Mathematical Methods for Physics and Engineering : A Comprehensive Guide

REPRESENTATION THEORY

(d) No explicit calculation is needed to see that ifi=j=k=l= 1, with

Dˆ(λ)=Dˆ(μ)=A 1 (or A 2 ), then each term in the sum is either 1^2 or (−1)^2

and the total is 6, as predicted by the right-hand side of (29.13) sinceg=6

andnλ=1.

29.6 Characters

The actual matrices of general representations and irreps are cumbersome to

work with, and they are not unique since there is always the freedom to change

the coordinate system, i.e. the components of the basis vector (see section 29.3),

and hence the entries in the matrices. However, one thing that does not change

for a matrix under such an equivalence (similarity) transformation – i.e. under

a change of basis – is the trace of the matrix. This was shown in chapter 8,

but is repeated here. The trace of a matrixAis the sum of its diagonal ele-

ments,

TrA=

∑n

i=1

Aii

or, using the summation convention (section 26.1), simplyAii. Under a similarity

transformation, again using the summation convention,

[DQ(X)]ii=[Q−^1 ]ij[D(X)]jk[Q]ki

=[D(X)]jk[Q]ki[Q−^1 ]ij

=[D(X)]jk[I]kj

=[D(X)]jj,

showing that the traces of equivalent matrices are equal.

This fact can be used to greatly simplify work with representations, though with

some partial loss of the information content of the full matrices. For example,

using trace values alone it is not possible to distinguish between the two groups

known as 4mmand ̄ 42 m,orasC 4 vandD 2 drespectively, even though the two

groups are not isomorphic. To make use of these simplifications we now define

the characters of a representation.

Definition.Thecharactersχ(D)of a representationDof a groupGare defined as

the traces of the matricesD(X), one for each elementXofG.

At this stage there will begcharacters, but, as we noted in subsection 28.7.3,

elementsA,BofGin the same conjugacy class are connected by equations of

the formB=X−^1 AX. It follows that their matrix representations are connected

by corresponding equations of the formD(B)=D(X−^1 )D(A)D(X),andsobythe

argument just given their representations will have equal traces and hence equal

characters. Thuselements in the same conjugacy class have the same characters,