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 4

The Iwahori-Hecke algebra $H_k\left(q\right)$ has generators $$T_i=\begin{array}{c} i i+1 \end{array},1\le i\le k+1,$$ and relations $$T_i{T}_{i+1}{T}_i=T_{i+1}{T}_i{T}_{i+1}\text{and}{T}_i-T_i^{-1}=(q-q^{-1})\text{.}$$ If $q=1$ then $H_k\left(q\right)$ is $ℂS_k,$ where $S_k$ is the symmetric group. If $$y^{{\epsilon }_i^{\vee }}=\begin{array}{c} \end{array}$$ then $$y^{{\epsilon }_i^{\vee }}{y}^{{\epsilon }_j^{\vee }}=y^{{\epsilon }_j^{\vee }}{y}^{{\epsilon }_i^{\vee }}$$ and $$y^{{\epsilon }_1^{\vee }}\cdots y^{{\epsilon }_n^{\vee }}=T_{w_0}^2\in 𝒵\left(H_k\left(q\right)\right),$$ where $𝒵\left(H_k\left(q\right)\right)$ is the center of $H_k\left(q\right)\text{.}$

We want to study the tower $$H_1\subseteq H_2\subseteq H_3\subseteq \cdots ,$$ where $$\begin{array}{ccc}{H}_{k-1}& \hookrightarrow & H_k\\ \begin{array}{c} b \end{array}& ⟼& \begin{array}{c} b \end{array}\end{array}$$ using ${\text{Res}}_{H_{k-1}}^{H_k}$ and ${\text{Ind}}_{H_{k-1}}^{H_k}\text{.}$

A partition is a collection of boxes in a corner $$\lambda =\begin{array}{c} \end{array}=(6,6,4,1,1)\text{.}$$

Let $${\hat{H}}_k=\left\{\text{partitions with}\hspace{0.17em}k\hspace{0.17em}\text{boxes}\right\}\text{.}$$ The Bratelli diagram of the tower $H_1\subseteq H_2\subseteq \cdots$ has $$\lambda \in {\hat{H}}_k\hspace{0.17em}\text{as vertices on level}\hspace{0.17em}k$$ $$\lambda -\mu \hspace{0.17em}\text{if}\hspace{0.17em}\mu \hspace{0.17em}\text{is obtained by adding a box to}\hspace{0.17em}\lambda \text{.}$$ $$\begin{array}{c} ∅ \end{array}$$ This means $$\left\{\text{Irreducible}\hspace{0.17em}{H}_k\text{-modules}\right\}\stackrel{1-1}{⟷}{\hat{H}}_k$$ and $${\text{Res}}_{H_{k-1}}^{H_k}\left(H_k^{\lambda }\right)=\sum _{\underset{\lambda /\mu =▫}{\mu \le \lambda }}{H}_{k-1}^{\mu }\text{.}$$ Since $${\text{Hom}}_{H_k}({\text{Ind}}_{H_{k-1}}^{H_k}\left(H_{k-1}^{\mu }\right),H_k^{\lambda })={\text{Hom}}_{H_{k-1}}(H_{k-1}^{\mu },{\text{Res}}_{H_{k-1}}^{H_k}\left(H_k^{\lambda }\right))$$ and $${\text{Hom}}_{H_k}(H_k^{\lambda },H_k^{\nu })=\{\begin{array}{ll}0,& \text{if}\hspace{0.17em}\lambda \ne \nu ,\\ ℂ·\text{Id},& \text{if}\hspace{0.17em}\lambda -\nu ,\end{array}$$ we get $${\text{Ind}}_{H_{k-1}}^{H_k}\left(H_{k-1}^{\mu }\right)=\underset{\underset{\lambda /\mu =▫}{\lambda \supseteq \mu }}{⨁}{H}_k^{\lambda }\text{.}$$ Note that $$\text{dim}\left(H_k^{\lambda }\right)=\text{\# of paths from}\hspace{0.17em}\varnothing –\cdots –\lambda$$

As vector spaces $$\begin{array}{rcl}{\displaystyle H_4^{ }}& =& {\displaystyle H_3^{ }\oplus H_3^{ }}\\ & =& {\displaystyle H_2^{ }\oplus H_2^{ }\oplus H^{ }}\\ & =& {\displaystyle H_1^{ }\oplus H_1^{ }\oplus H_1^{ }\text{.}}\end{array}$$ $\text{(}{H}_1\simeq M_1\left(ℂ\right)$ which has one 1-dimensional simple module.)

A standard tableau of shape $\lambda$ is a filling $T$ of the boxes of $\lambda$ with $1,2,\dots ,k$ such that

(a)	the rows increase left to right,
(b)	the columns increase top to bottom.

There is a bijection $$\begin{array}{ccc}\left\{\text{standard tableaux of shape}\hspace{0.17em}\lambda \right\}& \stackrel{1-1}{⟷}& \{\text{paths}\hspace{0.17em}\varnothing –\cdots –\lambda \}\\ \begin{array}{c} 1 3 2 4 5 \end{array}& ↤& \varnothing –\begin{array}{c} \end{array}⟶\begin{array}{c} \end{array}⟶\begin{array}{c} \end{array}⟶\begin{array}{c} \end{array}⟶\begin{array}{c} \end{array}\end{array}$$ so that $$\text{dim}\left(H_k^{\lambda }\right)=\text{\# of standard tableaux of shape}\hspace{0.17em}\lambda \text{.}$$

The irreducible $H_k\left(q\right)\text{-modules}$ are $$H_k^{\lambda }=\text{span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{a standard tealux of shape}\hspace{0.17em}\lambda \}$$ with $H_k\text{-action}$ given by $$\begin{array}{rcl}{\displaystyle y^{{\epsilon }_i^{\vee }}{v}_T}& =& {\displaystyle q^{c\left(T\left(i\right)\right)}{v}_T,}\\ {\displaystyle T_i{v}_T}& =& {\displaystyle \frac{q-q^{-1}}{1-q^{2\left(c\left(T\left(i\right)\right)\right)-c\left(T(i+1)\right)}}{v}_T+\left(q^{-1}\frac{q-q^{-1}}{1-q^{2\left(c\left(T\left(i\right)\right)\right)-c\left(T(i+1)\right)}}{v}_{s_{i}T}\right)}\end{array}$$ where $$\begin{array}{rcl}{\displaystyle T\left(i\right)}& =& {\displaystyle box\; containing\hspace{0.17em}i\hspace{0.17em}\text{in}\hspace{0.17em}T,}\\ {\displaystyle c\left(b\right)}& =& {\displaystyle s-r,\text{if}\hspace{0.17em}b\hspace{0.17em}\text{is in row}\hspace{0.17em}r\text{, column}\hspace{0.17em}s,}\\ {\displaystyle s_{i}T}& \text{is}& {\displaystyle T\hspace{0.17em}\text{with}\hspace{0.17em}i\hspace{0.17em}\text{and}\hspace{0.17em}i+1\hspace{0.17em}\text{switched,}}\\ {\displaystyle v_{s_{i}T}}& =& {\displaystyle 0\text{if}\hspace{0.17em}{s}_{i}T\hspace{0.17em}\text{is not standard.}}\end{array}$$

$H_5^{ }$ has basis $$v_{\underset{\underset{5\hspace{0.17em}\phantom{5}}{3\hspace{0.17em}4}}{1\hspace{0.17em}2}},v_{\underset{\underset{4\hspace{0.17em}\phantom{5}}{3\hspace{0.17em}5}}{1\hspace{0.17em}2}},v_{\underset{\underset{5\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}4}}{1\hspace{0.17em}3}},v_{\underset{\underset{4\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}3}},v_{\underset{\underset{3\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}4}}$$ and $$\begin{array}{rcl}{\displaystyle T_2{v}_{\underset{\underset{4\hspace{0.17em}\phantom{5}}{3\hspace{0.17em}5}}{1\hspace{0.17em}2}}}& =& {\displaystyle \frac{q-q^{-1}}{1-q^{2(1-(-1))}}{v}_{\underset{\underset{4\hspace{0.17em}\phantom{5}}{3\hspace{0.17em}5}}{1\hspace{0.17em}2}}+(q^{-1}+\frac{q-q^{-1}}{1-q^{2(1-(-1))}})v_{\underset{\underset{4\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}3}}}\\ {\displaystyle T_2{v}_{\underset{\underset{3\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}4}}}& =& {\displaystyle \frac{q-q^{-1}}{1-q^{2(-1-(-2))}}{v}_{\underset{\underset{3\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}4}}+(q^{-1}+\frac{q-q^{-1}}{1-q^2})v_{\underset{\underset{2\hspace{0.17em}\phantom{5}}{3\hspace{0.17em}5}}{1\hspace{0.17em}4}}}\\ & =& {\displaystyle -q^{-1}{v}_{\underset{\underset{3\hspace{0.17em}\phantom{5}}{2\hspace{0.17em}5}}{1\hspace{0.17em}4}}\text{.}}\end{array}$$ Since $$\begin{array}{ccc}{H}_k& \twoheadrightarrow & {TL}_k\\ T_i-q& ⟼& e_i\end{array}$$ is a surjective homomorphism every ${TL}_k\text{-module}$ is an $H_k\text{-module.}$

The Bratelli diagram for the tower ${TL}_1\subseteq {TL}_2\subseteq \cdots$ is $$\begin{array}{c} ∅ \end{array}=\begin{array}{c} ∅ ∅ ∅ \end{array}$$

Lie algebras

A Lie algebra is a vector space $𝔤$ with a bracket $[,]:𝔤\otimes 𝔤\to 𝔤$ such that

(a)	$[x,y]=-[y,x],$ for $x,y\in 𝔤,$
(b)	$[x,[y,z]]=[[x,y],z]+[y,[x,z]],$ for $x,y\in 𝔤\text{.}$

A Lie algebra is not an algebra.

The enveloping algebra of $𝔤$ is the algebra $U𝔤$ generated by the vector space $𝔤$ with relations $$yx=xy-[x,y],$$ for $x,y\in 𝔤\text{.}$

The Lie algebra ${𝔰𝔩}_2$ $$\begin{array}{rcl}{\displaystyle {𝔰𝔩}_2}& =& {\displaystyle \{x\in M_2\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}x=0\}}\\ & =& {\displaystyle \left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a+d=0\}}\end{array}$$ with $$[x,y]=xy-yx$$ (product on the right is matrix multiplication).

The vector space ${𝔰𝔩}_2$ has basis $$e=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),f=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ and $$[e,f]=h,[h,e]=2e,[h,f]=-2f\text{.}$$ The enveloping algebra $U{𝔰𝔩}_2$ is generated by $e,f,h$ with relations $$ef=fe+h,eh=he-2e,hf=fh-2f\text{.}$$ The algebra $U{𝔰𝔩}_2$ has basis $$\left\{f^{m_1}{h}^{m_2}{e}^{m_3}\hspace{0.17em}\right|\hspace{0.17em}{m}_1,m_2,m_3\in {\mathbb{Z}}_{\ge 0}\}\text{.}$$ Note: $U{𝔰𝔩}_2$ is not far from $ℂ[\epsilon ,\phi ,\eta ],$ the algebra generated by $\epsilon ,\phi ,\eta$ with relations $$\epsilon \phi =\phi \epsilon ,\epsilon \eta =\eta \epsilon ,\eta \phi =\phi \eta \text{.}$$

$U{𝔰𝔩}_2$ is a Hopf algebra

Let $M,N$ be $U\text{-modules.}$ $$\begin{array}{c}M\hspace{0.17em}\text{has basis}\hspace{0.17em}\{m_1,\dots ,m_r\}\\ N\hspace{0.17em}\text{has basis}\hspace{0.17em}\{n_1,\dots ,n_s\}\end{array}$$ The tensor product vector space is $M\otimes N$ with basis $$\{m_i\otimes n_j\hspace{0.17em}|\hspace{0.17em}1\le i\le r,1\le j\le s\}$$ so that $\text{dim}(M\otimes N)=r·s\text{.}$ (Note $M\oplus N$ has basis $\{m_1,\dots ,m_r,n_1,\dots ,n_s\}$ and $\text{dim}(M\oplus N)=r+s\text{).}$

$U$ is a Hopf algebra means that it comes with a map $$\mathrm{\Delta }:U⟶U\otimes U,$$ the coproduct, that tells me how to make $U$ act on $M\otimes N\text{.}$

For $U=U{𝔰𝔩}_2$ this map is $$\begin{array}{rcl}{\displaystyle \mathrm{\Delta }\left(e\right)}& =& {\displaystyle e\otimes 1+1\otimes e,}\\ {\displaystyle \mathrm{\Delta }\left(f\right)}& =& {\displaystyle f\otimes 1+1\otimes f,}\\ {\displaystyle \mathrm{\Delta }\left(h\right)}& =& {\displaystyle h\otimes 1+1\otimes h\text{.}}\end{array}$$

An ${𝔰𝔩}_2\text{-module}$ is a $U{𝔰𝔩}_2\text{-module.}$

Modules for $U{𝔰𝔩}_2$

$L\left(▫\right)$ has basis $\{v_1,v_{-1}\}$ with $$\begin{array}{ccc}U{𝔰𝔩}_2& ⟶& \text{End}\left(L\left(▫\right)\right)\\ e& ⟼& \left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\\ f& ⟼& \left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)\\ h& ⟼& \left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\text{.}\end{array}$$ $L\left(▫\right)\otimes L\left(▫\right)$ has basis $\{v_1\otimes v_1,v_1\otimes v_{-1},v_{-1}\otimes v_1,v_{-1}\otimes v_{-1}\}$ and $$\begin{array}{ccc}\begin{array}{rcl}{\displaystyle e(v_1\otimes v_1)}& =& {\displaystyle 0}\\ {\displaystyle v_0=f(v_1\otimes v_1)}& =& {\displaystyle v_{-1}\otimes v_1+r_1\otimes v_{-1}}\\ {\displaystyle 2v_{-2}=f{v}_0}& =& {\displaystyle 2(v_{-1}\otimes v_{-1})}\end{array}& & \begin{array}{rcl}{\displaystyle e{v}_0}& =& {\displaystyle 2(v_1\otimes v_1)=2v_2}\\ {\displaystyle e\left(v_2\right)}& =& {\displaystyle v_0=v_1\otimes v_1}\\ {\displaystyle e(v_{-1}\otimes v_{-1})}& =& {\displaystyle v_{-1}\otimes v_1+v_1\otimes v_{-1}}\end{array}\\ \begin{array}{rcl}& \uparrow & e\\ & & v_2\\ f& ⇵& e\\ & •& v_0\\ f& ⇵& e\\ & •& v_{-2}\\ f& \downarrow \end{array}& & \begin{array}{rcl}{\displaystyle h{v}_2}& =& {\displaystyle h{v}_1\otimes v_1=v_1\otimes v_1+v_1\otimes v_1=2v_1\otimes v_1}\\ {\displaystyle h{v}_0}& =& {\displaystyle 0}\\ {\displaystyle h{v}_{-2}}& =& {\displaystyle -2v_{-1}\otimes v_{-1}}\end{array}\end{array}$$ and $$v^0=v_1\otimes v_{-1}-v_{-1}\otimes v_1$$ has $$\begin{array}{rcl}{\displaystyle e{v}^0}& =& {\displaystyle 0,}\\ {\displaystyle f{v}^0}& =& {\displaystyle 0,}\\ {\displaystyle h{v}^0}& =& {\displaystyle 0\text{.}}\end{array}$$ So $$L\left(▫\right)\otimes L\left(▫\right)=L\left(\begin{array}{c} \end{array}\right)\oplus L\left(\varnothing \right)$$ where $$\begin{array}{c}L\left(\begin{array}{c} \end{array}\right)\hspace{0.17em}\text{has basis}\hspace{0.17em}\{v_2,v_0,v_{-2}\},\\ L\left(\varnothing \right)\hspace{0.17em}\text{has basis}\hspace{0.17em}\left\{v\right\}\text{.}\end{array}$$ $L\left(\varnothing \right)\otimes L\left(▫\right)$ has basis $\{v^0\otimes v_1,v^0\otimes v_{-1}\}$ $$\begin{array}{ccc}\begin{array}{rcl}{\displaystyle e(v^0\otimes v_1)}& =& {\displaystyle 0,}\\ {\displaystyle f(v^0\otimes v_1)}& =& {\displaystyle v^0\otimes v_{-1},}\\ {\displaystyle h(v^0\otimes v_1)}& =& {\displaystyle v^0\otimes v_1,}\end{array}& & \begin{array}{rcl}{\displaystyle e(e^0\otimes v_{-1})}& =& {\displaystyle v^0\otimes v_1,}\\ {\displaystyle f(v^0\otimes v_{-1})}& =& {\displaystyle 0,}\\ {\displaystyle h(v^0\otimes v_{-1})}& =& {\displaystyle -v^0\otimes v_{-1}\text{.}}\end{array}\end{array}$$ So $$\begin{array}{ccc}L\left(\varnothing \right)\otimes L\left(▫\right)& \stackrel{\sim }{⟶}& L\left(▫\right)\\ v^0\otimes v_1& ⟼& v_1\\ v^0\otimes v_{-1}& ⟼& v_{-1}\text{.}\end{array}$$ Then $$L\left(\begin{array}{c} \end{array}\right)\otimes L\left(▫\right)\hspace{0.17em}\text{has basis}\hspace{0.17em}\left\{\begin{array}{c}{v}_2\otimes v_1,v_2\otimes v_{-1}\\ v_0\otimes v_1,v_0\otimes v_{-1}\\ v_{-2}\otimes v_1,v_{-2}\otimes v_{-1}\end{array}\right\}$$ and $$\begin{array}{rcl}{\displaystyle e(v_2\otimes v_1)}& =& {\displaystyle 0}\\ {\displaystyle v_1=f(v_2\otimes v_1)}& =& {\displaystyle v_0\otimes v_1+v_2\otimes v_{-1}}\\ {\displaystyle 2v_{-1}=f\left(v_1\right)}& =& {\displaystyle v_{-2}\otimes v_1+v_0\otimes v_{-1}+v_0\otimes v_{-1}}\\ {\displaystyle 3v_{-3}=f{v}_{-1}}& =& {\displaystyle v_{-2}\otimes v_{-1}+v_{-2}\otimes v_{-1}+v_{-2}\otimes v_{-1}\text{.}}\end{array}$$ $$\begin{array}{ccc}\begin{array}{rcl}& \uparrow & e\\ & v_3& =v_2\otimes v_1\\ f& ⇵& e\\ & v_1& \\ f& ⇵& e\\ & v_{-1}& \\ f& ⇵& e\\ & v_{-3}& =v_{-2}\otimes v_{-1}\\ f& \downarrow & \end{array}& & \begin{array}{rcl}{\displaystyle h{v}_3}& =& {\displaystyle 3v_3}\\ {\displaystyle h{v}_1}& =& {\displaystyle v_1}\\ {\displaystyle h{v}_{-1}}& =& {\displaystyle -v_1}\\ {\displaystyle h{v}_{-3}}& =& {\displaystyle -3v_3}\end{array}\end{array}$$ and if $$\begin{array}{ccc}\begin{array}{rcl}{\displaystyle v^1}& =& {\displaystyle v_0\otimes v_1-2v_2\otimes v_1}\\ {\displaystyle v^{-1}}& =& {\displaystyle 2v_{-2}\otimes v_1+v_0\otimes v_{-1}}\\ {\displaystyle -2v_0\otimes v_{-1}}& =& {\displaystyle 2v_{-2}\otimes v_1-v_0\otimes v_{-1}}\end{array}& & \begin{array}{rcl}& \uparrow & e\\ & v^1& \\ f& \downarrow & \\ & v^{-1}& \\ f& \downarrow & \end{array}\end{array}$$ So $$L\left(\begin{array}{c} \end{array}\right)\otimes L\left(▫\right)=L\left(\begin{array}{c} \end{array}\right)\oplus L\left(▫\right)\text{.}$$ In general $$L\left(\stackrel{\stackrel{k}{⏞}}{\begin{array}{c} \end{array}}\right)\otimes L\left(▫\right)=L\left(\stackrel{\stackrel{k+1}{⏞}}{\begin{array}{c} \end{array}}\right)\oplus L\left(\stackrel{\stackrel{k-1}{⏞}}{\begin{array}{c} \end{array}}\right)$$ and $$\begin{array}{ccc}U{𝔰𝔩}_2& ⟶& \text{End}\left(L\left(\stackrel{\stackrel{k}{⏞}}{\begin{array}{c} \end{array}}\right)\right)\\ e& ⟼& \left(\begin{array}{cc}0& 1\\ & 0& 2\\ & & 0& 3\\ & & & 0\\ & & & & \ddots \\ & & & & & 0& k\\ & & & & & & 0\end{array}\right)\\ f& ⟼& \left(\begin{array}{c}0\\ 1& 0\\ & 2& 0\\ & & 3& \ddots \\ & & & & 0\\ & & & & k& 0\end{array}\right)\\ h& ⟼& \left(\begin{array}{ccccccc}k& & & & & & 0\\ & k-2\\ & & k-4\\ & & & \ddots \\ & & & & -(k-4)\\ & & & & & -(k-2)\\ 0& & & & & & -k\end{array}\right)\end{array}$$ The rule for $-\otimes L\left(▫\right)$ is given by $$\begin{array}{c} ∅ ∅ ∅ \end{array}$$ So $U{𝔰𝔩}_2$ should have something to do with ${TL}_1\subseteq {TL}_2\subseteq \cdots \text{.}$

Notes and References

These are a typed copy of Lecture 4 from a series of handwritten lecture notes for the class Representation Theory given on August 19, 2008.

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