Stable Sheaves on a Smooth Quadric Surface with Linear Hilbert Bipolynomials

Abstract

We investigate the moduli spaces of stable sheaves on a smooth quadric surface with linear Hilbert bipolynomial in some special cases and describe their geometry in terms of the locally free resolution of the sheaves.

1. Introduction

Throughout the paper, our base field is , the field of complex numbers.

By the work of Simpson [1], we can consider the moduli space of semistable sheaves on a smooth projective variety with a fixed Hilbert polynomial, which is itself a projective variety, and the moduli space has been studied quite intensively in the last decade for the case with linear Hilbert polynomial over projective spaces [2–5]. Our interest is on the moduli space over a smooth quadric surface.

Let be a smooth quadric surface in and let be the moduli space of semistable sheaves on with linear Hilbert polynomial with respect to the ample line bundle . Unlike the case of projective spaces, this moduli space is not irreducible in general. Indeed, for a purely -dimensional sheaf on , we can define a linear Hilbert bipolynomial such that for all . Then we can consider, due to [6], the moduli space of semistable sheaves on with linear Hilbert bipolynomial . The moduli space is a projective variety with a Zariski open subset consisting of stable ones, with dimension and the open set is nonempty if one of or is nonzero (see Proposition 7).

By its definition we have a natural decomposition Thus, the moduli is an irreducible component of Simpson's moduli space because the bidegree function is locally constant.

If is a stable sheaf in , then its schematic support is a curve of bidegree on and so a general sheaf is a line bundle over a smooth subcurve. Thus, the moduli space can be considered as an analogue of the universal line bundle of some fixed degree over the family of the bidegree -curves in .

Now, some simple observations lead us to consider only with due to proper twists. For small or , the moduli space is very simple. Indeed, is isomorphic to if and is empty otherwise. If or is equal to , say , then it is isomorphic to . These descriptions are quite simple from the definition of stability condition and so the first nontrivial case happens when . The main result of this paper is to describe the moduli spaces with .

Theorem 1. For , one obtains the following: (1) is isomorphic to and it is rational;(2) is birational to and it is unirational with degree .

In fact, we explicitly describe the sheaves in each moduli space in terms of their locally free resolution. Indeed, a sheaf is in if and only if it admits a resolution where the degeneracy locus of the first map is the support of . It enables us to identify with and show its rationality.

For , the situation is a bit more complicated; we can classify the sheaves in up to types, in terms of the short exact sequences they admit, and express the moduli as the union of subschemes In particular, we can show that every sheaf in is globally generated, from which we obtain a resolution that they admit: We investigate the property of the subvarieties and the relationship between them. We also construct a map from to , which is generically to and thus we obtain that is unirational of degree . We leave the rationality question of as a conjecture.

2. Preliminaries

Let be a smooth quadric surface isomorphic to for 2-dimensional vector spaces and , and then it is embedded into by the Segre map where . If we denote by the two projections from to each factor, then we will denote simply by . We also denote by for a coherent sheaf on and in particular the canonical sheaf of is .

Proposition 2. For a purely 1-dimensional sheaf on , there is a bipolynomial of degree 1 such that for all .

Proof. Let us assume that is the Hilbert polynomial of with respect to the ample line bundle . Let us take any , , and a smooth conic such that neither , , nor is contained in the -dimensional reduced curve .
The curves , and induce maps ,  , and . Since neither nor is contained in the -dimensional reduced curve , we have and . Since is pure, we obtain that , , and are injective. Thus, there are exact sequences
Let us set and . The sheaves , , and have finite supports and thus the dimensions of their cohomology do not change even if we twist them by a line bundle on . From (9), we get .
We claim that for all . If , then the claim is true. Now assume that , say . We use exact sequences like (8) with instead of with to get .

Definition 3. One defines the Hilbert bipolynomial of to be a linear bipolynomial such that In particular, the Hilbert polynomial of with respect to is defined to be .

We are mainly interested in the case when is a linear function, that is, for some .

Definition 4. Let be a pure sheaf of dimension 1 on with . The -slope of is defined to be . is called semistable (stable) with respect to the ample line bundle if (1) does not have any -dimensional torsion,(2)for any proper subsheaf , one has  where .

For every semistable -dimensional sheaf with , let us define to be its scheme-theoretic support and then we have . We often use slope stability and slope semistability instead of Gieseker stability or Gieseker semistability just to simplify the notation; they should be the same because the support is -dimensional, and from and , the inequality for Hilbert and slopes the same.

Definition 5. Let be the moduli space of semistable sheaves on with linear Hilbert bipolynomial .

We can define in a different way as a subvariety of , the moduli space of semistable sheaves on with linear Hilbert polynomial , which are -sheaves. To be precise, if is -sheaf, then all of its -subsheaves are also -sheaves. It implies that the notions of -stability and -stability of are the same and thus may be defined without using . Moreover, the sheaf with linear Hilbert bipolynomial has Hilbert polynomial with respect to and thus we have a natural decomposition In particular, is a subvariety of .

Remark 6. Let be any purely 1-dimensional coherent sheaf on with Hilbert polynomial . Assume that is not semistable and let be the Harder-Narasimhan filtration of (see page 55 in [1]). If is an -module, then each is an -module because it is a subsheaf of . Thus the Harder-Narasimhan filtration of as an -sheaf is the same as the one as an -sheaf.

Proposition 7. The moduli is a projective and irreducible scheme. If , then is a Zariski dense and open subset of with dimension .

Proof. The first assertion follows verbatim from the proof of Proposition 2.3 and Theorem 3.1 in [6], only when the assertion in Lemma 3.3 over holds. But it holds, using Castelnuovo-Mumford criterion with the Serre duality for and .
For the second assertion, let us consider a map defined by sending to , where and . Then the dimension of the image of this map is at least and it is at least if . In other words, general sheaf in is stable.

For any pure sheaf on with Hilbert bipolynomial , let us define to be the Grothendieck dual of . Since is pure, the natural map is injective. Since the support of is -dimensional, is bijective as in Remark 4 of [4]. Moreover, the support of is also -dimensional and so is also linear. By the Serre duality, we have for and, in particular, we have

Lemma 8. There is an isomorphism sending to .

Note also that . Since the map defined by , is an isomorphism, so we may assume that .

Lemma 9. For a not necessarily integral curve in , the sheaf is semistable. If is integral, then is stable.

Proof. We have the following sequence: In particular, we have and so . If is integral, then is stable since every line bundle on an integral curve is stable. In general, is semistable. Otherwise, there exists a semistable quotient sheaf such that the Hilbert bipolynomial satisfies and . By induction, we get that with is semistable and thus we have This is absurd since is a decreasing function on and .

Let us assume that , that is, with .

Proposition 10. One has In fact, each point in corresponds to an equivalence class , where is a line in .

Proof. Let us assume that and let us choose and then it fits into Thus we have Clearly, is stable. For a line , we have From the sequence for , we have and the map is a zero map. Thus, there exists a nontrivial extension of by and it is . In particular, and represent the same point in . In general, and with represent the same point in . Thus, with is strictly semistable if and only if . Conversely, let us choose a semistable sheaf with . In particular, the schematic support of is in . Since , there exists a nontrivial morphism and it induces an injection , where is a subscheme of . Here we have for some and so . Thus, the quotient is a semistable sheaf with . By induction, we have with . In particular, is an extension of by with and thus is equivalent to .
Now, let us assume that and fix with . Since , there is a nonzero map . Since is an -sheaf, induces a nonzero map . Since has slope and it is semistable, we get a contradiction. Alternatively, as in Lemma 4.10 of [6], we may first take the schematic support of and then use an injective map with with , and thus we have .

For the case of , that is, , it is enough to check the case since .

Proposition 11. consists of with . In particular, one has .

Proof. From the sequence we have and is semistable by Lemma 9. Conversely, let be a semistable sheaf with and so is a curve in . Since we have , there exists a nonzero map and it induces a nonzero map . Note that has no -dimensional torsion since is semistable. Since is also semistable, we have The map factors through an injection , where is a curve contained in . If is properly contained in , we have contradicting the semistability of and thus we have ; that is, is an isomorphism from to its image. Since and have the same Hilbert polynomial, we have .

3. Hilbert Bipolynomial

For the moduli space of semistable sheaves with linear Hilbert bipolynomial , it is enough to investigate the case when . Let us denote the moduli space by .

Proposition 12. The moduli space consists of the unique nontrivial extensions of by for each curve and a point , and one also has .

Proof. Since , there is a nonzero map , inducing a nonzero map , where . Since , we have . The map factors through an injection , where is a curve contained in . If is properly contained in , we have contradicting to the semistability of and thus we have ; that is, is an isomorphism from to its image, that is, we have where . In particular, we have , the skyscraper sheaf supported on a point . Since has no -dimensional torsion, the sequence does not split. Note that , and thus from the sequence of we have Here, the map is the transpose of which is given by the multiplication by the defining equation of . Since is a point on , the map is a zero map. In particular, the dimension of is and so corresponds to a unique nontrivial extension From the sequence (32), we have and that if and only if no injective map is an isomorphism at . This is certainly true if is not locally free of rank at . Note that is a line bundle at each point of and thus it is sufficient to prove when is a line bundle on the curve . In this case the nonexistence of a section of that does not vanish at is equivalent to the nonsplitting of (32). Thus, we have and so the point is uniquely determined by .
Conversely, let us assume that is a nontrivial extension of by , where is a point on . If is not semistable, then there exists a subsheaf with and so we have with and . If the composite is a zero map, then we have an injection , contradicting the semistability of . Thus, the composite is surjective and so we have the following diagram: (33) Here, is the kernel of the map and is the quotient . Since and is semistable, we have and thus with no constant term. Since is the quotient of , it must be for some curve contained in . But no such curves have the Hilbert polynomials with no constant term. Hence is semistable.

Remark 13. There is no strictly semistable sheaf in . Let us assume the existence of a polystable sheaf with . We have . If we let , then we have It implies that for all and thus we have , a contradiction.

Proposition 14. A sheaf is in if and only if it admits the following resolution: where and . Here, is a defining equation of .

Proof. Note that and so . If admits the sequence (32), then is globally generated outside and so is . Take any which is not contained in and with . The multiplication by an equation of gives an exact sequence where . Thus we have and . Together with the exact sequence (32) tensored by , we obtain that is globally generated at and so we have a surjection Let us set and then is a torsion-free sheaf of rank on with . By Theorem 19.9 in [7], the sheaf is locally free. Note that . Thus, we have and so we have an exact sequence where is a -dimensional subscheme of and . If , then we have and it is absurd since . Thus, we have and . Since , we have and the sequence (35). Note that the map is given by , where is a defining equation of .
The converse is trivial.

Remark 15. Using the proof of Lemma 5.3 in [2], we can obtain the same assertion of Proposition 14. Similarly, we also obtain that is globally generated and so a surjection . In this case, is no longer a direct sum of two line bundles.

Let us define a vector space to be and to be the set of such that . Then we have a surjective morphism

Let us choose with ; that is, we have the following diagram: where is an isomorphism. Since , we have a map associated with . Note that is given by , where and . Similarly, we have a map which is , where . In particular, we have In this equation, we can assume that . In other words, if and only if and are in the same orbit in under the action by

Theorem 16. is a geometric quotient map by the action of . In particular, one has and so is isomorphic to .

Proof. To get the assertion it suffices to prove that it has local sections as in Lemma 5.1 and Theorem 5.5 in [3].
Since every element of is stable, has a universal family on (see page 180 of [8] or Theorem of [9]). Since every semistable sheaf with bipolinomial is of the form for a unique , we also have a universal family on with as the fibre. Since and for all , the base change theorem gives that is a line bundle on , where is the first projection. Since and for all , the base change theorem gives that is a vector bundle of rank on by identifying with , where . For a fixed and a matrix , let us write , where is a defining equation of . Take an open neighborhood of in over which and are trivial. The matrix was constructed starting with a section of which spans together with the twist of a nonzero section of . Since and are trivial, there are maps and with and . Since and span , there is a neighborhood of in such that the sections and span every . The construction of gives that and induce a section of in a neighborhood of whose image by is .

As an automatic consequence, we obtain that is irreducible and unirational. In fact, we can prove more.

Theorem 17. is rational.

Proof. Let be the diagonal and denote its ideal sheaf by . Denoting by and the projection from to each factor, let us define a sheaf to be on . For each point , we have . Thus, we have and so is a vector bundle of rank on since . Let us consider the projective bundle By its definition, the fibre of over a point is the set of curves of type on passing through and so there is a natural map from to . In other words, is the universal curve of type on and it is isomorphic to . Since is locally trivial over , it is rational.

4. Hilbert Bipolynomial

Lemma 18. Any sheaf admits one of the following types:(A), where is a skyscraper -sheaf with degree 2,(B) with ,(C), where and with .

Proof. Since , we have . Thus, there exists a nonzero map and it induces a nonzero map , where . The map factors through an injection where is a curve contained in : (45) If we have , that is the map is an isomorphism from to its image in , then its cokernel is the skyscraper sheaf supported on two points, say . Thus we have the sequence
Let us assume that is properly contained in and then we obtain that has bidegree , or since and is semistable. Let be the only curve such that . Let be the quotient of by its torsion , that is, .
First, assume and so we have . Since is semistable, we get . Since every quotient of has the slope at least , the same is true for . Thus, is semistable and Proposition 11 gives .
Now, without loss of generality, let us assume that , that is, and so we have . If has -dimensional torsion with length , then the quotient is a quotient of with the -slope , contradicting the semistability of . Thus has no -dimensional torsion and so we have for a curve with .

Corollary 19. Every sheaf in is globally generated.

Proof. Let us take and then there is no nonzero map since is semistable. Thus we have and so . It is clear that of types and is globally generated and so we may assume that is of type , but neither of nor of .
Let be the image of the evaluation map and then is pure. Assume that . Since is of type , it is globally generated outside at most two points of . In particular, we have with and . Since , we have . Note that every nonzero section of vanishes at finitely many points since is neither of types nor . Since , we have and . A nonzero section of induces an exact sequence where for some . Since is pure, this exact sequence does not split. As in the proof of Proposition 12, we get a contradiction. Thus, we have and so is globally generated.

Lemma 20. is a sheaf in if and only if it admits a sequence where , such that is a defining equation of .

Proof. Let be a sheaf of type and then it is globally generated by Corollary 19. Since , we have a surjection Let us set and then it is a torsion-free sheaf of rank on with . By Theorem 19.9 in [7], is locally free. Note that . From the sequence we obtain that the map is an isomorphism and so . Similarly, we have and . By Remark 2.3 in [10], we obtain that is globally generated. Since or , we have and the resolution (48). The cases of the other types also work verbatim.

Definition 21. Let us define a subscheme as follows: Similarly, we define and for the semistable sheaves of types and , respectively. In particular, we have

Lemma 22. The sheaves of type are strictly semistable. In particular, they are contained in .

Proof. It is enough to check the semistability of . Let be a subsheaf of with and the quotient sheaf . If the composite map is a zero map, then is a subsheaf of , contradicting the semistability of . The sheaf is a subsheaf of and so we have . Similarly, the sheaf is a subsheaf of and so . From the exact sequence we have , a contradiction.

Let us denote by the closed subscheme of , consisting of the strictly semistable sheaves.

Corollary 23. One has where is the permutation group of order . In particular, is a rational variety.

Proof. Obviously, we have . Let be a strictly semistable sheaf and so it has a proper quotient sheaf with . From the semistability of , has no -dimensional torsion. From the equality , we obtain that is also semistable. Since , the Hilbert bipolynomial of is either , or . The first cases cannot happen due to Proposition 10. Thus, we have and so with and by Proposition 11. If is the kernel of the quotient map , then its p-slope is again and so is semistable. Similarly as before, we have with and . Hence, we have .
Let be a sheaf of type , that is, it corresponds to a pair of two curves . Let us assume that admits another sequence with . Note that is stable for all . Thus, the composite map is either a zero map or an isomorphism. In the former case, we have and so . In the latter case, we have and . Hence, the class of a strictly semistable sheaf corresponds to a uniquely determined pair of two curves in and we have . The second assertion follows from the fact that any symmetric product of any projective space is a rational variety (see Theorems 4.2.8 and 4.2.8′ in page 137 of [11, 12]).

Lemma 24. For two curves , one has

Proof. Note that we have Thus, if we apply the functor to the sequence of , we obtain We also have and so their dimensions are and , respectively. As -sheaves, we have for example, because and are reduced, and so the assertion is derived.

Lemma 25. The sheaves of type , but not of type , are stable. In particular, the sheaves of type are semistable.

Proof. As before let us assume the existence of a proper subsheaf of with and the quotient sheaf . Since the composite is not a zero map, thus we have for a -dimensional subscheme of with length . In particular, its Hilbert bipolynomial is . If we let , then we have . In particular, we have . If we define to be the kernel of the map , then it is a subsheaf of and thus we have . Combining the two inequalities, we have and so the map is surjective. Thus, we have . Note also that can be either or . If , then we have and so . In particular, we have and so , a contradiction. Now, assume and so . In particular, is not a -dimensional sheaf. Moreover, is a quotient sheaf of with constant term and so we have with and . For example, if , then we have and it contradicts the nontriviality of the extension . Thus, also admits the sequence where . Since is a subsheaf of with , we have , where is a subcurve of such that . Thus, is an extension of by . It is nontrivial, otherwise we would have as a direct factor of . Since there exists such a unique extension , admits an extension of by : (62) It implies that is of type .

Lemma 26. Let be a line bundle on a reduced curve with degree 2. (1) is semistable if and only if one has:(a) for all subcurves of in with ,(b) for each smooth subcurve of in with .(2) is stable if and only if for all subcurves of in with .

Proof. In both parts, the “only if” part is obvious. Assume that is not stable (resp., semistable) and take a proper subsheaf of with (resp. ). Taking a saturation of in , we may assume that is a pure sheaf. Call the scheme support of and the scheme support of . The definition of scheme support of a purely -dimensional sheaf gives as effective divisors. Thus has one of the types in the assertion. Since is reduced and is a line bundle on , the support of must be a proper subcurve of . If does not have a type of or , then we are done. But the case of having such types is excluded using the argument in the proof of Lemma 18.

Lemma 27. One has .

Proof. Let us set with and , and set to be smooth. For any extension of by , for example, , let be the kernel of the composition and then is a pure sheaf with as its scheme support and has Hilbert bipolynomial . Note that it has as its subsheaf.
To prove , it is sufficient to prove that is semistable. Suppose is not semistable and take a proper saturated stable subsheaf with . Its scheme support is contained in and it is of type . Without loss of generality, let us assume that . First, assume . In this case, we would have because and so we have , contradicting the fact that and that is globally generated. Assume . The map on must be just the inclusion , because is a line bundle. Thus either we have or is not saturated in . Hence the saturation of in has slope greater than , contradicting the semistability of . Now assume and , that is, . Since is smooth, is a line bundle on . If its degree is at least , then contradicts the semistability of . If , then we have , a contradiction. Hence is also contained in .

Lemma 28. For and , one has

Proof. Applying the functor to the sequence of , we obtain since we have and similarly . Note also that and . Thus we have the assertion.

Remark 29. When and meet transversally at two points, say and , then is the global sheaf of a sheaf with support on and with one copy of on each point , for the following reason.
Let be a regular local ring of dimension and take generators of its maximal ideal. All groups are with respect to . Since is Gorenstein, so the duality gives and for all . From the exact sequence in which is the multiplication by , we get that is the cokernel of the multiplication by in ; that is, we have . The same is true for extensions of by when and are transversal.

Lemma 30. Let be a sheaf of type with no 0-dimensional torsion. Then is semistable unless it admits the sequence where and with .

Proof. Let be a subsheaf with maximal -slope and so the quotient sheaf has no -dimensional torsion. Let us set with and . If the composite is a zero map, then is a subsheaf destabilizing , a contradiction. If is not surjective, for instance, , then is a subsheaf of with Hilbert bipolynomial . Thus we have and the quotient has Hilbert bipolynomial with zero constant term. Since has no -dimensional torsion, we have for a curve contained in . But the Hilbert bipolynomial of has nonzero constant term, a contradiction. Thus the map is surjective. Following the same argument before, we obtain that and . Without loss of generality, let us assume that . Then we have and thus we have , where is a curve contained in and . Since is a subsheaf of with , we have since has no -dimensional torsion. Thus fits into the sequence (68).

Remark 31. Applying the functor to the sequence of , we obtain Since the map is the dual of the map given by the multiplication by the defining equation of , the map is a zero map and thus we have . In particular its dimension is .

Lemma 32. Let be a sheaf of type fitting into an exact sequence with . Then is of type if and only if and have no common components; that is, has no multiple component.

Proof. If and have a common component, say , then has rank at the general point of and thus is not of type .
Conversely, assume that is finite. Since we have , the sequence (70) implies that and is globally generated. Let be a general section of and then it does not vanish at the general point of any of the components of . Since is reduced, induces an injective map and thus has type .

Lemma 33. Let be a sheaf of type . (1)If is not a component of , then is of type .(2)If is a double component of , that is, , then it is not of type .

Proof. Let us assume that .(1)Since is not a component of , is a line bundle on outside finitely many points of . Moreover, it is not an -sheaf. Note that is globally generated since and are globally generated with . Thus, a general section of does not vanish at a general point of and so it does not induce an injection . Hence, fits into some sequence (46).(2)Let us set and , where . Let be the projectivisation of and in particular we have by Lemma 28. We also know from Lemma 25 that any gives a semistable sheaf. Such a sheaf has rank at the points of and, in particular, it is not a line bundle over its support at a general point of . Thus, it never fits into an exact sequence (46). Otherwise it would be locally free of rank at each point of the support of but not in .

In general, the question whether the variety is rational is difficult as in the projective plane. We observed that is rational and so is . Below we give a partial answer to this question in the case of .

Theorem 34. is unirational with degree 4.

Proof. Let us fix a smooth curve of bidegree in and a point to consider a sheaf . If is the tangent plane of at , then we have for some points on since . It defines a rational map sending to . Note that . We claim that the map is generically to and so the assertion follows.
Let (resp., ) be the dense open subset of (resp., ) formed by the sheaves such that is smooth. Each element of (resp., ) is uniquely determined by a smooth and a degree one (resp., degree two) line bundle on . By Riemann-Roch, each degree one line bundle on is associated with a unique . Then the map sends to . Fix any degree two line bundle on . Since we are in characteristic zero, there are exactly four line bundles on such that . Hence, for each there are exactly points such that . Hence, is dominant and the preimage of each element of has cardinality .

We did not succeed in getting any smaller degree of unirationality of as of now, and we left the rationality question as a conjecture.

Conjecture 35. is rational.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

Sukmoon Huh is supported by Basic Science Research Program 2010-0009195 through NRF funded by MEST.