# Stable Sheaves on a Smooth Quadric Surface with Linear Hilbert Bipolynomials

Authors: Edoardo Ballico; Sukmoon Huh
Journal: The Scientific World Journal
 (2014)

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## 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.


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## Body

## 1. Introduction

Throughout the paper, our base field isℂ, 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 X 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.LetQ be a smooth quadric surface in ℙ3 and let MQ(μ,χ) be the moduli space of semistable sheaves on Q with linear Hilbert polynomial χ(m)=μm+χ with respect to the ample line bundle 𝒪Q(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 Q, we can define a linear Hilbert bipolynomial χℱ(x,y) such that
(1)χ(ℱ(x,y))=χℱ(x,y)
for all (x,y)∈ℤ⊕2. Then we can consider, due to [6], the moduli space M(m,n,t) of semistable sheaves on Q with linear Hilbert bipolynomial χ(x,y)=mx+ny+t. The moduli space is a projective variety with a Zariski open subset M∘(m,n,t) consisting of stable ones, with dimension 2mn+1 and the open set is nonempty if one of m or n is nonzero (see Proposition 7).By its definition we have a natural decomposition(2)MQ(m+n,t)=∐0≤a≤m+nM(a,m+n-a,t).
Thus, the moduli M(m,n,t) is an irreducible component of Simpson's moduli space because the bidegree function is locally constant.Ifℱ is a stable sheaf in M(m,n,t), then its schematic support Cℱ is a curve of bidegree (n,m) on Q 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 𝒫ic(n,m)d of some fixed degree d over the family of the bidegree (n,m)-curves in Q.Now, some simple observations lead us to consider onlyM(m,n,t) with 0≤t≤gcd(m,n) due to proper twists. For small m or n, the moduli space is very simple. Indeed, M(n,0,t) is isomorphic to ℙn if t=n and is empty otherwise. If m or n is equal to 1, say m=1, then it is isomorphic to ℙ2n+1. These descriptions are quite simple from the definition of stability condition and so the first nontrivial case happens when (m,n)=(2,2). The main result of this paper is to describe the moduli spaces M(2,2,t) with t=1,2.Theorem 1.
ForMt=M(2,2,t), one obtains the following: (1)
M1 is isomorphic to 𝒫ic(2,2)1 and it is rational;(2)
M2 is birational to 𝒫ic(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 sheafℱ is in M1 if and only if it admits a resolution
(3)0⟶𝒪Q(-2,-1)⊕𝒪Q(-1,-2)⟶𝒪Q⊕𝒪Q(-1,-1)⟶ℱ⟶0,
where the degeneracy locus of the first map is the support of ℱ. It enables us to identify M1 with 𝒫ic(2,2)1 and show its rationality.ForM2, the situation is a bit more complicated; we can classify the sheaves in M2 up to 3 types, in terms of the short exact sequences they admit, and express the moduli as the union of 3 subschemes
(4)M2=𝔄∪𝔅∪ℭ.
In particular, we can show that every sheaf in M2 is globally generated, from which we obtain a resolution that they admit:
(5)0⟶𝒪Q(-1,-1)⊕2⟶𝒪Q⊕2⟶ℱ⟶0.
We investigate the property of the subvarieties and the relationship between them. We also construct a map from M1 to M2, which is generically 4 to 1 and thus we obtain that M2 is unirational of degree 4. We leave the rationality question of M2 as a conjecture.


## 2. Preliminaries

LetQ be a smooth quadric surface isomorphic to ℙV1×ℙV2 for 2-dimensional vector spaces V1 and V2, and then it is embedded into ℙ3≅ℙV by the Segre map where V=V1⊗V2. If we denote by f1,f2 the two projections from Q to each factor, then we will denote f1*𝒪ℙ1(a)⊗f2*𝒪ℙ1(b) simply by 𝒪Q(a,b). We also denote ℰ⊗𝒪Q(a,b) by ℰ(a,b) for a coherent sheaf ℰ on Q and in particular the canonical sheaf ωQ of Q is 𝒪Q(-2,-2).Proposition 2.
For a purely 1-dimensional sheafℱ on Q, there is a bipolynomial χℱ(x,y)∈ℚ[x,y] of degree 1 such that
(6)χ(ℱ(u,v))=χℱ(u,v)
for all (u,v)∈ℤ⊕2.Proof.
Let us assume thatmt+c is the Hilbert polynomial of ℱ with respect to the ample line bundle 𝒪Q(1,1). Let us take any D∈|𝒪Q(0,1)|, T∈|𝒪Q(1,0)|, and a smooth conic C∈|𝒪Q(1,1)| such that neither D, T, nor C is contained in the 1-dimensional reduced curve Supp(ℱ).
The curvesD,T, and C induce maps jD:ℱ(t,t)→ℱ(t,t+1),  jT:ℱ(t,t)→ℱ(t+1,t), and jC:ℱ(t,t)→ℱ(t+1,t+1). Since neither D nor T is contained in the 1-dimensional reduced curve Supp(ℱ), we have jD≠0 and jT≠0. Since ℱ is pure, we obtain that jD, jT, and jC are injective. Thus, there are exact sequences
(7)0⟶ℱ(t,t)⟶ℱ(t,t+1)⟶ℱ(t,t+1)⊗𝒪D⟶0,(8)0⟶ℱ(t,t)⟶ℱ(t+1,t)⟶ℱ(t+1,t)⊗𝒪T⟶0,(9)0⟶ℱ(t,t)⟶ℱ(t+1,t+1)⟶ℱ(t+1,t+1)⊗𝒪C⟶0.
Let us seta:=h0(ℱ(t,t+1)⊗𝒪D) and b:=h0(ℱ(t+1,t)⊗𝒪T). The sheaves ℱ(t,t+1)⊗𝒪D, ℱ(t+1,t)⊗𝒪T, and ℱ(t+1,t+1)⊗𝒪C have finite supports and thus the dimensions of their cohomology H0(Q,-) do not change even if we twist them by a line bundle on Q. From (9), we get a+b=h0(ℱ(t+1,t+1)⊗𝒪C)=m.
We claim thatχ(ℱ(u,v))=av+bu+c for all (u,v)∈ℤ⊕2. If u=v, then the claim is true. Now assume that u≠v, say u>v. We use u-v exact sequences like (8) with ℱ(c,0) instead of ℱ with 0≤c<u-v to get χ(ℱ(u,v))=χ(v,v)+(u-v)b.Definition 3.
One defines the Hilbert bipolynomialχℱ(x,y)∈ℚ[x,y] of ℱ to be a linear bipolynomial such that
(10)χℱ(x,y)=χ(ℱ⊗𝒪Q(x,y)).
In particular, the Hilbert polynomial of ℱ with respect to 𝒪Q(1,1) is defined to be χℱ(t)=χℱ(t,t).We are mainly interested in the case whenχℱ(x,y) is a linear function, that is, χℱ(x,y)=mx+ny+t for some (m,n,t)∈ℤ⊕3.Definition 4.
Letℱ be a pure sheaf of dimension 1 on Q with χℱ(x,y)=mx+ny+t. The p-slope of ℱ is defined to be p(ℱ)=t/(m+n). ℱ is called semistable (stable) with respect to the ample line bundle 𝒪Q(1,1) if (1)
ℱ does not have any 0-dimensional torsion,(2)
for any proper subsheafℱ', one has
(11)p(ℱ′)=t′m′+n′≤(<)tm+n=p(ℱ),whereχℱ′(x,y)=m′x+n′y+t′.For every semistable1-dimensional sheaf ℱ with χℱ(x,y)=mx+ny+t, let us define Cℱ:=Supp(ℱ) to be its scheme-theoretic support and then we have Cℱ∈|𝒪Q(n,m)|. 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 mt+χ and m′t+χ′, the inequality for Hilbert and slopes χ/m the same.Definition 5.
LetM(m,n,t) be the moduli space of semistable sheaves on Q with linear Hilbert bipolynomial χ(x,y)=mx+ny+t.We can defineM(m,n,t) in a different way as a subvariety of MQ,ℙ3(m+n,t), the moduli space of semistable sheaves on ℙ3 with linear Hilbert polynomial χ(x)=mx+t, which are 𝒪Q-sheaves. To be precise, if ℱ is 𝒪Q-sheaf, then all of its 𝒪ℙ3-subsheaves are also 𝒪Q-sheaves. It implies that the notions of p-stability and μ-stability of ℱ are the same and thus MQ,ℙ3(m+n,t) may be defined without using ℙ3. Moreover, the sheaf with linear Hilbert bipolynomial χ(x,y)=ax+by+c has Hilbert polynomial χ(x)=(a+b)x+c with respect to 𝒪Q(1,1) and thus we have a natural decomposition
(12)MQ,ℙ3(m+n,t)=∐0≤a≤m+nM(a,m+n-a,t).
In particular, M(m,n,t) is a subvariety of MQ,ℙ3(m+n,t).Remark 6.
Letℱ be any purely 1-dimensional coherent sheaf on ℙ3 with Hilbert polynomial mx+ti. Assume that ℱ is not semistable and let
(13)0=ℱ0⊂ℱ1⊂⋯⊂ℱk=ℱ
be the Harder-Narasimhan filtration of ℱ (see page 55 in [1]). If ℱ is an 𝒪X-module, then each ℱi is an 𝒪X-module because it is a subsheaf of ℱ. Thus the Harder-Narasimhan filtration of ℱ as an 𝒪ℙn-sheaf is the same as the one as an 𝒪X-sheaf.Proposition 7.
The moduliM(m,n,t) is a projective and irreducible scheme. If mn>0, then M∘(m,n,t) is a Zariski dense and open subset of M(m,n,t) with dimension 2mn+1.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 Q holds. But it holds, using Castelnuovo-Mumford criterion with the Serre duality
(14)H1(ℰxt1(ℱ,ℱ(j,j)))≅Ext2(ℱ,ℱ(j,j))≅Hom(ℱ(j,j),ℱ(-2,-2))∨=0
for ℱ∈M(m,n,t) and j≥-1.
For the second assertion, let us consider a map(15)M(m′,n′,t′)×M(m′′,n′′,t′′)⟶M(m,n,t)
defined by sending (ℱ′,ℱ′′) to ℱ′⊕ℱ′′, where m=m′+m′′ and n=n′+n′′. Then the dimension of the image of this map is at least 2mn-2m′n′-2m′′n′′-1 and it is at least 1 if mn>0. In other words, general sheaf in M(m,n,t) is stable.For any pure sheafℱ on Q with Hilbert bipolynomial χℱ(x,y)=mx+ny+t, let us define
(16)ℱD:=ℰxtQ1(ℱ,ωQ)
to be the Grothendieck dual of ℱ. Since ℱ is pure, the natural map φℱ:ℱ→ℱDD is injective. Since the support of ℱ is 1-dimensional, φℱ is bijective as in Remark 4 of [4]. Moreover, the support of ℱD is also 1-dimensional and so χℱD(x,y) is also linear. By the Serre duality, we have
(17)Hi(ℱD(c,d))≅Hi((ℱ(-c,-d))D)≅H1-i(ℱ(-c,-d))∨
for i∈{0,1} and, in particular, we have
(18)χℱD(x,y)=-χℱ(-x,-y)=mx+ny-t.Lemma 8.
There is an isomorphism(19)M(m,n,t)⟶M(m,n,-t)
sending ℱ to ℱD.Note also thatχℱ(d,e)(x,y)=mx+ny+t+(md+ne). Since the map
(20)M(m,n,t)⟶M(m,n,t+md+ne),
defined by ℱ↦ℱ(d,e), is an isomorphism, so we may assume that 0<t≤gcd(m,n).Lemma 9.
For a not necessarily integral curveC in |𝒪Q(n,m)|, the sheaf 𝒪C is semistable. If C is integral, then 𝒪C is stable.Proof.
We have the following sequence:(21)0⟶𝒪Q(-n,-m)⟶𝒪Q⟶𝒪C⟶0.
In particular, we have χ𝒪C(x,y)=mx+ny+(m+n-mn) and so p(𝒪C)=1-1/(1/m+1/n). If C is integral, then 𝒪C is stable since every line bundle on an integral curve is stable. In general, 𝒪C is semistable. Otherwise, there exists a semistable quotient sheaf 𝒪C→ℱ→0 such that the Hilbert bipolynomial χℱ(x,y)=m′x+n′y+t′ satisfies m′+n′<m+n and p(ℱ)<p(𝒪C). By induction, we get that 𝒪C′ with C′:=Cℱ is semistable and thus we have
(22)p(𝒪C′)≤p(ℱ)<p(𝒪C).
This is absurd since p(𝒪C) is a decreasing function on m and n.Let us assume thatm=0, that is, HilbQ(ny+t) with 0<t≤n.Proposition 10.
One has(23)M(0,n,t)≅{(ℙ1)[n]≅ℙnift=n;∅if0<t<n.
In fact, each point in HilbQ(ny+n) corresponds to an equivalence class [𝒪L1⊕⋯⊕𝒪Ln], where Li is a line in |𝒪Q(1,0)|.Proof.
Let us assume thatt=n and let us choose L∈|𝒪Q(n,0)| and then it fits into
(24)0⟶𝒪Q(-n,0)⟶𝒪Q⟶𝒪L⟶0.
Thus we have
(25)χ𝒪L(x,y)=χ𝒪Q(x,y)-χ𝒪Q(-n,0)(x,y)=(x+1)(y+1)-(x-n+1)(y+1)=ny+n.
Clearly, 𝒪L is stable. For a line L∈|𝒪Q(1,0)|, we have
(26)χ𝒪2L(x,y)=χ𝒪L⊕𝒪L(x,y)=2y+2.
From the sequence for L, we have
(27)0⟶Hom(𝒪L,𝒪L)⟶Hom(𝒪Q,𝒪L)⟶fHom(𝒪Q(-1,0),𝒪L)⟶Ext⁡1(𝒪L,𝒪L)⟶0
and the map f is a zero map. Thus, there exists a nontrivial extension of 𝒪L by 𝒪L and it is 𝒪2L. In particular, 𝒪L⊕2 and 𝒪2L represent the same point in HilbQ(2y+2). In general, 𝒪L⊕k and 𝒪kL with k≥1 represent the same point in M(0,k,k). Thus, 𝒪L with L∈|𝒪Q(n,0)| is strictly semistable if and only if n≥2. Conversely, let us choose a semistable sheaf ℱ with χℱ(x,y)=ny+n. In particular, the schematic support L=Supp(ℱ) of ℱ is in |𝒪Q(n,0)|. Since χ(ℱ)=n>0, there exists a nontrivial morphism 𝒪Q→ℱ and it induces an injection 𝒪L1→ℱ, where L1 is a subscheme of L. Here we have L1∈|𝒪Q(s,0)| for some s≤n and so χL1(x,y)=sx+s. Thus, the quotient 𝒢=ℱ/𝒪L1 is a semistable sheaf with χ𝒢(x,y)=(n-s)y+(n-s). By induction, we have [𝒢]=[𝒪L2] with L2∈|𝒪Q(n-s,0)|. In particular, ℱ is an extension of 𝒪L2 by 𝒪L1 with L1+L2∈|𝒪Q(n,0)| and thus ℱ is equivalent to 𝒪L1⊕𝒪L2.
Now, let us assume that0<t<n and fix ℱ∈M(0,n,t) with C:=Cℱ∈|𝒪C(n,0)|. Since χ(ℱ)=t>0, there is a nonzero map f:𝒪Q→ℱ. Since ℱ is an OC-sheaf, f induces a nonzero map h:𝒪C→ℱ. Since 𝒪C has slope 1>t/n and it is semistable, we get a contradiction. Alternatively, as in Lemma 4.10 of [6], we may first take the schematic support T⊆C of Im⁡(h) and then use an injective map 𝒪T→ℱ with 𝒪T∈|𝒪Q(n′,0)| with 1≤n′<n, and thus we have μ(𝒪T)=1.For the case ofm=1, that is, χℱ(x,y)=x+ny+t, it is enough to check the case t=1 since gcd(1,n)=1.Proposition 11.
M
(
1
,
n
,
1
) consists of 𝒪C with C∈|𝒪Q(n,1)|. In particular, one has M(1,n,1)≅ℙ2n+1.Proof.
From the sequence(28)0⟶𝒪Q(-n,-1)⟶𝒪Q⟶𝒪C⟶0,
we have χ𝒪C(x,y)=x+ny+1 and 𝒪C is semistable by Lemma 9. Conversely, let ℱ be a semistable sheaf with χℱ(x,y)=x+ny+1 and so C:=Cℱ is a curve in |𝒪Q(n,1)|. Since we have χ(ℱ)=1, there exists a nonzero map 𝒪Q→ℱ and it induces a nonzero map h:𝒪C→ℱ. Note that Im⁡(h) has no 0-dimensional torsion since ℱ is semistable. Since 𝒪C is also semistable, we have
(29)p(𝒪C)≤p(Im⁡(h))≤p(ℱ).
The map h factors through an injection 𝒪D↪ℱ, where D is a curve contained in C. If D is properly contained in C, we have p(𝒪D)>p(ℱ) contradicting the semistability of ℱ and thus we have D=C; that is, h is an isomorphism from 𝒪C to its image. Since 𝒪C and ℱ have the same Hilbert polynomial, we have ℱ≅𝒪C.


## 3. Hilbert Bipolynomial2x+2y+1

For the moduli space of semistable sheaves with linear Hilbert bipolynomial2x+2y+t, it is enough to investigate the case when t=1,2. Let us denote the moduli space M(2,2,t) by Mt.Proposition 12.
The moduli spaceM1 consists of the unique nontrivial extensions ℱ of 𝒪P by 𝒪C for each curve C∈|𝒪Q(2,2)| and a point P∈C, and one also has h0(ℱ)=1.Proof.
Sinceχ(ℱ)=1, there is a nonzero map 𝒪Q→ℱ, inducing a nonzero map h:𝒪C→ℱ, where C:=Cℱ∈|𝒪Q(2,2)|. Since χ𝒪C(x,y)=2x+2y, we have p(𝒪C)=0<1/4=p(ℱ). The map h factors through an injection 𝒪D↪ℱ, where D is a curve contained in C. If D is properly contained in C, we have p(𝒪D)>p(ℱ) contradicting to the semistability of ℱ and thus we have D=C; that is, h is an isomorphism from 𝒪C to its image, that is, we have
(30)0⟶𝒪C⟶ℱ⟶𝒢⟶0,
where χ𝒢(x,y)=1. In particular, we have 𝒢≅𝒪P, the skyscraper sheaf supported on a point P∈C. Since ℱ has no 0-dimensional torsion, the sequence does not split. Note that Ext1(𝒪P,𝒪Q)≅H1(𝒪P)∨=0, and thus from the sequence of C we have
(31)0⟶Ext1(𝒪P,𝒪C)⟶Ext2(𝒪P,𝒪Q(-2,-2))⟶sExt2(𝒪P,𝒪Q).
Here, the map s is the transpose of Hom(𝒪Q(2,2),𝒪P)→Hom(𝒪Q,𝒪P) which is given by the multiplication by the defining equation of C. Since P is a point on C, the map s is a zero map. In particular, the dimension of Ext1(𝒪P,𝒪C) is 1 and so ℱ corresponds to a unique nontrivial extension
(32)0⟶𝒪C⟶ℱ⟶𝒪P⟶0.
From the sequence (32), we have h0(ℱ)≤2 and that h0(ℱ)=1 if and only if no injective map 𝒪C→ℱ is an isomorphism at P. This is certainly true if ℱ is not locally free of rank 1 at P. Note that ℱ is a line bundle at each point of C∖{P} and thus it is sufficient to prove h0(ℱ)=1 when ℱ is a line bundle on the curve C. In this case the nonexistence of a section of ℱ that does not vanish at P is equivalent to the nonsplitting of (32). Thus, we have h0(ℱ)=0 and so the point P is uniquely determined by ℱ.
Conversely, let us assume thatℱ is a nontrivial extension of 𝒪P by 𝒪C, where P is a point on C. If ℱ is not semistable, then there exists a subsheaf 𝒦⊂ℱ with p(𝒦)>p(ℱ)=1/4 and so we have χ𝒦(x,y)=m′x+n′y+t′ with (m′,n′)≤(2,2) and t′≥1. If the composite s:𝒦→ℱ→𝒪P is a zero map, then we have an injection 𝒦↪𝒪C, contradicting the semistability of 𝒪C. Thus, the composite is surjective and so we have the following diagram:
(33)
Here, 𝒦′ is the kernel of the map s and ℋ is the quotient ℱ/𝒦. Since χ𝒦′(x,y)=m′x+n′y+(t′-1) and 𝒪C is semistable, we have t′=1 and thus χℋ(x,y)=(2-m′)x+(2-n′)y with no constant term. Since ℋ is the quotient of 𝒪C, it must be 𝒪T for some curve T contained in C. But no such curves have the Hilbert polynomials with no constant term. Hence ℱ is semistable.Remark 13.
There is no strictly semistable sheaf inM1. Let us assume the existence of a polystable sheaf ℱ=ℱ1⊕⋯⊕ℱs with s≥2. We have χ(ℱ)=1=χ(ℱ1)+⋯+χ(ℱs). If we let χℱi(x,y)=aix+biy+ci, then we have
(34)c1+⋯+cs=1,ciai+bi=12,∀i.
It implies that ci>0 for all i and thus we have s=1, a contradiction.Proposition 14.
A sheafℱ is in M1 if and only if it admits the following resolution:
(35)0⟶𝒜⟶φ𝒪Q⊕𝒪Q(-1,-1)⟶ψℱ⟶0,
where 𝒜:=𝒪Q(-2,-1)⊕𝒪Q(-1,-2) and φ=(h1l1h2l2). Here, f:=h1l2-h2l1 is a defining equation of Cℱ.Proof.
Note thath0(ℱ)=1 and so h1(ℱ)=0. If ℱ admits the sequence (32), then ℱ is globally generated outside P and so is ℱ(1,1). Take any A∈|𝒪Q(1,1)| which is not contained in C and with P∉A. The multiplication by an equation of A gives an exact sequence
(36)0⟶ℱ⟶ℱ(1,1)⟶ℱ(1,1)|A⟶0,
where deg⁡(ℱ(1,1)|A)=deg⁡(A∩C)=4. Thus we have h1(ℱ(1,1))=0 and h0(ℱ(1,1))=5. Together with the exact sequence (32) tensored by 𝒪Q(1,1), we obtain that ℱ(1,1) is globally generated at P and so we have a surjection
(37)ψ:𝒪Q⊕𝒪Q(-1,-1)⟶ℱ⟶0.
Let us set ℋ:=ker⁡(ψ) and then ℋ is a torsion-free sheaf of rank 2 on Q with c1=(-3,-3). By Theorem 19.9 in [7], the sheaf ℋ is locally free. Note that χℋ(1,2)(x,y)=2xy+3x+y+1. Thus, we have h0(ℋ(1,2))>0 and so we have an exact sequence
(38)0⟶𝒪Q(a,b)⟶ℋ(1,2)⟶ℐZ(-1-a,1-b)⟶0,
where Z is a 0-dimensional subscheme of Q and (a,b)∈{(0,0),(0,1)}. If (a,b)=(0,1), then we have χℐZ(-1,0)(x,y)=xy+x+1 and it is absurd since χ𝒪Q(-1,0)(x,y)=xy+x. Thus, we have (a,b)=(0,0) and Z=∅. Since Ext1(𝒪Q(-1,1),𝒪Q)=0, we have ℋ(1,2)≅𝒪Q⊕𝒪Q(-1,1) and the sequence (35). Note that the map φ:ℋ→𝒪Q⊕𝒪Q(-1,-1) is given by (h1l1h2l2), where f:=h1l2-h2l1 is a defining equation of C=Cℱ.
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 ℱ(1,0) is globally generated and so a surjection φ′:𝒪Q⊕𝒪Q(-1,0)→ℱ. In this case, ker⁡(φ′) is no longer a direct sum of two line bundles.Let us define a vector spaceW to be
(39)W:=Hom(𝒜,𝒪Q⊕𝒪Q(-1,-1))
and W0⊂W to be the set of φ∈W such that h1l2-h2l1≠0. Then we have a surjective morphism
(40)π:W0⟶M1.Let us chooseφ1,φ2∈W0 with π(φ1)=π(φ2); that is, we have the following diagram:
(41)0⟶𝒜⟶φ1𝒪Q⊕𝒪Q(-1,-1)⟶ψ1ℱ⟶0↓f0⟶𝒜⟶φ2𝒪Q⊕𝒪Q(-1,-1)⟶ψ2ℱ⟶0,
where f is an isomorphism. Since Ext1(𝒪Q⊕𝒪Q(-1,-1),𝒜)=0, we have a map f1∈End(𝒪Q⊕𝒪Q(-1,-1)) associated with f. Note that f1 is given by (a0zb), where a,b∈ℂ× and z∈H0(𝒪Q(1,1)). Similarly, we have a map f2:𝒜→𝒜 which is (c100c2), where c1,c2∈ℂ×. In particular, we have
(42)φ2=(c100c2)-1φ1(a0zb).
In this equation, we can assume that c1=1. In other words, π(φ1)=π(φ2) if and only if φ1 and φ2 are in the same orbit in W0 under the action by
(43)G:=(Aut(𝒜)×Aut(𝒪Q⊕𝒪Q(-1,-1)))ℂ×={((100c),(a0zb))∣a,b,c∈ℂ×,z∈H0(𝒪Q(1,1))((100c),(a0zb))}.Theorem 16.
π
:
W
0
→
M
1 is a geometric quotient map by the action of G. In particular, one has M1≅W0/G and so M1 is isomorphic to 𝒫ic(2,2)1.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 ofM1 is stable, M1 has a universal family ℍ1 on M1×ℙ2 (see page 180 of [8] or Theorem 4.6.5 of [9]). Since every semistable sheaf with bipolinomial 2x+2y+5 is of the form ℱ(1,1) for a unique ℱ∈M1, we also have a universal family ℍ5 on M5×ℙ2 with ℱ(1,1) as the fibre. Since h0(ℱ)=1 and h1(ℱ)=0 for all ℱ∈M1, the base change theorem gives that u*(ℍ1) is a line bundle on M1, where u:M1×ℙ2→M1 is the first projection. Since h0(ℱ(1,1))=5 and h1(ℱ(1,1))=0 for all ℱ∈M1, the base change theorem gives that v*(ℍ5) is a vector bundle of rank 5 on M1 by identifying M5 with M1, where v:M5×ℙ2→M5. For a fixed ℱ∈M1 and a matrix φ∈π-1(ℱ), let us write φ=(h1l1h2l2), where f:=h1l2-h2l1 is a defining equation of Cℱ. Take an open neighborhood U of ℱ in M1 over which u*(ℍ1) and v*(ℍ5) are trivial. The matrix φ was constructed starting with a section σ of ℱ(1,1) which spans ℱ(1,1) together with the twist σ′ of a nonzero section of ℱ. Since u*(ℍ1)|U and v*(ℍ5)|U are trivial, there are maps e1:𝒪U→𝒪U and e2:𝒪U→𝒪U⊕5 with e1(ℱ)=σ′ and e2(ℱ)=σ. Since σ′ and σ span ℱ, there is a neighborhood V of ℱ in U such that the sections e1(𝒢) and e2(𝒢) span every 𝒢∈V. The construction of φ gives that e1 and e2 induce a section of π in a neighborhood of φ whose image by π is V.As an automatic consequence, we obtain thatM1 is irreducible and unirational. In fact, we can prove more.Theorem 17.
M
1 is rational.Proof.
LetΔ⊂Q×Q be the diagonal and denote its ideal sheaf by ℐΔ. Denoting by p1 and p2 the projection from Q×Q to each factor, let us define a sheaf 𝒰 to be p1*𝒪Q(2,2))⊠ℐΔ on Q×Q. For each point P∈Q, we have 𝒰|Q×{P}≅ℐP(2,2). Thus, we have h1(𝒰|Q×{P})=0 and so p2*𝒰 is a vector bundle of rank 8 on Q since h0(𝒰|Q×{P})=8. Let us consider the projective bundle
(44)𝒵=ℙ(p2*𝒰)⟶Q.
By its definition, the fibre of 𝒵 over a point P∈Q is the set of curves of type (2,2) on Q passing through P and so there is a natural map from 𝒵 to ℙH0(𝒪Q(2,2))≅ℙ8. In other words, 𝒵 is the universal curve of type (2,2) on Q and it is isomorphic to M1. Since 𝒵 is locally trivial over Q, it is rational.


## 4. Hilbert Bipolynomial2x+2y+2

Lemma 18.
Any sheafℱ∈M2 admits one of the following types:(A)
0→𝒪C→ℱ→η→0, where η is a skyscraper 𝒪C-sheaf with degree 2,(B)
0→𝒪T1→ℱ→𝒪T2→0 with T1,T2∈|𝒪Q(1,1)|,(C)
0→𝒪T1→ℱ→𝒪T2→0, where T1∈|𝒪Q(a,b)| and T2∈|𝒪Q(2-a,2-b)| with (a,b)∈{(1,2),(2,1)}.Proof.
Sinceχ(ℱ)=2, we have h0(ℱ)≥2. Thus, there exists a nonzero map 𝒪Q→ℱ and it induces a nonzero map h:𝒪C→ℱ, where C:=Cℱ∈|𝒪Q(2,2)|. The map h factors through an injection 𝒪T1↪ℱ where T1 is a curve contained in C:
(45)
If we have T1=C, that is the map h is an isomorphism from 𝒪C to its image in ℱ, then its cokernel ℋ is the skyscraper sheaf supported on two points, say P1,P2∈C. Thus we have the sequence
(46)0⟶𝒪C⟶ℱ⟶η⟶0.
Let us assume thatT1 is properly contained in C and then we obtain that T1 has bidegree (1,1),(1,2), or (2,1) since p(𝒪T1)≤p(ℱ)=1/2 and ℱ is semistable. Let T2⊂Q be the only curve such that T1+T2=C. Let ℋ' be the quotient of ℋ by its torsion τ, that is, ℋ′:=ℋDD.
First, assumeT1∈|𝒪Q(1,1)| and so we have χℋ′(x,y)=x+y+1-deg⁡(τ). Since ℱ is semistable, we get τ=0. Since every quotient of ℱ has the slope at least 1/2, the same is true for ℋ. Thus, ℋ is semistable and Proposition 11 gives ℋ≅𝒪T2.
Now, without loss of generality, let us assume thatT1∈|𝒪Q(1,2)|, that is, χ𝒪T1(x,y)=2x+y+1 and so we have χℋ(x,y)=y+1. If ℋ has 0-dimensional torsion 𝒯 with length k≥1, then the quotient ℋ/𝒯 is a quotient of ℱ with the p-slope 1-k≤0, contradicting the semistability of ℱ. Thus ℋ has no 0-dimensional torsion and so we have ℋ≅𝒪T2 for a curve T2∈|𝒪Q(1,0)| with C=T1+T2.Corollary 19.
Every sheaf inM2 is globally generated.Proof.
Let us takeℱ∈M2 and then there is no nonzero map ℱ→𝒪C≅ωC since ℱ is semistable. Thus we have h1(ℱ)=0 and so h0(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).
Letℋ⊆ℱ be the image of the evaluation map H0(ℱ)⊗𝒪Q→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 Cred. In particular, we have χℋ(x,y)=2x+2y+c with c≤1 and deg⁡(ℱ/ℋ)=2. Since h0(ℋ)=h0(ℱ)=2, we have h1(ℋ)=2-c. Note that every nonzero section of ℋ vanishes at finitely many points since ℱ is neither of types (B) nor (C). Since h0(𝒪C)<h0(ℋ), we have ℋ≠𝒪C and c=1. A nonzero section of ℋ induces an exact sequence
(47)0⟶𝒪C⟶ℋ⟶𝒢⟶0,
where 𝒢≅𝒪P for some P∈Cred. 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.Lemma 20.
ℱ is a sheaf in M2 if and only if it admits a sequence
(48)0⟶𝒪Q(-1,-1)⊕2⟶φ𝒪Q⊕2⟶ℱ⟶0,
where φ=(z11z12z21z22), zij∈H0(𝒪Q(1,1)) such that f:=z11z22-z12z21 is a defining equation of Cℱ.Proof.
Letℱ∈M2 be a sheaf of type (A) and then it is globally generated by Corollary 19. Since h0(ℱ)=2, we have a surjection
(49)ψ:𝒪Q⊕2⟶ℱ⟶0.
Let us set ℋ:=ker⁡(ψ) and then it is a torsion-free sheaf of rank 2 on Q with c1=(-2,-2). By Theorem 19.9 in [7], ℋ is locally free. Note that h0(ℋ(1,1))=2. From the sequence
(50)0⟶ℋ(1,0)⟶𝒪Q(1,0)⊕2⟶ℱ(1,0)⟶0,
we obtain that the map H0(𝒪Q(1,0)⊕2)→H0(ℱ(1,0)) is an isomorphism and so h1(ℋ(1,0))=0. Similarly, we have h1(ℋ(0,1))=0 and h2(ℋ)=h1(ℱ)=0. By Remark 2.3 in [10], we obtain that ℋ(1,1) is globally generated. Since c1(ℋ(1,1))=0 or h0(ℋ(1,1))=2, we have ℋ≅𝒪Q(-1,-1)⊕2 and the resolution (48). The cases of the other types also work verbatim.Definition 21.
Let us define a subscheme𝔄⊂M2 as follows:(51)𝔄:={ℱ∈M2∣ℱadmitsanontrivialextensionoftype(A)}.
Similarly, we define 𝔅 and ℭ for the semistable sheaves of types (B) and (C), respectively. In particular, we have
(52)M2=𝔄∪𝔅∪ℭ.Lemma 22.
The sheavesℱ of type (B) are strictly semistable. In particular, they are contained in 𝔅.Proof.
It is enough to check the semistability ofℱ. Let 𝒦 be a subsheaf of ℱ with p(𝒦)>1/2 and the quotient sheaf ℋ:=ℱ/𝒦. If the composite map s:𝒦↪ℱ↠𝒪T2 is a zero map, then 𝒦 is a subsheaf of 𝒪T1, contradicting the semistability of 𝒪T1. The sheaf Im⁡(s) is a subsheaf of 𝒪T2 and so we have p(Im⁡(s))≤1/2. Similarly, the sheaf ker⁡(s) is a subsheaf of 𝒪T1 and so p(ker⁡(s))≤1/2. From the exact sequence
(53)0⟶ker⁡(s)⟶𝒦⟶sIm⁡(s)⟶0,
we have p(𝒦)≤1/2, a contradiction.Let us denote by∂M2 the closed subscheme of M2, consisting of the strictly semistable sheaves.Corollary 23.
One has(54)∂M2=𝔅≅(ℙ3×ℙ3)𝔖2,
where 𝔖2 is the permutation group of order 2. In particular, ∂M2 is a rational variety.Proof.
Obviously, we have𝔅⊂∂M2. Let ℱ be a strictly semistable sheaf and so it has a proper quotient sheaf ℋ with p(ℋ)=1/2. From the semistability of ℱ, ℋ has no 0-dimensional torsion. From the equality p(ℱ)=p(ℋ), we obtain that ℋ is also semistable. Since p(ℋ)=1/2, the Hilbert bipolynomial of ℋ is either 2x+1, 2y+1 or x+y+1. The first 2 cases cannot happen due to Proposition 10. Thus, we have χℋ(x,y)=x+y+1 and so ℋ≅𝒪T2 with T2∈|𝒪Q(1,1)| and T2⊂Cℱ 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 𝒦≅𝒪T1 with T1∈|𝒪Q(1,1)| and Cℱ=T1+T2. Hence, we have ℱ∈𝔅.
Letℱ be a sheaf of type (B), that is, it corresponds to a pair of two curves {T1,T2}. Let us assume that ℱ admits another sequence
(55)0⟶𝒪T3⟶ℱ⟶𝒪T4⟶0
with T3,T4∈|𝒪Q(1,1)|. Note that 𝒪Ti is stable for all i. Thus, the composite map s:𝒪T3→ℱ→𝒪T2 is either a zero map or an isomorphism. In the former case, we have 𝒪T3≅𝒪T1 and so 𝒪T4≅𝒪T2. In the latter case, we have 𝒪T3≅𝒪T2 and 𝒪T4≅𝒪T1. Hence, the class of a strictly semistable sheaf ℱ corresponds to a uniquely determined pair of two curves in |𝒪Q(1,1)| and we have 𝔅≅(ℙ3×ℙ3)/𝔖2. The second assertion follows from the fact that any symmetric product Sd(ℙN) 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 curvesT1,T2∈|𝒪Q(1,1)|, one has
(56)dim⁡Ext⁡1(𝒪T2,𝒪T1)={3,ifT1=T2;2,ifT1≠T2.Proof.
Note that we have(57)Ext2(𝒪T2,𝒪Q(-1,-1))=H0(𝒪T2(-1,-1))∨=0.
Thus, if we apply the functor Hom(𝒪T2,-) to the sequence of T1, we obtain
(58)0⟶Hom(𝒪T2,𝒪T1)⟶Ext1(𝒪T2,𝒪Q(-1,-1))⟶Ext1(𝒪T2,𝒪Q)⟶Ext1(𝒪T2,𝒪T1)⟶0.
We also have
(59)Ext1(𝒪T2,𝒪Q(-1,-1))≅H1(𝒪T2(-1,-1))∨≅H0(𝒪T2)Ext1(𝒪T2,𝒪Q)≅H1(𝒪T2(-2,-2))∨≅H0(𝒪T2(1,1))
and so their dimensions are 1 and 3, respectively. As 𝒪Q-sheaves, we have
(60)h0(ℋom(𝒪T2,𝒪T1))={1,ifT1=T2;0,otherwise,
for example, because T1 and T2 are reduced, and so the assertion is derived.Lemma 25.
The sheavesℱ of type (C), but not of type (B), are stable. In particular, the sheaves of type (C) are semistable.Proof.
As before let us assume the existence of a proper subsheaf𝒦 of ℱ with p(𝒦)≥1/2 and the quotient sheaf ℋ:=ℱ/𝒦. Since the composite s:𝒦↪ℱ²↠𝒪T2 is not a zero map, thus we have Im⁡(s)≅𝒪T2(-Z) for a 0-dimensional subscheme Z of T2 with length k. In particular, its Hilbert bipolynomial is y+1-k. If we let χ𝒦(x,y)=m′x+n′y+t′, then we have p(𝒦)=t′/(m′+n′)≥1/2. In particular, we have t′≥1. If we define 𝒦′ to be the kernel of the map s, then it is a subsheaf of 𝒪T1 and thus we have p(𝒦′)=(t′-1+k)/(m′+n′-1)≤1/3. Combining the two inequalities, we have k=0 and so the map s is surjective. Thus, we have ℋ≅𝒪T1/𝒦′. Note also that t′ can be either 1 or 2. If t′=2, then we have m′=n′=2 and so χ𝒦′(x,y)=2x+y+1=χ𝒪T1(x,y). In particular, we have ℋ=0 and so 𝒦≅ℱ, a contradiction. Now, assume t′=1 and so m′+n′≤2. In particular, ℋ is not a 0-dimensional sheaf. Moreover, ℋ is a quotient sheaf of 𝒪T1 with constant term 1 and so we have ℋ≅𝒪T3 with T3⊂T1 and T3∈|𝒪Q(1,1)|. For example, if T3=T1, then we have 𝒦′=0 and it contradicts the nontriviality of the extension ℱ. Thus, ℱ also admits the sequence
(61)0⟶𝒦⟶ℱ⟶𝒪T3⟶0,
where χ𝒦(x,y)=x+y+1. Since 𝒦' is a subsheaf of 𝒪T1 with χ𝒦'(x,y)=x, we have 𝒦′≅𝒪T4(-1,0), where T4 is a subcurve of T1 such that T1=T3+T4. Thus, 𝒦 is an extension of 𝒪T2 by 𝒪T4(-1,0). It is nontrivial, otherwise we would have 𝒪T2 as a direct factor of ℱ. Since there exists such a unique extension 𝒪T2+T4, ℱ admits an extension of 𝒪T3 by 𝒪T2+T4:
(62)
It implies that ℱ is of type (B).Lemma 26.
Letℱ be a line bundle on a reduced curve C∈|𝒪Q(2,2)| with degree 2. (1)
ℱ is semistable if and only if one has:
(a)
deg⁡(ℱ|T)≥1 for all subcurves T of C in |𝒪Q(a,b)| with (0,0)⪇(a,b)⪇(1,1),(b)
deg⁡(ℱ|A)≥0 for each smooth subcurve A of C in |𝒪Q(a,b)| with (1,1)≤(a,b)⪇(2,2).(2)
ℱ is stable if and only if deg⁡(ℱ|T)≥1 for all subcurves T of C in |𝒪Q(u,v)| with (0,0)⪇(u,v)⪇(2,2).Proof.
In both parts, the “only if” part is obvious. Assume thatℱ is not stable (resp., semistable) and take a proper subsheaf ℋ of ℱ with p(ℋ)≥p(ℱ) (resp. p(ℋ)>p(ℱ)). Taking a saturation of ℋ in ℱ, we may assume that 𝒢:=ℱ/ℋ is a pure sheaf. Call A the scheme support of ℋ and T the scheme support of 𝒢. The definition of scheme support of a purely 1-dimensional sheaf gives A+T=C as effective divisors. Thus T has one of the types in the assertion. Since C is reduced and ℱ is a line bundle on C, the support of 𝒢 must be a proper subcurve T of C. If T does not have a type of 𝒪Q(1,2) or 𝒪Q(2,1), then we are done. But the case of T having such types is excluded using the argument in the proof of Lemma 18.Lemma 27.
One has𝔅∩ℭ≠∅.Proof.
Let us setB=B′+T2 with B′∈|𝒪Q(0,1)| and T2∈|𝒪Q(1,0)|, and set A∈|𝒪Q(1,1)| to be smooth. For any extension ℱ∈𝔅 of 𝒪B by OA, for example, ℱ=𝒪A⊕𝒪B, let ℋ be the kernel of the composition ℱ→𝒪B→𝒪T2 and then ℋ is a pure sheaf with T1:=A+B′ as its scheme support and has Hilbert bipolynomial χℋ=χ𝒪T1. Note that it has 𝒪A as its subsheaf.
To proveℋ≅𝒪T1, it is sufficient to prove that ℋ is semistable. Suppose ℋ is not semistable and take a proper saturated stable subsheaf 𝒢⊂ℋ with χ𝒢=ax+by+c. Its scheme support is contained in T1 and it is of type (b,a). Without loss of generality, let us assume that a≤b. First, assume (a,b)=(1,2). In this case, we would have c≥2 because p(𝒢)>p(ℋ) and so we have h0(ℋ)≥2, contradicting the fact that h0(ℱ)=2 and that ℱ is globally generated. Assume a=b=1. The map 𝒢→ℋ on A∖T2 must be just the inclusion 𝒪A→ℋ, because ℋ|A∖T2 is a line bundle. Thus either we have 𝒢=𝒪A or 𝒪A is not saturated in ℋ. Hence the saturation 𝒜 of 𝒪A in ℱ has slope greater than 1/2, contradicting the semistability of ℱ. Now assume a=0 and b=1, that is, C𝒢=B′. Since B′ is smooth, 𝒢 is a line bundle on B′. If its degree d is at least 1, then 𝒢 contradicts the semistability of ℱ. If d≤0, then we have p(𝒢)<p(ℋ), a contradiction. Hence ℱ is also contained in ℭ.Lemma 28.
ForT1∈|𝒪Q(1,2)| and T2∈|𝒪Q(1,0)|, one has
(63)dim⁡Ext1(𝒪T2,𝒪T1)=2.Proof.
Applying the functorHom(𝒪T2,-) to the sequence of T1, we obtain
(64)0⟶Ext⁡1(𝒪T2,𝒪Q)⟶Ext⁡1(𝒪T2,𝒪T1)⟶Ext⁡1(𝒪T2,𝒪Q(-1,-2))⟶0,
since we have
(65)Ext1(𝒪T2,𝒪Q(-1,-2))≅H1(𝒪T2(-1,0))∨≅H1(𝒪T2)∨≅H0(𝒪T2(-2))=0
and similarly Ext2(𝒪T2,𝒪Q)=0. Note also that Ext1(𝒪T2,𝒪Q)≅H0(𝒪T2) and Ext2(𝒪T2,𝒪Q(-1,-2))≅H0(𝒪T2)∨. Thus we have the assertion.Remark 29.
WhenT1 and T2 meet transversally at two points, say P1 and P2, then Ext1(𝒪T2,𝒪T1) is the global sheaf of a sheaf with support on P1 and P2 with one copy of ℂ on each point P1,P2,
(66)Ext1(𝒪T2,𝒪T1)≅ℂP1⊕ℂP2
for the following reason.
LetR be a regular local ring of dimension 2 and take x,y generators of its maximal ideal. All Exti groups are with respect to R. Since R/(y) is Gorenstein, so the duality gives ExtR1(R/(y),R)≅R/(y) and ExtRi(R/(y),R)=0 for all i≠1. From the exact sequence
(67)0⟶R⟶uR⟶R(x)⟶0,
in which u is the multiplication by x, we get that ExtR1(R/(y),R/(x)) is the cokernel of the multiplication by x in R/(y)→R/(y); that is, we have ExtR1(R/(y),R/(x))=ℂ. The same is true for extensions of 𝒪B by 𝒪A when A and B are transversal.Lemma 30.
Letℱ be a sheaf of type (A) with no 0-dimensional torsion. Then ℱ is semistable unless it admits the sequence
(68)0⟶𝒪T2⟶ℱ⟶𝒪T1⟶0,
where T1∈|𝒪Q(a,b)| and T2∈|𝒪Q(2-a,2-b)| with (a,b)∈{(1,2),(2,1)}.Proof.
Let𝒦 be a subsheaf ℱ with maximal p-slope p(𝒦)>1/2 and so the quotient sheaf ℋ:=ℱ/𝒦 has no 0-dimensional torsion. Let us set χ𝒦(x,y)=m′x+n′y+t′ with t′≥1 and (0,0)⪇(m′,n′). If the composite s:𝒦↪ℱ²↠η is a zero map, then 𝒦 is a subsheaf destabilizing 𝒪C, a contradiction. If s is not surjective, for instance, Im⁡(s)=𝒪P⊊η, then ker⁡(s) is a subsheaf of 𝒪C with Hilbert bipolynomial m′x+n′y+t′-1. Thus we have t′=1 and the quotient ℋ':=𝒪C/𝒦′ has Hilbert bipolynomial with zero constant term. Since ℋ′ has no 0-dimensional torsion, we have ℋ′≅𝒪D for a curve D contained in C. But the Hilbert bipolynomial of 𝒪D has nonzero constant term, a contradiction. Thus the map s is surjective. Following the same argument before, we obtain that t′=1 and m′+n′≤1. Without loss of generality, let us assume that (m′,n′)=(0,1). Then we have χℋ(x,y)=2x+y+1 and thus we have ℋ≅𝒪T1, where T1 is a curve contained in Cℱ and T1∈|𝒪Q(1,2)|. Since 𝒦 is a subsheaf of ℱ with χ𝒦(x,y)=y+1, we have 𝒦≅𝒪T2 since 𝒦 has no 0-dimensional torsion. Thus ℱ fits into the sequence (68).Remark 31.
Applying the functorHom(η,-) to the sequence of C∈|𝒪Q(2,2)|, we obtain
(69)0⟶Ext⁡1(η,𝒪C)⟶Ext⁡2(η,𝒪Q(-2,-2))⟶fExt⁡2(η,𝒪Q).
Since the map f is the dual of the map Hom(𝒪Q(2,2),η)→Hom(𝒪Q,η) given by the multiplication by the defining equation of C, the map f is a zero map and thus we have Ext1(η,𝒪C)≅H0(η)∨. In particular its dimension is 2.Lemma 32.
Letℱ be a sheaf of type (B) fitting into an exact sequence
(70)0⟶𝒪T1⟶ℱ⟶𝒪T2⟶0
with T1,T2∈|𝒪Q(1,1)|. Then ℱ is of type (A) if and only if T1 and T2 have no common components; that is, Cℱ has no multiple component.Proof.
IfT1 and T2 have a common component, say T, then ℱ has rank 2 at the general point of T and thus ℱ is not of type (A).
Conversely, assume thatT1∩T2 is finite. Since we have h1(𝒪T1)=0, the sequence (70) implies that h0(ℱ)=2 and ℱ is globally generated. Let σ be a general section of ℱ and then it does not vanish at the general point of any of the components of Cℱ. Since Cℱ is reduced, σ induces an injective map 𝒪Cℱ↪ℱ and thus ℱ has type (A).Lemma 33.
Letℱ be a sheaf of type (C). (1)
IfT2 is not a component of T1, then ℱ is of type (A).(2)
IfT2 is a double component of Cℱ, that is, T2⊂T1, then it is not of type (A).Proof.
Let us assume thatT2∈|𝒪Q(1,0)|.(1)
SinceT2 is not a component of T1, ℱ is a line bundle on C=Cℱ outside finitely many points of C. Moreover, it is not an 𝒪T1-sheaf. Note that ℱ is globally generated since 𝒪T1 and 𝒪T2 are globally generated with h1(𝒪T1)=0. Thus, a general section of ℱ does not vanish at a general point of T2 and so it does not induce an injection 𝒪T1↪ℱ. Hence, ℱ fits into some sequence (46).(2)
Let us setC=2T2+T3 and T1:=T2+T3, where T3∈|𝒪Q(0,2)|. Let Γ be the projectivisation of Ext⁡1(𝒪T2,𝒪T1) and in particular we have Γ≅ℙ1 by Lemma 28. We also know from Lemma 25 that any e∈Γ gives a semistable sheaf. Such a sheaf has rank 2 at the points of T2∖(T2∩T3) and, in particular, it is not a line bundle over its support at a general point of T2. 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 T3 but not in T2.In general, the question whether the variety𝒫ic(m,n)d is rational is difficult as in the projective plane. We observed that M1 is rational and so is 𝒫ic(2,2)1. Below we give a partial answer to this question in the case of 𝒫ic(2,2)2.Theorem 34.
M
2 is unirational with degree 4.Proof.
Let us fix a smooth curveC of bidegree (2,2) in Q and a point P∈C to consider a sheaf 𝒪C(P)∈M1. If 𝕋P is the tangent plane of Q at P, then we have 𝕋P∩C={2P,Q1,Q2} for some points Q1,Q2 on Q since deg⁡(C)=4. It defines a rational map
(71)Φ:M1⇢M2,
sending 𝒪C(P) to 𝒪C(Q1+Q2). Note that 𝒪C(Q1+Q2)=𝒪C(1,1)(-2P). We claim that the map Φ is generically 4 to 1 and so the assertion follows.
LetU (resp., V) be the dense open subset of M1 (resp., 𝒜⊂M2) formed by the sheaves ℱ such that Cℱ is smooth. Each element of U (resp., V) is uniquely determined by a smooth C∈|𝒪Q(2,2)| and a degree one (resp., degree two) line bundle on C. By Riemann-Roch, each degree one line bundle on C is associated with a unique P∈C. Then the map Φ sends 𝒪C(P) to ℛ:=𝒪C(1,1)(-2P). Fix any degree two line bundle ℳ on C. Since we are in characteristic zero, there are exactly four line bundles 𝒜 on C such that 𝒜⊗2≅𝒪C. Hence, for each ℛ∈Pic2(C) there are exactly 4 points P∈C such that ℛ≅𝒪C(1,1)(-2P). Hence, Φ is dominant and the preimage of each element of V has cardinality 4.We did not succeed in getting any smaller degree of unirationality ofM2 as of now, and we left the rationality question as a conjecture.Conjecture 35.
M
2 is rational.



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*Source: 346126-2014-02-11.xml*