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 3

Let ${ℂ}^d=\text{span}\left\{e_i\hspace{0.17em}\right|\hspace{0.17em}1\le i\le d\}$ where $$e_i=\left(\begin{array}{c}0\\ ⋮\\ 0\\ 1\\ 0\\ ⋮\\ 0\end{array}\right)$$ with $M_d\left(ℂ\right)\text{-action}$ by left multiplication.

(1)	If $M$ is a simple $M_d\left(ℂ\right)\text{-module}$ then $M\simeq {ℂ}^d\text{.}$
(2)	If $t:M_d\left(ℂ\right)\to ℂ$ is a trace then $t=\kappa ·\text{Tr},$ with $\kappa \in ℂ\text{.}$
(3)	$𝒵\left(M_d\left(ℂ\right)\right)=ℂ·\text{Id}\text{.}$
(4)	$M_d\left(ℂ\right)$ has one nonzero ideal.

		Proof.
	(1) Let $M$ be a simple $M_d\left(ℂ\right)\text{-module.}$ Let $m\in M$ be nonzero. Since $$m=1·m=\sum _{i=1}^d{E}_{ii}m,$$ then $E_{ii}m\ne 0$ for some $i\text{.}$ Let $m_j=E_{ji}m$ for $j=1,2,\dots ,d\text{.}$ Then $N=\text{span}\{m_1,\dots ,m_d\}$ is a submodule of $M$ such that $$E_{rs}{m}_j={\delta }_{sj}{m}_r\text{.}$$ Since $M$ is simple, $N=M$ and $$\begin{array}{ccc}M& ⟶& {ℂ}^d\\ m_j& ⟼& e_j\end{array}$$ is an $M_d\left(ℂ\right)\text{-module}$ isomorphism. (2) Let $t:M_d\left(ℂ\right)\to ℂ$ be a trace. then $$t\left(E_{ij}\right)=t\left(E_{i1}{E}_{1j}\right)=t\left(E_{1j}{E}_{i1}\right)={\delta }_{ij}t\left(E_{11}\right)\text{.}$$ So $t=t\left(E_{11}\right)·\text{Tr}\text{.}$ $\square$

The algebra $A={\displaystyle \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)}$

$A$ has basis $\left\{E_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le d_{\lambda },\lambda \in \hat{A}\}$ with $$E_{ij}^{\lambda }{E}_{rs}^{\mu }={\delta }_{\lambda \mu }{\delta }_{jr}{E}_{is}^{\lambda }\text{.}$$ Define $A^{\lambda }=\text{span}\left\{e_i^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}1\le i\le d_{\lambda }\}$ with $$E_{ij}^{\mu }{e}_r^{\lambda }={\delta }_{\lambda \mu }{\delta }_{jr}{e}_i^{\lambda }\text{.}$$ Define ${\text{Tr}}^{\lambda }:A\to ℂ$ by ${\text{Tr}}^{\lambda }\left(E_{ij}^{\mu }\right)={\delta }_{\lambda \mu }{\delta }_{ij}\text{.}$ Define $z_{\lambda }={\sum }_{i=1}^d{E}_{ii}^{\lambda },$ so that $z_{\lambda }^2=z_{\lambda }$ and $1=\sum _{\lambda \in \hat{A}}{z}_{\lambda }\text{.}$ Define $I^{\lambda }=z_{\lambda }A=\text{span}\left\{E_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le d_{\lambda }\}\text{.}$

(1)	$A^{\lambda },$ $\lambda \in \hat{A},$ are the simple $A\text{-modules.}$
(2)	If $t:A\to ℂ$ is a trace then $$t=\sum _{\lambda \in \hat{A}}{t}_{\lambda }{\text{Tr}}^{\lambda },$$ with $t_{\lambda }\in ℂ\text{.}$
(3)	$𝒵\left(A\right)=\text{span}\left\{z_{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A}\}\text{.}$
(4)	The minimal ideals of $A$ are $I^{\lambda },$ $\lambda \in \hat{A}\text{.}$ The ideals of $A$ are sums of $I^{\lambda }\text{.}$

The regular representation of $A$

$$A=\underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }}\text{since}A=\left\{\left(\begin{array}{c} 0 0 \end{array}\right)\right\}\text{.}$$ Note that $$\begin{array}{cccc}{\text{Tr}}^{\lambda }:& A& ⟶& ℂ\\ & a& ⟼& \text{Tr}\left(p^{\lambda }\left(a\right)\right)\end{array}$$ where $$\begin{array}{cccc}{\rho }^{\lambda }:& A& ⟶& M_{d_{\lambda }}\left(ℂ\right)\\ & a& ⟼& {\rho }^{\lambda }\left(a\right),\end{array}$$ where ${\rho }^{\lambda }\left(a\right)$ is the matrix of the action of $a$ on $A^{\lambda }\text{.}$ If $t:A\to ℂ$ is the trace of the regular representation then $$t=\sum _{\lambda \in \hat{A}}{d}_{\lambda }{\text{Tr}}^{\lambda }\text{.}$$ The dual basis to $\left\{E_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}$ is $$\left\{\frac{1}{d_{\lambda }}{E}_{ji}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},1\le i,j\le d_{\lambda }\}$$ since $t\left(E_{ij}^{\lambda }\frac{1}{d_{\mu }}{E}_{sr}^{\mu }\right)={\delta }_{ir}{\delta }_{\lambda \mu }{\delta }_{js}\text{.}$ Thus $$E_{ij}^{\lambda }=\sum _{\underset{1\le r,s\le d_{\mu }}{\mu \in \hat{A}}}{d}_{\mu }{A}^{\lambda }{\left(\frac{1}{d_{\mu }}{E}_{sr}^{\mu }\right)}_{ji}{E}_{rs}^{\mu }\text{.}$$ If $B=\left\{b\right\}$ is a basis of $A$ and $\left\{b^{*}\right\}$ is the dual basis with respect to $⟨,⟩$ defined by $t\text{.}$ Then $$E_{ij}^{\lambda }=\sum _{b\in B}{d}_{\lambda }{A}^{\lambda }{\left(b^{*}\right)}_{ji}b$$ and $$b=\sum _{\underset{1\le i,j\le d_{\lambda }}{\lambda \in \hat{A}}}{A}^{\lambda }{\left(b\right)}_{ij}{E}_{ij}^{\lambda }\text{.}$$

(Artin-Wedderburn) Let $A$ be a finite dimensional algebra such that the trace of the regular representation is nondegenerate. Then $$A\stackrel{\sim }{⟶}\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)$$ as algebras where $A=\underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }}$ as $A\text{-modules.}$

		Proof.
	By Maschke's theorem $$A\simeq \underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus d_{\lambda }}$$ as $A\text{-modules.}$ Since $$\begin{array}{ccc}A& ⟶& \text{End}\left(A\right)\\ a& ⟼& a_A\end{array}$$ is injective, $$\begin{array}{cccc}\rho :& A& ⟶& \text{End}\left(\underset{\lambda \in \hat{A}}{⨁}{A}^{\lambda }\right)\\ & a& ⟼& \underset{\lambda \in \hat{A}}{⨁}{\rho }^{\lambda }\left(a\right)\end{array}$$ is also injective and $\rho$ is an algebra homomorphism. $\square$

Temperley-Lieb ${TL}_3=\text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}$ has module $M=\text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}$ which we found was $M=N\oplus P$ where $$\begin{array}{rcl}{\displaystyle N}& =& {\displaystyle \text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{and}}\\ {\displaystyle P}& =& {\displaystyle \text{span}\{(1+(q+q^{-1}))\begin{array}{c}\end{array}-\begin{array}{c}\end{array}-\begin{array}{c}\end{array}\}\text{.}}\end{array}$$ We have $$\begin{array}{cccc}{\rho }^{}:& {TL}_3& ⟶& M_1\left(ℂ\right)\\ & \begin{array}{c}\end{array}& ⟼& 0\\ & \begin{array}{c}\end{array}& ⟼& 0\end{array}$$ and $$\begin{array}{cccc}{\rho }^{\varnothing }:& {TL}_3& ⟶& M_2\left(ℂ\right)\\ & \begin{array}{c}\end{array}& ⟼& \left(\begin{array}{cc}q+q^{-1}& 1\\ 0& 0\end{array}\right)\\ & \begin{array}{c}\end{array}& ⟼& \left(\begin{array}{cc}0& 0\\ 1& q+q^{-1}\end{array}\right)\end{array}$$ Then $$\begin{array}{ccc}{TL}_3& ⟶& {\displaystyle \underset{\lambda \in {TL}_3}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)}\\ \begin{array}{c}\end{array}& ⟼& \left(\begin{array}{ccc}q+q^{-1}& 1& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\\ \begin{array}{c}\end{array}& ⟼& \left(\begin{array}{ccc}0& 0& 0\\ 1& q+q^{-1}& 0\\ 0& 0& 0\end{array}\right)\end{array}$$ Let $G$ be a group. The group algebra of $G$ is the algebra $ℂG$ with basis $G$ and multiplication determined by the multiplication in $G\text{.}$ The map $$t:ℂG⟶ℂ\text{given by}t\left(a\right)=a{|}_1,$$ the coefficient of $1$ in $a,$ is a trace on $G\text{.}$ If $\text{Tr}$ is the trace of the regular representation of $G$ then $$\text{Tr}\left(g\right)=\sum _{h\in G}gh{|}_h=\{\begin{array}{cc}\left|G\right|,& \text{if}\hspace{0.17em}g=1,\\ 0,& \text{if}\hspace{0.17em}g\ne 1\end{array}=\left|G\right|·t\left(g\right)\text{.}$$

The braid group $B_n$ is the group of braids with $n$ strands with product $$b_1{b}_2=\begin{array}{c} b1 b2 \end{array}$$

(Artin) $B_n$ is presented by generators $g_1,\dots ,g_{n-1},$ where $$g_i=\begin{array}{c} 1 2 i i+1 n ⋯ ⋯ \end{array}$$ with relations $g_i{g}_{i+1}{g}_i=g_{i+1}{g}_i{g}_{i+1}\text{.}$

The symmetric group $S_n$ is the quotient of $B_n$ by the relations $g_i^2=1\text{.}$ $$(\text{write}\hspace{0.17em}{s}_i=\begin{array}{c} 1 2 i i+1 n ⋯ ⋯ \end{array}\hspace{0.17em}\text{since}\hspace{0.17em}\begin{array}{rcl}{\displaystyle g_i}& =& {\displaystyle \begin{array}{c} \end{array}}\\ {\displaystyle =g_i^{-1}}& =& {\displaystyle \begin{array}{c} \end{array}}\end{array}\hspace{0.17em}\text{in}\hspace{0.17em}{S}_n)$$

The Iwahori-Hecke algebra is the quotient of $ℂB_n$ by the relations $$(T_i-q)(T_i+q^{-1})=0\text{.}(T_i^2=(q-q^{-1})T_i+1)\text{.}$$ Let $$e_i=q-T_i\text{in}{H}_n\text{.}$$ Then $T_i{T}_{i+1}{T}_i=T_{i+1}{T}_i{T}_{i+1}$ becomes $$e_i{e}_{i+1}{e}_i-e_{i+1}{e}_i{e}_{i+1}=e_i-e_{i+1}\text{.}$$ The map $$\begin{array}{ccccc}ℂB_n& ⟶& H_n& ⟶& {TL}_n\\ & & e_i& ⟼& e_i\\ T_i& ⟼& T_i\end{array}$$ are surjective algebra homomorphisms.

Remark $$T_{w_0}^2=\begin{array}{c} \end{array}\in 𝒵\left(B_n\right)$$ and $$T_{w_0}^2=y^{{\epsilon }_1}\cdots y^{{\epsilon }_n},$$ where $$y^{{\epsilon }_i}=\begin{array}{c} i \end{array}$$ and $$y^{{\epsilon }_i}{y}^{{\epsilon }_j}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}=y^{{\epsilon }_j}{y}^{{\epsilon }_i},$$ for $1\le i,j\le n\text{.}$

So the image of $ℂ[y^{{\epsilon }_1},\dots ,y^{{\epsilon }_n}]$ is a large commutative subalgebra of $H_n$ (or ${TL}_n\text{).}$

Note: $$\begin{array}{ccc}{B}_n& \hookrightarrow & B_{n+1}\\ b& ⟼& \begin{array}{c} b \end{array}\end{array}$$ gives inclusions $$\begin{array}{rcl}{\displaystyle H_1}& \subseteq & {\displaystyle H_2\subseteq H_3\subseteq \cdots ,}\\ {\displaystyle {TL}_1}& \subseteq & {\displaystyle {TL}_2\subseteq {TL}_3\subseteq \cdots ,\text{and}}\\ {\displaystyle ℂS_1}& \subseteq & {\displaystyle ℂS_2\subseteq ℂS_3\subseteq \cdots \text{.}}\end{array}$$

Pullback functors

Suppose $A\stackrel{\phi }{\to }R$ is an algebra homomorphism. Then we get a functor $$\begin{array}{ccc}\left\{R\text{-modules}\right\}& ⟶& \left\{A\text{-modules}\right\}\\ M& ⟼& {\phi }^{*}\left(M\right)\end{array}$$ where ${\phi }^{*}\left(M\right)=M,$ as vector spaces and the $A\text{-action}$ is given by $$a·m=\phi \left(a\right)m,$$ for $a\in A,$ $m\in M\text{.}$

The map $H_n\stackrel{\pi }{\to }{TL}_n$ gives a functor $$\left\{{TL}_n\text{-modules}\right\}⟶\left\{H_n\text{-modules}\right\}$$ which takes simple modules to simple modules. The map $\pi$ is surjective and the map ${\pi }^{*}$ is injective. The map $${TL}_3\stackrel{\iota }{\hookrightarrow }{TL}_4$$ gives $$\left\{{TL}_4\text{-modules}\right\}\stackrel{{\iota }^{*}}{⟶}\left\{{TL}_3\text{-modules}\right\}\text{.}$$ The functor ${\iota }^{*}$ is Restriction, ${\text{Res}}_{{TL}_3}^{{TL}_4}\text{.}$

Adjoint functors

Let $F:\left\{A\text{-modules}\right\}\to \left\{B\text{-modules}\right\}$ be a functor. The adjoint functor $F^{\vee }:\left\{B\text{-modules}\right\}\to \left\{A\text{-modules}\right\}$ is determined by $${\text{Hom}}_{B\text{-mod}}(F^{\vee }M,N)\simeq {\text{Hom}}_{A\text{-mod}}(M,FM)\text{.}$$ The adjoint functor to ${\text{Res}}_A^B$ is induction ${\text{Ind}}_A^B\text{.}$ $${\text{Ind}}_A^B:\left\{A\text{-modules}\right\}⟶\left\{B\text{-modules}\right\}\text{.}$$ It is given explicitly by $${\text{Ind}}_A^B\left(M\right)=B{\otimes }_{A}M,$$ where $B{\otimes }_{A}M$ is generated by $b\otimes m,$ $b\in B,$ $m\in M,$ with relations $$ba\otimes m=b\otimes am$$ and bilinearity.

Bratelli diagrams

Let $$A_1\subseteq A_2\subseteq A_3\subseteq \cdots$$ be a sequence of inclusions of semisimple algebras. The Bratelli diagram for $A_1\subseteq A_2\subseteq \cdots$ is the graph with $$\text{vertices on level}\hspace{0.17em}k\text{:}\hspace{0.17em}{\hat{A}}_k,$$ where ${\hat{A}}_k$ is an index for the simple $A_k\text{-modules}$ and $$m_{\lambda \mu }\hspace{0.17em}\text{edges connecting}\hspace{0.17em}\lambda \hspace{0.17em}\text{and}\hspace{0.17em}\mu (\lambda \in {\hat{A}}_k,\mu \in {\hat{A}}_{k-1})$$ if $${\text{Res}}_{A_{k-1}}^{A_k}\left(A_k^{\lambda }\right)=\underset{\mu \in {\hat{A}}_{k-1}}{⨁}{\left(A_{k-1}^{\mu }\right)}^{\oplus m_{\lambda \mu }}\text{.}$$

The Bratelli diagram for $${TL}_1\subseteq {TL}_2\subseteq \cdots$$ has $$\text{vertices on level}\hspace{0.17em}k\text{:}\{\text{partitions of}\hspace{0.17em}k\hspace{0.17em}\text{with}\hspace{0.17em}\le 2\hspace{0.17em}\text{rows}\}={\widehat{T_L}}_k$$ and an edge $\lambda -\mu$ if $\mu$ is obtained from $\lambda$ by removing a box.

A partition is a collection of boxes in a corner. $$\lambda =\begin{array}{c} \end{array}=\left(442211\right)$$ Write $\lambda =({\lambda }_1,\dots ,{\lambda }_{\ell })$ with ${\lambda }_1\ge \cdots \ge {\lambda }_{\ell }$ and ${\lambda }_i=\text{\#}$ of boxes in row $i\text{.}$

(a)	The Bratelli diagram for $$H_1\subseteq H_2\subseteq H_3\subseteq \cdots$$ has $${\hat{H}}_k=\left\{\text{partitions with}\hspace{0.17em}k\hspace{0.17em}\text{boxes}\right\}\text{and}\lambda -\mu$$ if $\mu$ is obtained from $\lambda$ by removing a box.
(b)	The Bratelli diagram for $$ℂS_1\subseteq ℂS_2\subseteq ℂS_3\subseteq \cdots$$ has $${\hat{S}}_k=\left\{\text{partitions with}\hspace{0.17em}k\hspace{0.17em}\text{boxes}\right\}\text{and}\lambda -\mu$$ if $\mu$ is obtained from $\lambda$ by removing a box.

The algebra $U{𝔰𝔩}_2$

A Lie algebra is a vector space $𝔤$ with a bracket $[,]:𝔤\otimes 𝔤\to 𝔤$ such that

(1)	$[x,y]=-[y-x],$ for $x,y\in 𝔤,$
(2)	$[[x,y],z]+[[z,x],y]+[[y,z],x]=0,$ for $x,y,z\in 𝔤\text{.}$

A Lie algebra is not an algebra.

The enveloping algebra of $𝔤$ is the algebra $U𝔤$ generated by the vector space $𝔤,$ with the relations $$xy=yx+[x,y],$$ for $x,y\in 𝔤\text{.}$

The Lie algebra ${𝔰𝔩}_2$ is the vector space $${𝔰𝔩}_2=\{a\in M_2\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}a=0\}$$ with bracket $$[a,b]=ab-ba,$$ for $a,b\in {𝔰𝔩}_2$ (where the product on the RHS is matrix multiplication).

The enveloping algebra of ${𝔰𝔩}_2$ is the algebra $U{𝔰𝔩}_2$ generated by $x,y,k$ with relations $$xy=yx+k,kx=xk+2x,ky=yk-2y\text{.}$$

The Lie algebra ${𝔰𝔩}_2$ is presented by generators $$x=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),y=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),k=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ and relations $$[x,y]=k,[k,x]=2x,[k,y]=-2y\text{.}$$

HW: Show that $U{𝔰𝔩}_2$ has basis $$\left\{y^{m_1}{k}^{m_2}{x}^{m_3}\hspace{0.17em}\right|\hspace{0.17em}{m}_1,m_2,m_3\in {\mathbb{Z}}_{\ge 0}\}\text{.}$$ Hence $\text{dim}\left(U{𝔰𝔩}_2\right)=3\infty \text{.}$

Notes and References

These are a typed copy of Lecture 3 from a series of handwritten lecture notes for the class Representation Theory given on August 12, 2008.

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