Representation Theory Lecture 1

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

Last update: 26 May 2013

Lecture 1

A vector space is a set of linear combinations of a basis,

$$V=\text{span}\{b_1,\dots ,b_n\}\text{.}$$

Example: ${TL}_3$ has basis

$$, , , , \text{and}3 +4 +2·3 \in {TL}_3\text{.}$$

An algebra is a vector space $A$ with a product $A\otimes A\to A$ such that

(a)	$\left(a_1{a}_2\right)a_3=a_1\left(a_2{a}_3\right)$ for $a_1,a_2,a_3\in A,$ and
(b)	There exists $1\in A$ such that $$1·a=a·1=a,\text{for}a\in A\text{.}$$

Example: Define

$$\begin{array}{cc}{b}_1{b}_2={(q+q^{-1})}^{\text{\# of internal loops}}& b1 b2 \end{array}$$

An $A\text{-module}$ is a vector space $M$ with an action of $A,$ $A\otimes M\to M$ such that

(a)	$a_1\left(a_{2}m\right)=\left(a_1{a}_2\right)m,$ for $a_1,a_2\in A,$ $m\in M,$
(b)	$1·m,$ for all $m\in M\text{.}$

Note: $\otimes$ means that we require the distributive laws.

Example: Let $M$ be the vector space with basis $\{ , , \}$ with $b_m= b m {(q+q^{-1})}^{\text{\# of internal loops}}\text{.}$

A representation of $A$ is an $A\text{-module}$ $M\text{.}$

Given $M$ define

$$\begin{array}{cccc}\rho :& A& ⟶& \text{End}\left(M\right)\\ & a& ⟼& a_M\end{array}$$

where $a_M$ is the matrix describing the action of $A$ on $M\text{.}$

The map $\rho$ is a homomorphism of algebras.

Categories

A category $𝒞$

$$\text{objects}M\in 𝒞,\text{morphisms}\text{Hom}(M,N)\text{.}$$

Examples:

Objects	Morphisms
Vector spaces	Linear transformations
Algebras	Algebra homomorphisms
$A\text{-modules}$	$A\text{-module}$ homomorphisms
Sets	Functions

A simple module is a module $M$ such that if $N$ is a submodule of $M$ then

$$N=M\text{or}N=0\text{.}$$

A module is decomposable if

$$M\simeq N\oplus P$$

Definition: $N\oplus P$ has basis

$$\{n_1,\dots ,n_r,p_1,\dots ,p_5\}\text{with action}$$

$b_i{n}_j$ and $b_i{p}_k$ determined by $N$ and $P\text{.}$ Namely,

$$\begin{array}{ccc}A\otimes (N\oplus P)& ⟶& N\oplus P\\ an& ⟼& an\\ ap& ⟼& ap\text{.}\end{array}$$

Problem 1 Classify the simple modules.

Problem 2 Classify the indecomposable modules.

$$∅ ∅ 2 1 $ 2^2$+1^2=5$$

Consider $- - \text{.}$

Then

$$\begin{array}{ccccc} \hspace{0.17em}(2 - - )& =& 2 - - & =& 0\\ \hspace{0.17em}(2 - - )& =& 2 - - & =& 0\end{array}$$

So

$$\begin{array}{ccc}P& =& \text{span}\{2 - - \}\\ N& =& \text{span}\{ , \}\end{array}$$

are submodules and $M=N\oplus P\text{.}$

$P$ is a simple module since $\text{dim}\left(P\right)=1\text{.}$

Assume $Q\subseteq N$ is a submodule.

Assume $Q\ne 0\text{.}$ Let $a +b \in Q\text{.}$

Then

$$\begin{array}{ccc} ·(a +b )& =& a(q+q^{-1}) +b \\ ·(a +b )& =& a +b(q+q^{-1}) \end{array}$$

So $\in Q$ if $a(q+q^{-1})+b\ne 0$ and $\in Q$ if $a+b(q+q^{-1})\ne 0\text{.}$

If $a(q+q^{-1})+b\ne 0$ and $a+b(q+q^{-1})=0$ then $N=\text{span}\{ , \}$ and $P=\text{span}\left\{p\right\}$ with $p=p,$ $p=0,$ $p=0,$ $p=2 - -$ are simple modules.

If $a(q+q^{-1})+b=0$ and $a+b(q+q^{-1})=0$ then $b=-a(q+q^{-1})$ and $1-{(q+q^{-1})}^2=0\text{.}$

So that ${(q+q^{-1})}^2=1$ or $q^2+1+q^{-2}=0\text{.}$

So $(q+q^{-1})=\pm 1$ or $q=e^{\pm 2\pi i/3}\text{.}$

Generators and relations

Let $A$ be the algebra given by generators $e_1,e_2$ and relations

$$e_1{e}_2{e}_1=e_1,e_2{e}_1{e}_2=e_2,e_1^2=(q+q^{-1})e_1\text{and}{e}_2^2=(q+q^{-1})e_2\text{.}$$

Then $A$ contains

$$1,e_1,e_2,e_1{e}_2,e_2{e}_1$$

and the multiplication is determined.

Then

$$\begin{array}{ccc}A& ⟶& {TL}_3\\ e_1& ⟼& \\ e_2& ⟼& \end{array}$$

is an algebra isomorphism.

The Regular representation

$A$ is a vector space and $A$ acts on $A$ by multiplication.

A submodule of $A$ is a left ideal of $A\text{.}$

Example:

$${TL}_3=\text{span}\{ , , , , \}\text{.}$$

The left ideal generated by $$ is $I^{\left(1\right)}=\text{span}\{ , \}$ and $I^{\left(2\right)}=\text{span}\{ , \}$ are left ideals. Note: $I^{\left(1\right)}\simeq Q$ and $I^{\left(2\right)}\simeq Q$

Quotients

$\raisebox{1ex}{${TL}_3$}\!\left/ \!\raisebox{-1ex}{$⟨I^{\left(1\right)},I^{\left(2\right)}⟩$}\right.=\text{span}\left\{\overline{ }\right\}$ with

$$·\overline{ }=0\text{and} ·\overline{ }=0\text{.}$$

If $M$ is a module and $N$ is a submodule

$$\begin{array}{c}N\hspace{0.17em}\text{has basis}\{n_1,\dots ,n_e\}\\ M\hspace{0.17em}\text{has basis}\{n_1,\dots ,n_{\ell },p_1,\dots ,p_r\}\text{.}\end{array}$$

Then $\raisebox{1ex}{$M$}\!\left/ \!\raisebox{-1ex}{$N$}\right.$ has basis $\{{\bar{p}}_1,\dots ,{\bar{p}}_r\}$ and action

$$a={\bar{p}}_i={\overline{ap}}_i\text{where}{\bar{n}}_1=\dots {\bar{n}}_{\ell }=0\text{.}$$

The center

$$Z\left(A\right)=\{z\in A\hspace{0.17em}|\hspace{0.17em}za=az\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$

Example: Suppose

$$a +b +c +d \in Z\left(A\right)\text{.}$$

Then

$$(a(q+q^{-1})+b)+(c(q+q^{-1})+d) =(a(q+q^{-1})+c) +(b(q+q^{-1})+d)$$

so that $b=c$ and $c(q+q^{-1})+d=0\text{.}$

So that $b=c$ and $d=-c(q+q^{-1})\text{.}$

Similarly $a=-b(q+q^{-1})\text{.}$

The center

$$Z\left(A\right)=\{z\in A\hspace{0.17em}|\hspace{0.17em}az=za\hspace{0.17em}\text{for all}\hspace{0.17em}a\in A\}\text{.}$$

Example: Suppose

$$z=a +d +b +c \in Z\left(A\right)\text{.}$$

Then

$$\begin{array}{cccc}& ·z& =& (a(q+q^{-1})+b) +(d+c(q+q^{-1})) \\ =& z· & =& (a(q+q^{-1})+c) +(d+b(q+q^{-1})) \end{array}$$

so that $b=c$ and $d+c(q+q^{-1})=0\text{.}$ Also

$$\begin{array}{ccc} ·z& =& (a+c(q+q^{-1})) +(d(q+q^{-1})+b) \\ z· & =& (a+b(q+q^{-1})) +(d(q+q^{-1})+c) \end{array}$$

so that $b=c$ and $a+c(q+q^{-1})=0\text{.}$

So

$$z=-(q+q^{-1})( + )+( + )\text{.}$$

Then

$$\begin{array}{ccc}{z}^2& =& (+{(q+q^{-1})}^2-{(q+q^{-1})}^2-{(q+q^{-1})}^2+1)+\dots \\ & =& (1-{(q+q^{-1})}^2)z\text{.}\end{array}$$

So $z_{\varnothing }=\frac{1}{1-{(q+q^{-1})}^2}z$ satisfies $z_{\varnothing }^2=z_{\varnothing },$ ${(1-z_{\varnothing })}^2=1-z_{\varnothing }$ and $z_{ }+z_{ }=1\text{.}$

Semisimple algebras

An algebra is split semisimple if

$$A\simeq \underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\hat{\lambda }}}\left(ℂ\right)$$

for some index set $\hat{A}$ and some positive integers $d_{\lambda }\text{.}$ $\underset{\lambda \in \hat{A}}{⨁}{M}_{d_{\lambda }}\left(ℂ\right)$ has basis $\left\{E_{ij}^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda \in \hat{A},\hspace{0.17em}1\le i,j\le d_{\lambda }\}$ and multiplication

$$E_{ij}^{\lambda }{E}_{rs}^{\mu }={\delta }_{\lambda \mu }{E}_{is}^{\lambda }{\delta }_{jr}\text{.}$$

Example: Let

$$e_{11}^{\varnothing }= ·\frac{1}{q+q^{-1}}\text{and}{e}_{22}^{\varnothing }=z_{\varnothing }-e_{11}^{\varnothing }\text{and}{e}_{11}^{ }=z_{ }$$

so

$$(e_{11}^{\varnothing }=\frac{1}{q+q^{-1}} ,\hspace{0.17em}{e}_{22}^{\varnothing }=\frac{1}{1-{(q+q^{-1})}^2}( + )-\frac{(q+q^{-1})}{1-{(q+q^{-1})}^2} +? )$$ $$e_{12}^{\varnothing }=e_{11}^{\varnothing }\frac{1}{q+q^{-1}} e_{22}^{\varnothing },e_{21}^{\varnothing }=e_{22}^{\varnothing }\frac{1}{q+q^{-1}} e_{11}^{\varnothing }\text{.}$$

Claim:

$$\begin{array}{ccc}A& ⟶& M_2\left(ℂ\right)\oplus M_1\left(ℂ\right)\\ e_{11}^{\varnothing }& ⟼& E_{11}^{\varnothing }\\ e_{22}^{\varnothing }& ⟼& E_{22}^{\varnothing }\\ e_{12}^{\varnothing }& ⟼& E_{12}^{\varnothing }\\ e_{21}^{\varnothing }& ⟼& E_{21}^{\varnothing }\\ e_{11}^{ }& ⟼& E_{11}^{ }\end{array}$$

is an isomorphism. Here $\hat{A}=\{\varnothing , \}$ and $d_{\varnothing }=2,$ $d_{ }=1\text{.}$ Alternatively, action of ${TL}_3$ on $N$ is

$$\begin{array}{cc}\begin{array}{ccc} \left( \right)& =& (q+q^{-1}) \\ \left( \right)& =& (q+q^{-1}) \end{array}& \begin{array}{ccc} \left( \right)& =& \\ \left( \right)& =& \end{array}\end{array}$$

so

$$⟼\left(\begin{array}{cc}q+q^{-1}& 1\\ 0& 0\end{array}\right)\text{and} ⟼\left(\begin{array}{cc}0& 0\\ 1& q+q^{-1}\end{array}\right)\text{.}$$

So

$$\begin{array}{ccc}A& ⟶& M_r\left(ℂ\right)\oplus M_1\left(ℂ\right)\\ & ⟼& \left(\begin{array}{cc}\begin{array}{cc}1& 0\\ 0& 1\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}1\end{array}\end{array}\right)\\ & ⟼& \left(\begin{array}{cc}\begin{array}{cc}q+q^{-1}& 1\\ 0& 0\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}0\end{array}\end{array}\right)\\ & ⟼& \left(\begin{array}{ccc}\multicolumn{2}{c}{\begin{array}{cc}0& 0\\ 1& q+q^{-1}\end{array}}& \begin{array}{c}0\\ 0\end{array}\\ \\ \multicolumn{2}{c}{\begin{array}{cc}0& 0\end{array}}& \begin{array}{c}0\end{array}\end{array}\right)\end{array}$$

is an isomorphism.

Restriction and Inclusion

$${TL}_2=\text{span}\{ , \}\text{with}(q+q^{-1}) = · \text{.}$$

Simple modules: $ℂv_{\varnothing }$ and $ℂv_{ }$

$$v_{\varnothing }=(q+q^{-1})v_{\varnothing }\text{and} v_{ }=0\text{.}$$

Then

$$z_{\varnothing }=e_{11}^{\varnothing }=\frac{1}{q+q^{-1}} \text{and}{z}_{ }=1-z_{\varnothing }\text{.}$$ $$\begin{array}{ccc} & ⟼& \left(\begin{array}{cc}\begin{array}{cc}q+q^{-1}& 1\\ 0& 0\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}0\end{array}\end{array}\right),\end{array}\begin{array}{ccc} & ⟼& \left(\begin{array}{cc}\begin{array}{cc}0& 0\\ 1& q+q^{-1}\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}0\end{array}\end{array}\right)\text{.}\end{array}$$

Then

$$\begin{array}{ccc} & ⟼& \left(\begin{array}{cc}\begin{array}{cc}1& q+q^{-1}\\ 0& 0\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}0\end{array}\end{array}\right),\end{array}\begin{array}{ccc} & ⟼& \left(\begin{array}{cc}\begin{array}{cc}0& 0\\ q+q^{-1}& 1\end{array}& \begin{array}{c}0\\ 0\end{array}\\ \begin{array}{cc}0& 0\end{array}& \begin{array}{c}0\end{array}\end{array}\right)\text{.}\end{array}$$

So

$$\begin{array}{c}\begin{array}{ccc} + & ⟼& \left(\begin{array}{cc}1& q+q^{-1}\\ q+q^{-1}& 1\end{array}\right),\end{array}\begin{array}{ccc} + & ⟼& \left(\begin{array}{cc}q+q^{-1}& 1\\ 1& q+q^{-1}\end{array}\right),\end{array}\\ + -(q+q^{-1})( + )=\left(\begin{array}{cc}1-{(q+q^{-1})}^2& 0\\ 0& 1-{(q+q^{-1})}^2\end{array}\right)\text{.}\end{array}$$

Notes and References

This is a typed copy of handwritten notes by Arun Ram on 28/7/2008.

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