## Lectures in Representation Theory

Last update: 12 August 2013

## Lecture 4

Definition 1.38 An algebra $A$ is semisimple if it is isomorphic to a direct sum

$A≅⨁λ∈Aˆ Mdλ(ℂ)$

of matrix algebras where $\stackrel{ˆ}{A}$ is a finite index set and ${d}_{\lambda }$ is a positive integer for all $\lambda \in \stackrel{ˆ}{A}\text{.}$

Let $A$ be semisimple algebra, and let $\varphi :A⟶\underset{\lambda \in \stackrel{ˆ}{A}}{⨁}{M}_{{d}_{\lambda }}\left(ℂ\right)$ be an isomorphism. Then a basis of matrix units for $A$ is

${ eijλ= ϕ-1(Eijλ) | λ∈Aˆ,1≤ i,j≤dλ } .$

For $\lambda \in \stackrel{ˆ}{A},$ the minimal central idempotent ${z}_{\lambda }$ of $A$ indexed by $\lambda$ is given by

$zλ= ∑i=1dλ ei,iλ.$

The minimal ideal ${I}^{\lambda }$ of $A$ indexed by $\lambda$ is given by

$Iλ=ϕ-1 (Mdλ(ℂ)),$

and the irreducible representation ${W}^{\lambda }$ of $A$ indexed by $\lambda$ is given by $\lambda \in \stackrel{ˆ}{A}$ and given by

$Wλ: A ⟶ Mdλ(ℂ) a ⟼ ϕ(a)λ$

where $\varphi {\left(a\right)}_{\lambda }$ denotes the $\lambda \text{th}$ block of the matrix $\varphi \left(a\right)\text{.}$ In terms of modules, the irreducible module ${W}^{\lambda },$ is

$Wλ≅A eijλ.$

Since ${W}^{\lambda }\left(a\right)=\varphi {\left(a\right)}_{\lambda }$ for all $a\in A$ and $\lambda \in \stackrel{ˆ}{A},$ we let ${W}_{ij}^{\lambda }\left(a\right)$ denote the $i,j\text{-entry}$ of the matrix ${W}^{\lambda }\left(a\right)\text{.}$ Then the irreducible character ${\chi }^{\lambda }$ corresponding to $\lambda \in \stackrel{ˆ}{A}$ is given by

$χλ(a)= ∑i=1dλ Wiiλ(a).$

Moreover, it should be noted that

$Wijλ (ersμ)= δλ,μ δi,r ∂j,s.$

Let $\stackrel{\to }{t}$ be a trace on the semisimple algebra $A\text{.}$ Then $\stackrel{\to }{t}$ is determined by the weight vector ${\left({t}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}$ which satisfies the following:

 (a) $\stackrel{\to }{t}\left(a\right)=\sum _{\lambda \in \stackrel{ˆ}{A}}{t}_{\lambda }{\chi }^{\lambda }\left(a\right),$ for all $a\in A,$ (b) $\stackrel{\to }{t}\left({e}_{i}{i}^{\lambda }\right)={t}_{\lambda }\text{.}$

We suppose that $\stackrel{\to }{t}$ has the property that ${t}_{\lambda }\ne 0$ for all $\lambda \in \stackrel{ˆ}{A}$ (we will see that this means that $\stackrel{\to }{t}$ is non-degenerate) and define a bilinear form on $A$ by

$⟨a,b⟩= t→(ab), for a,b∈A.$

The form has the following properties:

 (a) $\text{symmetric}:\phantom{\rule{1em}{0ex}}⟨a,b⟩=\stackrel{\to }{t}\left(ab\right)=\stackrel{\to }{t}\left(ba\right)=⟨b,a⟩,$ (b) $\text{associative}:\phantom{\rule{1em}{0ex}}⟨ab,c⟩=\stackrel{\to }{t}\left(abc\right)=⟨a,bc⟩\text{.}$
If $A$ has a non-degenerate associative form, then $A$ is a Frobenius algebra. If $A$ has a symmetric, associative non-degenerate form, then $A$ is a symmetric algebra. Many of the results of this chapter hold in these more general algebras.

Let $\left\{{g}_{1},{g}_{2},\dots ,{g}_{d}\right\}$ be a basis for $A\text{.}$ A dual basis with respect to the bilinear form $⟨,⟩$ is the basis $\left\{{g}_{1}^{*},{g}_{2}^{*},\dots ,{g}_{d}^{*}\right\}$ having the property that

$⟨gi,gj*⟩ =δij.$

The basis $\left\{{e}_{ij}^{\lambda } | \lambda \in \stackrel{ˆ}{A},1\le i,j\le {d}_{\lambda }\right\}$ of matrix units in $A$ has as its dual basis the set $\left\{{e}_{ij}^{\lambda }/{t}_{\lambda } | \lambda \in \stackrel{ˆ}{A},1\le i,j\le {d}_{\lambda }\right\}\text{.}$ To verify this, we see that

$⟨eijλ,esrμ/tμ⟩ =tr(eijλesrμtμ)= δλ,μδjstr (eirλtλ) =δλ,μδjs δir.$

Theorem 1.39 Let $\stackrel{\to }{t}={\left({t}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}$ be a trace on $A$ such that ${t}_{\lambda }\ne 0$ for all $\lambda \in \stackrel{ˆ}{A}\text{.}$

 (a) Fourier Inversion Formula. $\phantom{\rule{1em}{0ex}}$ If $\lambda \in \stackrel{ˆ}{A}$ and $1\le i,j\le {d}_{\lambda }$ then $eijλ= ∑μ∈Aˆ1≤r,s≤dμ tλWijλ (esrμtμ) ersμ.$ (b) Central Idempotents. $\phantom{\rule{1em}{0ex}}$ If $\lambda \in \stackrel{ˆ}{A}$ then $zλ=∑μ∈Aˆ1≤r,s≤dμ tλχλ (esrμtμ) ersμ.$ (c) Orthogonality of Characters. $\phantom{\rule{1em}{0ex}}$ If $\lambda ,\nu \in \stackrel{ˆ}{A},$ then $∑μ∈Aˆ1≤r,s≤dμ χλ (esrμtμ) χν(ersμ) =δλ,ν dλtλ.$

 Proof. For (a) we have ${W}_{ij}^{\lambda }\left(\frac{{e}_{sr}^{\mu }}{{t}_{\mu }}\right)={\delta }_{\lambda ,\mu }{\delta }_{j,s}{\delta }_{i,r},$ and the result follows. For (b), we see that $zλ= ∑i=1dλ eiiλ = ∑i=1dλ ∑μ∈Aˆ1≤r,s≤dμ tλWiiλ (esrμtμ) ersμ ⏟Fourier inversion = ∑μ∈Aˆ1≤r,s≤dμ tλ ∑i=1dλ Wiiλ (esrμtμ) ⏟χλ(esrμtμ) ersμ$ Part (c) follows by taking the character ${\chi }^{\lambda }$ of each side the equation in (b). $\square$

Our next goal is to show that these formulas are independent of the basis used. To this end we let $𝒜=\left\{{g}_{1},\dots ,{g}_{d}\right\}$ and $ℬ=\left\{{h}_{1},\dots ,{h}_{d}\right\}$ be bases of $A$ and let ${𝒜}^{*}=\left\{{g}_{1}^{*},\dots ,{g}_{d}^{*}\right\}$ and ${ℬ}^{*}=\left\{{h}_{1}^{*},\dots ,{h}_{d}^{*}\right\}$ be their duals. Let $S=\left({s}_{ij}\right)$ be the transition matrix between $𝒜$ and $ℬ\text{.}$ That is

$hi=∑j=1d sijgj.$

If $T$ is the transition matrix between ${𝒜}^{*}$ and ${ℬ}^{*},$ then

$δim= ⟨hi,hm*⟩ = ⟨ ∑j=1d sijgj, ∑n=1d tmngn* ⟩ = ∑j,nsij tmn ⟨gj,gn*⟩ ⏟ =δj,n = ∑j=1d sijtmj.$

Therefore, $I=S{T}^{t}\text{.}$

Now consider the orthogonality of characters formula. We have

$∑jχλ (hj*)χν (hj) = ∑i=1d χλ (∑ntingn*) χν (∑msimgm) = ∑m,n ( ∑iti,nsi,m ⏟δm,n ) χλ(gn*) χν(gm) = ∑mχλ (gn*) χν(gm).$

Thus, orthogonality of characters holds for any basis. In fact, the same argument proves that all three formulas in 1.39 are basis independent. Therefore,

Theorem 1.40 Let A be a semisimple algebra with non-degenerate trace $\stackrel{\to }{t}={\left({t}_{\lambda }\right)}_{\lambda \in \stackrel{ˆ}{A}}$ $\text{(}{t}_{\lambda }\ne 0\text{)},$ and let $𝒜=\left\{{g}_{1},\dots ,{g}_{d}\right\}$ be a basis of $A\text{.}$ Then

 (a) Fourier Inversion Formula. $\phantom{\rule{1em}{0ex}}$ The elements $eijλ= ∑g∈𝒜tλ Wijλ(g*)g$ form a complete set of matrix units for $A\text{.}$ (b) Central Idempotents. $\phantom{\rule{1em}{0ex}}$ The central idempotents of $A$ are given by $zλ=∑g∈𝒜 tλχλ(g*) g.$ (c) Orthogonality of Characters. $\phantom{\rule{1em}{0ex}}$ If $\lambda ,\nu \in \stackrel{ˆ}{A},$ then $∑g∈𝒜χλ (g*)χν (g)=δλ,ν dλtλ,$ where ${d}_{\lambda }$ is the dimension of the irreducible $A\text{-representation}$ ${W}^{\lambda }\text{.}$

## 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.