29.7 COUNTING IRREPS USING CHARACTERS
Classes
Irrep IABCDE
A 1 11 1
A 2 11 − 1
E 2 − 10D 30 1Table 29.2 The characters of the irreps of the group 3mand of the represen-
tationD, which must be a superposition of some of them.29.7.1 Summation rules for irrepsThe first summation rule for irreps is a simple restatement of (29.14), withμset
equal toλ;itthenreads
∑
X[
χ(λ)(X)]∗
χ(λ)(X)=g.In words, the sum of the squares (modulus squared if necessary) of the characters
of an irrep taken over all elements of the group adds up to the order of the
group. For group 3m(table 29.1), this takes the following explicit forms:
for A 1 , 1(1^2 )+2(1^2 )+3(1^2 )=6;
for A 2 , 1(1^2 )+2(1^2 )+3(−1)^2 =6;
for E, 1(2^2 )+2(−1)^2 +3(0^2 )=6.We next prove a theorem that is concerned not with a summation within an irrep
but with a summation over irreps.
Theorem.Ifnμis the dimension of theμth irrep of a groupGthen
∑μn^2 μ=g,wheregis the order of the group.
Proof.Define a representation of the group in the following way. Rearrangethe rows of the multiplication table of the group so that whilst the elements in
a particular order head the columns, their inverses in the same order head the
rows. In this arrangement of theg×gtable, the leading diagonal is entirely
occupied by the identity element. Then, for each elementXof the group, take as
representative matrix the multiplication-table array obtained by replacingXby
1 and all other element symbols by 0. The matricesDreg(X) so obtained form the
regular representationofG;theyareeachg×g, have a single non-zero entry ‘1’
in each row and column and (as will be verified by a little experimentation) have