Lectures in Representation Theory

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 12 August 2013

Lecture 4

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

$$A\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)$$

of matrix algebras where $\hat{A}$ is a finite index set and $d_{\lambda }$ is a positive integer for all $\lambda \in \hat{A}\text{.}$

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

$$\{e_{ij}^{\lambda }={\varphi }^{-1}\left(E_{ij}^{\lambda }\right)\hspace{0.17em}|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}\text{.}$$

For $\lambda \in \hat{A},$ the minimal central idempotent $z_{\lambda }$ of $A$ indexed by $\lambda$ is given by

$$z_{\lambda }=\sum _{i=1}^{d_{\lambda }}{e}_{i,i}^{\lambda }\text{.}$$

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

$$I^{\lambda }={\varphi }^{-1}\left(M_{d_{\lambda }}\left(ℂ\right)\right),$$

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

$$\begin{array}{cccc}{W}^{\lambda }:& A& ⟶& M_{d_{\lambda }}\left(ℂ\right)\\ & a& ⟼& \varphi {\left(a\right)}_{\lambda }\end{array}$$

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

$$W^{\lambda }\cong A{e}_{ij}^{\lambda }\text{.}$$

Since $W^{\lambda }\left(a\right)=\varphi {\left(a\right)}_{\lambda }$ for all $a\in A$ and $\lambda \in \hat{A},$ 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 $\lambda \in \hat{A}$ is given by

$${\chi }^{\lambda }\left(a\right)=\sum _{i=1}^{d_{\lambda }}{W}_{ii}^{\lambda }\left(a\right)\text{.}$$

Moreover, it should be noted that

$$W_{ij}^{\lambda }\left(e_{rs}^{\mu }\right)={\delta }_{\lambda ,\mu }{\delta }_{i,r}{\partial }_{j,s}\text{.}$$

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

(a)	$\overrightarrow{t}\left(a\right)={\displaystyle \sum _{\lambda \in \hat{A}}}{t}_{\lambda }{\chi }^{\lambda }\left(a\right),$ for all $a\in A,$
(b)	$\overrightarrow{t}\left(e_i{i}^{\lambda }\right)=t_{\lambda }\text{.}$

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

$$⟨a,b⟩=\overrightarrow{t}\left(ab\right),\text{for}\hspace{0.17em}a,b\in A\text{.}$$

The form has the following properties:

(a)	$\text{symmetric}:⟨a,b⟩=\overrightarrow{t}\left(ab\right)=\overrightarrow{t}\left(ba\right)=⟨b,a⟩,$
(b)	$\text{associative}:⟨ab,c⟩=\overrightarrow{t}\left(abc\right)=⟨a,bc⟩\text{.}$

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 $\{g_1,g_2,\dots ,g_d\}$ be a basis for $A\text{.}$ A dual basis with respect to the bilinear form $⟨,⟩$ is the basis $\{g_1^{*},g_2^{*},\dots ,g_d^{*}\}$ having the property that

$$⟨g_i,g_j^{*}⟩={\delta }_{ij}\text{.}$$

The basis $\left\{e_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}$ of matrix units in $A$ has as its dual basis the set $\{e_{ij}^{\lambda }/t_{\lambda }\hspace{0.17em}|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}\text{.}$ To verify this, we see that

$$⟨e_{ij}^{\lambda },e_{sr}^{\mu }/t_{\mu }⟩=\text{tr}\left(\frac{e_{ij}^{\lambda }{e}_{sr}^{\mu }}{t_{\mu }}\right)={\delta }_{\lambda ,\mu }{\delta }_{js}\text{tr}\left(\frac{e_{ir}^{\lambda }}{t_{\lambda }}\right)={\delta }_{\lambda ,\mu }{\delta }_{js}{\delta }_{ir}\text{.}$$

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

(a)	Fourier Inversion Formula. $$ If $\lambda \in \hat{A}$ and $1\le i,j\le d_{\lambda }$ then $$e_{ij}^{\lambda }=\sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{t}_{\lambda }{W}_{ij}^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right)e_{rs}^{\mu }\text{.}$$
(b)	Central Idempotents. $$ If $\lambda \in \hat{A}$ then $$z_{\lambda }=\sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{t}_{\lambda }{\chi }^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right)e_{rs}^{\mu }\text{.}$$
(c)	Orthogonality of Characters. $$ If $\lambda ,\nu \in \hat{A},$ then $$\sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{\chi }^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right){\chi }^{\nu }\left(e_{rs}^{\mu }\right)={\delta }_{\lambda ,\nu }\frac{d_{\lambda }}{t_{\lambda }}\text{.}$$

		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 $$\begin{array}{ccc}{z}_{\lambda }=\sum _{i=1}^{d_{\lambda }}{e}_{ii}^{\lambda }& =& \sum _{i=1}^{d_{\lambda }}\underset{\underset{\text{Fourier inversion}}{⏟}}{\sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{t}_{\lambda }{W}_{ii}^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right)e_{rs}^{\mu }}\\ & =& \sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{t}_{\lambda }\underset{\underset{{\chi }^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right)}{⏟}}{\sum _{i=1}^{d_{\lambda }}{W}_{ii}^{\lambda }\left(\frac{e_{sr}^{\mu }}{t_{\mu }}\right)}{e}_{rs}^{\mu }\end{array}$$ 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 $𝒜=\{g_1,\dots ,g_d\}$ and $ℬ=\{h_1,\dots ,h_d\}$ be bases of $A$ and let ${𝒜}^{*}=\{g_1^{*},\dots ,g_d^{*}\}$ and ${ℬ}^{*}=\{h_1^{*},\dots ,h_d^{*}\}$ be their duals. Let $S=\left(s_{ij}\right)$ be the transition matrix between $𝒜$ and $ℬ\text{.}$ That is

$$h_i=\sum _{j=1}^d{s}_{ij}{g}_j\text{.}$$

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

$$\begin{array}{ccc}{\delta }_{im}=⟨h_i,h_m^{*}⟩& =& ⟨\sum _{j=1}^d{s}_{ij}{g}_j,\sum _{n=1}^d{t}_{mn}{g}_n^{*}⟩\\ & =& \sum _{j,n}{s}_{ij}{t}_{mn}\underset{\underset{={\delta }_{j,n}}{⏟}}{⟨g_j,g_n^{*}⟩}\\ & =& \sum _{j=1}^d{s}_{ij}{t}_{mj}\text{.}\end{array}$$

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

Now consider the orthogonality of characters formula. We have

$$\begin{array}{ccc}\sum _j{\chi }^{\lambda }\left(h_j^{*}\right){\chi }^{\nu }\left(h_j\right)& =& \sum _{i=1}^d{\chi }^{\lambda }\left(\sum _n{t}_{in}{g}_n^{*}\right){\chi }^{\nu }\left(\sum _m{s}_{im}{g}_m\right)\\ & =& \sum _{m,n}\left(\underset{\underset{{\delta }_{m,n}}{⏟}}{\sum _i{t}_{i,n}{s}_{i,m}}\right){\chi }^{\lambda }\left(g_n^{*}\right){\chi }^{\nu }\left(g_m\right)\\ & =& \sum _m{\chi }^{\lambda }\left(g_n^{*}\right){\chi }^{\nu }\left(g_m\right)\text{.}\end{array}$$

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 $\overrightarrow{t}={\left(t_{\lambda }\right)}_{\lambda \in \hat{A}}$ $\text{(}{t}_{\lambda }\ne 0\text{)},$ and let $𝒜=\{g_1,\dots ,g_d\}$ be a basis of $A\text{.}$ Then

(a)	Fourier Inversion Formula. $$ The elements $$e_{ij}^{\lambda }=\sum _{g\in 𝒜}{t}_{\lambda }{W}_{ij}^{\lambda }\left(g^{*}\right)g$$ form a complete set of matrix units for $A\text{.}$
(b)	Central Idempotents. $$ The central idempotents of $A$ are given by $$z_{\lambda }=\sum _{g\in 𝒜}{t}_{\lambda }{\chi }^{\lambda }\left(g^{*}\right)g\text{.}$$
(c)	Orthogonality of Characters. $$ If $\lambda ,\nu \in \hat{A},$ then $$\sum _{g\in 𝒜}{\chi }^{\lambda }\left(g^{*}\right){\chi }^{\nu }\left(g\right)={\delta }_{\lambda ,\nu }\frac{d_{\lambda }}{t_{\lambda }},$$ 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.

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