Mathematical Methods for Physics and Engineering - Matematica.NET

REPRESENTATION THEORY(d) No explicit calculation is needed to see that if i = j = k = l = 1, withˆD (λ) = ˆD (µ) =A 1 (or A 2 ), then each term in the sum is either 1 2 or (−1) 2and the total is 6, as predicted by the right-hand side of (29.13) since g =6and n λ =1.29.6 CharactersThe actual matrices of general representations and irreps are cumbersome towork with, and they are not unique since there is always the freedom to changethe 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 changefor a matrix under such an equivalence (similarity) transformation – i.e. undera change of basis – is the trace of the matrix. This was shown in chapter 8,but is repeated here. The trace of a matrix A is the sum of its diagonal elements,n∑Tr A =or, using the summation convention (section 26.1), simply A ii . Under a similaritytransformation, again using the summation convention,i=1A ii[D Q (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 withsome 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 groupsknown as 4mm and ¯42m, orasC 4v and D 2d respectively, even though the twogroups are not isomorphic. To make use of these simplifications we now definethe characters of a representation.Definition. The characters χ(D) of a representation D of a group G are defined asthe traces of the matrices D(X), one for each element X of G.At this stage there will be g characters, but, as we noted in subsection 28.7.3,elements A, B of G in the same conjugacy class are connected by equations ofthe form B = X −1 AX. It follows that their matrix representations are connectedby corresponding equations of the form D(B) =D(X −1 )D(A)D(X),andsobytheargument just given their representations will have equal traces and hence equalcharacters. Thus elements in the same conjugacy class have the same characters,1092