Representation theory Lecture Notes: Chapter 1

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

Last update: 28 October 2013

Algebras and representations.

Algebras.

An algebra is a vector space (over $ℂ\text{)}$ with a multiplication such that $A$ is a ring with identity, i.e. there is a map $A\times A\to A,$ $(a,b)\mapsto ab,$ which is bilinear and satisfies the associative and distributive laws. The following are examples of algebras:

(1)	The group algebra of a group $G$ is the vector space $ℂG$ with basis $G$ and with multiplication forced by the multiplication in $G$ (and the bilinearity).
(2)	If $M$ is a vector space (over $ℂ\text{)}$ then the space $\text{End}\left(M\right)$ of $ℂ\text{-linear}$ transformations of $M$ is an algebra under the multiplication given by composition of endomorphisms.
(3)	Given a basis $B=\{b_1,\dots ,b_d\}$ of the vector space $M$ the algebra $\text{End}\left(M\right)$ can be idenitified with the algebra $M_d\left(ℂ\right)$ of $d\times d$ matrices $T={\left(T_{ij}\right)}_{1\le i,j\le d}$ with entries in $ℂ$ via $$T{B}_i=\sum _{i=1}^d{b}_j{T}_{ji},\text{for}\hspace{0.17em}t\in \text{End}\left(M\right)\text{.}$$

Let $A$ be an algebra. An ideal in $A$ is a subspace $I\subset A$ such that $ar\in I$ and $ra\in I,$ for all $a\in A$ and $r\in I\text{.}$ A minimal ideal of $A$ is a nonzero ideal $I$ which cannot be written as a direct sum $I=I_1\oplus I_2$ of nonzero ideals $I_1$ and $I_2$ of $A\text{.}$ An idempotent is a nonzero element $p\in A$ such that $p^2=p\text{.}$ Two idempotents $p_1,p_2\in A$ are orthogonal if $p_1{p}_2=p_2{p}_1=0\text{.}$ A minimal idempotent is an idempotent $p$ that cannot be written as a sum $p=p_1+p_2$ of orthogonal idempotents $p_1,p_2\in A\text{.}$ The center of $A$ is $$Z\left(A\right)=\{z\in A\hspace{0.17em}|\hspace{0.17em}az=za\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$ A central idempotent is an idempotent in $Z\left(A\right)$ and a minimal central idempotent is a central idempotent $z$ that cannot be written as a sum $z=z_1+z_2$ of orthogonal central idempotents $z_1$ and $z_2\text{.}$

A trace on $A$ is a linear map $\overrightarrow{t}:A\to ℂ$ such that $$\overrightarrow{t}\left(a_1{a}_2\right)=\overrightarrow{t}\left(a_2{a}_1\right),\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$ A character of $A$ is a trace on $A\text{.}$ A trace $\overrightarrow{t}$ on $A$ is nondegenerate if for each $b\in A$ there is an $a\in A$ such that $\overrightarrow{t}\left(ba\right)\ne 0\text{.}$ The radical of a trace $\overrightarrow{t}$ is $$\begin{array}{cc}\text{rad}\hspace{0.17em}t=\{b\in A\hspace{0.17em}|\hspace{0.17em}\overrightarrow{t}\left(ba\right)=0\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}& \text{(1.1)}\end{array}$$

Every trace $\overrightarrow{t}$ on $A$ determines a symmetric bilinear form $⟨,⟩:A\times A\to ℂ$ given by $$\begin{array}{cc}⟨a_1,a_2⟩=\overrightarrow{t}\left(a_1{a}_2\right),\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}& \text{(1.2)}\end{array}$$ The form $⟨,⟩$ is nondegenerate if and only if the trace $\overrightarrow{t}$ is nondegenerate and the radical $$\text{rad}\hspace{0.17em}⟨,⟩=\{a\in A\hspace{0.17em}|\hspace{0.17em}⟨a,b⟩=0\hspace{0.17em}\text{for all}\hspace{0.17em}b\in A\}$$ of the form $⟨,⟩$ is the same as rad $\overrightarrow{t}\text{.}$

Let $\overrightarrow{t}$ be a trace on $A$ and let $⟨,⟩$ be the bilinear form on $A$ defined by the trace $\overrightarrow{t},$ as in ??. Let $B$ be a basis of $A\text{.}$ Let $G={\left(⟨b,b\prime ⟩\right)}_{b,b\prime \in B}$ be the matrix of the form $⟨,⟩$ with respect to $B\text{.}$ The following are equivalent:

(1)	The trace $\overrightarrow{t}$ is nondegenerate.
(2)	$\text{det}\hspace{0.17em}G\ne 0\text{.}$
(3)	The dual basis $B^{*}$ to the basis $B$ with respect to the form $⟨,⟩$ exists.

		Proof.
	(2) $\iff$ (1): The trace $\overrightarrow{t}$ is degenerate if there is an element $a\in A,$ $a\ne 0,$ such that $\overrightarrow{t}\left(ac\right)=0$ for all $c\in B\text{.}$ If $a_b\in ℂ$ are such that $$a=\sum _{b\in B}{a}_{b}b,\text{then}0=⟨a,c⟩=\sum _{b\in B}{a}_b⟨b,c⟩$$ for all $c\in B\text{.}$ So $a$ exists if and only if the columns of $G$ are linearly dependent, i.e. if an only if $G$ is not invertible. (3) $\iff$ (2): Let $B^{*}=\left\{b^{*}\right\}$ be the dual basis to $\left\{b\right\}$ with respect to $⟨,⟩$ and let $P$ be the change of basis matrix from $B$ to $B^{*}\text{.}$ Then $$d^{*}=\sum _{b\in B}{P}_{db}b,\text{and}{\delta }_{bc}=⟨b,d^{*}⟩=\sum _{d\in B}⟨b,c⟩={\left(G{P}^t\right)}_{b,c}\text{.}$$ So $P^t,$ the transpose of $P,$ is the inverse of the matrix $G\text{.}$ So the dual basis to $B$ exists if and only if $G$ is invertible, i.e. if and only if $\text{det}\hspace{0.17em}G\ne 0\text{.}$ $\square$

Let $A$ be an algebra and let $\overrightarrow{t}$ be a nondegenerate trace on $A\text{.}$ Define a symmetric bilinear form $⟨,⟩:A\times A\to ℂ$ on $A$ by $⟨a_1,a_2⟩=\overrightarrow{t}\left(a_1{a}_2\right),$ for all $a_1,a_2\in A\text{.}$ Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis to $B$ with respect to $⟨,⟩\text{.}$ Let $a\in A$ and define $$\left[a\right]=\sum _{b\in B}ba{b}^{*}\text{.}$$ Then $\left[a\right]$ is an element of the center $Z\left(A\right)$ of $A$ and $\left[a\right]$ does not depend on the choice of the basis $B\text{.}$

		Proof.
	Let $c\in A\text{.}$ Then $$c\left[a\right]=\sum _{b\in B}cba{b}^{*}=\sum _{b\in B}\sum _{d\in B}⟨cb,d^{*}⟩da{b}^{*}=\sum _{d\in B}da\sum _{b\in B}⟨d^{*}c,b⟩b^{*}=\sum _{d\in B}da{d}^{*}c=\left[a\right]c,$$ since $⟨cb,d^{*}⟩=\overrightarrow{t}\left(cb{d}^{*}\right)=\overrightarrow{t}\left(d^{*}cb\right)=⟨d^{*}c,b⟩\text{.}$ So $\left[a\right]\in Z\left(A\right)\text{.}$ Let $D$ be another basis of $A$ and let $D^{*}$ be the dual basis to $D$ with respect to $⟨,⟩\text{.}$ Let $P=\left(P_{db}\right)$ be the transition matrix from $D$ to $B$ and let $P^{-1}$ be the inverse of $P\text{.}$ Then $$d=\sum _{b\in B}{P}_{db}b\text{and}{d}^{*}=\sum _{\stackrel{\sim }{b}\in B}{\left(P^{-1}\right)}_{\stackrel{\sim }{b}d}{\stackrel{\sim }{b}}^{*},$$ since $$⟨d,{\stackrel{\sim }{d}}^{*}⟩=⟨\sum _{b\in B}{P}_{db}b,\sum _{\stackrel{\sim }{b}\in B}{\left(P^{-1}\right)}_{\stackrel{\sim }{b}\stackrel{\sim }{d}}{\stackrel{\sim }{b}}^{*}⟩=\sum _{b,\stackrel{\sim }{b}\in B}{P}_{db}{\left(P^{-1}\right)}_{\stackrel{\sim }{b}\stackrel{\sim }{d}}{\delta }_{b\stackrel{\sim }{b}}={\delta }_{d\stackrel{\sim }{d}}\text{.}$$ So $$\sum _{d\in D}da{d}^{*}=\sum _{d\in D}\sum _{b\in B}{P}_{db}ba\sum _{\stackrel{\sim }{b}\in B}{\left(P^{-1}\right)}_{\stackrel{\sim }{b}d}{\stackrel{\sim }{b}}^{*}=\sum _{b,\stackrel{\sim }{b}\in B}ba{\stackrel{\sim }{b}}^{*}{\delta }_{b\stackrel{\sim }{b}}=\sum _{b\in B}ba{b}^{*}\text{.}$$ So $\left[a\right]$ does not depend on the choice of the basis $B\text{.}$ $\square$

Representations.

An $A\text{-module}$ is a vector space $M$ (over $ℂ\text{)}$ with an $A\text{-action,}$ i.e. a map $A\times M\to M,$ $(a,m)\mapsto am,$ which is bilinear and such that $$1_{A}m=m\text{and}{a}_1\left(a_{2}m\right)=\left(a_1{a}_2\right)m,$$ for all $a_1,a_2\in A$ and $m\in M$ $\text{(}{1}_A$ denotes the identity in the algebra $A\text{).}$ A representation of $A$ is an $A\text{-module.}$ A representation of a group $G$ is a representation of the group algebra $ℂG\text{.}$ The character of an $A\text{-module}$ $M$ is the map ${\chi }^M:A\to ℂ$ given by $${\chi }^M\left(a\right)=\text{Tr}\left(M\left(a\right)\right),\text{for}\hspace{0.17em}a\in A,$$ where $M\left(a\right)$ is the linear transformation of $M$ determined by the action of $A$ and $\text{Tr}\left(M\left(a\right)\right)$ is the trace of $M\left(a\right)\text{.}$ An irreducible character of $A$ is the character of an irreducible representation of $A\text{.}$

An $A\text{-module}$ $M$ gives rise to a map $$\begin{array}{cc}\begin{array}{ccc}A& ⟶& \text{End}\left(M\right)\\ a& ⟼& M\left(a\right)\end{array}& \text{(1.5)}\end{array}$$ where $M\left(a\right)$ is the linear transformation of $M$ determined by the action of $a$ on $M\text{.}$ This map is linear and satisfies $$\begin{array}{ccc}M\left(1_A\right)& =& {\text{Id}}_M,\\ M\left(a_1{a}_2\right)& =& M\left(a_1\right)M\left(a_2\right),\end{array}$$ for all $a_1,a_2\in A,$ i.e. $A\to \text{End}\left(M\right)$ is a homomorphism of algebras. (Given a basis $B=\{b_1,\dots ,b_d\}$ of $M$ the map $A\to \text{End}\left(M\right)$ can be identified with a map $M:A\to M_d\left(ℂ\right)\text{.)}$ Conversely, an algebra homomorphism as in ??? and ??? determines an $A\text{-action}$ on $M$ by $$am=M\left(a\right)m,\text{for all}\hspace{0.17em}a\in A\hspace{0.17em}\text{and}\hspace{0.17em}m\in M\text{.}$$ Thus, the map $M:A\to \text{End}\left(M\right)$ and the $A\text{-module}$ $M$ are equivalent data. Historically, the map $M:A\to \text{End}\left(M\right)$ was the representation and $M$ was the $A\text{-module,}$ but now the terms representation and $A\text{-module}$ are used interchangeably. This is the reason for the use of the letter $M,$ both for the $A\text{-module}$ and the corresponding algebra homomorphism $M:A\to \text{End}\left(M\right)\text{.}$

A submodule of an $A\text{-module}$ $M$ is a subspace $N\subseteq M$ such that $an\in N,,$ for all $a\in A$ and $n\in N\text{.}$ An $A\text{-module}$ $M$ is simple or irreducible if it has no submodules except $0$ and itself. The direct sum of two $A\text{-modules}$ $M_1$ and $M_2$ is the vector space $M=M_1\oplus M_2$ with $A\text{-action}$ given by $$a(m_1,m_2)=(a{m}_1,a{m}_2),\text{for all}\hspace{0.17em}a\in A,m_1\in M_1\hspace{0.17em}\text{and}\hspace{0.17em}{m}_2\in M_2\text{.}$$ An $A\text{-module}$ $M$ is semisimple or completely decomposable if $M$ can be written as a direct sum of simple submodules. An $A\text{-module}$ $M$ is indecomposable if $M$ cannot be written as a direct sum $M=M_1\oplus M_2$ of nonzero submodules $M_1\subseteq M$ and $M_2\subseteq M\text{.}$

Here we need a reference to the reader to look at the examples in Chapter 2 etc.

Homomorphisms

Let $M$ and $N$ be $A\text{-modules.}$ Then define $${\text{Hom}}_A(M,N)=\{\varphi \in \text{Hom}(M,N)\hspace{0.17em}|\hspace{0.17em}a\varphi \left(m\right)=\varphi \left(am\right)\text{, for all}\hspace{0.17em}a\in A\hspace{0.17em}\text{and}\hspace{0.17em}m\in M\},$$ where $\text{Hom}(M,N)$ is the set of $ℂ\text{-linear}$ transformations from $M$ to $N\text{.}$ The proof of the following Proposition is identical to the proof of Proposition ??? except with $a$ replaced by $\varphi \text{.}$

Let $A$ be an algebra and let $\overrightarrow{t}$ be a nondegenerate trace on $A\text{.}$ Define a symmetric bilinear form $⟨,⟩:A\times A\to ℂ$ on $A$ by $⟨a_1,a_2⟩=\overrightarrow{t}\left(a_1{a}_2\right),$ for all $a_1,a_2\in A\text{.}$ Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis to $B$ with respect to $⟨,⟩\text{.}$ Let $M$ and $N$ be $A\text{-modules}$ and let $\varphi \in \text{Hom}(M,N)\text{.}$ Define $$\left[\varphi \right]=\sum _{b\in B}b\varphi b^{*}\text{.}$$ Then $\left[\varphi \right]\in {\text{Hom}}_A(M,N)$ and $\left[\varphi \right]$ does not depend on the choice of the basis $B\text{.}$

Direct sums of algebras

Let $A$ and $B$ be algebras and let $A^{\lambda },$ $\lambda \in \hat{A},$ and $B^{\mu },$ $\mu \in \hat{B},$ be the irreducible representations of $A$ and $B,$ respectively. The irreducible representations of $A\oplus B$ are $A^{\lambda },$ $\lambda \in \hat{A},$ with $A\oplus B$ action given by $$(a,b)m=am,\text{for}\hspace{0.17em}a\in A,b\in B,m\in A^{\lambda },$$ and $B^{\mu },$ $\mu \in \hat{B},$ with $A\oplus B$ action given by $$(a,b)n=bn,\text{for}\hspace{0.17em}a\in A,b\in B,\text{and}\hspace{0.17em}n\in B^{\mu }\text{.}$$

		Proof.
	The elements $(1,0)$ and $(0,1)$ in $A\oplus B$ are central idempotents of $A\oplus B$ such that $(1,0)(0,1)=(0,0)\text{.}$ If $P$ is an $A\oplus B\text{-module}$ then $$P=(1,0)P\oplus (0,1)P,$$ and this is a decomposition as $A\oplus B\text{-modules.}$ Since $$(a,b)(1,0)p=(a,0)(1,0)p,\text{and}(a,b)(0,1)p=(0,b)(0,1)p,$$ for all $a\in A,$ $b\in B,$ and $p\in P,$ the structure of $(1,0)P$ is determined completely by the $A\text{-action}$ and the structure of $(0,1)P$ is determined by the action of $B\text{.}$ If $P$ is a simple module then $P=(1,0)P$ or $P=(0,1)P\text{.}$ In the first case $P\cong A^{\lambda }$ for some $\lambda \in \hat{A}$ and in the second $P\cong B^{\mu }$ for some $\mu \in \hat{B}\text{.}$ $\square$

Similar arguments with the elements $(1,0)$ and $(0,1)$ in $A\oplus B$ yield the following.

(1)	If $A$ and $B$ are algebras then the ideals of $A\oplus B$ are all of the form $I\oplus J$ where $I$ is an ideal of $A$ and $J$ is an ideal of $B\text{.}$
(2)	If $A$ and $B$ are algebras then $Z(A\oplus B)=Z\left(A\right)\oplus Z\left(B\right)\text{.}$
(3)	If $A$ and $B$ are algebras and $\overrightarrow{t}$ is a trace on $A\oplus B$ then $\overrightarrow{t}$ is given by $$\overrightarrow{t}(a,b)={\overrightarrow{t}}_A\left(a\right)+{\overrightarrow{t}}_B\left(b\right),$$ where ${\overrightarrow{t}}_A$ is the trace on $A$ given by ${\overrightarrow{t}}_A\left(a\right)=\overrightarrow{t}(a,0)$ and ${\overrightarrow{t}}_B$ is the trace on $B$ given by ${\overrightarrow{t}}_B\left(b\right)=\overrightarrow{t}(0,b)\text{.}$

Tensor products

Let $M$ and $N$ be vector spaces and let $$B_m=\left\{m_i\right\}\text{and}{B}_n=\left\{n_j\right\}$$ be bases of $M$ and $N,$ respectively. The tensor product $M\otimes N$ is the vector space with basis $$B_{M\otimes N}=\{m_i\otimes n_j\hspace{0.17em}|\hspace{0.17em}{m}_i\in B_M,n_j\in B_n\}\text{.}$$ If $m={\sum }_i{c}_i{m}_i,$ and $n={\sum }_j{d}_j{n}_j,$ then write $$m\otimes n=\left(\sum _i{c}_i{m}_i\right)\otimes \left(\sum _j{d}_j{n}_j\right)=\sum _{i,j}{c}_i{d}_j(m_i\otimes n_j)\text{.}$$

If $A$ and $Z$ are algebras the tensor product is the vector space $A\otimes Z$ with multiplication determined by $$(a_1\otimes z_1)(a_2\otimes z_2)=a_1{a}_2\otimes z_1{z}_2,\text{for all}\hspace{0.17em}{a}_1,a_2\in A,z_1,z_2\in Z\text{.}$$ If $M$ and $N$ are vector spaces then $$\text{End}(M\otimes N)=\text{End}\left(M\right)\otimes \text{End}\left(N\right)\text{as algebras.}$$ This equality can be expressed in terms of matrices by choosing bases $\{m_1,\dots ,m_r\}$ and $\{n_1,\dots ,n_s\}$ of $M$ and $N,$ respectively. The $\text{End}\left(M\right)$ is identified with $M_r\left(ℂ\right)$ and $\text{End}\left(N\right)$ is identified with $M_s\left(ℂ\right)$ by $$E_{ij}{m}_j=m_i\text{and}{E}_{k\ell }{n}_{\ell }=n_k,\text{for}\hspace{0.17em}1\le i,j,\le r\hspace{0.17em}\text{and}\hspace{0.17em}1\le k,\ell \le s\text{.}$$ Then $$(E_{ij}\otimes E_{k\ell })(m_j\otimes n_{\ell })=E_{ij}{m}_j\otimes E_{k\ell }{n}_{\ell }=m_i\otimes n_k\text{.}$$ Use the (ordered) basis $$\{m_1\otimes n_1,\dots m_1\otimes n_s,m_2\otimes n_1,\dots ,m_2\otimes n_s,\dots ,m_r\otimes n_1,\dots ,m_r\otimes n_s\}$$ of $M\otimes N$ to identify $\text{End}(M\otimes N)$ with $M_{rs}\left(ℂ\right)\text{.}$ Then, if $a=\left(a_{ij}\right)\in M_r\left(ℂ\right)$ and $b=\left(b_{k\ell }\right)\in M_s\left(ℂ\right)$ then $a\otimes b$ is the $rs\times rs$ matrix $$a\otimes b=\left(\begin{array}{cccc}{a}_{11}b& a_{12}b& \cdots & a_{1r}b\\ a_{21}b& a_{22}b& \cdots & a_{2r}b\\ ⋮& & \ddots & ⋮\\ a_{r1}b& a_{r2}b& \cdots & a_{rr}b\end{array}\right)$$

Let $A$ and $B$ be algebras. Let $A^{\lambda },$ $\lambda \in \hat{A},$ be the simple $A\text{-modules}$ and let $B^{\mu },$ $\mu \in \hat{B},$ be the simple $B\text{-modules.}$ The simple $A\otimes B\text{-modules}$ are $$A^{\lambda }\otimes B^{\mu },\lambda \in \hat{A},\mu \in \hat{B},\text{where}(a\otimes b)(m\otimes n)=am\otimes bn,$$ for $a\in A,$ $b\in B,$ $m\in A^{\lambda },$ $n\in B^{\mu }\text{.}$

		Proof.
	There are two things to show: (1) $A^{\lambda }\otimes B^{\mu }$ is a simple $A\otimes B\text{-module,}$ (2) If $P$ is a simple $A\otimes B\text{-module}$ then $P\cong A^{\lambda }\otimes B^{\mu }$ for some $\lambda \in \hat{A}$ and $\mu \in \hat{B}\text{.}$ (1) By Burnside’s theorem $\text{End}\left(A^{\lambda }\right)=A^{\lambda }\left(A\right)$ and $\text{End}\left(B^{\mu }\right)=B^{\mu }\left(B\right)$ and therefore $$\text{End}(A^{\lambda }\otimes B^{\mu })=\text{End}\left(A^{\lambda }\right)\otimes \text{End}\left(B^{\mu }\right)=A^{\lambda }\left(A\right)\otimes B^{\mu }\left(B\right)=(A^{\lambda }\otimes B^{\mu })(A\otimes B)\text{.}$$ So $A^{\lambda }\otimes B^{\mu }$ has no submodules. So $A^{\lambda }\otimes B^{\mu }$ is simple. (2) Let $P$ be a simple $(A\otimes B)\text{-module.}$ Let $A^{\lambda }$ be a simple $A\text{-submodule}$ of $P$ and let $B^{\mu }$ be a simple $B\text{-submodule}$ of ${\text{Hom}}_A(A^{\lambda },P)\text{.}$ We claim that $A^{\lambda }\otimes B^{\mu }\cong P\text{.}$ Consider the $(A\otimes B)\text{-module}$ homomorphism $$\begin{array}{cccccc}\Phi :& A^{\lambda }\otimes B^{\mu }& \hookrightarrow & A^{\lambda }\otimes {\text{Hom}}_A(A^{\lambda },P)& ⟶& P\\ & & & m\otimes \varphi & ⟼& \varphi \left(m\right)\text{.}\end{array}$$ This map is nonzero since the injection $\varphi :A^{\lambda }\hookrightarrow P$ is a nonzero element of ${\text{Hom}}_A(A^{\lambda },P)\text{.}$ Since $A^{\lambda }\otimes B^{\mu }$ is simple $\text{ker}\hspace{0.17em}\Phi =0$ and since $P$ is simple $\text{im}\hspace{0.17em}\Phi =P\text{.}$ So $A^{\lambda }\otimes B^{\mu }\cong P\text{.}$ $\square$

The algebra $M_d\left(ℂ\right)\text{.}$

Let $A=M_d\left(ℂ\right)$ be the algebra of $d\times d$ matrices with entries from $ℂ\text{.}$ Set $$E_{ij}=\text{the matrix with}\hspace{0.17em}1\hspace{0.17em}\text{in the}\hspace{0.17em}(i,j)\hspace{0.17em}\text{entry and all other entries}\hspace{0.17em}0\text{.}$$ Then $\left\{E_{ij}\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le d\}$ is a basis of $A$ and $$E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il},1\le i,j,k,l\le d,$$ describes the multiplication in $A\text{.}$

Let $M_d\left(ℂ\right)$ be the algebra of $d\times d$ matrices with entries from $ℂ\text{.}$

(a)	Up to isomorphism, there is only one irreducible representation $M$ of $M_d\left(ℂ\right)\text{.}$
(b)	$\text{dim}\left(M\right)=d\text{.}$
(c)	The character ${\chi }^M:A\to ℂ$ of $M$ is given by $${\chi }^M\left(a\right)=\text{Tr}\left(a\right),\text{for all}\hspace{0.17em}a\in A,$$ where $\text{Tr}\left(a\right)$ is the trace of the matrix $a\text{.}$
(d)	The irreducible representation $M$ is the vector space $$M=\left\{{(c_1,\dots ,c_d)}^t\hspace{0.17em}\right|\hspace{0.17em}{c}_i\in ℂ\}$$ of column vectors of length $d$ with $A\text{-action}$ given by left multiplication, or, equivalently, $M$ is given by the map $$\begin{array}{cccc}M:& A& ⟶& M_d\left(ℂ\right)\\ & a& ⟼& a,\end{array}$$

		Proof.
	There are two things to show: (1) $M,$ as defined in (d), is a simple $A\text{-module,}$ and (2) If $C$ is a simple $A\text{-module}$ then $C\cong M\text{.}$ (1) Let ${\epsilon }_i$ be the column vector which has $1$ in the $i\text{th}$ entry and $0$ in all other entries. The set $\{{\epsilon }_1,\dots ,{\epsilon }_d\}$ is a basis of $M\text{.}$ Let $N\subseteq M$ be a nonzero submodule of $M$ and let $n={\sum }_{i=1}^d{n}_i{\epsilon }_i$ be a nonzero vector in $N\text{.}$ Then $n_j\ne 0$ for some $j$ and so $${\epsilon }_k=\frac{1}{n_j}{E}_{kj}n\in N,\text{for all}\hspace{0.17em}1\le k\le d\text{.}$$ Thus $N=M,$ since $N$ contains a basis of $M\text{.}$ (2) Let $C$ be a simple $A\text{-module}$ and let $c$ be a nonzero vector in $C\text{.}$ Since $c=\text{Id}·c={\sum }_{i=1}^d{E}_{ii}c\ne 0,$ $E_{jj}c\ne 0$ for some $j\text{.}$ Define an $A\text{-module}$ homomorphism by $$\begin{array}{cccc}\varphi :& M& ⟶& C\\ & {\epsilon }_k& ⟼& E_{kj}c\text{.}\end{array}$$ Since $\varphi \left({\epsilon }_j\right)\ne 0,$ $\text{ker}\hspace{0.17em}\varphi \ne 0\text{.}$ Since $M$ is simple, $\text{ker}\hspace{0.17em}\varphi =M$ and so $\varphi$ is injective. Since $\text{im}\hspace{0.17em}\varphi \ne 0$ and $C$ is simple, $\text{im}\hspace{0.17em}\varphi =C$ and so $\varphi$ is surjective. So $\varphi$ is an isomorphism and $C\cong M\text{.}$ $\square$

Let $M_d\left(ℂ\right)$ be the algebra of $d\times d$ matrices with entries from $ℂ\text{.}$

(1)	The only ideals of $M_d\left(ℂ\right)$ are $0$ and $M_d\left(ℂ\right)\text{.}$
(2)	$Z\left(M_d\left(ℂ\right)\right)=ℂ·\text{Id}$ and $\text{Id}$ is the only central idempotent in $M_d\left(ℂ\right)\text{.}$
(3)	Up to constant multiples, the trace $\text{Tr}:M_d\left(ℂ\right)\to ℂ$ given by $$\text{Tr}\left(a\right)=\sum _{i=1}^d{a}_{ii},\text{for all}\hspace{0.17em}a=\left(a_{ij}\right)\in M_d\left(ℂ\right),$$ is the unique trace on $M_d\left(ℂ\right)\text{.}$

		Proof.
	Let $E_{ij}$ denote the matrix in $M_d\left(ℂ\right)$ which has a $1$ in the $(i,j)$ entry and $0$ everywhere else. (1) Let $I$ be a nonzero ideal of $M_d\left(ℂ\right)$ and let $r=\left(r_{ij}\right)\in I,$ $r\ne 0\text{.}$ Let $r_{ij}$ be a nonzero entry of $r\text{.}$ Then $$\frac{1}{r_{ij}}{E}_{ki}r{E}_{jl}=E_{kl}\in I,\text{for all}\hspace{0.17em}1\le k,l\le d\text{.}$$ So $I$ contains a basis of $M_d\left(ℂ\right)\text{.}$ So $I=M_d\left(ℂ\right)\text{.}$ (2) Clearly $ℂ\text{Id}\subseteq Z\left(M_d\left(ℂ\right)\right)\text{.}$ Let $z=\left(z_{ij}\right)\in Z\left(M_d\left(ℂ\right)\right)\text{.}$ If $i\ne j$ then $$z_{ij}{E}_{ij}=E_{ii}z{E}_{jj}=z{E}_{ii}{E}_{jj}=0\text{.}$$ So $z_{ij}=0$ if $i\ne j\text{.}$ Further $$z_{ii}{E}_{ii}=E_{ii}z{E}_{ii}=E_{i1}z{E}_{1i}{E}_{ii}=z_{11}{E}_{ii},$$ so $z_{ii}=z_{11}$ for all $1\le i\le d\text{.}$ So $z=z_{11}\text{Id.}$ So $Z\left(M_d\left(ℂ\right)\right)\subseteq ℂ\text{Id}\text{.}$ So $Z\left(M_d\left(ℂ\right)\right)=ℂ\text{Id}\text{.}$ (3) Let $\chi :M_d\left(ℂ\right)\to ℂ$ be a trace on $M_d\left(ℂ\right)\text{.}$ If $a=\left(a_{ij}\right)\in M_d\left(ℂ\right)$ then $$\chi \left(E_{ii}a{E}_{jj}\right)=a_{ij}\chi \left(E_{ij}\right)=a_{ij}\chi \left(E_{i1}{E}_{1j}\right)=a_{ij}\chi \left(E_{1j}{E}_{i1}\right)=a_{ij}{\delta }_{ij}\chi \left(E_{11}\right)\text{.}$$ Thus $$\chi \left(a\right)=\chi \left(\left(\sum _{i=1}^d{E}_{ii}\right)a\left(\sum _{j=1}^d{E}_{jj}\right)\right)=\sum _{i,j=1}^d{a}_{ij}{\delta }_{ij}\chi \left(E_{11}\right)=\chi \left(E_{11}\right)\text{Tr}\left(a\right)\text{.}$$ So $\chi$ is a multiple of the trace Tr. $\square$

The algebra ${⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$

Let $\hat{A}$ be a finite set and let $d_{\lambda }$ be positive integers indexed by the elements of $\hat{A}\text{.}$ Let $$A=\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ be the algebra of block diagonal matrices with blocks $M_{d_{\lambda }}\left(ℂ\right)\text{.}$ Let $E_{ij}^{\lambda }$ be the matrix which has a $1$ in the $(i,j)$ entry of the $\lambda \text{th}$ block and $0$ everywhere else. Then $\left\{E_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}$ is a basis of $A$ and the relations $$E_{ij}^{\lambda }{E}_{kl}^{\mu }={\delta }_{\lambda \mu }{\delta }_{ij}{E}_{il}^{\lambda }$$ determine the multiplication in $A\text{.}$

The following theorems are consequences of Theorems ?? and Proposition ???.

Let $\hat{A}$ be a finite set and let $d_{\lambda }$ be positive integers indexed by the elements of $\hat{A}\text{.}$ Let $$A=\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ be the algebra of block diagonal matrices with blocks $M_{d_{\lambda }}\left(ℂ\right)\text{.}$

(1)	The irreducible representations $A^{\lambda }$ of $A$ are indexed by the elements of $\hat{A}\text{.}$
(2)	$\text{dim}\left(A^{\lambda }\right)=d_{\lambda }\text{.}$
(3)	The character ${\chi }^{\lambda }:A\to ℂ$ of $A^{\lambda }$ is given by $${\chi }^{\lambda }\left(a\right)=\text{Tr}\left(A^{\lambda }\left(a\right)\right),a\in A,$$ where $A^{\lambda }\left(a\right)$ is the $\lambda \text{th}$ block of the matrix $a\text{.}$
(4)	The irreducible representation $A^{\lambda }$ is given by the map $$\begin{array}{cccc}{A}^{\lambda }:& A& ⟶& M_{d_{\lambda }}\left(ℂ\right)\\ & a& ⟼& A^{\lambda }\left(a\right),\end{array}$$ where $A^{\lambda }\left(a\right)$ is the $\lambda \text{th}$ block of the matrix $A,$ or, equivalently, by the vector space $A^{\lambda }$ of column vectors of length $d_{\lambda }$ and $A\text{-action}$ given by $$am=A^{\lambda }\left(a\right)m,\text{for}\hspace{0.17em}a\in A\hspace{0.17em}\text{and}\hspace{0.17em}m\in A^{\lambda }\text{.}$$

Let $\hat{A}$ be a finite set and let $d_{\lambda }$ be positive integers indexed by the elements of $\hat{A}\text{.}$ Let $$A=\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ be the algebra of block diagonal matrices with blocks $M_{d_{\lambda }}\left(ℂ\right)\text{.}$ If $a\in A$ let $A^{\lambda }\left(a\right)$ denote the $\lambda \text{th}$ block of the matrix $a\text{.}$ Let $E_{ij}^{\lambda }$ be the matrix which has a $1$ in the $(i,j)$ entry of the $\lambda \text{th}$ block and $0$ everywhere else.

(1)	The minimal ideals of $A$ are given by $$I^{\lambda }=\{a\in A\hspace{0.17em}|\hspace{0.17em}{A}^{\mu }\left(a\right)=0\hspace{0.17em}\text{for all}\hspace{0.17em}\mu \ne \lambda \},\lambda \in \hat{A},$$ and every ideal of $A$ is of the form $I=\underset{\lambda \in S}{⨁}{I}^{\lambda },$ for some subset $S\subseteq \hat{A}\text{.}$
(2)	The minimal central idempotents of $A$ are $$z_{\lambda }=\sum _{i=1}^{d_{\lambda }}{E}_{ii}^{\lambda },\lambda \in \hat{A},$$ and $\left\{z_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A}\}$ is a basis of the center $Z\left(A\right)$ of $A\text{.}$
(3)	The irreducible characters ${\chi }^{\lambda },$ $\lambda \in \hat{A},$ of $A$ are given by $${\chi }^{\lambda }\left(a\right)=\text{Tr}\left(A^{\lambda }\left(a\right)\right),a\in A,$$ and every trace $\overrightarrow{t}:A\to ℂ$ on $A$ can be written uniquely in the form $$\overrightarrow{t}=\sum _{\lambda \in \hat{A}}{t}_{\lambda }{\chi }^{\lambda },t_{\lambda }\in ℂ\text{.}$$

Let $A$ be an algebra which is isomorphic to a direct sum of matrix algebras and fix an isomorphism $$\begin{array}{cc}\varphi :A\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}& \text{(3.3)}\end{array}$$ The elements $$e_{ij}^{\lambda }={\varphi }^{-1}{\left(E_{ij}\right)}^{\lambda },\lambda \in \hat{A},1\le i,j\le d_{\lambda },$$ are matrix units in $A,$ i.e. $\left\{e_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}$ is a basis of $A$ and $$e_{ij}^{\lambda }{e}_{kl}^{\mu }={\delta }_{\lambda \mu }{\delta }_{ij}{e}_{i_l}^{\lambda },$$ for all $\lambda ,\mu \in \hat{A},$ $1\le i,j\le d_{\lambda },$ $1\le k,l\le d_{\mu }\text{.}$ If $a\in A,$ let $A^{\lambda }{\left(a\right)}_{ij}\in ℂ$ be defined by the expansion $$a=\sum _{\lambda \in \hat{A}}\sum _{i,j=1}^{d_{\lambda }}{A}^{\lambda }{\left(a\right)}_{ij}{e}_{ij}^{\lambda }\text{.}$$ It follows from Theorem ??? that the maps $$\begin{array}{cccc}{A}^{\lambda }:& A& ⟶& M_{d_{\lambda }}\left(ℂ\right)\\ & a& ⟼& A^{\lambda }\left(a\right)=\left(A^{\lambda }{\left(a\right)}_{ij}\right)\end{array}\text{and}\begin{array}{cccc}{\chi }^{\lambda }:& A& ⟶& ℂ\\ & a& ⟼& \text{Tr}\left(A^{\lambda }\left(a\right)\right),\end{array}\lambda \in \hat{A},$$ are the irreducible representations and the irreducible characters of $A,$ respectively. The homomorphisms $A^{\lambda }$ depend on the choice of $\varphi$ but the irreducible characters ${\chi }^{\lambda }$ do not. The weights of a trace $\overrightarrow{t}$ on $A$ are the constants $t_{\lambda },$ $\lambda \in \hat{A},$ defined by the expansion in ???. The trace $\overrightarrow{t}$ is nondegenerate if and only if the $t_{\lambda }$ are all nonzero.

Let $A$ be an algebra which is isomorphic to a direct sum of matrix algebras, indexed by $\lambda \in \hat{A}\text{.}$ Let $\overrightarrow{t}$ be a nondegenerate trace on $A$ and let $⟨,⟩$ be the corresponding bilinear form. Let $B=\left\{b\right\}$ be a basis of $A$ and let $B^{*}=\left\{b^{*}\right\}$ be the dual basis to $B$ with respect to $⟨,⟩\text{.}$ Let ${\chi }^{\lambda },$ $\lambda \in \hat{A},$ be the irreducible characters of $A,$ $t_{\lambda }$ be the weights of $\overrightarrow{t},$ $d_{\lambda }$ the dimensions of the irreducible representations, $\left\{e_{ij}^{\lambda }\right\}$ a set of matrix units of $A,$ and $A^{\lambda }$ the corresponding irreducible representations of $A\text{.}$

(a)	(Fourier inversion formula) $$e_{ij}^{\lambda }=\sum _{b\in B}{t}_{\lambda }{A}_{ji}^{\lambda }\left(b^{*}\right)b\text{.}$$
(b)	The minimal central idempotent $z_{\lambda }$ in $A$ indexed by $\lambda \in \hat{A}$ is given by $$z_{\lambda }=\sum _{b\in B}{t}_{\lambda }{\chi }^{\lambda }\left(b^{*}\right)b\text{.}$$
(c)	(Orthogonality of characters) For all $\lambda ,\mu \in \hat{A},$ $$\sum _{b\in B}{\chi }^{\lambda }\left(b^{*}\right){\chi }^{\mu }\left(b\right)={\delta }_{\lambda \mu }\frac{d_{\lambda }}{t_{\lambda }}\text{.}$$

		Proof.
	(a) Since $\overrightarrow{t}$ is nondegenerate, the equation $\overrightarrow{t}\left(e_{ij}^{\lambda }\right)=\sum _{\mu \in \hat{A}}{t}_{\mu }{\chi }^{\mu }\left(e_{ij}^{\lambda }\right)=t_{\lambda }{\delta }_{ij}$ implies that $$\left\{\frac{e_{ji}^{\lambda }}{t_{\lambda }}\right\}\text{is the dual basis to}\left\{e_{ij}^{\lambda }\right\}\text{with respect to}\hspace{0.17em}⟨,⟩\text{.}$$ Thus, by (???), $A_{ij}^{\lambda }\left(a\right)=\frac{1}{t_{\lambda }}⟨a,e_{ji}^{\lambda }⟩,$ and so $e_{ij}^{\lambda }=\sum _{b\in B}⟨e_{ij}^{\lambda },b^{*}⟩b=\sum _{b\in B}{t}_{\lambda }{A}_{ji}^{\lambda }\left(b^{*}\right)b\text{.}$ (b) By part (a), $z_{\lambda }=\sum _{i=1}^{d_{\lambda }}{e}_{ii}^{\lambda }=\sum _{b\in B}{t}_{\lambda }\text{Tr}\left(A^{\lambda }\left(b^{*}\right)\right)b\text{.}$ (c) By part (b), $d_{\lambda }{\delta }_{\lambda \mu }={\chi }^{\mu }\left(z_{\lambda }\right)=\sum _{b\in B}{t}_{\lambda }{\chi }^{\lambda }\left(b^{*}\right){\chi }^{\mu }\left(b\right)\text{.}$ $\square$

Example 1. Let $A={⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$

(1)	As a left $A\text{-module}$ under the action of $A$ by left multiplication $$A\cong \underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }},$$ where $A^{\lambda }$ is the irreducible $A\text{-module}$ of column vectors of length $d_{\lambda }\text{.}$
(2)	As an $(A,A)$ bimodule under the action of $A$ by left and right multiplication $$A\cong \underset{\lambda \in \hat{A}}{⨁}{A}^{\lambda }\otimes {\overleftarrow{A}}^{\lambda },$$ where $A^{\lambda }$ is the left $A\text{-module}$ of column vectors of length $d_{\lambda }$ and ${\overleftarrow{A}}^{\lambda }$ is the right $A\text{-module}$ of row vectors of length $d_{\lambda }\text{.}$
(3)	Let $a,b\in A\text{.}$ If $a$ acts on $A$ by left multiplication and $b$ acts on $A$ by right multiplication then $$\text{Tr}(a\otimes b)=\sum _{\lambda \in \hat{A}}{\chi }^{\lambda }\left(a\right){\chi }^{\lambda }\left(b\right),$$ where ${\chi }^{\lambda },$ $\lambda \in \hat{A},$ are the irreducible characters of $A\text{.}$

Example 2. Let $G$ be a finite group and let $ℂG$ be the group algebra of $G\text{.}$ The trace of the regular representation of $ℂG$ is given by $$\text{tr}\left(g\right)=\sum _{h\in G}gh{|}_h=\{\begin{array}{cc}\left|G\right|,& \text{if}\hspace{0.17em}g=1,\\ 0,& \text{otherwise.}\end{array}$$ So, (provided $\left|G\right|\ne 0$ in $ℂ\text{)}$ the basis $${\left\{\frac{g^{-1}}{\left|G\right|}\right\}}_{g\in G}\text{is the dual basis to}G$$ with respect to the form $⟨,⟩$ defined by tr. Since tr is nondegenerate $$ℂG\cong \underset{\lambda \in \hat{G}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ for some set $\hat{G}$ and positive integers $d_{\lambda }\text{.}$ Then $$\text{tr}=\sum _{\lambda \in \hat{G}}{d}_{\lambda }{\chi }^{\lambda },$$ where ${\chi }^{\lambda },$ $\lambda \in \hat{G},$ are the irreducible characters of $G$ and, by (???), $$z_{\lambda }=\frac{1}{\left|G\right|}\sum _{g\in G}{d}_{\lambda }{\chi }^{\lambda }\left(g^{-1}\right)g,\lambda \in \hat{G},$$ are the minimal central idempotents in $ℂG\text{.}$ The orthogonality relation for characters of $G$ (???) is $$\frac{1}{\left|G\right|}\sum _{g\in G}{\chi }^{\lambda }\left(g^{-1}\right){\chi }^{\mu }\left(g\right)={\delta }_{\lambda \mu },\text{for}\hspace{0.17em}\lambda ,\mu \in \hat{G}\text{.}$$ If $G^{\lambda }:ℂG\to M_{d_{\lambda }}\left(ℂ\right)$ are the irreducible representations of $G$ then $$e_{ij}^{\lambda }=\frac{1}{\left|G\right|}\sum _{g\in G}{d}_{\lambda }{G}^{\lambda }{\left(g^{-1}\right)}_{ji}g,\lambda \in \hat{G},1\le i,j\le d_{\lambda },$$ are a set of matrix units in $ℂG,$ i.e. $$e_{ij}^{\lambda }{e}_{k\ell }^{\mu }={\delta }_{\lambda \mu }{\delta }_{kj}{e}_{\subset \ell }^{\lambda }$$ and $\left\{e_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{G},1\le i,j\le d_{\lambda }\}$ is a basis of $ℂG\text{.}$

Let $g,h\in G$ and let $g$ act on $ℂG$ by left multiplication and let $h$ act on $ℂG$ by right multiplication. Then $$\text{Tr}(g\otimes h)=\sum _{k\in G}gkh{|}_k=\sum _{k\in G}kh{k}^{-1}{|}_{g^{-1}}=\{\begin{array}{cc}\text{Card}\left({𝒞}_h\right),& \text{if}\hspace{0.17em}h\hspace{0.17em}\text{is conjugate to}\hspace{0.17em}{g}^{-1},\\ 0,& \text{otherwise,}\end{array}$$ where ${𝒞}_h$ is the conjugacy class of $h\text{.}$ Thus, by (???), $$\sum _{\lambda \in \hat{G}}{\chi }^{\lambda }\left(g\right){\chi }^{\lambda }\left(h\right)=\{\begin{array}{cc}\text{Card}\left({𝒞}_h\right),& \text{if}\hspace{0.17em}h\hspace{0.17em}\text{is conjugate to}\hspace{0.17em}{g}^{-1},\\ 0,& \text{otherwise,}\end{array}$$ which is the second orthogonality relation for characters of $G\text{.}$

The elements $$c_g=\sum _{x\in {𝒞}_g}x$$ are a basis of the center of $ℂG\text{.}$ Since $\left\{z_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{G}\}$ is also a basis of $Z\left(ℂG\right)$ we have that $$\text{Card}\left(\hat{G}\right)=\text{\# of conjugacy classes of}\hspace{0.17em}G,$$ though there is no (known) natural bijection between the irreducible representations of $G$ and the conjugacy classes of $G\text{.}$

It follows from ??? that $$\left|G\right|=\sum _{\lambda \in \hat{G}}{d}_{\lambda }^2\text{.}$$ Every trace $\overrightarrow{t}$ on $ℂG$ has a unique decomposition $$\overrightarrow{t}=\sum _{\lambda \in \hat{G}}{t}_{\lambda }{\chi }^{\lambda },t_{\lambda }\in ℂ\text{.}$$ So, since every $G\text{-module}$ is semisimple, its decomposition is determined by its character. So $$\text{Two}\hspace{0.17em}G\text{-modules are isomorphic if and only if they have the same character.}$$ and $$\begin{array}{ccc}\text{dim}\left(Z\left(ℂG\right)\right)& =& \left(\text{\# of irreducible representations of}\hspace{0.17em}G\right)\\ & =& \left(\text{\# of conjugacy classes of}\hspace{0.17em}G\right)\text{.}\end{array}$$

Centralizers.

Let $A$ be an algebra and let $M$ be an $A\text{-module.}$ The centralizer or commutant of M is the algebra $${\text{End}}_A\left(M\right)=\{T\in \text{End}\left(M\right)\hspace{0.17em}|\hspace{0.17em}Ta=aT\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$ If $M$ and $N$ are $A\text{-modules}$ then ${\text{Hom}}_A(M,N)$ is a left ${\text{End}}_A\left(M\right)\text{-module}$ and a right ${\text{End}}_A\left(N\right)\text{-module.}$

(Schur’s Lemma) Let $A$ be an algebra.

(1)	Let $A^{\lambda }$ be a simple $A\text{-module.}$ Then ${\text{End}}_A\left(A^{\lambda }\right)=ℂ·{\text{Id}}_{A^{\lambda }}\text{.}$
(2)	If $A^{\lambda }$ and $A^{\mu }$ are nonisomorphic simple $A\text{-modules}$ then ${\text{Hom}}_A(A^{\lambda },A^{\mu })=0\text{.}$

		Proof.
	Let $T:A^{\lambda }\to A^{\mu }$ be a nonzero $A\text{-module}$ homomorphism. Since $A^{\lambda }$ is simple, $\text{ker}\hspace{0.17em}T=0$ and so $T$ is injective. Since $A^{\mu }$ is simple, $\text{im}\hspace{0.17em}T=A^{\mu }$ and so $T$ is surjective. So $T$ is an isomorphism. Thus we may assume that $T:A^{\lambda }\to A^{\lambda }\text{.}$ When $A^{\lambda }$ is finite dimensional: Since $ℂ$ is algebraically closed $T$ has an eigenvector and a corresponding eigenvalue $\alpha \in ℂ\text{.}$ Then $T-\alpha ·\text{Id}\in {\text{Hom}}_A(A^{\lambda },A^{\lambda })$ and so $T-\alpha ·\text{Id}$ is either $0$ an isomorphism. However, since $\text{det}(T-\alpha ·\text{Id})=0$ $T-\alpha ·\text{Id}$ is not invertible. So $T-\alpha ·\text{Id}=0\text{.}$ So $T=\alpha ·\text{Id.}$ So ${\text{End}}_A\left(A^{\lambda }\right)=ℂ·\text{Id.}$ When $A^{\lambda }$ is countable dimensional: We shall show that there exists a $\lambda \in ℂ$ such that $T-\lambda ·\text{Id}$ is not invertible. Suppose $T-\lambda ·\text{Id}$ is invertible for all $\lambda \in ℂ\text{.}$ Then $p\left(T\right)$ is invertible for all polynomials $p\left(t\right)\in ℂ\left[t\right]\text{.}$ So $p\left(T\right)/q\left(T\right)$ is well defined for all $p\left(t\right),q\left(t\right)\in ℂ\left[t\right]\text{.}$ Let $v\in A^{\lambda }$ be nonzero. Then the map $$\begin{array}{ccccc}ℂ\left(t\right)& ⟶& \text{End}\left(V\right)& ⟶& V\\ \frac{p\left(t\right)}{q\left(t\right)}& ⟼& \frac{p\left(T\right)}{q\left(T\right)}& ⟼& \frac{p\left(T\right)}{q\left(T\right)}v\end{array}$$ is injective. Since $\text{dim}\hspace{0.17em}ℂ\left(t\right)$ is uncountable and $\text{dim}\hspace{0.17em}V$ is countable this is a contradiction. So $T-\lambda ·\text{Id}$ is invertible for some $\lambda \in ℂ\text{.}$ Then the same proof as in the finite dimensional case shows that $T=\lambda ·\text{Id.}$ If $A^{\lambda }$ is unitary: Let $$A=\frac{T+T^{*}}{2}\text{and}B=\frac{T-T^{*}}{2i}$$ where $T^{*}$ is defined by $⟨T{v}_1,v_2⟩=⟨v_1,T^{*}{v}_2⟩$ for all $v_1,v_2\in A^{\lambda }\text{.}$ Then $$A=A^{*},B=B^{*},T=A+iB,\text{and}A,B,T\in {\text{Hom}}_A(A^{\lambda },A^{\lambda })\text{.}$$ Then the spectral theorem for self adjoint operators says that $A$ and $B$ can be diagonalized [Rud1991, Thm. 12.22], $$A=\sum _i{\lambda }_i{P}_i\text{and}B=\sum _j{\mu }_j{Q}_j,\text{with}\hspace{0.17em}{P}_i^2=P_i,\hspace{0.17em}{Q}_j^2=Q_j,\hspace{0.17em}{P}_i,Q_j\in {\text{Hom}}_A(A^{\lambda },A^{\lambda }),\hspace{0.17em}{\lambda }_i,{\mu }_j\in ℂ\text{.}$$ Then $P_i{A}^{\lambda }$ is a submodule of $A^{\lambda }\text{.}$ So $P_i{A}^{\lambda }=A^{\lambda }\text{.}$ So $A=\lambda ·\text{Id.}$ $\square$

Suppose that $V$ is a unitary representation. Then $${\text{Hom}}_A(V,V)=ℂ·{\text{Id}}_V\text{implies that}V\hspace{0.17em}\text{is irreducible.}$$

		Proof.
	Suppose that $V$ is not irreducible. Then let $W\subseteq V$ be a submodule of $V\text{.}$ Let $$W^{\perp }=\{v\in V\hspace{0.17em}|\hspace{0.17em}⟨v,w⟩=0\text{, for all}\hspace{0.17em}w\in W\}\text{.}$$ Then $W^{\perp }$ is a submodule since, if $v\in W^{\perp }$ and $w\in W,$ then $⟨av,w⟩=⟨v,a^{*}w⟩=0$ because $a^{*}w\in W\text{.}$ Now, for Hilbert spaces, we have $V=W\oplus W^{\perp }$ and we can define a $$\begin{array}{ccc}V& \stackrel{p}{⟶}& V\\ w& ⟼& w,& \text{if}\hspace{0.17em}w\in W,\\ w^{\perp }& ⟼& 0,& \text{if}\hspace{0.17em}w\in W^{\perp },\end{array}$$ This map is a nonidentity $A\text{-module}$ homomorphism. So ${\text{Hom}}_A(V,V)\ne ℂ·\text{Id.}$ $\square$

Let $A$ be an algebra. Let $M$ be a semisimple $A\text{-module}$ and set $Z={\text{End}}_A\left(M\right)\text{.}$ Suppose that $$M\cong \underset{\lambda \in \hat{M}}{⨁}{\left(A^{\lambda }\right)}^{\oplus m_{\lambda }},$$ where $\hat{M}$ is an index set for the irreducible $A\text{-modules}$ $A^{\lambda }$ which appear in $M$ and the $m_{\lambda }$ are positive integers.

(a)	$Z=\underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(ℂ\right)\text{.}$
(b)	As an $(A\otimes Z)\text{-module}$ $$M\cong \underset{\lambda \in \hat{M}}{⨁}{A}^{\lambda }\otimes Z^{\lambda },$$ where the $Z^{\lambda },\lambda \in \hat{M},$ are the simple $Z\text{-modules.}$

		Proof.
	Index the components in the decomposition of $M$ by dummy variables ${\epsilon }_i^{\lambda }$ so that we may write $$M\cong \underset{\lambda \in \hat{M}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{A}^{\lambda }\otimes {\epsilon }_i^{\lambda }\text{.}$$ For each $\lambda \in \hat{M},$ $1\le i,j\le m_{\lambda }$ let ${\varphi }_{ij}^{\lambda }:A^{\lambda }\otimes {\epsilon }_j\to A^{\lambda }\otimes {\epsilon }_i$ be the $A\text{-module}$ isomorphism given by $${\varphi }_{ij}^{\lambda }(m\otimes {\epsilon }_j^{\lambda })=m\otimes {\epsilon }_i^{\lambda },\text{for}\hspace{0.17em}m\in A^{\lambda }\text{.}$$ By Schur’s Lemma, $$\begin{array}{ccc}{\text{End}}_{A\left(M\right)}={\text{Hom}}_A(M,M)& \cong & {\text{Hom}}_A(\underset{\lambda }{⨁}\underset{j}{⨁}{A}^{\lambda }\otimes {\epsilon }_j^{\lambda },\underset{\mu }{⨁}\underset{i}{⨁}{A}^{\mu }\otimes {\epsilon }_i^{\mu })\\ & \cong & \underset{\lambda ,\mu }{⨁}\underset{i,j}{⨁}{\delta }_{\lambda \mu }{\text{Hom}}_A(A^{\lambda }\otimes {\epsilon }_j^{\lambda },A^{\mu }\otimes {\epsilon }_i^{\mu })\\ & \cong & \underset{\lambda }{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}ℂ{\varphi }_{ij}^{\lambda }\text{.}\end{array}$$ Thus each element $z\in {\text{End}}_A\left(M\right)$ can be written as $$z=\sum _{\lambda \in \hat{M}}\sum _{i,j=1}^{m_{\lambda }}{z}_{ij}^{\lambda }{\varphi }_{ij}^{\lambda },\text{for some}\hspace{0.17em}{z}_{ij}^{\lambda }\in ℂ,$$ and identified with an element of ${\oplus }_{\lambda }{M}_{m_{\lambda }}\left(ℂ\right)\text{.}$ Since ${\varphi }_{ij}^{\lambda }{\varphi }_{kl}^{\mu }={\delta }_{\lambda \mu }{\delta }_{jk}{\varphi }_{il}^{\lambda }$ it follows that $${\text{End}}_A\left(M\right)\cong \underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(ℂ\right)\text{.}$$ (b) As a vector space $Z^{\mu }=\text{span}\left\{{\epsilon }_i^{\mu }\hspace{0.17em}\right|\hspace{0.17em}1\le i\le m_{\mu }\}$ is isomorphic to the simple ${\oplus }_{\lambda }{M}_{m_{\lambda }}\left(ℂ\right)$ module of column vectors of length $m_{\mu }\text{.}$ The decomposition of $M$ as $Z\otimes Z$ modules follows since $$(a\otimes {\varphi }_{ij}^{\lambda })(m\otimes {\epsilon }_k^{\mu })={\delta }_{\lambda \mu }{\delta }_{jk}(a\otimes {\epsilon }_i^{\mu }),\text{for all}\hspace{0.17em}m\in A^{\mu },a\in A\text{.}$$ $\square$

If $A$ is an algebra then $A^{\text{op}}$ is the algebra $A$ except with the opposite multiplication, i.e. $$A^{\text{op}}=\left\{a^{\text{op}}\hspace{0.17em}\right|\hspace{0.17em}a\in A\}\text{with}{a}_1^{\text{op}}{a}_2^{\text{op}}={\left(a_2{a}_1\right)}^{\text{op}},\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$ Let left regular representation of $A$ is the vector space $A$ with $A$ action given by left multiplication. Here $A$ is serving both as an algebra and as an $A\text{-module.}$ It is often useful to distinguish the two roles of $A$ and use the notation $\overrightarrow{A}$ for the $A\text{-module,}$ i.e. $\overrightarrow{A}$ is the vector space $$\overrightarrow{A}=\left\{\overrightarrow{b}\hspace{0.17em}\right|\hspace{0.17em}b\in A\}\text{with}\hspace{0.17em}A\text{-action}a\overrightarrow{b}=\overrightarrow{ab},\text{for all}\hspace{0.17em}a\in A,\overrightarrow{b}\in \overrightarrow{A}\text{.}$$

Let $A$ be an algebra and let $\overrightarrow{A}$ be the regular representation of $A\text{.}$ Then ${\text{End}}_A\left(\overrightarrow{A}\right)\cong A^{\text{op}}\text{.}$ More precisely, $${\text{End}}_A\left(\overrightarrow{A}\right)=\left\{{\varphi }_b\hspace{0.17em}\right|\hspace{0.17em}b\in A\},\text{where}\hspace{0.17em}{\varphi }_b\hspace{0.17em}\text{is given by}{\varphi }_b\left(\overrightarrow{a}\right)=a\overrightarrow{b},\text{for all}\hspace{0.17em}\overrightarrow{a}\in \overrightarrow{A}\text{.}$$

		Proof.
	Let $\varphi \in {\text{End}}_A\left(\overrightarrow{A}\right)$ and let $b\in A$ be such that $\varphi \left(\overrightarrow{1}\right)=\overrightarrow{b}\text{.}$ For all $\overrightarrow{a}\in \overrightarrow{A},$ $$\varphi \left(\overrightarrow{a}\right)=\varphi (a·\overrightarrow{1})=a\varphi \left(\overrightarrow{1}\right)=a\overrightarrow{b}=\overrightarrow{ab},$$ and so $\varphi ={\varphi }_b\text{.}$ Then ${\text{End}}_A\left(\overrightarrow{A}\right)\cong A^{\text{op}}$ since $$({\varphi }_{b_1}\circ {\varphi }_{b_2})\left(\overrightarrow{a}\right)=a\overrightarrow{b_2{b}_1}={\varphi }_{b_2{b}_1}\left(\overrightarrow{a}\right),$$ for all $b_1,b_2\in A$ and $\overrightarrow{a}\in \overrightarrow{A}\text{.}$ $\square$

Characterizing algebras isomorphic to ${⨁}_{\lambda }{M}_{d_{\lambda }}\left(ℂ\right)$

Suppose that $A$ is an algebra such that the regular representation $\overrightarrow{A}$ of $A$ is completely decomposable. Then $A$ is isomorphic to a direct sum of matrix algebras, i.e. $$A\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ for some set $\hat{A}$ and some positive integers $d_{\lambda },$ indexed by the elements of $\hat{A}\text{.}$

		Proof.
	If $\overrightarrow{A}$ is completely decomposable then, by Theorem ???, ${\text{End}}_A\left(\overrightarrow{A}\right)$ is isomorphic to a direct sum of matrix algebras. By Proposition ??, $$A^{\text{op}}\cong \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right),$$ for some set $\hat{A}$ and some positive integers $d_{\lambda },$ indexed by the elements of $\hat{A}\text{.}$ The map $$\begin{array}{ccc}{\left({⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\right)}^{\text{op}}& ⟶& {⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\\ a& ⟼& a^t,\end{array}$$ where $a^t$ is the transpose of the matrix $a,$ is an algebra isomorphism. So $A$ is isomorphic to a direct sum of matrix algebras. $\square$

Let $A={⨁}_{\lambda \in \hat{A}}{M}_{d_{\lambda }}\left(ℂ\right)\text{.}$ Then the trace $\text{tr}$ of the regular representation of $A$ is nondegenerate.

		Proof.
	As $A\text{-modules,}$ the regular representation $$\overrightarrow{A}\cong \underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }},$$ where $A^{\lambda }$ is the irreducible $A\text{-module}$ consisting of column vectors of length $d_{\lambda }\text{.}$ So the trace tr of the regular representation is given by $$\text{tr}=\sum _{\lambda \in \hat{A}}{d}_{\lambda }{\chi }^{\lambda },$$ where ${\chi }^{\lambda }$ are the irreducible characters of $A\text{.}$ Since the $d_{\lambda }$ are all nonzero the trace tr is nondegenerate. $\square$

(Maschke’s theorem) Let $A$ be an algebra such that the trace tr of the regular representation of $A$ is nondegenerate. Then every representation of $A$ is completely decomposable.

		Proof.
	Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis of $A$ with respect to the form $⟨,⟩:A\times A\to ℂ$ defined by $$⟨a_1,a_2⟩=\text{tr}\left(a_1{a}_2\right),\text{for all}\hspace{0.17em}{a}_1,a_2\in A\text{.}$$ The dual basis $B^{*}$ exists because the trace tr is nondegenerate. Let $M$ be an $A\text{-module.}$ If $M$ is irreducible then the result is vacuously true, so we may assume that $M$ has a proper submodule $N\text{.}$ Let $p\in \text{End}\left(M\right)$ be a projection onto $N,$ i.e. $pM=N$ and $p^2=p\text{.}$ Let $$\left[p\right]=\sum _{b\in B}bp{b}^{*},\text{and}e=\sum _{b\in B}b{b}^{*}\text{.}$$ For all $a\in A,$ $$\text{tr}\left(ea\right)=\sum _{b\in B}\text{tr}\left(b{b}^{*}a\right)=\sum _{b\in B}⟨ab,b^{*}⟩=\sum _{b\in B}ab{|}_b=\text{tr}\left(a\right),$$ So $\text{tr}\left((e-1)a\right)=0,$ for all $a\in A\text{.}$ Thus, since tr is nondegenerate, $e=1\text{.}$ Let $m\in M\text{.}$ Then $p{b}^{*}m\in N$ for all $b\in B,$ and so $\left[p\right]m\in N\text{.}$ So $\left[p\right]M\subseteq N\text{.}$ Let $n\in N\text{.}$ Then $p{b}^{*}n=b^{*}n$ for all $b\in B,$ and so $\left[p\right]n=en=1·n=n\text{.}$ So $\left[p\right]M=N$ and ${\left[p\right]}^2=\left[p\right],$ as elements of $\text{End}\left(M\right)\text{.}$ Note that $[1-p]=\left[1\right]-\left[p\right]=e-\left[p\right]=1-\left[p\right]\text{.}$ So $$M=\left[p\right]M\oplus (1-\left[p\right])M=N\oplus [1-p]M,$$ and, by Proposition ??, $[1-p]M$ is an $A\text{-module.}$ So $[1-p]M$ is an $A\text{-submodule}$ of $M$ which is complementary to $M\text{.}$ By induction on the dimension of $M,$ $N$ and $[1-p]M$ are completely decomposable, and therefore $M$ is completely decomposable. $\square$

Together, Theorems ???, ??? and Proposition ??? yield the following theorem.

(Artin-Wedderburn) Let $A$ be a finite dimensional algebra over $ℂ\text{.}$ The following are equivalent:

(1)	Every representation of $A$ is completely decomposable.
(2)	The trace of the regular representation of $A$ is nondegenerate.
(3)	The regular representation of $A$ is completely decomposable.

Example 1. Let $A$ be the algebra with basis $\{1,e\}$ and mulitplication given by $e^2=0\text{.}$ Then $$\overrightarrow{t}:A\to ℂ\text{given by}\overrightarrow{t}(a+be)=a+b$$ is a nondegenerate trace on $A\text{.}$ The regular representation of $A$ is given by $$\overrightarrow{A}\left(1\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\text{and}\overrightarrow{A}\left(e\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$$ and $ℂe$ is the only submodule of $\overrightarrow{A}\text{.}$ Thus, $\overrightarrow{A}$ is not completely decomposable. The trace tr of the regular representation of $A$ is given by $$\text{tr}(a+be)=2a,\text{for}\hspace{0.17em}a,b\in ℂ\text{.}$$

(Burnside’s Theorem) Let $A$ be an algebra and let $M:A\to \text{End}\left(M\right)$ be an irreducible representation of $A\text{.}$ Then $M\left(A\right)=\text{End}\left(M\right)\text{.}$

		Proof.
	Clearly, $M\left(A\right)\subseteq \text{End}\left(M\right)$ and $M$ is both a simple $M\left(A\right)\text{-module}$ and a simple $\text{End}\left(M\right)\text{-module.}$ As $\text{End}\left(M\right)\text{-modules}$ $$\overrightarrow{\text{End}\left(M\right)}\cong M^{\oplus d},$$ and so, by restriction, this is also true as an $M\left(A\right)\text{-module.}$ Thus, by Schur’s lemma, $${\text{End}}_{M\left(A\right)}\left(\overrightarrow{\text{End}\left(M\right)}\right)=M_d\left(ℂ\right)\text{.}$$ Let us label the summands in the decomposition by dummy variables ${\epsilon }_i,$ $$\overrightarrow{\text{End}\left(M\right)}=\underset{i=1}{\overset{d}{⨁}}M\otimes {\epsilon }_i,\text{so that}{E}_{ii}\left(\overrightarrow{\text{End}\left(M\right)}\right)=M\otimes {\epsilon }_i\text{.}$$ Now $\overrightarrow{M\left(A\right)}\subseteq \overrightarrow{\text{End}\left(M\right)}$ is an $M\left(A\right)$ submodule of $\overrightarrow{\text{End}\left(M\right)}\text{.}$ However, $$E_{ii}\left(\overrightarrow{\text{End}\left(M\right)}\right)\subseteq M\otimes {\epsilon }_i\text{and}\overrightarrow{M\left(A\right)}=E_{11}\overrightarrow{M\left(A\right)}\oplus \cdots \oplus E_{dd}\overrightarrow{M\left(A\right)}\subseteq M\otimes {\epsilon }_1\oplus \cdots \oplus M\oplus {\epsilon }_d\text{.}$$ Since $M$ is a simple $M\left(A\right)$ module, each $E_{ii}\overrightarrow{M\left(A\right)}$ is isomorphic to $M$ or $0\text{.}$ So $$\overrightarrow{M\left(A\right)}\cong M^{\oplus k},\text{for some}\hspace{0.17em}1\le k\le d\text{.}$$ So the regular representation of $M\left(A\right)$ is semisimple and $M\left(A\right)\cong M_k\left(ℂ\right)\text{.}$ Since $\text{dim}\left(M\right)=d$ and $M$ is a simple module for $M\left(A\right)$ we have $M\left(A\right)\cong M_d\left(ℂ\right)\text{.}$ So $M\left(A\right)=\text{End}\left(M\right)\text{.}$ $\square$

Remark 1. We used Schur’s lemma in a crucial way so we are assuming that $ℂ$ is algebraically closed. In general we can say: $$\text{If}\hspace{0.17em}M\hspace{0.17em}\text{is a simple}\hspace{0.17em}A\text{-module then}\hspace{0.17em}M\left(A\right)={\text{End}}_Z\left(M\right)\hspace{0.17em}\text{where}\hspace{0.17em}Z={\text{End}}_A\left(M\right)\text{.}$$ The proof is similar to that given above and is called the Jacobson density theorem.

Example. Assume that $A$ is a commutative algebra and let $M$ be a simple $A\text{-module.}$ Then $M\left(A\right)$ is commutative and $M\left(A\right)=\text{End}\left(M\right)\cong M_d\left(ℂ\right),$ where $d=\text{dim}\left(M\right)\text{.}$ However, $M_d\left(ℂ\right)$ is commutative if and only if $d=1\text{.}$ This shows that every irreducible representation of a commutative algebra is one dimensional.

Example 2. Explain what the error is in the following proof of Burnside’s theorem: If $M$ is an irreducible $A\text{-module}$ then $M\left(A\right)=\text{End}\left(M\right)\text{.}$

		Proof.
	Let $\{m_1,\dots ,m_d\}$ be a basis of $M\text{.}$ Since $M$ is irreducible, for any $i$ and $j$ there is an $a\in A$ such that $M\left(A\right)m_j=m_i\text{.}$ So the matrix $E_{ji}\in M\left(A\right)$ for all $1\le i,j\le n\text{.}$ So $\text{End}\left(M\right)\subseteq M\left(A\right)\text{.}$ So $M\left(A\right)=\text{End}\left(M\right)\text{.}$ $\square$

Notes and references

This is a typed exert of Representation theory Lecture notes: Chapter 1 by Arun Ram. Research supported in part by National Science Foundation grant DMS-9622985.

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