Rational Curves in Calabi-Yau Threefolds
#Trygve Johnsen1,* and Andreas Leopold Knutsen2
1Department of Mathematics, University of Bergen, Bergen, Norway
2Universitat Essen, Essen, Germany
ABSTRACT
We study the set of rational curves of a certain topological type in general members of certain families of Calabi-Yau threefolds. For some families we investigate to what extent it is possible to conclude that this set is finite. For other families we investigate whether this set contains at least one point representing an isolated rational curve.
Our study is inspired by Johnsen and Kleiman (Johnsen, T., Kleiman, S. (1996). Rational Curves of degree at most 9 on a general quintic threefold. Comm. Algebra 24(8):2721–2753).
Key Words: Rational curves; Calabi-Yau threefolds.
#Dedicated to Steven L. Kleiman on the occasion of his 60th birthday.
*Correspondence: Trygve Johnsen, Department of Mathematics, University of Bergen, Johs. Bruns gt 12, N-5008 Bergen, Norway; E-mail: johnsen@mi.uib.no, andreask@mi.uib.no.
Vol. 31, No. 8, pp. 3917–3953, 2003
3917
DOI: 10.1081/AGB-120022448 0092-7872 (Print); 1532-4125 (Online) Copyright#2003 by Marcel Dekker, Inc. www.dekker.com
1991 Mathematics Subject Classification: Primary 14J32;
Secondary: 14H45, 14J28.
1. INTRODUCTION
The famous Clemens conjecture says, roughly, that for each fixedd there is only a finite, but non-empty, set of rational curves of degree d on a general quintic threefoldFin complexP4. A more ambitious version is the following:
The Hilbert scheme of rational, smooth and irreducible curves C of degree d on a general quintic threefold inP4is finite, nonempty and reduced, so each curve is embedded with balanced normal bundleO(1)O(1).
Katz proved this statement ford7, and in Nijsse (1995) and Johnsen and Kleiman (1996) the result was extended tod9. An even more ambi- tious version also includes the statement that for generalF, there are no singular rational curves of degree d, and no intersecting pair of rational curves of degreesd1andd2withd1þd2¼d. This was proven (in Johnsen and Kleiman, 1996) to hold for d9, with the important exception for d¼5, where a general quintic contains a finite number of 6-nodal plane curves. This number was computed in Vainsencher (1995).
We see that, in rough terms, the conjecture contains a finiteness part and an existence part (existence of at least one isolated smooth rational curve of fixed degree d on general F, for each natural number d). For the quintics in P4, the existence part was proved for all d by S. Katz (1986), extending an argument from Clemens (1983), where existence was proved for infintely many d.
In this paper we will sum up or study how the situation is for some concrete families of embedded Calabi-Yau threefolds other than the quintics inP4.
In Sec. 2 we will briefly sketch some finiteness results for Calabi-Yau threefolds that are complete intersections in projective spaces. There are four other families of Calabi-Yau threefolds F that are such complete intersections, namely, those of type (2, 4) and (3, 3) in P5, those of type (2, 2, 3) in P6, and those of type (2, 2, 2, 2) in P7. For these families we have finiteness results comparable to those for quintics in P4involving smooth curves. The existence question has been answered in positive terms for curves of low genera, includingg¼0 and all positived, in these cases. See Kley (2000) and Ekedahl et al. (1999).
In Sec. 3 we study the five other families of Calabi-Yau threefoldsF that arecomplete intersections with Grassmannians G(k, n), namely, those of type (1, 1, 3) and (1, 2, 2) with G(1, 4), those of type (1, 1, 1, 1, 2) with
answered, in positive terms for curves withg¼0, for the types (1, 1, 3), (1, 2, 2), and (1, 1, 1, 1, 2). See Knutsen (2001b).
In Sec. 4 we will study rational curves in families of Calabi-Yau threefolds F on four-dimensional rational normal scrolls in projective spaces. These threefolds will then correspond to sections of the anti- canonical line bundle on the scroll, a very simple case of ‘‘complete intersection’’.
The main result of this paper, Theorem 4.3, ensures existence of at least one isolated smooth rational curve of given fixed topological type on generalF, for each topological type (a bidegree (d,a)) within a certain range. The main steps of the proof of Theorem 4.3 are sketched in Sec. 4.
There we also briefly investigate the possibilities for showing finiteness, applying similar methods developed to handle the complete intersection cases described above.
Sections 5 and 6 are devoted to the details of the proof of Theorem 4.3. In Sec. 5 we describe some general, useful facts about polarizedK3 surfaces, and we make some specific lattice-theoretical considerations that will be useful to us.
In Sec. 6 we prove Theorem 4.3 step by step. In Step (I) we prove Proposition 6.2, which describes curves on K surfaces, in Steps (II)–
(IV) we produce threefolds, which are unions of one-dimensional families ofK3 surfaces. We produce the desired rational curves on such threefolds and on smooth deformations of such threefolds.
1.1. Conventions and Definitions
The ground field is the field of complex numbers. We say a curve C in a variety V is geometrically rigid in V if the space of embedded deformations ofCinVis zero-dimensional. If, furthermore, this space is reduced, we say thatCisinfinitesimally rigidorisolatedinV.
Acurvewill always be reduced and irreducible.
2. COMPLETE INTERSECTION CALABI-YAU THREEFOLDS IN PROJECTIVE SPACES
In this section we will sketch the situation for the complete intersec- tion Calabi-Yau threefolds in projective spaces. In Johnsen and Kleimen (1996) one wrote, regarding the finiteness question for (smooth) rational curves: ‘‘The authors have checked the key details, and believe the follow- ing ranges come out:d7 for types (3, 3) and (2, 4), and d6 for types
(2, 2, 3) and (2, 2, 2, 2). In fact, except for the case d¼6 and F of type (2, 2, 2, 2), the incidence schemeIdof pairs (C,F) is almost certainly irre- ducible, generically reduced, and of the same dimension as the spacePof F.’’ Moreover, the ‘‘full theorem’’ for smooth rational curves in quintics, parallel to that of Nijsse (1995), was:
Theorem 2.1. Let d9, and let F be a general quintic threefold inP4. In the Hilbert scheme of F, form the open subscheme of rational, smooth and irreducible curves C of degree d. Then this subscheme is finite, nonempty, and reduced; in fact, each C is embedded in F with normal bundle OP1(1)OP1(1).
We are now ready to give corresponding results for the four other complete intersection cases:
Theorem 2.2. Assume that we are in one of the following cases:
(a) d7, and F is a general complete intersection threefold of type (2, 4) or (3, 3) in P5.
(b) d6, and F is a general complete intersection threefold of type (2, 2, 3) inP6.
(c) d5, and F is a general complete intersection threefold of type (2, 2, 2, 2) inP5.
In the Hilbert scheme of F, form the open subscheme of rational, smooth and irreducible curves C of degree d. Then this subscheme is finite, nonempty, and reduced; in fact, each C is embedded in F with normal bundle OP1(1)OP1(1). Moreover, in the cases (2, 2, 3) and d¼7, and in the case (2, 2, 2, 2) and d¼6, this subscheme is finite and non-empty, and there exists a rational curve C in F with normal sheafOP1(1)OP1(1).
Proof. In all cases of (a), (b), and (c) one proceeds as follows: LetIdbe the natural incidence of smooth rational curvesCand smooth complete intersection threefoldsFin question. One then shows thatIdis irreduci- ble, and dimId¼dimG, whereGis the parameter space of complete inter- section threefolds in question. This is enough to prove finiteness. In Jordanger (1999) not only the key details, but a complete proof of this result, was given.
Secondly, for each of the five intersection types (of CICY’s in some Pn) one has the following existence result, proven in Kley (2000) and Ekedahl et al. (1999). (Oguiso (1994) settled the (2, 4) case). It is also a special case of Knutsen (2001b, Thm. 1.1 and Rem. 1.2) and Kley (1999, Thm. 2.1).
Theorem 2.3. For all natural numbers d there exists a smooth rational curve C of degree d and a smooth CICY F, with normal sheaf NCjF¼ OP1(1)OP1(1) (which gives h0(NCjF)¼0).
Using these two pieces of information Theorem 2.2 follows as in Katz (1986, p. 152–153).
In the cases (2, 2, 3) and d¼7, and (2, 2, 2, 2) andd¼6, one proves dimId¼dimGand combines with Theorem 2.3. &
3. CICY THREEFOLDS IN GRASSMANNIANS
There are several ways of describing and compactifying the set of smooth rational curves of degree d in the Grassmann variety G(k, n).
See for example (Strømme, 1987). Let Md,k,ndenote the Hilbert scheme of smooth rational curves of degree d in G(k, n). It is well known that the dimension ofMd,k,nis (nþ1)dþ(kþ1)(nk)3.
Let each G(k, n) be embedded in PN, where N¼ nþ1kþ1 1, by the Plu¨cker embedding. LetGparametrize the set of smooth complete inter- section threefolds withG(k,n) by hypersurfaces of degrees (a1,. . .,as) in PN, where s¼dimG(k, n)3¼(kþ1)(nk)3, and a1þ þas¼ nþ1. Adjunction gives that the complete intersections thus defined have trivial canonical sheaves, and thus are Calabi-Yau threefolds. An easy numerical calculation gives that there are five families of Calabi-Yau threefoldsFthat are complete intersections with GrassmanniansG(k,n), beside those that are straightforward complete intersections of projective spaces PN (corresponding to the special case k¼0, n¼N). It will be natural for us to divide these five cases into two categories:
(a) Those whereai¼1, for alli. These are of type (1, 1, 1, 1, 1, 1, 1) in G(1, 6) in P20, or of type (1, 1, 1, 1, 1, 1) in G(2, 5) in P19. The dimensions of the parameter spacesG ofF in question, are 98 and 84, respectively.
(b) Those where ai2, for some i. These are of type (1, 1, 3) or (1, 2, 2) in G(1, 4) in P9, or of type (1, 1, 1, 1, 2) in G(1, 5) in P14. The dimensions of the parameter spaces of F in question, are 135, 95 and 109, respectively.
The existence question for rational curves of all degrees has been settled by the second author in Knutsen (2001b, Thm. 1.1 and Rem.
1.2), where it is concluded that for general F of types (1, 1, 3), (1, 2, 2),
or (1, 1, 1, 1, 2), and any integerd>0, there exists a smooth rational curve C of degree d in F with NCjF¼OP1(1)OP1(1) (which gives h0(NCjF)¼0). For the types (1, 1, 1, 1, 1, 1, 1) and (1, 1, 1, 1, 1, 1) we know of no such result.
The finiteness question seems hard to handle in all these cases. LetId be the incidence of Cand complete intersectionF. Denote by athe pro- jection to the parameter space Md,k,nof points [C] representing smooth rational curves Cof degreed in G(k,n), and byb the projection to the parameter space of complete intersections of the G(k, n) in question. A natural strategy is to look at each fixed rational curve Cand study the fibre a1([C]). If the dimension of all such fibres can be controlled, so can dimId. A way to gain partial control is the following: LetMbe a sub- scheme of Md,k,n, with dimM¼m, and assume that dima1([C])¼c is constant onM. Then of course this gives rise to a parta1(M) ofIdwhich has dimensioncþm. IfcþmdimG, one concludes that, for a general point [g] ofG, there is only a finite set of points froma1(M) inb1([g]).
More refined arguments may reveal that in many such cases a1(M) is irreducible.
An obvious argument shows that if M is the subscheme of M(d, k, n) corresponding to rational normal curves of degree d, then dima1([C]) is constant on M. Moreover the constant value is c¼dimGdimMd,k,n¼dimG((nþ1)dþ(kþ1)(nk)3). Of course the rational normal curves for fixedn,konly occur for (low) dNr, where G(k, n) is embedded in PN, andr is the number of i withai¼1.
(One observes that Nr¼maxfdjdimGdimMd,k,n0g for the cases in category (a), but Nr<maxfdjdimGdimMd,k,n0g for the cases in category (b).)
For d¼1, 2, 3 the only smooth rational curves of degreed are the rational, normal ones. In Osland (2001) it was shown for all five cases thatIdis irreducible ford¼1, 2, 3, and that on a generalFthere is no sin- gular (plane) cubic curve onF. In a similar way, one can show that on a general Fthere is no pair of intersecting lines, and no line intersecting a conic and no double line. For the complete intersection types of category (b), one then has:
Proposition 3.1. (i) Let d3, and let F be a general threefold of a given type as described above. In the Hilbert scheme of F, form the open sub- scheme of rational, smooth and irreducible curves C of degree d. Then this subscheme is finite, nonempty, and reduced; in fact, each C is embedded in F with normal bundle OP1(1)OP1(1). Moreover, there are no singular curves (reducible or irreducible) of degree d in F. We havedimId¼dimG, and Idis irreducible.
(ii) For all natural numbers d the incidence Idcontains a component of dimensiondimG, and this component dominates G by the second projection map b.
Proof. Part (i) follows from the irreducibility ofId, ford¼1, 2, 3, and the existence result in Knutsen (2001b), using Katz (1986, p. 152–153). Part (ii) follows from the same results, focusing only on one particular compo- nent ofId, corresponding to the curve found in Knutsen (2001b). &
In Batyrev et al. (preprint), one finds virtual numbers of rational curves of degree d on a generic threefold of each of the five types described. For the ones of category (b), there should be no problem in interpreting these numbers as actual numbers of smooth rational curves of degreedin a genericF, ford¼1, 2, 3, but it is a challenge to prove the analogue of Part (i) for higherd.
3.1. An Analysis of the IncidenceId
For each of the five types one might ask whether it is reasonable to believe that dimId¼dimG (or equivalently: dimIddimG) for many mored, or even for alld. We will point out below that the number of such d is very limited.
We recall that in the well known case of the hyperquintics inP4we have Id irreducible and of dimension 125¼dimG, for d9, reducible for d¼12, and reducible with at least one component of dimension at least 126 ford13. The cases d¼10, 11 seem to represent ‘‘open terri- tory’’, while it is also open whether some component has dimension at least 126 for d¼12. See Johnsen and Kleiman (1997). (None of these pieces of information contradict the Clemens conjecture, which predicts that all components of dimension at least 126 project to some subset of Gof positive codimension).
In each of the five types of complete intersection with Grassmannians G(k,n), a similar phenomenon occurs. The most transparent example is perhaps that of threefolds of intersection type (17) of G(1, 6) in P20. We now will show that in this case dimId>dimGfor alld4:
For all pointsPofP6, look atHP¼P5inG(1, 6) parametrizing all lines throughP. The subset ofMd,k,nparametrizing curves inHP, only spanning aP3(insideHPinsideG(1, 6) insideP20) has dimension 4dþ8. There is a 70-dimensional family of 13-planes in P20 containing a givenP3. Hence dimId6þ(4dþ8)þ70¼4dþ84. Clearly this exceeds 98 ford4.
Let J be the subset of Id thus obtained. The set of 3-spaces contained in some HP has dimension 6þdimG(3, 5)¼14, so
dimb(J)14þ70¼84<98. HenceJ, although big, gives no contradic- tion to the analogue of the Clemens conjecture.
To complete the picture we will also exhibit another part of the inci- dence Idof dimension 4dþ69. This is larger than 98 ford8. We will study the part ofIdthat arises from curvesCinG(1, 6), such that its asso- ciated ruled surface inP6only spans aP3inside thatP6. Each such curve is contained in aG(1, 3) in aP5insideP20, and a simple dimension count gives dimension 4dþ69.
On the other hand, it is for example clear that a generalP13insideP20 (corresponding to a general (17) ofG(1, 6)) does not contain aP5spanned by a sub-G(1, 3) ofG(1, 6). This means that the subsets ofId, correspond- ing to curvesCcontained in aG(1, 3), such that CandG(1, 3) span the same P5, project by b to subsets of G of positive codimension. Some Schubert calculus reveals that the same is true for the part of Idcorre- sponding to those Ccontained in a G(1, 3) and spanning at most a P4 also. Hence the ‘‘problematic’’ part ofIdin consideration here does not give a contradiction to the analogue of the Clemens conjecture.
Ford12, the part ofIdarising from curves such that its associated ruled surface spans a P4inside P6will have dimension at least 5dþ41, which is larger than 98. As above we see that generalP13insideP20does not contain aP9spanned by a sub-G(1, 4) ofG(1, 6).
For d15, the part of Id arising from curves such that their asso- ciated ruled surfaces spans aP5insideP6will have dimension at least 99.
LetJbe the subset ofIdthus obtained. The set of 3-spaces contained in some HP has dimension 6þdimG(3, 5)¼14, so dimb(J)14þ70¼ 84<98. HenceJ gives no contradiction to the analogue of the Clemens conjecture.
A similar phenomenon occurs for the case of threefolds of intersec- tion type (16) ofG(2, 5) inP19. We recall dimG¼84 in this case. For a given C in G(2, 5), the associated ruled threefold in P5 may span a 3-space, a 4-space, or all of P5. The former ones give rise to a part of the incidence of dimension 4dþ68. This is equal to dimG for d¼4, andIdis reducible then. Ford5 we see that dimId>dimG.
4. RATIONAL CURVES IN SOME CY THREEFOLDS IN FOUR-DIMENSIONAL RATIONAL
NORMAL SCROLLS
In this section we state the main result of this paper, Theorem 4.3, and we sketch the main steps of its proof. We also give some supplementary
results, and remark on the possibility of finding analogues of our main result.
We start by reviewing some basic facts about rational normal scrolls.
Definition 4.1. Let E¼OP1(e1) OP1(ed), with e1 ed0 and f¼e1þ þed2. Consider the line bundleL¼OP(E)(1) on the cor- responding Pd1-bundle P(E) over P1. We map P(E) into PN with the complete linear system jLj, where N¼fþd1. The image T is by definition a rational normal scroll of type (e1,. . ., ed). The image is smooth, and isomorphic toP(E), if and only if ed1.
Definition 4.2. LetTbe a rational normal scroll of type (e1,. . ., ed). We say thatTis a scroll of maximally balanced type if e1ed1.
Denote byHthe hyperplane section of a rational normal scrollT, and letCbe a (rational) curve inT. We say that thebidegreeofCis (d,a) if degC¼C.H¼d, considered as a curve on projective space, and CF¼a, whereFis the fiber of the scroll.
From now on we will letTbe a rational normal scroll of dimension 4 in PN, and of type (e1,. . ., e4), where the ei are ordered in an non- increasing way, ande1e31. Hence the subscrollP(OP1(e1)OP1(e2) OP1(e3)) is of maximally balanced type. We will show that for positivea, anddexceeding a lower bound depending ona, a general 3-dimensional (anti-canonical) divisor of type 4H(N5)F will contain an isolated rational curve of bidegree (d,a). To be more precise, we will show:
Theorem 4.3. LetTbe a rational normal scroll of dimension 4 inPNwith a balanced subscroll of dimension 3 as decribed. Assume this subscroll spans aPg(so g¼e1þe2þe3þ2) Let d1, and a1, be integers satisfying the following conditions:
(i) If g1(mod3), then either ðd;aÞ 2 fðg13 ;1Þ;ð2ðg1Þ=3;2Þg;
ord >ðg1Þa3 3a, (d, a)6¼(2(g1)=31, 2) and 3d6¼(g1)a.
(ii) If g2(mod3), then either (d, a)2 f(g1, 3), (2g2, 6)g;
or d >ðg1Þa3 3a, (d, a)62 f(2(g2)=3, 2), ((4g5)=3, 4), ((7g8)=3, 7)gand 3d6¼(g1)a.
(iii) If g0(mod3), then either (d, a)2 f((g3)=3, 1), ((2g3)=3, 2)g;
or dga=3.
Then the zero scheme of a general section of 4H(N5)F will be a smooth Calabi-Yau threefold and contain an isolated rational curve of bidegree (d, a).
This theorem will be proved in several steps. The fact that a general section is smooth follows from Bertini’s theorem, and since the divisor is anti-canonical and of dimension 3, it will be a Calabi-Yau threefold. Here are the main steps in the proof of the statement about the existence of an isolated curve as described:
(I) Set g:¼e1þe2þe3þ2. Using lattice-theoretical considerations we find a (smooth) K3 surface S in Pg with PicS’ZHZDZG, where H is the hyperplane section class, D is the class of a smooth elliptic curve of degree 3 and G is a smooth rational curve of bidegree (d,a). LetT¼TSbe the 3-dimensional scroll inPgswept out by the linear spans of the divisors injDjonS. The rational normal scrollTwill be of maximally balanced type and of degree e1þe2þe3.
(II) Embed T¼P(OP1(e1)OP1(e2)OP1(e3)) (in the obvious way) in a 4-dimensional scroll T¼P(OP1(e1) OP1(e4)) of type (e1,. . ., e4). HenceTcorresponds to the divisor classHe4FinT, andScorre- sponds to a ‘‘complete intersection’’ of divisors of type He4F and 3H(g4)F on T. We now deform the complete intersection in a rational family (i.e., parametrized by P1) in a general way. For ‘‘small values’’ of the parameter we obtain a K3 surface with Picard group of rank 2 and no rational curve on it.
(III) Take the union overP1of all the K3 surfaces described in (II).
This gives a threefoldV, which is a section of the anti-canonical divisor 4H(g4þe4)F¼4H(N5)F on T. For a general complete intersection deformation the threefold will have only finitely many singu- larities, none of them on G. Then Gwill be isolated onV.
(IV) DeformVas a section of 4H(g4þe4)F¼4H(N5)F FonT. Then a general deformationWwill be smooth and have an iso- lated curveGWof bidegree (d,a).
This strategy is analogous to the one used in Clemens (1983) to show the existence of isolated rational curves of infinitely many degrees in the generic quintic inP4, and in Ekedahl et al. (1999) to show the existence of isolated rational curves of bidegree (d, 0) in general complete intersection Calabi-Yau threefolds in some specific biprojective spaces.
Step (I) will be proved in Secs. 5 and 6, and Steps (II)–(IV) in Sec. 6.
4.1. Finiteness Questions
Let us say a few words about finiteness. LetTbe a rational, normal scroll of dimension 4 inPN. A divisor of type 4H(N5)Fcorresponds
to a quartic hypersurface Qcontaining N5 given 3-spaces in the F- fibration ofT. Then, for general suchQ, we see thatQ\Tis the union of a Calabi-Yau threefold and the N5 given 3-spaces. Each rational curveCof type (d,a) fora1 intersects each 3-space in at mosta, and hence a finite number of, points. For a given rational curveC0we want to study the following subset
p11 ð½C0Þ ¼ fð½C0;½FÞjC02Fg of the incidenceI¼Id,a.
Finding the dimension of p11 ([C0]) is essentially, as we shall see below, equivalent to findingh0(J(4)) (or (h1(J(4))), whereJis the ideal sheaf inPNof the unionXofC0and theN5 disjoint, linear 3-spaces, each intersecting C0 as described. We then have the following result, which is Corollary 1.9 of Sidman (2001):
Lemma 4.4. Let Jbe the ideal sheaf of a projective scheme X that con- sists of the union of d schemes X1,. . ., XdinPN, whose pairwise intersec- tions are finite sets of points. Let mi be the regularity of Xi. Then J is Pd
i¼1 mi-regular.
We will use this result. Look at the following exact sequence:
0!JX=Tð4HÞ !OTð4HÞ !OXð4HÞ !0:
This gives rise to the exact cohomology sequence 0!H0ðJX=Tð4HÞÞ !H0ðOTð4HÞÞ
!H0ðOXð4HÞÞ !H1ðJX=Tð4HÞÞ !0: This gives:
h0ðJX=Tð4HÞÞ
¼h1ðJX=Tð4HÞÞ þh0ðOTð4HÞÞ h0ðOXð4HÞÞ
¼h1ðJX=Tð4HÞÞ þ35ðN2Þ ð35ðN5Þ þ4dþ1 ðN5ÞaÞ
¼h1ðJX=Tð4HÞÞ þ105 ð4dþ1þ ð5NÞaÞ:
Hence we see that
dimp1ð½C0Þ ¼h1ðJX=Tð4HÞÞ þ104dimMd;a
¼h1ðJX=Tð4HÞÞ þdimGdimMd;a;
if 4e4(N5) 2. If we work with a stratum W of M¼Md,a
where h1(JX=T(4H)) is constant, say c, then the incidence stratum p1(W) has dimension cþdimGcodim(W,M). Now it is clear that h1(JX=T(4H))¼h1(JX=PN(4))¼h1(J(4)), sinceh1(JT=PN(4))¼0, which is true because rational normal scrolls are projectively normal. More- over h1(J(4))¼0 if X is 5-regular, by the definition of m-regularity in general. By Theorem 1.1 of Gruson et al. (1983) we have:
Lemma 4.5. A non-degenerate, reduced, irreducible curve of degree d inPr is (dþ2r)-regular.
Moreover in Corollary 1.10 of Sidman (2001) one has:
Lemma 4.6. The ideal sheaf of s linear k-spaces meeting (pairwise) in finitely many (or no) points is s-regular.
Putting these two results together, we observe that if C0 spans an r-space, thenXis (dþ2rþN5)¼(drþN3)-regular. In particu- lar, ifC0is a rational normal curve, thenXis (N3)-regular, and, in par- ticular, 5-regular ifNis 7 or 8 (and of coursedNthen). Also, curves spanning a (d1)-space are 5-regular ifN¼7 (ford8). We then have:
Corollary 4.7. On a general F inTof type (1, 1, 1, 1) inP7there are only finitely many smooth rational curves of degree at most 4. On a general F in T of scroll type (2, 1, 1, 1) in P8 there are only finitely many smooth rational curves of degree at most 3.
Proof. We deduce that allXin question are 5-regular, soh1(JX=PN(4H))¼0 for all X, and hence all non-empty incidence varietiesId,ahave dimen- sion equal to dimG, and hence the second projection mapp2has finite
fibres over general points of G. &
4.2. Analogous Questions for Other Threefolds
We would like to remark on the possibility of finding an analogue of Theorem 4.3. Is it possible to produce isolated, rational curves of bide- gree (d,a) for many (d,a), also on general CY threefolds of intersection type (2Hc1F, 3Hc2F) on five-dimensional rational normal scrolls in PN? Here we obviously look at fixed (c1,c2) such thatc1þc2¼N6.
A natural strategy, analogous to that in the previous section, would be to limit oneself to work with rational normal scrolls
P(OP1(e1) OP1(e5)) such that P(OP1(e1) OP1(e4)) is maximally balanced given its degree (that is:e1e41).
A natural analogue to Step (I) in the proof of Theorem 4.3 in the previous section is: Set g¼e1þ þe4þ3. Using lattice-theoretical considerations again, and with certain conditions on n, d and a, we find a (smooth) K3 surfaceS in Pg with PicS’ZHZDZG, where His the hyperplane section class,Dis the class of a smooth elliptic curve of degree 4 andCis a smooth rational curve of bidegree (d,a). In parti- cular,Shas Clifford index 2. (See Sec. 5.1 for the definition of the Clifford index of aK3 surface.) LetT¼TSbe the 4-dimensional scroll inPgswept out by the linear spans of the divisors injDjonS. The rational normal scroll T will be maximally balanced of degreee1þe2þe3þe4. In other words we should find an analogue of Proposition 6.2. It seems clear that we can do this.
The natural analogue of Step (II) in the previous section is:
Embed T¼P(OP1(e1) OP1(e4)) (in the obvious way) in a 5-dimensional scroll T¼P(OP1(e1) OP1(e5)) of type (e1,. . ., e5).
HenceTcorresponds to the divisor classHe5Fin T.
Ifgis odd, one would like to show thatSis a ‘‘complete intersection’’
of 2 divisors, both of type 2Hg52 Frestricted toT. Therefore, it is a
‘‘complete intersection’’ of three divisors Z5, Q1, Q2, the first of type He5F, and the two Qi of type 2Hg52 F on T. At the moment we have no watertight argument for this.
Ifg is even, one can show that Sis a ‘‘complete intersection’’ of 2 divisors, of types 2Hg42 Fand 2Hg62 F, restricted toT. Therefore, it is a ‘‘complete intersection’’ of three divisors Z5, Q1, Q2, of types He5F, 2Hg42 F, and 2Hg62 FonT.
If one really obtains a complete intersection as described above, one might deform it in a rational family (i.e., parametrized by P1). If S is given by equations Q1¼Q2¼0, one looks at deformations
Q1þsQ01¼Q2 ¼Z5sBðt;u;Z1;. . .;Z5Þ ¼0; or
Q1¼Q2þsQ02 ¼Z5sBðt;u;Z1;. . .;Z5Þ ¼0:
Here theBcorrespond to sections ofHe5F. For ‘‘small values’’ of the parameter one would like to obtain aK3 surface with Picard group of rank 2, Clifford index 1 (see Sec. 5.1 for the definition of the Clifford index of a K3 surface), and no rational curve on it. For g odd, there is no essential difference between the two types of deformations. Forgeven, the two deformations are different.
An analogue of Step (III) is: Eliminatingsfrom the first set of equa- tions, we obtain:
Q1BþZ5Q01¼Q2¼0:
Eliminatingsfrom the second set of equations, we obtain:
Q1¼Q2BþZ5Q02¼0:
Ifgis odd, we obtain in both cases a ‘‘complete intersection’’ three- fold of type
2Hg5
2 F;3H g5 2 þe5
F
:
Ifg is even, the first threefold is of type 2Hg6
2 F;3H g4 2 þe5
F
;
while the second is of type 2Hg4
2 F;3H g6 2 þe5
F
;
Since g¼N1e5, we see that in all cases we have intersection type (2Hc1F, 3Hc2H), such thatc1þc2¼N6.
The analogue of Step (IV) seems doable forgodd, but here Step (II), as remarked, is unclear. The details of this analogue for geven are also not quite clear to us.
5. K3 SURFACE COMPUTATIONS
The purpose of the section is to make the necessary technical pre- parations to complete Step (I) of the proof of Theorem 4.3. First we will recall some useful facts about K3 surfaces and rational normal scrolls.
In Lemma 5.3 we introduce a specific K3 surface which will be essential in the proof of Step (I). In the last part of the section we make some K3-theoretical computations related to the Picard lattice of this particular K3 surface.
5.1. Some General Facts AboutK3 Surfaces
Recall that aK3 surface is a (reduced and irreducible) surfaceSwith trivial canonical bundle and such thatH1(OS)¼0. In particularh2(OS)¼1 andw(OS)¼2.
We will use line bundles and divisors on aK3 surface with little or no distinction, as well as the multiplicative and additive notation, and denote linear equivalence of divisors by.
Before continuing, we briefly recall some useful facts and some of the main results in Johnsen and Knutsen (2001) which will be used in the proof of Theorem 4.3.
Let C be a smooth irreducible curve of genus g2 and A a line bundle on C. The Clifford index of A (introduced by Martens (1968) is the integer
CliffA¼deg A2ðh0ðAÞ 1Þ:
Ifg4, then theClifford index of Citself is defined as CliffC¼minfCliffAjh0ðAÞ 2;h1ðAÞ 2g:
Clifford’s theorem then states that CliffC0 with equality if and only ifCis hyperelliptic and CliffC¼1 if and only ifCis trigonal or a smooth plane quintic.
At the other extreme, we obtain from Brill-Noether theory (cf.
Arbarello et al., 1985, Chapter V) that CliffC bg12 c. For the general curve of genusg, we have CliffC¼ bg12 c.
We say that a line bundleAonC contributes to the Clifford index of C if h0(A), h1(A)2 and that it computes the Clifford index of C if in addition CliffC¼CliffA.
Note that CliffA¼CliffoC A1.
It was shown by Green and Lazarsfeld (1987) that the Clifford index is constant for all smooth curves in a complete linear systemjLjon aK3 surface. Moreover, they also showed that if CliffC<bg12 c (where g denotes the sectional genus ofL, i.e.,L2¼2g2), then there exists a line bundleMonSsuch thatMC:¼M OCcomputes the Clifford index ofC for all smooth irreducibleC2 jLj.
This was investigated further in Johnsen and Knutsen (2001), where we defined the Clifford index of a base point free line bundleLon aK3 surface to be the Clifford index of all the smooth curves in jLj and denoted it by CliffL. Similarly, if (S, L) is a polarized K3 surface, we defined the Clifford index ofS, denoted by CliffL(S) to be CliffL.
The following is a summary of the results obtained in Johnsen and Knutsen (2001) that we will need in the following. Since we only need those results for ample L, we restrict to this case and refer the reader to Johnsen and Knutsen (2001) for the results when Lis only assumed to be base point free.
Proposition 5.1. Let L be an ample line bundle of sectional genus g4 on a K3 surface S and let c:¼CliffL. Assume thatc<bg12 c. Then c is equal to the minimal integer k0 such that there is a line bundle D on S satisfying the numerical conditions:
2D2ðiÞL:D¼D2þkþ2ðiiÞ2kþ4
with equality in (i) or (ii) if and only if L2D and L2¼4kþ8. (In parti- cular, D2cþ2, with equality if and only if L2D and L2¼4cþ8, and by the Hodge index theorem
D2L2 ðL:DÞ2¼ ðD2þcþ2Þ2:Þ
Moreover, any such D satisfies (with M:¼LD and R:¼L2D):
(i) D.M¼cþ2.
(ii) D.LM.L (equivalently D2M2).
(iii) h1(D)¼h1(M)¼0.
(iv) jDj and jMj are base point free and their generic members are smooth curves.
(v) h1(R)¼0, R2 4, and h0(R)>0 if and only if R2 2.
(vi) If RR1þR2 is a nontrivial effective decomposition, then R1.R2>0.
Proof. The first statement is Knutsen (2001a, Lemma 8.3). The proper- ties (i)–(iv) are the properties (C1)–(C5) in Johnsen and Knutsen (2001, p. 9–10), under the additional condition that L is ample. The fact that h1(R)¼0 in (v) follows from Johnsen and Knutsen (2001, Prop. 5.5) (where D¼0 sinceLis ample), and the rest of (v) is then an immediate consequence of Riemann-Roch. Finally, (vi) follows from Johnsen and Knutsen (2001, Prop. 6.6) sinceLis ample. &
Now denote byfL the morphism fL:S!Pg
defined by jLj and pick a subpencil fDlg jDj ’P12D2þ1 generated by two smooth curves (so that, in particular, fDlg is without fixed compo- nents, and with exactly D2 base points). Each fL(Dl) will span a (h0(L)h0(LD)1)-dimensional subspace of Pg, which is called the linear span of fL(Dl) and denoted by Dl. Note that Dl¼Pcþ1þ12D2. The variety swept out by these linear spaces,
T ¼ [
l2P1
DlPg;
is a rational normal scroll (see Schreyer, 1986) of type (e1,. . .,ed), where
ei¼#fjjdjig 1; ð1Þ
with
d¼d0 :¼h0ðLÞ h0ðLDÞ;
d1 :¼h0ðLDÞ h0ðL2DÞ;
...
di:¼h0ðLiDÞ h0ðL ðiþ1ÞDÞ;
...
Furthermore, T has dimension dimT ¼d0¼h0ðLÞ h0ðFÞ ¼cþ 2þ12D2 and degree deg T¼h0ðFÞ ¼gc112D2.
We will need the following.
Lemma 5.2. Assume that L is ample, D2¼0, and h1(LiD)¼0 for all i0 such that LiD0. Then the scroll T defined by jDj as described above is smooth and of maximally balanced scroll type. Furthermore, dimT¼cþ2 anddegT¼gc1.
Proof. Let r:¼maxfijLiD0g. Then by Riemann-Roch, and our hypothesis thath1(LiD)¼0 for alli0 such thatLiD0, one easily findsr¼ bcþ2g cand
d0¼ ¼dr1 ¼L:D¼cþ2;
1dr¼gþ1 ðcþ2Þrcþ2; di¼0 for irþ1;
whence the scrollTis smooth and of maximally balanced scroll type. The assertions about its dimension and degree are immediate. &
5.2. Some SpecificK3 Surface Computations
In the following lemma we introduce a specific K3 surface with a specific Picard lattice, which will be instrumental in proving Theorem 4.3. The element G in the lattice will correspond to a curve of bidegree (d,a) as described in that theorem.
Lemma 5.3. Let n4, d>0 and a>0 be integers satisfying d >na3 3a. Then there exists an algebraic K3 surface S with Picard group PicS’ZHZDZGwith the following intersection matrix:
H2 H:D H:G D:H D2 D:G G:H G:D G2 2
4
3 5¼
2n 3 d
3 0 a
d a 2
2 4
3 5
and such that the line bundle L:¼H bn43 cDis nef.
Proof. The signature of the matrix above is (1, 2) under the given con- ditions. By a result of Nikulin (1980) (see also Morrison, 1984, Theorem 2.9(i)) there exists an algebraic K3 surface S with Picard group PicS¼ZHZDZGand intersection matrix as indicated.
Since L2>0, we can, by using Picard-Lefschetz tranformations, assume thatLis nef (see e.g., Oguiso, 1994 or Knutsen, 2002). &
Note now that
L2 ¼ 8 if n4 mod 3;
10 if n5 mod 3;
12 if n6 mod 3:
8<
: ð2Þ
We will from now on write L2¼2m, for m:¼n3bn4
3 c ¼4;5 or 6 ð3Þ
(in other wordsnm(mod 3)) and define d0 :¼G:L¼d bn4
3 ca>0: ð4Þ
Note that the conditiond >na3 3ais equivalent to d0 >ma
3 3
a: ð5Þ
Also note that PicS’ZLZDZG, and that
d:¼ jdiscðL;D;GÞj ¼ jdiscðH;D;GÞj
¼ j2að3dnaÞ þ18j ¼ j2að3d0maÞ þ18j ð6Þ divides jdisc(A,B,C)jfor anyA,B,C2PicS.
Now we will study the K3 surface defined in Lemma 5.3 in further detail.
Lemma 5.4. Let S, H, D,G and L be as in Lemma 5.3. Then L is base point free and CliffL¼1. Furthermore, L is ample (whence very ample) if and only if we are not in any of the following cases:
(i) ma¼3d0and 9jma,
(ii) m¼4 and (d0, a)¼(2, 2), (5, 4) or (9, 7), (iii) m¼5 and (d0, a)¼(2, 2), (6, 4) or (13, 8), (iv) m¼6 and (d0, a)¼(3, 2).
Moreover, if L is ample, thenjDjis a base point free pencil, LD is also base point free and h1(LD)¼h1(L2D)¼0.
Proof. SinceD.L¼3, we have CliffL1 by Proposition 5.1. To prove the two first statements, it suffices to show (by classical results on line bundles onK3 surfaces such as in Saint-Donat (1974) and by Proposition 5.1) that there is no smooth curveEsatisfyingE2¼0 and E.L¼1, 2.
SinceEis base point free, being a smooth curve of non-negative self- intersection (see Saint-Donat, 1974), we must haveE.D0. IfE.D¼0, then the divisor B:¼3E(E.L)D satisfies B2¼0 and B.L¼0, whence by the Hodge index theorem we have 3E(E.L)D, contradicting that Dis part of a basis of PicS. IfE.D2, the Hodge index theorem gives the contradiction
322L2ðE:DÞ ¼L2ðEþDÞ2 ðL:ðEþDÞÞ225: We now treat the caseE.D¼1. IfE.L¼1, we get 162L2¼L2ðEþDÞ2 ðL:ðEþDÞÞ2¼16;
whence by the Hodge index theorem L2(EþD), which is impossible, since L is part of a basis of PicS. So we have E.L¼2. Write ExLþyDþzG. From E.L¼2 andE.D¼1, we get
x¼1 3a
3zandy¼2ma3d0
9 z2ðm3Þ
9 : ð7Þ
Inserting into 12E2¼0¼mx2z2þ3xyþd0xzþayz, we find
½aðma3d0Þ 9z2 ¼m6: ð8Þ Ifm¼4, we get from (8) thatz¼±1 anda(4a3d0)¼7. Sinced0>0, we must have (d0,a)¼(9, 7), which is present in case (b). (From (7) we get the integer solution (x,y,z)¼(2, 3, 1).)
Ifm¼5, we get from (8) thatz¼±1 anda(5a3d0)¼8. Sinced0>0, we must have (d0,a)¼(2, 2), (6, 4) or (13, 8), which are present in case (c).
(We can however check from (7) that (2, 2) and (13, 8) do not give any integer solutions for x and y, whereas (6, 4) gives the integer solution (x,y,z)¼(1, 2, 1).)
Ifm¼6, we get from (8) that eitherz¼0 ora(2ad0)¼3. In the first case, we get the absurdity x¼1=3 from (7), and in the latter we get the only solution (d0, a)¼(5, 3), which inserted in (7) gives the absurdity x¼1=3z. We have therefore shown thatLis base point free and that CliffL¼1.
To show that L is ample we have to show by the Nakai criterion that there is no smooth curveEsatisfyingE2¼ 2 and E.L¼0.
By the Hodge index theorem again we have 2L2ðE:D1Þ ¼L2ðDEÞ2 ðL:ðDEÞÞ2¼9;
giving1E.D1. The casesE.D¼±1 are symmetric by interchanging EandE, so we can restrict to treating the casesE.D¼0 andE.D¼1.
We can writeExLþyDþzG. We get x¼ a
3zþE:D
3 : ð9Þ
Combining this withE.L¼0¼2mxþ3yþd0z, we get y¼2ma3d0
9 z2mðE:DÞ
9 : ð10Þ
Now we use 12E2¼ 1¼mx2z2þ3xyþd0xzþayz, and find
½aðma3d0Þ 9z2 ¼mðE:DÞ29: ð11Þ We first treat the caseE.D¼0. We get
½aðma3d0Þ 9z2 ¼ 9; ð12Þ which means thatz¼±1 orz¼±3.
Ifz¼±1, we find from (12) thatma3d0¼0, and from (9) we find x¼ a3, whence 3ja. If in addition 9jaorm¼6, then (x,y,z)¼±(a=3, ma=9, 1) defines an effective divisorEwithE2¼ 2 andE.L¼E.D¼0.
Ifz¼±3, we find from (10) that 3j2ma. But sincea(ma3d0)¼8, we get the absurdity 3j16.
We now treat the caseE.D¼1. Then we have
½aðma3d0Þ 9z2¼m9: ð13Þ We now divide into the three casesm¼4, 5, and 6.
Ifm¼4, then (13) reads
½að4a3d0Þ 9z2¼ 5; ð14Þ which means thatz¼±1 and a(4a3d0)¼4, in particular a¼1, 2 or 4.
Sinced0>0 we only get the solutions
ða;d0Þ ¼ ð2;2Þorða;d0Þ ¼ ð4;5Þ: ð15Þ From (9) we get x¼13ð1azÞ ¼13ð1aÞ, which means that we only have the possibilities (x, z, a, d0)¼(1, 1, 2, 2) and (x, z, a, d0)¼ (1, 1, 4, 5). Inserting into (10) we get
y¼1
9½ð8a3d0Þz8 ¼ 2 and 1; ð16Þ
respectively, whence (x,y,z, a,d0)¼(1, 2,1, 2, 2) and (1, 1, 1, 4, 5) are the only solutions.
Ifm¼5, then (13) reads
½að5a3d0Þ 9z2¼ 4; ð17Þ which means thatz¼±1 orz¼±2. Ifz¼±1, we havea(5a3d0)¼5, and since d0>0, there is no solution. Soz¼±2 and a(5a3d0)¼8. Again, sinced0>0, we only get the solutions
ða;d0Þ ¼ ð2;2Þ;ða;d0Þ ¼ ð4;6Þor ða;d0Þ ¼ ð8;13Þ: ð18Þ From (9) we get x¼13ð1azÞ ¼13ð12aÞ, which means that we only have the possibilities (x,z, a, d0)¼(1, 2, 2, 2), (3, 2, 4, 6) or (5, 2, 8, 13). Inserting into (10) we get
y¼1
9½ð10a3d0Þz10 ¼2;6 or 8; ð19Þ respectively, whence (x,y,z,a,d0)¼(1, 2, 2, 2, 2), (3,6,2, 4, 6) and (5, 8, 2, 8, 13) are the only solutions.
Ifm¼6, then (13) reads
½að6a3d0Þ 9z2¼3z2½að2ad0Þ 3 ¼ 3: ð20Þ We obtain z2[a(2ad0)3]¼ 1, which givesz¼±1 anda(2ad0)¼2.
Since d0>0, the latter yields (a, d0)¼(2, 3). If z¼1, then (9) gives the absurdity x¼13. For z¼ 1 we obtain (x, y)¼(1, 3) from (9) and (10). Hence (x,y,z,a,d0)¼(1,3,1, 2, 3) is the only solution form¼6.
So we have proved thatLis ample except for the cases (a)–(d).
It is well-known (see e.g., Saint-Donat, 1974 or Knutsen, 2001a) that an ample line bundle with CliffL¼1 is very ample.
IfLis ample, it follows from Proposition 5.1 thatjDjandjLDjare base point free andh1(LD)¼h1(L2D)¼0. &
We get the corresponding statement forH:
Lemma 5.5. Assume n, d and g do not satisfy any of the following condi- tions:
(i) na¼3d, with 9ja if n1, 2 (mod3), and 3ja if n0 (mod3).
(ii) n0 (mod3), a¼2 andd ¼3þ23ðn6Þ.
(iii) n1 (mod3) andd¼d0þ23ðn4Þ, for (d0, a)¼(2, 2), (5, 4) or (9, 7).
(iv) n2 (mod3) andd¼d0þ23ðn5Þ, for (d0, a)¼(2, 2), (6, 4) or (13, 8).
Then H is very ample and CliffH¼1. Moreover, h1(HiD)¼0 for all i0 such that HiD0.
Denote by T the scroll defined by D. Then T is smooth and of maxi- mally balanced scroll type.
Proof. The first two statements are clear since D.H¼3 and H ¼Lþ bn43 cD, with CliffL¼1 and D nef, since the cases (i), (ii), (iii) and (iv) are direct translations of the cases (a), (d), (b) and (c), respectively, in Lemma 5.4 above.
We now prove thath1(HiD)¼0 for alli0 such thatHiD0.
By Lemma 5.4 we have thath1(LD)¼h1(L2D)¼0. SinceDis nef we haveh1(LþiD)¼0 for alli0. Now letR:¼L2D. ThenR2¼ 4,2 or 0, corresponding to L2¼8, 10 or 12. Since we have h1(R)¼0 we get h0(R)¼0, 1 or 2 respectively. In the first case we are therefore done, and in the second, we clearly haveh0(RD)¼0, and now we also want
to show this forL2¼12. Assume thath0(RD)>0. Then, we have R¼DþD;
wherejDjis the moving part ofjRjandDis the fixed part. SinceR.D¼3, we get D.D¼3, and since D2¼0 we get D2¼ 6. Moreover D.L¼ (L3D).L¼3. This yields
D¼G1þG2þG3;
where the Gi’s are smooth non-intersecting rational curves satisfying Gi.L¼Gi.D¼1 for i¼1, 2, 3. Writing Gi¼xiLþyiDþziG, the three equationsGi.L¼1,Gi.D¼1 andG2i¼ 2 yield at most two integer solu- tions (xi,yi,zi), whence at least two of theGis have to be equal, a contra- diction.
So we have proved that h1(HiD)¼0 for all i0 such that HiD0. By Lemma 5.2 it follows that the scroll T is of maximally
balanced type, whence smooth. &
In the next section we will describe under what conditions G is a smooth rational curve. We end this section with two helpful lemmas.
Lemma 5.6. Assume L is ample. LetDxLþyDþzGbe a divisor on S such thatD2¼ 2 and setd0:¼ jdisc(L, D,D)j.
Thend0¼z2d, and is zero if and only ifDL2D and m¼5.
Proof. It is an easy computation to show thatd0¼z2d. Hence it is zero if and only ifz¼0. It is then an easy exercise to find that DL2Dand
m¼5. &
Lemma 5.7. Let B:¼3LmD. (We have B2¼0 and B.D¼9, whence by Riemann-Roch B>0.) IfD is a smooth rational curve satisfyingD2¼ 2 and D.B0, then we only have the following possibilities:
m¼4 and ðD:L;D:D;D:BÞ ¼ ð1;1;1Þ;ð4;3;0Þ;ð4;4;4Þ;
ð5;4;1Þ;ð6;5;2Þ;ð8;6;0Þ;
ð9;7;1Þ
m¼5 and ðD:L;D:D;D:BÞ ¼ ð1;1;2Þ;ð3;2;1Þ;ð4;3;3Þ m¼6 and ðD:L;D:D;D:BÞ ¼ ð1;1;3Þ;ð2;1;0Þ;ð3;2;3Þ;
ð4;2;0Þ:
Proof. Set as before R:¼L2D. We have 3D.R¼D.(3L6D) D.(3LmD)0, whence D.R0, with equality only ifm¼6.
If equality occurs, we have D<R (since R2¼0 and R>0 by Pro- position 5.1), whence we have a nontrivial effective decomposition RDþD0. Since R.L¼6 and L is ample, we have D.L5, whence (D.L,D.D)¼(4, 4) and (2, 1) are the only possibilities.
IfD¼R, thenR2¼ 2, whencem¼5 and (D.L,D.D,D.B)¼(4, 3,3).
So for the rest of the proof, we can assume thatD.R<0 withD6¼R.
If R2¼ 2 or 0 (i.e., m¼5 or 6), then R>0 by Proposition 5.1, whence D<R. If D.R 2, we get a nontrivial effective decomposition RDþD0 with D.D00. But this contradicts Proposition 5.1. So D.R¼ 1. Since R.L¼4 and 6 for m¼5 and 6 respectively, and L is ample, we have D.L3 and 5 respectively. Ifm¼6 andD.L¼5, we get D.D¼3 and we calculate jdisc(L, D, D)j ¼0, contradicting Lemma 5.6. This leaves us with the possibilities listed above form¼5 and 6.
Now we treat the case R2¼ 4, i.e., m¼4. Then we have h0(R)¼h1(R)¼0.
If D.R¼ 1, then, since D.B¼3D.Rþ2D.D¼ 3þ2D.D and D is nef, we get the only possibility (D.L,D.D)¼(1, 1).
IfD.R 2, we get (RD)2 2, whence by Riemann-Roch either RD>0 or DR>0. In the first case we get the contradiction R>D>0, so we must have L2D<D<3L4D (the latter inequality due to the fact thatB2¼0,B>0 andD.B0). SinceLis ample, we there- fore get
3D:L11; ð21Þ
and from the Hodge index theorem
16ðD:B1Þ ¼ ðBDÞ2L2 ððBDÞ:LÞ2¼ ð12D:LÞ2; that is,
D:B bð12D:LÞ2
16 þ1c bð123Þ2
16 þ1c ¼6: ð22Þ
IfD.B¼0, thenD:D¼3D:L4 , which means by (21) that (D.L,D.D)¼(4, 3) or (8, 6).
If D.B¼ 1, then D:D¼3D:Lþ14 , which means by (21) that (D.L, D.D)¼(5, 4) or (9, 7).
If D.B¼ 2, then (21) and (22) gives 3D.L8 and D:D¼3D:Lþ24 , which means that (D.L, D.D)¼(6, 5).
Continuing this way up to D.B¼ 6, one ends up with the choices
listed in the lemma. &
6. PROOF OF THEOREM 4.3
We will now complete the four steps of the proof of Theorem 4.3.
6.1. Proof of Step (I)
We start with some further investigations of theK3 surface with the Picard lattice introduced in Lemma 5.3:
Lemma 6.1. Assume L is ample. ThenGis a smooth rational curve if and only if none of these special cases occurs:
(a) m¼4, 3d0¼4a and a>9, in which case G(3L4D)þ (G3Lþ4D) is a nontrivial effective decomposition.
(b) m¼5 and 4<d0<2a, in which caseG(L2D)þ(GLþ2D) is a nontrivial effective decomposition.
(c) m¼6, d0¼2a and a>3, in which caseG(L2D)þ(GLþ2D) is a nontrivial effective decomposition.
Proof. SinceG2¼ 2 andG.H>0 we only need to show thatGis irre- ducible. ConsiderB¼3LmDas defined above. Then
d:¼ jdiscðL;D;GÞj ¼ j2ðG:DÞðG:BÞ þ18j: ð23Þ
Case I. G.B>0. Then d>18. Assume that G is not irreducible. Then there has to exist a smooth rational curve g<G such that g.BG.B.
We also haveg.DG.Dby nefness ofD. SetD:¼Gg>0. Ifg.B¼G.B, then D.B¼3D.LmD.D¼0, whence D.D>0, since L is ample, and henceg.D<G.D. In other words we always have (g.D)(g.B)<(G.D)(G.B), whence
discðL;D;gÞ ¼2ðg:DÞðg:BÞ þ18<d: ð24Þ If nowg.B<0, we get from Lemma 5.7 that (g.D)(g.B) 16, whence
discðL;D;gÞ ¼2ðg:DÞðg:BÞ þ18 32þ18 14>d: ð25Þ So we must have disc(L, D, g)¼0, whence by Lemma 5.6 we have m¼5 and gL2D¼:R. By ampleness of L we must have
0<D.L¼(GLþ2D).L¼d010þ6¼d04, whence d0>4. We will now show that d0<2aas well, so that we end up in case (b) above.
To get a contradiction, assume thatd02a. WriteGkRþDk, for an integer k1 such that Dk>0 and RäDk. By our assumption we have D2k¼ 2(k2þ1þk(d02a)) 2, so Dkmust have at least one smooth rational curve in its support. Since we have just shown that the only smooth rational curve g such that g.B0 is R, we have g0.B>0 for any smooth rational curve g0Dk. Pick one such g0Dk such that g0.BDk.B¼G.Bþ3k. Then, since also 0g0.DDk.D¼G.D3k, and G.B¼3d05aa¼G.Dby our assumptions, we get the contradic- tion
0<discðL;D;g0Þ ¼2ðg0:DÞðg0:BÞ þ182ðG:D3ÞðG:Bþ3Þ þ18<2ðG:DÞðG:BÞ þ18¼d:
So we are in case (b). Conversely, if m¼5 and 4<d0<2a, one sees that (GLþ2D)2 2 and (GLþ2D).L>0, whence by Riemann- Roch (GLþ2D)>0 and G(L2D)þ(GLþ2D) is a nontrivial effective decomposition.
Case II. G.B¼0. Thend¼18 and 3d0¼ma. Since we assume thatLis ample, we have that 9 does not divideaifm¼4 or 5 and 3 does not divide aifm¼6 by Lemma 5.4. Assume thatGis not irreducible. Then there has to exist a smooth rational curveg<Gsuch thatg.B0. Ifg.B<0, then g.D>0 by Lemma 5.7, and we can argue as in Case I above. We end up in the case m¼5 and g¼L2D, and since d0¼5a3 <2a this is a special case of (b).
So we can assume thatg.B¼0 for any smooth rational curve in the support ofB. By Lemma 5.7 again, for any suchgwe have the possibili- ties
ðm;g:L;g:DÞ ¼ ð4;4;3Þ;ð4;8;6Þ;ð6;2;1Þorð6;4;2Þ; ð26Þ whence the casem¼5 is ruled out. To prove that we end up in the cases (a) and (c) above, we have to show that a6¼3, 6 whenm¼4 anda6¼1, 2 whenm¼6.
So assumem¼4 and (d0,a)¼(4, 3) or (8, 6). Sinceg.L<G.L¼d, we see from (26) that (d0,a)¼(8, 6) and (g.L,g.D)¼(4, 3) for anyg<G. For any suchg, considerD:¼Gg>0. Then (D.L,D.D)¼(4, 3). IfD22, we get the contradiction from the Hodge index theorem:
64¼8L2L2ðDþDÞ2 ðL:ðDþDÞÞ2 ¼49:
IfD2¼0, we writeDxLþyDþzGand use the three equations D:L¼ 8xþ3yþ8z ¼4;
D:D¼ 3xþ6z ¼3;
D2¼ 8x22z2þ6xyþ16xzþ12yz ¼0:
to find the absurdity (x,y,z)¼(1, 4=3, 0).
SoD2 2, which means thatD2¼ 2, since for any smooth rational curve g0 in its support we must have (g0.L, g0.D)¼(4, 3). But then D.G¼g.G¼ 1, and D and ghave the same intersection numbers with all three generators of PicS. But theng¼Dand Gwould be divisible, a contradiction.
Assume now thatm¼6 and (d0,a)¼(2, 1) or (4, 4). As above we end up with the only possibility (d0, a)¼(4, 4) and (g.L, g.D)¼(2, 1). Now also (D.L,D.D)¼(2, 1), andD2¼ 2 and we reach the same contradiction as above.
So we have proved that we end up in cases (a) and (c) above. Con- versely, if m¼4 and d0 and a satisfy the conditions in (a), one easily checks thatG(3L4D)þ(G3Lþ4D) is a nontrivial effective decom- position, since both the components have self-intersection 2 and positive intersection with L. The same holds for the decomposition G(L2D)þ(GLþ2D) ifm¼6 andd0andasatisfy the conditions in (c).
Case III. G.B<0. As in the proof of Lemma 5.7 we haveG.R<0. If m¼4 and G.R¼ 1, we again end up with the possibility (G.L, G.D)¼ (1, 1), in which caseGis irreducible.
In all other cases, we have (RG)2 2, whence by Riemann-Roch eitherRG>0 or GR>0.
Case III(a). R>G. We must havem¼5 or 6, sinceh0(R)¼0 ifm¼4 by Proposition 5.1. We proceed as in the proof of Lemma 5.7, withGin the place ofD, and show thatG.R¼ 1, which gives the cases:
ðm;d0;aÞ ¼ ð5;1;1Þ;ð5;3;2Þ;ð5;4;3Þ;ð6;1;1Þorð6;3;2Þ:
We considerm¼5 first. Ifd0¼1, thenGis irreducible, and if (d0,a)¼ (4, 3), we getd¼0, whence the absurdityGL2Dby Lemma 5.6. So we must have (d0,a)¼(3, 2) and G.B¼ 1. If Gis reducible, there exists a smooth rational curve g<G such that g.B<0. Since g.L<G.L¼3, we get by looking at the list in Lemma 5.7 that (g.L, g.D,g.B)¼(1, 1,2).
Since then disc(L,D,g)¼disc(L,D,G)¼14, we get from Lemma 5.6 that
gxLþyDþzG, for z¼± 1. Using g.L¼1¼10xþ3yþ3z and g.D¼1¼3xþ2z, we find the integer solution (x,y,z)¼(1,2,1), which yieldsD2¼ 6, whereD:¼Ggas usual. HoweverD.L¼2, whenceDhas at most two components, contradicting its self-intersection number.
We next considerm¼6. If d0¼1, thenGis irreducible, so we must have (d0, a)¼(3, 2) and G.B¼ 3. This yields d¼6. If G is reducible, there exists a smooth rational curve g<G such that g.B<0. Since g.L<G.L¼3, we get by looking at the list in Lemma 5.7 that (g.L, g.D,g.B)¼(1, 1,3) or (2, 1, 0), yielding respectively disc(L, D, g)¼12 or 18, contradicting Lemma 5.6, which givesz2¼2 or 3 respectively.
Case III(b). G>R. We have 9
a<3d0ma¼G:B<0; and (GR).D¼a30, whence
3a8 and ma 3 3
a<d0 <ma
3 : ð27Þ
We leave it to the reader to verify that there are no integer solutions to (27) form¼6 and that the only solutions form¼4 and 5 are
m¼4: ðd0;aÞ ¼ ð5;4Þ;ð9;7Þ;
m¼5: ðd0;aÞ ¼ ð6;4Þ;ð8;5Þ;ð13;8Þ:
The cases withm¼5 belong to case (b) above. We now show that we can rule out the cases with m¼4.
Assume thatGis reducible. Then there has to exist a smooth rational curveg<Gsuch thatg.B<0, and we can use Lemma 5.7 again.
Assume first that (d0, a)¼(5, 4), which gives d¼10. Then g.L<5, whence by Lemma 5.7 we get the possibilities (g.L,g.D,g.B)¼(1, 1, 1) or (4, 4,4), yielding, respectively, disc(L,D,g)¼16 or 14, none of which are divisible byd¼10, a contradiction.
Assume now that (d0,a)¼(9, 7), which givesd¼4. Theng.L<9 and by Lemma 5.7 we get the possibilities (g.L,g.D,g.B)¼(1, 1,1), (5, 4,1), (6, 5,2) or (4, 4,4) yielding, respectively, disc(L,D,g)¼16, 10, 2 or 14.
By Lemma 5.6 the only possibility is therefore (g.L, g.D, g.B)¼ (1, 1,1) withgxLþyDþzG, forz¼± 2. Ifz¼2 we get the absurdity 1¼g.D¼3xþ14. If z¼ 2, we get from the two equations 1¼g.D¼ 3x14 and 1¼g.L¼8xþ3y18 the solution (x, y, z)¼(5, 7, 2), sog5L7D2G, which yieldsg.G¼0. Since we have just shown that
