Stable Sheaves on a Smooth Quadric Surface with Linear Hilbert Bipolynomials

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.


Introduction
Throughout the paper, our base field is C, 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][3][4][5]. Our interest is on the moduli space over a smooth quadric surface.
Let be a smooth quadric surface in P 3 and let M ( , ) be the moduli space of semistable sheaves on with linear Hilbert polynomial ( ) = + with respect to the ample line bundle O (1, 1). Unlike the case of projective spaces, this moduli space is not irreducible in general. Indeed, for a purely 1-dimensional sheaf F on , we can define a linear Hilbert bipolynomial F ( , ) such that (F ( , )) = F ( , ) for all ( , ) ∈ Z ⊕2 . Then we can consider, due to [6], Thus, the moduli M( , , ) is an irreducible component of Simpson's moduli space because the bidegree function is locally constant. If F is a stable sheaf in M( , , ), then its schematic support F 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 P ( , ) of some fixed degree over the family of the bidegree ( , )-curves in . Now, some simple observations lead us to consider only M( , , ) with 0 ≤ ≤ gcd( , ) due to proper twists. For small or , the moduli space is very simple. Indeed, M( , 0, ) is isomorphic to P if = and is empty otherwise. If or is equal to 1, say = 1, then it is isomorphic to P 2 +1 . These descriptions are quite simple from the definition of stability condition and so the first nontrivial case happens when ( , ) = (2,2). The main result of this paper is to describe the moduli spaces M(2, 2, ) with = 1, 2. (1) M 1 is isomorphic to P 1 (2,2) and it is rational; 2 The Scientific World Journal (2) M 2 is birational to P 2 (2,2) and it is unirational with degree 4.
In fact, we explicitly describe the sheaves in each moduli space in terms of their locally free resolution. Indeed, a sheaf F is in M 1 if and only if it admits a resolution where the degeneracy locus of the first map is the support of F. It enables us to identify M 1 with P 1 (2,2) and show its rationality.
For M 2 , the situation is a bit more complicated; we can classify the sheaves in M 2 up to 3 types, in terms of the short exact sequences they admit, and express the moduli as the union of 3 subschemes In particular, we can show that every sheaf in M 2 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 M 1 to M 2 , which is generically 4 to 1 and thus we obtain that M 2 is unirational of degree 4. We leave the rationality question of M 2 as a conjecture.

Preliminaries
Let be a smooth quadric surface isomorphic to P 1 × P 2 for 2-dimensional vector spaces 1 and 2 , and then it is embedded into P 3 ≅ P by the Segre map where = 1 ⊗ 2 . If we denote by 1 , 2 the two projections from to each factor, then we will denote * 1 O P 1 ( ) ⊗ * 2 O P 1 ( ) simply by O ( , ). We also denote E⊗O ( , ) by E( , ) for a coherent sheaf E on and in particular the canonical sheaf of is O (−2, −2).

Definition 3. One defines the Hilbert bipolynomial
In particular, the Hilbert polynomial of F with respect to O (1, 1) is defined to be F ( ) = F ( , ).
We are mainly interested in the case when F ( , ) is a linear function, that is, F ( , ) = + + for some ( , , ) ∈ Z ⊕3 . Definition 4. Let F be a pure sheaf of dimension 1 on with for any proper subsheaf F , one has where F ( , ) = + + .
For every semistable 1-dimensional sheaf F with F ( , ) = + + , let us define F := Supp(F) to be its scheme-theoretic support and then we have F ∈ |O ( , )|. 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 1-dimensional, and from + and + , the inequality for Hilbert and slopes / the same. We can define M( , , ) in a different way as a subvariety of M ,P 3 ( + , ), the moduli space of semistable sheaves on P 3 with linear Hilbert polynomial ( ) = + , which are O -sheaves. To be precise, if F is O -sheaf, then all of its O P 3subsheaves are also O -sheaves. It implies that the notions of -stability and -stability of F are the same and thus M ,P 3 ( + , ) may be defined without using P 3 . Moreover, the sheaf with linear Hilbert bipolynomial ( , ) = + + has Hilbert polynomial ( ) = ( + ) + with respect to O (1, 1) and thus we have a natural decomposition In particular, M( , , ) is a subvariety of M ,P 3 ( + , ).

Remark 6.
Let F be any purely 1-dimensional coherent sheaf on P 3 with Hilbert polynomial + . Assume that F is not semistable and let be the Harder-Narasimhan filtration of F (see page 55 in [1]). If F is an O -module, then each F is an O -module because it is a subsheaf of F. Thus the Harder-Narasimhan filtration of F as an O P -sheaf is the same as the one as an O -sheaf. 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 F ∈ M( , , ) and ≥ −1. For any pure sheaf F on with Hilbert bipolynomial F ( , ) = + + , let us define to be the Grothendieck dual of F. Since F is pure, the natural map F : F → F is injective. Since the support of F is 1dimensional, F is bijective as in Remark 4 of [4]. Moreover, the support of F is also 1-dimensional and so F ( , ) is also linear. By the Serre duality, we have for ∈ {0, 1} and, in particular, we have sending F to F .

Lemma 9. For a not necessarily integral curve in
Proof. We have the following sequence: In particular, we have O ( , ) = + + ( + − ) and so (O ) = 1 − 1/(1/ + 1/ ). If is integral, then O is stable since every line bundle on an integral curve is stable. In general, O is semistable. Otherwise, there exists a semistable quotient sheaf O → F → 0 such that the Hilbert bipolynomial F ( , ) = + + satisfies + < + and (F) < (O ). By induction, we get that O with := F is semistable and thus we have This is absurd since (O ) is a decreasing function on and .

Proposition 10. One has
In fact, each point in Hilb ( + ) corresponds to an

The Scientific World Journal
Proof. Let us assume that = and let us choose ∈ |O ( , 0)| and then it fits into Thus we have From the sequence for , we have Since O has slope 1 > / and it is semistable, we get a contradiction. Alternatively, as in Lemma 4.10 of [6], we may first take the schematic support ⊆ of Im(ℎ) and then use an injective For the case of = 1, that is, F ( , ) = + + , it is enough to check the case = 1 since gcd(1, ) = 1. Proof. From the sequence we have O ( , ) = + + 1 and O is semistable by Lemma 9. Conversely, let F be a semistable sheaf with F ( , ) = + + 1 and so := F is a curve in |O ( , 1)|. Since we have (F) = 1, there exists a nonzero map O → F and it induces a nonzero map ℎ : O → F. Note that Im(ℎ) has no 0-dimensional torsion since F is semistable. Since O is also semistable, we have The map ℎ factors through an injection O → F, where is a curve contained in . If is properly contained in , we have (O ) > (F) contradicting the semistability of F and thus we have = ; that is, ℎ is an isomorphism from O to its image. Since O and F have the same Hilbert polynomial, we have F ≅ O .

Proposition 12.
The moduli space M 1 consists of the unique nontrivial extensions F of O by O for each curve ∈ |O (2, 2)| and a point ∈ , and one also has ℎ 0 (F) = 1.
where is a curve contained in . If is properly contained in , we have (O ) > (F) contradicting to the semistability of F and thus we have = ; that is, ℎ is an isomorphism from O to its image, that is, we have where G ( , ) = 1. In particular, we have G ≅ O , the skyscraper sheaf supported on a point ∈ . Since F has no 0-dimensional torsion, the sequence does not split. Note that Ext 1 (O , O ) ≅ 1 (O ) ∨ = 0, and thus from the sequence of we have Here, the map is the transpose of Hom(O (2, 2), O ) → Hom(O , O ) 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 Ext 1 (O , O ) is 1 and so F corresponds to a unique nontrivial extension From the sequence (32), we have ℎ 0 (F) ≤ 2 and that ℎ 0 (F) = 1 if and only if no injective map O → F is an isomorphism at . This is certainly true if F is not locally free of rank 1 at . Note that F is a line bundle at each point of \ { } and thus it is sufficient to prove ℎ 0 (F) = 1 when F is a line bundle on The Scientific World Journal 5 the curve . In this case the nonexistence of a section of F that does not vanish at is equivalent to the nonsplitting of (32). Thus, we have ℎ 0 (F) = 0 and so the point is uniquely determined by F. Conversely, let us assume that F is a nontrivial extension of O by O , where is a point on . If F is not semistable, then there exists a subsheaf K ⊂ F with (K) > (F) = 1/4 and so we have K ( , ) = + + with ( , ) ≤ (2, 2) and ≥ 1. If the composite : K → F → O is a zero map, then we have an injection K → O , contradicting the semistability of O . Thus, the composite is surjective and so we have the following diagram: Remark 13. There is no strictly semistable sheaf in M 1 . Let us assume the existence of a polystable sheaf F = F 1 ⊕ ⋅ ⋅ ⋅ ⊕ F with ≥ 2. We have (F) = 1 = (F 1 ) + ⋅ ⋅ ⋅ + (F ). If we let F ( , ) = + + , then we have It implies that > 0 for all and thus we have = 1, a contradiction.
Since every element of M 1 is stable, M 1 has a universal family H 1 on M 1 × P 2 (see page 180 of [8] or Theorem 4.6.5 of [9]). Since every semistable sheaf with bipolinomial 2 +2 +5 is of the form F(1, 1) for a unique F ∈ M 1 , we also have a universal family H 5 on M 5 × P 2 with F(1, 1) as the fibre. Since ℎ 0 (F) = 1 and ℎ 1 (F) = 0 for all F ∈ M 1 , the base change theorem gives that * (H 1 ) is a line bundle on M 1 , where : M 1 × P 2 → 1 is the first projection. Since ℎ 0 (F(1, 1)) = 5 and ℎ 1 (F(1, 1)) = 0 for all F ∈ M 1 , the base change theorem gives that V * (H 5 ) is a vector bundle of rank Since and span F, there is a neighborhood of F in such that the sections 1 (G) and 2 (G) span every G ∈ . The construction of gives that 1 and 2 induce a section of in a neighborhood of whose image by is .
As an automatic consequence, we obtain that M 1 is irreducible and unirational. In fact, we can prove more.
Proof. Let Δ ⊂ × be the diagonal and denote its ideal sheaf by I Δ . Denoting by 1 and 2 the projection from × to each factor, let us define a sheaf U to be * 1 O (2, 2)) ⊠ I Δ on × . For each point ∈ , we have U| ×{ } ≅ I (2, 2). Thus, we have ℎ 1 (U| ×{ } ) = 0 and so 2 * U is a vector bundle of rank 8 on since ℎ 0 (U| ×{ } ) = 8. Let us consider the projective bundle By its definition, the fibre of Z over a point ∈ is the set of curves of type (2, 2) on passing through and so there is a natural map from Z to P 0 (O (2, 2)) ≅ P 8 . In other words, Z is the universal curve of type (2, 2) on and it is isomorphic to M 1 . Since Z is locally trivial over , it is rational.

Hilbert Bipolynomial 2 +2 +2
Lemma 18. Any sheaf F ∈ M 2 admits one of the following types: If we have 1 = , that is the map ℎ is an isomorphism from O to its image in F, then its cokernel H is the skyscraper sheaf supported on two points, say 1 , 2 ∈ . Thus we have the sequence Let us assume that 1 is properly contained in and then we obtain that 1 has bidegree (1, 1), (1, 2), or (2, 1) since (O 1 ) ≤ (F) = 1/2 and F is semistable. Let 2 ⊂ be the only curve such that 1 + 2 = . Let H be the quotient of H by its torsion , that is, H := H .
First, assume 1 ∈ |O (1, 1)| and so we have H ( , ) = + + 1 − deg( ). Since F is semistable, we get = 0. Since every quotient of F has the slope at least 1/2, the same is true for H. Proof. Let us take F ∈ M 2 and then there is no nonzero map F → O ≅ since F is semistable. Thus we have ℎ 1 (F) = 0 and so ℎ 0 (F) = 2. It is clear that F of types (B) and (C) is globally generated and so we may assume that F is of type (A), but neither of (B) nor of (C).
Let H ⊆ F be the image of the evaluation map 0 (F) ⊗ O → F and then H is pure. Assume that H ̸ = F. Since F is of type (A), it is globally generated outside at most two points of red . In particular, we have H ( , ) = 2 + 2 + with ≤ 1 and deg(F/H) = 2. Since ℎ 0 (H) = ℎ 0 (F) = 2, we have ℎ 1 (H) = 2 − . Note that every nonzero section of H vanishes at finitely many points since F is neither of types (B) nor (C). Since ℎ 0 (O ) < ℎ 0 (H), we have H ̸ = O and = 1.

A nonzero section of H induces an exact sequence
where G ≅ O for some ∈ red . Since H is pure, this exact sequence does not split. As in the proof of Proposition 12, we get a contradiction. Thus, we have H = F and so F is globally generated. Proof. Let F ∈ M 2 be a sheaf of type (A) and then it is globally generated by Corollary 19. Since ℎ 0 (F) = 2, we have a surjection
Similarly, we define B and C for the semistable sheaves of types (B) and (C), respectively. In particular, we have we have (K) ≤ 1/2, a contradiction.
Let us denote by M 2 the closed subscheme of M 2 , consisting of the strictly semistable sheaves.

Corollary 23. One has
where S 2 is the permutation group of order 2. In particular, M 2 is a rational variety.
Proof. Obviously, we have B ⊂ M 2 . Let F be a strictly semistable sheaf and so it has a proper quotient sheaf H with (H) = 1/2. From the semistability of F, H has no 0-dimensional torsion. From the equality (F) = (H), we obtain that H is also semistable. Since (H) = 1/2, the Hilbert bipolynomial of H is either 2 + 1, 2 + 1 or + +1. The first 2 cases cannot happen due to Proposition 10. Thus, we have H ( , ) = + + 1 and so H ≅ O 2 with 2 ∈ |O (1, 1)| and 2 ⊂ F by Proposition 11. If K is the kernel of the quotient map H → H, then its p-slope is again 1/2 and so K is semistable. Similarly as before, we have Let F be a sheaf of type (B), that is, it corresponds to a pair of two curves { 1 , 2 }. Let us assume that F admits another sequence ≅ O 1 . Hence, the class of a strictly semistable sheaf F corresponds to a uniquely determined pair of two curves in |O (1, 1)| and we have B ≅ (P 3 × P 3 )/S 2 . The second assertion follows from the fact that any symmetric product (P ) of any projective space is a rational variety (see Theorems 4.2.8 and 4.2.8 in page 137 of [11,12]).

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The Scientific World Journal Lemma 24. For two curves 1 , 2 ∈ |O (1, 1)|, one has Proof. Note that we have Thus, if we apply the functor Hom(O 2 , −) to the sequence of 1 , we obtain We also have and so their dimensions are 1 and 3, respectively. As Osheaves, we have for example, because 1 and 2 are reduced, and so the assertion is derived.
Lemma 25. The sheaves F 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 K of F with (K) ≥ 1/2 and the quotient sheaf H := F/K. Since the composite : is not a zero map, thus we have Im( ) ≅ O 2 (− ) for a 0dimensional subscheme of 2 with length . In particular, its Hilbert bipolynomial is + 1 − . If we let K ( , ) = + + , then we have (K) = /( + ) ≥ 1/2. In particular, we have ≥ 1. If we define K to be the kernel of the map , then it is a subsheaf of O 1 and thus we have (K ) = ( − 1 + )/( + − 1) ≤ 1/3. Combining the two inequalities, we have = 0 and so the map is surjective. Thus, we have H ≅ O 1 /K . Note also that can be either 1 or 2. If = 2, then we have = = 2 and so K ( , ) = 2 + + 1 = O 1 ( , ). In particular, we have H = 0 and so K ≅ F, a contradiction. Now, assume = 1 and so + ≤ 2. In particular, H is not a 0dimensional sheaf. Moreover, H is a quotient sheaf of O 1 with constant term 1 and so we have H ≅ O 3 with 3 ⊂ 1 and 3 ∈ |O (1, 1)|. For example, if 3 = 1 , then we have K = 0 and it contradicts the nontriviality of the extension F. Thus, F also admits the sequence where K ( , ) = + + 1.  Proof. In both parts, the "only if " part is obvious. Assume that F is not stable (resp., semistable) and take a proper subsheaf then H is a pure sheaf with 1 := + as its scheme support and has Hilbert bipolynomial H = O 1 . Note that it has O as its subsheaf.
To prove H ≅ O 1 , it is sufficient to prove that H is semistable. Suppose H is not semistable and take a proper saturated stable subsheaf G ⊂ H with G = + + . Its scheme support is contained in 1 and it is of type ( , ). Without loss of generality, let us assume that ≤ . First, assume ( , ) = (1,2). In this case, we would have ≥ 2 because (G) > (H) and so we have ℎ 0 (H) ≥ 2, contradicting the fact that ℎ 0 (F) = 2 and that F is globally generated. Assume = = 1. The map G → H on \ 2 must be just the inclusion O → H, because H| \ 2 is a line bundle. Thus either we have G = O or O is not saturated in H. Hence the saturation A of O in F has slope greater than 1/2, contradicting the semistability of F. Now assume = 0 and = 1, that is, G = . Since is smooth, G is a line bundle on . If its degree is at least 1, then G contradicts the semistability of F. If ≤ 0, then we have (G) < (H), a contradiction. Hence F is also contained in C.
Lemma 28. For 1 ∈ |O (1, 2)| and 2 ∈ |O (1, 0)|, one has Proof. Applying the functor Hom(O 2 , −) to the sequence of 1 , we obtain since we have and similarly Ext 2 (O 2 , O ) = 0. Note also that Thus we have the assertion. Remark 29. When 1 and 2 meet transversally at two points, say 1 and 2 , then is the global sheaf of a sheaf with support on 1 and 2 with one copy of C on each point for the following reason. Let be a regular local ring of dimension 2 and take , generators of its maximal ideal. All Ext groups are with respect to . Since /( ) is Gorenstein, so the duality gives Ext 1 ( /( ), ) ≅ /( ) and Ext ( /( ), ) = 0 for all ̸ = 1. From the exact sequence in which is the multiplication by , we get that Ext 1 ( /( ), /( )) is the cokernel of the multiplication by in /( ) → R/( ); that is, we have Ext 1 ( /( ), /( )) = C. The same is true for extensions of O by O when and are transversal.
Proof. Let K be a subsheaf F with maximal -slope (K) > 1/2 and so the quotient sheaf H := F/K has no 0dimensional torsion. Let us set K ( , ) = + + with ≥ 1 and (0, 0) ⪇ ( , ). If the composite : K → F 2 is a zero map, then K is a subsheaf destabilizing O , a contradiction. If is not surjective, for instance, Im( ) = O ⊊ , then ker( ) is a subsheaf of O with Hilbert bipolynomial + + − 1. Thus we have = 1 and the quotient H := O /K has Hilbert bipolynomial with zero constant term. Since H has no 0-dimensional torsion, we have H ≅ O for a curve contained in . But the Hilbert bipolynomial of O has nonzero constant term, a contradiction. Thus the map is surjective. Following the same argument before, we obtain that = 1 and + ≤ 1. Without loss of generality, let us assume that ( , ) = (0, 1). Then we have H ( , ) = 2 + + 1 and thus we have H ≅ O 1 , where 1 is a curve contained in F and 1 ∈ |O (1, 2)|. Since K is a subsheaf of F with K ( , ) = + 1, we have K ≅ O 2 since K has no 0dimensional torsion. Thus F fits into the sequence (68).
Remark 31. Applying the functor Hom( , −) to the sequence of ∈ |O (2, 2)|, we obtain   The Scientific World Journal Conversely, assume that 1 ∩ 2 is finite. Since we have ℎ 1 (O 1 ) = 0, the sequence (70) implies that ℎ 0 (F) = 2 and F is globally generated. Let be a general section of F and then it does not vanish at the general point of any of the components of F . Since F is reduced, induces an injective map O F → F and thus F has type (A). Lemma 33. Let F be a sheaf of type ( ).
(1) If 2 is not a component of 1 , then F is of type (A).
(2) If 2 is a double component of F , that is, 2 ⊂ 1 , then it is not of type (A).
(1) Since 2 is not a component of 1 , F is a line bundle on = F outside finitely many points of . Moreover, it is not an O 1 -sheaf. Note that F is globally generated since O 1 and O 2 are globally generated with ℎ 1 (O 1 ) = 0. Thus, a general section of F does not vanish at a general point of 2 and so it does not induce an injection O 1 → F. Hence, F fits into some sequence (46).
(2) Let us set = 2 2 + 3 and 1 := 2 + 3 , where 3 ∈ |O (0, 2)|. Let Γ be the projectivisation of Ext 1 (O 2 , O 1 ) and in particular we have Γ ≅ P 1 by Lemma 28. We also know from Lemma 25 that any ∈ Γ gives a semistable sheaf. Such a sheaf has rank 2 at the points of 2 \ ( 2 ∩ 3 ) and, in particular, it is not a line bundle over its support at a general point of 2 . Thus, it never fits into an exact sequence (46). Otherwise it would be locally free of rank 1 at each point of the support of 3 but not in 2 .
In general, the question whether the variety P ( , ) . We observed that M 1 is rational and so is P 1 (2,2) . Below we give a partial answer to this question in the case of P 2 (2,2) .
Proof. Let us fix a smooth curve of bidegree (2, 2) in and a point ∈ to consider a sheaf O ( ) ∈ M 1 . If T is the tangent plane of at , then we have T ∩ = {2 , 1 , 2 } for some points 1 , 2 on since deg( ) = 4. It defines a rational map sending O ( ) to O ( 1 + 2 ). Note that O ( 1 + 2 ) = O (1, 1)(−2 ). We claim that the map Φ is generically 4 to 1 and so the assertion follows. Let (resp., ) be the dense open subset of M 1 (resp., A ⊂ M 2 ) formed by the sheaves F such that F is smooth. Each element of (resp., ) is uniquely determined by a smooth ∈ |O (2, 2)| 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 Φ sends O ( ) to R := O (1, 1)(−2 ). Fix any degree two line bundle M on . Since we are in characteristic zero, there are exactly four line bundles A on such that A ⊗2 ≅ O . Hence, for each R ∈ Pic 2 ( ) there are exactly 4 points ∈ such that R ≅ O (1, 1)(−2 ). Hence, Φ is dominant and the preimage of each element of has cardinality 4.
We did not succeed in getting any smaller degree of unirationality of M 2 as of now, and we left the rationality question as a conjecture.