Lectures in Representation Theory

Last update: 19 August 2013

Lecture 8

 Proof of $\left(2\right)⇒\left(4\right)$ in Theorem 1.41. Let $\stackrel{\to }{A}$ be the left regular representation of $A\text{.}$ Assuming (2), we have $A→≅⨁λ∈Aˆ (Wλ)⊕mλ,$ where the ${W}^{\lambda }$ are irreducible representations. Let $a\in \stackrel{\to }{A}\text{.}$ If $a\ne 0,$ then $\stackrel{\to }{A}\left(a\right)\ne 0,$ since $\stackrel{\to }{A}$ is a faithful $A\text{-module.}$ Therefore, ${W}^{\lambda }\left(a\right)\ne 0$ for some $\lambda \in \stackrel{ˆ}{A},$ and there exists $w\in {W}^{\lambda }$ so that $aw\ne 0\text{.}$ Then $Aaw$ is a non-trivial submodule of the irreducible module ${W}^{\lambda },$ so $Aaw={W}^{\lambda }\text{.}$ There exists $b\in A$ such that $baw=w\text{.}$ This tells us that $ba$ is not nilpotent and thus $ba\notin \sqrt{A}\text{.}$ Since $\sqrt{A}$ is an ideal, $a\notin \sqrt{A},$ and therefore, $\sqrt{A}=0\text{.}$ $\square$

Let $A$ be a semisimple algebra with basis $G=\left\{{g}_{1},{g}_{2},\dots ,{g}_{d}\right\}\text{.}$ Let $\stackrel{\to }{t}={\left({t}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}$ be a non-degenerate trace on $A$ (i.e., ${t}_{\lambda }\ne 0\text{).}$ Let ${G}^{*}=\left\{{g}_{1}^{*},{g}_{2}^{*},\dots ,{g}_{d}^{*}\right\}$ be the dual basis with respect to $\stackrel{\to }{t},$ and for each $a\in A,$ let $[a]=∑g∈G gag*.$

(This is analogous to the sum over all conjugates of a in group theory.) We know that the sum $\sum _{g\in G}ga{g}^{*}$ is independent of the basis $G$ chosen, so we compute it for the basis ${\left\{{e}_{ij}^{\lambda }\right\}}_{\underset{1\le i,j\le {d}_{\lambda }}{\lambda \in \stackrel{ˆ}{A}}}$ of matrix units. Recall that the dual basis is ${\left\{{e}_{ji}^{\lambda }/{t}_{\lambda }\right\}}_{\underset{1\le i,j\le {d}_{\lambda }}{\lambda \in \stackrel{ˆ}{A}}},$ and note that $a$ can be written as $a=∑λ∈Aˆ1≤i,j≤dλ aijλeijλ, (1.56)$ for some ${a}_{ij}^{\lambda }\in ℂ\text{.}$ Thus, we have $[a] = ∑λ∈Aˆ1≤i,j≤dλ eijλa ejiλtλ = ∑λ∈Aˆ1≤i,j≤dλ ajjλ eiiλtλ By (1.56) = ∑λ∈Aˆ1≤i,j≤dλ ∑j=1dλajjλ⏟χλ(a) eiiλtλ = ∑λ∈Aˆ χλ(a)tλ ∑i=1dλeiiλ⏟zλ = ∑λ∈Aˆχλ (a)zλtλ,$ which shows that $\left[a\right]$ is a linear combination of the ${z}_{\lambda }$ and that $\left[a\right]\in Z\left(A\right)\text{.}$

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

 Proof. Let $G=\left\{{g}_{1},{g}_{2},\dots ,{g}_{d}\right\}\text{.}$ Then $∑g∈Gtλ2dλ χλ(g*)[g] = ∑g∈Gtλ2dλ χλ(g*)[g] ∑μ∈Aˆ χμ(g)tμ zμ = ∑μ∈Aˆtλ2dλ 1tμzμ ∑g∈Gχλ(g*)χμ(g)⏟orthogonality of characters = ∑μ∈Aˆ tλ2dλ zμtμδλ,μ dλtλ=zλ.$ $\square$

Therefore, the set $\left\{\left[g\right] | g\in G\right\}$ spans $Z\left(A\right)\text{.}$ Using the Gram-Schmidt process we can orthogonalize this set to compute the minimal central idempotents.

1.1$\phantom{\rule{1em}{0ex}}$Group Algebras

We now specialize this work to the case where $A$ is a group algebra. Let $G$ be a finite group, and let $A=ℂ\left[G\right]$ be its group algebra. We compute the trace of the regular representation for $h\in G$ as follows $tr(h) = ∑g∈Ghg|g = ∑g∈Gh|1 = |G|·h|1.$

Therefore, for $h,g\in G,$ we have $\text{tr}\left(h{g}^{-1}\right)={\delta }_{g,h}|G|,$ so the dual basis to $G$ is ${\left\{{g}^{-1}/|G|\right\}}_{g\in G}\text{.}$ This proves that the trace of the regular representation is non-degenerate, and $ℂ\left[G\right]$ is semisimple. The regular representation of $ℂ\left[G\right]$ decomposes as $ℂ[G]⟶≅ ⨁λ∈Aˆ (Wλ)⊕dλ.$ Therefore, $\text{tr}={\sum }_{\lambda \in \stackrel{ˆ}{A}}{d}_{\lambda }{\chi }^{\lambda },$ and so $\text{tr}={\left({d}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}\text{.}$ In this case 1.40 reads

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

The set of all traces $\stackrel{\to }{t}:A⟶ℂ$ forms a $ℂ\text{-vector}$ space. Since $t→=∑λ∈Aˆ tλχλ,$ the irreducible characters ${\chi }^{\lambda }$ span the spaces of traces on $A\text{.}$ Since $A$ is a group algebra, we have $t→(ghg-1) =t→(h),$ so $\stackrel{\to }{t}$ is constant on the conjugacy classes of $G\text{.}$ That is, $\stackrel{\to }{t}$ is a class function. Let $\Lambda$ index the conjugacy classes of $G,$ and for $\rho \in \Lambda ,$ let ${𝒞}_{\rho }$ denote the conjugacy class. Define ${c}_{\rho }:ℂ\left[G\right]⟶ℂ$ by $cρ(g)= { 1 if g∈𝒞ρ, 0 otherwise.$ Then $\stackrel{\to }{t}={\sum }_{\rho \in \Lambda }{T}_{\rho }{c}_{\rho }$ for some ${T}_{\rho }\in {𝒞}_{\rho },$ since $\stackrel{\to }{t}$ is a class function. Therefore, the set $\left\{{c}_{\rho } | \rho \in \Lambda \right\}$ spans the vector space of traces on $ℂ\left[G\right],$ and since the ${c}_{\rho }$ are characteristic functions, they are independent and form a basis. Thus, we can index the conjugacy classes of $G$ by elements of $\stackrel{ˆ}{A}\text{.}$

Now let ${\chi }^{\lambda }\left(\rho \right)$ denote the value of ${\chi }^{\lambda }$ on ${𝒞}_{\rho },$ and let $\rho \prime$ be the index of the conjugacy class that contains the inverses of the elements of ${𝒞}_{\rho }\text{.}$ (Note that $\rho$ and $\rho \prime$ may be the same). Then using 1.58, we get $1|G| ∑ρ∈Aˆ χλ(ρ′) χμ(ρ) |𝒞ρ|= δλ,μ.$ Let $\Xi$ be the matrix whose rows and columns are indexed by $\stackrel{ˆ}{A}$ and whose entry in the $\lambda \text{th}$ row and $\rho \text{th}$ column is ${\chi }^{\lambda }\left(\rho \right),$ and let $\Xi \prime$ be the matrix whose rows and columns are indexed by $\stackrel{ˆ}{A}$ and whose entry in the $\lambda \text{th}$ row and $\rho \text{th}$ column is ${\chi }^{\lambda }\left(\rho \prime \right)\text{.}$ Then we have just shown that $ΞΞ′=I.$ Moreover, $\Xi \prime \Xi =I,$ so $∑λ∈Aˆ χλ(ρ)χμ (τ′) |𝒞τ||G| =δλ,ρ,$ and we get the second orthogonality relation for finite groups $∑λ∈Aˆ χλ(ρ) χμ(τ′)= δλ,ρ |G||𝒞τ|. (1.59)$

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.