Representation Theory Lecture 1

Last update: 26 May 2013

Lecture 1

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

$V=span{b1,…,bn}.$

Example: ${TL}_{3}$ has basis

$, , , , and 3 +4 +2·3 ∈TL3.$

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 $1·a=a·1=a, fora∈A.$

Example: Define

$b1b2= (q+q-1) # of internal loops {b}_{1} {b}_{2}$

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 $\left\{ , , \right\}$ with ${b}_{m}= b {}_{m} {\left(q+{q}^{-1}\right)}^{\text{# of internal loops}}\text{.}$

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

Given $M$ define

$ρ:A⟶ End(M) a⟼aM$

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 $𝒞$

$objectsM∈𝒞, morphismsHom(M,N).$

Examples:

ObjectsMorphisms
Vector spacesLinear transformations
AlgebrasAlgebra homomorphisms
$A\text{-modules}$$A\text{-module}$ homomorphisms
SetsFunctions

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

$N=MorN=0.$

A module is decomposable if

$M≃N⊕P$

Definition: $N\oplus P$ has basis

${n1,…,nr,p1,…,p5} with action$

${b}_{i}{n}_{j}$ and ${b}_{i}{p}_{k}$ determined by $N$ and $P\text{.}$ Namely,

$A⊗(N⊕P) ⟶ N⊕P an ⟼ an ap ⟼ ap.$

Problem 1 Classify the simple modules.

Problem 2 Classify the indecomposable modules.

$\varnothing \varnothing 2 1 {2}^{2}+12=5$

Consider $- - \text{.}$

Then

$( 2 - - ) = 2 - - = 0 ( 2 - - ) = 2 - - = 0$

So

$P = span { 2 - - } N = span { , }$

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

$· ( a +b ) = a(q+q-1) +b · ( a +b ) = a +b(q+q-1)$

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

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

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

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

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

Generators and relations

Let $A$ be the algebra given by generators ${e}_{1},{e}_{2}$ and relations

$e1e2e1=e1, e2e1e2=e2, e12=(q+q-1)e1 and e22=(q+q-1)e2.$

Then $A$ contains

$1,e1,e2, e1e2, e2e1$

and the multiplication is determined.

Then

$A⟶TL3 e1⟼ e2⟼$

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:

$TL3=span { , , , , } .$

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

Quotients

${TL}_{3}}{⟨{I}^{\left(1\right)},{I}^{\left(2\right)}⟩}=\text{span}\left\{\stackrel{‾}{ }\right\}$ with

$· ‾ =0and · ‾ =0.$

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

$N has basis {n1,…,ne} M has basis { n1,…,nℓ, p1,…,pr } .$

Then $M}{N}$ has basis $\left\{{\stackrel{‾}{p}}_{1},\dots ,{\stackrel{‾}{p}}_{r}\right\}$ and action

$a=p‾i= ap‾i wheren‾1= …n‾ℓ=0.$

The center

$Z(A)= { z∈A | za=az for all a∈A } .$

Example: Suppose

$a + b + c + d ∈Z(A).$

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 $c\left(q+{q}^{-1}\right)+d=0\text{.}$

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

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

The center

$Z(A)= { z∈A | az=za for all a∈A } .$

Example: Suppose

$z=a +d +b +c ∈Z(A).$

Then

$·z = (a(q+q-1)+b) + (d+c(q+q-1)) = z· = (a(q+q-1)+c) +(d+b(q+q-1))$

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

$·z = (a+c(q+q-1)) +(d(q+q-1)+b) z· = (a+b(q+q-1)) +(d(q+q-1)+c)$

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

So

$z=-(q+q-1) ( + ) + ( + ) .$

Then

$z2 = ( +(q+q-1)2 -(q+q-1)2 -(q+q-1)2 +1 ) +… = (1-(q+q-1)2)z.$

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

Semisimple algebras

An algebra is split semisimple if

$A≃⨁λ∈A^ Mdλ^(ℂ)$

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

$Eijλ Ersμ= δλμ Eisλ δjr.$

Example: Let

$e11∅= ·1q+q-1 and e22∅=z∅- e11∅and e11 = z$

so

$( e11∅=1q+q-1 , e22∅= 11-(q+q-1)2 ( + ) - (q+q-1) 1-(q+q-1)2 + ? )$ $e12∅=e11∅ 1q+q-1 e22∅, e21∅= e22∅ 1q+q-1 e11∅.$

Claim:

$A ⟶ M2(ℂ)⊕ M1(ℂ) e11∅⟼E11∅ e22∅⟼E22∅ e12∅⟼E12∅ e21∅⟼E21∅ e11 ⟼ E11$

is an isomorphism. Here $\stackrel{^}{A}=\left\{\varnothing , \right\}$ and ${d}_{\varnothing }=2,$ ${d}_{ }=1\text{.}$ Alternatively, action of ${TL}_{3}$ on $N$ is

$( ) = (q+q-1) ( ) = (q+q-1) ( ) = ( ) =$

so

$⟼ ( q+q-11 00 ) and ⟼ ( 00 1q+q-1 ) .$

So

$A⟶ Mr(ℂ)⊕ M1(ℂ) ⟼ ( 1001 00 00 1 ) ⟼ ( q+q-1100 00 00 0 ) ⟼ ( 001q+q-1 00 00 0 )$

is an isomorphism.

Restriction and Inclusion

$TL2= span { , } with (q+q-1) = · .$

Simple modules: $ℂ{v}_{\varnothing }$ and $ℂ{v}_{ }$

$v∅= (q+q-1)v∅ and v =0.$

Then

$z∅=e11∅= 1q+q-1 and z =1-z∅.$ $⟼ ( q+q-1100 00 00 0 ) , ⟼ ( 001q+q-1 00 00 0 ) .$

Then

$⟼ ( 1q+q-100 00 00 0 ) , ⟼ ( 00q+q-11 00 00 0 ) .$

So

$+ ⟼ ( 1q+q-1 q+q-11 ) , + ⟼ ( q+q-11 1q+q-1 ) , + -(q+q-1) ( + ) = ( 1-(q+q-1)20 01-(q+q-1)2 ) .$

Notes and References

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