Rigidity of Bott-Samelson-Demazure-Hansen variety for PSO(2n+1,C)
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P SO (2 n + 1 , C ) S.SENTHAMARAI KANNAN AND PINAKINATH SAHA
Abstract.
Let G = P SO (2 n +1 , C )( n ≥
3) and B be the Borel subgroup of G containingmaximal torus T of G. Let w be an element of Weyl group W and X ( w ) be the Schubertvariety 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 reducedexpression i of w. In this article, we study the cohomology modules of the tangent bundle on Z ( w , i ) , where w is the longest element of the Weyl group W. We describe all the reduced expres-sions of w in terms of a Coxeter element such that all the higher cohomology modules ofthe tangent bundle on Z ( w , i ) vanish (see Theorem 7.1). 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 = N G ( T ) /T denote the Weyl group of G withrespect to T . We denote by R the set of roots of G with respect to T and by R + a set ofpositive roots. Let B + be the Borel subgroup of G containing T with respect to R + . Let w 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 = n w B + n − w , where n w is a representative of w in N G ( T ). Note that the roots of B is the set R − := − R + of negative roots. We use thenotation β < β ∈ R − . Let S = { α , . . . , α n } 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 simpleroot α is denoted by s α . For simplicity of notation, the simple reflection corresponding toa simple root α i is denoted by s i .For w ∈ W , let X ( w ) := BwB/B denote the Schubert variety in
G/B correspondingto w . Given a reduced expression w = s i s i · · · s i r of w , with the corresponding tuple i := ( i , . . . , i r ), 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 intro-duced 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, wewill 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 reducedexpression i of w . In [5], the automorphism groups of these varieties were studied. There,the following vanishing results of the tangent bundle T Z ( w,i ) on Z ( w, i ) were proved (see [5,Section 3]): (1) H j ( Z ( w, i ) , T Z ( w,i ) ) = 0 for all j ≥ G is simply laced, then H j ( Z ( w, i ) , T Z ( w,i ) ) = 0 for all j ≥ i of w . While computing thefirst cohomology module H ( Z ( w, i ) , T Z ( w,i ) ) for non simply laced group, we observed thatthis 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 cohomologymodule H ( Z ( w, i ) , T Z ( w,i ) ) does vanish ? In [4], a partial answer is given to this questionfor w = w when G = P Sp (2 n, C ). In this article, we give a partial answer to this questionfor w = w when G = P SO (2 n + 1 , C ).Recall that a Coxeter element is an element of the Weyl group having a reduced ex-pression of the form s i s i · · · s i n such that i j = i l whenever j = l (see [11, p.56, Section4.4]). Note that for any Coxeter element c , there is a decreasing sequence of integers n ≥ a > a > . . . > a k = 1 such that c = k Q j =1 [ a j , a j − − a := n + 1,[ i, j ] := s i s i +1 · · · s j for i ≤ j .In this paper we prove the following theorem. Theorem 1.1.
Let G = P SO (2 n + 1 , C ) ( n ≥ and let c ∈ W be a Coxeter ele-ment. Let i = ( i , i , . . . , i n ) be a sequence corresponding to a reduced expression of w ,where i r (1 ≤ r ≤ n ) is a sequence of reduced expressions of c (see Lemma 2.8). Then, H j ( Z ( w , i ) , T ( w ,i ) ) = 0 for all j ≥ if and only if c = k Q j =1 [ a j , a j − − , where a := n + 1 and a = n − . By the above vanishing results, we conclude that if G = P SO (2 n + 1 , C ) ( n ≥
3) and i = ( i , i , . . . , i n ) is a reduced expression of w as above, then the BSDH-variety Z ( w , i )is rigid.The main differences in the proof of main theorem between the case of type C n and thecase of type B n are as follows:When G is of type C n , there is only one long simple root, namely α n . Therefore, by[14, Corollary 5.6, p.778], we have H ( w, α j ) = 0 for any w ∈ W and for any j = n. But, if G is of type B n , α , α , . . . , α n − are all long simple roots. So, we can not apply[14, Corollary 5.6, p.778]. Hence, we need to study the cohomology modules H ( w, α j ) ( w ∈ W , j = n − H ( w, α j ) = 0 for any w ∈ W and for any j = n − s s · · · s n ) r − s s · · · s n − ( α j ) < ≤ r ≤ n and n + 1 − r ≤ j ≤ n − G = P SO (2 n + 1 , C ) ( n ≥
3) in the later sections 3, 4, 5, 6 and 7. In Section 3,we prove H ( w, α j ) = 0 for j = n − w ∈ W. In Section 4 (respectively, Section 5)
IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 3 we compute the weight spaces of H (respectively, H ) of the relative tangent bundle ofBSDH-varieties associated to some elements of the Weyl group. In Section 6, we provesome results on cohomology modules of the tangent bundle of BSDH varieties. In Section7, 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 = N G ( T ) /T denote the Weyl group of G with respect to T and we denote the setof 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 = n B + n − ,where n is a representative in N G ( T ) of the longest element w of W . Let R + ⊂ R bethe set of positive roots of G with respect to the Borel subgroup B + . Note that the set ofroots of B is equal to the set R − := − R + of negative roots.Let S = { α , . . . , α n } denote the set of simple roots in R + . For β ∈ R + , we also use thenotation β >
0. The simple reflection in W corresponding to α i is denoted by s α i . 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 = Hom R ( h R , R ),the dual of the real form of h . The positive definite W -invariant form on Hom R ( h R , R )induced by the Killing form of g is denoted by ( , ). We use the notation h , i to denote h µ, α i = µ,α )( α,α ) , for every µ ∈ X ( T ) ⊗ R and α ∈ R . We denote by X ( T ) + the set ofdominant characters of T with respect to B + . Let ρ denote the half sum of all positiveroots of G with respect to T and B + . For any simple root α , we denote the fundamentalweight corresponding to α by ω α . For 1 ≤ i ≤ n, let h ( α i ) ∈ h be the fundamental coweightcorresponding to α i . That is ; α i ( h ( α j )) = δ ij , where δ ij is Kronecker delta.For a simple root α ∈ S, let n α ∈ N G ( T ) be a representative of s α . We denote theminimal parabolic subgroup of G containing B and n α by P α . We recall that the BSDH-variety corresponds to a reduced expression i of w = s i s i · · · s i r defined by Z ( w, i ) = P α × P α × · · · × P α ir B × B × · · · × B , where the action of B × B × · · · × B on P α i × P α i × · · · × P α ir is given by( p , p , . . . , p r )( b , b . . . , b r ) = ( p · b , b − · p · b , . . . , b − r − · p r · b r ) , p j ∈ P α ij , b j ∈ B and i = ( i , i , . . . , i r ) (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 φ w , the natural birational surjective morphism from Z ( w, i ) to X ( w ) . Let f r : Z ( w, i ) −→ Z ( ws i r , i ′ ) denote the map induced by the projection P α i × P α i × · · · × P α ir −→ P α i × P α i × · · · × P α ir − , where i ′ = ( i , i , . . . , i r − ) . Thenwe observe that f r is a P α ir /B ≃ P -fibration.For a B -module V, let L ( w, V ) denote the restriction of the associated homogeneousvector bundle on G/B to X ( w ) . By abuse of notation, we denote the pull back of L ( w, V ) S.SENTHAMARAI KANNAN AND PINAKINATH SAHA via φ w 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 bundlefrom X ( w ) , we conclude that the cohomology modules H j ( Z ( w, i ) , L ( w, V )) ≃ H j ( X ( w ) , L ( w, V ))for all j ≥ i. Hence we denote H j ( Z ( w, i ) , L ( w, V )) by H j ( w, V ) . In particular, if λ is character of B, then we denote the cohomology modules H j ( Z ( w, i ) , L λ ) by H j ( w, λ ) . We recall the following short exact sequence of B -modules from [5], we call it SES :(1) H ( w, V ) ≃ H ( s γ , H ( s γ w, V )) . (2) 0 → H ( s γ , H ( s γ w, V )) → H ( w, V ) → H ( s γ , H ( s γ w, V )) → . Let α be a simple root and λ ∈ X ( T ) be such that h λ, α i ≥ . Let C λ denote onedimensional 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 h λ, α i ≥ . Let ev : H ( s α , λ ) −→ C λ be the evaluation map. Then we have (1) If h λ, α i = 0 , then H ( s α , λ ) ≃ C λ . (2) If h λ, α i ≥ , then C s α ( λ ) ֒ → H ( s α , λ ) , and there is a short exact sequence of B -modules: → H ( s α , λ − α ) −→ H ( s α , λ ) / C s α ( λ ) −→ C λ → . Further more, H ( s α , λ − α ) = 0 when h λ, α i = 1 . (3) Let n = h λ, α i . As a B -module, H ( s α , λ ) has a composition series ⊆ V n ⊆ V n − ⊆ · · · ⊆ V = H ( s α , λ ) such that V i /V i +1 ≃ C λ − iα for i = 0 , , . . . , n − and V n = C s α ( λ ) . We define the dot action by w · λ = w ( λ + ρ ) − ρ, where ρ is the half sum of positiveroots. 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 = ws α . Lemma 2.2. If l ( w ) = l ( v ) + 1 , then we have (1) If h λ, α i ≥ , then H j ( w, λ ) = H j ( v, H ( s α , λ )) for all j ≥ . (2) If h λ, α i ≥ , then H j ( w, λ ) = H j +1 ( w, s α · λ ) for all j ≥ . (3) If h λ, α i ≤ − , then H j +1 ( w, λ ) = H j ( w, s α · λ ) for all j ≥ . (4) If h λ, α i = − , then H j ( w, λ ) vanishes for every j ≥ . The following consequence of Lemma 2.2 will be used to compute the cohomology mod-ules in this paper. Now onwards we will denote the Levi subgroup of P α ( α ∈ S ) containing T by L α and the subgroup L α ∩ B by B α . Let π : ˜ G −→ G be the universal cover. Let ˜ L α (respectively, ˜ B α ) be the inverse image of L α (respectively, B α ). Lemma 2.3.
Let V be an irreducible L α -module. Let λ be a character of B α . Then wehave IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 5 (1) As L α -modules, H j ( L α /B α , V ⊗ C λ ) ≃ V ⊗ H j ( L α /B α , C λ ) . (2) If h λ, α i ≥ , then H ( L α /B α , V ⊗ C λ ) is isomorphic as an L α -module to the tensorproduct of V and H ( L α /B α , C λ ) . Further, we have H j ( L α /B α , V ⊗ C λ ) = 0 forevery j ≥ . (3) If h λ, α i ≤ − , then H ( L α /B α , V ⊗ C λ ) = 0 , and H ( L α /B α , V ⊗ C λ ) is isomorphicto the tensor product of V and H ( L α /B α , C s α · λ ) . (4) If h λ, α i = − , then H j ( L α /B α , V ⊗ C λ ) = 0 for every j ≥ .Proof. Proof (1). By [13, Proposition 4.8, p.53, I] and [13, Proposition 5.12, p.77, I], forall j ≥
0, we have the following isomorphism of L α -modules: H j ( L α /B α , V ⊗ C λ ) ≃ V ⊗ H j ( L α /B α , C λ ) . Proof of (2), (3) and (4) follows from Lemma 2.2 by taking w = s α and the fact that L α /B α ≃ P α /B . (cid:3) Recall the structure of indecomposable modules over B α and e B α (see [1, Corollary 9.1,p.130]). Lemma 2.4. (1)
Any finite dimensional indecomposable e B α -module V is isomorphicto V ′ ⊗ C λ for some irreducible representation V ′ of e L α and for some character λ of e B α . (2) Any finite dimensional indecomposable B α -module V is isomorphic to V ′ ⊗ C λ forsome irreducible representation V ′ of e L α and for some character λ of e B α . Now onwards we will assume that G = P SO (2 n + 1 , C )( n ≥ . Note that longestelement w of the Weyl group W of G is equal to − id. We recall the following Propositionfrom [15, Proposition 1.3, p.858].
Proposition 2.5.
Let c ∈ W be a Coxeter element, let ω i be the fundamental weightcorresponding to the simple root α i . Then there exists a least positive integer h ( i, c ) suchthat c h ( i,c ) ( ω i ) = w ( ω i ) . Lemma 2.6.
Let c ∈ W be a Coxeter element. Then we have (1) w = c n . (2) For any sequence i r (1 ≤ r ≤ n ) of reduced expressions of c ; the sequence i =( i , i , . . . , i r ) is a reduced expression of w . Proof.
Note that for n ≥ , there is an isomorphism of Weyl group of B n and Weyl groupof C n sending s i s i for (1 ≤ i ≤ n ) . Proof of the lemma holds in the case of type C n for( n ≥
3) (see [4, Lemma 4.2, p.441]). Therefore lemma holds for type B n ( n ≥ . (cid:3) Lemma 2.7.
Let n ≥ a > a > · · · > a r − > a r ≥ be a decreasing sequence of integers.Then, w = ( n Q j = a s j )( n Q j = a s j ) · · · ( n Q j = a r − s j )( n − Q j = a r s j ) is a reduced expression of w. S.SENTHAMARAI KANNAN AND PINAKINATH SAHA
Proof.
Note that for n ≥ , there is an isomorphism of Weyl group of B n and Weyl groupof C n sending s i s i for (1 ≤ i ≤ n ) . Proof of the lemma holds in the case of type C n for( n ≥
3) (see [4, Lemma 4.3,p.441]). Therefore lemma holds for type B n ( n ≥ . (cid:3) Let c be a Coxeter element in W. We take a reduced expression c = [ a , n ][ a , a − · · · [ a k , a k − − , where [ i, j ] = s i s i +1 · · · s j for i ≤ j and n ≥ a >a > · · · > a k = 1 . Then we have following.
Lemma 2.8. (1)
For all ≤ i ≤ k − ,c i = ( i Q l =1 [ a l , n ])( k Q l = i +1 [ a l , a l − i − i − Q l =1 [ a l k , a k − i + l − . (2) For all k ≤ j ≤ n,c j = ( k − Q l =1 [ a l , n ])([ a k , n ] j +1 − k )( k − Q l =1 [ a k , a l − . (3) The expressions of c i for ≤ i ≤ n as in (1) and (2) are reduced.Proof. Note that for n ≥ , there is an isomorphism of Weyl group of B n and Weyl groupof C n sending s i s i for (1 ≤ i ≤ n ) . Proof of the lemma holds in the case of type C n for( n ≥
3) (see [4, Lemma 4.4, p.442]). Therefore lemma holds for type B n ( n ≥ . (cid:3) cohomology modules H ( w, α j ) where j = n − and w ∈ W In this section, we prove that H ( w, α j ) = 0 for every w ∈ W and j = n − . Lemma 3.1.
Let v ∈ W and α ∈ S . Then H ( s j , H ( v, α )) = 0 for j = n. Proof.
By [14, Corollary 5.6, p.778] we have H ( w, α n ) = 0 . Therefore, we may assume that α is a long simple root. If H ( s j , H ( v, α )) µ = 0 , then there exists an indecomposable ˜ L α j -summand V of H ( v, α ) such that H ( s j , V ) µ = 0 . By Lemma 2.4, we have V ≃ V ′ ⊗ C λ for some character λ of ˜ B α j and for some irreducible ˜ L α j -module V ′ . Since H ( s j , V ) µ = 0from Lemma 2.3(3) we have h λ, α j i ≤ −
2. Since α is a long root, there exists w ∈ W such that w ( α ) = α . Thus H ( v, α ) ⊆ H ( vw, α ) . Again, since α is highest long root, H ( w , α ) = g −→ H ( vw, α ) is surjective. Let µ ′ be the lowest weight of V. Then bythe above argument µ ′ is a root. Therefore we have µ ′ = µ + λ, where µ is the lowestweight of V ′ . Hence, we have h µ ′ , α j i ≤ −
2. Since α j is a long root and µ ′ is a root, wehave h µ ′ , α j i = − , , . This is a contradiction. Thus we have H ( s j , H ( v, α )) µ = 0 . (cid:3) Lemma 3.2.
Let v ∈ W and α j ∈ S be such that j = n − . Then we have H ( s k , H ( v, α j )) = 0 , for every k = 1 , , . . . , n. Proof.
Step 1: H ( v, α j ) − ( α n − +2 α n ) = 0 . Case 1:
Assume that j = n, choose an element u ∈ W of minimal length such that u − ( α n ) = β , the highest short root. Then we have H ( v, α j ) ⊆ H ( vu, β ) . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 7 Since β is dominant weight the natural restriction map H ( w , β ) −→ H ( vu, β )is surjective.Hence H ( v, α j ) µ = 0 implies either µ = 0 or µ is a short root.Therefore, we have H ( v, α j ) − ( α n − +2 α n ) = 0 . Case 2:
Assume that 1 ≤ j ≤ n − . Note that if H ( v, α j ) µ = 0 then either µ = α j , µ ≤ − α j (see [4, Corollary 4.5, p.678]).Hence H ( v, α j ) − ( α n − +2 α n ) = 0 . Step 2: If H ( v, α j ) − ( β i +2 α n ) = 0 for some 1 ≤ i ≤ n − , then H ( v, α j ) − ( β i + α n ) = 0 . Proof of Step 2: If v = id, we are done. So choose 1 ≤ t ≤ n such that l ( s t v ) = l ( v ) − . Let v ′ = s t v. Then H ( v, α j ) = H ( s t , H ( v ′ , α j )) . Case 1:
Assume that t = n. In this case h− ( β i + 2 α n ) , α t i = − . If H ( v, α j ) − ( β i +2 α n ) =0, then there is an indecomposable B α n -summand V of H ( v ′ , α j ) with highest weight − β i . Since h− β i , α n i = 2 , we have H ( s t , V ) − ( β i + α n ) = 0 . Therefore we have H ( v, α j ) − ( β i + α n ) =0 . Case 2:
Assume that t = n − . In this case h− ( β i + 2 α n ) , α t i = 1 . If H ( v, α j ) − ( β i +2 α n ) =0, then there is an indecomposable B α n -summand V of H ( v ′ , α j ) with highest weight − ( β i + 2 α n ) . Thus by induction hypothesis we have H ( v ′ , α j ) − ( β i + α n ) = 0 . Since h− ( β i + α n ) , α t i = 0 , we have H ( v, α j ) − ( β i + α n ) = 0 . Case 3:
Assume that 1 ≤ t ≤ n − . In this case h− ( β i + 2 α n ) , α t i = − , . Assume that i = t. Then we have h− ( β i + 2 α n ) , α t i = − . If further H ( v, α j ) − ( β i +2 α n ) =0, then there is an indecomposable B α t -summand V of H ( v ′ , α j ) with highest weight − ( β i +1 + 2 α n ) . Therefore we have H ( v ′ , α j ) − ( β i +1 +2 α n ) = 0 . It is clear from Step: 1 that t + 1 ≤ n −
2. Therefore by induction H ( v ′ , α j ) − ( β t +1 + α n ) = 0 . Since h− ( β t +1 + α n ) , α t i = 1 , we have H ( v, α j ) − ( β t + α n ) = 0 . Assume that 1 ≤ t ≤ i − i + 1 ≤ t ≤ n − . Then we have h− ( β i + 2 α n ) , α t i = 0 . Thus H ( v, α j ) − ( β t +2 α n ) = H ( v ′ , α j ) − ( β t +2 α n ) = 0 . Therefore by induction H ( v, α j ) − ( β t + α n ) = 0 . Since h− ( β i + α n ) , α t i = 0 , we have H ( v, α j ) − ( β i +2 α n ) = 0 . Assume that i = t + 1 . Since h− ( β i + α n ) , α t i = 1 , then there is an indecomposable B α t -summand V of H ( v ′ , α j ) such that: V = C − ( β i +2 α n ) ⊕ C − ( β i − +2 α n ) or V = C − ( β i +2 α n ) . Then we have H ( v ′ , α j ) − ( β t +2 α n ) = 0 . Therefore by induction H ( v ′ , α j ) − ( β i + α n ) = 0 . Since h− ( β t + α n ) , α t i = 1, we have H ( v, α j ) − ( β i + α n ) = 0 Hence the proof of Step 2. Proof of Lemma :Case 1:
Assume that k = n. Then by Lemma 3.1 we have H ( s k , H ( v, α j )) = 0 . Case 2:
Assume that k = n. By Step 1 we see that H ( v, α j ) − ( α n − +2 α n ) = 0 . Note thatif β is a root such that H ( v, α j ) β = 0 and h β, α n i = − , then we have β = − ( β i + 2 α n )for some 1 ≤ i ≤ n − . S.SENTHAMARAI KANNAN AND PINAKINATH SAHA
By Step 2 if H ( v, α j ) − ( β i +2 α n ) = 0 for some 1 ≤ i ≤ n −
2, then H ( v, α j ) − ( β i + α n ) = 0 . Therefore C − ( β i + α n ) ⊕ C − ( β i +2 α n ) is an indecomposable B α n -summand of H ( v, α j ) . ByLemma 2.4, C − ( β i + α n ) ⊕ C − ( β i +2 α n ) is isomorphic to V ⊗ C − ω n , where V is an irreducible˜ L α n -module. Therefore by Lemma 2.3(4) we have H ( s k , C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ) = 0 . Thusour result follows. (cid:3)
Lemma 3.3.
Let w be an element of W and α j be an element of S such that j = n − . Then H ( w, α j ) = 0 . Proof.
We will prove by induction on length of w . If length of w is 0, then w = id. Thus itfollows trivially. Now suppose w ∈ W such that l ( w ) ≥ . Then there exists a simple root α ∈ S such that l ( s α w ) = l ( w ) − . Then using SES:0 −→ H ( s α , H ( s α w, α j )) −→ H ( w, α j ) −→ H ( s α , H ( s α w, α j )) −→ . From the above SES using induction hypothesis and Lemma 3.2, we get H ( w, α j )) = 0for j = n − . (cid:3) cohomology module H of the relative tangent bundle In this section we describe the weights of H of the relative tangent bundle.Notation:Let c be a Coxeter element of W. We take a reduced expression c = [ a , n ][ a , a − · · · [ a k , a k − − , where [ i, j ] = s i s i +1 · · · s j for i ≤ j and n ≥ a > a > · · · > a k = 1 . Let β i = α i + α i +1 + · · · + α n − for all 1 ≤ i ≤ n − . For 1 ≤ r ≤ k, let n ≥ a > a > a > · · · > a r ≥ w r = ( n Q j = a s j )( n Q j = a s j )( n Q j = a s j ) · · · ( n Q j = a r − s j )( n − Q j = a r s j )and let τ r = ( n Q j = a s j )( n Q j = a s j )( n Q j = a s j ) · · · ( n Q j = a r − s j )( n − Q j = a r s j ) . Note that l ( w r ) = l ( τ r ) + 1 . Lemma 4.1.
Assume that r ≥ . (1) Let v = s a r − s a r − +1 · · · s n s a r s a r +1 · · · s n − . Then we have H ( v, α n − ) = a r − − L i = a r ( C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕· · ·⊕ C − ( β i +2 α n + β ar − ) ) n − L i = a r − ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . (2) Let v ′ = s · · · s n s · · · s n − . Then we have H ( v ′ , α n − ) = n − L i =1 ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕· · ·⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 9 Proof.
Proof of (1): Let u = s a r s a r +1 · · · s n − . By using SES we have H ( u, α n − ) = C h ( α n − ) ⊕ ( n − L j = a r C − β j ) . Since h− β j , α n i = 2 for all a r ≤ j ≤ n − , by using SES we have H ( s n u, α n − ) = C h ( α n − ) ⊕ ( n − L j = a r C − β j ⊕ C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ) . (4.1.1)Since C h ( α n − ) ⊕ C − α n − is an indecomposable B α n − -module (see [5, p.11] and [14, p.8]),by Lemma 2.4, we have C h ( α n − ) ⊕ C − α n − = V ⊗ C − ω n − , where V is the two dimensional irreducible repres-entation of ˜ L α n − . Therefore by Lemma 2.3(4) we have H ( ˜ L α n − / ˜ B α n − , C h ( α n − ) ⊕ C − α n − ) = 0 . Moreover, we have h− ( β n − + α n ) , β n − i = − h− ( β n − + 2 α n ) , β n − i = 0 , h− β j , β n − i = − h− ( β j + α n ) , β n − i = 0 and h− ( β j + 2 α n ) , β n − i = 1 for all a r ≤ j ≤ n − . Therefore we have H ( s n − s n u, α n − ) = ( n − L j = a r C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ) ⊕ C − ( β n − +2 α n ) . (4.1.2) Claim:
For a r − ≤ k ≤ n − H ( s k s k +1 · · · s n u, α n − ) = k − L j = a r ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β k ) ) n − L j = k ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . Proof of the claim: We will prove by descending induction on k. By hypothesis we have H ( s k +1 · · · s n u, α n − ) = k L j = a r ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β k +1 ) ) n − L j = k +1 ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . (4 . . V = k − L j = a r ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β k +1 ) ) and V ′ = n − L j = k +2 ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . Then roots { ( β j + α n ) , ( β j +2 α n ) , ( β j +2 α n + β n − ) , . . . , ( β j +2 α n + β k +2 ) : a r ≤ j ≤ k − } , { ( β j + 2 α n ) , ( β j + 2 α n + β n − ) , . . . , ( β j + 2 α n + β j +1 ) : k + 2 ≤ j ≤ n − } and − ( β n − + 2 α n )are orthogonal to α k . Therefore V , V ′ are direct sums of irreducible ˜ L α k − modules. By Lemma 2.3(2), we have H ( ˜ L α k / ˜ B α k , V ) = V, H ( ˜ L α k / ˜ B α k , V ′ ) = V ′ and H ( ˜ L α k / ˜ B α k , C − ( β n − +2 α n ) ) = C − ( β n − +2 α n ) . Further the remaining roots of (4 . .
3) are {− ( β k + α n ) , − ( β k + 2 α n ) , − ( β k + 2 α n + β n − ) , . . . , − ( β k + 2 α n + β k +2 ) , − ( β k + 2 α n + β k +1 ) } , {− ( β k +1 + 2 α n ) , − ( β k +1 + 2 α n + β n − ) , . . . , − ( β k + 2 α n + β k +2 ) } and {− ( β j + 2 α n + β k +1 ) : a r ≤ j ≤ k } . Since h− ( β k + α n ) , α k i = − , by Lemma 2.3(4) we have H ( ˜ L α k / ˜ B α k , C − ( β k + α n ) ) = 0 . Since C − ( β k +2 α n ) ⊕ C − ( β k +1 +2 α n ) is the irreducible two dimensional ˜ L α k -module, by Lemma2.3(2) we have H ( ˜ L α k / ˜ B α k , C − ( β k +2 α n ) ⊕ C − ( β k +1 +2 α n ) ) = C − ( β k +2 α n ) ⊕ C − ( β k +1 +2 α n ) . Similarly for each k + 2 ≤ j ≤ n − , C − ( β k +2 α n + β j ) ⊕ C − ( β k +1 +2 α n + β j ) is the irreducibletwo dimensional ˜ L α k − module. Therefore by Lemma 2.3(2) we have H ( ˜ L α k / ˜ B α k , C − ( β k +2 α n + β j ) ⊕ C − ( β k +1 +2 α n + β j ) ) = C − ( β k +2 α n + β j ) ⊕ C − ( β k +1 +2 α n + β j ) ) for each k + 2 ≤ j ≤ n − . Moreover, h− ( β j + 2 α n + β k +1 ) , α k i = 1 for all a r ≤ j ≤ k −
1. Therefore by Lemma2.3(2) we have H ( ˜ L α k / ˜ B α k , C − ( β j +2 α n + β k +1 ) ) = C − ( β j +2 α n + β k +1 ) ⊕ C − ( β j +2 α n + β k ) for all a r ≤ j ≤ k − . Since h− ( β k + 2 α n + β k +1 ) , α k i = 0 , by Lemma 2.3(2) we have H ( ˜ L α k / ˜ B α k , C − ( β k +2 α n + β k +1 ) ) = C − ( β k +2 α n + β k +1 ) . From the above discussion, we have H ( s k s k +1 · · · s n u, α n − ) = k − L j = a r ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β k ) ) n − L j = k ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . Therefore the claim follows.The claim implies H ( v, α n − ) = H ( s a r − s a r +1 · · · s n v ′ r , α n − ) = a r − − L j = a r ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β ar − ) ) n − L j = a r − ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . Proof of (2): By (1) we have H ( s s · · · s n s s · · · s n − , α n − ) = ( C − ( β + α n ) ⊕ C − ( β +2 α n ) ⊕ C − ( β +2 α n + β n − ) ⊕ · · · ⊕ C − ( β +2 α n + β ) ) n − L j =2 ( C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕· · ·⊕ C − ( β j +2 α n + β j +1 ) ) ⊕ C − ( β n − +2 α n ) . (4 . . {− ( β + α n ) , − ( β + 2 α n ) , − ( β + 2 α n + β n − ) , . . . , − ( β + 2 α n + β ) , − ( β + 2 α n + β ) } , {− ( β + 2 α n ) , − ( β + 2 α n + β n − ) , . . . , − ( β + 2 α n + β ) } and − ( β + 2 α n + β ) . Since − ( β n − + 2 α n ) is orthogonal to α , by Lemma 2.3(2) we have H ( ˜ L α / ˜ B α , C − ( β n − + α n ) ) = C − ( β n − + α n ) . Since h− ( β + α n ) , α i = − , by Lemma 2.3(4) we have H ( ˜ L α / ˜ B α , C − ( β + α n ) ) = 0 . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 11 Since C − ( β +2 α n ) ⊕ C − ( β +2 α n ) is the irreducible two dimensional ˜ L α -module, by Lemma2.3(2) we have H ( ˜ L α / ˜ B α , C − ( β +2 α n ) ⊕ C − ( β +2 α n ) ) = C − ( β +2 α n ) ⊕ C − ( β +2 α n ) . Similarly for each 3 ≤ j ≤ n − , C − ( β +2 α n + β j ) ⊕ C − ( β +2 α n + β j ) is the irreducible twodimensional ˜ L α − module. Therefore by Lemma 2.3(2) we have H ( ˜ L α / ˜ B α , C − ( β +2 α n + β j ) ⊕ C − ( β +2 α n + β j ) ) = C − ( β +2 α n + β j ) ⊕ C − ( β +2 α n + β j ) ) for each3 ≤ j ≤ n − . Since h− ( β + 2 α n + β ) , α i = 0 , by Lemma 2.3(2) we have H ( ˜ L α / ˜ B α , C − ( β +2 α n + β ) ) = C − ( β +2 α n + β ) . From the above discussion, we have H ( v ′ , α n − ) = n − L i =1 ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕· · ·⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . (cid:3) Lemma 4.2.
Let ≤ r ≤ k and let v = s a r − s a r − +1 · · · s n s a r s a r +1 · · · s n − . Then we have (1) H ( s a r − · · · s n v, α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β ar − ) ) a r − − L i = a r − ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . (2) H ( w r , α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β ar − ) ) a r − − L i = a r − ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . (3) Let u = ( s a · · · s n )( s a · · · s n ) · · · ( s a k − · · · s n ) v ′ , where v ′ is defined as in Lemma4.1. Then we have H ( u , α n − ) = a k − − L i =1 ( C − ( β i +2 α n + β ak − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Proof.
Proof of (1): Since r ≥ , we have a r − < n. By Lemma 4.1(1), we have H ( v, α n − ) = a r − − L i = a r ( C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β ar − ) ) n − L i = a r − ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . Since {− ( β i + 2 α n + β n − ) , . . . , − ( β i + 2 α n + β a r − ) : a r ≤ i ≤ a r − − } are orthogonalto α n , by Lemma 2.3(2), we have H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n + β t ) ) = C − ( β i +2 α n + β t ) for all a r ≤ i ≤ a r − − a r − ≤ t ≤ n − . Since {− ( β i + 2 α n + β n − ) , . . . , − ( β i + 2 α n + β i +1 ) : a r − ≤ i ≤ n − } are orthogonal to α n , by Lemma 2.3(2) we have H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n + β l ) ) = C − ( β i +2 α n + β l ) for all i + 1 ≤ l ≤ n − , where a r − ≤ i ≤ n − . Since h− ( β i + 2 α n ) , α n i = − a r ≤ i ≤ n − , by Lemma 2.3(3) we have H ( ˜ L α n / ˜ B α n , C − ( β n − +2 α n ) ) = 0 for all a r − ≤ i ≤ n − . Moreover, for each a r ≤ i ≤ a r − − , C − ( β i + α n ) ⊕ C − ( β i +2 α n ) is an indecomposabletwo dimensioal B α n -module. Therefore by Lemma 2.4, we have C − ( β i + α n ) ⊕ C − ( β i +2 α n ) = V i ⊗ C − ω n , where V i is the irreducible two dimensional representation of ˜ L α n . By Lemma2.3(4) we have H ( ˜ L α n / ˜ B α n , C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ) = 0 for each a r ≤ i ≤ a r − − . From the above discussion, we have H ( s n v r , α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β ar − ) ) n − L i = a r − ( C − ( β i +2 α n + β n − ) ⊕· · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Since h− ( β i + 2 α n + β n − ) , α n − i = − a r ≤ i ≤ n − , by Lemma 2.3(4) wehave H ( ˜ L α n − / ˜ B α n − , C − ( β i + α n + β n − ) ) = 0 for each a r ≤ i ≤ n − . Moreover, {− ( β i + 2 α n + β n − ) , . . . , − ( β i + 2 α n + β a r − ) : a r ≤ i ≤ a r − − } , {− ( β i +2 α n + β n − ) , . . . , − ( β i + 2 α n + β i +1 ) : a r − ≤ i ≤ n − } , and β n − + 2 α n are orthogonal to α n − . Therefore we have H ( s n − s n v, α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β n − ) ⊕· · ·⊕ C − ( β i +2 α n + β ar − ) ) n − L i = a r − ( C − ( β i +2 α n + β n − ) ⊕· · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Proceeding recursively, we have H ( s a r − s a r − +1 · · · s n v, α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β ar − ) ) L a r − − L i = a r − ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Proof of (2):Since {− ( β i + 2 α n + β a r − − ) , . . . , − ( β i + 2 α n + β a r − ) : a r ≤ i ≤ a r − − } , {− ( β i +2 α n + β a r − − ) , . . . , − ( β i + 2 α n + β i +1 ) : a r − ≤ i ≤ a r − − } are orthogonal to α j for all a r − ≤ j ≤ n, by Lemma 2.3(2) we have H ( s a r − · · · s n s a r − · · · s n v, α n − ) = H ( s a r − · · · s n v, α n − ) . Proceeding recursively we have H ( w r , α n − ) = H ( s a r − · · · s n v, α n − ) = a r − − L i = a r ( C − ( β i +2 α n + β ar − − ) ⊕· · ·⊕ C − ( β i +2 α n + β ar − ) ) a r − − L i = a r − ( C − ( β i +2 α n + β ar − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Proof of (3):By Lemma 4.1(2) we have H ( v ′ , α n − ) = n − L i =1 ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . Since {− ( β i + 2 α n + β n − ) , . . . , − ( β i + 2 α n + β i +1 ) : 1 ≤ i ≤ n − } are orthogonal to α n ,by Lemma 2.3(2) we have H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n + β l ) ) = C − ( β i +2 α n + β l ) for all i + 1 ≤ l ≤ n − , where 1 ≤ i ≤ n − . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 13 Since h− ( β i + 2 α n ) , α n i = − ≤ i ≤ n − , by Lemma 2.3(3) we have H ( ˜ L α n / ˜ B α n , C − ( β n − +2 α n ) ) = 0 for all 1 ≤ i ≤ n − . From the above discussion, we have H ( s n v ′ , α n − ) = n − L i =1 ( C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Since h− ( β i + 2 α n + β n − ) , α n − i = − ≤ i ≤ n − , by Lemma 2.3(4) we have H ( ˜ L α n − / ˜ B α n − , C − ( β i +2 α n + β n − ) ) = 0 for each 1 ≤ i ≤ n − . Moreover, {− ( β i + 2 α n + β n − ) , . . . , − ( β i + 2 α n + β i +1 ) : 1 ≤ i ≤ n − } , and β n − + 2 α n are orthogonal to α n − . Therefore we have H ( s n − s n v ′ , α n − ) = n − L i =1 ( C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Proceeding recursively we have H ( s a k − s a k − +1 · · · s n v ′ , α n − ) = a k − − L i =1 ( C − ( β i +2 α n + β ak − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . Since {− ( β i + 2 α n + β a k − − ) , . . . , − ( β i + 2 α n + β i +1 ) : 1 ≤ i ≤ a k − − } are orthogonalto α j for all a k − ≤ j ≤ n , we have H ( s a k − · · · s n s a k − · · · s n v ′ , α n − ) = H ( s a k − · · · s n v ′ , α n − ) . Proceeding recursively we have H ( u , α n − ) = a k − − L i =1 ( C − ( β i +2 α n + β ak − − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) . (cid:3) Lemma 4.3.
Let ≤ r ≤ k. Then H ( w r − s n s a r − s a r − +1 · · · s n +2 − r , α n +2 − r ) µ = 0 if µ isof the form µ = − ( β j + α n ) for some a r − ≤ j ≤ a r − − . Proof.
By applying SES repeatedly, it is easy to see that H ( s n s a r − s a r − +1 · · · s n +2 − r , α n +2 − r ) = C h ( α n +2 − r ) ⊕ ( n +2 − r L j = a r − C − γ j,n +2 − r ) , where γ j,j ′ =( α j + · · · + α j ′ ) for j ′ ≥ j .Let V = H ( s n s a r − s a r − +1 · · · s n +2 − r , α n +2 − r ) . We next calculate H ( s a r − · · · s n − , V ) . Since n ≥ a > a > · · · > a k = 1 , we have a i ≤ n + 1 − i for all 1 ≤ i ≤ k. Assume l ≥ n + 4 − r, then h γ j,n +2 − r , α l i = 0 for all a r − ≤ j ≤ n + 2 − r. By Lemma 2.3(2) wehave H ( ˜ L α l / ˜ B α l , V ) = V for all l ≥ n + 4 − r. Therefore we have H ( s a r − · · · s n − , V ) = H ( s a r − · · · s n +2 − r s n +3 − r , V ) . Note that, since h− γ j,n +2 − r , α n +3 − r i = 1 for all a r − ≤ j ≤ n + 2 − r, by Lemma 2.3(2)we have H ( s n +3 − r , V ) = C h ( α n +2 − r ) ⊕ ( n +2 − r L j = a r − ( C − γ j,n +2 − r ⊕ C − γ j,n +3 − r )) . Since C h ( α n +2 − r ) ⊕ C − γ n +2 − r,n +2 − r is an indecomposable two dimensional B α n +2 − r -module,by Lemma 2.4, C h ( α n +2 − r ) ⊕ C − γ n +2 − r,n +2 − r = V ⊗ C − ω n +2 − r , where V is the irreducibletwo dimensional ˜ L α n +2 − r -module.Since h γ j,n +2 − r , α n +2 − r i = − , for all a r − ≤ j ≤ n + 1 − r, by Lemma 2.3(4) we have H ( ˜ L α n +2 − r / ˜ B α n +2 − r , C γ j,n +2 − r ) = 0 for all a r − ≤ j ≤ n + 1 − r. From the above discussion, we have H ( s n +2 − r , H ( s n +3 − r , V )) = n +1 − r L j = a r − C − γ j,n +3 − r . Since h γ n +1 − r,n +3 − r , α n +1 − r i = − , by Lemma 2.3(4), we have H ( ˜ L α n +1 − r / ˜ B α n +1 − r , C γ n +1 − r,n +3 − r ) = 0 . Moreover, h γ j,n +3 − r , α n +1 − r i = 0 for all a r − ≤ j ≤ n − r. Therefore we have H ( s n +1 − r , H ( s n +2 − r , H ( s n +3 − r , V ))) = n − r L j = a r − C − γ j,n +3 − r . Proceeding recursively we have H ( s a r − · · · s n − , V ) = a r − − L j = a r − C − γ j,n +3 − r and H ( s n s a r − · · · s n − , V ) = a r − − L j = a r − C − γ j,n +3 − r . Let V = H ( s n s a r − · · · s n − , V ) . Similarly, we have H ( s a r − · · · s n − , V ) = a r − − L j = a r − C − γ j,n +4 − r and H ( s n s a r − · · · s n − , V ) = a r − − L j = a r − C − γ j,n +4 − r . Proceeding recursively we have V r − = H ( s a · · · s n − , V r − ) = a r − − L j = a r − C − γ j,n − ,where V r − = H (( s n s · · · s n ) · · · ( s a r − · · · s n +2 − r ) , α n +2 − r ) . Note that γ j,n − = β j . Since h− α n − , α n i = 2 , by Lemma 2.3(2) and Lemma 2.4, wehave H ( s n , V r − ) = a r − − M j = a r − ( C − β j ⊕ C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ) . Moreover, h− β j , α n − i = − h− ( β j + α n ) , α n − i = 0 and h− ( β j + 2 α n ) , α n − i = 1 for all a r − ≤ j ≤ a r − − H ( s n − , H ( s n , V r − )) = a r − − M j = a r − ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ) . Proceeding recursively we have H ( s a , H ( s a +1 · · · s n , V r − )) = a r − − L j = a r − ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β a ) ) . Hence the proof of the lemma follows. (cid:3)
Lemma 4.4.
Let ≤ r ≤ k. Then H ( w r − s n , α n ) µ = 0 if and only if µ is of the form µ = − ( β j + α n ) , for some a r − ≤ j ≤ a r − − . Proof.
By applying SES repeatedly, it is easy to see that
IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 15 H ( s a r − · · · s n − s n , α n ) = n − L j = a r − C − ( β j + α n ) . Let V = H ( s a r − · · · s n − s n , α n ) . Since h− ( β j + α n ) , α n i = 0 , we have H ( s n , V ) = V . Since h− ( β n − + α n ) , α n − i = − h− ( β j + α n , α n − ) i = 0 , for all a r − ≤ j ≤ n − , by Lemma 2.3(2), Lemma 2.3(4) and Lemma 2.4 we have H ( s n − , V ) = n − M j = a r − C − ( β j + α n ) . Proceeding recursively we have H ( s a r − · · · s n − s n , V ) = a r − − M j = a r − C − ( β j + α n ) . Since n ≥ a > a > · · · > a k = 1, we see that h− ( β j + α n ) , α t i = 0, for all a r − ≤ j ≤ a r − − a r − ≤ t ≤ n, therefore by Lemma 2.3(2) and Lemma 2.4, we have H ( s a r − · · · s n s a r − · · · s n s a r − · · · s n − s n , α n ) = a r − − M j = a r − C − ( β j + α n ) . Since n ≥ a > a > · · · > a k = 1, we see that h− ( β j + α n ) , α t i = 0 for all a r − ≤ j ≤ a r − − a r − ≤ t ≤ n. By Lemma 2.3(2) and Lemma 2.4, we have H ( s a r − · · · s n s a r − · · · s n s a r − · · · s n s a r − · · · s n − s n , α n ) = a r − − M j = a r − C − ( β j + α n ) . Proceeding recursively we have H ( w r − s n , α n ) = a r − − M j = a r − C − ( β j + α n ) . Hence the lemma follows. (cid:3)
Lemma 4.5. If µ is of the form µ = − ( β j + α n ) for some ≤ j ≤ a k − − , then we have H ( w k − s n s s · · · s n +1 − k , α n +1 − k ) µ = 0 . Proof.
By applying SES repeatedly, it is easy to see that H ( s n s a s · · · s n +1 − k , α n +1 − k ) = C h ( α n +1 − k ) ⊕ ( n +1 − k L j =1 C − γ j,n +1 − k ) , where γ j,j ′ = ( α j + · · · + α j ′ ) for j ′ ≥ j .Let V = H ( s n s s · · · s n +1 − k , α n +1 − k ) . We next calculate H ( s a k − · · · s n − , V ) . Since n ≥ a > a > · · · > a k = 1 , we have a i ≤ n + 1 − i for all 1 ≤ i ≤ k. Moreover, h γ j,n +1 − k , α l i = 0 for all 1 ≤ j ≤ n + 1 − k and for all l ≥ n + 3 − k. Therefore by using Lemma 2.3(2) and Lemma 2.4, we have H ( s a k − · · · s n − , V ) = H ( s a k − · · · s n +1 − k s n +2 − k , V ) . Since h− γ j,n +1 − k , α n +2 − k i = 1 for all 1 ≤ j ≤ n + 1 − k, by using Lemma 2.3(2) andLemma 2.4, we have H ( s n +2 − k , V ) = C h ( α n +1 − k ) ⊕ ( n +1 − k L j =1 ( C − γ j,n +1 − k ⊕ C − γ j,n +2 − k )) . Since C h ( α n +1 − k ) ⊕ C − γ n +1 − k,n +1 − k is an indecomposable two dimensional B α n +1 − k -module,by Lemma 2.4 we have C h ( α n +1 − k ) ⊕ C − γ n +1 − k,n +1 − k = V ⊗ C − ω n +1 − k , where V is theirreducible two dimensional ˜ L α n +1 − k -module.By Lemma 2.3(4) we have H ( ˜ L α n +1 − k / ˜ B α n +1 − k , C h ( α n +1 − k ) ⊕ C − γ n +1 − k,n +1 − k ) = 0 . Since h γ j,n +1 − k , α n +1 − k i = − ≤ j ≤ n − k, by Lemma 2.3(4) we have H ( ˜ L α n +1 − k / ˜ B α n +1 − k , C γ j,n +1 − k ) = 0 for all 1 ≤ j ≤ n − k. From the above discussion, we have H ( s n +1 − k , H ( s n +2 − k , V )) = n − k L j =1 C − γ j,n +2 − k . Since h γ j,n +2 − k , α n − k i = 0 for all 1 ≤ j ≤ n − k − , and h γ n − k,n +2 − k , α n − k i = − , byusing Lemma 2.3(2), Lemma 2.3(2)(4) and Lemma 2.4, we have H ( s n − k , H ( s n +1 − k , H ( s n +2 − k , V ))) = n − k − M j =1 C − γ j,n +2 − k . Proceeding recursively we have H ( s n s a k − · · · s n − , V ) = a k − − M j =1 C − γ j,n +2 − k . Let V = H ( s n s a k − · · · s n − , V ) . Then similarly, we have H ( s a k − · · · s n − , V ) = a k − − M j =1 C − γ j,n +3 − k . Proceeding recursively we have V k − = H ( s a · · · s n − , V k − ) = a k − − M j =1 C − γ j,n − where V k − = H (( s n s · · · s n ) · · · ( s a k − · · · s n +1 − k ) , α n +1 − k ) . Note that γ j,n − = β j . Since h− α n − , α n i = 2 , by using Lemma 2.3(2) and Lemma 2.4we have H ( s n , V k − ) = a k − − L j =1 ( C − β j ⊕ C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ) . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 17 Moreover, h− β j , α n − i = − , h− ( β j + α n ) , α n − i = 0 and h− ( β j + 2 α n ) , α n − i = 1, forall 1 ≤ j ≤ a k − − . Therefore by using Lemma 2.3(2), Lemma 2.3(4) and Lemma 2.4 we have H ( s n − , H ( s n , V k − )) = a k − − M j =1 ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ) . Proceeding recursively we have H ( s a , H ( s a +1 · · · s n , V k − )) = a k − − L j =1 ( C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ⊕ C − ( β j +2 α n + β n − ) ⊕ · · · ⊕ C − ( β j +2 α n + β a ) ) . Hence the proof of the lemma follows. (cid:3)
Lemma 4.6. H ( w k s n , α n ) µ = 0 if and only if µ is of the form µ = − ( β j + α n ) for some ≤ j ≤ a k − − . Proof.
By applying SES repeatedly it is easy to see that H ( s · · · s n − s n , α n ) = n − M j =1 C − ( β j + α n ) . Let V = H ( s · · · s n − s n , α n ) . Since h− ( β j + α n ) , α n i = 0 , for all 1 ≤ j ≤ n, by Lemma2.3(2) and Lemma 2.4 we have H ( s n , V ) = V . Moreover, h− ( β n − + α n ) , α n − i = − h− ( β j + α n , α n − ) i = 0 for all 1 ≤ j ≤ n − . Therefore by using Lemma 2.3(4), Lemma 2.3(2) and Lemma 2.4, we have H ( s n − , V ) = n − M j =1 C − ( β j + α n ) . Proceeding recursively we have H ( s a k − · · · s n − s n , V ) = a k − − M j =1 C − ( β j + α n ) . Since n ≥ a > a > · · · > a k = 1, we have h− ( β j + α n ) , α t i = 0 for all 1 ≤ j ≤ a k − − a k − ≤ t ≤ n. By using Lemma 2.3(2), Lemma 2.4 we have H ( s a k − · · · s n s a k − · · · s n s · · · s n − s n , α n ) = a k − − M j =1 C − ( β j + α n ) . Similarly, since n ≥ a > a > · · · > a k = 1, we have h− ( β j + α n ) , α t i = 0 for all1 ≤ j ≤ a k − − a k − ≤ t ≤ n, therefore by using Lemma 2.3(2) and Lemma2.4 we have H ( s a k − · · · s n s a k − · · · s n s a k − · · · s n s · · · s n − s n , α n ) = a k − − M j =1 C − ( β j + α n ) . Proceeding recursively we have H ( w k s n , α n ) = a k − − M j =1 C − ( β j + α n ) . Hence the proof of the lemma follows. (cid:3) cohomology module H of the relative tangent bundle In this section, we describe the weights of H of the relative tangent bundle. Let n ≥ a > a > . . . > a k − > a k = 1 be a decreasing sequence of integers such that k ≥ . Fix3 ≤ r ≤ k. Lemma 5.1.
Let v r = s n s a r · · · s n − , v r − = s a r − · · · s n − and v r − = s a r − · · · s n − s n . Then we have (1) H ( v r − v r s n − , α n − ) = 0 . (2) Let w = v r − v r − v r s n − . H ( w, α n − ) µ = 0 if and only if µ is of the form µ = − ( β t + α n ) for some a r − ≤ t ≤ a r − − . In such a case,dim ( H ( w, α n − ) µ ) = 1 . Proof.
Proof of (1): Note that H ( s n − , α n − ) = C − α n − ⊕ C h ( α n − ) ⊕ C α n − (see [5, Corol-lary 2.5]). (5.1.1)Since h α n − , α n − i = 2 , we have H ( s n − , α n − ) = 0 . Since h α n − , α n − i = −
1, we have H ( s n − , H ( s n − , α n − )) = 0 . Therefore by using SES we have H ( s n − s n − , α n − ) = 0 . (5.1.2)Since h− α n − , α n − i = 1, by using (5.1.1) we have H ( s n − s n − , α n − ) = C h ( α n − ) ⊕ C − α n − ⊕ C − β n − . Since h− α n − , α n − i = 0 and h− β n − , α n − i = 1 we have H ( s n − , H ( s n − s n − , α n − )) = 0 . (5.1.3)Therefore by using SES together with (5.1.2), (5.1.3) we have H ( s n − s n − s n − , α n − ) = 0 . (5.1.4)Proceeding in this way we have H ( s a r · · · s n − s n − , α n − ) = 0 (5.1.5)and H ( s a r · · · s n − s n − , α n − ) = C h ( α n − ) ⊕ n − L j = a r C − β j . (5.1.6)Since h− β j , α n i > a r ≤ j ≤ n − , by using (5.1.6) we have H ( s n , H ( s a r · · · s n − s n − , α n − )) = 0 . (5.1.7)Therefore by using SES, (5.1.5) and (5.1.7) together we have H ( v r s n − , α n − ) = 0 . (5.1.8)In the proof of Lemma 4.1(1) we notice that H ( v r s n − , α n − ) = C h ( α n − ) ⊕ ( n − L j = a r C β j ⊕ C − ( β j + α n ) ⊕ C − ( β j +2 α n ) ) . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 19 Thus we have H ( s n − , H ( v r s n − , α n − )) = 0 (see lines from (4.1.1) to (4.1.2)). (5.1.9)Proceeding by similar arguments and using (5.1.8), (5.1.9) we have H ( v r − v r s n − , α n − ) = 0 . Proof of (2)From the Lemma 4.1(1) we have H ( v r − v r s n − , α n − ) = a r − − L i = a r ( C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕· · ·⊕ C − ( β i +2 α n + β ar − ) ) n − L i = a r − ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( α n − +2 α n ) . Notice that for each a r ≤ i ≤ a r − − C − ( β i + α n ) ⊕ C − ( β i +2 α n ) forms an indecomposabletwo dimensional B α n -module.Since h− ( β i + 2 α n ) , α n i = − , by Lemma 2.4 we have C − ( β i + α n ) ⊕ C − ( β i +2 α n ) = V i ⊗ C − ω n , where V i is the irreducible two dimensional ˜ L α n -module.By Lemma 2.3(4) we have H j ( ˜ L α n / ˜ B α n , C − ( β i + α n ) ⊕ C − ( β i +2 α n ) ) = 0 for all j ≥ a r ≤ i ≤ a r − − . Since h− ( β i + 2 α n + β t ) , α n i = 0 for each a r ≤ i ≤ a r − −
1, and a r − ≤ t ≤ n − , wehave H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n + β t ) ) = 0 for all a r ≤ i ≤ a r − − a r − ≤ t ≤ n − . Moreover, we have h− ( β i + 2 α n + β i +1 ) , α n i = 0 for each a r − ≤ i ≤ n − h− ( β i +2 α n ) , α n i = − a r − ≤ i ≤ n − H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n + β i +1 ) ) = 0 for all a r − ≤ i ≤ n − . By Lemma 2.3(3) we have H ( ˜ L α n / ˜ B α n , C − ( β i +2 α n ) ) = H ( ˜ L α n / ˜ B α n , s n − · − ( β i + 2 α n )) = C − ( β i + α n ) for all a r − ≤ i ≤ n − . From the above discussion, we have H ( s n , H ( v r − v r s n − , α n − )) = n − L i = a r − C − ( β i + α n ) . (5.2.1)By (1) and using SES we have H ( s n , H ( v r − v r s n − , α n − )) = H ( s n v r − v r s n − , α n − ) . (5.2.2)From (5.2.1) and (5.2.2) we have H ( s n v r − v r s n − , α n − ) = n − M i = a r − C − ( β i + α n ) . (5 . . H ( s n − , H ( s n v r − v r s n − , α n − )) = 0 . (5.2.4)Let v = v r − v r s n − . Therefore by using SES and (5.2.4) we have H ( s n − s n v, α n − ) = H ( s n − , H ( s n v, α n − )) . Since h− ( β n − + α n ) , α n − i = − h− ( β i + α n ) , α n − i = 0 for all a r − ≤ i ≤ n − , by using (5.2.3) we have H ( s n − s n v, α n − ) = n − L i = a r − C − ( β i + α n ) . Proceeding recursively we have H ( v r − v r − v r s n − , α n − ) = H ( s a r − · · · s n − s n v, α n − ) = a r − − L i = a r − C − ( β i + α n ) . (cid:3) Recall that w r = [ a , n ][ a , n ] · · · [ a r − , n ][ a r , n − , where 1 ≤ r ≤ k and n ≥ a > a >. . . > a k − > a k = 1 . Lemma 5.2. (1) H ( w , α n − ) = 0 . (2) If a = n − , then H ( w , α n − ) = 0 . (3) Let ≤ r ≤ k. Then, H ( w r , α n − ) µ = 0 if and only if µ = - ( β j + α n ) for some j such that a r − ≤ j ≤ a r − − . In such case dim ( H ( w r , α n − )) µ = 1 . Proof.
Proof of (1): Follows from proof of Lemma 3.1 and using SES.Proof of (2): By proof of (1), we have H ( s a · · · s n − , α n − ) = 0 . (5.2.1)Since a = n − , H ( s a · · · s n − , α n − ) = C h ( α n − ) ⊕ ( n − L j = a C − ( β j ) ) . Since h− β j , α n i ≥ a ≤ j ≤ n − , we have H ( s n , H ( s a · · · s n − , α n − )) = 0 . (5.2.2)By SES, (5.2.1), (5.2.2) we have H ( s n s a · · · s n − , α n − ) = 0 . By using Lemma 3.1 and SES respeatedly, we see that H ( w , α n − ) = H ( s a · · · s n s a · · · s n − , α n − ) = 0 . Proof of (3):Let v r = s n s a r · · · s n − , v r − = s a r − · · · s n − and v r − = s a r − · · · s n − s n . Then by Lemma5.1(2) we have H ( v r − v r − v r s n − , α n − ) = a r − − M j = a r − C − ( β j + α n ) . By Lemma 4.2(1) if ( H ( v r − v r − v r s n − , α n − )) µ = 0 then h µ, α n i ≥ . Therefore we have H ( s n , H ( v r − v r − v r s n − , α n − )) = 0 . (5.3.1)By using SES and (5.3.1) we have H ( s n v r − v r − v r s n − , α n − ) = H ( s n , H ( v r − v r − v r , α n − )) . (5.3.2)Since h− ( β j + α n ) , α n i = 0 for all a r − ≤ j ≤ a r − −
1, we have from (5.3.2) H ( s n v r − v r − v r s n − , α n − ) = a r − − L j = a r − C − ( β j + α n ) . By Lemma 3.1, we have H ( s n − , H ( s n v r − v r − v r s n − , α n − )) = 0 . (5.3.4)Therefore by using SES together with (5.3.4) we have H ( s n − s n v r − v r − v r , α n − ) = H ( s n − , H ( s n v r − v r − v r s n − , α n − )) . (5.3.5) IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 21 Since h− ( β j + α n ) , α n − i = 0 for all a r − ≤ j ≤ a r − − , using (5.3.5) we have H ( s n − s n v r − v r − v r s n − , α n − ) = a r − − L j = a r − C − ( β j + α n ) . Proceeding recursively we have H ( s a r − · · · s n − s n v r − v r − v r s n − , α n − ) = a r − − M j = a r − C − ( β j + α n ) . Since h− ( β j + α n ) , α t i for all a r − ≤ j ≤ a r − − a r − ≤ t ≤ n, using similararguments as above we have H ( w r , α n − ) = a r − − M j = a r − C − ( β j + α n ) . (cid:3) Corollary 5.3.
Let ≤ r ≤ k. If H ( w r , α n − ) µ = 0 , then H ( w r − s n s a r − · · · s n +2 − r , α n +2 − r ) µ = 0 . Proof.
Corollary follows from Lemma 4.3 and Lemma 5.2(2). (cid:3)
Corollary 5.4.
Let ≤ r ≤ k. If H ( w r , α n − ) µ = 0 , then H ( w r − s n , α n ) µ = 0 . Proof.
Corollary follows from Lemma 4.4 and Lemma 5.2(2). (cid:3)
Lemma 5.5.
Let u = w k s n [ a k , n − . Then H ( u , α n − ) µ = 0 if and only if µ is of theform, µ = − ( β j + α n ) for some ≤ j ≤ a k − − . Proof.
Let v k +1 = s n s · · · s n − , v k = s · · · s n − and v k − = s a k − · · · s n − s n . Step 1: H ( v k − v k v k +1 s n − , α n − ) = a k − − M j =1 C − ( β j + α n ) . Proof of Step 1: By Lemma 4.1(2), we have H ( s s · · · s n s · · · s n − , α n − ) = n − L i =1 ( C − ( β i +2 α n ) ⊕ C − ( β i +2 α n + β n − ) ⊕ · · · ⊕ C − ( β i +2 α n + β i +1 ) ) ⊕ C − ( β n − +2 α n ) . By Lemma 5.2(1), H ( s · · · s n s · · · s n − , α n − ) = 0 . Therefore by SES, we have H ( v k v k +1 s n − , α n − ) = H ( s , H ( s · · · s n s · · · s n − , α n − )) = 0 (by Lemma 3.1). (5.5.0)By SES and (5.5.0) we have H ( s n v k v k +1 s n − , α n − ) = H ( s n , H ( v k v k +1 s n − , α n − )) = n − L i =1 C − ( β i + α n ) . By Lemma 3.1, we have H ( s n − , H ( s n v k v k +1 s n − , α n − )) = 0 . Therefore by SES, wehave H ( s n − s n v k v k +1 s n − , α n − ) = H ( s n − , H ( s n v k v k +1 s n − , α n − )) . Since h− ( β n − + α n ) , α n − i = − h− ( β i + α n ) , α n − i = 0 for all 1 ≤ i ≤ n − , wehave H ( s n − s n v k v k +1 s n − , α n − ) = n − L i =1 C − ( β i + α n ) . Proceeding in this way recurssively, wesee that H ( v k − v k v k +1 s n − , α n − ) = a k − − L i =1 C − ( β i + α n ) . Hence Step 1 follows.By Lemma 4.2, we have H ( s n , H ( v k − v k v k +1 s n − , α n − )) = 0 . (5.5.1)Therefore by using SES and (5.5.1) we have H ( s n v k − v k v k +1 s n − , α n − ) = H ( s n , H ( v k − v k v k +1 s n − , α n − )) . (5.5.2)Since h− ( β j + α n ) , α n i = 0 for all 1 ≤ j ≤ a k − − , by using (5.5.2) we have H ( s n v k − v k v k +1 s n − , α n − ) = a k − − L j =1 C − ( β j + α n ) . By Lemma 3.1, we have H ( s n − , H ( s n v k − v k v k +1 s n − , α n − )) = 0 . (5.2.3)Therefore by using SES and (5.2.3) we have H ( s n − s n v k − v k v k +1 s n − , α n − ) = H ( s n − , H ( s n v k − v k v k +1 s n − , α n − )) . (5.2.4)Since h− ( β j + α n ) , α n − i = 0 for all 1 ≤ j ≤ a k − − , by using (6.2.4) we have H ( s n − s n v k − v k v k +1 s n − , α n − ) = a k − − L j =1 C − ( β j + α n ) . Proceeding recursively we have H ( s a k − · · · s n − s n v k − v k v k +1 s n − , α n − ) = a k − − M j =1 C − ( β j + α n ) . Using the similar arguments as above we have H ( u , α n − ) = a k − − M j =1 C − ( β j + α n ) . (cid:3) Corollary 5.6. If H ( u , α n − ) µ = 0 , then H ( w k − s n s s · · · s n +1 − k , α n +1 − k ) µ = 0 . Proof.
Corollary follows from Lemma 4.5 and Lemma 5.5. (cid:3)
Corollary 5.7. If H ( u , α n − ) µ = 0 , then H ( w k s n , α n ) µ = 0 . Proof.
Corollary follows from Lemma 4.6 and Lemma 5.5. (cid:3)
Let 3 ≤ r ≤ k. Let M r := { µ ∈ X ( T ) : H ( w r , α n − ) µ = 0 } and M := { µ ∈ X ( T ) : H ( u , α n − ) µ = 0 } . Then we have
Corollary 5.8. (1) M r ∩ M r ′ = ∅ whenever r = r ′ . (2) M ∩ M r = ∅ for every ≤ r ≤ k. Proof.
Proof of (1) and (2) follow from Lemma 5.2 and Lemma 5.5. (cid:3)
IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 23 Lemma 5.9.
Let c = s s · · · s n . Let T r = c r − s s · · · s n − , for all ≤ r ≤ n. Then, T r ( α j ) < for all n + 1 − r ≤ j ≤ n − . Proof.
Assume r = 2 . Then we see that T r ( α n − ) = − α . We assume that for 2 < l < n, we have T l ( α j ) < n + 1 − l ≤ j ≤ n − . Note that T l +1 = T l s n s · · · s n − . Then for all n − l ≤ i ≤ n − , we have s · · · s n − ( α i ) = α i +1 . Since n − ( i + 1) ≥ n + 1 − l ≤ i + 1 ≤ n − , we have T l +1 ( α i ) = T l ( α i +1 ) < s · · · s n − ( α n − ) = α n − , we have T l +1 ( α n − ) = T l ( s n ( − α n − )) . Since s n ( α n − ) = α n − +2 α n , we have T l s n ( α n − ) = T l ( α n − +2 α n ) . As s · · · s n − ( α n − +2 α n ) = β + 2 α n , we have T l ( α n − + 2 α n ) = T l − s n ( β + 2 α n ).Since s n s · · · s n − s n ( β + 2 α n ) = − α , we have T l − s n ( β + 2 α n ) = T l − ( − α ) . Since s n s · · · s n − ( − α ) = − α , we have T l − ( − α ) = T l − ( − α ) . Therefore by recursively wehave T l − ( − α ) = − α l − . Hence T l +1 ( α n − ) = − α l − < . Also it is clear that T l +1 ( α n − ) < . Therefore we have T l +1 ( α j ) < n − l ≤ j ≤ n − . Hence the result follows. (cid:3)
Lemma 5.10.
Let u, v ∈ W, let v := ( n Q j =1 s j ) l − s s · · · s n − for some positive integer l ≤ n, such that l ( uv ) = l ( u ) + l ( v ) . Let w = uv. If l ≥ , then H i ( w, α n − ) = 0 for all i ≥ . Proof.
We note that by SES, we have H ( w, α n − ) = H ( u, H ( v, α n − )) . We show that H ( v, α n − ) = 0 . By Lemma 5.9 we have for each 1 ≤ r ≤ n − , c r − s · · · s n − ( α j ) < n + 1 − r ≤ j ≤ n − . In particular, we have l ( vs n − ) = l ( v ) − . Therefore, by Lemma 2.2(4) and using SES, we have H ( v, α n − ) = H ( vs n − , H ( s n − , α n − )) = 0 . (5 . . H ( w, α n − ) = H ( u, H ( v, α n − )) . We will prove by induction l ( u ) . If l ( u ) = 0 then it follows trivially. Next suppose that l ( u ) > . Then there exists a simple root γ such that l ( s γ u ) = l ( u ) − . By using SES and(5.10.1) we have H ( s γ uv, α n − ) = 0 . Again, by induction hypotheses H ( s γ uv, α n − ) = H ( s γ u, H ( v, α n − )) . Therefore by SES, we have H ( w, α n − ) = H ( s γ , H ( s γ uv, α n − )) = H ( s γ , H ( s γ u, H ( v, α n − ))) = H ( u, H ( v, α n − )) .H ( v, α n − ) = 0 , follows from the fact that l ( vs n − ) = l ( v ) − H ( w, α n − ) = 0 . Therefore by [14, Corollary 6.4, p.780] we have H i ( w, α n − ) = 0for all i ≥ . (cid:3) cohomology modules of the tangent bundle of Z ( w, i )Let w ∈ W and let w = s i s i · · · s i r be a reduced expression for w and let i =( i , i , . . . , i r ). Let τ = s i s i · · · s i r − and i ′ = ( i , i , . . . , i r − ) . Recall the following long exact sequence of B -modules from [5] (see [5, Proposition 3.1,p.673]): → H ( w, α i r ) → H ( Z ( w, i ) , T ( w,i ) ) → H ( Z ( τ, i ′ ) , T ( τ,i ′ ) ) → H ( w, α i r ) → H ( Z ( w, i ) , T ( w,i ) ) → H ( Z ( τ, i ′ ) , T ( τ,i ′ ) ) → H ( w, α i r ) → H ( Z ( w, i ) , T ( w,i ) ) → H ( Z ( τ, i ′ ) , T ( τ,i ′ ) ) → H ( w, α i r ) → · · · By [14, Corollary 6.4, p.780], we have H j ( w, α i r ) = 0 for every j ≥ . Thus we have thefollowing exact sequence of B -modules:0 → H ( w, α i r ) → H ( Z ( w, i ) , T ( w,i ) ) → H ( Z ( τ, i ′ ) , T ( τ,i ′ ) ) → H ( w, α i r ) → H ( Z ( w, i ) , T ( w,i ) ) → H ( Z ( τ, i ′ ) , T ( τ,i ′ ) ) → w = s j s j · · · s j N be a reduced expression of w such that i = ( j , j , . . . , j r ) and letand j = ( j , j , . . . , j N ) . Lemma 6.1.
The natural homomorphism f : H ( Z ( w , j ) , T ( w ,j ) ) −→ H ( Z ( w, i ) , T ( w,i ) ) of B - modules is surjective.Proof. (see [4, Lemma 7.1, p.459]). (cid:3) Lemma 6.2.
Let J = S \ { α n − } . Let v ∈ W J and u ∈ W be such that l ( uv ) = l ( u ) + l ( v ) . Let u = s i · · · s i r and v = s i r +1 · · · s i t be reduced expressions of u and v respectively. Let i = ( i , i , . . . , i r ) and j = ( i , i , . . . , i r , i r +1 , . . . , i t ) . Then we have (1)
The natural homomorphism H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( u, i ) , T ( u,i ) ) of B -modules is surjective. (2) The natural homomorphism H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( u, i ) , T ( u,i ) ) of B -modules is an isomorphism.Proof. Let r + 1 ≤ l ≤ t. Let v l = us i r +1 · · · s i l and i l = ( i, i r +1 , . . . , i l ) . Proof of (1): By Lemma 3.3 we have H ( v t , α i t ) = 0 . Therefore, using LES the naturalhomomorphism H ( Z ( v t , i t ) , T ( v t ,i t ) ) −→ H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) )is surjective.By induction on l ( v ), the natural homomorphism H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) ) −→ H ( Z ( u, i ) , T ( u,i ) )is surjective. Hence we conclude that the natural homomorphism H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( u, i ) , T ( u,i ) ) IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 25 is surjective.Proof of (2): We will prove by induction on l ( v ) . By LES, we have the following exactsequence of B -modules:0 −→ H ( uv, α i t ) −→ H ( Z ( uv, j ) , T ( uv, j ) ) −→ H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) ) −→ H ( uv, α i t ) −→ H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) ) −→ . By induction on l ( v ) , the natural homomorphism H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) ) −→ H ( Z ( u, i ) , T ( u,i ) )is an isomorphism.By Lemma 3.3, H ( uv, α i t ) = 0 . Therefore, by the above exact sequence we have H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( v t − , i t − ) , T ( v t − ,i t − ) ) is an isomorphism. Hence we con-clude that the homomorphism H ( Z ( uv, j ) , T ( uv,j ) ) −→ H ( Z ( u, i ) , T ( u,i ) )of B -modules is an isomorphism. (cid:3) Recall that by Lemma 2.6(1) and Lemma 2.8(2) we have w = ( k − Q l =1 [ a l , n ])([ a k , n ] n +1 − k )( k − Q l =1 [ a k , a l − w . Let i be the tuple corresponding to this reduced of w . Let u = w k s n [ a k , n −
1] and i be the tuple corresponding to the reduced expression k Q l =1 [ a l , n ]([ a k , n − . Note that a k = 1 . With this notation, we have
Lemma 6.3. (1)
The natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) ) of B -modules is an isomorphism. (2) The natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) ) of B -modules is an isomorphism.Proof. For 1 ≤ n − k , let u j = w k s n [ a k , n ] j − s s · · · s n − and i j be the tuple correspondingto the reduced expression u j = ( k Q l =1 [ a l , n ])([ a k , n ]) j − [ a k , n −
1] (see Lemma 2.8(2)). Notethat w = u n − k s n ( k − Q l =1 [ a k , a l − . Case 1: a = n . In this case, we have s n ( k − Q l =1 [ a k , a l − ]) ∈ W J , where J = S \ { α n − } . Then by Lemma 6.2, the natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u n − k , i n − k ) , T ( u n − k ,i n − k ) ) (6 . . is surjectie and the natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u n − k , i n − k ) , T ( u n − k ,i n − k ) )of B -modules is an isomorphism. (6.3.2)If j ≥ , then by Lemma 5.10, we have H ( u j , α n − ) = 0 . Let u ′ j = u j s n − and let i ′ j bethe partial subsequence of u ′ j such that i j = ( i ′ j , n − . Hence by LES , we observe thatthe natural homomorphism H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u ′ j , i ′ j ) , T ( u ′ j ,i ′ j ) ) (6 . . H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u ′ j , i ′ j ) , T ( u ′ j ,i ′ j ) ) (6 . . H ( Z ( u ′ j , i ′ j ) , T ( u ′ j ,i ′ j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) ) (6 . . H ( Z ( u ′ j , i ′ j ) , T ( u ′ j ,i ′ j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) ) (6 . . H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) )is surjective and H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) )is an isomorphism.Proceeding recursively we get that the homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is surjective and homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is an isomorphism.Further since u − ( α ) < , by [5, Lemma 6.2, p.667], we have H ( Z ( u , i ) , T ( u ,i ) ) − α =0 . By [5, Theorem 7.1], H ( Z ( w , i ) , T ( w ,i ) ) is a parabolic subalgebra of g and hence thereis a unique B -stable line in H ( Z ( w , i ) , T ( w ,i ) ) , namely g − α . Thereore we conclude thatthe natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is an isomorphism. Case 2 : a = n Then by Lemma 6.2, the natural morphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u n +1 − k , i n +1 − k ) , T ( u n +1 − k ,i n +1 − k ) )is surjectie and the natural homomorphism IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 27 H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u n +1 − k , i n +1 − k ) , T ( u n +1 − k ,i n +1 − k ) )of B -modules is an isomorphism.If j ≥ , then by Lemma 5.10, we have H ( u j , α n − ) = 0 . Hence by LES, for each2 ≤ j ≤ n + 1 − k , we observe that the natural homomorphism H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) )is surjective and H ( Z ( u j , i j ) , T ( u j ,i j ) ) −→ H ( Z ( u j − , i j − ) , T ( u j − ,i j − ) )is an isomorphism. Therefore, the homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is surjective and the homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is an isomorphism.Further since u − ( α ) < , by [5, Lemma 6.2, p.667], we have H ( Z ( u , i ) , T ( u ,i ) ) − α =0 . By [5, Theorem 7.1], H ( Z ( w , i ) , T ( w ,i ) ) is a parabolic subalgebra of g and hence thereis a unique B -stable line in H ( Z ( w , i ) , T ( w ,i ) ) , namely g − α . Thereore we conclude thatthe natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) )of B -modules is an isomorphism. (cid:3) The following is a useful Corollary.
Corollary 6.4. If µ ∈ X ( T ) \ { } , then, we have dim ( H ( Z ( u , i ) , T ( u ,i ) ) µ ) ≤ . Proof.
By [5, Theorem 7.1], H ( Z ( w , i ) , T ( w ,i ) ) is a parabolic subalgebra of g . By Lemma6.3(1) we have H ( Z ( w , i ) , T ( w ,i ) ) ≃ H ( Z ( u , i ) , T ( u ,i ) ) . (as B -modules).Hence for any µ ∈ X ( T ) \ { } , we have dim ( H ( Z ( u , i ) , T ( u ,i ) ) µ ) ≤ . (cid:3) Let u ′ = w k s n [ a k , n − and let i ′ be the tuple corresponding to the reduced expression k Q l =1 [ a l , n ][ a k , n − . Let ≤ r ≤ k, and let j r = ( a , . . . , n ; a , . . . , n ; . . . ; a r − , . . . , n ; a r , . . . , n − and j ′ r = ( a , . . . , n ; a , . . . , n ; a r , . . . , n − . We now prove
Lemma 6.5.
Let µ ∈ X ( T ) \ { } . (1) If H ( u , α n − ) µ = 0 , then dim( H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ ) ≤ . (2) If H ( u , α n − ) µ = 0 , then dim( H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ ) = 2 , and the natural homo-morphism H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ −→ H ( u , α n − ) µ is surjective.Proof. By LES, we have the following long exact sequence of B -modules:0 −→ H ( u , α n − ) −→ H ( Z ( u , i ) , T ( u ,i ) ) −→ H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( u , α n − ) −→ · · · (6 . . H ( u , α n − ) µ = 0 and µ ∈ X ( T ) \ { } , then by theabove exact sequence the natural homoorphism (of T -modules) H ( Z ( u , i ) , T ( u ′ ,i ) ) µ −→ H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ is surjective. By Corollary 6.4, we have dim( H ( Z ( u , i ) , T ( u ,i ) ) µ ) ≤ . Proof of (2): Assume that H ( u , α n − ) µ = 0 . Then by Lemma 5.5, µ is of the form − ( β j + α n ) for some 1 ≤ j ≤ a k − − H ( u , α n − ) µ )=1. Hence by using Corollary6.4, we see that if H ( u , α n − ) µ = 0 , then dim ( H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ ) ≤ . (6.5.2)Let j ′′ k = ( j k , n ) be the tuple corresponding to the reduced expression w k s n = k Q l =1 [ a l , n ] . Then by Lemma 6.2 the natural homomorphism H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) )is surjective. By (6.5.2), we have dim( H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) µ ) ≤ . (6.5.3)Again by the Lemma 6.2 the natural homomorphism H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) (6.5.4)is surjective. Recall that u ′ t = u k s n − . By LES we have the following long exact sequenceof B -modules: 0 −→ H ( w k , α n − ) −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) −→ H ( Z ( u ′ k , i ′ k ) , T ( u ′ k ,i ′ k ) ) −→ H ( w k , α n − ) −→ · · · Since H ( u , α n − ) µ = 0 , by Corollary 5.8 we have H ( w k , α n − ) µ = 0 . Therefore we havean exact sequence 0 −→ H ( w k , α n − ) µ −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ −→ H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) µ −→ . Let σ k = w k − s n s · · · s n +1 − k and j ∗ k = ( j k − , n, , , . . . , n +1 − k ) be the tuple correspond-ing to this reduced expression of σ k . Then by using Lemma 6.2, the natural homomorphism
IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 29 H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) −→ H ( Z ( σ k , j ∗ k ) , T ( σ k ,j ∗ k ) ) (6.5.5)is surjective. Therefore the natural map H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ −→ H ( Z ( σ k , j ∗ k ) , T ( σ k ,j ∗ k ) ) µ (6 . . H ( u , α n − ) µ = 0 , by Corollary 5.6 we have H ( σ k , α n +1 − k ) µ = 0 . Therefore, H ( Z ( σ k , j ∗ k ) , T ( σ k ,j ∗ k ) ) µ = 0 . Thus from (6.5.6) we have H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ = 0 . Then we have an exact sequence of T -modules0 −→ H ( w k s n , α n ) µ −→ H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) µ −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ −→ H ( u , α n − ) µ = 0 , by Corollary 5.7 we have H ( w k s n , α n ) µ = 0 . Therefore dim( H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) µ ) ≥ . On the other hand by (6.5.3) we havedim( H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) µ ) ≤ . Hence we have dim( H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) ) µ )=2 . Since the natural homomorphism H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( Z ( w k s n , j ′′ k ) , T ( w k s n ,j ′′ k ) )is surjective, we have dim( H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) µ ))= 2 . (6.5.7)It is clear from (6.5.1),(6.5.7), and Corollary 6.4, that H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ −→ H ( u , α n − ) µ is surjective. (cid:3) Corollary 6.6.
The natural homomorphism H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( u , α n − ) is surjective.Proof. Note that by Lemma 5.5, if H ( u , α n − ) µ = 0 then µ ∈ X ( T ) \ { } . Now the prooffollows from Lemma 6.5. (cid:3)
Lemma 6.7. (1) If H ( w m , α n − ) µ = 0 for all r ≤ m ≤ k and H ( u , α n − ) µ = 0 , thendim ( H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ ) ≤ . (2) Let ≤ r ≤ k. If H ( w r , α n − ) µ = 0 , then dim ( H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ ) = 2 , andthe natural homomorphism H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ −→ H ( w r , α n − ) µ is surjective.Proof. Proof of (1): If H ( u , α n − ) µ = 0 , then by Lemma 6.5, we havedim( H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ ) ≤ . By Lemma 6.2, the natural homomorphism H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) µ −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ is surjective. If H ( w m , α n − ) µ = 0 for all r ≤ m ≤ k, by using LES, we see that thenatural homomorphism H ( Z ( w k , j k ) , T ( w k ,j k ) ) µ −→ H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ is surjective. Therefore, we have dim( H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ ) ≤ . Proof of (2): Assume that H ( w r , α n − ) µ = 0 . Then by Corollary 5.8, we have H ( w m , α n − ) µ =0 for all r + 1 ≤ m ≤ k and H ( u , α n − ) µ = 0 . Then by (1) , we havedim( H ( Z ( τ r +1 , j ′ r +1 ) , T ( τ r +1 ,j ′ r +1 ) ) µ ) ≤ . By Lemma 6.2, the natural homomorphism H ( Z ( τ r +1 , j ′ r +1 ) , T ( τ r +1 ,j ′ r +1 ) ) µ −→ H ( Z ( w r , j r ) , T ( w r ,j r ) ) µ is surjective. Hence, we have dim( H ( Z ( w r , j r ) , T ( w r ,j r ) ) µ ) ≤ . (6.7.1)By LES we have the following long exact sequence of B -modules:0 −→ H ( w r , α n − ) −→ H ( Z ( w r , j r ) , T ( w r ,j r ) ) −→ H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) −→ H ( w r , α n − ) −→ · · · Since dim( H ( Z ( w r , j r ) , T ( w r ,j r ) ) µ ) ≤ H ( w r , α n − ) µ )= 1 , we see thatdim( H ( Z ( τ r , j ′ r )) µ ) ≤ . (6.7.2)Let j ′′ r − = ( j r − , n ) be the tuple corresponding to the reduced expression w r − s n = r − Q l =1 [ a l , n ] . Then by Lemma 6.2 the natural homomorphism H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) −→ H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) )is surjective. By (6.7.2), we have dim( H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) µ )) ≤ . (6.7.3)Again by the Lemma 6.2 the natural homomorphism H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) ) −→ H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) (6.7.4)is surjective.By LES we have the following long exact sequence of B -modules:0 −→ H ( w r − , α n − ) −→ H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) −→ H ( Z ( τ r − , j ′ r − ) , T ( τ r − ,j ′ r − ) ) −→ H ( w r − , α n − ) −→ · · · Since H ( w r , α n − ) µ = 0, then by Corollary 5.8 we have H ( w r − , α n − ) µ = 0 . Thereforewe have an exact sequence0 −→ H ( w r − , α n − ) µ −→ H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) µ −→ H ( Z ( τ r − , j ′ r − ) , T ( τ r − ,j ′ r − ) ) µ −→ . . σ r − = w r − s n s a r − ··· s n +2 − r , and let j ∗ r − = ( j r − , n, a r − , a r − + 1 , . . . , n + 2 − r ) bethe reduced expression of σ r − . IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 31 Then by using Lemma 6.2 the natural homomorphism H ( Z ( τ r − , j ′ r − ) , T ( τ r − ,j ′ r − ) ) −→ H ( Z ( σ r − , j ∗ r − ) , T ( σ r − ,j ∗ r − ) ) (6 . . H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) µ −→ H ( Z ( σ r − , j ∗ r − ) , T ( σ r − ,j ∗ r − ) ) µ (6 . . H ( w r , α n − ) µ = 0, by Corollary 5.3 we have H ( σ r − , α n +2 − r ) µ = 0 . Therefore, H ( Z ( σ r − , j ∗ r − ) , T σ r − ,j ∗ r − ) ) µ = 0 . Thus from (6.7.7) we have H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) µ = 0 . Then we have an exactsequence of T -modules0 −→ H ( w r − s n , α n ) µ −→ H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) ) µ −→ H ( Z ( w r − , j r − ) , T ( w r − ,j r − ) ) µ −→ H ( w r , α n − ) µ = 0 , by Corollary 5.4 we have H ( w r − s n , α n ) µ = 0 . Therefore dim( H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) µ )) ≥ . Hence, by (6.7.3) we have dim( H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) ) µ )= 2 . Since the natural homomorphism H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) −→ H ( Z ( w r − s n , j ′′ r − ) , T ( w r − s n ,j ′′ r − ) )is surjective, we have dim( H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ )= 2 . Therefore by (6.7.1), H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) µ −→ H ( w r , α n − ) µ is surjective. (cid:3) main theorem In this section we prove the main theorem.Recall that G = P SO (2 n + 1 , C )( n ≥ , and let c be a Coxeter element of W. Thenthere exists a decreasing sequence n ≥ a > a > · · · > a k = 1 of positive integers suchthat c = [ a , n ][ a , a − · · · [ a k , a k − − , where [ i, j ] for i ≤ j denotes s i s i +1 · · · s j . Let i = ( i , i , . . . , i n ) be a sequence corresponding to a reduced expression of w , where i r (1 ≤ r ≤ n ) is a sequence of reduced expressions of c (see Lemma 2.8). Theorem 7.1. H j ( Z ( w , i ) , T ( w ,i ) ) = 0 for all j ≥ if and only if a = n − . Proof.
From [5, Proposition 3.1, p. 673], we have H j ( Z ( w , i ) , T ( w ,i ) ) = 0 for all j ≥ . Itis enough to prove the following: H ( Z ( w , i ) , T ( w ,i ) ) = 0 if and only if c is of the form[ a , n ][ a , a − · · · [ a k , a k − −
1] with a = n − . Proof of ( = ⇒ ): If a = n − , then a = n and c = s n s n − v, where v ∈ W J and J = S \ { α n − , α n } . Let u = s n s n − . Then c = uv. Let j = ( n, n −
1) be the sequencecorresponding to u. Then using LES, we have: −→ H ( u, α n − ) −→ H ( Z ( u, j ) , T ( u,j ) ) −→ H ( s n , α n ) −→ H ( u, α n − ) f −→ H ( Z ( u, j ) , T ( u,j ) ) −→ . We see that H ( s n s n − , α n − ) = C α n + α n − and H ( s n , α n ) α n + α n − = 0 . Hence f is nonzero homomorphism. Hence H ( Z ( u, j ) , T ( u,j ) )) = 0 . By Lemma 6.1, the natural homo-morphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u, j ) , T ( u,j ) )is surjective.Hence we have H ( Z ( w , i ) , T ( w ,i ) ) = 0 . Proof of ( ⇐ = ) : Assume that a = n − . Recall that w k = [ a , n ] · · · [ a k − , n ][ a k , n − ,u = w k s n [ a k , n − . By Lemma 6.3(2), the natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( u , i ) , T ( u ,i ) ) (7.1.1)of B -modules is an isomorphism.By using LES, we have the following exact sequence of B -modules: · · · −→ H ( Z ( u , i ) , T ( u ,i ) ) −→ H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) h −→ H ( u , α n − ) −→ H ( Z ( u , i ) , T ( u ,i ) ) −→ H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ . By Corollary 6.6, we see that the natural homomorphism h : H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( u , α n − ) is surjective. Therefore, the natural homomorphism H ( Z ( u , i ) , T ( u ,i ) ) −→ H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) (7 . . H ( Z ( u ′ , i ′ ) , T ( u ′ ,i ′ ) ) −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) (7 . . H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) (7 . . τ k = [ a , n ][ a , n ] · · · [ a k − , n ][ a k , n − . By using LES, we have the followingexact sequence of B -modules: · · · −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) −→ H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) h −→ IGIDITY OF BOTT-SAMELSON-DEMAZURE-HANSEN VARIETY FOR
P SO (2 n + 1 , C ) 33 H ( w k , α n − ) −→ H ( Z ( w k , j k ) , T ( w k ,j k ) ) h −→ H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) −→ . By Lemma 6.7(2), we see that the map h : H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) −→ H ( w k , α n − ) issurjective.Therefore, the map h : H ( Z ( w k , j k ) , T ( w k ,j k ) ) −→ H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) (7 . . H ( Z ( τ k , j ′ k ) , T ( τ k ,j ′ k ) ) −→ H ( Z ( w k − , j k − ) , T ( w k − ,j k − ) ) (7 . . B -modules: · · · −→ H ( Z ( w k − , j k − ) , T ( w k − ,j k − ) ) −→ H ( Z ( τ k − , j ′ k − ) , T ( τ k − ,j ′ k − ) ) → H ( w k − , α n − ) −→ H ( Z ( w k − , j k − ) , T ( w k − ,j k − ) ) → H ( Z ( τ k − , j ′ k − ) , T ( τ k − ,j ′ k − ) ) −→ . By Lemma 6.7(2), we see that the natural map H ( Z ( τ k − , j ′ k − ) , T ( τ k − ,j ′ k − ) ) −→ H ( w k − , α n − ) is surjective.Therefore, the natural map H ( Z ( w k − , j k − ) , T ( w k − ,j k − ) ) −→ H ( Z ( τ k − , j ′ k − ) , T ( τ k − ,j ′ k − ) ) (7.1.7)is an isomorphism.By using Lemma 6.2(2) and 6.7(2) repeatedly, we see that the natural map H ( Z ( τ k − , j ′ k − ) , T ( τ k − ,j ′ k − ) ) −→ H ( Z ( τ r , j ′ r ) , T ( τ r ,j ′ r ) ) (7 . . ≤ r ≤ k − . Therefore by ( (7.1.4), (7.1.5), (7.1.6), (7.1.7), (7.1.8) we have the natural homomorphism H ( Z ( w , i ) , T ( w ,i ) ) −→ H ( Z ( τ , j ′ ) , T ( τ ,j ′ ) ) (7 . . H ( Z ( τ , j ′ ) , T ( τ ,j ′ ) ) −→ H ( Z ( w , j ) , T ( w ,j ) )is an isomorphism.Since a = n − , by Lemma 5.2(2) we have H ( w , α n − ) = 0 . Note that by Lemma5.2(1) we have H ( w , α n − ) = 0 . By using Lemma 3.3 and using LES, we have H ( Z ( w , j ) , T ( w ,j ) ) = 0 . Hence weconclude that H ( Z ( w , i ) , T ( w ,i ) ) = 0 . This completes the proof of the theorem. (cid:3)
Corollary 7.2.
Let c be a Coxeter element such that c is of the form [ a , n ][ a , a − · · · [ a k , a k − − with a = n − and a k = 1 . Let ( w , i ) be a reduced expression of w interms of c as in Theorem 7.1. Then, Z ( w , i ) has no deformations.Proof. By Theorem 7.1 and by [5, Proposition 3.1, p.673], we have H i ( Z ( w , i ) , T ( w ,i ) ) =0 for all i > . Hence, by [12, Proposition 6.2.10, p.272], we see that Z ( w , i ) has nodeformations. (cid:3) Remark . Theorem 7.1 does not hold for
P SO (5 , C ) . Proof.
We take c = s s . Here a = 1 . Further, we have w = c = s s s s . Let i = (1 , , , . It is easy to see by using SES repeatedly that H ( s s s , α ) = C α + α ⊕ C α . Further, note that H ( Z ( s s , (1 , , T ( s s , (1 , ) α + α = 0 (see [5, Proposition 6.3(1),p.688]). Hence by using LES we have H ( Z ( s s s , (1 , , , T ( s s s , (1 , , ) = 0 . Therefore by using Lemma 6.1 we have H ( Z ( w , i ) , T ( w ,i ) ) = 0 . (cid:3) Acknowledgements
We thank the referee for the useful comments and suggestions.We thank the Infosys Foundation for the partial finance support.
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