Semisimple algebras

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

Last update: 6 November 2012

Semisimple algebras

An algebra $A$ is simple if $A\cong M_d\left(ℂ\right)\text{.}$

Suppose $\overrightarrow{t}$ is a trace on $M_d\left(ℂ\right)\text{.}$ Then

$$\begin{array}{ccc}\overrightarrow{t}\left(E_{ij}\right)& =& \overrightarrow{t}\left(E_{i1}{E}_{1j}\right)\\ & =& \overrightarrow{t}\left(E_{1j}{E}_{i1}\right)\\ & =& \overrightarrow{t}\left(E_{11}\right){\delta }_{ij}\text{.}\end{array}$$

If $a=\mid a_{ij}\mid \in M_d\left(ℂ\right)$ then

$$\begin{array}{ccc}\overrightarrow{t}\left(a\right)& =& \overrightarrow{t}\left(\sum _{i,j}{a}_{ij}{E}_{ij}\right)\\ & =& \sum _{i,j}{a}_{ij}\overrightarrow{t}\left(E_{ij}\right)\\ & =& \sum _{i,j}{a}_{ij}\overrightarrow{t}\left(E_{11}\right){\delta }_{ij}\\ & =& \overrightarrow{t}\left(E_{11}\right)\left(\sum _i{a}_{ii}\right)\text{.}\end{array}$$

So, up to a constant factor there is a unique trace function on $M_d\left(ℂ\right),$ that given by the standard trace on matrices.

Suppose $J$ is an ideal of $M_d\left(ℂ\right)$ and that $a=\mid a_{ij}\mid \in J,$ with $a\ne 0\text{.}$ So $a_{ij}\ne 0$ for some $(i,j)\text{.}$ Since $a\in J$ and $J$ is an ideal,

$$(1/a_{ij})\sum _{k=1}^d{E}_{ki}a{E}_{jk}=(1/a_{ij})\sum _k{a}_{ij}{E}_{kk}=I_d$$

is an element of $J\text{.}$ Thus $J=M_d\left(ℂ\right)\text{.}$ This shows that the only ideals of $M_d\left(ℂ\right)$ are the trivial ones, 0 and $M_d\left(ℂ\right)\text{.}$ It is an immediate consequence of §1 Ex. 1 that the center of $M_d\left(ℂ\right)$ is $I_d\left(ℂ\right)=ℂI_d\text{.}$ Furthermore, $I_d$ is the unique central idempotent in $M_d\left(ℂ\right)\text{.}$

(3.1) Proposition. There is a unique irreducible representation of $M_d\left(ℂ\right)$ given by the usual multiplication of $d\times d$ matrices on all column vectors of size $d\text{.}$

		Proof.
	Let $V$ be the $d$ dimensional vector space of column vectors of size $d\text{.}$ The standard basis of $V$ consists of the vectors $e_i={(0,\dots ,0,1,0,\dots ,0)}^1,$ $1\le i\le d$ where the 1 appears in the $i\text{th}$ spot. Suppose that $V^{\prime }$ is a nonzero invariant subspace of $V\text{.}$ Let $v=\sum _i{v}_i{e}_i,$ $v_i\in ℂ,$ be a nonzero element of $V^{\prime }\text{.}$ So $v_i\ne 0$ for some $1\le i\le d\text{.}$ Then $(1/v_i)E_{ji}v=e_j\text{.}$ Since $V^{\prime }$ is invariant we have that $e_j\in V^{\prime }$ for each $1\le j\le d\text{.}$ But since the $e_j$ are a basis of $V$ this implies that $V=V^{\prime }\text{.}$ So $V$ is an irreducible representation of $M_d\left(ℂ\right)\text{.}$ Now let W be an arbitrary irreducible representation of $A=M_d\left(ℂ\right)\text{.}$ There is some vector $w\in W$ and some $a\in A$ such that $aw\ne 0,$ otherwise $W$ would be the zero representation. If $a=\mid a_{ij}\mid$ then $aw=\sum _{i,j}{a}_{ij}{E}_{ij}w\ne 0$ implies that $E_{ij}w\ne 0$ for some pair $(i,j)\text{.}$ The space $M_d\left(ℂ\right)E_{ij}$ consists of all matrices that are 0 except in the $j\text{th}$ column and is isomorphic to $V\text{.}$ The map $$\begin{array}{cc}\begin{array}{cccc}\varphi :& M_d\left(ℂ\right)E_{ij}& ⟶& W\\ & a{E}_{ij}& ⟶& a{E}_{ij}\end{array}& \text{(3.2)}\end{array}$$ is an isomorphism since both $V$ and $W$ are irreducible. $\square$

So the regular representation of $M_d\left(ℂ\right)$ decomposes as a direct sum of $d$ copies of the unique irreducible representation $V$ of $M_d\left(ℂ\right),$ one copy for each column in $M_d\left(ℂ\right)\text{.}$

An algebra $A$ is semisimple if

$$\begin{array}{cc}A\cong {\oplus }_{\lambda \in \Lambda }{M}_{d_{\lambda }}\left(ℂ\right),& \text{(3.3)}\end{array}$$

where $\Lambda$ is a finite index set. The vector $\overrightarrow{d}=\left(d_{\lambda }\right),$ $\lambda \in \Lambda$ of positive integers is called the dimension vector of the algebra $A\text{.}$ We will use $M_{\bar{d}}\left(ℂ\right)$ as a shorthand notation for the algebra given by the right hand side of (3.3). We can view $M_{\bar{d}}\left(ℂ\right)$ as the full algebra of block diagonal matrices where the $\lambda \text{th}$ block is dimension $d_{\lambda }\text{.}$ We denote the matrix having 1 in the $(i,j)\text{th}$ position of the $\lambda \text{th}$ block and zeros everywhere else by $E_{ij}^{\lambda }\text{.}$ Denote the matrix which is the identity on the $\lambda \text{th}$ block and 0 everywhere else by $I_{\lambda }\text{.}$

Any trace on ${\oplus }_{\lambda \in \Lambda }{M}_{d_{\lambda }}\left(ℂ\right)$ is completely determined by a vector $\overrightarrow{t}=\left(t_{\lambda }\right)$ of complex numbers such that $\overrightarrow{t}\left(E_{11}^{\lambda }\right)=t_{\lambda }$ for each $\lambda$ in the finite index set $\Lambda \text{.}$ The vector $\overrightarrow{t}=\left(t_{\lambda }\right)$ is the trace vector of the trace $\overrightarrow{t}\text{.}$ A trace $\overrightarrow{t}$ on $M_{\bar{d}}\left(ℂ\right)$ is nondegenerate if and only if $t_{\lambda }\ne 0$ for all $\lambda \in \Lambda \text{.}$ The only ideals of $M_{\bar{d}}\left(ℂ\right)={\oplus }_{\lambda \in \Lambda }{M}_{d_{\lambda }}\left(ℂ\right)$ are of the form ${\oplus }_{\lambda \in {\Lambda }^{\prime }}{M}_{d_{\lambda }}\left(ℂ\right)$ where ${\Lambda }^{\prime }\subset \Lambda \text{.}$ The $I_{\lambda },$ $\lambda \in \Lambda$ form a basis of the center of $M_{\bar{d}}\left(ℂ\right)\text{.}$ Every central idempotent is a sum of some subset of the $I_{\lambda }\text{.}$ There is, up to isomorphism, one irreducible representation of ${\oplus }_{\lambda \in \Lambda }{M}_{d_{\lambda }}\left(ℂ\right)$ for each $\lambda \in \Lambda \text{.}$ It can be given by left multiplication on the space $M_{\bar{d}}\left(ℂ\right)E_{ij}^{\lambda },$ for any $i,j,$ $1\le i,j\le d_{\lambda }\text{.}$ The decomposition of the regular representation of ${\oplus }_{\lambda \in \Lambda }{M}_{d_{\lambda }}\left(ℂ\right)$ into irreducibles is given by

$$\begin{array}{cc}\stackrel{⟶}{M_{\stackrel{‾}{d}}\left(ℂ\right)}={\oplus }_{\lambda \in \Lambda }{W}_{\lambda }^{\oplus d_{\lambda }}& \text{(3.4)}\end{array}$$

where $W_{\lambda }$ denotes the irreducible representation corresponding to $\lambda \text{.}$

Matrix units and characters

Let $A$ be an algebra and $\hat{A}$ a finite index set such that $A\cong {\oplus }_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)$ under an isomorphism $\varphi :A\to {\oplus }_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$ (Let Md‾(ℂ) denote the algebra ${\oplus }_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right)\text{.)}$ Warning: The isomorphism $\varphi$ is not unique; nontrivial automorphisms of Md‾(ℂ) do exist, just conjugate by an invertible matrix. $z_{\lambda }={\varphi }^{-1}\left(I_{\lambda }\right)$ is an idempotent and an element of the center of $A\text{.}$ The zλ are the minimal central idempotents of $A\text{.}$ They are minimal in the sense that every central idempotent of $A$ is a sum of $z_{\lambda }\text{'s.}$ These elements are independent of the isomorphism $\varphi \text{.}$

A set of elements $e_{ij}^{\lambda }\in A,$ $\lambda \in \hat{A},$ $1\le i,j\le d_{\lambda }$ is the set of matrix units of $A$ if

$$\begin{array}{cc}{e}_{ij}^{\lambda }{e}_{rs}^{\mu }\{\begin{array}{cc}0,& \text{if}\lambda \ne \mu ;\\ 0,& \text{if}\lambda =\mu ,j\ne r;\\ e_{is}^{\lambda },& \text{if}\lambda =\mu ,j=r\text{.}\end{array}& \text{(3.5)}\end{array}$$

A complete set of matrix units of $A$ is a set of matrix units which forms a basis of $A\text{.}$ Let $E_{ij}^{\lambda }\in M_{\bar{d}}\left(ℂ\right)$ denote the matrix having 1 in the $(i,j)\text{th}$ position of the $A\text{th}$ block and zeros everywhere else. If $\left\{e_{ij}^{\lambda }\right\}$ is a set of matrix units of $A,$ the mapping $e_{ij}^{\lambda }⟼E_{ij}^{\lambda }$ determines explicitly an isomorphism $A\to M_{\bar{d}}\left(ℂ\right)\text{.}$ Conversely, an isomorphism $\varphi :A\to M_{\bar{d}}\left(ℂ\right)$ determines a set of matrix units $e_{ij}^{\lambda }={\varphi }^{-1}\left(E_{ij}^{\lambda }\right)\text{.}$ Note that the $e_{ii}^{\lambda }$ are minimal orthogonal idempotents in $A\text{.}$

Let $W_{\lambda },$ $\lambda \in \hat{A}$ denote the irreducible representations $A\text{.}$ By (3.1) and (3.2), for each $\lambda \in \hat{A},$

$$\begin{array}{cc}{W}_{\lambda }\cong A{e}_{ij}^{\lambda },& \text{(3.6)}\end{array}$$

for any $i,j,$ $1\le i,j\le d_{\lambda },$ where the action of $A$ on $A{e}_{ij}^{\lambda }$ is given by left multiplication. For each $\lambda \in \hat{A}$ denote the character of the irreducible representation $W_{\lambda }$ by ${\chi }^{\lambda }$ and for each $\lambda \in \hat{A}$ and $a\in A$ let $W_{ij}^{\lambda }\left(a\right)$ denote the $(i,j)\text{th}$ entry of the matrix $W_{\lambda }\left(a\right)\text{.}$ Note that we can view each $W_{\lambda }\left(a\right)$ as a matrix in $M_{\bar{d}}\left(ℂ\right)$ with all but the $\lambda \text{th}$ block 0.

Let $B$ be an arbitrary basis of $A\text{.}$ Let $\overrightarrow{t}=\left(t_{\lambda }\right)$ $\lambda \in \hat{A}$ be a nondegenerate trace on $A\text{.}$ For each $g\in B$ let $g^{*}$ denote the element of the dual basis to $B$ with respect to the trace $\overrightarrow{t}$ such that $\overrightarrow{t}\left(g{g}^{*}\right)=1\text{.}$

(3.7) Theorem. (Fourier inversion formula) The elements

$$e_{ij}^{\lambda }=\sum _{g\in B}{t}_{\lambda }{W}_{ji}^{\lambda }\left(g^{}\right)g$$

form a complete set of matrix units of $A\text{.}$

		Proof.
	Let $\varphi :A\to M_{\bar{d}}\left(ℂ\right)$ be given by $\varphi \left(a\right)={\oplus }_{\lambda }{W}_{\lambda }\left(a\right)\text{.}$ This is an isomorphism. For each $\lambda \in \hat{A}$ and $1\le i,j\le d_{\lambda }$ let $$e_{ij}^{\lambda }={\varphi }^{-1}\left(E_{ij}^{\lambda }\right)\text{.}$$ The set $B=\left\{e_{ij}^{\lambda }\right\}$ forms a basis of $A\text{.}$ The dual basis with respect to the trace $\overrightarrow{t}=\left(t_{\lambda }\right)$ is the basis $\left\{(1/t_{\lambda })e_{ji}^{\lambda }\right\}\text{.}$ $$\begin{array}{ccc}\sum _{g\in B}{t}_{\lambda }{W}_{ji}^{\lambda }\left(g^{*}\right)g& =& \sum _{k,l,\mu }{t}_{\lambda }{W}_{ji}^{\lambda }\left((1/t_{\mu })e_{lk}^{\mu }\right)e_{kl}^{\mu }\\ & =& \sum _{k,l,\mu }{t}_{\lambda }(1/t_{\mu }){\delta }_{jl}{\delta }_{ik}{\delta }_{\lambda \mu }{e}_{kl}^{\mu }\\ & =& e_{ij}^{\lambda }\text{.}\end{array}$$ Notice that $$k\text{th row of}\left(\sum _{g\in B}{W}_{ji}^{\lambda }\left(g^{*}\right)W_{\lambda }\left(g\right)\right)=j\text{th row of}\left(\sum _{g\in B}{W}_{\lambda }\left(g^{*}\right)E_{ik}^{\lambda }{W}_{\lambda }\left(g\right)\right)\text{.}$$ By § Ex. 1 we know that $\sum _{g\in B}{W}_{\lambda }\left(g^{*}\right)E_{ik}^{\lambda }{W}_{\lambda }\left(g\right)$ is independent of the basis $B\text{.}$ $\square$

(3.8) Theorem.

$$\sum _{g\in B}{t}_{\lambda }{\chi }^{\lambda }\left(g^{*}\right)g=z_{\lambda }\text{.}$$

		Proof.
	$$\begin{array}{ccc}{z}_{\lambda }& =& \sum _{i=1}^{d_{\lambda }}{e}_{ii}^{\lambda }\\ & =& \sum _i\sum _{g\in B}{t}_{\lambda }{W}_{ii}^{\lambda }\left(g^{*}\right)g\\ & =& \sum _{g\in B}\left(\sum _i{t}_{\lambda }{W}_{ii}^{\lambda }\left(g^{*}\right)\right)g\\ & =& \sum _{g\in B}{t}_{\lambda }{\chi }^{\lambda }\left(g^{*}\right)g\text{.}s\end{array}$$ $\square$

(3.9) Theorem.

$$\sum _{g\in B}{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{*}\right)(d_{\lambda }/t_{\lambda }){\delta }_{\lambda \mu }\text{.}$$

		Proof.
	$$\begin{array}{ccc}{d}_{\lambda }{\delta }_{\lambda \mu }& =& {\chi }^{\lambda }\left(z_{\mu }\right)\\ & =& {\chi }^{\lambda }\left(\sum _{g\in B}{t}_{\lambda }{\chi }^{\mu }\left(g^{*}\right)g\right)\\ & =& \sum _{g\in B}{t}_{\lambda }{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{*}\right)\text{.}\end{array}$$ $\square$

Examples

1. If $A$ is commutative and semisimple then all irreducible representations of $A$ are one dimensional. This is not necessarily true for algebras over fields which are not algebraically closed (since Schur's lemma takes a different form).
2. If $R$ is a ring with identity and $M_n\left(R\right)$ denotes $n\times n$ matrices with entries in $R$. the ideals of $M_n\left(R\right)$ are of the form $M_n\left(I\right)$ where $I$ is an ideal of $R.$
3. If $V$ is a vector space over $ℂ$ and $V*$ is the space of $ℂ$-valued functions on $V$ then $\dim V*=\dim V.$ If $B$ is a basis of $V$ then the functions ${\delta }_b,b\in B,$ determined by $${\delta }_b\left(b_i\right)=\left\{\begin{array}{rc}1,& \text{if }b=b_i,\\ 0,& \text{otherwise,}\end{array}\right\}$$ for $b_i\in B,$ form a basis of $V*.$ If $A$ is a semisimple algebra isomorphic to $M_d\left(ℂ\right)={\oplus }_{\lambda \in \tilde{A}}{M}_{d_{\lambda }}\left(ℂ\right),$ $\tilde{A}$ an index set for the irreducible representations $W_{\lambda }$ of $A,$ then $$\dim A=\sum _{\lambda \in \tilde{A}}{d}_{\lambda }^2,$$ and the functions $W_{ij}^{\lambda }(W_{ij}^{\lambda }\left(a\right)$ the $\left(i,j\right)$-th entry of the matrix $W_{\lambda }\left(a\right),a\in A)$ on $A$ form a basis of $A*.$ The $W_{ij}^{\lambda }$ are simply the functions ${\delta }_{e_{ij}^{\lambda }}$ for an appropriate set of matrix units $\left\{e_{ij}^{\lambda }\right\}$ of $A.$ Thi shows that the coordinate functions of the irreducible representations are linearly independent. Since ${\chi }^{\lambda }=\sum _i{W}_{ii}^{\lambda },$ the irreducible characters are also linearly independent.
4. Let $A$ be a semisimple algebra. Virtual characters are elements of the vector space $R\left(A\right)$ consisting of the $ℂ$-linear span of the irreducible characters of $A.$ We know that there is a one-to-one correspondence between the minimal central idempotents of $A$ and the irreducible characters of $A.$ Since the minimal central idempotents of $A$ form a basis of the center $Z\left(A\right)$ of $A,$ we ca define a vector space isomorphism $\phi :Z\left(A\right)\to R\left(A\right)$ by setting $\phi \left(z_{\lambda }\right)={\chi }^{\lambda }$ for each $\lambda \in \tilde{A}$ and extending linearly to all of $Z\left(A\right).$

   Given a nondegenerate trace $\overrightarrow{t}$ on $A$ with trace vector $\left(t_{\lambda }\right)$ it is more natural to define $\phi$ by setting $\phi \left(z_{\lambda }/t_{\lambda }\right)={\chi }^{\lambda }.$ Then, for $z\in Z\left(A\right),$ $$\phi \left(z\right)\left(a\right)=\overrightarrow{t}\left(za\right),$$ since $$\begin{array}{rcl}\phi \left(z_{\mu }/t_{\mu }\right)\left(a\right)& =& \overrightarrow{t}\left(z_{\mu }/t_{\mu }a\right)\\ & =& \overrightarrow{t}\left(\left(\frac{1}{t_{\mu }}\right)z_{\mu }a\right)\\ & =& \left(\frac{1}{t^{\mu }}\right)\left(t_{\mu }{\chi }^{\mu }\left(a\right)\right)\\ & =& {\chi }^{\mu }\left(a\right).\end{array}$$

5. If $A$ is a semisimple algebra isomorphic to $M_b\left(ℂ\right)={\oplus }_{\lambda \in \tilde{A}}{M}_{d_{\lambda }}\left(ℂ\right),\tilde{A}$ an index set for the irreducible representations $W_{\lambda }$ of $A,$ then the right regular representation decomposes as $$\overrightarrow{A}\cong {\oplus }_{\lambda \in A}{W}_{\lambda }^{\oplus d_{\lambda }}.$$ If matrix units $e_{ij}^{\lambda }$ are given by (3.7) then $$\mathrm{tr}\left(e_{ii}^{\lambda }\right)=\mathrm{tr}\left(d_{\lambda }{E}_{ii}^{\lambda }\right)=d_{\lambda }.$$ So the trace of the regular representation of $A,$$\mathrm{tr},$ is given by the trace vector $\overrightarrow{t}=\left(t_{\lambda }\right),$ where $t_{\lambda }=d_{\lambda }$ for each $\lambda \in \tilde{A}.$
6. Let $A$ be a semisimple algebra and let $B*=\left\{g*\right\}$ be a dual basis to $B=\left\{g\right\}$ of $A$ with respect to the tracde of the regular representation of $A.$ We can define an inner product on the space $R\left(A\right)$ of virtual characters, Ex 4, of $A$ by $$⟨\chi ,\chi \text{'}⟩=\sum _{g\in B}\chi \left(g\right)\chi \text{'}\left(g*\right).$$ The irreducible characters of $A$ are orthonormak with respect to this inner product. Nate that $\chi ,\chi \text{'}$ are characters of representations $V$ and $V\text{'}$ respectively, then, by Ex4 and Theorem 3.9, $$⟨\chi ,\chi \text{'}⟩=\dim {\mathrm{Hom}}_A\left(V,V\text{'}\right).$$ If ${\chi }^{\lambda }$ is the character of the irreducible representation $W_{\lambda }$ of $A$ then $⟨\chi ,\chi \text{'}⟩$ gives the multiplicity of $W_{\lambda }$ in the representation $V$ as in Section 1, Ex 3.
7. Let $A$ be a semisimple algebra and $\overrightarrow{t}=\left(t_{\lambda }\right)$ be a non-degnerate trace on $A.$ Let $B$ be a basis of $A$ and for each $g\in B$ let $g*$ denote the element of the dual basis to $B$ ith respect to the trace $\overrightarrow{t}$ such that $\overrightarrow{t}\left(gg*\right)=1.$ For each $a\in A$ define $$\left[a\right]=\sum _{g\in B}gag*.$$ By Section 2, Ex 1, the element $\left[a\right]$ is independent of the choice of the basis $B.$ By using a set of matrix units $e_{ij}^{\lambda }$ of $A$ we get $$\begin{array}{rcl}\left[a\right]& =& \sum _{i,j,\lambda }\left(\frac{1}{t_{\lambda }}\right)e_{ij}^{\lambda }a{e}_{ji}^{\lambda }\\ & =& \sum _{i,j,\lambda }\left(\frac{1}{t_{\lambda }}\right)a_{jj}^{\lambda }{e}_{ii}^{\lambda }\\ & =& \sum _{\lambda }\left(\frac{1}{t_{\lambda }}\right)\left(\sum _j{a}_{jj}^{\lambda }\left(\sum _i{e}_{ii}^{\lambda }\right)\right)\\ & =& \sum _{\lambda }\left(\frac{1}{t_{\lambda }}\right){\chi }^{\lambda }\left(a\right)z_{\lambda }.\end{array}$$ So ${\chi }^{\lambda }\left(\left[a\right]\right)=\left(\frac{d_{\lambda }}{t_{\lambda }}\right){\chi }^{\lambda }\left(a\right).$ By 3.9 $$\sum _{g\in B}\left(\frac{t_{\lambda }^2}{d_{\lambda }}\right){\chi }^{\mu }\left(g*\right)\left[g\right]=\sum _{\lambda }\sum _{g\in B}\left(\frac{t_{\lambda }^2}{d_{\lambda }}\right)\left(\frac{1}{t_{\lambda }}\right){\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g*\right)z_{\lambda }=\sum _{\lambda }{\delta }_{\lambda \mu }{z}_{\lambda }=z_{\mu }.$$ Thus the $\left[g\right],g\in B,$ span the center of $A.$
8. Let $G$ be a finite group and let $A=ℂG$. Let $\overrightarrow{t}$ be the trace on $A$ given by $$\overrightarrow{t}\left(a\right)=a{|}_1,$$ where 1 is the identity in $G.$ By Ex 5 and Section 2 Ex 3 the trace vector of $\overrightarrow{t}$ is given by $t_{\lambda }=\left(\frac{d_{\lambda }}{\left|G\right|}\right)$ where $d_{\lambda }$ is the dimnesion of the irreducible representation of $G$ corresponding to $\lambda .$

   If $h\in G,$ then the element $$\left[h\right]=\sum _{g\in B}ghg*=\sum _{g\in B}gh{g}^{-1}$$ is a multiple of the sum of the elements of $G$ that are conjugate to $h.$ Let $\Lambda$ be an index set of the conjugacy classes of $G$ and for each $\lambda \in \Lambda$, let $C_{\lambda }$ denote the sum of the elements in the conjugacy class indexed by $\lambda$. The $C_{\lambda }$ are linearly independent elements of $ℂG$. Furthermore by Ex 7 they span the center of $ℂG.$ Thus $\Lambda$ must also be an index set for the irreducible representations of $G.$ So we see that the irreducible representations of the group algebra of a finite group are indexed by the conjugacy classes.

9. Let $G$ be a finite group and let $C_{\lambda }$ denote the conjugacy classes of $G.$ Note that since $$\mathrm{tr}\left(V\left(hg{h}^{-1}\right)\right)=\mathrm{tr}\left(V\left(h\right)V\left(g\right)V{\left(h\right)}^{-1}\right)=\mathrm{tr}\left(V\left(g\right)\right)$$ for any representation $V$ of $G$ and all $g,h\in G,$ characters of $G$ are constant on conjugacy classes. Using theorem 3.8, $$\begin{array}{rcl}\left|G\right|{\delta }_{\lambda \mu }& =& \sum _g{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{-1}\right)\\ & =& \sum _{\rho }\sum _{g\in C_{\rho }}{\chi }^{\lambda }\left(g\right){\chi }^{\mu }\left(g^{-1}\right)\\ & =& \sum _{\rho }\left|C_{\rho }\right|{\chi }^{\lambda }\left(\rho \right){\chi }^{\mu }\left(\rho \text{'}\right),\end{array}$$ where $\rho \text{'}$ is such that $C_{\rho \text{'}}$ is the conjugacy class which contains the inverses of the elements in $C_{\rho }.$ Define matrices $\Xi =\parallel {\Xi }_{\lambda \rho }\parallel$ and $\Xi \text{'}=\parallel {\Xi \text{'}}_{\lambda \rho }\parallel$ by ${\Xi }_{\lambda \rho }={\chi }^{\lambda }\left(\rho \right)$ and ${\Xi }_{\lambda \rho }\text{'}=\left|C_{\rho }\right|{\chi }^{\lambda }\left({\rho }^{\text{'}}\right).$ By Ex 8 these matrices are square. In matrix notation the above is $$\Xi {\Xi }^{\text{'}t}=\left|G\right|I,$$ but then we also have that ${\Xi }^{\text{'}t}\Xi =\left|G\right|I,$ or equivalently that $$\sum _{\lambda }{\chi }^{\lambda }\left({\rho }^{\text{'}}\right){\chi }^{\lambda }\left(\tau \right)=\left(\frac{\left|G\right|}{\left|C_{\rho }\right|}\right){\delta }_{\rho \tau }.$$
10. This example gives a generalisation of the preceding example. Let $A$ be a semisimple algebra and suppose that $B$ is a basis of $A$ and that there is a partition of $B$ into classes such that if $b$ and $b\text{'}\in B$ are in the same classes then for every $\lambda \in \tilde{A}$, $${\chi }^{\lambda }\left(b\right)={\chi }^{\lambda }\left(b^{\text{'}}\right).$$ The fact that the characters are linearly independent implies that the number of classes must be the same as the number of irreducible characters ${\chi }^{\lambda }.$ Thus we can inbox the classes of $B$ by the elements of $\tilde{A}.$ Assume that we have fixed such a correspondence and denote the classses of $B$ by $C_{\lambda },\lambda \in \tilde{A}.$

    Let $\overrightarrow{t}$ be a nondegenerate trace on $A$ and let $G$ be the Gram matrix with respect to the basis $B$ and the trace $\overrightarrow{t}.$ If $g\in B,$ let $g*$ denote the element of the dual basis to $B$, with respect to the trace $\overrightarrow{t}$, such that $\overrightarrow{t}\left(gg*\right)=1.$ Let $G^{-1}=C=\parallel c_{gg\text{'}}\parallel$ and recall that $g*={\sum }_{g\text{'}\in B}{c}_{gg\text{'}}g\text{'}.$ Then Collecting $g,g\text{'}\in B$ by class size gives where ${\chi }^{\lambda }\left(\rho \right)$ denotes the value of the character ${\chi }^{\lambda }\left(\rho \right)$ at elements of the class $C_{\rho }.$ Now define a matrix $C=\parallel c_{\rho \tau }\parallel$ with entries $$c_{\rho \tau }=\sum _{g\in C_{\rho },g\text{'}\in C_{\tau }}{c}_{gg\text{'}},$$ and let $\Xi =\parallel {\Xi }_{\lambda \rho }\parallel$ and $\Xi \text{'}=\parallel {\Xi \text{'}}_{\lambda \rho }\parallel$ be matrices given by ${\Xi }_{\rho \lambda }={\chi }^{\lambda }\left(\rho \right)$ and ${\Xi \text{'}}_{\lambda \rho }=\left(\frac{t_{\lambda }}{d_{\lambda }}\right){\chi }^{\lambda }\left(\rho \right).$ Note that all of these matrices are square. Then the above gives that $$I=\Xi C{\Xi \text{'}}^t.$$ So $$I=C{\Xi \text{'}}^t\Xi ,$$ or equivalently that $$\begin{array}{rcl}{\delta }_{\rho \tau }& =& \sum _{\sigma ,\lambda }{c}_{\rho \sigma }\left(\frac{t_{\lambda }}{d_{\lambda }}\right){\chi }^{\lambda }\left(\sigma \right){\chi }^{\lambda }\left(\tau \right)\\ & =& \sum _{\sigma ,\lambda }\sum _{g\in C_{\rho },g\text{'}\in C_{\sigma }}{c}_{gg\text{'}}\left(\frac{t_{\lambda }}{d_{\lambda }}\right){\chi }^{\lambda }\left(\sigma \right){\chi }^{\lambda }\left(\tau \right)\\ & =& \sum _{\lambda }\sum _{g\text{'}\in B}\sum _{g\in C_{\rho }}{c}_{gg\text{'}}{\chi }^{\lambda }\left(\sigma \right){\chi }^{\lambda }\left(\tau \right)\\ & =& \sum _{g\in C_{\rho }}\left(\sum _{\lambda }{\chi }^{\lambda }\left(g*\right){\chi }^{\lambda }\left(\tau \right)\right).\end{array}$$

Notes and References

The Fourier Inversion formula for representations of finite groups appears in [Ser1977] p. 49. I must thank Prof. A. Garsia for suggesting the problem of finding a generalization. I know of no references giving a similar generalization. Theorems (3.7) and (3.8) are due to R. Kilrnoyer and appear in [CRe1981] (9.17) and (9.19). Ex. 3 is the Frobenius-Schur theorem. Ex. 9 is known as the second orthogonality relation for characters of finite groups (the first orthogonality relation being (3.8)), see [CRe1981] (9.26) or [Ser1977] Chap. 2, Prop. 7. The generalization given in Ex. 10 is new as far as I know. [Rl] shows that the Brauer algebra is an example of semsimple algebra that is not a group algebra with a natural basis that can be partitioned into classes such that (3.13) holds.

Notes and References

This is an excerpt from the unpublished first chapter of Arun Ram's dissertation entitled Representation Theory, written July 4, 1990.

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