This paper investigates the rigidity of Bott-Samelson-Demazure-Hansen varieties associated with the group PSO(2n+1, C), focusing on cohomology vanishing of tangent bundles for specific reduced expressions of the longest Weyl group element.
Contribution
It characterizes all reduced expressions of the longest Weyl group element for which the tangent bundle's higher cohomology vanishes, revealing rigidity conditions.
Findings
01
All reduced expressions with vanishing higher cohomology are described.
02
The study provides conditions for the rigidity of these varieties.
03
Cohomology modules of tangent bundles are explicitly computed.
Abstract
Let G=PSO(2n+1,C)(n≥3) and B be the Borel subgroup of G containing maximal torus T of G. Let w be an element of Weyl group W and X(w) be the Schubert variety in the flag variety G/B corresponding to w. Let Z(w,i) be the Bott-Samelson-Demazure-Hansen variety (the desingularization of X(w)) corresponding to a reduced expression i of w. In this article, we study the cohomology modules of the tangent bundle on Z(w0,i), where w0 is the longest element of the Weyl group W. We describe all the reduced expressions of w0 in terms of a Coxeter element such that all the higher cohomology modules of the tangent bundle on Z(w0,i) vanish (see Theorem \ref{theorem 8.1}).
No public reviews on file for this paper yet. If you reviewed it on a platform where reviews are public (OpenReview, ICLR, NeurIPS, ICML), you can paste yours below so the community can read it here.
Videos
No videos yet. Explain this paper in a talk, walkthrough, or lecture? Add one.
Taxonomy
TopicsAdvanced Algebra and Geometry · Advanced Combinatorial Mathematics · Algebraic structures and combinatorial models
Full text
rigidity of bott-samelson-demazure-hansen variety for PSO(2n+1,C)
S.Senthamarai Kannan and Pinakinath Saha
Chennai Mathematical Institute, Plot H1, SIPCOT IT Park,
Siruseri, Kelambakkam, 603103, India.
Let G=PSO(2n+1,C)(n≥3) and B be the Borel subgroup of G containing maximal torus T of G. Let w be an element of Weyl group W and X(w) be the Schubert variety in the flag variety G/B corresponding to w. Let Z(w,i) be the Bott-Samelson-Demazure-Hansen variety (the desingularization of X(w)) corresponding to a reduced expression i of w.
In this article, we study the cohomology modules of the tangent bundle on Z(w0,i), where w0 is the longest element of the Weyl group W. We describe all the reduced expressions of w0 in terms of a Coxeter element such that all the higher cohomology modules of the tangent bundle on Z(w0,i) vanish (see Theorem 7.1).
1. introduction
Let G be a simple algebraic group of adjoint type over the field C of
complex numbers. We fix a maximal torus T of G and
let W=NG(T)/T denote the Weyl group of G with respect to T.
We denote by R the set of roots of G with respect to T and by
R+ a set of positive roots. Let B+ be the Borel
subgroup of G containing T with respect to R+.
Let w0 denote the longest element of the Weyl group W. Let B be the
Borel subgroup of G opposite to B+ determined by T, i.e. B=nw0B+nw0−1, where nw0 is a representative of w0 in NG(T).
Note that the
roots of
B is the set R−:=−R+ of negative roots. We use the notation β<0 for
β∈R−.
Let S={α1,…,αn}
denote the set of all simple roots in R+, where n is the rank of G.
The simple reflection in the Weyl group corresponding to a simple root α is denoted
by sα. For simplicity of notation, the simple reflection corresponding to a simple root αi is denoted
by si.
For w∈W, let X(w):=BwB/B denote the Schubert variety in
G/B corresponding to w. Given a reduced expression
w=si1si2⋯sir of w, with
the corresponding tuple i:=(i1,…,ir), we denote by
Z(w,i) the desingularization of the Schubert variety X(w),
which is now known as Bott-Samelson-Demazure-Hansen variety. This was
first introduced by Bott and Samelson in a differential geometric and
topological context (see [2]). Demazure in [6] and
Hansen in [8] independently adapted the construction in
algebro-geometric situation, which explains the reason for the name.
For the sake of simplicity, we will denote any Bott-Samelson-Demazure-Hansen
variety by a BSDH-variety.
The construction of the BSDH-variety Z(w,i) depends on the
choice of the reduced expression i of w. In [5], the automorphism groups of these varieties were studied. There, the following vanishing results of the tangent bundle TZ(w,i) on Z(w,i) were proved
(see [5, Section 3]):
(1) Hj(Z(w,i),TZ(w,i))=0 for all j≥2.
(2) If G is simply laced, then Hj(Z(w,i),TZ(w,i))=0 for all
j≥1.
As a consequence, it follows that the BSDH-varieties are rigid
for simply laced groups and their deformations are unobstructed in general (see [5, Section 3] ).
The above vanishing result is independent of the choice of the reduced expression i of
w. While computing the first cohomology module H1(Z(w,i),TZ(w,i)) for non simply laced group, we observed that this cohomology module very much depend on the choice of a reduced expression i of w.
It is a natural question to ask that for which reduced expressions i of w,
the cohomology module H1(Z(w,i),TZ(w,i)) does vanish ?
In [4], a partial answer is given to this question for w=w0 when G=PSp(2n,C).
In this article, we give a partial answer to this question for w=w0 when G=PSO(2n+1,C).
Recall that a Coxeter element is an element of the Weyl group
having a reduced expression of the form si1si2⋯sin
such that ij=il whenever j=l (see [11, p.56, Section 4.4]). Note that for any Coxeter element c, there is a decreasing sequence of integers n≥a1>a2>…>ak=1 such that c=j=1∏k[aj,aj−1−1], where a0:=n+1, [i,j]:=sisi+1⋯sj for i≤j.
In this paper we prove the following theorem.
Theorem 1.1**.**
Let G=PSO(2n+1,C)(n≥3) and let c∈W be a Coxeter element.
Let i=(i1,i2,…,in) be a sequence corresponding to a reduced expression of w0, where
ir(1≤r≤n) is a sequence of reduced expressions of c (see Lemma 2.8).
Then, Hj(Z(w0,i),T(w0,i))=0 for all j≥1 if and only if c=j=1∏k[aj,aj−1−1], where a0:=n+1 and a2=n−1.
By the above vanishing results, we conclude that if G=PSO(2n+1,C)(n≥3) and i=(i1,i2,…,in)
is a reduced expression of w0 as above, then the BSDH-variety Z(w0,i) is rigid.
The main differences in the proof of main theorem between the case of type Cn and the case of type Bn are as follows:
When G is of type Cn, there is only one long simple root, namely
αn. Therefore, by [14, Corollary 5.6, p.778], we have H1(w,αj)=0 for any w∈W and for any j=n. But, if G is of type Bn,α1,α2,…,αn−1 are all long simple roots. So, we can not apply [14, Corollary 5.6, p.778]. Hence, we need to study the cohomology modules H1(w,αj) ( w∈W, j=n−1 ).
While studying these modules, we prove that H1(w,αj)=0 for any w∈W and for
any j=n−1 (see Lemma 3.3). Also, in this article, we need to prove an additional statement namely, (s1s2⋯sn)r−1s1s2⋯sn−1(αj)<0 for 2≤r≤n and n+1−r≤j≤n−1 (see Lemma 5.9).
The organization of the paper is as follows:
In Section 2, we recall some preliminaries on BSDH-varieties. We deal with the special case G=PSO(2n+1,C)(n≥3) in the later sections 3, 4, 5, 6 and 7. In Section 3, we prove H1(w,αj)=0 for j=n−1 and w∈W.
In Section 4 (respectively, Section 5) we compute the weight spaces of H0 (respectively, H1) of the relative tangent bundle of BSDH-varieties
associated to some elements of the Weyl group.
In Section 6, we prove some results on cohomology modules of the tangent bundle of BSDH varieties.
In Section 7, we prove the main result using the results from the previous sections.
2. preliminaries
In this section, we set up some notation and preliminaries. We refer to [3], [9], [10], [13] for preliminaries in algebraic groups and Lie algebras.
Let G be a simple algebraic group of adjoint type over C and T be a maximal torus of
G. Let W=NG(T)/T denote the Weyl group of G with respect to T and we denote
the set of roots of G with respect to T by R. Let B+ be a Borel subgroup of G
containing T. Let B be the Borel subgroup of G opposite to B+ determined by T.
That is, B=n0B+n0−1, where n0 is a representative in NG(T) of the longest element w0 of W. Let R+⊂R be
the set of positive roots of G with respect to the Borel subgroup B+. Note that the set of
roots of B is equal to the set R−:=−R+ of negative roots.
Let S={α1,…,αn} denote the set of simple roots in
R+. For β∈R+, we also use the notation β>0.
The simple reflection in W corresponding to αi is denoted
by sαi. Let g be the Lie algebra of G.
Let h⊂g be the Lie algebra of T and b⊂g be the Lie algebra of B. Let X(T) denote the group of all characters of T.
We have X(T)⊗R=HomR(hR,R), the dual of the real form of h. The positive definite
W-invariant form on HomR(hR,R)
induced by the Killing form of g is denoted by (\leavevmode,\leavevmode).
We use the notation ⟨\leavevmode,\leavevmode⟩ to
denote ⟨μ,α⟩=(α,α)2(μ,α), for every μ∈X(T)⊗R and α∈R.
We denote by X(T)+ the set of dominant characters of
T with respect to B+. Let ρ denote the half sum of all
positive roots of G with respect to T and B+.
For any simple root α, we denote the fundamental weight
corresponding to α by ωα. For 1≤i≤n, let h(αi)∈h be the fundamental coweight corresponding to αi. That is ; αi(h(αj))=δij, where δij is Kronecker delta.
For a simple root α∈S, let nα∈NG(T) be a representative of sα. We denote the minimal parabolic subgroup of G containing B and nα by Pα. We recall that the BSDH-variety corresponds to a reduced expression i of w=si1si2⋯sir defined by
[TABLE]
where the action of B×B×⋯×B on Pαi1×Pαi2×⋯×Pαir is given by
(p1,p2,…,pr)(b1,b2…,br)=(p1⋅b1,b1−1⋅p2⋅b2,…,br−1−1⋅pr⋅br),pj∈Pαij, bj∈B and i=(i1,i2,…,ir) (see [6, Definition 1, p.73], [3, Definition 2.2.1, p.64]).
We note that for each reduced expression i of w,Z(w,i) is a smooth projective variety. We denote by ϕw, the natural birational surjective morphism from Z(w,i) to X(w).
Let fr:Z(w,i)⟶Z(wsir,i′) denote the map induced by the projection
Pαi1×Pαi2×⋯×Pαir⟶Pαi1×Pαi2×⋯×Pαir−1, where i′=(i1,i2,…,ir−1). Then we observe that fr is a Pαir/B≃P1-fibration.
For a B-module V, let L(w,V) denote the restriction of the associated homogeneous vector bundle on G/B to X(w). By abuse of notation, we denote the pull back of L(w,V) via ϕw to Z(w,i) also by L(w,V), when there is no confusion. Since for any B-module V the vector bundle L(w,V) on Z(w,i) is the pull back of the homogeneous vector bundle from X(w), we conclude that the cohomology modules
[TABLE]
for all j≥0 (see [3, Theorem 3.3.4(b)]), are independent of choice of reduced expression i. Hence we denote Hj(Z(w,i),L(w,V)) by Hj(w,V). In particular, if λ is character of B, then we denote the cohomology modules Hj(Z(w,i),Lλ) by Hj(w,λ).
We recall the following short exact sequence of B-modules from [5], we call it SES :
Let α be a simple root and λ∈X(T) be such that ⟨λ,α⟩≥0. Let Cλ denote one dimensional B-module associated to λ. Here, we recall the following result due to Demazure [7, p.271] on short exact sequence of B-modules:
Lemma 2.1**.**
Let α be a simple root and λ∈X(T) be such that ⟨λ,α⟩≥0. Let ev:H0(sα,λ)⟶Cλ be the evaluation map. Then we have
(1)
If ⟨λ,α⟩=0, then H0(sα,λ)≃Cλ.
2. (2)
If ⟨λ,α⟩≥1, then Csα(λ)↪H0(sα,λ), and there is a short exact sequence of B-modules:
[TABLE]
Further more, H0(sα,λ−α)=0 when ⟨λ,α⟩=1.
3. (3)
Let n=⟨λ,α⟩. As a B-module, H0(sα,λ) has a composition series
[TABLE]
such that Vi/Vi+1≃Cλ−iα for i=0,1,…,n−1 and Vn=Csα(λ).
We define the dot action by w⋅λ=w(λ+ρ)−ρ, where ρ is the half sum of positive roots. As a consequence of exact sequences of Lemma 2.1, we can prove the following.
Let w∈W, α be a simple root, and set v=wsα.
Lemma 2.2**.**
If l(w)=l(v)+1, then we have
(1)
If ⟨λ,α⟩≥0, then
Hj(w,λ)=Hj(v,H0(sα,λ)) for all j≥0.
2. (2)
If ⟨λ,α⟩≥0, then Hj(w,λ)=Hj+1(w,sα⋅λ) for all j≥0.
3. (3)
If ⟨λ,α⟩≤−2, then Hj+1(w,λ)=Hj(w,sα⋅λ) for all j≥0.
4. (4)
If ⟨λ,α⟩=−1, then Hj(w,λ) vanishes for every j≥0.
The following consequence of Lemma 2.2 will be used to compute the cohomology modules in this paper.
Now onwards we will denote the Levi subgroup of Pα(α∈S) containing T by Lα and the subgroup Lα∩B by Bα. Let π:G~⟶G be the universal cover. Let L~α (respectively, Bα~) be the inverse image of Lα (respectively, Bα).
Lemma 2.3**.**
Let V be an irreducible Lα-module. Let λ
be a character of Bα. Then we have
(1)
As Lα-modules, Hj(Lα/Bα,V⊗Cλ)≃V⊗Hj(Lα/Bα,Cλ).
2. (2)
If
⟨λ,α⟩≥0, then
H0(Lα/Bα,V⊗Cλ)
is isomorphic as an Lα-module to the tensor product of V and
H0(Lα/Bα,Cλ). Further, we have
Hj(Lα/Bα,V⊗Cλ)=0
for every j≥1.
3. (3)
If
⟨λ,α⟩≤−2, then
H0(Lα/Bα,V⊗Cλ)=0,
and H1(Lα/Bα,V⊗Cλ)
is isomorphic to the tensor product of V and H0(Lα/Bα,Csα⋅λ).
4. (4)
If ⟨λ,α⟩=−1, then
Hj(Lα/Bα,V⊗Cλ)=0
for every j≥0.
Proof.
Proof (1).
By [13, Proposition 4.8, p.53, I] and [13, Proposition 5.12, p.77, I],
for all j≥0, we have the following isomorphism of Lα-modules:
[TABLE]
Proof of (2), (3) and (4) follows from Lemma 2.2 by taking w=sα and
the fact that Lα/Bα≃Pα/B.
∎
Recall the structure of indecomposable
modules over Bα and Bα (see [1, Corollary 9.1, p.130]).
Lemma 2.4**.**
(1)
Any finite dimensional indecomposable Bα-module V is isomorphic to
V′⊗Cλ for some irreducible representation
V′ of Lα and for some character λ of Bα.
2. (2)
Any finite dimensional indecomposable Bα-module V is isomorphic to
V′⊗Cλ for some irreducible representation
V′ of Lα and for some character λ of Bα.
Now onwards we will assume that G=PSO(2n+1,C)(n≥3).
Note that longest element w0 of the Weyl group W of G is equal to −id. We recall the following Proposition from [15, Proposition 1.3, p.858].
Proposition 2.5**.**
Let c∈W be a Coxeter element, let ωi be the fundamental weight corresponding to the simple root αi. Then there exists a least positive integer h(i,c) such that
ch(i,c)(ωi)=w0(ωi).
Lemma 2.6**.**
Let c∈W be a Coxeter element. Then we have
(1)
w0=cn.**
2. (2)
For any sequence ir(1≤r≤n) of reduced expressions of c; the sequence i=(i1,i2,…,ir) is a reduced expression of w0.
Proof.
Note that for n≥3, there is an isomorphism of Weyl group of Bn and Weyl group of Cn sending si↦si for (1≤i≤n).
Proof of the lemma holds in the case of type Cn for (n≥3) (see [4, Lemma 4.2, p.441]). Therefore lemma holds for type Bn(n≥3).
∎
Lemma 2.7**.**
Let n≥a1>a2>⋯>ar−1>ar≥1 be a decreasing sequence of integers. Then,
w=(j=a1∏nsj)(j=a2∏nsj)⋯(j=ar−1∏nsj)(j=ar∏n−1sj)*
is a reduced expression of w.*
Proof.
Note that for n≥3, there is an isomorphism of Weyl group of Bn and Weyl group of Cn sending si↦si for (1≤i≤n).
Proof of the lemma holds in the case of type Cn for (n≥3) (see [4, Lemma 4.3,p.441]). Therefore lemma holds for type Bn(n≥3).
∎
Let c be a Coxeter element in W. We take a reduced expression
c=[a1,n][a2,a1−1]⋯[ak,ak−1−1], where [i,j]=sisi+1⋯sj for i≤j and n≥a1>a2>⋯>ak=1.
The expressions of ci for 1≤i≤n as in (1) and (2) are reduced.
Proof.
Note that for n≥3, there is an isomorphism of Weyl group of Bn and Weyl group of Cn sending si↦si for (1≤i≤n).
Proof of the lemma holds in the case of type Cn for (n≥3) (see [4, Lemma 4.4, p.442]). Therefore lemma holds for type Bn(n≥3).
∎
3. cohomology modules H1(w,αj) where j=n−1 and w∈W
In this section, we prove that H1(w,αj)=0
for every w∈W and j=n−1.
Lemma 3.1**.**
Let v∈W and α∈S. Then H1(sj,H0(v,α))=0 for j=n.
Proof.
By [14, Corollary 5.6, p.778] we have H1(w,αn)=0. Therefore, we may assume that α is a long simple root. If H1(sj,H0(v,α))μ=0, then there exists an indecomposable L~αj-summand V of H0(v,α) such that H1(sj,V)μ=0. By Lemma 2.4, we have V≃V′⊗Cλ for some character λ of B~αj and for some irreducible L~αj-module V′.
Since H1(sj,V)μ=0 from Lemma 2.3(3) we have ⟨λ,αj⟩≤−2. Since α is a long root, there exists w∈W such that w(α)=α0. Thus H0(v,α)⊆H0(vw,α0). Again, since α0 is highest long root, H0(w0,α0)=g⟶H0(vw,α0) is surjective.
Let μ′ be the lowest weight of V. Then by the above argument μ′ is a root. Therefore we have μ′=μ1+λ, where μ1 is the lowest weight of V′. Hence, we have
⟨μ′,αj⟩≤−2. Since αj is a long root and μ′ is a root, we have ⟨μ′,αj⟩=−1,0,1. This is a contradiction. Thus we have H1(sj,H0(v,α))μ=0.
∎
Lemma 3.2**.**
*Let v∈W and αj∈S be such that j=n−1. Then we have
H1(sk,H0(v,αj))=0, for every k=1,2,…,n.*
Proof.
Step 1:** H0(v,αj)−(αn−1+2αn)=0.
Case 1:
Assume that j=n, choose an element u∈W of minimal length such that u−1(αn)=β0, the highest short root. Then we have H0(v,αj)⊆H0(vu,β0).**
Since β0 is dominant weight the natural restriction map
[TABLE]
is surjective.
**Hence H0(v,αj)μ=0 implies either μ=0 or μ is a short root.
Therefore, we have H0(v,αj)−(αn−1+2αn)=0.
Case 2:
Assume that 1≤j≤n−2. Note that if H0(v,αj)μ=0 then either μ=αj,0 or
μ≤−αj (see [4, Corollary 4.5, p.678]).**
Hence H0(v,αj)−(αn−1+2αn)=0.
Step 2:** If H0(v,αj)−(βi+2αn)=0 for some 1≤i≤n−2, then H0(v,αj)−(βi+αn)=0.**
**Proof of Step 2: If v=id, we are done.
So choose 1≤t≤n such that l(stv)=l(v)−1.
Let v′=stv. Then H0(v,αj)=H0(st,H0(v′,αj)).
**
**Case 1: Assume that t=n. In this case ⟨−(βi+2αn),αt⟩=−2.
If H0(v,αj)−(βi+2αn)=0, then there is an indecomposable Bαn-summand V of H0(v′,αj) with highest weight −βi. Since ⟨−βi,αn⟩=2, we have H0(st,V)−(βi+αn)=0.
Therefore we have H0(v,αj)−(βi+αn)=0.
Case 2: Assume that t=n−1. In this case ⟨−(βi+2αn),αt⟩=1. If H0(v,αj)−(βi+2αn)=0, then there is an indecomposable Bαn-summand V of H0(v′,αj) with highest weight −(βi+2αn).
Thus by induction hypothesis we have H0(v′,αj)−(βi+αn)=0.
Since ⟨−(βi+αn),αt⟩=0, we have
H0(v,αj)−(βi+αn)=0.
Case 3: Assume that 1≤t≤n−2.
In this case ⟨−(βi+2αn),αt⟩=−1,0 or 1.
**
Assume that i=t. Then we have ⟨−(βi+2αn),αt⟩=−1. If further H0(v,αj)−(βi+2αn)=0, then there is an indecomposable Bαt-summand V of H0(v′,αj)
with highest weight −(βi+1+2αn).
Therefore we have H0(v′,αj)−(βi+1+2αn)=0. It is clear from Step: 1 that t+1≤n−2. Therefore by induction
H0(v′,αj)−(βt+1+αn)=0.
Since ⟨−(βt+1+αn),αt⟩=1, we have H0(v,αj)−(βt+αn)=0.
Assume that 1≤t≤i−2 or i+1≤t≤n−2. Then we have ⟨−(βi+2αn),αt⟩=0. Thus H0(v,αj)−(βt+2αn)=H0(v′,αj)−(βt+2αn)=0. Therefore by induction
H0(v,αj)−(βt+αn)=0. Since ⟨−(βi+αn),αt⟩=0, we have H0(v,αj)−(βi+2αn)=0.
Assume that i=t+1. Since ⟨−(βi+αn),αt⟩=1, then there is an indecomposable Bαt-summand V of H0(v′,αj)
such that: V=C−(βi+2αn)⊕C−(βi−1+2αn) or V=C−(βi+2αn).
Then we have H0(v′,αj)−(βt+2αn)=0. Therefore by induction
H0(v′,αj)−(βi+αn)=0.
Since ⟨−(βt+αn),αt⟩=1, we have H0(v,αj)−(βi+αn)=0
Hence the proof of Step 2.
**Proof of Lemma :
Case 1: Assume that k=n. Then by Lemma 3.1 we have H1(sk,H0(v,αj))=0.**
Case 2: Assume that k=n.
By Step 1 we see that H0(v,αj)−(αn−1+2αn)=0. Note that if β is a root such that H0(v,αj)β=0 and ⟨β,αn⟩=−2, then we have β=−(βi+2αn) for some 1≤i≤n−2.
By Step 2 if H0(v,αj)−(βi+2αn)=0 for some 1≤i≤n−2, then H0(v,αj)−(βi+αn)=0.
Therefore C−(βi+αn)⊕C−(βi+2αn) is an indecomposable Bαn-summand of H0(v,αj). By Lemma 2.4, C−(βi+αn)⊕C−(βi+2αn) is isomorphic to V⊗C−ωn, where V is an irreducible L~αn-module.
Therefore by Lemma 2.3(4) we have H1(sk,C−(βi+αn)⊕C−(βi+2αn))=0.
Thus our result follows.
∎
Lemma 3.3**.**
Let w be an element of W and αj be an element of S such that j=n−1. Then H1(w,αj)=0.
Proof.
We will prove by induction on length of w. If length of w is [math], then w=id. Thus it follows trivially.
Now suppose w∈W such that l(w)≥1. Then there exists a simple root α∈S such that l(sαw)=l(w)−1.
Then using SES:
From the above SES using induction hypothesis and Lemma 3.2, we get H1(w,αj))=0 for j=n−1.
∎
4. cohomology module H0 of the relative tangent bundle
In this section we describe the weights of H0 of the relative tangent bundle.
Notation:
Let c be a Coxeter element of W. We take a reduced expression c=[a1,n][a2,a1−1]⋯[ak,ak−1−1], where [i,j]=sisi+1⋯sj for i≤j and n≥a1>a2>⋯>ak=1.
Let βi=αi+αi+1+⋯+αn−1 for all 1≤i≤n−1.
For 1≤r≤k, let n≥a1>a2>a3>⋯>ar≥1 be a decreasing sequence of integers.
Let V=j=ar⨁k−1(C−(βj+αn)⊕C−(βj+2αn)⊕C−(βj+2αn+βn−1)⊕⋯⊕C−(βj+2αn+βk+1)) and V′=j=k+2⨁n−2(C−(βj+2αn)⊕C−(βj+2αn+βn−1)⊕⋯⊕C−(βj+2αn+βj+1))⊕C−(βn−1+2αn).
Then roots {(βj+αn),(βj+2αn),(βj+2αn+βn−1),…,(βj+2αn+βk+2):ar≤j≤k−1},{(βj+2αn),(βj+2αn+βn−1),…,(βj+2αn+βj+1):k+2≤j≤n−2} and −(βn−1+2αn)
are orthogonal to αk.
Therefore V, V′ are direct sums of irreducible L~αk−modules. By Lemma 2.3(2), we have
Further the remaining roots of (4.1.3) are {−(βk+αn),−(βk+2αn),−(βk+2αn+βn−1),…,−(βk+2αn+βk+2),−(βk+2αn+βk+1)},{−(βk+1+2αn),−(βk+1+2αn+βn−1),…,−(βk+2αn+βk+2)} and {−(βj+2αn+βk+1):ar≤j≤k}.
Since ⟨−(βk+αn),αk⟩=−1, by Lemma 2.3(4) we have
H0(L~αk/B~αk,C−(βk+αn))=0.
Since C−(βk+2αn)⊕C−(βk+1+2αn) is the irreducible two dimensional L~αk-module, by Lemma 2.3(2) we have
The roots of (4.1.4) are {−(β1+αn),−(β1+2αn),−(β1+2αn+βn−1),…,−(β1+2αn+β3),−(β1+2αn+β2)},{−(β2+2αn),−(β2+2αn+βn−1),…,−(β1+2αn+β3)} and −(β1+2αn+β2).
Since −(βn−1+2αn)
is orthogonal to α1, by Lemma 2.3(2) we have
H0(L~α1/B~α1,C−(βn−1+αn))=C−(βn−1+αn).
Since ⟨−(β1+αn),α1⟩=−1, by Lemma 2.3(4) we have
H0(L~α1/B~α1,C−(β1+αn))=0.
Since C−(β1+2αn)⊕C−(β2+2αn) is the irreducible two dimensional L~α1-module, by Lemma 2.3(2) we have
Since {−(βi+2αn+βn−1),…,−(βi+2αn+βar−1):ar≤i≤ar−1−1} are orthogonal to αn, by Lemma 2.3(2), we have
H0(L~αn/B~αn,C−(βi+2αn+βt))=C−(βi+2αn+βt) for all ar≤i≤ar−1−1 and ar−1≤t≤n−1.
Since {−(βi+2αn+βn−1),…,−(βi+2αn+βi+1):ar−1≤i≤n−2} are orthogonal to αn, by Lemma 2.3(2) we have
H0(L~αn/B~αn,C−(βi+2αn+βl))=C−(βi+2αn+βl) for all i+1≤l≤n−1, where ar−1≤i≤n−2.
Since
⟨−(βi+2αn),αn⟩=−2 for all ar≤i≤n−1, by Lemma 2.3(3) we have
H0(L~αn/B~αn,C−(βn−1+2αn))=0 for all ar−1≤i≤n−1.
Moreover, for each ar≤i≤ar−1−1,C−(βi+αn)⊕C−(βi+2αn) is an indecomposable two dimensioal Bαn-module. Therefore by Lemma 2.4, we have C−(βi+αn)⊕C−(βi+2αn)=Vi⊗C−ωn,
where Vi is the irreducible two dimensional representation of L~αn.
By Lemma 2.3(4) we have
H0(L~αn/B~αn,C−(βi+αn)⊕C−(βi+2αn))=0 for each ar≤i≤ar−1−1.
Since ⟨−(βi+2αn+βn−1),αn−1⟩=−1 for each ar≤i≤n−2, by Lemma 2.3(4) we have H0(L~αn−1/B~αn−1,C−(βi+αn+βn−1))=0 for each ar≤i≤n−2.
Moreover, {−(βi+2αn+βn−2),…,−(βi+2αn+βar−1):ar≤i≤ar−1−1}, {−(βi+2αn+βn−2),…,−(βi+2αn+βi+1):ar−1≤i≤n−3}, and βn−1+2αn are orthogonal to αn−1.
Since {−(βi+2αn+βar−2−1),…,−(βi+2αn+βar−1):ar≤i≤ar−1−1},{−(βi+2αn+βar−2−1),…,−(βi+2αn+βi+1):ar−1≤i≤ar−2−2} are orthogonal to
αj for all ar−3≤j≤n, by Lemma 2.3(2) we have
Let 3≤r≤k. Then H0(wr−2snsar−1sar−1+1⋯sn+2−r,αn+2−r)μ=0 if μ is of the form μ=−(βj+αn) for some ar−1≤j≤ar−2−1.
Proof.
By applying SES repeatedly, it is easy to see that
H0(snsar−1sar−1+1⋯sn+2−r,αn+2−r)=Ch(αn+2−r)⊕(j=ar−1⨁n+2−rC−γj,n+2−r), where γj,j′=(αj+⋯+αj′) for j′≥j.
Let V1=H0(snsar−1sar−1+1⋯sn+2−r,αn+2−r). We next calculate H0(sar−2⋯sn−1,V1).
Since n≥a1>a2>⋯>ak=1, we have ai≤n+1−i for all 1≤i≤k. Assume l≥n+4−r, then ⟨γj,n+2−r,αl⟩=0 for all ar−1≤j≤n+2−r. By Lemma 2.3(2) we have H0(L~αl/B~αl,V1)=V1
for all l≥n+4−r.
Therefore we have H0(sar−2⋯sn−1,V1)=H0(sar−2⋯sn+2−rsn+3−r,V1).
Note that, since ⟨−γj,n+2−r,αn+3−r⟩=1 for all ar−1≤j≤n+2−r, by Lemma 2.3(2) we have H0(sn+3−r,V1)=Ch(αn+2−r)⊕(j=ar−1⨁n+2−r(C−γj,n+2−r⊕C−γj,n+3−r)).
Since Ch(αn+2−r)⊕C−γn+2−r,n+2−r is an indecomposable two dimensional Bαn+2−r-module, by Lemma 2.4,
Ch(αn+2−r)⊕C−γn+2−r,n+2−r=V⊗C−ωn+2−r, where V is the irreducible two dimensional L~αn+2−r-module.
Since ⟨γj,n+2−r,αn+2−r⟩=−1, for all ar−1≤j≤n+1−r, by Lemma 2.3(4) we have
H0(L~αn+2−r/B~αn+2−r,Cγj,n+2−r)=0 for all ar−1≤j≤n+1−r.
H0(sar−2⋯sn−1,V1)=j=ar−1⨁ar−2−1C−γj,n+3−r and H0(snsar−2⋯sn−1,V1)=j=ar−1⨁ar−2−1C−γj,n+3−r.
Let V2=H0(snsar−2⋯sn−1,V1). Similarly, we have H0(sar−3⋯sn−1,V2)=j=ar−1⨁ar−2−1C−γj,n+4−r and H0(snsar−3⋯sn−1,V2)=j=ar−1⨁ar−2−1C−γj,n+4−r.
Let V1=H0(sar−1⋯sn−1sn,αn).
Since ⟨−(βj+αn),αn⟩=0, we have H0(sn,V1)=V1.
Since ⟨−(βn−1+αn),αn−1⟩=−1
and ⟨−(βj+αn,αn−1)⟩=0, for all ar−1≤j≤n−2, by Lemma 2.3(2), Lemma 2.3(4) and Lemma 2.4 we have
[TABLE]
Proceeding recursively we have
[TABLE]
Since n≥a1>a2>⋯>ak=1, we see that ⟨−(βj+αn),αt⟩=0, for all ar−1≤j≤ar−2−1 and for all ar−3≤t≤n, therefore by Lemma 2.3(2) and Lemma 2.4, we have
[TABLE]
Since n≥a1>a2>⋯>ak=1, we see that ⟨−(βj+αn),αt⟩=0 for all ar−1≤j≤ar−2−1 and for all ar−4≤t≤n. By Lemma 2.3(2) and Lemma 2.4, we have
[TABLE]
Proceeding recursively we have
[TABLE]
Hence the lemma follows.
∎
Lemma 4.5**.**
If μ is of the form μ=−(βj+αn) for some 1≤j≤ak−1−1, then we have H0(wk−1sns1s2⋯sn+1−k,αn+1−k)μ=0.
Proof.
By applying SES repeatedly, it is easy to see that
Since Ch(αn+1−k)⊕C−γn+1−k,n+1−k is an indecomposable two dimensional Bαn+1−k-module, by Lemma 2.4 we have Ch(αn+1−k)⊕C−γn+1−k,n+1−k=V⊗C−ωn+1−k, where V is the irreducible two dimensional L~αn+1−k-module.
Similarly, since n≥a1>a2>⋯>ak=1, we have ⟨−(βj+αn),αt⟩=0 for all 1≤j≤ak−1−1 and for all ak−3≤t≤n, therefore
by using Lemma 2.3(2) and Lemma 2.4 we have
[TABLE]
Proceeding recursively we have
[TABLE]
Hence the proof of the lemma follows.
∎
5. cohomology module H1 of the relative tangent bundle
In this section, we describe the weights of H1 of the relative tangent bundle. Let n≥a1>a2>…>ak−1>ak=1 be a decreasing sequence of integers such that k≥3. Fix 3≤r≤k.
Lemma 5.1**.**
Let vr=snsar⋯sn−2,vr−1=sar−1⋯sn−1 and vr−2=sar−2⋯sn−1sn. Then we have
(1)
H1(vr−1vrsn−1,αn−1)=0.**
2. (2)
Let w=vr−2vr−1vrsn−1.H1(w,αn−1)μ=0 if and only if μ is of the form
μ=−(βt+αn)* for some ar−1≤t≤ar−2−1. In such a case,*
dim(H1(w,αn−1)μ)=1.**
Proof.
Proof of (1): Note that H0(sn−1,αn−1)=C−αn−1⊕Ch(αn−1)⊕Cαn−1 (see [5, Corollary 2.5]). (5.1.1)
Since ⟨αn−1,αn−1⟩=2, we have H1(sn−1,αn−1)=0. Since ⟨αn−1,αn−2⟩=−1, we have
H1(sn−2,H0(sn−1,αn−1))=0.
Therefore by using SES we have H1(sn−2sn−1,αn−1)=0. (5.1.2)
Since ⟨−αn−1,αn−2⟩=1, by using (5.1.1) we have
H0(sn−2sn−1,αn−1)=Ch(αn−1)⊕C−αn−1⊕C−βn−2.
Since ⟨−αn−1,αn−3⟩=0 and ⟨−βn−2,αn−3⟩=1 we have
H1(sn−3,H0(sn−2sn−1,αn−1))=0. (5.1.3)
Therefore by using SES together with (5.1.2), (5.1.3) we have
H1(sn−3sn−2sn−1,αn−1)=0. (5.1.4)
Proceeding in this way we have H1(sar⋯sn−2sn−1,αn−1)=0 (5.1.5)
and H0(sar⋯sn−2sn−1,αn−1)=Ch(αn−1)⊕j=ar⨁n−1C−βj. (5.1.6)
Since ⟨−βj,αn⟩>0 for all ar≤j≤n−1, by using (5.1.6) we have
H1(sn,H0(sar⋯sn−2sn−1,αn−1))=0. (5.1.7)
Therefore by using SES, (5.1.5) and (5.1.7) together we have
By Lemma 3.1, we have H1(sn−1,H0(snvkvk+1sn−1,αn−1))=0. Therefore by SES, we have H1(sn−1snvkvk+1sn−1,αn−1)=H0(sn−1,H1(snvkvk+1sn−1,αn−1)).
Since ⟨−(βn−1+αn),αn−1⟩=−1 and ⟨−(βi+αn),αn−1⟩=0 for all 1≤i≤n−2, we have H1(sn−1snvkvk+1sn−1,αn−1)=i=1⨁n−2C−(βi+αn).
Proceeding in this way
recurssively, we see that H1(vk−1vkvk+1sn−1,αn−1)=i=1⨁ak−1−1C−(βi+αn). Hence Step 1 follows.
By Lemma 4.2, we have H1(sn,H0(vk−1vkvk+1sn−1,αn−1))=0. (5.5.1)
Let 3≤r≤k. Let Mr:={μ∈X(T):H1(wr,αn−1)μ=0} and M0:={μ∈X(T):H1(u1,αn−1)μ=0}.
Then we have
Corollary 5.8**.**
Mr∩Mr′=∅* whenever r=r′.*
M0∩Mr=∅* for every 1≤r≤k.*
Proof.
Proof of (1) and (2) follow from Lemma 5.2 and Lemma 5.5.
∎
Lemma 5.9**.**
Let c=s1s2⋯sn. Let Tr=cr−1s1s2⋯sn−1, for all 2≤r≤n. Then, Tr(αj)<0 for all n+1−r≤j≤n−1.
Proof.
Assume r=2. Then we see that Tr(αn−1)=−α0.
We assume that for 2<l<n, we have Tl(αj)<0 for n+1−l≤j≤n−1.
Note that Tl+1=Tlsns1⋯sn−1. Then for all n−l≤i≤n−3, we have s1⋯sn−1(αi)=αi+1. Since n−(i+1)≥2 and n+1−l≤i+1≤n−1, we have Tl+1(αi)=Tl(αi+1)<0 (by assumption). Since
s1⋯sn−1(αn−2)=αn−1, we have Tl+1(αn−2)=Tl(sn(−αn−1)).
Since sn(αn−1)=αn−1+2αn, we have Tlsn(αn−1)=Tl(αn−1+2αn). As s1⋯sn−1(αn−1+2αn)=β1+2αn, we have
Tl(αn−1+2αn)=Tl−1sn(β1+2αn).
Since sns1⋯sn−1sn(β1+2αn)=−α1, we have Tl−1sn(β1+2αn)=Tl−2(−α1).
Since sns1⋯sn−1(−α1)=−α2, we have Tl−2(−α1)=Tl−3(−α2). Therefore by recursively we have
Tl−3(−α2)=−αl−1. Hence Tl+1(αn−2)=−αl−1<0.
Also it is clear that Tl+1(αn−1)<0.
Therefore we have
Tl+1(αj)<0 for all n−l≤j≤n−1. Hence the result follows.
∎
Lemma 5.10**.**
Let u,v∈W, let v:=(j=1∏nsj)l−1s1s2⋯sn−1 for some positive integer l≤n, such that l(uv)=l(u)+l(v). Let w=uv. If l≥3, then Hi(w,αn−1)=0 for all i≥0.
Proof.
We note that by SES, we have
[TABLE]
We show that H0(v,αn−1)=0.
By Lemma 5.9 we have for each 1≤r≤n−1,cr−1s1⋯sn−1(αj)<0 for all n+1−r≤j≤n−1. In particular, we have l(vsn−2)=l(v)−1.
We will prove by induction l(u). If l(u)=0 then it follows trivially. Next suppose that l(u)>1. Then there exists a simple root γ such that l(sγu)=l(u)−1. By using SES and (5.10.1) we have H0(sγuv,αn−1)=0. Again, by induction hypotheses H1(sγuv,αn−1)=H0(sγu,H1(v,αn−1)).
Therefore by SES, we have
[TABLE]
H1(v,αn−1)=0, follows from the fact that l(vsn−2)=l(v)−1 and Lemma 2.2(4). Thus, we have H1(w,αn−1)=0. Therefore by [14, Corollary 6.4, p.780] we have Hi(w,αn−1)=0 for all i≥0.
∎
6. cohomology modules of the tangent bundle of Z(w,i)
Let w∈W and let w=si1si2⋯sir be a reduced expression for w and let i=(i1,i2,…,ir). Let τ=si1si2⋯sir−1 and i′=(i1,i2,…,ir−1).
Recall the following long exact sequence of B-modules from [5] (see [5, Proposition 3.1, p.673]):
[TABLE]
[TABLE]
[TABLE]
By [14, Corollary 6.4, p.780], we have Hj(w,αir)=0 for every j≥2. Thus we have the following exact sequence of B-modules:
[TABLE]
[TABLE]
Now onwards we call this exact sequence by LES.
Let w0=sj1sj2⋯sjN be a reduced expression of w0 such that i=(j1,j2,…,jr) and let and j=(j1,j2,…,jN).
Let J=S∖{αn−1}. Let v∈WJ and u∈W be such that l(uv)=l(u)+l(v). Let u=si1⋯sir and v=sir+1⋯sit be reduced expressions of u and v respectively. Let i=(i1,i2,…,ir) and j=(i1,i2,…,ir,ir+1,…,it).
Then we have
(1)
The natural homomorphism
[TABLE]
of B-modules is surjective.
2. (2)
The natural homomorphism
[TABLE]
of B-modules is an isomorphism.
Proof.
Let r+1≤l≤t. Let vl=usir+1⋯sil and il=(i,ir+1,…,il).
Proof of (1): By Lemma 3.3 we have H1(vt,αit)=0. Therefore, using LES the natural homomorphism
By Lemma 3.3, H1(uv,αit)=0. Therefore, by the above exact sequence we have H1(Z(uv,j),T(uv,j))⟶H1(Z(vt−1,it−1),T(vt−1,it−1)) is an isomorphism. Hence we conclude that the homomorphism
H1(Z(uv,j),T(uv,j))⟶H1(Z(u,i),T(u,i))
of B-modules is an isomorphism.
∎
Recall that by Lemma 2.6(1) and Lemma 2.8(2) we have
is a reduced expression for w0. Let i be the tuple corresponding to this reduced of w0. Let u1=wksn[ak,n−1] and i1 be the tuple corresponding to the reduced expression l1=1∏k[al1,n]([ak,n−1]). Note that ak=1. With this notation, we have
Lemma 6.3**.**
(1)
The natural homomorphism
[TABLE]
of B-modules is an isomorphism.
2. (2)
The natural homomorphism
[TABLE]
of B-modules is an isomorphism.
Proof.
For 1≤n−k, let uj=wksn[ak,n]j−1s1s2⋯sn−1 and ij be the tuple corresponding to the reduced expression uj=(l1=1∏k[al1,n])([ak,n])j−1[ak,n−1] (see Lemma 2.8(2)). Note that
w0=un−ksn(l2=1∏k−1[ak,al2−1]).
Case 1: a1=n. In this case, we have sn(l2=1∏k−1[ak,al2−1])∈WJ, where J=S∖{αn−1}.
If j≥2, then by Lemma 5.10, we have H1(uj,αn−1)=0. Let uj′=ujsn−1 and let i′j be the partial subsequence of uj′ such that ij=(i′j,n−1). Hence by LES , we observe that the natural homomorphism
[TABLE]
is surjective and
[TABLE]
is an isomorphism.
Therefore by Lemma 6.2 we have the natural homomorphism
[TABLE]
is surjective and
[TABLE]
is an isomorphism. Therefore, by combining (6.3.3), (6.3.5) together and (6.3.4), (6.3.6) together we see that the homomorphism
[TABLE]
is surjective and
[TABLE]
is an isomorphism.
Proceeding recursively we get that the homomorphism
[TABLE]
of B-modules is surjective and homomorphism
[TABLE]
of B-modules is an isomorphism.
Further since u1−1(α0)<0, by [5, Lemma 6.2, p.667], we have H0(Z(u1,i1),T(u1,i1))−α0=0. By [5, Theorem 7.1], H0(Z(w0,i),T(w0,i)) is a parabolic subalgebra of g and hence there is a unique B-stable line in H0(Z(w0,i),T(w0,i)), namely g−α0. Thereore we conclude that the natural homomorphism
If j≥2, then by Lemma 5.10, we have H1(uj,αn−1)=0. Hence by LES, for each 2≤j≤n+1−k, we observe that the natural homomorphism
[TABLE]
is surjective and
[TABLE]
is an isomorphism. Therefore, the homomorphism
[TABLE]
of B-modules is surjective and the homomorphism
[TABLE]
of B-modules is an isomorphism.
Further since u1−1(α0)<0, by [5, Lemma 6.2, p.667], we have H0(Z(u1,i1),T(u1,i1))−α0=0. By [5, Theorem 7.1], H0(Z(w0,i),T(w0,i)) is a parabolic subalgebra of g and hence there is a unique B-stable line in H0(Z(w0,i),T(w0,i)), namely g−α0. Thereore we conclude that the natural homomorphism
[TABLE]
of B-modules is an isomorphism.
∎
The following is a useful Corollary.
Corollary 6.4**.**
If μ∈X(T)∖{0}, then, we have
[TABLE]
Proof.
By [5, Theorem 7.1], H0(Z(w0,i),T(w0,i)) is a parabolic subalgebra of g. By Lemma 6.3(1) we have
[TABLE]
(as B-modules).
Hence for any μ∈X(T)∖{0}, we have
[TABLE]
∎
Let u1′=wksn[ak,n−2] and let i′1 be the tuple corresponding to the reduced expression l1=1∏k[al1,n][ak,n−2]. Let 3≤r≤k, and let jr=(a1,…,n;a2,…,n;…;ar−1,…,n;ar,…,n−1) and j′r=(a1,…,n;a2,…,n;ar,…,n−2).
We now prove
Lemma 6.5**.**
Let μ∈X(T)∖{0}.
(1)
If H1(u1,αn−1)μ=0, then dim(H0(Z(u1′,i′1),T(u1′,i′1))μ)≤1.**
2. (2)
If H1(u1,αn−1)μ=0, then dim(H0(Z(u1′,i′1),T(u1′,i′1))μ)=2,* and the natural homomorphism*
[TABLE]
is surjective.
Proof.
By LES, we have the following long exact sequence of B-modules:
[TABLE]
[TABLE]
Proof of (1): Assume that H1(u1,αn−1)μ=0 and μ∈X(T)∖{0}, then by the above exact sequence the natural homoorphism (of T-modules) H0(Z(u1,i1),T(u1′,i1))μ⟶H0(Z(u1′,i′1),T(u1′,i′1))μ is surjective. By Corollary 6.4, we have dim(H0(Z(u1,i1),T(u1,i1))μ)≤1.
Proof of (2): Assume that H1(u1,αn−1)μ=0. Then by Lemma 5.5, μ is of the form −(βj+αn) for some 1≤j≤ak−1−1 and dim(H1(u1,αn−1)μ)=1. Hence by using Corollary 6.4, we see that if H1(u1,αn−1)μ=0, then
is surjective. Recall that ut′=uksn−1. By LES we have the following long exact sequence of B-modules:
0⟶H0(wk,αn−1)⟶H0(Z(wk,jk),T(wk,jk))⟶
H0(Z(uk′,i′k),T(uk′,i′k))⟶H1(wk,αn−1)⟶⋯
Since H1(u1,αn−1)μ=0, by Corollary 5.8 we have H1(wk,αn−1)μ=0.
Therefore we have an exact sequence
0⟶H0(wk,αn−1)μ⟶H0(Z(wk,jk),T(wk,jk))μ⟶
H0(Z(τk,j′k),T(τk,j′k))μ⟶0.
Let σk=wk−1sns1⋯sn+1−k and j∗k=(jk−1,n,1,2,…,n+1−k) be the tuple corresponding to this reduced expression of σk. Then by using Lemma 6.2, the natural homomorphism
Since H1(u1,αn−1)μ=0, by Corollary 5.6 we have H0(σk,αn+1−k)μ=0. Therefore, H0(Z(σk,j∗k),T(σk,j∗k))μ=0. Thus from (6.5.6) we have H0(Z(wk,jk),T(wk,jk))μ=0. Then we have an exact sequence of T-modules
Since H1(u1,αn−1)μ=0, by Corollary 5.7 we have H0(wksn,αn)μ=0.
Therefore dim(H0(Z(wksn,j′′k),T(wksn,j′′k))μ)≥2. On the other hand by (6.5.3) we have dim(H0(Z(wksn,j′′k),T(wksn,j′′k))μ)≤2. Hence we have dim(H0(Z(wksn,j′′k),T(wksn,j′′k))μ)=2.
is surjective. Therefore, we have dim(H0(Z(τr,j′r),T(τr,j′r))μ)≤1.
Proof of (2): Assume that H1(wr,αn−1)μ=0. Then by Corollary 5.8, we have H1(wm,αn−1)μ=0 for all r+1≤m≤k and H1(u1,αn−1)μ=0. Then by (1), we have
is surjective. Therefore, by (6.7.5) and (6.7.6) we have
H0(Z(wr−1,jr−1),T(wr−1,jr−1))μ⟶H0(Z(σr−1,j∗r−1),T(σr−1,j∗r−1))μ(6.7.7) is surjective. Since H1(wr,αn−1)μ=0, by Corollary 5.3 we have H0(σr−1,αn+2−r)μ=0.
Therefore, H0(Z(σr−1,j∗r−1),Tσr−1,jr−1∗))μ=0.
Thus from (6.7.7) we have H0(Z(wr−1,jr−1),T(wr−1,jr−1))μ=0. Then we have an exact sequence of T-modules
is surjective, we have dim(H0(Z(τr,j′r),T(τr,j′r))μ)=2.
Therefore by (6.7.1),
H0(Z(τr,j′r),T(τr,j′r))μ⟶H1(wr,αn−1)μ
is surjective.
∎
7. main theorem
In this section we prove the main theorem.
Recall that G=PSO(2n+1,C)(n≥3), and let c be a Coxeter element of W. Then there exists a decreasing sequence
n≥a1>a2>⋯>ak=1 of positive integers such that c=[a1,n][a2,a1−1]⋯[ak,ak−1−1], where [i,j] for i≤j denotes sisi+1⋯sj.
Let i=(i1,i2,…,in) be a sequence corresponding to a reduced expression of w0, where ir (1≤r≤n) is a sequence of reduced expressions of c (see Lemma 2.8).
Theorem 7.1**.**
Hj(Z(w0,i),T(w0,i))=0* for all j≥1 if and only if a2=n−1.*
Proof.
From [5, Proposition 3.1, p. 673], we have Hj(Z(w0,i),T(w0,i))=0 for all j≥2.
It is enough to prove the following:
H1(Z(w0,i),T(w0,i))=0 if and only if c is of the form [a1,n][a2,a1−1]⋯[ak,ak−1−1] with a2=n−1.
Proof of (⟹): If a2=n−1, then a1=n and c=snsn−1v, where v∈WJ and J=S∖{αn−1,αn}. Let u=snsn−1. Then c=uv.
Let j=(n,n−1) be the sequence corresponding to u. Then
using LES, we have:
[TABLE]
[TABLE]
We see that H1(snsn−1,αn−1)=Cαn+αn−1 and H0(sn,αn)αn+αn−1=0. Hence f is non zero homomorphism.
Hence H1(Z(u,j),T(u,j)))=0.
By Lemma 6.1, the natural homomorphism
[TABLE]
is surjective.
Hence we have
[TABLE]
Proof of (⟸): Assume that a2=n−1. Recall that wk=[a1,n]⋯[ak−1,n][ak,n−1],u1=wksn[ak,n−1]. By Lemma 6.3(2), the natural homomorphism
By using LES, we have the following exact sequence of B-modules:
[TABLE]
[TABLE]
By Corollary 6.6, we see that the natural homomorphism h1:H0(Z(u1′,i′1),T(u1′,i′1))⟶H1(u1,αn−1)
is surjective. Therefore, the natural homomorphism
By using LES, we have the following exact sequence of B-modules:
[TABLE]
[TABLE]
By Lemma 6.7(2), we see that the natural map H0(Z(τk−1,j′k−1),T(τk−1,j′k−1))⟶H1(wk−1,αn−1)
is surjective.
Therefore, the natural map
H1(Z(wk−1,jk−1),T(wk−1,jk−1))⟶H1(Z(τk−1,j′k−1),T(τk−1,j′k−1)) (7.1.7)
is an isomorphism.
By using Lemma 6.2(2) and 6.7(2) repeatedly, we see that the natural map
[TABLE]
is an isomorphism for all 3≤r≤k−2.
Therefore by ( (7.1.4), (7.1.5), (7.1.6), (7.1.7), (7.1.8) we have the natural homomorphism
[TABLE]
is an isomorphism.
Again from Lemma 6.2(2), we see that the natural map
[TABLE]
is an isomorphism.
Since a2=n−1, by Lemma 5.2(2) we have H1(w2,αn−1)=0. Note that by Lemma 5.2(1) we have H1(w1,αn−1)=0.
By using Lemma 3.3 and using LES, we have H1(Z(w2,j2),T(w2,j2))=0. Hence we conclude that H1(Z(w0,i),T(w0,i))=0. This completes the proof of the theorem.
∎
Corollary 7.2**.**
Let c be a Coxeter element such that c is of the form [a1,n][a2,a1−1]⋯[ak,ak−1−1] with a2=n−1 and ak=1.
Let (w0,i) be a reduced expression of w0
in terms of c as in Theorem 7.1. Then, Z(w0,i) has no deformations.
Proof.
By Theorem 7.1 and by [5, Proposition 3.1, p.673], we have Hi(Z(w0,i),T(w0,i))=0 for all i>0. Hence, by [12, Proposition 6.2.10, p.272], we see that Z(w0,i) has no deformations.
∎
Remark 7.3*.*
Theorem 7.1 does not hold for PSO(5,C).
Proof.
We take c=s1s2. Here a2=1.
Further, we have w0=c2=s1s2s1s2. Let i=(1,2,1,2). It is easy to see by using SES repeatedly that H1(s1s2s1,α1)=Cα1+α2⊕Cα2. Further, note that H0(Z(s1s2,(1,2)),T(s1s2,(1,2)))α1+α2=0 (see [5, Proposition 6.3(1), p.688]). Hence by using LES we have
Therefore by using Lemma 6.1 we have H1(Z(w0,i),T(w0,i))=0.
∎
Acknowledgements
We thank the referee for the useful comments and suggestions. We thank the
Infosys Foundation for the partial finance support.
Bibliography15
The reference list from the paper itself. Each links out to its DOI / PubMed record.
1[1] V. Balaji, S. Senthamarai Kannan, K.V. Subrahmanyam, Cohomology of line bundles on Schubert varieties-I, Transformation Groups 9 (2004), no.2, 105-131.
2[2] R. Bott and H. Samelson, Applications of the theory of Morse to symmetric spaces, Amer. J. Math. 80 (1958), 964-1029.
3[3] M. Brion, S. Kumar, Frobenius Splitting Methods in Geometry and Representation theory, Progress in Mathematics, Vol. 231 , Birkhäuser, Boston, Inc., Boston, MA, 2005.
4[4] B.Narasimha Chary, S.Senthamarai Kannan, Rigidity of Bott-Samelson-Demazure-Hansen variety for P S p ( 2 n , ℂ ) , 𝑃 𝑆 𝑝 2 𝑛 ℂ P Sp(2n,\mathbb{C}), Journal of Lie Theory 27 (2017), no. 2, 435-468.
5[5] B.Narasimha Chary, S.Senthamarai Kannan, A.J.Parameswaran, Automorphism group of a Bott-Samelson-Demazure-Hansen variety, Transformation Groups 20 (2015), no.3, 665-698.
6[6] M. Demazure, Desingularisation des varieties de schubert generalisees. Ann. Sci. Ecole Norm. Sup 7(1974), 53-88.
7[7] M. Demazure, A very simple proof of Bott’s theorem, Invent. Math. 33 (1976), 271-272.
8[8] H.C. Hansen, On cycles on flag manifolds, Math. Scand. 33 (1973), 269-274.
