[lambda]-rings and the representation theory of the symmetric group

Lecture Notes in Mathematics A collection of informal reports and seminars Edited by A. Dold, Heidelberg and B. Eckmann, ZUrich

308 Donald Knutson Columbia University in the City of New York, New York, NY/USA

Z-Rings and the Representation Theory of the Symmetric Group

Springer-Verlag Berlin. Heidelberg" New York 1973

A M S Subject Classifications (1970): 1 3 A 9 9 , 20-02, 2 0 C 3 0

I S B N 3-540-06184-3 S p r i n g e r - V e r t a g B e r l i n - H e i d e l b e r g - N e w Y o r k I S B N 0-387-06184-3 S p r i n g e r - V e r l a g N e w Y o r k • H e i d e l b e r g • B e r l i n

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 1973, Library of Congress Catalog Card Number 73-75663. Printed in Germany. Offsetdruck: Julius Beltz, Hemsbach/Bergstr.

CONTENTS

Introduction

Chapter

I:

.

k-Rings

.

i.

The Definition

2.

General

3.

Symmetric

4.

Adams

Chapter

II:

Operations

The

Irreducible

3.

Characters

4.

Permutation

i.

on k - R i n g s

.

46

Ring

Theory

of G r o u p s

of a F i n i t e

Representations

and

Group

Schur's

59

.

60

.

Lemma

76

. .

Representations

81

a n d the B u r n s i d e 104

The Group

III:

15 28

. Algebra

Approach

The Fundamental

The Fundamental Theory

of t h e

2.

Complements

3.

Schur

4.

and D e f i n i t i o n s

Representation

2.

5

.

Representation

Ring

k-Ring

Functions

The

Chapter

of

Constructions

1.

5.

5

Theorem

Symmetric

.115

.

Theorem

.

124

of the R e p r e s e n t a t i o n Group

124

and C o r o l l a r i e s

Functions

137

a n d the F r o b e n i u s

Formula

.

Methods

of C a l c u l a t i o n .

Character 155

Young

Diagrams

~

166

IV

Bibliography

Index

of

Index

.

.

Notation

194

.

198 200

INTRODUCTION

These are notes in the y e a r 1971-72. christened

from a seminar given The object

the F u n d a m e n t a l

of the Symmetric

Group.

isomorphism between

at C o l u m b i a U n i v e r s i t y

is to prove what

is herein

Theorem of the R e p r e s e n t a t i o n

This theorem

states that there is an

a ring c o n s t r u c t e d b y composing

way all the r e p r e s e n t a t i o n

Theory

rings of the symmetric

and the ring of all symmetric p o l y n o m i a l s

in a certain

groups

S , n

in an infinite number

of variables. This

isomorphism has b e e n more or less known

of the subject w i t h F r o b e n i u s reasons

it has b e e n expressed

through its relation linear group.

around 1900, (e.g.,

But for various

in Weyl's Classical

to the r e p r e s e n t a t i o n

The i s o m o r p h i s m

is given

The main technical

in its pure form seems not to

context,

in a b r i e f

in K-Theory,

introductory

in 1956

(~i~

first

in an a l g e b r a i c - g e o m e t r i c

and later used in group theory b y A t i y a h and Tall

The notion of k-ring T h e o r e m of Symmetric

is b u i l t upon the classical

Functions:

where

section.

tool is the notion of k-ring,

introduced b y G r o t h e n d i e c k

Groups)

theory of the general

have appeared until Atiyah's Power Operations a "dual" version

since the origins

(~43).

Fundamental

Theorem:

Let n~l

integer

coefficients

permutation

o6S

and

,

n

having f(X

("f is s y m m e t r i c " . ) F ( a l , a 2 ..... an), coefficients,

(the there

f ( X l , X 2 ..... Xn)

a2

=

XIX 2 + XlX 3 + X2X 3 +

a. l

=

~ jl<..<ji

an

=

XIX2..X n symmetric

some

of c l a s s i c a l

a symmetric

of t w o p o l y n o m i a l s

polynomial

a l , a 2 .... a , a n d w i t h n

XI + X2 +

and

for e a c h

integer

substituting ... + X n ... + X n - 1 X n

X. X ..X. 31 32 3i

functions"

o f X 1 ..... Xn)

identity

f ( X l , X 2 ..... X n ) ~

given

is a u n i q u e

=

(For a p r o o f ,

that,

with

(n) ) ~ f ( X l , X 2 ..... >in).

aI

is a p o l y n o m i a l

Much

there

in v a r i a b l e s

"elementary

a polynomial

the property

(1)'Xo(2) ..... X

Then

so that,

be

F (al . a. .2 . . .

applications,

algebra

function

of

- and asked

an)

see

is b u i l t

van

der Waerden

on t h i s

some variables to compute

the

theorem.

- e.g.,

the

associated

F46 ~

One

is

resultant F.

5

A k-ring R is a commutative operations function

kn:R---~R,

n=0,1 .....

F(a l,...,a

n

) associated

and kn(km(r)),

from scratch~

Several

Schur

in Chapter

relevant

functions

F(klr,l 2r, .... knr).

kn(rl+r2 ) • k n (rlr2),

kl(r)=r. ChapterII

is a discussion,

theory of finite groups.

III for the Fundamental

topics have been

in a more permanent

One notable

an element

find the polynomial

k0(r)=l,

of the representation

appear

the future.

first

and with

and any symmetric

X 1 .... Xn,

I, we study k-rings.

two are combined

rER,

the composites

and specifying

In Chapter

identity,

to f, and then evaluate

are given describing

we hope,

Given

f(Xl'''''Xn ) of variables

f(X 1 .... Xn) (r)ER is defined by,

Axioms

ring with

omission

Theorem.

omitted here and will,

version

of this work

is the algorithm

and its application

The

in

for multiplying

to the classical

Schubert

Calculus. One object

of these notes

from an elementary

point

Birkhoff/MacLane/van nothing

is assumed

theory). discussion

This

is to present

of view.

der Waerden

Beyond

reason

of the representation

Especially,

algebra background,

a little bit of category

for including

a complete

theory of groups.

It will be clear to m a n y people sources.

the standard

( ~ 7 ~ 6 J )

(except occasionally

is our main

all of this theory

this applies

h o w m u c h we owe to various

to Chapter

II vis-a-vis

Serre's

Representations

Lineares

des Groupes Finis,

slick enough that it is d i f f i c u l t While w o r k i n g

on these notes,

Science Foundation

with

a large number

especially, seminar

to make any improvement. I was

(Contracts GP-21341,

C o l u m b i a University.

supported b y the National GP-33019~),

and b y

I was supported m a t h e m a t i c a l l y b y c o n v e r s a t i o n s of p e o p l e

at C o l u m b i a

and elsewhere.

I want to thank several p e o p l e

- Ron Proulx,

p u t t i n g up with,

his exposition b e i n g

Evelyn Boorman,

and helping correct,

Here,

in the original

and Mel Brender m y first version

- for of

these ideas.

D.

Knutson

Columbia University New York City August

1972

CHAPTER

I

k-RINGS

:

i. _ T h e Definition of k-ring A k-ring R is a commutative certain additional n=0,1 . . . . .

ring with identity on which

unary operations

are defined,

satisfying certain axioms.

The simplest example is

the ring of integers ~ , for which In(m)= coefficient. however,

The general definition

kn:R~R,

(~), the binomial

is somewhat complicated,

and will be best understood by first analyzing one

manifestation

of the ring ~ , i t s

example of K-theory.

This

appearance

in the simplest

is the construction of ~ as the

ring of finite dimensional vector spaces over a given field F. Let F be a given field

(the real numbers will do).

Let

be the class of all finite dimensional vector spaces over F. Isomorphism of vector spaces gives an equivalence ~,

and we write

~ for the set of equivalence

~V3 for the isomorphism class of V° isomorphic

In

~ ~

classes,

writing

Since two vector spaces are

if and only if they have the same dimension and the

range of possible dimensions integers,

relation on

is exactly the set of non-negative

can be identified with this set, by

[V]~--> dim V.

there is an operation of addition: we define

~Vl~ = ~V2] + ~V3]

if Vl=V2~V 3 (the direct sum).

are introduced by enlarging

~

Additive

to the set K(F) :

inverses

we define K(F)

to b e the set consisting of all expressions -~ i ] - ~ 2 ~'

~V~

modulo the equivalence relation: ~ V l ~ - ~V2"] =

iVl3-

~V~

if and only if ~

~2 ~

~V~V2~.

6

Taking

[V7 = ['V~ - COT, "[~ is a subset

It is easy to show that K(F) identifying additive

~

with

group,

the positive positive

and this

integers.

integer

K(F)

An element

thought

corresponding

is an abelian group. is isomorphic

is just the usual

can be written

space V, and thus of K(F)

{0, i,2, ..},

of K(F). Indeed,

to Z, as an

construction

of K(F)

of ZZ from

corresponding

to a

as [U'J - [0,~; for some vector

of as an actual vector

to a negative

integer

space.

An element

is a "virtual

vector

space". The tensor product multiplication corresponds

in K(F),

ring with

The e~teMior

introduces

the isomorphism of integers.

Adv of a vector

corresponding

dimV. ( d ) in X.

for V, then one basis

products

under

spaces

Thus K(F)

Prploosition:

i)

Recall

to the formation that

of b i n o m i a l

if [Vl, V 2 ..... Vn]

is a

for Adv is the set of d-fold wedge

if 0 < d ~ n, none such if d b n .

AIv:

Anw = ~ i=0

(AiU ~ An-iv)

integers

(d)

By fiat we take

V

If W = U ® V, then for all positive

i) is obvious.

is

space V now give rise to

A0V = F.

Proof:

72.,

[ V.ll^V.12^.. ^V'id' i~_ 'II<... < id<_ n], and there are

such expressions

ii)

K(F) ~

a

identity.

powers

on K(F)

coefficients basis

which

to the usual m u l t i p l i c a t i o n

a commutative

operations

V 1 M V 2 of vector

n,

In ii),

let

{u 1 ..... Udl } be a b a s i s

a basis

of V,

so that

one basis

of U, and

{v 1 ..... Vd2} b e

of W = U @ V is

[u I .. u ,

A basis basis

of Anw is made

elements

the elements of

of W,

element

forms

Looking

part

the i d e n t i t y

Thus

the A

with

for all x,y

induce

of the f o l l o w i n g

A pre-k-ring

involved

on K(F)

i k :R----FR,

= x n

ki (x) xn-i (y)

3) k n ( x + y ) = i=O

}. d2

of of

one b a s i s

But such

an

I

in this proposition,

we

of p r e - k - r i n g

definition.

6R,

kl(x)

wedges

the order

Hence

the s t r u c t u r e

i) k0 (x) = 1 2)

v " .,

of Aqu@An-qv.

is a c o m m u t a t i v e

of o p e r a t i o n s

,v I, d I

of the form u. ll^'''^U~lq ^V.]I ^..^V,3n_q ,

of a b a s i s

operations

a series

changing

the w e d g e by +i.

at the d i m e n s i o n s

in the sense

are n-fold

d I, 1 4 j
have

n

which

in any such wedge,

of elements

.. < i q ~

Definition:

and

just m u l t i p l i e s

Anw consists

1 ~il<

up of symbols

,

ring R w i t h

i=0,1 ....

identity

, satisfying

I,

There

is an e q u i v a l e n t

formal p o w e r kt(x) Then

series

= X0(x)

in the v a r i a b l e

For x ER,

consider

the

t:

+ kl(x) t + X2 (x) t 2 + ...

the r e q u i r e m e n t s

Xt(x+y)

definition.

are

: Xt(x)kt(Y).

that k0(x)=l,

In this

form

xl(x)=x,and

it is e v i d e n t

that i t ( 0 ) = l

and

k t (-x)= I/k t(x) . As we h a v e is not with

Z is a k-ring,

the only p r e - k - r i n g

integer

coefficients

we can define, with

seen,

X-structures

kt(n)

later

on Z e x c e p t

Another

kt(r)

(l+t)

r

structure

on ~ .

Given

any p o w e r

series

1 + t + n2 t2 + n3t3 +

(i + t + n2t2 + we w i l l

the o r i g i n a l

. )n.

But

rule out all

one above.

pr~

we can d e f i n e

a X-structure

the b i n o m i a l

is o b t a i n e d by

will

This

taking

later pass m u s t e r

by

taking,

theorem.

kt(r)=

e

tr

as a k-ring,

(Again, but

the

won't.)

Let

R be

a pre-k-ring.

if k t ( x ) is a p o l y n o m i a l kin(x)=0,

for all m > n ,

O~ R is a d i f f e r e n c e under

(l+t) n.

, expanding by

l~re'k-structure

the first second

=

=

of h-ring,

On the real numbers ~, for r%R,

kt(n ) =

of the form

for all n,

the d e f i n i t i o n

structure

taking

Xt(n)=(l+t )

n

An

element

of d e g r e e

but

is finitary,

is of f i n i t e d e g r e e

n, or in o t h e r words,

kn(x)~0.

of e l e m e n t s

x%R

R is f i n i t a r y

of finite bur

degree.

R under

n

if

if each e l e m e n t (For eKample,

the same rule

isn't.)

In either Z5 or R, taking kt(x)=(l+t) x, we have kn(x) =

x(x-! ) (x-2)~. (x-n+l) n(n-l) (n-2).. (i)

In general,

we say an element

type if nkn(x)=xkn-l(x-l) element

Ixl =

/(x-l)(x-2).. (x-n+l)) ~x~kn-1 " <(n'l){n-2), [ (i) ] =(n] (x-l) .

in a pre-k-ring R is of binomial

for all n > 0 .

in R is of binomial

type.

R is binomial

if every

Thus b o t h ]% and Z are binomial

under kt(x)=(l+t) x, but not under any of the other pre-k-ring structures mentioned

above.

It turns out surprisingly are not binomial: representation next chapter)

that most of the interesting X-rings

In particular,

ring of G

for each finite group G, the

(which we shall study in detail in the

is a l-ring and is binomial

trivial group,

For a specific example,

if and only if G is the

the permutation group S 3

gives a ring R(S3) which as an abelian group is free on three generators.

One can choose these three generators

where 1 is the identity element,

a2=l, b2=l+a+b,

k t (a)=l+at,

In particular,

bk2-1(b-l)

and I t(b)=l+bt+at 2.

= bkl(b-l):b(b-l)=b2-b=

to be i, a, and b,

ab=b,

lt(1)=l+t,

2k 2 (b)=2a, while

(l+a+b)-b = l+a j 2a, so R(S3)

isn't binomial.

The definition of pre-k-ring kn(x+y). kn(xy) to be.

involves an expression

for

The definition of l-ring involves also expressions

and kn(lm(x)).

for

We now derive what these expressions ought

Suppose R is a pre-k-ring and xER is a sum of elements of

degree i:

x= x I + x 2 +

+ x n, kt(xi) = l+x~t. l

Then

I0 n

1 + il (x) t + X2 (x) t 2 + .

= k t (x)

: ~ k t (l+xit) i=l 2

1 +

(Xl+X2+..+Xn)t

.... + Thus iq(x)

=

1.

=

k2(xy)

taken over all different k2(xy)

= 2 ~ i
But ~" ~x i2 =

Hence

=

(. l

(~ X i ) 2

of the x.'s. 1

are two elements

in R, the product i.

Then

xy =

of degree

pairs +

(xi,Yj,)

(i,j),

of

(~xi) (~yj) I.

Hence

)

the sum b e i n g

(i',j').

~ xi2yjyj, j~" j'

(j j 'YJYJ'

,

+

Hence ~ xixi,Yj 2 i
Sx i 2) (j ~

+

(

j,YjYj

)

, ( 2 . xix i ) ~ Y j ) l' - 2 ~-.

k 2 (xy) : 2 k 2 x X 2 y : x 2 k2y

xixi,

+

:

x 2 - 2 k 2 x.

(x 2_ 2X2x)

+ y 2 X2x

-

k 2y +

2X2 x k 2 y.

Ditto

of

x i,yj

(i + xiYjt )

~(xiYj)

xixi,yjyj,

: 2 (i ~ i'xixi' )

+

that

a sum of elements

~ i,j

function

as the sum of elements

1 is again of degree

k t(xy)

kl(xy)=xy,

and y : y l + . . . , y m

suppose

~_ _ xiY j , b y a s s u m p t i o n

Hence

symmetric

to be expressible

Furthermore,

of degree

+

(XlX 2...x n)t n

x = Xl+...Xn,

R which b o t h h a p p e n

elements

(XlX2+XlX3+X2X3 + . . + x n _ l x n )t

is just the qth e l e m e n t a r y

Now suppose

of d e g r e e

+

for y.

(y2 - 2k2y)

X2x

11

A tedious but similar calculation gives

3 33 33 i (xy) = x k y + y k x +

+ 3k3xk3y + xyk2xk2y _ 3xk2xk3y - 3yk2yX3x.

Similarly,

since k t(x) = ~ ( l + x i t

),

I

kt(kqx)

= --~ l~il< • .
(i +

(x. x. ..x. )t) 11 12 1q

Hence k 2 (k2(x)) : k3 xk 1x - k 4 x, and k2 (k3(x)) = k6x-k5xklx+k4x~2x. Etc.

In the x-ring ~ , each of the expressions

calculated above

gives rise to an identity involving binomial coefficients.

and

<~))=

~ I -

(~]

Thus

Etc.

Thus the general pattern emerges.

The k-powers,

n k x, of an

element x, give the elementary symmetric functions of some variables xi, and then

(using the Fundamental Theorem on Symmetric Functions

- see p.2 and

~46~

can be expressed particular,

) any other symmetric

i in terms of the k x in a unique fashion.

n k (xy) and kn(km(x))

the variables

involved,

in terms of the k-powers Of course,

function of the x.!s i

so have,

are b o t h symmetric

In

funct#~ns of

uniquely by the Theorem,

expressions

of x and y.

all this holds a priori only if x and y are each

expressible as a finite sum of elements of degree l, and if the product of two elements of degree one is again of degree one.

But

12 we

can use this

involved

are,

definition be

and

example

of d e g r e e

function

a way

Let s. and i

o

given

a pre-~-xing

into the formal which

is k n o w n say,

in the second)

in the x.'s, 1

and

W e now give

the v a l i d i t i y

(with n ~ 5

calculating then p a s s i n g

the formal

number

in the first

the result

as

to a p o l y n o m i a l

definition

of this p r o c e d u r e

to

can be

that x is the sum of a finite

~i,~2 , .... ~q;DI, D 2 ..... ~r be 3

the identities

k 7 k3 ( (x)),

or

i: x = X l + X 2 + ' ' + X n

of x.

that

out what

identities

of k5(xy)

just p r e t e n d i n g

in the X-powers in such

all the

Hence

and n ~ 7-3 =21

a symmetric

to figure

any c a l c u l a t i o n

out b y

of elements

case

then b u i l d

of X-ring.

a X-ring,

carried

special

of X - r i n g

is b u i l t

indeterminahes.

in.

Define

by

(1 + sit + s2t2 +

..)

= ~

(I + ~i t) i

(! + Ol t + o2 t2 +

..)

= ~

s~s and o'~s are the e l e m e n t a r y l 3 ~3'.s.

Let Pn(Sl . s. .2 .

(i + ~jt)

symmetric

- in other words,

functions

.. s n ;~I,~2 ..... °n) be

of the

the c o e f f i c i e n t

the

~s i

and

of

t n in

~ (I + ~i~jt) and Pnd(Sl, S2 ..... Snd) b e the c o e f f i c i e n t i,j of t n in the p r o d u c t ~ (1 + By the l<~il< . .
Fundamental with

integer

as q ~ n

and

Theorem

on S y m m e t r i c

coefficients, r~n

in the

Functions,

Pn and

and are i n d e p e n d e n t ~irst

case~

Pnd are p o l y n o m i a l s

of q and r as long

and q 9 nd in the second.

13

Pn and

since over

Pnd h a v e

any ring with

universal

integer

coefficients,

identity,

and so are

they

sometimes

are well-defined referred

to as

polynomials.

Main Definition:

A k-ring

i)

X t (i) =

2)

For

R is a p r e - x - r i n g

in w h i c h

l+t

all x,y

in R,

n ~ 0,

xn(xy)

(xlx, x2x, • . , xn x , X 1 y, ..xny)

: P n

3)

For

a l l x in R,

and

n,m~0,

xm(xn(x) ) = p

(xlx ..... xmnx) mn

Note

the r e q u i r e m e n t

given

on Z and

zero,

since

i) a l r e a d y

~ above,

for any

l) a l s o

integer

m,

rules

requires

Xt(m)

out that

the exotic

X-structures

R is o f c h a r a c t e r i s t i c

= kt(l + 1 +

. +

i)

m t~mes = lt(1) m : zero

(l+t) m = 1 +

in a n y ring.

.

Hence

.

+ tm .

This

in a n y l - r i n g

last p o l y n o m i a l

1 +i +i +...+i

is n o t

~ 0.

m times In any k3(xX2y),

say,

through

as

little

more

P

n

or P

nm

X-ring,

the procedure

in t e r m s

indicated tha~

of

from scratch

Unfortunately,

the k-powers

above.

in e a c h

case

Indeed,

this

an expression

of x a n d y c a n b e

that procedure

calculating

and plugging

from

for r e d u c i n g

in v a l u e s

definition,

for

carried

amounts

the universal

to

polynomial

the s , ~. l 3

it is n o t

like

so c l e a r

14

how

to p r o v e

that a n y t h i n g

In the n e x t section, producing kt(n ) = involves

we will give a canonical

lots of k-rings,

(l+t)

n

is one.

the A d a m s

in p a r t i c u l a r , (Z, say)

and use

and

construction

it to s h o w that ~

The m o s t c o n v e n i e n t

operations

is a l-ring~

is g i v e n

criterion, On p.

49 .

for

under though,

15

2.

General

Constructions

Let A be 1 + A[[t~ with

+

a ring, denote

and D e f i n i t i o n s

commutative

the

with

on l-rings

identity.

~ a tn n ' n=0 in the t h e o r y of

set of all formal

a EA, for all n, and a0=l.

Let

power

(As usual

series

n

formal

power

series,

no q u e s t i o n s

infinite

sequence

to index

the c o e f f i c i e n t s

term

of a

n

of c o n v e r g e n c e

is allowed, a

n

.)

and the p o w e r s

Such

power

endow

l+A[[t~7 + w i t h

multiplication

".",

and

operations

i-powers

and b . l.+ b l .t + b .2 t 2 +

usual

product

power

series

a"+"b

with

serve

first

additive

= p

n

defined

unit a"."b

above

c = l+Clt

is then

unit

).

).

is the

P

n

series

w i t h Phi

l+0t+0t2+..

after

given

the first

is simply l+t.

with

is the p o l y n o m i a l

In particular,

"I" is the series

series

c=l+Clt+c2t2+..,

, where

are zero

is the power

Cn =Pni(al 'a2"'''ani)'

If

a"+"b

the power

series

and d= l+dlt , the p r o d u c t

c="Ai"(a)

12

12

coefficients

multiplicative

( page

"0"

is that power

( page

then

"+",

(a2+albl+b2) t2 + ( a 3 + a 2 b l + a l b 2 + b 3 ) t 3

(al,a2,..,an;bl,b2,..,bn)

series w h o s e

i~0,

element

of a d d i t i o n

of a and b:

1 + (al+bl) t +

The p r o d u c t n

of t just

"A n'' as follows:

a = l + a l t + a 2 t 2 +..,

c

series,

- any

i, are c a l l e d s p e c i a ! .

We

The

arise

two p o w e r term,

cd= l+Cldlt.

Finally,

c=l+Clt+c2t2+..,

the p o l y n o m i a l

say,

defined

The

for an integer where above

•

16

Proposition:

Under

the o p e r a t i o n s

defined

above,

l+[[t]] + is

a pre-k-ring.

Proof:

Let us check

the d i s t r i b u t i v e side

axiom:

is a power

equality,

series

polynomials

n

the i d e n t i t i e s

be

l+blt+b2t2+

independent

=~(l+~jt)

a

n

~ 0.

b=l+blt

integer

These

n,

coefficients

al,..,an,bl,..,bn,Cl,..,Cn.

axiom

i

of P

axioms

facts

the o p e r a t i o n s if and only

for each

agree.

Each

and to show the

to check

them

are

Hence in

independent.

So let

and let l + a l t + a 2 t 2 + . . = ~ ( l + ~ i t ) ,

and l+Clt+C2 t2 + . . = ~ ( l + ~ k

= I~(l+~ • l,~

pre-k-ring

- say

to be p r o v e d

t) .

is then

The

left

computed:

o~(l+~kt)

) "+"~l+~i~k

Several

sides

variables

• .="~( 1+~j t) ,

(here the d e f i n i t i o n

other

that

al, .... Cn are a l g e b r a i c a l l y

a" ." (b"+"c)

~(l+~i~jt

from a,b,c

it is s u f f i c i e n t

side of the d i s t r i b u t i v e b"+"e

to check

on b o t h

of p r e - X - r i n g s

= (a" ."b) "+" (a" ."c) .

in t, c o m p u t e d

in the q u a n t i t i e s

the case w h e r e ~i,~j,~k

of t

axiom

a"."(b"+"c)

it is s u f f i c i e n t

the c o e f f i c i e n t s

to check

some p a r t i c u l a r

n

~ t),~ 3 i,k

t)

which

follow

in l+A[[t]] +. th

If a is of d e g r e e is of d e g r e e

•

is u s e d

are simple

if a is an n

(l+~i~t)

This

is the right h a n d

is e x a c t l y

side.

The

I

similarly.

consequences Letting

degree n,

explicitly).

of the d e f i n i t i o n

a= ~ a n t

polynomial:

"An"a

= l+a t. n

i, a .... . b = ~ a n ( b l tn)

n

of

, a is of d e g r e e a =0, m > n, b u t m For

any a, if

n

17

Let A be a pre-k-ring.

kt: A -

Proposition:

~l+A[[t~

Consider

+

A is a k-ring

the m a p

defined by

a~kt(a)

.

if and only

if k t is a h o m o m o r p h i s m

of p r e - k - r i n g s . Proof:

By the d e f i n i t i o n

r i n g A.

The

operations

m a k e the p r o p o s i t i o n T h u s we h a v e it is s o m e w h a t

Proposition: i)

additive,

"A n,, h a v e b e e n

exactly

Each

for any defined

to

I

true.

an a l t e r n a t e

easier

definition

to p r o v e

of l-ring.

that c e r t a i n

rings

With

are

this

k-rings.

in w h i c h

= l+t element

finite element iii)

"." and

kt is always

Let R b e a p r e - k - r i n g

kt(1)

ii)

of "+",

sum in w h i c h a

of d e g r e e

l

The p r o d u c t degree

rCR is e x p r e s s i b l e

in the f o r m r = ~ + a i, a

each s u m m a n d

is p l u s - o r - m i n u s

an

1

of two e l e m e n t s

of d e g r e e

one is again

of

one.

T h e n R is a k-ring. Proof:

We

"i" = l+t.

are g i v e n

kt:R ~

We must check

l + R [ [ t ~ + additive,

that

kt(xY)=kt(x)kt(y)

and kt(1)

=

and

kt(kn(x)) = " A n " ( k t ( x )) • F i r s t we c h e c k sums

of + e l e m e n t s

do the case w h e r e

the m u l t i p l i c a t i v i t y . of d e g r e e

i, and

S i n c e x and y are each

kt is additive,

x and y are of d e g r e e

i.

Hence

we can

just

their product

18

is of degree i, which is to say, kt(xY) equals

But this latter

(l+xt) .... (l+yt) = kt(x)"."kt(y) -

The proof of kt(kn(x))="An"(kt(x)) identity holds for x=a, b Xt( ~

= l+xyt.

ki(a) kn-i(b))

We then compute

=

Suppose the

kt(kn(a+b)

= ,,~,, lt(ki(a) kn-i(b))

" ~ " ~kt(ki(a)) ....kt(k n-i(b)))

= "An"(xt(a)"+"kt(b))

is similar.

=

=

= " Z "(''Ai''(kt(a)''- ....An-l"(kt(b)))

"An"(lt(a+b)) .

So again, we can check the identity for x of degree one, which is straightforward.

Corollary:

~, under the X-operations

defined by ln(m)= (~) is

a k-ring.

J

(There is a converse to this proposition Principle

(L43)asserting

- the Splitting

every l-ring can be imbedded inside

a X-ring of the type described in the proposition

Theorem:

(Grothendieck)

the pre-x-ring

For any commutative

l+A[[t~]

+

is a k-ring.

Proof: We already know that rl"i" = l+t "An"(l+t)

=~"0"

1

n=0,1 n>~2

above.j

ring A with identity,

19

Thus we must

show,

for e a c h n~I,

that

"An"(x"."y)

p ~ x , " A 2 " x .... , " A n " x ; y , " A 2 " y .... ,"An"y), n iterated

k-powers.

again boils

But

as in the p r o o f

d o w n to p r o v i n g

certain

T o c h e c k e a c h of t h e s e i d e n t i t i e s , are the

"sum"

of a f i n i t e n u m b e r

l + A [ [ t ] ~ +.

Hence

concerned,

w e can

of t w o e l e m e n t s

this

previous

on p a g e

sub-pre-k-ring

proposition,

, this identities.

w e c a n a s s u m e t h a t x and y

of e l e m e n t s

of e l e m e n t s

of d e g r e e one

16

formal polynomial

j u s t l o o k at t h e e l e m e n t s

l + A [ [ t T ] + so t h e s e e l e m e n t s Since

and the same for

"of d e g r e e

as far as the t h i n g s w e h a v e

are sums a n d d i f f e r e n c e s

=

to c h e c k

i" in are

of l + A [ [ t ~ + w h i c h

of d e g r e e

i.

The product

is a g a i n of d e g r e e one in

form a sub-pre-k-ring

satisfies

the h y p o t h e s e s

it is a k-ring,

of l + A [ [ t ] ~ +. ot the

so the i d e n t i t i e s h o l d .

I

20 A map of k-rinqs RI---->R2, is a ring h o m o m o r p h i s m such that kn(f(r)):f(kn(r)) augmented ¢:R~

f:R 1

for all r6R I, and all n A 0.

..~... R2,

An

k-ring is a k-ring R together with a map of x-rings

2.

The class of all X-rings and maps of k-rings form the cateqory of k-rinqs, which we denote by

(k-rings). The special

power series ring construction provides

a functor from the

category of rings,

(Rings), to (k-rings): A ~

l+A[[t]] +. Let

U:(k-rings) ---~> (Rings) be the "forgetful functor"

assigning to

each k-ring R, its underlying ring R (with the k-structure Proposition:

Let R be a ring and S be a k-ring.

Then there is

a one-one correspondence between maps of k-rings ~:S ~ and maps of rings 9:U(S) ~ >

The first is the map

m R, by f(l+alt+a2t2+..)=al , which is a map of rings.

The second is the map of k-rings,

kt:S----~l+S[[t]] +.

Now for each map of x-rings ~:S ~ ] f gives a map of rings f ~ : U ( S ) ~ ) rings

%:S~R

l+R[[t]] +,

R.

Proof: W e make use of two maps. f:l+R[[t]] +

ignored).

R.

gives a map of k-rings

l+R[[t]] +, composing with

Convers~y,

a map of

~:l+S[[t]]+~

l+R[[t]] +,

~(l+Slt+s2t2+ . .)=l+%(sl)t+~(s2 )t2 +..

Composing this with

a map of x-rings

These two operations

~kt:S----~l+R[[t]] +.

kt gives

inverse,

i

From a cateqorical pdint of view*, this r e s u l t *The reader who is not categorically paragraph without loss of continuity.

are

is surprisinq.

inclined can skip this

21

It says that the forgetful a riqht adjoint.

functor

By far the usual

functor, (e.g., the one from its u n d e r l y i n g construction

(Groups)

out c o n s t r u c t i o n s

left exactness

a product

l-ring structure

as a set,

(in that case,

the

Of c o u r s ~ our functor Hence

fact can be used to carry For example,

to

it must be the case ~ y definition

then U(RxS)=U(R)xU(S) .

on the product

of Thus

of rings RxS.

M o t i v a t e d by these considerations,

W e define the product or R and S,

the c a r t e s i a n product

are d e f i n e d coordinate-wise:

= (i,i)

a group to

definitions.

(rl,Sl).(r2,s2)=(rlr2,sls2)

and kt(r,0)

taking

of R and S, one only need find the

Let R and S be l-rings.

operations

(sets)

of l-rings.

in (l-rings),

for tensor products.

to be,

has

is that a forgetful

of U that if the usual c a t e g o r i c a l

we make the following

RxS,

This

in the category

RxS ,is r e p r e s e n t a b l e

Sinilarly

(Rings)

(which the reader may construct).

define a product RxS of k-rings,

appropriate

to

of the free g r o u p on a set).

left and right exact.

to c o n s t r u c t

situation

set~ has only a left adjoint

U also has a left adjoint it is b o t h

(k-rings)~D

.

W e define

of R and S.

Ring

(rl,Sl)+(r2,s2)=(rl+r2,sl+s2) kt(r,s)=kt(r,0).kt(0,s),

+

(kn(r) ,0) t n, and similarly for kt(0,s) . n 1 The tensor product of two k-rings is c o n s t r u c t e d b y taking the tensor product R~S of the rings R and S, and defining kn(a)~l,

kn(l~b)=l~kn(b),

and kn(a~b)=kn(

a~l

. l~b).

kn(aMl) = Similarly,

22

the inverse

l i m i t of an i n v e r s e s y s t e m of k-rings,

o b t a i n e d b y t a k i n g the i n v e r s e obvious

k-structure.

but rather

tedious

l i m i t as rings

(The v e r i f i c a t i o n

so w e

L i m R,,

and i m p o s i n g

of all t h i s

l e a v e it to the reader.)

is the

is s t r a i g h t f o r w a r d

23 Proposition:

Let R be a k-ring and R[X] the ring of polynomials

in one variable X over R. k-ring on R[X] satisfying i)

Then there is a unique structure the three equivalent

of

requirements:

X is of degree 1

iX)

X n is of degree 1 for all n ~ l

iii)

kq(rx n) = kq(r)X nr for all integers n,q~/ 0, r6R.

Furthermore,

if R is augmented,

R[X] may be augmented

either by

e(X) =0 or ¢(X)=I.

Proof: The equivalence

of the three requirements

on an element X

in a k-ring has been noted previously. ( page 16 ) The uniqueness clear:

if f(x) = 7 a

l

x

i

6 R[X~ is any element,

to be a k-ring satisfying kt(f(]{)) = kt( ~

(aiXi))

the requirements, : ] ~ kt(aiXi)

Taking this as the definition

then for R[X]

it must be that

= l-~kxit(a )l "

of kt(f(X))

in R[X],

it is easy

to see that kt(f(X)+g(X))=kt(f(X)) kt(g(X)) =kt(f(X))"+"kt(g(X)) To check the multiplicativity,

it is sufficient

case of the product of monomials. xt(rs ) = it(r ) ....it(s)

iX)

kt(xmx n) = kt(X m) "."kt(X n)

definition

r,sCR

kt(rX n) = kt(r ) ....kt(X n)

i) is true by hypothesis

m,n~/ 0 r6R, n ~ 0 .

and iX) and iii)

given here of i t .

to consider the

We only need show

i)

iii)

-

follow easily from the

is

24

In c h e c k i n g sufficient,

that

since

kt is a d d i t i v e

for the two t r i v i a l E a c h of t h e s e

the

cases:

degree

of X = i.

~ gives

W e now c o n s t r u c t

in the p r e v i o u s

homomorphism

and f(X)=r,

to show this

r an e l e m e n t

the only c o n d i t i o n

to r e q u i r e

R[X~ --~ R, by f(X) ~

to b e

from

o

of R.

is a k-ring,

by

Thus

¢(X) =0.

is that

Then

For

with

¢(X)=I,

f(1) .

k-ring

n-i

¢(X)=0.

f(0) , c o m p o s e d

on R[X~ w i t h

f(X) ~ >

proposition.

~rn:Qr ~

to X,

Say we w i s h

the free

~ be c o n s t r u c t e d n

kt(~')=l+~'t'l 1

f(X)=X,

an a u g m e n t a t i o n

the f i r s t m a p is t a k e n

and

and m u l t i p l i c a t i v e ,

an a u g m e n t a t i o n

k-ring h o m o m o r p h i s m

¢:R~)

it is

is trivial.

In a s s i g n i n g ¢(X) ~

kt(kn(f(X)))="An"(kt(f(X))),

I

on one g e n e r a t o r .

adding

Let

an i n d e t e r m i n a t e

Q0 = ~' ~ , as n

Qn = ZZ [~i, ~2 ' "" ' ~n I, w i t h

and for any r > n ,

there

is a k - r i n g

Qn' ~([i ) = [i or O, d e p e n d i n g

on w h e t h e r

i ~ n or not. Let

Q = Lim

Qr"

(Recall

(...,an,an_l, ..,al,a0) Thus

an e l e m e n t

set of v a r i a b l e s polynomial set equal

of

with

if all b u t

is the

set of all s e q u e n c e s

aiEQ.1 and ~ n ~ i , n ( a n + l ) = a n

~ is an i n f i n i t e

~., h a v i n g 1

to zero.)

this

power

the p r o p e r t y

a finite

number

As r e m a r k e d

series that

in the i n f i n i t e

it r e d u c e s

of the v a r i a b l e s

above,

for all n.

(page 22)

to a

~. are l

Q is a l-ring.

25 th Let symmetric and a

n

an = an (?I'~2' .... ~r ) 6 Qr be the n function

of the

6 Q be the a s s o c i a t e d

kn(al)=an , for all n a's n

involving

some

a polynomial and this

Let comments,

of

identity

independent

involving

a I must

generated

contain

each

On the other hand,

f(al,a 2,..) EZ[al,a 2,..] coefficients.

In the next the

to m e n t i o n

section, free

by

ring

form

give

a I.

By the previous

in an infinite Indeed

any s u b - k - r i n g

polynomials,

in the al w i t h

is a s u b - k - r i n g

Three,

on one generator.

r

functions).

kn(f(al,a2,..)),

a polynomial

~ [al,a2,..]

a categorical

already

the u n i v e r s a l

and in C h a p t e r

k-ring

id~tity

an=kn(al ) , and so c o n t a i n

using

of the

as again Hence

Then

©r ) and the

would

A = ~ [ a l , a 2 .... ]-

any e x p r e s s i o n

if n ~ r)

limit.

any p o l y n o m i a l

of s y m m e t r i c

is a p o l y n o m i a l

of indeterminates:

we can e v a l u a t e

wish

inverse

in each

(since

to zero

al($1 '''' [ r ) ' ..,ar(~l,.. ' ~r ) 6

sub-k-ring

A, as a ring,

intensively

is true

out by the t h e o r e m

A c Q be the

Q containing

integer

this

in the

of a l , a 2 , . . , a r, for example,

is ruled

~ [ a l , a 2 , . . ].

(equal by d e f i n i t i o n element

(since

are a l g e b r a i c a l l y

number

~'s

elementary

we w i l l

of

study

H e r e we w i s h

application. A n a t u r a l

Q.

just

operation

26 on the category of ~-rings is a natural transformation from the underlying set identity functor to itself. That is, we have an assignment to each k-ring R, a map

(of sets)

~R:R ~ >

R such that for any map

of ~-rings f:R ---~ S, f~R = ~S f :R--~ S. operations

is defined b y

for multiplication

Proposition: isomorphic

A d d i t i o n of natural

(~R+~R) (r)= ~R(r)+~R(r)

and similarly

and k-operations.

The set of natural operations

is a i-ring,

to the free i-ring on one generator,

Proof: Let ~ be an operation.

and is

A-

Let alEA be the generator of A

and suppose uA(a I) = f(al,a 2 .... ) E A-

For any i-ring R, and rER,

let g:A---~R be the unique i-ring h o m o m o r p h i s m with g(al)=r. Then uR(r) Conversely, A-~A,

= uRg(a I) = g(uA(al))

= g(f(al,a2,..

given any element f(al,a2,..)EA,

taking a I to f(al,a2,..)

)) = f(r,~2(r),..) •

the unique map

extends uniquely to a natural

operation.

I

Hence given a natural operation, in the i-operations.

it is uniquely a polynomial

To check that a given polynomial f(il,i 2,..)

is equal to a given ~, one only need check u(al)=f(al,a 2 .... ). This being a proposed identity in to check in each Qn"

A c Q = Lim Q , it is sufficient n

This can be formally phrased as the

27

Verification uniquely

Principle:

a polynomial

If

in the l - o p e r a t i o n s

particular polynomial

f

(kl

s u f f i c i e n t to c h e c k t h a t of e l e m e n t s

of d e g r e e

I

2 ,l ,..),

and f o r m u l a t e

This process category.

such,

finite groups,or

compact Hausdorff

an o p e r a t i o n

on the c a t e g o r y principle

as f o l l o w s .

Let C be

The

K

0

the c a t e g o r y

s p a c e ~ , w e can d e f i n e f r o m C to the c a t e g o r y

functor C ~>

set of all s u c h o p e r a t i o n s

(l-rings) ~

any of

a K-theory of x - r i n g s . trans-

(Sets)

for a g i v e n C and K f o r m

a n d t h e r e is a n a t u r a l m a p of k - r i n g s

in g e n e r a l

for such.

of

in the K - t h e o r y K 0 is a n a t u r a l

f r o m the c o m p o s i t e

a k-ring/Op(K0~ Of course,

k - r i n g on k g e n e r a t o r s ,

a similar verification

to b e a c o n t r a v a r i a n t ~ n c t o r

to itself.

[l+[2+..+~r

B y a n a l o g y w i t h the c a s e w h e r e C is the c a t e g o r y

~ - m ~ u ~

formation

on a s u m

operations

can be generalized

~u~mmmmm~mmmmm~m~

Given

U = f, it is

E

construct the free

and d i s c u s s k - a r y n a t u r a l

of k-rings,

U is

and for any

to c h e c k t h a t

U = f, o p e r a t i n g

then

i, for all r > 0.

One can similarly any k ~

U is a x - r i n g o p e r a t i o n ,

A~OP(K0)

t h i s m a p n e e d b e n e i t h e r o n e - o n e n o r onto.

.

28

3.

Symmetric Functions

Let that

A b e the f r e e

l - r i n g on one g e n e r a t o r .

A as a r i n g is a p o l y n o m i a l

ov v a r i a b l e s : A was

Recall

this m e a n s

r i n g o v e r ~ in an i n f i n i t e n u m b e r

n [al,a 2 .... ], a n d k (al)=a n-

A = ~

constructed

as a s u b r i n g

of a k-ring

Q = Lim Q , < n

Qn = ~[[i' ~2'''" ~ ' and w e can sum up the r e l a t i o n b e t w e e n

g's by the

a's and the An

element

equation

--

if as a f u n c t i o n of the

of d e g r e e k.

of w e i g h t

Thus

a monomial

Let

of w e i g h t

{al,a2,... ] forms rI aI

r2 a2

a2

r

...an

n

has

n in A. addition,

n is a g a i n of w e i g h t a polynomial basis

n.

of

s i n c e the s u m of two S i n c e the set

A, the set of m o n o m i a l s

rn ..an

integers)

' (all n, all s e q u e n c e s

f o r m an a d d i t i v e b a s i s

of

r l , . . , r n of n o n - n e g a t i v e

A-

Hence rI

consists

_-

aI

r~

An d e n o t e the set of all e l e m e n t s

A is an a b e l i a n g r o u p u n d e r n elements

~ 's, it is

r]

rl+2r2e3r3 +'''+nrn"

isobaric

an tn = ~ (l+~.t)l " n=0 i=l of A is c a l l e d i s o b a r i c

1

homogeneous weight

k t (a I ) = ~

f ( a l , a 2 , . . , a n ) = F ( ~ l , ~2''')

of w e i < ~ t k in t h e a ' s --

the

of all m o n o m i a l s

~ ir. = n. l

The number

of t h e f o r m a I

of such m o n o m i a l s

one b a s i s

r2 a2

of

..an

with

is t h e n u m b e r

of p a r t i t i o n s

rl2r 2 of the n u m b e r n.

Indeed,

to e a c h p a r t i t i o n rI

w e can a s s o c i a t e be denoted

a .

set of all aT,

the m o n o m i a l Thus

aI

r2 a2

~ = 1

rn ...n

of n.

of n,

rn ..an

This monomial will

An is a free a b e l i a n g r o u p w i t h b a s i s

~ a partition

An

rn

the

29

At this p o i n t notation

it is c o n v e n i e n t

on p a r t i t i o n s .

Given

is any s u m n = n l + n 2 + . . . + n k ,

to i n t r o d u c e

a number

n!>0.

some g e n e r a l

n, a p a r t i t i o n

~ of n

If r I of the n's are equal

to l, r 2 are equal to 2, etc., this p a r t i t i o n is d e n o t e d r r r ~= 1 12 2 ...n n T w o p a r t i t i o n s are equal if and only if the corresponding write

rl,r2,..,

the p a r t s

in d e c r e a s i n g

n l ~ n 2 ~ ... >_nk. associated with n squares,

=

Given

~ =

in k rows,

any p a r t i t i o n

b e the lengths

graph

notation

or Y o u n q

(nl,n 2, .... n k)

with

and all of the rows

common

is to

(nl,n 2 .... ,nk) , w i t h

(Ferrar's

the p a r t i t i o n

(6,5,3,3,1,i,i)

Another

order:~=

The diagram

arranged

t h e squares,

are equal.

the i

th

diagram)

consists

of

row c o n t a i n i n g

lined up at the left.

n ! of

Thus

gives

~, w e can d r a w its g r a p h

of the c o l u m n s .

The

sequence

and let l l , 1 2 , . . . , 1 q ( l l , 1 2 , . . . , 1 q)

is /

also a p a r t i t i o n Its d i a g r a m diagonal. The

is o b t a i n e d b y f l i p p i n g Thus

~ E ~(n),

partition

of ~, d e n o t e d

~

the d i a g r a m

for ~ a l o n g

its

for ~ = ( 6 , 5 , 3 , 3 , 1 , I , i ) ,

set of all p a r t i t i o n s

(For a t a b l e For

of n, the c o n j u q a t e

another

common

~'= ( 7 , 4 , 4 , 2 , 2 , 1 ) .

of a n u m b e r n is d e n o t e d

of the size of H(n)

for n=l,2 ..... 200,

notation

.

is ~ ~ n.

see

H(n). [31~

.

30 Now back monomials gives

of

As

Since

the w e i g h t

An>~ik~---m>in+k,

of the p r o d u c t

of two

multiplication

for each n,k.

Hence

in A

h = ~ A n is n a 0

ring. shown

{aT I ~ ~n]. classical bases

An.

i is the sum of the weights,

a map

a graded

to

above, B u t this

theory

are also

The

first

In terms

each

An is a free

abelian

is not the o n l y " n a t u r a l "

of s y m m e t r i c

functions,

group w i t h basis.

several

other

From

"natural"

of t h e s e

is the [,

set of h o m o q e n e o u s

power

we can d e f i n e

l

hl

: ~i

h2

= i~z-j gi ~j

n

This

=

= Zan

k_t(al)

al =

a12-a2

~ ~ . • .. L_ i 11 1 2 ll~- 12L-n

definition

kt(al)

the

indicated.

of the v a r i a b l e s

h

a basis

? in

can also be w r i t t e n

tn

= ~

(l+~it) "

-

~ i (1-~l t)

=

~i t

nnl

•

as follows:

Thus

1 - ~i t

i

=

h nt n

n

We have

sums.

31

Hence

the a's and h ' s

are r e l a t e d b y the i d e n t i t y

~-.~ant

(-i) n h n t n

n

n=0 Equating a

~ /

J

coefficients

of t n g i v e s

H e n c e the set

we

+

n - hlan-I

A, w i t h

each h

.

{ hn

n

i

.

+

n ~0}

of w e i g h t

is

{ h

(-1)nh

n

1

=

0

n>~l

Given

a p o l y n o m i a l b a s i s of r r a p a r t i t i o n ~ = (l 12 2 ° - .)

Then another b a s i s for An as f r e e a b e l i a n

the n a t u r a l w a y to d e s c r i b e

on a k - r i n g R is to s p e c i f y n o t the o p e r a t i o n s r a t h e r the o p e r a t i o n s to the same t h i n g

n

's.

And,

•

{~ ~ n } .

In some cases,

h

.

also forms

n.

w r i t e h~= h l r l h 2 r2 . . . .

group

=

n=0

n

.

By the

formulas

k n, b u t

above,

• since the k n 's are e x p r e s s e d

of course,

c h o i c e of b a s e s

h

the k - s t r u c t u r e

this

in t e r m s of the

the same c o m m e n t w o u l d h o l d

of A , n~l. n

amounts

for any

32 Another basis is given by the monomial symmetric functions. rI r2 Let ~ = (i 2 ...) be a partition of a number n. We give two descriptions

of the monomial

symmetric

function

<~) associated

to ~.

One is a formula

.... ~ <~> =

~

2

'

2

2

{il~i 2 " ' [l r l { J l [ J 2" " []r2

3

3

~k l" "~k r3"'"

il< i 2 < ..
.. ~ k

r3

(all indices distinct) For a more abstract r1 11

definition,

consider the set of all

monomials

r

k all r > O, i > 0 "'" ik ' i ]

in ~. The infinite

symmetric

33

g r o u p S~

acts on this r 1

G i v e n now rI E1

~ = 1

set by,

~'

r 2

2

r ~(~°i..) lI

for ~6S

...

6 H(n),

consider

r = ~i11" )

the o r b i t

of

r2 ~2

..

under

the action.

sum in Q of all the e l e m e n t s

(~> is d e f i n e d in this

orbit.

as the formal In o l d - f a s h i o n e d rI

terminology,

(~> is o b t a i n e d b y

starting with

~i

r2 ~2

... and

then

symmetrizing. (~> is s y m m e t r i c b y the f u n d a m e n t a l

and h o m o g e n e o u s

theorem

in the a 's, and i n d e e d n O n the o t h e r hand, every

of s y m m e t r i c

an i n t e g r a l symmetric

a , is c l e a r l y w r i t e a b l e

functions. Thus

every

(Note,

form~a basis

of

Depending

combination function,

that

symmetric

n in the

functions,

in terms

for instance,

the set of m o n o m i a l

of d e g r e e

a

a polynomial

of the a , ~ ~ n .

and in p a r t i c u l a r

of m o n o m i a l n

=

~'s, h e n c e

(in>,

functions

symmetric

and h

{ (~>I

n

=

~

(~>) . ~n

~ ~n}

A • n

on the p r o b l e m

of

A may be easiest n

in

A, the a ' s

encountered,

to deal with.

For

one or a n o t h e r b a s i s

example,

when multiplying

or h ' s are m o s t c o n v e n i e n t : a .a = a where ~ ~i ~2 ~I'~2 ~i.~2 is the p a r t i t i o n o b t a i n e d b y a d j o i n i n g ~i and ~2" rI r2 s s r +s r +s (I.e., if ~l =(i 2 --I ~2 =(i 12 2..) , t h e n ~i.~2 =(i 1 12 2 2..) .) The multiplication

of m o n o m i a l

hand more reminiscent consider

the p r o d u c t

and 4 r e s p e c t i v e l y ,

symmetric

of c l a s s i c a l (133>.(122>.

so the p r o d u c t

functions

combinatorics. The

factors

is of w e i g h t

is on the o t h e r For

example,

are of w e i g h t i0.

Hence

6

34

<133).

<122) =

~

a <~ ~i0 ~

Let us find a 0 where ~0=(12224) .

for some integers a

a~o occurs as the coefficient of

2 2 4 ~i~2~3 ~4 ~5 and can be interpreted as the number of times this particular term arises as the product of terms of the forms 3 [il~i2 i3~i4

(il< i2 <~ i3, i4~ii,i2,i3)

and

2 ~Jl ~j2 <j3

3 j3~Jl,J2 ) . We first observe that the term ~i

(Jl~ J2"

can only contribute 4

4 to <5

so i4=5 and since 4

occurs in the product but only E35 in

the first term, either ~Jl or Since J2 ~Jl '~nO ~j' j ~ 5

~32 of

the second term must be ~5"

occurs in the result, it must be that

Thus the product must be of the form ~j2=~5" 3 ~ii~i2~i3~5 (Here i S

i2~ i3).

2 ~Jl~5~J3

"

Similar argument shows that J3=3 or 4.

made tha£ choice, Jl is determined. 2

2

4

EI£2E3 ~4 E5

2 2 4 ~i~2~3 ~4 ~5

=

arises is either as 2)

as (EI~2~4[53) o(E4E5~3

Similarly %6221

Etc.

0,

Thus the only way that

(~1E2~3~53) .(E3E5E42) or a

. Hence

Having

=

<12224)

2.

2, and a(1523)= 10.

(The eager reader

is invited to work out some of these products, and also to give for A3, say, the transformations between the various bases.)

35

T h e next

set of s y m m e t r i c

all b u t r a t h e r a vector

a "rational

space over

the p o w e r

sums.

functions

basis",

the r a t i o n a l

is not a real b a s i s

i.e.,

numbers

a basis These

@.

=

[i + [2 +

s2

=

2 2 ~i +~2 +

" "

= aI

s3

=

~13+~2 3 +

" "

= al 3 - 3 a l a 2 + 3 a a

kt = ~antn

~t

d -

A ~ ~ as

functions

are

We define

Sl

Letting

of

at

= aI

= ~

(l+~i t), we c o m p u t e

log ~t

d

dt

2 2a2

~log(l.~

=



=

<2>

=

<3>

Waring

s formula:

t) = ~

- dt

i

i i+~

kt

t

i co

2 -

(in terms

of

1 k_ t

~i

t

+

~i

_~hntn,

W e can solve this

kl t

3 2

=

t

n -

--- ~

n=O

(-i)

tn Sn+ 1

d_ ~t ~ n dt log( - ) = n=0° Sn+it ")

for the a ' s n

in terms

of the s 's as follows: n

i (-l) n s n + i t n n= 0

Hence ~--(n+l) an+itn n=0

Equating

coefficients

-- ~

. .n Sn+ 1 tn~/ (-i) n=0

of t n gives N e w t o n ' s

antn ~ n=0

Formulas

/

36

(_l) n

n-i Sn+ 1 +

(-i)

n-2 Sna I +

(-i)

Sn_la2

+

.

=

s1

=

s2

=

s3

=

s4

+

sla n =

(n+l) a n + I

Thus aI sla I -

2a 2

s2a I - Sla 2 + s3a I - s2a 2 ÷

Solving A ~ Q n

for and

an by

allow

3a 3 sla 3 -

Cramer's

rule

ourselves

to II

det

4a 4

(here

divide

by

we

step

out

integers)

0

of we

s1

sI

-2

0

s2

-s I

3

Sn_l-Sn_

s2 0

s3

2

sn

a

n

1

det

0

0

sI

-2

0

0

s2

-s I

3

0

s

(-i) n ~ l l n-1

1 n

det

sI

1

0

0

..

0

s2

sI

2

0

..

0

s3

s2

sI

3

..

0

ot

o

s

n

s

n-i

. .

s1

An have

into

37

(where

the last

operations

that

Given rI

evaluating

column

the denominator,

signs cancel.) r r ~ = (! 12 2...) of an integer

a partition

s2

.. .

b y n:,

involving,

Then

Is

the above

In ~n}

Three,

identities

expression

is a b a s i s

a number

element

and

n, we w r i t e

in 0~Q.

The

of i, and all e x p r e s s i o n s

A n element

in A ~i in each

[i~l = x ,l It(x)

as f(x,y).

modulo

A . n

of

will be required in~i n.

is thus

latter has {i~l,

a polynomial

We will

set of v a r i a b l e s .

= I-~(l+x t)l , kt(y) Thus

if

{r

~i,~2

In ~ n } ~n

and l~[j,

.

of

the power

series

Then

are two b a s e s

i .((l_x y. t) 1,3

so

f can be e x p r e s s e d

An ~A n -

~ h n (xy) t n = n=0

.

to w r i t e

and y = Yl + Y2 + "

[s~ I n ~ n }

} is a b a s i s

basis

[2MI,..;I~[I,..)

o~

Consider

verify

i,j=1,2,3,..

It is c o n v e n i e n t

:-~(l+yjt) and

a polynomial

f([iMl,

i ~ [ ] = y j , x : x I +x 2 +..,

{r i(x) s ~72(Y) I

that,

here.

consisting

symmetrical

of

shows

of i d e n t i t i e s

for each n, a c e r t a i n

AMA is c o n t a i n e d

that

elementary

all the minus

In C h a p t e r

these

by p e r f o r m i n g

r2

= sI

division

follows

on the numerator,

observing

s

step

i 3

of

An,

38 Calculating this product, we get

i I-~j(l-xiYj

t) I

= ~I I~0

hn(Y) (xit)n i

co

= ~ n=0 Thus h (xy) = n

~ ~h ~ t-n

By symmetry, h n(xy) =

i~

~ (-n

h (y) <~> (x) i tn

(y

<~>(x) ) 6 Anna / n

~

~<~> (y)h (x)~. ~n

Another expression for hn(xY) arises from the identity d d-~ log

1 k_t (x)

=

~ n=0

Sn+l

(x) tn

Hence log

so

i ~ (1-xit) 1

log ~-T (l-xiYjt) i,j

Hence

~ (l-xiYjt) 1,3

j

= ~

=

n=l

~ sn (x) t n=l n s n (x) (yjt)n n

h (xy)tn n

Calculating the exponential, we get

n

s n(x)s n(y)t n = ~ n= i n

/ ~ sn(x)sn (y)tn I e\n=l n /

39

Sl(X)Sl(Y)t

+ S2(X) S 2 ( Y ) t 2 2

e

+ S3(X) S 3 ( Y ) t 3 3

Sl(X) Sl (y) n=l

where

the

~+2~+...+ny Given

sum

)

.

2 (x) 2

1 ~ a.'

inside

+

n

2 ~ B'

is o v e r

all

tn nY

sequences

7-'

(~,8 .... y) w i t h

= n. a partition

~ =

(i~2 8 . . n 7) we d e f i n e

n:

With

this

notation,

the e x p r e s s i o n

!nside

the p a r e n t h e s e s

above

becomes ~n

s T (x) s T (Y)

n.'

Thus h

n

(xy)

Finally, Here

we

take

symmetric and

£

(y) =

we

=

c-T ~n

need

-7-r I ~ i<j

s

T

an e x p r e s s i o n

x = Xl+...+Xk,

functions

n:

T

due

(y)

originally

Y = Yl +'" "+ Yk

of k v a r i a b l e s .

(yi-Yj) .

(x)s

Define

to Cauchy.

and work A(x)

=

with

~ i<j

(xi-x j)

40 Lemma:

Proof:

1 ]-~ (l-xiYj) i,j

1 A (x) £(y)

det

1 l_xiY j

Consider the product

i,j Multiply the i

th

(l-xiYj)

1

I

row of the matrix i-l_xl--iYj ) bY the product~(l-xiYj),j

for each i = 1 ..... k .

The result is

(l-XlY j) j~l

det

• detl 1 |l-xiYj

~(l-x2Y j~l

j)

~ (l-XlY j) j~2

~(l-x2Y j~2

j)

Think of each entry ~--~j~j0(l-xiYj) as a polynomial fj0 (xi) in the variables x., with coefficients in the ring of polynomials 1 [Yl ..... Yk ]"

Observe f. (x) is of degree k-i in x. 30

Thus

the product is

det

and

fl (Xl)

f2 (Xl)

fl (x2)

f2 (x2)

fl(x) = Ax

powers of x)

k-i

+ (lower powers of x), f2 (x) = Bx

with A and B E ~ [Yl ......Yk ]"

k-2

+ (lower

Replacing the second

41

column b y -Bfl+Af 2 changes only b y a factor

of A, which does

reduces

the degree

Similar

column

x, without Thus

the value of the determinant

in x of the polynomials

operations

changing

the original

reduce

the degree

determinant

(k-1)+(k-2)+(k-3)+...+(1)+ hand,

the original

x. 's (i.e., l Hence

it is divisible

is independent by

£(y)

the 3d column

is of degree

by -[-[ (xi-x j) = £(x). i<j

in the x's,

the result

of the x's.

of the y's.

the constant,

consider

det)lq~j ~ (l-xiYq) )

~=i, ...,k.

=

of ~,

A simple

~

in

Etc.

On the other

function

Since

of the

for i~j).

h(x)

, the same product

is also of

divided

Hence ~

det)

1 l-xiY j

so equal

to a constant

the equation w r i t t e n

~. To

in the form

~(x). ~(y)

just evaluate both

computation

k-3

~(x)l ......7]- (l_xiYj)de t 1-xiYj

Similarly

of the x's and y's,

To find the value

in x.

column.

at most

is an alternating

1 ~ .~--~,(1-xiYj) (x) A (y) l, j

evaluate

to degree

¥fI~- 1} "~v'-~' in the x's. 2

(0) =

But it

in the second

sign if x. and x. are switched, l J

is independent

is independent

the x's.

of the determinant

determinant

it changes

degree k(k-1)2

not involve

involved

shows

1 sides setting x.= - - , l Yi

that ~=i.

42

We now write the above lemma,

substituting yt=Ylt+Y2t+..Ykt

for y = y l + Y 2 + . . . + y k , and calculate

ill kiiil detl xinyjntnl ~(l-xiYjt)

(x) A (y) t

1

1 k (k-l) t 2

(x) ~ (y)

2

n=0

,

o6S k

n •

= ~ 1 k(k_l) n=0 A(x) £ (y) t 2

x

~ t~

nI nl (1)Yl t / <

nI

sgn(~)(~ x \n I

n n> (k) Y k

sgn(~)x

nl n2 (1)x (2

nk)

nk n1 -x o (k)Yl "''Yk /

w h e r e the sum is over all sums nl+n2+..+nk=n of non-negative integers and all permutations

o6S k of the x's.

Note that since the terms in the determinant they are alternating, coefficient

so in any term where n =n , with i~j, l ]

the

is zero. Hence the sum can be taken over all ~6S k

and all sums nl~n2+...+nk:n, those expressions by adding

involve sgn(o),

ni~n j for i~j.

n=nl+n2+...+nk,

Hence we can sum

with the n.l in decreasing order,

in the extra summation over all permutations

The expression

over

of the y's.

in the parentheses becomes

sgn(~T - I ) x (1)Xc(2) " " X (k)YT (1)YT (2) " "YT (k) with the sum b e i n g taken over all sequences with nl+...+nk=n,

and all o, TCS k.

n l > n 2 > n3b..

>nk>0,

43 Finally,

define

kl~- k 2 >

kk=nk,

kk_l=nk_l-l,

.. m kk and k l + k 2 + . . . + k k = n 1

expression

., kl=nl+ (k-l) k (k-l) 2

then b e c o m e s

so that

The w h o l e

equal

to

- ~ (l-xiY j t)

n -

co

t n=O

k (k-l) 2

kk sgn(~)sgn(T)x

(i)

k l+k- i

• ''Xa(k)Y~(1)'"

kkh "y

(x) a (y) • .+X k = n

-

2

o, T6S k

W e define,

for a p a r t i t i o n

k =(kl, k 2 ..... k k ] the Schur

function

by kl+k-i ~6S k {X]

sgn (o) x

~(i)

k2+k-2 x

kk

(2)

"

x

g(k)

=

g(x)

Comparing

coefficients

of m = n -

k (k-l) 2

in the identity

above,

we get

h

(xy)

=

~

m

Notice

{k] (x) {k] (y)

I ~m that

{i] is g i v e n

variables

Xl, X 2 .... x k.

that b o t h

g(x)

functions

of xl,x2,..,Xk,

permutation

and

as a s y m m e t r i c

function

(The symmetry follows kl+k-i kk ~sgn(~)x ... x o(i) o(k)

of the x

l

's.

and so their Hence

ratio

{k] can be

from

of k

the fact

are a l t e r n a t i n g

is u n c h a n g e d

expressed

as a

b y any

Ill

44

polynomial

in the e l e m e n t a r y

{k] (x) = F l ( x , k 2 x ..... kkx)

coefficients.

Now,

integer k chosen Fl, k.

But

functions

a priori,

in C h a p t e r

F

k

a corollary III t h a t

(The r e a d e r

H e n c e F k is i n d e p e n d e n t any p a r t i c u l a r

of the x.'s: l

for s o m e p o l y n o m i a l

(k = the n u m b e r

it w i l l b e

F k , k = F k , k + I.

functions

2 k x : XlX2+XlX3+X2X3+...+Xk_iXk ,

x = X l + X 2 + . . + x k, Thus

symmetric

k ~ n,

F with

integer

c o u l d d e p e n d on the o r i g i n a l of x.! 's), of the

if k~n,

so w e s h o u l d w r i t e

interpretation

and k ~ n ,

of t h e s e

then

is i n v i t e d to p r o v e t h i s d i r e c t l y . )

of k,

even t h o u g h to c a l c u l a t e

a particular

it for

number k~n must be chosen

and fixed. Exercise: Schur been

{1 n} = a

functions,

n

, In] = h

.

also c a l l e d S - f u n c t i o n s

d e f i n e d and s t u d i e d

in a n u m b e r

of r e p r e s e n t a t i o n

theory

other

see L i t t l e w o o d

approaches,

n

as we

of ways,

in the l i t e r a t u r e , not

shall do in C h a p t e r (£29~), R e a d

just in t e r m s III.

(~4J),

For

and S t a n l e y

have

45

The S c h u r

functions

of m o n o m i a l

symmetric

calculation

shows:

{1}

=



{2]

=

<2>

[ii]

=

{3}

=

{21]

=

{lll]

=

{4]

:

{31]

=

{22 ]

=

{211} [iiii}

of course,

functions.

+

be expressed

Straightforward

in terms

but

tedious

<3>

+

<21)

+



<21>

+

2 lli> ill)

<4>

+

<31

+

<22

+

<31

+

<22

+ 2<211>

+ 3

<22>

+

(211>

+ 2

<211>

+ 3
=

<211>

+





=

In C h a p t e r

Three,

functions,

related

groups.

can,

we w i l l

glve a m o r e

natural

to the r e p r e s e n t a t i o n

definition

theory

of these

of the s y m m e t r i c

46

4.

Adams

Operations

Let A b e the of A.

Regarding

free k - r i n g on one g e n e r a t o r

A as the ring of s y m m e t r i c p o l y n o m i a l s

i n f i n i t e n u m b e r of v a r i a b l e s by giving

its d e s c r i p t i o n

A = ~[ I, a I, a 2 . . . . f u n c t i o n of the the a :i

with

7, w h e r e

['s,

(One e x c e p t i o n is p o s t p o n e d

These named)

)n n

- the o p e r a t i o n s

until Chapter

operations

in the c o n t e x t

n

power

(V) = S y m m n v

symmetric

as a p o l y n o m i a l

in

a natural

the o p e r a t i o n s

associated

m o s t n o t a b l y the p o w e r

rise to the A d a m s

associated with

operations.

the S c h u r

functions

easily be described

(and h e n c e

of the e x a m p l e of the k - r i n g R c o n s t r u c t e d vector

symmetric

to the e x t e r i o r p o w e r s : homogeneous

since

III.)

can m o s t

from f i n i t e d i m e n s i o n a l The e l e m e n t a r y

say,

~ ( r ) = F ( r , k2r, k3r,...)

functions,

which give

in an

Thus,

a l - r i n g R, ~ b e c o m e s

investigate

s e v e r a l t y p e s of s y m m e t r i c s

function.

expressible

for e a c h r£R,

s e c t i o n we w i l l

sum f u n c t i o n s

h

Given

an e l e m e n t

~ might be described,

a. is the i th e l e m e n t a r y 1

~ is u n i q u e l y

on R b y taking,

In this

~I,~2 .....

as a s y m m e t r i c

~ = F ( a 1,a 2, ...).

operation

and ~ b e

sum h

for V6R.

spaces

gives Just

a fixed

functions give

for V6R, n

over

we w r o t e

the n - f o l d

a

rise,

n

field F

as p e r design,

(V)= Anv.

symmetric

(p. 5)

The

power

as w i t h e x t e r i o r p o w e r s

there

is

47 an additivity formula n hn(Vl+V2 ) = ~ hi(Vl)hn_i(V 2) i=0

To see this, write ht(V) = ~ h gives

h

n

an identity

n

ht(V)k_t(V)=l

(V)t n. ,

Then the definition of h

so t h e

additivity

formula

n

for

follows from that for k n.

The monomial denoted

{~>.

<~>

symmetric

functions

<~> provid~ operations,

In this case there is an addition formula

(VI+V 2) =

To prove this,

~. <~i > (V 11 <~2 > (V 2) ~i~2 =~

it is enough,

using the Verification

assume V 1 and V 2 are sums of i st degree elements.

Principle,

powers

of <~> and

just as in the additivity formula for the exterior

(p.6).

Details

are left to the reader.

(Or see

[323, p.91,

Theorem 33). The main subject of this section is the set of operations associated with the power sums s . n Operation,

to

The argument

can then be carried out with the explicit definition the < ~ > ' s , l

also

Y? b y yn(v) = Sn(V),

We define the n

th

for V in a k-ring R.

Adams

48

Proposition :

Let

a,b be

elements

in a k-ring

R, and n,m i n t e g e r s

>~ 1

Then I)

yl(a)

= a

2)

~n (i) = 1

3)

~n(a+b)

4)

'~n(ab) = %+n(a)%+n(b)

5)

~n(km(a))

= km(+~n(a))

6)

Yn(ym(a))

= ynm(a)

= ~n(a)

Thus e a c h yn is a k-ring is a ring h o m o m o r p h i s m Proof:

+ ~n(b)

endomorphism

are each sums of e l e m e n t s of the p o w e r

The A d a m s elements

+~n

of d e g r e e

sums

operations

Principle,

s

n

we can

i.

(p.35)

also serve

assume

Then u s i n g all t h e s e

that a and b

the o r i g i n a l

m

are clear.

to d i s t i n g u i s h

binomial

of k-rings:

Proposition: ~n(a)=a

of R, and the map n ~ >

~-~-~ End R.

By the V e r i f i c a t i o n

definition

= Ym(~'n(a))

Let R be a k-ring

and aER.

Then

a is b i n o m i a l

iff

for all n > l . ~>

Proof: kt(a)

Use the i d e n t i t y =

d___log(kt(a)) dt

(l+t) a iff the r i g h t h a n d

Corollary:

For all m E2Z, ~'n(m)=m.

side

= ~

( - l ) n ~ n + l ( a ) t n.

Then

rl--o

is ~ ( - l ) n a t

n -

a l+t

I

49 Hence, sub-k-ring will

given

R 1 of R b y R 1 =

at least

The

any k-ring

include

Ix I ~n(x)=x,

the unit

first p r o p o s i t i o n

in v e r i f y i n g

that v a r i o u s

we n e e d

a definition:

element

r£R,

R, we can p i c k

pre-k-rings

include

This

binomial subring

a copy of ~.

a converse,

which

are in fact

a ring R is t o r s i o n - f r e e

and any integer

a maximal

all n ~ i}.

i, so will

above has

out

will be useful

k-rings.

First

if for any n O n z e r o

n ~ i, nr = r+...+r

(n summands)

is

Let o p e r a t i o n s

yn

also nonzero.

Theorem:

Let R be

a torsion-free

pre-k-ring. oo

be d e f i n e d by,

for a£R,

(So in particular, Suppose

yn(1)=l,

all a,b6R

pre-~-rinq

we h a v e

n,m.

the theorem,

~n:R~>R,

we m a k e

a general

for

definition:

unit,

n ~ i, s a t i s f y i n g

for all a,b£R,

is a p r e - ~ - r i n g

satisfying

Yn(ym(a))=ynm(a)

and Yn(ym(a))=ynm(a)

ring R w i t h

yn(a+b)=yn(a)+yn(b), also

all a,b,n.)

Then R is a k-ring.

R is a c o m m u t a t i v e

set of o p e r a t i o n s

n

'~'l(a)=a, "~n(a+b)=yn(a)+yn(b),

yn(ab)=yn(a)yn(b),

and integers

TO p r o v e

d ~ =~--~-l°gkt(a) = ~ ( - l ) n ~ n + l ( a ) t n=0

for all a,b6R

yn(1)=l,

together

yl(a)=a,

and integers

A with

and

n. A

~n(ab)=yn(a)~n(b),

and integers

n,mZ

a

i.

and

50

Given

any c o m m u t a t i v e

ring

R w i t h unit,

countable

sequences{(rl,r2,r3,..)

structure

by defining

For each yn(

integer

R

w

R w is g i v e n

I r i £ R ].

and m u l t i p l i c a t i o n

n ~ i, we d e f i n e

(rl,r2 .... )) =

Proposition:

addition

an o p e r a t i o n

a ring

coordinatewise.

yn:RW~R

w, b y

(rn,r2n,r3 n .... ).

is a Y-ring. if the m a p

If R is a p r e - Y - r i n g ,

Y-ring

if and only

Y(r)

(yl(r),y2(r) .... ), is a h o m o m o r p h i s m

=

let R w be the set of

then R is a

Y:R---~ R w, d e f i n e d b y of '~-rings.

D Proof:

Clear.

Note

that

since

1

is the i d e n t i t y map,

the m a p

Y:R ~

R ~ is

one-one.

Let R be

a torslon-free

kn : R - - ~ R, n>~O, so t h a t E.g.,

following

p.

pre-~-ring.

Sul]pose t h e r e

are o p e r a t i o n s

co

~d

logkt(x)

= ~ - ( - 1 ) n y n + l ( x ) t n, a l l x£R. n=l

, we can c a l c u l a t e

l(x) 1

,

I~ y2

n. k n(x)

= de

yl

(x)

(x)

yl(xI

n(x)

T h e n we s u p p o s e

that

b y the d e t e r m i n a n t so t h a t kn(x) that this

for each n > i, and each x 6 R the e l e m e n t

on the right h a n d

is d e f i n e d .

division

side

is d i v i s i b l e

(The t o r s i o n - f r e e

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

provision

if defined.)

Note

b y n:

defined in R,

guarentees

that

if

51

R contains possible

a field of c h a r a c t e r i s t i c

so the kn's

As usual,

automatically

the p o w e r

series

division

is always

exist.

k t =~knt n gives

R----~I+R[[t]] + and we can d e f i n e diagram

zero,

a map

L making

a map the

following

commute: 1+R[[t~ 3+ R //kt/~

L is

defined

by:

d -~log(t+alt+a2t

B(l+alt+a2t2+...) 2

Proposition:

~L

+...)

i)

=

(-1)nrn+l t

L(x"+"y)

2) L("I")

= (rl,r

= L(x)

2 ....

) if

n

+ L(y)

= 1

3) L(x .....y) = L(x)L(y) 4) L("Yn"(x))

= ynL(x)

5) If R is t o r s i o n - f r e e , 6)

If R c o n t a i n s onto,

Proof:

(Recall

operations there.)

a field of c h a r a c t e r i s t i c

and h e n c e

an i s o m o r p h i s m

first that the

"..."

in l + R [ [ t ~ + - thus

I) and 2) are easy.

easy observation

i),

notation

so the i n j e c t i v i t y

that the k e r n e l

refers

L is

to the k-ring

to the A d a m s

T h e n to c h e c k i, g i v i n g

zero,

of Y-rings.

,,yn,, refers

to take x and y to b e of d e g r e e F o r 5), we can use

L is one-one.

3),4)

again

operations

it is s u f f i c i e n t

an e a s y v e r i f i c a t i o n .

of L f o l l o w s

of L is trivial.

from the

For 6) the

52

inverse map L -I: RW--~ l + R [ [ t ~ + is easily c ~ I c u l a t e d n

-i

as

D

((bl,b 2 .... )) = exp(-g(t)) , where g ( t ) = ~ (-l)nbn tn" n+l

Corollary: zero

If R is an a~gebra over a field k of c h a r a c t e r i s t i c

(e.g., k = Q

Proof:

), then so is l + R [ [ t ~ +.

R w is c e r t a i n l y

isomorphic

a k-algebra,

and under the hypothesis,

is

to l + R [ [ t ~ +.

The Proof of the original R has k-operations, R is a k-ring

theorem is n o w accomplished.

and hence Y-operations,

iff R is a Y-ring.

A useful

If

and is torsion-free, restatement

of the

theorem is the following proposition.

Proposition:

Let R b e a t o r s i o n - f r e e

Suppose there

is a ring homomorphism.

There

and S b e any ring.

is given a map of sets ~:S--~> l + R [ ~ t ~ +. Then

is a ring h o m o m o r p h i s m

k-structure

ring,

iff the c o m p o s i t e map S --~ I + R [ [ t ~ + - ~ R If S is a pre-k-ring,

~ preserves

the

iff the c o m p o s i t e map L~ does.

is an interesting

a p p l i c a t i o n of this p r o p o s i t i o n

algebraic g e o m e t r y of varieties

over finite

fields

to the

(which the

n o n - g e o m e t e e r m a y ignore since it will not b e relevant to the sequel.)

w

Let k be a finite

field and S the G r o t h e n d i e c k

ring of

53

varieties

d e f i n e d o v e r k, w h e r e

of v a r i e t i e s and ~ : S ~ >

and p r o d u c t

~x(t).

S --~ X j

on X w i t h c o o r d i n a t e s

on e a c h v a r i e t y ,

proposition

implies

that

One m i g h t h o p e t h a t seems not

so.

the n - f o l d

for d e f i n i t i o n s . )

L ( ~ x ( t ) ) n is the n u m b e r of field k'

for p r o d u c t s ,

~ is a ring h o m o m o r p h i s m

the r e s t a t e d

R ---~ l + ~ [ t ] ] +.

{ is also a m a p of k - r i n g s .

symmetric power

, where

is the sum of the r a t i o n a l

and s i m i l a r l y

The o b v i o u s

of k

S i n c e the n u m b e r of r a t i o n a l p o i n t s

on a d i s j o i n t u n i o n of two v a r i e t i e s points

~44J

in the e x t e n s i o n

o v e r k is n.

Let R = Z,

to e a c h v a r i e t y X its

(See S w i n n e r t o n - D y e r

T h e n in the c o m p o s i t e m a p

the d e g r e e of k'

f r o m the d i s j o i n t u n i o n

f r o m the c a r t e s i a n p r o d u c t .

l + ~ [ [ t ] ] + b e the a s s i g n m e n t

zeta-function

points

sum c o m e s

k-structure

Alas,

it

to p u t on R w o u l d be to take

of a v a r i e t y X to b e h

(X), and d e f i n e n

the o t h e r o p e r a t i o n s 1-space A

1

is of d e g r e e

in a k-ring,

degree

in t h i s d e f i n i t i o n , We

accordingly.

s h o u l d remark,

i, b u t

U n d e r this d e f i n i t i o n

1 1 2 its s q u a r e A X A = A is not,

is m u l t i p l i c a t i v e .

powers,

seems n e a r l y

as p l a u s i b l e .

t h a t t h i s c a t e g o r y of v a r i e t i e s

is a g o o d e x a m p l e of a c a t e g o r y w h i c h has symmetric

but whose Grothendieck

and

H e n c e R is not a k - r i n g

and no o t h e r d e f i ~ i t i o n

however,

_affine

sums, ring

products,

is not

over k and

a k-ring.

54

W e n o w u s e the t h e o r e m to c o n s t r u c t k-rings. K(S)

Let S b e

c l a s s of

a set, K a field of c h a r a c t e r i s t i c

the set of all m a p s

two m a p s

a general

is d e f i n e d

as usual,

identity.

Suppose there

n = l , 2 .....

satisfying i)

from S to K.

o

The

m a k i n g K(S)

sum and p r o d u c t a commutative

is g i v e n on S a set of m a p s

is the

zero,

and of

ring with

~ :S---~S, n

identity map

i ii)

o ~ = n m

We t h e n d e f i n e o p e r a t i o n s 9n(f(s))

= f(On(S)).

possible

in K(S)

nm 9n:K(S)~)

K(S)

As in the theorem,

for e v e r y

by,

for f : S ~ )

K,

and sES,

s i n c e d i v i s i o n b y n:

is

i n t e g e r n, we can d e f i n e o p e r a t i o n s

in by

kn

_

1 n:

det

91

1

0

,2

~I

2

~n and w e can c o n c l u d e to the o p e r a t i o n s ~-ring,

k

t h a t the n

th

~n are the A d a m s o p e r a t i o n s c o r r e s p o n d i n g since K(S)

is c l e a r l y

a

it is a l s o a k-ring.

one is g i v e n

n

0

41

By the theorem,

This construction

of G.

/

a

n

power

applies

a g r o u p G,

is the m a p in G.

on G w i t h v a l u e s

in p a r t i c u l a r

and S is the set of c o n j u g a c y c l a s s e s

i n d u c e d on S b y

K(S) in ]~,

to the c a s e w h e n

the o p e r a t i o n of t a k i n g the

is t h e n c a l l e d the ~__~rin~ of c e n t r a l

functions

55

The context Let

L l + R [ [ t ] ] +~ P R

map

- that

R be

of u n i v e r s a l

a commutative

the moment)

and

of elements

of R.

different

ring

let W R b e

ring

Consider

classically

As

the map

rings

with

unit

the

a set,

structure

Witt

set

of

WR = R

w

appears

- which

we now

(assumed all

another

describe.

torsion-free

w-tuples

, but

in

we will

for

(Wl,W2,~.-) put

a rather

on W R.

M:WR----~ R ~ d e f i n e d

by n/d

M ( ( w ! , w 2 .... )) = Thus

(rl,r 2 .... )

where

r

= n

~dw dh

d

can

identify

rI = w 1 2 r2 = w I

+

2w 2

3 r3 = wI =

r4

+

3w 3

4

wI

+

2w22

+ 4w 4

etc. If R is t o r s i o n - f r e e ,

M

is

a one-one

map

and we

W R

with

its

image

Proposition: there

M(WR)

WR(M)

c R

.

is c l o s e d

are p o l y n o m i a l s

F.,

under

G. w i t h

sum

and product

integer

coefficients

l

F n depends

[wl I i d i v i d e s

on t w o

n])

sets

of variables:

Indeed

such

j

M ( ( W l , w 2 .... ) ) + M ( ( w l , w 2 .... ) )= M ( ( F I ( W l , W l (here

in R e.

that

e

)'F2(wI'w2'wI'w~)

[wi I i d i v i d e s

n]

.... ))

and

and

M ( (Wl,W 2, .. )) -M ( (wl,w ~, .. ))=M((G 1 (Wl,W i) ,G 2 (Wl, w2,w{, w ~, .. )) (where

similarly

Hence cation

using

in WR,

G

n

is

these

a function polynomials

W R becomes

of

the

two

to d e f i n e

a commutative

ring

sets

of v a r i a b l e s ) .

addition with

and multipli-

identity

- the

56 (Universal) ring of Witt vectors of R.

For the proof of the proposition,

we must construct the

polynomials F., G., which can be accomplished by just proving l ] the special case for the ring R = Z. polynomials,

For, once we have the

they define operations in the set W R, for any ring R,

(torsion-free or not).

The ring axioms, associativity ~£c., will

follow for an arbitrary ring because they are true over Z, hence are polynomial identities valid in all rings.

Actually,

to carry out the proof, we will just assume that

R is torsion-free

(an obvious property of Z).

WRM---~R w is injective.

Then the map

Also, the map l+R[[t]] + is injective.

The proposition follows by observing that these two inclusions give the same subset of RW:

Indeed, we can define f : W R ~ > l + R [ [ t ] ] +

by, for w = (Wl,W 2, ...)£W R, f(w)

Then L~f(w) -

=~(i d

d log ( ~ ( i dt d

1 d\ l-Wd(-t)d ]

= ~{d

= ~

Wd(-t)d-i

1

-

Wd (- t )

C¸-d

= ~ d

( - l ) ~ d ~ d wdn/d(-t)n I

(-l)

n

- Wd(-t) d)

n-i r

tn-i n

d)

)

Wd(-t) d t

)(i + Wd(-t) d + w d 2 j

= ~n(-l)n+/d~hd wdn/d I t n-I

(-t)

2d

+.

57

Hence,

b y the d e f i n i t i o n

M is one-one, via

so is f, so

f as a subset

it is n e c e s s a r y which

of the map L, L f(w) (for R t o r s i o n - f r e e )

of l+R[[t]] +.

once we w r i t e

(rl,r2,...).

Since

W R can be c o n s i d e r e d

To see that W R is all of l+R[[t]] +,

o n l y to s h o w that

is o b v i o u s

=

f is an i n v e r t i b l e

out the d e f i n i t i o n

map,

a fact

of f in m o r e

detail: Given w =

(Wl,W2,W3, ...)

f(w) = ~ ( i d =

£ W R,

- Wd(-t) d)

(l+Wlt) (l-w2 t2) (l+w3t3) (l-w4 t4) ...

= 1

+

+

(wl)t

(-w2)t 2

(w5-w4wl-w2w3)t5

The c o e f f i c i e n t sum is over

+

of t n is

w ..w nI n2 nk

(-w4+wlw3)t 4

where

the

of n into k d i s t i n c t

if l + a l t + a 2 t 2 + . . .

solve

+

( - w 6 + w 5 w l + w 4 w 2 - w 3 w 2 w l)t 6 +

n l > n 2 > .. > n k

Hence

the image of f, we m u s t

(w3-wlw2)t3

( - 1 ) n ~( - l ) k w < -

all p a r t i t i o n s

and all i n t e g e r s k.

+

+

El+R[ [t~ + is to be

the e q u a t i o n s

aI = w I a 2 = -w 2 a 3 = w3-wlw 2 etc. Since

in the n

coefficient w's

th

equation

w

n

occurs

+ l, the e q u a t i o n s

in terms of the a's.

in o n l y one term w i t h the

can b e

solved

inductively

for the

parts, in

58 The

first

few terms

are

w I = a1 w 2 = -a 2 w 3 = a 3 + ala 2 2 w 4 = -a 4 + a3a I + a 2 a I 2 w 5 = a 5 - a4a I - a3a 2 - ala 2

2 3 + aI a3 + aI a2

etc. H e n c e W R is i s o m o r p h i c

as a ring to l+R[[t]] +,

(It s h o u l d be m e n t i o n e d have

set up the i s o m o r p h i s m

(Wl,W 2 .... ) - ~ 2 ~ i d to take the to s a t i s f y

- wdtd ) .

(l+at) (l+bt)=(l+abt) Indeed our~-~ d

cited

degree - so

above.

( g~i

~4~, I~7

+ as

forces

them

of first d e g r e e Our c h o i c e

elements

authors

elements

in l+R[[ t ]I +

of f is g o v e r n e d b y

should m u l t i p l y

is to t h e i r ~--~ (l-wdtd) d n is to k_t(x)

properties

to the p - v e c t o r s

several

(l-at) (l-bt)=(l+(-a) (-b)t)=(l+abt).

(l-Wd(-t)d)

kt(x)=~kn(x)t

Additional relation

This

(1-at) (l-bt)=(1-abt). first

that

WR~)I+R[[t]]

"multiplication"

our rule that

expression

here

for any ring R.

=

( ~ h n ( x ) t n ) -I.

of this c o n s t r u c t i o n of W i t t can b e

as the

found

)

of W R and its in the r e f e r e n c e s

CHAPTER

The

II

following

:

notation

All

groups

are

All

vector

spaces

denotes If V

the

and W

linear

maps

S

Denote

these

o(1)=2,

relations

cons~[sts this

and

this

chapter:

numbers,

act

is t h e

are

over

= n,

picking ),

the

z*

is t h e

~ unless

complex

is

its

vector

space

a basis

the

group

numbers.

conjugate.

of Hom(V,V)

o(j)=j,

"column

vectors"

group

j>2.

7(n)=l. n

= i,

elements

Let

Then

S

and

1,7,T

2

otherwise

of

consisting

of V gives

of

invertible

specified.

left multiplication

1 , 2 , 3 ..... n.

~

z6C,

subset

~ Gl(n,~

on V by

symmetric

2 ~ =i,

V

AutV

for

Hom(V,W)

If d i m V

, are

of V

and

over

.

~

objects

of the

order.

~

to

the

o(2)=1,

i = i ; 2 .... n - 1 the

over

elements

OF G R O U P S

throughout

finite-dimensional

of groups

products,

denotes

n

are

complex

Aut V

THEORY

finite

elements.

matrices

so t h e

used

Aut

isomorphism

take

be

f r o m V t o W.

an

We

will

spaces,

invertible

Tensor

REPRESENTATION

are vector

of

n~fn

THE

n

rather

than

of permutations Let

oES

T be

the

n

be

oT = T

n-i

,o,o~,o7

2

~.

the

n-cycle

is g e n e r a t e d

by

(f(g(v))=(fg)

"row vectors".

of n objects. transposition T(i)=i+l o,

we

,

T subject

In p a r t i c u l a r ,

which

(v)

usually

to

S3 take

in

)

60

i.

The R e p r e s e n t a t i o n

A reprgsentation

~ :G ~ >

Rin 9 of a F i n i t e

Gzoup

of a g r o u p G, of d e g r e e ~,

is a h o m o m o r p h i s m

GI(n,C) .

For example,

if G = S 3, we can a s s i g n

-1

f:

9 :

-1

p Thus

~

for x , y ~ S 3, the m a t r i x p r o d u c t

of

~ (x) and

~(y)

is equal

to

(xy).

While take

this

is a nice c o n c r e t e

a coordinate-free

approach

Let V b e a v e c t o r

space,

in V is a h o m o m o r p h i s m is the d i m e n s i o n an i s o m o r p h i s m the c o n c r e t e

While the p a i r

of V.

of

and G a group.

f:G ~Aut (Of course,

to

V.

A represent gtion of G

The d e ~ r e e

picking

of the r e p r e s e n t a t i o n

a basis

of V gives

so this does g e n e r a l i z e

above.)

speaking,

(v, ~ ), we w i l l o f t e n

is c l e a r

we p r e f e r

to the problem.

A u t V w i t h Gl(n,~),

definition

strictly

definition,

from the context.

the r e p r e s e n t s t i o n speak of

of G is g i v e n b y

"the r e p r e s e n t a t i o n

In a s i m i l a r

way,

given

V"

if

p : G _ - ~ k-u-t j

V,

61

and g ~ G, we should and write g both

refer

to the a s s o c i a t e d

~(g) (v) for v 6V.

for the element

In other

common

of G and

of G

defined

Qr a c o m p l e x

the m o r e g e n e r a l arbitrary is said

Given G if g w ~ W

notion

of invariant

Zf gv=v

elements

vector

spaces

be

The set of all representation

for all in V.

is then exactly

a linear representation,

(to d i s t i n g u i s h

~GI(n,R), a

a subspace

where

(left)

it from

R is an

G-module

and

W of V is invariant

An element

g EG.

v ~ V is invariant

We w r i t e

V G is a s u b s p a c e

linear

maps

of G b y defining,

from V 1 for ~

is a linear

map of f(gv)=gf(v).

V 1 to V 2.

to V 2 , H o m ( V I , V 2) gives :V 1 -~) V 2 and g ~ G,

9Y(~j~)(~iy;jfor

Hom(Vl,V2)~ ~

, or

is invariant.

for any v e V # and g ~ G , of G - ~ o d u l e s

under

V G for the set

o~ V and

f: (VI, ~ ) ....) (V2, ~ )

the set of maps

V 1 --~ V 2 is the map

gv for F ( g ) (v).

of G on V.

f : V l - - ~ > V 2 satisfying,

Let HomG(VI,V2)

g~:

F:G

and w ~W.

A map of G-modules

called

to write

it from a p e r m u t a t i o n

V is called

of G on V,

for all g ~ G

and to write

representation

ring).

on V as /(g)

simpler

(V, p ) is also

of a map

an action

an action

a fixed point,

for F ( g ) ,

(to d i s t i n g u i s h

commutative

to give

it is u s u a l l y

terminology,

representation below)

But

operation

all v ~ V I.

H o m G ( V I , V 2)

a

62 A map of G-modules

f (VI, ~ ) --~ ( V 2 , ~ )

there is a map of G-modules composites (Vl~t)

f': ( V 2 , ~ )

if

(VI, ~j) so that the

ff' and f'f are the identity maps on V 2 and V I.

and

(V2j ~

) are isomorphic

there exists such an isomorphism. (V2, ~ ) ,

-~

is an isomorphism

(also called equivalent)

Note that, given

V 1 and V 2 can be isomorphic

if

(V I, ~,) and

(indeed identical)

as

vector spaces without being isom~Dphic as G-modules.

W e are interested really only in isomorphism classes of representations. of basis

Given a representation F

:G -m Aut V, each choice

for V gives a matrix representation of G, G --~ GI(n,C),

and all these are isomorphic. more natural than others,

Proposition:

Let

But some choices of basis of V are

as the following proposition shows.

~ : G ---~ Aut V be a representation of G.

there is an inner product on V, call it each g 6 G ,

and all Vl,V 2 in V, ¢ i , v 2 >

,

Then

such that for

= ~gvl,gv2>

.

Hence

every representation of G is isomorphic to a representation b y unitary matrices.

Proof:

Pick any basis e I .... e

n

of V and let

(-,-) be the

usual inner product with respect to this basis: (~a.e., z

z

~b.e.) 3

conjugation.

]

3

= ~a.b.*, ~

i

Now define,

the asterisk denoting complex

z

for Vl,V 2

in V,

63

_

This

1 S ( g v l , gV2 ) IGi g~G

is c l e a r l y b i l i n e a r ,

the u s u a l

inner p r o d u c t

VlhV2>

skew-symmetric is.

1

=

~

J GI

hg,

the m i d d l e

g ~G

lhgVl,hgv 2)

~

<~-g6 G

equality

the G r a m - S c h m i d

of V for the

(gvl,gV2)

holds

since

process, //

inner p r o d u c t

all the m a t r i c e s

assigned

Corollary

P:G ~

Let

b e an e l e m e n t

=

<-Vl,V 2 >

the set of all e l e m e n t s

of G.

<.-,->.

we can find an o r t h o n o r m a l With

respect

basis

to this basis,

to G are unitary.

Aut V b e a r e p r e s e n t a t i o n

Then there

is a b a s i s

of G, and

of V in w h i c h

let g

the m a t r i x

p (g) is d i a g o n a l .

Proof: way

for h £=G,

is the same as the set of all g, g (;G.

Using

of

since

g~ G

IG; - where

Also,

and n o n d e g e n e r a t e ,

Every u n i t a r y

to see this

for some n~/l, must have

matrix

is to o b s e r v e and h a v i n g

a diagonal

that every m a t r i x

entries

matrix

is d i a g o n a l i z a b l e . A

QED.

Another

satisfying

in a field of c h a r a c t e r i s t i c

for its J o r d a n

form.

An=I, zero,

64

We will matrices

see later

that we can s i m u l t a n e o u s l y

of G if and o n l y

Representations

arise

unique map

Less

g C G.

;0 : G ~ >

in m a n y ways.

For n=0,

The t r i v i a l

the

vector

A I , . . . , A n. and

~iAi

map with

by

n-dimensional

representation

the m a p S ~ n

n= i Gi

G---->S

Gl(n,C)

G ~>GI(n,C).

left m u l t i p l i c a t i o n ,

degree

, called

Generalizing

of the n e l e m e n t s

representation.

n with basis the b a s i s

of a g r o u p n

.

Given

given

elements

Let V

labeled

elements:

given

~6

Sn

a linear

the r e g u l a r

this process,

we can c o m p o s e

the g r o u p G acts

representation

representation

w e can start w i t h

G, and any g r o u p h o m o m o r p h i s m

maps

a linear r e p r e s e n t a t i o n

of H.

is b y this

to get a linear

In p a r t i c u l a r ,

giving

G, o_~f d e g r e e ~,

such,

above

of a g r o u p gives

the

e( ~'~iAi ) = ~iA~(i).

a homomorphism

representation

unit m a t r i x

is the zero r e p r e s e n t a t i o n :

Sn acts on V b y p e r m u t i n g

A permutation definition

of the n y n

of all p e r m u t a t i o n s

n

space of d i m e n s i o n

~ V,

representation

dimV=0.

the g r o u p S

A I , A 2 ..... A n has a c a n o n i c a l be

this

A u t V, w h e r e

trivially,

all the

if G is abelian.

o__ffd e g r e e ~ of a g r o u p G is the a s s i g n m e n t to each e l e m e n t

diagonalize

H ----2G.

on itself

of G of

of G.

any

linear

representatioz

Then composition

In p a r t i c u l a r ,

if H is a

of

65

subgroup

of G, each representation H of H, Res G ~

a representation

Conversely, cosets

if H is a subgroup

representation

representation generalized, trivial

of G.

V.

of G, G acts on the set of left

left multiplication,

of G, so b y the above,

Using a ~ e r m i n o l o g y

one-dimensional

~:G---~ A u t V

let H be the kernel

G/H~)

Aut V.

normal

subgroup

representation

of ~'

giving a

a linear

which will

we say that this representation

A representation not,

:H - - ~ A u t

of H in G, ~ g H ~ g 6 GI b y

permutation

~ :G---) Aut V gives by restriction

later be

of G is induced b y the

of H.

is faithful Then ~

gives

if

is one-one.

a faithful

representation

In this case, ~ will be said to be associated

G, the subgroup

H.

In particular,

if ~

K e r ( ~ ) is contained

with the

is induced b y a subgroup

in H and equals

If

H of

H if and only

if

H is normal.

Thus

already

for the group S 3, we have

(where we just indicate ~(~) a consequence

l)

and

~(t)

the following

since

representations

the other matrices

of these):

The trivial

representation

of degree

n:

the n • n

unit m a t r i x

are

66

2)

The canonical

representation:

p (~)= !

3)

o

0

1

r(~)=

The regular

representation:

the o r d e r

9, ~ ' .

0

0

1

0

0

0 i

0

0

0

0

1

0

0

0

0

0

1 0

0 1

0 0

0 0

0 0

The r e p r e s e n t a t i o n

I

o

o

0

1

0

,A2:

Let the e l e m e n t s

induced by

Cf (H) = H

1 0 0 0

Let V b e the v e c t o r

g

=

This

in

~H

0

0 1 0 0

The

=~T-~H : ~ / :]f~I, a n d

=

~

H

H

~ H

"3CH =

=

io0) 1 0

0 0 0 1

space with basis

1

0 1

0 0 0 0

the s u b g r o u p H : I I,@~of S 3.

T (v ~H) =

Then

taken

OOOOoo

=

x(H)

I°l

, then

of S 3 b e

o

~(~)

c o s e t s o f H in S 3 are H = S H - - ~( ~

H =

,A3=

~-~ ~'~ ~ ~, t h e n

~0 a (~,)' =

4)

o

Let A I =

and

~

is isomorphi:c to the c a n o n i c a l

=


(00j 1

0

0 1

0 0

representation but

this

is

67

not so easy to see. trivial

5)

Later,

the theory of characters will make it

to check such an isomorphism.

The r e p r e s e n t a t i o n

induced b y the s u b g r o u p K =

This r e p r e s e n t a t i o n K =[0) b e a basis,

is two-dimensional.

~ i!~

y ~~

of S 3.

Letting K =(l)and

we h a v e

f

6)

Another

representation,

the a l t e r n a t i n g =(-i)

and ~ =

permuta£ion

not given b y the above construction,

representation

.

It is of degree one,

(i), assigning

(~I) d e p e n d i n g

and assigns

on w h e t h e r

the

is even or odd.

Let G be a given group, representations

of G.

and consider

the class of all

There are a number of operations w h i c h can

be p e r f o r m e d on this class,

and our object

is to c o n s t r u c t out

of this class a ring R(G) w i t h all of these operations built the arithmetic

into

of R(G).

Let V and W be two r e p r e s e n t a t i o n s two r e p r e s e n t a t i o n s V~W

is

The sum

V~W

is c o n s t r u c t e d b y taking the vector space

and l~tting G act by,

In terms of matrices,

of G.

this

for g 6 G, is quite

of the sum

(v,w)(~: VG,~W, g(v,w)=(gv, gw). simple.

Given bases

v 1 ..... v n

68

of V and w I ..... w n of W, basis with

of V(gW. respect

G i v e n g 6 G,

to the b a s i s

Aut W w i t h r e s p e c t respect

the

set

if A is the n ~ n

v I ..... v n,

is ( O

s p a c e V ~ } W on w h i c h

v I ..... Vn

is a b a s i s

set of all operation

The

B0) w h e r e

s y m b o l s v i < ~ w j is a b a s i s on m a t r i c e s

i

of V and W l .... ,wm a b a s i s

th

exterior

is c a l l e d

power

of V ~ W .

the n ; < m

is t h e n the o n e - d i m e n s i o n a l

Given

G-modules

to be c o n s i d e r e d one-dimensional

t h e n the

i th

space V a n d l e t t i n g g 6-G act b y

representation.

~nv

If

~iv is o b t a i n e d b y t a k i n g the

one-dimensional

If n > d i m V,

of W,

is the

the K r o n e c k o r p r o d u c t .

be the t r i v i a l

the d e t e r m i n a n t

V and W

The a s s o c i a t e d

Note

V.

o den°tes

representations

g(vl, % ... ,\vi)=gVl,% .... '\gv i.

g 6G,

in

of g in Aut V ~ } W w i t h

g @ G acts b y g ( v d ~ w ) = g v ~ g w .

e x t e r i o r p o w e r of the v e c t o r

~nv

of G in Aut V

zero m~trices.

The p r o d u c t V o W of the two vector

matrix

is a

and B the m ~ < m m a t r i x

to w I ..... win, the m a t r i x

to the c o m b i n e d b a s i s

and m k n

v I ..... V n , W 1 ..... w m

~ I v = V.

We

representation

of the m a t r i x

assigned

take ~ 0 V

to

If n= d i m V, assigning

to e a c h

to g in the r e p r e s e n t a t i o n

is the z e r o r e p r e s e n t a t i o n .

V , W we have a l r e a d y d e s c r i b e d

as a G - m o d u l e . G-module,

In p a r t i c u l a r ,

Hom(V,W)

is the dual

how H o m ( V , W )

if W

is

is the t r i v i a l

V of the

69

representation ( g ~ ) (v) =

V.

G thus acts

~(g-lv).

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

dual b a s i s

a c t i n g on ~ is the

Let

inverse

complex number

Then

of ~.

transpose

Let

check

the c o n j u g a t e s

of V*,

the t r a n s i t i o n

on the c h o i c e

Proposition:

Proof: G-invariant

Proposition:

of V, and

assigned

to g

to g in V.

Pick a basis

is an n x n m a t r i x w i t h

of the entries.

&'~4~,

(Of c o u r s e

V / X ~ > V * does not depend,

obtained

9*(g)

denoted

of V.

By the p r e v i o u s inner product.

V, V

proposition,

V*,

e

Aut V,

and

one m u s t

up to i s o m o r p h i s m

is i s o m o r p h i c

matrices.

is i d e n t i c a l

to

to V*.

we can a s s u m e V has a

P i c k an o r t h o n o r m a l

basis

But a u n i t a r y m a t r i x

of V.

Then

is one w h o s e

its c o n j u g a t e .

Let 0 be the zero r e p r e s e n t a t i o n dimensional

the m a t r i x

a basis

of that a s s i g n e d

For any r e p r e s e n t a t i o n

transpose

we take

g e G,

of basis.)

G acts on V via u n i t a r y inverse

v 6 V,

be the n w n m a t r i x

the c o n j u g a t e r e p r e s e n t a t i o n that

~6V,

representation.

~*(g)

and th~s r * : G --~ Aut V is a n o t h e r called

Then

for each g &G, r (g)

entries.

r (g) b y t a k i n g

for

of matrices,

C :G --~ Aut V be a l i n e a r

v I ..... v n of V.

from

In terms

on ~ by,

of G and 1 the t r i v i a l

one-

representation.

Let U,V,

and W be G - m o d u l e s .

Then

t h e r e are

isomorphisms:

70

i)

(U(gV)~jW

ii)

VC~U ~

iii)

O~V

iv)

U-V

vi)

~ I~U ~

U~V

ix)

U

xi)

~

U~V<~;U-W

~ U~)V

-~u

Hom(U,V)

xii)

~

H o m ( V , W l ) ~ j H o m ( V , W 2)

xv)

Hom(Vl~

Hom(V!,W)(~Hom(V2,W)

~i=0

All of these

equivalence

is to c h e c k

Hom(V,W.~)W_) ~ ± z

~ Hom(W,V)

kn(v'iw)

Proof:

xiv)

this

Ai{v)'~n-i(w)

are c o n s i d e r e d

to be true

as v e c t o r

action

if b o t h

spaces.

of G on b o t h

sides

The only p r o b l e m

sides

is the

same.

is clear.

We now c o n s t r u c t set of all

n

~_" Hom(W,V)

are well k n o w n

that the d e f i n e d

case,

Vm,W)~

m U-V

Hom(V,W)

xiii)

the

u

U.V ~ U - V

x)

Take

-:~ V

~ U-(V-W)

U~fV~W)

viii)

In each

~ V<~O

~ V- U

u~l

vii)

Ue;(V~4)

U~V

(U-V)- W

v)

of each

~

I

the r e p r e s e n t a t i o n finite

formal

sums

r in~ R(G) ~

n~~ F V i J

of the g r o u p where

G.

V la r e

i representations

of G and n. are 1

integers.

This

set

is a group

under

7i

~n.[V,~l1

the o p e r a t i o n

+

~mifVi~

=

~(ni+mi)fVi~

"

We n o w require

that i)

If V and W are

ii) The

For all V and W,

resulting

Thus we can

isomorphic

set

i~an

representations,

~V~ + ~W~ = ~+WJ

fV3

is R(G).

element ~ni~i~

let W = ~ ) ( V i +

is then

of R(G)

has all

. +V.)

a representation

is c a l l e d

~nifVi~an ~ni~Vi] positive, referred sloppy

its c o e f f i c i e n t s

(where

there

are n.

"actual

and

say

= Sni~Vi~

summands

of V.) .

W 1 comes

Such

The general

representation

element

nl w h i c h

A general of G.

an element

representations:

from those

are negative.

1

.

of two actual

representation"

"the v i r t u a l

positive,

are

element

is often

Often we w i l l

WI-W 2

for

be

"the e l e m e n t

°f

Parts abelian

i~ii),iii)

group

under

s h o w that R(G) viii), ix),x)

of the above +.

show that

consequence

together

with

Defining

is a c o m m u t a t i v e

the Horn o p e r a t i o n

simple

, where

to as a "virtual

W

representation".

as a d i f f e r e n c e

- fW2]

n.1

1

for w h i c h

W 2 from those w h i c h

-

that

an

be w r i t t e n = ~WI~

fWJ.

.

l

W

=

of

ring.

does not ix),x),xi) on

imply

~WJ-~Uy = ~W-UJ,

is an

the c o n v e n t i o n s

proposition

Defining

involution

introduce

~0

and

parts ['VJ

~I

= ~v~

=~v*~

Part xi)

new.

later.

is an

iv),v),vi),vii)

on R(G).

anything

w i l l be u s e d

that R(G)

xii),

Finally,

imply that R(G)

parts

shows a xiii),

is a

72

pre-X-

ring.

dimension

Note

of V

R(G)

that

is e q u a l

for any a c t u a l to the d e g r e e of

is in fact a ~ - r i n g ,

later with

character

but

representation

V,

the

kt(V).

this w i l l be m o r e

easily seen

theory.

The e q u i v a l e n c e

of V and V* a l l o w s

us to c o n s t r u c t

the f o l l o w i n g

inner p r o d u c t on R(G) : (V,W) The

form

d i m c H o m G (V, W)

is b i l i n e a r b y xiv),xv)

HomG(V,V ) includes multiples of

=

of the

(-,-) .

at l e a s t

and n o n d e g e n e r a t e ,

the o n e - d i m e n s i o n a l

i d e n t i t y map.

The

A simple computation

since

space of

subtle point

for any V, scalar

is the s y m m e t r y

shows H O m G ( V * , W * )

=

(Hom(V*,W*)) G =

(Hom (U, ~) ) G = (Hom(V,W)) G = (Hom(W,V)) G = H o m G ( W , V ) .

Hence

sufficient

).

bases by:

to s h o w t h a t d i m c H o m G ( V , W )

of V and W.

given

f:V--~W,

conjugate

entries,

if a n d o n l y spaces. r(f*)

=

so h a v e

Consider write

f as a m a t r i x .

(rf)*. the

i n d e e d an

Hence

Let

as a m a p V*-~-~ W*.

is additive,

the spaces

same d i m e n s i o n

inner p r o d u c t .

are

defined

f* be the m a t r i x Then

of

f is G - i n v a r i a n t

setting up a one-one correspondence

The c o r r e s p o n d e n c e

Pick

the m a p H o m G ( V , W ) - - - ~ H O m G ( V * , W * )

considered

if f* is,

= dim~omG(V*,W*

it is

between

and for any r e a l n u m b e r

isomorphic

as c o m p l e x v e c t o r

as real v e c t o r

spaces.

Thus

these r,

spaces,

(-,-)

is

73

As the s i m p l e s t The r e s u l t i n g identity

R(G)

example

is then

of the above,

just

map, ~ n ( m ) = / m l ,

and

let G be the trivial

the ring of (n,m)=

nm,

integers, under

with

the

group.

the

isomorphism

~V] "vd2dim V.

Let G be a subgroup that e v e r y of G.

representation

It is not hard

-rings

with

particular,

remarked

by r e s t r i c t i o n

this gives

and a u g m e n t a t i o n subgroup,

if G is a s u b g r o u p

in this

a homomorphism

Res:R(H)~> R(G)'~ ~

case

a representation of

R(G).

In

so this

is the

the

induced

of H and V a r e p r e s e n t a t i o n

representation

on H as a set of p e r m u t a t i o n s

space w i t h b a s i s

the e l e m e n t s

right

G-module

~(gh)

= ~(h) (m (g) .) modulo

hg~ v - h~gv. h!(hg~

If several

already

R(H)---~ Z, ~ V ~ > d e g r e e V .

we can define

(since

have

of H gives

if G is the trivial

Conversely,

~H~V,

We

to see that

involution

augmentation

G acts

of H.

(i.e., ~ : G ~ Ind(V)

the s u b s p a c e H acts

groups

Ind:R(G) ----> R(H).

are

by g(h)=hg,

Aut ~4

is an a n t i - h o m o m o r p h i s m : defined

generated by,

= hlh~gv we

action

Let C H be the vector

This

is then

involved,

of H as follows:

of H.

on Ind(V)

v) = h h l g ~ v

Ind(V)

of G makes

as the vector

by all e l e m e n t s for h16 H,

write

~

a

space

of the form

hl(h~v)

= hl(h~gv), sometimes

of G,

=

(hlh)~ v-

this makes Ind H for G

sense.)

74 It is e a s y to see that gives

an a d d i t i v e

Ind(V)

m a p R(G) ~

= Ind(V).

= Ind(V)~)Ind(W)

R(H) .

To see this,

since H acts b y m a t r i c e s whence

Ind(V~W)

with

note

Slightly

less

so that

trivially,

that as an H-module,

real entries.

Hence

Ind

CH = ~£*,

Ind(V*)

=

(Ind(V))*,

the c o n c l u s i o n .

Ind

increases

degree(IndH(v))

These

Theorem:

operations

Let G c H.

dim~Oms(V,

2)

Ind(V~Res

In o t h e r words, spaces

Proof:

if n =

IHI/~G~

, then

= n d e g r e e V.

i)

product

the a u g m e n t a t i o n :

L e t V be a r e p r e s e n t a t i o n Res W) = d i m ~ o m H ( I n d

W)

=

There

of G, W of H.

V, W)

(Ind V ~ W

I) Res and Ind are a d j o i n t R(G)

Rec iproc ity:

Ind and Res are r e l a t e d by F r o b e n i u s

and R(H),

and

is a n a t u r a l

operators

2) I n d ( R ( G ~ C

imbedding

R(H)

on the is an

V---~ ~ H ~ V ,

inner

ideal.

v ~) l~v.

C3

Consider

the d i a g r a m

of v e c t o r f

V inclusion Ind V

Given

a G-map

diagram

determined

-'> Res W

1

: ~{~rg

f:V---~ Res W,

commute:

spaces

I

isomorphism

of v e c t o r

spaces

is a u n i q u e H - m a p

g, m a k i n g

this

--~

g there

take g ( h ~ v )

as the r e s t r i c t i o n

W

= hf(v).

Given

an H - m a p

of g to the s u b s p a c e

g,

f is u n i q u e l y

V of Ind V.

75 The isomorphism h~(v~gw)

~

Ind(V~Res W) ~ >

(h~v)~hw.

Ind(V)~W

is defined by

The inverse map is ( h ~ v ) ~ u

Each is an H-module homomorphism.

~> h ~ ( v ~ h - ! u ) .

76

2.

Irreducible

Representations

L e t V be a r e p r e s e n t a t i o n if t h e r e

is a s u b s p a c e W ~ V ,

w i t h W ~ <01! V. is b y d e f i n i t i o n

and Schur's

Lemma

of a g r o u p G.

V

with gweW

Otherwise V

V

to the sum of two r e p r e s e n t a t i o n s

Otherwise,

V

there

If V

is r e d u c i b l e ,

In o t h e r words,

is a n o t h e r G - s u b m o d u l e

choose a G-invariant W,

let W' be

if W W'

its o r t h o g o n a l

irreducible

one-dimensional

V

is

of G: V ~ V I @ g V 2.

is d e c o m p o s i b l e .

is a n o n t r i v i a l of V w i t h W~)W'

inner p r o d u c t on V

G-submodule ~ V.

of V,

To see this,

(as on p a g e 62

).

Given

complement.

L e t I r r e p G = t V I , V 2, of

if V

is i n d e c o m p o s i b l e .

(Maschke) :

Proof:

(The z e r o r e p r e s e n t a t i o n

is c a l l e d d e c o m p o s i b l e

isomorphic

Theorem

reducible

for all w e W , g ~ G and

is it_reducible.

reducible.)

is c a l l e d

representations representation,

"7 be the set of of G. which

( i s o m o r p h i s m classes)

W e t a k e V 1 to be the t r i v i a l is i r r e d u c i b l e

for any group

G.

Any representation irreducibles: follows:

If V

V =

V of G can be w r i t t e n

~ n.V.. 1 1 IrrepG is i r r e d u c i b l e ,

To a c c o m p l i s h

as a f i n i t e this,

sum of

proceed

V = V i , for s o m e V lo~ I r r e p G.

as If n o t i

77

let V . ~ V be a n o n t r i v i a l T h e n V.l is irreducible, is a G - m o d u l e

Schur's

number

as to w h e t h e r

V. is a G - m o d u l e 1 ~ such that

Proof: G-submodule

t h a n V.

L e t V . , V . ~ I r r e p G. l 3

or z e r o a c c o r d i n g

f(v)

Suppose

of V of m i n i m a l

=kv

Proceed by

i=j or not.

In p a r t i c u l a r ,

then there

V.. j

Then

the k e r n e l

0 or V.. 1

Hence

if f is n o t the z e r o map,

one-one

hence

an

since V

isomorphism. of f, w i t h

is a c o m p l e x v e c t o r

f-~I:V---~ V.

Since

irreducible,

f - ~ is the z e r o map.

Theorem:

irreducible

basis

The

its k e r n e l

product

the

it m u s t be b o t h

If V.=V. = V, l ]

space.)

r i n g RiG)

f:V---2 V,

(Such a l w a y s

Consider

the m a p

is n o n - z e r o , (it c o n t a i n s Hence

image

v)

and V

is

|

f = kI.

of G f o r m an o r t h o n o r m a l

with

respect

to the

inner

(-~-).

Proof: the c o m m e n t

R~G).

of f is a

eigenvalue ~

representations

of the r e p r e s e n t a t i o n

if

is a c o m p l e x

similarly,

of f is 0 o@ V.. 3

exists

is one

for all v E V.. 1

f:V.~> l

let v 6 V be an e i g e n v e c t o r

induction.

T h e n dim_HomG(Vi,Vj)~

homomorphism,

of V. so is e i t h e r 1

and onto,

dimension.

and b y the t h e o r e m V ~ V { ~ V 2 w h e r e V 2 c V

of s m a l l e r d i m e n s i o n

Lemma:

f:V. ~ > 1

G-submodule

Schur's

lemma gives

after Maschke's

the o r t h o n o r m a l i t y

Theorem

shows

t h a t the

immediately, irreducibles

and span

|

78 Corollary:

As an abelian

is isomorphic

to k ,

nlm I + n2m2+

Corollary:

under

+ with

inner product,

R(G)

k = IIrrep G I, and((nl, n 2 .... ), (ml, m 2 .... )) =

J

....

Let V 6 R(G)

representation

Corollary:

group

be any element.

if and only

Let

~

be

if

Then V is + an irreducible

J

(V,V)=I.

the regular

representation

of G.

Then

~ n.V. , w h e r e each V. E Irrep(G) appears, and n. = dim V.. 1 1 1 1 1 IrrepG In particular, the n u m b e r of irreducible r e p r e s e n t a t i o n s of G is =

f in ite~

Also

Proof: of the unit Reciprocity,

IGi~----~_~ rdim Vi) IrrepG

Let

1 0 be the trivial

subgroup for each

1

of G.

one-dimensional

representation

Then ~ = Ind(l O) so by F r o b e n i u s

irreducible

representation

V, of G, 1

n.l = ( ~ ,Vi ) = = dim

(Ind(lo) 'Vi)

=

(lo'Res

V i) = dim HOm{l ( (Io,V i)

B

V..

A representation it is of the

V of G is c a l l e d

form V = nV., 1

for some

is o t y p i c a l

irreducible

o f type ~i'

representation

if V.1

79

Proposition:

Let V. ~ Irrep G.

Given

any G - m o d u l e

V,

there

is a

1

unique maximal

G-submodule

W of V w h i c h

is i s o t y p i c a l

of type V

of V. l

It is c a l l e d

the v . - i s o t y p i c a l

component

of v.

v is the sum of

--1

its

isotypical

Proof:

components.

W r i t e V _~

n.V. as a sum of i r r e d u c ~ b l e s o This ! 1 IrrepG g i v e s an i s o m o r p h i s m V ~ = ( n i V i ) ~ i (~_~njVj). U n d e r this i s o m o r p h i s m j/i n.V. c o r r e s p o n d s to an i s o t y p i c a l G - s u b m o d u l e W of V, and ~ Q n.V. i i j~i 3 3 to a c o m p l e m e n t a r y submodule

submodule

of V of type V.

W' of V.

We c l a i m

must be c o n t a i n e d

that any

in W.

isotypical

If not,

there

1

w o u l d be a s u b m o d u l e nontrivially.

of V,

Since V.

isomorphic

t__ooVi,

is irreducible,

which

this m a k e s

intersects V.

1

of W'.

Hence

*~n.V. = Vi~complementary j~i ] ]

negative

In fact, V into

part)

so 0 =

components.

of

~ n.V., j~i 3 ]

( ~jnjVj, j~i

~c°mplement'Vi)

This c o n t r a d i c t i o n

one can be m o r e

isotypical

to a s u b m o d u l e

V i) = rV.,V.)l i +

integer).

a submodule

1

V. is i s o m o r p h i c l

cV i ~ c o m p l e m e n t ) ,

W'

explicit

so

V i) = = 1 +

(a n o n -

gives

the result.

about

the d e c o m p o s i t i o n

For e a c h V~ ~ Irrep G,

J

of

let

l

Vi~HOmG(Vi,V)

~--7 V be the m a p vi~ f ~c~ f(v) .

(Vi~HOmG(Vi,V)) ~ Irrep G d e c o m p o s i t i o n of G. Note we m a p

2Z~

R~G)

by n ~

V is an

isomorphism

that HomG(Vi, V) nV.,l this m a k e s

Then and

is the

is a t r i v i a l

G-module.

HomG(Vi, V) = n~.l

If

80 Proposition: Then

Let G be a s u b g r o u p

in w r i t i n g

of H, V. ~ Irrep io

Ind(V i ) = ~ m~W., o IrrepH 3 ]

the c o e f f i c i e n t s

m. and n, are equal. ]o 1o

Proof:

is just a r e s t a t e m e n t

This

and

G and W, ~ Irrep H. 3o

Res{W, ) = ~ n i V i, 3o IrrepG

of F r o b e n i u s

Reciprocity.

81

3~

Characters

Let G be a finite of sets

f:G~>C

group.

, such

(Equivalently,

that

f~a)=f(b)

A central f(ab)=f(ba)

whenever

be the set of all c e n t r a l

following

operations

0{a)

~a)

Conjugate:

f*(a)

(f(a))*

Inner

Product

~fl' f2 ) = degree(f)=

54, CF(G) defined

as follows:

also

the f u n c t i o n

~ G fl{g ) . f 2 ( g ) g6-

f(e)

via the Adams

For each

under

to o r d e r

sum,

product,

and the

operations.

over C.

conjugacy

on G, K ~ a ) = l

It is c u s t o m a r y

~1

is a ~ - r i n g

is also an a l g e b r a

basis

not.

=

( ~ n f ) (a) = f(a n)

CF(G)

the

= 1

Operations

-operations

We d e f i n e

+ f2~a)

Adams

per page

on G.

= f(a -1)

Augmentation As

functions

in G.)

(fl'f2) (a) = fl(a) f2(a)

l{a)

Dual:

G.

= 0

Multiplication: One:

for all a,b

on CF(G) :

(fl+f2) (a) = fl(a)

Zero:

on G is a map

a and b are c o n j u g a t e

Let CF~G)

Addition:

function

We can c o n s t r u c t class

K of G,

or 0, a c c o r d i n g the c o n j u g a c y

a natural

let K d e n o t e

to w h e t h e r

classes

a ~ K or

KI, K 2 .... ,K n

82

t a k i n g K 1 to be Clearly,

each

the

K. i s 1

f is of the form dimension

classes

Lemma:

(singleton)

of

a central

f =

CF~G)

as

class

containing

function,

~

and

every

c.K. , for u n i q u e i 1

i C-vector

space

is

the

the unit e l e m e n t central

c. C-~. l

number

of

Thus

of G.

function

the

conjugacy

of G.

For a n y p a i r of c o n j u g a c y

0 (Ki,Kj)

=

,

classes

K.,K. l ]

K, I ~ Kj

iKiI/LGl i--Kj Proof : A simple

Since form

the K. form an o r t h o g o n a l 1

(-,-),

it is an

Proposition: I)

For all

% 'IG~ ffa)

3) ~IG~-l(f)

Proof:

respect

to the b i l i n e a r

f•CF(G)

~n

f)

for all a L4G for all

integers

n~0

: T

Again a simple

for all a ~ G.

set w i t h

inner product.

= degree(f)

9) ~ IG~ + n ( f ) :

J

computation.

computation,

using

the fact that a I G I =

e

83 Before some facts

giving about

be a linear a basis

of V and let A = is d e f i n e d

i)

If

iii)

Tr~d]~

Given

Given

Tr{~&~

(&ij)

W and

we recall

Let o4:V---~ V

space V into

be the m a t r i x

of ~

itself,

pick

The trace

the sum of the d i a g o n a l

it does not d e p e n d

[ :W -~ V,

is the

of

entries.

on the

then T r ( ~

i d e n t i t y map,

) = Tr(~

Tr(~

)= dimV.

). If

Tr(~< )=0.

~ :V--~ V,

) = Tr(w

v)

of a v e c t o r

as ~

~:V~-~V

is the zero map, iv)

section,

of V

~:V~

If

in this

of an a u t o m o r p h i s m .

T r [ ~ ) is w e l l - d e f i n e d ;

of b a s i s

ii)

theorem

Aii, i properties:

the f o l l o w i n g

Lemma: choice

the trace

transformation

, Tr( ~ ), It has

the m a i n

)+ T r ( ~

~ :V~)

@ :W ---~ W so that

~(/+~ :V+~)W --~ V ~ W ,

7-

V,

[ :W ~

W so that

~_9~ :V(WW --~ V4£/W,

) = Tr( ~)Tr( ~ ) .

vi)

Given

~ :V---~ V, T r ( ~

vii)

Given

viii)

Let

~ :V--~ V, ~ :V ~ )

Let T be a formal ~The l e f t - h a n d

symbol

side here

taking

the m a t r i x

adding

the

result

)

for

V,

Tr(~

): T r [ % ~ )= Tr( ~

inducing

~i

)

(~ )

: transpose ~ (~

: ~i v ~

= conjugates ~iv,

i % i.

Ca " v a r i a b l e " ) . is e v a l u a t e d in that basis,

i d e n t i t y matrix,

and taking

)

D e t ( I + ~ T) = ~ T r ( ~l i b y p i c k i n g a b a s i s for V, multiplying

)T i,

each e n t r y by T,

the d e t e r m i n a n t

of the

84

ix]

If oK :V ~

V is idempotent

(i.e., ~ ~ = ~ )

then

t

Tr(o m) = d i m ( I m a g e ~ ) .

Let ~ : G ~ ) A u t

V be a representation

define a map of sets by ~V

7 ( V , ~ ):G-~) ~

~(V, ~ ) ~g): Tr( ~ (g)), g £ G. is a central

of a group G.

We

(usually denoted just 7 V )

By part ii) of the lemma above,

function so ~ V is an element of CF(G),

character of the representation

V.

Also by ii), 7 V

on the isomorphism type of {V, ~ ). Hence the assignment of characters

By iii), ~ V ~ W

the

depends only = 7V

to representations

+ ~W"

gives a map

R~G) ----~CF ~G) .

Theorem:

The map R!G)---> CF(G)

preserving the involution, augmentation.

is a homomorphism of ~ -rings

conjugation,

It is a one-one map,

inner product,

identifying R(G) with

image, which we call the character ring of G. irreducible representations,

and its

The images of the

called the irreducible characters,

form an orthonormal basis of CF(G).

Proof:

The fact that the map preserves

the involution~

the conjugate,

sum, product,

and the augmentation

i, 0,

follow

immediately from the lemma above. To show that the map preserves sufficient

the b-ring structure,

to show that it preserves

it is

the Adams operations.

Thus,

85

given a G-module V, if we take the element ~ k ( v ) & via the universal polynomial ~ k ( v ) = Q k ( then compute ~k(~/~V) .

~k(v ),

R(G), computed

k2v ..... kkv),

~IV,

and

the result must be the same as computing

Hence we mu~t~ show, for every g6 G , ~ k ( v

) (g)=~k(~v)(g).

Let g 6 G and assume a basis for V is picked so that the matrix for g is diagonal

(which is possible by

).

gl O}

g =

g2

0 gn

Then, using the lemma above: 1 + TrCg)T + Tr~ ~ 2 g ) T 2 +

=

Det II + gT)

~+gl T Detl

: \ Thus Tr~ i g )

~

l+g2T

~

(l+giT)

.th is the i elementary symmetric function of the gi s.

We now compute =

Qk I Tr ~g,

=

gl

k

Tr /~2g ..... Tr ~ k g )

k + g2

+

+ gn

k

(by definition of Qk )

= Tr(q k)

= ?~v(g k)

= VkNv~g) Hence the map R(G) ---9CF(G) preserves the

~ -rinq structure°

86 N e x t we show that will

have

two o t h e r

the set of

the m a p p r e s e r v e s

consequences:

irreducible

characters

F i r s t we n e e d a fact: 1

S

I GI Proof:

Then

=

lq

set

and

in CF(G).

V of G,

linear m a p d e f i n e d b y

g~G

gJ

Hence

so v G •

is equal

Image~J) d

Image{J) ~

-

to J for all g 6 G.

It is a s i m p l e

and e v e r y gd G acts

trivially

V G.

But also,

H e n c e V G = Image(J)

~ Tr (g) geG N o w take any two r e p r e s e n t a t i o n s d i m V G = Tr(J)

is an o r t h o n o r m a l

This

dim V G

that J is i d e m p o t e n t

image of J.

m a p on V G,

the m a p m u s t be one-one,

any r e p r e s e n t a t i o n

=

be the

the c o m p o s i t i o n

consequence the

~v(g )

inner product.

g (

Let J : V ~ ) V

J

given

the

1 ~GI

J is the

identity

so b y the lemma,

QED. V and W of G, w i t h

characters~v,

1 gEG 1

~

]IV, (g) 7 < w ( g )

iG% g ~ G

Pl

geG

1

~_

on

~TCgw(g

)

(GI g c G i ~i'/Hom~v, I Gl g C G =

dim C HomG(V,W)

=

~v,w)

wl ~g)

(using

the fact above)

~W"

87 Finally, we show that the set of irreducible characters span CFIG).

Let f~-CFIG) be any central function.

an irreducible representation of degree n.

Let

~:G~)Aut

Define ~f =

~

V be f(g) ~ (g),

gc~G a map from V to V. identity map, where

Then

I

~ f = ~I, a constant times the

k--
Proof:

Let g6 G.

Then

f(gl) ~ (g)-l~ (gl) f (g) gl & G f(gl ) ~ (g-lglg) glC-G (where g2=g- Igl g

f(gg2-1g) ~ (g2) g2 G G =

~ f(g2 ) ~(g2 ) g2 G G

(f is central so f(gg2

g)=f(g~

P

if Hence, by Schur's lemma, ~Of, is a constant multiple of the identity: ,~f = •I. =

Calculating the trace of both sides, we get

~ ffg)Tr( ~> {g)) giG

=

~f(g) 7 v ( g ) g4 G

= (f*,~(~V) .

Now let f be any central function on G. to all irreducible characters,

i.e.,

n /~ = Tr( k I) QED.

Suppose f is orthogonal

(f,'~V)=0 for all irreducible V.

Hence for every irreducible representation ~:G ---~Aut V, zero, by the above.

Since every representation

~ f,

~:G~

is zero for all

representation.

Aut V.

Gf* is

is a sum of irreducibles,

Now apply this to the regular

We calculate the transform of the basis vector e by

88 ~f,:

0 = ~f,(e)

~ fCg)*~g(e) = ~ f(g)*g. Thus f(g)*=0 g& G g~G since the set of all g 6 G form a basis of the regular

for all g & G ,

representation.

=

Hence f = 0.

Given any central function f, now, ff - ~ {f' ~ V ) ~ v . ~ is U IrrepG i l orthogonal to all irreducible characters, so zero. Hence f =

~ ~f, ~ V ) ~ V . IrrepG l 1

CorollarY:

RCG)

I

is a ~ - r i n g ~

Proof: We already knew R(G]

to be a pre- ~ -ring.

n o w is whether a number of identities hold. contained as a pre- k - r i n g CF(G)

the conclusion

in CF~G)

condition

~ v ( g -I) = ] ~ ( g ) identity,

But since R~G)

is

and the identities hold in

follows.

Since ~ = V* in R{G), sufficient)

In question

this gives a necessary

(but hardly

for a central function to be a character:

= ~V,(g)

=

take coordinates / gl / ~

~v(g)*.

For another proof of this

in V so that g is a diagonal matrix:

0

\i

gn Since g IGI = e, each gi must be a IGI th root of unity. each i, gi*=gi -1, so Trig*)

= Tr g-l) .

Hence,

for

@9

corollary:

The n u m b e r

to the n u m b e r

Corollary:

of

of c o n j u g a c y

G is a b e l i a n

of G has d i m e n s i o n

all the m a t r i c e s

one.

of G, n,

representations

Q

if e v e r y

irreducible

G is a b e l i a n

V of G, a b a s i s

that

the n u m b e r

representation

if and o n l y

for V can be

to G b y that b a s i s

know

of G is equal

of G.

Equivalently,

assigned

We a l r e a d y

classes

if and only

for e v e r y r e p r e s e n t a t i o n

proof:

irreducible

if

found so that

are diagonal.

of

irreducible

representations

of c l a s s e s of G, and the sum of the s q u a r e s n of the d e g r e e s is G: ~ d. 2 -- iG] , integer d.. H e n c e each d.=l l l l i=l if and o n l y if n = I GI~ The s e c o n d a s s e r t i o n of the c o r o l l a r y follows

is the n u m b e r

immediately

We n o w have

two n a t u r a l

characters

~V,

characters

by u p p e r

bases

and the class

of the t r i v i a l

indices:

The

symbol

~i

the sets

- the

K i.

letting

~i

representation ~j

the c h a r a c t e r ~ i

Since b o t h

for R!G)

functions

one-dimensional

for any g r o u p G, obtained when

from the first~

i

refers

index the

be the which

irreducible

character is i r r e d u c i b l e

to that c o m p l e x n u m b e r

is a p p l i e d

-'hi and

We

irreducible

to an e l e m e n t

K. are o r t h o g o n a l 3

of K.. 3

for the

inner

90 product,

we can e a s i l y

?<

Computing (K,,K,)

J

=

E

express

[Kj,

i

()
i (%(,Kj)

we find

in terms

of the other:

K.

J

i) * iKji }( = ~

:'K,

i *

ia}

J

J

Recall

=

J

lal

~,i

Hence

=

- i*

~j J

If we r e n o r m a l i z e

K. J

the K. by d e f i n i n g 3

also an o r t h o n o r m a l

0

('~

i

,Lj)

L

= J

IGI IKjl

K. J

the L. are 3

basis:

(L.,L.) l ]

and

~j) ]

j

either

=

i = ~ j.

i@

j

}el

<7
Lj _- S

i

J Proposition: and both

Z

i contained

Proof:

~ ! fg) *

i

(h)

in the class

-

IGi ]K~

K; zero,

if g and h are c o n j u g a t e otherwise.

Say q 6 K, and h { K.. D e f i n e L,,L. as above. Then l ] l j the sum in q u e s t i o n is ~ i k* ~ j k = (Li' Lj ) " k (Note that in general, K. and L. are only central f u n c t i o n s , n o t c h a r a c t e r ~ ) ] J i The m a t r i x of complex numbers~j__ is c a l l e d the c h a r a c t e r table

91

of the g r o u p G. across

It

the top a n d

is t r & d i t i o n a l l y w r i t t e n irreducible

K1

K2

characters

K3

listing

the c l a s s e s

at the left:

K

K.

n

E

I

n

The c o n s t r u c t i o n calculation

of this t a b l e

of the r e p r e s e n t a t i o n

w i l l n e e d to c o n s t r u c t to r e v i e w the m a i n

The t a b l e has two w a y s

linear

rows:

a number

features

two

1 V)

~g&G

t h e o r y of the group.

of e x a m p l e s

interpretations.

obey

As above,

between

it is w o r t h w h i l e

i

columns:

it r e p r e s e n t s

o r t h o g o n a l bases,

the o r t h o ~ o n a l i t y

~i(g)* ~ j(g)

l£irrepG

later,

Since we

of the table.

transformations

the rows and c o l u m n s

Also,

for a g i v e n g r o u p G is the b a s i c

=

:: ~ j k

relations

~i 3=

in

and thus

g i v e n above:

<~i

, kj)

;Gl

J •

t h e r e are as m a n y rows as columns,

n u m b e r of c o n j u g a c y c l a s s e s

in G.

the c o m m o n

number being

the

92

There for each

is a l s o the

irreducible

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

i n t e r p r e t a t i o n of the t a b l e representation

character.

the table are obvious. character,

the

t r i v i a l class, irreducible

the

the

representations. Theorem

is

of G { n e c e s s a r i l y ~

[G,G~ ,

irreducible

representation

character

here

/G I.

of the d e g r e e s

is that a o n e - d i m e n s i o n a l

We of

the c h a r a c t e r

for any n o r m a l t a b l e of G.

representation

is just a g r o u p h o m o m o r p h i s m ~

factors

through G modulo

representation

of G.

one-dimensional

Indeed,

matter G/H

of the

divide

inducing a representation

any o n e - d i m e n s i o n a l

G / G,G.

one-dimensional

is that the d e g r e e s

observation

irreducible)

subgroup

irreducible

of

S i n c e K 1 is the

sum of the s q u a r e s

Conversely,

of

of all ones.

features

is a l i s t of the d e g r e e s

S i n c e ~ is abelian,

commutator

certain

G

Another useful

p :G ~ 3

of this,

taken by

A fact w h i c h we w i l l not p r o v e

II.12)

that the

of the v a l u e s

7~ 1 is the t r i v i a l

first c o l u m n

already proved irreps

Since

first r o w c o n s i s t s

Cbut see Serre, have

Because

of G,

as the listing,

Hence,

characters

of G/~G, G J g i v e s

in p a r t i c u l a r ,

appears

, or for that as a p a r t

an

the n u m b e r

of G is the o r d e r of

t a b l e of G / G,G

s u b g r o u p H,

its

i o of G / ~ G , G ~ .

of the

95 since

the r e p r e s e n t a t i o n

the character -operations possible

ring,

the addition,

can all be carried

to be very explicit

As an example, three conjugacy Hence

of S 3 is [ I ~ ]

the arithmetic

R(S3)

and S3/IS 3,$3~

of these

down the character

table

already

have the first two rows of the table.

columns

gives

the last row. KI

K2

of R(G) .

Since

S 3 has

6,

IS31

irreps must be 6.

The commutator

is the cyclic

it is

, and K 3 = ~ , J T i ~ 7 ~ .

representations.

of the degrees

to

and

and

completely.

K 1 = [e ~, K 2 = ~'~ ~

must be i, 1 and 2.

It is easy to write

conjugacy,

out on the characters,

about

irreducible

the sum of the squares the degrees

multiplication,

we shall describe

classes:

S 3 has three

Hence

ring of the group G is isomorphic

subgroup

group of 2 elements

for ~Z 2"

Hence we

Orthogonality

of

K3 _

%2

~2

is the character

the character

of the alternating

of the matrix

Every representation three.

For example,

representation regular

representation

of S 3 is an

consulting

~has character

[l,q I of S 3 is (3,0,1)

of these

, the trivial

so is n ~ I.

The

so i s ] < i + ~ ( 2 + 2 ~ ,

The representation so it is ~ I +

~

is

60.

combination

(n,n,n)

(6,0,0)

of S 3. ~ 3

on page

the list on pp.65-7

w i t h the corollary

the subgroup

given

integral

of degree n has c h a r a c t e r

representation

agreement

representation

induced by 3

The

in

z2

94

canonical representation has character

(3,0,1)

so also is "~l+?c 3 , -

thus proving the isomorphism mentioned on page 7.

Etc.

Let us now write out explicitly the arithmetic of R(S3). ~dditively, ~i, ~ 2

R(S ) is the free abelian group on three generators 3 ~3. Since the characters take all real values, the

involution is trivial. (n i ~ l + n 2 0~2+n3 ~< 3

Degree(nl~l+n2~2+n3"~3)= nl+n2+2n 3.

ml~( l+m 2 ?<2+m 3 ~ 3) = nlml+n2m2+n3m3"

To compute the multiplication, each pair (i,j). For example,

we can just compute ~ i ~ j ,

for

the product of the two characters

~2 and ~3 is the character

{ 7 (-2 ?C3) (e)

: ?(~2(e).?C3(e)

('~27C3)

= ~2(-'r)'-'/,-3(~'-C)

=-i

= %2(~-)'Y3(~F)

= o

(~c)

(O(z~3)(~)

Hence ~ 2 ~ 3

is the character

(2,-1,0) so equal to ~ 3

The full multiplication table for

kl Si

.........~2

= 1,2 = 2

~ i 9fj is

73

72 ~~2

?<1

73

~f3 7i+7~2+73

~/3

)<3

The general product of two elements of R(G) follows by bilinearity.

95

~i

and ~ 2 are one-dimensional,

on them:

kt (~(I) = 1 + 71t,

dimensional

so

be computed

in two ways^

~ - operations are easy

= 1 +

~2t.

SO ~ ( ~ 3 )

for

~3

a can and a is

since they are 29<2.

by

~2(~3(e))

: ~3(e2)

klb2{'~3("Y)

='~3:("C2

0-)

is two-

from the second exterior power of these

in this case this is the determinant

bu2 (~3

~3

+ at 2 for some a 6R(G).

First we have matrices

Or, we can c o m p u t e ~ 2 ( ~ 3 )

~2(x)

~t (Q£2)

~ t ( ~ 3) = 1 + ~ 3 t

just the character obtained matrices;

so the

: ~<3 (

is the character

= (1/2)(~l(x)~l(x)

~

: ~3(e)

= 2

) = "~3('t" ) = -1

2)

=

<<~ ( £ )

= 2

(2,-i,2).

Now,

in any

h-ring,

- ~2(x)).

~ ' I ( 7 3 ) : X 3 : (2,-i,0).

It's square,

using the formula for product of two central functions,

is

Subtracting~2(~

(4,1,0).

result b y two yields

Exercise:

~(~%3)

3) = (2,-1,2) = (i,i,-i) = ~ 2 .

Carry out the same calculation

groups of order eight

and dividing the

for the two non-abelian

(the quaternion group and the dihedral group).

Show the resulting character rings are isomorphic as rings, but not isomorphic as

~-rings.

each case how many solutions

(Hint on the latter:

calculate

there are to the equation

in

k2(x) = 1

.

96

Let H b e of H. be

a subgroup

We c o m p u t e a conjugacy

function (X,L)

a formula class

(p. 90).

= X(~).

of G and X a c h a r a c t e r

of G,

Recall,

and L

character

the a s s o c i a t e d

for any c h a r a c t e r

Ind~ X-

Let

central

X of G, the dot p r o d u c t

ThUs

IndGx(~)

G

=

G ( I n d H X , L ~)

=

H (X, Res L ) 1

~_.

-

(here u s i n g

for the i n d u c e d

of a r e p r e s e n t a t i o n

the

(by F r o b e n i us

X(O)

fact that L

Res L

takes

Reciprocity)

(q)

real v a l u e s

so L

(O)=L

(0)*).

By definition ResL

(o)

G

SO

IndHX(~)

L

(~) =

]

-

OEHn[~] In p a r t i c u l a r ,

if k is the unit character,

G

Ind H 1 (4) -

(where, ~,

recall,the

and the n u m b e r

id

{HI

tHn[~]l [~]

notation

[~]

of e l e m e n t s

refers

to b o t h

of this class).

the c o n j u g a c y

class

97 The next computation is the representation ring R ( G ~ H) of the cartesian product of two groups.

Suppose P l : G . ~ ) A u t

and P2:H---~ Aut V 2 are given representations.

V1

Let

pl ~ P2:G~H---a A u t ( V I ~ V 2) be defined by

p l ~ P 2 ( ( g , h ) ) (Vl~)V2) = pl(g) (Vl)~P2(h) (v 2) This is a representation and its character XP I X P2 is given by ~pl N p2((g,h)) = ~01(g) Xp2(h) (since the trace of a tensor product is the product of the traces.)

Proposition: Conversely, Proof:

If Pl and P2 are irreducible,

so is Pl x P2"

every irrep of G × H is given in this way.

We have given

IG) geG' ~I

[H'J ....

Hence, using the formula for ]{p I×P2

h6H

above,

l

2 =

!G W ~

(g,h) 6G X H

1

O I X P2

so, by the corollary on p.78, X P I ~ P2 is irreducible. if pl,Pl I

,

are irreps of G, and p2,P2 I

ii

p2~P2 , then pl m p2~p I )< ~2 "

,

Similarly,

of H, and either p l#Oi' or

98

TO c h e c k

that

set of i r r e p s

{O I ~ p2 i P l E I r r e p G , P 2 E I r r e p H ]

of G wH,

is a c o m p l e t e

one o n l y n e e d c o m p u t e

(degree (01 x 02)) 2

:

i%~ pl,~ 2

Pl £ I r r e p G

(deg Pl" deg

Q2 )

P26IrrepH ~-

2

(deg

pl ) (deg p2 )

2

~-~-

= d

Pl" ~2

There

PI 5

are o t h e r w a y s

and c o r r e s p o n d i n g Given groups

theorems

rule

on the

(h,g) (h',g')

the c h a r a c t e r s

=

G and a g r o u p H.

(deg D2)

to g e t a third, rings.

(where Aut h e r e

the s e m i - d i r e c t p r o d u c t

( h , g ) £ H ~ G, w i t h the m u l t i p l i c a t i o n For the c o n s t r u c t i o n

( ~41~

in the

Here there

G on a set S w i t h n e l e m e n t s .

special case

of that

, p. II-18).

is the w r e a t h p r o d u c t

(H ~ .... ~H)

~

representation

G - ~ o Aut H

of a s e m i d i r e c t p r o d u c t

More complex

G---9 Aut

associated

(h ~(h') , gg') .

see S e r r e

2

~2

two g r o u p s

of H as a group)

is the set of p a i r s

H is abelian,

to c o m p o s e

G, H and a h o m o m o r p h i s m

r e f e r s to a u t o m o r p h i s m s H x G

(deg Ol)

is g i v e n This

(n factors)

of a p e r m u t a t i o n

an action,

induces

group

G - - D Aut S, of

a group homomorphism

b y g( .... h .... )=( .... h -i .... ) " s g (s)

99

The a s s o c i a t e d product,

written

the s p e c i a l

results

G[H~.

this

of this

section,

first

subgroups.

p' = R e s ~ Ind~

without

at least

- just

is M a c k e y ' s Let

- i.e.,

Theorem.

no a c c o u n t

standard While

of s u b j e c t

(The p r o o f s

here).

Let G be a g r o u p a representation

p b e the r e p r e s e n t a t i o n back

three

their mention.

irrelevant

p : H - - > Aut W b e

on n letters.

proof

obtained by

to K.

and H and K

and inducing

The p r o b l e m

a set of d o u b l e

coset

representatives

a set SC-G such that G is the d i s j o i n t

(A s t a n d a r d

notation

= sHs-IAK,

giving

group

later w i t h

p

is to

p'

Choose

Hs

arise

and some c o r o l l a r i e s .

for the sequal,

up to G, and then r e s t r i c t i n g compute

will

we g i v e w i t h o u t

theory,

is n e c e s s a r y

are not so d i f f i c u l t The

of H m n w i t h G is the w r e a t h

These products

in r e p r e s e n t a t i o n

would be complete

be

product

c a s e of G = S , the s y m m e t r i c n

To finish

none

semidirect

is S = K ~ G / H

a subgroup

a representation

of K. Ps:H

s

.)

Put,

For

for x6H

~ Aut W.

union

s£S, s

of G m o d

(H,K)

of KsH,

sES.

let

, Ps(X)=p(s-lxs),

H c K so s

Ind~

(ps)

is

s

defined.

Theorem:

R esGIn K d H(p) G

The r e p r e s e n t a t i o n

of the r e p r e s e n t a t i o n s

Ind~

(ps) s

is i s o m o r p h i c

for s £ K ~ G / H .

to the

sum

100

(For the proof,

and some c o r o l l a r i e s ,

For a c a t e g o r i c a l

generalization

G~R(G),

see A . D r e s s

Corollary:

Let H , K b e

T h e n the dot p r o d u c t cosets

of G m o d

The s e c o n d any g r o u p G, divides ( 41

be

deg

~ deg

of G.

of G.

is equal

,p. II-10).

than

Let ~H = Ind~l, to the n u m b e r

This

Let

~K = Ind~ 1 .

of d o u b l e

a finite

is an e l e m e n t

PG = I GI

for of p

in S e r r e

of the c h a r a c t e r s

A more natural

proof would

conjecture:

group

with

, hence

is true

(e.g.,

that the v a l u e s

and p£R(G)

be the r e g u l a r

~£R(G)

is that

the d e g r e e

is u s u a l l y p r o v e d

integers".

PG£R(G)

above,

Aut V of G,

to the f o l l o w i n g

Let G be

p = deg

p:G~)

via the t h e o r e m

This c o n j e c t u r e general.

functors

already mentioned

are " a l g e b r a i c

representation. Then t h e r e

(~H,~K)

and any irrep

as a c o r o l l a r y

Conjecture:

41

)) .

subgroups

theorem,

the order

gEG,

to other

(

(K,H).

, p. II.4))

Xp(g),

( 15

see S e r r e

an i r r e d u c i b l e

representation

~p = PG"

(In p a r t i c u l a r

the s t a t e m e n t

in m a n y

examples,

of G.

above.)

but unproved

in

I01 The t h i r d t h e o r e m Theorem:

Let G be

a finite group.

linear combination, induced

is B r a u e r ' s

with

integer

up from c h a r a c t e r s

(For the proof,

see S e r r e

theory:

this

is to p o i n t

the c h a r a c t e r

table

R(G) as

computable,

In p a r t i c u l a r ,

is e f f e c t i v e l y

an a b e l i a n group,

can g i v e

explicitly

and d e s c r i b e

We w i l l p r o v e describing

a

First of d e g r e e

observe

the

that,

X:G --~) ~

.

dimensional since b o t h

g r o u p ~n of n characters

groups

homomorphisms

are

between

table

of

ring

of g e n e r a t e r s ,

as f u n c t i o n s

the i n t e g e r s

algorithm

computable.

Xi:G

~ ,

Indeed

k rij

, Siq.

table by

the c h a r a c t e r s each

n so g =e,

such is a g r o u p all g£G,

= 1 so the image of X is c o n t a i n e d th

roots

of u n i t y

in ~.

are g r o u p h o m o m o r p h i s m s finite, them.

a computer

one

for its c o m p u t a t i o n .

group G,

If G has o r d e r n,

is,

rijXk,

of the c h a r a c t e r

for any finite

(x(g)) n= x(g n) = x(e) (finite)

giving

inefficient)

1 are e f f e c t i v e l y

number

XiXj =

is,

is e f f e c t i v e l y

Not only can one say R(G)

the p r o d u c t s

the e f f e c t i v i t y

of G.

on r e p r e s e n t a t i o n

that the c h a r a c t e r

a set of g e n e r a t e r s

(admittedly

homomorphism then

it f o l l o w s

kq (Xi) = ~ SiqXq,

subgroups

the m u l t i p l i c a t i o n

free on a c e r t a i n

explicitly

and k - o p e r a t i o n s

of a finite g r o u p

describeable.

on G is a

out a fact w h i c h

of the l i t e r a t u r e

given

characters.

of c h a r a c t e r s

1 on v a r i o u s

, p. II-29)).

in m o s t

the group.

coefficients,

([417

not e v i d e n t

("recursively")

on i n d u c e d

Every character

of d e g r e e

The reason we m e n t i o n at least,

Theorem

Hence

G -~) ~n"

can e a s i l y

oneand

list all

in

I02 N o w let G b e

any group.

all the o n e - d i m e n s i o n a l associated

induced

procedure.

Doing

characters

characters this

of r e p r e s e n t a t i o n s

of G.

of t h e s e Xq-

with

coefficients

coefficients The i)

of a b s o l u t e

important

By B r a u e r ' s on this

ii)

Given

Given to check

facts

two

again

Set the c o m p u t e r E.g.,

first

of a b s o l u t e

the

a mechanical

we get a finite to l i s t i n g

set

[Xq}

all

integer

list all such c o m b i n a t i o n s value

~i,

then t h o s e w i t h

v a l u e <--2, etc. are now:

theorem,

each

irrep of G w i l l

of the Xq,

eventually

appear

it is a m e c h a n i c a l

to see if X is an irrep of G:

irreps,

X1 and k2,

test

if

it is a m e c h a n i c a l

if they are isomorphic: in a d v a n c e

the n u m b e r

the c o m p u t e r

how many

of c o n j u g a c y

generates

to see if it is i r r e d u c i b l e , irreps

and then c o m p u t e

and x(e)>0.

It is k n o w n compute

Thus

IndGx of G,

any l i n e a r c o m b i n a t i o n

(X,X)=I

iv)

7. of H,

H of G, c o m p u t e

list.

computation

iii)

a subgroup

for e v e r y HOG,

combinations integer

Given

test

if

irreps classes

(XI,X2)=O

when

or i.

G has - namely, of G.

a list of r e p r e s e n t a t i o n s , and stops

matter

checks

the right n u m b e r

each of

is found.

It w o u l d b e n i c e to a s s e r t

that e v e r y

dimensional

if B r a u e r ' s

Theorem

could be

irrep of a g r o u p G is i n d u c e d

representation

of some

subgroup

of G.

strengthened from a oneGroups

having

103

this p r o p e r t y monomial.

are c a l l e d monomial.

An example

semidirect p r o d u c t

i,j,k.

of a n o n - m o n o m i a l

of the q u a t e r n i o n

w i t h the cyclic group

Alas,

~/3~

not all groups

group

group H =

are

is the {+l,+i,~j,~k}

acting on H b y c y c l i c l y p e r m u t i n g

This p r o d u c t has an irrep of degree 2, b u t no subgroup

at all of index 2.

104

4.

Permutation

Let G b e

Representations

a group

a c t i o n of G on S permutations

s.

of S.

For g£G

applying

For gl,g26G,

S is c a l l e d a G-map

(finite as always)

is a h o m o m o r p h i s m

of S o b t a i n e d b y element

a G-set.

and S a f i n i t e

of G to the g r o u p

and s£S, w e w r i t e g(s)

the p e r m u t a t i o n

If two G-sets, f:S~TT

f is an i s o m o r p h i s m

f is a G - m a p w h i c h

Ring

set.

An

of

for the e l e m e n t

associated

to g,

to the

we h a v e g l ( g 2 ( s ) ) = ( g l g 2) (s).

is a m a p of sets

all g£G.

and the B u r n s i d e

S and T,

such that

are given,

f(g(s))=g(f(s)),

and S and T are i s o m o r p h i c ,

is an i s o m p r p h i s m

for

if

as m a p of sets - i.e.,

one-

o n e and onto. If S has n e l e m e n t s , and so the

concept

representation we do not

require

section

same as t h a t of p e r m u t a t i o n

discussed

n

to b e

ring B(G)

above.

Note that

one-one:

is to m i m i c

and the c o n s t r u c t i o n

of the B u r n s i d e

representation

is the

the m a p G - - ~ S

of t h i s

of g r o u p s

construction

of G - s e t

of d e g r e e n of G,

The object theory

the m a p G - - 7 A u t S is a m a p G --~S n,

the

of R(G)

linear representation w i t h the

for the p e r m u t a t i o n

t h e o r y of g r o u p s .

An o r b i t of the a c t i o n of G on S is any s u b s e t of S of the form

{gs I g£G~

for some s£S.

An o r b i t

is c l e a r l y d e t e r m i n e d

105 b y any element The action

in it, and S is the disjoint

is transitive

S is one orbit. with

union of orbits.

, or S is a simple G-set,

Equivalently,

given

any Sl,S2£S,

if all of there

is a g6G

g(sl)=S 2.

For one example

of a G-set,

itself by inner m u l t i p l i c a t i o n confusion

here,

The action

lets write

consider

(=conjugation).

To avoid

~g for the permutation

is then defined by,

for g,h£G,

that ~g is a group homomorphism. called

the action of G on

~g(h)

associated

= g-lhg.

Such automorphisms

to g.

Note

of G are

inner automorphisms.

G also acts on the set of subgroups Given gEG,HC-G, given He-G,

= [ g-lhg I g£G,

[ gEG i ~g(H)CH]

[g£G I ~g(h)=h,

Earlier,

take ~g(H)

for all hEH]

a class

of G was described.

is called

h£H~.

Some terminology

the normalizer

is the centralizer

of examples Given

of G b y conjugation.

of H.

of permutation

a subgroup

of H, and

representations

H of G, G acts on the set

of left cosets G/H = [gH I gEG ~ of H in G by left multiplication: gl(gH)

=

converse:

(glg)H.

This action

is transitive

and we have the

-

106

proposition: of the

form G / H

Proof: H =

Every transitive permutation for some

G/H :

Consider

[gH I g E G ] - - - ~

Corollary: elements

of S d i v i d e s

this

Pick

and a G - m ~ p

the n u m b e r

: gs.

It is e a s i l y c h e c k e d

and an i s o m o r p h i s m ,

of e l e m e n t s

says a b o u t the a c t i o n of G on i t s e l f b y

c o n j u g a c y class,

and so we h a v e that the n u m b e r

the o r b i t of an e l e m e n t

element divides

Let S,T b e G - s e t s . disjoint union

Then under

the o b v i o u s

For each of T.

This

of T, T n,

is its

of e l e m e n t s

action,

c a l l e d the sum

the

of

S+T.

G a l s o acts on the c a r t e s i a n p r o d u c t , g(s,t)=(g(s),g(t)),

inner

the o r d e r of the group.

of S and T is a G-set,

S and T and w r i t t e n

I

of G.

In that case,

to a g i v e n

i

in S, t h e n the n u m b e r of

multiplication.

conjugate

sES and let

the G - s e t G / H and the m a p

If G acts t r a n s i t i v e l y

Note what

on S.

S g i v e n b y ~(gH)

t h a t ~ is w e l l - d e f i n e d

of G is

s u b g r o u p H of G.

S u p p o s e G acts t r a n s i t i v e l y

[g I g ( s ) = s ] .

representation

g i v i n g the p r o d u c t

i n t e g e r n~l,

SXT

of S and T.

G acts on the n - f o l d

set is c o n s t r u c t e d b y t a k i n g

and i d e n t i f y i n g

S>~T, b y

symmetric power

all n - t u p l e s

for any p e r m u t a t i o n

0£S

n

,

of e l e m e n t s

107

(tl't2' .. "' t n )~ (to(l) ..... t o (n) ). class containing

(tl,...,tn)

(g(t I) ..... g(tn)).

= the equivalence

We write SymmnT

As in the case of linear ring out of all permutation B(G),

of G-sets

in B(G),

Si, m o d u l o

is equal to

makes B(G)

a ring,

as operations

hn,

- i.e.,

symmetric proof

powers

- more

The product, power

structure

is a k-ring

of ordinary

be v i a the embedding the n e c e s s a r y yet known.

Given orbits. Maschke's

See b e l o w

any G-set

Hence

sets,

operation,

the sum above, interpreted

to B(G).

and not just a pre-k-ring

sums,

- follows

from the fact

products,

and

and hold "naturally". for R(G)

Another - would

into a "ring of characters".

lemma for "characters"

and B(G)

But

is not

(p.l13).

S, S is obviously

for permutation

Theorem

sums,

as defined

to our proof of k-hood

of B(G)

relevant

ring,

[SI]+[$2],

h o l d amongst

analogous

The Burnside

the relation:

and the symmetric

identities

a

formal

the truth of all the indentities

that these

we construct

of all finite

give a k-ring

of

for this G-set.

representations

[SI+$2].

(The fact that B(G)

equivalence

class

representations.

of a group G, consists

Ei[Si],

G acts b y g(the

is easy

the sum,

representations,

(making the analogy:

as G-sets,

of

the analog of irreducible~simple).

108 Proposition: transitive set

Every G-sets,

as a b e l i a n

is t h e n u m b e r we n e e d

Lemma:

the

analog

Let U and V b e

gEG

g

-i

Ug

~:G/U ~

B(G)

so g - l h g

G/V.

This different

This

is n o t

quite

irreducible

one c a n c h a r a c t e r i z e Corollary:

Suppose

inverse

Since

there

isomorphism

g-lh-iVhgC-g-lugcV. of e l e m e n t s

as V.

Then

is n o n e m p t y

e£G b e

for

G/U

as the

subgroups

gives But Hence

HomG(G/U,G/V), if and o n l y

identity

= h~(U)

with

and G / V

is a m a p

k

To c o m p u t e

k,

= hgV.

are

if

Since

Hence

Pick ~ is

g-lhgv

= V,

so g -1 UgcV.

Schur

Lemma

a nontrivial of

the

element.

h U = U = eU.

all hEU,

as n i c e G-sets

where

in V.

the

a n y hEU,

isomorphism

and V are c o n j u g a t e

Proof:

of G.

= ~(hU)

holds

order). I

to

of r a n k k, G-sets.

the

Lemma.

Let For

is u n i q u e ( u p

free

of U is c o n t a i n e d

gV = ~(eU)

E V.

is

sum of

Furthermore,

transitive

to G/V,

so t h a t g V = ~(eU).

well-defined,

U

group,

subgroups

from G/U

conjugate Let

appearing

of S c h u r ' s

as a f i n i t e

... + S n.

G-sets

of n o n - i s o m o r p h i c

set of G - m a p s

Proof:

S can be written

S = S1 + S2 +

IS ~ of t r a n s i t i v e 1

Hence,

some

G-set

simple

I

- there

can be

map between.

But

G-sets.

isomorphic

as G - s e t s .

Then

of G.

G/U

h-iVhCU,

(hg)-ivhg

>G/V,

g

-i

UgCV

some hEG. has

g-lh-ivhg=V,

for

some

gEG.

The

Hence

the

same

(finite:)

so g

-1U g = V .

number

I

109

Corollary: subgroups

Rank(B(G))

= k = the n u m b e r of c o n j u g a c y c l a s s e s

of G.

As in the

first s e c t i o n

of this chapter,

representation

of G, we can a s s o c i a t e

G.

a homomorphism

This gives

but

for G = S 3, t h e r e

Consider

is not u s u a l l y o n e - o n e .

of e l e m e n t s .

Hence,

as groups,

= ~ 3, and so the m a p h e r e c a n ' t b e o n e - o n e .

representation

of G,

the c h a r a c t e r

of the

linear representation.

Proposition: associated

> R(G).

the c o m p o s i t e m a p B(G)~--> R(G)----~>CF(G), a s s i g n i n g

to each p e r m u t a t i o n associated

B(G)

of

are four c o n j u g a c y c l a s s e s of s u b g r o u p s

only three conjugacy classes

B ( S 3) = Z~ 4 b u t R(S3)

to e a c h p e r m u t a t i o n

a linear representation

of k - r i n g s

It s h o u l d b e n o t e d that this m a p E.g.,

of

Let S b e

a G-set

and X the c h a r a c t e r

linear representation.

n u m b e r of p o i n t s integer-valued

of S left

central

Then

fixed b y g.

function

for gEG,

of the

x(g)

is the

In p a r t i c u l a r ,

on G, t a k i n g

X is an

only nonnegative

values. Proof:

Immediate

from the c o n s t r u c t i o n

of the c h a r a c t e r

representation, Matrices

matri~s

l

s u c h as o c c u r

the o b v i o u s b a s i s

in such r e p r e s e n t a t i o n s

of the v e c t o r

space

e a c h r o w and c o l u m n c o n t a i n

that entry being

of a

"i"

involved)

(choosing

are p e r m u t a t i o n

o n l y one n o n - z e r o

entry,

110

Hence the map B(G) valued characters

~ R(G)

of R(G),

has image at most the integer-

and so isn't onto in general.

it isn't even onto the set of integer-valued example,

the group discussed

So far, permutation

The first new ingredient the composite characters

B(G)~

of the theory

is the analog

R(G)~>CF(G)

do not suffice

characters

(see,

for

on p. 103).

all of this is a straightforward representations

In fact

generaliTation

for linear representations.

of character

is not one-one,

to distinguish

to

unequal

theory.

Since

usual

elements

of B(G).

So the notion must be extended.

Definition: ~:

A super central

(conjugacy

classes

The set of super central ~I([U~)+~2([U~),

Given

a G-set

In shorthand

on G

of subgroups functions

is a map of sets

of G)

is a ring

(~1~2) ([U~)=~I([U])~2([U~)

S, the super character

representation H of G, ~s(H)

function

is the function

notation,

~s(H)

= I HS I

(by (~i+~2) ([U]) = denoted

SCF(G).

of this permutation

%~S' given by,

= the number of elements

~

for each subgroup

Of ~sES I h(s)=s,

all h£H~.

111

Note that ~SI+S2(U) Hence the map B ( G ) P

= ~S I(U)+~S2(U ) and ~SIS2(U)=~S l(U)~S2 (U).

CSF(G)

is a ring homomorphism.

One interpretation of ~s(U) Hom

G

(G/U,S).

~S (T) =

The notation generalizes

I HomG(T,S)

i

all UCG.

Two G-sets S and T are isomorphic

associated super characters Proof:

to apply to any G-set T,

Of course it is easy to compute ~s(T),

given the values of ~s(U),

Theorem:

is the number of elements in

One direction

if and only if the

~S and ~T are equal.

is of course trivial.

To prove the other,

it

is convenient to set up first a partial ordering on the set of simple G-sets, G/U < G/V

We write U <

using any of the following equivalent definitions: if and only if

HOmG (G/U, G/V) ~

if and only if

g-lug c V

if and only if

~G/v(U) / 0

some g£G

V if G/U
NOW suppose S = ~

nu(G/U)

and T = ~ m u ( G / U )

are given

G-sets, where the sum ranges over a set of conjugacy classes of subgroups of G with integral coefficients nu,m U. Suppose for all V c G, ~s(V) = ~T(V). 0 = ~s(V) - ~T(V) = ~ (nu-mu)~G/u(V),

Then

for all VC-G.

Suppose for

112

some U c G, n u # m U.

Let V 0 b e

the a b o v e o r d e r i n g ) .

a maximal

Then nu-mu=O

for U>V 0.

u n l e s s UO <, V_

so the

sum b e c o m e s

~G/v0(V0)#0,

n

, a contradiction.

=m V0

such

0 =

(in the s e n s e of B u t ~G/u(V0)

(nv0-mvo)~G/V0(V0) .

= 0

Since

H e n c e n u = m U for all UCG,

V0

so S = T.

The c a t e g o r i c a l Lemma exactly determines

reader

s t a t e s that k n o w l e d g e

S up to i s o m o r p h i s m .

h a v e to k n o w the w h o l e

Corollary:

Given

s h o u l d n o t i c e h e r e that the Y o n e d a

functor,

The m a p B (G) ~ >

a supercentral

of the f u n c t o r T / ~ - ~ H o m ( T , S )

This theorem

says t h a t w e d o n ' t

just its e f f e c t on o b j e c t s :

m

SCF (G) is o n e - o n e .

f u n c t i o n ~,

and an e l e m e n t g£G,

can be

a p p l i e d to the c y c l i c

s u b g r o u p g e n e r a t e d b y G.

Lemma:

This a s s i g n m e n t

a ring h o m o m o r p h i s m

Theorem:

There

gives

is a c o m m u t a t i v e

B (G) ..............

SCF(G)

~

diagram

of m a p s

SCF(G)--~CF(G).

of r i n g s

R (G)

-~ CF(G)

a

113

What's m i s s i n g here is the k-ring aspect of this construction. The obvious

k-structure

operations,

Yn(~(H))

generated by n ring

th

to put on SCF(G)

= ~(Hn),

powers

remains to b e p r o v e d on SCF(G),

B(G)~R(G) product G-sets

also,

so that one cannot

to SCF(G) .

super character

classes

The map

find any dot

We leave it to the has to do w i t h the

involved.

Just as in linear r e p r e s e n t a t i o n

Let G b e

.

w o u l d define a dot product

reader to figure out what the map B ( G ) ~ b R ( G ) dot products

What

But taking the transitive

of B(G)

an easy extension

k-

is a k-homomorphism.

p r e s e r v e d b y the map.

having

is a

: with this k - s t r u c t u r e

have not b e e n m e n t i o n e d

as an orthonormal b a s i s

on B(G),

Then SCF(G)

line is a k-homomorphism.

SCF(G)

isn't one-one,

on B(G)

of H.

is the c o n j e c t u r e

the m a p B(G) ~

Dot products,

where H n is the subgroup of G

of elements

( p. 54) and the b o t t o m

is to take as Adams

tables. a

theory,

Here are a couple.

cyclic group of prime power order p.

of subgroups

of G are G itself,

and {e].

classes of subgroups ---I simple G-sets

one can construct

G/G G/{e}

~e]

G

1

1

IGI

0

Conjugacy

114-

Let G = S 3, w i t h C2, C 3 of orders

subgroups

2 and classes

{e]

simple

!

S 3,

[e],

and the cyclic

3. of s u b g r o u p s

C2

C3

S3

i

f

i

2

0

0

0

$3/C 3

2

0

$3/C 2

3

1

S3/{e]

6

j

G-sets

subgroups

115

5.

The Group Algebra Approach

Let G be a finite group. is the set of sums

, of dimension c ~Cgg ( ~c

=

~

~ g6G Cgg ,

The group algebra of G, ~ [G~, c 6C. g

IGi, u n d e r ~ C g g

+ ~dgg

(CCg)g , and an algebra,

g) (~dhh)

=

g

~ (Cgdh)k gh=k

C[G~

is a vector space over

= ~(Cg+dg)g

and

under

(the multiplication

induced by

the group operation.

Given a representation C-algebra homomorphism, by,

p (~Cgg)

Theorem:

= ~Cgp

p:G - ~ A u t

also denoted

V, there is induced a p, p:~ [ G 7

(g)

Let Pi:G ~ > Aut V.,1 i6I, be a complete

representations

~ End V,

of G.

Then the sum p = ~ p i : C ~ G ~

set of irreducible ~-~End i£Z

is

an

Proof:

isomorphism.

Suppose there is an element

Then p i ( S C g g ) map on V., 1

particular,

g£GCgg~ £C[G~ with p ( ~ C g g )

= 0 for each i, so that ~ C g P i ( g )

for each i.

irreps of G, ~ c In

V.1

g

= 0

is the zero

Since I runs through a set of all

g must induce the zero map on every G-module.

~ [G~ is itself a G-module

(by left multiplication)

116

so

(~c

g

g)y

= 0 for all y£C[G].

and the l i n e a r Hence

independence

p is o n e - o n e .

dimension

But

(Corollary,

This theorem the t r a d i t i o n a l

of the g's

since C[G]

p. 78)

reveals

N 2 w i t h M = N I ~ N 2.

of w h i c h

and o b s e r v e s

is,

S i n c e the r e p r e s e n t a t i o n is t r a n s p a r e n t

about r e p r e s e n t a t i o n s F r o m this p o i n t of G, G ~ >

Aut V,

homomorphism)

(over C:)

our a p p r o a c h

theory.

There,

and

one

t h a t it is s e m i - s i m p l e : is a s u b m o d u l e

Theorem,

the M a s c h k e

a special case

is a sum of m a t r i x

t h e o r y of a sum of m a t r i x

this produces

the m a i n

facts

of G. of view,

a representation

(group h o m o m o r p h i s m )

is the same as a r e p r e s e n t a t i o n

of C[GT,

a v o i d e d this a p p r o a c h f o r m i n g R(G),

so an i s o m o r p h i s m .

of course,

is that any s e m i s i m p l e ~ - a l g e b r a

all g.

V. h a v e the same l

and s u b m o d u l e N I, t h e r e

T h e n one i n v o k e s W e d d e r b u r n ' s

algebras. algebras

and ~ E n d

p is a l s o onto,

For ~ [G7 this

~-c g = 0, g

i m p l i e s t h a t c =0, g

a p p r o a c h to r e p r e s e n t a t i o n

for any ~ [ G ~ - m o d u l e M,

Then

the c o n n e c t i o n b e t w e e n

s t a r t s w i t h the a l g e b r a C[G~

Theorem.

T a k e y = le.

~[G~---~> End Vo

The r e a s o n

(aside from O c c ~ ' s

razor)

w e w a n t to t a k e t e n s o r p r o d u c t s

(~ - a l g e b r a t h a t we h a v e is t h a t

in

and e x t e r i o r

p o w e r s or r e p r e s e n t a t i o n s . Consider f,g,h:V ~gV

a vector inducing

f(vl)^...^f(Vn)

s p a c e V and l i n e a r t r a n s f o r m a t i o n s fn ,g n ,h n : Anv---~ Anv

and s i m i l a r l y

n for g , hn).

(by fn (v 1 ^...^Vn) Suppose

f = gh,

=

the

117

composite A

n

is not

of maps.

Then

"additive".

fn = g n h n .

Even

if f=g+h,

it can still h a p p e n

that

shows).

of groups,

groups

Thus

a map

doesn't

induce

the e x t e r i o r

Given

representations

there

is no n a t u r a l Another

[GI~ = C[G2~. the t h e o r e m

their

For example,

dihedral

4-group

b y the t h e o r e m groups,

their g r o u p

and the g r o u p

algebras

as k-rings,

they d i f f e r

is that

irreps G

to ~

group

of

? End V

a pair

(E.g.,

groups

algebras this

are

tables

though

is true

Again

of these

to

R(D4)=R(H) .

out on p. 95).

for

Or take the

u n i t s H.

isomorphic

In fact as rings

(as p o i n t e d

is, u s i n g

two a b e l i a n

of o r d e r 4.)

character

are each

C[G3

Hence group

of

algebras

of a b e l i a n

of q u a t e r n i o n

and the k n o w n

C G ~ G ~ ~> ~ • E n d ( ~ 2) .

~

isomorphic

rings m a y differ.

above,

a map

End(V~W).

isomorphic

and the c y c l i c

g r o u p D4,

induces

if G is abelian,

isomorphic

representation

any e x a m p l e

~ End Anv.

~I~

fact that

of the same order have

the K l e i n

map

of the p r o b l e m

above and the

all v£V,

are not the only p r o b l e m .

GI,G 2 may yield

all o n e - d i m e n s i o n a l , groups

~[

But

End V, ~_~--> End W of a C - a l g e b r a

C-algebra

groups

V,

of ~ - a l g e b r a s ,

powers

~i ~ >

indication

nonisomorphic

a map

f(v)=g(v)+h(v),

(as almost

G--TAut

a m a p of C - a l g e b r a s ,

Of c o u r s e

i.e.,

fn ~ gn + h n

G - - 7 Aut Anv, b u t

A n is a functor.

i.e.,

But

118

Let G b e

a group.

to a sum of m a t r i x

elements

By the t h e o r e m

rings,

e. 6 ~ [G],

above,

is i s o m o r p h i c

-[-7- End V i, we can ask for t h o s e i6IrrepG

i6IrrepG,

satisfying

e.=l on V.,

1

and v., ]

C[G]

1

j~i.

These

elements

are u n i q u e l y

0 on

1

characterized

by

the p r o p e r t i e s i)

e i y = y.e.

all 76(12 [G],

all i

(centrality)

1

ii)

e.-e. = 0 l 3

i#j

(orthogonality)

e.. e. = e~ 1

ill)

Theorem:

1

(idempotence)

1

~e. = i i i

Let X

i

(completeness,

.th • b e the i irrep of G.

positivity)

Then

i e.

=

1

Proof:

deg X

J

/Gi

ki(g),g

g£G

Let Y. E ~ [G] d e n o t e l

the e l e m e n t

For each G-module, p:G.--> Aut V,

Y.

g i v e n b y the a b o v e

acts on V by,

for v£V,

l

i Yi(v)

=

deg X

~

k l(g)*

pg(V)

IGI The t r a c e of this e n d o m o r p h i s m deg X T r y (Yi) = IG 1

can be c a l c u l a t e d

i

= degree

~_" xl(g) * Tr(Qg)

X

i

(kl,P)

as

formula.

119 Furthermore

the map Y1. : V ~ >

deg X

PhYi (v) =

V commutes

i

"

i GI

- ~ - x l (g) *Oh gEG

with

Ph for each h6G:

(pg(V))

i deg X

~

IGI

k I (h-lq) *p

q~G

(v)

(here taking

q = hg)

(here taking

q = gh)

q

and i YiPh(V)

deg X IGl

=

~ x!(g).p (Ph(V)) g gEG i

deg;Gl ~

~

x i ( q h - l ) p q (v)

q6G Since

X

i

is a c e n t r a l

H~nce, is a scalar

by

function,

Schur's

multiple

Lemma,

of the

these

two sums

are equal.

if V is irreducible,

identity:

Y. = rI.

Y~:V----~V 1

Applying

the

1

calculation

of the trace,

i V = X , r=l and Y.

Tr(Yi)

is the

' = r-deg V = deg x i °(XX,V). i If V ~ k , r=0.

identity.

Hence

If Y.

1

is the p r o j e c t i o n V of G, p r o j e c t s this

is what

operator

constructed

which a p p l i e d

onto y.i (V)cV,

Y., 1

operators them

t o any r e p r e s e n t a t i o n

i

the X - i s o t y p i c a l

the e. 's o b v i o u s l y l

The o p e r a t o r s Symmetrizing

1

do,

for i£Irrep

the two m u s t be

G,

are c a l l e d

(after the Rev.

in the algebras

component.

C[Sn]

Alfred - see

Since

identical.

Young

Y o u n g who

~39%~7J

first

120 One a p p l i c a t i o n of the

symmetrizer,

of these Y0'

operators

of the u n i t r e p r e s e n t a t i o n

symmetric

group

S . n

symmetric

power

of V has two e q u i v a l e n t

i)

Symmnv

Let V be

The n a t u r a l subspace

identification

of V ®n,

Isomorphism

Symmnv

Theorem

we have domain(Y0)

Image(Yo) = V ®n,

We next p r o v e later.

Given

a Young

for the a s s o c i a t e d

Lemma:

Let Y b e

G-modules the sum, Proof:

as l i n e a r

This

is u n i q u e l y

constructed

operator

is a simple

corollary

a sum of i s o t y p i c a l

th

exterior

out of an irrep of

w h i c h will be u s e f u l Y,

lets w r i t e

of Y : V

~V,

and

Y(V) operator

transformation

transformation,of

the n

and

of V ®n.

proposition

symmetrizing

as

By the F i r s t

Similarly

o n t o map V ~ >

V , W the l i n e a r

fixed b y

~6Sn]

= Domain(Yo)/Kernal(Yo)

g i v e n b y the action

a Young

left

from the fact that,

image of Y0"

symmetrizing

YV for the e n d o m o r p h i s m

the n - f o l d

factors).

[ (x-~x) I x6V ®n,

and q u o t i e n t

a technical

Then

of V ®n b y W, w h e r e W is the

Ker Y0=W.

as s u b s p a c e

of a

(n factors)

of these comes

is the

case

definitions:

(permuting

n

and any other type of p o w e r

S , appears both n

YV'

W =

space.

of V ® . . . ® V

of S

S y m m n v = the q u o t i e n t subspace

power,

any v e c t o r

= the s u b s p a c e the action

ii)

is in the s i m p l e s t

of G.

Given

YV~w:V@W----~V@W

Yv:V

-~V and Yw:W.

is ~ W

.

of the fact that e v e r y G - m o d u l e

components.

721

Lemma:

Given

a G-map T:V~>

linear transformation

T'

W of G - m o d u l e s ,

so t h a t the

V

there

is a u n i q u e

following commutes:

~W

YW

T'

Y(V)

Proposition: natural

.....~> Y(W)

Let HT-G and V b e an H - m o d u l e .

Then there

is a

isomorphism

(Ind~ V) G

vH--~

Proof:

R e c a l l V H d e n o t e s the e l e m e n t s

h6H,

thus the i m a g e of the Y o u n g

unit

representation

Consider

operator

Similarly

1 H of H.

of V left fixed b y

each

Y associated with

the

for the s u p e r s c r i p t G.

the c o m p o s i t e m a p VH

where

m

the

the second

2 V

~

first m a p

Y

Ind~ V

is i n c l u s i o n

is V--~ C[G]

®

V,

. dGv .G ~> (In H )

of an i s o t y p i c a l

v ~

l®v.

s u b s p a c e and

This composite map

is an

[H] isomorphism:

Indeed,

b y the f i r s t

lemma,

the f o r m a t i o n of i n d u c e d r e p r e s e n t a t i o n s , w e can r e s t r i c t

our a t t e n t i o n

V is n o t the u n i t (Ind V,

IG) I G =

it is a d d i t i v e

)i G =

of H, V H = 0 and also

of

in V,

to the c a s e of i r r e d u c i b l e V.

representation

(V,Res~l G~

and the a d d i t i v i t y

so

If

(Ind V) G =

(V, IH)! G = 0 1 G = 0 and any m a p

122 f r o m 0 to 0 is an i s o m D r ~ h i s m . (by F r o b e n i

reciprocity)

so the c o m p o s i t e

is a m a p

s h o w t h a t the m a p

In c a s e V = 1H, V

(IndV) G =

1 H - - - ~ l G.

isn't zero.

(i G + Hence

This

= V,

and

(other)) G = 1 G and it is s u f f i c i e n t to

follows easily by unravelling

the d e f i n i t i o n s

of i n d u c e d r e p r e s e n t a t i o n

this case.

final d e t a i l s

The

H

and Y o u n g o p e r a t o r

are left to the r e a d e r

in

as an

a exercise.

Here of g r o u p

is a final h i s t o r i c a l note. representation

algebras,

but with

one c o m p u t e s

theory were

done,

the s o c a l l e d g r o u p

its g r o u p d e t e r m i n a n t ,

variables

X , g£G. g

consider

the m a t r i x

representation

The o r i g i n a l

of G

® is the d e t e r m i n a n t

Given

the

assigned

®,

"generic

not w i t h

determinant. a polynomial element

this element

~ g£G

f r o m our p o i n t of view,

to the i r r e d u c i b l e

factor occuring

the n u m b e r of t i m e s

conjecture

t h a t an o r i g i n a l

a g r o u p G,

function

Xgg

in

in C[G~"

of t h a t space).

Looking

w e see t h a t the i r r e d u c i b l e

of ® c o r r e s p o n d

interesting

Given

The p r o b l e m w a s to r e l a t e

the s t r u c t u r e of G to the p r o b l e m of f a c t o r i n g ®. this

semisimple

in the r e g u l a r

(in t h e o b v i o u s c h o i c e of b a s i s of t h i s m a t r i x .

investigations

representations as its degree.

of G,

at

factors each

It is

i m p e t u s to the t h e o r y was D e d e k i n d ' s

( c o m m u n i c a t e d to F r o b e n i u s

in 1896)

that the n u m b e r of

123

linear

factors

commutator

of ® is equal

subgroup

For a h i s t o r y see H a w k i n s

([22]).

(a s i m p l e

to the index fact

in G of its

for us - see p.92).

of the early w o r k

in r e p r e s e n t a t i o n

theory

CHAPTER

III

The Fundamental

i.

of the

The

of k-rings certain rings

of all

Before

n letters,

the

say t h e

such that

asserts

X-ring

symmetric

o is a c y c l e

of t h e

the

constructed

about

combining

this

object,

symmetric

letters

o =

group

if t h e r e

- indeed

under

breaks

a number

of o r b i t s ,

=

structure

(k I ..... k n) w h e r e

different

structures:

=

(i,i,i)

=

(2,1)

~ =

(3)

and a

the

example,

corresponds corresponds

corresponds

Let

~ be

to the

o(m)=m, that

associated

(23) (i),

to the

elements

set

{I ..... n]

is a c y c l e .

e = o =

~ =

~ES n

o=(il,.,iq) (ji,.,jr)..

of S 3 h a v e

and

one

partition

element

OT =

{I .... n}

element

of the cycles

elements

of S n-

for m n o t

every

of w h i c h

the elements

to the

an e l e m e n t

of o, t h e

sizes

of

{i I .... i q ] C

cycles:

action each

to recall

of p e r m u t a t i o n s

n

of d i s j o i n t

t h e kl are t h e

For

S

Observe

of o is the

decomposition. cycle

I,

it is n e c e s s a r y

is a s u b s e t

" .. .,lq). (ll,

..(k I ..... k s )

The cycle

is an i s o m o r p h i s m

all t h e r e p r e s e n t a t i o n

..... O ( i q ) = i I a n d

as a p r o d u c t

into

there

in C h a p t e r

1 , 2 , 3 ..... n.

can be written

up

Theory

groups.

of length ~

Write

that

A discussed

by

~(ii)=i2,~(i2)=i3

ij .

THEOREM

Group

we construct

a few facts

FUNDAMENTAL

of t h e R e p r e s e n t a t i o n

in q u e s t i o n

between

k-ring

THE

Theorem

Symmetric

theorem

:

~

in t h e

one of three

(i) (2)(3)

(12) (3), 2

(123)

=

(13) (2) and

T 2=

(132)

125

Proposition: h a v e the

Proof:

Two elements

same c y c l e

Suppose

as a p r o d u c t

of S

are c o n j u g a t e

n

structure.

~ and J' h a v e the

of d i s j o i n t

a b o v e the other,

if and o n l y if t h e y

cycles

so t h a t c y c l e s

same c y c l e

in the

structure.

Write

following pattern:

of equal

each

one

length correspond:

(i I ..... iq) (Jl ..... Jr ) "'" (kl ..... ks)

(ll,

Let

T 6S

n

...

')...(k ,i ) (3' 1 ..... 3r

b e the c o r r e s p o n d e n c e

I ....

g i v e n b y the v e r t i c a l

T (il)-' --±i' '. .--' T (jl)-'' --31, ...etc. ' . . T. (lq)--l~ ~' are c o n j u g a t e .

Corollary:

Then

The c o n v e r s e p r o p o s i t i o n

The n u m b e r

of c o n j u g a c y

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

classes

representations

for the c o n j u g a c y c l a s s of S

-i

~ T=u so o and

|

is clear.

of S n,

and h e n c e

also

of S , is e q u a l to n

a

n

[~

, for ~ a p a r t i t i o n

of e l e m e n t s

The o t h e r c a l c u l a t i o n we w i l l n e e d (which n u m b e r w i l l

T

lines:

of n°

H e n c e f o r t h we a d o p t the n o t a t i o n

of [ ~

'k 's)

of

cycle

s t r u c t u r e ~.

is the n u m b e r

also be denoted by

[~.).

of n,

of e l e m e n t s

126

Proposition:

Suppose ~ =

(i~2 ~ . . . n ¥)

is a p a r t i t i o n

of n.

Then

n' t

Proof:

Given

the c y c l e

(-) (-) ... (-) ( - , - )

s t r u c t u r e ~:

(-,-)

let us c o u n t the n u m b e r n: w a y s to w r i t e one-cycles

can b e

rearranged

arise.

.........

to fill

(-,-,

. . . . -)

in the b l a n k s .

1,2 ..... n.

However

in ~: w a y s g i v i n g the

d i v i d e b y ~:

the ~: ..... y: t e r m s

have

of w a y s

in the n u m b e r s

of S , so we m u s t n

l e n g t h q,

... ( - , - )

to r e m o v e

Furthermore

are

the f i r s t same e l e m e n t

the d u p l i c a t i o n .

Similarly

in e a c h of the c y c l e s of

any one of the q e l e m e n t s can be l i s t e d

to d i v i d e b y q for e a c h c y c l e

There

of l e n g t h q.

first

- so we

H e n c e the

f a c t o r l ~ 2 ~ . . . n Y.

We now construct n=0,1,2,.., recall,

let R(Sn)

R(S0)=~).

X-structure

on e a c h R(Sn)

as follows. s u b g r o u p of S m

First n+m

of 1,2 ..... m.

is a p a i r i n g

observe that n

ring of Sn

(where,

f o r g e t the m u l t i p l i c a t i o n

and c o n s i d e r

by taking S

as p e r m u t a t i o n s

For e a c h i n t e g e r

b e the r e p r e s e n t a t i o n

F o r the while,

The o u t e r p r o d u c t

S

the k - r i n g R(S).

it as just an a b e l i a n group.

R(Sn)XR(Sm ) ~

R(Sn+m)

S ~ S can b e c o n s i d e r e d n m

to b e p e r m u t a t i o n s

of n + l , n + 2 ..... m,

and

and S

n+m

defined

as a

of 1,2 ..... n,

as p e r m u t a t i o n s

In d i v i d i n g up the n + m s y m b o l s p e r m u t e d b y S

n+m

127

into

one

set of n a n d a n o t h e r

the division

can be

of c o n s t r u c t i n g of S

done

set of m,

there

in ~ n ~ m ) w a y s , b u t

an i n j e c t i o n

S ~ n

S

m

is s o m e

any two

--> S

n+m

ambiguity

such

-

ways

give c o n j u g a t e subgroups

n+m Given

~n ~ ~ m

now

elements

£ R(Sn~

Sm)

~ n 6 R ( S n ), ~ m £ R ( S m ) , c o n s i d e r

(see d e f i n i t i o n ,

p.

).

the

The

element

outer

product

~n ~m o f ~n and ~m i s t h e e l e m e n t o f R(Sn+ m) g i v e n b y

~n~m (Note t h a t

S n+m = Ind S ~ S n m

since

any two of our ways

Sn+ m are conjugate, the way

chosen.

is j u s t

(n,~)~?

In t e r m s described and ~

m

:S

X

21

,X

41

F o r n=0, ~+~+..+~

32

,X

221,~nda (using

R(Sn)m

(n s u m m a n d s ) ,

n

~ ~

m

two

S < S n m

into

~ n ~ m £ R ( S n + m ) is i n d e p e n d e n t

the map

Given

R(Sm)---~

and ditto

R ( S n + m)

for m = 0 .

the product

representations

:S ~ S - - 7 A u t ( V ~ g W ) n m

of

~

n

can be :S--~ n

Aut(V)

and

~_~ (V~gW)) . -S U S n m

the X

of i m b e d d i n g

representations,

C[Sn+m]

an e x e r c l s e , ,X

element

of a c t u a l

~ Aut

respectively then

)

-----2A u t ( W ) ,

~ :S • n m n+m

As

the

as f o l l o w s .

m

(~n ~ ~ m )

311

reader denote

the notation

is i n v i t e d characters

to show that

2

•

~,~ of S 2 , S 3 , S 5 , S 5 , S 5 , a n u

of t h e c h a r a c t e r

2 21 41 32 221 311_~II~ X X = X + X + X + X

if k

tables

S5

128

Definition: R(S) =

- the sum as abelian for all n , m ~ 0 ,

Proposition:

Proof:

~=0

groups.

R(Sn )

The outer p r o d u c t s , R ( S n ) ~ R ( S m ) ~ >

induce a m u l t i p l i c a t i o n

R(S)

axioms

commutativity

for multiplication,

Commutativity

follows

follows

~

S

n+m

ring w i t h identity.

are a s s o c i a t i v i t y

and

and the d i s t r i b u t i v e

law.

from the fact that R ( S n ~ S m) is isomorphic

to R ( S m X Sn ) in the obvious way, S ~ S n m

on R(S).

is a graded c o m m u t a t i v e

The only n o n - o b v i o u s

R~Sn+m ~

and the two subgroups

and S ~ S c- S are conjugate. m n n+m

from the fact that the operation

Distributivity

of inducing

representations

is additive. To show associativity,

it is enough to show, given elements

~ n £ R ( S n ), ~ m £ R ( S m ), and ~p£R(Sp) S - d n+m+p in Sn~ Smx S

= (~n~m) ~p

We will

just p r o v e

Writing

this p r o p o s e d

S I d n+m+p n S xS n+m p

that

p

the left-hand

(~n~ ~m ~ ~p)

equality,

=

~n (~m~p)

the other b e i n g

equality out in more detail,

S _ n+m ( Inds x S n m

(~n ~ m

) x ~p)

=

similar.

we get

S Inds~s~sn+m+P ( ~ n m p

~m~p)

129

Let o6S (p. ~

n+m+p

and apply the formula for induced characters

) to both sides. (n+m+p):

i

n: m: p:

[o]

The right hand side yields

~i×~2~3 ~ S~Sm~S p

<(~n ((71)C~m(o2)C~p(o3 ) j

oi~a2~o3~ ~ in Sn+m+ p The left hand side yields (n+m+P): (n+m) : p'

1 [o]

~ I/ Sn+m 1 > Inds S (~n ~m ) ~p(°3) Oq~O3£Sn+m~< Sp n m / / oqXO3~o in Sn+m+ p

where

Sn+m, (n+m) : Inds m m ~ n~ ~m ) = n m n'. m:

1 [°q]

~-

/\

~n

(Ol)~m(a2)h/

al~ °2£Sn~Sm o l ~ 2 ~ q in Sn+m

Given a typical element o iAo2~c 3 6 Sn Sm Sp , the term ~n(Ol)~m(a2)~ (o3) is counted once in the right hand side, but on P the left side this term is counted once for each Oq~Ol~G 2 in Sn+m. Hence the factor [Oq] enters to eliminate the duplication. the two sides are indeed equal, proving associativity.

Thus

130

We n o w c o n s t r u c t fundamental given,

theorem.

and let V b e

on the v e c t o r O6Sn,

the map ® : R ( S ) ~

Let a r e p r e s e n t a t i o n any v e c t o r

space

o(W~Vl~>

space.

WimV~V~q..

~)V

...C~'~Vn)=O(w)C~v o

Let W(V)

be the s u b s p a c e W(V)

=

(WdgV C~n)

F o r example,

is the n - f o l d

alternating

representation W(V)

The c o n s t r u c t i o n Given

a m a p T:V I - ~

iwd~ T n : W d ~ V ~1 n action

The g r o u p

symmetric of S

n

W(V),

-7 W 4 ~ V 2~ ] n

acts

of V) by,

for

fixed u n d e r

one-dimensional

power

of V.

(the n o n t r i v i a l

this

representation

If W is the one-dimensional

power.

for fixed W,

This m a p

is f u n c t o r i a l

[W~VI6~n ) Sn

W(T):

W(VI) ~ W ( V 2 ) .

in V.

V2C~] n and h e n c e

is c o m p a t i b l e

: (W~JV 1~] n.) Sn~ 2

as we m i g h t write,

then

w i t h the

a map

IW ~9 TCW n

or,

n

n

V 2, w e h a v e T C ~ n : V l ~ n ~

of S , so g i v e s n

S

be

n

-i GO v -i ~j ...C~)v (i) o (2) ~-l(n)

is the e x t e r i o r

V ~

in the

Aut W of S

(n c o p i e s

if W is the t r i v i a l

of S , W(V) n

representation),

S ~> n

of ZU_~V ~-gn of v e c t o r s S

action:

A involved

(W~V2~n)

Sn

131

S u p p o s e that V 1 = V 2, and t h a t the t r a n s f o r m a t i o n diagonalizable

and has

eigenvalues

t h a t T r a c e ( T ) = t l + . . . + t k. monomials

t I ..... t k,

The e i g e n v a l u e s

in t I .... ,tk and s i m i l a r l y

(k=dim V)

T is so

of T ~gn are n t h - d e g r e e

for i w ~ f ~ / n .

Recall

the

elementary Lemma:

Let S:U----)U b e

a linear transformation.

subspace UI~ U satisfies the r e s t r i c t i o n a c t i n g on U. Hence

of d e g r e e n, w i t h

following

of W(T)

are m o n o m i a l s

reason:

to a c h a n g e

be unaffected by Finally,

of b a s i s

o f W(T)

of the s y m b o l s

of V.

But

Then,

iI ik tI ... t k has d e g r e e n and so c a n n o t t h a n n of the v a r i a b l e s a monomial would

t I ..... tk.

o c c u r in W(T)

of p o w e r s

of the

symmetric

a l = t i + . . + t k ..... a k = t l . . t k is i n d e p e n d e n t s u p p o s e k > n.

the a n s w e r w o u l d

change.

in t e r m s of e l e m e n t a r y

Indeed,

t I ..... t k

since the g i v e n d a t a

are c o o r d i n a t e - f r e e ,

such a c o o r d i n a t e

This

t I ..... t k for the

we c l a i m t h a t the e x p r e s s i o n

f u n c t i o n s W(T)

same p a t t e r n

of S

of d e g r e e n in

in t I ..... tko

in the v a r i a b l e s

any p e r m u t a t i o n

and the c a l c u l a t i o n

as k ~n.

of

is a h o m o g e n e o u s p o l y n o m i a l

integer coefficients,

is s y m m e t r i c

corresponds

the e i g e n v a l u e s

B

and so the t r a c e of W(T)

polynomial

Then

some

of S to U 1 are a s u b s e t of the e i g e n v a l u e s

the e i g e n v a l u e s

t I ..... ~ ,

S(U I) c~ U I.

Suppose

symmetric

functions

of k = d i m V,

still,

as long

any m o n o m i a l

involve nontrivially more

S i n c e W(T)

is s y m m e t r i c ,

such

if and o n l y if a m o n o m i a l w i t h

i I ..... ik o c c u r s

in W(T)

involving

just

the

132

the variables t I ..... tn. of monomial

symmetric

Hence the expression of W(T)

functions

is independent of k ~ n, and so

likewise for the elementary symmetric Thus,

function ®(W)

Lemma:

functions.

after all this work, we have a map:

representation

in terms

given a

S --9 w in R(S ), there is an associated n n

symmetric

£ A . n

®(Wl(*~W2) = ®(WI)

@(W2)

*

Proof: (Wl~9 W 2) Since S

n

=

(

(Wl~

in) ~

)

(W2~) V ~n)

for any V.

acts independently on the two factors

(WI~ v~n) S n ~ Hence,

v

( W 2 ~ V ~ n ) Sn

=

(

(WlfmvC~n) ~

(W2~Vd~n)) S n

for a linear transformation T,

Tr(WI+~JW2(T)

)

=

Tr(WI(T)(~gW2(T))

and this latter is

equal to Tr(WI(T))+Tr(W2(T ))

Corollary:

Lemma:

I

I

® gives a well-defined map R(Sn) --> A n .

® is multiplicative.

I.e.,

product of &n£R(Sn ) and ~m£R(Sm), product of symmetric

functions

if ~n+m E R(Sn+m)

is the outer

then ®(~n+m)=®(~n)G(~m ), where the

is taken in A = ~ A

n

.

133

Proof:

We can assume ~

~n:Sn ~

and ~

n

are actual

m

representations:

Aut W n, ~m:Sm---~ Aut Wm, ~ n + m : S n + m - o

V be any vector

Aut Wn+m.

Let

space.

W n + m ~ V ~w'n+m

=

= Ind

Ind

(Wn~ W m ) ~

~h+m

S ~ S

•

n

(Wn~WmC~)(ReSs

m

V ~n+m)

(by Froben ius

)

n+m C~n ~

Reciprocity)

v@Qm

= Ind

( Wn~Wm~)V

)

= Ind

((WnC~V CWn) ~_P(Wm~)v~)m))

Hence S v~m) i n+m

S (Wn+mC~ v~n+m)

n+m

=

v~n)

:

This

isomorphism

S

Wn+m(V)

if T is any linear it ind~ces

ind((Wn6~V~n

v~m)

Ca (Wm

= Wn(V)~_JWm(V )

operator

an equality

of a tensor p r o d u c t

n

) ~(Wm~

with

Wn+m(T)

S

m

is functorial

eigenvalues = Wn(T)~Wm(T)

is the p r o d u c t

t I ..... tq, q "

give a map of rings ® : R ( S ) ~ > A .

so

n+m,

Since the trace

of the traces,

this gives

I

®(Wn+ m) = ®(W n)®(W m) .

Hence the maps ® : R ( S n ) ~ >

in V,

An, n=0,1,2 .....

can be added up to

134 Lemma: Proof:

®:R(Sn)--->An One b a s i s

is onto,

for A

is

for each n~0.

{h

I ~ ~n].

Such an h

n S n is ~ ( I n d s k ~ s k ~

..~Sk

~

i) = ® ( t h e

outer product

= h k l h k ..h k 2 n

of the unit

n representations

Corollary: Proof:

®:R(S

of Skl ..... S kn )"

n

)~:)A

is o n e - o n e

n

An onto h o m o m o r p h i s m

the same

finite

rank m u s t

An i m m e d i a t e

corollary

(it is c o n v e n i e n t

between

two free a b e l i a n

of the

Indeed,

are c l e a r l y

integer-valued,

of

m

fact that

of the g r o u p s

functions.

groups

also be one-one.

for each n, R(Sn)

to think of ® as the i d e n t i t y map),

fact that the c h a r a c t e r s

basis

for all n~0.

the characters

S

-i

®

n

is the

are all i n t e g e r - v a l u e d S n

(hkl ""

and they give

hkn) = IndSkl''Sknl

an integral

for R(Sn).

Thus we h a v e rings.

Since

that ® : R ( S ) r - - ~ A

A is also a k-ring,

a corresponding additional

shown

k-structure

structure,

we have

is an i s o m o r p h i s m

the i s o m o r p h i s m

on R(S),

and t a k i n g

the m a i n

= A

result:

of

® induces

R(S)

with

this

n

135 Theorem:

The Fundamental

of the Symmetric

Theorem

Group).

of the Representation

The map ® : R ( S ) ~ > A

Theory

is an isomorphism

of k-rings.

The induced integers

k,n~l

k-structure

and ~6R(Sk),

these operations

plethysm).

applied

k-rings,

R(Sn),

on R(S),

to a representation the induced

guess

&:S k

2V

to

Classically

were referred

from the k-structure

n~0, which

is to describe

It is a reasonable

constructing

for example,

an element hn(~)E(Rnk).

(to be distinguished

One problem

explicitly.

assigns,

on A, so b y extension,

to as outer p l e t h y s m of the individual

on R(S)

is called

inner

the outer plethysm that the operation is performed

representation

h

n

by first

of the wreath product

S n [ S k ] ~ ? V ®n, and, using the n a t u r a l i n c l u s i o n Sn[Sk]CSnk, inducing verify

to a representation

this explicitly.

reasonable

algorithms

Few calculations

The point pass

Another

have b e e n made

and forth,

ring R(Sn)

on one hand.

and the group

(=the group

of k-operations

But we have been unable

outstanding

for computing

of the fundamental

freely b a c k

representation

of Snk"

outer

problem

(or inner)

is to find plethysm.

( see Littlewood ~ 2 7 ~ 8 ~

theorem

is that

of symmetric

functions

"of weight n")

).

it allows us to

for each integer n, between (=the ring of characters

to

the

of R(Sn)) of weight n

on the other.

On

136

each

s i d e of t h e

one c a n

do,

involved, relate this

equality

a number

and

these.

chapter.

some

there

of o b v i o u s

"canonical"

This project

are

a number

bases

for the

elements,

is c a r r i e d

of c a l c u l a t i o n s abelian

group

and the g a m e

is to

out

in t h e

rest

of

137 2.

Complements

Let n ~ l

and C o r o l l a r i e s

be

the i s o m o r p h i s m Given

a basis

a fixed i n t e g e r of a b e l i a n

[h

I ~n}

groups

for the g r o u p

for A , and v i c e - v e r s a . n are a b a s i s

and let ®:R(Sn)----~A n be given

R(Sn),

the s y m m e t r i c

and we h a v e

n

Theorem.

its image u n d e r ® is a b a s i s

F o r example, of A

in the F u n d a m e n t a l

functions

already noted

that,

if

is the p a r t i t i o n ®-l(h

where

~=(k I ..... kn) of n, S ) = Ind n S k 7< .. x S k 1 1 n

1 is the p r o d u c t

1 ~ ... ~ l

representations

of Skl .... Skn"

Littlewood

) this

~29~

it is n e c e s s a r y

associated identify

representation

h

is the

representation

R(Sn)

with

A

n

functions,

[a I ~ ~ n ] ,

S

(aT)

=

Inds

w h e r e ~=(k I ..... kn) , and representations and w r i t e

on the

"a °' not only

But

of A

n x

n

the n o t a t i o n

R(Sn)

by products

• .. × S k

~ .

When adhere

is the

it is e a s i e r

to

for b o t h objects.

of e l e m e n t a r y

) is t h e r e p r e s e n t a t i o n

(alt)

kl

n

(alt)

is the p r o d u c t

groups

and ~

in g e n e r a l

®-l(a

of

and A , we will n

function

v i a ® and use h

-i ®

between

of S . n

one-dimensional

is often d e n o t e d

symmetric

In the c a s e of the b a s i s symmetric

(Following

to d i s t i n g u i s h

to this n o t a t i o n :

of the t r i v i a l

of the a l t e r n a t i n g

Sk. As a b o v e , we w i l l 1 for the s y m m e t r i c f u n c t i o n a

be but

sloppy also

138

for the representation ®-l(a ), again treating ® as an identity map.

So far, only the abelian group structure of R(S n) has appeared. But R(Sn)

is also a k-ring with a dot product.

To avoid confusion

with the outer product defined above R(S n) ~

R(S m) - ~ f,g

R(Sn+m)

I~/<~O> fg

the usual representation-theoretic

R(S n) ~ will b e called, denoted f , g ~ >

n,m~ 0

R(S n) ~ _

product

J~ R(Sn)

in this chapter,

n~ 0

the inner product,

and

f*g.

The scalar product

R(S n) X will be called,

R(S n) ~

~

Z

in this chapter,

n~O the dot product,

and

denoted f,gw%~o)f.g . It is essential to keep these three products

straight,

and

we will adhere rigorously to these terms throughout this chapter.

As we have seen, the Fundamental Theorem corresponds outer p r o d u c t

to

since ®:R(Sn) ~

the

usual

product

of

An is an isomorphism,

symmetric

functions.

the But

we can induce an inner

product and a dot produc t on An from that on R(Sn).__

A bit later

139

we will g i v e e x p l i c i t

Using

the inner product,

en of the group all ~ 6 R ( S n ) , refer

R(Sn),

eniS

9(a )=h

for these.

the e l e m e n t

o.-~an*O.

an involution:

also to the i n d u c e d

the m a p of g r a d e d by

formulas

groups

, all ~

n,

Since

e2=l. n

involution

given

( t lq Y )

a £R(S n) g i v e s n an*an=hn,

We l e t

on A . n

by ~=t~ n.

all n~0

the

Let

symbol e

for

n

0:A - - ~ A be

also b y

characterization is given on p. 181

Another

and h n ~ O = o

e can also

and again,

a map

Note

be

specified

e(h )=a

.

e is not

a ring h o m o m o r p h i s m .

The m a i n Theorem:

result

The e l e m e n t

character

corresponds

Proof:

L

section

£ R(Sn)

is the f o l l o w i n g

defined(as

theorem•

previously)by

the

I:0 u n d e r ® to the p o w e r

F i r s t we

show the t h e o r e m

n into one part: only n e c e s s a r y

~ =

n

Then

to show that,

~

[(n)~

0 is an n - c y c l e in A, a n d t h e

(n).

of L k . . . . 1

outer p r o d u c t

Since

in this

- n,

or not.

immediate

sum f u n c t i o n

s

for the t r i v i a l

partition

of

since ® is m u l t i p l i c a t i v e ,

if ~ =

(k I .... kn),

it is

L~ is the

,L k n

Ln(O) s

n

= n or 0, d e p e n d i n g

is the

object

is

symmetric

to

show t h a t

on w h e t h e r

function

n n ~i+~2 +

®(Ln)=S n.

140

This will be n=l,

Ll=~l,

For n > l ,

accomplished

we use the N e w t o n

the t h e o r e m

the c h a r a c t e r

~£S

is true

associated

to s

®

-i

structure

®-l(srhn_r)

(o) =

Lr*~n_r

[o]

Given

the formula

dl£SrX

Sn_r,

" + Slhn-i

n

on R(S

n

In the case is trivial.

= nh n We will

1 6 r{ n-l.

(l~2$...nY).

(o)

in S

(Lr~ i) (Ol)

n

for induced

characters

(Lr~l) (ql) will be

of p.

zero u n l e s s

). o I is of the

form ~i

=

(

).(

an r-cycle c o n t a i n i n g the n u m b e r s i, ...,r in some order If o I is of this

form,

evaluate

).

= Lr*~n_r,

OlESrXSn_ r o ~ 1

(applying

.

(Srhn_r)

cycle

r

+

for Sl,S 2 .... Sn_ I.

have

n

on n.

formula

+ Sn-2h2

First we e v a l u a t e Let

induction

® ( ~ l ) = h l , and hl=S I, so the statement

s n + Sn-lhl Suppose

by

).(

) .....

(

)

other cycles i n v o l v i n g the numbers from r+l to n

(Lr~< I)(dl)=r.

141

Hence we want to count the ratio

iii[u[ilii ini:Snr ~

r

~ of / 01 in S n which a re ~ conjugate in S n to

(Srhn_r) =

. r . (this ratio)

G~ven that o has cycle structure is [l~2~..r~.]

(i~2 ~ . .r Z. .) , the denominator

(using the notation of p.

).

The numerator of

the ratio is then the number of ways of taking one r-cycle on the letters 1,2 ..... r times the number of permutations (l~2~..r ~-I..) on the remaining letters r+l .... . n. numerator is [r].[l~2 ~...r Z-I..~.

< rrl~. )

(

i

(n-r) '. I~:2~,...rZ-I(z_I)

n

of type

Hence the

The ratio is then

:...

)

i ~ : 2 ~ : . . . r ~ • ...

~ence~,~r~nr><~, ~r*~nr<~ rlnl~ ~r ~ has cycle structure

(l~2~..r~..).

Note that ~r is just the number of letters from 1,2 .... ,n contained in some r-cycle of ~.

142

Hence

@-l(Slhn_l+...+Sn_lhl)

= the n u m b e r

of e l e m e n t s

Applying ® -i

(o) = ~ + 2B +

o.. + r~ +

...

from 1,2 ..... n not in an n - c y c l e

of o.

N e w t o n 's formula, -i

(sn) (~) = ®

( n h n - S n _ l h l - . . . - S l h n _ l) (~) -i

=

nh

n

(<7)

= n -

-

®

(the n u m b e r of e l e m e n t s in an n - c y c l e of ~)

= the n u m b e r = ~n

L

of l e t t e r s

from 1,2 ..... n not c o n t a i n e d

in an n - c y c l e

of 0

if ~ is an n - c y c l e

Co =

(Sn_lhl + . . . + S l h n _ l ) (o)

n

if o is not

an n - c y c l e

(o).

Hence ®(Ln)

= Sn

all n >~ i.

N o w we m u s t p r o v e

~2=(i~22~2...)

where will

that,

of n 2, the outer p r o d u c t

Ul~2=(l&l+~2281+~2...) then

Given

for p a r t i t i o n s

follow by

induction

of L 1 and L 2

is a p a r t i t i o n on the n u m b e r

O £ S n l + n 2, w i t h c y c l e

Zl=(l~1281...)

structure

of nl+n 2. of p a r t s

of n 1

is L i~2

The t h e o r e m of n.

~=(l&28) , we c a l c u l a t e

143 S

Indsn~n Sn2(L ~'LI~2) (0) using the formula for induced characters:

S

IndSn~S (L~.~L 2 ) (o)= (nl+n2), , ' nI n2 ± nl.n 2 • [~] ( o i ~ 2)

Sn2

~i(~i )L~ 2 (0 2 )

oi~o2~o in S n (n=nl+n 2 ) Observe that

~/n i ,._ l

n2-

if each ~. has cycle structure 1 (l~i2Bi...)

L~ I(O1) L~ 2(°2) = "h

- so necessarily

o has cycle structure ~i~2 = (i ~1+~22~1+~2 ...)

otherwise <. Hence

So

S In d n L "~L (O) = S AS ~ n I n21 2

/ ~i~2

i

(nl+n2) :

1

n 1.

nl : n2 :

[~1~2]

[~l]

I

n 2. [~2] "N

~ = ~i~2

of cycle (oi~o2) of S ~S nI n2 This number is [~i][~2]. Hence

where N is the number of elements structure ~i~2 .

s

Ind n S ~S L~ix L~ 2 (~) nI n2

Hence L

=

/0 if ' I (nl+n2)_" [~i~2 ]

<

if

is the outer product of L ~i~2

2

=

(~)

L

~i~2

~ = TTI~

and L ~i

I

~2

I

144 We h a v e

so far four p r o p e r t i e s

®:R(Sn)__ >An, among

any one of w h i c h

all g r o u p

homomorphisms

i)

The o r i g i n a l

2)

®-l(h

3)

®-l(aM)

4)

®(L

The next

of the

isomorphism

characterize

R(S

n

)~>

it u n i q u e l y

A : n

definition S ) = IndSn 1 all w ~ n S = indsnal t

) = s

all ~ ~ n

all ~

few c o r o l l a r i e s

n

w i l l give

several m o r e ways

to d e f i n e

this

map.

First

is a w a y of d e f i n i n g

and is m e n t i o n e d recall

the d e f i n i t i o n

for ~6S n, K conjugacy

®(X)

=

class

using ~

Classical

of the c e n t r a l

~ or not.

X of Sn"

X =

the t h e o r e m

X(~) ~ ~n

n:

s

Thus L

goes b a c k

Groups

(~) is 1 or 0 a c c o r d i n g

any c h a r a c t e r

Hence,

in W e y l ' s

® which

(

functions

to F r o b e n i o u s

[45 ~i K

(n:/[~])K

First

on S : n

as to w h e t h e r =

p.215).

o is in the

.

Given

now

~ X(g)K , s o X = ~ ([ ] / n - ) x ( ~ ) L ~ ° ~ ~n ~ ~n

above,

and the

- clearly

fact that ® is additive,

a quite

explicit

recipe

for the

m a p ®.

This

leads

to the

of the dot p r o d u c t product

L~I.L

formula,

promised

and inner p r o d u c t

2 (respectively,

the

above,

in A . n

inner

for the c a l c u l a t i o n

Recall,

product

the dot

L~IL 2) i s

145

equal to (n:/[~l])

or zero

(respectively

according as to whether ~i=~2 or not.

[ s 1. s~ 2

i]

(n.'/[~l])L 1 or zero)

Hence

if ~i=~2

=

In

if ~ i ~ 2

*s S~l

=

I] s~l

if ~i=~2

~2 if ~ i ~ 2

These definitions can be found in some of the fairly ancient literature on the subject. using the symbols functions

Redfield

) in 1927 defined these

f ~ g for the inner product of two symmetric

f and g, and f ~ g

Another characterization Proposition:

( ~5J

for the dot product.

of ® is the following.

Let G be a subgroup of S

n

(In other words,

G be a faithful permutation group of degree n.)

let S Let x=IndGnl.

Then

1 ®(k)

-

~G~

~

(s~ s~..)

x£G

x = (1~2~..) all ~,~ .... Proof: We must check that

1

~-

L

x of cycle structure u and this follows easily from the formula for induced characters.

t46

This construction of a polynomial in the variables Sl,S2,... associated to a given faithful permutation group is of course, just the classical cycle index

of Polya ([337), (also called the

group reduction formula by Redfield (~35~) and Littlewood ( ~ 9 ~ ) I . A worthwhile example (of Redfield) in this context is the pair of subgroups of $6:

/(1)(2) (3)(4)(5)(6)

~i) (2) (3) (4) (5) (6)

(1)(2)(34)(56)

(1)(2)(34)(56) GI=

G2=

(12)(3)(4)(56)

(1)(2)(36)(45) (1)(2)(46) (35)

(12)(34)(5)(6)

6 2 2 These two subgroups have the same cycle index, namely Sl + 3SlS2 4 but are not conjugate as subgroups of S 6.

There are also

examples of non-isomorphic subgroups of some S

n

which yield the

same cycle index.

Another classical instance of the map ® is in the theory of immanents. (See Littlewood~9~ ) . Let there be given an n/.n matrix A:

I

All

Aln 1

e

A =

B

Anl

with entries in a given ring.

Ann

147 Let

J£S

n

be a permutation.

Write

PC7

Ang (n)

Let X£R(Sn) written

= be

IAI X

A l o (1) . . . . . .

The i m m a n e n t

a character.

of A a s s o c i a t e d t__qo],

is d e f i n e d b y

n

Thus the u s u a l

if ~ is the a l t e r n a t i n g determinant.

representation

of

S , n

and X is the u n i q u e

=

2AIIA22A33

In general,

If X is the t r i v i a l

2-dimensional

AI2 23A31

theorem,

I~ X = det A,

one-dimensiohal of A.

irreducible

Etc.

If n=3

representation

of S

3'

A13 21 32

A and B,

representation.

if A or B is a p e r m u t a t i o n

The relevance

we get

IAi X is the p e r m a n a n t

for m a t r i c e s

n o t the a l t e r n a t i n g

character,

iABI X ~

This

IBAI X , if X is

f o r m u l a does hold,

however

matrix.

of i m m a n e n t s

to our s u b j e c t

w h e r e t h e s. d e n o t e the p o w e r 1

is the f o l l o w i n g

sum s y m m e t r i c

functions.

148

Theorem:

For

X 6 R ( S n)

s1

®(x)

Proof:

=

Both the

given

partition

n'l

of n u n d e r the

element

~ =

~,

0

. .

s2

sI

Sn_ 1

Sn_ 2

.....

s

Sn_ 1

. . . . . . . .

n

are

additive

when

~.

=

Suppose

2

0 0

I_

identity

kAIX

0

n[

sides

prove

0

X is one

In t h i s

1

~ n-q--, o

in X,

sI

n-i!

Sll

so it is s u f f i c i e n t

of the

characters

to

L , for a

case

L(O)P

q

[11'] oE[~]

(Here

(nl,n 2 ..... nl).

in n o p a r t i c u l a r

1

-

the n

order).

l

p

are

Suppose

(1,2 ..... nl) (nl+l ..... n2) ( ....

just

the p a r t s

~6S n is

Then

po = 1 . 2 . 3 . . . (nl-i) .snl. (nl+l) ... (nl+n2-1) .Sn2 .....

n:

(S

n l ( n l + n 2) (nl+n2+n3) ( .....

If o is any

element

in the c l a s s

only

if no A j , j + 2 , A j , j +

term

in the

immanent

3 ....

comes

~,

appears.

.S

nI

there

will be

Hence

from permuting

....

)

n2

every

a nonzero other

n l , n 2 .... ,n I.

P

non~ero

149

If nl,n 2 ..... n Z are all different,

then

/ o

s

o£~

all permutations, • ,.. . ni I nl 2 'nl~ of

For the general

nl,n

2 .....

(nil) (nil+ni2) ( ....

n g

case when the n. aren't distinct,

we n e e d a

1

Lemma: 1

Z permutations

all n ,n , ..,n i I 12 l~

i

n i l ( n i l + n i 2 ) (...) ( n i l + . . + n i )

nln2...n~

of nl,n 2 .... nz Proof of Lemma:

By induction

it is true for all integers

i

on Z.

less than

all p e r m u t a t i o n s n. ,..,n. of lI iZ n I .... nz w i t h n appearing

=Z

last:

For £=i,

(nil) (n

Z.

nln2..ni_ini+l..n Z

i nln 2. .n~

~

n

I

Suppose

Then the left hand side is

+ni2)... ( n i l + . . . + n i ) ll

l

iz=i

(pulling the common last factor from each term and using the induction hypothesis.)

1 i

it is trivial.

nl+n2+-.+n Z

1 (nl+n2+..+nz)

1 nln 2 - .n Z

- end of proof of lemma. U

150

n' O

Hence

if n I ..... n Z are all d i f f e r e n t , ~

Pc -

nln2--n Z general,

though,

Z: p e r m u t a t i o n s permutations

if

- so the c o e f f i c i e n t

n'

n' i~2~., n

7_. P

but

(I~2B...),

~+~+..=Z,

are o n l y

in g e n e r a l

In

all the

(~:/&:$:..)

distinct

is not

n'

rather

- [~]

.

l&2~..~:~:..

1

-

Hence

L sI

_

0

1

e

- [7] ~ p o -

n'

S

This

is the c l a s s i c a l p r o o f

Littlewood

[29~

= s~ = s ( n ) .

and that the t h e o r e m

Hence

it w o u l d b e

sufficient

in X-

the m a t t e r

B

(taken f r o m

I X = ®(X)

for x=a

n

T h i s we k n o w

We k n o w

are a d d i t i v e

, n = 0 , 1 , 2 ....

to s h o w a l s o that b o t h for ®(X) -

seems a l i t t l e m o r e

is true:

alternate proof.

is true

fact

an e a s i e r p r o o f .

sides of - - i i ! n~

in X,

the theorem

of this

), b u t t h e r e m a y b e

to b e g i n w i t h t h a t b o t h

multiplicative

[~] s~

sI

n

side,

=

are n o t d i s t i n c t - t h e r e

nln2"'n ~

So

(nl,n 2 .... n~)

s . ?7

But

(p. q q

sides

).

are

for the left h a n d

s u b t l e - b u t of c o u r s e

We l e a v e it to the r e a d e r to w o r k

true s i n c e

out t h i s

151

Fix

the

integer

representations) write fact the

down

the

of S

resulting

so this

takes

care

irreps.)

since

the

labeled by partitions

Schur

Schur

of n,

methods

here

that

assume

in A

the

n

w e can

.

A non-obvious

n

- is t h a t

{ {~}

[In]=a

I~

this

n

and

n

associated

following

functions

so m u s t

in A

functions

functions

for the

(irreducible

to its p r o o f

showed

We will

the

above

functions

symmetric

mainly

irreps

devoted

(There w e

of the

section,

convenience:

symmetric

are

i.

the

any of the

is m a i n l y

functions

in C h a p t e r

of this

By

section

one-dimensional rest

.

n

associated

- the next

discussed

n and c o n s i d e r

fact

]

{n]=h

n

to the for the

notational are n a t u r a l l y

irreps

of S

also be

so

n labe l e d .

We will

write

in Thus

X

= alt

, and k

n

It is i n t e r e s t i n g one-one

correspondence

set

of c o n j u g a c y

for

any g r o u p

sets

have

natural

p.173

group

that

between

the

number

~ a partition

irrep

is c a l l e d (

to n o t e

classes

same

irrep

associated

to

{~}.

= i.

G, w e h a v e

the

for the

of S

n

shown

:

this set

labeling

of i r r e p s

gives of

S

X <>~-~k[ ° I~ of t y p e ( f. ~

) that

of e l e m e n t s ,

but

usually

a natural and t h e

n

~].

Of c o u r s e

these

there

two

is n o

such

correspondence.

Given of the

k

the

X

, which "hook

) for

of n, w e

we will

number"

can

denote

H

.

of ~, b e c a u s e

its c o m p u t a t i o n .

representations,

associate

one can w r i t e

This

to ~ ~ e

deg:~ee

positive

integer

of a s i m p l e

algorithm

From

the elementary

such

simple

theory

identities

as

of

152 H

=

i

=

1

=

n:

in

H

n

~" H2 ~Pn A slight b i t m o r e

(a I ) n

Lemma:

involved

~

=

is the

H

following

identity

in A : n

[~]

~n Proof:

Let U be

any r e p r e s e n t a t i o n

(Ui~

of a finite

Hom G ( U i t U))

~

group

G.

Then

U

U. 6I r r e p G 1 Here H°mG(U'1'U)

is as in p. ~I

is the obvious since

each

one:

side

u~f-u~>f(u)

is additive

U is i r r e d u c i b l e

The map U i ~ Q H o m G ( U i,U) - ~

(i.e.,

and the i s o m o r p h i s m

in U,

when

and the

U=U.,

some

identity

i).

U

follows

is true w h e n

Indeed,

this

is just

1

a restatement

of Schur's

Let n o w V b e on ~

a vector

by permutation

V

~n

~

lemma.

~

space

of factors.

(W~Hom

~-n But HornS

(W ,V ~ n )

=

and U = V ~i)n.

Then t h i s

S

Let G=S

identity

n

act

becomes

(W , ~ n ) )

n (Hom(W

, ~ n))

Sn =

(wdual S n ) n

S n = wdual (V)

n the last e q u a l i t y associated

being the definition

to a r e p r e s e n t a t i o n

dual

W~

.

of the operation

wdual(

)

153

N o w n o t e t w o things:

W

is an i r r e p

and t h e y b o t h h a v e t h e same d e g r e e . An i r r e l e v a n t

fact is t h a t

if and o n l y if W d u a l

(This is t r u e

is,

for any group.

for S , the i r r e p s W and W n ~

dual

are

isomorphic. ) Hence: V C~n

wdual~(~Dw~(v)

= ~i-n

This

is a f u n c t o r i a l

endomorphism

of V w i t h

t r a c e of b o t h

of b i n o m i a l

in V,

eigenvalues

so r e p l a c i n g V b y

t l , t 2 .... tn,

an

and t a k i n g the

sides we get

(tl+t2+''")n

Schur

isomorphism

=

~ H ~ n

functions

. {~] (t I .... t n)

interpreted

coefficients.

i n t e g e r n and let V b e

as o p e r a t i o n s g i v e

Namely,

a vector

let ~ b e

a generalization

any p a r t i t i o n

of any

s p a c e of d i m e n s i o n m.

Write (~> w h e r e ~'

=

dimension

is the p a r t i t i o n

If ~=(n), usual binomial

(~I

im) n)

, =

conjugate

=

to ~.

ira3 : dim ~in~v~: dim A n v the

coefficient.

The last p r o p o s i t i o n binomial

[~](V)

gives

an i d e n t i t y

coefficients mn

=

S ~n

H~

() m

for t h e s e g e n e r a l i z e d

154

(~1

In terms of k-rings, operations

[~/~ £ A to the e l e m e n t

We c l o s e this

identify group

(i.e.,

we have extends

t h e s e two

product,

a dot product,

b y a ). n

There

basis

{h ,~I-n},

{s , ~ n } .

of p a r t i t i o n s

natural bases

{<~>,~n},

Given

R(Sn) _ ~ > A

object

and an i n v o l u t i o n

are s e v e r a l

(binomial)

isomorphism

The r e s u l t i n g

the n u m b e r

of a p p l y i n g

This

integer matrix combinatorial

if the s theory

for c a l c u l a t i n g of the r e s u l t i n g

of the

numbers.

some of this theory.

For

which We

is a free a b e l i a n an inner

(as free a b e l i a n group):

integer matrix

and a r a t i o n a l (or one b a s i s matrix between

(or l/n:

times

an

and a large p a r t of the

symmetric

these matrices

~.

(inner m u l t i p l y i n g

{ [~}, ~I-n}

are chosen)

n

of n, w i t h

any two of t h e s e b a s e s

is an i n v e r t i b l e

k-ring

R(S)~>A).

and the set of s ) one can ask for the t r a n s i t i o n them.

the

w h a t we n o w have.

an i s o m o r p h i s m to a ring

sets.

of rank ~(n),

{a , ~ n } ,

m in the

section b y r e v i e w i n g

each i n t e g e r n ~0, is n a t u r a l

is the result

group consists

explicitly

In the n e x t

of m e t h o d s

and i n t e r p r e t a t i o n s

two s e c t i o n s w e e x p l o r e

155

Schur Functions

The m a i n functions

and the F r o b e n i u s C h a r a c t e r

object

of this

section

are the i m a g e s u n d e r

representations

the c o m p u t a t i o n

is to p r o v e

t h a t the S c h u r

the i s o m o r p h i s m ® of the i r r e d u c i b l e

of the s y m m e t r i c

m e t h o d of p r o o f g i v e s

Formula

groups.

It t u r n s

out t h a t the

a set of f o r m u l a s w h i c h w i l l b e b a s i c

of the c o m b i n a t o r i a l

t h e o r y of the

in

symmetric

group.

The p r o o f c o m e s

f r o m the

simple observation

in this c o n t e x t b y P h i l i p H a l l g r o u p of f i n i t e must be unique

rank,

with

[20~

) that in a free a b e l i a n

a dot p r o d u c t ,

(if it e x i s t s

at all)

This c o m e s

bases

the t r a n s i t i o n m a t r i x

be orthogonal with matrices which

integer

from the fact t h a t

entries.

are s i g n e d p e r m u t a t i o n

e a c h row and e a c h c o l u m n

entry,

that entry being ~

R(Sn)

= An has

irreducible

an o r t h o n o r m a l b a s i s

- at least,

sign and order. are given,

(pointed out

u n i q u e up to

if two o r t h o n o r m a l

from one to the o t h e r m u s t

The only o r t h o g o n a l

matrices contain

- i.e.,

square matrices

e x a c t l y one n o n - z e r o

i.

a dot p r o d u c t ,

representations.

Hence

and an o r t h o n o r m a l b a s i s , any o r t h o n o r m a l b a s i s

m u s t c o r r e s p o n d up to sign and o r d e r w i t h the irreps. l e m m a is the

following.

integral

the

in A

The k e y

n

in

156

Lemma:

Given

qk'

rk £An

[rk

I k~n]

under

any e x p r e s s i o n

(notation

facts

as in

are b o t h b a s e s

the dot p r o d u c t

The p r o o f about

product.

of this

satisfying, i)

I ~7

there

with

rk(y)

), the sets

[qk I k~n]

to each

and

other

n

lemma

requires

us

first

in free abelian

a free abelian

Thus,

-- ~. qk(x) XPn

of An and are dual

in A

dot p r o d u c t s

Let F be

hn(XY)

group

some

groups•

of finite

is a function

to recall

(-,-)

rank k w i t h

defined

a dot

on F ~ F

for all x,y, z6F

(x,x)

ii)

£

(x,x)~ 0

and e q u a l i t y

iii)

(x,y)

iv)

(x,y+z)

=

(x,y)

+

(x,z)

(x+y,z)

=

(x,z)

+

(y,z)

This gives

a map

:

holds

only

x.u~

(x,-)

if x=0

(y,x)

~ : F - - ~ Hom(F,~),

which

is n e c e s s a r i l y

one-one. Suppose, sense that,

now, if [rk]

det(

(rk,r~))

know

that

R(S

n

is any b a s i s

of F,

=~i.

(This is true

) has

an o r t h o n o r m a l

representations.) the map

that the dot p r o d u c t

An e q u i v a l e n t

8:F~-->Hom(F,Z)

given

is also normal

in the

the k × k d e t e r m i n a n t

in F = R(S n) = An since we basis,

definition

above,

the i r r e d u c i b l e of n o r m a l i t y

is onto

(hence

is that

an isomorphism).

157

Another

equivalent

is a u n i q u e (ri,s9)

statement

dual b a s i s

= 5ij

{sk}

(Kroneckor

Let us p r o v e to a g i v e n b a s i s

is that,

of F , dual

in the

implies

To find the

s k, one m u s t

for i n t e g e r s

equation

,r

by r kI

1

=

=

By Cramer's

all(rkl,r

this

the i n v e r s e map

There

,rk2)

+ a12(rki,rk2)

+

+

set of e q u a t i o n s

this

...

...

is s o l v a b l e

over Z since

in ~ .

n o w in the c a s e of F = A

n

, we w a n t to look at

~-I:Hom(F,Z)----->F.

is a n a t u r a l

for any free a b e l i a n is a n a t u r a l

Dotting

equations

kl

d e t ( ( r k . , r k ))= ~ 1 is i n v e r t i b l e i ]

Specifically

aij.

solve

of a dual

.... we get

) + al2(r

all(rki,rkl)

rule,

that

k2

~i

0

sense

[rl} of F, there

the e x i s t e n c e

like s k = a l l r k l + a 1 2 r k 2 + .... in s u c c e s s i o n

a basis

delta).

that n o r m a l i t y {rk}.

given

isomorphism

group

isomorphism

of finite

Hom(Hom(F,ZS),F)= rank F,

Hom(Hom(F,~),G)

F~F.

Indeed,

and any g r o u p G,

= F~G.

This m a p

is

there

158

given

by ~: F ~ G

by,

--~ H o m (Hom (F, ~) ,G)

for ~fiOgi

9( ~ Both

sides

to b e

show

entail dual

fic~}gi) (9)

are

~

-i

that

that

to one

Hence

{qk I k I-n] another.

and For,

it is an i s o m o r p h i s m

~-I =

[rk

simply

I kL--n]

chasing

B -I is o n t o

Furthermore,

for r6A

map

given

n

by

r = ~ - i B ( r ) = B-l(q)

are , the

just

so this enough

with

(r,rk)s k.

of l i n e a r

qk,rk£An

are b o t h

bases

isomorphism

), ~-l(q)

definition

dotting = ~

the

,

is e a s i l y

=

shows

algebra

must

a n d are

back,

(.]~rk(~sk) (q) = that

of them, of

shown

in g e n e r a l .

a matter

~ q~rk kl n

so for q E H o m ( A n , 2 Z

there

i 6 G

for F = ~ , t h i s

It is n o w

any e x p a n s i o n

Since

A --7 ~ n

S~(fi)g

in F and

A . n

= ~_ q ( r k) s k E An. k A . n

=

additive

£ A ~ n

~-i = ~ q k i ~ r k ,

span

and ~£Hom(F,Z)

an i s o m o r p h i s m .

Thus to

6 F~G

the

they

~ gives

r: q =

(r,-).

Let

r = sk,.

sl m u s t

form

q=8(r)

a basis. = the

Thus Then

k s k, = z~ ( s k , , r k i s k. S i n c e t h e s k are a b a s i s , (sk,,rk) = 1 or O, k a c c o r d i n g as to w h e t h e r k'=k or not. H e n c e the set of r k is a dual basis

to the

set of s k.

159 This argument dual basis r k ,

is reversible:

given any basis qk' and

o f An, ~-1 can be e x p a n d e d as

~_ qk@grk. k~-n

To identify ~ - I £ A ~ A , it would be sufficient n -~ n

everything with the rational find the element

~-i = k~--~nq ~

~

rk

-1

numbers

("introduce

in ( A / ~ Q ) ~ ) (An<~)~). n

to tensor

denominators")

and

Any expression

with qk'r x not only in An @ ~ but in

A ~ An~)~ would still give dual bases of A . n n

Now the argument at hand by noticing

h (xy) n

and,[([k]/n:)sx

to the case

that we have the expansion

=

i ~i-n}

Hence h (xy) = 6 -I. n

is brought back from generality

~kbn

Ix] n:

sk(x)

s (y) k

(p. ~ 9

and [sk} are dual bases of An~'¢

)

(p. I~5-- ). g

160

Corollary: i)

For

all n~l

{ I k ~ n}

ii)

{ [k] I k~- n}

iii) ®(X)£A

and

Let X b e is of the

n

[ hk I kSn}

are dual b a s e s

is an o r t h o n o r m a l b a s i s

an i r r e d u c i b l e form ®(k)

= +

of An.

of A

representation

n

of S . n

Then

{k} for some k m n.

J Proof:

See pp.

Later

(p. 169

fact a l w a y s k

k

for ®

on A

{k}).

)

) we will

a plus

-i(

conjugate

%[ /

sign. We w i l l

and m a p p i n g

comes

and i r r e p s of S

functions.

These

in the last

that,

if k'

the

of A b y S c h u r

of the c o r r e s p o n d e n c e is a n u m b e r

n

between

of f o r m u l a s

are d e d u c e d b y u s i n g the v a r i o u s section

for the m a p ®.

Thus k

[X]

n,

is the p a r t i t i o n

In o t h e r words,

the b a s i s

is in

endomorphism functions

IX' } .

Another consequence

given

also prove

from t a k i n g

[k} ~ >

in iii)

H e n c e we can write, for e a c h ~

to k, t h e n k k' = a * XX. n

(p.I~)

functions

s h o w t h a t the ! sign

sI

1

0

.......

s2

s1

2

0

s

s

...

=

n

(using the d e f i n i t i o n

sI

n-i

of ® b y

immanents)

X

Schur

for S c h u r definitions

161 and {k]

=

s

U

n:

k

(using the facts formulas

hold

The m o s t

that @(L

(p. @ 0

expansions bases.

) between

important

hn(XY)__ 6 And~ An c o m e s of h

(xy)

n

result

from the in the

functions

from the

fact that we have three

~ . [~] n: ~ n

s (x) s (y)

R(Sn),

R(Sn)~JA

gives

both

.

R(S

n

rise to f o r m u l a e

relating

comes =

from the i d e n t i t y

~ ~n

h

(x)<w>(y)

in A n ~ A n : Identifying

• ~ < ~ > (y)

~ P n S ) = IndSnl

= ®-l(h

n

different

we get

~kn

~

of

{qk],{r k] dual

qk~rk,

n'.

where

identification

involved.

derivation

with

and that t h e s e

k~ n

The e a s i e s t hn(xY ) =

{k],

the X k and the L .)

form

E a c h p a i r of e x p a n s i o n s

the s y m m e t r i c

An(X)

, ®(X k) :

)=s

) being

sides can be

, as usual.

Both

a ring of functions,

applied

to the c l a s s

sides class

D of S . n

are n o w in functions

The r e s u l t

on S is

n

I

162

s (y)

=

~

P

~(p)

<~>(y)

~n

Hence

the c h a r a c t e r s

basis

[ I ~ n ]

Conversely,

~(0)

of A

since

n

give

the t r a n s i t i o n

and the r a t i o n a l b a s i s

the s y m m e t r i c

functions

s

matrix between [ s

the

I p~n].

p

can e a s i l y b e P

expressed

in t e r m s

an a l g o r i t h m

for c o m p u t i n g

Similarly,

S

(x)

=

~-

functions,

<~>(p)

A

h

n

(y) w i t h

this g i v e s

~ •

R(Sn),

we have

(x)

v~n

that

symmetric

symmetric

the c h a r a c t e r s

identifying

O showing

of m o n o m i a l

the c h a r a c t e r s

functions

give

of S

n

corresponding

the t r a n s i t i o n

to the m o n o m i a l

matrix between

the

s

amd P

the h

It is w o r t h w h i l e functions -

see

~13~

symmetric [a

I ~l-n].

As above,

which

here

to m e n t i o n

have n o t b e e n m u c h

).

Let us call

functions".

these

another kind

studied if

T h e y are d e f i n e d

Equivalently,

f

h n(xy)

=

S (X) P

=

~ fk (x) a k (y) k~ n ~ ~n

(except q u i t e

I zl-n}, the

f (o)a re(x)

~Fn,

recently

"forgotten

as the dual b a s i s

= a * <~> w h e r e n

one d e d u c e s

of s y m m e t r i c

to

n = 0 , 1 , 2 ....

163

and

s (x) P

(where

=

in the second

character

equation,

associated

The next p a i r

h n (xy) =

~

~ a ~ n

to the

(p) f (x)

we have

symmetric

of e x p r e s s i o n s

n: [~]

Sr~(X ) S1~(y )

function

for hn(xY)

=

~

"ffVn

Identifying

denoted

by

f , the

f

).

gives

the i d e n t i t y

[k} (x) {k~ (y)

kPn

An(Y ) = R(Sn)

as above,

and e v a l u a t i n g

on the class

we get the Frobenius

Character

Formula

s (x)

=

P

It is w o r t h w h i l e variables m

sp

(x)

=

=

to w r i t e

X l , X 2 .... x n. m

s

m

definition

Recall

m

x 1 +x 2 +..+x

s .... spl P2

~ Xk(p) kPn

of Schur

this that,

(x)

out m o r e

a symmetric

function

was,

explicitly.

for an integer

For a p a r t i t i o n

n

again

{k}

p =

__

function

in the x. 's. z

for k =

(kl,k 2 .... kn),

a partition / kl+n-i k2+n-2 k ~sgn o ~ x x .. x n o£S o(i) 0(2) " o(n) n (Xl,X 2 ..... x n)

where

A ( X l , X 2 .... Xn)

=

~

i<j

m,

(pl,P2 .... ),

kl~k2~...~kn~0,

{k~

(xi-xj)

Take n

/

The

p,

164

Thus we get

the classical

expression

of t h e F r o b e n i u s

Character

Formula

s

(x) A 0

. (Xl,.

~ ,Xn)

+ XX kl+n-i _ (p) x I

=

k2+n-2 k x2 ... x n

k~-n

where,

following

indicates

that

permutations

the classical

the

two-dimensional consisting

of

n=3,

X

21

with

=

irreducible 3-cycles.

(Xl-X 2)

left

The

just

all

kl-n, b u t

also

sgn of the permutation

already

know

of t h e

of S 3 a c t i n g (p. q 3

) the

on t h e c l a s s

answer

side

right

k=(l,l,l) so

X

(Xl-X 3) <(x2-x 3)

is

hand

3

3,

( X l + X 2 + X 3 ) (Xl-X2) (Xl-X 3) (x2-x 3)

side

we get

this partition

13 (3)

is -i.

are Xl,X2,X 3

is,

first

o f all,

a sum over

all p a r t i t i o n s

k of n = 3 . For

all

included.

of t h e c h a r a c t e r

representation

We

on t h e r i g h t

3 3 3 so Sp = s 3 = x I + x 2 + x 3

. 3

so t h e

the

the notation

(3) - t h e v a l u e

so t h e v a r i a b l e s

p=(3)

style,

s u m is o v e r n o t

of t h e x., 1

L e t us c o m p u t e

n

(kl+n-l,k2+n-2,k3+n-3) contributes

=

to t h e t o t a l

(3,2,1) the

terms

, 32 32 32 32 32 32 , (x x x + x x x + x x x -x x x -x x x -x x x I) 123 231 31:2 132 213 32

165

For k=(2,1,0)

we get

(kl+n-l,k2+n-2,k3+n-3)

so this p a r t i t i o n 21

, 4 2

X

(3)

contributes

42

42

we get

4

2

XlX2.

(3) appears

Computation

coefficient

systematization, identity,

section.

contributes

on the right

of the

=

(5,1,0)

to the total

hand

left h a n d

one of the p r o b l e m s

to find a less m e s s y

technique

42

the terms

side

side

as the c o e f f i c i e n t

shows

that

of

this

is -i.

Obviously

final

42

the terms

, 5 +5 5 5 5 5 , tXlX 2 x 2 x 3 + x 3 x l - X l X 3 - X 2 X l - X 3 X 2 J

X 3(3)

21

to the total

(kl+n-l,k2+n-2,k3+n-3)

so this p a r t i t i o n

X

(4,2,0)

(XlX2+X2X3+x3Xl-XlX3-X2Xl-X3X2)

For k=(3,0,0)

Hence

42

=

systematization

together h n(xy)

=7

with

in c o m p u t i n g

characters

of this procedure.

the p r o c e d u r e

that

[4} (x) {~} (y) = ~ ( x ) < ~ > ( y )

of Y o u n g Diagrams,

which

is the

subject

arises

is

This from the is the

of the next

166

4.

Methods

of Calculation.

The object calculations, theory

of this chapter

and derive

of the symmetric

first compute

general

transition aT, <~>,

group

without

triangularity

between

are given,

about the representation

from them.

proof,

Frame's

kk(U)

the Schur

of characters

derived

in the last section:

~(x)=

sgn(G) o£S

Xk(p)

n

k~n where

of computing Finally,

the

and the h ,

the phenomenon

of symmetric

a simple way to use the Frobenius

(x)

result on

functions

finding

P

representations

is given.

and we discuss

we

of the

of these matrices.

The calculation

s

In particular,

Then the method

of the characters

matrices

and f

is to carry out several

of the irreducible

these via "hooks".

values

Diagrams.

some facts

the degrees

of S , and then give, n computing

Young

p ~ n, and ~(X) =

~ I (Xi-Xj)l<_i< j<__n

groups

Character

involves

Formula,

kl+n-i kn Xo(i) .... X (n)

167 Consider

a monomial X. X . . . . X with the subscripts 11 12 iq

chosen from the set {i .... n]. lattice permutation

Such a monomial

if in the first i factors,

is called a the number of

i

X 1 s occuring is ~ the number of X2's occuring ~ the number of X 3 s, etc.)

for all i.

permutations

of

For example,

the following are lattice

X13X2X3:

XI3X2X 3

X 1 2 X2XlX3

X 12X2X3XI

XlX2XI2X3

XlX2X3XI 2

XlX2XIX3X 1

and the following

X2X13X3

Recall

are not:

XI3X3X 2

(p.29) that, given a partition

kl~k2~...~kn~0,

XlX3XlX2Xl

k~-n,

its Young diagram is a pattern of boxes with

k I boxes in the first row, k 2 in the second,

x =

k = (kl,k 2 .... k n

(5,4,1,i)

t

etc.

168

A Young

tableau,

of shape

t a k i n g the Y o u n g the n u m b e r s

diagram

k =

(k i .... k n)

k and f i l l i n g

1,2 ..... n in some ord6r,

is o b t a i n e d b y

in the b o x e s w i t h

one n u m b e r

to e a c h box.

For example

I1

9 i ioi

6

i8

~ 7

!

-t I

and

li A standard Young column,

tableau

the e n t r i e s

tableaux

above

is one w h e r e

are i n c r e a s i n g .

are not

standard.

in e a c h r o w and e a c h F o r example,

But the

the t w o

f o l l o w i n g two are

standard:

iv,F-;-! ......

......................... 51 i001 i1!

111

Lemma:

Given

a monomial

kI X2 k Z = X 1 X 2 ...X n • k l ~ k 2 ~ . . . ~ k n , ~ k i n

the n u m b e r of l a t t i c e p e r m u t a t i o n s of s t a n d a r d t a b l e a u x of s h a p e Proof:

By example.

(k I ..... kn).

54 ~ = X I X 2 X 3.

Suppose

3 2 XIX2XIX2X3XIX2,

permutation

of ~ is e q u a l to the n u m b e r

label

Given

the t e r m s

1

2

3

4

5

6

7

8

9

i0

X1

X1

X1

X2

X1

X2

X2

X3

X1

X2

the l a t t i c e

in o r d e r

= n,

169

Fill out the tableau

(5,4,1) b y putting the numbers written above

the X 1 i s in the i st row, those above the X 2 i s in the 2nd, etc. Conversely,

given a standard tableau

F---~ I2 I v /-7 L~

|

/ 10!

8

,~ - - " ~

write out the monomial with an X I in the is~ 3d, 5 th; 6 th and 9 th position,

X 2 in 2 n d , 7 t h , 8 t h a n d

10 t h ,

and X 3 i n t h e 4 t h .

This

yields X X2X X X 2X22X X2 1 1 3 1 1 " These two processes between

Theorem: X

k

are inverse,

giving a one-one correspondence

standard tableaux and lattice permutations,

Let k ~ n .

The degree of the irreducible representation

is the number of standard tableaux of shape k.

Proof:

X The degree is X (In) -

k n o w so far that X

k

=

®-i

(More accurately,

([k]) is + an irrep,

the absolute value of XX(! n) .

By the Frobenius Character

(Xl+'''+Xn)n~i<j (Xi-Xj) = ~ES~

the degree is

for all k.)

formula

sgn(a)

k k l+n - i n XX(I n) X (i) ..... .X (n)

n k~n ~i<j

since we only

But it will follow from this

calculation that k (in) is positive

The product

i

(Xi-Xj) can be expanded ~ as

170

A (X) =

~

+

n-i X 1 ...... Xn_ 1

where the sum is over all permutations with the sgn(o)=+l

n

of the subscripts

attached.

We want to calculate the coefficient

oES

xk(1 n) which,

on the right side,

is

of, in particular,

X Ikl+n-1 X 2~2+n-2 . . .

.Xnkn = (X 111X22k_ . . .

So we may ask for the ways this monomial

xkn) n

n-I n-2.. . (Xl X2 "Xn-l)

appears on the left in

the product (Xl+...+ X )n ~ n

+ n-i n-2 -- Xl X2 "''Xn-1

Let us expand this product, multiplying by

(Xl+...+X) n

the function obtained

starting with &(X) and successively n times.

At each stage,

is still alternating

are permuted by an odd permutation

if the X's

~6S n, the sign changes by -i).

This occurs since &(X) is alternating symmetric

( i.e.,

(Xl+..+Xn)i~(x),

function by an alternating

and the product of a function

is alternating.

Hence writing

(Xl+...+Xn) i~(x) as a sum of monomials 11 in) ~integer coefficient)X 1 ... X in each monomial with nonzero n

coefficient,

the powers of different X's have to be different.

Jl J2 ' Let ~ = X 1 X 2 ...X 3n be one such term in (Xl+..+Xn) I&(X) n with distince (XI+...+Xn)

indices

Jl ..... Jn"

Multiply

to get to the next step

(XI+..+Xn) iA(x) by

(Xl+..+Xn)i+ig(x).

For T as i

above,

and any k, look at the resulting

term ~ 7

"

!

el 92 .x3n = X1 X 2 . n

171

Each 3q=]~ except for q=k where 3k=]k I. •

~

•

i

•

+

Then either two of

the j' are equal, in which case, since (XI+..+Xn) i+IA(x) is alternating,

the coefficient of

T is zero, or the jl,J2,..,]n i

I

'

i

still exhibit the same pattern as the jl,J2 ..... Jn in the I

sense that if Jk is the largest of Jl .... in" then j k is the largest of 31, .... 3n, and etc. for the second largest, third largest

...

In

particular,

to X 1kl+n-i ...X nkn with

(X~ -l...xn_l)

to

find

out

terms

contribute

in (X1+...+X n )nA(x), one only need start and ask how this term changes each time one

multiplies in a factor of (Xl+'''+Xn)

given

which

At each step, one is

Ul Z2 (x~-l...Xn_l) (X1 X 2 ...), ~lk~2k... After multiplying by

some X.,1 one gets

n-I ~{ ~2 (X1 ...Xn_l)(X 1 X 2 ...), U i ~ p ~ . . .

k n-i kl k2 in the resulting X (in) (X1 ...Xn_l) (Xl X 2 ...)

Hence

the coefficient

X (in) is the number of ways of arriving at the term

(x -i ...Xn_l)

x2

(XI±X 2 ..), which is the number of lattice

kI k2 k permutations of X 1 X 2 ...Xnn

(In particular xk(l n) is positive.) I

172

For example,

for n=3,

the s t a n d a r d t a b l e a u x are: D e g r e e of a s s o c i a t e d c h a r a c t e r

Standard Tableaux

Shape

(21)

deg X

(iii)

F o r n=4,

k =

deg X

2,2),

of this s h a p e w h i c h

say,

deg X

22

are s t a n d a r d

21

ill

= 2

= 1

= 2 since the o n l y two t a b l e a u x are

and

Given k

n, w e h a v e d e f i n e d the h o o k number,

b e this d e g r e e X method

k

-

This name comes

for c o m p u t i n g H k

diagram,

f r o m the f o l l o w i n g

(due to F r a m e

we a s s i g n to e a c h s q u a r e

( ~7~))-

directly below,

lying e i t h e r d i r e c t l y

the s q u a r e

Given

in it a p o s i t i v e

l e n g t h of the h o o k s u b t e n d e d b y the square. of all the s q u a r e s

H k, of k to

in q u e s t i o n ,

This

simple a Young

integer, "hook"

to the right,

the

consists

or

t o g e t h e r w i t h the s q u a r e

173

itself.

t

e

s

The length is the number of squares involved.

uare

su

which has length 7.

tends

t

e

For example,

hoe

In the following diagrams,

the hook numbers

are given in the squares.

Theorem:

Hk

n' 0

-

.~hij 1,3

Proof:

(See [ 8 ~ ,

[17], or ~ 2 ~

l

).

A "natural" proof of this has yet to be discovered, counts the number of all tableaux of the given shape, the number of standard tableaux.

n'

and H k

It should b e that ~ h . .

13

counts the number of some objects and there is an algorithm which to each pair consisting of one object and one standard tableau,

assigns a general tableau,

algorithm is now known. to three dimensional

and vice-versa.

No such

(Note that the obvious generalization

tableaux is false.)

~i~

174

To

calculate

a similar

but

characters

more

kk(p)

complicated

for g e n e r a l

use

of

the

X,P~

Frobenius

n

involves

Character

formula. Given Xl+n-i

s o A(X)

Let

p =

=

7

-+ kk(P)

(rl,r 2 .... ) so s

:

s

p

recall we

A(X)

start

=

~-

with

£(X)

Suppose

we

are

by

s

= X[ +

s

r. 1

=

r

alternating X's

with

the

term

at

s

...

..

rI r2

Xn_ 1

and multiply,

one

stage

£(X) S r l s r 2

°

°

different

so a s u m

subscripts

r.1

and

n

s

£(X), P

at

a time,

.Sri_l

The

by

and

•

the

product

in e a c h

different

terms

s

r~ 1

the multiplication

partial

of m o n o m i a l s have

r. 1 ... + X

In c a l c u l a t i n g

. .. + X r is n e x t . n

function,

Xnn r. 1 = X1 +

s ")

n-i n-2 X1 X2 ...

+

k

X1

is

an

of which,

exponents.

Consider

n-i n-2 ~i ~~ X1 X2 .. X n - I) (X 1 X 2 2 . . X q q x q +ql+ l "" .X q + i . . . X n) " q+i n where

~l~2k...~n.

which

arises

result

of

the

...s rI

multiplying

multiplying

this

the multiplication,

exponents,

A(X) s

by

When

coefficient

since r

this

of

is

an

given

by

s

term

r

consider

by

X

r q+i

two

different

X. e n d ]

the

resulting

term

alternating

term

If

as a

up with

will

function.

the

be

equal

zero

in

175

Hence,

for some q,

q Hence

+ n - q

>

in A(X)s

~

...s rI

. + n q+l

(q+i) + r

>

a

q+i

+ n -

(q+l)

, we get the t e r m r

(Xl-I ...A~n-q~An-q-l•t-_n-q-2 q q+l q+! " ° °Xn-l)

"

&l e a .+r-i+l a q + l + l & ~+i e . (X 1 ...X q x q+1 q+i-± x q+l+l q q+i Xq+l "''Xq+i-I --q+i+l "'')

There

is a c o r r e s p o n d i n g

increasing This

order

- with

term w h e r e a minus

is the term c o n s i d e r e d

have to k e e p

track

In this way,

the s u b s c r i p t s

sign in front

at the n e x t

stage,

if i is even. w h e r e we w i l l

of the + sign that has arisen we m a k e

the t r a n s i t i o n

are in

here.

from A(X) s

A(X) s

...s rI

s , and c a l c u l a t e ri_ 1 r

(X~-i

~i a ...Xn_ I) (X 1 "" ]X n n) '

end,

we have

can

all the w a y s

~l~&2~..~&n~0 that the t e r m

arise - and the c o e f f i c i e n t

is the n u m b e r w e w i s h e d

each w a y

of this

diagram

(Xl-i

- -

~i .Xn_l) (X 1

(~i ..... ~ )" n

ri_ 1

a t e r m of type

can arise.

At the

n-I kI k (X 1 ...Xn_l) (Xl ..Xnn) term is xk(p),

which

to c a l c u l a t e .

The idea n o w is to r e p r e s e n t

to the t e r m

to

...s rI

A(X)

- -

the p r o c e d u r e b y ~n) n

associating

~l~...~n~0

, the Y o u n g

is thus r e p r e s e n t e d

b y the e m p t y

.X

t

176

diagram.

Passing

from

A(X) Srl • .. s

is i n t e r p r e t e d

as a d d i n g

A(X) s

, r squares

..s rI

set of

to each

to A (X) Srl... s

ri_ 1

of the

(signed:)

"in all p o s s i b l e

s ri_ 1 r

diagrams

ways"

to g e t

of

a

ri_ 1 (signed)

in A(X) s

...s rI

diagrams

the

terms

of i n t e r e s t

s . ri_ I r

Specifically: transition

representing

is f r o m

in 5(X) Srl

Sri_l

multiplying

b y X qr+ i

the

a term

n-i ~i ~2 ~ ~ ~ ~ 2 (X 1 ...Xn_l) (X 1 X2 . . .Xq.q Xq+ .q+± 1 .X q +q+2

e " " ~ ~ ' " X q +q 1+.-l ±-.i X q +q+i l. X q +q1+.+l.±+.l.

X n n)

to a t e r m n-i ~i ~2 ~ ~ .+r-i+l ~ q + l + l x q+l-z x q+l-± (X 1 ...Xn_ I) (X 1 X 2 ...X q x q+l ~ . ^+i ~ . 4+1 q q+ 1 Xq+ 2 " ""-q+ i- 1 -'q+ i "

•X q + i + l q+i+l

So t h e

passage

is f r o m t h e Y o u n g

X n) n

diagram

(~i' ~2 ' ' ~ q ,6 q + l . . . . . . . . " ~ q + i ' ~ q + i + l ....... ,~ n ) to t h e Y o u n g

diagram

(~i' ~2 .... ~q' ~ q + i + r - i + l ' ~ q + l +l ..... ~q+ i-1 +I '~ q + i + l ..... ~n )

[77

The t r a n s i t i o n

from the old d i a g r a m

d e s c r i b e d as a d d i n g r s q u a r e s row,

add e n o u g h

~q+i_l+l,

in a c e r t a i n way:

s q u a r e s to c h a n g e

i.e.,

to the n e w can b e to the q+i

it from l e n g t h ~q+i to l e n g t h

fill out the r o w u n t i l

it e x c e e d s

the p r e v i o u s

r o w b y one square:

q+i-i q+i

row row

T h e n add e n o u g h to ~ q + i _ 2 + l

s q u a r e s to the q + i - i

- i.e.,

square.Keep

r o w to c h a n g e

e n o u g h to e x c e e d the n e x t h i g h e r r o w b y one

d o i n g this u n t i l you run out of s q u a r e s

Suppose this exhaustion

occurs

a Young diagram

the e x h a u s t i o n

the t i m e s

its l e n g t h

(i.e.,

when

Pictorially,

in the q

a row has one m o r e

a typical

addition

th

row and the r e s u l t

is

d o e s n o t o c c u r at one of

s q u a r e than a p r e v i o u s

row.)

is

q+i

This p r o c e s s

to add.

(here r=13 s q u a r e s are added)

is c a l l e d the r e g u l a r a d d i t i o n

Young diagram n u m b e r of rows

(~I ..... ~n )" involved

of rows i n v o l v e d

It is a p o s i t i v e

(= i+l)

is even.

is odd,

of r s q u a r e s to the application

negative

(This is to k e e p

if the

if the n u m b e r

t r a c k of the ~

signs).

178

Thus our algorithm

for c o m p u t i n g

Let k =

(k I ..... kn )' and p =

Compute

all w a y s

adding,

in a r e g u l a r

t h e n r3, .... t h e n

Xk(p)

can b e d e s c r i b e d :

(r I ..... rn ) b e p a r t i t i o n s

of s t a r t i n g w i t h the e m p t y Y o u n g

r

fashion,

f i r s t r i squares,

so t h a t the r e s u l t

of n.

d i a g r a m and

t h e n r 2,

is the Y o u n g d i a g r a m

k.

n

For e a c h

s~ch way,

count

negative

applications,

negative

applications.

(-i)

(+i)

if t h e r e

if t h e r e

are an odd n u m b e r

are an even n u m b e r

Add up the s i g n e d total,

of (or zero)

the r e s u l t

is

k

x (p). F o r example: H e r e k=(3,3,1) 2 squares, (3,3,1). the n e x t be

X331(2221) and

then

= -3

p=(2,2,2,1).

2 more,

The p r o c e d u r e

then 2 more,

If the f i r s t 2 s q u a r e s 2,

indicated

"c",

and the last,

then i, to get the p a t t e r n

are l a b e l e d

"d",

is first to add

the w a y s

"a",

the n e x t

2,

"b",

of d o i n g t h i s can

as follows: negative

Lb c I

app!ic ations

one

(c)

-i

one

(a)

-i

three

(a,b,c)

-1

t o t a l = -3 = X

331

(2221)

179

Another example:

X(832) (5422) = 2

Allowable patterns

Negative applicat.!ons

Sign

la lal a! al c Ic I d two

(a,b)

+i

[bj

zero

+l

Total = 2

From this algorithm we can deduce certain general results on characters.

a)

xn(p) = 1 for all p Proof:

There is exactly one way to regularly add

p= (rI ..... rn) squares to get the diagram

Namely

This involves /~ero negative applications.

I

In

b)

X

(P) = ~i

Proof:

for all p

There is only one way to regularly add

p=(r I ..... rn) squares to get a vertical column.

For each

r. which is even this involves a negative application. 1

180

Hence

for o6S n, if 0 i n v o l v e s an even of even l e n g t h

in

number

of c y c l e s

X if o i n v o l v e s an odd n u m b e r of e v e n l e n g t h

of c y c l e s

in Hence

{~ I k

(o) = i]

since

a cycle

permutations

Of course, fact

c)

we

is e x a c t l y

og e v e n

length

the

alternating

is a p r o d u c t

group

An ,

of an o d d n u m b e r

and vice-versa.

already

knOew

both

of t h e s e

facts.

But

the n e x t

is n ~ .

Xk(P)

Proof:

in k' 7. (P) X (P)

=

In i n t e r c h a n g i n g

diagram,

a regular

The number where and

number

are

to the c o n j u g a t e application

application

remains

turns

into

And

If r is even, a positive

r+l,

if r is odd,

to a diagram,

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

positive.

regular.

equals

Hence

is also n e g a t i v e .

of k)

of a Y o u n g

of c o l u m n s

added.

application

diagram

remains

of s q u a r e s

the n u m b e r

involved,

the c o n j u g a t e

and c o l u m n s

of s q u a r e s

a negative

of rows

rows

application

r is the n u m b e r

negative

the

of rows p l u s

one h a s

(k' b e i n g

so an e v e n

application a positive

then

a

application,

and

vice-versa. So

for e a c h w a y

can build

k'.

of b u i l d i n g

k from

If p is a p o s i t i v e

p = ( r I ..... rn),

class,

i.e.,

one

if p has

an

181

even number

of even cycles,

p is n e g a t i v e ,

there

there

is a c h a n g e

is no c h a n g e in sign.

An a l t e r n a t i v e w a y to w r i t e t h i s t h e o r e m all k p n.

Thus,

a

all ~ ~ n,

.~> h ,

all X F n,

all n.

explicitly, all n

the m a p (p. 139)

in sign.

0:A-->A,

is a *X = X n given by

is g i v e n b y X k ~

X

,

If

182

The third calculation

to be done is to determine

transition matrix between the bases of A . n

We start with the formula

h (xy) = n

~ k~n

{k] (x) {k] (y) =

Given a partition {k] = ~ _

rk h ~

r~ [% ] I ~ n ]

and { h

I ~n]

(pp.38,43)

~ ~n

h (x)<~>(y)

k | ~ n, consider

in An.

the

the problem of expressing

Unique integers

rk~ exist making

g~n

this an identity,

since the set [h _

I ~ n ] _ is a basis of An.

Inserting this into the formula above, we get ~k~n

~ ~n

~ ~n

rkrTh1~(X) {k] (Y)

Since the

{h ] form a basis,

particular

~ must be equal: %~-~" rk {k] (Y) = %Pn

h (X) <11">(y)

the coefficients

of

h , for a

(X)

So
~

-~-

k[-n

o£S

Fix k and ~ and let us compute coefficient problem

of, among others,

rk~ n

rk .

This integer is the

the term

is to compute the coefficient

<~>(Y) A(Y) =

kl+n-i k (sgn ~) Yo (i) n "''Y0(n)

~ YO I) .... YJ(n) °£Sn

kl+n-i kn Y1 .... Yn "

So the

of this term in

S

(sgn T)YT(1)...Y n

(n_

183

I.e.,

we have

to decide

terms

on the right

for w h i c h

~,~

side here m u l t i p l y

kl+n-i kn So we start w i t h Y1 .... Y and ask n it into

41 ~n (y (1)...YG(n))

a product

such w a y

a ~ sign

6S , the c o r r e s p o n d i n g n kl+n-i kn to give Y1 .... Yn

for all ways

of d e c o m p o s i n g

n-i 1 (Y (i) ...Y (n_l)).

(= sgn T) is attached,

To each

and rk~ is the

sum of

these +i. In terms

representing

the

first

of Y o u n g

kl+n-i kn Y1 ..... Yn "

• (2) row,

start w i t h

kI k the term Y1 ....... Ynn"

row,

subtracting

diagrams,

n-2 to the second,

Now,

n-1

etc.,

dividing

squares where

from w h i c h k squares

by Y

from the

Add

etc.,

in n-i

squares

n-I n-2 1 (1)YT (2) "''YT (n-l) ~(i)

(k I .... kn)

to

to get the d i a g r a m

row,

n-2

the only c o n s t r a i n t are to be

the d i a g r a m

subtracted

of

amounts

squares

to

from the

on T is that

a row

must

at least

contain

k squares. This formally

expression

for

as follows.

and a p e r m u t a t i o n (kl+n-~(n),

~6S

{k] in terms

Given

n

of h ' s

a partition

, write

o*k

k =

can be

expressed

(k I ..... kn)

of n

for the p a r t i t i o n

k2+(n-i ) -O(n-l) ..... kn+l-~(1))

of n.

(If any of

184 these numbers expression

are negative,

G*k is undefined.)

Then the

derived above is

sgn (~) ho, k n This looks like the expression fact it is:

of some determinant,

[k] = det I h k _s+t I . s hkl

{k] =

det

hkl+l

hx2-1 hk2 i ....

I.e.,

hkl+2

expression

"'"

hk2+l

hkl+n

i

hk2+n-11

....

i hk -n ~ n

(where, by definition,

and in

h 0 = l, and h

hk

q

for [k] is the Jacobi-Trudi

n

= 0 for q<0.) equation.

This

185

We carry out the computations

If k = (lll), we start with

~

for the case n=3.

, add in (2,1,0)

squares

and then there are four ways to subtract two

to get ~

squares from some row and one from another:

2 minus

[ I

1

yielding

(i,i,i)

210. sgn (210) =+±

yielding

(2, i, 0)

210 sgn (201) =-i

(2,1,0)

210 1 sgn (120)=-

0

minus

0 1

minus ~--~

1 2 0

u-'

yielding

i * ~ ....

0

~

minus

2

210

yielding :

i =

(3,0,0)

sgn (021) =+±

1

Hence

a(lll), (iii) = i,

a(lll), (210) = -2,

a(lll), (300)

1

186

Similarly,

starting with k = (2,1,0) =

(2,1,0) squares gives

~

, adding in

, and there are two ways

to subtract two squares from some row and one from some other:

1

1

minus

1

yielding

= (2,1,0)

sgn(210)=+l

[/= (3,0,0)

210 1 sgn(120)=-

0

1 ~ m i n u s

2

yielding~--~

0

so a(210), (300) = -i,

Similarly

a(2,1,0), (2,1,0) = i,

a(2,1,0), (3,0,0) = 0.

a(3,0,0), (3,0,0)=I, a(3,0,0), (2,1,0)=0,a(3,0,0), (1,1,1)=0.

Henc e [3}

=

lh 3

+

0h21

+

0hll 1

{21]

=

-lh 3

+

lh21

+

0hll 1

lh

-

2h21

+

lhll I

{ill? =

and

3

so the transition matrix is

-

1

-2

0

1

I87

Note in the calculation above, the final transition matrix in triangular:

It is a general fact that the transition matrices

relating the basis of Schur functions to any of the bases h ,a ,<~>, and f

are triangular and this follows from some

observations

on the algorithm above,

as we now show.

Definition:

On the set H(n) of partitions of an integer n, the

natural partial ordering is defined as the least ordering in which,

given ~l,~26H(n),

and GESn, with ~l*C = ~2" then ~ i ~ 2 .

We could also consider the two linear orderings defined by, for

I

(k I ..... Xn),(kl', .... kn ) £H(n),

(kl'''''kn)>l(kl"

. .,k~) . . i f . k l = k i ,.

k i = k [ , b u t ki+l>k'i+l

and (k I ..... Xn)>2(k 1 ..... kn) if kn=kn,..,ki+l-ki+ I -' Thus the first

is lexicographic b y largest part first,

second is lexicographic b y smallest part last carefully, kI ~ k2 ~

,but ki
all partitions

and the

(where, note

of n are taken to have n parts

..kkn~0 , and in general the smallest parts kn, kn_l,etc-

will be zero). It is straightforward to show that if ~i'o = ~2,for ~l,n2E~(n), and ~£S n, then b o t h ~i~i~2 and ~i~2~2 .

In particular this shows

that the binary relation on ~ i , ~ 2 : n l ~ = ~ 2 a partial ordering in w h i c h ~i<__~2 and ~ i

for some o, extends to implies ~i=~2 .

188

In the algorithm

=Z~ i k

above,

rk = 0 unless

rk h ~, and In] = hn,

if the b a s e s

[ [~] \ ~

any linear ordering transition

matrix

n} and

[n-l,l} [ h

which contains

is triangular.

(clear from the algorithm)

following

\ ~

the natural

~y k =

part,

ordering

Thus

the all k,

is of the form

is equivalent

"unipotent") .

to each of the

on U(n).

: Let 1 ~ i ~ j ~ n, and k 6 H(n) be a partition:

Raising

unit

orders

(kl,X2,..,kn).

Rji(k)

order,

In fact since rkk=l,

the matrix

etc.

n} are each ordered b y

(so-called

The natural

partial

Thus

= -hn +hn-l,l'

© Proposition:

k ~ ~.

Operator

Suppose

ki_l>k i and kj>kj+ I.

R.. associates 31

to k the partition

=(k!,k2,..,ki+l,ki+l,..,kj-l,kj+l,..kn

is removed the i

th

from a small part, We say,

Young Raising

That

is, one

, and added to a larger k ,k' of n, that

from k b y applying

the class

of convex

[i .... ,n] to itself.

is less than or equal a partition

).

a finite

series of

Operators.

: Consider integers

the j

th

for two partitions

k ~ y k' if k' is obtained

The Y o u n g

k =

to 2f(i),

functions

(Convex means for each

from the set of

that f(i+l)+f(i-l)

i = 2 ..... n-l.)

(k I ..... k ) of n, consider n

the convex

Given

function

189

fk given b y fk(i) function

f from

= k I + 12 +...+ k i.

{i ..... n~ to itself

the form fk for k =

(f(1),

two sets are in one-one define

functions,

Proof:

i.e.,

a convex

f(n)=n

f(2)-f(1) ..... n-f(n-l)).

k,k'

is of So the

and it makes

sense to

of n that I ~D k' if fk ~ fk'

fl(q) i fl,(q)

implies ~ ~ ' .

that if ~ *~=~', of raising n,n-l,.,

to get

distinct.

Let ~ =

from w h i c h the number

tracted

mutation,

a series

are now all

from the first number, Consider Suppose

rows.

Then ~*G = R

to a partition

the number

the number

it is larger Let p,q,

Let ~' be a followed qP

(~*~').

Proceeding

from which

But this can only happen so ~*~I = ~"

a series

of the form ~*al, where,

from which the number

is less than the number

for all i.

o(n)

Let i6{i .... n}.

(i,i+l).

is to show

we can assume ~*a is obtained b y applying

of raising operators ~*~i'

The numbers

of these

likewise

A d d in the numbers

i+l is to be subtracted.

be the indices

in this fashion,

(41 , .... ~n ) .

i is to be subtracted.

from w h i c h

b y the t r a n s p o s i t i o n

computing

etc.

essentially

from ~ b y applying

one now subtracts

from the second,

respectively,

remains

(~l+n,~2+n-l,..,~n).

To get ~*~,

than the number

What

then ~' is obtained

operators.

as

for all q=l,,,n.

It is easy to see that ~ y and ~ D are equivalent,

that ~ ~ y ~

g(n-l)

satisfying

correspondence

for two partitions

Conversely,

in

i is to be sub-

i+l is to be subtracted,

if o I is the identity perf

190 Another property of the natural ordering is that for ~,k 6 H(n), w ~ k if and only if, taking conjugate partitions, ~' < ~'. One might conjecture

that the natural ordering is also

obtained as the intersection of the two linear orderings ~i,~2 above.

But this is false,

(6,6,2,2)

as (7,4,4,1)

is greater than

in each of these, but the two are incomparable under

the natural order. This partial ordering on partitions has been investigated by, among others, Doubilet Inversion Theorem on it Liebler and Vitale

([14]) who has proved a Mobius

and also b y Brylawski

([9]) and

([25]) .

Young originally gave the algorithm in terms of his Raising operator,

and conversely the expression of the h

of the { ~ ' s

b y means of a lowering operator.

T = with

s in terms

(See

Now that we have the transition matrices R = {k} = ~ r k

!

[39]).

(rk~),

h ~, it is a simple matter to compute the matrix

(tk~),

tk~{~}.

To find tk~, dot this equation

[~}: = (,

[~)

~tk~({~},{~})

= t~

(using the orthonormality of the [~}'s)

191

But

{~] = ~ r w

since the h the m a t r i x

h

so

's and 's

Theorem.

the transitions

matrix U =

= T - R =

(and easy to prove) is a symmetric

Recall

gives

known

h ~ is just the product

and

{~} = ~ r h

R transp°se-

R.

of

so the

It is a general

fact

of a matrix b y its transpose

matrix W =

the conjugate

a *{k] = ~ w k ~ a n * a n

Hence wk~ = rk, ~. calculation

Hence tk =r k, so

= ~ u k

h p,

for all k,~.

a *{k] = [k'} n Thus

~k

Hence in the expression

is the transition

a *a =h n ~ ~

= r

T is also triangular.

that the product

matrix.

we have uk =u k

Recall

= ~ U k

= ~ t k ~ { W } ,

(uk~)

(<X>,h)

of R, a fact traditionally

In particular

N o w the transition

r

are dual bases .

T is the transpose

as Kostka's

Next

(, {~]) = ~ w

Similarly the Schur

that passing

turns

the partial

(rk~)

is triangular

f

(wkw),

partition, ~

from a partition

ordering implies

upside that

down. (wk~)

so this

in terms

a ~.

and an*h =a w and

so {k'] = ~

'

= an*,

functions

{k} = ~ w k

wkwh w.

same sort of

of the f .

to its conjugate Hence

is also.

just

the fact that

192

Notice k£H(n),

in p a r t i c u l a r

(ak,,h X) = i.

still g e t s

this b e c o m e s

=

This

is

in

(hk,ak,)=l

This

that

shape

n

(hk,ak,)=l

] - the Y o u n g

"positive

[~}.

1 ,

with

a , n

k' b y k , ~' b y ~,

Hence

there

correspondence

, (by u s i n g

is then

the M a c k e y

a standard

Theorem,

are b o t h

between

actual

exists say)

k X -

is first and

representations,

one and only one irrep

labeled

in

The m a i n p r o b l e m irrep

of S

n

arises

in

k. of A l f r e d Y o u n g

(for w h i c h

see L 3 9 J

c o m e s up in his c o n s t r u c t i o n symmetrizers.

is c o n s t r u c t e d

symmetric

Indeed,

) that the c o r r e s p o n d e n c e

since h k and ak,

for a u n i q u e

k~n,

r

is to show that e v e r y

In the a p p r o a c h

in ~[S

Relabel

{k} of S . n

that t h e y h a v e

irrep

of course,

fact that

{~'].

~Z~

entries

the inner p r o d u c t

the n a t u r a l

Ii2 J ~9J

fact i m p l i e s

this w a y

[k} +

k of n and irreps

then to o b s e r v e

common.

r

R one

([k], {k}) = I. QED.

(e.g.,

to c o m p u t e

this

ak, =

~ W>k

for each

the m a t r i x

diagonal

Taking

[~}

{k'} +

phenomenon:

inverting

matrix with

fact again gives

partitions proof

r

a~ = A

and the result

(hk,ak,)

Proof:

a triangular

thus h k = {k} + ~ "

the f o l l o w i n g

group

on the

For each Y o u n g

an e l e m e n t P(k)

of

), the

idempotents

tableau

of

£ ~[S n] - the

rows of k" and an e l e m e n t

193

N(k)

6 ~ IS ~ - the n

of k"

"negative

and then one shows

is an o r t h o g o n a l

symmetric

that

{

group

on the c o l u m n s

--~ P(k)N(k) all Y o u n g t a b l e a u x of shape k

I ki- n}

set in ~ IS I. n

The c o m b i n a t o r i a l

aspect

of the r e p r e s e n t a t i o n

theory

the s y m m e t r i c

group

is the s t u d y of t h e s e

transition

(rk~),

etc.

An e n t i r e l y

approach

(tk~),

subject

(in, e.g.,

symmetric <~> = ~

functions

Algorithm. properties

and the

R .

This

are derived.

representation

to us, m a y

bodily

elements

approach,

certain

in that context.

one has v e r y n a t u r a l

not h a v e o c c u r r e d

, wk to us,

It seems

has yet to b e w r i t t e n .

, etc.,

Knuth

thus

and their of

shortening

which now appear natural On the other hand,

interpretations

in

of the

w h i c h we c o u l d derive,

synthesis

(w~k),

the w h o l e n o t i o n

had not the c o m b i n a t o r i a l

the final

W =

functions

ignored,

notions

to the

R for w h i c h

is done b y the i n g e n i o u s

In this

However,

rk

the m a t r i x

Schur

matrices,

[ 3 8 J ) starts w i t h

of the m a t r i x

can be c o m p l e t e l y

look ad hoc

that approach,

found them.

symmetry

Then R is u s e d to d e f i n e

the e x p o s i t i o n .

matrix

['437 , and Rota

w kh k and c o n s t r u c t s

W = R transp°se"

group

Stanley

different

of

but might

approach

first

of the two a p p r o a c h e s

194

BIBLIOGRAPHY

Eli

Adams, J.F., ~ -Rinqs and lecture, 1961; Adams,

~-Operations,

J.F., Lectures on Lie Groups,

(unpublished

1969, Benjamin

Atiyah, M., Power O~erations in K-Theory, Quart. J. Math., (2) 17 (1966), 165-93. (Also reprinted in Atiyah, M., K-Theory, 1967, Benjamin [4~

Atiyah, M., and D.O.TalI, Group Representations, -Rings, and the J-homomorphism Topology, 8, 1969, 253-97

[9

Bergman, G.M., Ring Schemes: The Witt Scheme, Chapter 26 in D. Mumford Lectures on Curves on an Algebraic Surface, Princeton, 1966

L63

Berthelot, P., Generalities sur les ~-Anneaux. Expose V in the Seminaire de Geometrie Algebrique, Springer-Verlag Lecture Notes in Mathematics 225, 1972

[~

Bir~hoff,G., and S. MaeLane, A Surve Z of Modern Alqebra, 3 Edition, 1965, MacMillan Co. Boerner,

LgJ

H., Representations

of Groups,

1970, North-Holland

Brylawski, T., The Lattice of Integer Partition~s, of North Carolina Dept. of Math. report, 1972

university

Burroughs, J., Operations i__qnGrothendieck Rings and Grou~ RePresentations, Math. Dept. Preprint 228, State University of New York at Albany

LIO

Cartier, P., Groupes formels associes aux anneaux de Witt generalises, C.R. Acad. Sc. Paris, t.265, 1967, A-49-52

Coleman, A.J., Induced Representations with Applications to S and Gl(n), Queen's Papers in pure and Applied Math No.4, Queen's University, Kingston, Ontario, 1966

El%

Doubilet, P., Symmetric Functions through the Theorff o__ff Distribution and O c c u r , (No.VII of G.C. Rota's On the Foundations o__ffCombinatorial Theory) (to appear) Doubilet, P., An Inversion Formula (mimeographed notes, 1972)

involving

Partitions,

195

[19

Dress, A., R__epresentations of Finite Gro_~, Part i, The Burnside Ring, (mimeographed notes, Bielefeld, 1971)

[36]

Foulkes, H.O., O__nnRedfield's Grou~ Reduction F_unctions, Canadian J. Math, 15, 1963, 272-84 Frame, J.S., G.de B.Robinson, and R.M.Thrall, The Hook Lengths o f ~ , Canadian J. Math., 6, 1954, 316-325 n

L183

Grothendieck, A., La Theorie des Classes de Chern, Bull. Soc.Math.France, 86,1958,137-54 Grothendieck,A., Classes de Faisceaux et Theoreme de Riemann-Roch, (O, Appendix, in Seminaire de Geometrie Algebrique, Springer-Verlag Lecture Notes in Mathematics No. 225, 1972) Hall, P., The Algebra of Partitions, Proc. 4thcanad.Math. Congress, Banff, 1957 (1959) 147-159

[20

Harary, F., and E. Palmer, The Enumeration Methods of Redfield, Am.Journal Math., 89, 1967, 373-384 Hawkins, T., The Origins of the Theory of Group Characters, Archive for the History of the Exact Sciences,VI__II,2,1971, 142-70 Koerber,A., ReDresentatio~ o_f Permutation Groups~, 197~, Springer-Verlag Lecture Notes in Mathematics, No. 240

[20

[20

Lang, S., Algebra, 1965, Addison-Wesley Liebler,R.A. and M.R.Vitale, Ordering the Partition Characters of the Symmetric Group (to appear) Littlewood, D., A Universit~Algebra,

Heinemann, 1950

Littlewood, D., Plethysm and the Inner Product of S-Functions, J.London Math. Soc., 3_22,1957, 18-22

L20

Littlewood, D., The Inner P!ethysm of S-Functions, Canadian J. Math., i~, 1958, 1-16

[29

Littlewood, D., The Theory of Group Characters, 2nd Ed.,1958, Oxford

196

[30]

1 31]

MacMahon, P.A., Combinatory Analysis, Vol. I,II, Cambridge, 1915, Reprinted Chelsea, 1960 MacMahon, P.A., table of the number of partitions of n, n ~ 200, in G.H.Hardy and S.Ramanujan, ~ Formulae in Combinatory ~ , Proc.London Math. Soc.,17,1918, p.l14 Milnor, J., Lectures on Characteristic Classes, notes, Princeton, 1957)

(mimeographed

Polya,G., Kombinatorische Anzahlbestimmungen fu__~rGruppen, Graphen, und Chemische Verbindungen, Acta Math, 6_8, 1937, 145-254 Read, R.C., The Use of S-Functions in Combinatorial Analysis, Canadian J. Math, 22 , 1968, 808-841 Redfield, J.H., The Theory of GrouR Reduced Distributions, Am. J. Math., 49, 1928, 433-55 Robinson, G. de B., ReRresentation Theory of the Symmetric Group, 1961, University of Toronto Press Rota, G=C., Combinatorial Theory, Notes by L. Guibas, Bowdoin Summer Seminar in combinatorial Theory, 1971 Rota, GTC., The Combinatorics of the Symmetric Grou~ , Notes from a Conference at George Washington University, June 1972, (to appear) Rutherford, D.E., Substitutional Analysis, University Press Salmon, G., Modern Higher Algebra,

E4g E42]

1948, Edinburgh

1885, Cambridge

Serre, J.-P., Representations Lineares des Groupes Finis, 1967, Hermann Snapper, E., ~ Characters and Nonnegative Integral Matrices, J. Algebra, 19, 520~35, 1971 Stanley, R.P., Theory and A pplicatio n of Plane Partitions, Studies in Applied Math., 5__O0,1971, 167-88 Swinnerton-Dyer, H.P.F., ~plications of Algebraic Geometry to Number Theorz, Proc. Symp. Pure Math, A.M.S., 1969, Number Theory Institute

197

~

Weyl,

H.,

The Classical

~6~

van der Waerden,

~7~

Young, A., Q u a n t i t i v e S u b s t i t u t i o n a l A n a l y s i s , I - VII, ( p u b l i s h e d at v a r i o u s t i m e s 1 9 0 1 - 1 9 3 4 in Proc. L o n d o n M a t h . Soc. - s e e R u t h e r f o r d ~39~ )

B.L.,

Group_s, Modern

1949,

Algebra,

Princeton Ungar,

1950

198 INDEX

a

2

OF N O T A T I O N

55

W R

n

K(F)

6

S

A V

6

R(G)

70

kt(a)

8

CF(G)

81

P n ( S l .... s n ;c~n ~' "''~ n )

K, i

89

12

L.

90

n

59

n

1

[rq

96

P n d ( Sl .... Snd)

12

1 + A[[t]] +

15

A

25

B(G)

107

a

28

S C F (G)

ii0

28

~I~[G]

115

v) 29

R(S)

128 130

h n

TT'

(conjugate

of

99

~Wn

29

®

If(n}

29

S

17 1

137

*S

2

h

30

<~>

32

s

35

139

A(X)

39

145

{k]

43

~/n

47

H

151

R

5O

X

151

S

.S

rrl

~2

~Prr

laij Ik

137 137

147

t99

153

T =

(tk~)

190

162

U =

(uk~)

191

182

W =

(Wk~)

191

o*k

184

P(k),N(k)

192

R.

189

ITI

f

R

=

.

13

(rk~)

the n u m b e r in a set T

of e l e m e n t s passim

lO0

INDEX

Adams O p e r a t o r s

47

Ferrar's graph

algebraic g e o m e t r y

52

finite degree (of an element in a k-ring) 8

b i n o m i a l coefficient, generalized b i n o m i a l type

153

forgotten symmetric

29

function 162

9

Brauer's Theorem

i01

B u r n s i d e Ring

107

Frobenius Character Formula

163

Frobenius R e c i p r o c i t y C a u c h y ' s Lemma

40

central

81

function

centralizer

105

character

84

c h a r a c t e r ring

84

F u n d a m e n t a l Theorem, T h e o r y of S n

Rep. 135

F u n d a m e n t a l Theorem, symmetric functions G-module -

c h a r a c t e r table

74

- -, map of

2 61 61

90 - - -, i s o m o r p h i s m of

c h a r a c t e r of p r o d u c t of two g r o u p s G H 97 c h a r a c t e r s of S n' integrality

G-set; -

G-map

- -, sum of

104 106

134 - - -, p r o d u c t of

characters,computation

62

106

101 - - -, symmetric p o w e r 106

conjectures

i00,i13,135

conjugate partition

group a l g e b r a

115

group d e t e r m i n a n t

122

29

cycle

124

cycle index

146

group reduction formula 146 H o m o g e n e o u s p o w e r sum cycle structure

124

dot p r o d u c t

138

in R(S)=A

30

hook n u m b e r

172

immanent

147

201

indecomposible

76

induced character formula inner

automorphism

inner p r o d u c t (=A) irreducible

96 105

in R(S) 138

character

84

k-ring, n a t u r a l o p e r a t i o n on - - -, p r o d u c t

28

isotypical

78

- - -, t e n s o r p r o d u c t of two

21

-

,

lattice permutation

monomial

isotypical

component

Jacobi-Trudi

Equation

K-Theory Knuth

Theorem

k-ring ---, -

-

binomial

- -, c a t e g o r y

of

- -,of central functions

---,

definition

---,

finitary

-

-

184 27

Algorithm

Kostka's

79

21 15

-

Theorem

Maschke's isobaric

of two

special

-

Mackey's

irrep ( = i r r e d u c i b l e representation)

25

99

Theorem

76

group

103

natural ordering partitions

on 187

Newton' s F o r m u l a s normal

167

dot p r o d u c t

35 156

193

normalizer

105

191

orbit

104

5

orthogonality

9

outer product

relations

91 127

20

partition

54

partition, natural o r d e r i n g on set of

187

13

permutation

109

8

plethysm, outer

29

matrix

inner

and 135

- -, free on one generater

24

power

sums

35

- -, m a p

20

pre-k-ring

7

of

202

pre-~-ring

49

Schurfunctions

43

Y-ring

49

Schur's Lemma

77

semidirect p r o d u c t

98

regular addition of squares

177

representation,linear conjugat~

of

semi-simple

116

simple G-set

105

60 69 Splitting Principle

decomposible

76

degree of

60

dual of

68

18

standard Young tableau 168 super central function ii0

exterior power faithful

super c h a r a c t e r

ii0

super c h a r a c t e r table

113

68 65 symmetric

induced

73

inner p r o d u c t

72

irreducible

76

permutation

64

-

function

- -, e l e m e n t a r y

- - -, forgotten

-

-

p r o d u c t of

-

76

- -, regular

64

-

-

-

-, Sum of

-

S-functions

2

- -, hom. p o w e r sum

30

- -, m o n o m i a l

32

- -, p o w e r sum

35

- - -, Schur

43

symmetric p o w e r

46

torsion free ring

49

64

trace

83

67

transitive

105

70

t r i a n g u l a r i t y of transition m a t r i c e s

189

64 of S

- -, a l t e r n a t i n g

representation

162

68

- - -, trivial representation canonical

28

68

- - -, reducible

-

- -, Fund. T h e o r e m

2

ring

n'

44

203

Verification Waring

Principle

formula

Wedderburn's

27 35

Theorem

i16

Witt vectors

56

wreath

98

product

Young

diagram

Young

Raising

29 Operator

189

Young Symmetrizing Operator

119

Young

tableau

168

zeta-function

53