Harmonic Analysis on Compact Solvmanifolds

Lecture Notes in Mathematics Edited by A. Dold and B. Eckmann

602 Jonathan Brezin

Harmonic Analysis on Compact Solvmanifolds

Springer-Verlag Berlin-Heidelberg • New York 1977

Author Jonathan Paul Brezin Department of Mathematics University of North Carolina Chapel Hill, NC 27514 USA

Library of Congress Cataloging in Publication Data

Brezin, Jonathan Paul, 1943Harmonic analysis on compact solvmanifolds. (Lecture notes in mathematics ; 602) Bibliography: p. Includes index. 1. Harmonic analysis. 2. Homogeneous spaces. I. Title. II. Series. QA3.L28 no. 602 rQA~03~ 5101.8s ~5151.785j 77-221~2

AMS Subject Classifications (1970): 43A85 ISBN 3-540-08354-5 Springer-Verlag Berlin. Heidelberg. New York ISBN 0-38?-08354-5 Springer-Verlag New York • Heidelberg • Berlin This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically those of translation, reprinting, re-use of illustrations, broadcasting, reproduction by photocopying machine or similar means, and storage in data banks. Under § 54 of the German Copyright Law where copies are made for other than private use, a fee is payable to the publisher, the amount of the fee to be determined by agreement with the publisher. © by Springer-Verlag Berlin • Heidelberg 1977 Printed in Germany Printing and binding: Beltz Offsetdruck, Hemsbach/Bergstr. 2141/3140-543210

PREFACE

This project began at the kind s u g g e s t i o n of P r o f e s s o r R. Remmert that I consider w r i t i n g an Ergebnisse Bericht on the general theme of harmonic analysis on compact homogeneous spaces.

For a number of reasons,

treatment of the m a t e r i a l w o u l d be more appropriate. me was

I felt that a less formal Perhaps the crucial thing for

that I did not feel that there was a sufficient body of generally interesting

results in e s s e n t i a l l y final form to justify an Ergebnisse monograph.

As the

reader who follows through chapter I in detail will see, w h a t is available now is more a collection of experimental data--albeit, hard w o n - - t h a n a coherent theory. Thus the real report will have to w a i t - - p e r h a p s

five years or more.

I w o u l d like to

thank P r o f e s s o r Remmert for his encouragement, w h i c h led me to w o r k far harder to articulate my subject than I had ever thought I could. My o r i g i n a l vision was to present in as complete and elementary form as possible the work of R. Howe and of L. A u s l a n d e r and myself on inductive methods in harmonic analysis on compact h o m o g e n e o u s spaces, and compact solvmanifolds

in particular.

The b a s i c m a t e r i a l was to b e an u n p u b l i s h e d manuscript of Howe, the essence of w h i c h is §l_!, and another u n p u b l i s h e d manuscript,

this time by L. A u s l a n d e r and myself.

Our w o r k appears here as chapter II and there is some overlap with Howe's manuscript. W h e n all was said and done, original idea,

there was something vaguely dissatisfying about that

the p r o b l e m being that very little of the color and variety the field

has to offer showed through in the s o l u t i o n of the p a r t i c u l a r question w e w e r e attacking there.

I have tried to remedy this with an i n t r o d u c t o r y chapter entirely

devoted to examples and a s s u m i n g a m i n i m u m of prerequisites.

I hope the beginner

w i l l find this chapter a reasonably a c c e s s i b l e o v e r v i e w of the field, and that it will put the immense b a r r a g e of technical detail of the later sections into some perspective.

The "Afterword", w h i c h is section 12, w i l l give the b e g i n n e r a guide to

the literature,

some suggestions as to problems one might attack, and a very informal

account of who did what, when, where,...

IV

A few words on prerequisites are in oder.

The first, fifth, and sixth sections,

wherein the main examples are described, require little more than the basic vocabulary of distribution theory, representation theory, and Lie theory.

I don't believe

that any theorems of any depth from these areas are used--the text is essentially self-contained,

knowledge

except at one or two minor points.

Sections 2 and 4 require some

of basic number theory, but again little more than some vocabulary.

notion of algebraic group also crops up in section 4. R. Steinberg's volume in this series all that is needed and more.

The

The introductory sections of

(Springer Lecture Notes 366 (1974)) tell one

Beginning with section ~, we assume the reader is

familiar with the material on representation theory presented in the introductory chapter by L. Auslander and C. C. Moore in their memoir on the unitary representation theory of solvable Lie groups

(American Math. Soc. Memoir 62 (1966)).

It is

not assumed that the reader knows the proofs of those results, but merely that he has some acquaintance with Mackey's general inductive procedure.

The final two

sections of the text (iO and ii) assume a familiarity with nil- and solvmanifolds, the most convenient

(and completely adequate

pp 1-6 of L. Auslander et al.

for our purposes)

Flows on Homogeneous Spaces

source for which is

(Annals of Math Studies

53 (1963), Princeton) I would like to thank Louis Auslander and Roger Howe for their generosity in just about every aspect of this project.

In chapters II and III I am serving in

good part as a scribe for their ideas, and I hope they find the result pleasing. Richard Tolimieri and Richard Penney have been a real help by being as sanguine as I am pessimistic.

They both have made substantial progress on the problems set

forth here, in papers only just available in preprint

form as I write.

Finally, the

birth pangs of this project were suffered during a supremely productive summer semester at the University of Heidelberg.

I would like to thank again my host there,

Wolf Beiglbock, whose patient questioning and own insight I deeply appreciate.

The

beautiful example in section 2 of "bounded multiplicity" is his. Finally I want to thank my parents--for many things, but here for the enjoyment of science they helped me to have, and the endless pleasure that has brought. work is dedicated to them.

This

Acknowledgment

Some of the research reported on herein was conducted while the author was a Sloan Fellow and some was partially supported by grants from the National Science Foundation.

The author is grateful to both institutions for their generous support.

The excellent job of typing was done by Janet Farrell.

Chapel Hill, North Carolina

CONTENTS

CHAPTER

I

SOME

1. 2. 3. 4. 5. 6.

CHAPTER

II

THE

10. 11.

INDEX

THEORY

SOLVMANIFOLDS

Nilmanifolds Solvmanifolds

.

.

12.

BIBLIOGRAPHY

GENERAL

COMPACT

.

.

.

.

i

.

.

.

. . . . . .

.

.

110 126 134

149

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 164

171

Attribution

.

.

1 23 27 40 56 77

ii0

The theorem on discrete spectrum . . . . . . . . . . . . . . . . The restriction problem . . . . . . . . . . . . . . . . . . . . The induction problem . . . . . . . . . . . . . . . . . . . . .

III

AFTERWORD

EXAMPLES

The most primitive examples . . . . . . . . . . . . . . . . . . Some multiplicity computations . . . . . . . . . . . . . . . . Some an&lytic results . . . . . . . . . . . . . . . . . . . . . The arithmetic nature o f multiplicities . . . . . . . . . . . . Dimension four . . . . . . . . . . . . . . . . . . . . . . . . . M u l t i p l i c i t y computations in dimension four . . . . . . . . . .

7. 8. 9.

CHAPTER

BASIC

.

.

175

179

and speculation

. . . . . . . . . . . . . . . . . .

171

CHAPTER I

SOME BASIC EXAMPLES

i.

The most primitive examples

Although the methods we shall eventually develop can be used to study very general compact homogeneous spaces, w e ourselves are interested here primarily in a rather limited class of such spaces, the so-called compact solvmanifolds. solvmanifold is a compact and transitively.

C=

A compact

manifold on which a solvable Lie group acts smoothly

In this chapter, we are going to describe some examples, begin-

ning in this and the next few sections with the most primitive case, three dimensions.

Already in looking at three dimensional groups one encounters all but one of

the phenomena that distinguish genuinely solvable harmonic analysis from its classical abelian counterpart.

The one missing phenomenon appears in dimension four,

which we shall treat in §5.

It is a little hard to say just what these mysterious

phenomena are without having at hand some examples, so we won't belabor that point right now.

The examples will speak for themselves.

One final remark before we get down to serious business: Harmonic analysis can be done in many settings:

L2-spaces, more general

LP-spaces, Sobolev spaces, spaces of distributions...and the list goes on. all of what has been written about solvmanifelds has dealt with the which is undeniably easier to work with in general situations. possible to work with other spaces.

L2

Almost

theory,

It is, however,

We have worked in a variety of settings through-

out this chapter and hopefully have betrayed some of the special advantages and disadvantages of each. The three-dimensional examples arise from a semi-direct product construction. Let C~

t~--> o(t) manifold

be a one-parameter subgroup of

]R2 x ~

SL(2,~) .

We can then turn the

into a Lie group by defining the group operations to be

(~,t)(~,s) : (~ + o(t)~,

t + s)

(1.1) (~,t) -I : (-O(-t)~, -t)

W e shall use

G

to denote this Lie group.

o

by w r i t i n g

as a m a t r i x group, w h i c h can be done

~(t)

for all

subset

F°

o of

a subgroup of

G . o

y(t)

@(t)

0

u2

t c ~ .

0

0

1

t

0

0

0

1

G

Proof:

as

and

o(t)

as

,

is a solvable group and is n o n - a b e l i a n unless

o

In order to produce a compact solvmanifold, w e must o

that

n e ~

implies

o(n) E SL(2,~).

A glance at (i.i) shows that w h e n

consisting of those

(u,t) w i t h

H e n c e w e may form the quotient

Proposition:

(Ul,U 2)

o

correspond to the m a t r i x

u

sake, w e take to be cosets of the form

(~.2)

Go

0

integral. Go

of

~(t)

place the further condition on call such a

6(t)]

~(t)

It is easy to see that o(t) = 1

(~,t)

u ~ ~2

G

(t)]

y(t)

and then letting the element

It is o c c a s i o n a l l y h e l p f u l to v i e w

Go/F ~

~ c g2

o

Let us

is integral,

and

Go/Fo, which,

t ~ Z

the

is actually

for convenience's

Fog.

is a compact solvmanifold.

This is e s s e n t i a l l y obvious, but the compactness of

Go/F °

deserves some

comment :

Let

Q

denote the h a l f - o p e n unit cube

Then every element of with

~ c Po

jectively onto

and

G

: 0 ! Ul,U2,t < i}

in

Go .

can be w r i t t e n in precisely one w a y as a product

o

q £ Q.

The natural map

Go/Fo, w h i c h implies,

image of the closure of

{(~,t)

Q

in

G

o

G O ÷ Go/F °

in particular,

therefore carries

that

and hence is compact.

G /r °

~q Q

bi-

is the continuous D

It is perhaps w o r t h making a short topological because

Go

is the universal

covering transformations, because

covering

the fundamental

F ° is abelian only when

not homeomorphic,

space of

in general,

digression

Go/Fo, with

group of

o(n) = 1

G /F o o

for all

to the 3-torus

to point out that, Fa

is

F . o

ding left tions on

in doing so to identify

F -periodic o G /F

function on

as periodic

o

Given a function translate

R f g

F -periodic, o G /F o o

f

then

R f g

function spaces on

abelian--that

is, G

o

F

o

Go/F °

functions on

Go/F o.

with the correspon-

G . o g

h ~--> f(hg)

e

on

Go, we define the right-

G . o

If

f

happens

We can roughly describe harmonic

of the operators

Go/Fo,

= I~ 3,

a function on

on

will be also. analysis"

is

T 3.

and an element

to be the function

as the "spectral

various

Go

Go/F °

G --infact, we shall almost always treat funco

functions

on

In particular,

n c Z, the manifold

What we want to study are certain spaces of complex-valued It is convenient

as the group of

such as

{R

C ~ (Go/Fo).

= Z 3, G /F = T 3, and o o

g

: g e G } o In case

analysis

restricted happens

Go

o(t) = 1

to be left

for all

on

to to be

t c IR--

the story is a familiar one: Let

C°°(T 3)

be the space of functions

Rg, g ~ G O = i~3 . eigenvectors, C°°( T 3)

"Spectral

and of course,

analysis"

on which w e wish to study the operators Rg 's means

of the

up to constant multiples,

the only such elements

Since every function in formly c o n v e r g e n t

Coo( T 3)

can be w r i t t e n

sum o f t h e form

2n an en =

n E g3}

is,

in effect,

of the operators

R

g

Of course, w e ' r e and when

of

are the characters

en(~) = e _ ~ ( 2 ~ i ( n l x I + n2x 2 + n3x3)), =

=

finding their common

G

a basis as

g

for

traces

2

o

really interested

is n o n - ~ e l i ~

diagonalize

an e ¢ '

the family

that

{en : =

simultaneously

diagonalizes

all

= ~3 . in what happens w h e n R

on

C~(Go/F

G )

o

is not abelian,

are not all going to

g

commute w i t h one another, w h i c h means

C~(G /F ) oo

with

=

the operators

o

~ ~ ~3

in precisely one w a y as a uni-

=

C~(T3) G

~ £ Z3,

that

diagonalizes

all

that it is futile to look for a basis of

of these operators.

some of them to good effect:

One c a n , on t h e o t h e r h a n d ,

The group A = {(~,0)

G

decomposes

o

: u • ~2}

and

into a semi-direct

B = {(~,t)

: t • ~} .

nalilize

the operators

Carrying

out this plan wil take a while...

f c C~(Ga/Fa), periodic with

coming from the elements

the function respect

to

since

for each fixed

on

u • ~2,

~2

f(u,t)

--

= ~

_

in which

The functions

an(t) =

, and

A.

that whenever

t • ]R

t• ~,

imply

= f(~,t)

f(',t)

=

=

in

factor

in a Fourier

series

an(t)e() n e ~ 2

= nlu I + n2u 2

abelian

is, for each fixed 2

n= •

t e ~ , we can expand

(1.3)

Our strategy will be to diago-

of the normal

f(u + ~,t) = f((~,O)'(~,t))

Hence,

AB, in which

We begin by observing

~ ~--> f(~,t)

Z2,

product

and

e

-

=

,

is the function

(1.3) are not entirely

t ~--> exp(2~it)

arbitrary,

on

~.

as the following

propo-

sition shows.

(1.4)

Proposition:

The function

a

--

denoting ~o(t)

again the entries

= (nl~(t)

+ n2~(t),

(1.5)

a

--

Proof:

on n =

by

is in

C ~ n L=(~)

.

Further

~(t),

B(t), ~(t),

and 6(t),

and setting

nlB(t ) + n2~(t)) , one has

for all

k •~

and

t • IR

(k)

a

n=

in terms of

f

as an integral,

1 1 = S an(t)= 0 I 0 f(u't)e)

=

with

b (t) = a (s+t)e(<m,o(t)v>). m= m = =

The natural sequence

{f.} J

integers

topology on converges

the sequence

aa ~b ~cf. uI u2 t

to

The subspaces

f e C~(Go/Fo)

of partial derivatives

t j

Z(S)

are evidently

= ~

If

f

i.e.

a

of non-negative

converges

uniformly

to

closed in this topology.

Z(S), those of the form

C~(Go/Fo)

(a,b,c)

C~(Go/Fo)

Z(~)

Z(w) in the

for some

is written as in (1.3)

w £ ~

are

Ca-topology:

in the sense that every

c~n be written in precisely one way as a convergent

~ Z(~).

partial sum

"C -convergence,"

when for every triple

These minimal subspaces also span

(1.9) Proposition:

f

f

is that of

uI u2

Now, among the subspaces minimal.

C~(Go/Fo)

{~a ~b acf.}

'

with

=

then the summand

sum f

~ is the

fw

(i. i0)

[ne~ an(t)e()~n

Suppose

that

interval

an(t 0 )

I

is not zero.

about

to

on which

(~.17), we see that whenever = ~2 v • .

It follows

= ~o(t)

that

for all

==

Because a

an(t)e()

a

is smooth,

does not vanish.

t e I, we must have

w = ~q(t)

t • ~ .

for all

In particular,

•

Comparing

e()

t

there is a whole open coefficients

= e()

in some open interval,

~ = ~.

It follows

that

in

for all and hence

f(~,t)

=

a (t)e(). n =

If

g = (~,s)

and

(-o(-s)~ - o(-s-r)~ it follows

that

(~.18)

Proof:

on cosets

Use

Thus tors on

(1.16)

desirable

theory

of

on

Z(S) @ Z(~-S)

yields

may identify

Z(~-S) ±

subspace

then

V

of

C (Go/F o)

consists

that

of functions

[]

family

R(~,0 )

to compute

in

G

o

on

in

space

fact,

whenever

it is a Montel

/F ) oo

For example,

if

and therefore

space.

S _c ~, and if

with

= Z(S) ± @ Z(~-S)

the dual

Z(S)'

of

.

Notice,

Z(S),

it is

The key here is

Hence

faithfully

~'(Go/Fo) , then the decomposition

~'(Go/Fo)

of distributions

possible.

is mirrored

of opera-

of some infinite-

~'(Go/F o)

are diagonalizable,

~'(Go/F o)

C~(G

{R(N,0 ) : ~ • ~ i }

the appearance

It is only on the larger

~'(Go/Fo).

Z(S)

t • ~ ,

dim(V).

of even the commuting

for the action of

annililator

for all

to

O'. o

is reflexive--in

on

~2 },

= {R(~,0 ) : u •

that the operators

Go

is equal

D

cannot be achieved without

to he able

~ = ~o(t)

is a finite-dimensional

and induction

summands.

C~(Go/Fo)

action of

V

the family

C~(Go/Fo )

Go/F °

that

if

of

Because

g-lh-lgh

= f((~,t)g-lh-lgh).

the reduction

dimensional on

f(~,t)

under

then the commutator

+ o(-s-r)~ + o(-r)~,0).

Corollary:

is invariant constant

h = (~,r),

the reduction by that for the

Z(S) ± C=(Go/Fo)

denotes

the

=

in particular,

that we

and that each of the spaces

Z(S)

is reflexive. We are going to show eventually

Z(~)

is, except

for a very few

closed proper R-invariant

subspaces.

irreducibility

G

of

on

o

m's, irreducible,

subspaces

closed R-invariant

that the action of

in

Z(m)

G

on each of the spaces

in the sense that there are no

that are complemented

Of course, what we shall actually

Z'(w), where

in

Z(w)

by

do is prove

the

the heart of the matter is that the dis-

tributions

(1.19)

1 1 I I f(u't)e()

.

=

One sees from (1.20) hence

the

map

C

r

=

that the single coefficient

that

sends

f

to

a

r

=

is

a

=r

determines

an injection

of

Z(co)

f

completely, into

and

oo COO L n (JR).

10

Let

zt(w)

denote the image of

analytic

description

or not.

Let us look at the simpler

case,

k e I(~),

and hence

If the image

zt(~)

is actually equal

of

C . Our next major objective is to give a precise =r The answer depends strongly on w h e t h e r I(~) = {0}

zt(~).

then

~o(k) = ~

of

C

to

C ( ~ /I(~)).

f(~,t)

is a subspace

=r

of

depend on the choice of base-point

Notice,

r

in

=

I(~) = {0}, the sum in (1.20)

describing

zt(~)

C~(~/I(~)).

Indeed if

Z(~).

first:

a (t) = a

(k)(t) = a (t+k).

Thus

When

zt(~)

I(w) # {0},

a e C ( ~ /I(~)),

= ~ke~/l(~)a(t+k)e()

w h i c h is easily seen to lie in

When

I(~) # {0}

then

a = Crf

where

,

incidentally,

that

zt(w)

does not

~. will be over

g

lies in determining what precise decay

itself, a

and the key to

and its derivatives

r

=

must have at infinity in order to guarantee verges

to a

Ca

function on

G /F . o o

that the right-hand

For example,

the following

side of (1.20) conlemma comes essen-

tially free.

(!o21) Lermna:

Proof:

Let

If

a E C~(~)

l(m) = O, then

It

a

= ~ k=- a(t+k)e(
must satisfy

The following commutation relations must therefore hold:

I

[T,UI] = SlIUI + e21U2 , [T,U 2] = sI2U I + ~22U2

(1.23)

with

o'(0) = (sij).

[UI,U 2] = 0

Now let

monomial in the operators

X

UI, U2, and

linear combination of the form UI(U2T + [T,U2])

be the operator on T; then

X

~h~ j~k ~h,j,kCh,j,kUlU2 i .

C~(Go/Fo)

given by some

is also given by a finite real For example,

UITU 2 =

2 = UIU2T + cI2UI + g22UIU2 •

(!.24) Lemma:

For each triple

(h,j,k)

of non-negative

, h j k fll~. f e C ~ (Go/Fo), __let llfIlh,j,k = IIUIU2T converges to a function

f

in the

C~

integers and each

_Then a sequence

{fm }

topology if, and only if

in all of the norms

P r o o f :

$u2

=

m

- -

!

['llh,j,k"

The equations in (!.22) can be inverted to yield -y(t)U I + ~(t)U2,

C (Go/r o)

{f } converges to '

f

in

a t = T.

aUl

=

6(t)U I

B(t)U 2,

The assertion of the lemma now follows by an easy

induction argument.

We are now ready to look at on

zt(~).

Zt(~) .

Set

Direct computation wil show that if

r Rg=

equal to the operator a e zt(~), then

C ~ R g C -I

12

r= R (u,t)a(x)

(1--.25)

If we set

D.

r Dj, a = limx_+0(R=j(x)a U2, =r

and

-i

= C D.C

+ t)

for each one-parameter This enables

- a)/x.

Taking

Tr .

= e()a(x

subgroup

us to compute

j

of

G O , then

Ul,~'

the operators

r = (rl,r2) , as usual, we get:

lil;rll Ul,ra(t)

(1.26)

~

= 2~i(rla(t ) + r2Y(t))a(t)

t) = 2~i(rlB(t ) + r26(t))a(t)

(~t a) (t) Because

Ul,rhU2 =

JT ka c Zt(~) ,_r_

it follows

integers,

it to belong

to

the quantity

c L °° n C~(IR)

whenever

are non-negative

h,j,k

r

that a necessary

zt(~)

condition

is that whenever

h,j,

a c L

on a function and

k

n C ~ (~)

are non-negative

for

integers,

I iaI Ih,j,k =

(1.27)

suPtel~

I (rl~(t)

+ r2Y(t))h(rl B(T) + r26(t))j~ka(t)

I

should be finite.

(!.28) sisting

Proposition:

of those functions

Then

So(~)

equal

to

depends

a

S (~) o

denote

for which

the subspace

of

all of the norms

only on the choice of

o

(and not on

L ~ n C=(~)

Iiailh,j, k r

con-

are finite.

or even

~)

and is

Z#(m).

We remark mean,

Let

that in saying

of course,

subject

that

$

(~)

to the condition

does not depend on the choice

of

~, we

I(~) = {0}, which is in force throughout

this discussion.

Proof

(of (1.28)):_

prove

the proposition

C0(~)

is dense in

II-Ilh,j, k.

We already know that by showing S (~)

with

(i) that S

(~)

C0(~)

_c zt(~) _c $o (~) "

zt(~)

given the topology

We begin our attack by comparing

the norms on

zt(~): Let

f e Z(m),

and set

a = C f. =r

is c]osed

Recall

that

in

Hence we can

S (~), o

and

(2) that

from the norms Z(m)

with

those on

13

a(t) =

fill

f (u, t) e() dUldU 2

0 0 If in place of

f

in the integral we put

T []a[ ]h,j,k )0.

On the

=

tion

{e() =

Let

=

= e (x)8

~

from

X(u),

then

Because

~ (x)-la

=

eu(X)e= = 0, we have

l#(w)

# ]R

nuX(U)

the subset

If

is again in

= = 0. =

Now

for all choices of

u.

X(u)

a • Zi(w) Z*(~),

of

IR

has its

and since

U =

=

=

0

might be empty, but that won't matter.

disjoint

nuX(U).

must equal

that the distribu-

U =

X(u) = {t E JR: e (x) = 0}.

is discrete--it support

- e(<s,u>)}0

=

It follows

Hence

0__ is supported

in

=

will be empty unless

s = rd(t)

for some

t • IR , and in

=

that case

nuX(U) = t + I#(w).

It follows

that

0 = ~k£i#(~)/i(w)Ok

with

Ok

=

supported

at the single point

an eigendistribution, a scalar multiple

t + k + I(~)

of the corresponding We shall prove

~t+k"

that

•

that

Each

Ok

will again be

~

Thus w e may assume

$ = $0

is a constant multiple

of

=

= 0

One can extend

IR/ I(~).

and therefore we will be done if w e can show that each

ported at one point showing

in

c

implies

= 0. =

to an entire analytic

Ok

is

is sup~

by

=t

The function function of

c

is the key.

z = x + iy.

The assump-

U =

tion that

~ = ~(t)

implies

that

~

has a zero at

z = t.

Let us assume that

U =

can be chosen so that this zero has order one. w i t h a zero at about

t

function

t

and an element

and vanishes

is in

Z*(~).

given any polynomial

that agrees w i t h

outside of some small interval

a : x e--> Cu(X)-ib(x ) =

b E Z*(~)

Then,

about

P

P

in some interval

t, one has that the

On the other hand, w h e n e v e r

17

c E Zt(~), one has

<e c,e> = 0. U

on an arbitrary e

=

vanishes

Thus

=

=

=

e e Z*(~)

depends

= 0.

Since the value of

=

only on the germ of

c

at

t, it follows

on a dense subset of--and hence on all of--{a E Z#(~)

is precisely w h e r e It remains

~ =t

of

that

: a(t) = 0}, which

vanishes.

to prove that

the derivative

0 =

u

can be chosen so that

~, w e have

~xeu(t)

8x~u(t)

# 0.

= 2~ie().

Denoting by Recall

that

=

o(t) = ~(0)a(t), ro(x)

= 0

tion in then

and hence w e cannot have

for all Z(~)

x E I~ , w h i c h is contrary

be constant

~x~u(t)

ro(t)

on cosets of

G'.

= O, for if it were, we w o u l d have

to our assumption It follows

that not every func-

that if w e take

u = ro(t),

~ 0, as desired.

=

Let

w

and

~

Hon~(Z(~),Z(~)) Z(m)

to

be two not necessarily

to denote

Z(~)

f((~,t)-ig). Z(~)

onto

Proof: sum

Le___tt s c ~

operator

Then

L(~,t )

L(0,s )

for all

satisf X

f ~ Z(m)

is in

and

f

HomR(Z(~),Z(n))

We shall use T

g c G . o

that map One w a y

is to use left-translations:

~ = {~o(-s)

on functions

~.

on

Go

: n ~ w}. by

and yields

Define

L(~,t)f(g)

the

=

an isomorphism of

Z(n).

Fix a base point

f(~,t)

TR f = R Tf g g

of

linear operators

that commute with right-translations

(!.34) Proposition: left-translation

elements

the family of all bounded

and satisfy

to get operators

distinct

r

in

=

~.

Given

= ~kar(t+k)e().

f ~ Z(w), one can then expand

It follows

f

in a

that

=

(!.35) Set

L(~,s)f(~'t)

s = ro(-s), =

=

and set

= ~k dr(t= - s + k) e()

bs(t)

= ar(t - s).

One can then rewrite

.

(!.35) in the

=

familiar

form

L(0,s)f(~'t)

Since

~ E n, it follows

that

= ~k hs(t= + k)e(<s__o(k),__u>)

L(o,s)f

e Z(n).

It is now clear that -i

L(~,s ) c HomR(Z(m),Z(~)), carry

Z(~)

and because

isomorphically

onto

Z(n).

L(~,_s)

= L(~,s )

, the map

L(~,s ) must

18

One way of restating

the result of (~.34)

is a subset of the real orbit Z(~))

# {0}.

(1.36) and

n

Proof:

The converse

Proposition: determine

HomR(Z(~),Z(D))

HomR(Z(~),Z(n))

= {0):

v R(~,0),

furthermore,

if

seen

It remains

T £ HomR(Z(~),Z(n)).

operators

then

or

HomR(Z(~) ,

If

It is thus clear

must have

T'~ = 0.

then that both orbits

~o(R)

, then

# DO(R)

, then

is an eigendistribution

for the

will also be such an eigendistribution,

from proposition

Since

Z-orbits

of a single point is

w o ( R ) = no(~)

that if

T'~ # 0, then the eigenvalues

the same.

consists

We may suppose

@ e Z(n)' !

if, and only if, the two

~o(~)

that if

to prove

T'~ c Z(w)

their closed linear span,

then

~o(~) = ~a(~).

wa(~)

We have already # {0}.

Let

# {0}

the same R - o r b i t :

HomR(Z(~),Z(D))

: ~ c w, t £ ~ } ,

form is that if

is also true.

and will be left to the reader.

are non-trivial.

hence

w a ( ~ ) = {~a(t)

The case in which either

easily handled

in slightly weakened

(~.33)

the delta-functions and since

T'

v R(~,0 )

of

on

~

that when

{~t : t ~ ~ }

is bounded,

it follows

and

T'~

and must be

wo(R) # no(R)

, one

have

as

Z*(~) '

that

T' = 0, and

T = 0.

D

(~.37) Corollary: translation

are isomorphic

We shall distinct

The actions

plications

Z(e)

and

Z(n).

going to finish our discussion

in (~.34)

on --

G

This

case.

of

wo(~)

and

Z(~)

= Do(~)

.

defined by rightD

of isomorphisms

existing

between

turns out to be one of the principal Before we worry

about

that, however,

comwe are

HomR(Z(~),Z(~)).

HomR(Z(w),Z(~)) , which

that the operators

Z(~)

to the phenomenon

new to the non-abelian

Consider

G

if, and only if,

return in a moment

summands

of

is an algebra

L(o,t ) for

over

t e I#(e)

¢.

= {t e •

We have already

shown

: ~o(t)

any

= [, r

i

point in

w}

operators

all lie in this algebra.

{L(~,t ) : t E l#(e)}

(~.38) Proposition: when

I#(m)

in

Let

L(~)

generated

by the

HomR(Z(~) ,Z (~)).

L(~) = HomR(Z(w),Z(w))

# ~ , the algebra

be the algebra

HomR(Z(~),Z(~))

if

I#(~)

# ~ .

is commutative

In particular,

and has finite

19

dimension equal to the index of

Proof:

I(~)

The argument is essentially

and count eigenvectors. tributions

~

in

For each

Z(~)'

then

T R(~,s ) maps

T'

with every effect of

on eigenvectors

Since the eigenvectors

The spaces

span

onto

to itself.

v R(o,s),_ we see that

T'

it.

isomorphically E(t)

let

T'IE(t)

E(t-s).

(~.36): dualize

denote the set of all dis-

E(t)

for all

since

T'

also commutes Hence the

is entirely determined by the effect of determines

I(m)

in

and in

T E HomR(Z(m),Z(~)),

! = R(~,_t)o(T'IE(0))OR~o,t ).

A close look at the proof of proposition

£ ]R2--ci.

are all isomorphic, Now if

Furthermore,

Z'(~), T'IE(0)

mension equal to the index of

E(t)

! R(~,0)~ = e ()~

satisfying

E(t)

must map each

the same as that for proposition t e I#(~),

(~.33) and the remarks preceeding fact,

i__n_n I#(~).

T'

on

E(0).

T'

(!.33) will show that

I#(m), and furthermore,

E(0)

has di-

that the operators

{R~0,t ). : t e I#(~)} permute the basis for E(0) corresponding to the delta= functions transitively. It follows that these operators generate a maximal abelian subalgebra

R(~)

of the algebra of all linear operators on

sions

L(~)IE(O) ! H°mR(Z(~),Z(~))IE(O)

that

L(m) IE(0) = R(~), and we are done.

(~.39) Cor___ollary: I(~) = {0}--then R-invariant

Z(~)

If

! R(~)

I(~) = I#(m)--are

must hold.

Thus the inclu-

Direct computation

therefore in particular,

cannot be properly decomposed

shows

if

into a direct sum of closed

subspaces.

A somewhat stronger assertion would be that R-invariant

E(0).

subspace of

C=(Go/Fo).

Z(m)

is actually a minimal

closed

This seems to be a subtle matter to resolve,

not at all amenable to the natural direct attack, which is to use integration. reason for some form of integration Suppose that to prove minimal. X(f)

X

be the subspace of

translates

of

f.

entering naturally here is this:

is a closed R-invariant

Let

f

be an arbitrary, X

subspace of

fixed non-zero

What we want to show is that

Because a typical element of

C (Go/F o)

that one hopes

element of

spanned by the finite linear combinations

is to identify enough elements

in the closure of X(f)

The

has the form

X(f) X(f)

is dense in

X.

X, and let of rightThe problem

to check the desired density.

y~jRg(j)f, _

it is natural

to try to

20

produce elements of the closure of

/G $(g)Rgfdg. The nato ural m e a s u r e against w h i c h to integrate is Lebesgue m e a s u r e dUldU2dt = dg on Go, which, b e c a u s e

o

maps

right H a a r measure on

IR

X(f)

into

G . o

h a v i n g the form

SL(2,1R) , is easily seen to be both a left and a

The real q u e s t i o n is:

can w e make sense of the integral

/~(g)Rgfdg

this a touchy q u e s t i o n is that w h e n

for how broad a class of functions as an element of

I(~) = {0}, the subset

C°°(G /F ), b e c a u s e w h e n e v e r u u

invariant differential o p e r a t o r on

G , we w i l l h a v e

is not a b o u n d e d function of

serious one because,

in view of

g

in general.

/~(g)Rgfdg

W h a t makes

{R f : r ~ G } g

need not be b o u n d e d as a subset of

Ad(g)D

X?

D

of

X

is a left-

D R f = R [(Ad(g)D)f] g g

and

The d i f f i c u l t y here is a

arising as a limit of sums of trans-

lates, w e w o u l d like to b e able to say that w h e n e v e r

8

is a d i s t r i b u t i o n on

G /Fo,

we have

(1.40)

= /~(g)dg

whereas because



,

w i l l in general be an unbounded function of

going to h a v e to impose stringent decay conditions on

~

at infinity b e f o r e the

r i g h t - h a n d s i d e of (~.40) w i l l define a continuous f u n c t i o n of /~(g)Rgfdg

is in

functions of

X

~

X

and

(1.40) is to hold.

Of course, any restriction on what

In order to m a k e all of this a little more con-

crete, let us look at w h a t happens w h e n

~ ~ Co(Go):

~ £ C O (Go), there is no p r o b l e m i n v o l v e d in m a k i n g sense of

R f; w e simply set

R f(h) = /$(g)f(hg)dg.

It is a routine matter,

t r a n s l a t i o n - i n v a r i a n c e of the m e a s u r e

dg, to show that

itself; one simply argues by induction

that if

o p e r a t o r on that

O, as it must be if

w e may use is a serious matter w h e n w e are trying to represent as much

as p o s s i b l e as integrals.

When

g, we are

D

R

maps

/~(g)Rgfdg =

thanks to the C~(Go/Fo)

is a left-invariant differential

Go, then there exists a c o r r e s p o n d i n g r i g h t - i n v a r i a n t o p e r a t o r

D R f = R& f.

into

The s t i c k y point comes in e s t a b l i s h i n g

A

such

(~.40), w h i c h w e shall do

by using R i e m a n n sums: Let F K = G

o

K .

proximate

be a cube in Decomposing R f

K

G

u

that contains

the support of

~

and satisfies

into a disjoint union of cubes of volume

w i t h the R i e m a n n s u m

~, w e can ap-

21

S(~,~)f = ~ n ~ ( h n ) ~

f

(n

traces some finite set).

n Now it is clear that lish to

lim ÷ 0 < S ( ~ , ~ ) f , e >

(~.40), all we need to show is that as R f

in

C~(G /To).

(!.41)

What we w i s h

lim

for every differential

D.

Now

D

operator ~

to show,

has the advantage

that

of the function.

D

and hence,

g • Gg .

(~.42) Proposition:

Let

function with

also for

K.

converge to

argument now yields

(~.41) for

of

C (Go/F d)

RE

on

into itself.

C~(Go/F

)

In particular,

closed R-invariant

subspace

con-

f.

Let

X

the annihilator n

supported in

compact support

~ • C~(Ga) ; then the operator

lies in the smallest

Proof:

and the latter

$.

if

taining

tends

- DR f(g) I = 0

subspace

R f

S(~,v)f

Thus

maps every closed R-invariant f e C (G /Fo) , then

= S(~,~)Df(g),

h --> ~ ( h ) ~ D f ( g )

An easy u n i f o r m continuity

as noted above,

0, the sum

- ~R f(g) I = 0

function

lim ÷01DS(~,~)f(g)

for each fixed

in order to estab-

then, is that

DS(~,~)f(g)

the Riemann sums for a continuous

the integral

tends to

Thus,

~ = ~a ~b ~c. A r g u i n g as w e did in lemma (1.24), uI u2 t with an arbitrary right-invariant differential opera-

is a Riemann sum for the continuous However,

~

0 SUpgeKI~S(~,~)f(g)

w e see that w e can replace tor

= /~(g)dg.

= 0

whenever

b e a closed R-invariant of

X

whenever

f • X.

in the dual of f • X

Hence

and

subspace of

C~(G /F ) . (7 o

h ~ G~.

C~°(Go/Fo), and let

Then

It follows

<X,e> = {0} from (1.40)

e E xi =

implies

that

that

= 0

R X c X ±i = X.

Now that w e k n o w that R carries Z(~) into itself w h e n e v e r ~ ~ ~, w e can r pass to R = = C R C -I, which maps Zi(m) into itself. Suppose for simplicity's ~ ~r sake that

~(~,t) = ~(~)

(t) for some

~ c C0(~

)

and

0 ~ C0(m)

•

Then

22

Rr~a(x)= /G ~(u,t)e()a(x+t)dUldU2dt

(1.43)

d = iff~(u)e()dUldU2}

= ~^(ro(x))[

with

(x) =

(-x).

The subspace

cisely when for any non-zero {x~--> ~^(~o(x))[ stringent

,a](x)

,a](x)

condition that

Z(w)

choice of

,

will be minimal in

a

: ~ ~ C (~2) , ~

and

f (t-x)a(t) dt

Zt(~),

closed R-invariant

pre-

the closed linear span of

~ C~(IR)}

is all of

Z*(~).

The

have compact support make the determination

of

this closed linear span a delicate matter that we shall pursue no further here.

The

notes in section 12 contain some more commentary and references. Let us return now to the phenomenon sible existence of isomorphisms you will recall, and

~(~)

~n

if

~o(~)

many elements of

Let

~

~ ~, let

0 < t < i.

gent sequence

HomR(Z(w),Z(~))

(~.37),

the pos-

even when

~ # D.

exist precisely when the two real orbits

Let us denote by

~

the quotient of

~

If

~(~)

by the relation

Each equivalence

contains

class in

at most finitely

~.

then there is some so that

to

in corollary

= no(k).

(~.44) Proposition:

Proof:

belonging

such isomorphisms

are the same.

introduced

{tn }

t E ~

~ e ~, and let

r ~ ~.

satisfying

~ = ~o(t +

If follows that if in the interval

w=

If

N

g).

is another element of

~,

Clearly we may choose

t

were infinite, we could find a conver-

[0,i]

such that

=r°(tn) ~ Z 2, which is

absurd.

D

The number of elements of phism class Notation:

~

in

C~(Go/Fo),

mul(~IC~(Go/Fa))

will be called the

touch on in section 3.

in

~

is called the multiplicity

or more briefly,

or simply

~-primary

ful role when one worries

~

sum~nand of

mul(~).

the multiplicity The direct sum

C (Go/Fo).

The spaces

about deeper analytic questions,

of

of the isomor~

in

Z(~) = ~

Go/F O. E Z(~) =

Z(~) play a use-

some of which we shall

23

The possibility of having an

~

the uniqueness of the decomposition

with

mul(~) ~ 2

~@w£~Z(~).

has implications concerning

One might hope that

composes uniquely into a sum of irreducible R-invariant subspaces.

C~(G /Fo)

de-

However, suppose

mul(~) = 2: Then Set

Z(~) = Z(~ I) • Z(w2) , and there is an isomorphism

Z+(~) = {f + Lf : f E Z(~I)}

Z(~) = Z+(~) ~ Z-(~) Thus Z(~)

mul(~)

and

as well as

Z-(~) = {f - Lf : f • Z(Wl)}.

provides a measure of the non-uniqueness of the decomposition of Furthermore, since

~ # ~

implies

= {0}, the only sources of non-uniqueness be within the summands

Z(~) and do not result from any interaction between them. are ~oing to compute

2.

Clearly

Z(~ I) ~ Z(~2).

into irreducible R-invariant subspaces.

HomR(Z(~),Z(~))

L ~ HomR(Z(~I),Z(w2)).

mul(~)

In the next section, we

explicitly for several choices of

o.

Some multiplicity computations Let us begin by looking at the example in which

(2.1)

If

o(t) = (~

~ • ~ 2 , then its real orbit is

(rl,r 2 + rlt).

If

{(rl,r 2 + tr I) : t • ~ } ,

since

=r°(t) =

r I = 0, this orbit degenerates to a single point.

degenerate orbits, those for which

r I # 0, are straight lines. - {0}

orbits can be parametrized by line

~)

by letting

x E ~-{0}

The non-

The non-degenerate correspond to the

{(x,t) : t £ • } = ~(x). Suppose now that

r E Z2

and that

=

of the corresponding real orbit

~(rl).

r I # 0.

As we saw in section i, this multiplicity

is equal to the number of integral orbits in then the integral orbit through there are precisely

(rl,s)

We shall compute the multiplicity

is

~(r I) n Z 2

If

(rl,s) c w(r I) n Z

{(rl,s + rln) : n E ~}.

It follows that

IrlI distinct integral orbits, corresponding to

s = 0,i ..... IrlI - i.

Hence

mul(~(rl) ) = IrlI-

This computation shows that, not only do the multiplicities of the various orbits not all have to be

i, they may in fact be arbitrarily large.

2

This phenomenon

of unbounded multiplicities is more the rule than the exception when one goes to

24

deal with more general examples.

What is rather exceptional in this example is the

steady growth of

Irll

function

mul

mul(~(rl))

as

is much more erratic.

grows•

Ordinarily,

the behavior of the

Our next example is, in that sense, more

nearly typical. Consider now the example [cos(~t/2)

sin(~t/2)]

(S.2) o(t) =

The real orbit of the point the circle through

=r

=r £ ~ 2

the positive real numbers

o ( Z ).

r

2

we see that

and

g2,

and let

about

~o(3)

~(r)

I(~) = 4g .

~(r)

and since

r

under

=

~(4)

= ~,

Hence if we let

be the real orbit through

of integer lattice points lying on r

be the orbit of

are all distinct,

2), then the multiplicity of

ber of ways of writing

r : ~, or

Thus the non-degenerate orbits in

and let

consists of 4 points and that

2 2,1/2 r = (r I + r2J r

~ # 2"

if

~+.

r,ro(1), ~ ( 2 ) , ~

if

{~}

centered at zero, which can be parametrized by

be a non-zero element of

Since

of radius

~2

cos(~t/2)

is, in this case, either

with center

this case are the circles in

Let

-sin(~t/2)

~

(which is the circle

is one-fourth of the number

~(r), or in other words, one-fourth of the num-

as a sum of two squares

m

2

+ n

2

with

m

and

n c Z .

One can gain more insight into this result by recasting it in number theoretical terms• [i].

Letting

~

denote

-~fl, we identify

That done, the operator

two elements

~

and

¢

of

~ [i]

a(1)

Furthermore,

~

and

lie on the same

~

they have the same norm, the norm of that

mul(~(r))

PI' • "''P~ aj

~ [i]--i.e. when they generate the

lie on the same ~ = m + in

are the distinct primes congruent to pj

i; hence

~ ( ~ )-orbit precisely when they

~(~) -orbit precisely when

being

is equal to the number of ideals in

is the exponent of

with the Gaussian integers

corresponds to multiplication by

differ by multiplication by a unit in the ring same ideal.

g 2

1

m

2

g [i] mod 4

in the prime factorization of

2 + n .

We thus conclude

with norm and dividing r, then

r

r.

If

r, and if will be a

norm precisely when the remaining odd primes in its factorization all appear to even powers, and in that case there are exactly

~=l(ajj + i)

distinct ideals with norm

25

r.

One can now see clearly the erratic b e h a v i o r of

and

mul(~(5k))

= k + 1

whenever

k

m u l - - f o r example m u l ( ~ ( 2 k ) ) = i

is a positive integer.

E v e n though the actual b e h a v i o r of mul is erratic,

its average behavior is not.

The key is the density theorem of D i r i c h l e t - D e d e k i n d :

(--2.3) Theorem: of

Let

k

be a n algebraic number field, let

k, and for each p o s i t i v e integer

ideals in

I

w h o s e n o r m (relative to

n, let @)

M(n)

where

limn_~oM(n)/n = 2

r I = the n u m b e r of real places of

R = the regulator of

k,

When of n

r2 (2z)

k,

R/wl/dl,

r 2 = the number of complex places,

r I = 0, r 2 = i, R = I, w = 4

as the average m u l t i p l i c i t y out to radius

grows large, the average m u l t i p l i c i t y to radius

w i t h the analogous result for the choice of mul(~(n))

=

Inl

Then

k, and

d = the

k.

k = ~(i), w e get

M(n)/n

n.

w = the n u m b e r of roots of unity in

absolute discriminant of

b e an ideal class

denote the n u m b e r of i n t e g r a l

is at most rI

(2.4)

I

for all n o n - z e r o

o

n

and

d = 4.

We can think

n, w h e n c e

(2.4)

says that as

tends to

~/4.

C o n t r a s t this

in (-2.1), w h e r e w e found that

n E g , and hence the c o r r e s p o n d i n g average w o u l d

be (I

m u l ( ~ ( j ) ) ) / 2 n = (n + 1)/2

,

0 xc

and ~ [~]

the ring of We shall asstme ~ = 2 + /3, over

~ ,

of the underlying additive

26

group of

Z [~].

Our assumption

If

E = a + b~

E

has positive

and E -1, both of which

are positive.

o(1) = M.

that

, then

Consider

then

r ~

~d(n)

2 2 r I - Dr 2.

Z2

Set

correspond

if, and only if,

~ ( g )-orbits

with n o r m

Irl.

~

and of

in

~

PE n.

of

sI + s2~

number of

group

with the element If follows

2

n ~(~)

x~ - Dx~ = r

g2

n ~(r)

If we wish to apply Dirichlet's

case

shall have more

M

are

E

group o(t) with

p = rI + r2 ~

that if

x~ - Dx~ = r

generate is

on which

(-2.5) stems from the non-linearity

lies.

~ [#D].

mul(~(~))

ideals

to that for

(-2.4) becomes

of

o ( ~ )-orbit as

o

theorem in this context, we take

d = 4D, whence

~

the same ideal in

= the number of principal

s ~ - Ds~ =

o(~)-orbit

~(-~), w e see that

analogous

of ~ [ ~ ] ,

=s = r (n), then

in

as in

= the

~ [~]

(-2.2).

r I = 2, r 2 = 0,

limn_~oM(n)/n = l o g ( E ) / / D

It turns out that the erratic behavior of mul in the elliptic hyperbolic

of

theory is essen-

will lie on the same

rI + r2~)

This result is completely

R = log(E), w = 2, and

Go; its multiplicity

of the hyperbola

~

Since the other component

there is a one-parameter

It is not hard now to see that the

~(~)

an element

that the eigenvalues

is chosen as in (-2.2):

to

r = r~ - Dry.

will be the component Furthermore,

o

If w e identify

will

Hence

the corresponding

tially the same as that w h e n Let

norm implies

case

.

(2.2) and

of the orbits in each case.

to say about the bad effects of non-linearity

We

in the next section as

well. In all of our examples not be the case in general, (--2.2). Let

r e ~2,

2 2 2 x I + % x 2 = r. same

as we shall now show.

be a positive

real number,

Our

o

This need

is a slight variation

on

and set

[cos (2~t)

~sin(2~t)]

(-%-Isin (2~t)

cos (2~t) J

O(t)

(-2.6)

Let

so far, the function mul has been unbounded.

and let

2 2 r = r~ + ~ r2.

Suppose now that

o(IR)-orbit.

(Notice that

= r E

Then the g2,

o(n) = 1

o(~)-orbit

and suppose that for all

n ~ Z,

of

=r

= s

Z2

so that

is the ellipse lies on the =r

is its own

27

o(~)-orbit.) follows

that

Thus if

~

whenever

Then r~-

s 2 = %2(s~ - r~).

s~ = % 2 ( s ~ -

r22)

is any non-degenerate %

responding

r~-

is chosen so that G /P

Assume now that

can hold only if o(]R) -orbit,

%2

will be bounded.

is irrational.

rl = +_ s I

then

is irrational,

%2

mul(~)

and

= 4.

It

r2 = _+ s 2.

In other words,

the function mul for the cor-

It is w o r t h remarking

that for any value of

O

whatever,

the manifold

generate solvmanifold Go, namely M

IR 3 .

Go/F °

is diffeomorphic

now)

finite index.

3.

In general,

one expects M

that the mul function

is degenerate,

that the fundamental

group of

M

perhaps

for a solvmanifold

in the sense

(generally

has a normal abelian subgroup of

We shall have more to say about this matter later.

Some analytic

results

W e return now to the setting of section i, in which arbitrary

T 3, w h i c h is a de-

in the sense that if can already be gotten from an abelian

will be bounded only w h e n

speaking,

to the 3-torus

integral one-parameter

subgroup

be to study some analytic properties

o : ~

GO

÷ SL(2,~)

of the projection

is formed from an .

Our main object will

operators

E

: C~(Go/Fo)

÷

Z(~)

W

associated w i t h the direct-sum decomposition theory will play a more

central

peared only as a formal tool. Go/F o.

If you will recall,

invariant

role here W e begin,

~@

~Z(m) of

than in section

C~(Go/Fo).

i, where integrals

a measure

ordinary Lebesgue

gotten by first viewing

S

on

Go/F

ap-

then, by putting a "natural" measure on measure

duldU2dt = dg

under both left and right translation by elements of

sure also defines

Measure

G . o

, the measure of a subset

as a subset of

G

on

S

Go

is

Lebesgue mea-

of

Go/F °

being

and then taking the Lebesgue measure O

of the intersection

of

S

w i t h the h a l f - o p e n

unit cube

Q = {(N,t) e G --

u2,t

< l}--cf.

(1.2).

We shall refer

As the name Haar measure suggests, this measure invariant. invariant measure on

In fact,

G /F

to this measure

the r i g h t - t r a n s l a t i o n

R

g

act as isometries.

G /F °

--

as the Haar m e a s u r e

operators

R

all leave

g

the Haar measure is the unique right-translation

giving mass

1

to

O ~

that w h e n w e form the space

on

: 0 < u I, O

G /F

.

If follows,

in particular,

O ~

LP(Go/F o)

with respect

One of the m a i n problems

a problem by no means resolved

even in dimension

to Haar measure,

of harmonic

analysis

the operators on

3 where we are now--is

G /F -oo to give a

28

useful description the operators projection

of the bounded operators

Rg, g E G O .

operator

E

For example,

(m

bounded operator on all of answered,

on

LP(Go/r o)

that commute with all of

one might ask for which values of

being some fixed element of LP(G /Fo).

~)

extends

C~

p

the

to a

Even this question remains essentially

un-

although it does suggest something of what we meant when we spoke of

studying the analytic properties

of the operator

E , and we will try to give a

reasonable partial answer to the problem a little later,

toward the end of this sec-

tion. Let J

J : C~(G /F ) ÷ LP(G /F )

belongs

to the class

operators

Rg, g c G o )

linear operator on

Mp

maps

C (Go/Fo)

Let

J ~ Mp

into itself.

(depending only on

o

Let

G

and because

and

J)

J Lp

D.f J

Furthermore,

if

in the range UI, U 2

and

1 ! p ! ~T

Then

denote the basis

in (1.22), then there exists a constant

such that whenever

j : ~ ÷

Ga

C

f ~ C=(Go/F ), the inequality

and is equal to

JD.f. J we get an D

the distributional

derivative

DJf

a

Go/F o.

L p function on

the distributional

LPloc(IR3 ) .

norm,

Because

subgroup of as

x

tends to

J(Axf) = A (Jf) x '

the limit limx~OAx(Jf )

Thus, when we form the derivative LP-function,

namely

is equal to

JDf

that whenever

JD.f. J

differential

D.Tf in 3 Arguing by in-

operator on

and hence, in particular, h, j, and

k

G , o is

are non-negative

derivative

The assertions

familiar Sobolev inequality

Lp

is a left-invariant

It follows

converges,

LP(Go/Fo).

is bounded with respect to the

duction, we see that whenever

be a one-parameter

Ax f = [Rj(x)f - f]/x

in the topology of

the sense of distributions,

s in

p

f £ C~(Go/Fo) , and let

0, to the derivative

integers,

can be extended to a bounded

for some

introduced

o

Then the difference quotient

exists in

J

(i.e. commutes with all of the

h j k I IJfl I~ Jf(0,0)

g E Fo, we have

are the corresponding

corresponds

proposition

determines

g e F .

two operators b e i n g their composition.

f £ C ~ (Go/F ~)

G /F . o

=

with a certain space of distributions

of an operator

IoJ

on

f ~ C~(G /Fo).

when it is convenient.

~

distribution

Suppose

satisfying

will said to be a distribution

and

Go/F O.

~.

J ~ Mp

whenever

A distribution

Mp

on

the

o

(ii), when --

for all

identify

be the corresponding

case, w e shall call the function

f

It is clear that

for condition

is a distribution

from (-3.3) that the distribution

the convolution

Proof:

E

¢

J E M p,

leads to the relation:

By analogy with the abelian

(ii)

into

Jf(g) = RgJf(~,0)

It follows

(i)

(--3.2), w e have that w h e n e v e r

Mp

w i l l be negative.

things out on a positive note.

C~(Go/Fo)

onto

Z(~)

We recall

corresponding

that

Our next E

de-

to the direct sum de-

30

(_3.5) Proposition:

The operator

is in

E

M2

and its extension to

W

L2(Gcr/I'd)

is the orthogonal projection of

The distribution

s

corresponding to

E

6O

(-3.6)

Let

[~6oZ(6o). to

f.

0

0

f c C~(Go/Fo), and let

L2(G /FO)

[6of6o be the decomposition of

f6oif

in

f

determined by

[6of6o is absolutely and uniformly convergent

onto the L2-closure of

E

is simply the orthogonal pro-

6O

Z(6o), we need only show that if

L2(Go/Fo).

Let us write out

f6o as

Zkc~/i(6o)a(t+k)e()=

[h(g/i(n)b(t+h)e(<s__o(h),u>).

(3.7)

z(6o).

is given by the formula

It follows that, in order to prove that

6o # q, then

L2-closure of

= ~n~6o ill 1 f(~,0)e()dUldU 2

If you will recall, the sum

jection of

onto the

6O

=

Proof:

L2(Go/F )

and

f

as

Now formally at least,

fGo/Fo

f6o(x)f (x)-dx

= IQ[kZh a (t+k)b (t+h) -e (duldUmdt rl

rl rl

= Zk~hJ0a(t+k)b(t+h)-J0J0e()duldu2dt

=0

because

~o(k) - ~o(h) # 0

for every value of

are both non-zero, then the sums over

h

and

h k

and

k.

If

However, if

I(q) = {0}, we must justify the change in the order of summation. I(6o) = {0}, so that the sum over

k

and

I(q)

are finite, and these formal

manipulations are perfectly justified with no further ado.

that

l(w)

in (-3.7) is infinite.

I(6o) or

Let us suppose We can then use

the computations in proposition (~.28) to deduce that there is a constant

c > 0

such that

I(q),

unlike

]a(t+k) I ! c(l+k2) -2

0 < t < I.

I(6o), is non-zero, then the sum over

the bound

[a(t+k) ]

gent when

0 < t < i.

stant

whenever

c

so that

c(l+k2) -2, the sum If both

I(6o) and

]b(t+h) l ! c(l+h2) -2

h

If it happens that

in (-3.7) is finite, and in view of

~kZha(t+k)b(t+h) I(q)

is absolutely conver-

are zero, we can rechoose the con-

holds as well as

Ia(t+k) ] ! c(i+k2)-2; the

3~

absolute convergence of Fubini's

Ik~ha(t+k)b(t+h) -

in this case then follows immediately.

t h e o r e m can n o w he used to j u s t i f y the m a n i p u l a t i o n s in

It remains to prove that the d i s t r i b u t i o n

e

(~.7).

c o r r e s p o n d i n g to

E

has the

form d e s c r i b e d in (~.6), and this is easily done:

= E f(~,O) = f (~,0)

= ~k~/l(~)a(k)

ill

f j

f(~'0)e(

(ii)

Y

is a sequence in

~ ~ M p, it follows to

In other words, ~ (i).

that F.

is continuous.

the definition of measures:

c ~ 0, one has requiring

(i).

Suppose,

case it makes sense to talk about the graph C0(G /Fo).

f

Thus the only implications

(iii) and (iv) implies

Let us begin with

{fn*~}

on

are equivalent:

(ii)

Proof:

~

of the operator Y

convergent

to

f, whence

{fn,g} Hence

f~--> f*$

then, that (iv) holds, in which f~--> f*~

on

to some point {fn }

converges in

(f,F)

converges in Lp

to

f,~

F = f*~, which implies that is a bounded operator on

Lp But

Y

is

C0(Go/F ).

35

It remains to prove that (i) implies (iii). sure.

Co (Go/Fa).

functional on a constant p

~

is a mea-

In fact, we may assume without loss of generality that as a set-function,

is real-valued and non-negative.

-I

We assume, now that

C

P

> 0

Set

I I~II = /id~ = the norm of

~

as a linear

The estimate we are going to establish is that there is

such that whenever

f ~ C _ (G _ /F o)_

and

g ~ Lq(Go/F o)

q-1 = i, then + (3.14)__

ISg(y)If(xy)d~(x)dyl -- 0

gives rise to a

ew e Mp.

such that w h e n e v e r

Then w h e n e v e r

F E

LP(Ga/F ~) , we will

have

IIF,~oil ~ : fQIIPdudt i =

0 [ I (R(0, t)F)[ (A/Fo ~A)*c@I

i

C

0

It seems likely a bounded

that when

translation-invariant

but not for all

p

4.

is hyperbolic,

operator

in the open interval

p r o b l e m in abelian harmonic also be interesting

w=

llR(o,t)FII~dt =

analysis,

to know w h e t h e r

on

[pPdt

P ! C]IFII p

D

the distribution

LP(A/F

n A)

1 < p < 2.

for some

This, however,

g@

will define

p

other than 2, is really a

and we shall pursue it no further.

the converse of proposition

It w o u l d

(3.21) is true.

The arithmetic nature of multiplicities The computation

of section

2

showed that the behavior of multiplicities

intimately linked to number-theoretical same idea from a somewhat our aim is to reduce

different

the computation

W e are going to pursue the

point of view in this section. of multiplicities

that are local in the sense of algebraic more precisely,

considerations.

number

theory.

is

Roughly

speaking,

to a family of computations In order to put the matter

we are going to have to fill in some background.

Working with

G

41

for

o

arbitrary w o u l d clutter up this e x p o s i t i o n more a n n o y i n g l y than the gain in

generality is worth. two special cases.

Note:

T h e r e f o r e w e are going to work,

The first case is the "parabolic" case

~(I)

o(t) =

The second case is the "hyperbolic"

here w e assume equality, not similarity.

case in w h i c h

for the time being, only w i t h

is

[o: :] where

D

is a positive,

square-free integer ~ 2

the fundamental unit greater than a _+ b F D = s-+i . and hence

s

-i

(The c o n d i t i o n

1

in

det(o(1)) = 1

G

modulo

4, and

The eigenvalues of implies that

s

a + b/D

~(i)

is

will be

has positive norm,

can be v i e w e d - - w i t h a small caveat to be explained later--as

o

the real points of an a l g e b r a i c s u b g r o u p (~,t)

3

= a - bF~.)

The groups

let

~(~).

or

G

of

GL(3)

defined over

~ ; w e simply

correspond to

(!.l)

i

yit)

~(t)0

u:I

N o t i c e the difference b e t w e e n this r e a l i z a t i o n of ginning of section i.

G

o

and that given at the be-

There w e had to deal w i t h an arbitrary o

matrices to guarantee that w e got a faithful r e a l i z a t i o n of w h e n the image of

o

in

SL(2,~)

was compact.

Here our

Go o

and n e e d e d

4 x 4

in all cases, even is narrowly con-

s t r a i n e d - - t h e r e is no p r o b l e m about (4.1) b e i n g faithful h e r e - - a n d

(~.i) has the

advantage that it is "rational" in a sense we shall make clear in a moment. First let us look at the p a r a b o l i c case.

01t 0 01

There

(4.1) actually takes the form

42

Thus the algebraic

generic

group

element of

=oG

GL(3)

(4.2)

in this case is defined by letting

and t a k i n g

the equations

defining

(Xij)

G =~

denote a

to be

Xll = x22 = x33 = i, x21 = x31 = X32 = 0

In the hyperbolic

case,

things are somewhat more complicated.

The form taken

by (i.l) is

I and thus the equations

s(t) D B(t)

B(t) e(t)

0

0

defining

G=~

ui

are 2

(/,2')

2

X21 = DXl2 , X 1 ] -

Xll = X22,

DXl2 = 1

x31 = X32 = 0, x33 = 1

One can see that in this case we would be in trouble if we were using tion of section ~. from it: t

~(t)

is actually

appear in the

Thus

Xll

and

4 x 4

x34

not by algebraic

G . =o

~(t)

( t + -t)/2

realization

are related

is not a rational

with

function of

e = a + b~--whereas

of section ~,

on t h e

(i.e. polynomial)

W e mentionned points of

The problem is that

C~-variety

~(t)

as

defining

the realiza-

×ii

G~

both e(t)

and

in

t--far

t

as

and X34.

GL(4,N) , b u t

relations.

at the start that there was a caveat about

The p r o b l e m occurs in this case, namely

G

being

the real

that there is a solution

li0ii10 of the equations solutions

for

(~.2') not lying in

(4.2').

Then

indeed it is easy to see that factor b e i n g the spurious no immediate ~o"

importance

G

Go .

contains G

= G

Let G

G o, ~

denote

the family of all real

as a normal subgroup of index 2--

× ( ~ /2g ), the generator

solution just shown above.

This

of the

"expansion"

of

~/2g GO

is of

to us, as our main concern will be with the p-adic points of

We shall return to the matter later.

43

Let

p

be a prime in

hyperbolic; and let

G

Z;

let

o

denote the p-adic points of

o ,p

moment to examine the structure of Ap

denote the closure in

s e m i - d i r e c t product p

again be arbitrary, either parabolic or

Go,p

G

of

G . =O

for various choices of

o,p

{(u,0)= : =u ~ Q2}.

A

o

and

We can realize

A "B , in w h i c h the n o n - n o r m a l factor P P

in a more c o m p l i c a t e d w a y than

W e are going to take a

B

p.

Let

Go, p

as a

depends generally on P

does, at least in the h y p e r b o l i c case.

In

P the p a r a b o l i c case, the natural choice for

B

is the subgroup P

lit if01 H e r e the situation resembles

t • ~p

.

the real case as closely as one could hope.

an equally n a t u r a l choice for

B

in the h y p e r b o l i c case, namely,

There is

the matrices

P

(_4.4)

a

,

a,b • ~p

and

a 2 - Db 2

1

0 There are two quite different forms

can assume, depending on w h e t h e r or not

B P

lies in G

~p.

Suppose to b e g i n w i t h that

/D

does lie in

~p.

The w h o l e group Let

can be u p p e r - t r i a n g u l a r i z e d in this case, as w e shall now show.

~

de-

~,P note the m a t r i x

-~

(_4.5)

0 Notice that that in

det(~) = -2/D, and since

~ e GL(3,Qp). Qp

is for

lie in the group matrix

~

D

Actually,

is in

g x P

will

~p

by assumption, we conclude

one can say a little more.

to b e relatively prime to of units in

~

.

p; thus, if

The only w a y ~

is in

It follows that except w h e n

/D

can lie

~p, it must p = 2, the

P

is already in

havior at p = 2

/D

GL(3, ~ p )

and not just in

GL(3,~p).

The singular be-

cause us no real trouble, it w i l l merely require us to do some

special p l e a d i n g in one or two places. To return to the m a t r i x

Let us assume for the moment that

~, one can v e r i f y directly that

p

is odd.

44

(4.6)

with

~

1

t = a + b/D,

element given

t

of

Ii

~

1

i]

:i°

~

t

0

0

t -I = a - b/D, and

u I' and

u 2' some elements

from

that appears in (4.6) can in fact be arbitrary,

t, we can set

is in

a

a = (t + t-l)/2

9p, the similarity

and

x ~--> ~

-i

b = (t - t - l ) / 2 ~ .

• x • ~

defines

~p.

The

for if we are

It follows

that when

an isomorphism of

Bp

on-

X

to

~p. Let us assume now t h a t

The field t-

~p(/D)

to denote

~

does n o t l i e

is then a genuine

the conjugate

of

t

extension

element

side of

x

of

(4.6) will be in

must satisfy

satisfying

of

~p

of

t

t- = t -I

degree

except

in

We will use

(~) ~

.

GL(3,~p).

B

Now an

t £ ~p(/D)

It therefore

follows

for some element of

if we are given an element

t- = t -I , we can get an element of

Notice

and the

9p, can have some

is eigenvalue

Conversely,

2.

that

instead of in

as an eigenvalue.

t c 9p(/D)

p , a s a b o v e , i s od&

for all t c ~p

situation,

GL(3,gp(/D))

only if it also has

from (4.6)_ that if an element t

and t h a t

Go,p, being a matrix w i t h entries

as an eigenvalue

then

~p

in that extension,

that (4.6) remains valid even in our present right-hand

in

having

t,t

, and 1

Bp,

t £ ~p(~) as its

P eigenvalues Hence

B

just as before, by taking is isomorphic

a = (t + t-l)/2

to the closed subgroup

of

~

P

Furthermore--and n GL(3,~ )

relative

to

~p, is equal to

of

B

is dense in

B , an immediate P

P in

~

i.

this is crucial in the computations

P ~[~]

[~]x

b = (t - t - l ) / 2 { ~ consisting

.

of those

P

elements whose norm,

B

and

In particular, to follow--the consequence

Bp

is compact.

subgroup

of the density of

[~]. P

Consider n o w the case being

2 or 3 mod.

4,

D

p = 2.

Here we cannot have

is not even a square

Hence we need only worry about the case just as it did for ments apply.

p

odd.

In fact,

~

except

for a r b i t r a r ~

t

~p, because,

~ 92, and there everything for one point,

satisfying

D

mod. 4, let alone a square in

The one point is w h e r e w e want to set

b = (t - t - l ) / 2 ~

~e

exactly

a = (t + t-l)/2

t- = t -I

92.

turns out

the same arguand

The problem,

of course,

45

is that n e i t h e r

~

assertion,

B 2 n GL(3, ~)

that

a

that

and

Q2(¢~)

b

nor

b

obviously has to lie in is dense in

do indeed lie in

for some

and hence our density

B2, does not yet follow.

£ 2 ' and the gap w i l l be filled.

must be a r a m i f i e d e x t e n s i o n of

t = 1 + ~(t 0 + tl~)

£2'

to

~2' and hence

and

tI

in

t- = t -I

We w i l l show

The key is that implies that

~2' whence

t + t-I = t + t - = 2 + 2Dt 1 and t-

Hence

a = (t + t-l)/2

and

t

-i

= t - t-

2~t 0 .

=

b = (t - t - l ) / 2 ~

is compact, is i s o m o r p h i c to the subgroup of tive to

~2' and has

B 2 ~ GL(3, £ )

lie in g2[~]

£2"

x

We conclude that

B2

of elements of n o r m 1 rela-

as a dense subgroup.

That is about all we n e e d to k n o w about the structure of

G

o,p

.

We can now

turn to the real question, w h i c h is f o r m u l a t i n g the p-adic version, mul

, of the F

m u l t i p l i c i t y f u n c t i o n mul.

Our approach w i l l be to define

as a function on the orbit space

~ =~2/o(g).

-mul -p

Next we w i l l establish by direct

computation the b a s i c redation, w h i c h is the p r o d u c t formula for all

~ e ~.

at first formally

mul(m) = ~ mul (~), P P

That done, w e w i l l return to the p r o b l e m of i n t e r p r e t i n g

mul

(~) P

in terms of the sort of h a r m o n i c analysis w e did in s e c t i o n i. If y o u w i l l recall, the function mul w a s c o m p u t e d in the real case as follows: W e w e r e given at the outset the map the left on

~2 .

W e fixed a p a i r i n g

and got a right action of • ~

and an element

R

o, w h i c h w e used to make

_,_~> = UlV 1 + u2v 2 ~

is a homomorphism

from

SL(2,~)

into

Aut(N).

Proof:

Suppose that

G'

and

o" are the elements

[~' y'

of

SL(2,~)

, and

suppose

further

~B'' 1

that

and

[~" ¥,'

Go'

B"] ~,,j

= o".

We w i s h

to

prove

that

G ==o '

= G".

Now

(~)~'

with

= (xa" + yy", xB" + y~", z + f(x,y))

f(x,y) = qo(x,y) + qo,(xa + yy, xB + y~).

It thus follows

from (5.5) that

qo,,(x,y) = qo(x,y) + qo,(xa + yy, xB + y~) + rx + sy

for some real

~

and

qo,,(x,y) - qo(x,y) whereas

rx + sy

~.

What we must show is that

- qG,(x~ + yy, xB + y6) is homogeneous

r = s = 0.

is homogeneous

But this is clear:

of degree 2 in

x

and

of degree i; thus the two can be equal only

rx + sy ~ 0, i.e. r = s = 0.

Let

t~-->o(t)

e SL(2,~)

be a one-parameter

group that is integral in the

59

s e n s e of section ~, namely,

o(t) • SL(2, ~ )

whenever

t • ~.

just c o m p l e t e d above then yeilds a o n e - p a r a m e t e r s u b g r o u p Let

S

denote the group w h o s e u n d e r l y i n g m a n i f o l d is

o

The construction

t ~---> ~(t) of

N x ~

Aut(N).

and w h o s e group

o p e r a t i o n is

(5.7)

(n,t)(m,s) = (n(mo(t)),t + s).

One might be tempted now to believe that the subset points

((x,y,z),t)

w i t h all coordinates in

~

A

o

of

S

consisting of those

o

is a subgroup of

So, as before.

One must be a little careful, however, b e c a u s e of the (unavoidable) that appears in the form

qo(t)"

In order to guarantee that

must impose the additional condition that w h e n e v e r SL(2, ~ )

o

~B ~ y6 E 0 mod.

~(r)

Let us assume,

Specific examples w i l l

In fact, the "unit cube"

{((x,y,z),t)

R

o

coset in precisely one point.

of

So

on

(5.8) holds is

so that

A

'

The quotient

A

where

SL(2, £)

S . o

every right

SL(2, g )

is both integral and s u i t a b l e

then, that

indeed a discrete subgroup of

C=(So/Ao)

S /A --right cosets, as usual--is go ~ S

o

: 0 < x,y,z,t < i}

meets

Our object is to study the repre-

b y right translations.

trudes itself into the analysis critically from the start. duce analogues for

So/A °

of the models

p r o b l e m as w e did in sections 2 and ~. abelian subgroup

A

The general situation

Zt(m)

i.

in-

The p r o b l e m is to pro-

and then to solve the m u l t i p l i c i t y

W h e n w e dealt w i t h

Go, w e had the normal

for the commuting family of operators

{R

x

: x E A}.

So, we have available a b i g enough a b e l i a n subgroup only in

w h a t must b e c o n s i d e r e d the degenerate case: o(t) are

Su

to w o r k with, w h i c h had the effect of reducing our p r o b l e m to an

exercise in spectral theory Now in dealing w i t h

is o

is exactly like that of section -i, except that the c o m p l i c a t e d structure of

of

of

2 .

We do note that the subset of

a n o n - n o r m a l subgroup of i n d e x 3 in

sentation

r e ~ , the element

does have this property, w e shall call it suitable.

be considered later.

compact.

is a subgroup, w e

must satisfy the congruence conditions

(~.8) When

k°

1

factor of

where

for all t • ~ , b o t h eigenvalues

In all other cases, we must use an e n t i r e l y different a p p r o a c h to

60

the whole problem. Before stating the first main result, we have one reduction to make, and that we shall do now. A°

Notice that if

is central in

other words, f

So.

Hence whenever

which

Y(n)

f c C (So/Ao), we have

f

R (k)f = f"

in a Fourier series in the variable

denote the subspace of

C (So/A o)

f((x,y,z,t) = e(nz)f((x,y,0),t)

C~(So/Ao)

~(k) = ((0,0,k),0)

is periodic of period 1 as a function of its z-coordinate.

follows that we can expand let

k c g, then the element

= ~n=_=Y(n).

C=(So/Ao).

R-invariant subspaces of each So/{((0,0,z),0)

: z ~ ~}

naturally isomorphic to

It

Thus if we

for all

((x,y,z),t)

E

f

for

So, then Y(n)

is a closed

Our problem is to describe the minimal

Y(n).

One case we don't need to bother with:

is naturally isomorphic to C~(G/Fo),

In

consisting of those functions

Direct computation also shows that each

R-invariant subspace of

z.

of

G,

the subspace

Y(0)

since is

whose irreducible R-invariant subspaces we al-

ready know.

(~.9) Proposition: Y(~), n # 0.

Then

the restriction map

H

Let

H

be an irreducible closed R-invariant subspace of

is already irreducible under the action of f e--> fi N

carries

closed R-invariant (I) subspace of

H

N.

Furthermore

isomorphically onto an irreducible

C~(N/N n A ). G

This proposition and its proof are central in our analysis.

One should notice

the remarkable contrast between this result and the corresponding situation for G ,F°

and

A.

assertion for summands

If one views C~(Go/Fo)

A

as the counterpart of

would be that

A

N, then the analogous

acts %rreducibly on the non-trivial

Z(~), which, of course, is as far from the case as possible.

that the restriction map from

Z(~)

into

C~(A/A NFo)

Notice also

is not one-to-one either.

In fact the situations here and in section i are dual to one another in the sense of Frobenius reciprocity, which puts restriction and induction in duality. The proof of proposition

(5.9) is a bit involved and is perhaps best understood

by exploring some of the implications of its conclusion.

Let us assume for the

(i) R-invariant here and later simply means invariant under right translations by elements of whatever group is appropriate.

61

moment,

then, that the proposition is true, and let us denote by

map from

H

into

C~'(N/N nA ). o

ducible closed subspace on

H

and

also use

of

K--in fact, given J

J

carries

C (N/Nn Ao). f c H

and

Now

H

N

the restriction

injectively onto an irre-

acts by right translation both

n c N, one has

J°R

~, = R oJ.

to pull back the operators

R(o,t ) to K, where t ~ ~ and = We simply take the operator Pt on K to be JoR(~,t)oJ -I.

(0,0,0) c N.

k c ~, the operator k c g

K

The map

J

implies

Pk

can be computed explicitly.

We can 0 = When

First observe that, since

(~,k) c Ao, we have

R(o,k)f(n'O)

= f((n,O)(O,k)) : f ( ( O , - k ) (n,O) (O,k)) = f (no__(-k) ,0)

for all

~ e N

says about

and

K : K

f c H.

Hence if

F c K, then

Notice what this

must be invariant not only under the operators

all of the operators

Pk

as well.

In the proof of proposition

to turn this whole chain of reasoning around. the image in

Pk F = Fo~(-k).

C~(N/N nAo)

of

Y(n)

R , but under n = (5.9) we are going

We begin by showing that if

under the restriction map, then

Z(n)

Z(n)

is

decom-

poses into a direct sum of irreducible closed R-invariant subspaces each of which is also

Pk-invariant for every

spaces of

Z(n)

k e ~.

We next show how to construct from these sub-

certain irreducible closed R-invariant subspaces of

have the properties asserted in proposition

(5.9).

Y(n)

Finally we shall prove that this

construction yields all of the irreducible closed R-invariant subspaces of The subspace satisfying

Z(n)

of

C~(N/N p Ao)

f(x,y,z) = e(nz)f(x,y,O)

denote the subspace of

(~.i0)

Z(n)

containing Z(n,m)

n # O.

(x,y,z)

in

N.

f

f(x,y,z) = e(nz + my) ~ = _ = a ( x + k ) e ( n k y )

.

~ = (n,m).

Z(n,m)

would be

Z(e)

with

~

f

We recall that we

For each integer

consisting of those functions

In the notation of section ~,

Y(n).

consists precisely of those functions

for every

are only interested in the case in which

that do

m, let

Z(n,m)

that have the form

the element of

Combining the results of sections 1 and 2, we have that each

is an irreducible closed R-invariant subspace of

Z(n), and that

62

Z(n) = Z(n,0) the map

C

Z(n,m)

n,m

• Z(n,l)e...¢Z(n, lnl-l). that assigns

isomorphically

fer the action of

N

to

f

Furthermore,

in

(5.10) --

onto the Schwartz on

Z(n,m)

to

the Fourier

space

S(~)

by virtue of proposition

S(~)

.

coefficient

When we use

, we find that

a -C

(1.28),

carries

n,m

Cn,mR(x,y,z)Cn,m

to trans-i

is

the operator

(5.11) _ on

R m'n (x,y,z)a(t)

= e(n(z+yt)+my)a(t+x)

S(~). If you will recall,

invariant

subspaces

composition construct

of

~eZ(n,m)

now

Tka(x) and with

Z(n)

k e £ , let

U

from

and

Pk

Tk

f

Z(n)

{--el..... $1nl }

(5.13)

the effect of

TkRm'n

--

(x,y,z)

having a tractable S(~)

form.

given by the rule C

n,m

~fm

® !m+l

for

Un

on

=

R m-k'n

C Inl R(x,y,z)

(x,y,z)Tk

w e note that

"

now yields

UnR(x,y,z)Unl

Pk

that will carry

Z(n), we can w r i t e it uniquely as a sum

is the standard basis

In order to compute

The operators

S ( ~ ) ¢ C Inl

The de-

We are going to

is defined in terms of the operators

Unf = ~I~!ol[TmCn,mfm]

(5.14)_

onto

be the operator on

of

k ~ ~.

to this task.

to operators

U

for all

closed R-

We shall set

(-5.12)

Direct substitution

Pk-invariant

is not well suited

The operator

Given an element

fm E Z(n,m).

where

of

goal is to exhibit irreducible

that are also

R(x,y,z )

= a(x - k/n).

T k.

Z(n)

an isomorphism

both the operators For each

our immediate

= R0'n(x,y,z)

®

do not transform quite so simply, but we do have

the following

proposition:

(--5.15) Proposition: Wn

of

~

on

There exist

¢ Inl such that for all

representations k e £,

Qn

of

on

S (IR)

and

63

UnPkU~l = Qn ® n k Wk "

Furthermore,

the representation

Wn

is completely reducible.

Let us take a moment before proving the lemma to comment on its use. be a closed subspace of when

Z(n).

U X = S(~) ® X' n. .

By virtue of (5.14),

for some subspace . .

irreducible under the action of

X'

is

Wn

invariant.

decomposes into a direct sum

~_X'

the corresponding decomposition invariant,

and irreducible under

step in the proof of proposition Proposition z t (~).

Let

X'

non-zero vector the map

a-->

~kx"

~ k -x:

U~I(s(~) H

of

m

X

X

is, in addition,

dimcX' = i.

Proposition

is invariant under the operators

Since

Wn

is completely reducible,

of one-dimensional

~Xm

is R-invariant precisely

¢Inl~ and

N if, and only if,

completes this picture by showing that and only if,

X'

X

Let

of

Z(n)

(5.15) Pk

if,

¢ Inl

• W -invarlant subspaces, and n

has its surmmands R-invariant, P-

N--exactly what we need to accomplish the first (5.9).

(5.15) also provides us with a candidate for a model analogous to W n -invariant one-dimensional subspace of

be a

x ~ X'. a®x.

We can then identify

Since

X'

When we pull back to

S(~) @

is wn-invariant S(~)

,

X'

¢ Inl , and choose a

with

there is some

S(~) ~ ~

the action of the subgroup

® X'), we get a representation of

i N

--

on

S(~)

CX

by means of such that

i° N

of

So

on

that we shall call

o

vn($); it has the form

= ~-kRO,n

(5.i6)

vn(~)((x,y,z),k) a

(The minus signs appear because

Pk

N/N n A

o(k).)

defined by

to the the homeomorphism of

vn($) comes about as r = close as one could hope to having the simplicity of the canonical form R of section !.

o(-k), not

corresponds

n (x,y,z)Q-k a"

The representation

There are, however, some problems.

only represents

A N

and not all of

S :

First, of course, is that

we have an extension problem to solve.

That will be our next order of business after proving to compute Z(n)

Qn

(~.15).

(5.15).

explicitly enough to justify working in

where it all started.

vn(~)

S(~)

Another problem is rather than back in

This problem will be solved in the course of proving

One final problem is to compute the possible values of

~, once the

64

representation

Qn

has been determined.

mining precisely w h i c h irreducible The solution

This will

representations

turns out to be rather tacky;

turn out to be crucial in deterof

S

actually occur in

o

it will be discussed

The time has come to look at the proof of proposition is essentially proving

A

in

SL(2,£),

a more general result

SL(2,£ ) (5.8).

arbitrary

of matrices

It is an easy exercise

is generated

in section 6.

(~.15).

Because

o(k)

the proof of (5.15) is greatly simplified

that brings

satisfying

Y(n).

into play the whole subgroup

the congruence

in elementary

conditions

A

of

~B ~ y~ ~ 0 mod.

row and column operations

by

2

from

to show that

as a group by the two elements

Both these elements

turn out to be much easier

to handle

than the general element of

A, and we will only have to work with them in the end. In order to frame the general fines an automorphism and hence defines

=~

of

an oeprator

N

result, we note that each element

that carries Po

on

N N A

isomorphically

C~(N/N n A G )

by

Po(-k)

parameter

esch of the operators

S .)

Of course,

if

P f = fo~.

Pk of (5.15) is, in the present notation, group defining

o

t I

of

A

de-

onto itself (The operator

> o(t) is the oneP

maps

Z(n)

into itself.

(5.15') Proposition: Wn

of

A

on

¢ Inl

There exist representations

such that for all

there is an inner product on

¢ Inl

T ~ A,

with reject

Qn

of

A

UnP U~I = QTn ®

o_nn S ( ~ )

Wn .

and

Furthermore,

to w h i c h the operators

Wn T

are

of

Z(n).

all unitary.

Proof:

The operators

Hence w h e n e v e r

_rr E C Inl

up to scalar multiple, follows that mined by

P

o

permute the irreducible is non-zero,

such that

there exists

P U-I(s(I~) ~ n

U P u-l(a ® v) = b(o) ® w(o) n ~ n ---

a,v,o,

and the choice of

w_(o) (or, equivalently,

b(o))

R-invariant

w_(o).

with

subspaces

a non-zero

Cv) = U-I(s(IR) n b(o) e S(IR)

w_(~) £ C Inl, unique ® Cw_(o)).

It

completely

deter-

Our p r o b l e m is w h e t h e r we can choose

systematically

so that

Qon a = b(o)

and

wnv w(O) o-- = --

85

are representations because when Inl > 1

of

4.

n = ~ 1

It turns out to be simpler to normalize

we can see exactly what

~i

b(o)

first,

has to bej and generalizing

to

is easy.

Consider,

then, the case

n = ~ i.

Since

Z(! i) = Z( ! l , 0), we can define ±

= C±I,oP~C±-~ ,0

Q~I

for all

will be helpful in handling explicitly. n = -i m = 0

W 1 = 1

for all

~ c A.

the general case to compute the operators

In order to avoid confusion, we will work only with

being similar. here,

~ c 4, and of course

Our starting point is (~.i0), which, since

It

Q~I± more

n = ± I, the case n = 1

and

takes the form

f(x,y,z)

We can recover

a

from

(5.17)

f

= e(z)~k =_~a(x+k)e(ky).

by the formula

a(x) =

f(x,y,O)dy. 0

Now let

~ c 4.

Substituting

(5.18)

Written out, P f o

looks like

P f(x,y,z)

= f(x~+yy,

of

that

z+qo(x,y)).

into (5.17), we get 1 Q~a(x) = foPof(x,y,O)dy

The evaluation example,

xB+y6,

1 = f f(x~+Yy,o

xB+y~,O)e(q~(x,y))dy

of the final term in (5.18) is not difficult.

r e ~

and that

T(r)

.

Suppose,

for

is the element

A; then

(A.19)

Q$(r)a(x)

1 2 = Iof(X,y+2rx,O)e(rx )dy

= e(rx 2)

fl

f(x,y,O)dy

0

= e(rx2)a(x)

because

f(x,y,O)

is periodic of period i in the variable

y. Less obvious, but

66

perhaps more interesting, is the "Weyl" element

Here the situation is

Qla(x) = Iif(y,-x,0)e(-xy)dy

(5.20)

= Ii~k=_

a(y+k)e(-(y+k)x)dy

k+l = ~ a(y) e (-yx) dy k= -~ Jk

i

= ;_~: (y) e(-yx) dy ^ =

Thus

QI

a

(x)

.

is the ordinary Fourier transform.

Since

could, in principle, consider our computation of case it is easy to see what Wn

1 Qo

and

complete.

We remark that if

generate

y # O.

A, we

However, in this (It is only

~ ~ A, then either o = T(r),

o = -T(r) = ~2T(r), or

~

The final answer, when

y # O, is that there is a constant

i

~

is directly, and we shall do so now.

that is a computational horror.) has

Q1

T(1)

We are going to treat the case u(o)

y # 0 now.

of absolute value

such that

(5.21)

Q~a(x) = u(o) IYI2

a(x~+yy)e(q (x,y))dy .

The argument that leads to (5.21) is a variation on (5.20).

It begins with a slight

rewriting of the second line of (~.20):

Q~a(x) = ( l l l Y ] ~ a(xa+(y+k)y+j)e[(x~+y6)(ky+j)+qo(x,y)]dy JO j=l k = - ~ Using the same argument that got us from line 2 to line 4 in (5.20), we get here that

Qa ( x ) l

In the

j-th

= cj=l

~-~a(xa+yY+j)e[(xB+y6)J+q

s u m m a n d , we make t h e

change

of variables

(x,y)]dy

.

Y ~ Y _ y-lj,

which,

after

67

some arithmetic,

leads

Q~a(x)

This is essentially

to

= {~l~e(~j2/2y)}f_y(x

(~.21).

All that remains

has absolute

value i.

There

first approach

c A, the integers Hence the quantity known

~

are two ways

is through and

G

in (5.22)

the classical

are relatively

theorem for

L2(~)

(5.23)

that,

a'

extends

for all

morphism

of

sure on

N/N n A

operator on

analysis,

having

If follows

Combining

measure

tion

nn

in itself

ab-

A

on

then

.

Recall now from lemma CI, 0

satisfies P .

llCl,0f]12

Since

o

of

will satisfy

N/N n A a

the unitarity i.

=

is an auto-

is the only translation-invariant

Po

value

(3.9)

mea-

defined by

...IiPoflI2 = ..IifiI2 = IlalI2

o=

for all

for all

of the operator

in (~.23), we see that

1 Qo

operator

That

is a unitary

is of

and will be used later.

operator

the factor of of

2.

is classically

than simply having

IiQlaiI2 = IiCl,oP ClloaII2

in (5.22) has absolute

the dilation

include

y~ e 0 mod.

that the operator

i, the homeomorphism

Thus

that

this with

L2(~)

We are now ready to look at the case define

Since

in the form of the Planeherel

next the operator

total mass

measure.

on

N/N nAo,

Consider

N, and since Lebesgue

).

interest

sums.

a (xa+yy)e (qo (x,y)) dy

measure

f e Z(1).

f E C~(N/N N A

the quantity

sum whose value

is much better

is to use Fourier

to a unitary

Lebesgue

.

prime and satisfy

is indeed a Gaussian

> Iy] 2

must preserve

a c S(~)

theory of Gaussian

, which has as an easy consequence

if we use Lehesgue

.IifiI 2

this:

i!

The second approach

$(~)

to be seen is that

to approach

to be an eighth root of unity--which

solute value

on

.

1 i f12 IYI e .2 Y ~j=l (6j /2y)

(5.22)

This

+y~)e(qo(x,y))dy

t 1/2 S(~)

Dt

on

so that is defined

S(~)

n > i.

For each real number

by the formula

IIDtaII 2 = IlaiI 2 to be

Dta(x)

will hold.

1D-1

a ~--> D z-Q ~n ~

B

t > 0,

= tl/2a(tx)--we The representa-

More explicitly

68

n a(x) = e(nx2r)a(x), Qt (r)

and

(5.24) Qna(x)

Inll/21Yll/2u(~)I~ a(xa+yy)e(nq

=

-_co when

(For

exists a unique non-zero

the operator

a • S(~) w(o) e C n

The crucial point is that Wn

v~-->

w(o)

a • S(~)

amd non-zero

it contains when

v • Cn

~(i) and

o = T(1)

and

~.

Z

is a non-zero

u(O)

with

element of

u(o)-l.)

n C , there

for w h i c h

depends

only on

X

and

o.

Hence we can take as

o

in

A

for w h i c h

is a subgroup

of

(5.25) does hold for all

A and therefore computation

is equal to

that

A

if

(~.25) does hold

~.

there exist some

U-ib ® w. n --

and

that we replace

We shall show by direct

We saw at the beginning £ Cn

except

w(O).

The family of all elements

_v

y # 0 .

n ! -i, we w o u l d also use (5.24),

We will show that w h e n e v e r

(x,y))dy o

Written out,

of this proof that for every

b • S(~) U-ib ® w n --

and some looks

w • Cn

o e A, a e S ( ~ )

satisfying

[PUnl]a

and ® X

=

like

[n-i [C -l.T-Ib ] (x,y,z) j=0 Wj+l n,j j =

Z n-I

j =0 wj+le (nz+j y) [k=_ b (x+k+j/n) e (nky)

Therefore fl (5.26)

Wm+ib(x+m/n)

--

=

e(-my)[U-ib 0

n

® w](x,y,0)dy --

= ii e(-my) [PoUnla- ® v](x,y,0)dy. 0 We can write out

P U -I a ® v on --

explicitly,

just as we did

U-ib ® w. n --

The result is

n-i

(5.27)

~j=0Vj+le(n(z+qo(x,y))+j(xP+y6))

~k=_ a(x~+yy+k+j(n)e(nk(xB+yS)).

69

It should be appreciated that had we attacked proposition (5.15) in its original form, it is with (5.27) in all its glorious complexity that we would have had to contend. Let us look at what (~.26) and (~.27) have to say when

o = ~(1):

Wm+ib(x+m/n) 1 n-I = f0e(_my)lj= Vj+le(nx 2+j(2x+y))[k=_~a(x+k+j/n)e(nk(2x+y) )dy = Vm+le(nx2+2mnx)a(x+m/n)

Thus

w~ (1)z = Next comes

=

Vm+le(-m2/n)e(n(x+m/n)2)a(x+m/n)

=

Vm+le(-m2/n)[Qn~(1)a](x+m/n) •

with

Wm+ 1 = e(-m2/n)Vm+l .

~, which also goes smoothly:

Wm+l b (x+m/n) 1 n-i = /0e(-my)[j=0 Vj+le(-nxy-jx)[k=_~a(y+k+j/n)e(-nkx)dy =

~n-i

.

1

j=0 Vj+le(-3x)I0 [k

=-~ a(y+k+j/n)e(-(nx+m)(y+k))dy

f

Vj+le (-j x) _ a (y+j/n) e (-y (nx+m)) dy

{xn-1 j=O

Vj+le (jm/n) }

I ~ a (y)e (-ny (x+m/m)) dy ----¢o

= {n-i/2 [~.-i Vj+le(jm/n)}[Qn a](x+m/n).

It follows that

W n~ _v = _w with

win+1 = n-1/2 [n-lj=0Vj+le(jm/n)"

We have thus completely proved that mains to check that product. Cn

Wn o

Un Po U-In factors into

Qon ® W n.o It re-

can be made unitary by an appropriate choice of inner

In fact, we may take the inner product for which the standard basis for

is orthonormal: We use a variation on the second argument we gave to show that (~.22) had ab-

solute value i.

We showed there that

Q~

and thus our normalization of the operators a unitary operator on

defines a Dt

unitary operator on

guarantees that each

L2(~) . We have also seen that

Po

n Qo

L2(~), defines

is unitary on all of

70

L2(N/N nAo). U

n

Finally,

: Z(n) ÷ S(IR) ®

obvious and

inner

n Qc~

it follows

cn

is an isometry when

products.

are

Putting

unitary

from lemma

operators,

all

of

this

and

since

(3.9)

that the map

Z(n)

and

S(IR) ®

together, Un P d Un- 1

we s e e

Cn

are given the

that

since

U P U- 1 non

= Qon ® Wno' the operator

Wo n

must be unitary.

If you will recall, we defined

vn(g)

the representation

of A N on S(~)

by the formula

(~.28) (cf.

Vn(g)((x,y,z),k)

(5.16)).

Although we are w o r r i e d

n Wo(_l),

value of

the representation

(5.29) Proposition: vn[~]

of

the form

Proof:

S .

We are looking n = Qo(k)"

exponentiation, with.

2 in

N on

vn(~)

is w e l l - d e f i n e d

Vn(~)

extends

the only other extensions

--> e(mt)vn[~]((x,y,z),t)

for a suitable

for some

m c~.

to

group running

S

the representation

.

W o r k i n g with suitable

have

through

one-parameter SO

groups is to work

R 0'n, we see that it takes the Lie alge-

to the real Lie algebra spanned by the operators S(R)

~ £ cX.

to a representation

vn(~)

Since one easy way to manufacture

is an eigen-

for all

of

one-parameter

$

we are led to look for a copy of the Lie algebra of

Differentiating

bra of

Qn

(x,y,z) ~(k)

only about the case where

The representation

Furthermore,

((x,y,z),t)

V((0,0,0),k)

= g-kR0,n

one-parameter

can see that the Lie algebra of the semi-direct

D = d/dx,

subgroups

product

of

2 inx, and

SL(2,~),

N'SL(2,~)

one

is isomorphic

to the Lie algebra spanned by the six operators

D, ix, i, iD 2

Let us call this Lie algebra

L.

J

Notice

ix 2

,

that

1 + 2xD

L

•

consists

of b o u n d e d operators

on

S(IR) , and so there is no problem about exponentiating. Since

(5.30) --

N

is simply

connected,

exp (tZ)R 0'n

there exists an element

(x,y,z) e x p ( - t E ) =

R 0,n

Z

in

(x,y,z)~(t)

L

satisfying

71

Now consider the operator

exp(-k E ) Qq(k)" n

It follows

tor commutes with every one of the operators ducible, E

it follows

that there is some

is skew-symmetric

and hence

~

must actually

vn[~]((x,y,z),t)

~ ~ cx

lie in T .

of the extension

r • ~

S(~)

is irre-

got from

so that

is the desired extension of

vn[~]

But

L2(~),

e(r) = ~-i ; then

up to multiplication

from the same Shur's lemma argument we used to produce

We remark for future reference

R 0'n

n Qo(k) = ~kexp(k E ).

such that

Choose

that this opera-

Since

(x,y,z)"

to the inner product on

= e(rt)R0'n(x,y,z)eXp(tE)

The uniqueness follows

with respect

R 0'n

from (5.30)

by ~

vn(~).

e(mt),

m e ~ ,

above.

that Shur's lemma also yields

D

the following

proposition:

(~.31) Pr0posi.tion: of

S

are equivalent

if, and only if,

(5.32) Proposition: --

vn(~)

tending

•

value of

Proof:

W

n

Let

There is

RIH

the restriction

T_he two representations

is

a closed

and

to

vn[~]

subspace if,

D

of

H

S

of

a~--> a #

be the image of this map in be the restriction

R 0'n, the group

N

from

S(~)

element

v

C~(S /A ), and let

map.

o

~-1

on

H.

i s a n ei_gen_-

a e--> a #

Furthermore,

since

there is

then,

that

such that

)

a e S(I~) .

res

Hence

subspace of

¢ Inl such that

: C~(S /A ) ÷ of

vn[~]

res

Z(n).

res(H)

to

N

is

must map

H

iso-

It follows

= U-I(s(II) ®

that

Cv).

Thus

n

a E S(I~) there is a unique

U , res, and n

bounded.

such that

o

We assume,

--

for every

ex-

- -

the argument

C~(So/A

Since the restriction

R-invariant in

into

=s e S o'

already acts irreducibly

onto an irreducible

there is a non-zero

Since

C~(S /A )

and o n l y i f ,

how the other half ought to go.

Rs a # = (vn[~]sa) #

(5.33)

morphically

S(IR)

- -

(i).

w e are given an i s o m o r p h i s m

C~(N/N C A o)

on

(I

--

We are going to start with the "only if" part, because

H

e(m't)vn[~ ' ]

m = m'.

be a representation

R-invariant

equivalent

fairly natural and suggests

Let

vn[~]

~ = ~'

and

e(mt)vn[~]

a T c S(IR)

satisfying

are all bounded operators,

UnreS (a #) = a # ®

the map

a~--'> a %

is

v.

72

U res((R 0'n z)a) #) n (x,y, = U res(R

n

(a#))

(x,y,z),0)

by (5.33)

= UnR(x,y,z)res(a# )

= UnR(x,y,z)UnlUnreS(a#)

= (R0'n(x,y,z)(at))

we see that

R 0'n

there exists some

(x,y,z) ~ ~ ox

the definition of

at

by

(~.34)

commutes with such that

® Z

a~--> a #.

a t = ~a

by (!.14),

Hence, R 0'n

for all

~X, we can arrange that

a

being irreducible,

a E S(IR) . t

= a.

v

We thus get

is that

v

is an eigenvector

-

~-i.

rounding discussion

for

Wn ~(i)

To that end, we recall from proposition

(~.15) and the sur-

that

(a#)) res(R((0,0,0),l) = PO (-i) res (a #)

=

U~I(( Qo(1)a) n

® (W~(1)X)

•

On the other hand, we see from (5.33) that

res (R((0,0,0) ,i) (a#))

= res((vn(~) ((0,0,0) ,l) a)#) -i = U n ((vn(~)((O,O,O),l)a) ® _v)

= U-I n Hence

in

res(a #) = Unl{a ® ~}.

What we wish to show, you will remember, with eigenvalue

Replacing

n ( (Q (i) a) ® v)

.

n

Wo(1)_v = ~-iv,_ as desired. We must now prove the "if" half of the proposition.

Here we are given an

73

eigenvector

v in --

¢ Inl

duce a hounded operator subspace

H

of

for

W n~(i)

a +--> a #

C (So/A)

with eigenvalue

that (i) maps

and (ii) satisfies

-i,

and we are asked to pro-

bijectively

S(~)

onto a closed

The operator in question is

(5.33).

not hard to write down: Suppose element of

a • S(~) Z(n)

s •

and

and h e n c e ,

that function at

(0,0,0)

in

S

are given.

particular,

• N

to

get

Then

a function

U l{(vn[~]sa)-n on

N/N n A . o

Our construction

of

vn[~]

We e v a l u a t e

by exponentiation

.

(see the proof of proposition

makes it clear that

vn[~]s a, and hence a#(~), are C ~ functions of = is left A periodic, we need a slight strengthening

# to see that

a

It follows from proposition ((x,y,z),k)

is in

is an

a#(~):

a#(s) = [u~1{(vn[~]sa) ® v}](O,0,0)

(5.35)

® v}

(5.15') and the definition of

vn(~)

=s.

(5.29)) In order

of (~.34).

that if

~ =

A N, then

(~.36)

a#(~) = [Unl{a ® X}]((x,y,z)~(-k))

.

(We get (5.34) from (5.36) by taking

k = 0.)

One more remark we need is that if

one replaces

then it follows immediately

# satisfies and

~

in (5.35)

(5.33).

As for

with a#

~,

being left

A

periodic,

if

that

a ~--> a

r = ((a,b,c),k)

•

A

~ • So, then

a#(~s) = (vn[~]sa)#(~) =

(by (~.33))

[Unl{(vn[~]s a) ® v}]((a,b,c)~(-k)) =

(by (!.36))

[Unl{(vn[ a #

of

is mapped that

is a bounded operator.

H

is a

We begin

74

Let

{a } m

be a sequence in

to some function that

f.

S(~)

such that

We will show that

{a } m

{a #} m

converges in

converges

to some

C~(So/Ao)

a e S(~) , and

a # = f. The convergence

res(f).

of

{a #} m

to

f

implies the convergence

Since each of the functions

U-I(s(~)n

® Cv)_

of

res(a #) m

The function

u-l{amn ® --v}" The operator

Un

{res(a#)} m

to

lies in the closed subspace

C~(N/Nn Ao), there must be some

res(f) = u-l{an ® _v}-

of

a c S(~)

such that

res(a~) is, by virtue of (5.36)_ , equal to

being bounded,

it follows that

converges

{am ® _v} #

to

a ® ~, and hence

for each fixed

{a } m

converges

to

~ ~ So, the operator

a.

To prove that

vn[~]s

is bounded,

a

= f, we observe that

and therefore

=

{u-l{(vn[$]sam) n = space of

® v}} --

converges

U-I{( Vn [6] a) ® v}. N s -=

Thus,

H

is a closed sub-

C (So/Ao)-

In the course of proving from

to

H

to

S(~)

H

# a e--> a

closed, we showed that the inverse map

is a bounded operator.

It follows from the open mapping theorem

# that

a ~--> a

is also bounded.

It follows from proposition W~(1)

are,

we k n o w e x a c t l y

which

Of c o u r s e ,

determining

the

leave

to

section.

that

the

next

prove that the subspaces

(5.32) that, once we know what the eigenvalues irreducible

eigenvalues

of

Right

Y(n)

representations n Wo(1)

of

Here we face another problem of our approach: of what

H

occur

difficult;

a different

problem,

in proposition

in

Y(n).

we s h a l l which

(~.32) span

is

to

Y(n).

we don't really have a very good idea

is; we just know that it exists--it

around this by selecting certain

Su

may h e q u i t e

now we h a v e

constructed

of

of

is there, somewhere.

We shall get

H's to serve as a sort of base-point.

Here is the

story: Using the inner product on orthonormal,

¢ Inl

with respect to which the standard basis is

choose a complete orthonormal

system

{~i ..... Vlnl} i

mn G(1)"

for

Let

~j

an extension

~j

as for

a ® v . --

J

~k

-i ~j

denote the eigenvalue

vn[~j] whenever

of

n Wo,I~)

of

vn(~j), being careful

~j = ~k"

Now map

to the function whose value at

s =

S(~) is

®

on

of eigenvectors

L

--iv'" Choose for each

to take the same extension for ¢ [nl into

[u-l{(vn[~j]sa)n =

Y(n)

by sending

® v.}](O,O,O), --3

just

75

as in the proof of (~.32). Z#(n)

of

Y(n)

e(mt)Z#(n)

Let

Z#(n)

denote the image of this map.

will serve as the base-point.

to denote the subspace of

((x,y,z),t) e--> e(mt)f((x,y,z),t) Rle(mt)Z#(n)

Y(n)

for some

(5.31), RIe(mt)Z#(n)

m c ~, we will use

consisting of all functions of the form f e Z#(n).

Notice that the restriction

~ j =InI l e(mt)vn[~j],

is equivalent to the direct sum

noted in proposition

For each

The subspace

and

Rle(m't)Z#(n)

and hence, as we

are inequivalent if

m # m'.

(5.37) Proposition: element of with

f

in --

m

Proof:

Y(n)

Y(n) = ~@~=_ e(mt)Z#(n)

in the usual sense that every

can be written in precisely one way as a convergent sum

e(mt)Z#(n).

It is convenient to use some integration here.

on the cube

0 < x, y,z < 1

on

So/A o.

Let

res

the form

0 < x,y,z,t < 1

again denote restriction from

The crucial fact we shall need is that if The proof is not hard:

Recall that Lebesgue measure

defines an R-invariant measure on

lar fashion, Lebesgue measure on the "cube" measure

~mfm

we may assume

f

f c Z#(n), then

N/N N A .

In a simi-

defines an R-invariant So/A °

to

N/N nAo.

IifiI2 = IIres(f) Ii2.

comes from an element of

S(~)

® C Inl of

a ® --] v,, whence

2/

fI I2 =

I[Unl{(vn[~j]((0,0,0),t) a) ® _vj}](x,y,z)12dxdydzdt

=

l lUnl{(vn[$j]((0,0,0),t) a) ® vj} I I~ dt 0

IUn l{a ® ~ } I I ~

because

U

is an isometry and vn[~.] is unitary. It follows that n j 2 2 Iires(f) II2 = IiaiI2, since {~i ..... Vin I} is orthonormal. Let

Z#(n)

[']

denote

L2

to isometry from

struction of must map

Z#(n)

C~(So/Ao)

C~(N/N n Ao) n [Z(n)]

that

closure. [Z#(n)] res

N [Z#(n)]

We have just shown that

into maps

res

IifiI~ =

extends from

L2(N/N nA ), and it is clear from our cono [Z#(n)]

injectively into

is evidently just

onto

[Z(n)].

In particular,

C~(N/N n A ) p [Z#(n)].

Z(n), and

res

already maps

res

Now Z#(n)

onto

78

Z(n).

Hence

Z#(n) = [Z#(n)] n C~(So/Ao).

We come now to the body proper of the proof. e(mt)Z#(n)

and

[e(mt)Z#(n)]

e(m't)Z#(n)

and

are inequivalent when

[e(m't)Z#(n)]

of

L2(So/Ao)

Therefore the closed linear span in Hilbert-space

direct sum

lemma argument Let

(cf

E(m,n)

Each

E(m,n)

maps

C~(So/Ao)

[Y(n)]

of the subspaces

be the orthogonal projection of

is a bounded translation-invariant

Z#(n)].

We have therefore proved that whenever uniquely as a sum

words, the closed linear span of

[e(mt)Z#(n)]

L2

to

is the

:

L2(S /Ao)

onto

[e(mt)Z#(n)].

operator, which implies that E(m,n)

[~=_ E(m,n)f

f

f

C ~ topology.

the series

ally to

we can write

m # m'.

n [e(mt)Z#(n)] , and also that when

f • C (So/Ao) n [@~=_~[e(mt)Z#(n)], in the

to

We are going to use a little Sobolev

(3.1)) to go from

C ~ (So/Ao)

R

m # m', the subspaces

are orthogonal when

[@~=_~[e(mt)Z#(n)].

proposition

into

Since the restrictions of

converges uncondition-

As we just showed, e(mt)Z#(n) = C~(So/Ao) n [e(mt)

~=-~fm

f £ C~(So/Ao)

with each

{e(mt)Z#(n)

n ~@~=_~e(mt)Z#(n)],

fm e e(mt)Z#(n).

: m ¢ Z} in

Y(n)

In other

is the direct sum

[e~=_ooe(mt)Z#(n). In order to complete the proof, we need only show that What we will actually show is that if element

g e [@~=_~oe(mt)Z#(n)

f e Y(n)

and

that is not orthogonal

Y(n) ! [@~-__ ~e( mt)Z#(n)"

f # 0, then there exists an to

f

in

bined with the Sobolev lemma argument of the previous paragraph,

L2(S /A ).

Com-

this implies

Y(n) ! [@~=_ooe(mt)Z#(n) • Replacing arrange that

f

by a suitable translate

res(f) # 0.

Choose

g ~ Z (n)

R((0,0,0),t)f, so that

if need be, we can

res(g) = res(f).

We will

then have that the inner product

fxlzlz

{ r e s ( f ) ( x , y , z ) } { r e s ( g ) ( x , y , z ) } dxdydz ~O~OJO is strictly positive.

It follows that we can choose

plies that the real part of

i

f fill 0J0J0f ((x,y,z) ,t ) g( (x,y, z) ,t) dxdydz

c > 0

so that

It I < e

im-

77

is still strictly positive.

Let

~

in the interval

in

~/Z

and

0

(-E,e) + g

~(t) ! 1

for all

t.

Then

be a

, and which satisfies ~(t)g((x,y,z),t)

and the real part of its inner product with

(~.38) Corollary:

If

Y(n), there is a unique

Proof: maps

Because H

H

at most one at least one

f

is irreducible,

Proposition

the conditions

is in

%@~=_ e(mt)Z#(n), D

the projection

E(m,n)

is non-zero on

H.

either is zero on

H

RIe(m't)z#(n)

when

But surely in light of proposition

m # m', (5.37)

H.

D

(5.38) and hence is

Perhaps the most striking feature of the entire business is that

we still do not know exactly which irreducible representations of

S

appear in

RIY(n) , and furthermore even when we are given a irreducible representation of that does occur, we cannot say very precisely on what subspace of sentation occurs.

or

Since no subrepresentation of

(~.9) is an immediate consequence of corollary

completely proved.

~(0) = 1

H L e(mt)Z#(n) •

e(mt)Z#(n).

is non-zero on

E(m,n)

whose support lies

is positive.

is equivalent to a subrepresentation of E(m~n)

~/g

is an irreducible closed R-invariant subspace of

m E g such that

isomorphically into

RIe(mt)Z#(n)

H

function on

Ca

Y(n)

S

that repre-

One can convince himself of this second point by trying to write

out explicitly the orthogonal projection of

L2(So/Ao)

onto

Z#(n).

This lack

of

control is what we meant when, at the very beginning of section i, we referred to "one missing phenomenon" that would appear in dimension 4.

In dimension 3, things

may be a bit complicated, but the objects one is dealing with are all very precisely and perspicuously defined.

Not so in dimension 4, as this section shows.

Of course,

there remains the possibility that if we were to look at special cases in detail, these difficulties would evaporate.

We will take up this matter, along with ques-

tions analogous to those pursued in sections 2 and 4, in the next section.

6.

Multiplicity computations in dimension four

The multiplicity questions of interest for same as for

C~(Go/Fo),

C~(S /A ), while essentially

are somewhat different in their emphasis.

the

We begin this

78

section,

therefore, with a discussion in broad terms of what sorts of questions seem

natural to us in the present setting. Recall that Let

H

Y(n)

denotes

be an irreducible

{f E C~(So/Ao)

: f((x,y,z),t)

= e(nz)f((x,y,0),t}.

closed R-invariant subspace of some

Y(n)

with

n # 0o

Our analysis so far has shown (a)

that

H

is already irreducible under the action of

afortiori (b)

and hence

AoN;

under the action of

that the representation of

N

A N

on

H

is equivalent to one of the

on

S(~) , with

0

representations

vn(~)

of

A N

~

-i

an eigenvalue

U

of (c)

Wn

o(1);

that the multiplicity in all of

C~(So/Ao)

of the irreducible representation

RIH

of

of the equivalence class S

is equal to the 0

dimension of the space

E(~) = {v n )= = ~-i }; and there= E ¢ Inl : Wo_l.V(

fore

(d)

that if we can determine what the eigenvalues of the dimension of the corresponding eigenspaces

Wn o(I) E(~)

are, then we

will know exactly which irreducible representations of R

are and what

S

o

occur in

and how often each occurs.

To solve the problem posed in (d) turns out, as shall be seen, to be impossibly difficult in all but what must be regarded as rather degenerate cases.

One might

ask, then, whether it is at least possible to make some estimate of the asymptotic behaviour of the number

n Wo(1) • elgo(n )

as a function of

n, when

n

of distinct eigenvalues of

larger

eigo(n)

wn(1).

Can one be more precise?

is large. n Wo(1).

Consider, for example,

Roughly speaking,

the

is, the smaller the multiplicites of the individual eigenvalues of One might try to get some asymptotic estimate for

u

eigo(n),

for example some estimate for

Inl/eigo(n) , which one can interpret as the

expected multiplicity for each eigenvalue of

Wn o(1)"

One might also try to deter-

mine how greatly the actual multiplicities of the eigenvalues of from their expected value

Inl/eigo(n).

W ~(i) n

differ

The attraction of such estimates is that

one might hope to obtain them without having to know too precisely what the actual eigenvalues of

n Wo(1)

are.

There is, however, persuasive evidence that such hopes

79

are almost certainly contrast

doomed to disappointment.

to the situation in dimension

The crux of the matter is that, in

three w h e r e everything

well explored question of the distribution

revolved

around the

of primes in ideal classes,

the computa-

tions here lead one inevitably

into the terra incognita of the location of the gen-

erators

groups of finite fields.

for the multiplicative

when all our computations

The plain truth is that

are done, we shall find ourselves

precious

little better

off than we were before. Let us begin the computation by reviewing how we got If you will recall, we used the results

Un

an isomorphism

from the space i

f(x,y,O) }

S(~) ®

onto

Wn

in the first place:

of section 1 to show that there exists

Z(n) = {f e C~(N/N n A o)

: f(x,y,z)

= e(nz)

i

¢ Inl

with

the property

that if

R O'n

denotes

the repre-

sentation (R O'n

of

N

on

(x,y,z)a)(t)

=

e[n(z+yt)]a(x+t)

S(IR) , then -i RO,n UnN(x,y,z)Un = (x,y,z) ®

--see

(5.11) and (5.14).

representation

P

of

By direct computation we could then establish

A on

Z(n)

given by

Un P U-I n on = Qo The reduction of translation

L f n

of --

The operator

f

L

u

for some integers

Proof:

is accomplished

factors

as well:

Wn G by w o r k i n g w i t h left as well as right

operators.

(6.1) Lemma: tion

Wn

~

P f = foo=

that the

Direct

Given to

be

u E N the

carries a

and

and a function

function

Z(n)

f

v a~---> f ( u - l v )

on

N, define the left-transla-

on --

N.

into itself if, and only if,

b, and some real number

(Note

the

inverse.)

u = (an-l,bn-l,z)

z.

computation.

Left translations

D

enter the argument

with right-translations--see

proposition

for the obvious (--1.34).

reason that they commute

To be more precise,

let's let

80

E(n)

denote

whenever

u e £(n),

R0'n(x,y on

the subgroup

z) ® =1

~Inl;what

operators In

Lt u ¢ Inl

of

and

W n.

S(g/ng)

Lt u

(0,n-l,0),

between

L ut the

explicitly. given by the unit vectors

this basis, we shall identify

to the function

so that

for some operator

the interaction

~i,$2 ..... $1nl

of all complex-valued

¢ Inl corresponding

for

functions v

on

¢ Inl

g/n

on

~,

with

the element

defined by

g/ng

the

v(j) =

j = 0,i ..... Inl-l.

(6.2) Lemma: v E S(g/ng). = i ® L t. u

Using

1 ® L tu

is to study

basis

and

commute with all of the operators

the form

We begin by computing

we have the natural

space"

(n-l,0,0)

will

turns out to be instructive

"Schwartz

vj+ 1

by

-i U L U n u n

and therefore will have

directions.

of

generated

the operator

in the coordinate

~.v.e. JJ=J

N

Let

Define

Then

Lt u

u = (an-l,bn-l,cn -2)

the representation

Lt

be an element of --

~(n)

Notice

on --

E(n), and let

S(Z/ng)

by

U L U -I = nun

is the operator

(Ltv) (j) = e[-(c+bj)/n]_v(j+a) u=

(6.3)

of

the formal

similarity

between

Lt u

.

and the operator

R 0'n

(x,y,z)

on

S(~).

Proof

(of (6.2)):

(0,n-l,0), of

It suffices

because both

Lt

to check

and the right-hand

ments

S(~) ® f

equal

S(~/n~)

of the from

to the element

f(x,y,z)

If you go back to (5.10) Unl(~ ® ~)

side of

u = (n-l,0,0)

(6.3) define

Z(n),

of

Z(n)

the effect of the operator ~ ® ~. gotten

= e(nz)~=_~

and trace through

and

representations

To begin, from

~

-i U L U n u n

on ele-

let us write out Unl(~ ® ~).

by

~(x+k)e(nky)

the definition

of

Un, you will

find that

is the element

(f A V)(x,y,z)

of

cases

H(n) . What we need to do is to compute

Let

the special

the sum being over

= ~je~/ng v(j)e(jy)f(x+j/n,y,z)

j = 0,1,2,...,Inl-i

where

an integer

(and not an

81

element of case

£/n~)

is called for.

Now let us compute

Lu(f A ~)

in the special

u = (n-l,0,0):

[L(i/n,0,0)(f A V)](x,y,z) = [f A v]((n-l,0,0)-l(x,y,z)) = ~jv(j)e(jy)f(x+(j-l)n-l,y,z-yn -I) = ~jv(j)e ((j-l)y) f (x+(j-l)n-l,y, z) = ~j v(j+l) e (j y) f (x+j n-i ,y, z) f = [f A L (i/n,0,0)v](x,y,z) which shows that

(L # v)(j) = v(j+l). (i/n,0,0)= =

, u

The computation for

=

(0,n-l,0)

is

similar.

(6.4) Corollary:

Proof:

The representation

Lt

is irreducible.

If follows from (6.2) that the operator

has for its eigenvalues the n-th roots of unity.

Lt(0,1/n,0 )

is diagonalizable and

Since no eigenvalue occurs with

multiplicity more than

i, the operator is also cyclic, and hence whenever

operator on

S(~/n~)

that commutes with

tion

S(~/n~)

satisfying

all

t

=

in

j •

function

£/ng . =t

If the operator

also commutes with

for all

~ c S(~/n~)

L t (i/n,0,0)' then the

must be constant.

We are now ready to begin looking at the representation

W n.

what we already know from (5.24) e_ttseq.: In section 5 we worked directly with only two elements of

:I For

nd 10

T(1), we derived the formula

(6.5)

is an

Lt(0,1/n,0 ), there exists a unique func-

(T~)(j) = ~(j)~(j) T

T

[~(1)v](j)

= e(-j2/n)v(j)

A,

Let us recall

and

82

n WT(1)

Thus

problem.

comes to us diagonalized,

As for

and counting its eigenvalues presents no real

W n, b y i n t r o d u c i n g a H a a r measure on

£/n ~

n o r m a l i z e d so that

f__v(j)dj = inl -I/2 jv(j), w e force

(6.6)

= fv<m)e(mj/ )d

= v'(j)

,

^ b e i n g the o r d i n a r y Fourier t r a n s f o r m of used as base point

(instead of

j '

v

w i t h the character

j t---> e(-j/n)

> e(j/n), w h i c h is c u s t o m a r y - - t h e appearance of

the minus sign here is consistent w i t h its appearance in (6.3) and (6.5)). n WT(1)

analogy b e t w e e n the formulas for n QT(1)

n Qw

and

W n~

and

on the one hand and those for

on the other is evident, even if it might appear merely formal.

E v e n t u a l l y we are going to analyze the o p e r a t o r

Wn

in great detail--it is

one of a small h a n d f u l of cases that can be completely analyzed. want

to explore some general properties of the r e p r e s e n t a t i o n

obvious thing to do is to try to pursue the analogy between ple, our next lemma shows, in effect, that

(6.7) Lemma:

Proof:

If

o e A

and

We start by w o r k i n g in

S(£/ng)

The

via the o p e r a t o r

U

Wn

u • E(n), then

C ~ (N/A o n N ) , n

.

is to

Suppose,

in

L~

LgW n u o

W n. Wn

as

Of course, the

and Qn

Qn--for exam-

is to

RO'n:

= WnL t ~ NO=

to be precise; then w e get to

Z(n)

the, that

First, however, we

f

is in

Z(n).

W e have

[LuP f](v ) = [p f](u-lv) =

f((uz)-l(vz))

= [Luof](v~) = [PoLu f](v)

whenver

v • N.

n ® WnL* Qo o

u~

Hence

L P = P L . u o o U~ =

' w h i c h in turn implies

(6.8) Proposition: --from fying

SL(2, ~ )

to

[o] n = [o'] n. (i)

If

c (o,o')wno,

n .

Let

C o n j u g a t i n g by

u o

o ~---> [0]

SL(2, g / n g ) .

Un

yields

Qo ® L

=

L*W n = W n L #

Let

o

n~

denote the n a t u r a l m a p - - r e d u c t i o n mod.(n)

n o

and

o'

be two elements of

A

satis-

Then:

is odd, there is some constant

c(o,o') • T

satisfying

Wn = o

83

(ii)

_If _

n

c(o,o') e q" entries of

Proof:

is even, ._ then setting

satisfying 0" b e i n g

the proposition.

W n = c(o,o')Wn,L

~",B",y",

To b e g i n with,

let

n

and

with

u = (¥"6"/2n,-~"B"/2n,0),

the

~", as usual.

be arbitrary,

Our assumptions on

o

really must consider is the problem:

and set

and

[o"] n = $.

force

vo= ~ v mod. Ker L t

w i l l commute w i t h every

Lt v

The case of odd

Whatever

o

and

n

[0]

Thus w h a t w e

L t.

~.

and

(ii) of

The i d e a is to use lemma (6.7) together w i t h the i r r e d u c i b i l i t y of

ple of

o £ A

as in part

wn? o

if w e can show that

if

o'

o = 0'0"

= i, what is =

example, Wn

o = 0'0" , one can find a constant

n

holds for all

For

v c E(n), then

and h e n c e by Schur's lemma w i l l be a scalar multi-

n

goes exactly this way.

might be, it w i l l be true that

(i/n,0,0)~

=

(e/n,B/n,~B/2n

2)

(6.9) 0,1/n,0)~

Suppose now that

n

is odd and that

then be divisible by lie in

£ .

and

~8

y~/2n

whenever

and

B / n , ~ / 2 n 2 - B/n). is in

n

=

and

~.

The entries

y6

are divisible by

B

and

$/n

are divisible by

2, and therefore, n

Since any element

v z vo= -mod. Ker L -

v = (i/n,0,0). (a,b,c/n)

v-l'(v~)

of

c Ker L %.

holds for all

There ~(n)

y/n

o

will

will

being odd, aB/2n

v

v ~ v~ mod. Ker L t -i

with

"(v~) = ((~-l)/n, a, b, and

c

in

(0,1/n,0) simi-

The Schur's lemma argu-

W n = (constant)-l. o

is even, it may happen that

v ~v~

even w h e n

[o] n

=

!.

is that w e may have

~ ~ ~ 1 mod.(n)

> ~,6

odd because

n

and B z y ~ 2 mod.(4) whence neither

of

n; but b e c a u s e

One handles

v e E(n).

y

and

Now what w e want to show is that

C o n s i d e r the case

now applies to show that When

eB

K e r L t, it follows that

larly, w h e n c e ment

¥~

are both integers. v c Z(n).

[o] n

n, w h i c h means that in (6.9) the terms

It also follows that

o ~ A, both

(y/n,6/n,y6/2n 2)

~B/2n

nor

y6/2n

will lie in

g .

is even

The p r o b l e m

84

For example,

then Lf

suppose

n = 2

and

(i/n,0,0)-l[(i/n,0,0)~]

~ =

= (0,i,-i/4), w h i c h does not lie in

= -i.

(0,1,1/4)

Let us return now to the case w h e r e ever

[o] n = $, the element

conjugation,

n

is even but otherwise arbitrary.

u = (y6/2n,-~B/2n,0)

defines the same a u t o m o r p h i s m of

of

E(n), acting on

E(n) mod. Ker L t

(This a u t o m o r p h i s m w i l l be n o n - t r i v i a l p r e c i s e l y w h e n either mod.

(4).)

ccT

K e r r - - i n fact,

~

that or

When-

E(n) =o y6

by

does. is

2

The Schur's lemma argument shows that there must be some constant

such that

W n = cL t u

w h i c h proves

(ii).

The same technique of proof yields:

(6.10) Corollary:

The kernel of

Wn

lies in the k e r n e l of the n a t u r a l ma E

o D----> [O]n"

Proof:

If

[o] n # $, then

elements of

[O] n ~

Ker L t.

~

Thus the Schur's lemma a r g u m e n t yields that

Z(n) mod.

Wn # 1 whenever

~-

Our

aim

is to make a useful tool of p r o p o s i t i o n

complications in the even case.

(6.11) Proposition: that there is some to

defines a n o n - t r i v i a l a u t o m o r p h i s m of

[O]n.

Let

SL(2, ~ / n ~ ) .

k Then

p E A

Our next p r o p o s i t i o n suggests one technique.

Assume that such that

n

occurs w i t h m u l t i p l i c i t y

is an odd prime.

[0] n

denote the o r d e r of Wn

(6.8) despite the obvious

[O]n

has precisely

k

Let

o ~ A, and assume

is diagonal and is conjugate in

as an element of the finite group distinct eigenvalues, one of w h i c h

1 + (n-l)/k, the others w i t h m u l t i p l i c i t y

N o t i c e that w e do not say what the eigenvalues of there are and how often each occurs.

[A]n

Wn

(n-l)/k.

are, m e r e l y h o w many

If you w i l l recall, we m e n t i o n e d at the be-

ginning of the s e c t i o n that if all y o u care about is getting some q u a l i t a t i v e idea

85

of the behavior

of multiplicities,

values of the eigenvalues We see why here: [~]

n

= 1

to actually w r i t e

(6.8).

down the eigenvalues

because we do know

On the other hand,

largely an illusion, proposition

W no' and your p r o b l e m is there by somewhat

we don't know yet enough about the behavior

the proposition, sition

of

Wn o

of

the simplification

The stringent hypothesis

that

n

thus sometimes

affords

The other hypothesis,

SL(2, g / n g

)

gacy condition on nalizable

Our examples will show that

[ ~ In

cases.

n

is not prime,

down in a revealing way. n

the conjugacy

condition

on

is odd, w h e t h e r or not

n

n

(so that

the multiplicities The method of proof

is not prime--we

in connection with our analysis

of

It

of

shall see a good

W n.

[O]n' is easier to verify that

[A] n

is prime.

is in reality just the condition

that

is all of

Thus,

the conju-

[d]n

be diago-

in the usual matrix sense.

Proof

(of (6.11):

with

p e ~

constant

n

in

[O]n, but we give no hint

It turns out, as will be proved,

whenever

is

be an odd prime bears some explanation.

good results even when

example of this phenomenon

than it might appear.

to w r i t e

to propo-

All we are asserting

to carry out, except in very simple

(6. g)), but that when

are very much more difficult

{T £ A :

thanks

will be clear from the proof that all w e really need is the oddness we can apply proposition

on

achieved by the proposition

counts is the order of

is all but impossible

Wn

up to a constant multiple,

as to how one might compute this number.

this computation

of

simplified.

W n, but we can still prove o

and again the reason is easy to see.

(6.11) is that what

whatsoever

then you don't really need to know the actual

and

c ~

eigenvalues

of

By assumption, [P]n

diagonal.

, we have Wn o

there is some It follows

W n = cW n W n Tp

(wn) -I. T

T ~ A

satisfying

from proposition Hence,

we can work equally well w i t h

-i

]n

(6.8) that for some

for purposes

Wn ~ p

[o] n = [TpT

of counting

Thus w e shall assume

the o = p

no loss of generality b e i n g involved. We are now assuming

that

(~n-l,Bn-l,~B/2n 2) E (~n-l,0,0) By the same argument, Let

D

[o]

n

is diagonal,

mod. Ker L ~, since

(0,n-l,0)~

~ (0,6/n,0)

be the dilation operator on

which implies B/n

and

that

~B/2n

(n-l,0,0)~ = are both in

~.

(D v)(j) = v(~j)

on

mod. Ker L t.

S(~/n~)

given by

86

v c S(£/n~).

Using

=

holds for all for some

~6 ~ 1 mod.

u c S(n).

ccT.

G(e)

erated by

~.

It is at this point w h e r e

The order of

£/n~

that

is an eigenvalue

decomposes

if

now see easily for

plicity

into a disjoint D

and

% # i.

that each

union

group

the order of

~)

~ k E 1 mod.

of

G(~)

and

genas an

in ( ~ / n ~ ) X ,

D v = ~v.

of

Suppose Then

and further-

%k = i.

implies

Of course,

D

important.

[a]n

0,Sl,Sl,...,Sm,

k-th root of unity is an eigenvalue

=

m = (n-l)/k.

satisfies

(n)

u~

= cW n o

(£ /ng)X

k, and set

at the elements

m.

~

D

{0} u G(~)s IU...UG(~)sm.

v e SL(g/n

In addition,

= D Lt

~

the eigenvalues

from each of the cosets of

i, each occurs with multiplicity

One can

D , and that,

1 occurs with multi-

m + i.

When

n

(~/n~)X

is not prime, we can still form the group

of units of the ring

multiplication. {0}

of

u

being prime becomes

Call that order

determined by its values

~(0) = 0

except

n

LtD

so we must have

as a group is exactly

Sl,S2,...,s m

so that

more

G(e)

SL(2, £ / n g ) .

Choose representatives

is clearly

again,

denote the subgroup of the multiplicative

element of the group

~

lemma applies

Thus, all we have to do is to compute

their multiplicities. Let

Schur's

(n), one can easily check that

Of course, when

and the cosets of

structure

G(V)

in

~/n~ n

clear general

result.

We then let

is prime,

(£/n g)x.

can be much more complicated.

of orbits with given cardinality,

.

G(~) G(~)

in the group act on

~/nZ

by

the only orbits of this action are When

n

is not prime,

the orbit

Since what w e need to know is the number

one can see why in the non-prime

In any given case, however,

the analysis

case there is no

may be very straight-

forward to carry out.

(6--.12) Proposition:

Proof: Hence

Because [A ]n

n

is

If

odd,

n

there is some

k ~

satisfying

2k - 1 mod.

(n).

will contain

[T(1)k]n = Ii

It follows

[A] n = SL(2, g / n ~ ) .

is odd, then

that

[A]n

i]

is the same as

and

[~] =

[SL(2, g ) ] n "

We need only show,

therefore,

87

that

[SL(2,~)] Let

are going

of

= S L ( 2 , ~ /rig), w h i c h

n

~,B,y,

and

be any four integers

Suppose

is

first

[o]

that

ea + Bb = i.

for some

n

gcd(a,B) Let

= 1

in

SL(2, ~ ) ,

is odd:

~6 - By ~ 1 mod.

and

[°]n

B = rB', w e see

mod.

(n), so that if w e set

will

define

an e l e m e n t

of

~.

We can then find integers

is the element gcd(~,B)

that

~

= r

- By = i

s = ~'d - B'~,

SL(2, ~ / n ~ ) .

gcd(~',$')

= i, the a r g u m e n t

ry

is a c t u a l l y

This

We

in

completes

[SL(2, g ) ] n "

the proof.

a

and

The m a t r i x

(~.13)

of

SL(2, g / n ~ ) .

can be g r e a t e r mod.

(n)

implies

than

i.

the m a t r i x

Furthermore,

of the previous

r~

paragraph

n

As for the diagonal

factor,

Writing

r(~'~ - B'y)

Ii fill ilIrliiIm°dn Since

(n).

~-acn

Next c o n s i d e r t h e c a s e where and

in

e~ - By = 1 + cn.

y+bcn

= r~'

satisfying

n

o ~ SL(2,~).

i then l i e s

or not

to s h o w that the element

S L ( 2 , g /n g)

so that

~

is true w h e t h e r

shows

that

~ 1

b

88

With proposition odd n.

(6.12) we have finished our general treatment

We remark that the identity in (6.14) is useful in its on right, which is one

reason w e gave this particular

proof of (6.12).

want to give an example of how one can use of proposition group

(6.8) is that when

K(n) = {o c & : [O]n = ~}

whenever

o E K(n).

the case.

G(r,s)

One can use

in section

G(~,y)

~6 ~ 1 mod.

W2~(j)

As an example,

Wn

one upshot

to the sub-

K(n), namely,

W n = X~O)~

Xn ~ i, which, however,

and that is our next project. r

and

s, with

is not

A little

s # 0, w e set

sum"

Assume

(n), and let

= G(~,n)~(aj)

taken

= I e(-rj2/2s)dj JZ/sZ

u(~)

appearing in our computation

of

n Qo

n

is odd. o

(of (6o16)):

for all

direct computation

v ~ 9(~/~)

151 -56 that

and

that

6

be even integers

sat-

and

j c Z/nZ.

[o] 3 = 1 =

W 3 = -il.

There is n o t h i n g special about

Writing

and

26-ad2J

= -i, we see from (6.16)

character.

~

be the element

4 -15

G(4,3)

Let

n = 3, e = 6 = 4, so that

=

trivial

of

n, w e

5.

Then

Since

about even

If you w i l l recall,

a character of

Given any two integers

1-~ o_~f A.

(6.14).

like to have

as the factor

(6.16) Proposition: isfying

defines

]s]-i/2 ~]s]-i e(_rj2/2s) j=l

One recognizes

Before w o r r y i n g

is odd, the restriction

(6.14) to see why,

equal to the "Gaussian

(6.15) --

n

Ideally we w o u l d

notation will be helpful.

Proof

of the case of

T(k)

Hence

W3JK(3)

is not the

3; any odd prime w o u l d do.

for the k-th power

T(1) k

of

T(1), we see by

89

= mT (~/2)~-IT (e/2)~T(6/2)

(-6.17)

Hence

(-6.14) implies

for all

that there is some

v E S(Z/nTQ

any convenient

v

and

=

Multiplying

(_6.18)

and all j

ccT

j c ~/n~.

[W~](j)

such that

Thus,

to compute

= c%(~j)

holds

c, we are free to choose

we wish.

both sides

of (-6.17) by

~

-i

, we see that

W n

~ W n T(~/2)v](j)

[wnT(6/2)(wn)-IwnT(~/z)

= cv^ (-~j)

for all

v E S(~/n~E)

and all

j ~ g/nE.

The left-hand

side of (6.18) written

out

is

e (6j2/2n) I e (-~m2/2n) {le (-~h2/2n) v(h) e (mh/n) dh }e (-mj/n) dm

Now take is

~

0; for

the function whose value j, take

0.

That done,

at

0

is

Inl I/2

(-6.18) becomes

and whose value elsewhere

simply

/e(-~m2/2n)dm

= c, which

is what we set out to prove.

Let us now look at what happen when wit 2, seems Wn

to be giving us trouble,

in the sense of section ~:

tensor product written,

Wm ® Wn

but we are going

For the moment, than

n

speaking,

ged(m,n)

= i.

Since only one prime,

attack

to

is to try to "localize"

one wants

to factor

W rm~

into a

This is not quite possible

as just

to show now that one can come close. remains

an arbitrary

integer whose absolute

value is g r e a m r

i.

(-6.19) Proposition: is a unique and

is even.

the natural

Roughly

whenever

n

~

representation

Whenever W n'b

is an inte e ~ r e l a t i v e l y

b of

A

on

S(g/n~E)

by the formula [wn'bT(1)=v](j)

(-6.20)

I

[wn'b v](j)

=

= e(-bj2/n)v(j)

(

v(m)e(bjm/n)dm. 'Z/n~

prime

to

n, there

given on the generators

T(1)

90

Furthermore, if I

h

and

from

if k

m

is a third integer relatively prime both to

a r e integers s a t i s f y i n g

S~/mn~)

onto

(6.21)

S~/m~)

o

such that for all

~ m,bk . =

and

b, and

h m + kn = i, then there is an isomorphis m

® S ~ / n g ~)

iwmn,b i-i

--

n

(W

o)

o E A

n,bh ®

(W

o)

.

n,b WT(1)

The heart of the m a t t e r is easily seen to be showing that the operators and by

W n'b T(1)

shown in (6.20) actually satisfy all of the relations s a t i s f i e d in and

~.

A

C o n c e i v a b l y one can w o r k out a proof along that line, but it

seems easier to p r o c e e d indirectly w i t h the aid of (6.21), and the proof we give is by i n d u c t i o n on Of course, w e have

Inl

using (!.21).

(6.21) is of interest in its own right.

W mn'b = W mn'l = W mn, so that

(6.21) becomes

not quite the same as w h a t one might have expected, latter is false, W 5'I.

for one can show that

The i n t r o d u c t i o n of

W n'b

Notice that w h e n

W m n = W m ' k ® W n'h. W m n = W m ® W n.

W 15 = W 3'-I ® W 5'2

for general pairs

This is

Indeed,

whereas

(n,b)

b = 1

W 15 #

the W 3'I ®

is n e c e s s i t a t e d by

examples of this sort.

Proof

(of (6.19)):

lem about

W n'l

As we just noted, W n'l

b e i n g a representation.

is

Wn

and therefore there is no prob-

Similarly

we may assume for purposes of this proof that

W n'-I = W -n.

n > i, since

(In particular,

W n'b = W -n'-b

in

general.) S u p p o s e n o w that that teger

c

W n'b

is k n o w n to be a representation,

is an integer satisfying a.

Denote by

Da

gcd(n,c) = 1

the d i l a t i o n o p e r a t o r

and

c ~ a

2

and suppose further mod.

(n)

(D a= v)(j) = y(aj) -

on

for some inS(Z/n~).

Then

D W n'b D -I = W n'cb a T(1) a T(1)

It follows that fact, to

W n'cb

and

Da

W n'b D -I W n'cb = us

is also a w e l l - d e f i n e d r e p r e s e n t a t i o n of

A-equivalent,

W n'b.

Let us apply the o b s e r v a t i o n of the previous paragraph to the special case where

n

has the f o r m

p

r

with

p

a prime congruent to

3 mod.

(4).

If

in

9~

gcd(b,p) also

= i, then either

mod.

(pr).

is a well-defined

A#

b

or

-b

is a quadratic

residue

mod.

(p)

and hence

It follows from the argument of the previous paragraph

that

representation

.

of

A

and is equivalent

to

W n'-+l

=

W-

+n

In order to go farther we are going to need

(6.21).

The idea is this:

Using

denoted

wn'b--of

(6.20), we get a representation--also

generated by

T(1)

and

m.

If we can produce

I

so that

W n'b

the free group

(6.21) holds when #

= T(1) group

and

A #.

o = ~, then (6--.21) will hold for all elements

Let



transforms

prime

e(hja/n),

to

n

with

independently as W n'k

n

of

satisfies

±i k.

valid for an__n_y_odd

A.

Wn .

Of course,

the elements of W n'k

Applying

let's

A^

for the parameter

paragraph

one usually some

h

in

w i t h the maps

is just the Fourier

are taken on even functions,

the prime divisor conditions

as, for

The Fourier

to itself by choosing

W n'k

h. ±i

In particular, on

thus yields

of our computations,

odd functions, that, so long

the operator

this result, w e get the following

table,

n:

Eigenvalue

Inl ~ i mod. either

-i

group of

S(A)

-k

o, and

:

as a map from

The operator

The

of carrying

Instead,

to S(A^).

and then identifying

to

~.

of

~/n~

S(A)

Our remark in the previous

must be similar

wn'k

the character

to the choice of

for

of the ring

is a map from

j = 0,1,...,n-l.

transform corresponding the eigenvalues

denote

o,

is not divisible by any of

a moment w e ' l l work things out for

A^

and

[o] n.

Let's not worry about the details

in this instance,

the Fourier

of

(~/nZ)[in].

of

k

multiple--to

in some extension

, and let

n

generated by the eigenvalues

for counting the eigenvalues

A = g/n~

up to constant

n

n, but

only on

diagonalizes

out this generalization--in look at the upshot

~

on

on the

is diagonalizable,

the eigenvalues

same sort of argument applies more generally, w h e n

w h e n in addition,

o

up to constant multiple,

is similar--again

the ramified primes in the extension of

conditions

under these same conditions

If you will recall, we have seen that when

the diagonalization not

What about

Wn

Inl E 3 mo_~d. (4)

(4)

(InI+l)/4

(InI+3)/4 or (InI-l)/4

the other from above

i

(InI-l)/4

-i

(Inl-l)/4

Two things remain to b e done in our discussion

(InI+l)/4 either

(Inl-3)/4 or (InI+l)/4

the other from above

of

Wn :

w e must handle

the

97

case

n = 2 r, and we must show that w h e n the prime divisors of

mod.

(4)

to 3, the F o u r i e r t r a n s f o r m for the e x t e n s i o n

s i m i l a r to a d i l a t i o n operator. case

n = 2r When

n

are all congruent

~ / n Z ) [ i n]

of

~/nZ

is

There is actually very little w e can say about the

at the moment, so let us get that out of the way first:

n = 2 r, the m a t r i x

[~]n

cannot be diagonalized.

e s s e n t i a l l y u n i p o t e n t - - i t satisfies the e q u a t i o n ferent m e t h o d s from those for odd

n

In fact,

(! - [mn ]2)r = 0.

must therefore be used.

[m]n

is

E n t i r e l y dif-

We w i l l develop one

such method later in this section. Why one should encounter problems only w h e n n = 2r Q(i).

seems to come down to

2

b e i n g the only ramified prime in the e x t e n s i o n

One of the reasons for d e l a y i n g the d i s c u s s i o n of

n = 2r

is to enable us

to look first at some other examples that w i l l help clarify the effect of ramification. Let us now return to the case w h e r e

n

is odd.

the moment, let us put no further conditons on pose that of

A

B

W e are going to construct for

E(n)/Ker L , namely, w e let

E(B)

B:

x £ B

satisfying

on

as the c o n d i t i o n that

x(b) = e(xb)

E(z+yt)f(t+x).

L .

X

readily check that

b c B.

= (x+x',y+y',z+z' +xy'). i s o m o r p h i c to

E(B) ~

{(0,0,b)

sort of S t o n e - y o n N e u m a n n theorem.

~

of

on

S(B)

s

w i l l be obvious.

b y the formula

guarantees that the family

: b ~ B}

B

B, there is an element

should contain no n o n - t r i v i a l ideal of

is, up to equivalence,

w h o s e r e s t r i c t i o n to

Z(n)/Ker L .

(One can rephrase the condition

S(B), w h i c h in turn implies that L

1

an a n a l o g u e of the group

is another character of

Ker(e)

of

Sup-

Getting that requires us to make an addi-

for all

Our condition on

forms a basis for

E(B)

LB = L

and for

and h a v i n g the element

E(B) = E(A)

In all cases of interest to us, the existence of the r e p r e s e n t a t i o n

~/nZ

w e must assume that there exists a character

w i t h the p r o p e r t y that w h e n e v e r

e

B

A

(x,y,z)(x',y',z')

B = A, w e h a v e

Next w e w i l l w a n t the a n a l o g u e of tional assumption on

denote

denote the group w h o s e underlying set is

and w h o s e group o p e r a t i o n is

One should n o t i c e that w h e n

A

other than that it be odd.

is a finite cormnutative ring extending

as a unit.

B x B x B

n

Let

L

B.)

We n o w define

[L(x,y,z)f](t) = {s(yt)

: y ~ B}

is irreducible.

One can

the only i r r e d u c i b l e r e p r e s e n t a t i o n of is a m u l t i p l e of the character

~, a

98

W e are n o w ready to b r i n g £ Aut(E(B))

SL(2,B)

by the familiar rule

q (x,y) = ( ~ x 2 + 2 B y x y + y 6 y 2 ) / 2 . odd.

Since

(x,y,z) I

s t r i c t i n g to

s

Given

> L(x,y,z)~

2

with

is p e r m i s s i b l e b e c a u s e

is an irreducible r e p r e s e n t a t i o n of E(B), there exists an o p e r a t o r

such that for all

u c E(B)

w e have '

c o m p u t a t i o n reveals that w h e n

o £ SL(2,B), w e define

(x,y,z)~ = (x~ + y y , x B + y 6 , z + q o ( x , y ) )

The d i v i s i o n by

on the center of

to scalar multiple,

into play.

L WB u

o = r(1), we may :take

~

WB

is

E(B)

re-

, unique up

= WB L U

n

O

nO=

.

Direct

to be the o p e r a t o r

[wBT(1)f](t ) = E(t2)f(t)

and w h e n

o = m, w e get

[wB f](t) = /

f(x)E(-xt)dx

,

B the i n t e g r a t i o n b e i n g w i t h respect to H a a r m e a s u r e on unitary on

L2(B).

Furthermore,

if

~

is a unit in

B B

n o r m a l i z e d to make and

o

B W-

is the diagonal

matrix

[: :] then w e may take

WB

to be the dilation

o

[wBof](t) = f(~t)

Taking B = A[in] similar,

n

to h a v e only prime divisors

we see that since

~

congruent to

diagonalizes in

SL(2,B)

3 mod.

the o p e r a t o r

up to constant m u l t i p l e to a d i l a t i o n operator, namely,

Our d i s c u s s i o n of

Wn

is now complete,

(4) and taking ~

dilation by

except of course for

is i . n

r

n = 2 , to w h i c h

w e shall return later.

(6.27) Example.

W e are going to look next at some elements of

values are real quadratic irrational numbers. section 2 of the groups w h e r e the eigenvalues of rational.

A

w h o s e eigen-

If you w i l l recall our discussion in

Go, there was no essential d i f f e r e n c e b e t w e e n the cases o

w e r e imaginary and those w h e r e they w e r e real but ir-

The s i t u a t i o n for the groups

SO

in d i m e n s i o n 4 is entirely different.

99

As a first example,

take

(You will recognize

o

2 ~

in

Q(/3)

diagonalize

Wn

d

The primes in must avoid.

with respect to the basis

{i,/3}.)

for all integers

Inl > i.

~

Suppose,

holds either in SL(2,An[/~])

as the matrix of multiplication by a fundamental unit

n

with

that ramify in then, that

SL(2,A n)

if

otherwise.

wno, we find that when

~(/3)

are

gcd(6,n) = i.

3

Our problem once again is to

and

2

A

Set

n

3, so it is these two we = ~/n~

is a quadratic residue

mod.

.

Then

(n) or in

Hence, arguing as we did at the end of our discussion of gcd(6,n) = 1

'

we have

[O]n

diagonalizable and hence

Wn

o

similar to a constant multiple of the dilation operator coming from either of the eigenvalues of Wn

[o] n.

In particular,

is equal to the order of

[o]

n

the number

as an element of

only if the upper right-hand entry of if it is at least as big as at most

3k.

n.

It follows that

eig (n), which is 4 whenever at least logarithmically.

eigo(n) of distinct eigenvalues of

o

k

SL(2,~/nL~.

is divisible by

Now

[o] k = i n =

n, which can only happen

An easy estimate shows that this entry of

ok

is

eig (n) ~ log3(n), which means that, in contrast to Inl ~ 4, the quantity

Actually,

eigo(n)

is unbounded and grows

the growth is somewhat more rapid, as the fol-

lowing proposition shows:

(6.28) Proposition: order in

SL(2,~).

element of

SL(2,~/p~.

positive integer. SL (2 ,~/pJ ~

Proof:

Let

is

Let p

p E SL(2,g)

be an arbitrary element of infinite

be a prime, and let

Then

p

r

= $ +

Furthermore whenever

pkp

'

r

with

denote the order of p' ~ 0 mod.

j ~ k, the order of

[p]pj

rp j-k .

Using the binomial

theorem, we see that if

(p)

q = pJ, then

and in

[O]p k

as an some

100

O rq = $ + pj+kp ' + p"

with

0" z 0 mod (pk+j-l),

Applying the proposition cp pJ ' where

cp

to

o, we see that when

is some constant. When

multiplicity

is the constant

n

The behavior of

eigo(pJ)

c P

c = (eig (p)'pk)-i P

is the average

pJ, we see that the expected

elg (pj) = 4, so that

is the exact opposite of that of

with

k

eigo(pJ ) =

and therefore is independent of

~, where

Let's look more closely at the constant that

Inl/eigo(n )

is the prime power

Contrast this situation with that for pJ/4.

p > 6, we have

Recall now that

or expected multiplicity. pJ/eigo(pJ )

as asserted.

c . P

w• (pJ)

pJ/elg

eig (pJ).

We know from the proposition

some positive integer depending on

p.

There is

not much we can say about either, but here is what we do have, beginning with We know nothing about is

1

k

in general.

for almost all primes or not.

bounded as

p

varies.

So much for

primes divides

p

d

congruent to

We do not know, for example, whether

In fact, we do not known whether

eig (p), but only very little. SL(2,g/pg )

diagonalizes in

±i mod.

(12).

p - i, because the order of

[O]p

eigo(p)

is congruent to

the eigenvalues of

whenever

[o]

is

p

precisely for those eig (p)

is the same as the order of either of

group

p + 1

k

(~/pg)x.

are elements of norm 1

By similar reasoning, 5

or

7 mod.

(12), for then

in some quadratic extension of

P ~ / p g , and by Hilbert's theorem 90, there are quadratic extension of to any element of

A

~/pg .

whose eigenvalues

This estimate

the positivity of the entries of ments of for all

A? ~ > 0

p + 1

such elements in any given

Of course, this reasoning will apply equally well are not

ask what else there might be to say about eigo(p) ~ log3(p).

±i, mutatis mutandis.

eigd(p).

is peculiar to this o.

p

One might

We have already noted that o

insofar as it relies on

Is such a result true for more general ele-

On the other side, it would be nice to know that as

k

Using quadratic

For these primes, we have that

its eigenvalues in the multiplicative divides

k:

k.

We can say a little more about reciprocity, we see that

j.

goes to infinity.

eigo(p)

is

0(p s)

The reason this would be nice is that it

would imply that the average multiplicity which is as close to the known behavior for

p/eigo(p) p/eigw(p)

would be at least

0(pl-e),

as one could reasonably

101

hope,

given that one already knows

eigq(p) # O(i).

This sort of p r o b l e m is, super-

ficially at least, one of class field theory, in the following sense: Let

~p

denote one of the e i g e n v a l u e s of

could w r i t e

~

as

2 + ~,

[q]p--for this p a r t i c u l a r

q, w e

w i t h the u n d e r s t a n d i n g that the s q u a r e - r o o t is taken

P in

Z/p~

;

w h a t w e are about to say, however,

have seen, eigo(p) of the field

is the order of

Ap($p), w i t h

~p

Ap = ~ / p Z

r-th power residue for some

r

applies quite generally in A.

As we

as an element of the m u l t i p l i c a t i v e group

.

dividing

If

gp

a c t u a l l y lies in

Ap

p - i, then the order of

and is an

$

w i l l be at P

most

(p - l)/r.

If

~

fails to be an r-th power r e s i d u e for all

r

dividing

P p - i, then

~p

is primitive,

w e must have that

~p

and

eigo(p) = p - i.

is a square in

A(~p)

N o t e that since

whenever

must b e some r-th power residue for a larger

r

Sp

eigo(p) _< p,

fails to lie in

divisible b y

Ap, or

p - i, b e c a u s e

P the order of

~p

must b e a proper divisor of

that the p r o b l e m of d e t e r m i n i n g for w h i c h

r

the element

~p

eigo(p)

p + i.

In any event, it can be seen

comes down to the p r o b l e m of determining

is an r-th power residue in

Ap(~p), and this is a

p r o b l e m i n class field theory. Let us say a few w o r d s about the r a m i f i e d p r i m e s - - i n the case of this primes that

2 o

and 3. mod.

W e run into the same p r o b l e m here that w e did for

powers of

2

and

3

values of

Wn

and

2,

behaves e s s e n t i a l l y as if w e r e unipotent.

ever the p r o b l e m is not so easily resolved here as for our s o l u t i o n for

~

o, the

How-

~; this w i l l be clear from

~, the k e y to w h i c h w i l l be that w e k n o w precisely w h a t the eigenare, w h e r e a s in the case of

Wn

w e do not. O

One m i g h t ask w h e t h e r the prime 2 causes trouble even w h e n it is not ramified, as for e x a m p l e in

@(~).

If we are w o r k i n g in

there is no great problem, b e c a u s e w h e n tation of

SL(2,Z/n~,

SL(2,~_/n~). lem.

Let

n = 2r

and

with

d ~ i rood.(8), then

one can extend

Wn

to a represen-

and the congruence condition guarantees d i a g o n a l i z a b i l i t y in

On the other hand, the case w h e r e n = 2r

~(/d)

A

= Z/nZ

.

Then

d

d ~ 5 mod.

(8)

does present a p r o b -

is not a q u a d r a t i c residue in

n

n

once

n > 4.

Our p r o b l e m is to get an a n a l o g u e for

tion

~

SL(2,B)

of

here is the embedding

B = An(~)

that w e h a v e already d i s c u s s e d w h e n of

A

> =o

of

SL(2,B)

in

n

of the r e p r e s e n t a is odd.

W h a t w e lack

Aut(E(B)), w h i c h is unavailable

102

because it involves division by

2, a zero-divisor in

B.

One can partially remedy

this by lifting to characteristic zero: Let

B # = g[~],

elements of

N

and form

all of whose coordinates

E(B#), again because of

SL(2,B #)

Z(B#), which, if you wish, may be thought of as the

2

is not a unit in

answer is unique.

get

~

B#

does not act on

o

satisfying the congruence conditions o t

> ~, the point being that

2

the

and reduce it mod. 2rB #, from which point

Of course, this remedy is only partial, because we only

defined on part of

fined on that element of

SL(2,B #)

so that when division by 2 is possible,

We can take ~ +----> =o

as before.

Now

If, however, we look at the subgroup

(2B#), we do get a homomorphism

is at least not a zero-divisor in

~

B #.

B #.

consisting of those elements

~B ~ y~ ~ 0 mod.

we can get

lie in

SL(2,B), and there is no guarantee that

SL(2,B)

that conjugates

WB

is de-

the element you are interested

in to diagonal form. This completes our discussion of the case of real irrational eigenvalued elements of

A.

It is possible to take a different approach to the multiplicity problem that is not quite so dependent on the explicit analysis of the operators

Wn .

We propose

O

to describe that approach now, largely because it does shed some light on the ramified primes, particularly in simple cases like that of If you will recall, the group left translation on the space that satisfy

of all those functions for all

u e H(n), the isomorphism

to an operator in factored form U

1 @ L t.

(x,y,z) e N. U

Let

f

in

We showed in lemma

: Z(n) + S(~) ®

n

~t(n)

C~(N/nn A o)

S(~/n~)

carries

denote the quotient

U

~(n)/Ker(Lt). E(An),

As we saw in our discussion of

~, the group

An = ~ / n Z , the latter group, you will recall, being

operation identify to

£(n) = {(a/n,b/n,c/n 2) : a,b,c c ~} operates by

f(x,y,z) = e(nz)f(x,y,0)

(6.2) that whenever L

Z(n)

~.

Zt(n)

(x,y,z)(x',y',z') Et(n)

with

becomes

~(An).

= (x+x',y+y',z+z'+xy').

Et(n)

is isomorphic to

An x An x An

with the

When convenient, we shall

With this identification,

the natural map from

(a/n,b/n,c/n 2) e---> (a,b,c).

The following lemma is the main step in our second approach:

E(n)

103

(6.29) Lemma:

The maximal abelian subgroups

abelian subgroups of order whenever

0t

S(Z/nZ)

under the operators

Warning: if

n

Zt(n)

into one-dimensional

Zt(n) .

subspaces

all of which are invariant

composite--for

instance,

subgroups

defines an automorphism,

Zt(n)

also denoted

and

~

Ker L t

A.

variance

condition

EI,E 2

,En.

~''"

S(~/nZ)

Et(n)

The associated

onto themselves

q__, of the quotient

it actually leaves one of them--call it @t --fixed, of

in

as follows:

in

will, of course, permute the maximal abelian subgroups of

the unique decomposition

2

n = 8.

Suppose that we are interested in an element carries both

n

of order

One can use the lemma, which we will prove presently,

N

Furthermore,

Et(n), there exists a unique de-

there may well exist non-abelian

__q of

the

{L~ : u ~ 0t}.

is sufficiently

morphism

are precisely

n 2, and all such are normal in

is a maximal abelian subgroup of

composition of

of

Zt(n). Et(n).

and let

into eigenspaces

for

0to = 0 t implies that the operator

Wn

auto-

and therefore

This automorphism Let us suppose that

E 1 e E2e ...eEn {L t : u ~ st}. n permutes

be

The in-

the subspaces Wn

Already one can see that we are getting some picture of

o

after

the fashion of our original attack, a dilation operator being in essence a permutation of certain lines in

S(~/ng).

a litte more concrete by cbserving X_.j : @t + T

such that

further that of

~

E.]

W n E. = E k o ]

were

and their multiplicities.

that for each

is precisely Xk = X_oO.j =

on the character group of

=

What the permutation

@t

E. ]

{v= e S(g/n~Z)

looks like here can be made

there is a character : LtVu = = Xj(U)V V u

e St}, and

Thus the orbit structure of the action

can be used to compute the eigenvalues

of

Wn

We shall go into more detail later; as for now, let us

get the proof of (6.29) out of the way.

Proof

(of (6.29)):

First observe that the center

c e Z/ng}, which is also the commutator subgroup. in any maximal abelian subgroup of mal in Let

zZt(n) As

Et(n), it follows

of

zZt(n)

Et(n)

is

{(0,0,0)

is evidently

:

contained

that any such subgroup is nor-

E%(n). 0t

Zt(n)/zEt(n)

be any abelian subgroup of is isomorphic

to

Zt(n)

that contains

zZt(n).

The quotient

Z / n ~ × ~ / n Z in a natural way, which enables us to

104

pull back × ~.

Ot/zEt(n)

Applying

(a,b)

and

to

g

×

Let

the fundamental

(a',b')

in

@

On the other hand, (O,O,ab'

in

ab' - a'b ~ O mod.

denote the inverse image of

theorem of abelian

equal to the "determinant"

of

in

E#(n)--is

the commutator

(n), w h i c h means Since

0 # in

Et(n)

n2--and in particular

of

@t/zEt(n)

the elements

(a,b)

and

implies Since

@t

first,

maximal

that

b'

and

abelian--as (b',c) = i.

(a/d,b/d,0) @t

unless

Choose (a/c,y,0) with

of

(O,O,O)

0f--not in

@t

of @t

if

if

g.c.d

~

so that

Et(n).

0 # because

It follows c > 1--that

that the

in k ~ ~,

of the form

(a,b),

The condition

ab' = clnl

with

abelian: d.

Then

We may assume,

ab' = cln I

0#

0 #.

is not

then, that

!

divides

d

divides both

centralizing

a.

Hence b

@t

and i,

@t, but would not be

g.c.d. (b,c) = g.c.d. (b',c) = i.

@t

Its commutator with

that

for

@, an observation

b'x + cy = b, and consider

ba/c - ay = (a/c)(b-cy)

@

d > i, w e see that either

Et(n)

Hence

commutes w i t h every element of

Let us show that it centralizes ab' = cln I .

cn

Whenever

then simply that

(b,c) = i; for if

in

is

b' are positive.

ab' = cn, it will follow that

and

since

and

(b',c) = i.

Thus we may assume

y

also generate

is not maximal

Zt(n)

d = i.

since in

becomes

would be an element of

Et(n).

(O,b',O),

ay), which, to

x

a

is the

n2; we will prove that

(a',b').

have a common divisor

desired--or ~.c.d. Since

that

n2

c

does not lie in

@

ab' - a'b.

n 2.

has a pair of generators

(0,b'/d,O)

will only be maximal abelian

in

@

is less than

that the element (O,b'/d,O)

then

that

of

and

g × g--

is abelian.

once again the preimage

(a,b)

(a'+ka, b'+kb)

We are now ready to show that

Suppose

$.c.d.

Consider

Of course, we may also assume

that the order of c > i.

abelian.

yields

is at most

in

n 3, it follows

O t is actually less than

with its generators

that, used repeatedly, (O,b').

itself has order

O

(a',b',O)

that the index of

divides

× ~

with Ot

order of

Z

(a,b,0)

since

£t(n)

Suppose now that the order of

The index of

O

c.

cannot then be maximal

O.

zE(a), which must be

integer

0#

of

in

groups, w e can find elements

0#

some positive @t

@#/zEt(n)

for

Zt(n),

- a'b)

@

that form a basis

and hence also the index of

element

g.

the element

It obviously

(a,b,O)

is

= (a/c)b'x = nx ~ O mod. (a/c,y,O)

centralizes

@#.

is an element

commutes

(O,O,ba/c (n), is equal

of

~t(n)

not in

We have thus shown that under

105

no circumstances

can

0 t be both maximal abelian and of order less than

We now corse to the final assertion of the lemma, abelian in

Et(n),

sum of lines invariant under the operators

Lt , u

As before, we may take the generator of u = (a,b,O)

and

v = (O,b',O).

0 t is maximal

that whenever

there exists a unique decomposition

of

n 2.

into a direct

S(g/ng)

u e 0 t.

0 ¢mod.

zEt(n)

to have the form

ab' = ,Inf" For the operator

Maximality means that

Ltv, we find

f £ S(77/n~Z).

(L;f) (j) = e(-b'j/n)f(j)

Therefore

f

is an eigenvector

one of the subsets Fix one such

(j + a~)/n~

Lt

of

precisely when the support of

v

g/n~, where

j, and as a basis for the functions

the Dirac delta functions u = (a,b,O),

for

~k

concentrated

Lt

modulo scalars,

scales,

Lt u

the operator

operates

a unique basis

L# u

cyclically

is also diagonalizable,

u

supported in k

lies in

0,1,2 ..... Inl-1.

(j + aZ)/nZ, of that set.

take If

then

In o t h e r w o r d s , But

may vary over

at the points

Ltu Sk = e(-bk/n)~k-a

Therefore,

j

f

"

permutes

on t h e

the basis

functions

supported

and a cyclic diagonalizable

of eigenvectors.

This uniqueness

{~k }

is

in

transitively. (j + a ~ / n ~

operator has, modulo

exactly

w h a t we s e t

out

to prove.

D

Let us look in detail now at how lemma (6.29) can be used to calculate multiplicities.

Choose and fix a maximal abelian subgroup

E1 e E 2e ...~Eln I for

be the corresponding

@t, as described in the lemma.

et ÷

such that if

Xj(0,O,c)

= e(-c/n), whatever

E 1 ~ E m e ... e E l n I the same. tion to

f ~ E. J

and j

For each u ~ O #, then

may be.

is

E. J

of

of

S(g/n~

(O,O,c) - - >

Inl

e(-c/n),

Et(n).

L#uf = Xj(u)f.

Now the uniqueness

Let

into eigenspaces

there is a character

implies that no two of the characters

Since there are precisely zEt(n)

decomposition

et

Xj :

Notice that

of the decomposition

xj(J = 1,2 ..... Inl)

distinct characters

of

it follows that these

@t Inl

are

whose restriccharacters

106

are p r e c i s e l y the characters given

X c @tt, set

E

X1 .....

×Inl

Let

equal to the line

@tt = {Xj

{f ~ S ~ / n ~

A.

consider later conditions on

n

given

~.)

implies

Because

Wn

E o

o

in

@tt

X

= E

o c A, and assume that

whenever

X'~

a cyclic

Wn

o t = = @t.

(We w i l l

that guarantee the e x i s t e n c e of such a

L # Wn = Wn Lt u ~ o NO

=

whenever

X ~ @tt.

@t

u e Ef(n), we see that

Thus,

invariant subspace of

for a

@t o = @t =

there corresponds to each orbit of

S(Z/n~.

Actually, m o r e is true:

o

Fix

X ~ @tt

and let

be a n o n - z e r o element of fr-l" know Wn

Let

Inl}, and

: L t f = X(u)f ~ u E @t}. u

X We are now ready to bring in

: J = 1,2 .....

o

Because

X, X ° ~ , - . . , X ° ~ r EX,

and set

be the orbit of

f2 = wnofl,

w n o E x o o r = E , we must have = X

c(o) E ~I~

because

Wn

is unitary.

o

the e i g e n v a l u e s of

Wn

~

wnof r = c(o)f I

for some

'

Let

fl

fr = Who c(o)

(T--we

so that w e k n o w the e i g e n v a l u e s of c(O)•

versus other elements of

Wn

Wn

Yon can see one again A: k n o w i n g exactly w h a t

are narrows down considerably w h a t

one place w h e r e k n o w i n g the s p e c t r u m of

~.

f3 = wnof2, "''' and

there as p r e c i s e l y - - a n d only as p r e c i s e l y - - a s w e know the advantage in dealing with

under

It follows that the minimal p o l y n o m i a l of

X r - c(o)

on this cyclic subspace is

X

c(m)

can be.

Here is

only up to scalar multiples is really

o

harmful• To see how this m e t h o d w o r k s in practice,

let us look at

Wn

when

r n = 2 ,

the one case to w h i c h our previous m e t h o d did not apply:

(6.30) Example:

We consider,

then,

Wn

w h e n n = 2 r.

to produce the subgroup

@t

consider:

is even, w e may take

(0,2s,0), (2s,2S,o),

when

r = 2s

and (0,0,i); w h e n

invariant under

r = 2s + 1

sS (-2 ,2 ,O), and (0,O,l).

n e e d to k n o w are the orbit lengths length 4 unless X

Xo~

takes the values

h a v i o r of

X

2

= X. ±i

Now

for

to have generators

the orbits of

Let

(x,y,z)~ 2

X ~ =

@tt.

~

=

E(n)

to have generators

w

is really helpful.

in

O ft , and all w e really

The orbit of

(-x,-y,z), so

(2s,0,0),

@t

the s i m p l i c i t y of

Xo~

on the two n o n - c e n t r a l generators of

on the center of

four p o s s i b i l i t i e s

•

0t

there are actually two cases to

is odd, w e may take

Again,

W h a t w e have to do now is determine

~. =

Our first p r o b l e m is

2

X

will have

= X if, and only i ~

@t.

Since the be-

cannot be altered, it follows that there are

X, w h i c h w e can describe s c h e m a t i c a l l y as

(i,i),

(-i,-i),

107

(-i,i),

(i,-i).

The effect of

(i,i),

m

is to yield three orbits:

(-i,-i), and

(-i,i),

(i,-i),

{(-i,i),

and

(i,-i)}

when

r = 2s

(i,i)}

when

r = 2s + 1 .

{(-i,-i),

Hence we get two orbits of length

i, one of length 2 (once

orbits of length 4 (once

n h 4).

Each orbit of length 4 determines

space on w h i c h

take

presumably,

Wn

will

±I

and

track down the eigenvalues

but since we are mainly interested

±i of

n h 4), and

a cyclic sub-

each once as eigenvalues. Wn

on the remaining

in qualitative

results w h e n

2 r-2 - 1

One could,

three subspaces, n

is large, we

w o n ' t bother.

D

Although

as technique

for counting multiplicities,

the sort of orbit-analysis--

cyclic operator method just used is really not very good

(except in the one case we

just looked at), it does give one some insight even for general of R-invariant

subspaces

in

C (NAo/Ao).

For that reason,

spend a little more time on it in order to give s criterion o-invariant

D

any square-free

the simplest

integer

situation,

to

for the existence

of a

(positive

w h i c h is when

or negative).

absolute value at least 2 and want to know w h e t h e r mal abelian subgroup

0t

invariant

(6.31) Proposition:

Suppose

suppose

further that either

t0_o ~.

Then there does exist a

Proof:

By hypothesis

Let @t

be the subgroup

follows

from lemma

therefore,

it seems w o r t h w h i l e

@t.

We shall begin with

with

o's into the nature

that

n

under

that

or

-n

o

of

(6.29) that

@t~ = @t.

@t

and

n

of

a maxi-

are as discussed Oust above, of the n o r m map from

maximal abelian subgroup

~

generated by is maximal

Computing with

Zt(n)

n

~.

o-invariant

Ht(n)

has the form

We fix an integer

there exists in

i ss in theimage

there exist integers

o { A

o

and

b

satisfying

(a,b,0),

(Db,a,0),

abelian in acting on

Zt(n). Z

of

and

Z[~]

~t(n).

in = a 2 _ Db 2" and (0,0,i).

It

We need only show,

x Z, we get

108

(a,b)o = ~(a,b) + ~(Db,a) (6.32) (Db,a)o = DB(a,b)

It follows

from (6.32) that modulo _

equal to the product that

(a,b,0)~

z m-t (n), the element

(a,b,0)~(Db,a,0) B

is in

0 t.

are precisely

n = ±2 r special

and

o

of

lies in

shows that

(Db,a,0)~

D = -i.

r.

The

@t

(4), because

~[/D]

When

D E 1 mod.

is in

e t.

particular, (6.30) was a

The construction

is most ef-

is then the ring of integers

is the matrix with respect to the "natural" basis

plication by some unit.

is

In that case, the

w e used in example

just done.

E(n)

0 t, it follows

that are sums of two squares--in

construction

D E 2 or 3 mod.

(a,b,0)o =

zEt(n)

for example, when

integer

case of the general

fective w h e n Q(~)

the integers

for any positive

Since

The same argument

We can apply this proposition, norms

+ a(Db,a)

{i,~}

4, it is more convenient

in

of multi-

to w o r k with

o

in the form

D D B If

An

is the norm of

(0,0,i)

to generate

of

@t

B+a

a + b((l + /D)/2),

then we use

(a,b,0),

(D b,a+b,0),

and

@t.

The condition of proposition existence

= (D-I)/4

For example,

(6.31) is definitely

take

a = 26, ~ = 15,

not a necessary

one for the

D = 3, and n = 15.

Then

o

is 26 26

45 and we may take

@t

to be generated by

(i,0,0),

lem is that in this case the ring

~[a + B~,

g[/D].

B = ±i!)

(The latter only holds if

(0,15,0),

a - B/D]

and (0,0,i).

is not generally equal to

We have now said about all that there is to say about the computation plicities question. not clear.

in dimension

4.

Clearly much thought remains

W h e t h e r it is likely to be profitable

of multi-

to be given to this whole

to pursue

Nor is it clear that yet another description

The prob-

the class field theory is

of the multiplicities

will

109

likely simplify the computation

in any essential way.

gloomy picture by writing out explicitly

a formula for

We shall complete this Wn

valid for all

n

and

U

all

o

in which

y # 0.

(6.33) Proposition: v e S(Z/nZ);

=

Let

o ~ A

have its

K = -(j+hn)

f~/yZ

The argument is essentially

in section 5. omit them.

e(-(~m2+2mK+6K2)/2ny)dhdj

and

G(6,y) = IY1-112llVl-lj=0 e(-Sj2/2y) Proof:

and let

then

(wn°v)(m) = G(~'Y)-I 17Z/nZ v(j)

where

y-entry non-zero,

=

Jz[Iyz e(-Sj2/2y)dj

the same as that used to compute

Wn

As the details are horrible and not really very instructive,

earlier we shall

CHAPTER II

THE GENERAL THEORY

!.

The theorem on discrete spectrum Our emphasis now shifts to the abstract setting of general locally compact

groups.

This is both the natural setting for much of what we have to prove and the

only setting sufficiently general to cover all of the groups - - r a n g i n g

from Lie

groups to adelic groups - - that one necessarily encounters in any serious study of compact solvmanifolds.

Put briefly, our objective is to adapt and generalize the

inductive methods developed by G. W. Mackey in the 1950's so that they apply not just to groups, but to homogeneous spaces as well.

In particular, we want to be

able to reduce the study of harmonic analysis on compact solvmanifolds to a very limited class of examples - - e s s e n t i a l l y those discussed in the first chapter. Such a program requires more than merely recasting Mackey's work, for those results only give one an inductive method that may well be difficult to carry out successfully in any given situation.

In the case of compact solvmanifolds, the pattern

of attack is now well understood.

We shall describe it in detail in

§ii.

In this section we are going to review Mackey's work and some related results, after which we shall show how his main result, the so-called little group theorem, translates into the context of homogeneous spaces.

Throughout this chapter group, and quotient

K G/K.

G

will denote a separable unimodular locally compact

will denote a closed unimodular subgroup of

G

with compact

As a matter of convenience, we will always treat

sisting of cosets of the form

Kg, g g G.

homogeneous spaces we shall encounter. the compact space

G/K

G/K

as con-

This convention will apply to all the

Because both

G

and

K

are unimodular,

admits a unique measure that is invariant the action of

G

111

by translation and gives measure on

G/K.

G/K

total mass

Any integration over

i.

G/K

unless the contrary is explicitly stated. to Haar measure.

We shall call this measure the Haar will be with respect to Haar measure

Thus,

Again, these conventions on

L2(G/K)

G/K

means

L2

with respect

are meant to apply to all

homogeneous spaces to be considered. The action of L2(G/K):

for each

G

on

L2(G/K)

g ¢ G,

defines a unitary representation

the operator

R

acts on a typical

g

(Rf)(x) = f(xg)

F ~ LI(G),

the representation

R

(RFf)(x) = L2(G/K),

on

G.

Proof:

If

the norm topology for C0(G)

F e LI(G),

Because the set of

_F's

Ll(G),

in

K

RF

LI(G)

G K~

K(KN, Kg) =

One easily verifies that

F

with respect to a Haar measure

is a compact operator.

for which

R~

is compact is closed in

with compact support. on

G/K x G/K

in the space _F s C0(G/K),

by

[ F(h-lkg) dk J K

is continuous.

Given

F

b

Now whenever

~ c L2(G/K),

we have

K(Kh,Kg)f (Kg)d(Kg) .

= Thus, ~ Because

f G/K-

is the integral operator whose kernel is the continuous function G/K

G/K.

the operator

we need only verify the lemma for

K =

via

the regular representation of

course--being

then

of continuous functions on

one defines a new function

f ~ L2(G/K)

on

G

the i n t e g r a t i o n - - o f

(~.i) Lemma:

G

f(xg) F (g)dg

-on

R

associates with

f

of

(x ¢ G/K).

Slightly abusing the language, we shall call When

R

is compact, it follows that

RF

is a compact operator.

K_.

112

Because

~

is compact w h e n e v e r

F ¢ LI(G),

the r e p r e s e n t a t i o n

R

behaves

v e r y much like the regular r e p r e s e n t a t i o n of a compact group:

(~.2) The_____ore___mm:The r e p r e s e n t a t i o n

R

d e c o m p o s e s into a direct sum of

countably m a n y distinct i r r e d u c i b l e r e p r e s e n t a t i o n s ~

each o c c u r r i n $ w i t h finite

multiplicity.

This theorem is really an a s s e r t i o n about s e l f - a d j o i n t algebras of compact operators on a Hilbert space.

From this point of view, it can be thought of as a

n o n - c o m m u t a t i v e g e n e r a l i z a t i o n of the spectral theorem for compact s e l f - a d j o i n t operators.

Proofs for the theorem can be found in many places, one of the m o r e

accessible being Actually,

[Gel'fand et al. 1966, pp. 23-26].

(7_.2)

is not quite strong enough for our purposes.

In order, how-

ever, to formulate something satisfactory, we w i l l need to use some notions from r e p r e s e n t a t i o n theory, b a s i c a l l y some facts about

C*

algebras,

general r e f e r e n c e for w h i c h is [Dixmier esp. pp.

59-79,

The key ingredient is the

G,

ideal space of subset

S

of

{ P s G^ :

C*[G] G^

C*

group a l g e b r a of

w i l l be denoted

I c P } .

R^

A,

270-71].

denoted

C*[G].

I

in

C*[G]

denoted by

A ^,

A

G.

such that

This notion of closure defines a topology on If

The p r i m i t i v e

and called the dual space of

is closed if there is an ideal

shall call the h u l l - k e r n e l topology. hull of

G^

an excellent

is a r e p r e s e n t a t i o n of

that w e

G,

then the

:

ker A c p } of

Let us return n o w to the regular r e p r e s e n t a t i o n

R

of

G ^.

Instead of using

for the hull of

R,

w e shall usually use the m o r e suggestive n o t a t i o n

(~.3) Theorem:

If

P g G/K ~ ,

Furthermore,

G/K ^

then

{ P}

is a closed subset of

itself is a countable closed subset of

G^

S =

G^

is the subset { P ~ G ^

G/K.

A

G / K ^.

G^ .

w i t h o u t accumula-

tion points.

(This t h e o r e m m a y be found in

[Fell 1960].

We shall omit the proof.)

So far, w e h a v e b e e n talking in absolute terms about

G/K.

However,

as we

m e n t i o n e d at the outset, w e are really interested in an induction procedure. i n d u c t i o n means here is that we assume to be g i v e n a closed normal subgroup

What

113

H

of

G

such that

(i)

H/HnK

regular representation of

is compact, and

H/HAK

(2)

the problem of analyzing the

has been completely solved; with

are then asked to analyze the regular representation of available from

H/HNK.

In practice,

G/K

H

given, we

in terms of the data

one would not expect to accomplish very much

without making some sort of assumption to the effect that

H

For our purposes,

should be abelian.

Henceforth,

then,

the appropriate assumption is that H

G/H

is not too thin in

will denote a closed normal subgroup of

quotient

G/H.

In addition,

L

will denote

HAK,

compact.

One can picture the situation as follows:

G

G.

with abelian

and we assume that

H/L

is

/G\ H

K

\ / L

As the argument develops,

this diagram will grow somewhat more complex.

Our immediate problem is to relate the subset ponding subset

H/L ^

of

H^

G/K ^

of

G^

to the corres-

We will only afterward undertake a more precise

analysis of the regular representations

of

G/K

and

H/L.

The principal tool is

Mackey's work, which we are now going to formulate in terms suitable to the situation at hand.

Let

P e G/K ^ .

basic result in the theory of equivalence--

Because C*

{P}

unitary representations

C*[G]

P~

G,

we may view

it makes sense to restrict

P~

to

C*[H]

P

is being r e s t r i c t e d - - b y

of

C*[G]

to unitary

whose kernel is

P.

are in bijective correspondence with the

of

of

G ^, we know from a

algebras that there is a u n i q u e - - u p

irreducible representation

Since the representations of

is closed in

H.

pN

as a representation of

G.

Thus

This restriction defines a representation

whose hull we shall denote by res H (P).

res (P) o r - Let us look at

if it is not clear to what res (P)

a little more

closely: If

~

is an automorphism of

This automorphism of

C*[H]

will define a homeomorphism equivalent to

PNIH,

then

H,

then

e

defines an automorphism of

will permute the primitive ideals of ~^ ~^

of

H ^.

will map

If the composition res

(P)

C*[H]

(P~IH)oe

onto itself.

C*[H]. and hence

is unitarily

In particular,

if

114

for each

g e G

(P~IH)O~g

we let

~

denote

g

the automorphism

will indeed be equivalent

[ (P~IH)o~g

to

] (h) = P~

P~IH,

h ---> g-lhg

of

H,

then

because

(g-lhg) = p~(g)-i P~(h) P~(g)

= p~(g)-i [ p~iH ] (h) P~(g) ; thus

~

will map

g

tion on

H^

res

(P)

and leaves

in place of

~ ^ g

(Q)

res

self acts trivially. Q g H ^,

depends

then

G[Q'] = G[Q].

of

hand,

the orbit

Proof;

Since

to

K,

permutes

~k

of

H/L.

G G'Q

K

(!.4) Lemma:

If

that

of

G'Q

connected while will lie in

and

the right

The map

~k

L

K,

determines,

of

H/L ^

k'S ^ = S ^,

H ^

is that

it follows

that

contains

H

Q' ~ G.Q,

the groups

G[Q]

on

HA

are equivalent,

as asserted.

which,

that

is that the

under

G.

In fact,

is always discrete, G.Q = {Q}.

the

so that if On the other

invariant:

k.Q ~ H/L ^ •

~k '

when

k

belongs

and hence defines a h o m e o m o r p h i s m

and let

S

a unitary

operator

on

denote the regular repre-

Then

S

and

in a moment.

G

via composition,

A(k),

it-

Q - - that is, if

the automorphism H

H

it implies

then

(Soak)(h) = A(k)S h A(k) -I

and

on

is abelian,

only when

k s K,

cosets in

sentation

H/L ^

H/L ^

does leave

Q c H/L A

G

because

going to be invariant

Let us call that operator

S=e I

G/H

and not on

g'Q

H ^.

to make about the action of

G

acts by cenjuga-

we will simply write

This is very convenient

L2(H/L).

Hence

on

G

G[Q] = { g c G : g.Q = Q }

L = H n K is normal in

H/L.

G

We shall have more to say about

is not generally

the subgroup

Ordinarily

the action of

G°

only on the orbit

typical case will find Q ~ H/L ^,

the group

to make about the action of

The second observation H/L ^

invairant.

the isotropy group

G[Q]

In sum,

Since we are assuming

hence is a normal subgroup

subset

(P)

to denote

The first observation

whenever

onto itself.

(h ~ H) .

on passing

to hulls,

implies

that

115

~ . 5 ) Theorem:

Let

G[Q]/KnG[Q]

is compact~

Proof:

Q e H/L ^,

Let

continuous

and

group of

and

and define

H ^,

Then

K/KAG[Q]

G[Q] = q-l({Q}).

closed subset of prove that

Q e H/L ^ .

q : G-->H ^ Since

{kng n}

is

(~.3)

G[Q]

is a closed subgroup of

G.

KG[Q]

a

We shall

is also a closed sub-

in

G

in

G[Q]

to some element

g.

We wish to prove that

--> g.Q.

{kngn- Q}

belongs

to

H/L ^

is a discrete subset of

for all H^

H ^.

n,

k

{k n}

for all

n,

of lemma

(7.3),

the sequence

-i n

means that

g E G[Q]

in

K

kngn---> g,

by virtue

constant--which that

and

Since

gn.Q = Q

by theorem

It follows in particular

g c KG[Q]

in

On the other hand,

converge only if it is eventually

that

q

{gn }

Consider the sequence

kngn. Q

is large.

Then

is by virtue of theorem

is compact by proving that

converges

kngn. Q = kn.Q H/L ^

by q(g) = g.Q.

G:

g ~ KG[Q]. have

G,

{Q}

Suppose that we are given sequences that

is a closed subgroup of

is discrete.

it follows that

G[Q]/G[Q]nK

G[Q]

such

we must

so that (7.4).

Because

{kn'Q}

kn. Q = g.Q

for at least one

can once n,

n so

as desired.

A similar sort of argument shows that

The subgroup

G[Q]

K/KnG[Q]

is discrete.

plays a crucial role by virtue

D

of the following theorem

of Mackey:

(7.6) Theorem: representation

of

Let G

Q c H ^,

and suppose

whose restriction

to

H

Then there exists an irreducible representation (i) (ii)

{Q} P

is the hull of

by

is an irreducible

P

G.Q

has the orbit X

of

unitary

for its hull.

such that:

G[Q]

X H, and

is unitarily equivalent

G[Q]

that

to the representation

of

G

induced from

X.

A complete proof of this theorem can be found in [Mackey, 1958].

In order to

apply the theorem in our context, we need to show that all of the elements of "restrict

to orbits" in

H ^,

as in the hypothesis

on

P

in

(7.6).

G/K ^

116

(_7.7) Theorem: res

If

P e G / K ~,

then there exists

some

Q e H/L ^

such that

(P) = G'Q.

Our proof of (_7.7) will involve us for the first time in the serious use of induced representations,

so we are going to take a moment

tions and some terminology. we will also use tations of class.

G

P

To begin with,

P

will denote a primitive

or a single representation,

subgroup

of

and

U

representation--of

A,

representations

G

of

class, whichever I(U:A,G)

then

I(U:A,G)

induced from

as much as possible, of

G

I(U:A,G)

= I(I(U:A,B)

containing

A,

: B,G).

The subgroup

theorem asserts

restriction

g.Ul(g(AnC)g-l),

ACg,

depending

is an equivalence

is appropriate

subgroup

ideal,

H/L ^,

or the like),

class of represen-

element of that equivalence

an equivalence

class of represen-

on the context.

If

A

class of representations--or

will denote the equivalence

A

by U,

to

is a closed a single

class of

or some representative

from context. either

(or

equivalence

even to denote a particular

tations,

G

P s G/K ^

to denote the corresponding

or sometimes

Thus

if

to set up a few conven-

from that

Ordinarily we will try to abbreviate

I(U)

or

I(U:A).

If

B

is a closed

then the theorem on inducing by stages says that Suppose now that

that if for each then

C

is a normal

g ~ G

one sets

I(ug:g(A~C)g -I, C)

depends

subgroup Ug

of

G.

equal to the

only on the coset

and furthermore,

I(U:A,G) IC =

I

I(ug:g(AnC)g -I, C)d(ACg), G/AC

the integration being with respect

to any m e a s u r e on

mapped into themselves by the action of there is a v e r s i o n of the subgroup not, however,

need anything

even will be unimodular

theorem valid w h e n

plication

in the version of the subgroup

(7.7)):

about what measure on

Let

P ~ G/K ^ .

G/AC

We remark that

is not normal.

fact,

theorem may be taken,

of worrying

(for

C

so that there is a Haar m e a s u r e

annoyance

w h o s e null sets are

by right translations.

quite so complicated--in

which the integral in the subgroup

Proof

G

G/AC

on

in most of our w o r k G/AC

with respect

thereby avoiding

to use,

We shall AC to

even the

the one technical

com-

theorem we have stated.

We are going to begin by showing

that

P

lies

117

over some element and let R

is

1

of

HK/K ^.

I(I:K,G).

Let

to

runs over

HK/K ^

From theorem

and

R = I(R0) =

The subgroup

n(Q)

~Qn(Q)I(Q).

Since

Q0 ~ HK/K^

P

of the direct integral

The hull of

I0

G.Q 0

is already closed, because

in HK ^

so that the orbit

G.Q 0

is the hull of

the hull of

PIHK

I 0.

G orbits.

isomorphism

from

v from

is the function of

H/L

of

HK/K.

L2(HK/K) H/L

Then whenever

whence

res H (Q0) a single

P

of

says that

I 0.

H/L,

F E L 2(HK/K),

in

Q0

is a sub-

G.Q 0.

Actually HK leaves

of

G/HK.

I(Q0) ,

the hull of

PIHK = G.Q 0. lies over some H-->

F e L2(HK/K),

for all R0

HK defines

~R0(h)~-i = S(h),

It follows that Q0 IH ! H ^.

h s H.

Q01H

the hull of

K.

then

[~R 0(k)F] (Lh) = [R 0(k)_F] (Kh) = F(Khk)

.

S

~F de-

which says that is a subrepresentation S

=

H/L ^.

We want to show that

= [~F] (Lk-lhk)

Let

then

continue to denote the same

Let us look, then, at the action of

= F(Kk-lhk)

R 0.

I(Q0).

and because

then

is, if

and let

we have S.

Q

composition with which yields an isometric

res H (Q0) = the hull of

orbit.

of

On the other hand,

(~F)(Lh) = F(Kh)

of

is

the integration being

HK ^,

Q0 e HK/K^'

R

it follows

PIHK

resHK (P) = the hull of

is a discrete closed subset of KH

Q

R,

is a subrepresentation

L2(H/L)--that

h ~ H, to

of

is a subrepresentation

is closed in

HK/K,

onto

G/K,

where

is really swept out by the compact group

given by

is unitarily equivalent S,

the multiplicity

is the closure of the orbit

Hence

note the regular representation

R01H

P

But since

onto

of

class of

~Qn(Q)Q

The crucial observation here is that the inclusion map

a homeomorphism

for

integer,

R0 =

I0 = I g'Q0d(HKg)'

Next we are going to show that if Q ~ H/L ^.

Thus the equivalence

we know that

is a subset of the hull of

is a union of

K.

A = C = HK,

{Q0 }

G.Q0

of

is a subrepresentation

such that

G/HK.

fixed,

(2.2),

is a positive

over

PINK

denote the regular representation

By the theorem on inducing by stages,

theorem, applied here with

representation

Thus

R

R 0 = I(I:K,HK).

I(R0).

that there is some

Q0

Let

denote the trivial representation

equivalent

Now

Q

If

Thus

res H (Q0)

k e K

and

is

118

Thus, get

letting

A(k)

~R0(k)v-i

L2(HK/K)

= A(k)

is

that

~

for all

= Q0"

by the subspaces

HK

operates

A(k)L,

In other words,

Q0

lies over

P

Q0 c HK/K ^

lies over Recall

says that clearly

Q--that that

PI H

P

will be an

Since

~M

L2(H/L),

subspace

M

of

is both S-invariant

and

subspace

~0'

L2(HK/K)

S-invariant

closed

on

of

subspace

S-invariant

of

subspace

we see that

SIA(k)L = k.Q,

such

~

L.

Given

P e G/K ^,

in turn,

is spanned

the hull of

we know that

lies over some

is a subrepresentation

is a subrepresentation

G.Q

as its hull,

the hull of

PI H

is itself

Q01H

is

P

Q ~ H/L ^.

We claim that

of

of

it follows

a union of

I(Q0).

Hence

the subgroup

Since the latter

I g'[Q01H]d(HKg)" that the hull of

G orbits,

whence

theorem

PIH

lies in

it must equal

G.Q. G.Q,

theorems

lies over some

sentation

X

of

Q

there exists I(X:G[Q],G)

G[Q]

H/L ^

Let

an element

and we

(~.7),

and hence

we see that whenever

is induced

P c G/K ^.

by some irreducible

then

repre-

Q.

Then for some

X s G[Q]/KnG[Q] ^

P g G/K ^,

such that

Q ~ H/L ^ (i)

X

over which

lies over

Q

P

lies,

and

(2)

= P.

Let

X

Q0

be an element

of

of

H/L ^

over which

G[Q0]/KnG[Q0 ]^.

lies over an element

Q ~ H/L ^.

(_7.7), res H (P) = G.res H (X) = G'Q.

G[Q] = G[Q0]. gX

and

that also lies over

lies over some element X

proving

and

(~.6)

in

(~.8) Corollary:

see that

But

D

Combining

P

K'Q.

lies over

are done.

Proof:

Set

res H (P) = G.Q:

has

P

we

Q.

Q0'

is,

~O

irreducibly

done.

and that

R0-invariant

an irreducible

k e K.

We are now essentially some

on

that a closed

when its image

closed

As we just saw,

Because

It follows

= f(Lk-lhk)

k s K.

and thus will contain

Q = sIe.

(A(k)f)(Lh)

k c K.

precisely

be any irreducible

R01 ~

L2(H/L)

the operator

for all

Ro-invariant

A(k)-invariant Let

denote

Since

is equivalent

G[Q] to

lies.

Applying

By theorem

theorem

(__7.7),

(_7.7) again, we

Now, as we observed in the course Hence

= G[Q0],

we see that

X,

g'Q

then

P

G.Q0 = G.Q,

res H (X) = {Q}.

is equivalent

to

Q,

which means Hence,

whence

if

of that

g ~ G

g ~ G[Q].

119

In other words, theorem that

G[X] = G[Q].

(8.1)] G[X] = G[Q]

By Mackey's

implies

I(X:G[Q],G)

satisfies

(~.6)

res H (X0) = {Q}.

res (p)is the closure of G[Q] follows that

X = g.Xo,

that

Thus, G.X 0.

[Mackey 1958,

is irreducible.

corollary

(~.8).

group" theorem

Let us show

so that

X 0 s G[Q]

= I gX0d(G[Q]g)"

that

Hence

res (P) = G'X. G[Q]

It

I(X) = l(gXo) = I(Xo) = P.

G/H

is abelian,

it is possible for us to refine

[Mackey 1958, theorem

refinement

of Mackey's

"little-

(8.4)].

Q

be an irreducible

unitary representation

be a subgroup of

G

that contains

and is maximal with respect to the

that

(i)

Q

M

(ii)

is a closed normal subgroup of

If

Let Q

Q#

is a representation

M

is irreducible

be an subgroup

of

G

extends to a representation

Q(m-lhm)

= Q#(m-lhm)

equivalence

H

extends to a representation

I(Q#:M,G)

that

= I(X0)IG[Q]

for some

Let

property

Proof:

PIG[Q]

The key here is the following

(~.9) Theorem: M

P = I(X0:G[Q],G )

But we already know that

Because we can assume that

of

M

class of

Q

M.

G

Q#

of

of

H,

and

Then:

lying in the isotropy group

whose restriction

that contains M.

H,

If

to

fixed, which says that

m ~ M

and

i_!s Q,

then

G^

G[Q]

h ~ H, M

we have

all leave the

On the other hand, it is

M = G[Q]--in other words,

~nat Mackey has shown is that this obstruction

G[Q].

and that has the property

M ~G[Q].

Q# of

H

an element of

Hence the elements of

tion in general to finding a representation Q.

of

and hence determines

= Q#(m)-iQ(h)Q#(m).

not true in general that we can take

is

criterion

e = I(X:G[Q],G): We know from theorem

let

that

irreducibility

there is an obstruc-

whose restriction

to

can be realized as a group

extension

(_7.10)

in which

(i)

central in (_7.10)

i

--> T - - >

G#(Q)

G*(Q),

describes

and

Gt(Q)--f > G [ Q ] / H - - >

1

is a locally compact group, (3)

f

is a continuous

the obstruction

is this:

if

(2)

the circle group

epimorphism. M

H

T

is

The sense in which

is a subgroup of

G[Q]

that

120

contains

H,

then

(~.ii)

of

Q

1 -->

(~.i0)

splits.

extension

extends to

if

precisely when the subextension

> f-I(M/H)

Because

(~.Ii)

Conversely,

T

M

G/H

f > M/H

is abelian and

can split only when

f-I(M/H)

> i

is abelian,

T

f-I(M/H) then

is central in

is an abelian subgroup of

(2.11)

is injective, so that Choose

M

duality,

(_7.11) will have to split.

to be a subgroup of

G[Q]

that contains

respect to either of the two equivalent properties abelian subgroup of

G~(Q).

will be an extension in the

category of locally compact abelian groups, where by virtue of Pontryagin T

the

G#(Q),

Gt(Q),

or

(2)

that

M.

It is easy to see in terms of property

of

G[Q]

and hence also of

G.

Q

(i)

H

and is maximal with

that

f

-i

(M/H)

is an

extends to a representation

(i)

that

M

Q#

of

will be a closed subgroup

What remains to be shown is that

I(Q#:M,G)

is

irreducible. Now

Q#

I(Q#),

for that

itself is evidently irreducible, because

will be irreducible whenever the equivalence class of

fixed by no element of

G

Suppose, then, that g e M.

Let

want to prove that that

M[g] = M

g e G M[g]

and that

Q#

M[g]/M

g.Q#

Since

Q

Q#

from

M

to

T

Q#

M[g]

> N

is a cyclic group,

does indeed extend to

(!.12) Theorem: element of

H/L ^

is equivalent

N

to

generated by M

Q#

is left

Q#. M

We want to and

g.

We

but no farther, we can show

extends from

M

to

M[g].

The obstruc-

is a group extension

> M[g]/M

> i.

will be generated by the central subgroup Thus

N

will necessarily be abelian,

M[g].

If in theorem

(and not merely

G

extends to

together with a single other element.

whence

As

M.

be the subgroup of

M[g] = M.

i-->

T

not in

simply by showing that

tion to extending

Because

already is.

we are going to apply Mackey's irreducibility criterion, which says

I(Q#:M,G)

show that

Q#1H = Q

(7.9) H^

can be chosen so that in addition to

the equivalence class of

as assumed in (i)

and

(ii)

(!.9)), of

Q

defines an

then the subgroup

~.9),

it satisfies the

M

121

condition that

Proof:

Let

M/MnK

be compact.

J[Q] = {g g KH:g.Q

is equivalent to

Q},

let

group of

J[Q]

containing

H

to which

Q

extends, and let

group of

G[Q]

containing

J

to which

Q

extends.

M

satisfies the hypothesis of theorem

in

Since

J M

be a maximal subbe a maximal sub-

H ~ J,

the subgroup

(_7.9) and hence satisfies

(7__.9). What requires proof is that

M/MnK

is compact.

(i)

and

(ii)

Here is the cast of

characters:

G

G [Q ] ~

~

KH

~

j[Q]

K

~ j

K[Q]

We are going to begin by showing that the index of Set

K[Q] = KnJ[Q].

Then

J[Q] = K[Q]H,

induces a homeomorphism from (7_.7), onto

H/L

onto

J

in

J[Q]

and hence the inclusion map J[Q]/K[Q].

J[Q]/K[Q]

Let

S

and

T

denote the regular representations of

respectively, and let

by those closed J[Q] ~ G[Q] T-invariant.

L2(Q)

denote the subspace of

S-invariant subspaces on which

and hence fixes

Q

S

~ J[Q]

is equivalent to

(7_.2), L2(Q)

L2(j[Q]/K[Q]) H/L

L2(H/L)

up to equivalence, the inverse image

Furthermore, according to theorem

H

As in the proof of theorem

we use composition with this map to get an isomorphism ~ from

L2(H/L).

is finite.

Q.

and spanned Since

~-IL2(Q)

is

is a finite direct

122

sum of irreducible closed ~-IL2(Q) and

are both closed subspaces in their

L2(j[Q]/K[Q]).

irreducible closed

Because

~-IL2(Q)

L2(Q)

and

of irreducible subspaces.

IJ[Q]:JI

of

J

in

of

J

resh(Qt) Qt#

Now

is the singleton Q.

{Q},

Q#

Q.

res H (Q#%) = {Q} J

Q

Q%

res H (Q%)

extends to

lying over

Q

k

is

the index

only says that

Q%%I H

= I(Q##:J,J[Q]).

J,

J/JAK ^.

= {Q},

Because

which means that

and therefore every

actually restricts on

In particular,

is a maximal subgroup of

yields that

is an S-

lies over some irreducible representation

we must have

Now by hypothesis

representation equivalent to

Because

vL

J[Q].

J

(!.9)

Q# = TI~.

We are going to show that

irreducible representation of

that

Set

L2(H/L)

T-invariant, it contains an

whose equivalence class determines an element of

lies over

and

therefore is the direct sum of a finite number

Using theorem (!.7), we see that Q%%

L.

L2(Q)

respective ambient spaces

is closed and

T-invariant subspace

invariant subspace of k

S-invariant subspaces--in particular,

Q##I H

to which

to a

is equivalent to

is a multiple of

J[Q]

H

Q

Q--note

Q.

extends, theorem

Therefore,

kQ = Q%IH = I(QtT:J,J[Q])I H

= I J[Q]/J [(x'Qtt) IH]d(Jx)

= I J[Q]/J Qd(Jx) = IJ[Q]:JIQ Hence

k = IJ[Q]:J I as asserted. We are now ready to deal with

since

G[Q]/KAG[Q]

compact. M/JnK

b

But

M/(MNK).

is compact by theorem

J[Q]NK/JnK

Since

(~.5),

is finite, and therefore

is a closed and therefore compact subset of

quotient of

M/JAK,

We remark that actually equals

G[Q]nK = J[Q]AK = K[Q], we see that G[Q]/JNK G[Q]/JnK.

G[Q]/J[Q]NK is compact. As

M/MnK

and is Hence

is a

it is also compact.

JAK = MnK

by virtue

of

J's

maximiality, so that

M/JnK.

Putting everything together, we get our main result:

M/MAK

123

(~.13) Theorem:

Let

P s G/K ^.

together with a representation (i) (ii)

M/MnK M

of

M

contains

H,

I(Q0:M,G)

and the equivalence class of

Q0]H

of

G

Q0

is in

M/MnK ^.

Q01H

is in

H/L ^.

In

is irreducible.

is eRuivalent to

The effect of theorem

M

with the following three properties:

is compact, and the equivalence class of

particular, (iii)

Q0

Then there exists a closed subgroup

(~.13)

P.

is to introduce a sharp dichotomy into the

general problem of lifting results from

H

to

G.

We want to explain that now in

some detail by outlining how the rest of the computation will go.

The next few sec-

tions will fill in that outline. Recall that P

in the hull

R

denotes the regular representation of

G/K ^

of

R

P~

of

G.

We define the

ponent

L2(p)

closed

R-invariant subspaces on which the restriction of

lent to ~

P~.

p L 2 (P),

sition of

L2(G/K)

L2(G/K)

the sum being over

L2(p)

is unitarily equiva-

decomposes into a Hilbert space direct sum

P ~ G/K ^.

teed by theorem

R.

We shall call this the primary decompo-

L2(p)

Some authors prefer the term central decomposition. is a fixed primary component of

into a finite direct sum

closed irreducible

or in

LI~...~ L r

L2(G/K).

We can

of mutually orthogonal

R-invariant subspaces, the finiteness of the sum being guaran-

(~.2).

The number

r

of summands is a function only of

not of the particular choice of summands R,

R

L2(G/K)--a slight abuse of language, since we really should speak of the

Suppose now that

in

P-primary com-

to be the closed subspace spanned by those irreducible

The whole space

primary decomposition of

decompose

To each element

these corresponds, up to unitary equivalence, precisely

one irreducible unitary representation of

G/K.

L.. 3

One calls

r

P,

and

the multiplicity of

P

L2(G/K)--notations mul(P).

Our goal is to describe

L2(p)

and mul(P)

as precisely as possible.

In

practice, one attacks this problem by adopting a point of view that is "relative" in the following sense: Rather than trying from the outset to get observes that since

P

lies over some

Q

in

L2(p) H/L ^,

and

mul(P)

directly, one

one can try to solve a simpler

124

problem, namely, describing

L2(p)

and

mul(P)

in terms of

What makes the relative problem simpler is theorem effect,

that one need only consider two possibilities:

and that in which can use theorem case

PIH = Q

I(Q:H,G) = P; (~.13)

L2(Q)

(~.13),

and

mul(Q).

which says, in

the case in which

PI H = Q,

for if one can take care of those situations, one

to patch together the general case.

the restriction problem.

We shall call the

In terms of the examples in chapter I, the

restriction problem is what one encounters in dimension four--see proposition You will recall that the results there were not entirely complete.

(~.9).

It is probably

fair to say that the restriction problem is as completely solved in the general case as it is in the examples we discussed. known in principle, but not in practice. -that is, the case

One could say that what happens is

The situation with the induction problem-

P = I(Q:H,G)--is much better, since the problem is analogous to

our computations in dimension three, where our results were about as simple and complete as one could reasonably hope.

Let us look at both of these problems--

first the restriction problem and then the induction problem--in more detail: We want to think of

H

as known and

G

as unknown,

Q

in

so the restriction prob-

lem is properly posed as follows: Suppose that we are given an element some element of

G ^.

Does there then exist some

More precisely can one describe can one describe

H/L ^

L2(p)

and

P

in

G/K ^

{P s G/K^: P extends Q},

P

in

G/K ^

remaining questions, we have only partial answers.

L2(HK/K)

onto

L2(H/L),

H/L

onto

HK/K

As for the

Q

from HK

H

to

to HK G. H

is

Both probto HK.

The

defines an isomorphism ~ from

as we have discussed in several of the proofs above.

denotes the regular representation of

in

H^

an

Ro-invariant closed subspace of

Q

P in that set,

Q.

Let us look first at the passage from

R0

(otherwise

Q?

What happens is a dichotomy

very different from the problem of extending further from

natural homeomorphism from

and for

extending

in that what happens in extending

lems have their subtleties.

that extends

mul(P)?

Yes, there does exist at least one

within our dichotomy,

that is the restriction of

would not extend to

HK/K, G~),

L2(HK/K).

then since

HK

leaves

the inverse image Let

~

=

Q

v-IL2(Q)

If

fixed is

{W g HK/K : WIH = Q}.

125

Then

~-IL2(Q) = Z "@~2(W),

has to be non-empty, Unfortunately, numbers

Curiously,

Q#

W E ~

remain a complete mystery.

QT s ~ ,

HK

qt

then G/K ^.

up to

In particular,

HK/K ^ ~

L2(p)

This comes as no surprise af-

and every element of

every element of

G/K ^

group.

lying over

lying over

Qt

L2(G/K)--in fact, with multiplicity precisely

for

P

lying over

Qt.

If

G

happens to be a Lie

into

L2(p)

L2(Qt)nC~(HK/K) onto

L2(Q),

f -->

flHK

can be

in such a way as to define an

and the inverse can be explicitly

One would think that this result has nothing to do with

However,

G

to be a catch somewhere, and indeed there is, in

L2(p)nC=(G/K)

isometric isomorphism from computed.

Q.

in general, and the

group, the problem is not so bad, because the restriction map shown to carry

lying over

doesn't seem to create so many probG,

Furthermore,

There has, however,

computing the spaces

G

extends to

occurs with the same multiplicity in (Qt).

~.

§6.

climbing from

actually lies in

mul

in

there is very little more one can say about

mul(W),

If

W

so there is at least one element of

ter our struggles in

lems.

the sum being over all

G

being a Lie

the exact domain of its validity is not entirely clear.

The prob-

lem, of course, is to find a replacement in the general case for the notion of C -vector. The restriction problem is treated in full detail in have a look at the induction problem. told that

P = I(Q:H,G) ~ G ^.

if so, what

L2(p)

is the only element of

G^

are.

Right now, let us

We are now given an element

We want to know

and mul(P)

§8.

(i)

whether

Q s H/L ^

P ~ G/K ^,

and

and (2)

It is a consequence of Mackey's work that

lying over

Q,

P

so the ambiguity we faced in the re-

striction problem is not present here. Recall that respectively~ from

H

to

R

and

denote the regular representations

of

G/K

The key observation is that if we induce the restriction G,

we get a subrepresentation of

the equivalence class of mul(Q)P N R.

S

SIL2(Q)

In particular

stand what is going on with space on which to realize

is mul(Q)Q,

P ~ G/K ^

and

I(SIL2(Q)), I(SIL2(Q))

is

R : I(SIL2(Q)

: H,G) ~ R.

it follows that

mul(P) ~ mul(Q).

and

SIL2(Q) Since

muI(Q)I(Q:H,G)

In order to under-

we observe that one natural Hilbert L2(G/H : L2(Q)),

H/L

the space of all

=

126

square-integrable we can view

L2(Q)-valued functions on

L2(G/H : L2(Q))

recognizes the latter as

as a subspace of

L2(G/L).

Using

that intertwines T,

I(S:H,G)

we can transport

namely, getting irreducibility of

I(Q:H,G)

covering space of

G/K,

L2(Q)

lies in

L2(G/H : L2(H/L)).

L2(H/L),

One instantly

T

from

L2(G/H : L2(H/L))

onto

with the regular representation of

I(SIL2(Q))

I(SIL2(Q))

Since

In fact, the theorem on inducing by stages says

that there exists an isometric isomorphism L2(G/L)

G/H.

to

L2(G/L).

transported to implies that

We need a little more,

L2(G/K). K/L

is

G/L.

It turns out that the

countable, so that

G/L

is a

which suggests averaging over the group of covering trans-

formations as a way of pushing

I(SIL2(Q)) down from

L2(G/L)

to L2(G/K).

In con-

trast to the restriction problem where nothing seems to go right, here one can do no wrong, and indeed averaging does work. then

0f(Kg) =

Ix ~ K/L (Tf)(xg)

The operator the image of in general. P;

0

0

intertwines

Q

(G.Q)nH/L ^.

(~.4).

Hence

L2(p)

and

f g L2(G/H)

IIfII2 =

L2(p).

(G.Q)nH/L ^ K

II@fIl2.

induces

P, K

H/L ^

so L2(p)

that induces

and hence in particular on

is a union of

orbits in

R,

It will not be all of

is not the only element of G.Q

: L2(Q)),

with a subrepresentation of

Now the action of

that there are only finitely many QI,°..,Qr

L2(G/K)

I(SIL2(Q))

in fact every element of orbit

every element of

H^ K

leaves orbits.

(G.Q)nH/L ^,

H/L ^

invari-

It turns out

and that if

are a set of representatives, one from each orbit, then the images in

under the

@

map of the various subspaces

mutally orthogonal and span the construction of

0

are to be found in

§9.

8.

is in

will be a closed subspace of

The problem is that

ant, by lemma

To be precise, if

L2(p).

L2(G/H : L2(Qj))

In particular,

mul(P) =

will he

~j mul(Qj).

answers all of the questions about induction.

Hence

The details

The restriction problem

Because it presents fewer problems, we are going to begin with the case of extending from G

HK

to

G.

Since

G/H

is abelian, we can bring the characters of

into play, which turns out to be the crucial move.

127

Let that

X : G-->

X

T

be a character of

may he viewed as a function on

fines a unitary operator

M(X)

on

mutes the primary components in fact, if then

P e G/K:

M(X) L2(p)

G/K ^,

and

that is identically one on

G/K.

denotes,

One can easily see that

as usual, the

This observation,

so X

M(X)

for the regular representation

x®P-primary component.

and mul(P) = mul(x®P ).

K,

Pointwise multiplication by

L2(G/K).

L2(G/K)

L2(p)

is the

G

deper-

R.

In

P-primary component of

In particular,

x®P

R,

is in

together with Mackey's little-

group theorem, are the keys to proving our first basic result:

(_8.1) Theorem: extends to Q.

G,

Let

and let

Q

be a representation of

W

be an element of

HK/K ^

H

that lies in

H/L ^

whose restriction to

and H

is

Then: (i)

One can also extend

every representation of (2)

Proof:

If

P

and

G

P'

W

to a representation of

extending

W

is in

are two extensions of

G,

G/K ^. W

to

G,

We know that there is at least one representation W 0 = P01HK.

We now have two irreducible representations

of

W0

and the representation

1958, theorem

on

W

H.

(8.1)] that there is a character

acter

must extend to a character G

By virtue

that extends

HK,

namely

It follows from Mackey's little group theorem [Mackey

W = x0®W 0.

tion of

that extends

given to us in the hypothesis, both of which re-

such that X0

P0

mul(P) = mul(P').

G

Set

Q

then

of

Q.

strict to give

and furthermore,

W.

XO

of

KH

identically one on

H

of the Pontryagin duality theorem, the char-

Set

X

of

G.

Then

x®W 0

will be a representa-

P=x®W 0.

Applying the little-group theorem again, we see that every representation of G KH.

extending Since

W q

precisely when

has the form

q®P

is, in particular, P

with

identically one on K,

G

extending

occur with the same multiplicity in

R.

least one element of

W.

R0

a character of

G

q®P

identically one on will lie in

does and will occur with the same multiplicity as

Thus, if any representation of

Let

q

G/K ^

extends

W

is in

G/K ^,

P

G/K ^ in R.

they all are and all

Therefore, we need only prove that at

denote the regular representation of

HK/K.

By the theorem on in-

128

ducing by stages, representation

R

of

is equivalent to R.

Since

subgroup theorem yields that irreducible summand of

G,

I(Ro:HK,G).

acting on

I(W:HK,G) IHK

I(W:HK,G)

irreducible representation of

every irreducible summand of

has

I(W:HK,G)

W

W.

Since

W

W

W.

the

Therefore any

extends to

G~

every

is actually an extension of

I(W:HK,G)

I(W:HK,G)

is a sub-

as a fixed point,

is a multiple of

lying over

Putting everything together, we see that R,

HK ^,

lies over

G

Hence

W.

being a subrepresentation of

is an element of

G/K ^

extending

W.

We have left open the question of what elements of remarked in our outline in it later.

§!,

HK/K ^

extend

this is a rather thorny question.

One other disturbing feature of theorem

(8.1)

Q--as we

We will face

is the use of Pontryagin

duality, which suggests that the abelianness of

G/H

would be interesting to know how much of

survives if we drop abelianness

for

(8.1)

is really essential.

It

G/H. Our next result pins down the multiplicities a little more precisely than

theorem

(8.1).

(_8.2) Theorem: P s G/K ^

Proof: and

Let

lie over

Let

HK/K.

T @ mul(W)W

R

W.

and

W

be an element of

Then

RO,

The equivalence class of disjoint from

representation of

G

induced by

I(T) • mul(W)l(W).

R0 W.

RO,

It follows that

Therefore,

R

is, up to equivalence,

the equivalence class of

mul(W)l(W).

that

occurs with multiplicity one in PIHK = W,

and let

P

occurs in

R

W

R

the

breaks up into

exactly as often as it

in order to prove the theorem,

we have that

G/K

decomposes into a direct sum Since

does in

Because

G,

as usual, denote the regular representations of

T

P

that extends to

mul(P) = mul(W).

with

a sum

HK/K ^

it must be shown

I(W). occurs with multiplicity one in

PIHK.

If we had a version of Frobenius reciprocity that applied here, we could immediately conclude that

P

occurs with multiplicity one in

l(W)...q.e.d.

Unfortun-

ately, none of the standard versions of Frobenius reciprocity in the literature seems to say exactly what we need.

Rather than doing contortions to fit our prob-

129

lem to the literature, we are simply going to give an ad hoc argument directly aimed to prove what we need. Let

W~

space of ble

be a representation in the class

W~.

K-valued Let

One can realize functions on

s: G

> G

I(W:KH,G)

cross-sections, setting

KHg

G/KH--that is, on

(and not on

g

those satisfying

K

denote the Hilbert

L2(G/KH,K)--as

be a Borel cross-section to g ~ G

KH

in

itself).

follows:

G--that is,

an element

s(g)

s

of

is a

KHg

that

(We will always use normalized

S(iG) = 1G .)

b(KHg,h) = s(g)hs(gh) -I.

L2(G/HK,K)

and let

on the Hilbert space of square-integra-

Borel measurable map that assigns to each depends only on

W,

Define

b:

G/KH x G

Then we define a representation

> KH

U~

of

by

G

on

by setting

(U~(g)f)(x) = W~(b(x,g))f(xg),

for all is

x ~ G/KH,

g ~ G,

and

f s L2(G/KH,K).

UN

occurs only once in

I(W),

If you will recall,

R

to a subrepresentation of

U~

which is what we want. is induced by

R.

It follows immediately from this that

R0,

Thus theorem

and therefore

(~.2)

U~-invariant subspace

more, whenever

is a character of

subspace Clearly that

X : G

> T

is equivalent

L2(G/KH,K) =

representations Actually

~xH,

(2)

to

x®(U~IH).

the sum being over

x®(Q~IH)

G

in

(8.1),

Thus the irreducible subrepresentations of

are in

G/KH ^

and

U,

x®U 0

L2(G/KH,K).

FurtherKH,

U -invariant subspace.

What we are going to show is X

in

the

(G/KH)^;

and

X

(2)

traces

(i) that the

G/KH ^.

consequence of the little-group theorem.

the restriction

irreducible subrepresentation of

is equivalent

identically one on

are pairwise inequivalent when

is a straightforward

As we saw in proving

and ~

H

X H = {x~:f E H} is also an irreducible closed U~IxH

U~

implies that there exists at

least one irreducible closed

X

of

decomposes into a direct sum of pairwise

inequivalent irreducible subrepresentations.

W.

U

I (W). We are going to prove that

P

The equivalence class

UIKH = I(W) IKH U

is a multiple of

all lie over

then the little-group is equivalent to

n®U0,

W.

If

U0

is any

theorem says that if then

X=n.

130

The heart of the matter strictions xH's

orthogonal

from

to

H#.



G/KH

orthogonal.

to

Let

x -->

Hence



= 0 a.e.

in

K.

= 0

G/KH,

the set

Let

D

{h(x)

Therefore,

: h E H 0}

be a countable

dense subset of

H

and

dense subgroup of

Therefore,

for almost every

invariant

under the operators

of

x = KHg,

x

in

G/ICH,

{W-(b(x,d))

then

b(x,d)

K

: h ~ H 0}

ible,

in

x,

a closed

it must be true

G/HK.

Thus,

if

Suppose, that

H(x)

tion

h]~(O) = 0

H(x)

is either

0 is

a.e.

f ± H#,

we have

f = O,

we need only

such that for almost all

x

K. and let

H0

U~(D).

be any countable Since

as honest functions

H0 on

is countG/KH.

If

of

is

a.e.

the subset : d ¢ D}.

{h(x)

for all leaves

subspace 0

or

K

It follows

the closed linear span

invariant

: h ¢ H0}. of

d ¢ D.

argument n o w yields

: h ¢ H 0}

{h(x)

K

Going back to the definitions,

W~(HK)

{h(x)

: h ¢ H 0}

Since

K, and since

that

for almost all H(x)

W~

is, for

is irreduc,

itself almost everywhere

in

: H(x) = 0}, then what we wish to prove is that

zero.

on the contrary,

be

H

A simple continuity

W~-invariant

~(0) = {x e G/HK

I(0) has measure

hl~(0)

fixed.

H(x) = the closed linear span of

almost all

H0

of

leaves the closed linear span of

Set

in

= s(g)ds(g) -I

of

x.

defines a function

Because

KH,

= W~(b(x,d))h(x)

W-(s(g)Ds(g) -I)

W~(HK)

h ~ H

under the group

that the dense subgroup {h(x)

be

then

[U~(d)h](x)

we see that if

H0

is dense in

that is invariant

d ~ D,

Each

in order to prove

able, we are free to treat the elements h ~ H0

f a L2(G/KH,K)

X g G/KH^.

prove that there is a coutable dense subset in

we know that the various

and let

.

I G/KHX(X)dx

Because the re-

f = 0.

denote the inner product by sending

~®xH.

=

inequivalent,

H# = ~®xH,

We will show that

~

L2(G/KH,K)

that

U~IxH, x ~ G/KH ^, are pairwise

are pairwise

Let

is proving

everywhere 0

a.e.

for all

that in

~(0)

I(0)

has positive measure.

implies

that whenever

We are going to get a contradiction h s H

implies

I(0) = G/HK

The condition

h ~ H,

the restric-

by showing that

and hence

H

= O.

Let

131

g s G.

Since

vanishes of

H

of

H0

U~(g)h(x)

a.e.

in

= W (b(x,g))h(xg),

~(0)g-l.

must vanish on

However

~(0)g -I

vanish simultaneously

for almost all ~(0)g -I

x

in

we See that if

U~(g)

as well as

H = H, ~(0).

h s H,

we have

differ by a null. set.

Since

In particular,

H(x) = O.

G

U~(g)h

and therefore every element all the elements

off of some set of measure zero in

l~O)g -I,

then

~(0)g -I,

It follows that

acts transitively

on

so that

~(0)

G/HK,

and

~(0)

have positive measure and differ from its translates by null sets only if almost all of

G.

G,

and if

P

We now know that if

is any such extension,

then

P s G/K ^

That might seem to be the whole story, but it isn't. L2(p)

from

philosophy If

L2(W).

R0

R

and

As it happens,

K0

the intertwining

and mul(P) = mul(W).

We must still show how to get The general

to

(RIHK) IL2(p) HK/K

of

RoIL2(W)

and

= (RIHK) IL2(p),

and

HK/K,

then the

L2(p)

R01L2(W)

to

L2(W)

which strongly suggests

not possible.

ought to be the restriction map.

tend to a bounded operator on

G/K,

There is an obvious remedy:

"' L"(P)nC(G/K)

onto

L2(p)--a

L2(W)nC(HK/K)

But of

so restruction

irreducibility

of

the needed bound?

restrict into

in such a way as to ex-

little work with Shur's lemma will then

does not look promising.

L2(p),

at all?

An how is one to use the

What does work is the opposite tack, which is to produce

not vice-w~rsa.

to work with Lie groups,

get

there is a way of arguing in this vein, although it

a bounded inverse for the restriction into

L2(W)

But why

PIHK---the only visible special feature of this situation--to Perhaps

is,

prove that the re-

yield that this restriction operator is a scalar multiple of an isometry. L2(p)nC(G/K)

that

that sets up the equivalence

will generally be a set of measure zero in

striction map carries

G/K

HK when restricted to L2(p) is unitarily equivalent ,) L"(W)--this is exactly what mul(P) = mul(W) means. In

operator from

strictly speaking,

should

extends

of

symbols, what we have is

course

W s HK/K ^

this is a rather sticky point.

are the regular representations

RIHK

restricted

between

is

is certainly clear enough:

representation to

~(0)

The proof of the theorem is now complete.

Let us review our general situation. to

can

operator.

In other words, we map

L2(W)

Even here we hit a technical snag that requires us

or something very like a Lie group,

such as an adelic

132

group.

(8.3) Theorem: restriction to

Let

G

HI< is in

restriction map.

be a Lie group, let HK/K ^,

and let

Then the image of

p

W = PIHK

as before.

some isometric isomorphism we have G/K

~R(h) = R0(h)~ ,

and

HK/K.

acting on particular,

~

L2(p)

to maps

C~

~

from

where

R

L2(p) and

onto

R0

vectors for

L2(p)nC~(G/K)

G

denote the identity element of

fined on

f

c (L2(p)nC~(G/K)) s

intertwines

L2(W)

and

R 0.

Since

ex-

(8.2)

that there is

such that for all

must take the L2(W)

G,

and let

C~

h E HK, of

vectors for

by

s

~R(G)~ -I.

G

In

be the operator de-

Ef(Kg) = [~R(g)~-if](K!).

extends to an isometric isomorphism from R

p

L2(W)nC~(HK/K).

1

by

and

be the

L2(pIHK).

acting on

into

whose

are the regular representations ~

Let

show that

onto

G/K ^

--> C~(HK/K)

L2(p IHK)nC~(HK/K),

It follows from theorem

The crucial fact is that

L2(p)

be an element of

p : L2(p)nC~(G/K)

lies in

tends to an isometric isomorphism from

Proof: Set

P

-i p = ~

L2(W)

We are going to onto

L2(p)

that

will then hold, the theorem will be

proved. We begin by computing then

llfl12 =

isometries.

I lefl 12.

ll~R(g)~-ifl]2

= =

G/HK

I

because both

~

and

R(g)

are

I~R (hg) ~-if (K" I) 12d (Kh) d (HKg) HK/K

GIHK

KH/Kl~R(g)a-lf (Kh) 12d (Kh)d (MEg)

I GIHK

has total mas

isomorphism from

L2(W)

> ~R(g)~ -I Hence

i.

It follows that

onto some subspace of g e G, of

~

onto

L2(W)

e

G

on

does extend to an isometric

L 2(G/K).

we have

is a representation must map

2 = l lfl]2

~R(g)~-ifIl2d(HKg)

Next observe that for all

RIL2(p).

g e G,

G/K I~R(g)~-If(K'- _i) [2d(Kg)

I GIHK

g

for every

f ~ L2(W),

Thus : l l~fl 122 =

since

The key fact here is that if

R(g)~ = g~R(g)~ -I. L2(W)

L2(p),

Now

untarily equivalent to

and we are done.

D

133

As you can see, the Lie condition subspace all

H0

of

g s G.

L2(p)

such that

We can take

on the context.

H0

is used to guarantee

~(R(g)H 0)

that there is a dense

consists of continuous

to be Schwartz functions

or

C

functions

functions,

for

depending

Thus, we could work with the adeles of an algebraic group.

How-

ever, this proof does not seem to carry over to general locally compact groups, although it is difficult Theorems

(8.1),

to believe (8.2),

that the theorem itself is false.

and

of what happens in going from

(8.3)

HK/K

to

to look next at what goes on between what we were worrying than

§5

about in

§5.

give as simple and complete a picture G/K

H/L

as one could expect.

and

HK/K.

onto

homeomorphism Letting

R0

HK/K

gotten from the inclusion

yields an isometric and

S

isomorphism

we get the fundamental

k E K

and

A(k)

~R 0(k)

=

HK.

Let

equivalent

SO to

Q. Q.

~L2(W)

(8.4) Theorem

Let

is in

H/L ^

= ~(Q)

that lie over

the sum

~

- > HK.

from

Composition with this

L2(HK/K)

onto

of

and

HK/K

vR0(h ) = S(h)~

(A(k)f)(Lh)

and that

for all

= f(Lk-lhk)

L2(H/L). H/L

re-

h s H.

on

via

can be extended HK/K ^

From Mackey we know that if Hence

over

~-IL2(Q)

v.

L2(H/L),

If then

to a represen-

consisting of those is in

~,

then

WIH

of our comments on

W s ~.

mul(Q) = ~wg~

R01H of

Thus the equivalence

On the other hand, if L2(W),

W

is, by virtue

denote the representation S

W

denote the subset of

(Notation as above.)

variant subspace of Q.

Q ~

is actually equal to

is

H

isomorphism)

A(k)~.

tation of

mul(Q)Q.

relation

denotes the operator

Suppose now that

Proof:

(the "Noether"

denote the regular representations

spectively,

above,

simpler

because one is asking for far less.

H/L

elements

This problem is, in essence,

The general situation is somewhat

The starting point is the natural homeomorphism from

We are going

W ~ ~

then since

and

K

mul(W).

H

class of

on

L2(HK/K). S01~-IL2(Q)

is an irreducible

WIH = Q,

The desired formula now follows immediately

the equivalence from

Then

SO

is

closed

R0-in-

class of

S01K

v-iL2(Q) = Z@We~ L2(W)

is

134

Notice that, in particular, mul(W) multiplicities.

For instance,

if

~ mul(Q).

mul(Q) = i,

Thus,

then

~ = {W}

T h e r e is not very m u c h one can say about the set special cases.

extending tends to reduce

~

and

mul(W) = i.

except in some rather

What follows is a sketch of what one can say and some remarks on

the problems. Fix an element

W c ~.

the little-group t h e o r e m identically one on space of KH

on

L2(W),

Cm

one element sentation

and if

V.

and

9.

H

S

K

H

lies in the kernel of

Answering

V,

HK/H ^

(i)

however,

and

one must

> T

that is

R o - i n v a r i a n t sub-

(2) (i)

V

of

R0]v-IL2(Q)

is

k n o w at least

of the c o r r e s p o n d i n g repre-

depends very m u c h on what

is reasonably simple, (2),

: HK.

(i)

KH

presents a c o m p a r a t i v e l y

is difficult even in the simplest of §6.

W h e n we come to apply these

to s o l v a b l e Lie groups, we will prove a lemma that enables us to and

of

~,

compute the hull in

H

~'

then according to

then there exists a r e p r e s e n t a t i o n

In order to identify

(2)

~,

is an irreducible closed

as can be seen in some detail in

considerations

groups

(i)

So long as

m i n o r annoyance.

take b o t h

if

for some character

How p r e c i s e l y one can answer

to be.

situations,

is a second element of

m = mul(Q),

(R01K)®V. W,

W'

W' = W ® ~' Thus,

such that

equivalent to

happens

H.

If

KH

§5

at once as simple as p o s s i b l e - - e s s e n t i a l l y as simple as the

and

§6.

There seems to be no general approach to

(2)

at all.

The induction p r o b l e m

What we are about to do is at the opposite extreme from w h a t went on in the previous section, both in the n a r r o w e s t technical sense of the h y p o t h e s e s on the element

Q

of

H/L ^,

and in b r o a d e r sense of the general feeling of the success

of our attack. We are given an element

Q

of

H/L ^

such that

M a c k e y ' s c r i t e r i o n for the i r r e d u c i b i l i t y of must be

H

I(Q:H,G)

I(Q:H,G)

is irreducible.

asserts that

itself--that is to say, m u s t be as small as possible.

extend at all beyond then we shall compute

H.

We are going to show that

L2(I(Q))

and

lution have already b e e n outlined in

mul(l(Q)). §7,

I(Q:H,G)

{g e G : g - Q = Q }

Thus

is in

Q

G / K ~,

will not and

As the m a i n features of our so-

we shall turn f o r t h w i t h to the relentless

135

d e t a i l e d p u r s u i t of our p r o g r a m and shall eschew any further attempts at motivation. Let G/H

L2(G/H/L)

denote the Hilbert space of s q u a r e - i n t e g r a b l e functions on

w i t h values in

L2(G/H/L)

with

L2(H/L).

L2(G/L),

T h e r e is a by now standard m e t h o d for identifying

which, b e c a u s e it plays a central role in our calcula-

tions, we shall take a moment to review. T h e r e exists in

G

a Borel set

In fact,

E

and

G/H

be abelian--of course w h e n

want

E

G.

H

Let

H,

s : G

satisfying

U s u a l l y the set b : G

we can arrange that

> E

E

one likes to think of s

and

E

meets

s

b

G

b

H

s

b(g) = gs(g) -I. g s G,

as a b u n d l e over

G/H

w i t h fibre

as the v e r t i c a l coordinate.

from

E

by a

the unique element

and it w i l l be the map

and

hs(g)

G

at the identity element

h e H

(Hg,h) - - >

be normal in

Multiplying

g ~ G

as trivializing this bundle, w i t h

b a s e coordinate and

H

a Borel c r o s s - s e c t i o n to

denote the "defect" function:

noting for future r e f e r e n c e that if

t h i n k of

properly.

One calls

will go unmentioned,

> H

coset exactly once.

that

H

be the map that assigns to

Hg = Hs(g).

H

is not normal, w e m u s t specify w h e t h e r w e

to meet the right or left cosets of

s(g) e E

Let

that m e e t s each

w i l l exist even if w e drop the h y p o t h e s e s

suitable element of of

E

The general source is [Mackey 1957].

then

s

H

in

that is visible. It is w o r t h

b(hg) = hb(g). H,

G.

If

then one should

as the h o r i z o n t a l or

Thus the maps

G / H x H to G

and g

> (Hg,b(g))

from

are a pair a inverse Borel isomorphisms. of

H.

Thus

Lg

> (Hg,Lb(g))

isometric.

T

b a c k to

formula for

G/H x H

We can also w o r k m o d u l o the subgroup

o

from

If w e normalize

L2(G/H/L)

G/L

onto

onto

G / H x H/L.

L2(G/L)

the Haar measures sensibly,

L

via o

will be

A n a n n o y i n g p r o b l e m is that there is nothing f u n c t o r i a l or "natural"

about the operator Let

to

is a Borel i s o m o r p h i s m from

U s i n g this map, w e get an i s o m o r p h i s m [o~](Lg) = [f(Hg)](Lb(g)).

G

T

o,

w h i c h depends h e a v i l y on the choice of the c r o s s - s e c t i o n

denote the regular r e p r e s e n t a t i o n of L2(G/H/L). To(g)

L2(G/L).

We can use

Let the resulting r e p r e s e n t a t i o n be denoted

is derived as follows:

To.

o The

s.

to pull

136

If

f E L2(G/L),

then

o-lf

is given by

[(-if)(Hg)] Hence, whenever

F e L2(G/H/L),

(Lh) = f(Lhs(g)).

we have

[~- IT (g) oF (Hg') ] (eh) = [T(g)oF] (Lhs(g'))

= [oF] (Lhs (g')g)

= [_F(Hg'g)](Lb(hs(g')g))

Let us compute stant on

H

b(hs(g')g),

cosets,

which is

we have

representation have

S

of

H/L.

T

formula for

Notice that since and

(9.1)

and

I(S:H,G);

I(S:H,G)

Let

P = I(Q:H,G).

and

I(S), in

o

(9.1)

in terms of the regular

F ~ L2(G/H/L) = L2(G/H, L2(H/L)),

is "the" operator that sets up the equivalance

for the right-hand side of

mary component

P

L2(p:R).

of

L2(G/K).

Q

in

H/L ^

is irreducible.

oL2(G/H, L2(Q))

which we shall denote L2(p:R)

(9.1')

in terms of the choice of cross-section

By hypothesis

we see that

L2(G/L),

down into

b(hs(g')g) -

says that

If you will recall, it is the element

T

is con-

TO(g)F(Hg ') = S(b(g')-ib(g'g))F_(Hg'g).

(_9.1') shows is that

between

s

We therefore get for our final result

(and instructive) to rewrite

F(Hg'g) c L2(H/L),

(9.1')

~fhat

Thus

Since

[(T(~)F)(Hg')](Lh) = [F(Hg'g)](Lhb(g')-ib(g'g)).

It is convenient

we

hs(g')g[s(hs(g')g)] -I.

s(hs(g')g) = s(g'g),

hs(g')gs(g'g) -I = hb(g')-ib(g'g).

(9.1)

.

lies in the L2(p:T)

is precisely the s.

we are interested in. Since ~

intertwines

T

I(Q)-primary component of

to distinguish it from the

Our next goal is to map

P-pri

~L2(G/H, L2(Q))

137

(9.2) Lemma: subset of

Proof:

G/H.

Define

Since

H/L ^

that

j

Let

Furthermore,

j : J[Q]

it is a u n i o n of

> H/L ^

by

is a d i s c r e t e subset of

H ^,

that

Warning:

(_7.5). Hence

K.(H/L ^) = H/L ^,

J[Q]

countable.

index is finite, or

(ii)

i

w e can form the "average"

(9.3)

~f(Kg) =

H = {gg G: g'Q = Q},

w e see H --cf.

It follows from lemma

(i)

D

is a d i s c r e t e group and therefore

the index of L

in

K

G/K

I IK:LI I-I ~ x ~ K/L f(xg)

L

in

K

w h e n that

is infinite. G/L

Whenever

w i t h compact support,

given by

.

guarantees that the sum in

(9.3)

is finite

o

is an isometry,

The c o m p o s i t i o n

coo

the c o m p o s i t i o n

~oo

defines an isometry from

is, too:

L2(G/H, L2(Q))

L 2 ~:R).

Remark:

The map

~o~

was first introduced b y

text of theta functions. Let

G

A. Weil in [Weil 1964]

in the con-

Weil's special case is of great interest in its own right.

A ^ denote a (separable) locally compact abelian group that is isomorphic

to its dual group

is continuous.

G.

as the f u n c t i o n on

f

j

The n o r m a l i z i n g factor is d e s i g n e d so that w h e n Haar m e a -

sures are chosen so that

(9.4) Theorem:

The map

c o m p l e x - v a l u e d f u n c t i o n of

Notice that the compact support of g e G.

cosets.

KJ[Q] = J[Q].

if the index of

~f

is a d i s c r e t e

onto a d i s c r e t e subset of

K / K n H = K/L to be

J[Q]/H

is discrete.

and therefore

IIK:LII

is bounded, Borel m e a s u r a b l e ,

into

J[Q]/H

is not g e n e r a l l y a subgroup of

Let us define

for each fixed

KH/H

and since

J[Q]/H

It follows from the lemma that

Then

j(x) = x'Q.

defines a continuous map from

the proof of theorem (_7.4)

J[Q] = {xe G : x'Q g H/L^}.

A ,

and let

a

> a^

d e n o t e one such isomorphism.

We take for our

the locally compact group w h o s e u n d e r l y i n g t o p o l o g i c a l space is

and w h o s e group o p e r a t i o n is

(t,a,b)(u,c,d) = (tud^(a),a+c,b+d).

T x A x A

138

Suppose further that there is a closed subgroup pact.

Let

L 0 = {a ~ A : a^(x) = 1

{i} x L x L 0

of

G,

and let

closed normal subgroup of HnK = {i} x L x {0}. character of

H;

to see that

L2(Q)

for

L

and that

As for

Q,

G/H = A.

since

H

L2(G/K).

The image

with

L2(A).

~o~ (L2(A))

let

K

is abelian,

The map

A/L

H

Q

is an abelian

~oo

L

with

will necessarily be a

×: (t,a,0) ~X

is com-

be the subgroup

We shall identify

is the one-dimensional subspace

L2(G/H, L2(Q))

such that

Notice that

our choice will be the character

identify

A

all x c L},

H = T x A x {0}.

G,

of

of

> t.

It is easy

L2(H/L). Hence we can

will embed

L2(A)

in

are the generalized theta functions that Weil in-

troduced--more precisely they are one of two types of generalized theta functions, the other type living on a lattice in a symplectic group.

let

The proof of theorem

(9.3)

(9.5) Lemma:

Let

denote a Borel set in

L2([, L2(Q))

denote that subspace of

~

functions supported in T°(h), h s H.

I"

hinges on the following observation:

Then

Furthermore,

if

Proof:

T°

We recall that

G,

H

T°

g ~ G

and

we have

h E H,

(9.6)

under

T°(H),

end

and if

I' n

L2(~ ', L2(Q))

(9.1').

Let us take

g = h E H

To(h)F(Hg') = S(b(g')-ib(g'h))FHg'),

Hg'H = Hg'.

Now

since

H

b(g')-ib(g'h) =

Thus, if we set

g[h] = s(g)hs(g) -I

for

[T°(h)F](Hg) = S(g[h])F(Hg).

(9.6)

the same support in since

L2(Q)

It also follows from L2(~, L2 (Q))

G/H,

we get our formula in the compact form

It follows immediately from

T°(h)F have

o__n_n L2(~, L2(Q))

is given by formula

s(g')(g')-ig'hs(g'h) -I = s(g')hs(g') -I. all

is invariant under the operators

are disjoint.

in that formula, which then becomes being normal in

consisting of these

is another Borel set in

is empty, then the representations of gotten by restricting

having positive measure, and

L2(G/H, L2(Q))

L2(~, L2(Q)) I'

G/H

is

that whenever G/H,

c

H,

the functions

which proves the invariance of

is invariant under (9.6)

h

F

L2(~, L2(Q))

S(H).

that the equivalence class of

T°IH

and

on

139

s(g)-l.Q

IZ Now because

{g ~ G:g.Q = Q} = H,

~

Therefore, when

and

I'

f ~

d(~g)

we can have

.

g-Q = g'.Q

only when

Hg = Hg'.

are disjoint, the two representations

s(g)-l'Q d(Hg)

and

f i

s(g)-l'Q d(Hg)

are also disjoint.

Proof (of theorem (9.3)): L2(G/H, L2(Q)). ward.

KH/H

I

coset in

is a Borel set in G/H

F e L2(~, L2(Q)).

is an isometry.

Now

I,

G/H

precisely once.

necessarily have positive measure. Let

I,

and because

lies in the subset

Then

Since

KH/H

We will begin with oF ~ L2(G/L),

being a union of

H

~F(Lg) = [F(Hg)](Lb(g)), ~

of

when

Hg E I,

and

KH

on

L2(I, L2(Q)).

IIoF112 = I IFI 12

G/L.

(9.3)

that defines

collapses to the single term

e, oF

Because

F

(9.6),

is supported in

I

has positive measure in

G/K

G/K

and also carries the measure on

translation-invariant measure on

G/L

K

is the Haar measure of

is finite, then and H/L

G/L

that sum,

o

G/K I"

If

The natural map from

that gives L

all have total mass

i.

vol(~) = i.

are compact, and Then

total mass

has infinite index in

is an isometry and and G/H

G/K

G/L

o

will be a

G/K

G/H, it makes sense

to

bijectively on

~.

I

Hg ~

G/L

in

I,

we need to observe how the Haar measures on

Because

alize our measures so that

L

Hence

to restrict the Haar measure of

vol(1)

o

is supported

we see that because

IIK:LII-IoF(Lg).

In order to use

~

since

it follows that the support of o F

[~(oF)](Kg) = IIK:LII-IoF(Lg)

carries

will

cosets, is a f o r t i o r i a union of

coset exactly once, and because

are related.

~oo

and

(~.6)

G/L

is countable, ~

G/L.

Going back to the sum in meets every

is an isometry on

having compact closure and meeting

cosets and therefore may be viewed as a subset of in

~oo

Let us begin with a special case that is relatively straightfor-

Suppose that

each

What we wish to prove is that

G/L

onto

I

to a

vol(~), K,

where

we can norm-

If the index of

is an isometry if

L

G/H,

]IK:Lll-fold covering

140

of

G/K,

so that the fundamental domain

measure on (9.6),

G/K

is

IK:LII

Because

IIK:LII -I

times the measure pushed down from

~(°F)II22 =

I.

Hence Haar Going back to

oF

is supported in

I ~ IOF(Lg) 12 d(Lg) .

~,

it follows that

We are now going to use lemma is arbitrary. above.

k,k' E K, Lk = Lk'

If

kI

or not.

k'~

and Since

As we just noted,

~oo

to lemma

(9.5),

are disjoint.

R

L2(k~, L2(Q)).

it follows that

~L2(x~,

and

x'

~oo

L2(Q)),

L2(G/H, L2(Q))

the sum being over

H

on

we see that

RIH

(~oo)[L2(x~, L2(Q))]

is

x c K/L.

Hence we will be done if we K/L,

(~oo)[L2(x'~, L2(Q)].

T°

then

However, according

L2(x~, L2(Q))

takes the representation

G/K,

Notice that if

are distinct elements of

is orthogonal to

of

on the two images

has the same properties we needed of

G/H = Uk c K k~,

x

F e L2(G/H, L2(Q))

are either equal or disjoint depending on whether

the representations of

Since

representation

k~

is isometric on each summand.

can show that whenever (~oo)[L2(x~, L2(Q))]

;]~(oF)II 2 = IIoFII 2 = I;FII 2 .

to do the case where

is an isometry on

Therefore then

(9.5)

then the set

k c K,

equal to the orthogonal direct sum

H

will have mass

we get

(9.7)

I

~

and

L2(x'~, L2(Q))

in question to the regular

yields disjoint representations of

and

(~oo)[(L2(x'~, L2(Q)],

so these

images must be orthogonal to one another.

As we just observed, with

~oo

RI~ ( L2(G/H, L2(Q))).

mul(Q)P = muI(Q)I(Q:H,G), does occur in

T° IL2 (G/H, L2 (Q))

Since the equivalence class of

we see that

mul(P) > mul(Q) > 0.

T° IL2 (G/H, L2 (Q)) In particular,

is

P

R.

Let us agree to use

(9.8) Theorem:

Proof:

intertwines the restriction

If

G(Q)

to denote

then

G = HK,

As usual in passing from

v : L 2 (HK/K) - - >

L2(H/L).

and that

here, so that

G = HK

H/L

Recall that R

~(oL2(G/H, L2(Q))).

mul(P) = mul(Q)

to

HK/K, S

and

@(Q) = L2(p:R).

we use the natural isomorphism

is the regular representation of

is the regular representation of

H/L,

G/K = HK/K.

141

Let

A(k),

If

K

for

k g K,

is a closed

least

denote the operator

S-invariant subspace of

R-invariant subspace of

the inverse image under W E H/L ^

and

K

will lie over

L2(W),

L2(Q) t.

RIL2(Q) #

The only elements of

L2(Q) *.

K,

because

G^

Q',

Q'

we must have

Since

L2(k.Q) H/L ^

Since Q.

P

P = I(Q:H,G) Hence

If of

L2(Q) ~ ~ L2(p:R).

RIK ~

We are

That

K.Q,

then since

P

where

L2(k.Q) * =

L2(p:R).

L2(Q) t = L2(p:R)

the sum being over

k.Q,

we have

contribute nothing new to

L2(p:R) ± L2(Q') #.

On the

does not lie over now follows from

W e H/L. P

occurs

mul(Q)

times in

This will be done in two steps:

(I)

If

Kt

and

K

and [t

[

S-invariant subspaces of

are orthogonal closed

are also orthogonal.

and furthermore if to

is

is irreducible,

lies are the elements

We can now complete the proof by showing that

then

Kt

{A(k)K : k ~ K}.

L2(k.Q) = A(k)L2(Q),

but not in

L2(p:R) = ~{L2(W) * n L2(p:R)},

L2(Q) t.

v-iK--in other words

= L2(p:R).

over which

G = HK.

is in

Q.

lying over

L2(Q) ~ H^

Hence the spaces

other hand if

will denote the

then any irreducible subrepresentation

lies over

is the unique element of

traces

Kt

L2(H/L).

It follows from what we just noted that any irreducible sub-

going to show that in fact

k

containing

then

on

That is the crux.

representation of P

L2(H/L),

of the closed linear span of

lies in

W.

Consider

~

L2(G/K)

(A(k)~)(Lh) = f(Lk-lhk)

k,k' c K,

then

Indeed

A(k)K ± A(k)h

A(k)L2(Q)

whenever

L2(Q), k e K,

is either equal to or orthogonal

A(k' )L2 (Q). (II)

If

K

is an irreducible closed

RIK t

is irreducible.

then

~(7 ~)

W e K.Q. L2(Q)).

is

The

Indeed, if

often as

Q

7t

(I)

does in

and

L2(Q).

v(7 #) n

~(j#) n L2(Q) ~ K.

~(7') n L2(Q) = 0 (II)

R-invariant

~ { v ( 7 ~) n L2(Q)},

implies that

7* c K #, we must have

ible, it then follows that Combining

is a closed

S-invariant and equal to R-invariance of

Since

jt

S-invariant subspace of

we see that

or P

K,

then

subspace of

K #,

the sum being over

L2(k.Q) = A(k)(v(7 #) n But,

whence

occurs in

L2(Q),

K

being irreduc-

7t = 0

L2(Q) #

or

K*.

exactly as D

142

If you will recall,

(9.9) Theorem:

If

J[Q]

denotes

K c H,

then

{g s G : g.Q s H/L^}.

mul(P) = ~ mul(W),

the sum being over

W e J[Q].Q.

Proof:

Since

situation, where

W

K ~ H,

we have

an isometry from traces

J[Q]Q.

R

is equivalent

to

we get that

L2(G/H, L2(H/K))

the operator

onto

that we are assuming

here--one

sees that

I(sIK:H,G ).

RIL2(p:R)

Therefore

L2(G/K).

o is, in the present Let

K =

~L2(W),

If one looks closely at the remarks surrounding

being careful to remember there is just

K = L.

Since

so that the representation

L2(p:R) = oL2(G/H,K), SIK = %

is equivalent

K=L,

to

~

(9.1')--

mul(W)W,

and that

where

muI(W)I(W:H,G)

W

T

RIL2(p:R)

traces

J[Q]'Q,

= (~ mul(W))P,

as

desired.

Putting theorems

(9.8)

and

(9.9)

together, we get our final rule for com-

puting multiplicies:

(_9.10) Theorem: is irreducible, of

I(Q)

every

in

K

In general,

if

Q

and if ~ = G'Q n H/L ^,

L2(G/K)

is an element of then

is computed as follows:

orbit exactly once; then

I(Q:H,G) Let

muI(I(Q:H,G))

H/L ^

s G/K ^

~*

such that

and the multiplicity

be a subset of ~ that meets

= ~W c ~,mul(W).

"In general" here means simply that only our usual assumptions namely,

that it is normal in

intersection with

Proof:

I(Q:H,G)

about

H

G with abelian quotient and is compact modulo

hold, its

K.

The theorem follows immediately

from

(9.8),

(9.9),

and the theorem on

inducing by stages.

If we were only interested and furthermore, purposes

we would not have needed theorem

of actually doing analysis on

adequate. spaces

in multiplicities,

G/K,

constructed

in theorem

(9.3) at all.

theorem

What we need is to show how the spaces

@(Q)

we would be done at this point,

(9_.10)

L2(p)

However,

for

is hopelessly

in-

are built out of the

(_9.3). That will be our next goal.

Before

143

we start that, however, that in

(9.10)

a brief remark on

the abelianness

Let us go back now to {g s G:g.Q s H/L^}.

section

s

space"

close examination will reveal

is not needed. L2(Q)),

and the subset

What we are going to show is that if HKj,

subspaces

of

that defines

and furthermore

L2(p:R),

o

(9.11) Theorem:

Let

= ~W

under the condition

E ~#@(W),

so lon$ as the cross-section

and

~

is acceptable

The reason we need some condition

on

s

that the Borel cross-

(~.i0):

@(W)'s

(9.10).

Then

are mutually ortho$onal,

in the sense defined below.

is that we are going to, in effect,

reprove the theorem on inducing by stages, which requires cross-sections subgroups Let

simultaneously. sI : G

the corresponding

that

> G

be a Borel cross-section

"defect" function,

g as

H

s,

to

H

does is acceptable The cross-sections

in

s2

As for

> s2(bl(g))sl(g) s

to

the image of

s2(HK ) ~ K.

to several

Here is the precise situation:

be a Borel cross-section is countable,

@(j'Q)

cosets yield mutu-

be as in theorem

and the various s

then

Our main result is the "Hilbert

result in theorem

G, H, Q,

J[Q] =

j e J[Q],

that distinct

is chosen properly.

version of the numerical

L2(I(Q:H'G):R)

G/H

0(Q) = e(oL2(G/H,

depends only on the coset ally orthogonal

of

(2.10):

to

HK

bl(g) = gsl(g) -I. HK.

Since

G,

will be countable.

and let

Next let

I(Q) e G/K ^

s 2 : HK

implies that

If one wishes,

bI

be > HK

HK/H

one may assume

we take it to be the cross-section in

G.

Any cross-section

for the purposes of theorem sI

in

and

: L 2 (G/HK,

s2

to

H

in

G

that factors

(9.11).

defines operators

L 2 (HK/K))

> L 2 (G/K)

i and ~2 : L2(HK/H'

just as in our discussion

> L2(HK/L)

leading up to theorem

the averaging operator

e2

L2(HK/L)

to

just as

whenever

j ~ J[Q],

L2(HK/K)

L2(H/L))

from bounded,

the operator

e

(9.3).

compactly

supported functions

was defined before. ~2 o °2

maps

In addition, we can define

Theorem

L2(HK/H,

(~.3)

L2(j°Q))

in says that

isometrically

144

onto a closed subspace

02(j.Q)

of

L2(I(j-Q:H,HK))

denotes the regular representation of ®2(j.Q) map

is a closed

into

under

(9.12) Theorem:

L2(G/K).

gl'

L2(HK/K).

so that if

R0-invariant subspace of

L2(G/HK, 02(j.Q))

L2(G/HK, @2(j'Q))

HK/K,

in

Set

Q0 = I(Q:H,HK),

L2(j'Q0:R0).

@I2(J'Q)

and recall that

As usual,

Now using

R0

then ~i'

we

equal to the image of

@(j.Q) = ~(oL2(G/H, L2(j'Q)).

O(j.Q) = OI2(J.Q) .

This theorem is a mildly disguised version of the theorem on induction by stages, the two stages being from

Proof (for theorem (9.12):

to

HK

Using the map

62 : L2(G/HK, L2(HK/H, L2(Q))) ~2(o2(F(HKg))).

H

and from

~2 o °2'

T I : L2(G/HK, L2(HK/H, L 2 ( Q ) ) ) ) - - > L 2 ( G / H [F (HKg) ] (Hb l(g)).

Sl,

to G.

we define an operator

> L2(G/HK, @2(Q))

Next, using the cross-section

HK

by

(B2F)(HKg) =

we define an operator

: L2(Q))

by

(~IF_)(Hg) =

We then have the following diagram: 62

L2(G/HK, L2KH/H, L2(Q)))

L2(G/HK,O2(Q))

~i ]

ali

L 2(G/H, n 2(Q))

~ o o

L 2(G/K)

What we shall prove is that this diagram is commutative. will work with Let

j'Q

u ~ L2(Q)

L2(G/HK), L2(HK/~,

in place of and L2(Q))

Q,

Since the same argument

the theorem will therefore be proved.

f ~ L2(G/HK).

Define

f A u

to be the element of

given by the rule:

{ f(HKg)H

if

x E H

[f ^ u (HKg)](lix) = 0

otherwise .

Then one easily checks by direct substitution that if

(9.13)

F = f A u,

one has

(~ o TIF)(Kg) = I IK:L I l-lf(HKg)u(Ls2(bl(g)) -I bl(g))

= (OlB2F)(Kg) •

145

It remains to show how we can shift the support of is simple enough: presentation

Z

result if we use Let

S (2)

R

we just pull back the representation on

L2(G/HK, L2(KH/H, L2(H/L))),

f ^ u

around.

The idea

via ~ ~ T 1

to a re-

and show that we get the same

~IB2 . denote the representation of

KH

on

L2(HK/H, L2(H/L))

defined

by the rule:

(S (2)(x)F)(Hy) = S(b2(y)-ib2(yx)).(F(Hyx)),

where

b2(Y) = Ys2(y)-i

L2(HK/H, L2(H/L)))

Next define the representation

Z

of

G

on

L2(G/HK,

by

(Z(g)F)(HKx) = S (2)(bl(x)-Ibl(xg)) "(F(HKxg)) .

What we seek to prove is that

~ o ~i A(g) = R(g)~ o T I

and

OlB2Z(g) = R(g)ol$ 2

Before doing so, let us see that this is what we need: Suppose that recalling that

s2

k ~ K.

Then

Z(k)

has its image in

shifts the support of K,

we can choose

k

f ^ u. so that

In fact, s2(k) = k,

in which case I

f (HKg) S (k-ls l(g)ks I (g)-lu, _

x ~ Hk -I

[(Z(k)f ^ H)(HKg)](Hx) = 0

Since

L2(G/HK, L2(HK/H, L2(H/L))

k s s2(K), f s L2(G/HK)

and

Let us look first at

is spanned by the elements

~ g L2(H/L),

intertwining relations between

Z

otherwise

and

R

we see that

G

on

~oT°(g)F = R(g)~oF densely defined. sides make sense.

(9.3)

together with the

L2(G/H, L2(H/L))

Recall that

T°

de-

defined by the rule

.

yields that, whenever both sides make sense,

--we have to be a little careful because

However, if

with

will prove the theorem.

(T°(g)F(Hx) = S(b(x)-ib(xg))(F(Hxg))

Our proof of theorem

(9.13)

~6~i Z(g) = R ( g ) ~ r I.

notes the representation of

Z(k) f A u

F g L2(G/H, L2(Q)),

~

is only

there is no problem and both

Thus, it is enough for our purposes to prove that

146

•iZ(g) = T°(g)TI , for then the relation on

L2(G/HK, L2(HK/H, L2(Q))).

~ o ~l Z(g) = R(g)

Our choice of

s

~ ~ ~i

would hold

proves crucial here, because

b(g) = gs(g) -I = gsl(g) -I s2(bl(g)) -I

= bl(g)s2(bl(g)) -I = b2(bl(g)) • It follows that

TIZ (g)F (Hx) = [Z (g)_F(HKx) ] (Hb I (x))

= [S (2) (b I (x)-ibl (xg))_F(HKxg)] (Hb I (x))

= S(b2(bl(X))-ib2(bl(Xg)))[F(Hkxg) (Hbl(Xg)) = S (b (x)-ib (xg)) [F (nKxg) (Hb I (xg)) ]

= S (b (x)-ib (xg)) [~IF(HKxg) ]

= [TO(g)TI F] (Hx) . Next let us look at two parts.

a I B2 Z(g).

Recall that

representation

R (I)

of

R0 G

As just above, we break the computation up into

is the regular representation of on

L2(G/HK, L2(HK/K))

HK/K.

by

(R (1)(g)F)(HKx) = R0(bl(X)-ibl(Xg))(F(HKxg))

One then easily checks that

The crucial point here is that the operator

L2(HK/H, L2(Q))

L2(HK/K)

the relation

,

OlR(1) (g) = R(g) °l " Thus what we must show is that

R(1)(g)B2 = B2Z(g). to

We define the

e2o2

from

satisfies--by virtue of the proof of theorem

~2a2S(2)(x) = R0(x)~2o 2

for all

x e HK.

Thus

(B2Z(g)F)(HKx) = ~2o2[(Z(g)F)(HKx)] = ~,2o2 S (2)

(bl(X)-ibl(Xg))[F(HKxg)]

= R 0 (b I (x)-ibl (xg))e2o2 [F(HKxg) ]

=

R0(bl(X)-ibl(Xg ))[ (B2F_)(HKxg)]

(9.3)--

147

= (R (I) (g) B21) (XKx) The proof of theorem

Theorem

(9.12)

(9.12),

is now complete.

whatever

the proof of the main result ~*

of

G.Q n H/L ^

L2(I(Q:H,G):R).

not in

W s G.Q n H/L ^,

Since

k c K. K.W.

G

Then

KH.

Hence Thus

oI

so also are

occurs in

Therefore

{012(W)

of

W'

direct sum

we see that

to

where

g'W 0

is not equivalent

but is

= I(Q:H,G) = P

whenever

to

W0,

g

is

which means

are mutually orthogonal. onto L2(G/K

{@2(W)

: W ~ ~t}

Hence the span in ~I2(W).

carries

L2(I(Q:H,G):R).

L2(G/HK,

are mutually orthoL2(G/K)

According

of {@ 12(W)

to theorem R

:

(9.10),

as a whole.

Finally we use theorem

(9.12),

and the proof is complete.

(9.10),

(9.11) and (9.12)

we do not use the

G/H.

and (9.11)

H

theorems,

n -----~ n+l

interested

in the

so we shall push on, even though theorems

are enough for solvmanifold

as the step

condition on

L2(W)),

@2(W) = 02(k.W)

I(W0:HK,G)

exactly as often as it does in

must equal

dernier cri in induction

or (9.11)

=

G'Q n H/L ^,

Anyone who is still reading at this point is undoubtedly

(9.10)

~(W)

~2(o2L2(HK/H,

is in

Since

L2(HK/K))

Since

We remark again that in theorems abelianness

in a subset

@2(W) = L2(W 0 : R0) ,

k E K,

: W ~ ~#}

: W s ~#} .

0(W) = 012(W),

denote

is inequivalent

L2(G/HK,

R I [~I2(W)

~I2(W)

which says that

that

j ~ KH.

I(W':H,HK)

@I2(W).

is their orthogonal

I(Q:H,G)

02(W)

whenever

W0

{@2(W)

from

onto the subspace

W e ~*}

(9.8)

for some

we must have that

The isometry

gonal,

W

orbit exactly once, we have

and let

k.W 0 = W 0

W' = j.W

02(W ) ± ®2(W').

02(W))

that, summing over

Suppose on the other hand that

but not in

that

K

It follows from theorem

is irreducible, in

(9.11)),

that meets each

W 0 = I(W : H,Hk). whenever

(theorem

We are going to prove this assertion next.

Suppose that as before.

its own interest may be, is in essence a lemma in

problems.

If one thinks of (9.10)

in an induction argument,

is not that it be normal in

G,

which is to say that there should exist a tower

the natural "global"

but rather that it he sub____normal, H = H o _c H I c...c_ _

Hr = G

of

148

closed subgroups

in

G

with each

to frame a theorem that relates duces

P c G ~

Hi

normal in

mul(P)

Hi+ I.

to mul(Q)

but that does not involve

when

the tower.

The "global" Q

is in

problem is

H/L ^

and in-

There is an obvious

assertion

to try: Let pact,

~

(2)

W

gH'g -I = H, of

H.

the set of all pairs

is in

(H'/KnH') ^,

and such that

The subgroup

where

K

K

in

~t,

of

W

Q

of

V ( k h ' k -I) = W(h')

each

and

denote

in

G

o,

operates

for all

h' g H'.

=

Also,

~mul(W),

of course,

this conjecture

lems that come up.

For example,

compact?

there exists some

~

via the rule

Let

~*

where

gHlg-i/KngHl g-I

is compact,

Try to relate the family

~i

know what can be said,

g ~ G

is that,

to

h -->

k.(H',W)

meeting (H',W)

of the operators

(9.11). general

set of con-

There are any number of prob-

does it follow that HI

=(kH'k-l,v)

the m u l t i p l i c i t y

the first step up the tower,

for

W(g-lhg)

summing over

means

there are analogues

is com-

such that

be a subset of ~

mul(W)

can be verified.

consider

H'/KnH'

to the representation

HI.

If

gHg-i/KngHg -I

is

to the family

you will see that this problem is a major annoyance.

folds.

(I)

there does not seem to be any reasonable

ditions under w h i c h

and

in which

conjecture

so one can frame a result analogous

Unfortunately,

g E G

on

The obvious

muI(I(Q:H,G))

L2(H'/H'nK).

(3)

is equivalent

orbit exactly once. we get

and

(H',W)

~

for

H,

It would be interesting

and to

even--as we shall see--in the context of compact nilmani-

C H A P T E R III

COMPACT SOLVMANIFOLDS

i0.

Nilmanifolds. As one w o u l d expect, h a r m o n i c analysis on compact n i l m a n i f o l d s is c o n s i d e r a b l y

simpler than h a r m o n i c analysis on compact s o l v m a n i f o l d s in general. special features in the nilpotent general solvable case.

case that have no counterpart at all in the

One therefore t r a d i t i o n a l l y develops the theory first in

detail for nilmanifolds, to that s p e c i a l case.

There are also

and then one reduces,

as best one can, the general theory

That is also the tack we shall follow here.

Let us b e g i n by reviewing a few b a s i c facts about nilpotent Lie groups and their u n i t a r y representations.

We shall use

connected n i l p o t e n t Lie group.

Let

N

N

denote the Lie algebra of

tial and l o g a r i t h m give inverse d i f f e o m o r p h i s m s , The group denoted by space of

N

acts on

Ad, or by N

N

Ad N

: N ~ N

N

N ^.

Ad

or

ordinate to

f

with f

(denoted:

R = {x e N : f([x,y])

= 0

N--

Ad N.

(and not one into

(cf__)(x) : e 2~i£(x)

now that

The dual vector

and as such

"Character" here is u n d e r s t o o d to m e a n unitary

that is, a h o m o m o r p h i s m into * that assigns to f in N the character

Suppose

log : N ~ N.

and the c o r r e s p o n d i n g r e p r e s e n t a t i o n of

character,

N

The exponen-

as a locally compact abelian group under addition,

it has a character group

w e can i d e n t i f y

and

if it is n e c e s s a r y to be so precise.

the so-called "coadjoint" r e p r e s e n t a t i o n - - b y N

exp

N.

via the adjoint representation, w h i c h will be

w i l l be denoted by

We can v i e w

to denote a connected, simply

~x

cx).

Using the rule

£ N ,

N ^. is

in

N .

H < f) for all

A subalgebra

if

f

H

vanishes on

of

N

[H,H].

is said to be subIf

y e N}, then the largest p o s s i b l e d i m e n s i o n for

150

a s u b a l g e b r a subordinate to

(dim(N)

+ dim(R))/2,

To say that

then

H < f

f H

is

(dim(N) + dim(R))/2.

is called a p o l a r i z a t i o n for

is equivalent

c h a r a c t e r of the connected subgroup character from

H

to

to saying that

H

of

N

h~

depends only on

f

H

and

H.

N, namely

l(f,H).

is a p o l a r i z a t i o n for

is a

Inducing this

One shows that l(f,H)

f, and in that case

and not on the choice of the p o l a r i z a t i o n

K i r ( f ) - - o r if necessary,

dim(H) =

> Ef(log(h))

N, w e get a unitary r e p r e s e n t a t i o n of

is i r r e d u c i b l e if, and only if,

use

H < f ~.

underlying

l((sf) olog : H,N), w h i c h w e shall denote simply as

l(f,H)

If

H.

We shall

K i r N ( ~ ) - - t o denote the i r r e d u c i b l e r e p r e s e n t a t i o n

I(f,H), H any p o l a r i z a t i o n for

f.

Thus

Kir is a map from

N

to

N .

One shows

that this map is surjective. * Let

~

m o r p h i s m of Kir(f) oe Hence

be an a u t o m o r p h i s m of N .

Kir

N /Ad N

{~' ~ N Proof:

onto

N

~

denote the c o r r e s p o n d i n g auto-

is natural in the sense that N .

is constant on orbits of

(i0.i) L e m m a H

Kir

define the same element of

jection from

let

Then the map

N, and let

If

Ad (N)

~

is inner,

in

N .

then

Kir(~ f)

and

K i r ( f ) o ~ = Kir(f).

If fact, Kir

defines a b i -

.

(Pukanszky):

Let

f e N , let

b e the connected subgroup of

N

H

be a p o l a r i z a t i o n for

underlying

H.

Then

(Ad~)~

f, and

=

:~'IH:~I~}.

The proof goes by induction o n the d i m e n s i o n of

First w e observe that if we may assume that

H

N, as one w o u l d expect.

H = N, then the a s s e r t i o n is certainly true.

is a proper s u b g r o u p of

a c o n n e c t e d normal subgroup

M

of

N

Therefore

N, w h i c h implies that there exists

that contains

H

and has c o d i m e n s i o n one in

N. The i n d u c t i o n h y p o t h e s i s

AdM(H)'(L]M)

= {£' { M

W e w i s h to p r o v e that

asserts that the lemma is true for

: £ ' I H = f]H}. ~' = AdN(h) ~

allows us to do is to find an element w e may assume that Because

Suppose now that

for some h c H

h c H. such that

£' { N

M.

Thus

and

_

=

What our i n d u c t i o n h y p o t h e s i s f'IM = (AdN(h)f)IM.

Thus

f'IM = flM.

% ' I H = %IH,

the two r e p r e s e n t a t i o n s

I(%',H)

and

l(f,H)

are equal.

151

H

Since I(f',H)

is a p o l a r i z a t i o n is irreducible,

I(f',H)

= KirN(f').

f,

no e N

usual one.

Since

also polarizes

n O c M.

~',

Hence

and that

can hold only if

f' ~ AdN(N)" ~.

What we are going to prove

By virtue of the naturality

= n0"KirM(flM),

[(AdNn0)!]IM

criterion

= !'IM = flM,

requires

that

the action of

it follows

= I(KirM(flM)

N

that

M^

KirM, being

K~!M(flM)

: M,N),

{x ~ N : x'KirM(fIM)

on

of

the

=

so that Mackey's

= KirM(flM)}

= M.

n o E M. We conclude

(AdMn0)'(flM)

the proof by showing

= flM.

g £ M , the identity underlying

It follows component

{x~ M : g([x,y])

{x ~ M : (AdMX)g = g}

exp(R(fIM)),

N/F.

of

{x e M

= 0

N~

N

N = N~IR.

Campbell-Hausdorff

implies

of

N

for all

N*

f

_fIN~

Since an element

Applying

of

implies

exp(N~)

N*

of

that = N~

M

if

subgroup

is nilpotent,

Furthermore, w e see that

subgroup

F

log(F) ~

with compact contains a basis

generated by

f (N)

F) if

of

N.

H n N~

~, whence

A closed subis dense in

~ ~, or in other terms,

is determined by its restriction

the luxury of identifying

N~

with

for

log(F) ! N.

is a Lie algebra over

to

no

H, which is what we wanted.

is a subgroup

(with respect

flM

that

that whenever

is the connected

this to

that

N~

Notice

formula

as one can easily see.

a discrete

N/F

n O E H.

y E M} = R(g), and since

contains

is called rational

are going to permit ourselves ments of

g.

One shows

that

An element

c N*~.

of

is called rational of

: (AdMX)g = g}

to denote the vector space over

We will then have

H

implies

which in turn must lie in

The compactness

We shall use

nO E M

is itself connected,

Let us assume now that quotient

that

from the Campbell-Hausdorff

will lie in any polarization

must lie in

to

H.

if

N , we

the rational

ele-

N .

A subalgebra H.

is irreducible.

f' = AdN(n0)" ~.

On the other hand, KirN(f)

irreducibility

group

H

= KirN(f')

satisfying

KirM([(AdNn0)~]IM)

n0"Kir(fIM).

N.

= KirN(f)

n O £ H.

we see that

R(g)

that

KirN(f)

Let us begin by showing that

Hence

I(f,H)

w h i c h means

However

Thus there is some is that

for

or subspace

An induction argument

rational polarization

for

H

of

N

is called rational

that is by now standard f.

if

shows that if

One other rationality

H n N~

is dense in

f ~ N~, there exist

result is worth

recalling:

152

If

H

is a rational

and the subgroup and

n e N.

[H,N]

subgroup

of

N, then so are the centralizer

generated by the commutators

In particular,

(10.2) Theorem [Ad (N)'f]

Proof:

N N ~

F

If

dim(N) N

may happen

(Howe-Moore-Richardson):

If

upper central,

H

in

with

N

h { H

and lower cen-

to be.

f ~ N

and

Kir(f)

E N/F

, then

is non-empty.

Again the proof is by induction

in which

[h,n] = h-ln-lhn

the terms of the derived,

tal series are all rational, w h a t e v e r

of

= 1

on the dimension

of

N.

to the reader and proceed immediately

itself polarizes

f, then

Kir(f)

is simply

We leave the case

to the induction

the character

step.

(sf)'log

^ of

N, which is in

!Z,

N/F

only when it is identically

which surely implies

~, in which H

Now

Notice

Let us assume

then that

case there is a rational p o l a r i z a t i o n

is contained that

f ~ N .

KirN(f)

in a rational

ideal

= I(K~M(fI~)

: M,N).

We are now going to use the results

M

of

one on

H N

for

F, whence N

f

f(log(F))

does not polarize that is proper in

that has codimension

from ~!.

By hypothesis

N.

one in

K~[N(f)

N.

lies in

^ N/F

, and we constructed

be induced from ~ M

M.

such that

M

so that

M/M n F

Theorem KirM(~)

(7.13) applied ^ ~ M/M n F and

would be compact and

KirN(f)

would

to this case says that there is some l(KirM(~)

duction hypothesis says that we can choose g * any element of N that extends g, and let K

: M,N) = KirN(f). * so that it lies in M~. polarize

g.

Since

The in-

# Let

g

KirM(g)

be in-

# duces an irreducible that

representation

_KirN(g# _ ) = KirN(f) We can actually

Given

: M,N).

f { N

along with obtaining and a p o l a r i z a t i o n

equal to the set of all pairs C(H,flN)

N, we see that Hence

K

to be the quotient

(Ad(n)H,

(Ad(n)H, of

for

theorem than formula

Ad(n)H = Ad(m)H

n

traces

by the equivalence

(Ad(m)H, and

g

and

(10.2).

This

for the multipli-

f, we shall set

Ad (n)f), where

C+(H,flN)

Ad (n)f) ~

a detailed H

also polarizes

g # E N~* n Ad * (N)f.

give a much more precise occurence

will be our next goal, cities.

= I(K~!M(g)

of

C+(H,flN) N.

We define

relation

Ad (m)f) (Ad (n)f)IAd(n)H

= (Ad (m)f)IAd(rr~)H.

153

Pukanszky's from

N/H

lemma implies that the map onto

(K,g)

in

g(K n log(P))

defines a bijection

(Ad(n)H, Ad (n)f)

c(H,flN).

We shall use pairs

n -->

C(H,f/F)

to denote the subset of

c+(H,fIN)

c

in which

K

c(H,flN)

defined by those

is rational with respect to

r

and

f

in

Z.

(10.3) Theorem:

Let

f • N~, and let

be a ~olarization

H

for

N

that

^ is rational with respect t~o

F.

Then

c(H,f/F)

is non-empty.

Furthermore,

precisely

the number of

F

occurs in

Kir(f)

N/F

of Howe and at the same time, essentially

that Richardson

in

L2(N/F)

is

C(H,flF).

L. Richardson

The theorem as it is stated here is due to R. Howe. independently

Kir(f)

the multi~licit~ of

orbits in

if, and only if,

required some (relatively innocuous)

proved,

the same theorem,

conditions

except

on

H.

The proof

on the dimension of

N.

We begin by

we are about to give is basically Howe's.

Proof:

The proof goes, as usual, by induction

remarking

that the theorem is evident

is no difficulty

in handling

dim(N) > 1

H #

and

N.

rational with respect to

the case

Let

M

respect to

F

(Ad (x(t))f)M

and complementary is in for all

F

to

dim(N) = i.

subgroup of M

in

if, and only if, t • IR.

N.

N.

N

N.

c(H(t),ftlM ) ÷ c(H,flN ).

subset of

C(H,flN)

H, is

In addition,

let

whose image is rational with

We assume that

t c Z.

there

then, that

that contains

one in N

In particular,

We can assume,

Set

x

has been chosen

H(t) = Ad(x(t))H

By virtue of Pukanszky's

injection

(10.4)

is equal to

F, and has codimension

denote a one-parameter

x(t)

H

denote an ideal in

x : IR ÷ N

so that

if

lemma,

Therefore we shall view

and

f

= t

there is a natural

C(H(t),ftIM)

as a

That done, we get

C(H,flN)

One should observe that when

= ot•]Rc(H(t),ftlM).

s # t, the sets

disjoint, which is a consequence

of Mackey's

I(KirM(f t) : M,N) = KirN(f), we can have

c(H(t),ftlM) irreducibility

x(t)'KirM(f0)

and

c(H(s),fsIM)

criterion:

= x(s)K~M(f0)

are

since only when

154

x(t)-ix(s) of

~ M--that

C(H,fIN)

over

is, when ]R

s = t.

What

corresponding

(1--0.4) yields,

to the fibration

of

therefore,

is a fibration

N/H

]R

over

with fibre

M/H. It follows {s ~ IR

from (10.4)

: c(H(s),fsIF

and only if t = s + k

n M)

that

C(H,f[F)

is non-emEty

t ¢ ~, one has that with

s E S

and

c(H(s),fs[Fn

and

0 ! s < i}.

c(H(t),ftlF

k ¢ ~.

seen to b e the sum, over all

= ut¢iRC(H(t),fti£

Hence

n M)

that

is non-empty

s ¢ S, of the number

Let

Because

the number of

F

orb(s)

x(t)

c F

if,

if, and only if,

orbits in of

S =

F R M

c(H,fI£) orbits

is

in

M).

Let us look next at some multiplicities. 0 ~ s < 1

n M).

and

x(s)KirM(fIM)

S ! S# : when

s

~ M/£ n M^}.

is in

Set

S#

equal to

{s ¢ IR

:

One can see from the induction hypothesis

S, the set

C(H(s),fsiF

n M)

is non-empty,

and hence

^ KirM(~),

w h i c h is equal to

orb(s).

To pass from

theorem

(9.10), which,

in

the multiplicity

S#

plicity of

KirN(f )

x(s)'KirM(f[F) , occurs in

M/M n F

to

N/F, we use the results

interpreted

in

of

in the present

KirM(f s)

L2(N/F)

(1--0.5)

in

S = S #, we w o u l d have

showing

C(H,fIF).

that

Suppose is then in element. both

H

that

f

mul(s)

Hence we can complete

s

is in

S#

is mul(s),

= orb(s)

s

then the multi-

and the sum in (1--0.5)

paragraph,

be the number of

£

the proof of the theorem simply by

and is a rational number.

s

is rational,

are rational,

that

(H(s),~),

KirM(f s)

it follows

is in

we see that

Our argument in the previous

C(H(s),f

the element

(Ad (x(s))'f)IM , are both

S #, w h i c h means to the pair

says that if for each

.

What we must show is that

Because

f , which is

of §9, particularly

S # = S.

S.

and

with multiplicity

is

would, by virtue of the argument of the previous orbits in

context,

L2(M/M n F)

~seS # mul(s)

If we knew that

M/M n £

that

x(s)

C(H(s),~/F

paragraph

contains

is rational

in

H(s), which is just

rational. M/M n F ^.

IF n M)

We shall prove that

Now

at least one N.

Since

Ad(x(s))H,

s, you will recall,

Applying

s

and

is in

the induction hypothesis

n M) is not empty, whence

shows that we can prove

s E S.

S # ! S--and

IB5

thus

S # = S--simply b y showing that

direct w a y of proving p r o b l e m is that shows that

H

H

c o r r e s p o n d i n g set

Unfortunately,

there seems to be no

S # ! Q--it may not even be true in this generality.

and

and

S # ! ~.

M

M

are "too general".

The actual p r o o f is indirect:

can b e replaced by another pair

S#

can b e provcd rational.

first to give conditions on

H

and

M

The

H'

and

Our program,

M'

one

for w h i c h the

therefore, w i l l be

that will guarantee

S# Z ~, and second, to

show that we can always arrange to w o r k w i t h such pairs. It is h e l p f u l at this point to reduce the p r o b l e m to the case in w h i c h the center

zN

of

N

has dimension one and

is to show that if

K = zN n ker(f),

f

is faithful on

then we can factor out

R e p e a t e d use of this o b s e r v a t i o n yields that whence

problem.

to show,

then, that

tion for

f.

f

K

may b e assumed faithful on

in

rational.

f, so in particular,

N/K

N.

K

ponent of

KirN(f).

L2(N/F)

KirN(f) in

w i l l lie in any polariza-

H/K

zN

and

f

with

that

Next set

K = exp(K)

must be constant on right

FK

is the same as that of

dim(zN) = 1

and that

f

and is to be denoted by

the c e n t r a l i z e r y

Z(y)

is rational.

d i m e n s i o n on e in

of

y

in

N

cosets.

is

If

KirN/K(f)

y

is in

in

L2(N/FK).

in

K.

z2N

does.

lies in com-

Putting

We may assume,

zN.

z2N N

{x ~ N :

but not in

zN, then

that is rational w h e n -

dim(zN) = I, the ideal M

K

Hence the m u l t i p l i c i t y

is a proper ideal in

N, just as the ideal y

K

KirN(f)-primary

N, w h i c h is the ideal

z2N.

In fact, b e c a u s e

w e can find a rational element

a polariza-

and o b s e r v e that

is faithful on

C o n s i d e r next the "second center" of [x,N] ~ zN}

N/K,

does not affect w h a t the rational p o l a r i z a t i o n of

e v e r y t h i n g together, w e see that it does no h a r m to factor out therefore,

H

are rational,

If follows that any function in the

L2(N/F)

zN

is viewed as a linear functional on

Furthermore, b e c a u s e both

Thus, factoring out

the kernel of

f

K

are p r e c i s e l y the subalgebras

are in any s u b s t a n t i a l way.

ever

zN,

can be factored out w i t h o u t affecting the

It follows too that if

its p o l a r i z a t i o n s in

of

N/K.

and w o r k w i t h

It is an easy consequence of the definition of p o l a r i z a t i o n s that

lies in every p o l a r i z a t i o n for

f

f

K

The idea

dim(zN) = i.

We w a n t

tion

log(zN) = zN.

Z(y)

w i l l h a v e co-

Let us assume for the moment that

such that

Z(y)

actually equals

M.

156

We shall prove that under these conditions, Let Y

denote the subspace of

is an ideal in

F n Y. and N

Y

N. H e n c e

z

is in

zN.

~

x(t) E £

-I -i x 7£ xy

commutator

~

in

Recall now that

precisely when

t ~ ~.

is

_zr

y

Q:

together w i t h

is a n o r m a l subgroup of

and

w h o s e image is c o m p l e m e n t a r y to

does lie in

spanned by

Y = exp(Y)

W e can choose elements

(2)

N

S#

Y

so that

x : IR + N

N.

x = x(1).

for some integer

Clearly

Consider

(i) they generate

F n Y,

is a o n e - p a r a m e t e r subgroup of

M, w h i c h is the c e n t r a l i z e r of Set

zN .

Since r.

_x

and

Y.

y

In terms of

We also have

are in

F, the

Ad*, w h a t this

means is that

(1--0.6)

[Ad (x(t))f](log X) = f ( l o g y) + rt f ( l o g z).

We note that the value of Kir(f)

b e i n g in

i, w h e n c e

N/F ^

f ( l o g z)

tral in

on

implies that w h e n

W e w a n t to prove M, the character

Q.

Taking

f(log(~))

As

s

on

t = s

F n zN, w e must have

We w i l l use

~.

But

f

we see that

zN

but does satisfy

Z(y). Let

y

assume that

H ! Z(y), then

(H,f).

is cen-

F n Y, w h e n c e f(log(y))

f(log(y))}/r-

s = {f (log(y)) -

W e shall use

y

in

z2N

that does not lie

and hence, by induction,

The p r o b l e m is that in general, H

H

does not

lie

in

the

need not sit in any

Z(y).

Since

z2N

not lying in

Z(y)

zN.

We may

has c o d i m e n s i o n one in

H + Z(y) = N, and therefore we choose a o n e - p a r a m e t e r group is complementary

has c o d i m e n s i o n one in y

S# ! Q

be an arbitrary rational element of

x(IR)

H n Z(y) element

on

Y

C o m p l e t i n g the induction therefore requires some stealth.

follows that so that

Since

w h i c h is rational, as desired.

t h e o r e m is true for such

1

is in

is rational, w h e n c e

We have now shown that if there is some rational in

(10.6).

[ef](v) =

zN, this

S#--that is, K i r M ( ~ )

w i l l be i d e n t i c a l l y

is in

in (10.6)

is in

Q.

Y

is in

the point b e i n g that

is k n o w n to be faithful on

s

is in

af

v f

Suppose now that

f (log y) ~ [Ad*(x(s))f](log y) is in

log(z) cannot be arbitrary,

is an integer.

integer is not zero. M/F n M ^.

f

cannot lie in K

to denote

H.

to H.

Hence

IR

exp(Z(y)),

x(IR) n F

Furthermore, b e c a u s e IRy +

(H n Z(y))

+ (H n Z(y)). Y

and

Because

= x~).

x : JR÷ H

Notice that

[exp(y),x(~)]

= zN, the

is a p o l a r i z a t i o n for y, Z(y), and

N, it

H

f.

are all

157

rational,

K

is a rational

same number of Set

F

orbits in

c(K,flN)

[ = H n K, and define

quotient of

{(AdN(n)L, AdN(n) ~)

two such ordered pairs blA.

subalgebra of

The equivalence

(A,a)

can get a natural map Let

{H,f,n}

p

from

as in

We shall prove that there are the C(H,f]N).

c(L,flN)

by analogy with

: n ~ N}

by the equivalence

relation under which

are equivalent

A = B

and

class of

Z(y).

(B,b)

(A~(n)L,

A~(n)f)

C(ff,f[N)

to

denote the equivalence

The "obvious" way to get the map

p

c(H,flN)

if

will be denoted

C(/,flN)

class of

to be the

and

alA =

{/,!,n}.

One

as follows:

(AdN(n)ff , AdN(n)f)

is simply to set

in

C(H,f,N).

p{H,f,n} = {/,f,n};

the only

question is whether this map is well defined.

It is, and the key is lemma (i0.i),

a careful look at which yields that if

and

(Ad

AdN(n)H = AdN(m)H

and

n c exp(H).

It follows

that if

(Ad* N(m)f)IAdN(n)H , then p

n e N

N(n)!)IH = ilH,

then

(Ad N ( n ) f ) I A ~ ( n ) H

=

AdN(n)/ = AdN(m) L , which is precisely what one needs for

to be well-defined. The map

C(H,!IF) {L,f,n} ~

p

is equivariant with respect to the action of

into the subset

C(/,!IF)

which have a representative

Hence

p

carries

F

orbits in

of

c(L,flN)

(A,!)

Of course, we also get a natural map carries

c(K,!IF)

into

any element

{L,~,n}

P-l({L,!,n})

n C(ff,!iP)

The action of n.{/,f,m}

c(L,!IF).

of

N

= {L,f,nm},

C(/,fl2),

q

c(L,f N)

and as

n

HK

operates

F

HK.

transitively

there are exactly as many q-l({L,!,n})

N, we see that N

for

We are interested

As for the proof:

log(2))

c(L,flF). c(L,~ N)

that

F

orbits meeting

n C(K,!IF):

p-l({/,f,l})

{L,f,n}

traces

{L,f,l}, which is H = exp(H)

and

C(L,f N). {n ~ N :

K = exp(K),

in this fact because it implies and on

of which is that in the end we will be able to take as we shall see.

!(An

is given in our present notation by

traces

on

orbits in

C(K,f N) to

What we shall prove is that if

then the isotropy group is that

to

rational and

The final step in the proof is to show that for

We begin by computing the isotropy group in {£,f,n} = {L,f,l}}.

A

from

as there are meeting on

and carries

consisting of those classes

with

C(H,flF)

N

q

-i

({L,f,l}),

N = HK, an enormous

the upshot convenienc%

158

Because

H

plarizes

formula implies that if whenever

f, we have h ~ H, then

h ~ H, it follows that

Similar reasoning applies to {L,f,l}

in

If say:

N.

Since

zN

Now

is equal

w e can conclude from lemma

{L,f,l}.

lies in the isotropy group for HK

contains that isotropy group: must "normalize"

6 + ~y,

where

Next consider

Ad N ( n ) ~

(i0.i) that

n = hk

Our problem,

(Ad N ( n ) ~ ) I L = flL.

Since

[x,IRy]

HK

such that

are trying to show.

y.

to

A d N ( n ) K = K.

h E H

by its value on

AdN(h) L = L

y

[, that is to

is some element of

6, anything that normalizes [ w i l l n o r m a l i z e

implies

we can produce some

Since

lies in the isotropy group of

then in particular, n

K

lies in

{L,f,n} = {L,f,l}

K

H

K, and thus

{/,f,n} = {L,f,l},

Hence the C a m p b e l l - H a u s d o r f f

[Ad N ( h ) f ] I L = flL.

The hard part is to show that

L = AdN(n)/.

z2N.

f([H,L]) = 0.

K =

If

x

therefore,

and

Ad N ( h ) ~

for some

is any element of

H

Suppose that

agree on

h.

K.

Then

Now b y assumption,

Ad N(n)f

is determined on

not lying in

z = [x,y], and if

Thus

k ~ K, w h i c h is w h a t we

is to p r o d u c e

L + IRy, we see that

= z N, w h i c h means that if

Ad N(n)~.

K.

K, then

t ~ ~ , then

×

lAd N(exp(tx))!](y)

therefore, element

since

exp(tx)

f

is faithful on

-i

({L,f,l}).

the element

zN, we see that we can take for our

for some suitable v a l u e of

W e n o w k n o w that q

= f(y) + tf(z);

HK

Since

~

{L,f,l}

p-l({L,~,l})

only serves as a base point for d e s c r i b i n g an

can be v i e w e d as the typical point in C(L,~IF)

the

t.

operates t r a n s i t i v e l y on both

assume that it is the element of

h

and N

orbit,

C(L,fl N), so we may

w h o s e preimages we are trying to study.

What we must show is that there are e x a c t l y as many

F n HK

orbits in

p -i ({L,!,l}) n c(L,!IF)

as there are in

ments of

enter not at all into this computation, we may assume

N = HK

N

not in

HK

Since the ele-

from now on.

C o n s i d e r the s u b s p a c e arize

q -i ((L,f,l}) n c(K,~IF).

f, w e have

N = H + K

f([H,L0])

b y assumption,

normal subgroup of

L0 = ~ c

N.

L0

L : f(1) = 0}

in

= 0 = f([K,L0]).

Hence

is an ideal in

N, and

The rationality of

L0

L. [H +

Because

ff

K,Lo] ! LO"

exp(Lo) = L 0

and

K

pol-

Since

is a connected

w i t h respect to F follows from that

159

of

L

and

f.

W e are going to show that

affecting our computation. vanishes on

Clearly if

L0, since it agrees w i t h

n ~ L0, then

plays no role in any action. in order to avoid

can be factored out of

{H,f,n} ~ p-l({L,f,l}), f

we are i n t e r e s t e d in are w e l l defined on ula implies that if

L0

on

L.

N/L O.

Ad N ( n ) f = f

then

Hence all of the linear func~onals Also,

the C a m p b e l l - H a u s d o r f f

on all of

N = HK.

Hence

W e can therefore safely factor it out of

F

N

so that

K n £ , and

is n o w three-dimensional,

z

generates

{x,y,z}

zero i n t e g e r

N.

form-

L0 However,

h a v i n g to create more n o t a t i o n - - t h a t may not even be possible,

and both

sional, rational, connected, normal subgroups. in

without

Ad N(n)f

as w e are close to the theoretical limit--let u s simply assume that The group

N

generates

r, which,

zN nP,{x,z} P.

H

and

K

L 0 = {0}.

are two-dimen-

Let us choose elements

generates

We w i l l then have

H n £, {y,z}

x,y, and z

generates

x-i y -i xy = z r

for some non-

as in (1--0.6) yields

f ( A d ( e x p ( - t log y ) ) l o g x) = f ( l o g x) + rt f(log z). Thus,

the n u m b e r of

0 ~ t < 1

and

P

orbits in

C(H,fIF)

f ( l o g x) + rtf(log z) ~ ~ .

of the value of

is the n u m b e r of

s a t i s f y i n g both

This n u m b e r clearly is a function only

rf(log z), and hence w i l l be the same for

D e s p i t e the s e e m i n g l y endless proof,

t

the m u l t i p l i c i t y

K

as for

H.

formula is just the be-

ginning of the study of h a r m o n i c analysis on N/F. At the very least, one w a n t s to ^ know for each P in N/F a precise d e s c r i p t i o n Of the P - p r i m a r y summand L2(p) in

L2(N/F).

This p r o b l e m was first solved c o m p l e t e l y by L. Richardson,

and the

d i s c u s s i o n that follows is e s s e n t i a l l y his. Because w e are dealing here e x c l u s i v e l y w i t h induced representation, n a t u r a l i n c l i n a t i o n is to try to use the results from §9.

one's

P r e s u m a b l y one could do

so, albeit only w i t h some special p l e a d i n g - - t h e p r o b l e m is that one must cope w i t h i n d u c t i o n not from normal, but rather subnormal subgroups.

Actually, what enables

one to carry the m a t t e r through is the k n o w l e d g e about m u l t i p l i c i t i e s (10.3).

A n o t h e r special feature, perhaps less critical,

gained in

is that we are only in-

ducing characters. W h e r e the approach of §9

comes a cropper here is in the dependence on a choice

160

of Borel cross-section if

s

(the maps

is a Borel cross-section

not be a cross-section H

normal,

for

cross-sections,

keeping.

As it happens,

realization

that

H = exp(H),

and

_f

in

(2)

~

(i)

IFI c L2(N/H)°

[J(n)F](m)

The equivalence

N*

and that

x

is in

x-iHx = H. H

N, then

s

will

In §9, we assumed

is only subnormal,

the character

we could

complicating

our book-

J

for all

J

is

embedding then,

H/F n H

L2(N/H,f)

that

w e simply sum over

in the

KirN(f)

Proof:

Let

R

and

~

in

is in

N.

H of

in

and

f.

H.

n

on

(We shall use

The

is in L2(N/H,f)

J(H,f)

on

H

and

f.)

f(H n log F) ! ~ ,

L2(N/£). or equivalently,

that

will be constant

on

of

L2(N/H,f)

L2(N/H,f) To get to

~ ( H A F) = i . £ n H

cosets

will define an ele-

with a subspace of L2(N/F)

Given a bounded ~F

for

on

N

We shall show that if

of §9.

we define

h

about the dependence

L2(N/H,f)

F/F r, H, as in §9.

x

in

every element

reminiscent

supported,

(10.7) Lemma:

when

f ( H n log £) ! ~ , of

~(h) = (E~)(log h)

the representation

Thus, we can identify

n F)--a situation

the sum being over

m

KirN(~).

being compact,

L2(N/H N F).

that is compactly

denote

polarization

the space of all functions

= #(h-l)F(n)

~(F N H) = i, every element

and hence,

is a rational

L2(N/H,f),

F(hn) Let

H

if we need to be specific

class of

Let us assume,

for all

unless

Here, where

denote

= F(mn)

J

there is a natural

L2(N/H

N

N, and if

there is a simple remedy at hand, which is to use Mackey's

and let

instead of simply

ment of

in

The problem is simple enough to see:

but only at the cost of immensely

is in

that satisfy

given by

Since

x-iHx

H

space we shall use will be

: N ÷ C N,

for

b).

of induced representations:

Suppose

Hilbert

and

so there was no problem.

work with

Set

s

from

function

to be the function

F

~F(n)

L2(N/H n £), in

L 2 ( N / H A P)

= [xFF(xn),

F/F n H.

The operator

~

primar X component o f

defines

an isometric

embedding o f

L2(N/H,f)

L2(N/F).

denote the regular representation

of

N/F.

Since

~J(n) = R(n)e

n ~ N, one sees that in order to prove the lemma, we need only prove that

is an isometry. results of §9.

The strategy

is to introduce

a cross-section

and then to use the

161

Let s($) = ~

s : N ÷ N and

be a Borel cross-section

s(F) ! F-

As before, we let

We then get an isometry

~ : L2(N/H

to b

for

o

Using these L2(N/H

formulas,

: C~)

of

usual notation, is enough

of

L2(H/H

to show that

on the dimension of H

is then normal Let

dimension

M

N.

duction hypothesis H

in

onto

~

carries

L2(N/H,f).

as the

It follows

the subspace

(If you like, you can

~-primary

component,

that to prove

a

L2(~)

in our

is an isometry,

it

The latter will be proved by induction

dim(N/H)

= 1

is taken care of in §9, because

connected

subgroup of

N

that contains

H

and has co-

We are going to begin by w r i t i n g out in detail what the in-

implies

M, n o r m a l i z e d

via

Now for the induction:

be a rational,

one in

C~

The case

N.

b(n) = ns(n) -I.

to be given by

is an isometry.

N/H.

in

space

n F)).

a~

so that

o r)h) = £ ( H n r.hs(n)).

: L2(H/H o F))

think of the one-dimensional

normalized

n F)b(n)).

one can readily verify that

L2(N/H

N

: L2(H/H n F)) ÷ L2(N/H n F)

is easily calculated [o-IG(Hn)]((H

in

denote the "defect"

[oF]((F n H)n) = [F(Hn)]((h

The inverse

H

about

so that

M.

Choose a Borel cross-section

t(~) = 1

and

t(F N M) c £ n M.

t : M + M

Let

for

c : M ÷ H

=

be the corresponding T : L2(M/H,

defect:

L2(H/H n F)) ÷ L 2 ( M / M n F)

we define the averaging pact support x

in

to

L2(M/H

The induction §9

is designed

complemented cross-section to

U

in

: C¢)

by

by setting

into the

M

in

N

defines an isometry [!(Hm)](F n Hc(m)),

functions

on

M/F o H

hypothesis, KirM(fiM )

with com-

then, is that primary

M/F n M

B~

component

up to

defines of

N/F, w h i c h is exactly what

is particularly one-dimensional

We get the usual isometry

n e U, we have

q : L2(U

: L2(M/F

an iso-

L2(M/M n F).

simple, because subgroup

U.

can be taken to be the unique h o m o m o r p h i s m

such that w h e n e v e r

and

[BF](M N Fm) = [xFF(xm), the s u m b e i n g over

Here our situation

M

c

[TF](F n Hm) =

by a connected rational to

As usual,

from bounded

step is to climb from

N u

~

The induction

to do.

with kernel

nu(n) -I.

operator

L2(M/F N M)

F n M/F n H.

metry from

c(m) = mt(m) -I.

u(n) = n.

Set

n M)) ÷ L2(N/F n M)

M

is

Thus, from

our N

d(n) = from

d,

162

namely

[nF](M n Fn) = [F(u(n))](M

is transverse

to

functions

N/F n M

on

being over L2(U

M n F

in

with

j ~ U n F.

: BT(L2((M/H)

compact

clearly

they can be--so

[pF](Hn) p

: L2(M/H

x ~ U.

yq(BT)#p -I

so

that

R-invariant We shall use

> L2(U

BT

on bounded

of

L2(N/F).

and

yq.

What we u

are

then the diagram

: ~))

L2(N/F)

(B~) #

defined

by

we use the operator

As we saw above,

from

s, t, and

: BTL2(M/H

•>

that

U n F

yq(B~) #

[(BT)#F](x) p

defined by

is as isometry,

the diagram commutes,

subspace @(f,H)

(10.7)

is to assign

~L2(N/H,f)

of the

to denote

If

X : L2(N/H,~)

and clearly

whence

~

=

of

and

g

by setting

@(g,K),

(f,H)

primary

an irreducible

component

of

L2(N/F).

~L2(N/H,f).

@((Ad x)f,

K = exp(K),

÷ L2(N/K,~)

to the pair

KirN(~)

the subspace

x ~ F, then

K = (Adx)H,

X~, we get an element

(Adx)H)

= @(_f,H).

= (Ad x)f. xF(n)

Then we can create

= F(x-lnx).

Applying

an iso~

to

namely:

(exF)(Fn) = ~y~F/Kr/FXF(yn) = ~ycr/KnFF(x-lynx)

=

Set

=

is an isometry.

(1--0.8) Lemma:

Set

component

in terms of

: ¢*)

checks

y

is an is0metry

s(n) = t(d(n))u(n),

: C~))

One readily

The effect of lemma

metry

y~

primary

For the left column,

= [F(u(n))](Hd(n)).

is an isometry.

Proof:

that

For the upper row, we use the operator for all

operator

so that if the cross-section

L2(N/H

Bz(F(x))

was chosen so that

[~F](Fn) = ~j F(M n Fjn), the sum

by

shows

is to write

operators

L2(U

commutes.

support

U

the averaging

into the KirN(f)

The heart of the problem

chosen--as

Since

F, we can define

Our work in §9

: ¢~)))

shall do is to produce

n Fd(n)).

G(n) = F ( n x ) .

Clearly

~v~r/HnrF(vx-lnx)

G is also in

•

L2(N/H,~).

Furthermore,

~G = ~xFF:

163

(~G)(Fn) = Zy~F/FnHF(ynx) -i = ~y~F/FnHF(yxx

nx)

= ~y~r/rnHF(yx-lnx)

Since some

~G =

in

~X_~, and since every element of

L2(N/H,f),

@(g,K)

has the form

C(H,f/F)

(1--0.9) Lemma: different

Proof:

F

that is constant on

If

(K,g)

orbit from

dimensions one and two.

H

and

Suppose, H

K.

(H,f), then

K

@

is a well-defined

c(H,f/F)

that lies on a

@(K,g) i 0(H,f).

N.

The result is obvious in

In order to carry out the induction step, we need to know M

in

N

that has codimension one in

N

and con-

We begin by establishing the existence of such an ideal.

to the contrary, no rational ideal of

and

for

orbits.

represents an element of

that there is a rational ideal

Since

F

1~e proof is by induction on the dimension of

tains both

~xF

the lemma is proved.

Recalling the notation of theorem (10.3), we see that function on

= (~xF)(Fn).

are both rational,

N

contains both

the subalgebra the

H

and

K.

span is rational and hence

cannot be proper, for if it were proper, it would lie in a proper rational ideal of N.

in

It follows that the vector space sum N.

But there is some

so that H +

H + K

+

n e N

[N,N] = N implies

there can't be two

H = N.

F

N.

Set

If

K = (Adn)H.

IN,N] = N.

H = N, then

H + K

M = exp M, and let

rational subgroup of

H +

L2(M/F n M)

N

H = K , mod. N

[N,N],

is nilpotent,

is a single element, so

lies in some codimension one ideal

U,

analogous to

M

rational

as in the proof of lemma (10.7), be a connected

complementary to

~(h) = (~f)(log h)

cial fact is that

Hence

Because

C(H,f_/F)

M.

We are going to use the cummutative

diagram introduced at the end of the proof of (10.7). character

[N,N]

contains a complement to

orbits to worry about.

We may assume then that in

such that

implies

[N,N] = N

H + K

Let

BTL2(M/H,¢~)

#~ as in (10.7), be the

of

H.

Then

is the subspace of

@(f,H)

in

L2(N/r); let us call it

@M(f,H).

The cru-

@(f,H) = Yn L2(U : @M(f,H)), because it enables us to use the

164

theorem (9.11). When we restrict to

KirM(!IM)

M, it may happen that there is no

= x.KirM(glM).

lie in the

KirM(flM)

Let us assume so.

and

K~M(glM)

Then, since

Since

that

@(f,H)

depends only on the

KirM(flM) = KirM(glM).

g = (Ad m)f. lie in

@M(g,K)

F n M.

Our original hypothesis on

C(H,flF)

@M(~, K)

L2(M/F n M)

x F

in F

f

for which

orbit of

and

re-

Since Yn

maps

L2(U : ~)

that meets each

(i0.i0) Theorem:

The

F

KirM(flM)

= x'KirM(g I

(H,f), we may actually assume

m e M g

K~M(flM)

such that

K = (Adm)H

guarantees that

Therefore, the induction hypothesis implies that

Choose a set of representatives in

and

L2(U : @M(f,H)) ± y~L2(U : @M(~,

Hence there is some

are orthogonal subspaces of the

L2(M/M n F).

F for which

@(f,H) i @(g,K).

We may now assume that there is some M).

in

@M(!,H)

primary components of

spectively, it follows from theorem (9.1!] that K)), whence

x

{(Hl,f I) .... ( H r , ~ ) }

does not

@M(f,H)

primary component

isometrically into

m

and

#

and of

L2(N/F), we are done.

for a family of elements

orbit exactly once.

KirN(~) primary component of

L2(N/F)

is

@(Hl,fl)~...m@(Hr,f_r) • Proof:

Combine (!O. 3) with (10.7) and (10.9).

We have now obtained a decomposition of ducible R-invariant subspaces. lation invariant operators on

L2(N/F)

into a direct sum of irre-

The next step is the detailed analysis of the transL2(N/F) , or more importantly, on

C (N/F).

The sort

of analysis we have in mind is that carried out in §3 for groups of dimension 3. Although we shall not carry out the analysis here, we remark that an essentially complete theory has been worked out by R. Penney, whose papers are about to appear.

ii.

Solvmanifolds Ironically, the theory parallel to that in §i0 for nilmanifolds becomes simpler

for solvmanifolds in general, for the simple reason that one cannot hope for results

165

nearly so complete.

We will set forth in this section what is known.

The essential

reduction was first carried out by R. Howe whose argument we shall follow with only minor modifications, As usual, let us begin with a barrage of notation.

We denote by

connected, but not necessarily connected, solvable Lie group. contains a discrete subgroup

F

such that

G/F

is discrete.)

G/N

The nil-radical of

G

a simply

We assume that

G

is both connected and compact.

(Actually, it is probably enough to assume that F

G

F

is closed in

will be denoted

N.

G, and not that

We shall assume that

is abelian. ^

Let

P e G/F .

representation ponent

L2(p)

R

We want to know about the multiplicity of

of

in

G

on

P

in the regular

L2(G/F), and we want to describe the P-primary com-

L2(G/F).

Our work in chapter II enables us to make several re-

ductions immediately: Since

N/N n F

is compact, we know that

P

lies over some element

P'

of

^

N/N n F . compact,

We can find a closed subgroup P'

extends to

F

F

containing

N

such that

but no further, and (assuming we chose

F/F n F P'

is

correctly

^

to begin with) G/F

P'

is abelian,

extends to an element

the results of §9

the multiplicity of we may assume that

P

and

G = F.

P"

of

F/F n F

G/F

L2(G/F)

L2(p)

from corresponding data in

L2(F/F n F).

Hence

^

and has the property that

P/N

is in

N/N n F , then

with exactly the same multiplicity as does G

to

PIFN

in

FN, we lose control of

therefore make do with multiplicity computations alone.

P

occurs in

L2(FN/F).

as yet no good general results on how to compute

Unfor-

L2(p) and must

We have already encountered

this phenomenon in our analysis of four dimensional groups in §5 and §6.

G

Since

We next use the results of §8, which show that if

tunately, in the transition from

when

P.

tell us all we need to know about how to compute

^

is in

that induces

There are

the primary components of

L2(G/F)

is solvable.

Accepting our fate gracefully, we shall give up on computing the multiplicity of

P

assume that

PIN

the kernel of

G = FN P.

and that Let

K

in

L2(G/F). is in

L2(p) and concentrate on

With this narrow aim in mind, we may

N/N n F ^.

Our next aim is to get rid of

denote the identity component of the kernel of

P.

If

166

K of

is non-trivial, N

then because

non-trivially.

K

is normal in

Furthermore,

because

N, it must meet the center

F n zN

acts trivially on

zN

L2(N/N n F),

^

the condition

that

P/N

and hence is closed in L2(G/F)

lie in zN.

is constant on

F(K n zN)

implies

cosets

(in fact,

G, we may replace

G

with

without affecting our multiplicity

assume that the kernel of

K n zN

Since every element of the

F(K n zN)

is closed in

F/F n K n zN

N/N n F

P

is discrete,

is compact modulo F n zN

P

on

primary component of FK

cosets),

G/K n zN

and

computation.

F

and since with

Therefore,

and that the center of

N

we may

is one-

dimensional. We have one final reduction

to make, for which the following short argument

lays the groundwork: Suppose that we can find a closed subgroup

H

in

G

that has the following

four properties: (i)

H/H n F

(£)

HN

(3)

H n N

(4)

P

=

is compact.

G.

is normal in

N, and

N/N n H

is abelian. ^

is induced from

H

by some element of

H/H n F .

^

Let P.

H If

yields

denote the subset of Q

is in

PIN.

Q e ~}.

H/H n F

~, then because

consisting of all the elements

HN = G, we see that

Hence the analysis of §9

Since

H n N

is normal in

applies to

QIN n H

PIN

N, we see that

and

F n N

that induce

induced to

N

HIN n H={QIN n H : acts on

^

(N n H)/(N n H n F) such that

and stabilizes

{QIIN n H ..... QrlN o H}

We saw in ~9 determines

an isomorphism L2(pIN),

induces a natural embedding in

L2(N/F n N).

%

from

of

to

H n N

Since

because ~

of

be a subset of

orbit exactly once in in

N

is chosen, n H))

G = HN

L2(p), in and

H

HIN n H. it

onto the

FN = G, the inclusion map

of the P-primary summand,

get a corresponding natural embedding

the same image in

n F).

In addition,

L2(N/N n H : ~¢j=l L r 2 (QjlN n H)).

F

{QI ..... Qr }

L2(N/N n H : ~¢;=IL2(QjlN

L2(N/N v

Let

meets every

that once a Borel cross-section

primary summand,

L2(pIN)

HIN n H.

L2(G/F)

PIN N + G inLo

H = (F n H)(N n H), we

r 2 L2(G/G n H : ~¢j=l L (Qj)) into

The crucial point here is that

L2(pIN), as one can see by noting that

%o~

~o~

and

intertwines

v the

have

167

r ]L2(Qj)) , S ~@j=II(S

induced representation

H/H n F, with the regular representation

being the regular representation

of

N/N n F.

of

We thus arrive at the first

basic result:

(ii.i) Lem~na: Qj

(Notation as above.)

Let

L2 (H/H n F); then the multiplicity

in

mul(Qj)

of

P

in

Thus, insofar as computing the multiplicity strict our attention fundamental

to the subgroup

H

if no such proper subgroup

precise structure of those groups

G

of H

L2(G/F)-

of

G.

denote the multiplicity o f

P

r ~j=imul(Qj).

is

is concerned, we can re-

We shall call the pair

can be found.

{G,P}

Our next theorem gives the

that can occur in a fundamental

pair

{G,P}.

In order to state the result cleanly, we will first describe our stituation somewhat more precisely. If you will recall, we have gotten ourselves identity component of center

zN

of

N

crete, satisfies can use

a dense subset of rational points. G, and hence acts on

(11.2) Theorem: (i) (ii)

N/zN

G

Q.

is compact, we

log(N n F)

N

a "rational" struc-

A subspace of

N

or equivalently,

Finally we note that This action leaves

Suppose that

G

zN

acts on pointwise

of

is called if it contains N

by conjuga-

fixed and in-

{G,P}

is fundamental.

Then:

is abelian.

The action of

Whenever containing

Proof:

N/N n £

N/zN.

on

fixed-point (iii)

N.

in

to give

F, it is dis-

if it is a finite linear combination

with coefficients

N

Since

As for

N

being "rational"

of

G.

N

log(N n F)

duces an action of

[F,F] c N.

We also have that the

in the Lie algebra

rational if there is a basis for it in

tion in

N.

and is central in

FN = G, and satisfies

log(N n F)

from

is the nilpotent Lie group

is one dimensional

ture, an element of elements

G

down to the point where the

U

G

on

N/zN

is reductive

and, except for

0, is

free. is a G-invariant zN, one has

rational subspace of

N

properly

[U,U] = zN.

The proof will be broken down into two steps, which,

for convenience's

sake,

168

we shall isolate as lemmas.

(i_ii.3) Lemma:

Suppose that

rational subspace of

Remark:

N

{G,P}

is fundamental.

that properly contains

[U,U] = zN, as in part (iii).

Proof:

Because

U

is G-invariant,

N.

non-trivially.

Applying this observation to

U/zN

invariant, U n zN.

Ad(N)

invariant.

U

z2N

zN.

z2N

= {X c N :

G

clearly leaves

is a G-invariant rational subspace of

N

that properly con-

U c z2N.

z2N

[U n z2N,U n z2N] = zN .

The lemma will be proved if we can show that

Furthermore, we might as well assume that no

G-invariant rational subspace of

U

contains

zN

properly, since the

lemma will fail in that case if it is going to fail at all.

[U,U] = 0.

Hence

Since

Hence we shall assume

We will also assume ~ a t

The problem is to get a contradiction.

There is some linear functional has a discrete kernel,

f

contains a polarization of exp(Z).

f

on

N

must be faithful on

Kirillov's method to show that if

from

zN.

N/zN, we see that

sitting in

N, which is the ideal

[X,N] ! zN}, in a subspace properly containing

P

contains

Now every ideal in a nilpotent Lie algebra meets the center

must meet the "second center" of

proper

is a G-invariant

[U,U]

zN, then

it is a f o r t i o r i

is an ideal in

tains

U

Since we don't yet know that part (i) of (1__1.2) is true, we can't quite

assert

U

If

f

such that zN.

PIN

is

Kir(f).

Since

We are going to use

[U,U] = O, then the centralizer

and hence is big enough that

PIN

With a little more work, we will then show that

Z

of

U

in N

can be induced {G,P}

is not funda-

mental, contradiction. Step I:

Produce a polarization for

complementary to in

N.

in

z2N, and let

We are first going to show that

exist bases dhj"

zN

{ ~ ..... ~ }

in

U0

and

f V0

in

Z.

Let

U0

be a subspace of

dim(U0) = dim(V0) {YI ..... Yk }

in

V0

be a subspace of N

U

complementary to

Z

and further that there such that

![~,Yj]=

A dimension check will then yield, with a moment's work, the desired polariza-

tion. If

X

is any non-zero element of

U0, then

[X,V0] = zN, since otherwise

X

169

would lie in

zN.

Similarly,

zN, since otherwise

Y

if

Y

is any non-zero element of

would lie in

degenerate pairing of

Z,

is a non-

• , whence the existence of the desired bases.

Consider now the bilinear

form

B(X,Y) = f[X,Y]

B, then any polarization of

If we restrict

B

to

Z, then

previous paragraph shows that neither N.

UO U0

on f

nor

V0

B

that, since

Show that U

{G,P}

N.

Hence, the subgroup

proper, normal subgroup of

~IZ

N.

Since

H0/F n N

Z

(dim(N)

B

on all

will have dimension

dim(U 0) = dim(U O) = k

and

N.

As Kirillov has shown,

in

Z.

Thus step I is done.

The first observation is

(F n N)[N,N]exp(Z)

= H0

By virtue of step I, we know that

PIN

is the dimen-

zN, we must have that

duced by a character from a connected subgroup of theorem on inducing by stages

in

cannot be fundamental.

is rational and not equal to

and rational in

r

will have dimension

flZ

as a form on all of

there does exist at least one polarization for Step II:

If

meets the radical of

(dim(Z) + r + k)/2 = (dim(N) + r)/2, since

contains the radical of

N.

will lie in its radical, whereas the

Hence, the dimension of a polarization for

at least

H 0.

> f[X,Y]

into

+ r)/2.

Z

(X,Y) !

[Y,U O] =

U 0 × V0

sion of the radical of

of

Hence

V0, then

H 0.

Z

is proper

is a closed PIN

can be in-

Hence, by virtue of the

can be induced by some representation

is compact, we can use the general results of §~

Q0

of

to see that

^

we can choose

QO

Because normalizes

U H0

so that it lies in is

F-invariant,

in

G.

Set

Ho/F n N .

the subspace

H = FH 0.

Z

is

F-invariant, and hence

We will show that

P

is induced from

F H,

^

contrary to

{G,P}

being fundamental.

Then the representation

P1

Let of

Q1

over

Q0"

PIN

by virtue of Mackey's subgroup theorem.

group theorem that if we multiply

Q1

element of

itself.

H/F ^

that induces

Call a subspace

U

of

z2N

P

G

be any element of induced by

satisfies

lying PIIN =

It follows from Mackey's little

by a suitable character of

H, we will get an

essential if it is rational, G-invariant, and con-

tains no G-invariant rational subspace other than essential in case

Q1

H/F

N = zN , which can happen.)

spanned by the essential subspaces.

Let

zN. W

(We allow zN

itself to be

denote the subspace of

The easiest way to visualize

V

z2N

is to pass to

170

the complexification with the pre-image

in

common eigenvectors have no non-zero lated.

Thus,

¢ ® z2N

of

G.

then since

W

is G-invariant

V/zN, since otherwise

rational

and rational,

non-degenerate

subspace

symplectic

U

of

form on

We shall

Z n V = zN. on

Z n z2N

is minimal.

generated by the

on

lemma

V.

Also,

G

can

(11.3) would be vio-

of

V

ideal in

Furthermore,

V, the codimension

that

Z n z2N # zN.

that contain

zN

Such a subspace,

form

R.

(X,Y) ~

> f[X,Y]

In fact, since

By lemma

satisfying

(11.3),

[U,U] = 0

Hence the bracket

f

V,

is faith-

the only

is

in

on

N

zN.

Since

defines a

V/zN. in

because

of

N, w h i c h will be denoted

N, so is

Z

Z.

Also,

the bracket

in

finish the proof by showing that

Suppose of

rational

z2N R = zN.

Consider next the centralizer is a G-invariant

so is

R = {X ~ V : [X,V] = 0}.

JR,R] c [R,V] = O, it follows that

degenerate

acts reductively

the radical of the symplectic

zN, we see that

see that

G

~ ® (z2N/zN)

i ® z2N

of

V = N.

denotes

V

as the intersection

the theorem will be proved once we show:

R

G-invariant

appears

Notice that

If

ful on

V

of the subspace of

fixed points in

(11.4) Lemma:

Proof:

C ® z2N, w h e r e

N

in

is dim(V)

using N - i.

however,

Since

(1--1.3) again, we

is essentially

non-

Hence

Z + V = N.

rational

subspaces

Z n z2N = zN.

Among all of the G-invariant

properly,

Z.

there will be at least one w h o s e dimension

is essential

and hence must lie in

V, w h i c h

is absurd.

D

It is instructive w o r k i n g on in §6. a solvmanifold §6.

to compare

G

in theorem

One way of viewing theorem

is not significantly

different

D e p e n d i n g on your outlook on life,

W h i c h was it? The lady?,

Or the tiger?

(11.2)

(11.3)

with

the groups we were

is that the worst behavior

from what we already encountered

for in

this news is either very good or very bad.

Afterword

12.

Attribution and speculation Section ~ amounts to an exposition of Mackey's little group method in the

simplest special case.

We chose to work with

C , rather than with

L 2 ' in order

to avoid the measure theoretical problems that lie at the heart of Mackey's work. Of course, Bruhat's thesis [1956] does similar results in greater generality.

If

there is a difference between Bruhat's point of view and my own, it is undoubtedly a consequence of our being interested in very different sorts of groups.

In gen-

eral, when dealing with specific examples I think that one is better off working with

C=

and distributions, rather than with

to the fact that one can restrict

C

C~

approach (thanks

functions to submanifolds of positive co-

dimension) tends to simplify the algebra. §~ are a good example of what I mean:

L 2, because the

The eigendistribution computations of

because one can really find eigenvectors for

the translation operators in the space of distributions, one can avoid a lot of tricky analysis in checking irreducibility.

I make these comments, which must seem

obvious, only because I am surprised at how little use Bruhat's variation on Mackey has gotten in the sort of harmonic analysis we are doing here.

I think that this

has been a serious mistake, particularly in dealing with difficult multiplicity questions such as those in §!. The question of indecomposability versus irreducibility seems to be a theory one.

[Poulsen, 1972 ] discusses the general situation.

nilpotent group, the computations at the end of §i ability.

For the three-dimensional

suffice to check idecompos-

For the genuinely solvable groups the situation is less clear, essentially

because of the too rapid decay of the functions in the associated Schwartz space

S (m). The first paper that I am aware of that treats multiplicity problems for solvmanifolds is folds.

[C. C. Moore, 1965], where a partial result is obtained for nilmani-

The connection between multiplicities and ideals in number fields, as

172

illustrated in §2, must have occurred to everyone who looked at the problem. R. Tolimieri [1975] has worked out these computations more thoroughly and from a somewhat different point of view.

As I mentioned in the preface, the example where

the multiplicities stay bounded is due to W. Beiglbock (unpublished).

The idea of

looking at average growth of multiplicities is my own, and it remains a pet projec~ The real question is:

If you believe that the Kirillov orbit picture really tells

you all about the irreducible representations, how do you use it to gain insight into the qualitative behaviour of multiplicities?

The quantitative behaviour is

obviously too complicated to say anything at all about, as the Howe-Richardson results of §i0 show only too plainly.

At any rate, one "qualitative" feature is

average growth of multiplicities as one goes to infinity

in the "dual"

G/F ^.

Can one find a consistent averaging method that, for a large class of groups, will give estimates analogous to those in §2? in the short note [Brezin, 1975]

Some thoughts on the matter can be found

where some reasonably precise conjectures are made.

Section 3 describes a different sort of qualitative behavior, namely the order of the distribution giving projection onto the P-primary component of

C (G/F)

as

^

P

varies over

G/F .

The first significant step was taken be Richardson [1976],

whoshowedwhy the order need not be zero.

That what was involved was just orbit

geometry was pointed out in [Brezin, 1975], at least in what could be termed the generic case.

The complete problem was solved by R. Penney in a series of papers

about to appear.

He also showed the relation of the problem to two superficially

very different notions, square-intergrability in the sense of [Moore-Wolf, 1973] and distinguishability of subspaces in the sense of [Auslander-Brezin, 1973]. The computations of sections 5 and 6 grew out of a study of [Weil, 1964].

It

is probably fair to say that the multiplicity problem in the hyperbolic case cannot be solved more explicitly than here, at least without some number theory of truly staggering d~fficulty being overcome first. example, in the notation used in 5),

In the elliptic case (o = ~, for

the situation is more promising.

The computa-

tions in [Auslander-Tolimieri, 1975] relating classical theta functions to func~ons on nilmanifolds has been shown recently by Tolimieri to provide an alternative approach to the case

o = ~

in sections 5 and 6--his results should be appearing

173

in the form of an announcement in the Bulletin of the Amer. Math. Soc. soon.

This

new approach looks very promising as a tool for attacking groups of the sort described in theorem (11.2) in the case analogous to algebraic group of automorphisms of from conjugation by elements of

G

N

o = ~, which is when the least

containing all the automorphisms coming

has compact reductive factor.

Work on the

question is in progress, but no early resolution is expected. Much more should be said about the projections onto the primary summands for the various four-dimensional solvmanifolds of section 5. results to those in section 3 for these larger groups?

What are the analogous Two other references on

related matters are [L. Corwin, 1976] and some recent work of R. Penney available in preprint form as of this writing. Sections 7, 8, and 9 originally form a paper done jointly by L. Auslander and the author.

The results in section ~, at least in part, had been done previously

in [Howe, 1971].

Aside from the questions already raised in section 8, there seems

to be little room for improvement here, although a generalization of section 9 to cover subnormal subgroups would be very nice.

The crucial role played by the

multiplicity formula, theorem (10.3), in proving Richardson's completeness theorem ((i0,i0)) suggests that the problem is not easy. Section i0 summarizes two papers part of Howe's work appears.

[Howe, 1971] and [Richardson, 1971].

Only

Whether one should be content with these results as

they stand is a matter of disagreement.

My personal feeling--and it is nothing

more than that--is that it would be very instructive to have some less precise results about some broad class of groups, results whose implications are a little easier to perceive than (10.3).

An outstanding example of what I mean is [Moore-

Wolf, 1973]. Section ii consists of some previously unpublished results of R. Howe. are many things one might try to do to elucidate this scene.

There

Perhaps the first

question that one might ask is this: Suppose

G

is a connected, simply connected solvable Lie group that is also

an algebraic group defined over

~.

tional on the Lie algebra of

for which there exists a rational polarization.

G

Suppose that

f

is a rational linear func-

174

Finally,

let

r

and suppose that

be a discrete subgroup of G/F

is compact.

G

consisting of rational points of

G,

To what extent can the results of section i0

be pushed through for the irreducible representation of

G

determined by

f?

This problem is immensely complicated by the failure of its hypotheses to be stable under the operations required to argue inductively. The second obvious question is, for Howe's reciprocity formula

[Howe,

there probably is no problem.

G

1971]?

as above, how does one generalize

When there is a rational polarization,

In general,

the computations of section 6 suggest

that a new idea may be needed to get a clean general result.

If worst comes to

worst, one will even have to cope with non-principal ideles, which could prove rather messy. Finally, here are some general bibliographical notes. discussion

(with a very good bibliography)

of the general representation theory of

solvable Lie groups can be found in [C. C. Moore, 1973]. solvmanifolds is treated most thoroughly

that this paper is difficult in parts--solvmanifolds therefore,

needed to be tackled.

For many purposes,

1963] is sufficient.

The algebraic theory of

(and in the manner best suited for most

applications of which I am aware) in [Auslander, 1973].

objects--and probably,

An excellent general

I would warn the beginner

are unfortunately complicated

should be treated as a reference, only those parts the short summary in [Auslander et al,

The treatment in [Auslander, 1973] is much more powerful.

other main tool we use, Mackey's little group method,is developed in a series of papers

[Mackey, 1952, 1957, 1958].

A beautiful and readily accessible example of

how to use Mackey's method is the famous thesis of Kirillov

[1962].

The

Bibliography

Auslander, L. [1973]

"An exposition of the structure of solvmanifolds, part I: theory." Bull. Amer. Math. Soc. 79 (1973) 227-261.

algebraic

Auslander, L. and Brezin, J. [1973]

"Translation-invariant subspaces in Inv. Math. 20 (1973) 1-14.

L2

of a compact nilmanifold, I."

Auslander, L. ; Green, L. W.; et al. [1963]

Flows on Homogeneous Spaces, Princeton, N.J. (Annals of Math. Studies 53).

Auslander, L. and Moore, C. C. [1966]

"Unitary representations of solvable Lie groups." Soc. 62.

Mem. Amer. Math.

Auslander, L. and Tolimieri, R. [1975]

Abelian Harmonic Analysis, Theta Functions, and Function Algebras on a Nilmanifold, Heidelberg, B.R.D. (Springer Lecture Notes in Mathematics

436). Brezin, J. [1975]

"Geometry and the method of Kirillov," in Non-Commutative Harmonic Analysis, Heidelberg, B.R.D. (Springer Lecture Notes in Mathemmtics 466).

Bruhat, F. [1956]

"Sur les representations induites des groupes de Lie." Math. France 84 (1956) 97-205.

Bull. Soc.

Corwin, L.

[1976 ]

"Decomposition of representations induced from uniform subgroups and the "Mackey mmchine." Jour. Func. Anal. 22 (1976) 39-57.

Dixmier, J. [1964]

Les C*-algebres et leurs representations, Paris, France (GauthierVillars).

Fell, J. M. G. [1960]

"The dual spaces of C* algebras." Trans. Amer. Math. Soc. 94 (1960) 365-403.

Gel'fand, I. M. et al. [1969]

Representation Theor[ and Autemorphic Functions, Philadelphia, PA (W. B. Saunders and Co.).

176

Howe, R. [1971]

"On Frobenius reciprocity for unipotent algebraic groups over Amer. Jour. Math. 93 (1971) 163-172.

Q.,,

Kirillov, A. A. [1962]

"Unitary representations of nilpotent Lie groups." Russ. Math. Surveys 17 (1962) 53-104.

Mackey, G. W. [1952]

"Induced representations of locally compact groups, I." Ann. of Math. 55 (1952) 101-139.

[1957]

"Borel structures in groups and their duals." Trans. Amer. Math. Soc. 85 (1957) 134-165.

[1958]

"Unitary representations of groups extensions, I." Acta Math. 99 (1958) 265-311.

Moore, C. C. [1965]

"Decomposition of unitary representations defined by discrete subgroups of nilpotent groups." Ann. of Math. 82 (1965) 146-182.

[1973]

"Representations of solvable and nilpotent Lie groups and harmonic analysis on nil and solv manifolds," in Proceedings of Symposia in Pure Mathematics, XXVI, Providence, RI (Amer. Math. Soc.).

Moore, C. C. and Wolf, J. [1973]

"Square-integrable representations of nilpotent Lie groups." Amer. Math. Soc. 185 (1973) 445-462.

Trans.

Poulsen, N. S. [1972]

"On C ~ vectors and intertwining bilinear forms for representations of Lie groups." Jour. Func. Anal. 9 (1972) 87-120.

Pukanszky, L. [1967]

"On the theory of exponential groups." Trans. Amer. Math. Soc. 126 (1967) 487-507.

Richardson, L. [1971]

"Decomposition of the L 2 space of a general compact, nilmanifold." Amer. Jour. Math. 93 (1971) 173-189.

[1975]

"A class of idempotent measures on compact nilmanifolds" Acta Math. 135 (1975) 129-154.

Rudin, W. [1962]

Fourier Analysis on Abelian Groups, New York, NY (Wylie-Interscience).

Steinberg, R. [1974]

Conjugacy Classes in Alsebraic Groups, Heidelberg, B.R.D. Lecture Notes in Mathematics 366).

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Tolimieri, R. [1975]

"Solvable groups and quadratic forms." Trans. Amer. Math. Soc. 201 (1975) 329-345.

Weil, A. [1964]

"Sur certaines groupes d'operateurs unitaires." Acta Math. iii (1964) 143-211.

INDEX

Compact homogeneous spaces, examples G/K H/L Compact solvmanifolds, Go/F ° So/A ° N/F G/F

ii0

i13 examples 2-3 59 151 165

Defect function

135

Dirichlet-Dedekind Theorem

25-6

Function spaces Z(S) D ' ( G / F o)

5 8

z'(s) = z(s)' zT(s)

8 lO

S

12

M P Y(n)

60

Z (n)

61

z (n,m)

61

Z#(n)

75

S (TZ/nY0

28

80

Induction by stages

ii0

Kirillov's correspondence, Kir

150

Little-group Theorem

119

Multiplicity

22, §2, §4, §6, 142 153

p-Primary component Product formula

45-8, 52-4

Representations R = right translation 3, 59~ Iii r R= ii, 12, 15 P 61, 64 R m'n 62 Vn

63

Qn

62-4

Wn

62-4

~[~]

71

L

79

LI

80

wn, b

89

I(U : A,G) S

116 117

Subgroup theorem

116

Theta functions

137-8