Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 19 August 2013

Lecture 8

Proof of (2)(4) in Theorem 1.41.

Let A be the left regular representation of A. Assuming (2), we have AλAˆ (Wλ)mλ, where the Wλ are irreducible representations.

Let aA. If a0, then A(a)0, since A is a faithful A-module. Therefore, Wλ(a)0 for some λAˆ, and there exists wWλ so that aw0. Then Aaw is a non-trivial submodule of the irreducible module Wλ, so Aaw=Wλ.

There exists bA such that baw=w. This tells us that ba is not nilpotent and thus baA. Since A is an ideal, aA, and therefore, A=0.

Let A be a semisimple algebra with basis G={g1,g2,,gd}. Let t=(tλ)λAˆ be a non-degenerate trace on A (i.e., tλ0). Let G*={g1*,g2*,,gd*} be the dual basis with respect to t, and for each aA, let [a]=gG gag*.

(This is analogous to the sum over all conjugates of a in group theory.) We know that the sum gGgag* is independent of the basis G chosen, so we compute it for the basis {eijλ}λAˆ1i,jdλ of matrix units. Recall that the dual basis is {ejiλ/tλ}λAˆ1i,jdλ, and note that a can be written as a=λAˆ1i,jdλ aijλeijλ, (1.56) for some aijλ. Thus, we have [a] = λAˆ1i,jdλ eijλa ejiλtλ = λAˆ1i,jdλ ajjλ eiiλtλ By (1.56) = λAˆ1i,jdλ j=1dλajjλχλ(a) eiiλtλ = λAˆ χλ(a)tλ i=1dλeiiλzλ = λAˆχλ (a)zλtλ, which shows that [a] is a linear combination of the zλ and that [a]Z(A).

Theorem 1.57 If G is a basis of A, then zλ=gG tλ2dλχλ (g*)[g].


Let G={g1,g2,,gd}. Then gGtλ2dλ χλ(g*)[g] = gGtλ2dλ χλ(g*)[g] μAˆ χμ(g)tμ zμ = μAˆtλ2dλ 1tμzμ gGχλ(g*)χμ(g)orthogonality of characters = μAˆ tλ2dλ zμtμδλ,μ dλtλ=zλ.

Therefore, the set {[g]|gG} spans Z(A). Using the Gram-Schmidt process we can orthogonalize this set to compute the minimal central idempotents.

1.1Group Algebras

We now specialize this work to the case where A is a group algebra. Let G be a finite group, and let A=[G] be its group algebra. We compute the trace of the regular representation for hG as follows tr(h) = gGhg|g = gGh|1 = |G|·h|1.

Therefore, for h,gG, we have tr(hg-1)=δg,h|G|, so the dual basis to G is {g-1/|G|}gG. This proves that the trace of the regular representation is non-degenerate, and [G] is semisimple. The regular representation of [G] decomposes as [G] λAˆ (Wλ)dλ. Therefore, tr=λAˆdλχλ, and so tr=(dλ)λAˆ. In this case 1.40 reads

(a) eijλ= gGdλ Wijλ (g-1|G|) g,
(b) zλ=gG dλχλ (g-1|G|) g,
(c) gGdλ χλ(g-1|G|) χμ(g)=δλ,μ,
and therefore we have 1|G|gG χλ(g-1) χμ(g)=δλ,μ. (1.58)

The set of all traces t:A forms a -vector space. Since t=λAˆ tλχλ, the irreducible characters χλ span the spaces of traces on A. Since A is a group algebra, we have t(ghg-1) =t(h), so t is constant on the conjugacy classes of G. That is, t is a class function. Let Λ index the conjugacy classes of G, and for ρΛ, let 𝒞ρ denote the conjugacy class. Define cρ:[G] by cρ(g)= { 1 ifg𝒞ρ, 0 otherwise. Then t=ρΛTρcρ for some Tρ𝒞ρ, since t is a class function. Therefore, the set {cρ|ρΛ} spans the vector space of traces on [G], and since the cρ are characteristic functions, they are independent and form a basis. Thus, we can index the conjugacy classes of G by elements of Aˆ.

Now let χλ(ρ) denote the value of χλ on 𝒞ρ, and let ρ be the index of the conjugacy class that contains the inverses of the elements of 𝒞ρ. (Note that ρ and ρ may be the same). Then using 1.58, we get 1|G| ρAˆ χλ(ρ) χμ(ρ) |𝒞ρ|= δλ,μ. Let Ξ be the matrix whose rows and columns are indexed by Aˆ and whose entry in the λth row and ρth column is χλ(ρ), and let Ξ be the matrix whose rows and columns are indexed by Aˆ and whose entry in the λth row and ρth column is χλ(ρ). Then we have just shown that ΞΞ=I. Moreover, ΞΞ=I, so λAˆ χλ(ρ)χμ (τ) |𝒞τ||G| =δλ,ρ, and we get the second orthogonality relation for finite groups λAˆ χλ(ρ) χμ(τ)= δλ,ρ |G||𝒞τ|. (1.59)

Notes and References

This is a copy of lectures in Representation Theory given by Arun Ram, compiled by Tom Halverson, Rob Leduc and Mark McKinzie.

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