Andreas Leopold Knutsen
On k th-order embeddings of K 3 surfaces and Enriques surfaces
Received: 28 March 2000 / Revised version: 20 October 2000
Abstract. We give necessary and sufficient conditions for a big and nef line bundleLof any degree on aK3 surface or on an Enriques surfaceSto bek-very ample andk-spanned.
Furthermore, we give necessary and sufficient conditions for a spanned and big line bundle on aK3 surfaceSto be birationallyk-very ample and birationallyk-spanned (our definition), and relate these concepts to the Clifford index and gonality of smooth curves in|L|and the existence of a particular type of rank 2 bundles onS.
1. Introduction
LetLbe a line bundle on a smooth connected surfaceSover the complex numbers.
Recall [Be-So2] thatL is calledk-very ample, for an integerk ≥ 0, if for any 0-dimensional subscheme(Z,OZ)of lengthh0(OZ)=k+1, the restriction map H0(L) → H0(L⊗OZ)is surjective. This definition is a natural generalization of the notions of spannedness and very-ampleness of line bundles. In fact, by definition,Lis 0-very ample if and only ifLis generated by its global sections, andLis 1-very ample if and only ifLis very ample.
There are various geometrical interpretations of the notion ofk-very ampleness.
Denoting byS[r]the Hilbert scheme of 0-dimensional subschemes ofSof lengthr, and by Grass(r, H0(L))the Grassmannian of allr-dimensional quotients ofH0(L), then the rational map
φk :S[k+1]−→Grass(k+1, H0(L)),
sending(Z,OZ)∈S[k+1]into the quotientH0(L)→H0(L⊗OZ)is a morphism ifLisk-very ample and an embedding if and only ifLis(k+1)-very ample [C-G].
Also, ifSis embedded in Ph0(L)−1via a very ample line bundleL, thenLis k-very ample if and only ifShas no(k+1)-secant(k−1)-planes.
There is also a slightly weaker condition thank-very ampleness as follows [B-F-S]:L is calledk-spanned, for an integerk ≥ 0, if for any curvilinear 0- dimensional subscheme(Z,OZ)of lengthh0(OZ)=k+1, the restriction map H0(L)→H0(L⊗OZ)is surjective. Recall that a 0-dimensional scheme(Z,OZ) is called curvilinear if dimTxZ≤ 1 for everyx ∈Zred. On a smooth irreducible A. L. Knutsen: Department of Mathematics, University of Bergen, Johs. Brunsgt 12, 5008 Bergen, Norway. e-mail: [email protected]
Mathematics Subject Classification (2000): 14J28
curve, the two notions of k-very ampleness and k-spannedness coincide for all k ≥ 0. On a smooth connected surface they coincide fork ≤ 2 [B-F-S, Lemma 3.1]. In the sequel we will show that forK3 and Enriques surfaces these two notions are equivalent for allk≥0.
In recent years a lot of work has been done in the study ofk-very ample and k-spanned line bundles on surfaces (see e.g. [Ba-So], [B-F-S], [Be-So1], [Be-So2], [Be-So3], [Be-So4], [DR1], [DR2], [Te2], [Be-Sz]), and in particular, in the classi- fication of pairs(S, L), whereSis a surface andLis ak-very ample ork-spanned line bundle onS.
In [B-F-S] (resp. [Be-So2]) Beltrametti, Francia and Sommese (resp. the first and third author) showed that ifLis nef andL2≥4k+5, andKS+Lis notk-very ample (resp.k-spanned), then there exists an effective divisorDsuch that
L.D−k−1≤D2< L.D/2< k+1.
Recently, Terakawa [Te2] showed that for line bundles of degree>4k+4 on surfaces of Kodaira dimension zero, these conditions are also sufficient. SinceKS
is numerically equivalent to zero, in particular necessary and sufficient conditions for a nef line bundleLto bek-very ample andk-spanned were granted. In fact these conditions are equivalent for thek-very ample and thek-spanned case.
IfLis a bigk-spanned line bundle on a smooth surfaceSof Kodaira dimension zero andL2≤4k+4, then eitherSis aK3 surface andL2=4k, 4k+2 or 4k+4, orSis an Enriques surface andL2=4k+4 (see Proposition 2.4 below).
In this paper we complete the description ofk-very ample andk-spanned line bundles on surfaces of Kodaira dimension zero. Our approach holds for allL2≥4k onK3 surfaces (resp. allL2≥4k+4 on Enriques surfaces), so no condition that L2≤4k+4 will be imposed, since they would not make the proofs easier. Thus, we give a unified presentation of the casesL2 ≥ 4k+5, already treated by the mentioned authors, and the cases with low values ofL2, where the results are new.
We will need a different approach than in [B-F-S] and [Be-So2], where Bogomolov stability is used, making the assumptionL2≥4k+5 necessary.
The author would like to mention that at the same time that an earlier version of this paper was written, and independently, T. Szemberg [Sz] treated the Enriques case. In particular, he showed that the caseL2=4k+4 andL k-very ample only occurs ifk=0.
For spanned and big line bundles of any degree on aK3 surface our approach also makes it possible to give a characterization of birationalk-very ampleness and birationalk-spannedness. A big and globally generated line bundleLwill be called birationallyk-very ample (resp. birationallyk-spanned), if there exists a non-empty Zariski-open subset ofSwhereLisk-very ample (resp.k-spanned). These concepts (which on aK3 surface turn out to be equivalent as well) are interesting not only in their own rights, but also because they are connected to the Clifford index of smooth curves in|L|, the minimal gonality of smooth curves in|L|[C-P] and the existence of a certain type of rank 2 vector bundle onS.
The following three theorems are the main results in this paper.
Theorem 1.1. LetLbe a big and nef line bundle on aK3 surface andk ≥ 0 an integer. The following conditions are equivalent:
(a)Lisk-very ample, (b)Lisk-spanned,
(c)L2 ≥ 4k and there exists no effective divisorD satisfying the conditions(∗) below:
2D2(≤i)L.D≤D2+k+1(≤ii)2k+2
(∗) with equality in (i) if and only ifL∼2DandL2≤4k+4, and equality in (ii) if and only ifL∼2DandL2=4k+4. Furthermore, if Lis notk-very ample (equivalentlyk-spanned), then among all divisors satisfying (∗), we can always find a smooth curve, and all smooth curves satisfying(∗)will contain a 0-dimensional scheme of degreek+1 where thek-spannedness fails (more preciselyL|Dis notk-spanned).
The corresponding result for Enriques surfaces is the following, which in the k-very ample case is similar to a result obtained independently by Szemberg in [Sz]. Our approach is slightly different.
Theorem 1.2. LetLbe a big and nef line bundle on an Enriques surface andk≥0 an integer. The following conditions are equivalent:
(a)Lisk-very ample, (b)Lisk-spanned,
(c) there exists no divisorD >0 satisfyingD2= −2,D.L≤k−1, orD2=0, D.L≤k+1.
So, unlike in theK3 case, thek-very ampleness ofLis governed by divisors with self-intersection 0 and−2. However, we will see that divisors satisfying similar conditions as the conditions(∗)play an important role also for Enriques surfaces.
We will say that a divisorDon aK3 surface satisfies the conditions(∗∗)if it satisfies the conditions(∗)and in additionD2≥0 and(L2, D2)=(4k+2, k).
Also recall that by a result of Green and Lazarsfeld [G-L] all smooth curves in a base point free linear system on aK3 surface have the same Clifford index (see Sect. 8 for more details). The same is not true for the gonality (see Remark 1.7 below).
Theorem 1.3. LetLbe a globally generated, big line bundle on aK3 surface and k ≥1 an integer. Denote bycthe Clifford index of all smooth curves in|L|. The following conditions are equivalent:
(a)Lis birationallyk-very ample, (b)Lis birationallyk-spanned,
(c)L2≥4kand there exists no effective divisorDsatisfying the conditions(∗∗), (d)L2≥4kand there exists no smooth curveDsatisfying the conditions(∗∗), (e) the minimal gonality of a smooth curve in|L|is≥k+2,
(f) c≥k, and there exists a smooth curve in|L|having gonalityc+2; orc=k−1 and all smooth curves in|L|have gonalityc+3 (in which case,L∼2D+, whereDandare smooth curves satisfyingD2=k,2= −2 andD.=1),
(g) there exists no smooth curveCin|L|containing a 0-dimensional subscheme of degreek+1 where thek-spannedness ofL|C ωCfails,
(h) there exists no rank 2 vector bundleEonS generated by its global sections satisfyingH1(E)=H2(E)=0, detE=Landc2(E)≤k+1.
The paper is organized as follows.
In Sect. 2 we fix notation and gather some results of Saint-Donat that will be needed in the rest of the paper. Then we prove in Sect. 3 that ifL2 ≥ 4kandS isK3, orL2 ≥4k+4 andSis Enriques, andL+KSis notk-very ample (resp.
notk-spanned), there exists a divisorDcontaining some (resp. some curvilinear) 0-dimensional scheme Z of degree≤ k+1 where thek-very ampleness (resp.
k-spannedness) fails and satisfying certain numerical conditions (which are(∗)in theK3 case).
In Sect. 4 we treat the Enriques case and give the proof of Theorem 1.2.
The rest of the paper will deal withK3 surfaces and the proofs of Theorems 1.1 and 1.3.
We first show, in Sect. 5, that it is always possible to find a smooth curve among all divisors satisfying(∗).
In Sect. 6 we conclude the proof of Theorem 1.1 by using such a smooth curve to show that L is notk-spanned (and hence not k-very ample). We also investigate more closely the case L2 < 4k. In any case we explicitly construct 0-dimensional schemes of degreek+1 where thek-spannedness ofLfails. This explicit construction will be needed in Sect. 7, where we show that the existence of any divisor satisfying(∗∗)will in fact imply thatL fails to bek-spanned on any Zariski-open subset ofS, but if the only divisorsD satisfying(∗)are those satisfying the special conditions D2 ≤ −2 or (L2, D2) = (4k+2, k), thenL is in fact birationallyk-spanned, even though it is notk-spanned. This shows the equivalence of parts (a)–(d) in Theorem 1.3.
In Sect. 8 we discuss the Clifford index and gonality of curves in|L|, relying upon results in [G-L] and [C-P], and finish the proof of Theorem 1.3.
Remark 1.4. Note that in [DR2] Di Rocco showed that ifLis ak-very ample line bundle of degree≤4k+4 on a surfaceSandk≥2, then(S, L)belongs to a certain list of pairs ([DR2, Table 2]1), and all the line bundles in the list are proved to be k-very ample, except for theK3 and Enriques cases. Thus, the results in this paper also complete the description ofk-very ample line bundles of degree≤4k+4 on a surface, fork≥2.
Remark 1.5. Conditions for birationalk-very ampleness and birationalk-spanned- ness can most probably be found for the other surfaces of Kodaira dimension zero, but as far as we can see, the connections to Clifford index and gonality of smooth curves in|L|are not obtained as easily as in theK3 case. In general, it would be interesting to know whether there are connections between birational k-very ampleness and birational k-spannedness and the Clifford index and gonality of smooth curves in|L|for other surfaces thanK3s.
1 Note that in this list the case of Enriques surfaces and line bundles of degrees 4k+4 are missing [DR3].
Remark 1.6. Note that fork = 0 and 1 we retrieve the special results of Saint- Donat concerning criteria for spannedness and very ampleness of line bundles on K3 surfaces. Denoting byLthe morphism defined by the complete linear system
|L|, whenLis spanned, we get the well-known result thatLis not birational if and only if all smooth curves in|L|are hyperelliptic.
Remark 1.7. The only example known where the smooth curves in a base point free linear system|L|on aK3 surface do not have constant gonality, is the famous Donagi–Morrison example [Do-Mo, (2.2)]. IfLis ample, Ciliberto and Pareschi [C-P] have showed that this is indeed the only such example, but the question remains open for the cases whereLis not ample.
In addition to the Donagi–Morrison example, the only other example known of exceptional curves in a base point free linear system on aK3 surface is an example of Eisenbud, Lange, Martens and Schreyer (see Remark 8.7). This example appears in a natural way in our treatment of birationalk-very ampleness and birationalk- spannedness (it is the second case of part (f) in Theorem 1.3), and in Sect. 8 we show that this is the only example of a base point free linear system on aK3 surface where all smooth curves are exceptional.
Unfortunately, we are not able, by our treatment, to “explain” the Donagi–
Morrison example in terms of birationalk-very ampleness and birationalk-spanned- ness, nor to treat the question of the constancy of gonality of the smooth curves in
|L|whenLis not ample. Those would be very interesting questions to treat.
2. Notation and background material
We use standard notation from algebraic geometry.
The ground field is the field of complex numbers. All surfaces are smooth algebraic surfaces.
By a curve on a surfaceSis always meant an irreducible curve (possibly sin- gular), i.e. a prime divisor. Line bundles and divisors are used with little or no distinction, as well as the multiplicative and additive notation. Linear equivalence of divisors is denoted by∼, and numerical equivalence by≡. Note that on aK3 surface linear and numerical equivalence is the same.
IfLis any line bundle on a surface,Lis said to be numerically effective, or simply nef, ifL.C ≥ 0 for all curvesC onS. In this caseLis said to be big if L2>0.
IfFis any coherent sheaf on a varietyV, we shall denote byhi(F)the complex dimension ofHi(V,F), and byχ(F)the Euler characteristic
(−1)ihi(F). If D is any divisor on a surfaceS, Riemann–Roch for D is χ(OS(D)) =
1
2D.(D−KS)+χ(OS), whereKSis the canonical bundle ofS.
IfDis any effective divisor onS, andLany line bundle onD, Riemann–Roch yieldsχ(L)=degL+χ(OD)=degL−12D(D+KS).
By an Enriques surface is meant a surfaceS withH1(OS)=0 and such that the canonical bundleKS satisfiesKS OS, andKS2 OS. Recall that we also haveh0(KS)=h1(KS)=h2(OS)=0,h2(KS)=h0(OS)=1 andχ(OS)=1.
By aK3 surface is meant a surfaceS with trivial canonical bundle and such thatH1(OS)=0. In particularh2(OS)=1 andχ(OS)=2.
We will make use of the following results of Saint-Donat on line bundles on K3 surfaces. The first result will be used repeatedly, without further mention.
Proposition 2.1. [SD, Cor. 3.2] A complete linear system on aK3 surface has no base points outside of its fixed components.
Proposition 2.2. [SD, Prop. 2.6] LetLbe an invertible sheaf on aK3 surfaceS such that|L| = ∅and such that|L|has no fixed components. Then either
(i) L2 > 0 and the generic member of|L|is an irreducible curve of arithmetic genusL2/2+1. In this caseh1(L)=0, or
(ii)L2=0, thenLOS(kE), wherekis an integer≥1 andEis an irreducible curve of arithmetic genus 1. In this caseh1(L)=k−1 and every member of
|L|can be written as a sumE1+...+Ek, whereEi ∈ |E|fori=1, ..., k. Note that by Bertini the generic members in|L|and|E|are smooth.
Lemma 2.3. [SD, 2.7] LetLbe a nef line bundle on aK3 surfaceS. Then|L|is not base point free if and only if there exist smooth irreducible curvesE,and an integerk≥2 such that
L∼kE+, E2=0, 2= −2, E.=1.
In this case, every member of|L|is of the formE1+...+Ek+, whereEi ∈ |E|
for alli.
The following result was mentioned in the introduction. Fork-very ampleness andk≥2, it is proved in [Ba-So, Cor. 3.2]. The proof for thek-spanned case and k≥0 is not much more involved and therefore left to the reader.
Proposition 2.4. Let L be a big and nef line bundle on a smooth surface S of Kodaira dimension zero. IfLisk-spanned andL2 ≤ 4k+4, then eitherSis a K3 surface andL2 = 4k, 4k+2 or 4k+4, or S is an Enriques surface and L2=4k+4.
3. Numerical conditions ifk-very ampleness ork-spannedness is not fulfilled We will now consider the case whereS is a K3 surface andL2 ≥ 4k, orS is an Enriques surface andL2 ≥ 4k+4, andKS⊗Lfails to bek-very ample or k-spanned.
The following result is due to Beltrametti, Francia and Sommese.
Proposition 3.1. LetLbe a nef and big line bundle on a surfaceS and letZ be any 0-dimensional subscheme ofSsuch that degZ =k+1. Assume that the map
H0(KS⊗L)−→H0(KS⊗L⊗OZ)
is not onto, and for any proper subschemeZofZ, the map H0(KS⊗L)−→H0(KS⊗L⊗OZ) is onto.
Then there exists a rank 2 vector bundleEonSfitting into the exact sequence
0−→OS −→E−→L⊗JZ −→0, (3.1)
and such that the coboundary map of the exact sequence tensorized withKS, δ:H1(L⊗KS⊗JZ)−→H2(KS)H0(OS)C, is an isomorphism.
Proof. This follows from the first part of the proof of [Be-So2, Thm. 2.1] and from [Ty, (1.12)].
Corollary 3.2. We haveh1(E⊗KS)≤h1(OS)andh2(E⊗KS)=0. In particular, ifSisK3 or Enriques, we haveh1(E⊗KS)=h2(E⊗KS)=0 (withKSOS in theK3 case).
Proof. The first assertion is immediate from the short exact sequence above (ten- sorized withKS), the second follows from the short exact sequence
0−→L⊗KS⊗JZ −→L⊗KS−→L⊗KS⊗OZ −→0, and the fact thath1(L⊗KS⊗OZ)=h2(L⊗KS)=0.
Note that from the sequence (3.1) we getc1(E)2=L2andc2(E)=degZ= k+1.
The approach in [B-F-S] and [Be-So2] is now based upon the fact that when L2≥4k+5, we havec1(E) >4c2(E). By the well-known Bogomolov stability criterion ([Bo], [Re]) one can then putEin a suitable exact sequence. Since in our casesL2 ≤ 4k+4, we need a different approach. Note that by our assumptions that L2 ≥ 4k in the K3 case and L2 ≥ 4k+4 in the Enriques case, we have c1(E)2−4c2(E)≥ −4 andc1(E)2−4c2(E)≥0 in theK3 and Enriques case, respectively.
We will need the following result by Donagi and Morrison.
Lemma 3.3. LetEbe a nonsimple, rank 2 vector bundle on a surface S. There exist line bundlesM,Nand a zero-dimensional subschemeA⊂Ssuch thatEfits in an exact sequence
0−→N −→E−→M⊗JA−→0 and either
(a)N ≥M, or
(b)A= ∅and the sequence splits.
Proof. This follows the lines of the proof of [Do-Mo, Lemma 4.4], by noting that the assumption thatSbe aK3 is unnecessary.
Now we prove the result which enables us to avoid the use of Bogomolov stability.
Proposition 3.4. LetEbe a vector bundle of rank two on aK3 surface (resp. an Enriques surface) satisfyingc1(E)2−4c2(E)≥ −4 (resp.c1(E)2−4c2(E)≥0).
Then there exist line bundlesM,Nand a zero-dimensional subschemeA⊂Ssuch thatEfits in an exact sequence
0−→N −→E−→M⊗JA−→0 (3.2)
and either (a)N ≥M, or
(b)A= ∅and the sequence splits.
Proof. For a vector bundle of rankeon a surfaceSRiemann-Roch gives χ(E⊗E∗)=(e−1)(c1(E))2−2ec2(E)+e2χ(OS).
From our assumptions we getχ(E⊗E∗)≥4 for both theK3 and the Enriques case, whenceh0(E⊗E∗)+h0(KS⊗E⊗E∗)≥4. In theK3 case, sinceKS is trivial, we geth0(E⊗E∗)≥2, and we are done by Lemma 3.3. So we will stick to the Enriques case.
If h0(E⊗E∗) ≥ 2, we are again done by Lemma 3.3. So we can assume h0(KS⊗E⊗E∗)≥3 for the rest of the proof.
Pick any non-zero section α : E → KS ⊗E in H0(KS ⊗E ⊗E∗) Hom(E, KS⊗E). Defineα:=KS⊗αand consider the composite morphism
α◦α:E−→E⊗KS2E.
Then we have three possibilities:
(i) α◦αis zero,
(ii) α◦αis a non-zero multiple of the identity, (iii) α◦αis not a multiple of the identity.
In case (iii),α◦αgives a non-trivial endomorphism ofE, whenceEis non- simple and we are done again.
In case (ii) bothαandαhave constant rank two, whenceE E⊗KS and h0(E⊗E∗)=h0(KS⊗E⊗E∗)≥3, and we are done again.
In case (i),αmust drop rank. Since detα∈Hom(L, L)H0(L)C andα is not zero,αhas constant rank equal to one, whence kerα=:N is a line bundle and we can write imα=M⊗JA, whereM:=(imα)∗∗is a line bundle andAis a zero-dimensional subscheme ofS. In other words we have an exact sequence
0−→N −→E−→M⊗JA −→0.
Sinceα◦α=0, we have imα⊆kerαN⊗KS, whenceN⊗KS⊗M∨ has a section, andN +KS ≥ M. From the short exact sequence above we get c1(E)=M+Nandc2(E)=M.N+degA, whence
(M−N)2=c1(E)2−4M.N ≥c1(E)2−4c2(E)≥0.
By Riemann–Roch eitherN−M≥0 orM−N ≥0. IfM−N >0, we would get the absurdityKS ≥M−N >0, whenceN−M≥0, and we are done.
Remark 3.5. Actually, one can get a similar result in the casec1(E)2−4c2(E)=
−2 on an Enriques surface. In fact, the whole argument goes through, except if h0(E⊗E∗)=0,h1(E⊗E∗)=1 andh2(E⊗E∗)=0. Such a vector bundle is called exceptional and by H. Kim [Ki] it sits in a non-split short exact sequence
0−→OS(D)−→E−→OS(D++KS)−→0,
for some divisorsDandsuch that2= −2,h0()=1 andh0(+KS)=0.
IfEis a vector bundle obtained as in Proposition 3.1, we haveD >0.
We leave to the reader to work out the details.
Note that sinceLis nef, we haveN.L≥ M.Lin case (a), and in case (b) we can also assume this by symmetry. We will use this in the proof of the next result, which is parallel to [B-F-S, Prop. 1.4], but slightly different, due to the different hypotheses.
Lemma 3.6. With the same assumptions and notation as in Proposition 3.1, assume furthermore thatSis aK3 surface andL2≥4k, or thatSis an Enriques surface andL2 ≥ 4k+4. LetEbe the rank 2 vector bundle andM,N the line bundles obtained as above and fitting in the sequences (3.1) and (3.2). Then the following conditions are satisfied (withKS =0 in theK3 case):
(i) The sequenceN →E→L⊗JZobtained from (3.1) and (3.2) is nontrivial.
(ii) |M|contains an effective divisor D containingZ.
(iii) N > 0 andH1(M +KS) = H2(M) = H2(M +KS) = H2(N) = 0.
FurthermoreM2=D2≥ −2.
(iv) N−M≥0 ifSis Enriques, or ifSisK3 and eitherL2≥4k+2 orL2=4k andA= ∅.
(v) (L−2D).L≥0.
(vi) L.D−degZ≤D2, with equality if and only ifA= ∅. (vii) D2≤ 12L.D, with equality if and only ifL≡2D.
(viii)D2≤degZ, with equality if and only ifL≡2DandL2=4k+4.
(ix) L.D≤2 degZ, with equality if and only ifL≡2DandL2=4k+4.
(x) If(L.D)2=L2D2, thenL≡λD, for someλ∈Q, 2≤λ≤1+degZ/D2. Proof. (i) First we need to show thath0(−N)=0. Note thatN is nontrivial, since otherwiseN.L = M.L = L2 = 0. Assume, to get a contradiction, that | −N|contains an effective memberN0. Then, since|L|contains an effective member (because it is big and nef) andL∼M−N0,|M|contains
an effective memberM0. But thenL.M0≤L.N = −L.N0≤0 contradicts thatLis big and nef.
If the compositionN →E→L⊗JZis zero, thenN⊆ker(E→L⊗JZ), soN =OS(−D)for someD≥0, contradicting the fact thath0(−N)=0.
(ii) Tensorizing the sequences (3.1) and (3.2) withN−1and taking cohomology, we get 0=H0(E⊗N−1)⊆H0(M⊗JZ).
(iii) We first prove thatN > 0. This is clear if we are in case (a) of Lemma 3.4, since we have proved that M > 0. So we can assume thatA = ∅ in sequence (3.2). SinceL ∼ M+N by (3.1) andN.L ≥ M.L, we find N2≥ M2. From (3.1) and (3.2) we findN.M =c2(E)=k+1, soN2≥
1
2(L2−2M.N) ≥ −1 ifS isK3 andN2 ≥ 0 , ifS is Enriques. In both cases, sinceNis nontrivial, by Riemann–Roch either|N|or| −N|contains an effective member. SoN >0, since we have proved thath0(−N)=0. By Serre dualityh2(N)=h0(KS−N)=0.
The exact sequence (3.4) gives rise to an exact sequence
0−→N⊗KS −→E⊗KS−→M⊗KS −→τ −→0,
where τ is a torsion sheaf supported on a finite set. Taking cohomology and using that H1(E ⊗KS) = H2(E ⊗KS) = H2(N ⊗KS) = 0, it follows that H1(M⊗KS) = H2(M ⊗KS) = 0. Clearly we also have h2(M)=h0(KS−M)=0. By Riemann–Roch
χ(M⊗KS)=h0(M⊗KS)= 1
2M2+χ(OS), whenceM2=D2≥ −2 in both theK3 and Enriques case.
(iv) Since this is clear when we are in case (a) of Lemma 3.4, we can assume we are in case (b) of that lemma, and hence that degA=0. From (3.2) we get (N−M)2 =L2−4k−4. By our assumptions, we get(N−M)2 ≥ 0 if Sis Enriques and(N−M)2≥ −2 ifSisK3, so by Riemann–Roch either N−M≥0 orM−N ≥0. By symmetry, we can assumeN−M≥0.
(v) This is(N−M).L≥0 rewritten.
(vi) By (3.1) and (3.2) we havec2(E)=(L−D).D+degA=degZ. (vii) By Hodge Index TheoremL2D2 ≤ (L.D)2, with equality if and only if
(D.L)L ≡ (L2)D, which means thatD ≡ aL for somea ∈ Q+ (since L2= 0 andD >0, one hasD.L =0). But sinceL≡D+N, this gives N ≡bLfor someb ∈ Q+. Combining this with (v) we get 2(D.L)D2 ≤ (L.D)2, with equality if and only ifD ≡ aL,N ≡ bLandD.L =N.L. This means thata =b=1/2 andL≡2D.
(viii) Immediate from (vi) and (vii).
(ix) Immediate from (vi) and (viii).
(x) This follows from (v) and (vi).
As a consequence, we have
Proposition 3.7. With the same assumptions and notation as in Proposition 3.1, assume furthermore thatSis aK3 surface andL2≥4k, or thatSis an Enriques surface andL2≥4k+4.
Then there exists an effective divisorDpassing throughZand such that L.D−k−1≤D2(≤i) 1
2L.D(≤ii)k+1 (3.3) with equality in (i) if and only ifL≡2DandL2≤4k+4, and equality in (ii) if and only ifL≡2DandL2=4k+4.
FurthermoreL−2D≥0 ifL2≥4k+2.
As a special case, we get that ifL+KSis notk-very ample, then there exists an effective divisorDas above.
The numerical conditions in (3.3) can also be formulated as 2D2(≤i)L.D≤D2+k+1(≤ii)2k+2
(∗) with equality in (i) if and only ifL∼2DandL2≤4k+4, and equality in (ii) if and only ifL∼2DandL2=4k+4, ifSis aK3 surface, and as
(#) 2D2(≤i)L.D≤D2+k+1(≤ii)2k+2
with equality in (i) or (ii) if and only ifL≡2DandL2=4k+4, ifSis an Enriques surface.
4. k-spannedness failing on an Enriques surface
In this section we will study the Enriques case. Given a divisor satisfying(#), we will give an explicit construction of 0-dimensional schemes where thek-spannedness ofL+KSfails.
First of all, note (in both the Enriques andK3 cases) that ifDis an effective divisor satisfying the conditions(#)or(∗)fork=k0, and the middle inequality is strict, thenDwill satisfy the same conditions fork=k0−1. So ifDis a divisor satisfying the conditions(#)for the smallest integerk0, thenDwill have to satisfy D.L=D2+k0+1.
Recall the following result due to Cossec and Dolgachev [C-D, p. 134]:
Theorem 4.1. LetDbe an effective divisor withD2>0 on an Enriques surface.
There exists a divisorf >0 satisfyingf2=0 andf.D≤√ D2. As a consequence, we get:
Lemma 4.2. AssumeDis an effective divisor satisfying the numerical conditions (#)for an integerk=k0and that there are no divisors satisfying the conditions for any integerk < k0.
ThenDis of one of the following types:
(i) D2= −2,D.L=k0−1, (ii) D2=0,D.L=k0+1, (iii)D2=2,L≡2D,k0=1, (iv) D2=4,L≡2D,k0=3,
and if D is as in (iii) or (iv), then D ∼ E1+E2, where E12 = E22 = 0 and E1.L=E2.L=k0+1.
Proof. We clearly must haveD2≥ −2. Also note that the result is clear fork0≤1, so we will assumek0≥2.
The divisorDhas to be nef ifD2≥0. Indeed, if there existed a curvesuch that 2= −2 and.D <0, then.L=0 (see the proof of Claim 2 in Proposition 5.1 below) andwould satisfy the conditions(#)fork=1, a contradiction.
IfD2 ≥2, then by Theorem 4.1 we can find anf >0 such thatf2 =0 and such that(D−f )2≥D2−2√
D2 ≥0 and(D−f ).D≥D2− √
D2>0.
By Riemann-Roch and the nefness ofD, we have thatA := D−f > 0. Since L.D = L.f +L.A≤ 2k0+2, we must haveL.f =L.A = k0+1,A2 =0, D2=2 or 4 andL.D=D2+k0+1=2k0+2, whenceL≡2Dandk0=1 or 3 respectively.
We will now prove
Proposition 4.3. Let L be a big and nef line bundle on an Enriques surface S satisfyingL2≥4k0+4. Assume there exists a divisorDsatisfying the numerical conditions(#)for an integerk =k0and that there are no divisors satisfying the conditions for any integerk < k0(so that, in particular,Dmust be as in (i)–(iv) in Lemma 4.2 above).
Define F := L−D and FD := F ⊗OD. Then the generic section of FD will define a zero dimensional curvilinear scheme of degreek0+1 where thek0- spannedness of(L+KS)D:=(L+KS)⊗ODfails.
Proof. The numerical conditions above giveD2≤k0+1, with equality if and only ifL≡2D. Note also thatF.D=k0+1
SinceD2≤k0+1, we get by Riemann–Roch onD, h0(FD)≥k0+1−1
2D2≥ 1 2k0+1
2 >0.
Any non-zero sectionsinH0(FD)gives a short exact sequence (after tensorising with the dualising sheafω◦D:=KS⊗OD(D))
0→ωD◦ →(L+KS)D →G→0,
whereG is a torsion sheaf supported on the zero scheme of s, so lengthG = degFD =k0+1.
Now h1((L+KS)D) = h0(−FD) = 0 and h1(ω◦D) = h0(OD) > 0, so (L+KS)Dfails to bek0-very ample on the zero scheme ofs. To show that(L+KS)D is not evenk0-spanned, we can assumek0≥2.
IfDis as in case (iv) of Lemma 4.2, thenF ≡D≡12Lis nef. If|F|is not base point free, then sinceF2 =D2 =4, there exists by Proposition 3.7 an effective divisorBsatisfyingB2=0 andB.F =1. But this would implyB.L=2, andB would satisfy the conditions(#)fork = 1 < k0, a contradiction. So|F|is base point free and the zero scheme of a generic section ofFDis curvilinear. Hence we are done.
By Lemma 4.2, what remains are the cases where D is as in (i) or (ii). In particular,L.D≤k0+1.
We will show thatDis reduced and that FD is base point free onD. It then follows that the zero scheme of a generic section ofFD is curvilinear, which will complete the proof.
IfD is reduced and irreducible, then since degFD = k0+1 ≥ 2pa(D) = D2+2, we have thatFDis base point free onD(see e.g. [C-F, Prop. 2.3]).
So assume D = n
i=1Di, for an integer n ≥ 2, with all Di > 0. Then L.D=n
i=1L.Di ≤k0+1, and since none of theDi can satisfy the conditions (#)for anyk < k0by assumption, we easily findk0 =2 or 3,n ≤ 3,D2 =0, L.D = k0+1,D2i = −2 andL.Di ≥ 1 for alli. SinceD2 = 0, we have that Dis nef, as in the proof of Lemma 4.2, whenceD.Di = 0 for alli. This gives degFDi =(L−D).Di ≥ 1 > 2pa(Di). By [C-F, Prop. 2.3] again,FDis base point free onD.
SinceDi2= −2 for alli(and any decomposition),Dis reduced.
Remark 4.4. Note that one can argue in the same way (i.e. show thatL+KSfails to bek-very ample on any section ofFD and show that a generic such section is curvilinear) in the caseL2 ≥ 4k+5 for surfaces of Kodaira dimension zero in general, and get a simpler proof of [Te2, Thm. 5]. Actually, this approach gives the stronger result that ifLis a big and nef divisor on any surfaceS satisfying L2 ≥ 4k+5 and there exists an effective divisorDsatisfyingL.D−k−1 ≤ D2 < L.D/2 < k+1 andD.KS ≤ 0, thenKS +Lis notk-spanned (here we leave some verifications to the reader). So Del Pezzo surfaces will for instance also be included. Note thatk-very ample line bundles on Del Pezzo surfaces have been completely characterized in [DR1].
SinceLandL+KSare numerically equivalent, we have shown forL2≥4k+4, thatLisk-very ample if and only if it isk-spanned if and only if there are no divisors satisfying the conditions(#). And among these divisors, we can always find one satisfying the conditions (i) and (ii) in Lemma 4.2, by the last statement in that lemma.
To finish the proof of Theorem 1.2, it suffices to note that ifL2≤4k+2, there exists by Theorem 4.1 anf >0 satisfyingf2=0 andf.L≤ √
4k+2 ≤k+1.
Note that we have also proved the following:
Corollary 4.5. IfLis a line bundle on an Enriques surface satisfyingL2≥4k+4 andLis(k−1)-very ample but notk-very ample, then any 0-dimensional scheme Z where thek-very ampleness of Lfails is contained in an effective divisor D satisfying one of the conditions (i)–(iv) in Lemma 4.2, and any such divisorDwill contain 0-dimensional schemes where thek-spannedness ofLfails.
As a consequence of Theorem 4.1, a line bundle of degree 4k+4 cannot be k-very ample unlessk=0, as remarked also by T. Szemberg in [Sz].
However, our approach shows that also for L2 = 4k+4 and k ≥ 1, any (minimal, in the sense of Proposition 3.1) 0-dimensional schemeZ where thek- very ampleness ofL fails, is contained in a divisorD satisfying the conditions above. (By Remark 3.5 a similar result holds forL2=4k+2.)
As an easy example, a line bundle of degree 8 is not very ample (as was already known to Okonek [Ok]), and Corollary 4.5 shows that (ifL is base point free) each length 2 schemeZ thatLfails to separate actually lies on a divisorD >0 satisfying (i)D2 = −2,D.L = 0, or (ii)D2 = 0,L.D = 2, or (iii)D2 = 2, L≡2D. Conversely, given such a divisor, Proposition 4.3 shows how to explicitly find 0-dimensional schemes of length 2 thatLdoes not separate. By Theorem 4.1 a divisor of type (ii) above always exists. The existence of a divisor of type (i) corresponds toL contracting rational curves, whereas the existence of a divisor of the type (iii) is crucial to determine whetherφLis birational or not (see [Co, Lemma 5.2.7]).
The next step in the paper is to prove the corresponding result to Proposition 4.3 whenSis aK3 surface andL2≥4k. Because of the weaker conditions, we will use a different approach and start by finding smooth curves satisfying the conditions (∗).
5. Smooth curves satisfying the conditions(∗)
For the rest of the paper, we will consider the case whereSis aK3 surface.
We will now show that we can always find a smooth irreducible curve among the divisors satisfying(∗). We will actually prove a stronger result, which we will need in the next sections.
Recall that we commented in the beginning of the previous section that ifk0is the smallest integer such that there exists a divisorDsatisfying(∗), thenDwill have to satisfyL.D=D2+k0+1.
In exactly the same way it follows that ifDis any divisor withD2≥0 satisfying the numerical conditions(∗)fork=k0and there are no divisors with non-negative self-intersection satisfying the numerical conditions(∗)for k < k0, thenDwill have to satisfyL.D=D2+k0+1.
With this in mind we prove
Proposition 5.1. Let S be aK3 surface andL a big and nef line bundle onS, withL2 ≥4k0for an integerk0 ≥0. Assume there is a divisorD satisfying the numerical conditions(∗)fork=k0.
(a) IfD2≤ −2, there exists a smooth rational curveD0satisfying the conditions (∗)fork=k0.
(b) If D2 ≥ 0 and furthermore there exists no divisor with non-negative self- intersection satisfying(∗)fork < k0(so that, in particular,L.D=D2+k0+1), there exists a smooth curveD0satisfying 0≤D20≤D2,L.D0=D20+k0+1 and the conditions(∗)fork =k0, and furthermore such thatF := L−D0
satisfiesh0(F )≥h0(D0).
Proof. If D2 ≤ −2,D must have at least one smooth rational curve D0 as its component. SinceLis nef, this curve will clearly satisfyL.D0≤L.D, so we get
−4=2D02< L.D0≤L.D≤D2+k+1≤D02+k+1=k−1, soD0is the desired curve. This proves (a).
In case (b), we first show with the help of three claims that we can reduce to the case whereL2≥2D.L,H1(D)=0 andDis nef2.
Claim 1. We can assumeL2≥2D.L(equivalentlyF2≥D2orF.L≥D.L).
Proof of claim. From(∗)we have 2D.L≤4k0+4. If equality holds, thenL∼2D, so we haveL2=2D.L. If 2D.L≤4k0, orL2≥4k0+2, we also haveL2≥2D.L. So we only have the case 2D.L=4k0+2 andL2=4k0left. SinceLis big, we must have k0 ≥ 1, and fromL.D = D2+k0+1, we getD2 = k0, whence k0 ≥ 2. We have L2−2D.L= −2, which is equivalent toD.L =F.L+2 or L2 =2F.L+2, forF :=L−D. SinceL∼D+F we also getD2=F2+2.
In particularF2 ≥ 0. By Hodge Index F2L2 ≤ (F.L)2, whence 2F2 < F.L. Furthermore
F.L=D.L−2=D2+k0−1=F2+k0+1≤2k0, so we can interchangeDwithF.
Claim 2. We can assumeDis nef.
Proof of claim. Ifis a smooth rational curve (necessarily contained in the base locus of|D|) such that.D < 0, defineD := D−. Then we haveD.L ≤ D.L≤ 12L2, whence by Hodge index 2D2 ≤L.D, with equality if and only if L∼2D.
Furthermore we have
D2=D2−2D.+2=D2−2D.−2≥D2, whence
D.L≤D.L=D2+k0+1≤D2+k0+1.
From the last equation combined with 2D2 ≤L.D, we getD2 ≤k0+1, with equality if and only ifL∼2D.
By the assumption that there exists no divisor with non-negative self-intersection satisfying(∗)fork < k0, we see that we must haveD.L=D.LandD2 =D2. Combining all this, we see that we can exchangeDwithD, and since|D|has one base divisor less than|D|, we are done by induction on the number of base components of|D|, counted with multiplicities.
Claim 3. We can assume thath1(D)=0.
2 Note that the two first properties are fulfilled ifDis a divisor obtained as in Lemma 3.6.
Proof of claim. By choosingDas in Claim 2, it follows by Proposition 2.2 and Lemma 2.3 thath1(D)=0, unless whenD ∼lE, for an integerl ≥ 2 andEa smooth elliptic curve. But in this caseL.E = 1lL.D = 1l(k0+1), soEwould satisfy the conditions(∗)for somek < k0, contrary to our assumptions.
By Proposition 2.2 and Lemma 2.3, choosingDaccording to the claims above, the generic member of|D|is a smooth curve of genus≥1, unless
D∼lE+,
whereEandare smooth irreducible curves satisfyingE2 =0,2 = −2 and E.=1 andl≥2 an integer.
By the conditions(∗), we haveL.D≤2k0+2, so we get E.L≤ 1
lD.L≤ 2(k0+1)
l ≤k0+1,
andEwill satisfy the conditions(∗)for somek≤k0, so we either have a contra- diction, orEis the desired smooth curve.
We now choose a smooth curveD0as above and we must show thatF :=L−D0
has the desired property.
Since F2 ≥ D02 ≥ 0, by Riemann–Roch either F ≥ 0 or−F ≥ 0. Since F.L ≥ D0.L = k0+1 > 0, we have F > 0. By Riemann–Roch again, using h1(D0)=0:
h0(F )=1
2F2+2+h1(F )≥ 1
2D2+2=h0(D).
This concludes the proof of Proposition 5.1.
6. k-spannedness failing onK3 surfaces
We now conclude the proof of Theorem 1.1 by explicitly constructing, in the next two propositions, 0-dimensional curvilinear schemes where thek-spannedness of Lfails, given the assumptions thatL2<4kor thatL2≥4kand the existence of a divisor satisfying the conditions(∗)above.
We will make use of the following result by Beltrametti and Sommese:
Theorem 6.1. [Be-So1, (1.2)] IfLis a specialk-spanned line bundle on a smooth curve of genusg, theng≥2k+1.
We first study the caseL2<4kmore thoroughly.
Proposition 6.2. LetLbe a globally generated, big line bundle on aK3 surface SsatisfyingL2 <4kfor an integerk ≥ 1. Then any smooth curveC ∈ |L|will contain a base point free complete linear system|A|of dimension≥1 and of degree l+1, for an integer 1≤l≤k, such that thel-very ampleness ofLfails on each member of|A|.
Proof. LetC be any smooth curve in|L|. Theng(C) = 12L2+1 ≤ 2k, so by Theorem 6.1,LC := L⊗OC = ωC is not k-spanned on C. So there exists a 0-dimensional subschemeZofCof degreel+1, for some integer 1≤l≤k, such that the map
H0(ωC)−→H0(ωC⊗OZ) arising from the short exact sequence
0−→ωC⊗OC(−Z)−→ωC −→ωC⊗OZ −→0 is not surjective.
Taking cohomology and using Serre duality, this is equivalent to h1(ωC⊗OC(−Z))=h0(OC(Z))≥2.
It is also clear that we can pick a minimal suchZ, i.e. with the property that the map
H0(ωC)−→H0(ωC⊗OZ)
is surjective for all proper subschemesZinZ. Then|OC(Z)|will be base point free, for if not, we would by removing base points, get a contradiction to the minimality ofZ.
Proposition 6.3. With the same assumptions and notation as in Proposition 5.1, defineF := L−D0. Then, for all smooth curves D0 in|D0| the line bundle FD0 :=F ⊗OD0 has the property that thek0-spannedness ofLD0 :=L⊗OD0 fails on any member of|FD0|.
Proof. This follows the first lines of the proof of Proposition 4.3. SinceD0 is smooth, any member of|FD0|will automatically be a curvilinear scheme where the k0-spannedness ofLfails.
This concludes the proof of Theorem 1.1.
In the sequel we will need the following observation.
Lemma 6.4. With the same notation and assumptions as in Proposition 6.3, assume furthermore thatLis generated by its global sections (so that in particulark0≥1) and thatD2≥0 (and henceD02≥0).
Thenh0(FD
0)≥2 for any smooth curveD0 in|D0|.
If furthermore(L2, D2)=(4k0+2, k0), then|FD0|is base point free for any smooth curveD0in|D0|.
Proof. First note that if(L2, D2) = (4k0+2, k0), then the smooth curve D0
constructed as in Proposition 5.1 will also satisfy(L2, D02)=(4k0+2, k0). To alleviate notation, we will work withD0. We first prove the statement that h0(FD0)≥2.
IfL∼2D0andL2=4k0+4, thenFD0 ωD0 andh0(ωD0)=12k0+32 ≥2.
