Introduction to admissible representations of p-adic groups. Chapter III. Induced representations and the Jacquet module

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1 Last revised 6:45 p.m. February 16, 2016 Introduction to admissible representations of p-adic groups Bill Casselman University of British Columbia Chapter III. Induced representations and the Jacquet module Inthischapter,let Gbethe k rationalpointsonareductivegroupdefinedover k. I llintroducehererepresentations induced from parabolic subgroups to G, as well as a related construction going from representations of G to those of parabolic subgroups. These constructions will lead eventually to a classification of irreducible admissible representations of G. arabolic induction is a classic technique in representation theory, but the adjoint construction has its origins in [Jacquet Langlands:1970] and [Jacquet:1971]. Contents 1. Representations induced from parabolic subgroups 2. Admissible representations of parabolic subgroups 3. The Jacquet module 4. Iwahori factorizations 5. Admissibility of the Jacquet module 6. The canonical pairing 7. References SUMMARY. If = M is a parabolic subgroup of G and V a smooth representation of G, its Jacquet module is its maximal trivial quotient V. This is a smooth representation of M = /, and if V is an admissible representation of G the Jacquet module is an admissible representation of M. It is almost by definition that the Jacquet module controls equivariant maps from V into representations induced from to G,butitalsocontrolstheasymptoticbehaviourofmatrixcoefficientsof V. Thisdualroleoriginatesinwork of Harish Chandra on representations of real reductive groups, and is the basis for harmonic analysis on G. 1. Representations induced from parabolic subgroups Suppose tobeaparabolic subgroupof G,and (σ, U)an admissiblerepresentationof. The(normalized) representation induced from σ is the right regular representation of G on the space Ind(σ, G) = {f C (G, U) f(pg) = σ(p)δ 1/2 (p)f(g)for all p, g G}. Here δ (p) = det n (p). Since \G is compact, according to roposition II.9.1 this representation is admissible. Bruhat and Tits define a goodcompact subgroup of G to be a compact open K such that G = K for all parabolic subgroups. Theyprovethatsuchexist. If K isgood thentherestrictionof Ind(σ, G) to K isa K isomorphism of Ind(σ, G) with Ind(σ K, K).

2 Chapter III. Induced representations and the Jacquet module 2 An unramified characterof isone that is trivial on K. Itfollows from the observation abovethat if χ is unramified then Ind(σ, G) and Ind(σχ, G) are canonically isomorphic as K representations. Another way of puttingthisisthat all theserepresentationsmaybe definedon thesame space. roposition II.9.3 implies: III.1.1. roposition.if (π, V )isanadmissiblerepresentationof Gand (σ, U)oneof,thencompositionwith the map Ω 1 : f f(1) induces an isomorphism Hom G ( V, Ind(σ, G) ) = Hom (V, U). If ϕliesin Ind(δ 1/2, G)the integral K ϕ(k) dk defines a G invariant linear functional. Since G/ is compact, roposition II.9.4 implies: III.1.2. roposition. The contragredient of Ind(σ, G) is isomorphic to Ind( σ, G). If f lies in the first and F the second, then the product pairing f(x), F(x) lies in Ind(δ 1/2, G). The pairing can thenbe chosento be f(k), F(k) dk. III.1.3. roposition. The representation Ind(σ, G) is unitary if σ is. K This choice of invariant integral is by no means canonical, and in other contexst other choices are natural. There is a great deal more to be said about these representations, but first we need to investigate admissible representations of. 2. Admissible representations of parabolic subgroups Let = M = M be a parabolic subgroup of G, A = A the split centre of M. There exists a basis of neighbourhoods of of the form U M U where U M is a compact open subgroup of M, U is one of, and U M conjugates U toitself. The group M may be identified with a quotient of, and therefore the admissible representations of M may beidentifiedwiththose of trivial on. Ithappens that thereare no others: III.2.1. roposition. Every admissible representation of is trivial on. roof. Let (π, V ) be an admissible representation of over the ring R, and suppose v in V. We want to show that π(n)v = v for all n in. So suppose n in, and suppose U = U M U fixes v. We can find a in A such that a 1 na U as well as a 1 Ua U. Then π(a)takes V U toitself, sincefor uin U wehave π(u)π(a)v = π(a)π(a 1 ua)v = π(a)v. The operator π(a) is certainly injective. I shall prove in a moment that it is bijective on V U. Assuming this, wecan find v such that π(a)v = v, and then: π(n)v = π(n)π(a)π(a 1 )v = π(n)π(a)v = π(a)π(a 1 na)v = π(a)v = v.

3 Chapter III. Induced representations and the Jacquet module 3 If Risafield,theclaimthat π(a)issurjectiveisistrivial,since V U isfinite dimensionaland π(a)isinjective. Butif Risanarbitrarycommutativeoetherianring,onehastobeabitmorecareful. Thefollowingwilltell uswhatisneeded: III.2.2. Lemma. Suppose B to be a finitely generated module over the oetherian ring R. If f: B B is an R injection with the property that for each maximal ideal m of R the induced map f m : B/mB B/mB is also injective, then f is itself an isomorphism. roof. Let C bethe quotient B/f(B). The exactsequence inducesfor each man exactsequence 0 B f B C 0 0 B/mB fm B/mB C/mC 0. It is by assumption that the left hand map is injective. Since F = R/m is a field and B is finitely generated, the space B/mB is a finite dimensional vector space over F, and therefore f m an isomorphism. Hence C/mC = 0 for all m. The module C is oetherian, which means that if C 0, it possesses at least one maximal proper submodule D. The quotient C/D must be isomorphic to R/m for some maximal ideal m. Butthen C/mC 0,acontradiction. Therefore C = 0and f an isomorphism. 3. The Jacquet module Suppose = M to be a parabolic subgroup of G. (π, V ) an admissible representation of G, and (σ, U) one of. Frobenius reciprocity(roposition II.9.3) tells us that Hom G ( V, Ind(σ, G) ) = Hom (V, U), while the results of the previous section tell us that σ is trivial on and factors through the canonical projection M. Inthissection weexplorethe consequencesof joining thesetwo facts. III.3.1. Lemma. If is a p adic unipotent group, it possesses arbitrarily large compact open subgroups. roof. It is certainly true for the group of unipotent upper triangular matrices in GL n. Here, if a is the diagonalmatrixwith a i,i = i thenconjugationbypowersof awillscaleanygivencompactopensubgroup toanarbitrarilylargeone. Butanyunipotentgroupcanbeembeddedasaclosedsubgroupinoneofthese. Fix the parabolic subgroup = M. If (π, V ) is any smooth representation of, define V () to be the subspace of V generatedbyvectorsof theform π(n)v v as nrangesover. Thegroup acts triviallyonthe quotient It is universal with respect to this property: V = V/V () III.3.2. roposition.theprojectionfrom V to V inducesforeverysmooth R representation (σ, U)onwhich acts trivially an isomorphism Hom (V, U) = Hom R (V, U). III.3.3. Lemma. For v in V the following are equivalent:

4 Chapter III. Induced representations and the Jacquet module 4 (a) v liesin V (); (b) v liesin V (U)for some compact open subgroupof ; (c) wehave π(u)v du = 0 for somecompact open subgroup U of. U roof. The equivalence follows immediately from Lemma III.3.1. That of(b) and(c) follows from roposition II.1.2. III.3.4. roposition. If 0 U V W 0 isan exact sequenceof smoothrepresentationsof,thenthesequence is also exact. 0 U V W 0 roof. That the sequence U V W 0 is exact follows immediately from the definition of V (). The only non trivial point is the injectivity of U V. If u in U lies in V () then it lies in V (S) for some compact open subgroup S of. According to roposition II.1.2, the space V has a canonical decomposition and v liesin V (S)if and onlyif V = V S V (S), S π(s)v ds = 0 Butthislast equationholds in U aswell,since U is stableunder S,so v mustliein U(S). If σistrivialon,any mapfrom V to U factorsthrough V. Thespace V ()isstableunder,andthere ishenceanaturalrepresentationof M on V. The Jacquet moduleof π isthisrepresentationtwistedbythe character δ 1/2. Inotherwords, if uliesin V/V ()and v in V has image u,then π (m)u istheimage of δ 1/2 (m)π(m)v for m in M. This isdesignedexactly toallow thesimplest formulation of this: III.3.5. roposition. If (π, V ) is any smooth representation of G and (σ, U) one of M then evaluation at 1 induces an isomorphism Hom G ( π, Ind(σ, G) ) = HomM (π, σ)

5 Chapter III. Induced representations and the Jacquet module 5 4. Iwahori factorizations Suppose = M to be a parabolic subgroup of G, the corresponding opposite one, with = M. If K is a compact open subgroup, it is said to have an Iwahori factorization with respect to if (a) the product map from K M K K to K is a bijection, where M K = M K etc. and (b) a K a 1 K, a 1 K a K for every ain A. III.4.1. Lemma. Let be a minimal parabolic subgroupof G. There exists a basis {K n } of neighbourhoods of {1}in G such that (a) each K n isanormal subgroupof K 0 ; (b) If is a parabolic subgroupof G containing then K n has an Iwahori factorization with respect to ; (c) if = M is a parabolic subgroup containing then M K n has an Iwahori factorization with respectto M. roof. Keep in mind that the statement depends only on the conjugacy class of the parabolic group. Assume first that G is split over k. According to [Iwahori Matsumoto:1967], there exists a split group scheme G o defining Gover kbybaseextension. Wemaychoose alsotobeobtainedbybaseextensionfromasmooth parabolic subgroup defined over o. The sequence of congruence subgroups G(p n ) for n 1 satisfies the conditions of the Lemma. owsuppose Garbitrary. Let l/kbeafinitegaloisextension l/koverwhich Gsplits. Let K l,n beasequence satisfying the roposition for a minimal parabolic subgroup contained in k l, and let K n = K l,n G. Galois theory together with uniqueness of the Iwahori factorizations allows us to conclude. 5. Admissibility of the Jacquet module owfixan admissible representation (π, V )of G. Let, bean opposing pair of parabolic subgroups, K 0 tobeacompactopensubgrouppossessing aniwahorifactorization K 0 = 0 M 0 0 withrespecttothispair. For each a in A let T a bethe smooth distribution µ K0aK 0/K 0 on G. For any smooth representation (π, V ) and v in V K0 let τ a betherestrictionof π(t a )to V K0. Thusfor v in V K0 τ a (v) = π(t a )v = π(g)v K 0aK 0/K 0 = π(k)π(a)v. K 0/K 0 ak 0a 1 Thisisvalidsincetheisotropysubgroupof aintheactionof K 0 actingon K 0 ak 0 /K 0 is ak 0 a 1 K 0,hence isabijectionof K 0 /K 0 ak 0 a 1 with K 0 ak 0 /K 0. k kak 0 III.5.1. Lemma.If v liesin V K0 withimage uin V,theimage of τ a v in V isequal to δ 1/2 (a)π (a)u. roof. Since K 0 = 0 M 0 0, ak 0 a 1 = (a 0 a 1 )M 0 (a 0 a 1 ). Since aa 1, the inclusion of 0 /a 0 a 1 into K 0 /(ak 0 a 1 K 0 )isinturnabijection. Sincetheindexof a 0 a 1 in 0 or,equivalently, thatof 0 in a 1 0 ais δ 1 (a): τ a (v) = π(k)π(a)v K 0/aK 0a 1 K 0 = π(n)π(a)v 0/a 0a 1 =π(a) π(n)v. a 1 0a/ 0

6 Chapter III. Induced representations and the Jacquet module 6 Since π(n)v and v havethesame image in V, and [a 1 0 a: 0 ] = δ 1 (a), thisconcludestheproof. III.5.2. Lemma.For every a, bin A, τ ab = τ a τ b roof. We have T a T b = 0/a 0a 1 = = 0/a 0a 1 π(n 1 π(a)π(n 2 )π(b)v 0/b 0b 1 π(n 1 )π(an 2 a 1 )π(ab)v 0/b 0b 1 0/ab 0b 1 a 1 π(n)π(ab)v = T ab since as n 1 ranges over representatives of 0 /a 0 a 1 and and n 2 over representatives of 0 /b 0 b 1, the products n 1 an 2 a 1 range over representativesof 0 /ab 0 b 1 a 1. III.5.3. Lemma. For any a in A the subspace of V K0 on which τ a acts nilpotently coincides with V K0 V (). roof. Since Risoetherianand V K0 finitelygenerated,theincreasingsequence ker(τ a ) ker(τ a 2) ker(τ a 3)... iseventuallystationary. Itmustbe shownthat itisthesame as V K0 V (). Choose nlarge enoughso that V K0 V () = V K0 V (a n 0 a n ). Let b = a n. Since τ b v = π(b) π(n)v, b 1 0b/ 0 and τ b v = 0if and only if b 1 0b/ 0 π(n)v = 0,and again ifand onlyif v liesin V (). The canonical map from V to V takes V K0 to V M0. The kernelof thismapis V V (), whichbylemma III.5.3isequalto thekernelof τ a n for large n. III.5.4. Lemma.Theimageof τ a n in V K0 isindependentof nif nislargeenough. Themap τ a isinvertibleon it. The intersectionof itwith V ()istrivial. roof. Choose n so large that ker(τ a n) = ker(τ a m) for all m n. By Lemma III.5.3 this kernel coincides with V K0 V (). Let U be the image of τ a n. If u = τ a nv and τ a nv = 0 then τ a 2nv = 0, which means by assumption that in fact u = τ a nv = 0. Therefore the intersection of U with V () is trivial, the projection from V to V is injective on U, and τ a is also injective on it. If m is a maximal ideal of R, this remains true for U/mU, and therefore by Lemma III.2.2 τ a is invertible on U. This implies that U is independent of the choiceof n. Let V K0 V K0 to V M0 bethiscommonimageof the τ an for large n. Thepointisthatitsplitsthecanonicalprojectionfrom, whichturnsoutto beasurjection. III.5.5. roposition.the canonical projection from V K0 M0 to V isan isomorphism. roof. Suppose given u in V M0. Since M 0 is compact, we can find v in V M0 whose image in V is u. Suppose that v is fixed also by for some small. If we choose b in A such that b 0b 1, then v = δ 1/2 (b)π(b)v is fixed by M 0 0. Because K 0 = 0 M 0 0, the average of π(n)v over 0 is the same as the average of π(k)v over K 0. This average lies in V K0 and has image π (b)u in V. But then τ a v has

7 Chapter III. Induced representations and the Jacquet module 7 image δ 1/2 (a)π (ab)u in V and also liesin V K0. Since τ ab acts invertiblyon V K0, we can find v in V K0 such that τ ab v = τ a τ b v = τ a v,and whoseimage in V is u. As a consequence: III.5.6. Theorem.If (π, V )isanadmissiblerepresentationof Gthen (π, V )isanadmissiblerepresentation of M. Thuswhenever K 0 isasubgrouppossessing aniwahorifactorizationwithrespectto,wehaveacanonical subspace of V K0 projecting isomorphically onto V M0. For a given M 0 there may be many different K 0 suitable;howdoes thespace V K0 vary with K 0? III.5.7. Lemma.Let K 1 K 0 betwocompactopensubgroupsof Gpossessing aniwahorifactorization with respect to. If v 1 in V K1 and v 0 in V K0 havethesame imagein V, then π(µ K0/K 0 )v 1 = v The canonical pairing Continue to let K 0 be a compact open subgroup of G possessing an Iwahori factorization 0 M 0 0 with respect totheparabolic subgroup, (π, V )an admissible representationof G. III.6.1. Lemma.For v in V K0, ṽ in Ṽ K0 Ṽ (), ṽ, v = 0. roof. Fix a in A for themomentand choose v 0 in V K0 with τ av 0 = v. Then v, ṽ = τ a v 0, ṽ = v 0, τ a 1ṽ. AccordingtoLemmaIII.5.3, τ a isnilpotenton Ṽ K0 Ṽ (), soifwechoose asuitablytheright handside is 0. III.6.2. Theorem. If (π, V is an admissible representation of G, then there exists a unique pairing between V and Ṽ with the property that whenever v has image u in V and ṽ has image ũ in Ṽ, then for all a in A near enoughto 0 ṽ, π(a)v = δ 1/2 (a) ũ, π (a)u. Similarly withtherolesof V and Ṽ reversed. roof. Let u in V and ũin Ṽ be given. Suppose that u and ũ are bothfixed by elementsof M 0. Let v be a vectorin V K0 withimage u,and similarly for ṽ and ũ. Definethepairing bytheformula ũ, u can = ṽ, v. ItfollowsfromLemmaIII.6.1andLemmaIII.5.7thatthisdefinitiondependsonlyon uand ũ,andnotonthe choicesof v and ṽ. That ṽ, π(a)v = δ 1/2 (a) ũ, π (a)u can also follows from Lemma III.6.1 and Lemma III.5.7. That this property characterizes the pairing follows from theinvertibilityof τ a on V K0. This pairing is called the canonical pairing. III.6.3. Corollary. If R is a field, the canonical pairing defines an isomorphism of (π, Ṽ ) with the contragredientof (π, V ),as arepresentationof M. roof. The canonical pairing is invariant under M because for m in M the pairing π (m)u, π (m)ũ can also satisfies the conditions characterizing the canonical pairing. For non degeneracy, suppose u in V such that u, ũ can = 0 for all ũ Ṽ. Let v be a canonical lift of u. Let ṽ bearbitrary in Ṽ K0. Suppose v = τ a v 0 for v 0 also in V K0. Then v, ṽ = τ(a)v 0, ṽ = v 0, τ(a) 1 ṽ. But if we choose a suitably then τ a 1v lies in Ṽ, so this last is a canonical a pairing, hence 0. Therefore v, ṽ = 0for all ṽ in Ṽ K0,and v = 0.

8 Chapter III. Induced representations and the Jacquet module 8 7. References 1. H. Jacquet, Représentations des groupes linéaires p adiques, in Theory of group representations and Fourier analysis,(proceedings of a conference at Monecatini), C. I. M. E. Edizioni Cremonese, Rome, and R.. Langlands, Automorphic forms on GL(2), Lecture otes in Mathematics 114, Springer, 1970.