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 2

The Temperley-Lieb algebra ${TL}_k$ is $${TL}_k=\text{span}\left\{\text{noncrossing diagrams with}\hspace{0.17em}k\hspace{0.17em}\text{top dots and}\hspace{0.17em}k\hspace{0.17em}\text{bottom dots}\right\}\left(\text{generators}\hspace{0.17em}A\right)$$ with product $$b_1{b}_2={(q+q^{-1})}^{\text{\# of internal loops}}\begin{array}{c} b1 b2 \end{array}\left(\text{relations}\hspace{0.17em}A\right)\text{.}$$

$$\begin{array}{rcl}{\displaystyle {TL}_1}& =& {\displaystyle \text{span}\left\{\begin{array}{c}\end{array}\right\},}\\ {\displaystyle {TL}_2}& =& {\displaystyle \text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle {TL}_3}& =& {\displaystyle \text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{and}}\\ {\displaystyle {TL}_4}& =& {\displaystyle \text{span}\left\{\begin{array}{ccc}\begin{array}{c}\end{array},& \begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},& \begin{array}{c}\end{array},\begin{array}{c}\end{array},\\ & \begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},& \begin{array}{c}\end{array},\begin{array}{c}\end{array},\\ & \begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\end{array}\right\}\text{.}}\end{array}$$ Let $$e_i=\begin{array}{c} i i+1 ⋯ ⋯ \end{array},\text{for}i=1,\dots ,k-1\left(\text{generators}\hspace{0.17em}B\right)\text{.}$$

${TL}_k$ is presented by generators $e_1,\dots ,e_{k-1}$ and relations $$e_i^2(q+q^{-1})e_i\text{and}{e}_i{e}_{i\pm 1}{e}_i=e_i\left(\text{relations}\hspace{0.17em}B\right)\text{.}$$

		Proof.
	To show: (a) Generators $A$ can be written in terms of generators $B\text{.}$ (b) Relations $A$ can be derived from relations $B\text{.}$ (c) Generators $B$ can be written in terms of generators $A\text{.}$ (d) Relations $B$ can be derived from relations $A\text{.}$ $\square$

Homework

(1)

(a) Define the symmetric group $S_k$ (via permutations). (b) Let $$s_i=\begin{array}{c} i i+1 ⋯ ⋯ \end{array},i=1,2,\dots ,k-1\text{.}$$ Show that $S_k$ is presented by generators $s_1,\dots ,s_{k-1}$ and relations $$s_i^2=1\text{and}{s}_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}\text{.}$$ (c) Define the Young lattice and show that it is the Bratelli diagram for the tower $$ℂS_1\subseteq ℂS_2\subseteq ℂS_3\subseteq \cdots$$ (d) Let $$m_i=s_{1i}+s_{2i}+s_{3i}+\cdots +s_{i-1,i},$$ where $$s_{ij}=\begin{array}{c} i j ⋯ ⋯ \end{array}$$ for $1\le i<j\le k\text{.}$ Let $m_1=0\text{.}$ Show that $$m_i{m}_j=m_j{m}_i,$$ for $1\le i<j\le k\text{.}$ (e) Show that each irreducible $S_k\text{-module}$ $S^{\lambda }$ has a basis of simultaneous eigenvectors $v_T$ for $m_1,\dots ,m_k,$ $$\text{i.e.}{m}_i{v}_T=c\left(T\left(i\right)\right)v_T\text{.}$$ (f) Find the eigenvalues $c\left(T\left(i\right)\right)\text{.}$

(2)

Following the work of R. Block, classify the simple modules of $U_q{𝔰𝔩}_2,$ where $U_q{𝔰𝔩}_2$ is the algebra generated by $E,F,K^{\pm 1}$ with relations $$\begin{array}{c}K{K}^{-1}=K^{-1}K=1,\\ KE{K}^{-1}=q^{2}E,KF{K}^{-1}=q^{-2}F,\\ {\displaystyle EF-FE=\frac{K-K^{-1}}{q-q^{-1}}\text{.}}\end{array}$$

Traces

Let $A$ be an algebra. A trace on $A$ is a linear transformation $t:A\to ℂ$ such that $$t\left(a_1{a}_2\right)=t\left(a_2{a}_1\right),$$ for $a_1,a_2\in A\text{.}$

Let $$\begin{array}{cccc}{\rho }_M:& A& ⟶& \text{End}\left(M\right)\\ & a& ⟼& a_M\end{array}$$ be a representation of $A\text{.}$

The character of $M$ is the trace $$\begin{array}{cccc}{\chi }_M:& A& ⟶& ℂ\\ & a& ⟼& \text{Tr}\left(a_M\right)\end{array},$$ where $$\text{Tr}\left(a_M\right)=\sum _{m\in B}am{|}_m$$ for a basis $B$ of $M,$ with $$am{|}_m=\text{coefficient of}\hspace{0.17em}m\hspace{0.17em}\text{in}\hspace{0.17em}am$$ (expanded in the basis $B\text{).}$

Given a trace $t:A\to ℂ$ define $$⟨,⟩:A\otimes A⟶ℂ\text{by}⟨a_1,a_2⟩=t\left(a_1{a}_2\right),$$ for $a_1,a_2\in A\text{.}$ Then $$⟨a_1,a_2⟩=⟨a_2,a_1⟩\text{and}⟨a_1{a}_2,a_3⟩=⟨a_1,a_2{a}_3⟩,$$ for $a_1,a_2,a_3\in A\text{.}$ The radical of $⟨\hspace{0.17em}⟩$ is $$\text{Rad}\left(⟨,⟩\right)=\{r\in A\hspace{0.17em}|\hspace{0.17em}⟨r,a⟩=0\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$

Let $B=\{b_1,\dots ,b_n\}$ be a basis of $A\text{.}$ The dual basis to $B$ with respect to $⟨,⟩$ is $B^{*}=\{b_1^{*},\dots ,b_n^{*}\}$ such that $$⟨b_i,b_j^{*}⟩={\delta }_{ij}\text{.}$$ The Gram matrix of $⟨,⟩$ is $$G={\left(⟨b_i,b_j⟩\right)}_{b_i,b_j\in B}\text{.}$$

HW: Show that $$\begin{array}{rcl}{\displaystyle \text{Rad}\left(⟨,⟩\right)=0}& ⟺& {\displaystyle G\hspace{0.17em}\text{is invertible}}\\ & ⟺& {\displaystyle \text{det}\hspace{0.17em}G\hspace{0.17em}\text{is invertible}}\\ & ⟺& {\displaystyle The\; dual\; basis$ B^{*}$exists}\end{array}$$ A nondegenerate trace is a trace $t:A\to ℂ$ such that $\text{Rad}\left(⟨,⟩\right)=0\text{.}$

HW: Show that $\text{Rad}\left(⟨,⟩\right)$ is an ideal of $A\text{.}$

$$\begin{array}{rcl}{\displaystyle {TL}_3}& =& {\displaystyle \text{span}\{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\},}\\ {\displaystyle B}& =& {\displaystyle \{\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array},\begin{array}{c}\end{array}\}\text{.}}\end{array}$$ Define a trace on ${TL}_3$ by $$t\left(b\right)={(q+q^{-1})}^{\text{\# of cycles in cl}\left(b\right)},$$ where $$\text{cl}\left(b\right)=\begin{array}{c} b \end{array}\text{.}$$

Commuting operators

Let $A$ be an algebra, $M$ an $A\text{-module.}$ The commutant or centralizer algebra is $${\text{End}}_A\left(M\right)=\{\phi \in \text{End}\left(M\right)\hspace{0.17em}|\hspace{0.17em}{a}_{\bar{M}}\phi =\phi a_{\bar{M}},\hspace{0.17em}\text{for}\hspace{0.17em}a\in A\}\text{.}$$ Recall that $$\begin{array}{ccc}A& ⟶& \text{End}\left(M\right)\\ a& ⟼& a_M\end{array}$$ is an algebra homomorphism. Let $M$ and $N$ be simple $A\text{-modules}$ and $\phi :M\to N$ and $A\text{-module}$ homomorphism (i.e. $\phi a_M=a_N\phi ,$ for $a\in A\text{).}$ Then $$\text{ker}\hspace{0.17em}\phi \hspace{0.17em}\text{and}\hspace{0.17em}\text{im}\hspace{0.17em}\phi \hspace{0.17em}\text{are submodules}$$ of $M$ and $N,$ respectively. So $\text{ker}\hspace{0.17em}\phi =0$ or $\text{ker}\hspace{0.17em}\phi =M$ and $\text{im}\hspace{0.17em}\phi =0$ or $\text{im}\hspace{0.17em}\phi =N\text{.}$ So $\phi =0$ or $\phi$ is a bijection (and $M\simeq N\text{).}$

Let $\lambda$ be an eigenvalue of $\phi \text{.}$ Then $\phi -\lambda \in {\text{End}}_A\left(M\right)\text{.}$ So $\phi -\lambda =0$ or $\phi -\lambda$ is invertible. Since $\text{det}(\phi -\lambda )=0,$ $\phi -\lambda$ is not invertible. So $$\phi =\lambda ·\text{Id}\text{.}$$

(Schur's Lemma) Let $M$ be a simple module. Then $${\text{End}}_A\left(M\right)=ℂ·{\text{id}}_M\text{.}$$

Let $A$ be a finite dimensional algebra. Let $t:A\to ℂ$ be a nondegenerate trace on $A\text{.}$ Let $B$ be a basis of $A$ and let $B^{*}$ be the dual basis with respect to $⟨,⟩\text{.}$

(Maschke's theorem).

(a)	Let $M,N$ be $A\text{-modules}$ and let $\phi :M\to N$ be a vector space morphism. Then $$\left[\phi \right]=\sum _{b\in B}b\phi b^{*}$$ is an $A\text{-module}$ homomorphism, i.e. $\left[\phi \right]\in {\text{Hom}}_A(M,N)\text{.}$
(b)	Assume $t$ is the trace of the regular representation. Every finite dimensional $A\text{-module}$ $M$ is completely decomposable.

		Proof.
	(a) Let $a\in A\text{.}$ Then $$\begin{array}{rcl}{\displaystyle a\left[\phi \right]}& =& {\displaystyle \sum _{b\in B}ab\phi b^{*}}\\ & =& {\displaystyle \sum _{b,c\in B}⟨ab,c^{*}⟩c\phi b^{*}}\\ & =& {\displaystyle \sum _{b,c\in B}c\phi ⟨ab,c^{*}⟩b^{*}}\\ & =& {\displaystyle \sum _{b,c\in B}c\phi ⟨c^{*}a,b⟩b^{*}}\\ & =& {\displaystyle \sum _{c\in B}c\phi c^{*}a}\\ & =& {\displaystyle \left[\phi \right]a\text{.}}\end{array}$$ HW: Show that $\phi$ does not depend on the choice of the basis $B\text{.}$ (b) Let $M$ be a finite dimensional $A\text{-module.}$ Assume $N\subseteq M$ is a nonzero submodule. Let $\pi :M\to M$ be a vector space homomorphism such that $\text{im}\hspace{0.17em}\pi =N$ and $\pi \left(n\right)=n,$ for $n\in N\text{.}$ (i.e. define $\pi \left(n_i\right)=n_i,$ $\pi \left(m_j\right)=0$ for a basis $\{n_1,\dots ,n_r\}$ of $N$ and a basis $\{n_1,\dots ,n_r,m_1,\dots ,m_s\}$ of $M\text{.)}$ To show: (ba) $\text{im}\left[\pi \right]=N$ and $\left[\pi \right]n=n$ for $n\in N\text{.}$ (bb) $M=\left[\pi \right]M\oplus (1-\left[\pi \right])M,$ and $\left[\pi \right]M=N$ and $(1-\left[\pi \right])M$ are submodules. (ba) If $n\in N$ $$\left[\pi \right]n=\sum _{b\in B}b\pi b^{*}n=\left(\sum _{b\in B}b{b}^{*}\right)n\text{.}$$ Since $$⟨\sum _{b\in B}b{b}^{*},a⟩=\sum _{b\in B}⟨ab,b^{*}⟩=\text{Tr}\left(a_A\right)=⟨1,a⟩$$ for all $a\in A,$ it follows that $\sum _{b\in B}b{b}^{*}=1\text{.}$ So $\left[\pi \right]n=1·n=n\text{.}$ If $m\in M$ then $\left[\pi \right]m=\sum _{b\in B}b\pi b^{*}m\in N$ since $\pi b^{*}m\in N$ and $N$ is a submodule. (bb) If $m\in M$ then $$m=(\left[\pi \right]+(1-\left[\pi \right]))m=\left[\pi \right]m+(1-\left[\pi \right])m\text{.}$$ So $$M=\left[\pi \right]M+(1-\left[\pi \right])M\text{.}$$ If $m\in \left[\pi \right]M\cap (1-\left[\pi \right])M$ then $$m=\left[\pi \right](1-\left[\pi \right])m=(\left[\pi \right]-\left[\pi \right])m=0\text{.}$$ $\square$

The regular representation

The regular representation of $A$ is the vector space $A$ with $A\text{-action}$ given by left multiplication. $$\begin{array}{cccc}{\rho }_A:& A& ⟶& \text{End}\left(A\right)\\ & a& ⟼& a_A\end{array}$$ is injective because $a·1=a$ implies $\text{ker}\hspace{0.17em}{\rho }_A=0\text{.}$

Identify $A$ with $\text{im}\hspace{0.17em}{\rho }_A,$ so that $A$ "is" a set of matrices. A matrix $a$ is nilpotent if $a^k=0,$ for some $k\in {\mathbb{Z}}_{>0}\text{.}$

Let $$\begin{array}{cccc}t:& A& ⟶& ℂ\\ & a& ⟼& \text{Tr}\left(a_A\right)\end{array}$$ be the trace of the regular representation. Then $\text{Rad}\left(⟨,⟩\right)$ is the largest ideal of $A$ such that every element is nilpotent.

		Proof.
	To show: (a) Every element of $\text{Rad}\left(⟨,⟩\right)$ is nilpotent. (b) If $J$ is an ideal of $A$ and every element of $J$ is nilpotent then $J\subseteq \text{Rad}\left(⟨,⟩\right)\text{.}$ (a) Let $r\in \text{Rad}\left(⟨,⟩\right)\text{.}$ Then $$\text{Tr}\left(r^k\right)=⟨r^k,1⟩=⟨r,r^{k-1}⟩=0,$$ for all $k\in {\mathbb{Z}}_{>0}\text{.}$ Let ${\lambda }_1,\dots ,{\lambda }_n$ be the eigenvalues of $r\text{.}$ Then $$p_k({\lambda }_1,\dots ,{\lambda }_n)={\lambda }_1^k+\cdots +{\lambda }_n^k=\text{Tr}\left(r^k\right)=0,$$ for all $k\in {\mathbb{Z}}_{>0}\text{.}$ Since $p_k({\lambda }_1,\dots ,{\lambda }_n)=0,$ for $k\in {\mathbb{Z}}_{>0},$ $$\begin{array}{cc}{e}_k({\lambda }_1,\dots ,{\lambda }_n)=0& (*)\end{array}$$ for $k\in {\mathbb{Z}}_{>0}\text{.}$ So $$\prod _{i=1}^n(z+{\lambda }_i)=\sum _{k=0}^n{e}_k({\lambda }_1,\dots ,{\lambda }_n)z^{n-k}=z^n\text{.}$$ So ${\lambda }_1=\cdots ={\lambda }_n=0\text{.}$ Thus, by Jordan normal form, $r$ is nilpotent (conjugate to a strictly upper triangular matrix). (b) Assume $J$ is an ideal of $A$ and all elements of $J$ are nilpotent. Let $r\in J$ and $a\in A\text{.}$ Then $ra\in J$ and, since $ra$ is nilpotent, $$0=\text{Tr}\left(ra\right)=⟨r,a⟩\text{.}$$ So $r\in \text{Rad}\left(⟨,⟩\right)\text{.}$ $\square$

Expansion of $(*)\text{:}$ Since $p_k({\lambda }_1,\dots ,{\lambda }_n)=0$ for $k\in {\mathbb{Z}}_{>0}$ $$\begin{array}{rcl}{\displaystyle 1}& =& {\displaystyle e^{-\sum _{k\in {\mathbb{Z}}_{>0}}\frac{p_k({\lambda }_1,\dots ,{\lambda }_n)}{k}{(-z)}^k}}\\ & =& {\displaystyle e^{-\sum _{i=1}^n\sum _{k\in {\mathbb{Z}}_{>0}}\frac{{(-{\lambda }_{i}z)}^k}{k}}}\\ & =& {\displaystyle \prod _{i=1}^n{e}^{-\text{ln}\left(\frac{1}{1+{\lambda }_{i}z}\right)}}\\ & =& {\displaystyle \prod _{i=1}^n{e}^{\text{ln}(1+{\lambda }_{i}z)}}\\ & =& {\displaystyle \prod _{i=1}^n(1+{\lambda }_{i}z)}\\ & =& {\displaystyle \sum _{k\in {\mathbb{Z}}_{\ge 0}}{e}_k({\lambda }_1,\dots ,{\lambda }_n)z^k,}\end{array}$$ where the 3rd equality follows from $$\frac{1}{1-x}=1+x+x^2+\cdots ,$$ so that $$\frac{1}{1+x}=1+(-x)+{(-x)}^2+\cdots$$ and $$\begin{array}{rcl}{\displaystyle \text{ln}(1+x)}& =& {\displaystyle \int \frac{1}{1+x}dx}\\ & =& {\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots }\\ & =& {\displaystyle -(-x)-\frac{{(-x)}^2}{2}-\frac{{(-x)}^3}{3}-\frac{{(-x)}^4}{4}-\cdots }\\ & =& {\displaystyle -\sum _{k\in {\mathbb{Z}}_{>0}}\frac{{(-x)}^k}{k}\text{.}}\end{array}$$

Notes and References

These are a typed copy of Lecture 2 from a series of handwritten lecture notes for the class Representation Theory given on August 5, 2008.

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