Characters

Let T be a representation of a group G over a field K and M the corresponding KG-module. Then the trace TrM(a) with respect to the module M is defined for every element a E KG (Sect. 6.3); it is the trace of the matrix T(a) (in any basis). In particular, for every element x E G, we get the field element X(x)

=

TrM(x). The function X : G -+ K is called the character of the representation T. If T is irreducible, then X is called an irreducible character.

The character of the regular representation is called the regular character and is denoted by Xreg •

Proposition 7.3.1.

() { n for x = 1,

Xreg x = 0 for x

¥=

1.

The proof is obvious.

Proposition 7.3.2. For every character, X(gxg- l )

=

X(x). In other words, a character is constant on each conjugacy class.

Proof. For every representation T, T(gxg- l ) = T(g)T(x)T(g)-l, and the similar matrices T(x) and T(gxg- l ) have the same trace. 0

Observe also that, as an immediate consequence of Corollary 2.6.3, we get the following theorem.

Theorem 7.3.3. Let K be a field of characteristic O. Then every representa- tion is determined uniquely by its character, i. e. equality of characters implies similarity of representations.

Now, let the field K be algebraically closed and Xl, X2, ... ,Xs be all the irreducible characters of the group G over the field K. Denote by Xii the element Xi(9i), where gi E Ci (Cl, C2 , ••• , Cs are the conjugacy classes of the group G). The square matrix X

=

^(Xii) is called the character table of the group G over the field K. Let us remark that KG ~ EEl s diMi, where Mi is the module of the ith irreducible representation and di=l i = [Mi : Kj; hence,

s Xreg

= E

diXi.

i=l

As we have already pointed out, the elements Ci =

E

^{x, i = 1,2, ... s,}

xEGj

form a basis of the center C of the group algebra KG. On the other hand, C ~ KS and therefore, if 1 = el

+

e2

+ ... +

es is a decomposition of the iden- tity of the algebra C, then {el' e2, ... , es } is also a basis of C. Consequently, there are elements Qij and fiii in the field K such that Ci =

E

s ^{Qijej and}

i=l

7.3 Characters 121

s

ei =

I:

f3ijCj; and thus the matrices A = (Qij) and B = (f3ij) are reciprocal.

It turns out that the coefficients j=1 Qij and f3ij are closely related to the character table.

Proposition 7.3.4. Denote by di the dimension of the irreducible represen- tation with character Xi and hj the number of elements in the class Cj. Then

di -1

f3ij

=

-Xi(gj ), where gj E Cj.

n

Proof. Observe that the element ej acts on the jth irreducible representation as identity, while the elements ek (k =I-j) act on it trivially. Therefore Xj(ek) = 0 for k =I-j and Xj(ej) = dj. From here,

8 8

Xj(Ci) = Xj(I>ikek) = LQikXj(ek) = djQij.

k=1 k=1

On the other hand, Xj(c;) = hiXji and the formula for Qij follows.

In order to compute f3ij, we use the fact that Xreg =

I:

s diXi. Ob- serve that Xreg(Ckg)

=

0 if g-1

1.

Ck and Xreg(Ckg)

=

n if g-1 i=1 E Ck (this follows from Corollary 7.3.1). Therefore, if gj E Cj, then Xreg(ei9;1) = Xreg (

t

f3ikCk9;1)

=

nf3ij . On the other hand, Xreg( eig;1)

= t

dkXk(eig;1)

k=1 k=1

= diXi(9;1) because Xk( eig;1) = 0 for k =I- i and Xi( eig;1) = Xi(g;1). The

formula for f3ij follows. 0

Taking into account that the matrices A and B are reciprocal, we obtain immediately the following "orthogonality relations" for characters.

Theorem 7.3.5.

for i =I-j, fori=ji for i =I-j, for i = j.

Corollary 7.3.6. A representation T of a group G over an algebraically closed field of characteristic 0 is irred1lcible if and only if its character X satisfies

122 7. Representations of Finite Groups

Proof. Decompose the representation T into a direct sum of irreducible repre-

s

sentations. Correspondingly, the character X can be expressed as X =

I:

^{miXi ,}

where XI, X2,· .. ,Xs are irreducible characters. But then i=1

and this sum is equal to 1 if and only if X

=

Xi for some i, i. e., in view of Theorem 7.3.3, if T is an irreducible representation. D If ]{ = <C is the field of complex numbers, then the orthogonality relations can be given a slightly different form. To that end, we introduce the following lemma.

Lemma 7.3.7. If X is the character of a d-dimensional representation of a group G over the field of complex numbers, then, for every 9 E G, X(g) is a sum of d n-th roots of unity and X(g-I) = x(g ), where as usual,

z

is the complex conjugate of ihe number z.

Proof. Since gn

=

^e, we get (T(g)r

=

^E for every element 9 E G. Since the polynomial xn - 1 has no multiple roots, it follows that the matrix T(g) is similar to the diagonal matrix

(

CI C2

0)

T(g) '"

o

^{Cd where}

^ci

^{= 1.}

From here, X(g) = CI

+

^C2

+ ... +

^{Cd and}

o o

This results in

( -I) -1

+

^-I

+ +

^{-I - -}

+ - -()

X 9 = cI c2 . . . Cd = CI

+

^C2

+ . . .

^{cd = X 9 .}

D In particular, Xi(gjl)

rem 7.3.5 take the form

Xij and the orthogonality relations of Theo-

1

^s

{O

;: L

^{hkXikXjk = 1}

1

k=1 ⁸

{O

;: L

^{XkiXkj =} Ilh i k=1

7.4 Algebraic Integers

7.4 Algebraic Integers 123 for i =1= j,

for i

=

j;

for i =1= j, for i

=

^j.

In this section we shall need some properties of algebraic integers. Recall that an algebraic integer is, by definition, a (complex) root of an equation xm

+

alXm-1

+ ... +

am = 0 with integral coefficients ai.

Proposition 7.4.1. A rational number which is an algebraic integer zs an integer.

Proof. Let z be a root of an equation Xffi

+

alXm-1

+ ... +

am = 0 with integers ai and z = pi q with relatively prime integers p and q

>

1. Passing to a common denominator, we get pm = - a1Qpm-l - azqZpm-Z - ... - amqm.

This is impossible because p and q are relatively prime. 0 The following lemma provides a convenient criterion for a number z to be an algebraic integer.

Lemma 7.4.2. In order that z be an algebraic integer, it is necessary and suffi- cient that there exist complex numbers Yl, Y2, ... ,Yt such that ZYi =

L

t aijYj , where all aij are integers and not all Yi are zero. j=1

Proof. If z is a root of an integral equation xm

+

alXm-1

+ ... +

am = 0, then we may take, trivially, Yl = 1, Y2 = Z, ... , Ym = zm-l.

Conversely, let Yl, Yz , ... ,Yt have the required property. Denote by A the matrix (aij) and by Y the column vector whose coordinates are Yl, Y2, ... , Yt . Then (zE - A)Y

=

0 and thus det (zE - A)

=

O. However, the determinant det (zE - A) = zt +alzt-1

+ ... +

at, where ai are integral linear combinations of products of elements of the matrix A and thus integers. We conclude that

z is an aigebraic integer. 0

Corollary 7.4.3. The sum and product of algebraic integers are algebraic integers. In other words, the algebraic integers form a ring.

t

Proof. Let Yl, Yz, .. ·, Yt be complex numbers such that ZYi =

L

aijYj (with

r j=1

integers aij) and y~, y~, ... ,y~ such that Zl yi =

L

^{a~j yj} (with integers a~j)' j=1

124 7. Representations of Finite Groups

Then one can see easily that the numbers {Yiyj

I

^{i =} 1,2, ... , t; ^J 1,2, ... ,r} satisfy similar conditions for the numbers z

+

^Zl and ^ZZI. 0

Since the roots of unity are obviously algebraic integers, we obtain the following corollary of Lemma 7.3.7.

Corollary 7.4.4. If X is a character of a group G over the field of complex numbers, then X(x) is an algebraic integer for every x E G.

We shall now employ the notation of the previous section. In particular, let X =

(X

^ij) be the character table of a group G over the field of complex numbers.

Theorem 7.4.5. All numbers D:ij =

d:

hi Xji are algebraic integers.

J

Proof. Note that, for all i and j, CiCj is an element of the center of the algebra {;G. On the other hand, CiCj is an integral linear combination of the elements of the group G. It follows that CiCj =

L:

rijkCk , where rijk are integers. Besides,

k

CiCj =

(2::

^D:ipep)

(2::

^{D:jqeq) =}

2::

D:;pD:jpep

p q p

and Ck =

L:

D:kpep ; thus CiCj =

L:

rijkD:kpep and D:ipD:jp =

L:

rijkD:kp' Writing

p k,p k

Z

=

^{D:ip, Yj}

=

^D:jp (for a fixed p), we can apply Lemma 7.4.2 and conclude

that D:ip is an algebraic integer. 0

Corollary 7.4.6. The dimensions di of irreducible complex representations divide the order of the group.

Proof. Rewrite the list of the orthogonal relations of Theorem 7.3.5 to the form

~

^hkXik ,'( -1) _

~

L d. X, gk - d'

k=l I ,

S· lnce -d-'-^hkXik

,

= D:ki an d Xi gk (-1) are age ratc mtegers, a so t e num er l b " 1 h b -d' n.

,

IS

an algebraic integer. As a rational number, it must be an integer, as required.

o

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