Syzygies of curves via K3 surfaces

Michael Kemeny

Humboldt-Universitat zu Berlin

August 28, 2015.Haeundae, Busan, Korea

Notation

I L denotes a globally generated line bundle on a smooth, projectivecurve C .

I For M any line bundle Ki,j(C ,M; L) denotes the middle cohomologyin the sequence

i+1∧H0(C , L)⊗ H0(C , (j − 1)L + M)→

i∧H0(C , L)⊗ H0(C , jL + M)

→i−1∧

H0(C , L)⊗ H0(C , (j − 1)L + M)

I Ki,j(C , L) is short for Ki,j(C ,OC ; L)

I The line bundle L is said to satisfy property (Np) if Ki,j(C , L) = 0for i ≤ p, j ≥ 2.

p-very ampleness

A line bundle L is called p-very ample iff for any effective divisor D ofdegree p + 1

ev : H0(C , L)→ H0(D, L|D )

is surjective. Equivalently, L is not p v. a. iff φL : C → Pr 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 ofgenus g such that

d ≥ 2g + p + 1− 2h1(C , L)− Cliff(C ).

Then L fails property Np if and only if L is not p + 1-very ample.

The line bundle L is not p + 1-very ample iff φL admits a (p + 2)-secantp plane.

Special case: p = 0

Firstly the direction “L not (p + 1) v.a. =⇒ L fails Np” isstraightforward and was shown by GL when they formulated theconjecture. So the conjecture concerns the other direction.

For L very ample, condition N0 is equivalent to φL : C → Pr beingprojectively 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) ≥ 2g + 1− 2h1(C , L)− Cliff(C ).

Then φL : C → Pr is projectively normal.

Special case L = ωC

We next consider the special case L = ωC .Assume p < Cliff(C ). Then ωC is p + 1-very ample.Indeed, otherwise ∃D ∈ Cp+2 such that

H0(ωC )→ H0(D, ωC |D )

is non-surjective. Equivalently H0(D, ωC |D )→ H1(C , ωC (−D)) isnonzero, so the surjection

H1(C , ωC (−D)) H1(C , ωC )

is non-injective. Thus h0(C ,O(D)) = h1(C , ωC (−D)) ≥ 2 soCliff(C ) ≤ p which is a contradiction.

Green’s conjecture

Thus the Secant Conjecture becomes, in case L = ωC :

Conjecture (Green’s Conjecture, 1984)The canonical bundle ωC satisfies (Np) for p < Cliff(C ).

K3 Surfaces

A K3 surface is a smooth, projective surface with h1(X ,OX ) = 0 andωX ' OX . By the adjunction formula, if C ⊆ X is smooth curve thenO(C )|C ' ωC .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 iseffective and base point free and either:

1. L the trivial line bundle OX , or

2. (L · H) > 0.

In addition assume H1(X , qH − L) = 0 for all q ≥ 0. Then, for anysmooth D ∈ |H|, restriction gives an isomorphism

Kp,q(X ,−L;H) ' Kp,q(D,−L;ωD).

Proof of Lefschetz Thm

Proof.By the assumptions we have, for each q ∈ Z, a short exact sequence

0→ H0(X , (q − 1)H − L)s→ H0(X , qH − L)→ H0(D, qωD − L|D )→ 0,

where the first map is multiplication by the section s ∈ H0(X ,H)defining D.Thus we have a s.e.s of graded SymH0(X ,H) modules

0→⊕q

H0(X , (q−1)H−L)→⊕q

H0(X , qH−L)→⊕q

H0(D, qωD−L|D )→ 0.

Taking Koszul cohomology leads to the l.e.s

→ Kp,q−1(X ,−L,H)s→ Kp,q(X ,−L,H)→ Kp,q(B,H0(X ,H))

→ Kp−1,q(X ,−L,H)s→

where B is the graded SymH0(X ,H) module⊕

q H0(D, qωD − L|D ).

Proof Cont.

Proof (Cont.)But multiplication by s induces the zero map on cohomology so we have

Kp,q(B,H0(X ,H)) ' Kp,q(X ,−L,H)⊕ Kp−1,q(X ,−L,H).

Next we have

0→ H0(X ,OX )s→ H0(X ,H)→ H0(D,KD)→ 0

so choose a splitting H0(X ,H) ' H0(D,KD)⊕ Cs. This gives

p∧H0(X ,H) '

p∧H0(D,KD)⊕

p−1∧H0(D,KD).

A diagram chase then givesKp,q(B,H0(X ,H)) ' Kp,q(D,−L, ωD)⊕ Kp−1,q(D,−L, ωD). Inductionon p now gives the claim.

Consequences

The Lefschetz Theorem has some interesting consequences for sections ofK3 surfaces. For example, if H is very ample then it implies

Kp,q(D1, ωD1) ' Kp,q(X ,H) ' Kp,q(D2, ωD1)

for any two smooth sections D1,D2 ∈ |H|. Thus, Green’s conjecturewould imply that Cliff(D) is constant for all D ∈ |H|. This was verifiedby Green and Lazarfseld in 87.

In fact, one can always compute the Clifford index of sections of a veryample 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 pointfree and (H)2 > 0. Assume there is no decomposition H = L1 ⊗ L2 forline bundles Li with h0(Li ) ≥ 2 for i = 1, 2. Then a general smooth curveC ∈ |H| is Brill–Noether–Petri general.

In particular, Cliff(C ) = b g(C)−12 c in the conclusion of the theorem above.

Green’s conjecture for curves on K3

The results on the previous slides suggest that it might be a good idea totry and prove Green’s conjecture for a curve C on a K3 surface. This wasdone, by C. Voisin in 2002 and 2005, in the case Pic(X ) ' Z[C ] and byAprodu–Farkas (2011) in general:

Theorem (Voisin, Aprodu–Farkas)Green’s Conjecture holds for a smooth curve C ⊆ X lying on a K3surface X .

In particular we deduce by semicontinuity the fact that Green’sconjecture holds for general curves. But we also get even more, e.g. byapplying the Thm to K3s with Picard rank two generated by classes Cand E with (E )2 = 0, (C · E ) = k one sees that Green’s conjecture holdsfor general curves of each gonality (warning: this is not how this fact wasfirst proven!).

Kernel Bundles

We will need the kernel bundle description of Koszul cohomology. Let Lbe a globally generated line bundle on a variety. Set

ML := Ker(H0(X , L)⊗OX X ),

where the morphism is evaluation of sections. For any second line bundleF (or coherent sheaf), we have

Kp,q(X ,F ; L)

' coker(

p+1∧H0(L)⊗ H0(X ,F ⊗ Lq−1)→ H0(

p∧ML ⊗ F ⊗ Lq))

' ker(H1(

p+1∧ML ⊗ F ⊗ Lq−1))→

p+1∧H0(L)⊗ H1(F ⊗ Lq−1))

Kernel Bundles: Hyperelliptic curves

In particular, if either H0(∧p ML ⊗ F ⊗ Lq)) = 0 or

H1(∧p+1 ML ⊗ F ⊗ Lq−1)) = 0 then Kp,q(X ,F ; L) = 0.

One more fact will be useful. If C is a hyperelliptic curve andE ∈ Pic2(C ) denotes the g1

2 then

ωC ' (g − 1)E .

Further, the canonical morphism φωC: C → Pg−1 is the composite of

φE : C → P1 with the Veronese v : P1 → Pg−1.

Kernel Bundles: Hyperelliptic curves

The Euler sequence can be written

0→ ΩPg−1(1)→ H0(O(1))⊗OPg−1 → O(1)→ 0.

ThusMωC

' φ∗E (ΩPg−1(1)).

Further,v∗(ΩPg−1(1)) ' OP1(−1)⊕g−1.

So we haveMωC

' OC (−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 ageneral line bundle L of degree d on C .

The word general means that the property holds on a dense open subsetof the appropriate moduli space.

Proof of Generic Secant Conj.

Proof.We first state some simple reductions. In the case h1(C , L) ≥ 1, theclaim reduces to generic Green’s Conj, by an straightforward argument ofKoh–Stillman from 1989. So we may assume h1(C , L) = 0.

Assume L is p + 1 very ample with h1(L) = 0 andd ≥ 2g + p + 1− Cliff(C ). We need to show L satisfies (Np). In fact, Lis projectively normal under this bound on d (and h1(L) = 0), so by aresult of GL we only need to prove Kp,2(C , L) = 0.

As h1(C , L) = 0, then L fails to be p + 1 very ample iff there existsD ∈ Cp+2 such that h0(C , L(−D)) ≥ d − g − p.By Riemann–Roch this is equivalent to h1(C , L(−D)) ≥ 1 or−(L− D − KC ) effective.

Proof of Secant Conj.

Proof cont.In other words, if L is p + 1 very ample then

L− KC /∈ Cp+2 − C2g−d+p.

It is an easy exercise to prove that if a + b ≥ g then the difference map

Ca × Cb → Pica−b(C )

is surjective. Thus we must have (p + 2) + (2g − d + p) ≤ g − 1 or

d ≥ g + 2p + 3 (1)

in addition to

d ≥ 2g + p + 1− Cliff(C ). (2)

Proof of Secant Conj.

Proof cont.For simplicity, assume g = 2i + 1 is odd, the even case is similar.

It is easy to see h1(C , L) = 0 and Kp,2(C , L) = 0 impliesKp−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 ≥ 2p + 2i + 4.

Using the projective normality result (from two slides back) we see that itsuffices to assume

d = 2p + 2i + 4.

Proof of Secant Conj.

Proof cont.Thus we need to show that, for any p ≥ i − 1, there exists a line bundleL of degree 2p + 2i + 4 on a curve C of genus 2i + 1 satisfyingKp,2(C , L) = 0. To this end, consider a K3 surface X with Picard groupgenerated by two classes H, η giving the intersection matrix(

4p + 4 2p − 2i2p − 2i −4

).

We let C be a smooth curve in the linear system H − η, and will showKp,2(C ,H|C ) = 0.Following Green, from the sequence of graded Sym(H0(X ,H)) modules

0→⊕q

H0(X , qH − C )→⊕q

H0(X , qH)→⊕q

H0(C , qH)→ 0

we deduce the long exact sequence

→ Kp,2(X ,H)→ Kp,2(C ,H)→ Kp−1,3(X ,−C ,H)→

Proof of Secant Conj. cont.

Proof cont.Using Lazarsfeld’s theorem, one shows D is BNP general for D ∈ |H|general and thus Cliff(D) = p + 1 for any D ∈ |H| smooth. By theLefschetz hyperplane theorem and Green’s Conj for K3 surfacesKp,2(X ,H) = 0. Thus we need to show Kp−1,3(X ,−C ,H) = 0 orequivalently, Kp−1,3(D,OD(−C ), ωD) = 0, by the Lefschetz theorem. Forthis, using Lazarsfeld’s kernel bundle description of Koszul cohomology itis in turn sufficient to show

H1(D,

p∧MKD

⊗ KD ⊗ η|D ) = 0

where MKDis the kernel of

ev : H0(D,KD)⊗OD KD .

Proof of Secant Conj. cont.

Proof cont.For the last vanishing H1(D,

∧p MKD⊗ KD ⊗ η|D ) = 0, we further

degenerate so that D becomes hyperelliptic, by adding an elliptic curve Ewith (E · C ) = 2, (E · η) = 0 into the Picard group of the K3.

As D is hyperelliptic KD = (g(D)− 1)E . Further, MKD' OD(−E )⊕g(D)

soj∧MKD

' OD(−jE )⊕(g(D)−1j ).

This reduces the statement to the easy verification

H1(D, (p + 2)E + η) = 0.

Note that ((p + 2)E + η)|D has degree 2(g(D)− 1)− 2i , so thisvanishing is expected so long as 2i ≤ g(D)− 1. Since g(D) = 2p + 3,this is just the equality p ≥ i − 1 we started with.

The specific Secant Conj.

In some special cases we can prove effective versions of the secantconjecture. Consider the ‘extremal case’, where g = 2i + 1 is odd andd = 2g = 4i + 2, p = i − 1. This is the case where the two inequalities(1), (2) in the proof become equalities. In this case we have an optimalresult.

Theorem (Farkas–K.)The GL Secant Conjecture holds for every smooth curve C of odd genusg and every line bundle L ∈ Pic2g (C ), that is, one has the equivalence

K g−32 ,2(C , L) 6= 0 ⇔ Cliff(C ) <

g − 1

2or L− KC ∈ C g+1

2− C g−3

2.

(the condition L− KC ∈ C g+12− C g−3

2is the same as saying L is not g−1

2

very ample)

The specific Secant Conj.

The extremal case of the Secant conjecture is proven by a divisorcalculation on Mg ,2g . Let π :Mg ,2g →Mg be the forgetful morphism.Define three divisors:

1. The Hurwitz divisor M1g ,i+1 := [C ] ∈Mg : W 1

i+1(C ) 6= ∅ ofcurves with non-maximal Clifford index and its pull-backHur := π∗(M1

g ,i+1).

2. The syzygy divisor Syz :=

[C , x1, . . . , x2g ] ∈Mg ,2g :

Ki−1,2(C ,OC (x1 + · · ·+ x2g )

)6= 0.

3. The secant divisor Sec :=

[C , x1, . . . , x2g ] ∈Mg ,2g :

OC (x1 + · · ·+ x2g ) ∈ KC + Ci+1 − Ci−1

.

The specific Secant Conj.

One now calculates the class of Syz in CH1(Mg ,2g ). To do this, onefirst writes it as a degeneracy locus and then usesGrothendieck–Riemann–Roch. On the other hand, the classes of Sec andHur have already been computed in previous work of Farkas andHarris–Mumford. Comparing the formulae, one sees

[Syz] = [Sec] + i · [Hur] ∈ CH1(Mg ,2g )

where g = 2i + 1. This gives the claim.

The specific Secant Conj.

Using the extremal case we can deduce some other statements, such asthe extremal case for even genus , i.e. where deg(L) = 2g + 1, we provethe expected statement for BNP general curves and arbitrary line bundleson them:

TheoremThe Green-Lazarsfeld Conjecture holds for a Brill–Noether–Petri generalcurve C of even genus and every line bundle L ∈ Pic2g+1(C ), that is,

K g2−1,2(C , L) 6= 0 ⇔ L− KC ∈ C g

2+1 − C g2−2.

The specific Secant Conj.

Another result which follows from the extremal cases is:

TheoremLet C be a BNP general curve of genus g . For p ≥ 0, let

d = 2g + p + 1− Cliff(C ).

If L ∈ Picd(C ) is a non-special line bundle such that the secant varietyV g−p−4g−p−3 (2KC − L) has the expected dimension d − g − 2p − 4, then

Kp,2(C , L) = 0.

Here V g−p−4g−p−3 (2KC − L) denotes the locus of g − p − 3 secant g − p − 5

planes.

Natural resolutions

Let L be a globally generated line bundle on a curve C of genus g . Wesay that the resolution of the graded SymH0(C , L) moduleΓC (L) := ⊕qH

0(C , qL) is natural if, for every j at most one of the spacesKp,j−p(C , L) is non-zero for all p. Equivalently, at most one term in eachdiagonal of the Betti table is non-zero.

If h1(C , L) = 0 and d := deg(L), r := h0(C , L)− 1 then

dimKp,1(C , L)− dimKp−1,2(C , L) = p

(d − g

p

)(d + 1− g

p + 1− d

d − g).

So in fact one can determine the whole Betti table for ΓC (L) if oneknows that it has a natural resolution (by Green’s theorem the spacesKp,q(C , L) = 0 for q ≥ 3 and by the Vanishing theorem Kp,q(C , L) = 0for q ≤ 0 unless p = q = 0).

Prym–Green conjecture

Note that the generic Green’s conjecture implies that the resolution ofΓC (ωC ) is natural for the general curve C . The Prym–Green conjecturepredicts that this generalises to the general paracanonical curve, i.e. itstates that for the general element [C , η] in the moduli space Rg ,l

parametrising curves with a torsion line bundle η ∈ Pic0(C ) of order l ,the graded module ΓC (ωC ⊗ η) has a natural resolution.

The Prym–Green conjecture has been shown been tested probabilisticallyvia Macaulay and to very high likelihood fails for g = 8 and g = 16 andl = 2. This suggests one should amend the conjecture appropriately, e.g.by insisting g 6= 2k for l = 2. Nevertheless, we have . . .

Prym–Green conjecture

Theorem (Farkas, K.-)The Prym–Green conjecture holds for g odd and l = 2 or g ≥ 11,

l ≥√

g+22 . I.e. for such values ΓC (ωC ⊗ η) has a natural resolution for

the general [C , η] ∈ Rg ,l .

Proof in case l = 2

Proof.Write g = 2i + 1. Then naturality of ΓC (ωC ⊗ η) reduces to twovanishings, namely

Ki−1,1(C , ωC ⊗ η) = 0

andKi−3,2(C , ωC ⊗ η).

We will proceed by finding a K3 surface X together with two primitiveclasses C , η ∈ Pic(X ) with C integral of genus g and with at worstnodes, with η2 = −4, (η · C ) = 0 and η|C torsion of order l .

Proof of PG

Proof Cont.Assume such a K3 exists. Let H = C + η. If we can check thatH0(X ,H − C ) = 0 and H1(X , qH − C ) = 0 for all q (by setting q = 1and using RR for surfaces, this forces η2 = −4), then we get

→ Kp,q(X ,H)→ Kp,q(C ,H)→ Kp−1,q+1(X ,−C ,H)→

The proof now follows that of the GL Secant conjecture. Namely, onefirst shows that, on the given K3s, that D ∈ |H| is BNP general for Dgeneral, using Lazarsfeld’s theorem.This gives Ki−1,1(X ,H) = 0 and Ki−3,2(X ,H) = 0, via Green’sconjecture for K3s. The vanishing Ki−2,2(X ,−C ,H) = 0 andKi−4,3(X ,−C ,H) is then dealt with via a hyperelliptic degeneration.

Proof of PG

Proof Cont.It remains to construct such K3s.

For l = 2, use a classical construction of Nikulin. Namely, one takes a K3surface with symplectic involution i together with a big and nef linebundle L which is invariant under i and (L)2 = 2g − 2. There exist suchK3s of Picard rank 9 and containing a class η with the desired properties,where is C can be any smooth element of |L|.

The construction in the case l ≥√

g+22 is given by the resolution of the

following conjecture due to Barth–Verra.

Case l ≥√

g+22

Let Υg be the rank two lattice generated by elements L, η with

(L)2 = 2g − 2, η2 = −4, (L · η) = 0.

Theorem (Barth–Verra 88, Farkas–K.)Fix and l ≥ 2 and assume 2l2 − 2 ≥ g . Let Xg be a general Υg -polarised

K3 surface. Then there exist precisely

(2l2 − 2

g

)integral divisors C ∈ |L|

such that η|C ∈ Pic0(C ) is nontrivial and ηl|C ' OC . All such divisors Chave at worst nodal singularities.

Proof of Barth–Verra Conj

The count

(2l2 − 2

g

)from the previous slide does not give the number

of hyperplane sections C such that η|C has order exactly l unless l isprime; nevertheless this number is positive and is given by an explicitformula via subtracting the counts of sections with order dividing l .

The key step to proving the Barth–Verra conjecture is to degenerate to aK3 with rank three Picard lattice described in terms of the ordered basisE , Γ, η with intersection matrix: 0 1 0

1 −2 00 0 −4

Proof of Barth–Verra Conj cont.

We degenerate L to Γ + gE . The proof now runs by observing thefollowing two facts. Firstly, each divisor in |Γ + gF | has the formΓ + Fi + . . .+ Fg , for Fi ∈ |E | elliptic fibres.

Secondly, using the theory of elliptic surfaces one can show that thereexists 2l2 − 2 fibres F ∈ |E | such that ηl|F ' OF . This leads to the

desired formula (modulo a transversality statement which requiresadditional work).