29.7 COUNTING IRREPS USING CHARACTERS
that
∑
μn^2 μ=g. (29.21)This completes the proof.
As before, our standard demonstration group 3mprovides an illustration. Inthis case we have seen already that there are two one-dimensional irreps and one
two-dimensional irrep. This is in accord with (29.21) since
12 +1^2 +2^2 =6, which is the ordergof the group.Another straightforward application of the relation (29.21), to the group withmultiplication table 29.3(a), yields immediate results. Sinceg=3,noneofits
irreps can have dimension 2 or more, as 2^2 = 4 is too large for (29.21) to be
satisfied. Thus all irreps must be one-dimensional and there must be three of
them (consistent with the fact that each element is in a class of its own, and that
there are therefore three classes). The three irreps are the sets of 1×1 matrices
(numbers)
A 1 ={ 1 , 1 , 1 } A 2 ={ 1 ,ω,ω^2 } A∗ 2 ={ 1 ,ω^2 ,ω},whereω=exp(2πi/3); since the matrices are 1×1, the same set of nine numbers
would be, of course, the entries in the character table for the irreps of the group.
The fact that the numbers in each irrep are all cube roots of unity is discussed
below. As will be noticed, two of these irreps are complex – an unusual occurrence
in most applications – and form a complex conjugate pair of one-dimensional
irreps. In practice, they function much as a two-dimensional irrep, but this is to
be ignored for formal purposes such as theorems.
A further property of characters can be derived from the fact that all elementsin a conjugacy class have the same order. Suppose that the elementXhas order
m,i.e.Xm=I. This implies for a representationDof dimensionnthat
[D(X)]m=In. (29.22)Representations equivalent toDare generated as before by using similarity
transformations of the form
DQ(X)=Q−^1 D(X)Q.In particular, if we choose the columns ofQto be the eigenvectors ofD(X) then,
as discussed in chapter 8,
DQ(X)=
λ 1 0 ··· 00 λ 2..
.
..
.... 0
0 ··· 0 λn