Functional Equations-Langlands, R.P.on the Functional Equations Satisfied by Eisenstein Series

8/12/2019 Functional Equations-Langlands, R.P.on the Functional Equations Satisfied by Eisenstein Series

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On the Functional Equations Satisfied by Eisenstein Series

Robert P. Langlands

Appeared as vol. 544 of SpringerVerlag Lecture Notesin Math., SpringerVerlag, BerlinHeidelberg,

New York, 1976, pp. 1337.


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Eisenstein series ii

because it is the adelic form of the theorems which is most frequently applied. I caution the reader that

he will not appreciate the relation between 7 and this appendix until he has an intimate understanding

of7. The appendix should be read first however.

It is also difficult to come to terms with 7 without a feeling for examples. Some were given in my

lecture on Eisenstein series in Algebraic Groups and Discontinous Subgroups. Others exhibiting the

more complicated phenomena that can occur are given in the third appendix, whose first few pages

should be glanced at before 7 is tackled.

The last appendix has nothing to do with 7. It is included at the suggestion of Serge Lang, and is

an exposition of the Selberg method in the context in which it was originally discovered.

In the introduction I thank those who encouraged me during my study of Eisenstein series. Here

I would like to thank those, Godement and HarishChandra, who encouraged me after the notes

were written. HarishChandras encouragement was generous in the extreme and came at what was

otherwise a difficult time. Its importance to me cannot be exaggerated.

It has been my good fortune to have had these notes typed by Margaret (Peggy) Murray, whose

skills as a mathematical typist are known to all visitors to the IAS. I thank her for another superb job.


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Eisenstein series iii

TABLE OF CONTENTS

1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2. Statement of assumptions. Some properties of discrete groups satisfying

the assumptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3. Definition of a cusp form (after Gelfand). Basic properties of cusp forms . . . . . . . . 29

4. Definition of Eisenstein series. Investigation of the constant term in the

Fourier expansion of an Eisenstein series. A variant of a formula of Selberg

(Lemma 4.6(ii)) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

5. Some lemmas used in Sections 6 and 7 . . . . . . . . . . . . . . . . . . . . . . . 78

6. Proof of the functional equations for the Eisenstein series associated to cusp forms . . . 93

7. Proof of the functional equations for all Eisenstein series. Statement of theorem . . . . . 127

References for 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

Appendix I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184

Appendix II . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

Appendix III . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222

Appendix IV . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249


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1. Introduction.

One problem in the theory of automorphic forms that has come to the fore recently is that of

explicitly describing the decomposition, into irreducible representations, of the regular representation

of certain topological groups on Hilbert spaces of the form L2(\G)when is a discrete subgroup of

G. Usually is such that the volume of\Gis finite. Except for some abelian groups, this problem

is far from solved. However, Selberg has discovered that the gross features of the decompositionare determined by simple properties of the group and this discovery has led to the development,

mostly by Selberg himself, of the theory of Eisenstein series. Of course he has preferred to state the

problems in terms of eigenfunction expansions for partial differential equations or integral operators.

At present the theory is developed only for the connected reductive Lie groups which, without real loss

of generality, may be assumed to have compact centres. Even for these groups some difficulties remain.

However, some of the problems mentioned in [19] are resolved in this paper, which is an exposition of

that part of the theory which asserts that all Eisenstein series are meromorphic functions which satisfy

functional equations and that the decomposition ofL2(\G) is determined by the representations

occurring discretely in L2(\G)and certain related Hilbert spaces. For precise statements the reader

may refer to Section 7.

At present it is expected that the main assertions of this paper are true if the volume of\Gis

finite. It is of course assumed thatGis a connected reductive Lie group. Unfortunately not enough is

known about the geometry of such discrete groups to allow one to work with this assumption alone.

However, the property which is described in Section 2 and which I thereafter assume possesses is

possessed by all discrete groups known to me which have a fundamental domain with finite volume.

Indeed it is abstracted from the results of Borel [2] on arithmetically defined groups. Section 2 is

devoted to a discussion of the consequences of this property. In Section 3 the notion of a cusp form

is introduced and some preliminary estimates are derived. In Section 4 we begin the discussion of

Eisenstein series, while Section 5 contains some important technical results. In Section 6 the functional

equations for Eisenstein series associated to cusp forms are proved. For series in one variable the

argument is essentially the same as one sketched to me by Professor Selberg nearly two years ago, but

for the series in several variables new arguments of a different nature are necessary. In Section 7 the

functional equations for the remaining Eisenstein series are derived in the course of decompositionL2(\G)into irreducible representations.

I have been helped and encouraged by many people while investigating the Eisenstein series

but for now I would like to thank, as I hope I may without presumption, only Professors Bochner


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Introduction 2

and Gunning for their kind and generous encouragement, three years ago, of the first results of this

investigation.


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Chapter 2 3

2. The assumptions.

LetG be a Lie group with Lie algebra g. It will be supposed thatG has only a finite number

of connected components and that g is the direct sum of an abelian subalgebra and a semi-simple Lie

algebra gs. It will also be supposed that the centre ofGs, the connected subgroup ofG with Lie algebra

gs, is finite. Supposea is a maximal abelian subalgebra ofgs whose image inad g is diagonalizable.

Choose an order on the space of real linear functions on a and let Qbe the set of positive linear functions

on a such that there is a non-zero element Xin g such that[H, X] =(H)Xfor all Hin a. Q is called

the set of positive roots ofa. Supposea is another such subalgebra andQ is the set of positive roots

ofa with respect to some order. It is known that there is some ginGs such thatAdg(a) =a and such

that if Q then the linear function defined by(H) =(Adg(H))belongs toQ. Moreover any

two elements ofGs with this property belong to the same right coset of the centralizer ofa in Gs. G

itself possesses the first of these two properties and it will be assumed that it also possesses the second.

Then the centralizer ofameets each component ofG.

For the purposes of this paper it is best to define a parabolic subgroup PofG to be the normalizer

in G of a subalgebra p ofg such that the complexification pc = p RC ofp contains a Cartan subalgebra

jc ofgc together with the root vectors belonging to the roots ofjc which are positive with respect to

some order onjc. It is readily verified that the Lie algebra ofP ispso thatPis its own normalizer. Let

nbe a maximal normal subalgebra ofps =ps gs which consists entirely of elements whose adjoints

are nilpotent and letm be a maximal subalgebra ofpwhose image inad gis fully reducible. It follows

from [16] thatp = m +n and that m contains a Cartan subalgebra ofg. Leta be a subalgebra of the

centre ofm gs whose image inad gis diagonalizable. Ifm is the orthogonal complement ofa, with

respect to the Killing form ong, inm thena m= {0}. There is a setQ of real linear functions on a

such thatn =Qnawhere

na = {Xn [H, X] =(H)Xfor allHin a}.

aorA, the connected subgroup ofPwith the Lie algebraa, will be called a split component ofPif the

trace of the restriction ofad Y to nis zero for any Y in m and any in Q. There is a Cartan subalgebra

j ofg and an order on the real linear functions onjcsuch thata j m and such thatQ consists of the

restrictions of the positive roots to aexcept perhaps for zero. Let Qbe the set of positive roots whose

restriction toaequals; then

1/ dim nQ


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Chapter 2 4

is zero onj mand equalsona. Thus ifQ c= 0andc 0for allthen

Q

Q

(dim n)1c

= 0

which implies thatc= 0for all. In particular zero does not belong toQ so thatm is the centralizer

and normalizer ofaing.

Since m contains a Cartan subalgebra it is its own normalizer. Let us show that ifM is the

normalizer ofm inPthen the connected component ofM is of finite index inM. M is the inverse

image in G of the intersection of an algebraic group with AdG. Since AdG contains the connected

component, in the topological sense, of the group of automorphisms ofg which leave each element of

the centre fixed the assertion follows from Theorem 4 of [23]. Since the Lie algebra ofM is m it follows

from Lemma 3.1 of [16] thatM is the inverse image inGof a maximal fully reducible subgroup of the

image ofP inAdG. LetNbe the connected subgroup ofGwith the Lie algebra n. Since the image of

Nin AdGis simply connected it follows readily from [16] thatM andNare closed, thatP=M N,

and thatM N={1}.

We must also verify thatM is the centralizer ofa inG. M certainly contains the centralizer of

ainG. Let bbe a maximal abelian subalgebra ofgs which containsasuch that the image ofbinad gis

diagonalizable. Certainly m contains b. Let b= b1 +b2where b1is the intersection ofb with the centre

ofm and b2is the intersection ofb with the semi-simple part ofm. b2is a maximal abelian subalgebra

of the semi-simple part ofm whose image in ad m is diagonalizable. It may be supposed (cf. [11],

p. 749) that the positive roots ofbare the roots whose root vectors either lie in nc, the complexification

ofn, or lie in mc and belong to positive roots ofb2. Ifm lies in M thenAdm(b1) = b1. Moreover

replacing if necessarym by mm0 wherem0 lies in the connected component ofM and hence in the

centralizer ofa we may suppose thatAdm(b2) =b2and thatAdm takes positive roots ofb2to positive

roots ofb2. ThusAdm(b) =b andAdm leaves invariant the set of positive roots ofb; consequently, by

assumption, m lies in the centralizer ofb and hence on a. It should also be remarked that the centralizer

ofAmeets each component ofPandGandPmeets each component ofG.

IfMis the group of allm in M such that the restriction ofAdmto n has determinant1for

alla thenm is closed; sinceQ contains a basis for the space of linear functions on a the intersection

A M is {1}. Let1, , pbe such a basis. To see thatAM =M introduce the groupM1of allm

inM such that the restriction ofAdm toni has determinant 1for1 i p. CertainlyAM1=M.

So it has merely to be verified thatM, which is contained inM1, is equal toM1. Since the Lie algebra

of bothMandM1 is m the groupMcontains the connected component ofM1. SinceA M1 = {1}


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Chapter 2 5

the index[M1: M]equals[AM1 :AM]which is finite. It follows readily thatM=M1. It is clear that

MandS=M Nare uniquely determined byPandA. The pair(P, S)will be called a split parabolic

subgroup withA as split component. Its rank is the dimension ofA. Observe thatA is not uniquely

determined by the pair(P, S).

The next few lemmas serve to establish some simple properties of split parabolic subgroups

which will be used repeatedly throughout the paper. If(P, S)and(P1, S1)are any two split parabolic

subgroups then(P, S)is said to contain(P1, S1)ifPcontainsP1and ScontainsS1.

Lemma 2.1. Suppose(P, S) contains(P1, S1). LetAbe a split component of(P, S)andA1 a split

component of (P1, S1). There is an elementp in the connected component ofP such thatpAp1

is contained inA1.

SinceSis a normal subgroup ofP,pAp1 will be a split component of(P, S). According to

Theorem 4.1 of [16] there is a p in the connected component ofPsuch thata1+ m Adp(a + m). Thus

it suffices to show that ifa1+ m1 is contained ina+m thena is contained in a1. Ifa1+ m1 a+m

thena and a1 commute so thata is contained ina1+ m1; moreoverm containsm1 becausem s1 =

(a + m) s s1 (a1+ m2) s1 = m1. Consequently a is orthogonal tom1with respect to the Killing

form and hence is contained ina1.

Lemma 2.2. SupposePis a parabolic subgroup andais a split component ofP. Let{1,, , p,}

be a minimal subset ofQ such that any inQ can be written as a linear combinationp

i=1 mii

with non-negative integersmi. Then the set{1,, , p,} is linearly independent.

This lemma will be proved in the same manner as Lemma 1 of [13]. Let, be the bilinear

form on the space of linear functions ona dual to the restriction of the Killing form to a. It is enough

to show that ifi and j are two distinct indices that i, j, neither equals zero nor belongs to Q and

that ifandbelong toQ neither nor belongs toQ or is zero then , 0. If this is so

andpi=1 aii,= 0letF ={i

ai 0} andF ={i ai


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Chapter 2 6

which implies that = 0. As a consequence of a previous remark ai = 0, 1 i p. Certainly i, j,

is not zero ofi =j ; suppose thati, j,= belongs toQ. Then

i, j,=

pk=1

mkk,

or

(mi 1)i,+ (mj+ 1)j,+ k=i,j

mkk, = 0

so thatmi 1< 0. Hencemi = 0and

i,= (mj+ 1)j,+k=i,j

mkk,

which is a contradiction. Supposeandbelong toQand neither nor belongs toQor is

zero. Choose the Cartan subalgebraj as above and let(, )be the bilinear form on the space of linear

functions on the complexification ofj gs dual to the Killing form. Ifis the restriction of toathen

, = 1/ dim nQ

(, )

In particular of belongs toQ then

, = 1/ dim nQ

(, )

Because of the assumptions on and , is neither a root nor zero; thus (cf. [15], Ch. IV) each

term of the sum and hence the sum itself is non-positive. It is clear that the set {1,, , p,} is unique

and is a basis for the set of linear functions on a; it will be called the set of simple roots ofa. It is also

clear that ifP1containsPandA1is a split component ofP1contained in A then the set of simple roots

a1is contained in the set of linear functions on a1obtained by restricting the simple roots ofatoa1.

Lemma 2.3 Suppose

P=P1 P2 Pk

is a sequence of parabolic subgroups with split components

A1 A2 Ak

and

dim Ai+1 dim Ai = 1, 1 i < k


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Chapter 2 7

If{1,, , p,}is the set of simple roots ofa and{j,, , p,}restricted to aj is the set of simple

roots foraj, 1 j k, then

aj ={Hai,(H) = 0, i < j}

and if

Qj ={ Q (H)= 0 for some Hin aj}then

nj =Q

j

n, 1 j k.

Conversely ifF is a subset of{1, , p}; if

a= {Hai,(H) = 0 for all i F};

if

Q ={ Q(H)= 0 for some H a};

if

n=Q

n;

and if m is the orthogonal complement of a in the centralizer of a ing then p= a+m+n is

the Lie algebra of a parabolic subgroup P ofG which containsP and has a as a split component.

In the discussions above various objects such as A, Q, n have been associated to a parabolic

subgroupP; the corresponding objects associated to another parabolic group, sayP1, will be denoted

by the same symbols, for exampleA1,Q1, n1, with the appropriate indices attached. It is enough to

prove the direct part of the lemma fork = 2. SinceP2 properly contains P1 and since, as is readily

seen, Pjis the normalizer ofnj ,j = 1, 2the algebra n2must be properly contained in n1. Consequently

there is an Q whose restriction toa2 is zero and =pi=1 mii, with non-negative integersmi.

Let i,be the restriction ofi, to a2and let 1,=nj=2 njj,; then

0 =

p

j=2(mj+ m1nj)j;somj = 0,j 2and = m11,. Sincedim a2 dim a1 = 1the direct part of the lemma is proved.

Proceeding to the converse we see that ifPis taken to be the normalizer ofp in G then Pis parabolic

by definition. Pcontains the connected component ofPand the centralizer ofA in G; so it contains


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Chapter 2 8

all ofP. Moreover the image ofa + minad g is fully reducible and n is a normal subalgebra ofp; so

to prove thatais a split component ofPit has to be shown that ifbelongs to Q and

Q={Q(H) =(H) for allHin a},

then the trace of the restriction ofad X to

Q

n is zero for all Xinm. It is enough to show this

when Xbelongs to the centre of

m. But then Xcommutes witha and so lies ina + m; say X=Y+ Z.Ifi belongs to Fthe trace of the restriction ofad X toni, isi,(Y)dim ni, ; on the other hand it is

zero becauseni, belongs tom. Thusi,(Y) = 0for alli F, so that Y belongs to

aand hence is

zero. Since the assertion is certainly true forZit is true forX.

There are some simple conventions which will be useful later. If jc and jc are two Cartan

subalgebras ofgc and an order is given on the set of real linear functions on jc and jc then there is

exactly one map fromjcto jcwhich takes positive roots to positive roots and is induced by an element

of the adjoint group ofgc. Thus one can introduce an abstract Lie algebra which is provided with a set

of positive roots and a uniquely defined isomorphism of this Lie algebra with each Cartan subalgebra

such that positive roots correspond to positive roots. Call thistheCartan subalgebra ofgc. Suppose

(P, S)is a split parabolic subgroup with A and A as split components. Letj be a Cartan subalgebra

containing a and let j be a Cartan subalgebra containing a. Choose orders on jc and jc so that the

root vectors belonging to positive roots lie inpc. There is a p1 in Psuch that Adp1(a) = a; since the

centralizer ofA meets each component ofP there is ap in the connected component ofPsuch that

Adp(H) = Adp1(H)for allHin a. LetAdp(j) =j. There is an elementmin the adjoint group ofmc

such thatAdp Adm(a) =a,Adp Adm(j) =j, andAdp Adm takes positive roots ofj to positive roots

ofj. The maps ofa j jcand a j jcdetermine maps ofa and a

intotheCartan subalgebra of

gcand ifHbelongs toathenHand Adp1(H)have the same image. The image ofawill be called the

split component of(P, S). Usually the context will indicate whether it is a split component or the split

component which is being referred to. IfFis a subset of the set of simple roots of the split component it

determines a subset of the set of simple roots of any split component which, acccording to the previous

lemma, determines another split parabolic subgroup. The latter depends only on Fand will be called

simply the split parabolic subgroup determined by F; such a subgroup will be said to belong to(P, S).

If(P, S)is a split parabolic subgroup with the split component a let ,1, , ,p be the linearfunctions ona such that,i, j,= ij ,1 i,j p. Of course1,, , p, are the simple roots ofa.

If c1 < c2 let

a+(c1, c2) ={Ha c1< j,(H)< c2, 1 i p}


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Chapter 2 9

and let

+a(c1, c2) ={Ha c1< ,i(H)< c2, 1 i p}

It will be convenient to set a+(0, ) = a+ and +a(0, ) = +a. IfA is the simply-connected abstract

Lie group with the Lie algebra a then A will also be called the split component of(P, S). The map

Hexp His bijective; if is a linear function on a set(exp H) = exp((H)). If0 c1 c2

we let

A+(c1, c2) ={a A c1< i,(a)< c2, 1 i p}

and

+A(c1, c2) ={a A c1< ,i(a)< c2, 1 i p}

We shall make frequent use of the two following geometrical lemmas.

Lemma 2.4. For each s < there is a t < such that a+(s, ) is contained in +a(t, ). In

particulara+ is contained in+a.

For each s there is an element Hin a such that a+(s, )is contained in H+ a+; thus it is enough

to show that a+ is contained in +a. Suppose we could show thati, j 0, 1 i, j p. Then

,i =p

j=1 aijj,with a

ij 0 and it follows immediately that a

+ is contained in+a. Since i,, j, 0

ifi=j this lemma is a consequence of the next.

Lemma 2.5. SupposeVis a Euclidean space of dimensionnand1,, , n, is a basis forV such

that(i,, j,) 0 if i=j . If,i, 1 i n, are such that(,i, j,) =ij then either there are two

non-empty disjoint subsetsF1 andF2 of{1, , n}such thatF1 F2= {1, , n}and(i,, j,) = 0

if i F1, jF2 or(,i, ,j)> 0 for all i andj.

The lemma is easily proved ifn 2so suppose thatn >2 and that the lemma is true forn 1.

Suppose that, for some iandj ,(,i, ,j) 0. Choosek different fromiandj and project {, =k}

on the orthogonal complement ofk, to obtain {, = k}. Certainly for = k the vector, is

orthogonal to,kand (,, m,) =m. Moreover

(,, m,) = (,, m,) (,, k,)(m,, k,)/(k,, k,) (,, m,)

with equality only if,or m,is orthogonal to k,. By the induction assumption there are two disjoint

subsets ofF1and F2of{

1 n, =k} such that(,, m,) = 0if F1and m F2. For sucha pair(,, m,) = (,, m,); so either(,, k,) = 0for all F

1 or (m,, k,) = 0for allm F

2.

This proves the assertion.


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Chapter 2 10

Suppose thata is just a split component ofP andFis a subset ofQ. Letc = {H a(H) =

0 for all F}; ifFis a subset of the set of simple roots c is called a distinguished subspace ofa. Let

gcbe the orthogonal complement ofcin the centralizer ofcingand letGcbe any subgroup ofGwith

the Lie algebragcwhich satisfies the same conditions as G. Then p gcis the Lie algebra of a parabolic

subgroupP ofGcand b, the orthogonal complement ofc in a, is a split component ofP. We regard

the dual space ofb as the set of linear functions ona which vanish onc. Let1,, , q, be the simpleroots ofb; {1,, , q,} is a subset ofQ. There are two quadratic forms on the dual space ofb, namely

the one dual to the restriction of the Killing form on g to b and the one dual to the restriction of the

Killing form ongcto b. Thus there are two possible definitions of,1, , ,q and hence two possible

definitions of+b. In the proof of Theorem 7.7 it will be necessary to know that both definitions give the

same+b. A little thought convinces us that this is a consequence of the next lemma.

The split parabolic subgroup(P, S)will be called reducible ifgcan be written as the direct sum

of two ideals g1and g2in such a way thatp = p1+ p2withpi= p giand s= s1+s2withsi = s gi.

Thenn = n1+ n2withni =n gi. Ifais a split component of(P, S)andm is the centralizer ofaing

thenm =m1+ m2 withm

i = m

gi. Sincem = m s, it is also the direct sum ofm1and m2and a,

being the orthogonal complement ofminm, is the direct sum ofa1anda2. If(P, S)is not reducible it

will be called irreducible.

Lemma 2.6. Suppose that the split parabolic subgroup (P, S) is irreducible and suppose that is a

representation ofg on the finite-dimensional vector spaceV such that if is a linear function on

a and

V = {v V (H)v= (H)v for all H in a},then the trace of the restriction of(X) to V is zero for allX inm. Then there is a constant c

such thattrace{(H1) (H2)}= cH1, H2 for allH1 andH2 ina.

Ifg contains a non-trivial centre theng is abelian anda = {0}; so there is nothing to prove. We

suppose then that g is semi-simple, so that the Killing form X, Yis non-degenerate. Consider the

bilinear form trace (X) (Y) = (X, Y)ong. It is readily verified that

([X, Y], Z) + (Y, [X, Z]) = 0

LetTbe the linear transformation on g such that (X, Y) = T X , Y ; then T X , Y = X , T Y and

T([X, Y]) = [X , T Y ]. IfHbelongs toaandXbelongs tomthe assumption of the lemma implies that


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Chapter 2 11

(H, X) = 0. Moreover choosing a basis forVwith respect to which the transformations(H),Ha,

are in diagonal form we see that (X, Y) = 0ifXbelongs topandYbelongs ton. Ifbelongs toQ let

n ={Xg [H, X] =(H)Xfor allHin g}

and letn =

Qn

. IfXbelongs toa+ m+n

andYbelongs ton then(X, Y) = 0. Thus ifH

belongs toathen T H , Y = 0forY inm + n + n, so thatT Hlies ina. For the same reasonTm m

and Tn n. Let 1, , rbe the eigenvalues ofTand let gi = {Xg (T i)nX= 0 for somen}.

gi,1 i r , is an ideal ofg andg = gi,p = (p gi), ands = (s gi). We conclude thatr = 1.

The restriction ofT toa is symmetric with respect to the restriction of the Killing form to a. Since the

latter is positive definite the restriction ofT toais a multiple of the identity. This certainly implies the

assertion of the lemma.

Now suppose is a discrete subgroup ofG. When describing the conditions to be imposed on

we should be aware of the following fact.

Lemma 2.7. If is a discrete subgroup of G, if (P1, S1) and (P2, S2) are two split parabolic

subgroups, if Pi Si, i= 1, 2, if the volume of Si\Si is finite fori= 1, 2, and ifP1 P2,

thenS1 S2.

Let S = S1S2; then P2 S. S is a normal subgroup ofS2 and S\S2 is isomorphic

to S1\S1S2 and is consequently abelian. It follows readily from the definition of a split parabolic

subgroup that the Haar measure onS2is left and right invariant. This is also true of the Haar measure

onS\S2

and hence it is true of the Haar measure onS. ThusS2\S2

ds2 =

S\S2

ds2

S2\S

ds= (S\S2) ( S\S)

Consequently (S\S2) is finite and S\S2is compact. Since the natural mapping from S\S2to S1\S1S2

is continuousS1\S1S2 is also compact. ButS1\S1S2 is a subgroup ofS1\P1 which is isomorphic to

A1and A1contains no non-trivial compact subgroups. We conclude that S2 is contained inS1.

If is a discrete subgroup ofG and (P, S) is a split parabolic subgroup then (P, S) will be

called cuspidal if every split parabolic subgroup(P, S)belonging to(P, S)is such that P S,

N\N is compact, and S\S has finite volume. A cuspidal subgroup such that S\S is

compact will be called percuspidal. Since the last lemma implies thatSis uniquely determined byP

andwe will speak ofPas cuspidal or percuspidal. IfPis a cuspidal subgroup the groupN\Swhich

is isomorphic toMsatisfies the same conditions asG. It will usually be identified withM. The image


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Chapter 2 12

of Sin Mis a discrete subgroup ofM. If(P,S)is a split parabolic group belonging to(P, S)

then(P, S) = (N\P S, N\S)is a split parabolic subgroup ofM. If(P, S)is a cuspidal group

ofGthen(P, S)is a cuspidal subgroup ofMwith respect to the group .

Once we have defined the notion of a Siegel domain we shall state the condition to be imposed

on. Fix once and for all a maximal compact subgroupKofG which contains a maximal compact

subgroup ofG0. If(P, S)is a split parabolic subgoup, ifc is a positive number, and if is a compact

subset ofSthen a Siegel domainS = S(c, )associated to(P, S)is

{g= sak s , a A+(c, ), k K}

Ais any split component of(P, S).

A set Eof percuspidal subgroups will be said to be complete if when (P1, S1) and (P2, S2)

belong to Ethere is ag in G such thatgP1g1 =P2 and gS1g

1 = S2and when(P, S)belongs toE

andbelongs tothe pair(P 1,S1)belongs to E.

Assumption. There is a complete set ofEof percuspidal subgroups such that ifP is any cuspidal

subgroup belonging to an element ofE there is a subset{P1, , Pr} ofE such thatP belongs to

Pi, 1 i r, and Siegel domainSi associated to (N\P1 S, N\Si) such that M =ri=1Si.

Moreover there is a finite subsetF ofE such thatE=

PFP

1.

Henceforth a cuspidal subgroup will mean a cuspidal subgroup belonging to an element ofE

and a percuspidal subgroup will mean an element ofE. It is apparent that the assumption has been

so formulated that ifPis a cuspidal subgroup then it is still satisfied if the pair , Gis replaced by

the pair ,M. Let us verify that this is so ifEis replaced by the set of subgroups N\PSwhere

P belongs to Eand P belongs toP. It is enough to verify that if(P,S)is a split parabolic group

belonging to (P1, S1) and to (P2, S2) and gP1g1 = P2, gS1g

1 = S2 then g lies in P. Let a be

a split component of(P,S); let a1 be a split component of(P1, S1) containing a; and let b be a

maximal abelian subalgebra ofgs containinga1whose image inad g is diagonalizable. Choosep inP

so that(pP2p1, pS2p

1)has a split componenta1which containsaand is contained inband so that

Adpg(a1) =a2and Adpg(b) = b. Replacingg bypg if necessary we may suppose thatp = 1. Choose

an order on the real linear functions onbso that any root whose restriction to a1lies inQ1 is positive.

If the restriction of the positive root to alies in Qthen the restriction toa1of the root defined by

(H) =

Adg(H)


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Chapter 2 13

lies in Q1 and is thus positive. The roots whose restrictions toa are zero are determined by their

restrictions to the intersection ofbwith the semi-simple part ofm. It is possible (cf. [11]) to choose an

order on the linear functions on this intersection so that the positive roots are the restrictions of those

roots ofb such that , with(H) = (Adg(h)), is positive. It is also possible to choose an order

so that the positive roots are the restrictions of the positive roots ofb. Consequently there is an m in

Msuch that Admg(b) = b and Admgtakes positive roots to positive roots. Thenmg belongs to thecentralizer ofband hence to P; so the assertion is proved.

Some consequences of the assumption which are necessary for the analysis of this paper will

now be deduced. If(P, S)is a split parabolic subgroup ofG the map(p, k) pk is an analytic map

from P Konto G. IfA is a split component of(P, S)then every element p ofPmay be written

uniquely as a product ofp = aswitha in A and s in S. Although in the decompositiong = pk the

factorp may not be uniquely determined by g the factora = a(g)of the product p = as is. In fact

the image ofa(g)in the split component of(P, S) is. Henceforth, for the sake of definiteness,a(g)

will denote this image. Every percuspidal subgroup has the same split component which will callh.

Suppose the rank ofhispand,1, , ,pare the linear functions onh dual to the simple roots.

Lemma 2.8. IfP is a percuspidal subgroup there is a constant such that,i

a()

, 1 i

p, for all in.

IfC is a compact subset ofG so is K C and {a(h)h KG} is compact; thus there are two

positive numbers1and 2with1< 2 such that it is contained in+A(1, 2). Ifg = ask withain

A,sinS, andk inKand hbelongs toCthena(gh) =a(kh) a(g), so that

1,i(a)< ,i

a(gh)

< 2,i(a), 1 i p,

In particular in proving the lemma we may replace by a subgroup of finite index which will still

satisfy the basic assumption, and hence may suppose thatG is connected. IfPis a cuspidal subgroup

let = S. IfP is percuspidal there is a compact set in Ssuch that every left cosetof in

contains an element =sa()kwiths in. For the purposes of the lemma only thosesuch that

= need be considered. It is not difficult to see (cf. [22], App. II) that there is a finite number of

elements1, , n in Nsuch that the connected component of the centralizer of{1, , n}in

NisNc, the centre ofN. A variant of Lemma 2 of [18] then shows that Nc\Nc is compact so

that, in particular, there is an element = 1in Nc. IfQc is the set of in Q such thatn nc ={0}


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Chapter 2 14

there is a constantsuch that, for allin ,

a()

for at least one in Qc. If not there would

be a sequence {} with {1

a()

} converging to zero for allinQc, so that

1= k1

a1() (s1 s) a()

k

would converge to1which is impossible.

Ifg = ri=1gi with gi simple, then p = ri=1p gi so that nc = ri=1nc gi. Ifj is a Cartan

subalgebra containinga, if an order is chosen onjas before, and ifwis a subspace ofnc giinvariant

under Adpforp Pthen the complexification ofw contains a vector belonging to the lowest weight

of the representationg Ad(g1)ofGongi. Since this vector is uniquenc gi ni for someiin

Q. The lowest weight is the negative of a dominant integral function; so i =p

j=1 bji,j withb

ji 0.

Thus there is a constant such that, for allin ,

minijp

,j

a()

In any case Lemma 2.8 is now proved for percuspidal subgrouops of rank one.

IfG0 is a maximal compact normal subgroup ofG then in the proof of the lemma G may be

replaced by G0\G and by G0\G0. In other words it may be supposed that G has no compact

normal subgroup. LetZbe the centre ofG. If we could show that Z\Zwas compact we could

replaceG by Z\Gand by Z\Zand assume thatG has no centre. We will show by induction on

the dimension ofG that if\Ghas finite volume then Z\Zis compact. This is certainly true ifG

is abelian and, in particular, if the dimension ofG is one. Suppose then thatG is not abelian and of

dimension larger than one. Because of our assumptions the groupG has a finite covering which is a

product of simple groups. We may as well replaceG by this group and by its inverse image in this

group. LetG =ni=1 GiwhereGi is simple,1 i r . We may as well assume thatGiis abelian for

somei. Choosein but not inZ. It follows from Corollary 4.4 in [1] that the centralizer ofin is

not of finite index in and hence that for some in 11 = does not lie in the centre ofG.

Let =ni=1 iand suppose thatidoes not lie in the centre ofGifor1 i m where1 m < n. It

follows as in [20] that the projection of on G =m

i=1 Giis discrete and that the volume of G\G

is finite ifG =ni=m+1 Gi. Since G

contains a subgroup ofZwhich is of finite index in Z and

otherwise satisfies the same conditions as Gthe proof may be completed by induction.

IfG has no centre and no compact normal subgroup is said to be reducible if there are two

non-trivial closed normal subgroups G1 and G2 such that G1 G2 = {1}, G = G1G2, and is

commensurable with the product of1 = G1 and2 = G2. is irreducible when it is not


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Chapter 2 15

reducible. Since if one of a pair of commensurable groups satisfies the basic assumption so does the

other, it may be supposed when is reducible that it is the product of1 and 2. If we show that 1

and2 satisfy the basic assumption we need only prove the lemma for irreducible groups. IfP is a

cuspidal subgroup forthen P = P1P2 with

Pi =P Giand

N = N1N2 with

Ni =N Gi.

Since N\Nis thus the product ofN1\N1and

N2\N2both factors are compact. Moreover

if

Si =

S Githeni

Pi

Siand

P

S1

S2. If

Ais a split component of(

P,

S)and

Aiis the projection ofAonGithen

A1A2is a split component of

Pand determines the split parabolic

subgroup(P, S1S2). Since the measure of

S1S2\

S1S2is clearly finite Lemma 2.7 implies that

S= S1S2. It follows readily that(

Pi,Si)is a cuspidal subgroup fori, i= 1, 2. Once this is known

it is easy to convince oneself thati,i= 1, 2, satisifes the basic assumption.

To make use of the condition that is irreducible another lemma is necessary.

Lemma 2.9. Suppose is irreducible. If P is a cuspidal subgroup of and ,1, , ,q are the

linear functions ona dual to the simple roots then,i

, ,j

> 0 for alli andj.

To prove this it is necessary to show that if the first alternative of Lemma 2.6 obtains then

is reducible. LetF1 andF2 be the two subsets of that lemma and letP1 andP2 be the two cuspidal

subgroups determined by them. We will show in a moment that n is the Lie algebra generated byqi=1 nai, and that if i F1, j F2 then [ni, , nj, ] = 0. Thusn = n1 n2 ifni is the algebra

generated byjFi

nj . Moreoverni is the maximal normal subalgebra ofpi gs containing only

elements whose adjoints are nilpotent. The centralizer ofa1 is a fully reducible subalgebra ofg and

lies in the normalizer ofn1. The kernel of the representation of this algebra on n1 is a fully reducible

subalgebrag2. The normalizerg ofg2is the sum of a fully reducible subalgebra g1and g2. g

contains

n1 in the centralizer ofa1 and thus containsp1. Sincep1 is parabolicg = g. Sinceg2 containsn2 and

Ni ={1} fori = 1, 2it follows from Theorem 1 of [20] thatis reducible.

To begin the proof of the first of the above assertions we show that ifis inQthen , j,> 0

for somej . If this were not so then, since=qj=1 mjj,,

0< , =

q

j=1mjj,, 0

Choose a Cartan subalgebraj ofg containing a and choose an order as before on the real linear functions

onjc. If is a positive root and the restriction of toa is neither zero nori,,1 i q, then for


20/261

Chapter 2 16

somej there is a inQj, such that is a positive root. Indeed if this were not so then, since

is not a root for any such, we would have(, ) 0. Consequently,

, i, 0, 1 i q,

which is impossible. Letn be the algebra generated by

qj=1 nj, ; it is enough to show that n

c, the

complexification ofn, equalsnc. We suppose that this is not so and derive a contradiction. Order the

elements ofQ lexicographically according to the basis {1,, , q,}and let be a minimal element

for which there is a root inQsuch thatX , a root vector belonging to, is not innc. Choose aj

and a in Qj, so that is a root. The root vectors ofXand Xboth belong to n

cand thus

X which is a complex multiple of[X , X ]does also. As for the second assertion we observe

that if

i F1, j F2,

and Qi, ,

Qj, ,

then is neither a root nor zero. Moreover

0 =

Qj,

(, ).

So each term is zero and for no inQj, is + a root. This shows that

[ni, , nj, ] = 0.

Suppose that is irreducible and that the assertion of Lemma 2.8 is not true for the percuspidal

subgroupP. There is a sequence {j} and ak,1 k p, such that

limj

,k

a(j)

=

It may be supposed that k = 1. Let P be the cuspidal subgroup belonging to Pdetermined by

{i, i= 1}. Letj =njajmjkj withnj in

N,aj inA,mj in

M, andkj inK. Replacingj byjj

with j = Sand choosing a subsequence if necessary we may assume that {nj}belongs

to a fixed compact set and that {mj}belongs to a given Siegel domain associated to the percuspidal

subgroupP = N\P SofM. P is a percuspidal subgroup ofG to whichPbelongs. IfA is


21/261

Chapter 2 17

the split component ofP andA =Ais the split component ofP thena(j) =aja(mj). There is

a constantc such that

,i

a(j)

c ,i(aj)

This follows immediately ifi = 1since

,1a(j)= ,1(a)and from Lemma 2.5 ifi >1. Since

,i, ,1> 0, 1 i q

there is a positive constant r such that

r,i(aj) ,1(aj).

However

,1(aj) =,1

a(j)

;

so

limj

,i

a(j)

= , 1 i p,

which we know to be impossible.

The next lemma is a simple variant of a well known fact but it is best to give a proof since it is

basic to this paper. Suppose Pis a parabolic with split componenta. Letjbe a Cartan subalgebra such

thata j p and choose an order on the real linear functions onjc as before. Let,1, , ,q be the

linear functions ona dual to the simple roots and let ,i be the linear function onj which agrees with

,i on aand is zero on the orthogonal complement ofa. There is a negative numberdisuch thatdi,i

is the lowest weight of a representationiofG0, the connected component ofG, acting on the complex

vector spaceVito the right.

Lemma 2.10. If is a linear function on a such that there is a non-zero vector v in Vi with

vi(a) =(a) for alla inA then

= di,i+

qj=1

njj,

withnj 0, 1 j q. Moreover ifvi is a non-zero vector belonging to the lowest weight then

{g G0 vii(g) =viwith C}


22/261

Chapter 2 18

is the intersection withG0 of the split parabolic subgroup Pi determined by{j,j =i}.

Let

ai = {Haj,(H), j =i}

and letQibe the set of positive roots ofawhich do not vanish onai. Set

ni = Q

i

n;

then

g= ni +ai+ mi+ ni.

Let

Vi ={v vi(X) = 0 forXni}.

IfWis a subspace ofVi invariant and irreducible under ai+ mithen the vector belonging to the lowest

weight of the representation ofai+mi on Wmust be a multiple ofvi and the lowest weight must be

dii. ConsequentlyVi is the set of multiples ofvi. LetV

(n)i be the linear space spanned by

{vii(X1) i(Xk)Xj g andk n}

and let (n)Vibe the linear space spanned by

{vii(X1) i(Xk)Xj ni and k n}.

We show by induction that V(n)i

(n)Vi. This is certainly true for n = 1 since k may be zero. If

X1, , Xn1belong toni andXnbelongs togthen

vii(X1) i(Xn)

is equal to

vii(X1) i(Xn2) i([Xn1, Xn]) + vii(X1) i(Xn2) i(Xn) i(Xn1).

Applying induction to the two terms on the right we are finished. The first assertion of the lemma

follows immediately. Let

Pi ={g G0 vii(g) =viwith C}

The intersection ofPi with G0 is just the normalizer ofni in G0. Thus it leavesVi invariant and is

contained inPi . To complete the proof we need only show that piis contained in pi. Ifm

iis a maximal


23/261

Chapter 2 19

fully reducible subalgebra ofpicontaining ai+mithen {Xmi

vi i(X) = 0} is a normal subalgebraofmi and its orthogonal complement in m

i with respect to the Killing form lies inai+ mi because it

commutes withai. Thus its orthogonal complement isaiand [ai, mi] = 0; so m

i =a +mi. Letn

ibe a

maximal normal subalgebra ofpi such thatad Xis a nilpotent for allXinni. Thenn

i is contained in

niand pi= m

i+ n

i. It follows thatp

i = pi.

Before stating the next lemma we make some comments on the normalization of Haar measures.

We suppose that the Haar measure on G is given. The Haar measure onKwill be so normalized that

the total volume ofKis one. IfPis a cuspidal subgroup the left-invariant Haar measure onPwill be

so normalized that G

(g) dg=

P

K

(pk) dpdk.

Let be one-half the sum of the elements ofQ and ifa= exp Hbelongs to A let (a) = exp

(H)

.

LetdHbe the Lebesgue measure a normalized so that the measure of a unit cube is one and let da be

the Haar measure onAsuch thatd(exp H) =dH. Choose, as is possible, a Haar measure on Sso thatP

(p) dp=

S

A

(sa) 2(a) ds da.

Choose the invariant measure on N\Nso that the volume of N\Nis one and choose the Haar

measure onNso that N

(n) dn=

N\N

N

(n) dn.

Finally choose the Haar measure onMso that

S

(s) ds= NM

(nm) dndm

Lemma 2.11 LetPbe a percuspidal subgroup and a compact subset ofS. There are constantsc

andr such that for any t 1 and anyg inG the intesection ofg and the Siegel domainS(, t)

associated to p has at mostctr elements.

It is easy to convince oneself that it is enough to prove the lemma whenGis connected. In this

case the representationsiintroduced before Lemma 2.10 are representations ofG. Choose a norm on

Viso thati(k)is unitary for allk inK. Ifg = sa(g)kthen

vi i(g) =di,i

a(g)

vi i(k),

so that

vi i(g) =di,i

a(g)

vi.


24/261

Chapter 2 20

IfT is a linear transformation then T denotes as usual the norm ofT. Choosing a basis ofvij ,

1 j ni, forVisuch that

viji(a) =ij(a) vij

for allainA, we see that there is a constant c1such that, for allv in Viand allainA,

vi(a) c1 min1jni ij (a)v.Moreover, it follows from Lemma 2.10 that there is a constant ssuch that, for allainA+(t, ),

min1jni

ij (a) ts di,i(a).

Letc2be such that, for allsin and allv in Vi,

vi(s) c2v.

Supposegand g =g, within , both belong to S(, t). Certainly

vi (g) =di,i

a(g)

vi.

On the other hand

vi i(g) c1c2 ts di,i

a(g)

vii()

and

vii() =di,ia() vi.

It follows from Lemma 2.8 that there are constants c3and c4and s1such that

,i

a(g)

c3 ts1 ,i

a()

,i

a(g)

c4 t

s1 ,i

a(g)

.

Sinceg = 1g, the argument may be reversed. Thus there are constantsc5,c6, ands2 such that

c5> ,i

a()

> c6 ts2 , 1 r p.

Let us estimate the order of

U(t) ={= sak s 1, a+ A(c6ts2 , c5), k K}


25/261

Chapter 2 21

with1a compact subset ofS. There are certainly constantsb1,b2andr1, r2such that+A(c6 t

s2 , c5)is

contained inA+(b1tr1 , b2t

r2). Choose a conditionally compact open set in G such that1U 2U=

implies1 = 2; thenb1can be so chosen that U(t)implies

U2(t) A+(b1t

r1 , b2tr2)K,

where2(t) ={s1as2a

1 s11, s2 2, a A+(b1tr1 , b2tr2}

and2 is the projection ofK UonS. Consequently the order ofU(t)is at most a constant times the

product of A+(b1tr2 ,b2tr2)

2(a) da

and the volume of2(t). A simple calculation, which will not be given here, now shows that the order

ofU(t)is bounded by a constant times a power oft. If it can be shown that for eachg in S(, t)and

each in the number of elements in = Psuch that g belongs toS(, t)is bounded bya constant independent oft,, andg then the lemma will be proved. If g = sak thensmust be in

. If there is no suchthe assertion is true; if there is one, say 0, then any other equals0 with

=.

Corollary. Let P1 and P2 be percuspidal subgroups and let P be a cuspidal subgroup belonging

to P2. Let S1 be a Siegel domain associated to P2, let S2 be a Siegel domain associated to

P2 = N\P2

S, let be a compact subset of N, and let b, s, and t be positive numbers. Let

a2 be the split component ofP2. There is a constantr, which depends only onG ands, and a

constantc such that ifg S1, , andg = namk withn in , a in A+(t, ), m inS2, k

inK, and(a) bsas(m) then

a1(g)

cra2(m)

Moreover if P =G the constantr can be taken to be1.

If1,, , p,are the simple roots ofa1, then

a1(g)= sup1ip i,a1(g);similarly, if1,, , q, are the simple roots of

a2,

a2(m)= sup

1iqi,

a2(m).


26/261

Chapter 2 22

Suppose that

i,a2(m), 1 i q,

for allmin S. Ifm is given as in the lemma, letM=(a2(m)); then

log i,a2(m) log M, 1 i q.

Since

log t log i,(a) log b + s log M

and sincea2(g) =aa2(m)there is a constant r1, which depends only on Gands, and a constantr2

such that

| log i,

a2(g)

| r1log M+r2, 1 i p.

In particular there is a constant r3, which depends only on Gands, and two positive constantsc1and

c2such that

i,a2(g)

c1Mr3

and

,i

a2(g)

c2Mr3,

for1 i p. Chooseu so that uP2u1 = P1; then a1(ugu

1) = a2(g). Letvi have the same

significance as above except that the group Pis replaced by P1. Then there is a constantr4, which

depends only onGands, and a constantc3such that

di,i

a1(g)

vi= vi i

1u1(ugu1)u

c3 Mr4di,i

a2(g)

vi.

Thus there is a constant r5, which depends only on Gands, and a constantc4such that

,i

a1(g) c4 Mr5 .

Appealing to Lemma 2.4 we see that ,i

a1(g)

is bounded away from zero for 1 i p. Since

log j,

a1(g)

is a linear combination oflog ,i

a1(g)

,1 i p the first assertion of the lemma is

proved.

To complete the proof of the lemma we have to show that ifS1 and S2 are Siegel domains

associated to P1 and P2 respectively then there is a constant c such that ifg belongs to S1 and g

belongs toS2then

a1(g) ca2(g)Using Lemma 2.10 as above we see that,i

a1(g) a

12 (g)

is bounded away from zero and infinity

for1 i p. Thusi,

a1(g) a12 (g)

must be also.

The next lemma will not be needed until Section 5.


27/261

Chapter 2 23

Lemma 2.12. SupposeP andP are two percuspidal subgroups andS andS are associated Siegel

domains. LetF andF be two subsets, with the same number of elements, of the set of simple roots

ofhand let P and P be the cuspidal subgroups belonging to P andP respectively determined by

F andF. If0 b t

when does not belong to F andb

a(g)

>

a(g)

whenbelongs to F and does not, ifg

belongs to S

and satisifes the corresponding conditions, and if belongs to F andg = g

thenP 1 = P. Moreover ifP = P and, for someg inG, gP g1 = P andgSg1 = S then

P = P, S= S.

Suppose, for the moment, merely that g belongs toS,g belongs to S, andg = g . Chooseu

so thatubelongs toG0and so thatuPu1 =P. Choosingviin Vias above we see that

di,i

a(g)

vi= di,i

a(1u1)

wii(ug) ci

di,i

a(1u1)

d,i

a(g)

vi

ifwiis such thatdi,i

a(1u1)

wi = vi i(

1u1).

Of course a similar inequality is valid ifg andg are interchanged. Since umay be supposed to lie in a

finite set independent ofwe conclude as before that a1(g) a(g)lies in a compact set. Moreover, as

in the proof of Lemma 2.11, 1 must belong to one of a finite number of left-cosets of. Consequently

wi,1 i p, must belong to a finite subset ofViand there must be a constant csuch that

wi i(ugu1) c d,i

a(g)

Moreover, it follows from the proof of Lemma 2.10 that there are positive constants b andrsuch that if

wiis not a multiple ofvithen

wi i(ugu1) b dii,

a(g)

di,i

a(g)

.

Chooset so large thatbt > cand choosetin an analogous fashion. Ifg andg satisfy the conditios of

the lemma then1u1 must belong toi,F

Pi, wherePi is defined as in Lemma 2.10. It is easily

seen that

P = i,F

Pi

so1u1 belongs to P. Index the system of simple roots so that

1,

a(g)

2,

a(g)

p,

a(g)


28/261

Chapter 2 24

There is an integerqsuch thatF ={q+1,, , p,}. Ift is very large then

1,

a(g)

> j,

a(g)

ifi q < j. Thus if1,, , p,is the system of simple roots indexed so that

1,a(g) 2,a(g) p,a(g),then

{1,, , q,}= {1,, , q,}

Since {q+1,, , p,}= Fthe setsFandF are equal anduPu1 = P. Then

1P=1 u1P u= P .

To prove the second assertion we observe that (P, S) belongs to (P, S) and to

(gP g1,gSg1). We have proved while discussing the basic assumption that this implies that g

belongs to P.

The next lemma will not be needed until Section 6 when we begin to prove the functional

equations for the Eisenstein series in several variables. LetPbe a cuspidal subgroup of rank qwith

a as split compoment. A set{1,, , q,}of roots ofais said to be a fundamental system if every

other root can be written as a linear combination of1,, , q,with integral coefficients all of the same

sign. It is clear that ifP1 and P2 are two cuspidal subgroups, g belongs to G, Adg(a1) = a2, and

B={1,, , q,} is a fundamental system of roots for a2, then

g1B= {1, Adg, , q, Adg}

is a fundamental system. The Weyl chamberWB associated to a fundamental system is

{Ha i,(H)> 0, 1 i q},

so that

Ad(g1

)WB =Wg1B

It is clear that the Weyl chambers associated to two distinct fundamental systems are disjoint. The only

fundamental system immediately at hand is the set of simple roots and the associated Weyl chamber

isa+. IfP1andP2are as above we defined(a1, a2)to be the set of all linear transformations from a1


29/261

Chapter 2 25

toa2 obtained by restricting Adgto a1 ifg in G is such that Adg(a1) = a2. P1 andP2 are said to be

associate if(a1, a2)is not empty.

Suppose P0 is a percuspidal subgroup and P is a cuspidal subgroup belonging to P0. Let

{1,, , p,}be the set of simple roots for h and suppose that P is determined by{q+1,, , p,}.

If1 j q let Pj be the cuspidal subgroup determined by {j,, q+1,, , p,}. Suppose aj is

contained in a. To prove the next lemma it is necessary to know that for eachP and j there is an

element gin Mj such thatAdg(a mj)is the split component of a cuspidal subgroup which belongs

toP0Mj and such that if is the unique simple root ofa

mj then Adg1 is a negative root

ofAdg(a mj). Unfortunately the only apparent way to show this is to use the functional equations

for the Eisenstein series. Since the lemma is used to prove some of these functional equations a certain

amount of care is necessary. Namely ifq >1one needs only the functional equations of the Eisenstein

series for the pairs(j,Mj)and since the percuspidal subgroups have rank less than those for (, G)

one can assume them to be proved. On the other hand ifq= 1the lemma is not used in the proof of

the functional equations. In any case we will take this fact for granted and prove the lemma. Everyone

will be able to resolve the difficulty for himself once he has finished the paper.

Lemma 2.13. Let P0 be a percuspidal subgroup and let F = {P1, , Pr} be a complete family

of associate cuspidal subgroups belonging to P0. If P belongs to F and E is the collection of

fundamental systems of roots of a then a is the closure ofBEWB. If B E then there is a

unique i, 1 i r, and a uniques in(ai, a) such thatsa+i =WB.

Suppose as before that P is determined by {q+1,, , p,}. If1 j qletgj be one of the

elements ofMj whose existence was posited above. Denote the restriction ofAdgj toa bysj and let

sj(a) = bj . Denote the restriction of1,, , q, to a also by1,, , q,. Thenj, s1j restricted

tobj mj is the unique simple root. Thus the simple roots1,, , q, ofbj can be so indexed that

j, s1j =j,and i, s

1j =i,+ bijj,withbij 0ifi =j . More conveniently,j, sj =j,

andi, sj =i,+ bijj,. To prove the first assertion it is enough to show that ifH0belongs toaand

(H0)= 0for all rootsthen there is someiand somesin(ai, a)such thats1(H0)belongs toa

+i .

There is a pointH1in a+ such that the line throughH0and H1intersects none of the sets

{Ha (H) =(H) = 0}where and are two linearly independent roots. If no suchi and s exist letH2 be the point closest

toH0 on the segment joining H0 andH1 which is such that the closed segement from H1 to H2 lies


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Chapter 2 26

entirely in the closure ofri=1

s(ai,a)

s(a+i ).

H2is notH0. Let H2lie in the closure ofta+k witht in(ak, a). ReplacingH0by t

1(H0)andPby Pk

if necessary it may be supposed that H2lies in the closure ofa+. Choosej so that

,(H2)> 0, 1 q, =j,

and j,(H2) = 0. Thenj,(H0) < 0, so that ifH lies on the segment joining H0 and H2 and is

sufficiently close toH2thensjHlies inb+j ; this is a contradiction.

It is certainly clear that ifB belongs to E then there is an i and an s in (ai, a) such that

sa+i = WB . Suppose thatt belongs to (ak, a) and ta+k = WB . Thens

1t(a+k) = a+i . Ifs is the

restriction ofAdh to a+i and t is the restriction ofAdg to a+k then h

1gPkg1h = Pi. The previous

lemma imples thati = k and thath1gbelongs toPi. Since the normalizer and centralizer ofaiin Pi

are the same it follows thats1tis the identity.

Ifa is as in the lemma the transformations s1, , sqjust introduced will be called the reflections

belonging, respectively, to 1,, , q,. We have proved that ifa and b belong to {a1, , ar} then

every element of(a, b)is a product of reflections; ifs is the product ofnbut no fewer reflections then

nis called the length ofs. Two refinements of this corollary will eventually be necessary; the first in

the proof of Lemma 6.1 and the second in the proof of Lemma 7.4.

Corollary 1. Every s in (a, b) can be written as a product sn s2s1 of reflections in such a

way that ifsk lies in(aik , ajk) and belongs to the simple rootk ofaik , thensk1 s1(a+i1

) is

contained in

{Haikk(H)> 0}.

Of course n is not necessarily the length ofs. Let WB = sa+. Take a line segment joining

a point in the interior ofWB to a point in the interior ofb+ which does not meet any of the sets

{H b(H) = (H) = 0} where and are two linearly independent roots. If the segment

intersects only one Weyl chamber the result if obvious. The lemma will be proved by induction on

the number, m, of the Weyl chambers which it intersects. Ifm is greater than one, let the segment

intersect the boundary ofb+ atH0. Index the simple roots1,, , q, ofb so that1,(H0) = 0and

j,(H0) > 0 ifj > 1. Then ifHbelongs to b+ the number 1,(sH) is negative, so that ifr is the

reflection belonging to1,the number(1, r1)(rsH)is positive. Let t= rs; ifrbelongs to(b, c)


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Chapter 2 27

then tbelongs to(a, c). Since there is a line segment connecting WBand r1(c+)which meets only in

m 1Weyl chambers, there is a line segment connectingc+ andta+ =rWB which meets onlym 1

Weyl chambers. If the corollary is true fort, sayt = sn1 s1 and sn = r1 thens = sn s1 and

this product satisfies the conditions of the corollary.

Supposea1, , ar are, as in the lemma, split components ofP1, , Pr respectively. Suppose

that, for1 i r,Si is a collection ofm-dimensional affine subspaces of the complexification ofai

defined by equations of the form(H) = whereis a root andis a complex number. Ifsbelongs

to Si and tbelongs to Sj we shall define (s, t) as the set of distinct linear transformations from sto

{H Ht} obtained by restricting the elements of(ai, aj) tos. Suppose that eachs in S= ri=1 Si

is of the formX(s) + swheresis the complexification of a distinguished subspace ofh and the point

X(s)is orthogonal tos; suppose also that for each s in Sthe set(s,s)contains an element s0 such

that

s0X(s) + H

= X(s) + H

for allH ins. Then ifr (r,s)and t (s, t) the transformation ts0r belongs to (r, t). Every

element s of(s, t)defines an element of(s, t) in an obvious fashion.s is called a reflection belonging

to the simple root ofs if the element it defines in (s, t)is that reflection. It is easy to convince oneself

of the following fact.

Corollary 2. Suppose that for every s inS and every simple root ofS there is a t in S and areflection in(s, t) which belongs to . Then ifs and tbelong to S ands belongs to (s, t) there

are reflectionsrn, , r1 such that ifrk belongs to (s,sk) andsk in(sk,sk) defines the identity

in(sk,sk) the transformations equals the productrnsn1rn1 r2s1r1.

As before the minimal value fornis called the length ofs.


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Chapter 3 28

3. Cusp forms.

As usual the invariant measure on\Gis normalized by the condition thatG

(g) dg=

\G

(g)

dg

Ifis a locally integrable function on\G,Pis a cuspidal subgroup, andT =N, then

(g) =\T

(tg) dg=N\N

(ng) dn

is defined for almost all g . A function in L(\G), the space of square-integrable functions on\G,

such that (g)is zero for almost allg and all cuspidal subgroups exceptG itself will be called a cusp

form. It is clear that the space of all cusp forms is a closed subspace ofL(\G)invariant under the

action ofGon L(\G); it will be denoted by L0(\G). Before establishing the fundamental property

ofL0(\G)it is necessary to discuss in some detail the integral

(f) (g) = G

(gh) f(h) dh

whenis a locally integral function on \Gandfis a once continuously differentiable function on G

with compact support.

SupposePis a percuspidal subgroup ofG and F is a subset of the set of simple roots ofh. Let

P1be the cuspidal subgroup belonging toPdetermined by the setF. Let

2(g) =

N1\N1(ng) dn

and let1= 2. Then

(f)

(g)equals

(3.a)

G

(h) f(g1h) =

G

1(h) f(g1h) dh +

G

2(h) f(g1h) dh.

The second integral will be allowed to stand. The first can be written asN1(N)\G

N1\N

N1\N1

1(nh)

1N1

f(g11nh) dn

dh.

If we make use of the fact that1(nh) =1(nh)

and N1\N1

1(nh) dn= 0,


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Chapter 3 29

this repeated integral can be written asN1(N)\G

N1\N1

1(nh) f(g,nh) dn

dh

with

f(g, h) =

Nf(g1h)

N1\NN1f(g1nh) dn

=N1\N1

N

f(g1h) f(g1nh)

dn

It should be recalled thatN1is a normal subgroup ofNand N1a normal subgroup of N.

IfS= S(t, )is a Siegel domain associated toPit is necessary to estimatef(g, h)wheng is in

S(t, ). It may be supposed that (N) contains N. Since f(g,h) =f(g, h) if Nwe can take

h= n1a1m1k1withn1in N,a1inA,m1in M, andk1inK. Supposeg = sak = a(a1sa)k= au

with s in and a in A+(t, ); then u lies in a compact set U1which depends on and t. The integrand

in the expression forf(g, h)equalsN

f(u1a1ah1) f(u

1a1nah1)

withh1 = a1h. If1 is a compact subset ofN1 such that( N1)1 = N1 it is enough to estimate

this sum fornin1. LetUbe a compact set containing the support off. If a given term of this sum is

to be different from zero eithera1ah1ora1nah1must belong toU1U. Then either

(a1aa1n1a)(a1m1a1)

or

(a1naa1n1a)(a1m1a)

belongs to P U1UK. It follows that there is a compact setV inNdepending only onS andUsuch

thata1abelongs to V. Choose a conditionally compact open set V1in Nso that ifbelongs to N

and V1 V1is not empty then = 1; there is a compact set V2in Nsuch that a1V1a is contained in V2

ifa belongs toA+(t, ). Ifa1a belongs toV thenV1 is contained inaV V2a1. Consequently the

number of terms in the above sum which are different from zero is at most a constant times the measureofaV V2a

1 and a simple calculation shows that this is at most a constant times2(a). Finally there

is a compact subset2ofAMsuch that every term of the above sum vanishes unlessm1a1belongs to

a2.


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Chapter 3 30

If{Xi}is a basis ofg there is a constant such that |(Xi) f(g)| for alli and g . IfX g

then(X) f(g)is defined to be the value of dfdt

(g exp tX)att= 0. Then

|f(u1a1ah1) f(u1a1nah1)|

is less than or equal to

10

Ad(h11 )Ad(a1)X f(u1a1exp tXah1) dtifn= exp X. Sincenlies in a fixed compact set so does X. Moreover

h1= a1n1aa

1m1a1k1

lies in a compact set depending only on S and U. Consequently the right hand side is less than a

constant, depending only on S, U, and , times the largest eigenvalue ofAd(a1) on N1. In conclusion

there is a constantc, depending only onS,U, and, such that for allg in S and allh

|f(g, h)| c2

a(g)

mini,F

i,

a(g)1

.

Moreover the first integral in the expression for (f) (g)is equal to

a2K

2(b)

N1(N)\N

N1\N1

1(n1nbmk) f(g, n1nbmk) dn1

dn

dbdmdk

or, as is sometimes preferable,

a2K

2(b)

N\N

1(nbmk) f(g, nbmk) dk

dbdmdk.

The absolute value of the first integral is at most

c2

a(g)

min1,F

i,

a(g)1

a2K

2(b)

N\N

|1(nbmk)| dn

dbdmdk.

If3is a compact subset ofNsuch that( N)3= Nthis expression is at most

(3.b) c

2a(g) mini,F i,a(g)1

3a2K |1(h)| dhFor the same reasons the absolute value of the second integral is at most

(3.c) c2

a(g)

miniF

i

a(g)1

3a2K

|(h)| dh.


35/261

Chapter 3 31

Lemma 3.1. Let belong to L0(\G), let f be a once continuously differentiable function with

compact support, and letPbe a percuspidal subgroup. IfS= S(t, )is a Siegel domain associated

to P there is a constantc depending only onS andfsuch that forg inS

|(f) (g)| c1

a(g)

1

a(g)

.

Here is the norm ofin L(\G)and ifabelongs toAthen

(a) = max1ip

i,(a)

It is enough to establish the inequality on each

Si =

g S i,a(g) j,a(g), 1 j p

For simplicity takei = 1. In the above discussion takeF = {j,

j = 1}. The second term in (3.a) is

zero; so to estimate(f) (g)we need only estimate (3.b). The integral is at most3a2K

dh 1

2

3a2K

|(h)|2 dh 1

2

Since3a2Kis contained in a fixed Siegel domainS(t, )for allainA+(t, ), the second integral

is at most a constant times 2. If4 and 5 are compact subsets ofA and Mrespectively such that

2 is contained in45the first integral is at most

3dn

42(ab) db

5dm

.

Since

mini,F

i(a) =1,(a) =

a(g)

ifg = sak is inS1, the lemma follows.

It is a standard fact that(f)is a bounded linear operator onL(\G). It is readily seen to leave

L0(\G)invariant.

Corollary. Iffis once continuously differentiable with compact support then the restriction of(f)

to L0(\G) is a compact operator.

Since 1(a) 1(a) is square integrable on any Siegel domain the corollary follows immediately

from Ascolis lemma, the above lemma, and the fact that\Gis covered by a finite number of Siegel

domains. The significance of the corollary is seen from the following lemma.


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Chapter 3 32

Lemma 3.2. LetG be a locally compact group and a strongly continuous unitary representation

ofG on the separable Hilbert spaceV. Suppose that for any neighbourhoodU of the identity inG

there is an integrable functionf onG with support inU such that

f(g) 0, f(g) = f(g1),

G

f(g) dg= 1

and(f) is compact; thenV is the orthogonal direct sum of countably many invariant subspaces

on each of which there is induced an irreducible representation of G. Moreover no irreducible

representation ofG occurs more than a finite number of times inV.

Of course(f)is defined by

(f)v=

F

f(g) (g) v dg

ifv belongs to V. Consider the families of closed mutually orthogonal subspaces ofV which are

invariant and irreducible under the action ofG. If these families are ordered by inclusion there will be

a maximal one. Let the direct sum of the subspaces in some family beW. In order to prove the first

assertion it is necessary to show that WequalsV. Suppose the contrary and letW be the orthogonal

complement ofWinV. Choose a vin W with v= 1and choose Uso that v (g)v< 12 ifgis in

U. Choose fas in the statement of the lemma. Then (f)v v < 12 so that (f)v= 0. The restriction

of(f)to W is self-adjoint and thus has a non-zero eigenvalue. LetWbe the finite-dimensional

space of eigenfunctions belonging to the eigenvalue . Choose from the family of non-zero subspaces

ofWobtained by intersecting Wwith closed invariant subspaces ofW

a minimal one W0. Take the

intersectionV0of all closed invariant subspaces ofW containingW0. Since V0 ={0} a contradiction

will result if it is shown that V0 is irreducible. IfV0 were not then it would be the orthogonal direct

sum of two closed invariant subspacesV1and V2. SinceVi Wis contained inV0 W

= W

0for

i= 1and2, the spaceVi Wis either {0} orW

0. But(f) Vi Viso

W0= (V1 W) (V2 W

)

and, consequently, Vi W = W

fori equal to 1 or 2. This is impossible. The second assertion

follows from the observation that if some irreducible representation occurred with infinite multiplicity

then, for somef,(f)would have a non-zero eigenvalue of infinite multiplicity.

Before proceeding to the next consequence of the estimates (3.b) and (3.c) we need a simple

lemma.


37/261

Chapter 3 33

Lemma 3.3. Let S(1), , S(m) be Siegel domains, associated to the percuspidal subgroups

P(1), , P(m) respectively, which cover \G. Suppose c and r are real numbers and (g) is a

locally integrable function on\G such that

|(g)| c r

a(i)(g)

ifg belongs to S(i). If P is a cuspidal subgroup and

(a,m,k) =

N\N

(namk1) dn

fora in A, m in M andk inK then there is a constantr1, which does not depend on, such

that for any compact setC in A, any percuspidal subgroup P of M, and any Siegel domanS

associated to P there is a constantc1, which does not depend on, such that

|(a,m,k)| c1r1

a(m)

ifa belongs to C andm belongs to S. In particular if P =G thenr1 can be taken equal to r.

If is a compact subset ofNsuch that( N)= Nthen

|(a,m,k)| supn

|(namk1)|

Ifg = namk1 choose in so that g belongs to S(i) for some i. According to the corollary to

Lemma 2.11 there is a constant r2such that for anyC,P, and Sthere is a constantc2such that

a(i)(g)

c2a(m)r2

Since

a(i)(g)

is bounded below on S(i) for eachi, it can be supposed for the first assertion that

r 0. Then take r1=rr2and c1= ccr2. If

P is G the lemma also asserts that ifS is any Siegel domain

associated to a percuspidal subgroupPthen there is a constant c1such that |(g)| c1r

a(g)

onS.

Giveng in S again choosein so that gbelongs toS(i) for somei. The corollary to Lemma 2.11

asserts that there is a number c2independent ofiandg such that

c12 1a(g) a(i)(g) c2

Takec1 = ccr2ifr 0 and takec1=cc

r2 ifr


38/261

Chapter 3 34

Lemma 3.4. SupposeS(1), , S(m) are Siegel domains, associated to percuspidal subgroups, which

cover\G. Supose that(g) is a locally integrable function on\G and that there are constantsc

andr such that|(g)| cr

a(i)(g)

ifg belongs to S(i). LetUbe a compact subset ofG, let be

a constant, let{Xi} be a basis ofg, and letf(g) be a once continuously differentiable function on

G with support inU such that|(Xi) f(g)| for allg andi. IfS is a Siegel domain associated

to the percuspidal subgroup P and ifk is a non-negative integer there is a constant c1, dependingonc, r, U, , S, andk but not on orf, such that

(k)(f) (g) k(f)i(g) c1 rka(g)on

S1= {g S i,a(g) j,a(g), 1 j p}.

In accordance with our notational principles

i(g) = Ni\Ni

(ng) dn

ifPi is the percuspidal subgroup belonging to P determined by {j,j = i}. The assertion of the

lemma is certainly true fork = 0. The proof for generalk will proceed by induction. For simplicity

takei= 1. Since k(f)

i

=k(f)i,

it will be enough to show that if there is a constant s such that for any Siegel domainS associated toP

there is a constantc such that |i(g)| csa(g) onS1then for anyS there is a constantc1so that

|(f) (g) (f)1(g)| c1 s1

a(g)

onS1. Of course it will also have to be shown that the constantsc1 do not depend onfor. Indeed

we apply this assertion first to withs = r and then in general tok(f) withs = r k. Since

(f) (g) (f)1(g)

is nothing but the first term on the right side of (3.a) it can be estimated by means of (3.b). Thus

(f) (g) (f)1(g) c2 2a(g) 11,a(g) 3a2K

|1(h)| dh

ifgbelongs toS1. First observe that ifg belongs toS1then

a(g)

= 1,

a(g)


39/261

Chapter 3 35

There is a Siegel domain S such that whena = a(g)and g belongs toS the set3a2Kbelongs to

S. Let4and5be compact subsets ofA andMrespectively such that2is contained in45; then

the integral is less than or equal to a constant, which does not depend on , times4

2(ab) s(ab) db

which is certainly less than a constant times 2(a) s(a).

Corollary. Suppose V is a finite-dimensional subspace of L0(\G) invariant under (f) for f

continuous with compact support and such thatf(kgk1) = f(g) for all g in G and all k in K.

Then given any real number r and any Siegel domainS associated to a percuspidal subgroup P

there is a constantc such that, for all inV and allg inS,

|(g)| cr

a(g)

(g).

Since for a givent there are constantsc1and r1 such that1(a) c1r1(a)fora inA+(t, ),

the corollary will follow from Lemmas 3.1 and 3.3 if it is shown that there is a once continuously

differentiable functionf0 satisfying the conditions of the lemma such that(f0) = for all in V.

Let {1, , n}be an orthonormal basis for Vand let Ube a neighbourhood of the identity in G such

that

(g)i i< (2n)1

ifgbelongs toUand1 i n. Then, for anyinV,

(g) 12

ifgis in U. Choose fto be a non-negative function, once continuously differentiable with support in u,

such thatG

f(g) dg= 1 and f(kgk1) =f(g) for all gin G and all k in K. Then the restriction of(f)

tov is invertible. Thus there is a polynomialp with no constant term such thatp

(f)

is the identity

onV . In the group algebrap(f)is defined; setf0 = p(f). IfV was not a space of square-integrable

functions but a space of continuous functions and otherwise satisfied the conditions of the lemma then

a simple modification of the above argument would show the existence of the functionf0.

IfP is a cuspidal subgroup then the pair M , satisfies the same conditions as the pair G, . It

will often be convenient not to distinguish between functions on \M,T\S, andAT\P. Also every

function on G defines a function on P Kby(p, k) = (pk1). SinceG = P K, functions onG

may be identified with functions on P Kwhich are invariant under right translation by (k, k)ifk


40/261

Chapter 3 36

belongs toK P. IfVis a closed invariant subspace ofL(\M)let E(V)be the set of measurable

functionsonAT\Gsuch that(mg)belongs toVas a function ofmfor each fixedgin Gand\MK

|(mk)|2 dmdk = 2


41/261

Chapter 3 37

isomorphism ofE(V)with a space of functions on M K. Indeed let P = N\P Sthen P is a

cuspidal subgroup ofMand Mis the same asM. AlsoP Kis a cuspidal subgroup ofM K

for the group {1}. IfL is the space of square integrable functions onKthen the image ofE(V)

is the set of all functions in E(V L)which are invariant under right translations by(k, k)wherek

belongs to K Pand kis the projection ofk on M. Denote the group of such elements byK0and

let

Kbe the projection ofK

P on

M. The group

Kplays the same role for

MasKdoes forG.Supposebelongs toE(V, W). Then

(mk1, kk2) = (mk1k

12 k

1).

For fixed m and kthis function belongs to the space of functions on K Kof the form (k1k12 )with

inW. A typical element ofWis of the formij , that is, the matrix element of a representation in F.

Since

ij(k1k12 ) = i(k1) j(k

12 )

it belongs to the spaceW ifW is the space of functions on K Kspanned by the matrix elements

of those irreducible representations ofK Kobtained by taking the tensor product of an irreducible

representation ofK P contained in the restriction toK P, which is isomorphic to K, of one of

the representations inFwith a representation ofKcontragredient to one of the representations in F.

Thus the image ofE(V, W)is contained inE(V L, W); indeed it is readily seen to be contained in

E(V W, W)and to be the space of all functions inE(V W, W)invariant under right translation

by elements ofK0. On occasion it will be convenient to identify E(V, W)with this subspace.

Since the representation ofM on L(\M) is strongly continuous there is associated to each

element Xin the centreZ of the universal enveloping algebra ofm a closed operator (X) on L(\M).

Indeed if is any strongly continuous representation ofMon a Hilbert space L there is associated to

eachXin Z a closed operator(X). IfL is irreducible then

L= nj=1Lj

where each Lj is invariant and irreducible under the action ofM0, the connected component ofM.

The restriction of(X)to Lj is equal to a multiple, j(X)I, of the identity. The mapX j(X)is

a homomorphism ofZ into the complex numbers. Let us say that the representation belongs to the

homomorphismj . Suppose the closed invariant subspaceVis a direct sum Vi of closed, mutually


42/261

Chapter 3 38

orthogonal subspaces Vieach of which is invariant and irreducible under the action ofM. As we have

just remarked each Viis a direct sum

nij=1Vij

of subspaces invariant and irreducible under the action ofM0. SupposeVij belongs to the homomor-

phism ij .Vwill be called an admissible subspace ofL(\M)ifVis contained in L(\M)and there

are only a finite number of distinct homomorphisms in the set {ij}.

Lemma 3.5. IfV is an admissible subspace ofL0(\M) andW is an admissible subspace of the

space of functions onK thenE(V, W) is finite dimensional.

In the discussion above take P equal to P. Then E(V, W) is isomorphic to a subspace of

E(V W, W). It is readily seen that V Wis an admissible subspace ofL0( {1}\M K)and

thatW is an admissible subspace of the space of functions on K K. Since it is enough to show that

E(V W, W)is finite dimensional we have reduced the lemma to the case that P andMare equal

to G. SupposeV = Vi. IfV = Vi where the second sum is taken over those Vi which contain

vectors transforming according to one of the representations inF thenE(V, W) =E(V, W). In other

words it can be supposed that eachVicontains vectors transforming underKaccording to one of the

representations inF. For eachilet

Vi =nij=1Vij

where each Vijis invariant and irreducible under the action ofG0, the connected component ofG, and

belongs to the homomorphismij . It is known ([10], Theorem 3) that there are only a finite number

of irreducible unitary representations ofG0 which belong to a given homomorhpism ofZ, the centre

of the universal enveloping algebra ofg, and which contain vectors transforming according to a given

irreducible representation ofK G0. Thus there is a finite set Eof irreducible representations ofG0

such that for eachi there is aj such that the representation ofG0 onVij is equivalent to one of the

representations inE. As a consequence of Lemma 3.2 applied toG0there are only a finite number of

Vi. It is known however (cf. [10], Theorem 4) that for each i the space of functions in Vi transforming

according to one of the representations in F is finite dimensional. This completes the proof of the

lemma. Since E(V, W)is finite-dimensional it follows from the proof of the corollary to Lemma 3.4 that

it can be considered as a space of continuous functions.

Suppose(g)is a continuous function on T\Gsuch that for eachg in Gthe function(mg)on

\Mbelongs toVand the function(gk)onKbelongs toW. For eachain A consider the function

(sak)onT\S Kor onAT\P K. Ifk0belongs toK P =K Sthen(sk10 ak0k) =(sak)


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Chapter 3 39

sincesk10 ak0a1s1 is inN. Thus it defines a function (a)on AT\Gwhich is seen to belong to

E(V, W). The space of all such functions for which(), which is a function on A with values in

E(V, W), has compact support will be called D(V, W).

Lemma 3.6. SupposeV is an admissible subspace ofL0(\M) andWis an admissible subspace

of functions on K. If belongs to D(V, W) then\ (g) is absolutely convergent; its sum(g) is a function on\G. IfS0 is a Siegel domain associated to a percuspidal subgroup P0 andifr is a real number there is a constantc such that |(g)| c r

a0(g)

forg inS0.

There is one point in Section 6 where we will need a slightly stronger assertion than that of the

lemma. It is convenient to prove it at the same time as we prove the lemma.

Corollary. Let(g)be a function onT\Gand suppose that there is a constanttsuch that(namk) =

0 unless a belongs to A+(t, ). Let P1, , Pm be percuspidal subgroups to whichP belongs and

suppose that there are Siegel domains S1, ,Sm associated to

Pi = N\Pi S which cover

\M. Suppose that there is a constant s such that given any constant r1 there is a constant c1

such that, for1 i m,

|(namk)| c1 s(a) r1

ai(m)if m belongs to Si. Finally suppose that there are constants u and b with 1 b > 0 such that

(namk) = 0 if(a)> u and the projection ofm on\Mbelongs to the projection on\M of

{mSi

ai(m)

< b(a)}

for somei. Then \

(g) =(g)

is absolutely convergent and ifS0 is a Siegel domain associated to a percuspidal subgroup P0 and

r is a real number there is a constantc such that |(g)| cr

a0(g)

forg inS0.

It is a consequence of Lemma 3.5 and the corollary to Lemma 3.4 that the function of the lemma

satisfies the conditions of the corollary. Let be a compact subset ofNsuch that( N)= N. Ifgis

inS0let Ube the set of all elements in such thatg= namkwithnin,a inA+(t, ),min Si

for some i, andk in K. Since any left coset ofincontains an elementsuch thatg= namkwith

8/12/2019 Functional Equations-Langlands, R.P.on the Functional Equa

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Functional Equations-Langlands, R.P.on the Functional Equations Satisfied by Eisenstein Series