## Quantum ${SL}_{2}$

Last update: 25 September 2012

## Review

The Lie algebra

$𝔰𝔩2= { ( a b c d ) ∣a+d=0 }$

with bracket $\left[x,y\right]=xy-yx,$ for $x,y\in {𝔰𝔩}_{2}$ is presented by generators

$e= ( 01 00 ) , f= ( 00 10 ) , h= ( 10 0-1 )$

and relations

$[e,f]=h, [h,e]=2e, [h,f]=-2f.$

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

${ fm1hm2 em3∣ m1,m2,m3∈ ℤ≥0 } .$

If $M=\text{span}\phantom{\rule{0.2em}{0ex}}\left\{{m}_{1},\dots ,{m}_{r}\right\}$ and $N=\left\{{n}_{1},\dots ,{n}_{s}\right\}$ are $U{𝔰𝔩}_{2}$–modules, then

$M⊗N=span { mi⊗nj∣ 1≤i≤r,1≤j≤s }$

is a $U{𝔰𝔩}_{2}$–module, with

$e(mi⊗nj) = emi⊗nj+mi⊗ enj f(mi⊗nj) = fmi⊗nj+mi⊗ fnj h(mi⊗nj) = hmi⊗nj+mi⊗ hnj.$

## The quantum group ${U}_{q}{𝔰𝔩}_{2}$

${U}_{q}{𝔰𝔩}_{2}$ is generated by $E,F,{K}^{±1}$ with relations

$KEK-1=q2E, KFK-1=q-2F, EF=FE+ K-K-1 q-q-1 .$

The map $\Delta :\phantom{\rule{0.2em}{0ex}}U\to U\otimes U$ given by

$Δ(E) = E⊗K+1⊗E, Δ(F) = F⊗1+K-1⊗F, Δ(K) = K⊗K,$

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

$V=L(▫)=span {v1,v-1}$

with

$Kv1=qv1, Ev1=0, Fv1=v-1, Kv-1=q-1v-1, Ev-1=v1, Fv-1=0.$

So ${\rho }^{▫}:\phantom{\rule{0.2em}{0ex}}U\to \text{End}\phantom{\rule{0.2em}{0ex}}\left(L\left(▫\right)\right)$ has

$ρ▫(K)= ( q0 0q-1 ) , ρ▫(E)= ( 01 00 ) , ρ▫(F)= ( 00 10 )$

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

$V⊗2=V⊗V= span { v1⊗v1, v1⊗v-1, v-1⊗v1, v-1⊗v-1 }$

with

$E(v1⊗v1) = 0, E(v1⊗v-1) = v1⊗v1, K(v1⊗v1) = q2v1⊗v1, K(v1⊗v-1) = v1⊗v-1, F(v1⊗v1) = v-1⊗v1+ q-1v1⊗v-1, F(v1⊗v-1) = v-1⊗v-1, E(v-1⊗v1) = qv1⊗v1, E(v-1⊗v-1) = q-1v1⊗v-1 +v-1⊗v1, K(v-1⊗v-1) = q-2v-1⊗ v-1, K(v-1⊗v1) = v-1⊗v1, F(v-1⊗v1) = qv-1⊗v-1, F(v-1⊗v-1) = 0,$

or, equivalently,

$( ρ▫⊗ ρ▫ ) (E) = ρ▫(E)⊗ ρ▫(K)+ ρ▫(1)⊗ ρ▫(E) = ( 01 00 ) ⊗ ( q0 0q-1 ) + ( 10 01 ) ⊗ ( 01 00 ) = ( 0· ( q q-1 ) 1· ( q q-1 ) 0· ( q q-1 ) 0· ( q q-1 ) ) + ( 1· ( 01 00 ) 0· ( 01 00 ) 0· ( 01 00 ) 1· ( 01 00 ) ) = ( aa aa-1 qa aq-1 aa aa aa aa ) + ( 01 00 aa aa aa aa 01 00 ) = ( 01q0 000q-1 0001 0000 )$

In general, if $A=\left(\begin{array}{ccc}{a}_{11}& \dots & {a}_{1r}\\ ⋮& & \\ {a}_{r1}& \dots & {a}_{rr}\end{array}\right)$ and $B=\left(\begin{array}{ccc}{b}_{11}& \dots & {b}_{1s}\\ ⋮& & \\ {b}_{s1}& \dots & {b}_{ss}\end{array}\right)$ acting on $M=\text{span}\phantom{\rule{0.2em}{0ex}}\left\{{m}_{1},\dots ,{m}_{r}\right\}$ and $N=\text{span}\phantom{\rule{0.2em}{0ex}}\left\{{n}_{1},\dots ,{n}_{s}\right\}$ respectively then, if

$(A⊗B) (mi⊗nj)= Ami⊗Bnj,$

then the matrix of $A\otimes B$ in the basis

$m1⊗n1,…, m1⊗ns, m2⊗n1,…, m2⊗ns,…, mr⊗n1,…, mr⊗ns$

is

$A⊗B= ( a11B a12B … a1rB ⋮ ⋮ ar1B … arrB )$

## Decomposing ${V}^{\otimes 2}$

$v1⊗v+1 ↓F F(v1⊗v1)= v-1⊗v1+ q-1v1⊗ v-1 v-1⊗v1+ q-1v1⊗v-1 ↓F F ( v-1⊗v1+ q-1v1⊗v-1 ) =[2]v-1⊗v-1. [2]v-1⊗v-1. v-1⊗v1-q v1⊗v-1 E ( v-1⊗v1-q v1⊗v-1 ) =0 F ( v-1⊗v1-q v1⊗v-1 ) =0$

Let

$b1=v1⊗v1, b2=v-1⊗v1 +q-1v1⊗v-1, b3=v-1⊗ v-1,b4= v-1⊗v1-q v1⊗v-1.$

Then

$V⊗2=L(▫) ⊗L(▫)=L ( ▫ ▫ ) ⊕L(∅)$

where

$L ( ▫ ▫ ) =span{b1,b2,b3} andL(∅)= span{b4}$

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

$ρ ▫ ▫ ⊕∅ (E)= ( 0[2] 01 0 0 ) , ρ ▫ ▫ ⊕∅ (F)= ( 0 10 [2]0 0 ) , ρ ▫ ▫ ⊕∅ (K)= ( q2 q0 q-2 1 )$

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

$L(▫)⊗ L(▫)⊗ L(▫)= span { v1⊗v1⊗v1, v1⊗v1⊗ v-1 v1⊗v-1⊗ v1, v1⊗v-1 ⊗v-1 v-1⊗v1 ⊗v1, v-1⊗v1⊗ v-1 v-1⊗v-1 ⊗v1, v-1⊗v-1 ⊗v-1 }$

Another basis of ${V}^{\otimes 3}$ is

${ b1⊗v1, b1⊗v-1, b2⊗v1, b2⊗v-1, b3⊗v1, b3⊗v-1, b4⊗v1, b4⊗v-1 } ,$

i.e.

$V⊗3= ( L ( ▫ ▫ ) ⊕L(∅) ) ⊗V= ( L ( ▫ ▫ ) ⊗V ) ⊕ ( L(∅)⊗V )$

$L\left(\varnothing \right)\otimes V=\text{span}\phantom{\rule{0.2em}{0ex}}\left\{{b}_{4}\otimes {v}_{1},{b}_{4}\otimes {v}_{-1}\right\}$ with

$E(b4⊗v1)=0, F(b4⊗v1)= b4⊗v-1, K(b4⊗v1)= qb4⊗v1 E(b4⊗v-1)= b4⊗v1, F(b4⊗v-1) =0, K(b4⊗v-1) =q-1b4⊗v-1,$

So

$L(∅)⊗V ≅ L(▫) b4⊗v1 ⟼ v1 b4⊗v-1 ⟼ v-1$

Then

$L ( ▫ ▫ ) ⊗V=span { b1⊗v1, b1⊗v-1, b2⊗v1, b2⊗v-1, b3⊗v1, b3⊗v-1 }$

with

$b1⊗v1 ↓F F(b1⊗v1)= b2⊗v1+ q-2b1⊗ v-1 b2⊗v1+q-2 b1⊗v-1 ↓F [2]b3⊗v1+ b2⊗v-1+ q-2b2⊗ v-1+q-4b2 ⊗0=[2] ( b3⊗v1+ q-1b2⊗v-1 ) ↓F 0+[2]q2b3⊗ v-1+[2]b3 ⊗v-1+0+q-2 [2]b3⊗v-1 +0=[2][3]b3 ⊗v-1$

So, if

$c1=b1⊗v1, c2=b2⊗v1+ q-2b1⊗v-1, c3=b3⊗v1 +q-1b2⊗v-1 ,c4=b3⊗v-1$

then the action of $F$ on

$L ( ▫ ▫ ▫ ) =span {c1,c2,c3,c4} is given by$ $ρ ▫ ▫ ▫ (F)= ( 0 10 [2]0 [3]0 ) , ρ ▫ ▫ ▫ (E)= ( 01 1[2] 1[3] 0 ) , ρ ▫ ▫ ▫ (K)= ( q3 q1 q-1 q-3 )$

Note that

$0 ↑E b2⊗v1-qb1⊗v-1 F↓ [2]b3⊗v1 +b2⊗v-1- q[2]b2⊗ v-1=[2] b3⊗v-1- q2b2⊗v-1 F↓ [2]q2b3⊗ v-1-q2 [2]b3⊗v-1 =0$

So that, if

$c5=b2⊗v1- qb1⊗v-1, c6=[2]b3⊗ v-1-q2b2⊗ v-1and L∼(▫)=span {c5,c6}$

then

$L∼(▫)≅ L(▫)$

So

$V⊗3 = L(▫)⊗ L(▫)⊗ L(▫)≅ ( L ( ▫ ▫ ) ⊗L(∅) ) ⊗L(▫) = ( L ( ▫ ▫ ) ⊗V ) ⊕(L(∅)⊗V) ≅ L ( ▫ ▫ ▫ ) ⊕L(▫)⊕ L(▫)$ $∅ ∅ ∅ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ 1·2=2 1·1+1·3=4 2·2+1·4=8 2·1+3·3+1·5=16 5·2+4·4+1·6=32 1 2 1 2 5 1 3 1 4 1 1$

## What is the connection between $T{L}_{k}$ and ${V}^{\otimes k}$ for ${U}_{q}{𝔰𝔩}_{2}$?

Define an action of $T{L}_{2}=\text{span}\phantom{\rule{0.2em}{0ex}}\left\{,\right\}$ on

$V⊗2=span { v1⊗v1, v1⊗v-1, v-1⊗v1, v-1⊗v-1 }$

by

$(v1⊗v1)=0, (v-1⊗v-1)=0,$ $(v1⊗v-1)=qv1⊗ -v-1⊗v1, (v-1⊗v1)=q-1 v-1⊗v1-v1⊗v-1$

In matrices we have

$ρ⊗2 () = ( 0000 0q-10 0-1q-10 0000 ) .$

Note that

$( ρ⊗2 () ) 2 = ( 0000 0q2+1-(q+q-1)0 0-(q+q-1)1-q-20 0000 ) =[2]ρ⊗2 ()$

so that this is an action of $T{L}_{2}$ on ${V}^{\otimes 2}\text{.}$

The $T{L}_{2}$ action commutes with the ${U}_{q}{𝔰𝔩}_{2}$ action on ${V}^{\otimes k},$ i.e.

$ρ⊗2 (TL2)⊆EndU (V⊗2)$

The Temperley-Lieb algebra $T{L}_{k}$ is generated by

$ej= 1 2 … j j+1 … k , 1≤j≤k-1.$

Define an action of $T{L}_{k}$ on

$V⊗k=span { vi1⊗…⊗ Vik∣ i1,…,ik∈ {±1} }$

by

$ej ( vi1⊗…⊗ vik ) =vi1⊗…⊗ vij-1⊗ ( vij⊗ vij+1 ) ⊗vij+2 ⊗…⊗vik$

Claim

1. This defines a $T{L}_{k}$–action on ${V}^{\otimes k}$
2. This $T{L}_{k}$–action commutes with the ${U}_{q}{𝔰𝔩}_{2}$–action on ${V}^{\otimes k}\text{.}$

Let $A$ be an algebra and let $M$ be a semisimple $A$ module,

$M=⨁λ∈A^ (Aλ)⊕mλ .$

Let $Z={\text{End}}_{A}\phantom{\rule{0.2em}{0ex}}\left(M\right)\text{.}$ Then

$Z=⨁λ∈M^ Mmλ(ℂ),and M≅⨁λ∈M^ Aλ⊗Zλ$

as an $\left(A,Z\right)$ bimodule, where $\stackrel{^}{M}\subseteq \stackrel{^}{A}$ is an index set for the simple $A$–modules appearing in $M\text{.}$

 Proof. $Z = HomA(M,M)= HomA ( ⨁λ∈M^ ⨁i=1mλ Aiλ, ⨁μ∈M^ ⨁j=1mλ Ajμ ) = ⨁λ∈M^ ⨁i,j=1mλ HomA ( Aiλ,Ajλ ) ,$ by Schur's Lemma. Hence $Z=span { eijλ∣ λ∈M^,1≤ i,j≤mλ } whereeijλ :Aiλ→ Ajλ$ (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 ${e}_{ij}^{\lambda }{e}_{ji}^{\lambda }={e}_{ii}^{\lambda }\right)\text{.}$ $\square$