Representation Theory

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

Last updated: 2 October 2014

Lecture 5

Review

The Lie algebra $$\begin{array}{rcl}{\displaystyle {𝔰𝔩}_2}& =& {\displaystyle \left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a+d=0\}}\end{array}$$ with bracket $$[x,y]=xy-yx,$$ for $x,y\in {𝔰𝔩}_2$ is presented by generators $$e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ and relations $$[e,f]=h,[h,e]=2e,[h,f]=-2f\text{.}$$ The enveloping algebra $U{𝔰𝔩}_2$ is generated by $e,f,h$ with relations $$ef=fe+h,eh=he-2e,hf=fh-2f$$ and has basis $$\left\{f^{m_1}{h}^{m_2}{e}^{m_3}\hspace{0.17em}\right|\hspace{0.17em}{m}_1,m_2,m_3\in {\mathbb{Z}}_{\ge 0}\}\text{.}$$ If $M=\text{span}\{m_1,\dots ,m_r\}$ and $N=\{n_1,\dots ,n_s\}$ are $U{𝔰𝔩}_2\text{-modules,}$ then $$M\otimes N=\text{span}\{m_i\otimes n_j\hspace{0.17em}|\hspace{0.17em}1\le i\le r,1\le j\le s\}$$ is a $U{𝔰𝔩}_2\text{-module,}$ with $$\begin{array}{rcl}{\displaystyle e(m_i\otimes n_j)}& =& {\displaystyle e{m}_i\otimes n_j+m_i\otimes e{n}_j,}\\ {\displaystyle f(m_i\otimes n_j)}& =& {\displaystyle f{m}_i\otimes n_j+m_i\otimes f{n}_j,}\\ {\displaystyle h(m_i\otimes n_j)}& =& {\displaystyle h{m}_i\otimes n_j+m_i\otimes h{n}_j\text{.}}\end{array}$$

The quantum group $U_q{𝔰𝔩}_2$

$U_q{𝔰𝔩}_2$ is generated by $E,F,K^{\pm 1}$ with relations $$KE{K}^{-1}=q^{2}E,KF{K}^{-1}=q^{-2}F,EF=FE+\frac{K-K^{-1}}{q-q^{-1}}\text{.}$$ The map $\mathrm{\Delta }:U\to U\otimes U$ given by $$\begin{array}{rcl}{\displaystyle \mathrm{\Delta }\left(E\right)}& =& {\displaystyle E\otimes K+1\otimes E,}\\ {\displaystyle \mathrm{\Delta }\left(F\right)}& =& {\displaystyle F\otimes 1+K^{-1}\otimes F,}\\ {\displaystyle \mathrm{\Delta }\left(K\right)}& =& {\displaystyle K\otimes K,}\end{array}$$ is a coproduct.

$U=U_q{𝔰𝔩}_2$ at $q=1$ is $U{𝔰𝔩}_2\text{.}$

$U=U_q{𝔰𝔩}_2$ has a 2-dimensional simple module $$V=L\left(▫\right)=\text{span}\{v_1,v_{-1}\}$$ with $$\begin{array}{ccccc}\begin{array}{rcl}{\displaystyle K{v}_1}& =& {\displaystyle q{v}_1,}\\ {\displaystyle K{v}_{-1}}& =& {\displaystyle q^{-1}{v}_{-1},}\end{array}& & \begin{array}{rcl}{\displaystyle E{v}_1}& =& {\displaystyle 0,}\\ {\displaystyle E{v}_{-1}}& =& {\displaystyle v_1,}\end{array}& & \begin{array}{rcl}{\displaystyle F{v}_1}& =& {\displaystyle v_{-1},}\\ {\displaystyle F{v}_{-1}}& =& {\displaystyle 0\text{.}}\end{array}\end{array}$$ So ${\rho }^{▫}:U\to \text{End}\left(L\left(▫\right)\right)$ has $${\rho }^{▫}\left(K\right)=\left(\begin{array}{cc}q& 0\\ 0& q^{-1}\end{array}\right),{\rho }^{▫}\left(E\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),{\rho }^{▫}\left(F\right)=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\text{.}$$

Computing $V\otimes V=L\left(▫\right)\otimes L\left(▫\right)=V^{\otimes 2}$

$$V^{\otimes 2}=V\otimes V=\text{span}\{v_1\otimes v_1,v_1\otimes v_{-1},v_{-1}\otimes v_1,v_{-1}\otimes v_{-1}\}$$ with $$\begin{array}{ccc}\begin{array}{rcl}{\displaystyle E(v_1\otimes v_1)}& =& {\displaystyle 0,}\\ {\displaystyle E(v_1\otimes v_{-1})}& =& {\displaystyle v_1\otimes v_1,}\\ {\displaystyle K(v_1\otimes v_1)}& =& {\displaystyle q^2{v}_1\otimes v_1,}\\ {\displaystyle K(v_1\otimes v_{-1})}& =& {\displaystyle v_1\otimes v_{-1},}\\ {\displaystyle F(v_1\otimes v_1)}& =& {\displaystyle v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1},}\\ {\displaystyle F(v_1\otimes v_{-1})}& =& {\displaystyle v_{-1}\otimes v_{-1},}\end{array}& & \begin{array}{rcl}{\displaystyle E(v_{-1}\otimes v_1)}& =& {\displaystyle q{v}_1\otimes v_1,}\\ {\displaystyle E(v_{-1}\otimes v_{-1})}& =& {\displaystyle q^{-1}{v}_1\otimes v_{-1}+v_{-1}\otimes v_1,}\\ {\displaystyle K(v_{-1}\otimes v_{-1})}& =& {\displaystyle q^{-2}{v}_{-1}\otimes v_{-1},}\\ {\displaystyle K(v_{-1}\otimes v_1)}& =& {\displaystyle v_{-1}\otimes v_1,}\\ {\displaystyle F(v_{-1}\otimes v_1)}& =& {\displaystyle q{v}_{-1}\otimes v_{-1},}\\ {\displaystyle F(v_{-1}\otimes v_{-1})}& =& {\displaystyle 0,}\end{array}\end{array}$$ or, equivalently, $$\begin{array}{rcl}{\displaystyle ({\rho }^{▫}\otimes {\rho }^{▫})\left(E\right)}& =& {\displaystyle {\rho }^{▫}\left(E\right)\otimes {\rho }^{▫}\left(K\right)+{\rho }^{▫}\left(1\right)\otimes {\rho }^{▫}\left(E\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\otimes \left(\begin{array}{cc}q& 0\\ 0& q^{-1}\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\otimes \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)& 1·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)\\ 0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)& 0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)\end{array}\right)+\left(\begin{array}{cc}1·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)& 0·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\\ 0·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)& 1·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cccc}& & q& \\ & & & q^{-1}\\ \phantom{q^{-1}}& & & \\ \phantom{a}& \phantom{a}& & \end{array}\right)+\left(\begin{array}{cccc}0& 1& & \\ 0& 0& & \\ & & 0& 1\\ & & 0& 0\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cccc}0& 1& q& 0\\ 0& 0& 0& q^{-1}\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right)\text{.}}\end{array}$$ In general, if $$A=\left(\begin{array}{ccc}{a}_{11}& \cdots & a_{1r}\\ ⋮& & \\ a_{r1}& \cdots & a_{rr}\end{array}\right)\text{and}B=\left(\begin{array}{ccc}{b}_{11}& \cdots & b_{1s}\\ ⋮& & \\ b_{s1}& \cdots & b_{ss}\end{array}\right)$$ acting on $M=\text{span}\{m_1,\dots ,m_r\}$ and $N=\text{span}\{n_1,\dots ,n_s\}$ respectively then, if $$(A\otimes B)(m_i\otimes n_j)=A{m}_i\otimes B{n}_j,$$ then the matrix of $A\otimes B$ in the basis $$m_1\otimes n_1,\dots ,m_1\otimes n_s,m_2\otimes n_1,\dots ,m_2\otimes n_s,\dots ,m_r\otimes m_1,\dots ,m_r\otimes n_s$$ is $$A\otimes B=\left(\begin{array}{cccc}{a}_{11}B& a_{12}B& \cdots & a_{1r}B\\ ⋮& & & ⋮\\ a_{r1}B& \cdots & & a_{rr}B\end{array}\right)\text{.}$$

Decomposing $V^{\otimes 2}$

$$\begin{array}{ccc}\begin{array}{c}{v}_1\otimes v_1\\ \downarrow F\\ v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1}\\ \downarrow F\\ \left[2\right]v_{-1}\otimes v_{-1}\end{array}& & \begin{array}{rcl}{\displaystyle F(v_1\otimes v_1)}& =& {\displaystyle v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1},}\\ {\displaystyle F(v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1})}& =& {\displaystyle \left[2\right]v_{-1}\otimes v_{-1}\text{.}}\end{array}\\ v_{-1}\otimes v_1-q{v}_1\otimes v_{-1}& & \begin{array}{rcl}{\displaystyle E(v_{-1}\otimes v_1-q{v}_1\otimes v_{-1})}& =& {\displaystyle 0,}\\ {\displaystyle F(v_{-1}\otimes v_1-q{v}_1\otimes v_{-1})}& =& {\displaystyle 0\text{.}}\end{array}\end{array}$$ Let $$b_1=v_1\otimes v_1,b_2=v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1},b_3=v_{-1}\otimes v_{-1},b_4=v_{-1}\otimes v_1-q{v}_1\otimes v_{-1}\text{.}$$ Then $$V^{\otimes 2}=L\left(▫\right)\otimes L\left(▫\right)=L\left(\begin{array}{c} \end{array}\right)\oplus L\left(\varnothing \right)$$ where $$L\left(\begin{array}{c} \end{array}\right)=\text{span}\{b_1,b_2,b_3\}\text{and}L\left(\varnothing \right)=\text{span}\left\{b_4\right\}\text{.}$$ In the basis $b_1,b_2,b_3,b_4$ the matrices for the action of $E,F,K$ on $V\otimes V$ are $$\begin{array}{rcl}{\displaystyle {\rho }^{\oplus \varnothing }\left(E\right)}& =& {\displaystyle \left(\begin{array}{cccc}0& \left[2\right]& & \\ & 0& 1& \\ & & 0& \\ & & & 0\end{array}\right),}\\ {\displaystyle {\rho }^{\oplus \varnothing }\left(F\right)}& =& {\displaystyle \left(\begin{array}{cccc}0& & & \\ 1& 0& & \\ & \left[2\right]& 0& \\ & & & 0\end{array}\right),}\\ {\displaystyle {\rho }^{\oplus \varnothing }\left(K\right)}& =& {\displaystyle \left(\begin{array}{cccc}{q}^2& & & \\ & q^0& & \\ & & q^{-2}& \\ & & & 1\end{array}\right)\text{.}}\end{array}$$

Decomposing $V^{\otimes 3}=L\left(▫\right)\otimes L\left(▫\right)\otimes L\left(▫\right)\text{.}$

$$L\left(▫\right)\otimes L\left(▫\right)\otimes L\left(▫\right)=\text{span}\left\{\begin{array}{cc}{v}_1\otimes v_1\otimes v_1,& v_1\otimes v_1\otimes v_{-1},\\ v_1\otimes v_{-1}\otimes v_1,& v_1\otimes v_{-1}\otimes v_{-1},\\ v_{-1}\otimes v_1\otimes v_1,& v_{-1}\otimes v_1\otimes v_{-1},\\ v_{-1}\otimes v_{-1}\otimes v_1,& v_{-1}\otimes v_{-1}\otimes v_{-1}\end{array}\right\}\text{.}$$ Another basis of $V^{\otimes 3}$ is $$\{b_1\otimes v_1,b_1\otimes v_{-1},v_2\otimes v_1,v_2\otimes v_{-1},v_3\otimes v_1,v_3\otimes v_{-1},b_4\otimes v_1,b_4\otimes v_{-1}\},$$ i.e. $$V^{\otimes 3}=(L\left(\begin{array}{c}\end{array}\right)\oplus L\left(\varnothing \right))\otimes V=(L\left(\begin{array}{c}\end{array}\right)\otimes V)\oplus (L\left(\varnothing \right)\otimes V)$$ $$L\left(\varnothing \right)\otimes V=\text{span}\{b_4\otimes v_1,b_4\otimes v_{-1}\}$$ with $$\begin{array}{ccccc}\begin{array}{rcl}{\displaystyle E(b_4\otimes v_1)}& =& {\displaystyle 0,}\\ {\displaystyle E(b_4\otimes v_{-1})}& =& {\displaystyle b_4\otimes v_1,}\end{array}& & \begin{array}{rcl}{\displaystyle F_4(b_4\otimes v_1)}& =& {\displaystyle b_4\otimes v_{-1},}\\ {\displaystyle F(b_4\otimes v_{-1})}& =& {\displaystyle 0,}\end{array}& & \begin{array}{rcl}{\displaystyle K(b_4\otimes v_1)}& =& {\displaystyle q{b}_4\otimes v_1,}\\ {\displaystyle K(b_4\otimes v_{-1})}& =& {\displaystyle q^{-1}{b}_4\otimes v_{-1}\text{.}}\end{array}\end{array}$$ So $$\begin{array}{ccc}L\left(\varnothing \right)\otimes V& \simeq & L\left(▫\right)\\ b_4\otimes v_1& ⟼& v_1\\ b_4\otimes v_{-1}& ⟼& v_{-1}\end{array}$$ Then $$L\left(\begin{array}{c}\end{array}\right)\otimes V=\text{span}\{b_1\otimes v_1,b_1\otimes v_{-1},b_2\otimes v_1,b_2\otimes v_{-1},b_3\otimes v_1,b_3\otimes v_{-1}\}$$ with $$F(b_1\otimes v_1)=b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1},$$ $$\begin{array}{c}{b}_1\otimes v_1\\ \downarrow F\\ b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1}\\ \downarrow F\\ \left[2\right]b_3\otimes v_1+b_2\otimes v_{-1}+q^{-2}{b}_2\otimes v_{-1}+q^{-4}{b}_2\otimes 0=\left[2\right](b_3\otimes v_1+q^{-1}{b}_2\otimes v_{-1})\\ \downarrow F\\ 0+\left[2\right]q^2{b}_3\otimes v_{-1}+\left[2\right]b_3\otimes v_{-1}+0+q^{-2}\left[2\right]b_3\otimes v_{-1}+0=\left[2\right]\left[3\right]b_3\otimes v_{-1}\end{array}$$ So, if $$\begin{array}{rcl}{\displaystyle c_1}& =& {\displaystyle b_1\otimes v_1,}\\ {\displaystyle c_2}& =& {\displaystyle b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1},}\\ {\displaystyle c_3}& =& {\displaystyle b_3\otimes v_1+q^{-1}{b}_2\otimes v_{-1},}\\ {\displaystyle c_4}& =& {\displaystyle b_3\otimes v_{-1}}\end{array}$$ then the action of $F$ on $$L\left(\begin{array}{c}\end{array}\right)\otimes V=\text{span}\{c_1,c_2,c_3,c_4\}$$ is given by $${\rho }^{\begin{array}{c}\end{array}}\left(F\right)=\left(\begin{array}{c}0\\ 1& 0\\ & \left[2\right]& 0\\ & & \left[3\right]& 0\end{array}\right)$$ and $$\begin{array}{rcl}{\displaystyle {\rho }^{\begin{array}{c}\end{array}}\left(E\right)}& =& {\displaystyle \left(\begin{array}{cc}0& 1\\ & 0& \left[2\right]\\ & & 0& \left[3\right]\\ & & & 0\end{array}\right),}\\ {\displaystyle {\rho }^{\begin{array}{c}\end{array}}\left(K\right)}& =& {\displaystyle \left(\begin{array}{c}{q}^3\\ & q^1\\ & & q^{-1}\\ & & & q^{-3}\end{array}\right)\text{.}}\end{array}$$ Note that $$\begin{array}{c}0\\ \uparrow E\\ b_2\otimes v_1-q{b}_1\otimes v_{-1}\\ F\downarrow \\ \left[2\right]b_3\otimes v_1+b_2\otimes v_{-1}-q\left[2\right]b_2\otimes v_{-1}=\left[2\right]b_3\otimes v_{-1}-q^2{b}_2\otimes v_{-1}\\ F\downarrow \\ \left[2\right]q^2{b}_3\otimes v_{-1}-q^2\left[2\right]b_3\otimes v_{-1}=0\end{array}$$ So that, if $$\begin{array}{rcl}{\displaystyle c_5}& =& {\displaystyle b_2\otimes v_1-q{b}_1\otimes v_{-1},}\\ {\displaystyle c_6}& =& {\displaystyle \left[2\right]b_3\otimes v_{-1}-q^2{b}_2\otimes v_{-1}}\end{array}$$ and $$\stackrel{\sim }{L}\left(▫\right)=\text{span}\{c_5,c_6\}$$ then $$\stackrel{\sim }{L}\left(▫\right)\simeq L\left(▫\right)\text{.}$$ So $$\begin{array}{rcl}{\displaystyle V^{\otimes 3}}& =& {\displaystyle L\left(▫\right)\otimes L\left(▫\right)\otimes L\left(▫\right)}\\ & \cong & {\displaystyle (L\left(\begin{array}{c}\end{array}\right)\otimes L\left(\varnothing \right))\otimes L\left(▫\right)}\\ & =& {\displaystyle (L\left(\begin{array}{c}\end{array}\right)\otimes V)\oplus (L\left(\varnothing \right)\otimes V)}\\ & \simeq & {\displaystyle L\left(\begin{array}{c}\end{array}\right)\oplus L\left(▫\right)\oplus L\left(▫\right)}\end{array}$$ $$\begin{array}{ccc}\begin{array}{c} ∅ ∅ ∅ 1 1 1 2 1 2 3 1 5 4 1 \end{array}& & \begin{array}{rcl}{\displaystyle 1·2}& =& {\displaystyle 2}\\ {\displaystyle 1·1+1·3}& =& {\displaystyle 4}\\ {\displaystyle 2·2+1·4}& =& {\displaystyle 8}\\ {\displaystyle 2·1+3·3+1·5}& =& {\displaystyle 16}\\ {\displaystyle 5·2+4·4+2·6}& =& {\displaystyle 32}\end{array}\end{array}$$

What is the connection between ${T_L}_k$ and $V^{\otimes k}$ for $U_q{𝔰𝔩}_2\text{?}$

Define an action of ${TL}_2=\text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}$ on $$V^{\otimes 2}=\text{span}\{v_1\otimes v_1,v_1\otimes v_{-1},v_{-1}\otimes v_1,v_{-1}\otimes v_{-1}\}$$ by $$\begin{array}{rcl}{\displaystyle \begin{array}{c}\end{array}(v_1\otimes v_1)}& =& {\displaystyle 0,}\\ {\displaystyle \begin{array}{c}\end{array}(v_{-1}\otimes v_{-1})}& =& {\displaystyle 0,}\\ {\displaystyle \begin{array}{c}\end{array}(v_1\otimes v_{-1})}& =& {\displaystyle q{v}_1\otimes v_{-1}-v_{-1}\otimes v_1,}\\ {\displaystyle \begin{array}{c}\end{array}(v_{-1}\otimes v_1)}& =& {\displaystyle q^{-1}{v}_{-1}\otimes v_1-v_1\otimes v_{-1}\text{.}}\end{array}$$ In matrices we have $${\rho }^{\otimes 2\left(\begin{array}{c}\end{array}\right)}=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& q& -1& 0\\ 0& -1& q^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)\text{.}$$ Note that $${\left({\rho }^{\otimes 2\left(\begin{array}{c}\end{array}\right)}\right)}^2=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& q^2+1& -(q+q^{-1})& 0\\ 0& -(q+q^{-1})& 1-q^{-2}& 0\\ 0& 0& 0& 0\end{array}\right)=\left[2\right]{\rho }^{\otimes 2\left(\begin{array}{c}\end{array}\right)}$$ so that this is an action of ${TL}_2$ on $V^{\otimes 2}\text{.}$

The ${TL}_2$ action commutes with the $U_q{𝔰𝔩}_2$ action on $V^{\otimes 2},$ i.e. $${\rho }^{\otimes 2}\left({TL}_2\right)\subseteq {\text{End}}_U\left(V^{\otimes 2}\right)\text{.}$$

The Temperley-Lieb algebra ${TL}_k$ is generated by $$e_j=\begin{array}{c} 1 2 ⋯ j j+1 ⋯ k \end{array},1\le j\le k-1\text{.}$$ Define an action of ${TL}_k$ on $$V^{\otimes k}=\text{span}\{v_{i_1}\otimes \cdots \otimes v_{i_k}\hspace{0.17em}|\hspace{0.17em}{i}_1,\dots ,i_k\in \{\pm 1\}\}$$ by $$e_j(v_{i_1}\otimes \cdots \otimes v_{i_k})=v_{i_1}\otimes \cdots \otimes v_{i_{j-1}}\otimes \begin{array}{c}\end{array}(v_{i_j}\otimes v_{i_{j+1}})\otimes v_{i_{j+2}}\otimes \cdots \otimes v_{i_k}\text{.}$$

Claim

(a) This defines a ${TL}_k\text{-action}$ on $V^{\otimes k}\text{.}$ (b) This ${TL}_k\text{-action}$ commutes with the $U_q{𝔰𝔩}_2\text{-action}$ on $V^{\otimes k}\text{.}$

Let $A$ be an algebra and let $M$ be a semisimple $A$ module, $$M=\underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus m_{\lambda }}\text{.}$$ Let $𝒵={\text{End}}_A\left(M\right)\text{.}$ Then $$𝒵=\underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(ℂ\right),\text{and}M\simeq \underset{\lambda \in \hat{M}}{⨁}{A}^{\lambda }\otimes {𝒵}^{\lambda }$$ as an $(A,𝒵)$ bimodule, where $\hat{M}\subseteq \hat{A}$ is an index set for the simple $A\text{-modules}$ appearing in $M\text{.}$

		Proof.
	$$\begin{array}{rcl}{\displaystyle 𝒵}& =& {\displaystyle {\text{Hom}}_A(M,M)}\\ & =& {\displaystyle {\text{Hom}}_A(\underset{\lambda \in \hat{M}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{A}_i^{\lambda },\underset{\mu \in \hat{M}}{⨁}\underset{j=1}{\overset{m_{\lambda }}{⨁}}{A}_j^{\mu })}\\ & =& {\displaystyle \underset{\lambda \in \hat{M}}{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}{\text{Hom}}_A(A_i^{\lambda },A_j^{\lambda }),}\end{array}$$ by Schur's Lemma. Hence $$𝒵=\left\{e_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{M},1\le i,j\le m_{\lambda }\}$$ where $e_{ij}^{\lambda }:A_i^{\lambda }\to A_j^{\lambda }$ (choose $e_{ii}^{\lambda }$ so that ${\left(e_{ii}^{\lambda }\right)}^2=e_{ii}^{\lambda }$ and $e_{ij}^{\lambda }$ and $e_{ji}^{\lambda }$ so that $e_{ij}^{\lambda }{e}_{ji}^{\lambda }=e_{ii}^{\lambda }\text{).}$ $\square$

Notes and References

These are a typed copy of Lecture 5 from a series of handwritten lecture notes for the class Representation Theory given on August 26, 2008.

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