PARABOLIC INDUCTION AND JACQUET MODULES OF REPRESENTATIONS OF O(2n, F)

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1 PARABOLIC INDUCTION AND JACQUET MODULES OF REPRESENTATIONS OF O(2n, F) DUBRAVKA BAN Abstract. For the sum of the Grothendieck groups of the categories of smooth finite length representations of O(2n, F) (resp., SO(2n, F)), n 0, (F a p-adic field), the structure of a module and a comodule over the sum of the Grothendieck groups of the categories of smooth finite length representations of GL(n, F), n 0, is achieved. The multiplication is defined in terms of parabolic induction, and the comultiplicitation in terms of Jacquet modules. Also, for even orthogonal groups, the combinatorial formula, which connects the module and the comodule structures, is obtained. In this paper, we deal with 1. Introduction R(O) = n 0 R n (O) where R n (O) denotes the Grothendieck group of the category of smooth finite length representations of O(2n, F), F a p-adic field. R(O) is a module and a comodule over the Hopf algebra R = n 0 R n ; here R n denotes the Grothendieck group of the category of smooth finite length representations of GL(n, F). The structure of R was described by Zelevinsky in [Z1]. The definition of the multiplication m : R R R and the comultiplication m : R R R is based on the fact that for 0 k n there exists a standard parabolic subgroup of GL(n, F) whose Levi factor is isomorphic to GL(k, F) GL(n k, F). The multiplication is defined using parabolic induction, and the comultiplication by Jacquet modules (see the third section of this paper). The structure of a Hopf algebra on R 1991 Mathematics Subject Classification. 20 G 05, 22 E 50. Key words and phrases. representations of p-adic groups, even orthogonal groups, special even orthogonal groups, parabolic induction, Jacquet modules. 1

2 2 DUBRAVKA BAN includes the Hopf axiom; it is the property that m is a ring homomorphism. For O(2n, F) and 0 k n, there is a standard parabolic subgroup whose Levi factor is isomorphic to GL(k, F) O(2(n k), F). So, there is a natural way to define (using parabolic induction) the action of R on R(O), and (using Jacquet modules) the mapping µ : R(O) R R(O). This is done in the sixth section. There is also a connection between the module and the comodule structures on R(O). Let M = (m 1) ( m ) s m, where s : R R R R is the homomorphism determined by s(r 1 r 2 ) = r 2 r 1, r 1, r 2 R. Then we have (*) µ (π σ) = M (π) µ (σ), so R(O) is an M -Hopf module over R (see [T1] for the definition). The formula (*) can be used to find a composition series for Jacquet modules of parabolically induced representations. The kind of work we have done for even orthogonal groups was first done by Tadić; in [T1] he introduced such a structure in the cases of symplectic and special odd-orthogonal groups, and he proved the combinatorial formula (*) for those groups. He also raised the question of the existence of such a structure for other series of classical p-adic groups. We now give a short summary of the paper. In the second section, we give the definitions and some results of Bernstein and Zelevinsky, and Casselman, about parabolic induction and Jacquet modules. The third section describes the structure of R, as it is done in [Z1]. The fourth section is about standard parabolic subgroups of SO(2n, F) and about R(S) (the definition is analogous to R(O)). R(S) is an R-module and R-comodule. The fifth section contains calculations in the root system for the case of D n, i.e., for the group SO(2n, F). This is used in the sixth section to find double cosets of O(2n, F). In this section we also define the module and the comodule structures for even orthogonal groups. In the seventh section, we have applied the proof of the combinatorial formula from [T1] to our case. I would like to close the introduction by thanking Marko Tadić, who suggested this project and helped its realisation. Also, I am very grateful to the referee for his valuable comments and English corrections.

3 INDUCTION AND JACQUET MODULES 3 2. Preliminaries In this section, we shall introduce some basic notation and recall some results that will be needed in the rest of the paper. Our presentation follows the papers [BZ2] and [C]. A Hausdorff topological group G is called an l-group if any neighbourhood of the identity contains an open compact subgroup. Let G be an l-group, M, U closed subgroups, such that M normalises U, M U = e and the subgroup P = MU G is closed; let θ be a character of U normalised by M. In such a situation, we define the functors I U,θ, i U,θ : AlgM AlgG, r U,θ : AlgG AlgM. (Here AlgG denotes the category of algebraic (=smooth) representations of G.) (a) Let (ρ, L) AlgM. Denote by I(L) the space of functions f : G L satisfying the following conditions: 1. f(umg) = θ(u) 1/2 U (m)ρ(m)(f(g)), u U, m M, g G. (Here U denotes the modular character.) 2. There exists an open subgroup K f G such that f(gk) = f(g), for g G, k K f. Define the representation (δ, I(L)) AlgG by (δ(g)f)(g ) = f(g g). We call δ an induced representation and denote it by I U,θ (ρ). Denote by i(l) the subspace of I(L) consisting of all functions compactly supported modulo the subgroup P = MU. The restriction of δ to the space i(l) is called compactly induced and is denoted by i U,θ (ρ). (b) Let (π, E) AlgG. Denote by E(U, θ) E the subspace spanned by the vectors of the form π(u)ξ θ(u)ξ, u U, ξ E. The quotient space E/E(U, θ) is called the θ-localisation of the space E and is denoted by r U,θ (E). Define the representation (δ, r U,θ (E)) AlgM by δ(m)(ξ + E(U, θ)) = 1/2 U (m)(π(m)ξ + E(U, θ)), m M, ξ E; it is easily verified that δ is well-defined. Call the representation δ the θ-localisation of π and denote it by r U,θ (π).

4 4 DUBRAVKA BAN We shall now state a result of Bernstein and Zelevinsky (Theorem 5.2 of [BZ2]). Let G be an l-group, P, M, U and Q, N, V be closed subgroups, θ be a character of U and ψ be a character of V. Suppose that (1) MU = P, NV = Q, M U = N V = e, M normalises U and θ, N normalises V and ψ. Then there are defined functors i U,θ : AlgM AlgG and We want to compute the functor r V,ψ : AlgG AlgN. F = r V,ψ i U,θ : AlgM AlgN. It requires some complementary conditions. Suppose that (2) The group G is countable in infinity, and U, V are limits of compact subgroups. Consider the space X = P \G with its quotient-topology and the action δ of G on X defined by δ(g)(ph) = Phg 1, g, h G, Ph X. Suppose that (3) The subgroup Q has a finite number of orbits on X. Acording to ([BZ1],1.5), one can choose a numbering Z 1,..., Z k of the Q-orbits on X such that all sets Y 1 = Z 1, Y 2 = Z 1 Z 2,..., Y k = Z 1... Z k = X are open in X. In particular, all Q-orbits on X are locally closed. Fix a Q-orbit Z X. Choose w G such that P w 1 Z and denote by w the corresponding inner automorphism of G: w(g) = wg w 1. Call a subgroup H G decomposable with respect to the pair (M, U), if H (MU) = (H M)(H U). Suppose that (4) The groups w(p), w(m) and w(u) are decomposable with respect to (N, V ); the groups w 1 (Q), w 1 (N) and w 1 (V ) are decomposable with respect to (M, U). If the conditions (1)-(4) hold, we define the functor Φ Z : AlgM AlgN. Consider the condition (*) The characters w(θ) and ψ coincide when restricted to the subgroup w(u) V. If (*) does not hold, set Φ Z = 0. If (*) holds then define the functor Φ Z in the following way. Set M = M w 1 (N), N = w(m ) = w(m) N, V = M w 1 (V ), ψ = w 1 (ψ) V, U = N w(u), θ = w(θ) U.

5 INDUCTION AND JACQUET MODULES 5 It is clear that the following functors are defined r V,ψ : AlgM AlgM, i U,θ w : AlgM AlgN, : AlgN AlgN. Let ε 1 = 1/2 U 1/2 U w 1 (Q) be a character of M, ε 2 = 1/2 V 1/2 V w(p) be a character of N and ε = ε 1.w(ε 2 ) be a character of M. We define Φ Z by Φ Z = i U,θ w ε r V,ψ : AlgM AlgN (here ε is considered as a functor, see [BZ2] 1.5). In a more symmetric form, Φ Z = i U,θ ε 2 w ε 1 r V,ψ. Theorem 2.1. Under the conditions (1)-(4) the functor F = r V,ψ i U,θ : AlgM AlgN is glued from the functors Φ Z where Z runs through all Q-orbits on X. More precisely, if orbits Z 1,..., Z k are numerated so that all sets Y i = Z 1... Z i (i = 1,..., k) are open in X, then there exists a filtration 0 = F 0 F 1... F k = F such that F i /F i+1 Φ Zi. (Let A be an abelian category and C 1, C 2,..., C k A. We say that the object D A is glued from C 1, C 2,..., C k if there is a filtration 0 = D 0 D 1 D k = D in D, such that the set of quotients D i /D i 1 is isomorphic after a permutation to the set C i.) Let F be a locally compact nonarchimedean field. By an algebraic F-group we mean the group of F-points of some algebraic group, defined over F. In a natural locally compact topology such groups are l-groups. Let G be a connected (in an algebraic sense) reductive F-group. Fix from now on a minimal parabolic subgroup P 0 G and a maximal split torus A 0 P 0. Let P be a parabolic subgroup containing P 0. We call such a group a standard parabolic subgroup. Let U be the unipotent radical of P. There exists a unique Levi subgroup in P containing A 0 ; denote it by M (it is a connected reductive F-group). It is known that P normalises U and has the Levi decomposition P = MU, M U = e. We define the functors i G,M : AlgM AlgG and r M,G : AlgG AlgM

6 6 DUBRAVKA BAN by i G,M = i U,1, r M,G = r U,1. For σ AlgM we call i G,M (σ) the parabolically induced representation of G by σ from P, and for π AlgG we call r M,G (π) the Jacquet module of π with respect to P. Denote by Σ the set of (reduced) roots of G relative to A 0. The choice of P 0 determines a basis of Σ (which consists of simple roots). It also determines a set of positive roots Σ +. Denote by W the Weyl group of G. For θ, we denote by W θ the subgroup of W generated by all reflections w α α θ. If P = P θ = MU is the standard parabolic subgroup of G determined by θ, then W θ is also denoted by W M. Let Ω, θ. Now, we shall describe the set [W θ \W/W Ω ], a set of representatives of W θ \W/W Ω, defined in [C]. For α, set We have W α = w W wα > 0, [W/W Ω ] = α Ω W α, [W θ \W/W Ω ] = [W θ \W] [W/W Ω ]. α W = w W w 1 α > 0. [W Ω \W] = α Ω α W, If P = P θ = MU and Q = P Ω = NV are standard parabolic subgroups of G, then we have a bijection W M \W/W N = P \G/Q (see [BT], 5.15,5.20). From this relation and Theorem 1.1 Bernstein and Zelevinsky obtained the geometric lemma ([BZ2]). The same result was obtained independently by Casselman in [C]. Theorem 2.2 (Geometric lemma). Let G be a connected reductive p- adic group, P = P θ = MU, Q = P Ω = NV parabolic subgroups. Let σ be an admissible representation of M. Then r N,G i G,M (σ) has a composition series with factors i N,N w 1 r M,M(σ) where M = M w(n), N = w 1 (M) N and w [W θ \W/W Ω ]. Let π be a smooth finite length representation of G. We identify it canonically with an element of the Grothendieck group of the category of all smooth finite length representations of G. We denote this element by s.s.(π) and call this map semi-simplification.

7 INDUCTION AND JACQUET MODULES 7 3. General linear group In this section, we shall recall some results of the representation theory of p-adic general linear groups. The proofs can be found in [BZ2] and [Z1]. Fix the minimal parabolic subgroup of GL(n, F) which consists of all upper triangular matrices in GL(n, F). The standard parabolic subgroups of GL(n, F) can be parametrized by ordered partitions of n: for α = (n 1,..., n k ) there exists a standard parabolic subgroup (denote it in this section by P α ) of GL(n, F) whose Levi factor M α is naturally isomorphic to GL(n 1, F) GL(n k, F). Denote by R n the Grothendieck group of the category of smooth representations of GL(n, F). R n is a free abelian group; it has a basis consisting of equivalence classes of irreducible smooth representations of GL(n, F). Let R = n 0 R n. We shall define a multiplication and a comultiplication on R. Let π 1, π 2 be admissible representations of GL(n 1, F), GL(n 2, F), resp., n 1 + n 2 = n. Define π 1 π 2 = i GL(n,F),M(n1,n 2 )(π 1 π 2 ). Now, for irreducible smooth representations π, τ R, we put π τ = s.s.(π τ). We extend Z-bilinearly to R R. The induced mapping R R R, π τ π τ is denoted by m. Let π be a smooth representation of GL(n, F) of finite length. For α = (n 1,..., n k ) we define r α,(n) (π) = r Mα,GL(n,F)(π). This is a representation of M α = GL(n1, F) GL(n k, F), so we may consider s.s.(r α,(n) (π)) R n1 R nk. Now we define m (π) = n s.s.(r (k,n k),(n) (π)) R R. k=0 We extend m Z-linearly to all R. With the multiplicationm and the comultiplication m, R is a graded Hopf algebra. This means that R is Z + -graded as an abelian group, m and m are Z + -graded, R has an algebra and coalgebra structure, and the comultiplication m : R R R is a ring homomorphism. Let g GL(n, F). We denote by t g the transposed matrix of g, and by τ g the matrix of g transposed with respect to the second diagonal.

8 8 DUBRAVKA BAN 4. Special orthogonal group SO(2n, F) From now on, F will be a fixed local non-archimedean field of characteristic different from two. The special orthogonal group SO(2n, F), n 1, is the group For n = 1, we get SO(2n, F) = X SL(2n, F) SO(2, F) = τ XX = I 2n. [ λ 0 0 λ 1 ] λ F = F. SO(0, F) is defined to be the trivial group. Denote by A 0 the maximal split torus in SO(2n, F) which consists of all diagonal matrices in SO(2n, F). Hence, A 0 = diag(x 1,..., x n, x 1 n,..., x 1 1 ) x i F = (F ) n. Denote by a the natural isomorphism of (F ) n to A 0 defined by a(x 1,..., x n ) = diag(x 1,..., x n, x 1 n,..., x 1 1 ). We fix the minimal parabolic subgroup P 0 which consists of all upper triangular matrices in SO(2n, F). The root system is of type D n : the roots: ±e i ± e j, 1 i < j n, the positive roots: e i e j, 1 i < j n, e i + e j, 1 i < j n, the simple roots: α i = e i e i+1, 1 i n 1, α n = e n 1 + e n. The set of simple roots is denoted by. The action of the simple roots on A 0 is given by α i (a(x 1,..., x n )) = x i x 1 i+1, 1 i n 1, α n (a(x 1,..., x n )) = x n 1 x n. Let us describe the standard parabolic subgroup P θ = M θ U θ, θ. For i = 1,..., n we define \αi, i n 1, Ω i = \α n, α n 1, i = n 1. For i = 0, we put Ω 0 =. If θ can be written in the form θ = i I Ω i, I = i 1,..., i k, i 1 < i 2 <... < i k, then M θ = diag(g 1,..., g k, h, τ g 1 k,..., τ g 1 1 ) g i GL(n i, F), h SO(2(n m), F),

9 INDUCTION AND JACQUET MODULES 9 where n 1 = i 1, n 1 + n 2 = i 2,..., n n k = i k = m. Put α = (n 1,..., n k ). M θ is also denoted by M α. In this case we have M θ = GL(n1, F) GL(n 2, F) GL(n k, F) SO(2(n m), F). If θ cannot be written in such a form (this happens when α n 1 / θ, α n θ), then we have M θ = s M Ω s 1, where Ω = (θ\α n ) α n 1, I s = I Note that the presentation of θ in the form θ = i I Ω i is not always unique. Namely, when α n 1 / I and α n / I, we may take n 1 I, n I or n 1 I, n / I. In that case we have M θ = diag(g 1,..., g k, h, τ g 1 k,..., τ g 1 1 ) g i GL(n i, F), h = diag(x, x 1 ), x F, so we may consider or M θ = GL(n1, F) GL(n 2, F) GL(n k, F) GL(1, F), M θ = GL(n1, F) GL(n 2, F) GL(n k, F) SO(2, F). For us, it will be important that for any ordered partition α = (n 1,..., n k ) of a non-negative integer m n we have a standard parabolic subgroup of SO(2n, F) whose Levi factor M α is isomomorphic to GL(n 1, F) GL(n 2, F) GL(n k, F) SO(2(n m), F). Now, take smooth finite length representations π of GL(n, F) and σ of SO(2m, F). Let P (n) = M (n) U (n) be a standard parabolic subgroup of SO(2(m + n), F). Hence, M (n) = GL(n, F) SO(2m, F), so π σ can be taken as a representation of M (n).define π σ = i M(n),SO(2(m+n),F)(π σ). Proposition 4.1. Let π, π 1 and π 2 be finite length smooth representations of the groups GL(n, F), GL(n 1, F) and GL(n 2, F) respectively, and let σ be a finite length smooth representation of SO(2m, F). Then (i) π 1 (π 2 σ) = (π 1 π 2 ) σ, (ii) (π σ) = π σ.

10 10 DUBRAVKA BAN (Here π denotes the contragredient representation of π.) Proof. The proof is straightforward and follows from [BZ2], Proposition 2.3. Denote by R n (S) the Grothendieck group of the category of all finite length smooth representations of SO(2n, F). Define R(S) = n 0 R n (S). The multiplication of representations we introduced above gives rise to a multiplication : R R(S) R(S). For irreducible smooth representations π R and σ R(S), we put π σ = s.s.(π σ), and extend Z-bilinearly to R R(S). Now, we can get a Z-linear mapping, denote it by µ : R R(S) R(S), which satisfies µ(π σ) = s.s.(π σ) for π R and σ R(S). Proposition 4.2. (R(S), µ) is a Z + -graded module over R. Proof. See [Sw] for the definition of a module over a Hopf algebra. We are interested in the property of associativity, i.e., that the following diagram commutes: id µ R R R(S) R R(S) m id µ R R(S) µ R(S). The proof of this property relies on the previous proposition. Let σ be a finite length smooth representation of SO(2n, F). Let α = (n 1,..., n k ) be an ordered partition of a non-negative integer m n. Define s α,(0) (σ) = r Mα,SO(2n,F)(σ). This is a representation of M α = GL(n1, F) GL(n 2, F) GL(n k, F) SO(2(n m), F), so we may consider s.s.(s α,(0) (σ)) R n1 R nk R n m (S). Now we shall define a Z-linear mapping µ : R(S) R R(S). For an irreducible smooth representation σ R(S), we define n µ (σ) = s.s.(s (k),(0) (σ)). k=0 We extend µ Z-linearly to µ : R(S) R R(S). Proposition 4.3. (R(S), µ ) is Z + -graded R comodule.

11 INDUCTION AND JACQUET MODULES 11 Proof. The definition of a comodule over a Hopf algebra can be found in [Sw]. We are interested in coassociativity, i.e., that the following diagram commutes: µ R(S) R R(S) µ id µ R R(S) m id R R R(S). The proof follows from [BZ2], Prop.2.3. The above construction is analogous those Tadić did in [T1] for Sp(n, F) and SO(2n + 1, F). 5. Calculations in the root system, the case of D n In this section we shall make the calculations in the Weyl group we need for the geometric lemma. Precisely, for i 1, i 2 1, 2,..., n we shall find [ W Ωi1 \W/W Ωi2 ] and for w [ WΩi1 \W/W Ωi2 ], determine Ω i1 w(ω i2 ). First, we shall describe the Weyl group: W = ±1 n 1 Sym(n), where ±1 n 1 = (ǫ 1,..., ǫ n ) ±1 n i ǫ i = 1. Sym(n) acts on the roots ±e i ±e j by permutations of the set e 1,..., e n, and (ǫ 1,..., ǫ n ) acts as sign changes ( 1 in the i-th place of ǫ = (ǫ 1,..., ǫ n ) denotes the interchange of e i and e i ). For p Sym(n) and (ǫ 1,..., ǫ n ) ±1 n 1, we have It follows that p(ǫ 1,..., ǫ n )p 1 = (ǫ p 1 (1),..., ǫ p 1 (n)). [p(ǫ 1,..., ǫ n )] 1 = p 1 (ǫ p 1 (1),..., ǫ p 1 (n)), [(ǫ 1,..., ǫ n )p] 1 = (ǫ p(1),..., ǫ p(n) )p 1. Now we shall use the formulas from [C] for [W Θ \W/W Ω ] we listed before. The beginning of our calculation is almost the same as in [T1], and the first four lemmas are very similar.

12 12 DUBRAVKA BAN By the definition of the action of W on roots, for p Sym(n) and (ǫ 1,..., ǫ n ) ±1 n we have pǫ(α i ) = pǫ(e i e i+1 ) = p(ǫ i e i ǫ i+1 e i+1 ) = ǫ i e p(i) ǫ i+1 e p(i+1), 1 i n 1, pǫ(α n ) = pǫ(e n 1 + e n ) = p(ǫ n 1 e n 1 + ǫ n e n ) = ǫ n 1 e p(n 1) + ǫ n e p(n). As we said, W α i = w W wα i > 0. If we check when pǫ(α i ) > 0, 1 i n, then we easily get the following lemma. Lemma 5.1. a) For 1 i n 1, W α i is the disjoint union of the following three sets: (i) pǫ W ǫ i = ǫ i+1 = 1, p(i) < p(i + 1); (ii) pǫ W ǫ i = 1, ǫ i+1 = 1; (iii) pǫ W ǫ i = ǫ i+1 = 1, p(i) > p(i + 1) b) W αn is the disjoint union of the following three sets: (i) pǫ W ǫ n 1 = ǫ n = 1; (ii) pǫ W ǫ n 1 = 1, ǫ n = 1, p(n 1) < p(n); (iii) pǫ W ǫ n 1 = 1, ǫ n = 1, p(n 1) > p(n). In the same way, we can compute α i W = w W w 1 α i > 0. Lemma 5.2. a) For 1 i n 1, α i W is the disjoint union of the following three sets: (i) pǫ W ǫ p 1 (i) = ǫ p 1 (i+1) = 1, p 1 (i) < p 1 (i + 1) ; (ii) pǫ W ǫ p 1 (i) = 1, ǫ p 1 (i+1) = 1 ; (iii) pǫ W ǫ p 1 (i) = ǫ p 1 (i+1) = 1, p 1 (i) > p 1 (i + 1). b) αn W is the disjoint union of the following three sets: (i) pǫ W ǫ p 1 (n 1) = ǫ p 1 (n) = 1 ; (ii) pǫ W ǫ p 1 (n 1) = 1, ǫ p 1 (n) = 1, p 1 (n 1) < p 1 (n) ; (iii) pǫ W ǫ p 1 (n 1) = 1, ǫ p 1 (n) = 1, p 1 (n 1) > p 1 (n). In the next lemma, we shall use the formula [W/W Ω ] = α Ω W α, for Ω. Lemma 5.3. Let 1 i n and let 0 j i. Denote by Yj i the set of all pǫ W such that the following six conditions are satisfied: (i) ǫ k = 1, for 1 k j; (ii) p(k 1 ) < p(k 2 ), for 1 k 1 < k 2 j; (iii) ǫ k = 1, for j + 1 k i; (iv) p(k 1 ) > p(k 2 ), for j + 1 k 1 < k 2 i;

13 INDUCTION AND JACQUET MODULES 13 (v) ǫ k = 1, for i + 1 k n 1; (vi) p(k 1 ) < p(k 2 ), for i + 1 k 1 < k 2 n. Denote by Ȳj n the set of all pǫ W which satisfy the same conditions (for i = n), but instead of (iii), the condition (iii ) ǫ k = 1, for j + 1 k n 1, ǫ n = 1. Then Here Ω n = \α n 1. [W/W Ωi ] = [W/W Ω n ] = 0 j i 0 j n 1 Y i j, Ȳ n j. Proof. Take pǫ [W/W Ωi ] = α Ω i W α. If i < n 1, then pǫ W αn W α n 1. From Lemma 5.1 a) for n 1 and Lemma 5.1 b) for n we get or ǫ n 1 = 1, ǫ n = 1, ǫ n 1 = 1, ǫ n = 1, p(n 1) < p(n), p(n 1) < p(n). Anyway, ǫ n 1 = 1, p(n 1) < p(n). Further, Lemma 5.1.a) implies ǫ i+1 = ǫ i+2 = = ǫ n 1 = 1, Now, we have p(i + 1) < p(i + 2) < < p(n). ǫ k = 1, for i + 1 k n 1, and p(k 1 ) < p(k 2 ), for i + 1 k 1 < k 2 n. This condition is also satisfied for i = n 1 or i = n, because in those cases it is empty. Since pǫ W α k, k 1,..., i 1, Lemma 5.1 implies that for any k 1,..., i 1, we have ǫ k = ǫ k+1 = 1 or ǫ k = 1, ǫ k+1 = 1 or ǫ k = ǫ k+1 = 1. We cannot have ǫ k = 1, ǫ k+1 = 1. So we conclude that there exists j 0, 1,..., i such that ǫ j = 1 for 1 k j 1 and ǫ k = 1 for j+1 k i 1. Lemma 5.1 also implies p(k) < p(k+1) for 1 k j 1 and p(k) > p(k +1) for j +1 k i 1. Hence, pǫ Yj i where 0 j i. If pǫ 0 j i Y j i,then we see from Lemma 5.1 that pǫ W α l for l i, in the case i n 1, and pǫ W α l for l n 1, l n in the case i = n 1. This proves the other inclusion.

14 14 DUBRAVKA BAN Let pǫ [W/W Ω n ] = l =n 1 W α l. Suppose that ǫ n 1 = 1. Then by Lemma 5.1 a), we get ǫ 1 = ǫ 2 = = ǫ n 1 = 1, p(k 1 ) < p(k 2 ), for 1 k 1 < k 2 n 1. The condition ǫ i = 1 gives us ǫ n = 1. Put j = n 1. Then, the conditions (i), (ii) i and (iii ) are satisfied, and the others are empty. Let ǫ n 1 = 1. Then by Lemma 5.1 b) we get ǫ n = 1, p(n 1) > p(n).it follows from Lemma 5.1.a) that there exists j 0, 1,..., n 2 such that and ǫ k = 1, for 1 k j, p(k 1 ) < p(k 2 ), for 1 k 1 < k 2 j, ǫ k = 1, for j + 1 k n 1, p(k 1 ) > p(k 2 ), for j + 1 k 1 < k 2 n 1. Together with the first condition, we get ǫ k = 1, for j + 1 k n 1, p(k 1 ) > p(k 2 ), for j + 1 k 1 < k 2 n. Therefore, the conditions (i), (ii), (iii ) and (iv) are satisfied, and the others are empty. The other inclusion can be proved as before. Remark 5.1. If pǫ W, ǫ = (ǫ 1,..., ǫ n ), then n ǫ i = 1. Thus we have i=1 for i < n : pǫ Yj i implies ǫ n = ( 1) i j, for i = n : if n j odd, then Yj n =, if n j > 0 even, then Ȳj n =. If j = n, then Ȳ n n = id Ȳ n 1, n so we can write [W/W Ωn ] = Ȳj n. 0 j n For the set [W Ω \W], we can simply use the relation [W/W Ω ] 1 = [W Ω \W] and the previous lemma to obtain the following: Lemma 5.4. Let 1 i n and let 0 j i. Denote by Xj i the set of all pǫ W such that the following six conditions are satisfied: (i) ǫ p 1 (k) = 1, for 1 k j; (ii) p 1 (k 1 ) < p 1 (k 2 ), for 1 k 1 < k 2 j; (iii) ǫ p 1 (k) = 1, for j + 1 k i; (iv) p 1 (k 1 ) > p 1 (k 2 ), for j + 1 k 1 < k 2 i; (v) ǫ p 1 (k) = 1, for i + 1 k n 1; (vi) p 1 (k 1 ) < p 1 (k 2 ), for i + 1 k 1 < k 2 n.

15 INDUCTION AND JACQUET MODULES 15 Denote by X n j the set of all pǫ W which satisfy the same conditions (for i = n), but instead of (iii), the condition (iii ) ǫ p 1 (k) = 1, for j + 1 k n 1, ǫ p 1 (n) = 1. Then, [W Ωi \W] = [W Ωn \W] = 0 j i 0 j n 1 X i j, X n j. Let i 1, i 2 1,..., n. For integers d, k such that 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d, we define a permutation p n (d, k) i1,i 2 in the same way as in [T1]: j, for 1 j k; j + i 1 k, for k + 1 j i 2 d; p n (d, k) i1,i 2 (j) = (i 1 + i 2 d + 1) j, for i 2 d + 1 j i 2 ; j i 2 + k, for i j i 1 + i 2 d k; j, for i 1 + i 2 d k + 1 j n. The conditions on d and k imply that p = p n (d, k) i1,i 2 is well-defined. For k 0, we set 1 k = 1, 1,..., 1 and 1 k = 1, 1,..., 1. a) If i 1, i 2 n, 0 d mini 1, i 2, d even, max0, (i 1 + i 2 n) d k mini 1, i 2 d, then we define q n (d, k) (0,0) = p n (d, k) i1,i 2 (1 i2 d, 1 d,1 n i2 ). If i 1, i 2 < n, 0 d mini 1, i 2, d odd, max0, (i 1 +i 2 n) d+ 1 k mini 1, i 2 d, then we define q n (d, k) (0,0) = p n (d, k) i1,i 2 (1 i2 d, 1 d,1 n i2 1, 1). b) If i 1, i 2 < n, 0 d mini 1, i 2, d even, k = i 1 +i 2 n d 0, then we define q n (d, k) (1,1) = p n (d, k) i1,i 2 (1 i2 d 1, 1 d+1,1 n i2 1, 1).

16 16 DUBRAVKA BAN c) If i 1 n, i 2 < n, 0 d mini 1, i 2, d odd, k = i 1 +i 2 n d 0, then we define q n (d, k) (1,0) = p n (d, k) i1,i 2 (1 i2 d, 1 d,1 n i2 1, 1). d) If i 1 < n, i 2 n, 0 d mini 1, i 2, d odd, k = i 1 +i 2 n d 0, then we define q n (d, k) (0,1) = p n (d, k) i1,i 2 (1 i2 d 1, 1 d+1,1 n i2 ). q n (d, k) (0,0), q n (d, k) (1,1), q n (d, k) (1,0) and q n (d, k) (0,1) are elements of W. Lemma 5.5. Let i 1, i 2 1,..., n. Suppose that integers j 1 and j 2 satisfy 0 j 1 i 1 and 0 j 2 i 2. If X i 1 j 1 Y i 2 j 2, then one of the following three conditions is satisfied: (i) i 1 j 1 = i 2 j 2 ; (ii) i 1 j 1 = i 2 j even; (iii) i 2 j 2 = i 1 j even. In that case, we have: (a) If i 1 j 1 = i 2 j 2 is even, then X i 1 j 1 Y i 2 j 2 = q n (d, k) (0,0) d = i 1 j 1, max0, (i 1 + i 2 n) d k mini 1, i 2 d (b) If i 1 j 1 = i 2 j 2 is odd, then X i 1 j 1 Y i 2 j 2 = q n (d, k) (0,0) d = i 1 j 1, max0, (i 1 + i 2 n) d + 1 k mini 1, i 2 d q n (d, k) (1,1) d = i 1 j 1 1, k = i 1 + i 2 n d 0. (c) If i 1 j 1 = i 2 j is even, then X i 1 j 1 Y i 2 j 2 = q n (d, k) (1,0) d = i 2 j 2, k = i 1 + i 2 n d 0. (d) If i 2 j 2 = i 1 j is even, then X i 1 j 1 Y i 2 j 2 = q n (d, k) (0,1) d = i 1 j 1, k = i 1 + i 2 n d 0 Proof. Let pǫ W. Then pǫ X i 1 j 1 Y i 2 j 2 if and only if the following twelve conditions are satisfied: (1) ǫ p 1 (l) = 1, for 1 l j 1 ; (2) p 1 (l 1 ) < p 1 (l 2 ), for 1 l 1 < l 2 j 1 ;.

17 INDUCTION AND JACQUET MODULES 17 (3) ǫ p 1 (l) = 1, for j l i 1 ; (4) p 1 (l 1 ) > p 1 (l 2 ), for j l 1 < l 2 i 1 ; (5) ǫ p 1 (l) = 1, for i l n 1; (6) p 1 (l 1 ) < p 1 (l 2 ), for i l 1 < l 2 n; (7) ǫ l = 1, for 1 l j 2 ; (8) p(l 1 ) < p(l 2 ), for 1 l 1 < l 2 j 2 ; (9) ǫ l = 1, for j l i 2 ; (10) p(l 1 ) > p(l 2 ), for j l 1 < l 2 i 2 ; (11) ǫ l = 1, for i l n 1; (12) p(l 1 ) < p(l 2 ), for i l 1 < l 2 n. Suppose that there exists pǫ X i 1 j 1 Y i 2 j 2. Then conditions (1),(3) and (5) give that the number of 1 s which appear in ǫ must be i 1 j 1 if i 1 j 1 is even, or i 1 j if i 1 j 1 is odd. Conditions (7),(9) and (11) give that the number of 1 s which apear in ǫ must be i 2 j 2 if i 2 j 2 is even, or i 2 j if i 2 j 2 is odd. We conclude that the difference between i 1 j 1 and i 2 j 2 is at most 1, and, if they are not equal, the bigger one is even.thus, we get conditions (i), (ii) and (iii) from the lemma. a) If i 1 j 1 = i 2 j 2 even, then ǫ n = 1, ǫ p 1 (n) = 1, so pǫ satisfies conditions (1)-(12) from Lemma 4.5 [T1], which gives the statement. b) Let i 1 j 1 = i 2 j 2 odd. If i 1 = n or i 2 = n, then there is no pǫ W which satisfies conditions (1)-(12), so X i 1 j 1 Y i 2 j 2 =. Suppose i 1, i 2 < n. From (7),(9) and (11), we conclude that ǫ = (1 j2, 1 i2 j 2,1 n i2 1, 1). Conditions (3),(7),(9) and (11) imply p([j 2 + 1, i 2 ] N n) = [j 1 + 1, i 1 ] N n. If p(n) = n, then conditions (1)-(12)restricted to the set 1,..., n 1 are the same as in Lemma 4.5[T1]. It follows that p = p n (d, k) i1,i 2 and i 1 + i 2 d k + 1 n, i.e., k i 1 + i 2 n d + 1. If p(n) n, then from (4) and (10) we see that p(j 2 + 1) = n, p 1 (j 1 + 1) = n, p([j 2 + 2, i 2 ] N ) = [j 1 + 2, i 1 ] N.

18 18 DUBRAVKA BAN Set d = i 2 j 2 1. By (10), p is order-reversing as a mapping p : [j 2 + 2, i 2 ] N [j 1 + 2, i 1 ] N. It follows that p(j) = i 1 (j j 2 2) = (i 1 + i 2 d + 1) j, for i 2 d + 1 j i 2. From p(j 2 + 1) = n we have p 1 (n) = j Together with (6), this implies In the same, way we get p 1 ([i 1 + 1, n 1] N ) [1, j 2 ] N. p([i 2 + 1, n 1] N ) [1, j 1 ] N. Let K = p 1 ([i 1 +1, n 1] N ), L = [1, j 2 ] N \K. Suppose that L. Since p(k j 2 + 1) = [i 1 + 1, n] N, we have p(k) > p(l), and from (8) we see that K > L, (i.e., p > q, p K, q L). Thus, there exists k 1,..., j 2 such that p 1 ([i 1 + 1, n 1] N ) = [k + 1, j 2 ] N. If L =, we put k = 0, so the above condition is satisfied. Now, we have (*) p 1 ([i 1 + 1, n] N ) = [k + 1, j 2 + 1] N = [k + 1, i 2 d] N, In the same way, we get From (12), we obtain n i 1 1 = i 2 d k 1, k = i 1 + i 2 n d 0. p([i 2 + 1, n] N ) = [k + 1, i 1 d] N. p(j) = k j i 2 1 = j i 2 + k, By ( ), we have and from (6) we see that p([k + 1, i 2 d] N ) = [i 1 + 1, n] N, i j n p(j) = i j k 1 = j + i 1 k, k + 1 j i 2 d. It remains to determine p on [1, k] N. From the above observations, we get so by (8), we have We conclude that p = p n (d, k) i1,i 2. p([1, k] N ) = [1, k] N, p(j) = j, for 1 j k.

19 INDUCTION AND JACQUET MODULES 19 It remains to prove that q = q n (d, k) (0,0) X i 1 j 1 Y i 2 j 2 when d = i 1 j 1 = i 2 j 2 and max0, (i 1 + i 2 n) d < k mini 1, i 2 d, and q = q n (d, k) (1,1) X i 1 j 1 Y i 2 j 2 when d = i 1 j 1 1 = i 2 j 2 1 and k = i 1 + i 2 n d 0. One sees directly from the definition of q and q that conditions (7)- (12) are satisfied. In the same way, one sees that conditions (1)-(6) are satisfied. c) Let i 1 j 1 = i 2 j even. If i 2 = n, then there is no pǫ W which satisfies conditions (1)-(12), so X i 1 j 1 Y i 2 j 2 =. Suppose that i 2 < n. Set d = i 2 j 2. From (1),(3),(5),(7) and (9), we see that From (4), we get ǫ n = 1, ǫ p 1 (n) = 1, ǫ = (1 i2 d, 1 d,1 n i2 1, 1), p([j 2 + 1, i 2 ] N n) = [j 1 + 1, i 1 ] N. p 1 (j 1 + 1) = n, p([j 2 + 1, i 2 ] N ) = [j 1 + 2, i 1 ] N, p(j) = i 1 (j j 2 1) = (i 1 + i 2 d + 1) j, i 2 d + 1 j i 2. In the same way as in (b), it follows from p(n) = j that p([i 2 + 1, n 1] N ) = [k + 1, j 1 ] N, where k = i 1 + i 2 n d 0, and We conclude that From (8), we have and p(j) = j, 1 j k, p(j) = j i 2 + k, i j n. p([1, i 2 d] N ) = [1, k] N [i 1 + 1, n] N. p([1, k] N ) = [1, k] N, p([k + 1, i 2 d] N ) = [i 1 + 1, n] N, p(j) = i j k 1 = j + i 1 k, k + 1 j i 2 d. Therefore, p = p n (d, k) i1,i 2. The rest of proof is same as in (b).

20 20 DUBRAVKA BAN (d) Analogous to (c). For i 1 = n, i 2 n, 0 d i 2, d odd, k = i 2 d, we define q n (d, k) ( 1, 1) n,i 2 = p n (d, k) i1,i 2 (1 i2 d+1, 1 d 1,1 n i2 ). For i 1 = n, i 2 < n, 0 < d i 2, d even, k = i 2 d, we define q n (d, k) (±1, 1) n,i 2 = p n (d, k) i1,i 2 (1 i2 d+1, 1 d 1,1 n i2 1, 1). For i 1 = n, i 2 < n, d = 0, k = i 2, we define q n (d, k) (±1, 1) n,i 2 = p n (d, k) i1,i 2. For i 1 = i 2 = n, 0 < d n, d even, k = n d, we define For i 1 = i 2 = n, q n (d, k) ( 2, 2) n,n = p n (d, k) n,n (1 n d+1, 1 d 2, 1). d = 0, k = n, we define q n (d, k) ( 2, 2) n,n = p n (d, k) n,n. An argument analogous to that for Lemma 5.5 gives Lemma 5.6. Let i 2 1,..., n. Suppose that integers j 1 and j 2 satisfy 0 j 1 n 1 and 1 j 2 i 2. If Xn j1 Y i 2 j 2, then one of the following two conditions is satisfied: (i) (n 1) j 1 = i 2 j 2 even, (ii) (n 1) j 1 = i 2 j even. In that case, we have: (a) If (n 1) j 1 = i 2 j 2 > 0 is even, then X n j 1 Y i 2 j 2 = q n (d, k) ( 1, 1) n,i 2 d = n j 1, k = i 2 d and for (n 1) j 1 = i 2 j 2 = 0, we have X j n 1 Y i 2 j 2 = q n (d, k) ( 1, 1) n,i 2 d = 1, k = i 2 d q n (d, k) (±1, 1) n,i 2 d = 0, k = i 2 d., (b) If (n 1) j 1 = i 2 j is even, then n 1 j 1 > 0 and X j n 1 Y i 2 j 2 = q n (d, k) (±1, 1) n,i 2 d = n j 1 1, k = i 2 d.

21 INDUCTION AND JACQUET MODULES 21 Proposition [ 5.7. Let i 1, i 2 1,..., n. Then, \W/W ] WΩi1 Ω i2 = = = = 0 d min d even 0 d min d odd q n (d, k) (0,0) i 1,i 2 max0, (i 1 + i 2 n d) k mini 1, i 2 d q n (d, k) (1,1) k = i 1 + i 2 n d 0 q n (d, k) (0,0) max0, (i 1 + i 2 n d) + 1 k mini 1, i 2 d Particularly: (a) [ If i 1 = n, ] i 2 < n, then WΩn \W/W Ωi2 = 0 d i 2 d even q n (d, k) (0,0) n,i 2 k = i 2 d [(b) If i 1 < n, i 2 = n, then \W/W ] WΩi1 Ω n = 0 d i 1 d even q n (d, k) (0,0) i 1,n k = i 1 d (c) If i 1 = i 2 = n, then [W Ωn \W/W Ωn ] = q n (d, k) (1,0) k = i 1 + i 2 n d 0 q n (d, k) (0,1) k = i 1 + i 2 n d 0 0 d n d even 0 d i 2 d odd 0 d i 1 d odd q n (d, k) (0,0) n,n q n (d, k) (1,0) n,i 2 k = i 2 d. q n (d, k) (0,1) i 1,n k = i 1 d. k = n d. Proof. We know that [W Θ \W/W Ω ] = [W Θ \W] [W/W Ω ], for Θ, Ω. From Lemmas 5.3 and 5.4, we have [ \W/W WΩi1 Ω i2] = [ W Ωi1 \W ] [ ( ) ( ) ] W/W Ωi2 = X i1 j 1 Y i2 j 2 0 j 1 i 1 0 j 2 i 2 ( ) = X i1 j 1 Y i 2 j 2. 0 j 2 i 2 0 j 1 i 1

22 22 DUBRAVKA BAN Now, Lemma 5.5 tells us when X i 1 j 1 Y i 2 j 2 proposition. is nonempty and gives the (a) If i 1 = n, i 2 < n, then q n (d, k) (1,1) n,i 2 and q n (d, k) (0,1) n,i 2 are not defined and q n (d, k) (0,0) n,i 2 is not defined for d odd. For d even, the inequality max0, (i 1 +i 2 n) d k mini 1, i 2 d becomes max0, i 2 d k mini 1, i 2 d, and its only solution is k = i 2 d. (b),(c) Analogously. In the same way, we get Proposition [ 5.8. Let] i 2 1,..., n. Then, (i) W Ωn \W/W Ωi2 = q n (d, k) ( 1, 1) n,i 2 k = i 2 d 0 d i 2 d odd (ii) [W Ω n \W/W Ω n ] = 0 d n d even In particular, for i 2 = n (i) reduces to [W Ω n \W/W Ωn ] = 0 d n d odd 0 d i 2 d even q n (d, k) (±1, 1) n,i 2 k = i 2 d. q n (d, k) ( 2, 2) n,n k = n d qn (d, k) ( 1, 1) n,n k = n d.. Lemma 5.9. Fix i 1, i 2 1, 2,..., n. Suppose that integers d, d and k, k satisfy the following conditions: Then, 0 d, d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d, max0, (i 1 + i 2 n) d k mini 1, i 2 d. (i) (p n (d, k) i1,i 2 ) 1 = p n (d, k) i2,i 1. ( ) 1 (ii) q n (d, k) (0,0) = qn (d, k) (0,0) i 2,i 1, ( ) 1 q n (d, k) (1,1) = qn (d, k) (1,1) i 2,i 1, ( ) 1 q n (d, k) (1,0) = qn (d, k) (0,1) i 2,i 1. (iii) Let w = q n (d, k) ( ), w = q n (d, k ) ( ), where ( ), ( ) (0, 0), (1, 1), (1, 0), (0, 1). If w = w, then ( ) = ( ), d = d, k = k.

23 INDUCTION AND JACQUET MODULES 23 Proof. The proofs of (i) and (ii) are straightforward calculations (also, cf. Lema 4.7 [T1]). (iii) Write w = pǫ and w = p ǫ, where p = p n (d, k) i1,i 2 and p = p n (d, k ) i1,i 2. Suppose that ( ) = ( ), and that d and d are both odd or both even. If we compare the numbers of 1 s which appear in ǫ and ǫ, we get d = d. Therefore, p n (d, k) i1,i 2 = p n (d, k ) i1,i 2. The definition of p n (d, k) i1,i 2 implies that k is the maximal integer which satisfies 0 k mini 1, i 2 d and p n (d, k) i1,i 2 (l) = l for all 1 l k. This implies k = k. We are now going to prove that in other cases we cannot have w = w. a) Let ( ) = (0, 0), d even. Suppose that ( ) = (0, 0), d odd. Then w = w implies n i 2 = 0, n i 2 1 = 0, which is impossible. We can use the same reasoning in the cases ( ) = (1, 1) and ( ) = (1, 0). In the case ( ) = (0, 1), we consider w 1 and (w ) 1. They are of type (0, 0) and (1, 0), so w 1 (w ) 1, which implies w w. b) Let ( ) = (0, 0), d odd. Then i 1, i 2 < n. Suppose that ( ) = (1, 1). Then w = w implies d = d + 1, k = k = i 1 + i 2 n d. Now we have i 1 + i 2 d k = i 1 + i 2 d (i 1 + i 2 n d + 1) = n 1, i 1 + i 2 d k = n, and by definition p n (d, k) i1,i 2 (n) = n, p n (d, k) i1,i 2 (n) = n i 2 + k = i 1 d < n, which contradicts the assumption w = w. If we suppose ( ) = (1, 0), then we get d = d, k = k. But the condition for w is k > i 1 + i 2 n d, and for w k = i 1 + i 2 n d. This is again a contradiction. In the case ( ) = (0, 1), the equality w = w implies n i 2 = 0 and n i 2 1 = 0, which is impossible. c) Let ( ) = (1, 1), d even. Suppose that ( ) = (1, 0). Then w = w implies d = d + 1, k = k. But we have k = i 1 +i 2 n d and k = i 1 +i 2 n d = i 1 +i 2 n d 1, which is impossible. The assumption ( ) = (0, 1) gives n i 2 1 = 0 and n i 2 = 0, a contradiction.

24 24 DUBRAVKA BAN d) Let ( ) = (1, 0), ( ) = (0, 1). Then w = w implies n i 2 1 = 0, n i 2 = 0, which is impossible. Define q n (d, k) i1,i 2 = p n (d, k) i1,i 2 (1 i2 d, 1 d,1 n i2 ). This is an automorphism of Σ. If d is even, then q n (d, k) i1,i 2 is an element of W. Recall that for i 1,..., n, we defined \αi, i n 1, Ω i =, Ω \α n, α n 1, i = n 1, n = \α n 1, Ω 0 =. Lemma Let w = q n (d, k) i1,i 2. Then, Ω i1 w(ω i2 ) = Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k. Proof. The conditions on d and k imply Set 0 k i 1 d i 1 i 1 + i 2 d k n, 0 k i 2 d i 2 i 1 + i 2 d k n. β i = α i, i = 1,..., n 1, β n = e n. Then Γ = β 1,..., β n is the set of simple roots of the root system of type B n, into which our root system embeds. We shall use the following formula, proved in [T1]: (*) (Γ \ β i1 ) w(γ \ β i2 ) = Γ \ β l l k, i 1 d, i 1, i 1 + i 2 d k \ 0. Since we have it follows that Γ \ β n = \ α n, (Γ \ β j ) \ β n = Ω j \ α n, (for j 0), (**) (Ω i1 w(ω i2 )) \ α n, w(α n ) = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n, w(α n ). We consider several cases. (a) Let i 1, i 2 < n 1. First, suppose that i 1 +i 2 d k < n 1. Then k, i 1 d < n 1. From the definition of w = q n (d, k) i1,i 2, we get w(e n 1 ) = e n 1, w(e n ) = e n,

25 INDUCTION AND JACQUET MODULES 25 so w(α n 1 ) = α n 1, w(α n ) = α n. We apply this to formula (*) and we get ( \ α i1 ) w( \ α i2 ) = \ α l l k, i 1 d, i 1, i 1 + i 2 d k \ 0. Since i 1, i 2, k, i 1 d, i 1 + i 2 d k < n 1, this is exactly the formula from the lemma. Next, we consider the case when i 1 + i 2 d k = n 1. Then, w 1 (α n 1 ) = w 1 (e n 1 e n ) = p n (d, k) i2,i 1 (e n 1 e n ) = e i2 d e n. This is an element of if i 2 d = n 1, which is impossible since i 2 < n 1. So, w 1 (α n 1 ) / and α n 1 / w(ω i2 ). In the same way, we see that α n / w(ω i2 ). Hence, Ω i1 w(ω i2 ) \ α n 1, α n = Ω n 1. Similarly, w(α n ) /. Now from (**), we have (Ω i1 w(ω i2 )) \ α n = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n. Since Ω i1 w(ω i2 ) Ω n 1 and i 1 + i 2 d k = n 1, if we intersect the above equality with Ω n 1, we get the formula we need. It remains to consider the case when i 1 + i 2 d k = n. We have w 1 (α n ) = e i2 d 1 + e i2 d, and this is not in because i 2 < n 1. Hence, α n / w(ω i2 ). Also, we have w(α n ) = e i1 d 1 + e i1 d, and again this is not in. Now, the relation (**) becomes Ω i1 w(ω i2 ) = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n since Ω i1 +i 2 d k = Ω n = \ α n. = Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k, (b) Let i 1 n 1. If i 1 + i 2 d k = n 1 (this is possible for i 1 = n 1), then we have If i 1 + i 2 d k = n, then w(α n ) = w(e n 1 + e n ) = e n 1 i2 +k + e n. w(α n ) = w(e n 1 + e n ) = e n 1 i2 +k + e n i2 +k. Anyway, w(α n ) implies w(α n ) = α n, and the relation (**) becomes (Ω i1 w(ω i2 )) \ α n = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n. Since α n / Ω i1, the relation we want follows immediately.

26 26 DUBRAVKA BAN (c) Now, consider the case when i 1 < n 1, i 2 = n 1. First, suppose that i 1 + i 2 d k = n 1. If d = 0, then w(α n ) = w(e n 1 + e n ) = e n 1 + e n = α n, so α n / w(ω i2 ) and w(α n ) = α n / Ω i2. If d > 0, then w 1 (α n ) = w 1 (e n 1 + e n ) = p n (d, k) i2,i 1 (e n 1 + e n ) = e i2 d + e n /, w(α n ) = w(e n 1 + e n ) = p n (d, k) i1,i 2 ( e n 1 + e n ) = e i1 d+1 + e n /. Anyway, the relation (**) becomes Ω i1 w(ω i2 ) = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n, and since i 1 + i 2 d k = n 1, we are done. Now, suppose that i 1 + i 2 d k = n.then w 1 (α n ) = p n (d, k) i2,i 1 (e n 1 + e n ) = e i2 d 1 + e i2 d, and this is not in since i 2 d n 1. If d = 0, then w(α n ) = w(e n 1 + e n ) = p n (d, k) i1,i 2 (e n 1 + e n ) = e n + e i1 /, and the relation (**) gives the result. If d > 0, then w(α n ) = w(e n 1 + e n ) = p n (d, k) i1,i 2 ( e n 1 + e n ) = e i1 d+1 + e i1 d = α i1 d. Now from (**), we have (Ω i1 w(ω i2 )) \ α i1 d = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n, α i1 d. But α i1 d / w(ω i2 ) since α i1 d = w(α n ). Also, α n / Ω i1 +i 2 d k and α i1 d / Ω i1 d. The result follows. (d) It remains to consider the case i 1 < n 1, i 2 = n. Then i 1 + i 2 d k = n and we have w 1 (α n ) = p n (d, k) i2,i 1 (e n 1 + e n ) = e n 1 i1 +k + e n i1 +k = e n 1 d + e n d. We see that w 1 (α n ) /, for d > 0, and w 1 (α n ) = α n, for d = 0. Since i 2 = n, in both cases we have α n / w(ω i2 ). Now for d > 1, we have and for d = 1 w(α n ) = p n (d, k) i1,i 2 ( e n 1 e n ), w(α n ) = p n (d, k) i1,i 2 (e n 1 e n ) = e n e i1, and in both cases w(α n ) /. Hence, relation (**) becomes Ω i1 w(ω i2 ) = (Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k) \ α n, and the result follows from the condition i 1 + i 2 d k = n.

27 INDUCTION AND JACQUET MODULES 27 Lemma (i) If w = q n (d, k) (0,0) or w = q n (d, k) (1,0), then Ω i1 w(ω i2 ) = j ΠΩ j where Π = k, i 1 d, i 1, i 1 + i 2 d k\0, or, equivalently, Ω i1 w(ω i2 ) = Ω k Ω i1 d Ω i1 Ω i1 +i 2 d k. (ii) If w = q n (d, k) (1,1) or w = q n (d, k) (0,1), then or, equivalently, ( ) Ω i1 w(ω i2 ) = s Ω j Ω i1 w(ω i2 ) = Ω k Ω i1 d Ω i1 Ω n, j Π Π = k, i 1 d, i 1, i 1 + i 2 d k = n, where s = (1 n 1, 1) denotes the automorphism of Σ which interchanges α n 1 and α n. Proof. (a) Let w = q n (d, k) (0,0). If d is even, then w = q n (d, k) i1,i 2, and the statement follows from Lemma If d is odd, then i 1, i 2 < n. Now w = w s, where w = q n (d, k) i1,i 2. Note that s(ω i2 ) = Ω i2 for i 2 < n, so we have Ω i1 w(ω i2 ) = Ω i1 w s(ω i2 ) = Ω i1 w (Ω i2 ). The result follows from Lemma (b) Let w = q n (d, k) (1,1). Then, i 1, i 2 < n, i 1 + i 2 d k = n and w = sw s, where w = q n (d, k) i1,i 2. Now, we have Ω i1 w(ω i2 ) = Ω i1 sw s(ω i2 ) = s(ω i1 w s(ω i2 )) = = s(ω i1 w (Ω i2 )) = (Lemma 5.10) = s(ω k Ω i1 d Ω i1 Ω n ) = = Ω k Ω i1 d Ω i1 Ω n. (c) Let w = q n (d, k) (1,0). Then, i 2 < n and w = w s, where w = q n (d, k) i1,i 2. It follows that Ω i1 w(ω i2 ) = Ω i1 w s(ω i2 ) = Ω i1 w (Ω i2 ) and Lemma 5.10 gives the result.

28 28 DUBRAVKA BAN (d) Let w = q n (d, k) (0,1). Then, i 1 < n and w = sw, where w = q n (d, k) i1,i 2. Now, we have Ω i1 w(ω i2 ) = Ω i1 sw (Ω i2 ) = s(ω i1 w (Ω i2 )) = In the same way, we get = (Lemma 5.10) = s(ω k Ω i1 d Ω i1 Ω n ) = = Ω k Ω i1 d Ω i1 Ω n. Lemma Let w = q n (d, k) ( 1, 1) n,i 2 Ω n w(ω i2 ) = s(ω k Ω n d Ω n ). or w = q n (d, k) (±1, 1) n,i 2. Then, 6. Orthogonal group O(2n, F) The orthogonal group O(2n, F), n 1, is the group O(2n, F) = X GL(2n, F) τ XX = I 2n. O(2n, F) has two connected components. The first is SO(2n, F) = X O(2n, F) detx = 1, and the second is X O(2n, F) detx = 1. We have where O(2n, F) = SO(2n, F) s SO(2n, F), s = I Let denote the set of simple roots of SO(2n, F), W the Weyl group. Let α = (n 1,..., n k ) be an ordered partition of m n. Denote by P α = M α U α the standard parabolic subgroup of SO(2n, F) with Levi factor M α = GL(n1, F) GL(n k, F) SO(2(n m), F). We shall consider the following subgroups of O(2n, F): Pα sp Q α = α, for m < n, P α, for m = n. I. It follows that Q α = N α U α, where Mα sm N α = α, for m < n, M α, for m = n. We have N α = diag(g 1,..., g k, h, τ g 1 k,..., τ g 1 1 ) g i GL(n i, F), h O(2(n m), F),

29 INDUCTION AND JACQUET MODULES 29 so N α = GL(n1, F) GL(n k, F) O(2(n m), F). Let α = (i). The subgroups N = N α and V = U α are closed, N normalises V and N V = e, so by the first section, we have functors i V,1 and r V,1. Define i G,N = i V,1 and r N,G = r V,1. Hence, i G,N : AlgN AlgG, r N,G : AlgG AlgN. Let α = (i 1 ), β = (i 2 ), P = Q α = MU, Q = Q β = NV.Let σ be an admissible representation of O(2n, F). We consider r N,G i G,M (σ). By Theorem 2.1, we can find a composition series of r N,G i G,M (σ). We need to calculate representatives of P \ O(2n, F) / Q. Lemma 6.1. Let i 1, i 2 1,..., n, α = (i 1 ), β = (i 2 ), P = Q α = MU, Q = Q β = NV. (i) q n (d, k) i1,i 2 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d is a set of representatives of P \ O(2n, F) / Q. (ii) Let w = q n (d, k) i1,i 2. The groups w 1 (P), w 1 (M) and w 1 (U) are decomposable with respect to (N, V ), and the groups w(q), w(n) and w(v ) are decomposable with respect to (M, U). Proof. (a) Suppose that i 1, i 2 < n. Then, P = Q α = P α sp α, Q = Q β = P β sp β. Let x SO(2n, F). Then, PxsQ = (P α sp α ) x (sp β P β ) = PxQ, so [x] = [xs]. Analogously, [x] = [sx]. Thus, we can choose representatives from SO(2n, F). Let x, y SO(2n, F) with [x] = [y]. Now, we have It follows that PxQ = PyQ, (P α sp α ) x (sp β P β ) = (P α sp α ) y (sp β P β ), (P α xp β ) (sp α xp β ) (P α xsp β ) (P α sxs 1 P β ) = (P α yp β ) (sp α yp β ) (P α ysp β ) (P α sys 1 P β ). (P α xp β ) (P α sxs 1 P β ) = (P α yp β ) (P α sys 1 P β ),

30 30 DUBRAVKA BAN so P α xp β = P α yp β or P α xp β = P α sys 1 P β. We know that [ ] W Ωi1 \W/W Ωi2 is a set of representatives of Pα \ SO(2n, F) / P β. By the above considerations, a set of representatives of P \ O(2n, F) / Q can be chosen from the set [ ] W Ωi1 \W/W Ωi2 in the following way: we take all elements which satisfy P α wp β = P α sws 1 P β, and from the remaining set we choose w or a representative of P α sws 1 P β. For w = q n (d, k) (0,0) [ ] W Ωi1 \W/W Ωi2 and i1 + i 2 d k < n, we have w = sws 1. For i 1 + i 2 d k = n, we have sq n (d, k) (0,0) s 1 = q n (d, k) (1,1), if d is even, sq n (d, k) (1,0) s 1 = q n (d, k) (0,1), if d is odd. We conclude that q n (d, k) (0,0) 0 d mini 1, i 2, for d even max0, (i 1 + i 2 n) d k mini 1, i 2 d, max0, (i 1 + i 2 n) d < k mini 1, i 2 d for d odd q n (d, k) (1,0) 0 d mini 1, i 2, d odd, k = i 1 + i 2 n d 0 is a set of representatives of P \ O(2n, F) / Q. We have q n (d, k) (0,0), for d even, q n (d, k) i1,i 2 = q n (d, k) (0,0) s, for d odd and i 1 + i 2 d k < n, q n (d, k) (1,0) s, for d odd and i 1 + i 2 d k = n. Then, from the relation [x] = [sx], it follows that the set q n (d, k) i1,i 2 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d is a set of representatives of P \ O(2n, F) / Q. Let w = q n (d, k) i1,i 2. We shall show that the group w(q) is decomposable with respect to (M, U). If i 1 + i 2 d k < n, then w and s commute, so Then, w(s) = wsw 1 = s. w(q) (MU) = wqw 1 (MU) = w(p β sp β ) (M α U α sm α U α ) = = [w(p β ) M α U α ] s[w(p β ) M α U α ] = (because w(p β ) is decomposable with respect to (M α, U α )) = [(w(p β ) M α )(w(p β ) U α )] s[(w(p β ) M α )(w(p β ) U α )] = [(w(p β ) M α ) s(w(p β ) M α )] [w(p β ) U α ].

31 On the other side, INDUCTION AND JACQUET MODULES 31 (w(q) M)(w(Q) U) = = [(w(p β ) s w(p β )) (M α sm α )] [(w(p β ) s w(p β )) U α ] = [(w(p β ) M α ) s (w(p β ) M α )] [w(p β ) U α ], so w(q) is decomposable with respect to (M, U). If i 1 + i 2 d k = n, then w(q) P = w(p β sp β ) (P α sp α ) = (w(p β ) P α ) (w(sp β ) sp α ). It can be shown that w(sp β ) sp α =, which implies w(sp β ) sm α =. It follows that w(q) P = w(p β ) P α, w(q) M = w(p β ) M α. If w [ W Ωi1 \W/W Ωi2 ], we have w(q) MU = w(p β ) P α = w(p β ) M α U α = (since w(p β ) is decomposable with respect to (M α, U α )) = (w(p β ) M α )(w(p β ) U α ) = (w(q) M)(w(Q) U). If w / [ W Ωi1 \W/W Ωi2 ],then w = ws [ W Ωi1 \W/W Ωi2 ] and Now, we have w (P β ) = w P β (w ) 1 = wsp β sw 1 = wp β w 1. w(q) MU = w(p β ) P α = w (P β ) M α U α = = (w (P β ) M α )(w (P β ) U α ) = (w(q) M)(w(Q) U). Hence, w(q) is decomposable with respect to (M, U). For the other groups, the proof is similar. (b) Let i 1 = n, i 2 < n. For x SO(2n, F), we have PxsQ = P α x (sp β P β ) = PxQ, so [x] = [xs] (but the classes [x] and [sx] are not the same in general). Hence, we can choose representatives from SO(2n, F) again. Let x, y SO(2n, F) and [x] = [y]. Now, P α x (sp β P β ) = P α y (sp β P β ), (P α xp β ) (P α xsp β ) = (P α yp β ) (P α ysp β ).

32 32 DUBRAVKA BAN It follows that P α xp β = P α yp β, so [ ] W Ωi1 \W/W Ωi2 is a set of representatives of P \ O(2n, F) / Q. Recall that [ \W/W WΩi1 Ω i2] = q n (d, k) (0,0) n,i 2 k = i 2 d Since q n (d, k) n,i2 = 0 d i 2 d even 0 d i 2 d odd the equality [x] = [xs] implies that q n (d, k) (1,0) n,i 2 k = i 2 d. q n (d, k) (0,0) n,i 2, for d even, q n (d, k) (1,0) n,i 2 s, for d odd, q n (d, k) n,i2 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d is a set of representatives of P \ O(2n, F) / Q. Let w = q n (d, k) n,i2. Then, w [ W Ωi1 \W/W Ωi2 ] or ws [ WΩi1 \W/W Ωi2 ]. Since i 2 < n, we have (ws)(p β ) = w(p β ). Now, w(q) MU = w(p β sp β ) M α U α = w(p β ) M α U α = = (w(p β ) M α )(w(p β ) U α ) = (w(q) M)(w(Q) U). The proof for w(n) and w(v ) is similar. If w [ W Ωi1 \W/W Ωi2 ], then it is easy to show the statement for w 1 (P). If ws [ W Ωi1 \W/W Ωi2 ], then w 1 (P) NV = w 1 (P α ) (P β sp β ) = w 1 (P α ) (P β ) = = s (sw 1 (P α ) (P β )) s = (since (sw 1 )(P α ) is decomposable with respect to (M β, U β )) = s ((sw 1 )(P α ) M β )((sw 1 )(P α ) U β ) s = (w 1 (P α ) M β )(w 1 (P α ) U β ) = (w 1 (P α ) N)(w 1 (P α ) V ). Analogously for w 1 (M), w 1 (U). (c) For i 1 < n, i 2 = n, the argument is analogous to (b).

33 INDUCTION AND JACQUET MODULES 33 (d) Let i 1 = i 2 = n. For x, y SO(2n, F), we have [x] = [y] P α xp β = P α yp β, so elements of [ W Ωi1 \W/W Ωi2 ] represent different classes. Moreover, so P α xp β SO(2n, F), P α sxp β O(2n, F)\SO(2n, F), [x] [sx], Let x, y SO(2n, F), [sx] = [sy]. Then, [sx] [y]. P α sxp β = P α syp β, sp α sxp β = sp α syp β. We conclude that the elements sw, where w [W Ωn \W/W Ωn ], represent all the classes of type [sx], x SO(2n, F). Now, we get the following set of representatives: 0 d n d even Since qn (d, k) (0,0) n,n k = n d it follows that q n (d, k) n,n = 0 d n d odd q n (d, k) (0,0) n,n, for d even, sq n (d, k) ( 1, 1) n,n, for d odd, sqn (d, k) ( 1, 1) n,n k = n d. q n (d, k) n,n 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d is a set of representatives of P \ O(2n, F) / Q. For w [W Ωn \W/W Ωn ], we know by [BZ2] that w 1 (P), w 1 (M) and w 1 (U) are decomposable with respect to (N, V ), and that w(q), w(n) and w(v ) are decomposable with respect to (M, U). Let w [W Ω n \W/W Ωn ]. Then, (sw) 1 (P) NV = (sw) 1 (P α ) M β U β = w 1 sp α sw M β U β = (since w 1 (sp α s) is decomposable with respect to(m β, U β )) = (w 1 sp α sw M β )(w 1 sp α sw U β ) = [ (sw) 1 (P) N ][ (sw) 1 (P) V ].

34 34 DUBRAVKA BAN The arguments for (sw) 1 (M) and (sw) 1 (U) are similar. For (sw)(q), we have (sw)(q) MU = (sw)(p β ) M α U α = s (w(p β ) sm α ssu α s) s = (since w(p β ) is decomposable with respect to (sm α s, su α s)) = s (w(p β ) sm α s)(w(p β ) su α s) s = [(sw)(p β ) M α ] [(sw)(p β ) U α ] = [(sw)(q) M] [(sw)(q) U]. The arguments for (sw)(n) and (sw)(v ) are similar. It can be easily verified that, in our case, the character ε from Theorem 2.1 is equal to 1. Now, by Theorem 2.1 and Lemma 6.1, we have Lemma 6.2. Let i 1, i 2 1,..., n, α = (i 1 ), β = (i 2 ), P = Q α = MU, Q = Q β = NV. Let σ be an admissible representation of M. Then r N,G i G,M (σ) has a composition series with factors i N,N w 1 r M,M(σ), where N = w 1 (M) N, M = M w(n) and w q n (d, k) i1,i 2 0 d mini 1, i 2, max0, (i 1 + i 2 n) d k mini 1, i 2 d. The following lemma describes M and N from Lemma 6.2. Lemma 6.3. Let w = q n (d, k) i1,i 2, α = (i 1 ), β = (i 2 ). Then, N α w(n β ) = N γ, where γ = (k, i 1 d, i 1, i 1 + i 2 d k). Proof. Recall from [C] (Proposition 1.3.3) that for θ, Ω and w [W θ \W/W Ω ] we have M θ w(m Ω ) = M θ w(ω). a) Let i 1, i 2 < n. If d is even, then w [ W Ωi1 \W/W Ωi2 ] and we have N α w(n β ) = (M α sm α ) w(m β sm β ) = (M α w(m β )) (sm α w(sm β )). If i 1 + i 2 d k < n, then s and w commute, so N α w(n β ) = (M α w(m β )) s (M α w(m β )) = M γ sm γ = N γ. If i 1 + i 2 d k = n, then by the proof of Lemma 6.1 we have sm α w(sm β ) =, so N α w(n β ) = M γ.