Let be a finite group and let be functions. On page 13 of the book, Diaconis proves the following `Plancherel’ formula

where are the irreducible representations of and .

First of all I will give a fairly short proof of the formula using the group algebra . Define

On the representation , the element acts as . As a representation of , the group algebra affords the regular representation of . Hence, by Corollary 1 on page 12, has exactly irreducible summands on which acts according to . Therefore, the trace of acting on is

On the other hand, the character of the regular representation was found in Proposition 5 to be on non-identity elements, and on the identity. Hence, by linearity, the trace of acting on the regular representation is

This completes the proof of the Plancherel formula. One might object to this name, on the grounds that the bilinear forms that appear on either side are not inner products. To get round this, we note that if is defined by then

Hence the original formula implies that

where we now assume that the representations are unitary. (The above formula appears in Remark 3 on page 14.) The form on matrices defined by is an inner product, while on the left-hand side we have an inner product provided we restrict to real valued functions. This restriction seems a little surprising, but does appear to be necessary.

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