This paper investigates the moduli space of stable sheaves supported on genus three curves within a smooth quadric surface, demonstrating its rationality and computing its Betti numbers through a detailed geometric analysis.
Contribution
It provides a comprehensive description of the moduli space, including its rationality, Betti numbers, classification of sheaves, and a global geometric construction involving flips and blow-downs.
Findings
01
The moduli space is rational.
02
Betti numbers are explicitly computed.
03
Stable sheaves are classified via resolutions and extensions.
Abstract
We study the moduli space of stable sheaves of Euler characteristic 1 supported on curves of arithmetic genus 3 contained in a smooth quadric surface. We show that this moduli space is rational. We compute its Betti numbers by studying the variation of the moduli spaces of alpha-semi-stable pairs. We classify the stable sheaves using locally free resolutions or extensions. We give a global description: the moduli space is obtained from a certain flag Hilbert scheme by performing two flips followed by a blow-down.
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Full text
Moduli of stable sheaves supported on curves of genus three contained in a quadric surface
Mario Maican
Institute of Mathematics of the Romanian Academy, Calea Grivitei 21, Bucharest 010702, Romania
We study the moduli space of stable sheaves of Euler characteristic 1
supported on curves of arithmetic genus 3 contained in a smooth quadric surface.
We show that this moduli space is rational.
We compute its Betti numbers by studying the variation of the moduli spaces of α-semi-stable pairs.
We classify the stable sheaves using locally free resolutions or extensions.
We give a global description: the moduli space is obtained from a certain flag Hilbert scheme by performing two flips
followed by a blow-down.
Key words and phrases:
Moduli spaces, Semi-stable sheaves, Wall crossing
2010 Mathematics Subject Classification:
Primary 14D20, 14D22
1. Introduction
Let P1 be the projective line over C and consider the surface P1×P1 with fixed polarization
O(1,1)=OP1(1)⊗OP1(1).
For a coherent algebraic sheaf F on P1×P1, with support of dimension 1, the Euler characteristic
χ(F(m,n)) is a polynomial expression in m, n, of the form
[TABLE]
where r, s, t are integers depending only on F. This is the Hilbert polynomial of F.
The slope of F is
[TABLE]
Let M(P) be the coarse moduli space of S-equivalence classes of sheaves on P1×P1
that are semi-stable with respect to the fixed polarization and that have Hilbert polynomial P.
We recall that F is semi-stable, respectively, stable, if it is pure and for any proper subsheaf F′⊂F we have p(F′)≤p(F),
respectively, p(F′)<p(F).
According to [10], M(P) is projective, irreducible, and smooth at points given by stable sheaves.
Its dimension is 2rs+1 if r>0 and s>0.
The spaces M(rm+n+1), M(2m+2n+1) and M(2m+2n+2) were studied in [1].
In fact, it is not difficult to see that M(rm+n+1) consists of the structure sheaves of curves of degree (1,r),
so it is isomorphic to P2r+1.
The space M(3m+2n+1) was studied in [4] and [12].
We refer to the introductory section of [12] for more background information.
This paper is concerned with the study of M=M(4m+2n+1).
The closed points of M are in a bijective correspondence with the isomorphism classes [F] of stable sheaves F
supported on curves of degree (2,4) and satisfying the condition χ(F)=1.
As already mentioned, M is a smooth irreducible projective variety of dimension 17.
For any t∈Z, twisting by O(t,t) gives an isomorphism M≃M(4m+2n+6t+1).
According to [12, Corollary 1], M≃M(4m+2n−1).
In the following theorem we classify the sheaves in M.
Theorem 1.1**.**
The variety M can be decomposed into an open subset M0, two closed irreducible subsets M2, M2′,
each of codimension 2, a locally closed irreducible subset M3 of codimension 3,
and a locally closed irreducible subset M4 of codimension 4.
These subsets are defined as follows: M0 is the set of sheaves F having a resolution of the form
[TABLE]
where the entries φ12 and φ22 are linearly independent and the maximal minors of the matrix (φij)i=1,2,j=1,2,3,
describing the corestriction of φ to the first two summands, have no common factor;
M2 is the set of sheaves F having a resolution of the form
[TABLE]
with φ11=0, φ12=0;
M2′ is the set of sheaves F having a resolution of the form
[TABLE]
with φ11=0, φ12=0;
M4 is the set of extensions of the form
[TABLE]
satisfying the condition H0(F)≃C,
where Q⊂P1×P1 is a quintic curve of degree (2,3) and L⊂P1×P1 is a line of degree (0,1);
M3 is the set of extensions of the form
[TABLE]
where OQ(p) is a non-split extension of Cp by OQ for a point p∈Q, and satisfying the condition H0(F)≃C.
Moreover, M2 is the Brill-Noether locus of sheaves for which H1(F)={0}.
The proof of Theorem 1.1, given in Section 3, relies on the Beilinson spectral sequence,
which we recall in Section 2.
The varieties X that appear in this paper have no odd homology, so we can define the Poincaré polynomial
[TABLE]
Theorem 1.2**.**
The Euler characteristic of M is 288.
The Poincaré polynomial of M is
[TABLE]
The proof of this theorem rests on the wall-crossing method of Choi and Chung [3].
In Section 4 we investigate how the moduli spaces Mα(4m+2n+1) of α-semi-stable pairs
with Hilbert polynomial 4m+2n+1 change as the parameter α varies.
In Theorem 4.11 we find that Mα(4m+2n+1) are related by two explicitly described flipping diagrams.
Combining this with Proposition 4.12, yields a global description: M is obtained from the flag Hilbert scheme
of three points on curves of degree (2,4) in P1×P1 by performing two flips followed by a blow-down
with center the Brill-Noether locus M2.
The total space X of ωP1×P1 is a Calabi-Yau threefold.
For a homology class β=(r,s)∈H2(P1×P1)⊂H2(X) let
Nβ(X) be the genus zero Gromov-Witten invariant of X
and let nβ(X) be the genus zero Gopakumar-Vafa invariant of X, as introduced in [8].
It was noticed in [4] that, up to sign, the latter is the Euler characteristic of a moduli space:
Our main technical tool in Section 3 will be the Beilinson spectral sequence.
Let F be a coherent sheaf on P1×P1. According to [2, Lemma 1], we have a spectral sequence
converging to F, whose first level E1 has display diagram
[TABLE]
where E1ij={0} if i∈/{−2,−1,0} or if j∈/{0,1,2} and
[TABLE]
If F has support of dimension 1, then the first row of (1) vanishes and
the convergence of the spectral sequence forces θ2 to be surjective and yields the exact sequence
[TABLE]
An application of the Beilinson spectral sequence is the following lemma that will be used in Section 3.
Lemma 2.1**.**
Let Z⊂P1×P1 be a zero-dimensional subscheme of length 3
that is not contained in a line of degree (1,0) or (0,1).
Then the ideal of Z has resolution
[TABLE]
where the maximal minors of ζ have no common factor.
The dual of the structure sheaf of Z has resolution
[TABLE]
Proof.
We apply the spectral sequence (1) to the sheaf F=IZ(1,1).
By hypothesis, H0(IZ(1,0))={0} and H0(IZ(0,1))={0} hence, from (2), we obtain the vanishing of E1−1,0.
Since H0(IZ)={0}, also E1−2,0 vanishes.
From the short exact sequence
[TABLE]
we obtain the vanishing of H2(IZ). Analogously, H2(IZ(1,0)), H2(IZ(0,1)) and H2(IZ(1,1)) vanish.
The first row of (1) vanishes.
Denote d=dimCH1(IZ(1,1)). Display diagram (1) now takes the simplified form
[TABLE]
From the convergence of the spectral sequence we see that θ2 is surjective.
There is no surjective morphism θ2:O(0,−1)⊕O(−1,0)→dO for d≥1, hence d=0.
Thus, Ker(θ1) is a subsheaf of O. We claim that Ker(θ1)={0}.
Indeed, if Ker(θ1) were non-zero, then O/Ker(θ1) would be a torsion subsheaf of IZ(1,1).
Combining the exact sequences
[TABLE]
[TABLE]
yields the resolution
[TABLE]
Applying Hom(−,O(−1,−3)), we obtain resolution (4).
If the maximal minors of the matrix representing ζ had a common factor f,
then the reduced support of Coker(ζ) would contain the curve {f=0}.
But this is impossible because Ext2(OZ,O) has support of dimension zero.
∎
Lemma 2.2**.**
Let S be a smooth projective surface and let C⊂S be a locally Cohen-Macaulay curve.
Let Z be a coherent sheaf on S with support of dimension zero.
Let F be an extension of Z by OC without zero-dimensional torsion.
Then F is uniquely determined up to isomorphism,
meaning that if F′ is another extension of Z by OC without zero-dimensional torsion, then F′≃F.
Moreover, Z≃ExtOS2(OZ,OS) for a subscheme Z⊂C of dimension zero,
so we have the exact sequence
[TABLE]
Proof.
This lemma is a direct consequence of [13, Proposition B.5].
Indeed, given an exact sequence
[TABLE]
in which F has no zero-dimensional torsion, then the pair (OC,F) is a stable pair supported on C, in the sense of [13].
By [13, Lemma B.2], we have ExtOC1(F,OC)={0}.
Applying HomOC(−,OC) to (6), yields the exact sequence
[TABLE]
Thus, ExtOC1(Z,OC) is the structure sheaf OZ of a zero-dimensional subscheme Z⊂C.
Under the bijection of [13, Proposition B.5] between stable pairs supported on C
and zero-dimensional subschemes of C, the pair (OC,F) corresponds to Z,
so it is uniquely determined, up to isomorphism.
Tensoring (7) with the dualising line bundle ωC on C, yields the exact sequence
[TABLE]
We claim that Hom(F,ωC)≃Ext1(F,ωS).
This follows by applying Hom(F,−) to the exact sequence
[TABLE]
We obtain the exact sequence
[TABLE]
The last morphism is locally multiplication with an equation f defining C.
But C=supp(F), hence f annihilates F,
and hence f annihilates Ext1(F,ωS). This proves the claim.
According to [11, Remark 4], Ext1(Ext1(F,ωS),ωS)≃F.
Clearly,
[TABLE]
Applying Hom(−,ωS) to (8) yields extension (5).
Comparing with (6), we see that Z≃Ext2(OZ,OS).
∎
Crucial for our classification of semi-stable sheaves
is the following vanishing result that should be compared with [12, Proposition 4].
We fix vector spaces V1 and V2 over C of dimension 2 and we identify P1×P1
with P(V1)×P(V2). Let {x,y} be a basis of V1∗ and let {z,w} be a basis of V2∗.
A morphism O(i,j)→O(k,l) will be represented by a form in Sk−iV1∗⊗Sl−jV2∗.
Proposition 2.3**.**
Assume that the sheaf F gives a point in M.
(i)
We have H0(F(−1,−1))={0} and H0(F(−1,0))={0}.
2. (ii)
If F satisfies the vanishing condition H0(F(0,−1))={0}, then H1(F)={0}.
Proof.
(i) The vanishing of H0(F(−1,−1)) follows from [12, Proposition 2(i)].
To prove the vanishing of H0(F(−1,0)) we can argue as in the proof of [12, Proposition 3].
(ii) Assume now that H0(F(0,−1))={0}.
From (2) and part (i) of the proposition, we deduce that E1−1,1≃O(0,−1)⊕3O(−1,0).
Denote d=dimCH1(F).
There is no surjective morphism
[TABLE]
for d≥4, hence d≤3. Assume that d=3. The maximal minors for a matrix representation of θ2 have no common factor,
otherwise θ2 would not be surjective.
Thus, Ker(θ2)≃O(−3,−1), hence θ1=0, and hence,
from the exact sequence (3), we obtain a surjective morphism
F→O(−3,−1). This is absurd. Thus, the case when d=3 is unfeasible.
Consider now the case when d=2. If θ2 is represented by a matrix of the form
[TABLE]
then Ker(θ2)≃O(0,−1)⊕O(−3,0), hence O(−3,0) is a direct summand of Ker(θ2)/Im(θ1),
and hence, from the exact sequence (3), we obtain a surjective morphism F→O(−3,0). This is absurd.
If θ2 is represented by a matrix of the form
[TABLE]
then Ker(θ2)≃O(−2,−1)⊕O(−1,0), hence O(−2,−1) is a direct summand of Ker(θ2)/Im(θ1),
and hence we obtain a surjective morphism F→O(−2,−1). This is absurd.
If θ2 is represented by a matrix of the form
[TABLE]
then Ker(θ2)≃O(−1,−1)⊕O(−2,0) and we obtain a surjective morphism F→O(−2,0). This is absurd.
We claim that, if θ2 is not of the form A, B or C, then θ2 is represented by a matrix of the form
[TABLE]
with v=0. Indeed, since θ2≁A and θ2≁B, we may write
[TABLE]
with u=0, v=0. Since θ2≁B, v1 and v2 cannot be both zero.
If v1 and v2 are linearly independent, then θ2∼D.
If v1 and v2 span a one-dimensional vector space,
then, since θ2≁B, we may write
[TABLE]
Since θ2≁C, we have u1=0, forcing θ2∼D.
In the case when θ2=D, it is easy to see that the morphism
[TABLE]
is represented by a matrix of the form
[TABLE]
hence Ker(θ1)≃3O(−1,−1), and hence Coker(θ5) has Hilbert polynomial 3m+3n+3.
But then, in view of the exact sequence (3),
Coker(θ5) is a destabilizing subsheaf of F. Thus, the case when d=2 is also unfeasible.
It remains to examine the case when d=1. Recall that θ2 is surjective, hence it can have two possible forms.
Firstly, if
[TABLE]
then Ker(θ2)≃O(0,−1)⊕O(−2,0)⊕O(−1,0) and we obtain a surjective morphism
F→O(−2,0), which is absurd. The second form is
[TABLE]
If θ1 is represented by a matrix having two zero columns, then Ker(θ1)≃2O(−1,−1), hence Coker(θ5)
has Hilbert polynomial 2m+2n+2, and hence Coker(θ5) is a destabilizing subsheaf of F.
Thus, we may write
[TABLE]
hence Ker(θ1)≃O(−1,−2)⊕O(−1,−1), and hence Coker(θ5) has Hilbert polynomial 3m+2n+2.
But then Coker(θ5) is a destabilizing subsheaf of F.
We deduce that the case when d=1 is also unfeasible.
∎
3. Classification of sheaves
We begin our classification of semi-stable sheaves by examining the Brill-Noether locus of sheaves that do not satisfy
the first vanishing condition in Proposition 2.3(ii).
Proposition 3.1**.**
The sheaves F in M satisfying the condition H0(F(0,−1))={0} are precisely the non-split extension sheaves of the form
[TABLE]
where C⊂P1×P1 is a curve of degree (2,4) and p is a point on C.
Moreover, the sheaves from (9) are precisely the sheaves F having a resolution of the form
[TABLE]
with φ11=0, φ12=0.
Let M2⊂M be the subset of sheaves F from (9).
Then M2 is closed, irreducible, of codimension 2, and is isomorphic to the universal curve of degree (2,4)
in P1×P1.
Thus, M2 is a fiber bundle with fiber P13 and base P1×P1.
Proof.
Let F give a point in M and satisfy H0(F(0,−1))={0}. As in the proof of [12, Proposition 2],
there is an injective morphism OC→F(0,−1) for a curve C of degree (s,r), 0≤s≤2, 0≤r≤4, 1≤r+s≤6.
From the stability of F we have the inequality
[TABLE]
which has the unique solution (s,r)=(2,4).
We obtain extension (9).
Conversely, let F be given by the non-split extension (9).
As in the proof of [12, Proposition 3], we can show that OC(0,1) is stable,
from which it immediately follows that F gives a point in M and that H0(F(0,−1))={0}.
Choose φ11∈V1∗⊗C and φ12∈C⊗V2∗ defining p.
Since p∈C, we can find φ21∈S2V1∗⊗S3V2∗ and φ22∈V1∗⊗S4V2∗
such that the polynomial φ11φ22−φ12φ21 defines C.
Consider the morphism
[TABLE]
[TABLE]
From the snake lemma we see that Coker(φ) is an extension of Cp by OC(0,1).
Since Coker(φ) has no zero-dimensional torsion, we can apply Lemma 2.2 to deduce that F≃Coker(φ).
Thus, [F]∈M2 if and only if F has resolution (10).
∎
In the remaining part of this section we will assume that F satisfies both vanishing conditions from Proposition 2.3(ii).
The exact sequence (3) takes the form
[TABLE]
where
[TABLE]
Proposition 3.2**.**
Assume that [F]∈M and that H0(F(0,−1))={0}.
Assume that the maximal minors of θ1 have a common factor. Then F is an extension of the form
[TABLE]
for a quintic curve Q⊂P1×P1 of degree (2,3) and a line L⊂P1×P1 of degree (0,1),
or is an extension of the form
[TABLE]
where OQ(p) is a non-split extension of Cp by OQ for a point p∈Q.
Conversely, any extension F as in (12) or (13) satisfying the condition H0(F)≃C is semi-stable.
Let M4⊂M be the subset of sheaves F as in (12) satisfying the condition H0(F)≃C.
Let M3⊂M be the subset of sheaves F as in (13) satisfying the condition H0(F)≃C.
Then M3 and M4 are locally closed, irreducible subsets, of codimension 3, respectively, 4.
Proof.
Let ηi be the maximal minor of a matrix representing θ1 obtained by deleting column i.
Denote g=gcd(η1,…,η5). Let (s,r)=(2,4)−deg(g).
It is easy to check that the sequence
[TABLE]
[TABLE]
is exact.
From (11) we see that Coker(θ5) is a subsheaf of F,
hence we have the inequality
[TABLE]
forcing (s,r)=(2,3) or (s,r)=(2,2).
If (s,r)=(2,2), then PCoker(θ1)=2m+1 and Coker(θ1) is semi-stable, which follows from the semi-stability of F.
But, according to [1, Proposition 10], M(2m+1)=∅.
This contradiction shows that (s,r)=(2,2), hence (s,r)=(2,3).
From (11) we obtain the extension
[TABLE]
If Coker(θ1) has no zero-dimensional torsion, we obtain extension (12).
Otherwise, the zero-dimensional torsion has length 1, its pull-back in F is a semi-stable sheaf OQ(p),
and we obtain extension (13).
Conversely, let F be an extension as in (12) satisfying H0(F)≃C.
Assume that F had a destabilizing subsheaf F′. Let G be the image of F′ in OL(1,0).
According to [12, Proposition 1], OQ is stable, hence χ(F′∩OQ)≤−1.
Since χ(F′)≥1, we see that χ(G)≥2, hence G=OL(1,0) and OQ⊈F′.
Thus H0(F′∩OQ)={0}, hence the map H0(F′)→H0(OL(1,0)) is injective.
But this map factors through H0(F)→H0(OL(1,0)), which, by hypothesis, is the zero map.
We deduce that H0(F′)={0}, which yields a contradiction. Thus, there is no destabilizing subsheaf.
The same argument applies for extensions (13) satisfying H0(F)≃C.
By Serre duality
[TABLE]
From the short exact sequence
[TABLE]
we obtain the long exact sequence
[TABLE]
Thus Ext1(OL(1,0),OQ)≃C2, hence M4 is isomorphic to an open subset of a P1-bundle over P11×P1.
By Serre duality we have
[TABLE]
Using Lemma 2.2, it is easy to see that the sheaves OQ(p)
are precisely the sheaves having a resolution of the form
[TABLE]
where φ11=0, φ12=0 (cf. Proposition 3.1).
From resolution (14) we obtain the long exact sequence
[TABLE]
Thus, Ext1(OL,OQ(p))≃C2, hence M3 has dimension 14. The other claims about M3 are obvious.
∎
Lemma 3.3**.**
Assume that [F]∈M and H0(F(0,−1))={0}.
Assume that the maximal minors of θ1 have no common factor.
Then Ker(θ1)≃O(−2,−4) and Coker(θ1)≃Ext2(OZ,O) with Z described below.
We have an extension
[TABLE]
where C is a curve of degree (2,4) and Z⊂C is a subscheme of dimension zero and length 3.
Moreover, Z is not contained in a line of degree (0,1).
Proof.
The fact that Ker(θ1)≃O(−2,−4) is well-known.
The Hilbert polynomial of Coker(θ1) is 3, hence Coker(θ1) has dimension zero and length 3.
From (11), we obtain the exact sequence
[TABLE]
We can now apply Lemma 2.2 to obtain the extension (15) and the isomorphism
Coker(θ1)≃Ext2(OZ,O).
Assume that Z is contained in a line L of degree (0,1). Then OZ≃Ext2(OZ,O).
Choose φ11∈C⊗V2∗ defining L.
Choose φ12∈S3V1∗⊗C such that φ11 and φ12 define Z.
If L⊈C, then L.C=2, which contradicts the fact that Z⊂L∩C.
Thus L⊂C, so there is φ22∈S2V1∗⊗S3V2∗ such that φ11φ22 is a defining polynomial of C.
Consider the exact sequence
[TABLE]
[TABLE]
Then F′ is an extension of OZ by OC without zero-dimensional torsion.
Since, from the exact sequence (15),
F is also an extension of OZ by OC without zero-dimensional torsion,
we can apply Lemma 2.2 to deduce that F≃F′.
We obtain a contradiction from the isomorphisms C≃H0(F)≃H0(F′)≃C3.
∎
Proposition 3.4**.**
Let M0⊂M be the subset of sheaves F for which H0(F(0,−1))={0}, Ker(θ1)≃O(−2,−4)
and supp(Coker(θ1)) is not contained in a line of degree (1,0) or (0,1).
Then M0 is open and can be described as the subset of sheaves F having a resolution of the form
[TABLE]
where φ12 and φ22 are linearly independent and the maximal minors of the matrix (φij)i=1,2,j=1,2,3
have no common factor.
Proof.
Let F give a point in M0. Let Z and C be as in Lemma 3.3.
By hypothesis Z is not contained in a line of degree (1,0) or (0,1),
hence Ext2(OZ,O)≃Coker(ζ) as in (4).
Let ζ1, ζ2, ζ3 be the maximal minors of ζ.
They are the defining polynomials of Z, hence we can find
φ31∈V1∗⊗S3V2∗, φ32∈C⊗S3V2∗, φ33∈V1∗⊗S2V2∗
such that ζ1φ31−ζ2φ32+ζ3φ33 is the polynomial defining C.
Let
[TABLE]
Then Coker(φ) is an extension of Ext2(OZ,O) by OC without zero-dimensional torsion and,
by Lemma 3.3, the same is true of F.
From Lemma 2.2 we deduce that F≃Coker(φ).
By Proposition 2.3, H0(F)≃C, hence the map H1(O(0,−3))→H1(2O(0,−2)) is injective,
which is equivalent to saying that φ12 and φ22 are linearly independent.
We have shown that F has resolution (16).
Conversely, assume that F has resolution (16).
Then H0(F)≃C because φ12 and φ22 are linearly independent.
From the snake lemma we see that F is an extension of Ext2(OZ,O) by OC,
where Z is the zero-dimensional scheme of length 3 given by the maximal
minors of the matrix obtained by deleting the third row of φ,
and C is the curve of degree (2,4) defined by det(φ).
Thus, H0(F) generates OC.
We will show that F is semi-stable. Assume that F had a destabilizing subsheaf F′.
Then χ(F′)>0 and χ(F′)≤dimCH0(F)=1, hence χ(F′)=1, forcing H0(F′)≃C.
Thus H0(F′)=H0(F), hence OC⊂F′, and hence F′ has multiplicity 6.
There are no destabilising subsheaves of F of multiplicity 6.
Thus, F gives a point in M.
Since φ12 and φ22 are linearly independent, we have H0(F(0,−1))={0}.
Since H0(F) generates OC, Ker(θ1)≃O(−2,−4) and Coker(θ1)≃Ext2(OZ,O).
Note that Z is not contained in a line of degree (1,0) or (0,1).
In conclusion, F gives a point in M0.
∎
Proposition 3.5**.**
The variety M is rational.
Proof.
By Lemma 2.1, Lemma 2.2, Lemma 3.3 and Proposition 3.4,
the open subset of M0, given by the condition that Z consist of three distinct points,
is a P11-bundle over an open subset of HilbP1×P1(3), so it is rational.
∎
Proposition 3.6**.**
Let F be an extension as in (15) without zero-dimensional torsion, for a curve C of degree (2,4)
and a subscheme Z⊂C that is the intersection of two curves of degree (1,0), respectively, (0,3).
Then F gives a point in M. Let M2′⊂M be the subset of such sheaves F.
Then M2′ is closed, irreducible, of codimension 2, and can be described as the set of sheaves F having a resolution of the form
[TABLE]
with φ11=0, φ12=0.
Proof.
Note that OZ≃Ext2(OZ,O).
Let F be an extension of OZ by OC without zero-dimensional torsion.
Let φ11∈V1∗⊗C and φ12∈C⊗S3V2∗ be the defining polynomials of Z.
We can find φ21∈S2V1∗⊗V2∗ and φ22∈V1∗⊗S4V2∗ such that φ11φ22−φ12φ21
is the defining polynomial of C.
Then the cokernel of φ=(φij)1≤i,j≤2 is an extension of OZ by OC without zero-dimensional torsion,
hence, by Lemma 2.2, F≃Coker(φ).
Conversely, arguing as in Proposition 3.4, we can show that any sheaf of the form Coker(φ), with φ as in (17),
is semi-stable.
∎
Proof of Theorem 1.1. By Propositions 3.1, 3.2, 3.4 and 3.6,
M is the union of the subvarieties M0, M2, M2′, M3, M4.
For [F]∈M2, we have H0(F)≃C2, whereas, for [F] in any of the other subvarieties, we have H0(F)≃C.
Thus, M2 is disjoint from the other subvarieties.
For [F]∈M0∪M2′, H0(F) generates the structure sheaf of a curve C of degree (2,4),
whereas, for [F]∈M3∪M4, H0(F) generates the structure sheaf of a curve Q of degree (2,3).
Thus, M0∪M2′ is disjoint from M3∪M4.
For [F]∈M0, the support of F/OC is not contained in a line of degree (1,0),
whereas, for [F]∈M2′, the support of F/OC is contained in a line of degree (1,0).
Thus, M0 is disjoint from M2′.
For [F]∈M3, F/OQ has zero-dimensional torsion, whereas, for [F]∈M4, F/OQ is pure.
Thus, M3 is disjoint from M4.
In conclusion, the subvarieties in question form a decomposition of M.
∎
4. Variation of the moduli spaces of α-semi-stable pairs
A coherent systemΛ=(Γ,F) on P1×P1
consists of a coherent algebraic sheaf F on P1×P1 and a vector subspace Γ⊂H0(F).
Let α be a positive real number and let PF(m,n)=rm+sn+t be the Hilbert polynomial of F.
We define the α-slope of Λ as the ratio
[TABLE]
We say that Λ is α-semi-stable, respectively, α-stable, if F is pure and for any proper
coherent subsystem Λ′⊂Λ we have pα(Λ′)≤pα(Λ), respectively,
pα(Λ′)<pα(Λ).
According to [9] and [7],
for fixed positive real number α, non-negative integer k and linear polynomial P(m,n),
there is a coarse moduli space, denoted Syst(P1×P1,α,k,P),
which is a projective scheme whose closed points
are in a bijective correspondence with the set of S-equivalence classes of α-semi-stable coherent systems (Γ,F)
on P1×P1 for which dimΓ=k and PF=P.
When k=0 this space is M(P).
A coherent system for which dimΓ=1 will be called a pair.
Our main concern is with the moduli space of α-semi-stable pairs Mα(P)=Syst(P1×P1,α,1,P).
It is known that there are finitely many positive rational numbers α1<…<αn, called walls,
such that the set of α-semi-stable pairs with Hilbert polynomial P remains unchanged as α varies in one of the intervals
(0,α1), or (αi,αi+1), or (αn,∞).
In fact, from the definition of α-semi-stability, we can see that, if α is a wall, then there is a strictly α-semi-stable pair,
i.e. a pair Λ for which there exists a subpair or quotient pair Λ′, such that pα(Λ)=pα(Λ′). This equation has only rational solutions in α.
For α∈(αn,∞) we write M∞(P)=Mα(P).
For α∈(0,α1) we write M0+(P)=Mα(P).
If gcd(r+s,t)=1, then, from the definition of α-semi-stability, we see that (Γ,F)∈M0+(P) if and only if
F is semi-stable. At the other extreme we have the following proposition due to Pandharipande and Thomas.
Proposition 4.1**.**
For α≫0, a pair Λ=(Γ,F) is α-semi-stable if and only if F is pure and F/OC has dimension zero or is zero,
where OC is the subsheaf of F generated by Γ. In particular, t≥r+s−rs.
The scheme M∞(rm+sn+t) is isomorphic to the relative Hilbert scheme of zero-dimensional schemes of length t−r−s+rs
contained in curves of degree (s,r).
Proof.
Assume that (Γ,F) is α-semi-stable for α≫0. If POC(m,n)=r′m+s′n+t′ with r′+s′<r+s, then
[TABLE]
which contradicts our hypothesis. Thus, POC(m,n)=rm+sn+r+s−rs.
Conversely, assume that OC has this Hilbert polynomial and that F is pure.
Let Λ′=(Γ′,F′)⊂Λ be a proper coherent subsystem with PF′(m,n)=r′m+s′n+t′.
If Γ′={0}, then
[TABLE]
If Γ′=Γ, then OC⊂F′, hence r′=r, s′=s, t′<t, and we have
[TABLE]
The isomorphism between M∞(P) and the relative Hilbert scheme is a particular case of [13, Proposition B.8].
As a map, it is given by (Γ,F)↦(Z,C), where Z⊂C is the subscheme introduced at Lemma 2.2.
∎
Corollary 4.2**.**
The scheme M∞(4m+2n+1) is isomorphic to a fiber bundle with fiber P11 and base the Hilbert scheme of three points
in P1×P1, so it is smooth.
Proof.
The relative Hilbert scheme of pairs (Z,C), where C⊂P1×P1 is a curve of degree (2,4) and Z⊂C
is a subscheme of dimension zero and length 3, has fiber P(H0(IZ(2,4))) over Z.
If Z is not contained in a line of degree (0,1) or (1,0), then, from Lemma 2.1, we deduce that
H0(IZ(2,4))≃C12. If Z is contained in such a line, then it is straightforward to check that H0(IZ(2,4))≃C12.
∎
Lemma 4.3**.**
Assume that Mα(rm+sn+t)=∅. Then t≥r+s−rs.
For r, s non-negative integers, not both zero, and α∈(0,∞), we have
[TABLE]
Proof.
We use induction on r+s. If r+s=1, or if there is no wall in [α,∞),
then Mα(rm+sn+t)=M∞(rm+sn+t) and the conclusion follows from Proposition 4.1.
Assume that r+s>1 and that there is a wall α′∈[α,∞).
There is a pair Λ∈Mα′(rm+sn+t) and a subpair or quotient pair Λ′∈Mα′(r′m+s′n+t′),
such that pα′(Λ)=pα′(Λ′). We have 0≤r′≤r, 0≤s′≤s, 1≤r′+s′<r+s,
[TABLE]
hence
[TABLE]
If t=r+s−rs, then there is no wall in [α,∞), hence we have an isomorphism as in the lemma.
∎
Proposition 4.4**.**
With respect to P(m,n)=4m+2n+1 there are only two walls at α1=5 and α2=11.
Proof.
Assume that α is a wall. Then there are pairs Λ∈Mα(4m+2n+1) and Λ′∈Mα(rm+sn+t)
such that Λ′ is a subpair or a quotient pair of Λ and
[TABLE]
Here 0≤r≤4, 0≤s≤2, 1≤r+s≤5. By Lemma 4.3, we also have t≥r+s−rs.
Assume that r=3, s=2, t≥−1. Equation (18) has solutions
α1=5 for t=0 and α2=11 for t=−1. Assume that r=2, s=2, t≥0.
Equation (18) has solution α=2 for t=0.
In this case either Λ∈Ext1(Λ′,Λ′′) or Λ∈Ext1(Λ′′,Λ′)
for some Λ′′∈M(2m+1).
However, according to [1, Proposition 10], M(2m+1)=∅.
Thus, there is no wall at α=2.
For all other choices of r and s equation (18) has no positive solution in α.
∎
Denote Mα=Mα(4m+2n+1).
For α∈(11,∞), write Mα=M∞.
For α∈(5,11), write Mα=M5+=M11−.
For α∈(0,5), write Mα=M0+.
The inclusions of sets of α-semi-stable pairs induce the birational morphisms
[TABLE]
In view of Theorem 4.11, the above are flipping diagrams
(consult [12, Remark 5] for details).
Remark 4.5**.**
From the proof of Proposition 4.4, we see that an S-equivalence class of
strictly α-semi-stable elements in M11
consists of (split or non-split) extensions of (Γ1,E1) by (0,OL(1,0)),
together with the extensions of (0,OL(1,0)) by (Γ1,E1).
Here (Γ1,E1) lies in M11(3m+2n−1) and
L⊂P1×P1 is a line of degree (0,1).
We say, for short, that the strictly α-semi-stable elements of M11 are of the form
(Γ1,E1)⊕(0,OL(1,0)).
According to Lemma 4.3 and Proposition 4.1,
E1≃OQ for a quintic curve Q⊂P1×P1 of degree (2,3).
Thus, M11(3m+2n−1)≃P11.
Again from the proof of Proposition 4.4, we see that the
strictly α-semi-stable elements in M5 are of the form
(Γ,E)⊕(0,OL), where (Γ,E)∈M5(3m+2n).
We claim that M5(3m+2n)≃M∞(3m+2n).
To see this, we will show that there are no walls relative to the Polynomial P(m,n)=3m+2n.
As in the proof of Proposition 4.4, we attempt to solve the equation
[TABLE]
with 0≤r≤3, 0≤s≤2, 1≤r+s≤4, t≥r+s−rs.
For all choices of r and s we have t≥0, hence the above equation has no positive solutions in α.
From Proposition 4.1 we see that M5(3m+2n) isomorphic to the universal quintic of degree (2,3),
so it is a P10-bundle over P1×P1.
More precisely, the elements in M5(3m+2n) are of the form (H0(OQ(p)),OQ(p)),
where OQ(p) is a non-split extension of Cp by OQ.
Proposition 4.6**.**
Let Q⊂P1×P1 be a quintic curve of degree (2,3), let p∈Q be a point, let OQ(p) be a non-split
extension of Cp by OQ, and let L⊂P1×P1 be a line of degree (0,1).
Then any non-split extension sheaf F as in (13) is semi-stable.
The set of such sheaves is the closure of M3 in M.
The boundary M3∖M3 is contained in M2, more precisely,
it consists of extensions as in (9) in which C=Q∪L and p∈Q.
Proof.
The case when H0(F)≃C was examined at Proposition 3.2,
so we need only consider the case when H0(F)≃C2.
In this case the canonical morphism O→OL lifts to a morphism O→F, hence we can combine resolution (14)
with the standard resolution of OL to obtain the resolution
[TABLE]
[TABLE]
where φ11=0, φ12=0, and φ23 and φ33 are linearly independent.
Note that p is given by the equations φ11=0, φ12=0.
From the snake lemma, we obtain an extension
[TABLE]
where F′ is given by the resolution
[TABLE]
[TABLE]
We claim that F′≃OC(0,1), where C=Q∪L.
In view of Proposition 3.1, the claim implies that F is semi-stable, in fact [F]∈M2.
It remains to prove the claim. Let K be the kernel of the canonical morphism OC→OQ.
Since K has no zero-dimensional torsion and PK=m−1, K≃OL(−2,0).
Applying Hom(−,ω) to the exact sequence
[TABLE]
yields the exact sequence
[TABLE]
which is the same as the exact sequence
[TABLE]
Since H0(OC(0,1))≃C2, the canonical morphism O→OL lifts to a morphism O→OC(0,1),
hence the canonical resolutions of OQ and OL can be combined into a resolution of the form
[TABLE]
[TABLE]
Since OC(0,1) is a non-split extension of OL by OQ, ψ12 and φ33 are linearly independent.
It is clear now that the matrices representing φ′ and ψ are equivalent under elementary row and column operations.
We conclude that F′≃OC(0,1).
∎
The preimages of the sets of strictly semi-stable elements are the flipping loci:
[TABLE]
Proposition 4.7**.**
Consider Λ1∈M11(3m+2n−1), Λ2∈M(m+2), Λ3∈M5(3m+2n),
and Λ4∈M(m+1).
(i)
Over a point (Λ1,Λ2), F∞ has fiber P(Ext1(Λ1,Λ2)).
2. (ii)
Over a point (Λ1,Λ2), F11 has fiber P(Ext1(Λ2,Λ1)).
3. (iii)
Over a point (Λ3,Λ4), F5 has fiber P(Ext1(Λ3,Λ4)).
4. (iv)
Over a point (Λ3,Λ4), F0 has fiber P(Ext1(Λ4,Λ3)).
(ii) Assume that Λ=(Γ,F)∈F11 lies over (Λ1,Λ2).
Then Λ is a non-split extension of Λ1 by Λ2, or, vice versa, of Λ2 by Λ1.
If Λ2⊂Λ, then
[TABLE]
which violates the semi-stability of Λ. Thus Λ∈P(Ext1(Λ2,Λ1)).
Conversely, given such Λ, we need to show that Λ∈Mα for α∈(5,11).
Write Λ1=(Γ1,OQ), Λ2=(0,OL(1,0)).
We have a non-split extension of sheaves
[TABLE]
Let Λ′=(Γ′,F′) be a proper coherent subsystem of Λ.
Let G be the image of F′ in OL(1,0).
If F′∩OQ={0}, then G=OL(1,0), forcing χ(F′)=χ(G)≤1.
If F′∩OQ={0}, then χ(F′∩OQ)≤−1 because, by virtue of [1, Lemma 9], OQ is semi-stable.
We have in this case χ(F′)=χ(F′∩OQ)+χ(G)≤−1+2=1.
If Γ′={0}, then
[TABLE]
Assume now that Γ′={0}. Then Γ′=Γ=H0(OQ), hence OQ⊂F′.
If OQ=F′, then
[TABLE]
If OQ⫋F′, then r(F′)+s(F′)=6, hence χ(F′)≤0, and hence
[TABLE]
In all cases we have the inequality pα(Λ′)<pα(Λ), hence Λ∈Mα, for α∈(5,11).
(iii) We will show that every Λ=(Γ,F)∈P(Ext1(Λ3,Λ4)) gives a point in Mα
for α∈(5,11). Write Λ3=(Γ3,OQ(p)), Λ4=(0,OL).
We have a, possibly split, extension of sheaves
[TABLE]
Let Λ′=(Γ′,F′) be a proper coherent subsystem of Λ.
Let G be the image of F′ in OQ(p). Using the fact that OQ is semi-stable, it is easy to see that OQ(p) is semi-stable, as well.
Thus, χ(G)≤0, hence χ(F′)=χ(F′∩OL)+χ(G)≤1+0=1.
If Γ′={0}, then
[TABLE]
Assume now that Γ′={0}, i.e. Γ′=Γ. Then OQ⊂G.
If F′∩OL={0}, then F′≆OQ(p), otherwise Λ≃Λ3⊕Λ4.
In this case F′≃OQ, hence
[TABLE]
Assume now that F′∩OL={0}. Then r(F′)+s(F′)=6, hence χ(F′)≤0, and hence
pα(Λ′)<pα(Λ).
(iv) If (Γ,F)∈P(Ext1(Λ4,Λ3)), then we have the non-split extension (13),
hence, by Proposition 4.6, F is semi-stable.
Thus (Γ,F)∈M0+, i.e. (Γ,F)∈F0.
∎
Proposition 4.8**.**
([7, Corollaire 1.6])*
Let Λ=(Γ,F) and Λ′=(Γ′,F′) be two coherent systems on a separated scheme of finite type over C.
Then there is a long exact sequence*
[TABLE]
Proposition 4.9**.**
The flipping loci F∞, F11, F5, F0 are smooth bundles with fibers P3, P1, P2, respectively, P1.
Proof.
We need to determine the extension spaces of pairs occurring at Proposition 4.7.
(i) Choose Λ1=(Γ1,OQ) and Λ2=(0,OL(1,0)).
From Proposition 4.8 we have the long exact sequence
[TABLE]
For α≫0, Λ1 and Λ2 are α-stable coherent systems of different slopes, hence
Hom(Λ1,Λ2)={0}.
From the short exact sequence
[TABLE]
we obtain the long exact sequence
[TABLE]
Combining the last two long exact sequences, we obtain the isomorphism Ext1(Λ1,Λ2)≃C4.
(ii) From Proposition 4.8, we have the exact sequence
[TABLE]
From resolution (19), we obtain the exact sequence
[TABLE]
Combining the last two exact sequences, we obtain the isomorphism Ext1(Λ2,Λ1)≃C2.
(iii) Choose Λ3=(Γ,OQ(p)) and Λ4=(0,OL).
From Proposition 4.8, we have the long exact sequence
Combining the last two exact sequences, it follows that Ext1(Λ4,Λ3)≃C2.
∎
Lemma 4.10**.**
(i)* For Λ∈F11 we have Ext2(Λ,Λ)={0}.
(ii) For Λ∈F0 we have Ext2(Λ,Λ)={0}.*
Proof.
(i) In view of the exact sequence
[TABLE]
it is enough to show that Ext2(Λi,Λj)={0} for i,j=1,2.
From Proposition 4.8, we have the exact sequence
[TABLE]
The group on the right vanishes because H0(OQ(−3,−2))={0}.
Thus, Ext2(Λ1,Λ2)={0}.
From the exact sequence
[TABLE]
we obtain the vanishing of Ext2(Λ2,Λ1).
From the exact sequence
[TABLE]
we obtain the vanishing of Ext2(Λ2,Λ2).
From Proposition 4.8, we have the exact sequence
[TABLE]
According to [7, Théorème 3.12], Ext1(Λ1,Λ1) is isomorphic to the tangent space of M11(3m+2n−1)≃P11
(see Remark 4.5) at Λ1, so it is isomorphic to C11.
From resolution (19), we obtain the exact sequence
[TABLE]
Combining the last two exact sequences we obtain the vanishing of Ext2(Λ1,Λ1).
(ii) As above, we need to prove that Ext2(Λi,Λj)={0} for i,j=3,4.
From Proposition 4.8, we have the exact sequence
[TABLE]
Thus, Ext2(Λ3,Λ4)={0}.
From the exact sequence
[TABLE]
we obtain the vanishing of Ext2(Λ4,Λ3).
From the exact sequence
[TABLE]
we obtain the vanishing of Ext2(Λ4,Λ4).
From Proposition 4.8, we have the exact sequence
[TABLE]
From resolution (14), we obtain the exact sequence
[TABLE]
Since Hom(OQ(p),OQ(p))≃C, it follows that
[TABLE]
According to [7, Théorème 3.12], Ext1(Λ3,Λ3) is isomorphic to the tangent space at Λ3 of M5(3m+2n),
which, according to Remark 4.5, is smooth of dimension 12.
We obtain the vanishing of Ext2(Λ3,Λ3).
∎
Theorem 4.11**.**
Let Mα be the moduli space of α-semi-stable pairs on P1×P1 with Hilbert polynomial
P(m,n)=4m+2n+1. We have the following blowing up diagrams
[TABLE]
Here β∞ is the blow-up along F∞
and β11 is the contraction of the exceptional divisor F∞ in the direction of P3,
where we view F∞ as a P3×P1-bundle with base M11(3m+2n−1)×M(m+2).
Likewise, β5 is the blow-up along F5 and β0 is the contraction of the exceptional divisor
F5 in the direction of P2, where we view F5 as a P2×P1-bundle over
M5(3m+2n)×M(m+1).
Proof.
A birational morphism β11:M∞→M11− can be constructed as at [3, Theorem 3.3]
such that β11 contracts F∞ in the direction of P3, β11 is an isomorphism outside F11,
and β11−1(x)≃P3 for any x∈F11.
We now apply the Universal Property of the blow-up [6, p. 604] to deduce that β11 is a blow-up with center F11.
For this we need to know that M11− and F11 are smooth.
By Corollary 4.2, M∞ is smooth,
by Proposition 4.9, the blowing up center F∞ is smooth, hence M∞ is smooth, too.
Since β11 is an isomorphism outside F11, M11−∖F11 is smooth.
Since all points of M11− are α-stable, we can apply the Smoothness Criterion [7, Théorème 3.12],
which states that Λ∈M11− is a smooth point if Ext2(Λ,Λ)={0}.
Thus, in view of Lemma 4.10(i), M11− is smooth at every point of F11.
The smoothness of F11 was proved at Proposition 4.9.
For the second blow-up diagram we reason analogously, using the facts that F5 and F0 are smooth, and using Lemma 4.10(ii).
∎
According to [7, Théorème 4.3], there is a universal family (Γ,F)
of coherent systems on M0+×P1×P1.
In particular, F is a family of semi-stable sheaves on P1×P1 with Hilbert polynomial 4m+2n+1,
which is flat over M0+. It induces the so called forgetful morphismϕ:M0+→M.
We have ϕ(Γ,F)=[F].
Proposition 4.12**.**
The forgetful morphism ϕ:M0+→M is a blow-up with center the Brill-Noether locus M2.
Proof.
According to Proposition 2.3(ii), for [F]∈M∖M2 we have H0(F)≃C,
hence ϕ−1([F])=(H0(F),F) is a single point. Thus, ϕ is an isomorphism away from M2.
According to Proposition 3.1, for [F]∈M2 we have H0(F)≃C2, hence ϕ−1([F])≃P1.
Taking into account that M and M2 are smooth, we can apply the Universal Property of the blow-up [6, p. 604]
to conclude that ϕ is a blow-up with center M2.
∎
Proof of Theorem 1.2.
By virtue of Proposition 4.12, we have the relation
[TABLE]
According to Proposition 3.1, we have the relation
In view of Corollary 4.2 and Remark 4.5, we have the relation
[TABLE]
According to [5, Theorem 0.1], we have the equation
[TABLE]
The final result reads
[TABLE]
Acknowledgement.
The author would like to thank Jean-Marc Drézet for several helpful discussions.
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