On the vanishing ranges for the cohomology of finite groups of Lie type II
aa r X i v : . [ m a t h . G R ] D ec Proceedings of Symposia in Pure Mathematics
On the vanishing ranges for the cohomology of finite groups of Lie type II
Christopher P. Bendel, Daniel K. Nakano, and Cornelius Pillen
Abstract.
The computation of the cohomology for finite groups of Lie type in the describing characteristicis a challenging and difficult problem. In [
BNP ], the authors constructed an induction functor which takesmodules over the finite group of Lie type, G ( F q ), to modules for the ambient algebraic group G . In particularthis functor when applied to the trivial module yields a natural G -filtration. This filtration was utilized in[ BNP ] to determine the first non-trivial cohomology class when the underlying root system is of type A n or C n . In this paper the authors extend these results toward locating the first non-trivial cohomology classesfor the remaining finite groups of Lie type (i.e., the underlying root system is of type B n , C n , D n , E , E , E , F , and G ) when the prime is larger than the Coxeter number.
1. Introduction1.1.
Let G be a simple algebraic group scheme over a field k of prime characteristic p which is definedand split over the prime field F p , and F : G → G denote the Frobenius map. The fixed points of the r th iterate of the Frobenius map, denoted G ( F q ), is a finite Chevalley group where F q denotes the finitefield with p r elements. An elusive problem of major interest has been to determine the cohomology ringH • ( G ( F q ) , k ). Until recently, aside from small rank cases, it was not even known in which degree the firstnon-trivial cohomology class occurs.This present paper is a sequel to [ BNP ] where we began investigating three related problems of increasinglevels of difficulty:(1.1.1) Determining Vanishing Ranges: Finding
D > i ( G ( F q ) , k ) = 0 for0 < i < D .(1.1.2) Locating the First Non-Trivial Cohomology Class: Finding a D satisfying (1.1.1) such thatH D ( G ( F q ) , k ) = 0 . A D satisfying this property will be called a sharp bound .(1.1.3) Determining the Least Non-Trivial Cohomology: For a sharp D as in (1.1.2) compute H D ( G ( F q ) , k ).Vanishing ranges (1.1.1) were found in earlier work of Quillen [ Q ], Friedlander [ F ] and Hiller [ H ] . Sharpbounds (1.1.2) were later found by Friedlander and Parshall for the Borel subgroup B ( F q ) of the GL n ( F q ),and conjectured for the general linear group by Barbu [ B ]. A more detailed discussion of these results canbe found in [ BNP , Section 1.1].In [
BNP ], for simple, simply connected G and primes p larger than the Coxeter number h , we proved thatH i ( G ( F p r ) , k ) = 0 for 0 < i < r ( p − H ]. For a group with underlying root system of type C n , we demonstrated that D = r ( p −
2) is in fact asharp bound, answering (1.1.2). The first non-vanishing cohomology, as in (1.1.3), was also determined. Fortype A n , questions (1.1.2) and (1.1.3) were also answered, where the r > Mathematics Subject Classification.
Primary 20J06; Secondary 20G10.Research of the first author was supported in part by NSF grant DMS-0400558.Research of the second author was supported in part by NSF grant DMS-1002135. c (cid:13) at least twice the Coxeter number. Our methods also yielded a proof of Barbu’s Conjecture [ B , Conjecture4.11].In this paper, we continue these investigations in two directions. First we consider the case that G is agroup of adjoint type (as opposed to simply connected). For such G with p > h , when the root system issimply laced, one obtains a uniform sharp bound of r (2 p −
3) answering (1.1.2) (cf. Corollary 3.3.1). Thesame uniform bound also holds for the adjoint versions of the twisted groups of types A , D , and E when p > h .We then consider the remaining types in the simply connected case. For G being simple, simply connectedand having root system of type D n with p > h , (1.1.2) and (1.1.3) are answered (cf. Theorem 4.5.2). Fortype E n , (1.1.2) is answered for all primes p > h (with the exceptions of p = 17 ,
19 for type E ); cf.Theorems 5.1.3, 5.2.3, and 5.3.1.The calculations for the non-simply-laced groups are considerably more complicated. For type B weanswer (1.1.2) when r = 1 and p > h , see Theorem 6.7.1. Some discussion of the situation for types G and F is given in Sections 7 and 8 respectively. For r = 1 and p > h , we find improved answers to (1.1.1); cf.Theorem 7.5.1 and Theorem 8.1.1. Finding an answer to (1.1.2) and (1.1.3) continues to be elusive in thesetypes although some further information towards answering these questions is obtained. Throughout this paper, we will follow the notation and conventions given in the stan-dard reference [
Jan ]. G will denote a simple, simply connected algebraic group scheme which is defined andsplit over the finite field F p with p elements (except in Section 3.3 where G is assumed to be of adjoint typerather than simply connected). Throughout the paper let k be an algebraically closed field of characteristic p . For r ≥
1, let G r := ker F r be the r th Frobenius kernel of G and G ( F q ) be the associated finite Chevalleygroup. Let T be a maximal split torus and Φ be the root system associated to ( G, T ). The positive (resp.negative) roots are Φ + (resp. Φ − ), and ∆ is the set of simple roots. Let B be a Borel subgroup containing T corresponding to the negative roots and U be the unipotent radical of B . For a given root system of rank n ,the simple roots will be denoted by α , α , . . . , α n (via the the Bourbaki ordering of simple roots). For type B n , α n denotes the unique short simple root and for type C n , α n denotes the unique long simple root. Thehighest (positive) root will be denoted ˜ α , and for root systems with multiple root lengths, the highest shortroot will be denoted α . Let W denote the Weyl group associated to Φ, and, for w ∈ W , let ℓ ( w ) denote thelength of the element w (i.e., number of elements in a reduced expression for w ).Let E be the Euclidean space associated with Φ, and the inner product on E will be denoted by h , i . Let α ∨ = 2 α/ h α, α i be the coroot corresponding to α ∈ Φ. The fundamental weights (basis dualto α ∨ , α ∨ , . . . , α ∨ n ) will be denoted by ω , ω , . . . , ω n . Let X ( T ) be the integral weight lattice spannedby these fundamental weights. The set of dominant integral weights is denoted by X ( T ) + . For a weight λ ∈ X ( T ), set λ ∗ := − w λ where w is the longest word in the Weyl group W . By w · λ := w ( λ + ρ ) − ρ wemean the “dot” action of W on X ( T ), with ρ being the half-sum of the positive roots. For α ∈ ∆, s α ∈ W denotes the reflection in the hyperplane determined by α .For a G -module M , let M ( r ) be the module obtained by composing the underlying representation for M with F r . Moreover, let M ∗ denote the dual module. For λ ∈ X ( T ) + , let H ( λ ) := ind GB λ be the inducedmodule and V ( λ ) := H ( λ ∗ ) ∗ be the Weyl module of highest weight λ .
2. General Strategy and Techniques2.1.
We will employ the basic strategy used in [
BNP ] in addressing (1.1.1)-(1.1.3) which uses effectivetechniques developed by the authors which relate H i ( G ( F q ) , k ) to extensions over G via an induction functor G r ( − ). When G r ( − ) is applied to the trivial module k , G r ( k ) has a filtration with factors of the form H ( λ ) ⊗ H ( λ ∗ ) ( r ) [ BNP , Proposition 2.4.1]. The G -cohomology of these factors can be analyzed by usingthe Lyndon-Hochschild-Serre (LHS) spectral sequence involving the Frobenius kernel G r (cf. [ BNP , Section3]). In particular for r = 1, we can apply the results of Kumar-Lauritzen-Thomsen [ KLT ] to determinethe dimension of a cohomology group H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ), which can in turn be used to determineH i ( G ( F p r ) , k ). The dimension formula involves the combinatorics of the well-studied Kostant PartitionFunction. This reduces the question of the vanishing of the finite group cohomology to a question involvingthe combinatorics of the underlying root system Φ.For root systems of types A and C the relevant root system combinatorics was analyzed in [ BNP , Sections5-6]. In the cases of the other root systems ( B , D , E , F , G ) the combinatorics is much more involved and N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 3 we handle these remaining cases in Sections 4-8. In this section, for the convenience of the reader, we statethe key results from [
BNP ] which will be used throughout this paper.
We first record here a formula for − w · BNP ,Observation 2.1]:
Observation . If w ∈ W admits a reduced expression w = s β s β . . . s β m with β i ∈ ∆ and m = ℓ ( w ), then − w · β + s β ( β ) + s β s β ( β ) + · · · + s β s β . . . s β m − ( β m ) . Moreover, this is the unique way in which − w · Let G r ( k ) := ind GG ( F q ) ( k ). The functor G r ( − ) isexact and one can use Frobenius reciprocity to relate extensions over G with extensions over G ( F q ) [ BNP ,Proposition 2.2].
Proposition . Let
M, N be rational G -modules. Then, for all i ≥ , Ext iG ( F q ) ( M, N ) ∼ = Ext iG ( M, N ⊗ G r ( k )) . In order to make the desired computations of cohomology groups, we will make use of Proposition 2.3.1(with M = k = N ). In addition, we will use a special filtration on G r ( k ) (cf. [ BNP , Proposition 2.4.1]).
Proposition . The induced module G r ( k ) has a filtration with factors of the form H ( λ ) ⊗ H ( λ ∗ ) ( r ) with multiplicity one for each λ ∈ X ( T ) + . The filtration from Proposition 2.3.2 allows one to obtain a conditionon G -cohomology which leads to vanishing of G ( F p r )-cohomology (cf. [ BNP , Corollary 2.6.1]).
Proposition . Let m be the least positive integer such that there exists λ ∈ X ( T ) + with H m ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 . Then H i ( G ( F q ) , k ) ∼ = H i ( G, G r ( k )) = 0 for < i < m . While the identification of an m satisfying Proposition 2.4.1 gives a vanishingrange as in (1.1.1), it does not a priori follow that H m ( G ( F q ) , k ) = 0. The following theorem providesconditions which assist with addressing (1.1.2) or (1.1.3) [ BNP , Theorem 2.8.1].
Theorem . Let m be the least positive integer such that there exists ν ∈ X ( T ) + with H m ( G, H ( ν ) ⊗ H ( ν ∗ ) ( r ) ) = 0 . Let λ ∈ X ( T ) + be such that H m ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 . Suppose H m +1 ( G, H ( ν ) ⊗ H ( ν ∗ ) ( r ) ) = 0 for all ν < λ that are linked to λ . Then (a) H i ( G ( F q ) , k ) = 0 for < i < m ; (b) H m ( G ( F q ) , k ) = 0 ; (c) if, in addition, H m ( G, H ( ν ) ⊗ H ( ν ∗ ) ( r ) ) = 0 for all ν ∈ X ( T ) + with ν = λ , then H m ( G ( F q ) , k ) ∼ = H m ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) . From the filtration on G r ( k ) in Proposition 2.3.2, H i ( G ( F q ) , k ) ∼ = H i ( G, G r ( k )) can be decomposed asa direct sum over linkage classes of dominant weights. For a fixed linkage class L , let m be the leastpositive integer such that there exists ν ∈ L with H m ( G, H ( ν ) ⊗ H ( ν ∗ ) ( r ) ) = 0. Let λ ∈ L be suchthat H m ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0. Suppose H m +1 ( G, H ( ν ) ⊗ H ( ν ∗ ) ( r ) ) = 0 for all ν < λ in L . Thenanalogous to Theorem 2.5.1, it follows that H m ( G ( F q ) , k ) = 0. See [ BNP , Theorem 2.8.2]. G -cohomology. From Sections 2.4 and 2.5, the key to understanding the vanishingof H i ( G ( F p r ) , k ) is to understand H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) for all dominant weights λ . For r = 1, thesegroups can be related to G -cohomology groups (cf. [ BNP , Lemma 3.1]).
Lemma . Suppose p > h and let ν , ν ∈ X ( T ) + . Then for all j H j ( G, H ( ν ) ⊗ H ( ν ∗ ) (1) ) ∼ = Ext jG ( V ( ν ) (1) , H ( ν )) ∼ = Hom G ( V ( ν ) , H j ( G , H ( ν )) ( − ) . We remark that the aforementioned lemma would hold for arbitrary r th-twists if it was known that thecohomology group H j ( G r , H ( ν )) ( − r ) admits a good filtration, which is a long-standing conjecture of Donkin.For p > h , this is known for r = 1 by results of Andersen-Jantzen [ AJ ] and Kumar-Lauritzen-Thomsen [ KLT ].In that case, the lemma is only needed when ν = ν . For arbitrary r we can often work inductively fromthe r = 1 case. This requires slightly more general Ext-computations and the possibility that ν = ν . CHRISTOPHER P. BENDEL, DANIEL K. NAKANO, AND CORNELIUS PILLEN r = 1 . From Lemma 2.6.1, for ν ∈ X ( T ) + , the cohomology group H i ( G, H ( ν ) ⊗ H ( ν ∗ ) (1) ) can be identified with Hom G ( V ( ν ) , H i ( G , H ( ν ) ( − ). It is well-known that, from block consid-erations, H i ( G , H ( ν )) = 0 unless ν = w · pµ for w ∈ W and µ ∈ X ( T ). For p > h , from [ AJ ] and [ KLT ],we have(2.7.1) H i ( G , H ( ν )) ( − = ( ind GB ( S i − ℓ ( w )2 ( u ∗ ) ⊗ µ ) if ν = w · pµ u = Lie( U ). Note also that, since p > h and ν is dominant, µ must also be dominant.For a dominant weight ν = pµ + w ·
0, observe that, from Lemma 2.6.1 and (2.7.1), we haveH i ( G, H ( ν ) ⊗ H ( ν ∗ ) (1) ) ∼ = Hom G ( V ( ν ) , H i ( G , H ( ν )) ( − ) ∼ = Hom G ( V ( ν ) , ind GB ( S i − ℓ ( w )2 ( u ∗ ) ⊗ µ )) ∼ = Hom B ( V ( ν ) , S i − ℓ ( w )2 ( u ∗ ) ⊗ µ ) . Hence, if H i ( G, H ( ν ) ⊗ H ( ν ∗ ) (1) ) = 0, then ν − µ = ( p − µ + w · i − ℓ ( w )) / ν and n ≥
0, let P n ( ν ) denote the dimension of the ν -weight space of S n ( u ∗ ). Equivalently,for n > P n ( ν ) denotes the number of times that ν can be expressed as a sum of exactly n positiveroots, while P (0) = 1. The function P n is often referred to as Kostant’s Partition Function . By using[ AJ , 3.8], [ KLT , Thm 2], Lemma 2.6.1, and (2.7.1), we can give an explicit formula for the dimension ofH i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) (cf. [ BNP , Proposition 3.2.1, Corollary 3.5.1]).
Proposition . Let p > h and λ = pµ + w · ∈ X ( T ) + . Then dim H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P i − ℓ ( w )2 ( u · λ − µ ) . The following gives a fundamental constraint on non-zero i such thatH i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = Ext iG ( V ( λ ) (1) , H ( λ )) = 0 . It is stated in a more general Ext-context as it will also be used in some inductive arguments for r >
BNP , Proposition 3.4.1]).
Proposition . Let p > h with γ , γ ∈ X ( T ) + , both non-zero, such that γ j = pδ j + w j · with δ j ∈ X ( T ) + and w j ∈ W for j = 1 , . Assume Ext iG ( V ( γ ) (1) , H ( γ )) = 0 . (a) Let σ ∈ Φ + . If Φ is of type G , assume that σ is a long root. Then p h δ , σ ∨ i − h δ , σ ∨ i + ℓ ( w ) + h w · , σ ∨ i ≤ i. (b) If ˜ α denotes the longest root in Φ + , then p h δ , ˜ α ∨ i − h δ , ˜ α ∨ i + ℓ ( w ) − ℓ ( w ) − ≤ i. Equalityrequires that γ − δ = (( i − ℓ ( w )) / α and h− w · , ˜ α ∨ i = ℓ ( w ) + 1 . (c) If γ = γ = pδ + w · , then i ≥ ( p − h δ, ˜ α ∨ i − . Proposition 2.8.1 can be generalized to the following (cf. [
BNP , Proposition 4.3.1]).
Proposition . Let p > h , = λ ∈ X ( T ) + and i ≥ . If H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 , then thereexists a sequence of non-zero weights λ = γ , γ , . . . , γ r − , γ r = λ ∈ X ( T ) + such that γ j = pδ j + u j · forsome u j ∈ W and nonzero δ j ∈ X ( T ) + . Moreover, for each ≤ j ≤ r , there exists a nonnegative integer l j with Ext l j G ( V ( γ j ) (1) , H ( γ j − )) = 0 and P rj =1 l j = i . Furthermore, (2.8.1) i ≥ r X j =1 ( p − h δ j , ˜ α ∨ i − r. Equality requires that pδ j − δ j − + u j · l j − ℓ ( u j − )) / α and that h− u j · , ˜ α ∨ i = ℓ ( u j ) + 1 for all ≤ j ≤ r . Note that the assumption H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 in the proposition can be replaced byExt kG/G ( V ( λ ) ( r ) , H l ( G , H ( λ ))) = 0 , where k + l = i . In that case one arrives at the same conclusions with l = l . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 5
3. Vanishing Ranges in the Simply Laced Case
In this section we obtain some general vanishing information for those cases when the root system Φis simply laced. In such cases, the longest root ˜ α and the longest short root α coincide. Following thediscussion in Section 2, we want to consider whenH i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0for i > λ ∈ X ( T ) + . To gain information on such cohomology groups, we will use Lemma 2.6.1 and (2.7.1). Thefollowing proposition will aid us in showing that certain cohomology groups are non-zero.
Proposition . Let ˜ α denote the longest root of Φ and l be a nonnegative integer. Then Hom G ( V (( l + 1)˜ α ) , ind GB ( S l ( u ∗ ) ⊗ ˜ α )) ∼ = k. Proof.
The claim follows from the diagram below and the fact that all modules in the commutativediagram below have a one-dimensional highest weight space with weight ( l + 1)˜ α . The first embedding is aconsequence of the fact that the module V (˜ α ) ⊗ · · · ⊗ V (˜ α ) | {z } ( l +1) times has a Weyl module filtration. The Weyl module V (˜ α ) is isomorphic to the dual of the adjoint representation, g ∗ . Clearly, g ∗ maps onto u ∗ and V (˜ α ) mapsonto φ − ˜ α (of weight ˜ α ) as B -modules. Hence, we obtain the two B -surjections in the first line of the diagram.The remaining maps and the commutativity of the diagram arise via the universal property of the inductionfunctor. V (( l + 1)˜ α ) ֒ → V (˜ α ) ⊗ · · · ⊗ V (˜ α ) | {z } ( l +1) times u ∗ ⊗ · · · ⊗ u ∗ | {z } l times ⊗ ˜ α ։ ։ S l ( u ∗ ) ⊗ ˜ α ind GB ( S l ( u ∗ ) ⊗ ˜ α ) ↑ ❳❳❳❳❳❳❳❳❳❳③ (cid:3) For a G -module V and a dominant weight γ let [ V ] γ denote the unique maximal summand of V whose composition factors have highest weights linked to γ . Lemma . Assume that the root system Φ of G is simply laced. Let ˜ α denote the longest root anddefine λ = p ˜ α + s ˜ α · p − h + 1)˜ α. Then (a) for any non-zero dominant weight µ linked to zero we have H i ( G, H ( µ ) ⊗ H ( µ ∗ ) ( r ) ) = 0 whenever i < r (2 p − ; (b) for any non-zero dominant weight µ linked to zero we have Ext kG/G ( V ( µ ) ( r ) , H l ( G , H ( µ ))) = 0 whenever k + l < r (2 p − ; (c) [H i ( G , H ( λ )) ( − ] ∼ = ( H ( λ ) if i = 2 p − if < i < p − . Proof.
We apply Proposition 2.8.2. Note that µ being linked to zero forces all the weights δ j ofProposition 2.8.2 to be in the root lattice. This forces h δ j , ˜ α ∨ i ≥
2. Parts (a) and (b) now follow fromequation (2.8.1) and the remark in Section 2.8.For part (c) we make use of Proposition 3.1.1 with l + 1 = p − h + 1 and conclude thatHom G ( V ( λ ) , ind GB ( S ( p − h ) ( u ∗ ) ⊗ ˜ α )) ∼ = k. Note that in the simply laced case ℓ ( s ˜ α ) = 2 h − GB ( S p − h ( u ∗ ) ⊗ ˜ α )) ∼ =H p − ( G , H ( λ ) ( − ) . The weight λ is the smallest non-zero dominant weight in the zero linkage class. Any other non-zeroweight µ in the linkage class will be of the form µ = λ + σ = ( p − h + 1)˜ α + σ , where σ is a non-zero sum ofpositive roots. Clearly µ cannot be a weight of S m ( u ∗ ) ⊗ ˜ α whenever m ≤ p − h. Hence,Hom G ( V ( µ ) , H i ( G , H ( λ )) ( − ) ∼ = Hom G ( V ( µ ) , ind GB ( S i − ℓ ( s ˜ α )2 ( u ∗ ) ⊗ ˜ α )) = 0 CHRISTOPHER P. BENDEL, DANIEL K. NAKANO, AND CORNELIUS PILLEN for all 0 < i ≤ p − . Since H p − ( G , H ( λ )) ( − has a good filtration one obtains[H p − ( G , H ( λ )) ( − ] ∼ = H ( λ ) . Part (b) now implies that [H i ( G , H ( λ ))] = 0 whenever 0 < i < p − . (cid:3) Theorem . Assume that the root system of G is simply laced. Then H r (2 p − ( G ( F q ) , k ) = 0 . Proof.
Let µ be a weight in the zero linkage class. From Lemma 3.2.1(a), we know that H i ( G, H ( µ ) ⊗ H ( µ ∗ ) (1) ) = 0 for i < r (2 p − λ = ( p − h + 1)˜ α . We next show by induction on r thatH r (2 p − ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0. For r = 1, this follows from Lemma 2.6.1 and Lemma 3.2.1(c).For r >
1, we look at the Lyndon-Hochschild-Serre spectral sequence E k,l = Ext kG/G ( V ( λ ) ( r ) , H l ( G , H ( λ ))) ⇒ Ext k + lG ( V ( λ ) ( r ) , H ( λ )) . Lemma 3.2.1(b) implies that the E k,l = 0 for k + l < r (2 p − E k,l = Ext kG ( V ( λ ) ( r − , H l ( G , H ( λ )) ( − ) . Lemma 3.2.1(c) implies that E k,l = 0 for l < p −
3, and, moreover, that H ( λ ) is a summand ofH p − ( G , H ( λ )) ( − . Hence, (cf. [ BNP , Lemma 5.4]), E ( r − p − , p − = Ext ( r − p − G ( V ( λ ) ( r − , H p − ( G , H ( λ )) ( − )has a summand isomorphic to Ext ( r − p − G ( V ( λ ) ( r − , H ( λ )). By induction, this Ext-group is non-zero,and hence E ( r − p − , p − = 0 and transgresses to the E ∞ -page, which implies thatExt r (2 p − G ( V ( λ ) ( r ) , H ( λ )) = 0 . Since λ is the lowest non-zero dominant weight in the zero linkage class, the claim now follows by applyingthe argument given in Section 2.5 to the weight λ and the zero linkage class. (cid:3) In this section we assume that G is simply laced and of adjointtype. The fixed points of the r th iterated Frobenius map on G will again be denoted by G ( F q ). For example,if G is the adjoint group of type A then G ( F q ) is the projective linear group with entries in the field with q elements. Propositions 2.3.1 and 2.3.2 can also be applied to groups of adjoint type. Note that the rootlattice and the weight lattice coincide in this case. Therefore all dominant weights of the form pδ + w · Corollary . Assume that the root system Φ of G is simply laced and that G is of adjoint type.Then (a) H i ( G ( F q ) , k ) = 0 for < i < r (2 p − ; (b) H r (2 p − ( G ( F q ) , k ) = 0 . Note that, for type A n and q − n + 1 being relatively prime, the adjoint and the universal types ofthe finite groups coincide. In these cases the above claim was already observed in Theorems 6.5.1 and 6.14.1of [ BNP ]. Remark . Let G be of type A , D , E and of adjoint type. Let σ denote an automorphism ofthe Dynkin diagram of G . Then σ induces a group automorphism of G that commutes with the Frobeniusmorphism, which we will also denote by σ . Let G σ ( F q ) be the finite group consisting of the fixed points of σ composed with F . Note that σ fixes the maximal root ˜ α . Therefore the discussion in this section alsoapplies to the twisted groups G σ ( F q ) of adjoint type. In particular, Corollary 3.3.1 holds for these groups aswell.
4. Type D n , n ≥ D n , n ≥
4, and that p > h = 2 n −
2. Following Section2, our goal is to find the least i > i ( G, H ( λ ) ⊗ H ( λ ∗ )) = 0 for some λ . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 7
Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some i > λ = pµ + w · µ ∈ X ( T ) + and w ∈ W . From Proposition 2.8.1(c), i ≥ ( p − h µ, ˜ α ∨ i − ω j , h ω j , ˜ α ∨ i = ( j = 1 , n − , n ≤ j ≤ n − . Therefore, if µ = ω , ω n − , ω n , we will have h µ, ˜ α ∨ i ≥ i ≥ p −
3. This reduces us to analyzing thecases when µ = ω , ω n − , ω n , ω . We consider first the case that λ = pω + w · Lemma . Suppose Φ is of type D n with n ≥ and p > n − . Suppose λ = pω + w · ∈ X ( T ) + with w ∈ W . Then (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − n ; (b) if H p − n ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = pω − (2 n − ω = ( p − n + 2) ω . Proof.
Following the discussion in Section 2.7, λ − ω = ( p − ω + w · S j ( u ∗ )for j = i − ℓ ( w )2 . Recall that ω = α + α + · · · + α n − + α n − + α n . Consider the decomposition of − w · ℓ ( w ) distinct positive roots (see Observation 2.2.1). Write ℓ ( w ) = a + b where a is the numberof positive roots in this decomposition that contain α and b is the number of roots in this decompositionthat do not contain α . Then λ − ω contains p − − a copies of α . Since any root contains at most onecopy of α , we have i − ℓ ( w )2 = j ≥ p − − a. Replacing ℓ ( w ) by a + b and simplifying gives i ≥ p − − a + b. The total number of positive roots containing an α is 2 n −
2. Since we necessarily then have a ≤ n − i ≥ p − − (2 n −
2) + b = 2 p − n + b ≥ p − n since b ≥
0. This proves part (a). Furthermore, we see that i = 2 p − n if and only if b = 0 and a = 2 n − − w · n − α . That is, − w · n − ω , which gives part (b). (cid:3) ω continued. We will show in Proposition 4.3.2 thatH p − n ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0for λ = ( p − n + 2) ω . To do this, we will make use of Proposition 2.7.1. We first make some observationsabout relevant partition functions. Note that ω = ǫ . For Φ of type D n , with n ≥
4, and integers m, k , we set P ( m, k, n ) := P u ∈ W ( − ℓ ( u ) P k ( u · mǫ ) if m ≥ , k ≥ , m = 0 , k = 0 , P ( m, k, n ) = dim Hom G ( V ( mǫ ) , H ( G/B, S k ( u ∗ ))) = [ch H ( G/B, S k ( u ∗ )) : ch H ( mǫ )] , when m ≥ , k ≥ , n ≥ Lemma . Suppose Φ is of type D n with n ≥ and m ≥ . (a) P ( m, k, n ) = 0 whenever k < m . (b) P u ∈ W ( − ℓ ( u ) P k ( u · mǫ − ǫ ) = P ( m − , k, n ) . (c) P u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) = P ( m + 1 , k, n ) − P ( m + 1 , k, n − , for n ≥ . CHRISTOPHER P. BENDEL, DANIEL K. NAKANO, AND CORNELIUS PILLEN (d) P u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) = P ( m − , k − n + 2 , n ) . (e) P ( m, k, n ) = P ( m, k, n −
1) + P ( m − , k − n + 2 , n ) , for m ≥ , n ≥ . (f) P ( m, m, n ) = 1 , for n ≥ and m even. Proof. (a) The weight ω = ǫ = ( α + ˜ α ), written as a sum of simple roots, contains one copyof α . Assume that 0 ≤ k < m and n ≥
4. Recall that P ( m, k, n ) = P u ∈ W ( − ℓ ( u ) P k ( u · mǫ ).Note that u · mǫ = u ( mǫ ) + u · u ( ǫ ) = ǫ . Therefore, P ( m, k, n ) = P { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ ). If u ( ǫ ) = ǫ then u · mǫ = mǫ + u · . In addition − u ·
0, written as a sum of simple roots, contains no α . Hence, u · mǫ , written as a sum of simple roots,contains exactly m copies of α . Each positive root of Φ contains at most one copy of α . Therefore at least m positive roots are needed to sum up to u · mǫ . One concludes that P k ( u · mǫ ) = 0 for k < m .(b) Again u · mǫ − ǫ is a sum of positive roots only if u ( ǫ ) = ǫ . Therefore, X u ∈ W ( − ℓ ( u ) P k ( u · mǫ − ǫ ) = X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ − ǫ )= X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k (( m − ǫ + u · X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · ( m − ǫ )= P ( m − , k, n ) . (c) For the expression u · mǫ + ǫ to be a sum of positive roots one needs either u ( ǫ ) = ǫ or u ( ǫ ) = ǫ and u ( ǫ ) = ǫ . Set A = P u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) Then A = X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ + ǫ ) + X { u ∈ W | u ( ǫ )= ǫ ,u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ + ǫ )= X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k (( m + 1) ǫ + u · X { u ∈ W | u ( ǫ )= ǫ ,u ( ǫ )= ǫ } ( − ℓ ( u )+1 P k ( s α mǫ + u · − α + ǫ )= X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · ( m + 1) ǫ ) − X { u ∈ W | u ( ǫ )= ǫ ,u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · ( m + 1) ǫ )= P ( m + 1 , k, n ) − P ( m + 1 , k, n − . (d) We make use of the fact that ω = ǫ is a minuscule weight and obtain: A = [( X i ≥ ( − i ch H i ( G/B, S k ( u ∗ ) ⊗ − ǫ )) : ch H ( mǫ )]= [ch H ( G/B, S k ( u ∗ ) ⊗ − ǫ )) : ch H ( mǫ )]= [ch H ( G/B, S k − n +2 ( u ∗ ) ⊗ ǫ )) : ch H ( mǫ )] (by [ KLT , Lemma 6])= X u ∈ W ( − ℓ ( u ) P k − n − ( u · mǫ − ǫ ) (by [ AJ , 3.8])= P ( m − , k − n + 2 , n ) (by (b)) . Part (e) now follows directly from (c) and (d).(f) If n ≥
5, it follows from (e) and (a) that P ( m, m, n ) = P ( m, m, n − . So the claim holds if it holds for
N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 9 n = 4. If n = 4 and k = m , then (e) has to be replaced by P ( m, m,
4) = X { u ∈ W | u ( ǫ )= ǫ ,u ( ǫ )= ǫ } ( − ℓ ( u ) P m ( u · mǫ ) . Note that the both sides of the equation are zero unless m is even. For even m , a direct computation similarto that in the proof of [ BNP , Lemma 6.11] shows that X { u ∈ W | u ( ǫ )= ǫ ,u ( ǫ )= ǫ } ( − ℓ ( u ) P m ( u · mǫ ) = 1 . (cid:3) Proposition . Suppose Φ is of type D n with n ≥ . Assume that p > n − . Let λ = ( p − n +2) ω . Then H p − n ( G, H ( λ ) ⊗ H ( λ ) (1) ) = k. Proof.
From the previous discussion we know that λ = ( p − n + 2) ω is of the form pω + w · ℓ ( w ) = 2 n −
2. Set k = ( i − l ( w )) /
2. From Proposition 2.7.1 and Lemma 4.3.1(b), one concludesdim H i ( G, H ( λ ) ⊗ H ( λ ) (1) ) = [ch H ( G/B, S k ( u ∗ ) ⊗ ω ) : ch H ( λ )] = P ( p − n + 1 , k, n ) . By Lemma 4.3.1(a), this expression is zero unless k ≥ p − n + 1 and Lemma 4.3.1(d) it follows that P ( p − n + 1 , p − n + 1 , n ) = 1. Replacing k by ( i − n + 2) / i yields the claim. (cid:3) ω n − and ω n . We now consider the case that λ = pω n − + w · λ = pω n + w · w ∈ W with λ ∈ X ( T ) + . Lemma . Suppose Φ is of type D n with n ≥ and p > n − . Suppose λ = pω n − + w · ∈ X ( T ) + or λ = pω n + w · ∈ X ( T ) + with w ∈ W , and H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for i = 0 . Then (a) i ≥ ( p − n ) n ; (b) if n ≥ , then i ≥ p − n + 2 . Proof.
We consider the case of ω n . By symmetry, the case of ω n − can be dealt with in a similarmanner. Following the discussion in Section 2.7, λ − ω n = ( p − ω n + w · S j ( u ∗ ) for j = i − ℓ ( w )2 . Recall that ω n = 12 ( α + 2 α + 3 α + · · · + ( n − α n − ) + ( n − α n − + n α n . Consider the decomposition of − w · ℓ ( w ) = a + b where a is the number of positive roots in this decomposition which contain α n and b is thenumber of roots in this decomposition which do not contain α n . Then λ − ω n contains ( p − n − a copies of α n . Since any root contains at most one copy of α n , we have i − ℓ ( w )2 = j ≥ ( p − n − a. Substituting ℓ ( w ) = a + b , rewriting, and simplifying, we get i ≥ ( p − n − a + b. The total number of positive roots containing α n is ( n − n . Since we necessarily have a ≤ ( n − n and b ≥ i ≥ ( p − n − ( n − n b = ( p − n ) n b ≥ ( p − n ) n For part (b), assume that n ≥
5. We want to show that( p − n ) n ≥ p − n + 2 . This is equivalent to showing that ( p − n ) n ≥ p − n + 4. Consider the left hand side:( p − n ) n = np − n = 4 p + ( n − p − n . Hence the problem is reduced to showing that ( n − p − n ≥ − n + 4 or ( n − p − n + 4 n − ≥
0. Since p ≥ n −
1, we have ( n − p − n + 4 n − ≥ ( n − n − − n + 4 n − n − n = n ( n − ≥ n ≥
5. Part (b) follows. (cid:3)
Note that if n = 4, ( p − n ) n p − p − p − n. D . The following two theorems summarize our findings when the root systemis of type D n . Theorem . Suppose Φ is of type D n with n ≥ . Assume that p > n − . Then (a) H i ( G ( F p ) , k ) = 0 for < i < p − n ; (b) H p − n ( G ( F p ) , k ) = ( k if n ≥ k ⊕ k ⊕ k if n = 4 . Proof.
Part (a) follows from Section 4.1, Lemma 4.2.1(a), Lemma 4.4.1, and Proposition 2.4.1.For part (b), when n ≥
5, it follows from Section 4.1, Lemma 4.2.1, Proposition 4.3.2 and Lemma 4.4.1that λ = (2 p − n + 2) ω is the only dominant weight with H p − n ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0. Since λ isthe lowest weight in its linkage class, the claim follows from Theorem 2.5.1. For n = 4 the symmetry ofthe root system yields H p − n ( G, H ( λ ) ⊗ H ( λ ) (1) ) = k for the weights λ = ( p − ω , λ = ( p − ω and λ = ( p − ω , and those are the only weights with non-zero G -cohomology in degree 2 p − n . Each weightis minimal in its own linkage class. The claim follows. (cid:3) Working inductively from the r = 1 case, we can obtain sharp vanishing bounds for arbitrary r . Theorem . Suppose Φ is of type D n with n ≥ . Assume that p > n − . Then (a) H i ( G ( F q ) , k ) = 0 for < i < r (2 p − n ) ; (b) H r (2 p − n ) ( G ( F q ) , k ) = ( k if n ≥ k ⊕ k ⊕ k if n = 4 . Proof.
For part (a), we need to show that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 for 0 < i < r (2 p − n ) and λ ∈ X ( T ) + . If that is true, then the claim follows from Proposition 2.4.1. For part (b), we require preciseinformation on those dominant weights λ for which H r (2 p − n ) ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0. The root lattice Z Φ has four cosets within X ( T ): Z Φ, { ω + Z Φ } , { ω n − + Z Φ } , and { ω n + Z Φ } . If λ is a weight in the rootlattice claim (a) follows from Lemma 3.2.1(a). Furthermore, no such weights can contribute to cohomologyin degree r (2 p − n ).Assume that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 for some i > λ = pδ + u · δ = δ r ∈ { ω + Z Φ } . For each 1 ≤ j ≤ r , in order to have Ext l j G ( V ( γ j ) (1) , H ( γ j − )) = 0,then γ j − δ j − = pδ j + u j · − δ j − must be a weight in S lj − ℓ ( uj − ( u ∗ ). This implies that pδ j − δ j − must lie inthe positive root lattice. Since δ r ∈ { ω + Z Φ } , we necessarily have pδ r ∈ { ω + Z Φ } . Since pδ r − δ r − ∈ Z Φ,it then follows that δ r − ∈ { ω + Z Φ } . Inductively one concludes that δ j ∈ { ω + Z Φ } for all j .For a weight γ , when expressed as a sum of simple roots, let N ( γ ) denote the number of copies of α that appear. Since a positive root contains at most one copy of α , we have l j − ℓ ( u j − )2 ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 11
From Observation 2.2.1, we know that − u j · ℓ ( u j ) distinct positive roots.Write ℓ ( u j ) = a j + b j where a j denotes the number of roots containing an α and b j the number that donot. Then N ( u j ·
0) = − a j , and rewriting the above gives l j ≥ pN ( δ j ) − N ( δ j − ) − a j + a j − + b j − . Hence, summing over j gives i = r X j =1 l j ≥ r X j =1 (2 p − N ( δ j ) − r X j =1 a j + r X j =1 b j . Recall that the δ j are non-zero dominant weights. By the assumption on δ j , N ( δ j ) ≥
1. The total numberof positive roots containing an α is 2 n −
2. Hence, a j ≤ n −
2. With this, we get(4.5.1) i ≥ r (2 p − − r (2 n −
2) + r X j =1 b j = r (2 p − n ) + r X j =1 b j ≥ r (2 p − n ) , since b j ≥
0. This gives the necessary condition for part (a) for the coset { ω + Z Φ } . Before considering theremaining two cosets, towards addressing part (b), we consider when equality can hold in (4.5.1).As in Section 4.2 we see that equality holds in (4.5.1) if and only if N ( δ j ) = 1 and ℓ ( u j ) = 2 n − , which forces λ = γ j = ( p − n + 2) ω for all j . Moreover, one obtains from the discussion above andProposition 4.3.2 for λ = ( p − n + 2) ω that(4.5.2) [ H i ( G , H ( λ )) ( − ] λ ∼ = ( H ( λ ) if i = 2 p − n < i < p − n. Using the spectral sequence argument in the proof of Theorem 3.2.2 (see also the proof of [
BNP , Lemma5.4]), we can show that H r (2 p − n ) ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = k . We prove this by induction on r with the r = 1 case being Theorem 4.5.1. Consider the Lyndon-Hochschild-Serre spectral sequence E k,l = Ext kG/G ( V ( λ ) ( r ) , H l ( G , H ( λ ))) ∼ = Ext kG ( V ( λ ) ( r − , H l ( G , H ( λ )) ( − ) ⇒ Ext k + lG ( V ( λ ) ( r ) , H ( λ )) . From the remarks in Section 2.8 and the discussion above, E k,l = 0 for k + l < r (2 p − n ). Furthermore,from (4.5.2), E k,l = 0 for l < p − n . Finally, if E k,l = 0 and k + l = r (2 p − n ), then, from the aboveconclusion that γ j = λ for each j , we must have l = 2 p − n . Hence, the E ( r − p − n ) , p − n -term survives to E ∞ and is the only term to contribute in degree r (2 p − n ). Hence, by (4.5.2) and our inductive hypothesis,Ext r (2 p − n ) G ( V ( λ ) ( r ) , H ( λ )) ∼ = Ext ( r − p − n ) G ( V ( λ ) ( r − , H p − n ( G , H ( λ )) ( − ) ∼ = Ext ( r − p − n ) G ( V ( λ ) ( r − , H ( λ )) ∼ = k. To complete the proof of (a) we need to consider the case that λ = pδ + u · δ = δ r ∈{ ω n − + Z Φ } ∪ { ω n + Z Φ } . As above, pδ j + u j · − δ j − must lie in the positive root lattice. Since u j · pδ j − δ j − must lie in the positive root lattice. When expressed as a sum of simple roots ω n − = α + · · · (as does ω n ). Whereas, for 1 ≤ j ≤ n − ω j = α + · · · . Since δ ∈ { ω n − + Z Φ }∪{ ω n + Z Φ } ,for pδ − δ to lie in the positive root lattice, when expressed as a sum of fundamental weights, pδ mustcontain an odd number of copies of ω n − and ω n in total. Since p is odd, this also holds for δ . Inductively,every δ j has this property.We may assume therefore that each δ j contains at least one copy of ω n or one copy of ω n − . Proceed asabove, but let N α n ( γ ) and N α n − ( γ ) denote the number of copies of α n and α n − , respectively, appearingin γ . Set N ( γ ) = max { N α n ( γ ) , N α n − ( γ ) } . Note that, for the weights γ that appear in what follows, both N α n ( γ ) and N α n − ( γ ) are nonnegative. Again, a positive root contains at most one copy of α n or α n − .Just as above, we get l j ≥ pN α n ( δ j ) − N α n ( − u j · − N α n ( δ j − ) + ℓ ( u j − )and the corresponding dual statement for α n − . By choosing the appropriate root we obtain l j ≥ pN ( δ j ) − N α n ( − u j · − N α n ( δ j − ) + ℓ ( u j − ) or l j ≥ pN ( δ j ) − N α n − ( − u j · − N α n − ( δ j − ) + ℓ ( u j − ) . Either one will result in l j ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) + ℓ ( u j − ) . From earlier arguments we know that ℓ ( u j − ) ≥ N ( − u j − · l j ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) + N ( − u j − · . Summing over j , one obtains i = r X j =1 l j ≥ r X j =1 [(2 p − N ( δ j ) − N ( − u j · . Clearly, N ( − u j · ≤ N ( − w ·
0) = n ( n − . Moreover, we can say that N ( δ j ) ≥ n . Substituting this gives i ≥ r (2 p − (cid:16) n (cid:17) − rn ( n − r (cid:18) ( p − n )2 − n ( n − (cid:19) ≥ r (2 p − n ) , where the last inequality follows as in the proof of Lemma 4.4.1. Thus part (a) follows. For n ≥
5, the lastinequality is strict. Hence, λ = ( p − n + 2) ω is the only dominant weight for which H r (2 p − n ) ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0. As in the proof of Theorem 4.5.1, since λ is minimal in its linkage class, part (b) follows.Similarly, for n = 4, by symmetry, part (b) follows. (cid:3)
5. Type E E . Assume for this subsection that Φ is of type E with p > h = 12 (so p ≥ µ with h µ, ˜ α ∨ i < ω and ω . One concludes from Proposition 2.8.1 andProposition 2.4.1 that H i ( G ( F p ) , k ) = 0 for all 0 < i < p − λ of the form pω + w · pω + w · i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some 0 < i < p − . Lemma . Suppose Φ is of type E , p ≥ and λ ∈ X ( T ) + is of the form pω + w · or pω + w · with w ∈ W . Assume in addition that p = 13 , . Then H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for all < i < p − . Proof.
We prove the assertion for λ = pω + w · , w ∈ W. Let N denote the number of times that α appears in − w · α at most once. This implies that N ≤ ℓ ( w ). Moreover, there are exactly 16 distinct positive rootscontaining α . Hence, N ≤ ω = 1 / α + 3 α + 5 α + 6 α + 4 α + 2 α ), we note that λ − ω , written as a sum of simple rootscontains at least 4 / p − − N copies of α . From Section 2.7 we know that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0and i > λ − ω is a sum of ( i − ℓ ( w )) / p −
1) is divisible by 3. Again using the fact that α appears at most once in each positive root, one obtainsthe inequality: 43 ( p − − N ≤ i − ℓ ( w )2 . Solving for i yields i ≥
83 ( p − − N + ℓ ( w ) ≥ p −
1) + 23 ( p − − N ≥ p −
1) + 23 ( p − − . Note that equality holds if and only if N = ℓ ( w ) = 16.One obtains the desired claim i ≥ p −
1) for all primes except those of the form p = 3 t + 1 with13 ≤ p ≤ , i.e., the primes p = 13 and p = 19 . (cid:3) Theorem . Suppose Φ is of type E and p ≥ . (a) If p = 13 , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 13 (ii) H p − ( G ( F p ) , k ) = 0 . (b) If p = 13 , then (i) H i ( G ( F p ) , k ) = 0 for < i < ; (ii) H ( G ( F p ) , k ) = 0 . Proof.
For p = 19, part (a) follows immediately from Lemma 5.1.1, Proposition 2.4.1, and Theo-rem 3.2.2.For the proof of part (b), set p = 13. Part (i) follows from the proof of Lemma 5.1.1. Let W I denotethe subgroup of W generated by the simple reflections s α , ..., s α and let w denote the distinguished repre-sentative of the left coset w W I . Then ℓ ( w ) = 16 and − w · α , which equals the weight 12 ω . Let λ = pω + w · ω . Clearly, k ∼ = Hom G ( V ( λ ) , H ( λ )) ∼ = Hom G ( V ( λ ) , ind GB ( S ( u ∗ ) ⊗ ω )) ∼ = Hom G ( V ( λ ) , ind GB ( S (16 − ℓ ( w )) / ( u ∗ ) ⊗ ω )) ∼ = Hom G ( V ( λ ) , H ( G , H ( λ )) ( − ) ∼ = H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) . Since λ is the smallest dominant weight in its linkage class the assertion follows from the remarks in Sec-tion 2.5.For p = 19, part (a)(ii) follows from Theorem 3.2.2. It remains to show part (a)(i). If H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for i <
35 = 2 p − λ ∈ X ( T ) + , then λ = 19 ω + w · λ = 19 ω + w · w ∈ W . From the proof of Lemma 5.1.1, one can see that i ≥
32. Consider the case that λ = 19 ω + w · ω case is dual and analogous. One can explicitly, with the aid of MAGMA, identify all w ∈ W suchthat λ ∈ X ( T ) + and λ − ω lies in the positive root lattice. By considering the number of copies of α appearing in λ − ω (as in the proof of Lemma 5.1.1), one can identify the least k such that λ − ω is aweight in S k ( u ∗ ), and hence the least possible value of i . The three weights which can give a value of i < k . λ ℓ ( w ) k i = 2 k + ℓ ( w )7 ω + ω
14 10 347 ω + ω
15 9 337 ω
16 8 32For these weights, one can use MAGMA to explicitly compute X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω )in order to apply Proposition 2.7.1. For λ = 7 ω , one finds that in factdim H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P ( u · λ − ω ) = 0and dim H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P ( u · λ − ω ) = 0 . So, for λ = 7 ω , we have i ≥ λ = 7 ω + ω one findsdim H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P ( u · λ − ω ) = 0 . Therefore, i ≥
35 in this case.Finally, for λ = 7 ω + ω , one findsdim H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P ( u · λ − ω ) = 0 . Therefore, i ≥
36 in this case and the claim follows. (cid:3)
We now consider the situation for arbitrary r . Sharp vanishing can be obtained for primes about twicethe Coxeter number. Theorem . Suppose Φ is of type E and p ≥ . (a) If p = 13 , or p = 17 when r is even, then (i) H i ( G ( F q ) , k ) = 0 for < i < r (2 p − ; (ii) H r (2 p − ( G ( F q ) , k ) = 0 . (b) If p = 13 , then (i) H i ( G ( F q ) , k ) = 0 for < i < r ; (ii) H r ( G ( F q ) , k ) = 0 . (c) If p = 17 and r is even, then (i) H i ( G ( F q ) , k ) = 0 for < i < r ; (ii) H r ( G ( F q ) , k ) = 0 . (d) If p = 19 , then (i) H i ( G ( F q ) , k ) = 0 for < i < r ; (ii) H r ( G ( F q ) , k ) = 0 . Proof.
The validity of parts (a)(ii), (c)(ii), and (d)(ii) follows from Theorem 3.2.2. For part (a)(i),we need to show that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 for all dominant weights λ and all 0 < i < r (2 p − Z Φ has three cosetswithin X ( T ): Z Φ, { ω + Z Φ } , and { ω + Z Φ } . If λ is a weight in the root lattice, claim (a)(i) follows fromLemma 3.2.1(a).Assume that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 for some i > λ = pδ + u · δ = δ r ∈ { ω + Z Φ } ∪ { ω + Z Φ } . For each 1 ≤ j ≤ r , in order to haveExt l j G ( V ( γ j ) (1) , H ( γ j − )) = 0, then γ j − δ j − = pδ j + u j · − δ j − must be a weight in S lj − ℓ ( uj − ( u ∗ ). Thisimplies that pδ j − δ j − must lie in the positive root lattice. Since δ r ∈ { ω + Z Φ } ∪ { ω + Z Φ } , we necessarilyhave pδ r ∈ { ω + Z Φ } ∪{ ω + Z Φ } . Since pδ r − δ r − ∈ Z Φ, it then follows that δ r − ∈ { ω + Z Φ } ∪{ ω + Z Φ } .Inductively one concludes that δ j ∈ { ω + Z Φ } ∪ { ω + Z Φ } for all j .Before continuing, we investigate this condition on δ j a bit further. Recall that ω = α + · · · + α and ω = α + · · · + α . Suppose that δ j ∈ { ω + Z Φ } and δ j − ∈ { ω + Z Φ } . In order for pδ j − δ j − to lie in the root lattice, p − = ( p −
1) would need to be an integer. In other words, p − δ j and δ j − lie in { ω + Z Φ } . On the otherhand, suppose that δ j ∈ { ω + Z Φ } and δ j − ∈ { ω + Z Φ } (or vice versa). Then p − = (2 p −
1) (or p − = ( p − p − p is aprime greater than three, either p − p − | ( p − δ j ∈ { ω + Z Φ } or each δ j ∈ { ω + Z Φ } . We refer to this as the “consistent” case. Whereas,if 3 | ( p − δ j s alternately lying in { ω + Z Φ } or { ω + Z Φ } .Note further that since δ = δ r , the alternating case can only occur if r is even .Consider first the consistent case (when 3 | ( p − δ j ∈{ ω + Z Φ } . For a weight γ , when expressed as a sum of simple roots, let N ( γ ) denote the number of copiesof α that appear. Since a positive root contains at most one copy of α , we have l j − ℓ ( u j − )2 ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) . Rewriting this and using the fact that (see Observation 2.2.1) ℓ ( u j − ) ≥ N ( − u j − ·
0) gives l j ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) + ℓ ( u j − ) ≥ pN ( δ j ) − N ( − u j · − N ( δ j − ) + N ( − u j − · . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 15
Therefore, i = r X j =1 l j ≥ r X j =1 (2 pN ( δ j ) − N ( − u j · − N ( δ j − ) + N ( − u j − · r X j =1 ((2 p − N ( δ j ) − N ( − u j · . There are only 16 positive roots which contain an α . Hence, N ( − u j · ≤
16. Since N ( δ j ) ≥ , we get i ≥ r X j =1 (cid:18)
43 (2 p − − (cid:19) = r (cid:18)
43 (2 p − − (cid:19) = r (cid:18) p − p − − (cid:19) . For p ≥
25, we get i ≥ r (2 p −
2) as desired. Note that for p = 17 and p = 23, 3 ∤ ( p − p = 13 and p = 19. For p = 13, we conclude that i ≥ r , and for p = 19, we conclude that i ≥ r .Now consider the alternating case (which requires p − D n case, for a weight γ , let N α ( γ ) (or N α ( γ )) denote the coefficient of α (or α ) when γ is expressed as a sum of simple roots. And then set N ( γ ) = max { N α ( γ ) , N α ( γ ) } (wherethe max is considered only in cases where the quantities involved are nonnegative). Then we reach the sameconclusion on i as above. In this case, p = 13 and p = 19 cannot occur. Moreover, p = 17 and p = 23 arepotentially “bad.” However, for p = 23, since i is an integer, we still conclude that i ≥ r (2 p −
3) as needed.For p = 17, we conclude that i ≥ r .That completes the proof of all parts except for part (b)(ii) with p = 13. This follows inductivelyfrom the r = 1 case by using the the spectral sequence argument as in the proofs of Theorem 3.2.2 andTheorem 4.5.2. (cid:3) For p = 17 when r is even and p = 19, the theorem does not give a sharp vanishing bound. E . Assume for this subsection that Φ is of type E with p > h = 18 (so p ≥ µ with h µ, ˜ α i < ω . Again we conclude from Proposition 2.8.1 and Proposition 2.4.1that H i ( G ( F p ) , k ) = 0 for all 0 < i < p − λ of the form pω + w · i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some 0 < i < p − . Lemma . Suppose Φ is of type E , p ≥ and λ ∈ X ( T ) + is of the form pω + w · with w ∈ W .Assume in addition that p = 19 , . Then H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for all < i < p − . Proof.
Assume λ = pω + w · , w ∈ W. Let N denote the number of times that α appears in − w · α at mostonce. This implies that N ≤ ℓ ( w ). Moreover, there are exactly 27 distinct positive roots containing α .Hence, N ≤ ω as a sum of simple roots the coefficient for α is 3 /
2. Therefore λ − ω , writtenas a sum of simple roots contains at least 3 / p − − N copies of α . From Section 2.7, we know thatH i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 and i > λ − ω is a sum of ( i − ℓ ( w ) / α appears at most once in each positive root, one obtains the inequality:32 ( p − − N ≤ i − ℓ ( w )2 . Solving for i yields i ≥ p − − N + ℓ ( w ) ≥ p −
1) + p − − N ≥ p −
1) + p − − . Note that equality holds if and only if N = ℓ ( w ) = 27.Hence, i ≥ p −
1) for all primes except for 18 < p ≤ , i.e., the primes p = 19 and p = 23 . (cid:3) Theorem . Suppose Φ is of type E and p ≥ . (a) If p = 19 , , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; (ii) H p − ( G ( F p ) , k ) = 0 . (b) If p = 19 , then (i) H i ( G ( F p ) , k ) = 0 for < i < ; (ii) H ( G ( F p ) , k ) = 0 . (c) If p = 23 , then (i) H i ( G ( F p ) , k ) = 0 for < i < ; (ii) H ( G ( F p ) , k ) = 0 . Proof.
Part (a) follows from Lemma 5.2.1, Proposition 2.4.1, and Theorem 3.2.2.For the proof of part (b), set p = 19. Part (i) follows from the proof of Lemma 5.2.1. Let W I denotethe subgroup of W generated by the simple reflections s α , ..., s α and let w denote the distinguished repre-sentative of the left coset w W I . Then ℓ ( w ) = 27 and − w · α , which equals the weight 18 ω . Let λ = pω + w · ω . Using the same argument as for E ,we obtain H ( G , H ( λ )) ( − ∼ = H ( λ ) and hence H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) ∼ = k. Again λ is the smallestdominant weight in its linkage class and the assertion follows from the remarks in Section 2.5.For part (c), set p = 23. If H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for i <
43 = 2 p − λ ∈ X ( T ) + , then λ = 23 ω + w · w ∈ W . From the proof of Lemma 5.2.1, one can see that i ≥
39. One canexplicitly, with the aid of MAGMA, identify all w ∈ W such that λ ∈ X ( T ) + and λ − ω lies in the positiveroot lattice. By considering the number of copies of α appearing in λ − ω (as in the proof of Lemma 5.2.1),one can identify the least k such that λ − ω is a weight in S k ( u ∗ ), and hence the least possible value of i .The four weights which can give a value of i <
43 are listed in the following table along with the minimumpossible value of k . λ ℓ ( w ) k i = 2 k + ℓ ( w )5 ω + ω
24 9 425 ω + ω
25 8 415 ω + ω
26 7 405 ω
27 6 39For these weights, one can use MAGMA to explicitly compute X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω )in order to apply Proposition 2.7.1. For λ = 5 ω , one finds that in factdim H ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P ( u · λ − ω ) = 1 . Since there are no weights less than λ which can give cohomology in degree 40, H ( G ( F p ) , k ) = 0. (cid:3) We next consider the situation for arbitrary r . Theorem . Suppose Φ is of type E and p ≥ . (a) If p = 19 , , then (i) H i ( G ( F q ) , k ) = 0 for < i < r (2 p − ; (ii) H r (2 p − ( G ( F q ) , k ) = 0 . (b) If p = 19 , then (i) H i ( G ( F q ) , k ) = 0 for < i < r ; (ii) H r ( G ( F q ) , k ) = 0 . (c) If p = 23 , then (i) H i ( G ( F q ) , k ) = 0 for < i < r ; (ii) H r ( G ( F q ) , k ) = 0 . Proof.
The validity of part (a)(ii) follows from Theorem 3.2.2. For part (a)(i), we need to show thatH i ( G, H ( λ ) ⊗ H ( λ ∗ ) ( r ) ) = 0 for all dominant weights λ and all 0 < i < r (2 p − Z Φ has two cosets within X ( T ): Z Φand { ω + Z Φ } . If λ is a weight in the root lattice claim (a)(i) follows from Lemma 3.2.1(a). N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 17
Consider then the case that λ ∈ { ω + Z Φ } and apply Proposition 2.8.2. As before, one finds that each δ j ∈ { ω + Z Φ } . Further, if we let N ( γ ) denote the coefficient of α , when γ is expressed as a sum of simpleroots, then we again conclude that i ≥ r X j =1 ((2 p − N ( δ j ) − N ( − u j · . Here, since ω = α + · · · + α , we have N ( δ j ) ≥ . Furthermore, there are 27 positive roots which containan α , and so N ( − u j · ≤
27. Therefore, we get i ≥ r (cid:18)
32 (2 p − − (cid:19) = r (2 p − p − . For p ≥
27, we have i ≥ r (2 p −
3) as needed, which completes part (a).For p = 19 and p = 23, we conclude only that i ≥ r or i ≥ r , respectively, which gives parts (b)(i)and (c)(i). Parts (b)(ii) and (c)(ii) again follows inductively from the r = 1 case by the spectral sequenceargument in Theorem 3.2.2 and Theorem 4.5.2. (cid:3) E . Assume for this subsection that Φ is of type E with p > h = 30 (so p ≥ Theorem . Suppose Φ is of type E and p ≥ . Then (a) H i ( G ( F q ) , k ) = 0 for < i < r (2 p − ; (b) H r (2 p − ( G ( F q ) , k ) = 0 .
6. Type B n , n ≥ B n , n ≥
3, and that p > h = 2 n . Note that type B is equivalent to type C which was discussed in [ BNP ]. However, for certain inductive arguments, at pointswe will allow n = 1 ,
2. Following Section 2, our goal is to find the least i > i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some λ . From Proposition 2.8.1, we know that i ≥ p − Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some i > λ = pµ + w · µ ∈ X ( T ) + and w ∈ W . In this case, the longest root ˜ α = ω . From Proposition 2.8.1, i ≥ ( p − h µ, ˜ α ∨ i − ω j , h ω j , ˜ α ∨ i = ( j = 1 , n ≤ j ≤ n − . Therefore, if µ = ω , ω n , we will have h µ, ˜ α ∨ i ≥ i ≥ p − n is sufficiently large, and λ = pω n + w ·
0, then one also has i ≥ p − Lemma . Suppose Φ is of type B n with n ≥ and p > n . Suppose λ = pω n + w · ∈ X ( T ) + with w ∈ W and H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 . Then i > p − . Proof.
Following the discussion in Section 2.7, λ − ω n = ( p − ω n + w · S j ( u ∗ )for j = i − ℓ ( w )2 . Recall that 2 ω n = α + 2 α + 3 α + · · · + nα n . Consider the decomposition of − w · ℓ ( w ) = a + b + c where a is the number ofpositive roots in this decomposition which contain 2 α n , b is the number of roots in this decomposition whichcontain α n , and c is the number of roots in this decomposition that do not contain α n . Then λ − ω n contains (cid:18) p − (cid:19) n − a − b copies of α n . Since any root contains at most 2 copies of α n , we have i − ℓ ( w )2 = j ≥ (cid:18)(cid:18) p − (cid:19) n − a − b (cid:19) . Replacing ℓ ( w ) by a + b + c and simplifying gives i ≥ (cid:18) p − (cid:19) n − a + c. The total number of positive roots which contain 2 α n is n ( n − /
2. Hence, a ≤ n ( n − / c ≥
0. Sowe have(6.1.1) i ≥ (cid:18) p − (cid:19) n − (cid:18) n − (cid:19) n = (cid:18) p − n (cid:19) n = 2 p + (cid:16) n − (cid:17) p − n . Finally, using the assumption that p ≥ n + 1, one finds i ≥ p + (cid:16) n − (cid:17) (2 n + 1) − n p + n − n − . For n ≥
7, we have i ≥ p − (cid:3) For 3 ≤ n ≤
6, the lemma is in fact false. These cases will be discussed in Sections 6.3 - 6.6. ω . In this section we investigate the case that λ = pω + w ·
0. Throughout thissection Φ is of type B n . In order to make use of some inductive arguments we allow n ≥
1. We will frequentlyswitch between the bases consisting of the simple roots { α , . . . , α n } , the fundamental weights { ω , . . . , ω n } ,and the canonical basis { ǫ , . . . , ǫ n } of R n . Following [ Hum ] we have α i = ǫ i − ǫ i +1 , for 1 ≤ i ≤ n −
1, and α n = ǫ n . Note that ǫ = α is the maximal short root. The fundamental weights are ω j = ǫ + · · · + ǫ j , for1 ≤ j ≤ n −
1, and ω n = 1 / ǫ + · · · + ǫ n ). In particular, ω = ǫ . Definition . For Φ of type B n , with n ≥
1, we define P ( m, k, j, n ) := P u ∈ W ( − ℓ ( u ) P k ( u · ( mǫ + ( ǫ + · · · + ǫ j ))) if m ≥ , k ≥ , and 1 ≤ j ≤ n, m = − , k = 0 , and j = 1 , T ( m, k, j, n ) := dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) , H ( G/B, S k ( u ∗ )) ⊗ H ( ǫ ))if m ≥ , k ≥ , and 1 ≤ j ≤ n, p > n , P ( m, k, j, n ) = dim Hom G ( V ( mǫ + ( ǫ + ... + ǫ j )) , H ( G/B, S k ( u ∗ ))) , which equals the multiplicity of H ( mǫ + ( ǫ + · · · + ǫ j )) in a good filtration of H ( G/B, S k ( u ∗ )) (cf. [ AJ ,3.8]). In particular, P ( m, k, j, n ) ≥ m, k, j , and n . Lemma . Suppose Φ is of type B n with n ≥ and p > n . If m ≥ , k ≥ , and ≤ j ≤ n , then (a) P u ∈ W ( − ℓ ( u ) P k ( u · ( mǫ + ( ǫ + ... + ǫ j )) − ǫ ) = P ( m − , k, j, n ) + P ( m, k, j − , n − ; (b) P u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) = P ( m, k, , n ) − P ( m, k, , n − ; (c) P ( m, k, j, n ) = 0 whenever k < m + 1 ; (d) T ( m, k, j, n ) = P ni =1 P ( m − , k − i + 1 , j, n ) + P ni =1 P ( m, k − i + 1 , j − , n −
1) + P ( m, k − n, j, n ) ; (e) for n ≥ and ≤ j ≤ n − , T ( m, k, j, n ) ≥ P ( m − , k, j, n ) + P ( m, k, j + 1 , n ) + P ( m, k, j − , n ) ; (f) T ( m, k, n, n ) ≥ P ( m − , k, n, n ) + P ( m, k, n, n ) + P ( m, k, n − , n ) ; (g) for l ≥ , P (2 l, k, , n ) = P (2 l − , k − n, , n ) ; (h) for l ≥ , P u ∈ W ( − ℓ ( u ) P k ( u · (2 l + 1) ǫ + ǫ ) = P (2 l, k − n, , n ) . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 19
Proof. (a) P ( m − , k, j, n ) = P u ∈ W ( − ℓ ( u ) P k ( u · (( m − ǫ + ( ǫ + · · · + ǫ j ))). The expression u · (( m − ǫ + ( ǫ + · · · + ǫ j )) = u (( m − ǫ + ( ǫ + · · · + ǫ j )) + u · u ( ǫ ) = ǫ or u ( ǫ ) = ǫ . If u is of the second type, then us α stabilizes ǫ . Setting v = us α , oneobtains P ( m − , k, j, n ) = X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · (( m − ǫ + ( ǫ + · · · + ǫ j )))+ X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · (( m − ǫ + ( ǫ + · · · + ǫ j )))= X u ∈ W ( − ℓ ( u ) P k ( u · ( mǫ + ( ǫ + · · · + ǫ j )) − ǫ ) − X v ∈ W ( − ℓ ( v ) P k ( v · ( mǫ + ( ǫ + · · · + ǫ j ))) . The second term is just P ( m, k, j − , n −
1) as claimed. Note that the formula also holds for m = 0.(b) The expression P u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) will be a sum of positive roots only if either u ( ǫ ) = ǫ or u ( ǫ ) = ǫ . Arguing as above one obtains X u ∈ W ( − ℓ ( u ) P k ( u · mǫ + ǫ ) = X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ + ǫ )+ X { u ∈ W | u ( ǫ )= ǫ } ( − ℓ ( u ) P k ( u · mǫ + ǫ )= X u ∈ W ( − ℓ ( u ) P k ( u · ( mǫ + ǫ )) − X v ∈ W ( − ℓ ( v ) P k ( v · ( mǫ + ǫ )) . The first term is P ( m, k, , n ) and the second term is just P ( m, k, , n − ≤ k < m + 1. Part (a) implies that P ( m, k, j, n ) ≤ X u ∈ W ( − ℓ ( u ) P k ( u · (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) − ǫ ) . Note that u · (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) − ǫ is a sum of positive roots if and only if u ( ǫ ) = ǫ . If u ( ǫ ) = ǫ then u · (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) − ǫ = ( m + 1) ǫ + u · ( ǫ + · · · + ǫ j ) and − u · ( ǫ + · · · + ǫ j ), writtenas a sum of simple roots, contains no α . However, ( m + 1) ǫ , written as a sum of simple roots, containsexactly m + 1 copies of α . Each positive root of Φ contains at most one copy of α . Therefore at least m + 1 positive roots are needed to obtain the weight u · (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) − ǫ . One concludesthat P k ( u · (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) − ǫ ) = 0 and the assertion follows.(d) For a simple root α , let P α denote the minimal parabolic subgroup corresponding to α , and let u α denote the Lie algebra of the unipotent radical of P α . From the short exact sequence0 → α → u ∗ → u ∗ α → → S k − ( u ∗ ) ⊗ α → S k ( u ∗ ) → S k ( u ∗ α ) → . Tensoring with a weight µ yields0 → S k − ( u ∗ ) ⊗ α ⊗ µ → S k ( u ∗ ) ⊗ µ → S k ( u ∗ α ) ⊗ µ → . Induction from B to G yields the long exact sequence(6.2.1) · · · → H i ( G/B, S k − ( u ∗ ) ⊗ α ⊗ µ ) → H i ( G/B, S k ( u ∗ ) ⊗ µ ) → H i ( G/B, S k ( u ∗ α ) ⊗ µ ) → · · · . We apply (6.2.1) with α = α j = ǫ j − ǫ j +1 and µ = − ǫ j , where 1 ≤ j ≤ n −
1, giving(6.2.2) · · · → H i ( G/B, S k − ( u ∗ ) ⊗ − ǫ j +1 ) → H i ( G/B, S k ( u ∗ ) ⊗ − ǫ j ) → H i ( G/B, S k ( u ∗ α ) ⊗ − ǫ j ) → · · · . Note that, h− ǫ j , α ∨ j i = −
1, forces H i ( P α /B, − ǫ j ) = 0 for all i . The spectral sequence H r ( G/P α , S k ( u ∗ α )) ⊗ H s ( P α /B, µ ) ⇒ H r + s ( G/B, S k ( u ∗ α ) ⊗ µ )yields H i ( G/B, S k ( u ∗ α ) ⊗ − ǫ j ) = 0 for all i . Therefore, from (6.2.2), one obtains for 1 ≤ j ≤ n − i ≥ H i ( G/B, S k − ( u ∗ ) ⊗ − ǫ j +1 ) ∼ = H i ( G/B, S k ( u ∗ ) ⊗ − ǫ j ) . Iterating this process yields(6.2.4) H i ( G/B, S k ( u ∗ ) ⊗ − ǫ i ) ∼ = H i ( G/B, S k − n + i ( u ∗ ) ⊗ − ǫ n ) . Note that if k < n − i , the right hand side is identically zero, and the isomorphism still holds.Next we apply (6.2.1) with α = α n = ǫ n and µ = − ǫ n in order to obtain(6.2.5) · · · → H i ( G/B, S k − ( u ∗ )) → H i ( G/B, S k ( u ∗ ) ⊗ − ǫ n ) → H i ( G/B, S k ( u ∗ α ) ⊗ − ǫ n ) → · · · . Here h− ǫ n , α ∨ n i = −
2. Using the spectral sequence as above, one obtains H i ( G/B, S k ( u ∗ α ) ⊗ − ǫ n ) ∼ = H i − ( G/P α , S k ( u ∗ α )) ⊗ H ( P α /B, − ǫ n ) ∼ = H i − ( G/P α , S k ( u ∗ α )) ∼ = H i − ( G/B, S k ( u ∗ α )) . (6.2.6)Since H ( G/B, S k ( u ∗ α ) ⊗ − ǫ n ) = 0, one obtains via by (6.2.5),(6.2.7) H ( G/B, S k − ( u ∗ )) ∼ = H ( G/B, S k ( u ∗ ) ⊗ − ǫ n ) . From H i ( G/B, S k ( u ∗ )) = 0 for i ≥
1, using (6.2.5), one concludes(6.2.8) H i ( G/B, S k ( u ∗ ) ⊗ − ǫ n ) ∼ = H i − ( G/B, S k ( u ∗ α )) for i ≥ . Next, we apply (6.2.1) with α = α n = ǫ n and µ = 0 in order to obtain(6.2.9) · · · → H i ( G/B, S k − ( u ∗ ) ⊗ ǫ n ) → H i ( G/B, S k ( u ∗ )) → H i ( G/B, S k ( u ∗ α )) → · · · . From H i ( G/B, S k ( u ∗ )) = 0 and H i ( G/B, S k ( u ∗ ) ⊗ ǫ n ) = 0 for i ≥ KLT , 2.8], one concludes(6.2.10) 0 → H ( G/B, S k − ( u ∗ ) ⊗ ǫ n ) → H ( G/B, S k ( u ∗ )) → H ( G/B, S k ( u ∗ α )) → , and(6.2.11) H i ( G/B, S k ( u ∗ ) ⊗ − ǫ n ) ∼ = H i − ( G/B, S k ( u ∗ α )) = 0 for i ≥ . Similarly, apply (6.2.1) with α = α j = ǫ j − − ǫ j and µ = ǫ j , where 2 ≤ j ≤ n , to obtain · · · → H i ( G/B, S k − ( u ∗ ) ⊗ ǫ j − ) → H i ( G/B, S k ( u ∗ ) ⊗ ǫ j ) → H i ( G/B, S k ( u ∗ α ) ⊗ ǫ j ) → · · · . As before this yields(6.2.12) H i ( G/B, S k − ( u ∗ ) ⊗ ǫ j − ) ∼ = H i ( G/B, S k ( u ∗ ) ⊗ ǫ j ) . Iterating this process results in(6.2.13) H i ( G/B, S k − i +1 ( u ∗ ) ⊗ ǫ ) ∼ = H i ( G/B, S k ( u ∗ ) ⊗ ǫ i ) . Again, this isomorphism holds even if k < i − B -module M , we denote its Euler characteristic by χ ( M ) = X i ≥ ( − i ch H i ( G/B, M ) . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 21
From the above, one obtainsch( H ( G/B, S k ( u ∗ )) ⊗ H ( ǫ )) = χ ( S k ( u ∗ ) ⊗ H ( ǫ ))= n X i =1 χ ( S k ( u ∗ ) ⊗ ǫ i ) + χ ( S k ( u ∗ )) + n X i =1 χ ( S k ( u ∗ ) ⊗ − ǫ i )= n X i =1 χ ( S k − i +1 ( u ∗ ) ⊗ ǫ ) + χ ( S k ( u ∗ )) + n X i =1 χ ( S k − i +1 ( u ∗ ) ⊗ − ǫ n ) (by 6.2.4, 6.2.13)= n X i =1 ch H ( G/B, S k − i +1 ( u ∗ ) ⊗ ǫ ) + ch H ( G/B, S k ( u ∗ )) (by [ KLT , 2.8])+ n X i =1 ch H ( G/B, S k − i +1 ( u ∗ ) ⊗ − ǫ n ) − n X i =1 ch H ( G/B, S k − i +1 ( u ∗ ) ⊗ − ǫ n )(by 6.2.8, 6.2.11)= n X i =1 ch H ( G/B, S k − i +1 ( u ∗ ) ⊗ ǫ ) + ch H ( G/B, S k ( u ∗ ))+ n X i =1 ch H ( G/B, S k − i ( u ∗ )) + n X i =1 ch H ( G/B, S k − i − n +1 ( u ∗ ) ⊗ ǫ ) − n X i =1 ch H ( G/B, S k − i +1 ( u ∗ )) (by 6.2.7, 6.2.8, 6.2.10, 6.2.13)= n X i =1 ch H ( G/B, S k − i +1 ( u ∗ ) ⊗ ǫ ) + ch H ( G/B, S k − n ( u ∗ )) . The last equality yields T ( m, k, j, n ) = dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) , H ( G/B, S k ( u ∗ )) ⊗ H ( ǫ ))= n X i =1 dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) , H ( G/B, S k − i +1 ( u ∗ ) ⊗ ǫ ))+ dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) , H ( G/B, S k − n ( u ∗ )) . The assertion follows now from part (a).(e) A direct computation shows thatch( V ( mǫ + ( ǫ + · · · + ǫ j )) ⊗ V ( ǫ )) = ch V (( m − ǫ + ( ǫ + · · · + ǫ j ))+ ch V ( mǫ + ( ǫ + · · · + ǫ j +1 )) + ch V ( mǫ + ( ǫ + · · · + ǫ j − ))+ ch V (( m + 1) ǫ + ( ǫ + · · · + ǫ j )) + ch V (( m − ǫ + ( ǫ + ǫ ) + ( ǫ + · · · + ǫ j )) . It follows that T ( m, k, j, n ) = dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) , H ( G/B, S k ( u ∗ )) ⊗ H ( ǫ ))= dim Hom G ( V ( mǫ + ( ǫ + · · · + ǫ j )) ⊗ V ( ǫ )) , H ( G/B, S k ( u ∗ ))) ≥ P ( m − , k, j, n ) + P ( m, k, j + 1 , n ) + P ( m, k, j − , n ) , as claimed.Part (f) follows in similar fashion.(g) It is well-known that, for m ≥
2, ch( H ( mω )) is equal to the difference of the m th and the ( m − l thsymmetric power equals the dimension of the zero weight space of the (2 l + 1)st symmetric power. The sameis true for the pair H (2 lω ) and H ((2 l + 1) ω ). It follows from Kostant’s Theorem [ Hum , 24.2] that(6.2.14) X k ≥ P (2 l − , k, , n ) = X k ≥ P (2 l, k, , n ) . From (6.2.10) and (6.2.13) one obtains0 → H ( G/B, S k − n ( u ∗ ) ⊗ ǫ ) → H ( G/B, S k ( u ∗ )) → H ( G/B, S k ( u ∗ α n )) → . All three modules have good filtrations. Moreover, by part (i)dim Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k − n ( u ∗ ) ⊗ ǫ )) = P (2 l − , k − n, , n ). Hence for l ≥ P (2 l, k, , n ) = P (2 l − , k − n, , n ) + dim Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k ( u ∗ α n ))) . Summing over all k ≥ X k ≥ P (2 l − , k, , n ) = X k ≥ P (2 l, k, , n ) + X k ≥ dim Hom G ( V (2 lǫ + ǫ )) , H ( G/B, S k ( u ∗ α n ))) . Comparing with (6.2.14) yields dim Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k ( u ∗ α n ))) = 0, which forces P (2 l, k, , n ) = P (2 l − , k − n, , n ), for all k ≥ AJ , 3.8], the multiplicity of ch V (2 lǫ + ǫ ) in χ ( S k ( u ∗ ) ⊗ − ǫ ) equals X u ∈ W ( − ℓ ( u ) P k ( u · ((2 l + 1) ǫ ) + ǫ ) . Moreover, by (6.2.11) and (6.2.4), χ ( S k ( u ∗ ) ⊗ − ǫ ) = ch H ( G/B, S k ( u ∗ ) ⊗ − ǫ ) − ch H ( G/B, S k ( u ∗ ) ⊗ − ǫ )) . In addition, from (6.2.8) and (6.2.4), the vanishing of Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k ( u ∗ α n ))) forces thevanishing of Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k ( u ∗ ) ⊗ − ǫ )), for all k . Hence, P u ∈ W ( − ℓ ( u ) P k ( u · ((2 l +1) ǫ ) + ǫ ) = dim Hom G ( V (2 lǫ + ǫ ) , H ( G/B, S k ( u ∗ ) ⊗ − ǫ )), which equals P (2 l, k − n, , n ) by (6.2.4) and(6.2.7). (cid:3) Proposition . Suppose Φ is of type B n with n ≥ and p > n . For m ≥ , k ≥ , and ≤ j ≤ n ,define (6.2.15) t ( m, j, n ) = m + j +12 if j is odd and m is odd ,m + j if j is even and m is even ,m + 1 + n − j if j is even and m is odd ,m + 1 + n − j − if j is odd and m is even . Then P ( m, k, j, n ) = 0 whenever k < t ( m, j, n ) . Proof.
By Lemma 6.2.2(c), P ( m, k, j, n ) = 0 if k < m +1. We will prove the slightly stronger statementin the proposition inductively. To do so, we make some general observations. Define T ′ ( m, k, j, n ) = n X i =2 P ( m − , k − i + 1 , j, n ) + n X i =1 P ( m, k − i + 1 , j − , n −
1) + P ( m, k − n, j, n ) . Observe that by Lemma 6.2.2(d), T ′ ( m, k, j, n ) = T ( m, k, j, n ). Note further that if r is the smallest valueof k for which T ′ ( m, k, j, n ) = 0, then P ( m − , r − , j, n ) + P ( m, r, j − , n −
1) + P ( m, r − n, j, n ) = 0.Suppose that P ( m − , k, j, n ) = 0 whenever k < t ( m − , j, n ) and that P ( m, k, j − , n −
1) = 0 whenever k < t ( m, j − , n − T ′ ( m, k, j, n ) = 0 whenever k < min { t ( m − , j, n ) + 1 , t ( m, j − , n − , m + 1 + n } . Moreover, parts (d) and (e) of Lemma 6.2.2 would imply that, for 2 ≤ j ≤ n − P ( m, k, j + 1 , n ) + P ( m, k, j − , n ) = 0 whenever T ′ ( m, k, j, n ) = 0 , and from Lemma 6.2.2(f)(6.2.18) P ( m, k, n, n ) + P ( m, k, n − , n ) = 0 whenever T ′ ( m, k, n, n ) = 0 . In order to prove the proposition, we will use induction on n and or j . If n = 1 the claim follows frompart (c) of Lemma 6.2.2. Moreover, parts (c) and (d) of the Lemma 6.2.2 imply that the claim holds for j = 1 and n ≥ Step 1:
Here we will show that P ( m, k, j, n ) = 0, whenever k < t ( m, j, n ) and m + j is odd. We will useinduction on j . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 23
Assumption: P ( m, k, l, n ) = 0, whenever k < t ( m, l, n ), m + l is odd, and l ≤ j .Suppose that m + j + 1 is odd. Then m + j − P ( m, k, j + 1 , n ) = 0 whenever k < min { t ( m − , j, n ) + 1 , t ( m, j − , n − , m + 1 + n } . It suffices therefore to verify that(6.2.19) t ( m, j + 1 , n ) ≤ min { t ( m − , j, n ) + 1 , t ( m, j − , n − , m + 1 + n } . From (6.2.15) it follows that t ( m, j + 1 , n ) = ( m + 1 + n − ( j +1) − = m + n − j if j is even and m is even ,m + 1 + n − ( j +1)2 = m + n − j − if j is odd and m is odd , while t ( m − , j, n ) + 1 = ( m + n − j + 1 = m + n − j + 1 if j is even and m is even ,m + n − j − + 1 = m + n − j − if j is odd and m is odd , and t ( m, j − , n −
1) = ( m + 1 + n − − ( j − − = m + n − j if j is even and m is even ,m + 1 + n − − ( j − = m + n − j − if j is odd and m is odd . Inequality (6.2.19) indeed holds and Step 1 is complete.
Step 2:
Here we will show that P ( m, k, n, n ) = 0, whenever k < t ( m, n, n ) and m + n is even.Suppose that m + n is even. Step 1 implies that (6.2.16) holds for j = n . Together with (6.2.18) oneobtains P ( m, k, n, n ) = 0 whenever k < min { t ( m − , n, n ) + 1 , t ( m, n − , n − , m + 1 + n } . It suffices therefore to verify that t ( m, n, n ) ≤ min { t ( m − , n, n ) + 1 , t ( m, n − , n − , m + 1 + n } . This can easily be done by looking at (6.2.15). It is left to the interested reader.
Step 3:
Here we will show that P ( m, k, j, n ) = 0 whenever k < t ( m, j, n ) and m + j is even. We useinduction on n and on j . For j we work in decreasing order. The case j = n was settled above. Assumption:
We assume that P ( m, k, l, n −
1) = 0 whenever k < t ( m, l, n − P ( m, k, l, n ) = 0 whenever k < t ( m, l, n ), m + l is even, and l ≥ j .Suppose that m + j − P ( m, k, j − , n ) = 0 whenever k < min { t ( m − , j, n ) + 1 , t ( m, j − , n − , m + 1 + n } . It suffices therefore to verify that(6.2.20) t ( m, j − , n ) ≤ min { t ( m − , j, n ) + 1 , t ( m, j − , n − , m + 1 + n } . From (6.2.15) one obtains: t ( m, j − , n ) = ( m + j − if j is odd and m is even ,m + j if j is even and m is odd . while t ( m − , j, n ) + 1 = ( m + j − + 1 if j is odd and m is even ,m + j if j is even and m is odd , and t ( m, j − , n −
1) = ( m + j − if j is odd and m is even ,m + j if j is even and m is odd . This proves inequality (6.2.20). (cid:3)
Theorem . Suppose Φ is of type B n with n ≥ . Assume that p > n . Let λ = pω + w · be adominant weight. Then (a) H i ( G, H ( λ ) ⊗ H ( λ ) (1) ) = 0 for < i < p − , whenever ℓ ( w ) is even; (b) H i ( G, H ( λ ) ⊗ H ( λ ) (1) ) = 0 for < i < p − , whenever ℓ ( w ) is odd; (c) H p − ( G ( F p ) , k ) = 0 . Proof.
The set of dominant weights of the form λ = pω + w ·
0, written in the ǫ -basis, are( p − ℓ ( w ) − ǫ + ( ǫ + · · · + ǫ ℓ ( w )+1 ) , with 0 ≤ ℓ ( w ) ≤ n − , and ( p − ℓ ( w ) − ǫ + ( ǫ + · · · + ǫ n − ℓ ( w ) ) , with n ≤ ℓ ( w ) ≤ n − . Using Proposition 6.2.3 and Lemma 6.2.2(a) , a direct computation shows that X u ∈ W ( − ℓ ( u ) P k ( u · (( p − ℓ ( w ) − ǫ + ( ǫ + · · · + ǫ ℓ ( w )+1 )) − ǫ ) = 0 whenever k < t, where t = ( ( p − − ℓ ( w )2 for 0 ≤ ℓ ( w ) ≤ n − ℓ ( w ) even , ( p − − ℓ ( w )+12 for 0 ≤ ℓ ( w ) ≤ n − ℓ ( w ) odd , and X u ∈ W ( − ℓ ( u ) P k ( u · (( p − ℓ ( w ) − ǫ + ( ǫ + .... + ǫ n − ℓ ( w ) )) − ǫ ) = 0 whenever k < t, where t = ( ( p − − ℓ ( w )2 for n ≤ ℓ ( w ) ≤ n − ℓ ( w ) even , ( p − − ℓ ( w )+12 for n ≤ ℓ ( w ) ≤ n − ℓ ( w ) odd . Parts (a) and (b) follow from Proposition 2.7.1. Note that i = 2 k + ℓ ( w ).Let λ be the lowest dominant weight of the form pω + w ·
0. Then λ = ( p − n + 1) ǫ and ℓ ( w ) = 2 n − X u ∈ W ( − ℓ ( u ) P p − n − ( u · (( p − n + 1) ǫ ) − ǫ ) = 0 . By Lemma 6.2.2(a) this is equivalent to showing that P ( p − n − , p − n − , , n ) is not zero. Lemma6.2.2(b) and (h) imply that(6.2.22) P (2 l + 1 , k, , n ) = P (2 l, k − n, , n ) + P (2 l + 1 , k, , n − . Note that (6.2.22) also holds for l = −
1. Obviously P (2 l − , l, ,
1) = 1. It follows inductively from (6.2.22)that P (2 l − , l, , n ) = 0 , for all n ≥ , l ≥ . From Lemma 6.2.2(g) one obtains now that P (2 l, l + n, , n ) =0 . Setting 2 l = p − n − P ( p − n − , p − n − , , n ) = 0. Hence, (6.2.21) holds. In Proposition2.7.1, i = 2 k − ℓ ( w ) = 2( p − n −
1) + 2 n − p − p − ( G, H ( λ ) ⊗ H ( λ ) (1) ) = 0.The weight λ is the lowest non-zero weight in its linkage class. Part (c) of the theorem follows now from thediscussion after Theorem 2.5.1. (cid:3) B . Let Φ be of type B with p > h = 6 (so p ≥ i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for 0 < i < p −
3, we must have λ = pω + w · w ∈ W . With the aid of MAGMA [ BC , BCP ] or other software, one can explicitly compute all w · λ are dominant. Further, λ − ω must be a weight of S i − ℓ ( w )2 ( u ∗ ). By directcomputation, one can determine the least possible value of k for which λ − ω can be a weight of S k ( u ∗ ).The following table summarizes the weights which can give a value of i < p − λ = pω + w · ℓ ( w ) k i = 2 k + ℓ ( w )( p − ω + 2 ω p − p − p − ω + ω p − p − p − ω p − p − Lemma . Suppose that Φ is of type B with p ≥ . Let λ = pµ + w · ∈ X ( T ) + with µ ∈ X ( T ) + and w ∈ W . (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − . (b) If H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 25 (c) H p − ( G, H (( p − ω ) ⊗ H (( p − ω ∗ ) (1) ) = k. (d) If H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω + ω or λ = ( p − ω + 2 ω . (e) H p − ( G ( F p ) , k ) = k . Proof.
Parts (a), (b), and (d) follow from the discussion preceding the lemma. Part (c) follows fromProposition 2.7.1 and Lemma 6.3.3 below with m = p −
7. Since the weights in part (d) are larger than( p − ω , by Theorem 2.5.1 and Theorem 6.2.4, we obtain part (e). (cid:3) Remark . The weights in part (d) also appear to give cohomology classes as verified for p = 7 , , p = 7, λ = ( p − ω + ω gives a one-dimensional cohomology group. But for p = 11 , λ = ( p − ω + 2 ω gives aone-dimensional cohomology group. Lemma . Suppose that Φ is of type B . Let m ≥ be an even integer. Then X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = 1 . Proof.
Let n be such that m = 2 n . For n = 0, the claim readily follows, so we assume that n ≥ We work with the epsilon basis for the root system. Then the positive roots are ǫ , ǫ , ǫ , ǫ + ǫ , ǫ + ǫ , ǫ + ǫ , ǫ − ǫ , ǫ − ǫ , and ǫ − ǫ . Further ω = ( ǫ + ǫ + ǫ ). Relative to the ǫ basis, for any u ∈ W , u ( ǫ i ) = ± ǫ j . That is, u permutes the ǫ i up to a sign.For u ∈ W , set x u := u · (( m + 1) ω ) − ω . Using the fact that 2 ρ = 5 ǫ + 3 ǫ + ǫ , one finds that(6.3.1) x u = u (( n + 3) ǫ + ( n + 2) ǫ + ( n + 1) ǫ ) − ǫ − ǫ − ǫ . By direct calculation, one can identify all u ∈ W for which x u is a positive root. There are twelve suchelements which are summarized in the following table (using permutation notation) along with the parityof their lengths. An element marked with a superscript negative sign denotes the operation which consistsof the given permutation of the ǫ i s followed by sending ǫ to − ǫ . For example, let u = (123) − . Then u ( ǫ ) = ǫ , u ( ǫ ) = − ǫ , and u ( ǫ ) = ǫ . u (1) (12) (13) (23) (123) (132) (1) − (12) − (13) − (23) − (123) − (132) − ℓ ( u ) even odd odd odd even even odd even even even odd oddFor these twelve u , using (6.3.1), one can explicitly compute x u . The values are summarized in thefollowing table. Recall that m = 2 n . For small values of n , some of these cannot be sums of positive roots.The necessary condition on n is given in the third column. u x u := u · (( m + 1) ω ) − ω positive root sum(1) nǫ + nǫ + nǫ n ≥ n − ǫ + ( n + 1) ǫ + nǫ n ≥ n − ǫ + nǫ + ( n + 2) ǫ n ≥ nǫ + ( n − ǫ + ( n + 1) ǫ n ≥ n − ǫ + ( n + 1) ǫ + ( n + 1) ǫ n ≥ n − ǫ + ( n − ǫ + ( n + 2) ǫ n ≥ − nǫ + nǫ − ( n + 2) ǫ n ≥ − ( n − ǫ + ( n + 1) ǫ − ( n + 2) ǫ n ≥ − ( n − ǫ + nǫ − ( n + 4) ǫ n ≥ − nǫ + ( n − ǫ − ( n + 3) ǫ n ≥ − ( n − ǫ + ( n + 1) ǫ − ( n + 3) ǫ n ≥ − ( n − ǫ + ( n − ǫ − ( n + 4) ǫ n ≥ P n ( x u ) for these twelve valuesof u . We show below that there are four pairs of u s for which the lengths have opposite parity and thevalues of P n ( x u ) are the same. Hence those cancel each other out. We will further show that there is also Indeed, for small values of n the claim can be verified by hand, and it has been verified for n ≤ a relationship between the remaining partition functions that will lead to the desired claim. To see theserelationships, we make a few observations whose proofs are left to the interested reader. Observation . Let x = a ǫ + a ǫ + a ǫ with a + a + a = 3 n . Suppose that x is expressed asa sum of 2 n positive roots.(a) At least n of the roots have the form ǫ i + ǫ j (not necessarily all the same).(b) For any pair i, j ∈ { , , } (with i = j ), if a i + a j = 2 n + c , then the root sum decompositioncontains at least c copies of ǫ i + ǫ j . Observation . Let x = a ǫ + a ǫ − a ǫ with 1 ≤ a , a < a and a + a > n . Suppose that x is expressed as a sum of 2 n positive roots. Then at least one of the roots is ǫ − ǫ and at least one is ǫ − ǫ .We now identify the four pairs (of opposite parity) where P n ( x u ) is the same. Case 1. (13) and (132).If n = 1, as noted above, x (13) = − ǫ + ǫ + 3 ǫ cannot be expressed as a sum of positive roots. On theother hand, x (123) = 3 ǫ can be. However, it cannot be expressed as a sum of 2 n = 2 positive roots. So weassume n ≥
2. If x (13) is expressed as a sum of 2 n positive roots, by Observation 6.3.4(b), at least one ofthe roots is ǫ + ǫ (in fact, at least two). Hence, P n ( x (13) ) = P n − ( x (13) − ( ǫ + ǫ )) = P n − (( n − ǫ +( n − ǫ + ( n + 1) ǫ ). Similarly, if x (132) is expressed as a sum of 2 n positive roots, at least one of the rootsis ǫ + ǫ (and one is ǫ + ǫ ). Hence, P n ( x (132) ) = P n − (( n − ǫ + ( n − ǫ + ( n + 1) ǫ ) = P n ( x (13) ).For the remaining three pairs, if n is not sufficiently large for x u to admit a positive root sum decompo-sition, then P n ( x u ) = 0 in both cases. So we assume in what follows that n is sufficiently large to admit aroot sum decomposition. Case 2. (1) − and (12) − .If x (1) − is expressed as a sum of 2 n positive roots, by Observation 6.3.5, at least one of the roots is ǫ − ǫ . Hence, removing this root, P n ( x (1) − ) = P n − (( n − ǫ + nǫ − ( n + 1) ǫ ). Similarly, again byObservation 6.3.5, if x (12) − is expressed as a sum of 2 n positive roots, then at least one of the roots is ǫ − ǫ .Hence, P n ( x (12) − ) = P n − (( n − ǫ + nǫ − ( n + 1) ǫ ) = P n ( x (1) − ) . Case 3. (13) − and (132) − .Similar to Case 2, by removing an ǫ − ǫ for (13) − and removing an ǫ − ǫ for (132) − , one finds that P n ( x (12) − ) = P n − (( n − ǫ + ( n − ǫ − ( n + 3) ǫ ) = P n ( x (132) − ). Case 4. (23) − and (123) − .Again, similar to Case 2 with a slight generalization of Observation 6.3.5, by removing two copies of ǫ − ǫ for (23) − and two copies of ǫ − ǫ for (123) − , one finds that P n ( x (23) − ) = P n − (( n − ǫ + ( n − ǫ − ( n + 1) ǫ ) = P n ( x (123) − ).From Cases 1-4, we have that(6.3.2) X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = P n ( x (1) ) − P n ( x (12) ) − P n ( x (23) ) + P n ( x (123) ) . We now deduce several relationships among the terms on the right hand side. If n = 1, only the first threeterms can be non-zero, and one can readily check that the claim holds. So we assume that n ≥
2. Note thatthe following argument does still hold even when n = 1.Consider the identity element (1). Write P n ( x (1) ) = M + M + M where M denotes the number ofroot sum decompositions which contain an ǫ + ǫ , M denotes the number which contain an ǫ + ǫ butnot an ǫ + ǫ , and M denotes the number which contain neither an ǫ + ǫ nor an ǫ + ǫ . For M , byassumption, the decomposition contains an ǫ + ǫ . Removing this root gives(6.3.3) M = P n − (( n − ǫ + ( n − ǫ + nǫ ) . For M , by assumption, the decomposition contains an ǫ + ǫ . Removing this root gives(6.3.4) M = P ∗ n − (( n − ǫ + nǫ + ( n − ǫ ) , where P ∗ denotes the fact that we are only counting decompositions which contain no copies of ǫ + ǫ . Byassumption, M is the number of root decompositions of nǫ + nǫ + nǫ (into 2 n positive roots) which donot contain an ǫ + ǫ nor an ǫ + ǫ . By Observation 6.3.4(a), any such decomposition contains at least n N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 27 copies of ǫ + ǫ . Removing those leaves nǫ which must be expressed as a sum of n positive roots withoutusing ǫ + ǫ nor ǫ + ǫ . There is clearly only one such decomposition (using n copies of ǫ ). Hence, M = 1.Consider now the word (12). Write P n ( x (12) ) = N + N where N denotes the number of root sumdecompositions which contain at least one copy of ǫ + ǫ and N denotes the number which do not containan ǫ + ǫ . In the first case, by removing an ǫ + ǫ , we have(6.3.5) N = P n − (( n − ǫ + nǫ + nǫ ) . In the second case (as well as the first case), by Observation 6.3.4(b), any decomposition must include an ǫ + ǫ . Removing that, we see that N = P ∗ n − (( n − ǫ + nǫ + ( n − ǫ ) = M from (6.3.4).From Observation 6.3.4(b), by removing an ǫ + ǫ , P n ( x (23) ) = P n − (( n − ǫ + ( n − ǫ + nǫ ) = M , where the second equality follows from (6.3.3). From Observation 6.3.4(b), by removing an ǫ + ǫ , P n ( x (123) ) = P n − (( n − ǫ + nǫ + nǫ ) = N , where the second equality follows from (6.3.5) .From (6.3.2) and the preceding relationships, we have X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = P n ( x (1) ) − P n ( x (12) ) − P n ( x (23) ) + P n ( x (123) )= M + M + M − N − N − P n ( x (23) ) + P n ( x (123) )= M + M + 1 − N − M − M + N = 1as claimed. (cid:3) B . Let Φ be of type B with p > h = 8 (so p ≥ B , in order to have H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for 0 < i < p −
3, we must have λ = pω + w · w ∈ W . Again, by direct computation with MAGMA, the following table summarizes the weights which cangive a value of i < p − λ = pω + w · ℓ ( w ) k i = 2 k + ℓ ( w )( p − ω + 2 ω p − p − p − ω + ω p − p − p − ω p − p − Lemma . Suppose that Φ is of type B with p ≥ . Let λ = pµ + w · ∈ X ( T ) + with µ ∈ X ( T ) + and w ∈ W . (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − . (b) If H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω . (c) H p − ( G, H (( p − ω ) ⊗ H (( p − ω ∗ ) (1) ) = k . (d) If H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω + ω or λ = ( p − ω + 2 ω . (e) H p − ( G ( F p ) , k ) = k . Proof.
Parts (a), (b), and (d) follow from the discussion preceding the lemma. Part (c) follows fromProposition 2.7.1 and Lemma 6.4.2 below with m = p −
9. Since the weights in part (d) are larger than( p − ω , by Theorem 2.5.1 and Theorem 6.2.4, we obtain part (e). (cid:3) If n = 1, x (123) cannot be expressed as a sum of positive roots. However, in that case, N is necessarily zero, and so westill have P n ( x (123) ) = N . Lemma . Suppose that Φ is of type B . Let m ≥ be an even integer. Then X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = 1 . Proof.
The arguments to follow are quite similar to those in the proof of Lemma 6.3.3. Let n be suchthat m = 2 n . For n = 0, the claim readily follows, so we assume that n ≥
1. As with the proof of Lemma6.3.3, we work with the epsilon basis for the root system. Then the positive roots are ǫ , ǫ , ǫ , ǫ , ǫ + ǫ , ǫ + ǫ , ǫ + ǫ , ǫ + ǫ , ǫ + ǫ , ǫ + ǫ , ǫ − ǫ , ǫ − ǫ , ǫ − ǫ , ǫ − ǫ , ǫ − ǫ , and ǫ − ǫ . Further ω = ( ǫ + ǫ + ǫ + ǫ ). Relative to the ǫ basis, for any u ∈ W , u ( ǫ i ) = ± ǫ j . That is, u permutes the ǫ i up to a sign.For u ∈ W , let x u := u · (( m + 1) ω ) − ω . Using the fact that 2 ρ = 7 ǫ + 5 ǫ + 3 ǫ + ǫ , one finds that(6.4.1) x u = u (( n + 4) ǫ + ( n + 3) ǫ + ( n + 2) ǫ + ( n + 1) ǫ ) − ǫ − ǫ − ǫ − ǫ . By direct calculation, one finds that if u sends any ǫ i to − ǫ j (any j ), then x u is either not expressible as asum of positive roots or requires at least 2 n + 1 roots to do so. Therefore, the only u that can contribute tothe alternating sum under consideration are those u for which u ( ǫ i ) = ǫ j . That is, u is simply one of the 24permutations of the ǫ i s.Let u ∈ S ⊂ W . From (6.4.1), one finds that x u = a ǫ + a ǫ + a ǫ + a ǫ where a + a + a + a = 4 n .Since the positive roots are of the form ǫ i , ǫ i + ǫ j , or ǫ i − ǫ j , for this to be expressed as a sum of 2 n roots,each such root must be of the form ǫ i + ǫ j . That is the other two types of roots are not allowable. Similarto the arguments in the proof of Lemma 6.3.3, one can further see the following. Observation . Suppose that a ǫ + a ǫ + a ǫ + a ǫ is expressed as a sum of 2 n positive rootswhere a + a + a + a = 4 n . For any pair i, j ∈ { , , , } (with i = j ), if a i + a j = 2 n + c , then the rootsum decomposition contains at least c copies of ǫ i + ǫ j .Using Observation 6.4.3, by direct calculation, one can show that the 18 permutations u for which u ( ǫ ) = ǫ can be separated into 9 pairs of opposite parity having equal values of P n ( x u ). Hence theterms for those values of u cancel in the alternating sum. For example, consider the permutations (12) and(12)(43) of opposite parity. From (6.4.1), x (12) = ( n − ǫ + ( n + 1) ǫ + nǫ + nǫ . By Observation 6.4.3, adecomposition of x (12) must contain at least one copy of ǫ + ǫ (as well as a copy of ǫ + ǫ ). Subtractingthat root shows that P n ( x (12) ) = P n − (( n − ǫ + nǫ + ( n − ǫ + nǫ ) . On the other hand, x (12)(34) = ( n − ǫ + ( n + 1) ǫ + ( n − ǫ + ( n + 1) ǫ . Here, x (12)(34) must contain acopy of ǫ + ǫ (in fact, at least two copies). Subtracting this root gives P n ( x (12)(34) ) = P n − (( n − ǫ + nǫ + ( n − ǫ + nǫ ) = P n ( x (12) ) . The eight other pairings (which may not be unique) are (13) with (13)(24); (14) with (14)(23); (123) with(1243); (132) with (1342); (124) with (1234); (142) with (1432); (134) with (1324); and (143) with (1423).We leave the details to the interested reader.That leaves the six values of u for which u ( ǫ ) = ǫ : (1), (23), (24), (34), (234), and (243). However, asabove, one can show that P n ( x (24) ) = P n ( x (243) ). So those terms cancel as well and we are reduced to X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = P n ( x (1) ) − P n ( x (23) ) − P n ( x (34) ) + P n ( x (234) ) . From (6.4.1), we have x (1) = nǫ + nǫ + nǫ + nǫ . Write P n ( x (1) ) = M + M + M where M denotesthe number of root sum decompositions which contain at least one copy of ǫ + ǫ , M denotes the numberwhich contain no copies of ǫ + ǫ but contain at least one copy of ǫ + ǫ , and M denotes the number whichcontain neither an ǫ + ǫ nor an ǫ + ǫ . By assumption, subtracting a copy of ǫ + ǫ , we have(6.4.2) M = P n − ( nǫ + ( n − ǫ + ( n − ǫ + nǫ ) . For M , subtracting a copy of ǫ + ǫ gives(6.4.3) M = P ∗ n − (( n − ǫ + ( n − ǫ + nǫ + nǫ ) , where the P ∗ denotes the fact that the sum is only over those decompositions which do not contain a copyof ǫ + ǫ . For M , in order to get the nǫ appearing in x (1) , there must be exactly n copies of ǫ + ǫ . Butthen the remaining n factors must all be ǫ + ǫ . In other words, M = 1. N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 29
From (6.4.1), we have x (23) = nǫ + ( n − ǫ + ( n + 1) ǫ + nǫ . Write P n ( x (23) ) = N + N where N denotes the number of root sum decompositions which contain at least one copy of ǫ + ǫ and N denotesthe number which contain no copies of ǫ + ǫ . Subtracting a copy of ǫ + ǫ , we have(6.4.4) N = P n − ( nǫ + ( n − ǫ + nǫ + nǫ ) . For N , by Observation 6.4.3, any decomposition of x (23) contains at least one copy of ǫ + ǫ (as well as acopy of ǫ + ǫ ). Subtracting the ǫ + ǫ gives N = P ∗ n − (( n − ǫ + ( n − ǫ + nǫ + nǫ ) = M from (6.4.3).From (6.4.1), we have x (34) = nǫ + nǫ +( n − ǫ +( n +1) ǫ . From Observation 6.4.3, any decompositionof x (34) contains at least one copy of ǫ + ǫ (as well as a copy of ǫ + ǫ ). Subtracting the ǫ + ǫ gives P n ( x (34) ) = P n − ( nǫ + ( n − ǫ + ( n − ǫ + nǫ ) = M from (6.4.2).From (6.4.1), we have x (234) = nǫ + ( n − ǫ + ( n + 1) ǫ + ( n + 1) ǫ . From Observation 6.4.3, anydecomposition of x (234) contains at least one copy of ǫ + ǫ (in fact, at least two copies). Subtracting thisgives P n ( x (234) ) = P n − ( nǫ + ( n − ǫ + nǫ + nǫ ) = N from (6.4.4).In summary, we have X u ∈ W ( − ℓ ( u ) P m ( u · (( m + 1) ω ) − ω ) = P n ( x (1) ) − P n ( x (23) ) − P n ( x (34) ) + P n ( x (234) )= M + M + M − N − N − P n ( x ) + P n ( x )= M + M + 1 − N − M − M + N = 1as claimed. (cid:3) B . Let Φ be of type B with p > h = 10 (so p ≥ B , in order to have H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for 0 < i < p −
3, we must have λ = pω + w · w ∈ W . Specifically, substituting n = 5 into (6.1.1) gives(6.5.1) i ≥ p − p − . We obtain the following.
Lemma . Suppose that Φ is of type B with p ≥ . Let λ = pω + w · ∈ X ( T ) + with w ∈ W . (a) If p = 17 or p ≥ , then H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i ≤ p − . (b) Suppose p = 11 . Then (i) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (ii) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω = ω ; (iii) H p − ( G, H ( ω ) ⊗ H ( ω ∗ ) (1) ) = k ; (iv) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ω + ω or λ = 2 ω + ω ; (v) H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = k for λ = ω + ω or λ = 2 ω + ω ; (vi) H p − ( G ( F p ) , k ) = k . (c) Suppose p = 13 . Then (i) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (ii) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω = 3 ω ; (iii) H p − ( G, H (3 ω ) ⊗ H (3 ω ∗ ) (1) ) = k ; (iv) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ω + 3 ω or λ = 2 ω + 3 ω ; (v) dim H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 2 for λ = ω + 3 ω ; (vi) H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = k for λ = 2 ω + 3 ω ; (vii) H p − ( G ( F p ) , k ) = k . (d) Suppose p = 19 . Then (i) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (ii) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω = 9 ω ; (iii) dim H p − ( G, H (9 ω ) ⊗ H (9 ω ∗ ) (1) ) = 15 . Proof.
For p ≥
23, part (a) follows from (6.5.1). Parts (b)(i)-(v), (c)(i)-(vi), and (d) as well as part (a)for p = 17 follow by explicitly computing (with the aid of MAGMA) all possible w ·
0, and then computingpartition functions by hand or with the aid of MAGMA. For p = 11, since the weights in part (b)(iv) arelarger than that in (b)(ii), part (b)(vi) follows from Theorem 2.5.1 and Theorem 6.2.4. Similarly, part (c)(vii)follows. (cid:3) B . Let Φ be of type B with p > h = 12 (so p ≥ B , in order to have H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for 0 < i < p −
3, we must have λ = pω + w · w ∈ W . Recall the arguments in Section 6.1. Specifically, substituting n = 6 into (6.1.1) gives(6.6.1) i ≥ p + ( p − . We obtain the following.
Lemma . Suppose that Φ is of type B with p ≥ . Let λ = pω + w · ∈ X ( T ) + with w ∈ W . (a) If p ≥ , then H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i ≤ p − . (b) Suppose p = 13 . Then (i) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (ii) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ( p − ω = ω ; (iii) H p − ( G, H ( ω ) ⊗ H ( ω ∗ ) (1) ) = k ; (iv) if H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 , then λ = ω + ω or λ = 2 ω + ω ; (v) H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = k for λ = ω + ω or λ = 2 ω + ω ; (vi) H p − ( G ( F p ) , k ) = k . Proof.
Part (a) follows from (6.6.1). Parts (b)(i)-(v) follow by explicitly computing (with the aid ofMAGMA) all possible w ·
0, and then computing partition functions by hand or with the aid of MAGMA.Since the weights in part (b)(iv) are larger than that in part (b)(ii), part (b)(vi) follows from Theorem 2.5.1and Theorem 6.2.4. (cid:3) B . Theorem . Suppose Φ is of type B n with n ≥ . Assume that p > n . (a) If n ≥ or p > when n ∈ { , } , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; (ii) H p − ( G ( F p ) , k ) = 0 . (b) If n ∈ { , } and p = 13 , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; (ii) H p − ( G ( F p ) , k ) = k . (c) If n = 5 and p = 11 , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; (ii) H p − ( G ( F p ) , k ) = k . (d) If n ∈ { , } , then (i) H i ( G ( F p ) , k ) = 0 for < i < p − ; (ii) H p − ( G ( F p ) , k ) = k . Proof.
This follows from the discussion in Section 6.1, Theorem 6.2.4, Lemma 6.3.1, Lemma 6.4.1,Lemma 6.5.1, and Lemma 6.6.1. (cid:3)
7. Type G Assume throughout this section that Φ is of type G and that p > h = 6 (so p ≥ i > i ( G, H ( λ ) ⊗ H ( λ ∗ )) = 0 for some λ ∈ X ( T ) + . N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 31
Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some i > λ = pµ + w · µ ∈ X ( T ) + and w ∈ W . From Proposition 2.8.1(c), i ≥ ( p − h µ, ˜ α ∨ i −
1. Consider the two fundamentalweights ω and ω . Note that ω = α and ω = ˜ α . Furthermore, we have h ω , ˜ α ∨ i = 1 and h ω , ˜ α ∨ i = 2.Therefore, unless µ = ω = α , we have h µ, ˜ α ∨ i ≥ i ≥ p − λ = pω + w · w ∈ W . In order to have H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0,as discussed in Section 2.7, λ must be dominant and λ − ω must be a weight of S j ( u ∗ ) for some j . Inother words, λ − ω must be expressible as a non-negative linear combination of positive roots. By directcalculation (by hand or with the aid of MAGMA), one can identify all possible λ . These are listed in thefollowing table. As usual, s i := s α i and e is the identity element. w ℓ ( w ) λ = pω + w · e pω s p − ω + ω s s p − ω + 2 ω s s s p − ω + 2 ω s s s s p − ω + ω s s s s s p − ω Note that each λ has the form λ = aω + bω for a ≥ ≤ b ≤
2. From Proposition 2.7.1, we knowthat for λ = pµ + w · i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = X u ∈ W ( − ℓ ( u ) P i − ℓ ( w )2 ( u · λ − µ ) . In the next section, we consider such partition functions. Since the prime p does not per se play a role inthe partition function computations, we will work in a general setting. Let λ = aω + bω for a ≥ ≤ b ≤
2. From the previous section, our goal isto make computations of(7.2.1) X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω ) . In particular, we will identify the least value of k for which this sum is non-zero along with the value ofthe sum in that case. See Proposition 7.3.3 and Proposition 7.4.4.In order for P k ( u · λ − ω ) to be non-zero, u · λ − ω must lie in the positive (more precisely, non-negative)root lattice. By direct computation, one finds that there are only four elements u ∈ W for which this is true(under our assumptions on a and b above). This is summarized in the following table. The value of u · λ − ω is given in the root basis. u ℓ ( u ) u · λ − ω e a + 3 b − α + ( a + 2 b − α s a + 3 b − α + ( a + 2 b − α s a + 3 b − α + ( a + b − α s s a − α + ( a + b − α Note that in some of the cases a must be sufficiently large in order for u · λ − ω to lie in the positiveroot lattice. Specifically, for s , one needs a ≥ b ≥
1; for s , one needs a ≥ a ≥ b ≥ b ≥
2; and for s s , one needs a ≥ As noted in Section 7.2, our goal is to find the least value of k such that the sum(7.2.1) is non-zero. In this section, we notice some relationships among the partition functions which willallow us to identify a range under which the sum is zero. Lemma . Let λ = aω + bω with a ≥ and ≤ b ≤ . Suppose that k ≤ a + b − . Then P k ( λ − ω ) = P k ( s · λ − ω ) . Proof.
Recall the table in Section 7.2, and set γ := λ − ω = (2 a + 3 b − α + ( a + 2 b − α ,γ := s · λ − ω = (2 a + 3 b − α + ( a + b − α . Consider a decomposition of γ into k not necessarily distinct positive roots. Since a + 2 b − a + b −
2) +( b + 1) and k ≤ a + b −
2, at least b + 1 of those roots must contain 2 α . However, the only root containing2 α is ˜ α = 3 α + 2 α . Hence, any decomposition of γ into k roots must contain at least b + 1 copies of ˜ α .Therefore(7.3.1) P k ( γ ) = P k − b − ( γ − ( b + 1)(3 α + 2 α )) = P k − b − ((2 a − α + ( a − α ) . Now consider γ and the difference between the number of α s and α s appearing. Suppose γ = P ( m i α + n i α ) is expressed as a sum of k positive roots. Then X ( m i − n i ) = X m i − X n i = (2 a + 3 b − − ( a + b −
2) = a + 2 b. Note that for each i , m i − n i ∈ {− , , , } . Since k ≤ a + b −
2, for at least b + 1 values of i (in fact, atleast b + 2 values), we must have m i − n i = 2. However, the only root where that occurs is 3 α + α . Hence,any decomposition of γ into k roots must contain at least b + 1 copies of 3 α + α . Therefore, P k ( γ ) = P k − b − ( γ − ( b + 1)(3 α + α )) = P k ((2 a − α + ( a − α ) . Combining this with (7.3.1) gives the claim. (cid:3)
Lemma . Let λ = aω + bω with a ≥ . Suppose that k ≤ a + b − . Then P k ( s · λ − ω ) = P k ( s s · λ − ω ) . Proof.
Recall the table in Section 7.2, and set γ := s · λ − ω = ( a + 3 b − α + ( a + 2 b − α ,γ := s s · λ − ω = ( a − α + ( a + b − α . For γ , the same argument as in the preceding lemma gives(7.3.2) P k ( γ ) = P k − b − ( γ − ( b + 1)(3 α + 2 α )) = P k − b − (( a − α + ( a − α ) . Now consider γ and the difference between the number of α s and α s appearing. Suppose γ = P ( m i α + n i α ) is expressed as a sum of k positive roots. Then X ( m i − n i ) = X m i − X n i = ( a − − ( a + b −
2) = − b − . Note that for each i , m i − n i ∈ {− , , , } . For at least b + 1 values of i (in fact, at least b + 4 values), wemust have m i − n i = −
1. However, the only root where that occurs is α . Hence, any decomposition of γ into k roots must contain at least b + 1 copies of α . Therefore P k ( γ ) = P k − b − ( γ − ( b + 1) α ) = P k − b − (( a − α + ( a − α ) . Combining this with (7.3.2) gives the claim. (cid:3)
With the two aforementioned lemmas we can now prove the following proposition.
Proposition . Let λ = aω + bω for a ≥ and ≤ b ≤ . For k ≤ a + b − , X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω ) = 0 . Proof.
From the discussion in Section 7.2, X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω ) = P k ( λ − ω ) − P k ( s · λ − ω ) − P k ( s · λ − ω ) + P k ( s s · λ − ω ) . For a ≥
6, the claim follows from Lemma 7.3.1 and Lemma 7.3.2 above. For a <
6, one can see from theproof of Lemma 7.3.2 that the 2nd and fourth terms are zero. Hence, for 3 ≤ a ≤
5, the result follows fromLemma 7.3.1. For 1 ≤ a ≤
2, one can see from the proof of Lemma 7.3.1 that both the first and third termsvanish, and so the result follows. When a is small, the claim could also be readily verified by hand. (cid:3) N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 33
Let λ = aω + bω for a ≥ ≤ b ≤
2. The goal of this section is to determine X u ∈ W ( − ℓ ( u ) P a + b − ( u · λ − ω ) . See Proposition 7.4.4.From the discussion in Section 7.2 we need to consider the following weights (with notation followingSection 7.3): γ := λ − ω = (2 a + 3 b − α + ( a + 2 b − α ,γ := s · λ − ω = (2 a + 3 b − α + ( a + b − α ,γ := s · λ − ω = ( a + 3 b − α + ( a + 2 b − α ,γ := s s · λ − ω = ( a − α + ( a + b − α . More precisely,(7.4.1) X u ∈ W ( − ℓ ( u ) P a + b − ( u · λ − ω ) = P a + b − ( γ ) − P a + b − ( γ ) − P a + b − ( γ ) + P a + b − ( γ ) . We first make some reduction observations as done in the proofs in Section 7.3. Note that when a is small,some of the statements are trivially true since both sides are zero. But we include them here (and in thefollowing statements) for simplicity of exposition. Observe also that the right hand side is independent ofthe value of b . Lemma . Let λ , γ , γ , γ , and γ be as above, and let k = a + b − . Then (a) P k ( γ ) = P a − (2( a − α + ( a − α ) ; (b) P k ( γ ) = P a − ((2 a − α + ( a − α ) ; (c) P k ( γ ) = P a − (( a − α + ( a − α ) ; (d) P k ( γ ) = P a − (( a − α + ( a − α ) . Proof. (a) Suppose that γ is decomposed as a sum of k positive roots. Similar to the argument inLemma 7.3.1, since a + 2 b − a + b −
1) + b , at least b of those roots must (contain 2 α and hence) be˜ α = 3 α + 2 α . Hence, P k ( γ ) = P k − b ( γ − b ˜ α ), and the claim follows.(b) Again, as in the proof of Lemma 7.3.1, since the difference in the number of α s and α s appearingin γ is a + 2 b , if γ is expressed as k roots, then at least b + 1 of them must be 3 α + α . Hence, P k ( γ ) = P k − b − ( γ − ( b + 1)(3 α + α )), and the claim follows.(c) As in part (a), we must have P k ( γ ) = P k − b ( γ − b ˜ α ), and the claim follows.(d) As in part (b), similar to the proof of Lemma 7.3.2, we consider the difference in the number of α sand α s appearing in γ . Since this number is − b −
4, we can in particular assume that if γ is decomposedinto k roots, then at least b + 1 of them are α . Hence, P k ( γ ) = P k − b − ( γ − ( b + 1) α ), and the claimfollows. (cid:3) With the aid of Lemma 7.4.1, we now observe that there are some relationships among the P k ( γ i ). Tothis end, we introduce a bit of notation. Consider an arbitrary integer k ≥ γ = cα + dα for c, d ≥
0. Any decomposition of γ into a sum of k positive roots is of one of two types: either the sumcontains at least one copy of ˜ α or it does not contain any copies of ˜ α . Correspondingly, let P k, ˜ α ( γ ) and P k, ˜ α ( γ ) denote the number of such root sums. Then P k ( γ ) = P k, ˜ α ( γ ) + P k, ˜ α ( γ ). Observe that(7.4.2) P k, ˜ α ( γ ) = P k − ( γ − ˜ α ) . Lemma . Let λ , γ , γ , γ , and γ be as above, and let k = a + b − . Then (a) P k ( γ ) = P k ( γ ) + P a − , ˜ α (2( a − α + ( a − α ) ; (b) P k ( γ ) = P k ( γ ) + P a − , ˜ α (( a − α + ( a − α ) . Proof. (a) We have P k ( γ ) = P a − (2( a − α + ( a − α ) (by Lemma 7.4.1(a))= P a − , ˜ α (2( a − α + ( a − α ) + P a − , ˜ α (2( a − α + ( a − α )= P a − ((2 a − α + ( a − α ) + P a − , ˜ α (2( a − α + ( a − α ) (by (7.4.2))= P k ( γ ) + P a − , ˜ α (2( a − α + ( a − α ) (by Lemma 7.4.1(b)) . (b) We have P k ( γ ) = P a − (( a − α + ( a − α ) (by Lemma 7.4.1(c))= P a − , ˜ α (( a − α + ( a − α ) + P a − , ˜ α (( a − α + ( a − α )= P a − (( a − α + ( a − α ) + P a − , ˜ α (( a − α + ( a − α ) (by (7.4.2))= P k ( γ ) + P a − , ˜ α (( a − α + ( a − α ) (by Lemma 7.4.1(d)) . (cid:3) From (7.4.1), Lemma 7.4.1 and Lemma 7.4.2, we see that(7.4.3) X u ∈ W ( − ℓ ( u ) P a + b − ( u · λ − ω ) = P a − , ˜ α (2( a − α + ( a − α ) − P a − , ˜ α (( a − α + ( a − α ) . Lemma . Let c ≥ . Then P c, ˜ α (2 cα + cα ) − P c, ˜ α (( c − α + cα ) = (cid:24) c + 13 (cid:25) , where ⌈ x ⌉ denotes the least integer greater than or equal to x . Proof.
Let η := 2 cα + cα and η := ( c − α + cα . Observe first that if c <
2, then P c, ˜ α ( η ) = 0.On the other hand, we have P , ˜ α (0) = 1 and P , ˜ α (2 α + α ) = 1, and so the claim holds for c <
2. Assumefor the remainder of the proof that c ≥ η i is expressed as a sum of c positive roots, none of which are ˜ α , then each root isnecessarily of the form aα + α for a ∈ { , , , } . So the question of possible decompositions involveslooking only at the coefficients of α . For nonnegative integers m, n , let P m ( n ) denote the number of waysthat n can be expressed as a sum of m integers n = n + n + · · · + n m where n i ∈ { , , , } . With this notation, P c, ˜ α ( η ) = P c (2 c ), P c, ˜ α ( η ) = P c ( c − P c (2 c ) − P c ( c −
2) (when c ≥ m , n as above, let S m ( n ) denote the set of such partitions of n into m integers. We first show thatthere is an injection ϕ : S c ( c − → S c (2 c ). Let τ ∈ S c ( c − τ : c − τ + τ + · · · + τ c , where τ i ∈ { , , , } . Let s denote the number of τ i s which equal 3. The remaining c − s values mustsum to c − − s , and hence at most c − − s of those terms can be non-zero. In other words, at least( c − s ) − ( c − − s ) = 2 s + 2 of the remaining terms are zero. That is, we may assume that τ has the form: c − · · · + 3 | {z } s times + 0 + · · · + 0 | {z } (2 s +2) times + τ s +3 + · · · + τ c , where, for (3 s + 3) ≤ i ≤ c , τ i ∈ { , , } . Let ϕ ( τ ) be the partition:2 c = 3 + · · · + 3 | {z } s times +2 + 2 + 0 + · · · + 0 | {z } s times +( τ s +3 + 1) + ( τ s +4 + 1) + · · · + ( τ c + 1) . In words, the map ϕ leaves the initial s copies of 3 fixed, sends s of the zeros to 3, sends two of the zeros to2, leaves the other s zeros fixed, and adds one to the unknown integers at the end. Note that those unknownintegers are each at most 2, so adding one is allowable. One can also readily check that the new sum doesindeed add up to 2 c . It is clear that ϕ is an injection, but we will explicitly construct an inverse below.Observe that the resulting partition of 2 c contains 2 at least twice. We claim that the image of ϕ isin fact precisely the subset X ⊂ S c (2 c ) consisting of those partitions where 2 appears two or more times.Indeed, we can define a function ψ : X → S c ( c −
2) as follows. Let ξ ∈ X and s denote the number of timesthat zero appears in ξ . The remaining c − s values in ξ must sum to 2 c . We know that at least two of thosehave value 2. The remaining c − s − c −
4. Since 2( c − s −
2) = 2 c − − s , at least2 s of those terms must have value 3. In other words, ξ has the form:2 c = 0 + · · · + 0 | {z } s times +2 + 2 + 3 + · · · + 3 | {z } s times + ξ s +3 + · · · + ξ c , N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 35 where (for (3 s + 3) ≤ i ≤ c ) ξ i ∈ { , , } . Let ψ ( ξ ) be the partition: c − · · · + 0 | {z } (2 s +2) times + 3 + · · · + 3 | {z } s times +( ξ s +3 −
1) + ( ξ s +4 −
1) + · · · + ( ξ c − . In words, the map ψ leaves the zeros fixed, sends the two 2s to zero, sends s copies of 3 to zero, leaves theother s copies of 3 fixed, and subtracts one from each of the remaining integers. Clearly ψ is an inverse to φ . Hence, P c ( c −
2) = | X | .It remains to compute P c (2 c ) − | X | . That is, we need to count the number of partitions2 c = n + n + · · · + n c , where n i ∈ { , , , } but for which at most one value of n i = 2. Write c = 3 m + t for m ≥ t <
3. Thenit is a straightforward (but somewhat lengthy) computation to show that the number of such partitions is m + 1. This is left to the interested reader. The lemma follows. (cid:3) Applying Lemma 7.4.3 with c = a −
1, we obtain the following from (7.4.3).
Proposition . Let λ = aω + bω for a ≥ and ≤ b ≤ . Then X u ∈ W ( − ℓ ( u ) P a + b − ( u · λ − ω ) = l a m . Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some λ ∈ X ( T ) + and i > i < p −
3, then λ must be of the form λ = pω + w ·
0, andmore precisely, that it must be one of the weights listed in Table 7.1. For each such λ , from Proposition 7.3.3and Proposition 7.4.4, we can identify the least value of k such that X u ∈ W ( − ℓ ( u ) P k ( u · λ − ω ) = 0 , and moreover, identify the value of the sum. Further, from Proposition 2.7.1 (with k = ( i − ℓ ( w )) / i with H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 along with the dimension ofthe cohomology group. This information is summarized in the following table. Here k and i are minimumpossible values, and dim gives the dimension of the cohomology group (equivalently the value of (7.2.1)). w ℓ ( w ) λ = pω + w · k i dim e pω p − p − (cid:6) p (cid:7) s p − ω + ω p − p − (cid:6) p (cid:7) − s s p − ω + 2 ω p − p − (cid:6) p (cid:7) − s s s p − ω + 2 ω p − p − (cid:6) p (cid:7) − s s s s p − ω + ω p − p − (cid:6) p (cid:7) − s s s s s p − ω p − p − (cid:6) p (cid:7) − Theorem . Suppose Φ is of type G and p ≥ . (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for i < p − . (b) dim H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = ((cid:6) p (cid:7) − if λ = ( p − ω + ω else. (c) dim H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = ((cid:6) p (cid:7) − if λ = ( p − ω or ( p − ω + 2 ω else. (d) H i ( G ( F p ) , k ) = 0 for < i < p − . Proof.
Part (a) follows from Proposition 2.7.1 and Proposition 7.3.3. Parts (b) and (c) follow from thepreceding table and discussion. Part (d) follows from part (a) and Proposition 2.4.1. (cid:3)
One would like to apply Theorem 2.5.1 to conclude that H p − ( G ( F p ) , k ) = 0. However, the weight( p − ω is less than and linked to the weight ( p − ω + ω and so the Theorem is not applicable. Thenon-zero cohomology from the weight ( p − ω in degree 2 p − p − ω + ω . We refer the interested reader to [ BNP , Section 2.7] for discussionof this interplay.In a similar manner, cohomology in degree 2 p − p − ω + ω could cancelthat in degree 2 p − p − ω + 2 ω . So it is not even possible to conclude thatH p − ( G ( F p ) , k ) = 0. In summary, alternate methods are needed to determine the precise vanishing bound.
8. Type F Assume throughout this section that Φ is of type F and that p > h = 12 (so p ≥ i > i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 forsome λ ∈ X ( T ) + . Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for some i > λ = pµ + w · µ ∈ X ( T ) + and w ∈ W . From Proposition 2.8.1, i ≥ ( p − h µ, ˜ α ∨ i −
1. For 1 ≤ i ≤
3, we have h ω i , ˜ α ∨ i ≥ h ω , ˜ α ∨ i = 1. Therefore, unless µ = ω = α , we have h µ, ˜ α ∨ i ≥ i ≥ p − λ = pω + w · w ∈ W . With the aid of MAGMA, one can identify all w forwhich λ is in fact dominant. From Proposition 2.8.1(a), with σ = α and λ = pω + w ·
0, since h ω , α ∨ i = 2,we have(8.1.1) i ≥ p −
1) + ℓ ( w ) + h w · , α ∨ i . By checking all possible cases, one finds that ℓ ( w ) + h w · , α ∨ i ≥ −
7. Combining this with (8.1.1), weconclude that i ≥ p −
9. From Proposition 2.4.1, we get the following.
Theorem . Suppose Φ is of type F and p ≥ . Let λ ∈ X ( T ) + . Then (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (b) H i ( G ( F p ) , k ) = 0 for < i < p − . Based on the preceding discussion, the weights which could give H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0for i ≤ p − λ = pω + w · ℓ ( w ) h w · , α ∨ i i ( p − ω + ω −
20 2 p − p − ω + ω −
21 2 p − p − ω −
22 2 p − p − ω + 3 ω −
16 2 p − p − ω + 2 ω + ω −
17 2 p − p − ω + ω + ω −
18 2 p − p − ω + 2 ω + ω −
14 2 p − p − ω + ω + ω + ω −
15 2 p − p − ω + 2 ω −
16 2 p − λ − ω must be a weight of S i − ℓ ( w )2 ( u ∗ ), and hence i is congruent to ℓ ( w ) mod 2. Itfollows that some of the above degree bounds are even higher. For example, consider λ = ( p − ω + ω = N THE VANISHING RANGES FOR THE COHOMOLOGY OF FINITE GROUPS OF LIE TYPE II 37 pω + w ·
0. Since ℓ ( w ) = 14 but 2 p − i could take would be 2 p −
8. A similar situationholds for λ = ( p − ω + 2 ω + ω and λ = ( p − ω + ω + ω + ω . Similarly, for the other weights inthe above list, if the cohomology vanishes in the degree i listed, then the next possible non-vanishing degreeis i + 2. We summarize this in the following lemma. Lemma . Suppose Φ is of type F , p ≥ and λ ∈ X ( T ) + . Suppose that H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) =0 . Then (a) i ≥ p − ; (b) if i = 2 p − , then λ = ( p − ω + ω or ( p − ω ; (c) if i = 2 p − , then λ = ( p − ω + ω , ( p − ω + 3 ω , or ( p − ω + ω + ω ; (d) if i = 2 p − , then λ = ( p − ω + ω , ( p − ω , ( p − ω + 2 ω + ω , ( p − ω + 2 ω + ω ,or ( p − ω + 2 ω . In principle, one could use Proposition 2.7.1 to compute the dimension ofH i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) )in terms of partition functions for the weights in Lemma 8.2.1. For small p , one can use MAGMA to makethis computation. For p = 13 , , or 19, one finds that the two candidates in degree 2 p − p −
7. And in degree 2 p −
8, the only one weight (of thethree) which has cohomology is ( p − ω + ω . We make the following Conjecture . Suppose that Φ is of type F , p ≥ , and λ = pµ + w · ∈ X ( T ) + . Then (a) H i ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for < i < p − ; (b) H p − ( G, H ( λ ) ⊗ H ( λ ∗ ) (1) ) = 0 for λ = ( p − ω + ω . If part (a) of the conjecture holds, then H i ( G ( F p ) , k ) = 0 for 0 < i < p − p − ( G ( F p ) , k ) = 0. Analogous to the situation for type G (cf. Section 7.5), cohomology in degree 2 p − p − ω could cancel out the cohomology in degree 2 p − p − ω + ω .Conjecture 8.3.1 is a special case of a more general conjecture on partition functions (known to hold forsmall values of m ). Conjecture 8.3.1(a) would follow from parts (a) and (b) while Conjecture 8.3.1(b) wouldfollow from part (c). Conjecture . Suppose that Φ is of type F and m ≥ . Then (a) X u ∈ W ( − ℓ ( u ) P m +1 ( u · ( mω + ω ) − ω ) = 0 ; (b) X u ∈ W ( − ℓ ( u ) P m − ( u · ( mω ) − ω ) = ( if m is even , if m is odd ;(c) X u ∈ W ( − ℓ ( u ) P m +1 ( u · ( mω + ω ) − ω ) = 1 . References [ AJ]
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