lectures on representations of p-adic groups 1

LECTURES ON REPRESENTATIONS OF p-ADIC GROUPS

GORDAN SAVIN

1. Introduction

This text contains an elementary introduction, with exercises, to representations of p-adicreductive groups. The text is intended to be more accessible then the standard references,such as the paper of Bernstein and Zelevinsky [1], and (unpublished) notes of Casselman.The material presented here has been known for the past quarter century, and has undergonemany modifications in that period. Thus, it would be hard (if not impossible) to acknowledgeproperly contributions of many people who have worked on the subject. However, it is safe tosay that Bernstein, Casselman, Harish-Chandra, Howe and Jacquet (in alphabetical order)are some of the most prominent contributors to the subject.

The text is organized as follows. Sections 2-4 contain definitions and preliminary results onp-adic fields, structure of GLn(F ) over a p-adic field F , and smooth representations. Then, inorder to keep the exposition as simple as possible, we restrict ourselves to GL2(F ). However,the topics and their proofs are chosen so that they easily generalize to GLn(F ) and otherreductive groups.

In Section 5 and 6 we introduce induced and cuspidal representations, and prove thatirreducible smooth representations are admissible. Sections 7 and 8 are devoted to describingthe composition factors of induced representations. In section 9 we discuss unitarizablerepresentations, and construct the complementary series for GL2(F ). Sections 10 and 11 aredevoted to two examples. In section 10 we show that the Steinberg representation is squareintegrable, and in section 11 we construct one cuspidal representation.

Finally, in Sections 12 and 13, we go back to GLn(F ) and describe the composition factorsof (regular) induced representations. This result, due to Rodier [4], is based on the combi-natorics of the root system, and the reduction to the special case of GL2(F ) (obtained inSection 7). As such, it gives a good introduction to further, more advanced topics.

Acknowledgments: This material was presented during a week long tutorial in the Institutefor Mathematical Sciences at the National University of Singapore. The author would like tothank the people associated with the institute, especially Eng Chye Tan and Cheng Bo Zhu,for their hospitality and support. Thanks are also due to Jian Shu Li, Allen Moy, and MarkoTadic for useful comments and suggestions at various stages.

2. The field QpThis is a crash course on the field of p-adic numbers. Let Z denote, as usual, the ring of

integers. Define the p-adic absolute value on Z as follows. Let x 6= 0 be an integer. Write1

2 GORDAN SAVIN

x = pam where m is relatively prime to p. Then

jxjp = 1

pa.

We also put j0jp = 0. This absolute value satisfies the usual properties: the multiplicativity,jxjp ¢ jyjp = jxyjp, and the triangle inequality. In fact we have an even stronger property:

jx+ yjp · maxfjxjp, jyjpg.It follows that d(x, y) = jx¡ yjp defines a metric on Z. The ring of p-adic integers Zp is thecompletion of Z with respect to this metric.To understand this ring we can proceed as follows. Let B(x, r) be the (closed) ball of

radius r ¸ 0, centered at x. Note that Z can be writen as a union of p disjoint balls of radius1/p, centered at 0, 1, . . . , p¡ 1:

Z = B(0,1

p) [B(1, 1

p) [ . . . [B(p¡ 1, 1

p).

Note that B(i, 1p) is simply i+pZ, a coset of the maximal ideal (p). Similarly, Z can be writtenas a union of pn disjoint balls of radius 1/pn, centered at all reminders modulo pn. Thereare two consequences of this: First, Z is totally bounded, so its completion Zp is compact.Second, the completion of Z is the union of completion of individual balls of radius 1/pn,which implies that

Z/pnZ = Zp/pnZp.

Exercise. Let x be an integer relatively prime to p. Show that the multiplicative inverse of xexists in Zp. Hint: for every n, there exist integers yn and zn such that xyn + pnzn = 1.The ring Zp has (p) as unique maximal ideal. The field of fractions Qp is obtained by

adjoining 1/p. It follows that

Qp = [−∞<i<∞piZp.Next, we shall describe smooth additive characters of Qp. A character ψ is smooth if it is

trivial on piZp for some some i. The minimal such i is called the conductor of ψ. We shallnow construct a canonical character of conductor 0. Note that Qp/Zp is naturally isomorphicto the p-torsion part of Q/Z. The group Q/Z has a natural additive character given by

ψ(x) = e2πix.

The restriction of ψ to Qp/Zp is the additive character of conductor 0. Any other characteris of the form ψa(x) = ψ(ax) for an element a in Qp.We shall now say a couple of words about characters of Q×p . Note that Q×p = hpi £ Z×p ,

where hpi »= Z is the free group generated by p. The group Z×p has a filtration by subgroups1 + piZp with i ¸ 1. A character of Q×p is smooth if it is trivial on some 1+ piZp. Moreover,it is called unramified, if it is trivial on Z×p . Every unramified character χ is completelydetermined by its value χ(p). In particular, the group of unramified characters is isomorphicto C×.Any other p-adic field F is simply a finite extension of Qp (see [3]). The integral closure of

Zp will be denoted by O. It has a unique maximal ideal ($), which is the radical of the ideal

LECTURES ON REPRESENTATIONS OF p-ADIC GROUPS 3

(p). Let q be the order of the residue field O/$O. We shall normalize the absolute value onF so that

j$j = 1

q.

3. Structure of GLn(F )

Let F be a p-adic field, and G = GLn(F ). Let K = GLn(O). Then K is a (maximal)open, compact subgroup of G. It has a filtration by the principal congruence subgroups Kiof G is defined by

Ki = fg 2 G j g ´ 1 mod $ig.These groups are normal in K, and provide a fundamental system of neighborhoods of 1 inG, defining a topology of G. We have the following important decompositions in G:

3.1. Cartan decomposition. Let Λ »= Zn be the group of diagonal matrices $m1

. . .

$mn

where m1 . . . mn are integers. Let Λ

+ be the subset of Λ consisting of diagonal matrices suchthat m1 ¸ . . . ¸ mn. Then

G = KΛ+K.

3.2. Bruhat-Tits decomposition. Let B denote the group of upper triangular matrices inG, and N the subgroup of B consisting of matrices with 1 on the diagonal. Let W be thegroup of permutation matrices, which is isomorphic to the permutation group Sn. The groupW is called the Weyl group of G. Then

G = NWB.

3.3. Iwasawa decomposition.G = BK.

Each of the three decompositions given here can be proved by a variant of row-column re-duction. This is deceptively simple and somewhat misleading. An entirely different approachis needed for general reductive groups. Thus, we give only a hint here, in form of an exercise.

Exercise. Prove the Cartan decomposition. Hint: Let g be in GLm(F ). Let gij be the entryof g with the largest absolute value. Multiplying g by elementary matrices with coefficientsin O (so they are in K) turn all entries in the i-th row and j-th column (except gij) into 0.Proceed by induction on n.

A critical difference between real and p-adic groups is that N is a union of open compactsubgroups. For example, if G = GL2(F ), then N is a union of

Ni = fµ1 x0 1

¶j where x 2 $iOg.

As we shall see later, one consequence of this fact is that SLn(F ) does not have non-trivialfinite dimensional representations.

4 GORDAN SAVIN

3.4. Haar Measure. Recall that the congruence subgroups Ki give a system of open neigh-borhoods of 1 which defines the topology of G. Let C∞0 (G) be the set of locally constant,compactly supported functions on G. In particular, for every function f there exist a suffi-ciently small congruence subgroup Ki, such that f can be written as a linear combination

f =Xj

cj ¢ Char(gjKi)

where Char(gjKi) is the characteristic function of the right coset gjKi. Now put

µ(f) =

ZGf(g) dg =

1

[K : Ki]

Xj

cj.

This defines a (unique) left G-invariant measure on G such that vol(K) = 1.

Exercise. Show that the measure is right G-invariant. Hint: Let g be in G, and define a leftG-invariant measure µg by

µg(f) = µ(Rg(f))

where Rg(f) is the right translate of f by g. By uniqueness of the left G-invariant measure,µg = c(g)µ for some positive constant c(g). Show that c(g) is a character, and then use that[G,G] ¢ Z (Z is the center of G) is of finite index in G.

4. Smooth representations

Let V be a vector space over a complex (or real field). A smooth representation of G onV is a homomorphism

π : G! GL(V )

such that every vector v in V is fixed by a (sufficiently small) congruence subgroup. Weshall refer to V as a (smooth) G-module. For every congruence subgroup Ki, let V

Ki be thesubspace of all vectors fixed by Ki. A representation is admissible if V

Ki is finite dimensionalfor every i. It turns out that any irreducible smooth module is in fact admissible. We shallgive a proof of this fact for GL2(F ).

Proposition 4.1. (Schur’s Lemma.) Let V be an irreducible smooth representation of G,and A : V ! V be a non-trivial homomorphism. Then A = µ ¢ I for some µ in mathbbC.(Here I is the identity operator on V ).

Proof. First of all note that V has countable dimension. Indeed, V is spanned by π(G)v forany non-zero vector v in V . Since v is fixed by some Ki, the statement follows, as G/Ki iscountable (from the following exercise, for example).

Exercise. Show that for every λ in Λ+, the index of Kλ = λKλ−1\K in K is sandwiched by

δ(λ)−1 · [K : Kλ] · [K : K1] ¢ δ(λ)−1

where δ(λ) =Qi<j q

mi−mj .

Assume that A6= µ ¢ I for every µ in C. Then irreducibility of V implies that A¡ µ ¢ I isbijective. In particular, Rµ = 1/(A¡µ ¢ I) is well defined for every µ in C. We now need thefollowing:


Exercise. Let v be a non-zero vector in V . Assuming that A ¡ µ ¢ I are bijective, show thatRµv are linearly independent vectors in V.

But we know that V has countable dimension. This is a contradiction, and the propositionis proved. ¤

Proposition 4.2. Let V be an smooth representation. Then:

² If the representation V is finitely generated, then there exists an irreducible quotient.² Otherwise, there exits two submodules V 00 ½ V 0 such that V 0/V 00 is irreducible.

Proof. The second statement clearly follows from the first. To prove the first statement, letE be a finite dimensional subspace generating V . If U is a proper submodule, then E \U isa proper subspace of E. Since E is finite dimensional, there exists a maximal subspace Emaxof E which is an intersection with a proper invariant subspace. Let U be the family of allinvariant subspaces U such that E \ U = Emax. By the Zorn’s lemma, the family U has amaximal element Umax. Clearly, V/Umax is irreducible. The proposition is proved. ¤

Proposition 4.3. Let V be a finitely generated smooth G-module. Then V is also finitelygenerated over B.

Proof. Fix a finite set of generators. Since V is smooth, there exists a congruence subgroupKi fixing all generators. Since Ki is of finite index in K, the proposition easily follows fromthe Iwasawa decomposition. ¤

An interesting feature of smooth representations it that the trivial representation is theunique finite-dimensional representations of SLn(F ).

Proposition 4.4. Let (π, V ) be a finite dimensional irreducible smooth representation ofGLn(F ). Then V »= C, and π(g) = χ(det) for some character χ of F×.

Proof. Since V is smooth and finite dimensional, there exists a sufficiently small open compactsubgroup Nc ½ N acting trivially on V .

Exercise. Let n be any element in N . Then there exits an element t in T such that tnt−1 = ncis in Nc.

Since π(n)v = π(t−1)π(nc)π(t)v = v, it follows that N acts trivially on V . The sameargument shows that N acts trivially on on V , and the proposition follows from the followingexercise.

Exercise. SLn(F ) is generated by N and N .Hint: Use row-column reduction. ¤

For any smooth G representation of V , let V ∗ be the space of all linear functionals on V .Let hv, v∗i be the usual pairing between V and V ∗. Then G acts on V ∗ by

hv,π∗(g)v∗i = hπ(g−1)v, v∗i.Let V be the set of smooth vectors in V ∗, and let π denote the restriction of π∗ to V . Therepresentation V is called the contragredient dual of V . A priori, it may not be clear that Vis non-trivial. However, we have the following simple proposition:

6 GORDAN SAVIN

Proposition 4.5. Let V be a smooth representation of G. Then V Ki is isomorphic to theset of linear functionals on V Ki. In particular, V is admissible, if and only if V is so.

Proof. Let v be any vector in V . Then v is fixed by Kj for some sufficiently small j whichwe can assume to be less then i. Define

P (v) =1

[Ki : Kj ]

Xk∈Ki/Kj

π(k)v.

Clearly, P (v) does not depend on the choice of j, and P is a Ki-invariant projection from

V onto V Ki . Now any functional on V Ki composed with P defines an element in V Ki .Conversely, any functional in V Ki clearly factors through P . The proposition is proved. ¤

5. Induced representations, case of GL2(F )

Let χ1 and χ2 be two characters of the multiplicative group F×. Let χ be the character

of T , the group of diagonal matrices in G, defined by

χ

µa1 00 a2

¶= χ1(a1)χ2(a2).

We shall now describe the so-called parabolic induction, which produces a smooth represen-tation of G from a character of T . Since B/N = T , the character χ can be pulled back to B,and we can consider the smooth induced representation

IndGB(χ) = ff : G! C j f(ntg) = ja1/a2j1/2χ(t)f(g)g∞,where here, of course, we have considered only the smooth functions on G. The meaning (and

necessity) of the term δ1/2 = ja1/a2j1/2 will become more clear as we develop more theory.Basically, this factor compensates for the choice of the Borel subgroup B containing T , as wecould have defined the induction using B, the group of lower triangular matrices.

The induced representations are admissible. Indeed, the Iwasawa decomposition impliesthat

dim IndGB(χ)Ki · [K : Ki].

Our goal is to understand the composition factors of the induced representations. To do so,we need to introduce an important functor which is dual, in a certain sense, to the parabolicinduction.

5.1. Jacquet Functor. We first define a functor form the category of smooth N -modulesto the category of vector spaces. Let V be a smooth N -module, and put

VN = V/V (N)

where V (N) is the linear span of π(n)v ¡ v for all n 2 N and v 2 V . The functor V Ã VNis exact, which will be of fundamental importance:

Theorem 5.1. Let 0! V 0 ! V ! V 00 ! 0 be an exact sequence of N -modules. Then

0! V 0N ! VN ! V 00N ! 0

is exact.


Proof. The right exactness is clear. We only need to check that V 0N ! VN is injective.This is equivalent to V 0 \ V (N) = V 0(N), which will follow from the following importantcharacterization of V (N). ¤

Lemma 5.2. Let V be a smooth N -module. A vector v is in V (N) if and only ifZNj

π(n)v dn = 0

for a sufficiently large open compact subgroup Nj. In particular, V (N) = [j∈ZV (Nj), whereV (Nj) is the subspace of all v in V such that the above integral vanishes.

Proof. If v is in V (N), then v is a finite linear combination of terms π(n)u ¡ u. Let Nj besufficiently large so that it contains all n appearing in that linear combination. Then theintegral over Nj will be 0. ¤

Exercise. Prove the other direction of the Lemma. Hint: the vector v is invariant under someNi.

Now assume that V is a smooth G-module. Then VN is naturally a smooth T -module, andV Ã VN is called the Jacquet functor. The Jacquet functor and the parabolic induction arerelated by the Frobenius reciprocity:

HomG(V, IndGB(χ)) = HomT (VN , δ

1/2χ)

Exercise. The Frobenius reciprocity is a tautology. Prove it!

Proposition 5.3. Let V be an irreducible smooth G-module such that VN 6= 0. Then V is asubmodule of Ind(χ) for some character χ of T .

Proof. By Proposition 4.3 we know that V is finitely generated overB. Thus VN is also finitelygenerated over B, and it has an irreducible quotient by Proposition 4.2. The propositionfollows from the Frobenius reciprocity. ¤

6. Cuspidal representations, case of GL2(F )

We have just seen that a smooth irreducible representation V such that VN 6= 0, embeddsinto a principal series representation. Our concern in this section will be the representationssuch that VN = 0; so called cuspidal representations. Since the center Z of G is not compact,so it will be more convenient to work with a smaller group

G = fg 2 GL2(F ) j det(g) 2 O×g.Note that ZG is of index 2 in G, so representations of G can be easily related to those ofG.

Exercise. Let (π, V ) be an irreducible smooth G-module. Then the restriction of V to Gis either irreducible, or a direct sum of two non-isomorphic G-modules. Moreover, therestriction reduces if and only if π »= π − γ where γ is the unique non-trivial character of Gtrivial on ZG.Definition 6.1. A smooth representation (π, V ) of G is called cuspidal if VN = 0.

8 GORDAN SAVIN

We shall now show that the matrix coefficients of cuspidal representations are compactlysupported. A matrix coefficient of V is a (smooth) function

ϕv,v(g) = hπ(g)v, viwhere v is in V , and v in V , the contragredient of V .

Proposition 6.2. Let (π, V ) be a cuspidal (not necessarily irreducible) representation of G.Then the matrix coefficients of π are compactly supported.

Proof. Let U and U be the linear span of π(K)v and π(K)v, respectively. Both spacesare finite dimensional, since the representations are smooth and K is compact. Since U iscuspidal, by Lemma 5.2, we can find a sufficiently large i such that such thatZ

N−iπ(n)u dn = 0

for all u 2 U , and U is contained in V Ki . Let

λ =

µ$a 00 $−a

¶.

Then, for any u in U , and any u in U ,ZNi

hπ(λ)u, π(n)ui dn =ZNk

hπ(λ)π(n)u, ui dn,

where k = i ¡ 2a. Thus, if a ¸ i then the right hand side is 0. Since u is in V Ki , the lefthand side is equal to hπ(λ)u, ui, up to a non-zero factor (the measure of Ni). Therefore,ϕv,v(kλk

0) = 0 for any two elements k and k0 in K. The proposition follows form the Cartandecomposition. ¤

Proposition 6.3. Let V be an irreducible cuspidal representation of G, and P : U ! Va non-trivial homomorphism of smooth G modules. Then the above sequence splits, that isthere exists a homomorphism s : V ! U such that P ± s(v) = v for all v in V .Proof. Let V be the contragredient dual of V . Pick v any vector in V . We first define askew-linear (with respect to the complex conjugation) map i : V ! V by

i(v) =

ZG

ϕv,v(g)π(g)v dg.

Since ϕv,π(h)v = ϕv,v(h−1g), the map i is a skew-linear homomorphism. Since V is irreducible,

i must be onto. Finally, since

hi(v), vi =ZGjϕv,v(g)j2 dg > 0,

we see that i is injective, as well. Now pick any u in U such that P (u) = v. Define, analogously

to i, a skew-linear map from V to U by

j(v) =

ZG

ϕv,v(g)π(g)u dg.


(Here π denotes the action of G on U). Now put s = j ± i−1. Clearly, s is a linear homo-morphism from V to U , and by Schur’s Lemma P ± s has to be a multiple of the identityoperator. Since P (s(v)) = v, for our fixed vector v, the proposition follows. ¤

Exercise. The proof of the above proposition shows that smooth irreducible cuspidal represen-tations are admissible. Hint: the map i defines a skew-linear isomorphism of V and V . Nowuse the fact that V Ki is equal to the linear dual of V Ki (Proposition 4.5).

Corollary 6.4. Every irreducible smooth representation V of GL2(F ) is admissible.

Proof. We have just seen that cuspidal representations are admissible. Otherwise, the repre-sentation is a submodule of a principal series representation by Proposition 5.3, so it is againadmissible. ¤

7. Decomposing principal series

In this section we shall describe the composition series of IndGB(χ). Some critical calcula-tions will be performed in Section 8.

Proposition 7.1. Let V be an irreducible subquotient of IndGB(χ). Then VN 6= 0.Proof. If VN = 0, then V is cuspidal, so it has to be a submodule as well. But then byFrobenius reciprocity VN 6= 0. Contradiction. ¤

Since VN 6= 0 for any subquotient, the following proposition shows that the length ofa composition series of IndGB(χ) is at most two. The proof is based on the Bruhat-Titsdecomposition B [BwN of G, where

w =

µ0 11 0

¶.

Proposition 7.2. Let χ be a character of T , and χw be the character obtained by conjugatingχ by w, which simply means permuting χ1 and χ2. Then

0! δ1/2χw ! IndGB(χ)N ! δ1/2χ! 0.

Proof. Let IndGB(χ)w be the B-submodule of functions in IndGB(χ) supported on the big cell

BwN . We shall use the filtration

IndGB(χ)w µ IndGB(χ)and the exactnes of V Ã VN to calculate IndGB(χ).Define a functional α on IndGB(χ) by

α(f) = f(1).

Sinceα(π(nt)f) = δ1/2χ(t) ¢ α(f)

we have shown that δ1/2χ is a quotient of IndGB(χ)N . Clearly, IndGB(χ)w is the kernel of α.

Define a functional αw on IndGB(χ)w by

αw(f) =

ZNf(wn) dn.

10 GORDAN SAVIN

Exercise. Show that the kernel of αw is precisely IndGB(χ)w(N). Hint: note that Ind

GB(χ)w

is isomorphic to C∞0 (N), as an N -module, and apply Lemma 5.2.

An easy calculation shows that

αw(π(nt)f) = δ1/2χw(t) ¢ αw(f).The proposition is proved. ¤

We now note a very important consequence of the above result. If χ6= χw (such χ is calledregular) then the above exact sequence splits. This implies that αw extends to Ind

GB(χ), and

thus by the Frobenius reciprocity defines a non-zero intertwining operator

Aw(χ) : IndGB(χ)! IndGB(χ

w)

Proposition 7.3. The following gives a complete answer to decomposing principal seriesrepresentations of GL2(F ).

² The induced representation IndGB(χ) is irreducible if χ1/χ26= j ¢ j±1.² Otherwise, the composition series is of length two, with one of the subfactors a one-dimensional representation.

Proof. The singular case χ = χw will be dealt with later. Assume that χ 6= χw. Then theFrobenius reciprocity implies that

Hom(IndGB(χ), IndGB(χ))

is one-dimensional, so every intertwining operator is a multiple of the identity I . In particular,it follows that

Aw(χw) ± Aw(χ) = c(χ) ¢ I

for some scalar c(χ), (called c-function).

Lemma 7.4. If χ6= χw, then IndGB(χ) reduces if and only if c(χ) = 0.

To verify this lemma, recall that IndGB(χ) can have at most two subquotients. Let V be a

submodule. Then, V is a proper submodule if and only if VN = δ1/2χ. In that case IndGB(χ)

has a proper quotient U and UN = δ1/2χw. The Frobenius reciprocity implies that Aw(χ)annihilates V . Likewise, Aw(χ

w) annihilates U , which is a submodule of IndGB(χw). It follows

that Aw(χw) ± Aw(χ) = 0. The lemma is proved.

The first part of the proposition follows as soon as we calculate c(χ). This will be done inthe next section. Thus, assume that χ1/χ2 = j ¢ j±1. In fact, since subquotients of IndGB(χ)are the same as subquotients of IndGB(χ

w), it suffices to consider χ1/χ2 = j ¢ j−1. In this caseγ(det) appears as a submodule, where

γ = χ1j ¢ j1/2 = χ2j ¢ j−1/2.If γ is trivial, then the other (infinite-dimensional) subquotient is called the Steinberg repre-sentation. Obviously, for other γ, the infinite dimensional subquotient is is just the twist ofthe Steinberg representation by γ(det). ¤


8. c-function

The purpose of this section is to obtain an explicit formula for the c-function µ(χ). We alsoobtain a result which plays a crucial role in proving that IndGB(χ) is irreducible for singular

χ. In order to keep notation simple, we shall restrict ourselves to the family I(s) = IndGB(χ)where

χ

µa1 00 a2

¶= ja1/a2js/2.

If s6= 0 letAw(s) : I(s)! I(¡s)

be the intertwining operator which, via the Frobenius reciprocity, corresponds to the func-tional αw. Recall the definition of αw. On the subspace of functions f such that f(1) = 0, itis defined by

αw(f) =

ZNf(wn) dn,

and, if s6= 0, then αw is the unique extension to I(s) . In order to calculate Aw(¡s) ±Aw(s),we need to obtain an explict formula for αw on the whole I(s). In other words, our task isto regularize this integral for any f in I(s).

To do so, we shall now present some key technical ingredients. Let dx be the Haar measureon F , and for every s in C, let λs be the functional on C∞0 (F×) defined by

λs(f) =

ZF×f(x)jxjs dxjxj .

Let π denote the (right) regular representation of F× on C∞0 (F×). The functional λs isequivariant with respect to π in the sense that

λs(π(a)f) = jaj−sλs(f).Proposition 8.1. The functional λs extends to an equivariant functional on C

∞0 (F ) if and

only if s6= 0. The extension is unique if s6= 0.Proof. Let Λs be an extension, if any. Let f be in C

∞0 (F ). Then the function f ¡ π(ω)f is

in C∞0 (F×). In particular,

Λs(f ¡ π($)f) =

ZF×[f(x)¡ f($x)]jxjs dxjxj .

On the other hand, since Λs is equivariant, we have Λs(f ¡ π($)f) = (1 ¡ j$j−s)Λs(f).Putting things together,

(1¡ qs)Λs(f) =ZF×[f(x)¡ f($x)]jxjs dxjxj .

This formula shows that the extension exists and is unique if 1 ¡ qs is not zero. Otherwise,which happens precisely when s = 0, the left hand side is always zero, while the right handside can be easily arranged to be non-zero (take, for example, the characteristic function ofO). In particular, the functional λ0 does not extend. The proposition is proved. ¤

12 GORDAN SAVIN

We shall now apply this ideas to our situation. For any f in I(s) define

f 0(g) = j$j−(s+1)/2 ¢ πµ$ 00 1

¶f(g).

Next note that f¡f 0 vanishes at 1, so it is supported on the big double coset NwB. Arguingexactly as in the proof of the previous proposition we see that an extension of αw to I(s)exists if and only if s6= 0, in which case it is given by

αw(f) = (1¡ q−s)−1ZN(f ¡ f 0)(wn) dn.

Corollary 8.2. HomG(I(0), I(0)) = C.This corollary follows from the Frobenius reciprocity, and the fact that αw does not extend

to I(0). Combined with the fact that I(0) is completely decomposable, which will be shownin the next section, it will imply that I(0) is irreducible.

Proposition 8.3. Let L(s) = 1/(1¡ q−s). Then

c(s) =L(¡s)L(s)

L(1¡ s)L(1 + s) .

In particular, c(s) = 0 if and only if s = §1.Proof. In order to calculate Aw(¡s) ± Aw(s), we need to make the formula for αw(f) moreexplicit. Clearly, f is Ki-invariant for some i, and we shall show that f(1) determines f(wn)for all n in N nN−i+1. Indeed, note the following relation in SL2(F ).µ

1 ¡x−10 1

¶µ1 0x 1

¶µ1 ¡x−10 1

¶=

µ0 ¡x−1x 0

¶Let j · ¡i, and assume that x is in $jOn$jO. Since f is Ki-invariant, the relation impliesthat

f(w

µ1 x0 1

¶) = jxjs+1f(1).

In particular, it follows that f ¡ f 0 is supported on N−i.Next, note that the Iwasawa decomposition implies that I(s) has a unique element fs such

that fs(gk) = fs(g) for all k in K, and fs(1) = 1. The above calculation can be made veryprecise for fs:

Exercise. Show that

αw(fs) = L(s)/L(1 + s).

Since Aw(fs) is K-invariant, it must be a multiple of f−s. The above exercise shows thatthe multiple is L(s)/L(1 + s). The proposition follows.

¤


9. Unitary representations

A smooth representation of G is called unitarizable if it admits a positive definite G-invariant hermitian form. In this section we shall first show that IndGB(χ) is unitarizable ifχ is a unitary character (χ ¢ χ = 1). Combined with Corollary 8.2 this will (finally) showthat I(0) is irreducible. We finish this section by showing that the representations I(s) with0 < s < 1 (the so called complementary series) are also unitarizable.

The Iwasawa decomposition implies that each representation I(s) has a unique (up to anon-zero constant) K-invariant functional given by

f 7!ZKf(k) dk.

Since the trivial representation of G is a quotient of I(1), the K-invariant functional is alsoG-invariant if s = 1. Now we can easily construct a positive definite G-invariant hermitianform on IndGB(χ). Let f and g be any two functions in Ind

GB(χ). If χ is unitary, then f ¢ g is

in I(1), so the hermitian form

hf, gi =ZKf(k)g(k) dk

is clearly positive definite and G-invariant.

Proposition 9.1. The representation I(0) is irreducible.

Proof. Since I(0) is unitarizable, and of finite length, it will decompose as a direct sum ofirreducible constituents. In particular, Hom(I(0), I(0)) is isomorphic to a direct sum ofmatrix algebras Mn(C), one for each irreducible constituent of multiplicity n. On the otherhand, we know that Hom(I(0), I(0)) = C. The proposition follows. ¤

Our last topic for GL2(F ) will be a construction of the unitary structure on I(s) for¡1 < s < 1. We first need to renormalize the intertwining operators Aw(s), so that they aredefined at the singular point s = 0 as well. Recall that Aw(s) corresponds, via the Frobeniusreciprocity, to the functional

αw(f) = L(s)

ZN(f ¡ f 0)(wn) dn.

This is well defined only if s6= 0 since L(s) has a pole at s = 0. Let Aw(s) be the operatorwhich corresponds to the functional αwL(1 + s)/L(s). Then Aw(s) is well defined for every¡1 < s < 1, and

Aw(s)fs = f−s.

We can now easily define a G-invariant hermitian form on I(s) as follows. Let f and g betwo functions in I(s). Then Aw(s)f is in I(¡s), and as above,

hf, gi =ZK(Aw(s)f)(k)g(k) dk

defines a G-invariant sesquilinear form on I(s). Apriori, it is not clear that this form ishermitian. However, Schur’s Lemma implies that an irreducible smooth G-module can haveonly one sesquilinear G-invariant form, up to a non-zero scalar. Thus the form hg, fi must

14 GORDAN SAVIN

be proportional to the form hf, gi. Since hfs, fsi = 1, it follows that they are equal, so ourform is hermitian, after all.Next, note that Aw(0) is the identity operator on I(0). Indeed, we know that it is a

multiple of the identity operator, but the formula for the spherical vector shows that it is infact equal to the identity operator. It follows that the form is positive definite on I(0). Weshall use this observation to prove the following proposition.

Proposition 9.2. If 0 < s < 1, then the G-invariant hermitian form h¢, ¢i on I(s) is positivedefinite.

Proof. Since I(s) is irreducible for ¡1 < s < 1 the G-invariant form must be non-degenerate(otherwise the kernel of the form would be an invariant subspace). Moreover, as the fol-lowing exercise shows, the restriction of this form to every finite-dimensional subspace ofKi-invariants is also non-degenerate.

Exercise. Let V be a smooth G-module with a non-degenerate G-invariant hermitian form.Then the restriction of the form to V Ki is also non-degenerate. Hint: Use the Ki-invariantprojection operator P from V onto V Ki .

Every element in I(s) is completely determined by it restriction to K. In this way, we canidentify, in a K-invariant fashion, I(s)Ki with the (finite-dimensional) space Ei of functionson B\KnK/Ki. By the above exercise, the invariant form on I(s) induces a non-degenerateform h¢, ¢is on Ei, which is positive definite for s = 0. The proposition follows from thefollowing exercise.

Exercise. Let E be a finite dimensional vector space, and h¢, ¢is a family of non degeneratehermitian symmetric forms, depending continuously on the parameter s in a connected openset in R. If h¢, ¢is0 is positive definite for some s0 in the open set, then it is positive definitefor all s. ¤

10. Steinberg representation

By Proposition 7.3 we now know that the principal series I(s) decomposes if ond only ifs = §1, in which case it has a composition series of length 2. One subquotient is the trivialrepresentation, and the other is the Steinberg representation V . Recall that

VN = δ.

We shall show that the Steinberg representation is square integrable, which means that thematrix coefficients of V restricted to G0 are square integrable. The results of this section are,in large part, due to Casselman.

Proposition 10.1. Let V be the Steinberg representation of G. Then the matrix coefficientsof V are square integrable, when restricted to G0.

Proof. To prove this proposition, we need a result on the asymptotics of matrix coefficientsof V . Recall that

λa =

µ$a 00 $−a

¶.


Lemma 10.2. Let V be the Steinberg representation of G. Then, for every v in V , and v inV , there exists a positive integer i such that

hπ(kλa+ik0)v, vi = δ(λa)hπ(kλik0)v, vi.for every non-negative integer a, and any two elements k and k0 in K.

Proof. The proof of this proposition is analogous to the proof of Proposition 6.2. Let U andU be the finite dimensional subspaces generated by v and v over K, respectively. Let u bein U , and for every positive integer a, define

ua = π(λa)u¡ δ(λa)u.

Since VN = δ it follows that u1 is in V (N). Recall that V (N) is a union of V (Nj), whereV (Nj) is the kernel of the operator

Pj(u) =

ZNj

π(n)u dn.

Take i large enough so that every u1 is contained in V (N−i) and U is contained in V Ki . Thisis our choice of i.

Exercise: Show that π(λ1)V (N−i) ½ V (N−i+2). In particular, π(λ1)V (N−i) is again con-tained in V (N−i).

Since

ua = π(λ1)ua−1 ¡ δ(λ1)ua−1,

an induction argument based on the previous exercise implies that ua is in V (N−i) for allpositive integers a. Thus, for every u in U ,

P−i(ua) = 0

and if tildeu is in U ,ZNi

hπ(λi)ua, π(n)ui dn =ZN−ihπ(λi)π(n)u, ui dn = 0.

It follows that hπ(λi)ua, ui = 0, as desired. The lemma is proved.¤

We can now finish the proof of the proposition. Recall the Cartan decomposition

G0 = [∞a=0KλaK.The measure of KλaK is equal to the index of K\KλaK in K, which is essentially δ(λa)

−1 =q2a. This and Lemma 10.2 imply thatZ

G0jϕv,vj2 dg · C

∞Xa=0

q−2a,

for some positive constant C. This series clearly converges, and the proposition is proved. ¤

16 GORDAN SAVIN

11. A cuspidal representation of G0

In this section we present an example of a cuspidal representation. To make the exampleas simple as possible, the trick (which the author learned from Allen Moy) is to restrict tothe case p = 2. Thus, in this section only, our groups are defined over the diadic field Q2.Let F2 = f0, 1g be the residual field of Q2. Consider the projective line

P(2) = f(1, 0), (1, 1), (0, 1)gover F2. The action of K/K1 = GL2(2) on P(2) gives an isomorphism of GL2(2) with S3,the symmetric group on 3 letters. Let ² be the usual sign character on S3. Then one easilychecks that

²

µ1 10 1

¶= ¡1.

Pull ² back to K, and define V = indG0

K (²). The symbol ind stands for the induction withcompact support. In particular, V consists of compactly supported functions on G0 suchthat f(kg) = ²(k)f(g) for all k in K and g in G0.

Proposition 11.1. The representation V is an irreducible cuspidal representation of G0.

Proof. We shall first prove that VN = 0. First of all, note thatZN∩K

²(n) dn = 0.

Let eK in V be the function supported on K, such that eK(k) = ²(k) for all k in K. TheIwasawa decomposition implies that any element in V is a linear combination of π(b)eK whereb is in B. Let Nb = b

−1(N \K)b. The vanishing of the integral above is implies thatZNb

π(n)π(b)eK dn = 0.

It follows that V is cuspidal by Lemma 5.2. It remains to prove that it is irreducible.Let H²(G

0//K) be the Hecke algebra of compactly supported functions on G0 such that

f(kgk0) = ²(k)f(g)²(k0).

for all k and k0 in K. As usual, the multiplication in H²(G0//K) is defined by convolutionof functions

(f1 ¤ f2)(g) =ZG0f1(h)f2(h

−1g) dg.

Note that H²(G0//K) is contained in V , and equal to V K,², the subspace consisting of all

functions f in V such that π(k)f = ²(k)f for all k in K. The element eK defined above isthe identity element in H²(G

0//K).

Lemma 11.2. There is a natural isomorphism HomG0(V, V ) »= H²(G0//K).Proof. Let A be in HomG0(V, V ). Since V is generated by eK , the operator A is completelydetermined by its value A(eK). Since A(eK) lies in V

K,², which is the same as H²(G0//K),

the map

A7! A(eK)


furnishes a canonical injection from HomG0(V, V ) to H²(G0//K). Conversely, an element T

in H²(G0//K) defines an element A in HomG0(V, V ) by

A(f) = T ¤ f.Since A(eK) = T ¤ eK = T , we see that A 7! A(eK) is surjective as well. The lemma isproved. ¤

Lemma 11.3. The algebra H²(G0//K) is one-dimensional.

Proof. Recall the Cartan decomposition

G0 = [∞a=0KλaKwhere

λa =

µ2a 00 2−a

¶.

Let f be in H²(G0//K). We shall show that f is a multiple of eK . Clearly, f is determined

by its values at all λa. Let

n =

µ1 2a

0 1

¶and n0 =

µ1 10 1

¶.

If a ¸ 1, then n is in K1 and ²(n) = 1. On the other hand, recall that ²(n0) = ¡1. It followsthat

f(λa) = f(nλa) = f(λan0) = ¡f(λa).

Thus, f(λa) = 0 if a ¸ 1, and f must be a multiple of eK . The lemma is proved. ¤

We can now finish the proof of proposition. Since V is generated by eK , Proposition 4.2implies that there exists an irreducible quotient V 0 of V . Let P be the projection from Vonto V 0. Since V 0 is cuspidal, by Proposition 6.3 there is a splitting s : V 0 ! V such thatP ± s = IdV 0 . But

s ± P 2 HomG0(V, V ) = C ¢ IdVso s ± P must be equal to IdV . It follows that V = V 0, and the proposition is proved.

¤

12. The root system for GLn(F )

The purpose of this section is to introduce the root system for GLn. It is a combinatorialobject which provides us with a language to study representations of GLn(F ). (For moreinformation on root systems see [2]).

Let R be the field of real numbers, and consider the space Rn with the standard basise1, . . . , en. With respect to this basis, we shall identify every element x in Rn with an n-tuple(x1, . . . , xn). The group Sn acts on Rn by permuting the entries of (x1, . . . , xn). This actionpreserves the inner product

hx, yi =nXi=1

xiyi.

18 GORDAN SAVIN

Let

Ω = fx 2 Rn jXi

xi = 0g.

Note that the group Sn preserves Ω. In fact, it is an irreducible representation. We shall nowdescribe the action of Sn by means of euclidean reflections. Let

Ψ = fei ¡ ej j i6= jg,be a finite set in Ω. Its elements are called roots. Any root α defines a reflection w by

w(x) = x¡ hα, xi ¢ x.Exercise. Let α = ei ¡ ej. Show that the corresponding w is simply the permutation of i andj.

It follows from the exercise that the reflections w define the representation of Sn on Ω.Clearly, the fixed points of the reflection corresponding to the root α = ei¡ej is the hyperplanefxi = xjg, defined by the equation xi = xj . Consider the open subset obtained by removingall hyperplanes fxi = xjg

Ω = Ω n ([fxi = xjg).The connected components of Ω are called chambers. We shall point out a particular chambergiven by

C+ = fx 2 Ω j x1 > . . . > xng.Next, note that the entries of any x in Ω are pair-wise different, so there exists a uniquepermutation which puts them in a decreasing order. This means that Ω is a disjoint unionof w(C+) as w runs thru Sn. In particular, every chamber C is equal to

C = fx 2 Ω j xi1 > . . . > xingfor some permutation (i1, . . . , in).

The choice of C+ gives us a set of generators of Sn as follows. Note that the closure of C+

is obtained by adding parts (called walls) of hyperplanes fxi = xi+1g. The correspondingroots

fe1 ¡ e2, . . . , en−1 ¡ engare called simple. Note that the simple roots form a basis of Ω. Let wi be the (simple)reflection defined by ei ¡ ei+1. We claim that the reflections wi generate Sn. Of course,since wi is nothing but the permutation (i, i + 1) this is a well known fact. We shall give aproof based on the fact that Sn acts simply transitively on the set of chambers. Let w be anelement in Sn. Pick x

+ in C+ and x in w(C+) such that the segment between them avoidsthe singularities of ΩnΩ. Since those singularities form a codimension 2 subvariety, a genericchoice of x+ and x will do. The number of hyperplanes intersected by the segment is clearlyindependent of the particular choices, is the length `(w) of w. The chambers containing thesegment form a gallery

C+, C1, . . . C`(w) = w(C+).

Any two consecutive chambers in this gallery share a wall. In particular, there exists asimple reflection wi1 such that C1 = wi1(C

+). Next, note that the walls of C1 correspond toreflections wi1 (the shared wall) and wi1wjw

−1i1for all j 6= i1. In particular, there exists i2


such that C2 = wi1wi2w−1i1(C1), which implies C2 = wi2wi1(C

+). Continuing in this fashion,we can obtain wi1 , . . . , wi`(w) such that

wi`(w) ¢ . . . ¢ wi1(C+) = w(C+).

Exercise. Make a picture of the root system for GL3(F ). The roots form a regular hexagon.Let w be the element in W such that w(C+) = C−. Express w as a product of simplereflections by picking a gallery between C+ and C−.

13. Induced representations for GLn(F )

Let e1, e2, . . . , en be the standard basis of Fn. Let Vr be the span of e1, . . . , er. Consider

a partial flagVi1 ½ Vi2 ½ . . . ½ Vim

and let P be its stabilizer in G. Then P is called a parabolic subgroup of G, and it admits theLevi decomposition P =MN , where M is isomorphic to the product of GL(Vi/Vi−1), and Uis the unipotent radical of P . Note that the minimal parabolic subgroup B corresponds tothe full flag.

If (τ, E) is a smooth representation of M , then we can induce it to G by

IndGP (τ) = ff : G! τ j f(umg) = δ1/2P τ(m)f(g)g∞.

The character δP of M is defined so that for every locally constant, compactly supportedfunction h on U we have Z

Uh(mum−1) du = δP (m)

ZUh(u) du.

As a particular case of interest, consider the minimal parabolic subgroup B = TU . Notethat any smooth character χ of T is given by

χ

a1. . .

an

= χ1(a1) ¢ . . . ¢ χn(an)

for some smooth characters χ1 . . .χn of F×. Assume that χ is regular, which means that

χ6= χw for any w in W . In the case of GLn(F ) this simply means that χi 6= χj if i6= j. Asin the case of GL2(F ), the Bruhat-Tits decomposition can be used to show that

IndGB(χ)N =Mw∈W

δ1/2χw.

It turns out that VN 6= 0 for any irreducible subquotient, just as in the case of GL2(F ).In particular, each irreducible subquotient is completely determined by VN . Let S be the setof all fi, jg such that χi/χj = j ¢ j±1. Let

Ωχ = Ω n ([Sfxi = xjg)where fxi = xjg is the hyperplane in Ω defined, of course, by the given equation. A result ofRodier states that the irreducible subquotients corresponds to the connected components ofΩχ. We shall here give a special case of that result, describing VN for the unique irreducible

20 GORDAN SAVIN

submodule V . (Since any irreducible subquotient of IndGB(χ) is a submodule of IndGB(χ

w)for some w in W , the general case easily follows.)Let Ω+χ be the connected component containing the positive chamber C

+. Define Wχ tobe the set of all Weyl group elements w such that

w(C+) ½ Ω+χ .Proposition 13.1. Let χ be a regular character. Let V be the unique irreducible submoduleof IndGB(χ) (normalized induction). Then

VN =Mw∈Wχ

δ1/2χw.

Proof. The proof of this statement is combinatorial, and based on the special case of GL2(F ),proved in Proposition 7.3. Let Pi be the parabolic subgroup corresponding to the flag withVi ommited. Note that its Levi factor Mi has exactly one factor isomorphic to GL2, and letwi (= w) be the permutation matrix in that factor. Assume that χi/χi+1 6= j ¢ j±1, so wi isin Wχ. Then, by Proposition 7.3

IndPiB (χ)»= IndPiB (χwi)

Inducing both representation further up to G, this implies IndGB(χ)»= IndGB(χ

wi). In par-ticular, χwi is a summand of VN . For a general element w of length m in Wχ we have toconsider a gallery

C+, C1, . . . , Cm = w(C+)

which is contained in Ω+χ (since Ω+χ is convex), and repeat the above argument m-times. This

shows that

VN µMw∈Wχ

δ1/2χw.

The opposite inclusion will follow from the following lemma:

Lemma 13.2. Assume that χi/χj = j ¢ j±1 for some i < j. If χw is a summand of VN thenC+ and w(C+) are on the same side of the hyperplane fxi = xjg.To be specific, assume that χi/χj = j ¢ j. Let w0 be the permutation (i+1, j), and χ0 = χw

0

be the character of T obtained by permuting χj and χi+1. Then χ0i/χ

0i+1 = j ¢ j. Inducing in

stages (first from B to Pi), and using the second part of Proposition 7.3 (GL2(F )-reducibility)we get an inclusion

IndGPi(η) ½ IndGB(χ0),where η is a character of Mi, which on the GL2(F )-factor is given by composing the deter-minant with the character

χij ¢ j1/2 = χi+1j ¢ j−1/2.A more general form of the Bruhat-Tits decomposition implies that

IndGMi(η)N =

Mw∈Wi

δ1/2(χ0)w


where Wi is the set of all w such that w(i) < w(i + 1). This, however, is equivalent toww0(i) < ww0(j), which happens if and only if C+ and ww0(C+) are on the same side of thehyperplane fxi = xjg. Summarizing, we have

IndGMi(η)N =

Mw(C+)⊆xi>xj

δ1/2χw.

¤

Corollary 13.3. Let χ be a regular character, and let Ω1, . . .Ωm be the connected componentsof Ωχ. Then Ind

GB(χ) has m irreducible subquotients V1, . . . , Vm so that

(Vi)N =M

w(C+)⊆Ωiδ1/2χw.

Exercise. Show that IndGB(δ1/2) has 2n subquotients. Hint: Try first the special case of

GL3(F ).

References

[1] J. Bernstein and A. Zelevinsky, Representations of the group GLn(F ), where F is a non-archimedeanlocal field, Russian Math. Surveys 31 (1976), 1-68.

[2] J. Humphreys, Introduction to Lie Algebras and representation theory, Graduate Texts in Mathematics9, Springer-Verlag, 1978.

[3] N. Koblitz, p-adic Numbers, p-adic Analysis, and Zeta Functions, Second Edition, Graduate Texts inMathematics 58, Springer-Verlag, 1984.

[4] F. Rodier, Decomposition de la serie principale des groupes reductifs p-adiques, 408-424, Lecture Notesin Mathematics 880, Springer-Verlag, 1981.

[5] M. Tadic, Representations of classical p-adic groups, 129-204, Pitman Res. Notes Math. Ser. 311, Long-man, Harlow 1994.

Department of Mathematics, University of Utah, Salt Lake City, UT 84112E-mail address : [email protected]

lectures on representations of p-adic groups 1

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