Lectures in Representation Theory

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

Last update: 19 August 2013

Lecture 8

		Proof of $\left(2\right)\Rightarrow \left(4\right)$ in Theorem 1.41.
	Let $\overrightarrow{A}$ be the left regular representation of $A\text{.}$ Assuming (2), we have $$\overrightarrow{A}\cong \underset{\lambda \in \hat{A}}{⨁}{\left(W^{\lambda }\right)}^{\oplus m_{\lambda }},$$ where the $W^{\lambda }$ are irreducible representations. Let $a\in \overrightarrow{A}\text{.}$ If $a\ne 0,$ then $\overrightarrow{A}\left(a\right)\ne 0,$ since $\overrightarrow{A}$ is a faithful $A\text{-module.}$ Therefore, $W^{\lambda }\left(a\right)\ne 0$ for some $\lambda \in \hat{A},$ 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 $G=\{g_1,g_2,\dots ,g_d\}\text{.}$ Let $\overrightarrow{t}={\left(t_{\lambda }\right)}_{\lambda \in \hat{A}}$ be a non-degenerate trace on $A$ (i.e., $t_{\lambda }\ne 0\text{).}$ Let $G^{*}=\{g_1^{*},g_2^{*},\dots ,g_d^{*}\}$ be the dual basis with respect to $\overrightarrow{t},$ and for each $a\in A,$ let $$\left[a\right]=\sum _{g\in G}ga{g}^{*}\text{.}$$

(This is analogous to the sum over all conjugates of a in group theory.) We know that the sum ${\displaystyle \sum _{g\in G}}ga{g}^{*}$ is independent of the basis $G$ chosen, so we compute it for the basis ${\left\{e_{ij}^{\lambda }\right\}}_{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}$ of matrix units. Recall that the dual basis is ${\{e_{ji}^{\lambda }/t_{\lambda }\}}_{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}},$ and note that $a$ can be written as $$\begin{array}{cc}a=\sum _{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}{a}_{ij}^{\lambda }{e}_{ij}^{\lambda },& \text{(1.56)}\end{array}$$ for some $a_{ij}^{\lambda }\in ℂ\text{.}$ Thus, we have $$\begin{array}{cccccc}\left[a\right]& =& \sum _{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}{e}_{ij}^{\lambda }a\frac{e_{ji}^{\lambda }}{t_{\lambda }}& =& \sum _{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}{a}_{jj}^{\lambda }\frac{e_{ii}^{\lambda }}{t_{\lambda }}& \text{By (1.56)}\\ & =& \sum _{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}\underset{\underset{{\chi }^{\lambda }\left(a\right)}{⏟}}{\sum _{j=1}^{d_{\lambda }}{a}_{jj}^{\lambda }}\frac{e_{ii}^{\lambda }}{t_{\lambda }}& =& \sum _{\lambda \in \hat{A}}\frac{{\chi }^{\lambda }\left(a\right)}{t_{\lambda }}\underset{\underset{z_{\lambda }}{⏟}}{\sum _{i=1}^{d_{\lambda }}{e}_{ii}^{\lambda }}\\ & =& \sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(a\right)\frac{z_{\lambda }}{t_{\lambda }},\end{array}$$ 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 $$z_{\lambda }=\sum _{g\in G}{t}_{\lambda }^2{d}_{\lambda }{\chi }^{\lambda }\left(g^{*}\right)\left[g\right]\text{.}$$

		Proof.
	Let $G=\{g_1,g_2,\dots ,g_d\}\text{.}$ Then $$\begin{array}{ccc}\sum _{g\in G}\frac{t_{\lambda }^2}{d_{\lambda }}{\chi }^{\lambda }\left(g^{*}\right)\left[g\right]& =& \sum _{g\in G}\frac{t_{\lambda }^2}{d_{\lambda }}{\chi }^{\lambda }\left(g^{*}\right)\left[g\right]\sum _{\mu \in \hat{A}}\frac{{\chi }^{\mu }\left(g\right)}{t_{\mu }}{z}_{\mu }\\ & =& \sum _{\mu \in \hat{A}}\frac{t_{\lambda }^2}{d_{\lambda }}\frac{1}{t_{\mu }}{z}_{\mu }\underset{\underset{\text{orthogonality of characters}}{⏟}}{\sum _{g\in G}{\chi }^{\lambda }\left(g^{*}\right){\chi }^{\mu }\left(g\right)}\\ & =& \sum _{\mu \in \hat{A}}\frac{t_{\lambda }^2}{d_{\lambda }}\frac{z_{\mu }}{t_{\mu }}{\delta }_{\lambda ,\mu }\frac{d_{\lambda }}{t_{\lambda }}=z_{\lambda }\text{.}\end{array}$$ $\square$

Therefore, the set $\left\{\left[g\right]\hspace{0.17em}\right|\hspace{0.17em}g\in G\}$ spans $Z\left(A\right)\text{.}$ Using the Gram-Schmidt process we can orthogonalize this set to compute the minimal central idempotents.

1.1$$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 $$\begin{array}{ccc}\text{tr}\left(h\right)& =& \sum _{g\in G}hg{|}_g\\ & =& \sum _{g\in G}h{|}_1\\ & =& \left|G\right|·h{|}_1\text{.}\end{array}$$

Therefore, for $h,g\in G,$ we have $\text{tr}\left(h{g}^{-1}\right)={\delta }_{g,h}\left|G\right|,$ so the dual basis to $G$ is ${\{g^{-1}/\left|G\right|\}}_{g\in G}\text{.}$ 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 $$\stackrel{⟶}{ℂ\left[G\right]}\cong \underset{\lambda \in \hat{A}}{⨁}{\left(W^{\lambda }\right)}^{\oplus d_{\lambda }}\text{.}$$ Therefore, $\text{tr}={\sum }_{\lambda \in \hat{A}}{d}_{\lambda }{\chi }^{\lambda },$ and so $\text{tr}={\left(d_{\lambda }\right)}_{\lambda \in \hat{A}}\text{.}$ In this case 1.40 reads

(a)	$$e_{ij}^{\lambda }=\sum _{g\in G}{d}_{\lambda }{W}_{ij}^{\lambda }\left(\frac{g^{-1}}{\left|G\right|}\right)g,$$
(b)	$$z_{\lambda }=\sum _{g\in G}{d}_{\lambda }{\chi }^{\lambda }\left(\frac{g^{-1}}{\left|G\right|}\right)g,$$
(c)	$$\sum _{g\in G}{d}_{\lambda }{\chi }^{\lambda }\left(\frac{g^{-1}}{\left|G\right|}\right){\chi }^{\mu }\left(g\right)={\delta }_{\lambda ,\mu },$$

and therefore we have $$\begin{array}{cc}\frac{1}{\left|G\right|}\sum _{g\in G}{\chi }^{\lambda }\left(g^{-1}\right){\chi }^{\mu }\left(g\right)={\delta }_{\lambda ,\mu }\text{.}& \text{(1.58)}\end{array}$$

The set of all traces $\overrightarrow{t}:A⟶ℂ$ forms a $ℂ\text{-vector}$ space. Since $$\overrightarrow{t}=\sum _{\lambda \in \hat{A}}{t}_{\lambda }{\chi }^{\lambda },$$ the irreducible characters ${\chi }^{\lambda }$ span the spaces of traces on $A\text{.}$ Since $A$ is a group algebra, we have $$\overrightarrow{t}\left(gh{g}^{-1}\right)=\overrightarrow{t}\left(h\right),$$ so $\overrightarrow{t}$ is constant on the conjugacy classes of $G\text{.}$ That is, $\overrightarrow{t}$ 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 $$c_{\rho }\left(g\right)=\{\begin{array}{cc}1& \text{if}\hspace{0.17em}g\in {𝒞}_{\rho },\\ 0& \text{otherwise.}\end{array}$$ Then $\overrightarrow{t}={\sum }_{\rho \in \Lambda }{T}_{\rho }{c}_{\rho }$ for some $T_{\rho }\in {𝒞}_{\rho },$ since $\overrightarrow{t}$ is a class function. Therefore, the set $\left\{c_{\rho }\hspace{0.17em}\right|\hspace{0.17em}\rho \in \Lambda \}$ 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 $\hat{A}\text{.}$

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 $$\frac{1}{\left|G\right|}\sum _{\rho \in \hat{A}}{\chi }^{\lambda }\left(\rho \prime \right){\chi }^{\mu }\left(\rho \right)\left|{𝒞}_{\rho }\right|={\delta }_{\lambda ,\mu }\text{.}$$ Let $\Xi$ be the matrix whose rows and columns are indexed by $\hat{A}$ 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 $\hat{A}$ 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 $$\Xi \Xi \prime =I\text{.}$$ Moreover, $\Xi \prime \Xi =I,$ so $$\sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(\rho \right){\chi }^{\mu }\left(\tau \prime \right)\frac{\left|{𝒞}_{\tau }\right|}{\left|G\right|}={\delta }_{\lambda ,\rho },$$ and we get the second orthogonality relation for finite groups $$\begin{array}{cc}\sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(\rho \right){\chi }^{\mu }\left(\tau \prime \right)={\delta }_{\lambda ,\rho }\frac{\left|G\right|}{\left|{𝒞}_{\tau }\right|}\text{.}& \text{(1.59)}\end{array}$$

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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