradiction. (cid:3)
Lemma 4.6.
For ˆ Z g as above, the class B := H − (2 p + 2 − j ) E − η is not effective for j ≤ p . Proof.
We calculate ( B ) = 4( i + j ) − p − ≤ −
6. If B is effective, then there exists a basecomponent R of B with ( R · B ) < R ) = −
2. Write R = aH + bE + cη for a, b, c ∈ Z .Since R is effective, ( R · E ) ≥
0, so a ≥
0. We have a = 0 because ( R ) = −
2. Since B − R iseffective, ( B − R · E ) ≥ a = 1. Then B − R = − (2 p + 2 − j + b ) E − (1 + c ) η .Applying Lemma 4.2, we see c = −
1, and we have 2 p + 2 − j + b ≤
0, so b ≤ − p −
2. But then R = H + bE − η and one calculates( R ) = 4( b + i ) − ≤ i − p − − ≤ − , which is a contradiction. (cid:3) Corollary 4.7.
Let ˆ Z g be as above and let D ∈ | H | be an integral, smooth curve. Then H (cid:16) D, V j M K D ( K D + η D ) (cid:17) = 0 for j ≤ p .Proof. Essentially identical to that of Lemma 3.5. (cid:3)
Theorem 4.8.
Let ˆ Z g be as above and Z g be a generic Ξ g,p -polarized K3 surface, which isdeformation equivalent to ˆ Z g . Let C ∈ | L | be a smooth, integral curve, where L = H − η . Then K j, ( C, H C ) = 0 for j ≤ p .Proof. On the surface Z g , we choose a general divisor D ′ ∈ | H | . By semicontinuity and Corollary4.7, we have H (cid:16) D ′ , V j M K D ′ ( K D ′ + η ) (cid:17) = 0, for j ≤ p . Thus K j − , ( Z g , − L, H ) = 0 for j ≤ p .We further have h ( Z g , qH − L ) = 0 for q ≥ h ( Z g , H − L ) = 0 from Lemma 4.2. Thus Proposition 2.3 applies, and it suffices to show K j, ( Z g , H ) = 0 for j ≤ p . For this it suffices to show Cliff( D ′ ) ≥ p + 1 by [AF1, Thm. 1.3]. Toestablish this, it suffices in turn to show there is no decomposition H = A + A for divisors A i on Z g with h ( Z g , A i ) ≥ i = 1 , H = A + A is such a decomposition and write A i = a i H + b i η for a i , b i ∈ Z , when i = 1 ,
2. By semicontinuity, A i must deform to effective divisors on ˆ Z g , and then intersectingwith E shows that a i ≥ i = 1 ,
2. Since a + a = 1, we have either a = 0 or a = 0. Weassume a = 0, so A = b i η is effective. By semicontinuity and Lemma 4.2, we see b i = 0, so A is trivial and the claim holds. (cid:3) We thus derive the main result of this section:
Theorem 4.9.
Let C be a general curve of genus g = 2 i ≥ and H C ∈ Pic( C ) be a general linebundle of degree p + 2 i + 3 , where p + 1 ≥ i . Then K j, ( C, H C ) = 0 for j ≤ p .Proof. This follows from the above by setting H C := H | C . Note that we have h ( C, H C ) = h ( Z g , H ) = 2 p + 4 (as h ( Z g , η ) = h ( Z g , η ) = 0 by deforming to ˆ Z g ), which is the expectednumber of sections for a line bundle of degree 2 p + 2 i + 3. (cid:3) The Prym–Green conjecture for curves of odd genus
In this section we prove the odd genus case of the Prym-Green Conjecture formulated in[CEFS]. We start by recalling a few things about polarized Nikulin surfaces.The
Nikulin lattice N is the even lattice of rank 8 generated by elements N , . . . , N and e := P j =1 N j , where N j = − j = 1 , . . . , N ℓ · N j ) = 0 for ℓ = j . For an integer g ≥
2, following [GS, Definition 2.1] and [FV, Definition 1.2], we define a
Nikulin surface of thefirst kind to be a K3 surface X together with a primitive embedding Z · L ⊕ N ֒ → Pic( X ) suchthat L is a big and nef class with ( L ) = 2 g − HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 15
Nikulin surfaces of the first kind when g is odd). By definition, a Nikulin surface X containsan even set of eight disjoint smooth rational curves N , . . . , N which generate N over Z (butnot over Z ). Nikulin surfaces of the first kind form an irreducible 11-dimensional moduli space F N g . We refer to [Dol] and [vGS] for the construction of F N g via period domains and to [FV] fora description of its birational geometry for small genus.A general element (cid:2) X, Z · L ⊕ N ֒ → Pic( X ) (cid:3) ∈ F N g corresponds to a surface having Picardlattice equal to Λ g := Z · L ⊕ N . Set H = L − e ∈ Pic( X ), where 2 e = N + . . . + N . For asmooth curve C ∈ | L | , since e ⊗ C = 0, the pair [ C, e C ] is an element of the Prym moduli space R g and φ H C : C → P g − is a Prym-canonical curve of genus g .Suppose now that g = 2 i + 3 is an odd genus. The Prym-Green Conjecture (1) predicts that K i, ( C, H C ) = 0 and K i − , ( C, H C ) = 0 . This is equivalent to determining the shape of the resolution the Prym-canonical curve φ H C ( C ). Lemma 5.1.
Let g = 2 i +3 for i ≥ . There exists a Nikulin surface X g with Pic( X g ) ≃ Λ g suchthat L is base point free, H is very ample and the general smooth curve D ∈ | H | is Brill-Noethergeneral.Proof. Consider a general Nikulin surface X g of the first kind and let π : X g → ¯ Y be the mapwhich contracts all the exceptional curves N j . The base point freeness of L is a consequenceof [GS, Proposition 3.1]. Since g ≥
6, from [GS, Lemma 3.1] we obtain that H = L − e isvery ample. The fact that a general smooth curve D ∈ | H | is Brill–Noether general relies oncemore on showing that there is no decomposition H = A + A , for divisors A i on X g with h ( X g , A i ) ≥
2, for i = 1 ,
2. Suppose there is such a decomposition and write A i = a i L + b i m i ,where a i , b i ∈ Z and m i ∈ N , for i = 1 ,
2. By intersecting with the nef class L , we have that a i ≥ i = 1 ,
2. Since a + a = 1, we may assume a = 0 and a = 1; thus A ∈ N isorthogonal to L . This implies that π ( A ) is a finite sum of points (as π ( L ) is ample and in factgenerates Pic( ¯ Y )). This in turn forces A to be a sum of the disjoint ( −
2) curves N , . . . , N ,so that h ( X g , A ) = 1. (cid:3) Corollary 5.2.
Let X g and H be as above. Then K i, ( X g , H ) = K i − , ( X g , H ) = 0 .Proof. Let D ∈ | H | be a general divisor, hence g ( D ) = 2 i + 1. From the previous lemma,Cliff( D ) = i . Thus the result follows from [AF1, Theorem 1.3]. (cid:3) Lemma 5.3.
Assume g ≥ is odd and let X g be a general Nikulin surface with Picard lattice Λ g . Then H ( X g , qH − L ) = 0 for q ≥ and H ( X g , H − L ) = 0 .Proof. As already pointed out, H is very ample for g ≥
6. From [GS, Proposition 3.5], the class2 H − L is big and base point free for g ≥
10; thus H ( X g , qH − L ) = 0 for q ≥
2. Furthermore, H ( X g , H − L ) = H ( X g , H − L ) = 0 from [FV, Lemma 1.3] and H ( X g , L ) = 0 since L is bigand nef. (cid:3) From the above results and Proposition 2.3, to establish the Prym-Green Conjecture for ageneral curve C ∈ | L | , it suffices to show that for a smooth curve D ∈ | H | K i − , ( D, − L D , K D ) = 0 and K i − , ( D, − L D , K D ) = 0 . Observe that both these statements would follow from the Minimal Resolution Conjecture[FMP], provided L D ∈ Pic( D ) was a general line bundle in its respective Jacobian, which isof course not the case. However, one can try to follow the proof of [FMP, Lemma 3.3] andspecialize to hyperelliptic curves while still retaining the embedding of the Prym curve C in aNikulin surface. We carry this approach out below. Let T g be the rank two lattice with ordered basis { L, E } and with intersection form (cid:18) g − (cid:19) . Consider the lattice T g ⊕ N , where N is the Nikulin lattice. This rank 10 lattice is even, ofsignature (1 , K X g with Picard lattice isomorphic to T ⊕ N . After applying Picard-Lefschetz reflectionsand replacing L with − L if necessary, we may assume L is nef. Since we have a primitiveembedding Λ g ֒ → T g ⊕ N , we conclude that ˆ X g is a Nikulin surface of the first kind. Lemma 5.4.
Let ˆ X g be as above with g ≥ . Then H = L − e ∈ Pic( ˆ X g ) is big and nef.Proof. As ( H ) >
0, we need to show that H is nef. The class H is effective, as ( H · L ) > −
2) curve Γ with (Γ · H ) <
0. We may writeΓ = aL + bE + X j =1 c j N j , for a, b ∈ Z and c j ∈ Z for j = 1 , . . . ,
8. Since ( N j · H ) = 1 for all j , we have Γ = N j , sothat ( N j · Γ) ≥ c j ≤ j . Set k := (Γ · L ) = a (2 g −
2) + 2 b ; since L isbig and nef, k ≥
0. From (Γ · H ) < k < P j =1 | c j | and thus k < P j =1 c j , by theCauchy–Schwarz inequality. Further, since (Γ) = −
2, we find that ( aL + bE ) = 2 P j =1 c j − L ) ( aL + bE ) ≤ (cid:0) L · ( aL + bE ) (cid:1) , or equivalently(2 g − (cid:0) P j =1 c j − (cid:1) ≤ k . Putting these two inequalities together, we obtain P j =1 c j < g − g − ,and so P j =1 (2 c j ) ≤
4, for g ≥ c j ∈ Z , there are two cases. In the first case, c ℓ = − ℓ and c j = 0 for j = ℓ .From k < P j =1 | c j | , we then have k = 0 and further ( aL + bE ) = 0. Putting these togethergives 2 ab = 0 and thus a = 0 or b = 0. Using again that k = (cid:0) L · ( aL + bE ) (cid:1) = 0, we must have a = b = 0 so Γ = − N ℓ , which contradicts the effectiveness of Γ. In the second case | c j | ≤ for all j , and there are at most four values of j such that c j = 0. But since N is generated by N , . . . , N and e with 2 e = N + . . . + N , the only such elements in T g ⊕ N must have c j = 0for all j . But this implies k < (cid:3) Lemma 5.5.
Let ˆ X g and H be as above, with g ≥ . Then H is base point free.Proof. From Proposition 2.4 and the previous lemma, it suffices to show that there is no smoothelliptic curve F with ( F · H ) = 1. Suppose such an F exists and write F = aL + bE + P j =1 c j N j for a, b ∈ Z and c j ∈ Z for j = 1 , . . . ,
8. We have c j ≤ F · N j ) ≥ k := ( F · L ) = a (2 g −
2) + 2 b ≥
0. Since 1 = ( F · H ), we find k = 1 + | c | + · · · + | c | . Notethat ( aL + bE ) = 2 P j =1 c j . The Hodge Index Theorem applied to L and aL + bE gives(2 g − (cid:16) X j =1 c j (cid:17) ≤ k = (cid:16) X j =1 | c j | (cid:17) . One has (1 + P i =1 | c i | ) ≤ P j =1 | c j | + 8 P j =1 c j ≤ P j =1 c j , using the fact that2 c j ∈ Z . Combining this with the above inequality, we write (cid:0) g − − (cid:1) P j =1 c j ≤ P j =1 (2 c j ) = 0 for g ≥
6. Thus c j = 0 for all j and k = 1. But k = a (2 g −
2) + 2 b iseven, which is a contradiction. (cid:3) We now show that the E is the class of an irreducible, smooth elliptic curve. HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 17
Lemma 5.6.
Let ˆ X g and H be as above with g ≥ . Then E is the class of an irreducible,smooth elliptic curve.Proof. As E is primitive with ( E ) = 0, it suffices to show that E is nef, [H, Proposition 2.3.10].The class E is effective as ( E ) = 0 and ( E · L ) >
0. If E is not nef, then there exists a smoothrational curve R with ( R · E ) <
0. Obviously, R must be a component of the base locus of | E | . Write R = aL + bE + P j =1 c j N i , for a, b ∈ Z , c j ∈ Z . From ( R · E ) <
0, we have a <
0. Since L is nef, ( R · L ) ≥ b ≥ − a ( g − E − R is effective, so (cid:0) L · ( E − R ) (cid:1) ≥
0, giving b ≤ − a ( g − b = − a ( g − b = 1 − a ( g − R · L ) = 2( a ( g −
1) + b ) ≤
2. The Hodge Index Theorem applied to L and aL + bE yields(2 g − (cid:16) X j =1 c j − (cid:17) ≤ ( R · L ) ≤ . Thus P j =1 (2 c j ) ≤ g ≥
6, as 2 c j ∈ Z for all j . Since N is generated by N , . . . , N , and e ,the integers 2 c j all have the same parity for j = 1 , . . . ,
8, and so there are only two possibilities,namely c j = 0, for all j , or there is an index ℓ with c ℓ = − c j = 0, for j = ℓ (intersectingwith N j , shows that c j ≤ j as in the previous lemmas). In the case c = . . . = c = 0,we have either R = a ( L − ( g − E ) or R = aL + (1 − a ( g − E . Then ( R ) = − a ( g −
1) = 1 which is obviously impossible, or 1 = a ( a ( g − − a <
0. In the case that there exists ℓ with c ℓ = 1 and c j = 0, for j = ℓ , we have either R = a ( L − ( g − E ) − N j or R = aL + (1 − a ( g − E − N j . Then ( R ) = − a ( g −
1) = 0 or 2 a (2 − a ( g − a = 0, contradicting a < (cid:3) As a consequence of the above, we see that any smooth curve D ∈ | H | is hyperelliptic, with O D ( E ) defining a pencil g . The next two technical lemmas will be needed later. Lemma 5.7.
Let ˆ X g and H be as above with g ≥ odd. Then L − g − E is base point free.Proof. We have ( L − g − E ) = 0 and L · ( L − g − E ) = g − >
0, so L − g − E is effective andit suffices to show that it is nef. Suppose by contradiction that there is a smooth rational curve R with R · ( L − g − E ) < R = aL + bE + P j =1 c j N j as above. By intersecting with N j , we see that c j ≤ j = 1 , . . . ,
8. Next, since R is a component of L − g − E , the class L − g − E − R must be effective, and intersecting it with the nef class E gives a ≤
1. Further, R · ( L − g − E ) < R · L ) ≥
0, so ( R · E ) > a = 1. Intersecting the effective class L − g − E − R with the nef class L now yields b ≤ − g − . Hence ( L + bE ) = 2 g − b ≤ R ) = − L + bE ) = 2 P j =1 c j −
2. Thus P j =1 (2 c j ) ≤
4. Using that, asin the previous lemma 2 c j ∈ Z have the same parity, we distinguish two possibilities: either all c j = 0, or else, there exists ℓ with c ℓ = − c j = 0 for j = ℓ .In the former case, R = L + bE , with b ≤ − g − , and ( R ) = − − g −
1) + 4 b . As g is odd, 2( g −
1) + 4 b is divisible by 4, a contradiction. In the latter case R = L + bE − N ℓ and( R ) = − − g −
1) + 4 b −
2, which produces b = − g − . But then ( L − g − E ) · R =( L − g − E ) · ( L − g − E − N ℓ ) = 0, contradicting the assumptions. (cid:3) Lemma 5.8.
Let ˆ X g and H be as above with g ≥ odd. Then the class cE − e ∈ Pic( ˆ X g ) isnot effective for any c ∈ Z .Proof. By intersecting with L , we see that cE − e is not effective if c <
0. Suppose there existsan integer c > cE − e is effective and we choose c minimal with this property. Since( cE − e ) = −
4, there is an integral component R of cE − e with ( R · cE − e ) <
0. Necessarily,( R ) = −
2. Write R = aL + bE + P j =1 c j N j , as above. We have ( N j · cE − e ) = 1, so R = N j and ( R · N j ) ≥
0, implying c j ≤
0, for j = 1 , . . . ,
8. Intersecting R with the nef class E yields a ≥
0. Since R is a component of cE − e , we have that cE − e − R is an effective class which weintersect with E , forcing a = 0. From ( R ) = −
2, we have P j =1 c j = 1. As the integers 2 c j allhave the same parity, the only possibility is that there exists ℓ with c ℓ = − c j = 0 for j = ℓ .Then R = bE − N ℓ , and intersecting with L shows b ≥
0. We have b >
0, for − N ℓ is not effective.But then cE − e − R = ( c − b ) E − e + N ℓ is effective. Since N ℓ · (( c − b ) E − e + N ℓ ) = − <
0, wenecessarily have that N ℓ is a component of ( c − b ) E − e + N ℓ , so that ( c − b ) E − e is effective.This contradicts the minimality of c . (cid:3) We are now in a position to show that for the hyperelliptic Nikulin surface ˆ X g constructedbefore, the vanishing statements (1) hold. Corollary 5.9.
Let ˆ X g and H be as above with g = 2 i + 3 ≥ and D ∈ | H | be smooth andirreducible. Then H (cid:16) D, V i − M K D ⊗ K ⊗ D ⊗ L ∨ D (cid:17) = 0 and H (cid:16) D, V i − M K D ⊗ K ⊗ D ⊗ L ∨ D (cid:17) = 0 .Proof. For the first vanishing, we need to show that H (cid:16) D, V i − M K D ⊗ K D ⊗ e ∨ D (cid:17) = 0. As in[FMP, Prop. 3.5], the kernel bundle M K D splits into a direct sum of line bundles and we have i − ^ M K D ≃ O D (cid:16) − ( i − E D (cid:17) ⊕ ( g − i − ) . Using that K D ≃ ( g − E D , it suffices to show h ( D, O D (( i + 1) E D − e D )) = 0. From the abovelemmas, h ( ˆ X g , O ˆ X g (( i + 1) E − e )) = 0 and the primitive class L − ( i + 1) E is base point free,that is, it is represented by a smooth elliptic curve. Thus H (cid:0) ˆ X g , O ˆ X g (( i + 1) E − L ) (cid:1) = H (cid:0) ˆ X g , O ˆ X g (( i + 1) E − e − H ) (cid:1) = 0and the claim follows.The second vanishing boils down to h (cid:0) D, O D (( i + 2) E D − e D ) (cid:1) = 0. The class ( i + 2) E − e hasself-intersection − H ( ˆ X g , O ˆ X g (( i + 2) E − e ) (cid:1) = 0 by the above lemma. Moreover H (cid:0) ˆ X g , O ˆ X g (( i + 2) E − e ) (cid:1) = 0, since e − ( i + 2) E is not effective, having negative intersectionwith L . Also H (cid:0) ˆ X g , O ˆ X g (( i + 2) E − e ) (cid:1) = 0 and H (cid:0) ˆ X g , O ˆ X g (( i + 2) E − e − H ) (cid:1) = 0, since H − ( i + 2) E + e is not effective. Indeed, otherwise the smooth elliptic curve E would be asubdivisor of L − ( i + 1) E , which is also the class of a smooth elliptic curve. (cid:3) Proof of Theorem 1.2.
By deformation to the hyperelliptic case ˆ X g , it then follows that there isa nonempty, open subset of the moduli space F N g of Nikulin surfaces X g , such that H (cid:16) D, i − ^ M K D ⊗ K ⊗ D ⊗ L ∨ D (cid:17) = 0 and H (cid:16) D, i − ^ M K D ⊗ K ⊗ D ⊗ L ∨ D (cid:17) = 0 , for a general D ∈ | H | . In particular, K i − , ( D, − L D , K D ) = 0 and K i − , ( D, − L D , K D ) = 0.Putting everything together, we obtain K g − , ( C, K C ⊗ e C ) = K g − , ( C, K C ⊗ e C ) = 0, for ageneral curve C ∈ | L | general on such X g . (cid:3) The syzygy divisor on M g, g and the divisorial case of the Green-LazarsfeldConjecture The goal of this section is to prove Theorem 1.6. We use the convention that if M is a Deligne-Mumford stack, then we denote by M its coarse moduli space. All Picard and Chow groupsof stacks and coarse moduli spaces considered in this section are with rational coefficients. Werecall [AC] that the coarsening map M g,n → M g,n induces an isomorphism between (rational)Picard groups Pic( M g,n ) ∼ = → Pic( M g,n ). In particular, we shall stop distinguishing betweendivisor classes on M g,n and on M g,n respectively. HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 19
Recall that we have set g = 2 i + 1 and d = 2 g and defined Syz to be the divisor in M g, g ofpointed curves [ C, x , . . . , x g ] such that K i − , (cid:16) C, O C ( x + · · · + x g ) (cid:17) = 0 . To ease notation, we set L := O C ( x + · · · + x g ). We first express determinantally over modulithe condition that a point belong to the divisor Syz . Let M P g := Ω P g (1) be the universal rank g kernel bundle over P g . Then K i − , ( C, L ) = 0 if and only if H (cid:0) P g , V i − M P g ⊗ I C/ P g (2) (cid:1) = 0,or equivalently, the morphism between the following vector spaces of the same dimension ϕ [ C,L ] : H (cid:16) P g , i − ^ M P g (2) (cid:17) −→ H (cid:16) C, i − ^ M L ⊗ L ⊗ (cid:17) is an isomorphism. The statement that the two vector spaces above have the same dimension,follows because on one hand, it is well-known that h (cid:16) P g , i − ^ M P g (2) (cid:1) = h (cid:16) P g , Ω i − P g ( i + 1) (cid:17) = i (cid:18) i + 3 i (cid:19) , on the other hand, it is also known [La2] that M L is a stable vector bundle on C , hence H (cid:0) C, V i − M L ⊗ L ⊗ (cid:1) = 0, for slope reasons; since µ ( M L ) = −
2, one computes that h (cid:16) C, i − ^ M L ⊗ L ⊗ (cid:17) = (cid:18) i + 1 i − (cid:19)(cid:16) ( i − µ ( M L ) + 2 d + 1 − g (cid:17) = i (cid:18) i + 3 i + 1 (cid:19) . To express
Syz as a degeneracy locus of two vector bundles over the stack M g, g , we firstintroduce the following diagram of moduli stacks of pointed curves X v −−−−→ M g, g y f y π C g u −−−−→ M g where u : C g → M g is the universal curve. For j = 1 , . . . , g , let q j : M g, g −→ X be thesection of the universal family v defined by q j ([ C, x , . . . , x g ]) := (cid:0) [ C, x , . . . , x g ] , x j (cid:1) . We set E j := Im( q j ). For 1 ≤ j ≤ g , we denote by ψ j ∈ Pic( M g, g ) the cotangent class correspondingto the j -th marked point, that is, characterized by fibres ψ j (cid:0) [ C, x , . . . , x g ] (cid:1) := T ∨ x j ( C ). Finally, λ := c ( u ∗ ( ω u )) ∈ Pic( M g ) denotes the Hodge class.For ℓ ≥
1, we set F ℓ := v ∗ (cid:0) O X ( ℓE + · · · + ℓE g ) (cid:1) . By Grauert’s Theorem it follows that F ℓ isa vector bundle of rank (2 i + 1)(2 ℓ −
1) + 1. We define the kernel vector bundle over X via theevaluation sequence: 0 −→ M −→ v ∗ F −→ O X ( E + · · · + E g ) −→ . Clearly, M | v − ([ C,x ,...,x g ]) = M L , where we recall that we have set L = O C ( x + · · · + x g ).Next, for integers p ≥ q ≥
2, we introduce the vector bundle G p,q := v ∗ (cid:16) p ^ M ⊗ O X ( qE + · · · + qE g ) (cid:17) . Observe that G ,q = F q . Using the vanishing H (cid:0) C, V p M L ⊗ L ⊗ q (cid:1) = 0 valid for each line bundle L ∈ Pic g ( C ) and integer q ≥
2, we conclude that G p,q is locally free over M g, g . Furthermore,there are exact sequences of vector bundles(11) 0 −→ G p,q −→ p ^ G , ⊗ G ,q −→ G p − ,q +1 −→ , globalizing the corresponding exact sequences at the level of each individual curve. Followingthe path indicated in [Fa1], for each p ≥ q ≥
2, we define a vector bundle H p,q over M g, g such that H p,q (cid:0) [ C, x , . . . , x g ] (cid:1) := H (cid:16) P g , p ^ M P g ( q ) (cid:17) , where the last identification takes into account the embedding φ L : C ֒ → P g . To define thebundles H p,q , we proceed inductively. First, we set H , := G , and then H ,q := Sym q H , .Then, having defined H j,q for all j < p , we define H p,q via the following exact sequence:(12) 0 −→ H p,q −→ p ^ H , ⊗ H ,q −→ H p − ,q +1 −→ . There exist vector bundle morphism ϕ p,q : H p,q → G p,q , which over each fibre corresponding toan embedding C | L | ֒ → P g are the restriction maps at the level of twisted holomorphic forms: H (cid:16) P g , p ^ M P g ( q ) (cid:17) → H (cid:16) C, p ^ M L ⊗ L ⊗ q (cid:17) . Note that rk( H i − , ) = rk( G i − , ) and ϕ = ϕ i − , : H i − , → G i − , is the morphism whosedegeneracy locus is precisely the divisor Syz . The following formulas are standard, see [HM]:
Lemma 6.1.
Keeping the notation from above, the following identities hold: (1) (i) v ∗ ( f ∗ c ( ω u ) ) = 12 λ . (2) (ii) v ∗ ( v ∗ λ · f ∗ c ( ω u )) = (2 g − λ . (3) (iii) v ∗ ([ E j ] · v ∗ λ ) = λ . (4) (iv) v ∗ ([ E j ] · f ∗ c ( ω u )) = ψ j , for ≤ j ≤ g . (5) (v) v ∗ ([ E j ] ) = − ψ j , for ≤ j ≤ g . Using the exact sequence (11) and (12), we shall reduce the calculation of the first Chernclass of G p,q and H p,q respectively to the case p = 0. We have the following result in that case: Proposition 6.2.
For ℓ ≥ , the following relation holds in CH ( M g, g ) : c ( G ,ℓ ) = λ − (cid:18) ℓ + 12 (cid:19) g X j =1 ψ j . Proof.
We apply Grothendieck-Riemann-Roch to the proper morphism of stacks v : X → M g, g and to the sheaf L := O X ( E + · · · + E g ). Note that R v ∗ ( L ⊗ ℓ ) = 0, therefore we can write: c ( G ,ℓ ) = v ∗ h(cid:16) ℓ g X j =1 [ E j ] + ℓ (cid:16) g X j =1 [ E j ] (cid:17) + · · · (cid:17) · (cid:16) − f ∗ c ( ω u )2 + f ∗ c ( ω u )12 + · · · (cid:17)i . Applying repeatedly Lemma 6.1, we get the claimed formula. (cid:3)
We now prove Theorem 1.6. The morphism ϕ : H i − , → G i − , degenerates along Syz . In thecourse of proving Theorem 1.3, we have exhibited a pair [
C, L ] ∈ Pic gg with K i − , ( C, L ) = 0.Therefore ϕ is generically non-degenerate. In the next proof, we shall use the formulas c (cid:0) Sym n ( E ) (cid:1) = (cid:18) r + n − r (cid:19) c ( E ) and c (cid:0) n ^ E (cid:1) = (cid:18) r − n − (cid:19) c ( E )valid for a vector bundle E of rank r on any stack or variety. Proof of Theorem 1.6.
From the discussion above, [
Syz ] = c ( G i − , − H i − , ) ∈ CH ( M g,g ).We compute both Chern classes via the exact sequences (11) and (12) respectively, coupled with HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 21
Proposition 6.2 and write c ( G i − , ) = i − X ℓ =0 ( − ℓ c (cid:16) i − ℓ − ^ G , ⊗ G ,ℓ +2 (cid:17) = i − X ℓ =0 ( − ℓ h rk( G ,ℓ +2 ) · (cid:18) i + 1 i − ℓ − (cid:19) c ( G , ) + (cid:18) i + 2 i − ℓ − (cid:19) c ( G ,ℓ +2 ) i = i − X ℓ =0 ( − ℓ h(cid:18) i + 1 i − ℓ − (cid:19)(cid:16) (2 i + 1)(2 ℓ + 3) + 1 (cid:17) c ( G , ) + (cid:18) i + 2 i − ℓ − (cid:19)(cid:16) λ − (cid:18) ℓ + 32 (cid:19) g X j =1 ψ j (cid:17)i == (cid:18) ii (cid:19)(cid:16) i + 5 i − i − i + 1)( i + 2) · λ − i + 13 i − i − i + 1)( i + 2) · g X j =1 ψ j (cid:17) . On the other hand, recalling that H ,ℓ +2 = Sym ℓ +2 H , and that H , = G , , we have c ( H i − , ) = i − X ℓ =0 ( − ℓ c (cid:16) i − ℓ − ^ H , ⊗ H ,ℓ +2 (cid:17) = i − X ℓ =0 ( − ℓ h(cid:18) i + 2 i − ℓ − (cid:19) c ( H ,ℓ +2 ) + (cid:18) i + 1 i − ℓ − (cid:19) · rk( H ,ℓ +2 ) · c ( H , ) i = i − X ℓ =0 ( − ℓ h(cid:18) i + 2 i − ℓ − (cid:19)(cid:18) i + 3 + ℓℓ + 1 (cid:19) c ( H , ) + (cid:18) i + 1 i − ℓ − (cid:19)(cid:18) i + 3 + ℓℓ + 2 (cid:19) c ( H , ) i = (cid:18) ii (cid:19) i (2 i + 1)(2 i + 3)( i + 1)( i + 2) · (cid:16) λ − g X j =1 ψ j (cid:17) . Computing the difference c ( G i − , − H i − , ), we obtained the claimed formula for the class[ Syz ]. (cid:3) Finally, we explain how the divisorial case d = 2 g of the G-L Secant Conjecture (Theorem1.4) implies the conjecture for line bundles of extremal degree d = 2 g + p + 1 − Cliff( C ) on ageneral curve C . The argument proceeds in two steps. Proof of Theorem 1.7 in the case d = 2 g + 1 . We begin with the case of odd genus and take asmooth curve C of genus g = 2 i + 1 of Clifford index i , then set d := 3 i + p + 3, where i ≥ p + 1.In particular, inequality (2) is an equality, while inequality (4) is satisfied as well.We start with a line bundle L ∈ Pic d ( C ) and let us assume that K p, ( C, L ) = 0. We set ξ := L − K C ∈ Pic p − i +3 ( C ). By using Lemma 2.1, for every effective divisor D ∈ C i − p − , weobtain that K i − , ( C, L ( D )) = 0. Theorem 1.4 implies L + D ∈ K C + C i +1 − C i − , that is, ξ + C i − p − ⊂ C i +1 − C i − . This implies that dim V g − p − g − p − (2 K C − L ) ≥ i − p −
1, which is contradiction, hence K p, ( C, L ) = 0.Assume now that C is a curve of even genus g = 2 i and Cliff( C ) = i −
1. We grant Theorem1.5. Out of this, we derive Theorem 1.7 for all remaining cases. Assume d = 3 i + p + 2, where i > p + 1 (the case i = p + 1 corresponds to d = 2 g + 1, which is our hypothesis).If L ∈ Pic d ( C ) satisfies K p, ( C, L ) = 0, then by Lemma 2.1, we obtain K i − , ( C, L ( D )) = 0,for each divisor D ∈ C i − p − . Since deg( L ( D )) = 4 i + 1 we can apply Theorem 1.5 in that caseand conclude that L − K C + C i − p − ⊂ C i +1 − C i − , that is, dim V g − p − g − p − (2 K C − L ) ≥ i − p − (cid:3) Theorem 1.5, that is, the situation when g is even and d = 2 g + 1 can be viewed as a limitcase of the divisorial case dealt with by Theorem 1.4. Proof of Theorem 1.5.
Let C be a general curve of genus g = 2 i and L ∈ Pic i +1 ( C ) a linebundle with K i − , ( C, L ) = 0. Write L = O C ( x + · · · + x i +1 ), for distinct points x j ∈ C , andaim to show that L − K C ∈ C i +1 − C i − . We fix general points x, y ∈ C and let X := C ∪ E be the nodal curve of genus 2 i + 1 obtained byattaching to C at the points x and y a rational curve E . Since C has only finitely many pencils g i +1 and x and y are chosen generically, gon( X ) = i + 2. On X , we consider the line bundle L X of bidegree (4 i + 1 , L X | C = L and L E = O E (1) respectively. Choosinga point x i +2 ∈ E − { x, y } , note that [ X, L X ] corresponds to the (2 g + 2)-pointed stable curve[ X, x , . . . , x i +2 ] ∈ M g +1 , g +2 under the Abel-Jacobi map M g +1 , g +2 Pic g +2 g +1 .We show that K i − , ( X, L X ) = 0. To that end, we write the duality theorem [G] K i − , ( X, L X ) ∨ ∼ = K i +1 , ( X, ω X , L X ) . Using the isomorphisms H ( X, L X ) ∼ = H ( C, L ) and H ( X, ω X ) ∼ = H ( C, K C ( x + y )) respec-tively, we write the following commutative diagram, where the vertical arrows are both injective: V i +1 H ( C, L ) ⊗ H ( C, K C ) d i +1 , −−−−→ V i H ( C, L ) ⊗ H ( C, L ⊗ K C ) y yV i +1 H ( X, L X ) ⊗ H ( X, ω X ) d i +1 , −−−−→ V i H ( X, L X ) ⊗ H ( X, L X ⊗ ω X )This yields an injection 0 = K i +1 , ( C, K C , L ) ֒ → K i +1 , ( X, ω X , L X ). This implies the non-vanishing K i − , ( X, L X ) = 0, that is, [ X, x , . . . , x g +2 ] ∈ Syz , the closure being taken inside M g +1 , g +2 . Recall that we have established the following equality of closed sets in M g +1 , g +2 : Syz = Sec ∪ Hur . As already pointed out, X has maximal gonality, that is, [ X, x , . . . , x g +2 ] / ∈ Hur , thereforenecessarily [
X, x , . . . , x g +2 ] ∈ Sec . Rather than trying to understand the meaning of thesecant condition on the singular curve X , we recall the identification [FMP] Y i +1 − Y i − = Θ V i − Q Y ⊂ Pic ( Y ) , valid for any smooth curve Y of genus 2 i + 1. The right-hand-side of this equality makes sensefor semi-stable curves as well, in particular for X . Thus L X ⊗ ω ∨ X ∈ Θ V i − Q ωX , and by duality,(13) H (cid:16) X, i − ^ M ω X ⊗ ω ⊗ X ⊗ L ∨ X (cid:17) = 0 . Observe that M ω X | C = M K C ( x + y ) and M ω X | E = O ⊕ gE , hence we have the Mayer-Vietoris sequence0 −→ i − ^ M ω X ⊗ ω ⊗ X ⊗ L ∨ X −→ (cid:16) i − ^ M K C ( x + y ) ⊗ (cid:0) K C − L + 2 x + 2 y (cid:1)(cid:17) M(cid:16) O E ( − ⊕ ( gi − ) (cid:17) −→ (cid:16) i − ^ M ω X ⊗ ω ⊗ X ⊗ L ∨ X (cid:17)(cid:12)(cid:12) x,y → . Condition (13) yields H (cid:16) C, V i − M K C ⊗ (2 K C − L + x + y ) (cid:17) = 0. Using again [FMP], thisnon-vanishing translates into the condition(14) L − K C − x − y ∈ (cid:0) C i − C i − (cid:1) + (cid:0) C i +1 − C i − − x − y (cid:1) ⊂ Pic ( C ) , HE GENERIC GREEN-LAZARSFELD SECANT CONJECTURE 23 which holds for arbitrary points x, y ∈ C . If, for some x, y ∈ C , the bundle L − K C − x − y belongsto the second component of the cycle in (14), we are finished. Else, if L − K C − C ⊂ C i − C i − ,then we find that dim V i +1 i +2 ( L ) ≥
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Humboldt-Universit¨at zu Berlin, Institut f¨ur Mathematik, Unter den Linden 610099 Berlin, Germany
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