This paper investigates the properties of limit linear series on curves of compact type with three components, revealing new conditions for unique extensions and providing a constructive method to find all such extensions.
Contribution
It introduces a characterization for the uniqueness of exact extensions of refined Eisenbud-Harris limit linear series on three-component curves and develops a constructive approach to find all exact extensions.
Findings
01
Not all refined limit linear series have unique exact extensions.
02
A new condition characterizes when a unique extension exists.
03
Every refined limit linear series admits at least one exact extension.
Abstract
Our aim in this work is to study exact Osserman limit linear series on curves of compact type X with three irreducible components. This case is quite different from the case of two irreducible components studied by Osserman. For instance, for curves of compact type with two irreducible components, every refined Eisenbud-Harris limit linear series has a unique exact extension. But, for the case of three irreducible components, this property is no longer true. We find a condition characterizing when a given refined Eisenbud-Harris limit linear series has a unique exact extension. To do this, it is necessary to understand how to construct exact extensions. We find a constructive method, which describes how to construct all exact extensions of refined limit linear series. By our method, we get that every refined limit linear series has at least one exact extension.
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Full text
Limit linear series for curves of compact type
with three irreducible components
Gabriel Muñoz
Abstract
Our aim in this work is to study exact Osserman limit linear series on curves of compact type X with three irreducible components. This case is quite different from the case of two irreducible components studied by Osserman. For instance, for curves of compact type with two irreducible components, every refined Eisenbud-Harris limit linear series has a unique exact extension. But, for the case of three irreducible components, this property is no longer true. We find a condition characterizing when a given refined Eisenbud-Harris limit linear series has a unique exact extension. To do this, it is necessary to understand how to construct exact extensions. We find a constructive method, which describes how to construct all exact extensions of refined limit linear series. By our method, we get that every refined limit linear series has at least one exact extension.
1 Introduction
In Algebraic Geometry, the theory of linear series on smooth curves is closely related to that of Abel maps. The fibers of Abel maps consist precisely of complete linear series.
For curves of compact type, (Eisenbud and Harris 1986) developed the theory of limit linear series as an analogue of linear series. This theory is very powerful for degeneration arguments on curves. The idea is to analyze how linear series degenerate when a family of smooth curves degenerates to a compact type curve. Eisenbud and Harris approached this situation by considering only the possibe limit line bundles with nonnegative multidegree and degree d on one irreducible component of the curve.
(Osserman 2006a) developed a new and more functorial construction for the theory of limit linear series. The basic idea is to consider all possible limit line bundles with nonnegative multidegree. Thus, Osserman limit linear series carry more information about limit line bundles. This new theory has a generalization to higher rank vector bundles (Osserman 2014).
Abel maps for curves of compact type have been studied by (Coelho and Pacini 2010). Recently, for curves of compact type X with two irreducible components, (Esteves and Osserman 2013) related limit linear series to fibers of Abel maps via the definition of limit linear series by (Osserman 2006a). They studied the notion of exact limit linear series. These contain in particular all limits of linear series on the generic fiber in a regular smoothing family.
Also, for curves of compact type X with two irreducible components, (Osserman 2006b) studied the space of limit linear series corresponding to a given Eisenbud-Harris limit linear series. He obtained an upper bound for the dimension of that space. Using this result, he also obtained a simple proof of the Brill-Noether theorem using only the limit linear series theory.
Our aim in this work is to study exact limit linear series on curves of compact type X with three irreducible components. This case is quite different from the case of two irreducible components. For instance, for curves of compact type with two irreducible components, (Osserman 2006a) showed that every refined Eisenbud-Harris limit linear series has a unique exact extension. But, for the case of three irreducible components, this property is no longer true.
We will study the case of exact limit linear series which are obtained as the unique exact extension of a refined Eisenbud-Harris limit linear series. For curves X consisting of a chain of smooth curves, (Osserman 2014) studied the notion of chain adaptable refined limit linear series. He showed that every chain adaptable refined limit linear series has a unique extension. In particular, for our case of curves of compact type X with three irreducible components, chain adaptable refined limit linear series are an example of refined Eisenbud-Harris limit linear series with a unique extension. It turns out that, by our Theorem 5.3, every refined Eisenbud-Harris limit linear series having a unique extension is chain adaptable. We should mention that, for our case of curves of compact type with three irreducible components, it is easy to see that every chain adaptable refined limit linear series has a unique extension (we give a simple proof of that in Theorem 5.3), but proving that the unique extension property implies the chain adaptable condition is not easy at all. We really need to understand how to construct extensions (in Section 4, we describe a method for the construction of all exact extensions).
We now explain the contents of the paper in more detail, especially the statement of our main theorem, which is Theorem 5.3. We begin with the notation of a limit linear series. In (Esteves and Osserman 2013), a limit linear series of degree d and dimension r on a curve of compact type X with two irreducible components Y and Z, meeting transversally at a point P, is a collection g:=(L,V0,…,Vd), where L is an invertible sheaf on X of degree d on Y and degree [math] on Z, and Vi is a vector subspace of H0(X,Li) of dimension r+1, for each i=0,…,d, where Li is the invertible sheaf on X with restrictions \mathcal{L}\big{|}_{Y}(-iP) and \mathcal{L}\big{|}_{Z}(iP), and these vector subspaces are linked by certain natural maps between the sheaves Li. Thus, a limit linear series is defined by a collection of pairs (Li,Vi), for each i=0,…,d, and we can use the notation {(Li,Vi)}i, where 0≤i≤d. For each i=0,…,d, the invertible sheaf Li has multidegree d:=(d−i,i). So, for each i=0,…,d, setting d:=(d−i,i), Ld=Li and Vd=Vi, a limit linear series can also be denoted by a collection {(Ld,Vd)}d, where d≥0 of total degree d. We will use this notation for the case of three irreducible components.
In this work, X will denote the union of three smooth curves X1,X2 and X3, such that X1 and X2 meet transversally at a point A, and X2 and X3 meet transversally at a point B, with A=B. A limit linear series of degree d and dimension r on X is a collection g:={(Ld,Vd)}d, where d≥0 of total degree d, each Ld is an invertible sheaf of multidegree d, and Vd is a vector subspace of H0(X,Ld) of dimension r+1, for each d, where Ld is the invertible sheaf on X with restrictions \mathcal{L}\big{|}_{X_{1}}(-(d-i)A), \mathcal{L}\big{|}_{X_{2}}((d-i)A-lB) and \mathcal{L}\big{|}_{X_{3}}(lB), with d=(i,d−i−l,l) and L:=L(d,0,0), and the subspaces Vd are linked by certain natural maps between the sheaves Ld; see Subsection 2.1 for the precise definition of limit linear series.
We restrict our attention to the case of exact limit linear series which are obtained as the unique exact extension of a refined Eisenbud-Harris limit linear series. Given a refined limit linear series, for any i≥0 and l≥0 such that i+l≤d there is a natural space, here denoted Kil, satisfying that, for every extension g:={(Ld,Vd)}d, Kil contains V(i,d−i−l,l). It turns out that these spaces are very important to understand when the extension is unique (spaces with the same property as the Kil appeared in (Osserman 2006a) and (Osserman 2014), for curves of compact type with two irreducible components and for curves of compact type with more than two components, respectively).
We recall the analogous spaces for the case of curves of compact type X with two components. Given an invertible sheaf L on X of degree d on Y and degree [math] on Z, for each nonnegative multidegree d:=(i,d−i), let Ld be the invertible sheaf on X with restrictions \mathcal{L}\big{|}_{Y}(-(d-i)P) and \mathcal{L}\big{|}_{Z}((d-i)P), and let VY and VZ be vector subspaces of H0(L(d,0)) and H0(L(0,d)), respectively, of dimension r+1, such that \{(\mathcal{L}_{(d,0)}\big{|}_{Y},V_{Y}\big{|}_{Y}),(\mathcal{L}_{(0,d)}\big{|}_{Z},V_{Z}\big{|}_{Z})\} is a refined Eisenbud-Harris limit linear series. In this situation, (Osserman 2006a) showed that the unique exact extension is given as follows:
(∗) For each nonnegative multidegree d:=(i,d−i), Vd is the space of sections s of H0(X,Ld) whose image in H0(X,L(d,0)) belongs to VY and vanishes at P with order at least d−i, and whose image in H0(X,L(0,d)) belongs to VZ and vanishes at P with order at least i.
Now, in our case of three irreducible components, consider the analogous situation: given an invertible sheaf L on X of multidegree (d,0,0), for each nonnegative multidegree d:=(i,d−i−l,l), let Ld be the invertible sheaf on X with restrictions \mathcal{L}\big{|}_{X_{1}}(-(d-i)A), \mathcal{L}\big{|}_{X_{2}}((d-i)A-lB) and \mathcal{L}\big{|}_{X_{3}}(lB), and let VX1, VX2 and VX3 be vector subspaces of H0(L(d,0,0)), H0(L(0,d,0)) and H0(L(0,0,d)), respectively, of dimension r+1, such that \{(\mathcal{L}_{(d,0,0)}\big{|}_{X_{1}},V_{X_{1}}\big{|}_{X_{1}}),(\mathcal{L}_{(0,d,0)}\big{|}_{X_{2}},V_{X_{2}}\big{|}_{X_{2}}),(\mathcal{L}_{(0,0,d)}\big{|}_{X_{3}},V_{X_{3}}\big{|}_{X_{3}})\} is a refined Eisenbud-Harris limit linear series. For each i,l, define Kil as the natural generalization of (∗). More specifically:
For each nonnegative multidegree d:=(i,d−i−l,l), Kil is the space of sections s of H0(X,Ld) whose image in H0(X,L(d,0,0)) belongs to VX1 and vanishes at A with order at least d−i, whose image in H0(X,L(0,d,0)) belongs to VX2 and vanishes at A with order at least i and vanishes at B with order at least l, and whose image in H0(X,L(0,0,d)) belongs to VX3 and vanishes at B with order at least d−l; see Subsection 3.1 for the precise definition of the spaces Kil.
The difference with the case of two irreducible components is the fact that in our case the spaces defined above does not necessarily have dimension r+1.
In Section 4, we describe a method for the construction of all exact extensions of refined limit linear series. As a consequence, we get that every refined limit linear series has at least one exact extension. We use the method of Section 4 to understand when a refined limit linear series has a unique exact extension. Our Theorem 5.3 says that, for a refined limit linear series:
*There is a unique exact extension if and only if *
dim Kil=r+1ifi+l≤d, bj−1<i≤bj, bk−1′<l≤bk′andj+k≤r+1,
where b0,…,br are the orders of vanishing at A and b0′,…,br′ are the orders of vanishing at B, all orders correspond to the linear series on X2.
Moreover, this condition is also equivalent to the existence of a unique extension.
It follows from the exact sequence defining Kil (see Subsection 3.1), that the condition in Theorem 5.3 is equivalent to the following condition:
dim VX2(−iA−lB)=r+1−j−k
if i+l≤d, bj−1<i≤bj, bk−1′<l≤bk′ and j+k≤r+1.
This condition means exactly that our refined limit linear series is chain adaptable.
For compact type curves with two irreducible components, given an exact limit linear series, (Esteves and Osserman 2013) associated a closed subscheme of the fiber of the corresponding Abel map. In a subsequent work, for compact type curves with three irreducible components, we will give a description of this closed subscheme when the underlying exact limit linear series is the unique extension of a refined limit linear series.
Our techniques can be generalized to the case of compact type curves with an arbitrary number of irreducible components (work in progress).
2 Preliminaries
2.1**.**
(Limit linear series)* *Throughout this article, X will denote the union of three smooth curves X1,X2 and X3, such that X1 and X2 meet transversally at a point A, and X2 and X3 meet transversally at a point B, with A=B. If Y is a reduced union of some components of X, we get the following exact sequence
[TABLE]
for any invertible sheaf L on X.
Let L be an invertible sheaf on X of degree d on X1 and degree [math] on X2 and X3. For any i≥0 and l≥0 such that i+l≤d, let L(i,d−i−l,l) be the invertibe sheaf on X with restrictions \mathcal{L}\big{|}_{X_{1}}(-(d-i)A), \mathcal{L}\big{|}_{X_{2}}((d-i)A-lB) and \mathcal{L}\big{|}_{X_{3}}(lB). Note that L(i,d−i−l,l) has multidegree (i,d−i−l,l). For any i≥0 and l≥0 such that i+l≤d, let d:=(i,d−i−l,l) and set
[TABLE]
Whenever d≥0, there are natural maps
[TABLE]
[TABLE]
where the first map in each composition is the restriction map and the last maps are the natural inclusions. Note that the compositions φd,dφd,d and φd,dφd,d are zero.
If Y is a subcurve of X, for any d and for any subspace V⊆H0(X,Ld), we denote by VY,0 the subspace of V of sections that vanish on Y. If Y is an irreducible component of X, we denote by LY the invertible sheaf Ld, where the component of d corresponding to Y is equal to d and the other components of d are [math]. Also, to ease notation, let Lil:=L(i,d−i−l,l).
Fix integers d and r. A limit linear series on X of degree d an dimension r is a collection consisting of an invertible sheaf L on X of degree d on X1 and degree [math] on X2 and X3, and vector subspaces Vd⊆H0(X,Ld) of dimension r+1, for each d:=(i,d−i−l,l)≥0, such that φd,d(Vd)⊆Vd and φd,d(Vd)⊆Vd for each d≥0, whenever d≥0.
Given a limit linear series, if Y is an irreducible component of X, we denote by VY the corresponding subspace of H0(X,LY). Also, we denote by Vil the corresponding subspace of H0(X,Lil).
The conditions φd,d(Vd)⊆Vd and φd,d(Vd)⊆Vd are called the linking condition, and we say that Vd and Vd are linked by the maps φd,d and φd,d.
Note that φd,d:Vd→Vd has kernel VdXqc,0 and image contained in VdXq,0. Analogously, the map φd,d:Vd→Vd has kernel VdXq,0 and image contained in VdXqc,0.
A limit linear series {(Ld,Vd)}d is called exact if
Im (φd,d:Vd→Vd)=VdXq,0 and
Im (φd,d:Vd→Vd)=VdXqc,0
for each d:=(i,d−i−l,l)≥0, whenever d≥0.
Remark 2.2**.**
Note that, from the construction of the invertible sheaves Lil and the maps φd,d, we have that, for q=q, s∈H0(Ld) vanishes on Xq if and only if φd,d(s)∈H0(Ld) vanishes on Xq. In particular, given a limit linear series {(Ld,Vd)}d, we get natural inclusions, for q=q
[TABLE]
Also, if s∈H0(Ld) and φd,d(s)∈H0(Ld) vanishes on Xq, then φd,d(s) vanishes on Xq∪Xqc=X, as φd,d(H0(Ld)) is contained in the kernel of the map H^{0}(\mathcal{L}_{\underline{d}})\rightarrow H^{0}(\mathcal{L}_{\underline{d}}\big{|}_{X_{q}^{c}}), and hence φd,d(s)=0, which implies that s vanishes on Xq. Thus, given a limit linear series {(Ld,Vd)}d, we get the natural inclusion
[TABLE]
3 The kernel Kil
3.1**.**
(The kernel Kil)* *Let L be an invertible sheaf on X of degree d on X1 and degree [math] on X2 and X3. For any i≥0 and l≥0 such that i+l≤d, recall that Lil denotes the invertibe sheaf on X with restrictions \mathcal{L}\big{|}_{X_{1}}(-(d-i)A), \mathcal{L}\big{|}_{X_{2}}((d-i)A-lB) and \mathcal{L}\big{|}_{X_{3}}(lB).
Note that
[TABLE]
Then, we get the natural exact sequence:
[TABLE]
[TABLE]
where the first summand in k⊕k corresponds to the point A and the second summand corresponds to the point B. The last map in the exact sequence will be denoted evil.
Let VX1,VX2,VX3 be r+1-dimensional subspaces of H0(LX1),H0(LX2) and H0(LX3), respectively, such that they satisfy the linking condition. Assume that the associated Eisenbud-Harris limit linear series on X is refined. Call h this limit linear series.
For any i≥0 and l≥0 such that i+l≤d, we define Kil by the exact sequence:
[TABLE]
Thus
[TABLE]
where d:=(i,d−i−l,l), and the natural map
αqd:H0(Ld)→H0(LXq) is the restriction to Xq,
for each q=1,2,3.
Denote by b0,…,br the orders of vanishing of VX2 at A, and denote by b0′,…,br′ the orders of vanishing of VX2 at B. Also, let a0,…,ar denote the orders of vanishing of VX1 at A, and c0,…,cr the orders of vanishing of VX3 at B. Throughout this article, the data of this subsection will remain fixed.
**
Remark 3.2**.**
Note that, if {(Ld,Vd)}d is a limit linear series which is an extension of h, then Vil⊆Kil for any i≥0 and l≥0 such that i+l≤d.
Indeed, let d:=(i,d−i−l,l). By the linking condition, we have that α1d(Vd)⊆VX1. Since Im(\alpha_{1\underline{d}})\subseteq H^{0}(\mathcal{L}_{X_{1}}\big{|}_{X_{1}}(-(d-i)A)), \alpha_{1\underline{d}}(V_{\underline{d}})\subseteq H^{0}(\mathcal{L}_{X_{1}}\big{|}_{X_{1}}(-(d-i)A)), and hence α1d(Vd)⊆VX1(−(d−i)A). Thus Vd⊆(α1d)−1(VX1(−(d−i)A)). Analogously, we have Vd⊆(α2d)−1(VX2(−iA−lB)) and Vd⊆(α3d)−1(VX3(−(d−l)B)). It follows that Vd⊆Kil.
**
Remark 3.3**.**
By abuse of notation, denote the restriction of evil to the vector subspace
VX1(−(d−i)A)⊕VX2(−iA−lB)⊕VX3(−(d−l)B)
by evil as well. Notice that
[TABLE]
and
[TABLE]
Analogously, we have that
[TABLE]
and
[TABLE]
Remark 3.4**.**
Let C be a smooth curve, L an invertible sheaf on C of degree d, and V⊆H0(L) a linear series. Let r+1:=dim V, and let Q1,Q2∈C distinct points. Let e1,…,er be the orders of vanishing of V at Q1, and e0′,…,er′ the orders of vanishing of V at Q2. Then ej+ek′≤d if j+k≤r. Furthermore, dimV(−ejQ1−ek′Q2)≥r+1−(j+k) for any j,k.
Indeed,
[TABLE]
Thus, if j+k≤r, then dimV(−ejQ1−ek′Q2)≥1, which implies h0(L(−ejQ1−ek′Q2))≥1, and hence deg(L(−ejQ1−ek′Q2))≥0, i.e. ej+ek′≤d.
**
Proposition 3.5**.**
The following statements hold:
1. dim Kil≥r+1. Furthermore, dim Kil=r+1 if i≤b0 or l≤b0′.
2. The subspaces Kil⊆H0(Lil) satisfy the linking condition.
Proof. We will first prove that dim Kil≥r+1. There are five cases to consider.
Case 1: If i=bj and l=bk′ for some j,k.
Consider the exact sequence
[TABLE]
Since
VX1(−(d−i)A)=VX1(−(d−i+1)A) and VX3(−(d−l)B)=VX3(−(d−l+1)B),
as d−i=d−bj=ar−j is an order of vanishing of VX1 at A and d−l=d−bk′=cr−k is an order of vanishing of VX3 at B, we have that Im(evil)=k⊕k, by Remark 3.3, and hence
as ar−j<d−i<ar+1−j is not an order of vanishing of VX1 at A, bj−1<i<bj is not an order of vanishing of VX2 at A and d−l=cr−k is an order of vanishing of VX3 at B, we have that Im(evil)={0}⊕k, by Remark 3.3, and hence
[TABLE]
where in the last equality we used that VX2(−iA)=VX2(−bjA), as bj−1<i<bj, and in the last inequality we used Remark 3.4.
Case 3: If i=bj and bk−1′<l<bk′ for some j,k.
This case is analogous to Case 2.
Case 4: If bj−1<i<bj and bk−1′<l<bk′ for some j,k.
VX2(−iA−lB)=VX2(−iA−(l+1)B) and VX3(−(d−l)B)=VX3(−(d−l+1)B),
as d−i is not an order of vanishing of VX1 at A, i is not an order of vanishing of VX2 at A, l is not an order of vanishing of VX2 at B and d−l is not an order of vanishing of VX3 at B, we have that Im(evil)={0}⊕{0}, by Remark 3.3, and hence
[TABLE]
where in the last inequality we used Remark 3.4, and in the last equality we used that VX2(−iA)=VX2(−bjA) and VX2(−lB)=VX2(−bk′B).
Case 5: If i<b0 or i>br or l<b0′ or l>br′.
We will only prove the stated inequality in the case i<b0, as the other cases are analogous. Suppose i<b0. Then d−i>ar, and hence VX!(−(d−i)A)=0. Also, we have VX2(−iA−lB)=VX2(−(i+1)A−lB), as i is not an order of vanishing of VX2 at A.
Suppose first that l=bk′ for some k. Then VX3(−(d−l)B)=VX3(−(d−l+1)B), and it follows from the exact sequence defining Kil that
[TABLE]
We used above that VX2(−iA)=VX2, as i<b0.
Suppose bk−1′<l<bk′ for some k. Then VX3(−(d−l)B)=VX3(−(d−l+1)B) and VX2(−iA−lB)=VX2(−iA−(l+1)B), and hence
[TABLE]
Now, suppose l<b0′. Then VX2(−iA−lB)=VX2(−iA−(l+1)B). On the other hand, d−l>cr, and hence VX3(−(d−l)B)=0. It follows that
[TABLE]
We used above that VX2(−iA)=VX2 and VX2(−lB)=VX2.
Finally, suppose l>br′. Then
VX2(−iA−lB)=VX2(−iA−(l+1)B) and VX3(−(d−l)B)=VX3(−(d−l+1)B).
On the other hand, since l>br′, we have d−l<c0, and hence VX3(−(d−l)B)=VX3. Then
[TABLE]
We used above that VX2(−lB)=0, as l>br′. This finishes the proof of the stated inequality.
Now, we will prove that dim Kil=r+1 if i≤b0 or l≤b0′. We will only prove the stated equality in the case i≤b0, as the other case is analogous. Notice that, in Case 5, we saw dim Kil=r+1 if i<b0. Thus, it remains to show the stated equality in the case i=b0. Assume i=b0.
Suppose first that l=bk′ for some k. Notice that, in Case 1, for j=0, the equality holds in dimVX2(−iA−lB)≥r+1−(j+k), as VX2(−iA−lB)=VX2(−lB). Thus dim Kil=r+1.
An analogous reasoning works for the case bk−1′<l<bk′. Now, suppose l<b0′. In Case 5 we saw dim Kil=r+1 if i<b0. Analogously, we can show that dim Kil=r+1 if l<b0′.
Finally, suppose that l>br′. Then
VX3(−(d−l)B)=VX3(−(d−l+1)B), VX2(−iA−lB)=0 and VX3(−(d−l)B)=VX3.
On the other hand, since i=b0, we have d−i=ar. It follows that
[TABLE]
This finishes the proof of the stated equality.
Now, we will prove the statement 2 of the proposition. Keep the notation of multidegrees d and d used in Section 2. We will only prove the linking condition for q=1, as the proofs for q=2,3 are analogous. We will first prove that φd,d(Kil)⊆Ki−1,l. (Recall that, for q=1, d=(i−1,d−i−l+1,l).) Let s∈Kil. We have
[TABLE]
On the other hand, s∈(α2d)−1(VX2(−iA−lB))⊆(α2d)−1(VX2(−(i−1)A−lB)), as s∈Kil. Then
[TABLE]
Also, since s∈Kil, s∈(α3d)−1(VX3(−(d−l)B)), and hence
[TABLE]
It follows from (1), (2) and (3) that φd,d(s)∈Ki−1,l. This proves that φd,d(Kil)⊆Ki−1,l.
Now, we will prove that φd,d(Ki−1,l)⊆Kil. Let s∈Ki−1,l. Then
It follows from (4), (5) and (6) that φd,d(s)∈Kil. This proves that φd,d(Ki−1,l)⊆Kil, which finishes the proof of the proposition. □
Proposition 3.6**.**
The following statements hold:
1. For any i≥1 and l≥1 such that i+l≤d,
φd,d′(Kil)=Ki−1,l−1X2c,0,
where d:=(i,d−i−l,l) and d′:=(i−1,d−i−l+2,l−1).
2. For any i≥1 and l≥0 such that i+l≤d,
φd,d′′(Kil)=Ki−1,lX1,0,
where d:=(i,d−i−l,l) and d′′:=(i−1,d−i−l+1,l).
3. For any i≥0 and l≥1 such that i+l≤d,
φd,d(Kil)=Ki,l−1X3,0,
where d:=(i,d−i−l,l) and d:=(i,d−i−l+1,l−1).
Proof. We will first see how the statements 2 and 3 imply the statement 1. Let i≥1 and l≥1 such that i+l≤d. Let s′∈Ki−1,l−1X2c,0. Then s′∈Ki−1,l−1X3,0. But, by the statement 3 of the proposition, φd′′,d′(Ki−1,l)=Ki−1,l−1X3,0, so s′=φd′′,d′(s′′) for some s′′∈Ki−1,l. As φd′′,d′(s′′)=s′∈Ki−1,l−1X2c,0⊆Ki−1,l−1X1,0, it follows from Remark 2.2 that s′′∈Ki−1,lX1,0. Then by the statement 2 of the proposition, s′′=φd,d′′(s) for some s∈Kil. Thus
This proves that Ki−1,l−1X2c,0⊆φd,d′(Kil). But, it follows from Proposition 3.5, item 2, that φd,d′(Kil)⊆Ki−1,l−1X2c,0, so φd,d′(Kil)=Ki−1,l−1X2c,0.
It remains to show the statements 2 and 3. We will only prove the statement 2, as the statement 3 is analogous.
By abuse of notation, we denote the restriction of evil to the vector subspace
VX1(−(d−i)A)⊕VX2(−iA−lB)⊕VX3(−(d−l)B)
by evil as well. It follows from the exact sequence defining Kil that
[TABLE]
On the other hand, the exact sequence defining Kil induces the following exact sequence
[TABLE]
Then KilX1c,0≅VX1(−(d−i+1)A), so dim φd,d′′(Kil)=dim Kil−dim VX1(−(d−i+1)A).
Thus
[TABLE]
On the other hand, the exact sequence
[TABLE]
implies
[TABLE]
where ev1i−1,l is the restriction of evi−1,l to {0}⊕VX2(−(i−1)A−lB)⊕VX3(−(d−l)B). But, it follows from Proposition 3.5, item 2, that φd,d′′(Kil)⊆Ki−1,lX1,0, so from (3) and (8), we have that φd,d′′(Kil)=Ki−1,lX1,0 if and only if
[TABLE]
By checking cases i=bj, i=bj, l=bk′ and l=bk′, we see that both sides of (3) are equal to dimVX3(−(d−l)B)−dimVX3(−(d−l+1)B). Thus (3) is true, and hence φd,d′′(Kil)=Ki−1,lX1,0, proving the statement 2 of the proposition. This finishes the proof of the proposition. □
Proposition 3.7**.**
The following statements hold:
1. For any i≥1 and l≥0 such that i+l≤d, the following statements are equivalent:
(i) φd′′,d(Ki−1,l)=Ki,lX1c,0, where d:=(i,d−i−l,l) and d′′:=(i−1,d−i−l+1,l).
(ii) i−1 is an order of vanishing of VX2 at A and i−1 is not an order of vanishing of VX2(−lB) at A.
2. For any i≥0 and l≥1 such that i+l≤d, the following statements are equivalent:
(i) φd,d(Ki,l−1)=Ki,lX3c,0, where d:=(i,d−i−l,l) and d:=(i,d−i−l+1,l−1).
(ii) l−1 is an order of vanishing of VX2 at B and l−1 is not an order of vanishing of VX2(−iA) at B.
Proof. We will only prove the statement 1. (The statement 2 is analogous.) Suppose first that (i) holds. The exact sequence
[TABLE]
implies that \alpha_{1\underline{d}}\big{|}_{K_{il}^{X_{1}^{c},0}}:K_{il}^{X_{1}^{c},0}\rightarrow V_{X_{1}}(-(d-i+1)A) is an isomorphism. On the other hand, it follows from Proposition 3.5, item 2, that φd′′,d(Ki−1,l)⊆KilX1c,0. Then, we have that φd′′,d(Ki−1,l)=KilX1c,0 if and only if α1d(φd′′,d(Ki−1,l))=α1d(KilX1c,0), i.e., if and only if α1d′′(Ki−1,l)=VX1(−(d−i+1)A). By hypothesis, φd′′,d(Ki−1,l)=KilX1c,0, so α1d′′(Ki−1,l) is a proper subspace of VX1(−(d−i+1)A). The exact sequence defining Ki−1,l induces the following exact sequence
[TABLE]
By abuse of notation, we denote the restriction of evi−1,l to the vector subspace
VX1(−(d−i+1)A)⊕VX2(−(i−1)A−lB)⊕VX3(−(d−l)B)
by evi−1,l as well, and let evi−1,l be the restriction of evi−1,l to the vector subspace α1d′′(Ki−1,l)⊕VX2(−(i−1)A−lB)⊕VX3(−(d−l)B). We have
[TABLE]
and also
[TABLE]
Therefore
[TABLE]
and since dimIm(evi−1,l)−dimIm(evi−1,l)≤dimIm(evi−1,l)≤2, it follows that
as α1d′′(Ki−1,l) is a proper subspace of VX1(−(d−i+1)A). Thus, there are two cases to consider.
Case 1: If dimα1d′′(Ki−1,l)=dimVX1(−(d−i+1)A)−1.
It follows from (10) that dimIm(evi−1,l)=dimIm(evi−1,l)−1. We will first prove that i−1=bj for some j. Suppose by contradiction that i−1 is not an order of vanishing of VX2 at A. Then dimIm(evi−1,l)≤1, and hence dimIm(evi−1,l)≤0. So dimIm(evi−1,l)=0 and dimIm(evi−1,l)=1. Now, since i−1 is not an order of vanishing of VX2 at A, dimIm(evi−1,l)=1 implies that l=bk′ for some k, which implies that Im(evi−1,l)⊇{0}⊕k, and hence dimIm(evi−1,l)≥1, a contradiction. Thus i−1=bj for some j.
Now, we will prove that i−1 is not an order of vanishing of VX2(−lB) at A. Suppose first that l=bk′ for some k. Since i−1=bj and l=bk′, we have dimIm(evi−1,l)=2. Then dimIm(evi−1,l)=1. Since l=bk′, Im(evi−1,l)⊇{0}⊕k, and it follows from dimension considerations that Im(evi−1,l)={0}⊕k. This implies that all sections of α1d′′(Ki−1,l)⊆VX1(−(d−i+1)A) and all sections of VX2(−(i−1)A−lB) vanish at A, i.e.,
α1d′′(Ki−1,l)⊆VX1(−(d−i+2)A) and
VX2(−(i−1)A−lB)=VX2(−iA−lB).
Thus i−1 is not an order of vanishing of VX2(−lB) at A. In addition, since α1d′′(Ki−1,l) is contained in VX1(−(d−i+2)A), it follows from dimension considerations that α1d′′(Ki−1,l) is equal to VX1(−(d−i+2)A).
Now, assume l is not an order of vanishing of VX2 at B. Then dimIm(evi−1,l)=1, as i−1=bj, and hence Im(evi−1,l)={0}⊕{0}. This implies that all sections of α1d′′(Ki−1,l)⊆VX1(−(d−i+1)A) and all sections of VX2(−(i−1)A−lB) vanish at A. It follows that α1d′′(Ki−1,l)=VX1(−(d−i+2)A) and i−1 is not an order of vanishing of VX2(−lB) at A.
Case 2: If dimα1d′′(Ki−1,l)=dimVX1(−(d−i+1)A)−2.
It follows from (10) that dimIm(evi−1,l)=dimIm(evi−1,l)−2. As dimIm(evi−1,l)≤2, we have Im(evi−1,l)={0}⊕{0} and dimIm(evi−1,l)=2. Since dimIm(evi−1,l)=2, it follows that i−1=bj and l=bk′ for some j,k. Now, as l=bk′, we get dimIm(evi−1,l)≥1, a contradiction. Thus, the only case can happen is Case 1, and hence (ii) holds.
Suppose now that (ii) holds. Define K′⊆H0(Ld′′) by the exact sequence
[TABLE]
By abuse of notation, we denote the restriction of evi−1,l to the vector subspace
VX1(−(d−i+1)A)⊕VX2(−(i−1)A−lB)⊕VX3(−(d−l)B)
by evi−1,l as well, and let evi−1,l be the restriction of evi−1,l to the vector subspace VX1(−(d−i+2)A)⊕VX2(−(i−1)A−lB)⊕VX3(−(d−l)B). We have that K′⊆Ki−1,l, as VX1(−(d−i+2)A)⊆VX1(−(d−i+1)A). On the other hand, it follows from the definition of K′ that α1d′′(K′)⊆VX1(−(d−i+2)A), and hence α1d′′(K′)=VX1(−(d−i+1)A). Now, recall that φd′′,d(Ki−1,l)=KilX1c,0 if and only if α1d′′(Ki−1,l)=VX1(−(d−i+1)A). Therefore, to prove (i), we need only show that α1d′′(Ki−1,l)=VX1(−(d−i+1)A). For this, it suffices to show that Ki−1,l=K′.
Since K′⊆Ki−1,l, we need only prove that dim K′=dim Ki−1,l. We have
[TABLE]
and also
[TABLE]
Therefore dim K′=dim Ki−1,l if and only if
[TABLE]
i.e., if and only if dimIm(evi−1,l)=dimIm(evi−1,l)−1, as i−1=bj for some j. There are two cases to consider.
Case I: If l=bk′ for some k.
Since i−1=bj and l=bk′, dimIm(evi−1,l)=2. On the other hand, since i−1 is not an order of vanishing of VX2(−lB) at A, we have VX2(−(i−1)A−lB)=VX2(−iA−lB), i.e., all sections of VX2(−(i−1)A−lB) vanish at A. Then Im(evi−1,l)⊆{0}⊕k. But, since l=bk′, Im(evi−1,l)⊇{0}⊕k, and hence Im(evi−1,l)={0}⊕k. Therefore dimIm(evi−1,l)=1, and thus dimIm(evi−1,l)=dimIm(evi−1,l)−1.
Case II: If l is not an order of vanishing of VX2 at B.
Since i−1=bj, dimIm(evi−1,l)=1. As in Case I, Im(evi−1,l)⊆{0}⊕k. But, since l is not an order of vanishing of VX2 at B, Im(evi−1,l)⊆k⊕{0}, and hence we have Im(evi−1,l)={0}⊕{0}. So dimIm(evi−1,l)=0=dimIm(evi−1,l)−1. This finishes the proof of the proposition. □
Remark 3.8**.**
Let V1,V2 and V3 vector subspaces of a N-dimensional vector space V. We will say that V1distributes overV2 and V3 if V1∩(V2+V3)=V1∩V2+V1∩V3. Note that this notion is symmetric on V1,V2 and V3.
Indeed, since V1∩V2+V1∩V3⊆V1∩(V2+V3), we have that V1∩(V2+V3)=V1∩V2+V1∩V3 is equivalent to dim V1∩(V2+V3)=dim (V1∩V2+V1∩V3). We have
dimV1∩(V2+V3)=dim(V1∩V2+V1∩V3) if and only if
Now, just notice that (3.8) is symmetric on V1,V2 and V3. Thus, the following statements are equivalent:
V1 distributes over V2 and V3.
V2 distributes over V1 and V3.
V3 distributes over V1 and V2.
**
Proposition 3.9**.**
For any i≥0 and l≥0 such that i+l≤d:
KilXq1,0∩(KilXq2,0+KilXq3,0)=KilXq3c,0+KilXq2c,0* for distinct q1,q2,q3.*
Proof. By Remark 3.8, it is enough to prove the case qm=m for m=1,2,3. Via the injective map in the exact sequence defining Kil, we can see Kil as a subspace of VX1(−(d−i)A)⊕VX2(−iA−lB)⊕VX3(−(d−l)B). It follows from the exact sequence defining Kil that
KilX2,0=VX1(−(d−i+1)A)⊕{0}⊕VX3(−(d−l+1)B),
KilX3c,0={0}⊕{0}⊕VX3(−(d−l+1)B),
KilX2c,0={0}⊕VX2(−(i+1)A−(l+1)B)⊕{0}
andKilX1c,0=VX1(−(d−i+1)A)⊕{0}⊕{0}.
Also, by checking cases i=bj, i=bj, l=bk′ and l=bk′, we get
In particular, dim(KilX2,0+KilX3,0)≥dimKil−1, and equality holds if and only if l is an order of vanishing of VX2(−iA) at B. On the other hand, we have
so, by dimension considerations, we have that KilX3c,0+KilX2c,0=KilX1,0 if and only if VX2(−(i+1)A−lB)=VX2(−(i+1)A−(l+1)B). In this case, the statement of the proposition holds.
Suppose now that VX2(−(i+1)A−lB)=VX2(−(i+1)A−(l+1)B). Then, we have VX2(−iA−lB)=VX2(−iA−(l+1)B), and hence dim(KilX2,0+KilX3,0)=dimKil−1. Notice that
[TABLE]
Now, since KilX3c,0+KilX2c,0⊆KilX1,0∩(KilX2,0+KilX3,0) and dim(KilX2,0+KilX3,0)=dimKil−1, it suffices to show that KilX1,0 is not contained in the space KilX2,0+KilX3,0. Suppose by contradiction that KilX1,0⊆KilX2,0+KilX3,0. Notice that
a contradiction. So the statement of the proposition is shown. □
Proposition 3.10**.**
For any i≥0 and l≥0 such that i+l≤d, the following statements hold:
1. dim(KilXq1,0+KilXq2,0)≥ dim Kil−1 for any q1=q2.
2. dim KilX1,0+dim KilX2,0+dim KilX3,0≥2(dim Kil−1).
Proof. By the proof of Proposition 3.9, the statement 1 holds for q1=2 and q2=3. The proofs of the other cases are analogous. As for the statement 2, putting together the following equalities
we get dim KilX1,0+dim KilX2,0+dim KilX3,0≥2(dim Kil−1), and equality holds if and only if the equality holds in the two inequalities above. □
Proposition 3.11**.**
For any i≥0 and l≥0 such that i+l≤d:
dim KilX1,0+dim KilX2,0+dim KilX3,0=2(dim Kil−1)* if and only if i is an order of vanishing of VX2(−lB) at A and l is an order of vanishing of VX2(−iA) at B. In this case, we have that KilX1,0,KilX2,0 and KilX3,0 are proper subspaces of Kil.*
Proof. We have seen in the proof of Proposition 3.10 that
i.e., if and only if i is an order of vanishing of VX2(−lB) at A and l is an order of vanishing of VX2(−iA) at B.
Now, suppose that i is an order of vanishing of VX2(−lB) at A and l is an order of vanishing of VX2(−iA) at B. By the proof of Proposition 3.9, KilX2,0+KilX3,0 has dimension dimKil−1 if and only if l is an order of vanishing of VX2(−iA) at B. So, by the hypothesis on l, we have dim(KilX2,0+KilX3,0)=dimKil−1, and hence KilX2,0 and KilX3,0 are proper subspaces of Kil. Analogously, since i is an order of vanishing of VX2(−lB) at A, we have dim(KilX2,0+KilX1,0)=dimKil−1, and hence KilX1,0 is a proper subspace of Kil. □
4 Constructing exact extensions
We will describe a method for the construction of exact extensions. Furthermore, this method allows us to construct any exact extension. The main result of this section is the following proposition, which is the fundamental statement for our method.
Proposition 4.1**.**
For any i≥0 and l≥0 such that i+l≤d, let d:=(i,d−i−l,l), and let d′′:=(i−1,d−i−l+1,l) if i>0. Then, the following statements hold:
*1. Let i,l be positive integers such that i+l=d.
Let d′:=(i−1,d−i−l+2,l−1). Let Vd′ and Vd′′ be r+1-dimensional subspaces of Ki−1,l−1 and Ki−1,l, respectively, such that*
φd′,d′′(Vd′)=Vd′′X3c,0* and φd′′,d′(Vd′′)=Vd′X3,0.*
Set β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). Then, for any linearly independent elements u1,…,uβ∈Vd′′X1,0 such that Vd′′X1,0=(Vd′′X2c,0⊕Vd′′X3c,0)⊕⟨u1,…,uβ⟩, and for any elements v1,…,vβ∈Kil such that φd,d′′(v1)=u1,…,φd,d′′(vβ)=uβ, the subspace
Vd:=φd′,d(Vd′)+⟨v1,…,vβ⟩⊆Kil**
is r+1-dimensional, and
φd′,d(Vd′)=VdX2,0, φd,d′(Vd)=Vd′X2c,0 and
*2. Let i,l be positive integers such that i+l≤d−1.
Let d′:=(i−1,d−i−l+2,l−1) and d′′′:=(i,d−i−l−1,l+1). Let Vd′, Vd′′ and Vd′′′ be r+1-dimensional subspaces of Ki−1,l−1, Ki−1,l and Ki,l+1, respectively, such that*
φd′,d′′(Vd′)=Vd′′X3c,0, φd′′,d′(Vd′′)=Vd′X3,0, φd′′,d′′′(Vd′′)=Vd′′′X2,0 and φd′′′,d′′(Vd′′′)=Vd′′X2c,0.
Set β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). Then, for any linearly independent elements u1,…,uβ∈Vd′′X1,0 such that Vd′′X1,0=(Vd′′X2c,0⊕Vd′′X3c,0)⊕⟨u1,…,uβ⟩, and for any elements v1,…,vβ∈Kil such that φd,d′′(v1)=u1,…,φd,d′′(vβ)=uβ, the subspace
Let d′′′:=(i,d−i−l−1,l+1). Let Vd′′ and Vd′′′ be r+1-dimensional subspaces of Ki−1,l and Ki,l+1, respectively, such that*
φd′′,d′′′(Vd′′)=Vd′′′X2,0* and φd′′′,d′′(Vd′′′)=Vd′′X2c,0.*
Set β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). Then, for any linearly independent elements u1,…,uβ∈Vd′′X1,0 such that Vd′′X1,0=(Vd′′X2c,0⊕Vd′′X3c,0)⊕⟨u1,…,uβ⟩, and for any elements v1,…,vβ∈Kil such that φd,d′′(v1)=u1,…,φd,d′′(vβ)=uβ, the subspace
Vd:=φd′′′,d(Vd′′′)+⟨v1,…,vβ⟩⊆Kil**
is r+1-dimensional, and
φd′′′,d(Vd′′′)=VdX3,0, φd,d′′′(Vd)=Vd′′′X3c,0 and
Proof. We will first prove the statement 2. Consider the following diagram
[TABLE]
Notice that, by Proposition 3.6, item 2, elements v1,…,vβ∈Kil exist satisfying that φd,d′′(v1)=u1,…,φd,d′′(vβ)=uβ. Since all sections of φd′,d(Vd′)⊆H0(Ld) vanish on X2,
where in the first equality we used Remark 2.2 and in the second equality we used that φd′′,d′′′(Vd′′)=Vd′′′X2,0. On the other hand, φd′′,d(Vd′′)=φd′,d(φd′′,d′(Vd′′))⊆φd′,d(Vd′), as φd′′,d′(Vd′′)⊆Vd′. Analogously, φd′′,d(Vd′′)=φd′′′,d(φd′′,d′′′(Vd′′))⊆φd′′′,d(Vd′′′), as φd′′,d′′′(Vd′′)⊆Vd′′′. It follows that φd′′,d(Vd′′)⊆φd′,d(Vd′)∩φd′′′,d(Vd′′′), and hence φd′′,d(Vd′′)=φd′,d(Vd′)∩φd′′′,d(Vd′′′). Then
[TABLE]
On the other hand, since φd′′,d′′′(Vd′′)=Vd′′′X2,0, we have
where in the last equality we used that Vd′′X1,0=(Vd′′X2c,0⊕Vd′′X3c,0)⊕⟨u1,…,uβ⟩. Therefore φd,d′′((φd′,d(Vd′)+φd′′′,d(Vd′′′))∩⟨v1,…,vβ⟩)=0. On the other hand, as u1,…,uβ are linearly independent and φd,d′′(v1)=u1,…,φd,d′′(vβ)=uβ, it follows that
is r+1-dimensional. Since Vd′ and Vd′′′ are subspaces of Ki−1,l−1 and Ki,l+1, respectively, we have φd′,d(Vd′)⊆Kil and φd′′′,d(Vd′′′)⊆Kil. Thus, as ⟨v1,…,vβ⟩⊆Kil as well, it follows that Vd⊆Kil.
where the second equality follows from (18), and in the fourth equality we used that φd′′,d′(Vd′′)=Vd′X3,0. Since φd′,d(Vd′)⊆Vd, it follows that
dimVd′X2c,0+dimφd′,d(Vd′)=dimVd′=r+1,
and hence
[TABLE]
where in the last equality we used that φd,d′(Vd)=Vd′X2c,0. Since φd′,d(Vd′)⊆VdX2,0, we get φd′,d(Vd′)=VdX2,0. The proof of the equalities φd′′′,d(Vd′′′)=VdX3,0 and φd,d′′′(Vd)=Vd′′′X3c,0 is analogous to that of φd′,d(Vd′)=VdX2,0 and φd,d′(Vd)=Vd′X2c,0. This proves statement 2.
Now, we will prove the statement 1. Notice that (14) holds, as φd′′,d′(Vd′′)=Vd′X3,0. Then
[TABLE]
Since i+l=d, we have d′′=(i−1,1,l), and hence Vd′′X2c,0=0. Thus
[TABLE]
It follows from (4) and (20) that dimφd′,d(Vd′)=r+1−β. To prove that dimVd=r+1, it suffices to show that
φd′,d(Vd′)∩⟨v1,…,vβ⟩=0.
We have
[TABLE]
Then
[TABLE]
where in the last equality we used (20). Thus φd,d′′(φd′,d(Vd′)∩⟨v1,…,vβ⟩)=0. Reasoning as in the proof of the statement 2, we get φd′,d(Vd′)∩⟨v1,…,vβ⟩=0, and hence
Vd=φd′,d(Vd′)⊕⟨v1,…,vβ⟩
is r+1-dimensional. Reasoning as in the proof of the statement 2, we get Vd⊆Kil.
The proofs of the equalities φd′′,d(Vd′′)=VdX1c,0, φd′,d(Vd′)=VdX2,0 and φd,d′(Vd)=Vd′X2c,0 are the same as in the proof of the statement 2. So the statement 1 is shown.
Now, we will prove the statement 3. Notice that (13) holds, as φd′′,d′′′(Vd′′)=Vd′′′X2,0. Therefore
[TABLE]
Since l=0, we have d′′=(i−1,d−i+1,0), and hence Vd′′X3c,0=0. Then
[TABLE]
It follows from (4) and (23) that dimφd′′′,d(Vd′′′)=r+1−β. To prove that dimVd=r+1, it suffices to show that
φd′′′,d(Vd′′′)∩⟨v1,…,vβ⟩=0.
We have
[TABLE]
Then
[TABLE]
where in the last equality we used (23). So φd,d′′(φd′′′,d(Vd′′′)∩⟨v1,…,vβ⟩)=0, and reasoning as in the proof of the statement 2, we get φd′′′,d(Vd′′′)∩⟨v1,…,vβ⟩=0, and hence
Vd=φd′′′,d(Vd′′′)⊕⟨v1,…,vβ⟩
is r+1-dimensional. Reasoning as in the proof of the statement 2, we get Vd⊆Kil as well.
Reasoning as in the proof of the statement 2, we get the remaining equalities. So the statement 3 is shown, proving the proposition. □
Now, we describe the method for the construction of exact extensions {(Ld,Vd)}d. According to Remark 3.2, we necessarily have to do the construction in such a way that Vil to be contained in Kil for any i≥0 and l≥0 such that i+l≤d. The idea is first to construct the subspaces Vil for i=0, then for i=1, and so on, until i=d−1. For each i≥1, we first construct the subspaces Vil for l=d−i, then for l=d−i−1, and so on, until l=0.
Suppose inductively that, for i≥1, the r+1-dimensional subspaces
Vi−1,l⊆Ki−1,l
have been constructed for l=d−(i−1),…,l=0 in such a way that
φd′,d′′(Vd′)=Vd′′X3c,0 and φd′′,d′(Vd′′)=Vd′X3,0 for l=1,…,d−(i−1),
where, as usual, d′:=(i−1,d−(i−1)−l+1,l−1) and d′′:=(i−1,d−(i−1)−l,l). We say that the subspaces Vi−1,l satisfy the vertical exactness property.
Then we will construct r+1-dimensional subspaces Vil⊆Kil for l=d−i,…,l=0 in such a way that
(i) The subspaces Vil satisfy the vertical exactness property.
(ii)
φd′,d(Vd′)=VdX2,0, φd,d′(Vd)=Vd′X2c,0 for l=1,…,d−i, and
φd′′,d(Vd′′)=VdX1c,0, φd,d′′(Vd)=Vd′′X1,0 for l=0,…,d−i,
where d:=(i,d−i−l,l), d′:=(i−1,d−(i−1)−l+1,l−1) and d′′:=(i−1,d−(i−1)−l,l).
We inductively do the construction as follows:
Step 1. For l=d−i, the subspace Vil is the subspace Vd defined in Proposition 4.1, item 1.
Step 2. For l=d−i−1,…,l=1, the subspace Vil is the subspace Vd defined in Proposition 4.1, item 2.
Step 3. For l=0, the subspace Vil is the subspace Vd defined in Proposition 4.1, item 3.
By Proposition 4.1, the subspaces Vil satisfy the properties (i) and (ii). Thus, it remains to construct the r+1-dimensional subspaces V0l⊆K0l to satisfy the vertical exactness property, and verify that the exact limit linear series {(Ld,Vd)}d that we construct is in fact an extension.
By Proposition 3.5, item 1, dim Kil=r+1 if i≤b0 or l≤b0′. Since VX1,VX2 and VX3 are linked, we have VX1⊆Kd0, VX2⊆K00 and VX3⊆K0d. It follows from dimension considerations, that VX1=Kd0, VX2=K00 and VX3=K0d. On the other hand, dim K0l=r+1 if 0≤l≤d. Thus, we define V0l:=K0l for any nonnegative integer l≤d. (Note that, for l=0 and l=d, the definition coincides with our fixed subspaces VX2 and VX3.)
Now, we will prove that
φd′′′,d(Vd′′′)=VdX3,0, φd,d′′′(Vd)=Vd′′′X3c,0 for l=0,…,d−1,
where d:=(0,d−l,l) and d′′′:=(0,d−l−1,l+1). (Observe that this notation corresponds to the notation in Proposition 4.1 for i=0.)
Let l be a nonnegative integer such that l≤d. By Proposition 3.6, item 3, we have φd′′′,d(K0,l+1)=K0lX3,0. Then dim K0,l+1X3c,0+dim K0lX3,0=dim K0,l+1. On the other hand, since dim K0,l+1=r+1=dim K0,l, we have dim K0,l+1X3c,0+dim K0lX3,0=dim K0l. Now, dim K0lX3,0+dim φd,d′′′(K0l)=dim K0l. Hence
dim φd,d′′′(K0l)=dim K0,l+1X3c,0,
and since φd,d′′′(K0l)⊆K0,l+1X3c,0, we have φd,d′′′(K0l)=K0,l+1X3c,0, proving that the subspaces V0l=K0l satisfy the vertical exactness property.
Thus, we construct subspaces Vil⊆Kil for i=0,…,d−1 and l=0,…,d−i. Now, since dim Ki0=r+1 for i=0,…,d−1, we have that, by dimension considerations, Vi0=Ki0 for i=0,…,d−1. On the other hand, the subspaces {Ki0}i=0,…,d satisfy the following exactness property analogous to that of the subspaces K0l
φd′′,d(Ki−1,0)=Ki0X1c,0 and φd,d′′(Ki0)=Ki−1,0X1,0 for i=1,…,d,
where d:=(i,d−i,0) and d′′:=(i−1,d−i+1,0).
Thus, since VX1=Kd0, we get an exact limit linear series {(Ld,Vd)}d which is an extension of h.
Now, let {(Ld,Vd)}d be any exact extension. We will prove that the subspaces Vd are constructed by our method. Since dim K0l=r+1 if 0≤l≤d, we have V0l=K0l for each integer l such that 0≤l≤d. Let 0<i<d, 0≤l≤d−i, and d:=(i,d−i−l,l). We will show that Vd is constructed by our method if l>0 and i+l≤d−1. (The proofs of the cases i+l=d, l=0 are analogous.) Keep the notation of multidegrees used in Proposition 4.1. By Remark 3.2, we have that Vd, Vd′, Vd′′ and Vd′′′ are r+1-dimensional subspaces of Kil, Ki−1,l−1, Ki−1,l and Ki,l+1, respectively. Set β:=dimVd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). As
Vd⊇φd′,d(Vd′) and Vd⊇φd′′′,d(Vd′′′), it follows that Vd⊇φd′,d(Vd′)+φd′′′,d(Vd′′′). By the proof of the statement 2 of Proposition 4.1, we have
dim(φd′,d(Vd′)+φd′′′,d(Vd′′′))=r+1−β, so
This proves that our method constructs any exact extension.
5 Unique exact extension
In this section, we will show the conditions under which the exact extension is unique, and in this case, we will describe the scheme P(g) for such a unique extension. Keeping the notation of multidegrees used in Proposition 4.1, we have the following lemma.
Lemma 5.1**.**
If φd′′,d(Ki−1,l)=KilX1c,0, then dim Kil=dim Ki−1,l.
Proof. It follows from the hypothesis that
dimKi−1,lX1,0+dimKilX1c,0=dimKi−1,l.
On the other hand, by Proposition 3.6, item 2, we have φd,d′′(Kil)=Ki−1,lX1,0, and hence
dimKilX1c,0+dimKi−1,lX1,0=dimKil.
Thus dim Kil=dim Ki−1,l. □
Lemma 5.2**.**
Let i≥0 and l≥0 such that i+l≤d. Then, the following statements hold:
1. If i>0, then dim Kil≥ dim Ki−1,l.
2. If l>0, then dim Kil≥ dim Ki,l−1.
Proof. We will only prove the statement 1, as the proof of the statement 2 is analogous. Let d:=(i,d−i−l,l) and d′′:=(i−1,d−i−l+1,l). By Proposition 3.6, item 2, we have φd,d′′(Kil)=Ki−1,lX1,0. It follows that
dimKilX1c,0+dimKi−1,lX1,0=dimKil.
On the other hand, since φd′′,d(Ki−1,l)⊆KilX1c,0, we have
It follows that dim Kil≥ dim Ki−1,l, proving the statement 1 of the lemma. □
Theorem 5.3**.**
*The following statements are equivalent:
h has a unique exact extension.
dim Kil=r+1 if i+l≤d, bj−1<i≤bj, bk−1′<l≤bk′ and j+k≤r+1.
h has a unique extension.*
Proof. First, for a fixed integer j such that 1≤j≤r, let us consider the following statement:
dim Kil=r+1 if i+l≤d, bj−1<i≤bj, bk−1′<l≤bk′ and j+k≤r+1.
We will prove that, for that fixed integer j, the statement 4 implies the following statement:
Let ej0,…,ej,r−j be the orders of vanishing of VX2(−bjA) at B. Then ej,r−j=br−j′.
Indeed, assume statement 4 holds. by the proof of the first four cases of Proposition 3.5, we have that, for bj−1<i≤bj and bk−1′<l≤bk′ such that i+l≤d, dimKil=r+1 if and only if dimVX2(−iA−lB)=r+1−j−k. On the other hand, by Remark 3.4, bj+br−j′≤d. Thus dimKbj,br−j′=r+1, and it follows that
dimVX2(−bjA−br−j′B)=r+1−j−(r−j)=1.
If bj+br−j′+1≤d, then, by hypothesis, dimKbj,br−j′+1=r+1, as br−j′<br−j′+1≤br+1−j′. So dimVX2(−bjA−(br−j′+1)B)=r+1−j−(r+1−j)=0, and it follows that br−j′=er−j. Now, if bj+br−j′+1>d, then VX2(−bjA−(br−j′+1)B)=0 as well, and hence br−j′=ej,r−j. So statement 5 holds.
Now, we will prove that the statement 1 implies the statement 2. Assume statement 1 holds. Let {(Ld,Vd)}d be the unique exact extension. We claim that, for 0<i<d and 0≤l≤d−i,
KilX1c,0=φd′′,d(Vd′′) if β>0,
where d:=(i,d−i−l,l), d′′:=(i−1,d−i−l+1,l) and β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0).
Indeed, we will prove the claim for l>0 and i+l≤d−1, as the remaining cases are analogous. Assume β>0. In Section 4 we saw that
where d′:=(i−1,d−i−l+2,l−1), d′′′:=(i,d−i−l−1,l+1) and v1,…,vβ∈Kil satisfy that
Vd′′X1,0=(Vd′′X2c,0⊕Vd′′X3c,0)⊕⟨u1,…,uβ⟩, where u1:=φd,d′′(v1),…,uβ:=φd,d′′(vβ). Suppose that KilX1c,0 is not contained in Vd. Let v∈KilX1c,0∖Vd, and set
We have that Vd⊆Kil and φd,d′′(v1+v)=u1,…,φd,d′′(vβ)=uβ. Now, as v1∈Vd and v∈/Vd, it follows that v1+v∈/Vd, and hence Vd=Vd. However, by the method of Section 4, this allows us to construct an exact extension which is different from the unique exact extension, a contradiction. Thus KilX1c,0⊆Vd, and hence KilX1c,0=VdX1c,0=φd′′,d(Vd′′). So our claim is established.
Now, we will prove the statement 2 by induction on j. Let l0 be the largest order of vanishing of VX2(−b1A) at B. Notice that b1+l0≤d, as VX2(−b1A−l0B)=0. Also, we have l0≤br′, as l0 is necessarily an order of vanishing of VX2 at B. By definition of l0, and since VX2(−(b0+1)A)=VX2(−b1A), we get
dim VX2(−(b0+1)A−l0B)=1 and dim VX2(−(b0+1)A−(l0+1)B)=0.
Thus, since KilX2c,0≅VX2(−(i+1)A−(l+1)B) for any i≥0 and l≥0 such that i+l≤d, we get
[TABLE]
Now, set i:=b0+1 and l:=l0, keep the notation of multidegrees used in Proposition 4.1 and set β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). We will prove that β>0. Suppose first that l>0. It follows from the proofs of the statements 1 and 2 of Proposition 4.1 that β=dim Vd′X2c,0−dim Vd′′X2c,0. On the other hand, dim Kb0,l0=r+1 and dim Kb0,l0−1=r+1, so Vd′′=Kb0,l0 and Vd′=Kb0,l0−1. Therefore, by (25),
so β>0. Suppose now that l=0. It follows from the proof of the statement 3 of Proposition 4.1 that β=dim Vd′′X1,0−dim Vd′′X2c,0. On the other hand, as we saw, Vd′′=Kb0,l0. Since
dimKb0,l0X2c,0=0 and
Thus, in any case, β>0. It follows from the claim that KilX1c,0=φd′′,d(Vd′′)⊆φd′′,d(Ki−1,l), and hence KilX1c,0=φd′′,d(Ki−1,l). Then, by lemma 5.1, we get
[TABLE]
On the other hand, notice that, for l~>l0, VX2(−b0A−l~B)=VX2(−(b0+1)A−l~B) if and only if VX2(−l~B)=VX2(−(b0+1)A−l~B), i.e., if and only if VX2(−l~B)=0, that is, l~>br′. Thus, b0 is an order of vanishing of VX2(−l~B) if l0<l~≤br′. Then, by Proposition 3.7, item 1, if l0<l~≤br′ and i+l~≤d, Kil~X1c,0=φd′′,d(Ki−1,l~), where d:=(i,d−i−l~,l~) and d′′:=(i−1,d−i−l~+1,l~). It follows from lemma 5.1 that
[TABLE]
Therefore, by (26), (27), lemma 5.2, item 2 and Proposition 3.5, item 1,
[TABLE]
On the other hand, by Proposition 3.7, item 1, if b0+1<i~≤b1, l~≥0 and i~+l~≤d, Ki~l~X1c,0=φd′′,d(Ki~−1,l~), where d:=(i~,d−i~−l~,l~) and d′′:=(i~−1,d−i~−l~+1,l~). Then, it follows from lemma 5.1 that
[TABLE]
ifb0+1<i~≤b1,0≤l~≤br′andi~+l~≤d. Thus, (28) and (29) prove the case j=1. Now, suppose by induction that, for a certain 2≤j<r,
dim Ki~l~=r+1 if i~+l~≤d, bj−1<i~≤bj, bk−1′<l~≤bk′ and j+k≤r+1.
Then, since the statement 4 implies the statement 5, lj−1:=br−j′ is the largest order of vanishing of VX2(−bjA) at B. Let lj be the largest order of vanishing of VX2(−bj+1A) at B. Notice that bj+1+lj≤d, as VX2(−bj+1A−ljB)=0. Also, we have lj≤lj−1, i.e., lj≤br−j′. By definition of lj, and since VX2(−(bj+1)A)=VX2(−bj+1A), we get
dim VX2(−(bj+1)A−ljB)=1 and dim VX2(−(bj+1)A−(lj+1)B)=0,
and hence
[TABLE]
Now, we proceed as in the first case. Set i:=bj+1 and l:=lj, keep the notation of multidegrees used in Proposition 4.1 and set β:=dim Vd′′X1,0−dim(Vd′′X2c,0⊕Vd′′X3c,0). We will show that β>0. Suppose first that l>0. As we saw, β=dim Vd′X2c,0−dim Vd′′X2c,0. On the other hand, by induction, dim Kbj,lj=r+1 and dim Kbj,lj−1=r+1, as bj+lj≤d and lj≤br−j′, so Vd′′=Kbj,lj and Vd′=Kbj,lj−1. Therefore, by (30),
Thus, in any case, β>0. Reasoning as in the case j=1, we get
[TABLE]
On the other hand, notice that, for l~>lj, VX2(−bjA−l~B)=VX2(−(bj+1)A−l~B) if and only if VX2(−bjA−l~B)=0, i.e., if and only if l~>lj−1=br−j′. Thus, bj is an order of vanishing of VX2(−l~B) if lj<l~≤br−j′. Then, by Proposition 3.7, item 1, if lj<l~≤br−j′ and i+l~≤d, Kil~X1c,0=φd′′,d(Ki−1,l~), where d:=(i,d−i−l~,l~) and d′′:=(i−1,d−i−l~+1,l~). By induction and lemma 5.1,
Now, we will prove that the statement 2 implies the statement 3. Assume statement 2 holds. Let {(Ld,Vd)}d be an extension. Then, since bj+br−j′≤d, we have
Vil=Kil if 0<j≤r, bj−1<i≤bj and 0≤l≤br−j′.
Also,
Vil=Kil if i≤b0 or l≤b0′.
On the other hand, since the statement 4 implies the statement 5, for each 0<j≤r, br−j′ is the largest order of vanishing of VX2(−bjA)=VX2(−(bj−1+1)A) at B. Then VX2(−(bj−1+1)A−(br−j′+1)B)=0, and hence VX2(−(i+1)A−(l+1)B)=0 if i≥bj−1 and l≥br−j′. It follows that KilX2c,0=0 if i≥bj−1, l≥br−j′ and i+l≤d. Thus, for i>bj−1, l>br−j′ and i+l≤d, Ki−1,l−1X2c,0=0, and hence Vi−1,l−1X2c,0=0, implying that
φ(i−1,d−i−l+2,l−1),(i,d−i−l,l)(Vi−1,l−1)=Vil if i>bj−1, l>br−j′ and i+l≤d.
(In this case, Vil=VilX2,0.) It follows that the extension is unique, proving the statement 3.
Finally, by the method of Section 4, there exists at least one exact extension, so the statement 3 implies the statement 1. This finishes the proof of the theorem. □
Remark 5.4**.**
Suppose that h has a unique exact extension and let {(Ld,Vd)}d be its unique exact extension. Note that, in the proof of Theorem 5.3, we saw that
Vd=VdX2,0 for any d=(i,d−i−l,l) with i>bj−1, l>br−j′ and i+l≤d.
**
Remark 5.5**.**
Suppose that h has a unique exact extension and let {(Ld,Vd)}d be its unique exact extension. It follows from the proof of Theorem 5.3 that, for 0≤j≤r, b0′,…,br−j′ are the orders of vanishing of VX2(−bjA) at B. Analogously, for 0≤k≤r, b0,…,br−k are the orders of vanishing of VX2(−bk′B) at A.
**
Acknowledgements. The author would like to thank Eduardo Esteves for several helpful discussions.
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