Mathematical Methods for Physics and Engineering - Matematica.NET

29.7 COUNTING IRREPS USING CHARACTERSthat∑n 2 µ = g. (29.21)µThis completes the proof.As before, our standard demonstration group 3m provides an illustration. Inthis case we have seen already that there are two one-dimensional irreps and onetwo-dimensional irrep. This is in accord with (29.21) since1 2 +1 2 +2 2 =6, which is the order g of the group.Another straightforward application of the relation (29.21), to the group withmultiplication table 29.3(a), yields immediate results. Since g =3,noneofitsirreps can have dimension 2 or more, as 2 2 = 4 is too large for (29.21) to besatisfied. Thus all irreps must be one-dimensional and there must be three ofthem (consistent with the fact that each element is in a class of its own, and thatthere 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},where ω =exp(2πi/3); since the matrices are 1 × 1, the same set of nine numberswould 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 discussedbelow. As will be noticed, two of these irreps are complex – an unusual occurrencein most applications – and form a complex conjugate pair of one-dimensionalirreps. In practice, they function much as a two-dimensional irrep, but this is tobe 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 element X has orderm, i.e.X m = I. This implies for a representation D of dimension n that[D(X)] m = I n . (29.22)Representations equivalent to D are generated as before by using similaritytransformations of the formD Q (X) =Q −1 D(X)Q.In particular, if we choose the columns of Q to be the eigenvectors of D(X) then,as discussed in chapter 8,⎛⎞λ 1 0 ··· 0D Q (X) =0 λ. 2 ⎜ .⎝. ⎟. .. 0 ⎠0 ··· 0 λ n1099