Quantum ${SL}_2$

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

Last update: 25 September 2012

Review

The Lie algebra

$${𝔰𝔩}_2=\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\mid a+d=0\}$$

with bracket $[x,y]=xy-yx,$ for $x,y\in {𝔰𝔩}_2$ is presented by generators

$$e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

and relations

$$[e,f]=h,[h,e]=2e,[h,f]=-2f\text{.}$$

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

$$\{f^{m_1}{h}^{m_2}{e}^{m_3}\mid m_1,m_2,m_3\in {\mathbb{Z}}_{\ge 0}\}\text{.}$$

If $M=\text{span}\{m_1,\dots ,m_r\}$ and $N=\{n_1,\dots ,n_s\}$ are $U{𝔰𝔩}_2$–modules, then

$$M\otimes N=\text{span}\{m_i\otimes n_j\mid 1\le i\le r,1\le j\le s\}$$

is a $U{𝔰𝔩}_2$–module, with

$$\begin{array}{ccc}e(m_i\otimes n_j)& =& e{m}_i\otimes n_j+m_i\otimes e{n}_j\\ f(m_i\otimes n_j)& =& f{m}_i\otimes n_j+m_i\otimes f{n}_j\\ h(m_i\otimes n_j)& =& h{m}_i\otimes n_j+m_i\otimes h{n}_j\text{.}\end{array}$$

The quantum group $U_q{𝔰𝔩}_2$

$U_q{𝔰𝔩}_2$ is generated by $E,F,K^{\pm 1}$ with relations

$$\begin{array}{c}KE{K}^{-1}=q^{2}E,KF{K}^{-1}=q^{-2}F,\\ EF=FE+\frac{K-K^{-1}}{q-q^{-1}}\text{.}\end{array}$$

The map $\Delta :U\to U\otimes U$ given by

$$\begin{array}{ccc}\Delta \left(E\right)& =& E\otimes K+1\otimes E,\\ \Delta \left(F\right)& =& F\otimes 1+K^{-1}\otimes F,\\ \Delta \left(K\right)& =& K\otimes K,\end{array}$$

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

$$V=L\left(▫\right)=\text{span}\{v_1,v_{-1}\}$$

with

$$\begin{array}{ccc}K{v}_1=q{v}_1,& E{v}_1=0,& F{v}_1=v_{-1},\\ K{v}_{-1}=q^{-1}{v}_{-1},& E{v}_{-1}=v_1,& F{v}_{-1}=0\text{.}\end{array}$$

So ${\rho }^{▫}:U\to \text{End}\left(L\left(▫\right)\right)$ has

$${\rho }^{▫}\left(K\right)=\left(\begin{array}{cc}q& 0\\ 0& q^{-1}\end{array}\right),{\rho }^{▫}\left(E\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),{\rho }^{▫}\left(F\right)=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$$

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

$$V^{\otimes 2}=V\otimes V=\text{span}\{v_1\otimes v_1,v_1\otimes v_{-1},v_{-1}\otimes v_1,v_{-1}\otimes v_{-1}\}$$

with

$$\begin{array}{cc}\begin{array}{ccc}E(v_1\otimes v_1)& =& 0,\\ E(v_1\otimes v_{-1})& =& v_1\otimes v_1,\\ K(v_1\otimes v_1)& =& q^2{v}_1\otimes v_1,\\ K(v_1\otimes v_{-1})& =& v_1\otimes v_{-1},\\ F(v_1\otimes v_1)& =& v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1},\\ F(v_1\otimes v_{-1})& =& v_{-1}\otimes v_{-1},\end{array}& \begin{array}{ccc}E(v_{-1}\otimes v_1)& =& q{v}_1\otimes v_1,\\ E(v_{-1}\otimes v_{-1})& =& q^{-1}{v}_1\otimes v_{-1}+v_{-1}\otimes v_1,\\ K(v_{-1}\otimes v_{-1})& =& q^{-2}{v}_{-1}\otimes v_{-1},\\ K(v_{-1}\otimes v_1)& =& v_{-1}\otimes v_1,\\ F(v_{-1}\otimes v_1)& =& q{v}_{-1}\otimes v_{-1},\\ F(v_{-1}\otimes v_{-1})& =& 0,\end{array}\end{array}$$

or, equivalently,

$$\begin{array}{ccc}({\rho }^{▫}\otimes {\rho }^{▫})\left(E\right)& =& {\rho }^{▫}\left(E\right)\otimes {\rho }^{▫}\left(K\right)+{\rho }^{▫}\left(1\right)\otimes {\rho }^{▫}\left(E\right)\\ & =& \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\otimes \left(\begin{array}{cc}q& 0\\ 0& q^{-1}\end{array}\right)+\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\otimes \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)& 1·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)\\ 0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)& 0·\left(\begin{array}{cc}q& \\ & q^{-1}\end{array}\right)\end{array}\right)+\left(\begin{array}{cc}1·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)& 0·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\\ 0·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)& 1·\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\end{array}\right)\\ & =& \left(\begin{array}{cc}\begin{array}{cc}\phantom{a}& \phantom{a}\\ \phantom{a}& \phantom{a^{-1}}\end{array}& \begin{array}{cc}q& \phantom{a}\\ \phantom{a}& q^{-1}\end{array}\\ \begin{array}{cc}\phantom{a}& \phantom{a}\\ \phantom{a}& \phantom{a}\end{array}& \begin{array}{cc}\phantom{a}& \phantom{a}\\ \phantom{a}& \phantom{a}\end{array}\end{array}\right)+\left(\begin{array}{cc}\begin{array}{cc}0& 1\\ 0& 0\end{array}& \begin{array}{cc}\phantom{a}& \phantom{a}\\ \phantom{a}& \phantom{a}\end{array}\\ \begin{array}{cc}\phantom{a}& \phantom{a}\\ \phantom{a}& \phantom{a}\end{array}& \begin{array}{cc}0& 1\\ 0& 0\end{array}\end{array}\right)=\left(\begin{array}{cccc}0& 1& q& 0\\ 0& 0& 0& q^{-1}\\ 0& 0& 0& 1\\ 0& 0& 0& 0\end{array}\right)\end{array}$$

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 $M=\text{span}\{m_1,\dots ,m_r\}$ and $N=\text{span}\{n_1,\dots ,n_s\}$ respectively then, if

$$(A\otimes B)(m_i\otimes n_j)=A{m}_i\otimes B{n}_j,$$

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

$$m_1\otimes n_1,\dots ,m_1\otimes n_s,m_2\otimes n_1,\dots ,m_2\otimes n_s,\dots ,m_r\otimes n_1,\dots ,m_r\otimes n_s$$

is

$$A\otimes B=\left(\begin{array}{cccc}{a}_{11}B& a_{12}B& \dots & a_{1r}B\\ ⋮& & & ⋮\\ a_{r1}B& \dots & & a_{rr}B\end{array}\right)$$

Decomposing $V^{\otimes 2}$

$$\begin{array}{cc}{v}_1\otimes v_{+1}& \\ \downarrow F& F(v_1\otimes v_1)=v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1}\\ v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1}& \\ \downarrow F& F(v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1})=\left[2\right]v_{-1}\otimes v_{-1}\text{.}\\ \left[2\right]v_{-1}\otimes v_{-1}\text{.}& \\ & \\ v_{-1}\otimes v_1-q{v}_1\otimes v_{-1}& \begin{array}{c}E(v_{-1}\otimes v_1-q{v}_1\otimes v_{-1})=0\\ F(v_{-1}\otimes v_1-q{v}_1\otimes v_{-1})=0\end{array}\end{array}$$

Let

$$b_1=v_1\otimes v_1,b_2=v_{-1}\otimes v_1+q^{-1}{v}_1\otimes v_{-1},b_3=v_{-1}\otimes v_{-1},b_4=v_{-1}\otimes v_1-q{v}_1\otimes v_{-1}\text{.}$$

Then

$$V^{\otimes 2}=L\left(▫\right)\otimes L\left(▫\right)=L\left( ▫ ▫ \right)\oplus L\left(\varnothing \right)$$

where

$$L\left( ▫ ▫ \right)=\text{span}\{b_1,b_2,b_3\}\text{and}L\left(\varnothing \right)=\text{span}\left\{b_4\right\}$$

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

$${\rho }^{ ▫ ▫ \oplus \varnothing }\left(E\right)=\left(\begin{array}{cc}\begin{array}{ccc}0& \left[2\right]& \\ & 0& 1\\ & & 0\end{array}& \\ & 0\end{array}\right),{\rho }^{ ▫ ▫ \oplus \varnothing }\left(F\right)=\left(\begin{array}{cc}\begin{array}{ccc}0& & \\ 1& 0& \\ & \left[2\right]& 0\end{array}& \\ & 0\end{array}\right),{\rho }^{ ▫ ▫ \oplus \varnothing }\left(K\right)=\left(\begin{array}{cc}\begin{array}{ccc}{q}^2& & \\ & q^0& \\ & & q^{-2}\end{array}& \\ & 1\end{array}\right)$$

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

$$L\left(▫\right)\otimes L\left(▫\right)\otimes L\left(▫\right)=\text{span}\left\{\begin{array}{cc}{v}_1\otimes v_1\otimes v_1,& v_1\otimes v_1\otimes v_{-1}\\ v_1\otimes v_{-1}\otimes v_1,& v_1\otimes v_{-1}\otimes v_{-1}\\ v_{-1}\otimes v_1\otimes v_1,& v_{-1}\otimes v_1\otimes v_{-1}\\ v_{-1}\otimes v_{-1}\otimes v_1,& v_{-1}\otimes v_{-1}\otimes v_{-1}\end{array}\right\}$$

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

$$\{b_1\otimes v_1,b_1\otimes v_{-1},b_2\otimes v_1,b_2\otimes v_{-1},b_3\otimes v_1,b_3\otimes v_{-1},b_4\otimes v_1,b_4\otimes v_{-1}\},$$

i.e.

$$V^{\otimes 3}=(L\left( ▫ ▫ \right)\oplus L\left(\varnothing \right))\otimes V=(L\left( ▫ ▫ \right)\otimes V)\oplus (L\left(\varnothing \right)\otimes V)$$

$L\left(\varnothing \right)\otimes V=\text{span}\{b_4\otimes v_1,b_4\otimes v_{-1}\}$ with

$$\begin{array}{ccc}E(b_4\otimes v_1)=0,& F(b_4\otimes v_1)=b_4\otimes v_{-1},& K(b_4\otimes v_1)=q{b}_4\otimes v_1\\ E(b_4\otimes v_{-1})=b_4\otimes v_1,& F(b_4\otimes v_{-1})=0,& K(b_4\otimes v_{-1})=q^{-1}{b}_4\otimes v_{-1},\end{array}$$

So

$$\begin{array}{ccc}L\left(\varnothing \right)\otimes V& \cong & L\left(▫\right)\\ b_4\otimes v_1& ⟼& v_1\\ b_4\otimes v_{-1}& ⟼& v_{-1}\end{array}$$

Then

$$L\left( ▫ ▫ \right)\otimes V=\text{span}\{b_1\otimes v_1,b_1\otimes v_{-1},b_2\otimes v_1,b_2\otimes v_{-1},b_3\otimes v_1,b_3\otimes v_{-1}\}$$

with

$$\begin{array}{cc}{b}_1\otimes v_1& \\ \downarrow F& F(b_1\otimes v_1)=b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1}\\ b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1}& \\ \downarrow F& \\ \left[2\right]b_3\otimes v_1+b_2\otimes v_{-1}+q^{-2}{b}_2\otimes v_{-1}+q^{-4}{b}_2\otimes 0=\left[2\right](b_3\otimes v_1+q^{-1}{b}_2\otimes v_{-1})\\ \downarrow F& \\ 0+\left[2\right]q^2{b}_3\otimes v_{-1}+\left[2\right]b_3\otimes v_{-1}+0+q^{-2}\left[2\right]b_3\otimes v_{-1}+0=\left[2\right]\left[3\right]b_3\otimes v_{-1}\end{array}$$

So, if

$$c_1=b_1\otimes v_1,c_2=b_2\otimes v_1+q^{-2}{b}_1\otimes v_{-1},c_3=b_3\otimes v_1+q^{-1}{b}_2\otimes v_{-1},c_4=b_3\otimes v_{-1}$$

then the action of $F$ on

$$L\left( ▫ ▫ ▫ \right)=\text{span}\{c_1,c_2,c_3,c_4\}\text{is given by}$$ $${\rho }^{ ▫ ▫ ▫ }\left(F\right)=\left(\begin{array}{c}0\\ 1& 0\\ & \left[2\right]& 0\\ & & \left[3\right]& 0\end{array}\right),{\rho }^{ ▫ ▫ ▫ }\left(E\right)=\left(\begin{array}{cc}0& 1\\ & 1& \left[2\right]\\ & & 1& \left[3\right]\\ & & & 0\end{array}\right),{\rho }^{ ▫ ▫ ▫ }\left(K\right)=\left(\begin{array}{c}{q}^3\\ & q^1\\ & & q^{-1}\\ & & & q^{-3}\end{array}\right)$$

Note that

$$\begin{array}{c}0\\ \uparrow E\\ b_2\otimes v_1-q{b}_1\otimes v_{-1}\\ F\downarrow \\ \left[2\right]b_3\otimes v_1+b_2\otimes v_{-1}-q\left[2\right]b_2\otimes v_{-1}=\left[2\right]b_3\otimes v_{-1}-q^2{b}_2\otimes v_{-1}\\ F\downarrow \\ \left[2\right]q^2{b}_3\otimes v_{-1}-q^2\left[2\right]b_3\otimes v_{-1}=0\end{array}$$

So that, if

$$c_5=b_2\otimes v_1-q{b}_1\otimes v_{-1},c_6=\left[2\right]b_3\otimes v_{-1}-q^2{b}_2\otimes v_{-1}\text{and}\stackrel{\sim }{L}\left(▫\right)=\text{span}\{c_5,c_6\}$$

then

$$\stackrel{\sim }{L}\left(▫\right)\cong L\left(▫\right)$$

So

$$\begin{array}{ccc}{V}^{\otimes 3}& =& L\left(▫\right)\otimes L\left(▫\right)\otimes L\left(▫\right)\cong (L\left( ▫ ▫ \right)\otimes L\left(\varnothing \right))\otimes L\left(▫\right)\\ & =& (L\left( ▫ ▫ \right)\otimes V)\oplus (L\left(\varnothing \right)\otimes V)\\ & \cong & L\left( ▫ ▫ ▫ \right)\oplus L\left(▫\right)\oplus L\left(▫\right)\end{array}$$ $$∅ ∅ ∅ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ ▫ 1·2=2 1·1+1·3=4 2·2+1·4=8 2·1+3·3+1·5=16 5·2+4·4+1·6=32 1 2 1 2 5 1 3 1 4 1 1$$

What is the connection between $T{L}_k$ and $V^{\otimes k}$ for $U_q{𝔰𝔩}_2$?

Define an action of $T{L}_2=\text{span}\{,\}$ on

$$V^{\otimes 2}=\text{span}\{v_1\otimes v_1,v_1\otimes v_{-1},v_{-1}\otimes v_1,v_{-1}\otimes v_{-1}\}$$

by

$$(v_1\otimes v_1)=0,(v_{-1}\otimes v_{-1})=0,$$ $$(v_1\otimes v_{-1})=q{v}_1\otimes -v_{-1}\otimes v_1,(v_{-1}\otimes v_1)=q^{-1}{v}_{-1}\otimes v_1-v_1\otimes v_{-1}$$

In matrices we have

$${\rho }^{\otimes 2}\left(\right)=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& q& -1& 0\\ 0& -1& q^{-1}& 0\\ 0& 0& 0& 0\end{array}\right)\text{.}$$

Note that

$${\left({\rho }^{\otimes 2}\left(\right)\right)}^2=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& q^2+1& -(q+q^{-1})& 0\\ 0& -(q+q^{-1})& 1-q^{-2}& 0\\ 0& 0& 0& 0\end{array}\right)=\left[2\right]{\rho }^{\otimes 2}\left(\right)$$

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.

$${\rho }^{\otimes 2}\left(T{L}_2\right)\subseteq {\text{End}}_U\left(V^{\otimes 2}\right)$$

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

$$\begin{array}{ccc}{e}_j=& 1 2 … j j+1 … k ,& 1\le j\le k-1\text{.}\end{array}$$

Define an action of $T{L}_k$ on

$$V^{\otimes k}=\text{span}\{v_{i_1}\otimes \dots \otimes V_{i_k}\mid i_1,\dots ,i_k\in \{\pm 1\}\}$$

by

$$e_j(v_{i_1}\otimes \dots \otimes v_{i_k})=v_{i_1}\otimes \dots \otimes v_{i_{j-1}}\otimes (v_{i_j}\otimes v_{i_{j+1}})\otimes v_{i_{j+2}}\otimes \dots \otimes v_{i_k}$$

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,

$$M=\underset{\lambda \in \hat{A}}{⨁}{\left(A^{\lambda }\right)}^{\oplus m_{\lambda }}\text{.}$$

Let $Z={\text{End}}_A\left(M\right)\text{.}$ Then

$$Z=\underset{\lambda \in \hat{M}}{⨁}{M}_{m_{\lambda }}\left(ℂ\right),\text{and}M\cong \underset{\lambda \in \hat{M}}{⨁}{A}^{\lambda }\otimes Z^{\lambda }$$

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

		Proof.
	$$\begin{array}{ccc}Z& =& {\text{Hom}}_A(M,M)={\text{Hom}}_A(\underset{\lambda \in \hat{M}}{⨁}\underset{i=1}{\overset{m_{\lambda }}{⨁}}{A}_i^{\lambda },\underset{\mu \in \hat{M}}{⨁}\underset{j=1}{\overset{m_{\lambda }}{⨁}}{A}_j^{\mu })\\ & =& \underset{\lambda \in \hat{M}}{⨁}\underset{i,j=1}{\overset{m_{\lambda }}{⨁}}{\text{Hom}}_A(A_i^{\lambda },A_j^{\lambda }),\end{array}$$ by Schur's Lemma. Hence $$Z=\text{span}\{e_{ij}^{\lambda }\mid \lambda \in \hat{M},1\le i,j\le m_{\lambda }\}\text{where}{e}_{ij}^{\lambda }:A_i^{\lambda }\to A_j^{\lambda }$$ (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 $e_{ij}^{\lambda }{e}_{ji}^{\lambda }=e_{ii}^{\lambda })\text{.}$ $\square$

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