Representation theory Lecture Notes: Chapter 2

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

Last update: 29 October 2013

The groups $G(r,p,n)$

Let $r,p,n$ be positive integers such that $p$ divides $r\text{.}$ The group $G(r,p,n)$ is the group of $n\times n$ matrices such that

(a)	There is exactly one nonzero entry in each row and each column,
(b)	The nonzero entries are $r\text{th}$ roots of unity,
(c)	The $(r/p)\text{th}$ power of the product of the nonzero entries is $1\text{.}$

The group $G(r,1,n)$ has order $r^{n}n!,$ since condition (c) is trivially satisfied, there are $r$ choices for each of $n$ roots of unity and there are $n!$ permutation matrices. The group $G(r,p,n)$ is the normal subgroup of $G(r,1,n)$ of index $p$ given by the exact sequence $$\begin{array}{cccccccc}1& G(r,p,n)& \hookrightarrow & G(r,1,n)& ⟶& \mathbb{Z}/p\mathbb{Z}& ⟶& 1\\ & & & t_{\lambda }w& ⟼& {\lambda }_1+\cdots +{\lambda }_n\end{array}$$ Some special cases of these groups are:

(1)	$G(r,1,1)=\mathbb{Z}/r\mathbb{Z},$ the cyclic group of order $r,$
(2)	$G(r,r,2)=W{I}_r,$ the dihedral group of order $2r,$
(3)	$G(1,1,n)=S_n=W{A}_{n-1},$ the symmetric group of $n\times n$ permutation matrices,
(4)	$G(2,1,n)=W{B}_n,$ the hyperoctahedral group, or Weyl group of type $B_n,$
(5)	$G(2,2,n)=W{D}_n,$ the Weyl group of type $D_n,$
(6)	$G(r,1,n)=(\mathbb{Z}/r\mathbb{Z})\wr S_n={(\mathbb{Z}/r\mathbb{Z})}^n⋉S_n,$ where $S_n$ acts on ${(\mathbb{Z}/r\mathbb{Z})}^n$ by permuting the factors.

Let $\mathbb{Z}/r\mathbb{Z}=\{0,1,\dots ,r-1\}$ and let $\xi =e^{2\pi i/r}\text{.}$ For each $\lambda =({\lambda }_1,\dots ,{\lambda }_n)\in \mathbb{Z}/r\mathbb{Z},$ let $t_{\lambda }$ be the diagonal matrix with diagonal entries ${\left(t_{\lambda }\right)}_{ii}={\xi }^{{\lambda }_i}\text{.}$ Then $$\begin{array}{ccc}G(r,1,n)& =& \left\{t_{\lambda }\sigma \hspace{0.17em}\right|\hspace{0.17em}\lambda \in {(\mathbb{Z}/r\mathbb{Z})}^n,\sigma \in S_n\}\text{and}\\ G(r,p,n)& =& \{t_{\lambda }\sigma \in G(r,1,n)\hspace{0.17em}|\hspace{0.17em}{\lambda }_1+\cdots +{\lambda }_n=0\hspace{0.17em}\text{mod}\hspace{0.17em}p\}\text{.}\end{array}$$ where the multiplication in $G(r,1,n)$ is determined by the relations $$\begin{array}{ccc}{t}_{\lambda }{t}_{\mu }& =& t_{\lambda +\mu },\lambda ,\mu \in {(\mathbb{Z}/r\mathbb{Z})}^n,\\ w{t}_{\lambda }& =& t_{w\lambda }w,\lambda \in {(\mathbb{Z}/r\mathbb{Z})}^n,w\in S_n,\end{array}$$ where $w\lambda =({\lambda }_{w\left(1\right)},\dots ,{\lambda }_{w\left(n\right)})$ if $\lambda =({\lambda }_1,\dots ,{\lambda }_n)\text{.}$

The groups $G(r,p,n)$ are complex reflection groups. The reflections in $G(r,1,n)$ are the elements $$\begin{array}{c}{t}_i^k{t}_j^{-k}(i,j),k\in (\mathbb{Z}/r\mathbb{Z}),1\le i<j\le n,\text{and}\\ t_i^k=t_{(0,\dots ,0,k,0,\dots ,0)}=\begin{array}{c} ξk ith ⋯ ⋯ \end{array},k\in (\mathbb{Z}/r\mathbb{Z}),1\le i\le n\text{.}\end{array}$$ The reflections in $G(r,p,n)$ are those reflections in $G(r,1,n)$ which are also in $G(r,p,n)\text{.}$ These are $$t_i^{kp},0\le k\le (r/p)-1,\text{and all}{t}_i^k{t}_j^{-k}(i,j),k\in (\mathbb{Z}/r\mathbb{Z}),1\le i<j\le n\text{.}$$

Define $$\begin{array}{c}{t}_1=\begin{array}{c} ξ ⋯ \end{array},s_1=\begin{array}{c} ξ ξ-1 ⋯ \end{array},\\ s_i=(i-1,i)=\begin{array}{c} ⋯ ⋯ \end{array},2\le i\le n\text{.}\end{array}$$

The group $G(r,1,n)$ can be presented by generators $t_1,s_1,\dots ,s_{n-1}$ and relations $$\begin{array}{cccc}{s}_i{s}_j& =& s_j{s}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ s_i{s}_{i+1}{s}_i& =& s_{i+1}{s}_i{s}_{i+1},& 2\le i\le n-1,\\ t_1{s}_1{t}_1{s}_1& =& s_1{t}_1{s}_1{t}_1,\\ \multicolumn{3}{c}{t_1^r=1\text{and}{s}_i^2=1,}& 2\le i\le n\text{.}\end{array}$$ (b) The group $G(r,p,n)$ has a presentation by generators $t_1^p,s_1,s_2,\dots ,s_n$ and relations $$\begin{array}{ccc}{t}_1^{r/p}& =& 1\text{and}{s}_i^2=1,1\le i\le n,\\ s_i{s}_j& =& s_j{s}_i,\text{for}\hspace{0.17em}|i-j|>1\hspace{0.17em}\text{and}\hspace{0.17em}i,j\ge 2,\\ s_i{s}_{i+1}{s}_i& =& s_{i+1}{s}_i{s}_{i+1},\text{for}\hspace{0.17em}2\le i\le n-1,\\ s_1{s}_3{s}_1& =& s_3{s}_1{s}_3,\text{and}{s}_1{s}_j=s_j{s}_1,\text{for}\hspace{0.17em}j>3,\\ t_1^p{s}_2{t}_1^p{s}_2& =& s_2{t}_1^p{s}_2{t}_1^p,\text{and}{s}_0{t}_1^p=t_1^p{s}_0,\text{for}\hspace{0.17em}j>2,\\ \underset{\underset{2(r/p)\hspace{0.17em}\text{factors}}{⏟}}{t_1^p{s}_1{t}_1^p{s}_1\cdots }& =& \underset{\underset{2(r/p)\hspace{0.17em}\text{factors}}{⏟}}{s_1{t}_1^p{s}_1{t}_1^p\cdots },\text{and}\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{s_1{s}_2{s}_1\cdots }=\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{s_2{s}_1{s}_2\cdots },\end{array}$$

Note that only the groups ???? can be generated by $n$ reflections.

Let $\lambda =({\lambda }^{\left(0\right)},{\lambda }^{\left(1\right)},\dots ,{\lambda }^{(r-1)})$ be a $r\text{-tuple}$ of partitions with $n$ boxes total. A standard tableau of shape $\lambda$ is a filling of the boxes of $\lambda$ with $1,2,\dots ,n$ such that, in each partition ${\lambda }^{\left(j\right)},$

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

The rows and columns of each partition ${\lambda }^{\left(j\right)}$ are numbered as for matrices and $$\begin{array}{c}T\left(i\right)\hspace{0.17em}\text{is the box containing}\hspace{0.17em}i\hspace{0.17em}\text{in}\hspace{0.17em}T,\\ c\left(b\right)=j-i,\hspace{0.17em}\text{if}\hspace{0.17em}b\hspace{0.17em}\text{is in position}\hspace{0.17em}(i,j)\text{, and}\\ s\left(b\right)={\xi }^j,\hspace{0.17em}\text{if}\hspace{0.17em}b\hspace{0.17em}\text{is in}\hspace{0.17em}{\lambda }^{\left(j\right)},\end{array}$$ where $\xi =e^{2\pi i/r}\text{.}$ The numbers $c\left(b\right)$ and $s\left(b\right)$ are the content and the sign of the box $b,$ respectively.

(a)	The irreducible representations $S^{\lambda }$ of the group $G(r,1,n)$ are indexed by $r\text{-tuples}$ of partitions $\lambda =({\lambda }^{\left(0\right)},\dots ,{\lambda }^{(r-1)})$ with $n$ boxes total.
(b)	$\text{dim}\hspace{0.17em}{S}^{\lambda }=$ # of standard tableaux of shape $\lambda \text{.}$
(c)	The irreducible $G(r,1,n)\text{-module}$ $S^{\lambda }$ is given by $$S^{\lambda }=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}\lambda \}$$ with basis $\left\{v_T\right\}$ and with $G(r,1,n)$ action given by $$\begin{array}{ccc}{s}_1{v}_T& =& s\left(T\left(1\right)\right)v_T,\\ s_i{v}_T& =& {\left(s_i\right)}_{TT}{v}_T+(1+{\left(s_i\right)}_{TT})v_{s_{i}T},i=2,3,\dots ,n,\end{array}$$ where $${\left(s_i\right)}_{TT}=\{\begin{array}{cc}\frac{1}{c\left(T\left(i\right)\right)-c\left(T(i-1)\right)},& \text{if}\hspace{0.17em}s\left(T\left(i\right)\right)=s\left(T(i-1)\right),\\ 0,& \text{if}\hspace{0.17em}s\left(T\left(i\right)\right)\ne s\left(T(i-1)\right),\end{array}$$ $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$ $s_{i}T$ is the same as $T$ except that $i$ and $i-1$ are switched, $v_{s_{i}T}=0$ if $s_{i}T$ is not standard.

For each $0\le k\le r-1$ the elements $t_i^k,$ $1\le i\le n,$ form a conjugacy class in $G(r,1,n)$ and the elements $t_i^k{t}_j^{-k}(i,j),$ $1\le i<j\le n,$ $0\le k\le r-1,$ form another conjugacy class in $G(r,1,n)\text{.}$ Thus the elements $$z_s\left(k\right)=\sum _{i=1}^n{t}_i^k,1\le k\le r-1,\text{and}{z}_{\ell }=\frac{1}{r}\sum _{\underset{1\le k\le r-1}{i<j}}{t}_i^k{t}_j^{-k}(i,j),$$ are elements of $Z\left(ℂG(r,1,n)\right)\text{.}$ So $z_s\left(k\right)$ and $z_{\ell }$ must act by a constant on any irreducible representation $S^{\lambda }$ of $G(r,1,n)\text{.}$ Define $x_1=0,$ $$\begin{array}{ccc}{x}_k& =& \left(\sum _{\underset{0\le \ell \le r-1}{1\le i<j\le k}}{t}_i^{\ell }{t}_j^{-\ell }(i,j)\right)-\left(\sum _{\underset{0\le \ell \le r-1}{1\le i<j\le k-1}}{t}_i^{\ell }{t}_j^{-\ell }(i,j)\right)=\frac{1}{r}\sum _{1\le i<k,\hspace{0.17em}0\le \ell \le r-1}{t}_i^{\ell }{t}_j^{-\ell }(i,j),\text{for}\hspace{0.17em}2\le k\le n,\hspace{0.17em}\text{and}\\ y_k& =& \left(\sum _{i=1}^k{t}_i\right)-\left(\sum _{i=1}^{k-1}{t}_i\right)=t_k,\text{for}\hspace{0.17em}1\le k\le n\text{.}\end{array}$$

The elements $x_1,\dots ,x_n$ and $y_1,\dots ,y_n$ all commute with each other and the action of these elements on the irreducible representation $S^{\lambda }$ of $G(r,1,n)$ is given by $$y_k{v}_T=s\left(T\left(k\right)\right)v_T\text{and}{x}_k{v}_T=c\left(T\left(k\right)\right)v_T,$$ for all standard tableaux $T\text{.}$

		Proof.
	The proof is by induction on $k$ using the relations $$x_k=s_k{x}_{k-1}{s}_k+\sum _{\ell =0}^{r-1}{y}_{k-1}^{\ell }{s}_k{y}_{k-1}^{-\ell }\text{and}{y}_k=s_k{y}_{k-1}{s}_k\text{.}$$ The base cases $$x_1{v}_T=0=c\left(T\left(1\right)\right)v_T\text{and}{y}_1{v}_T=s\left(T\left(1\right)\right)v_T$$ are immediate from the definitions. Then $$\begin{array}{ccc}{y}_k{v}_T& =& s_k{y}_{k-1}{s}_k\\ & =& s_k(s\left(T(k-1)\right){\left(s_k\right)}_{TT}{v}_t+(1+{\left(s_k\right)}_{TT})s\left(T\left(k\right)\right)v_{s_{k}T})\\ & =& s_k(s\left(T\left(k\right)\right)s_k{v}_T+(s\left(T(k-1)\right)-s\left(T\left(k\right)\right)){\left(s_k\right)}_{TT}{v}_T)\\ & =& s\left(T\left(k\right)\right)v_T+0=s\left(T\left(k\right)\right)v_T,\text{and}\\ x_k{v}_T& =& (s_k{x}_{k-1}{s}_k+\frac{1}{r}\sum _{\ell =0}^{r-1}{s}_k{y}_{k-1}^{-\ell }{y}_k^{\ell })v_T\\ & =& s_k(c\left(T(k-1)\right){\left(s_k\right)}_{TT}{v}_T+c\left(T\left(k\right)\right)(1+{\left(s_k\right)}_{TT})v_{s_{k}T}+\frac{1}{r}\sum _{\ell =0}^{r-1}s{\left(T(k-1)\right)}^{-\ell }s{\left(T\left(k\right)\right)}^{\ell }{v}_T)\\ & =& s_k(c\left(T\left(k\right)\right)s_k{v}_T+(c\left(T(k-1)\right)-c\left(T\left(k\right)\right)){\left(s_k\right)}_{TT}{v}_T+\frac{1}{r}\sum _{\ell =0}^{r-1}{\left(s{\left(T\left(k\right)\right)}^{-1}s\left(T\left(k\right)\right)\right)}^{\ell }{v}_T)\\ & =& \{\begin{array}{cc}{s}_k\left(c\left(T\left(k\right)\right)\right)s_k{v}_T+((-1)+1)v_T,& \text{if}\hspace{0.17em}s\left(T\left(k\right)\right)=s\left(T(k-1)\right),\\ s_k\left(c\left(T\left(k\right)\right)\right)s_k{v}_T+(0+0)v_T,& \text{if}\hspace{0.17em}s\left(T\left(k\right)\right)\ne s\left(T(k-1)\right),\end{array}\\ & =& c\left(T\left(k\right)\right)v_T\text{.}\end{array}$$ $\square$

The cyclic group of order $r$

The cyclic group $G(r,1,1)$ of order $r$ is the group generated by a single element $g$ with the relation $g^r=1\text{.}$ The group $G(r,1,1)=\{1,g,g^2,\dots ,g^{r-1}\}$ can be realized as the group $\mathbb{Z}/r\mathbb{Z}=\{0,1,\dots ,r-1\}$ or as the group of $r\text{th}$ roots of unity in $ℂ,$ $$G(r,1,1)=\left\{e^{2\pi ik/r}\hspace{0.17em}\right|\hspace{0.17em}k=0,1,\dots ,r\}\text{.}$$ Since $G(r,1,1)$ is abelian every element is in a conjugacy class by itself. Since $ℂG(r,1,n)$ is commutative every irreducible representation is one dimensional. A representation $\rho :ℂG(r,1,n)\to ℂ$ is completely determined by the value $\rho \left(g\right)$ and the relation $g^r=1$ forces $\rho {\left(g\right)}^r=1\text{.}$ So $\rho$ is an irreducible representation of $\mathbb{Z}/r\mathbb{Z}$ if and only if $\rho \left(g\right)$ is an $r\text{th}$ root of unity. This proves the following theorem.

(a)	The irreducible representations $S^{\lambda }$ of the cyclic group $G(r,1,1)$ are indexed by $0\le \lambda \le r-1\text{.}$
(b)	$\text{dim}\hspace{0.17em}{S}^{\lambda }=1,$ for all $\lambda =0,1,\dots ,r-1\text{.}$
(c)	The irreducible representations (and the irreducible characters) of $G(r,1,1)$ are given by the maps $$\begin{array}{cccc}{\rho }^k:& ℂG(r,1,1)& ⟶& ℂ\\ & g& ⟼& e^{2\pi i/r}\end{array}0\le k\le r-1\text{.}$$
(d)	The irreducible $G(r,1,1)\text{-module}$ $S^{\lambda }$ is given by the one dimensional vector space $S^{\lambda }=ℂv_{\lambda }$ with $G(r,1,1)\text{-action}$ given by $$g^k{v}_{\lambda }={\xi }^{k\lambda }{v}_{\lambda },\text{where}\xi =e^{2\pi i/r}\text{.}$$

Fix $u_0,\dots ,u_{r-1}\in ℂ\text{.}$ The cyclic algebra $H_{r,1,1}=H_{r,1,1}(u_0,\dots ,u_{r-1})$ is the algebra given by a single generator $X$ which satisfies the relation $$\begin{array}{cc}(X-u_0)(X-u_1)\cdots (X-u_{r-1})=0,& \text{(1.2)}\end{array}$$ i.e. $H_{r,1,1}$ is the quotient of the polynomial ring $ℂ\left[X\right]$ by the ideal generated by the polynomial in ???. The algebra $H_{r,1,1}$ has basis $\left\{X^k\hspace{0.17em}\right|\hspace{0.17em}0\le k\le r-1\}$ and, if $u_k={\xi }^k,$ where $\xi =e^{2\pi i/r},$ then $H_{r,1,1}=ℂG(r,1,1)$ is the group algebra of the cyclic group.

The proof of the following theorem is exactly the same as the proof of Theorem ???. The hypotheses in the statement are exactly what is needed to guarantee that the maps ${\rho }^k$ defined in part (c) are distinct (and nonzero???)

Assume that $u_0,\dots ,u_{r-1}\in ℂ$ are all distinct (and nonzero???).

(a)	The irreducible representations $H^{\lambda }$ of the cyclic algebra $H_{r,1,1}$ are indexed by $0\le \lambda \le r-1\text{.}$
(b)	$\text{dim}\hspace{0.17em}{H}^{\lambda }=1,$ for all $\lambda =0,1,\dots ,r-1\text{.}$
(c)	The irreducible representations (and the irreducible characters) of $H_{r,1,1}(u_0,\dots ,u_{r-1})$ are given by the maps $$\begin{array}{cccc}{\rho }^k:& H_{r,1,1}(u_0,\dots ,u_{r-1})& ⟶& ℂ\\ & X& ⟼& u_i\end{array}0\le k\le r-1\text{.}$$
(d)	The irreducible $H_{r,1,1}\text{-module}$ $H^{\lambda }$ is given by the one dimensional vector space $H^{\lambda }=ℂv_{\lambda }$ with $H_{r,1,1}\text{-action}$ given by $$X{v}_{\lambda }=u_{\lambda }{v}_{\lambda }\text{.}$$

The dihedral group $G(r,r,2)$ of order $2r$

The dihedral group $G(r,r,2)$ is the group of $2\times 2$ matrices given by $$G(r,r,2)=\{\left(\begin{array}{cc}{\xi }^k& 0\\ 0& {\xi }^{-k}\end{array}\right),\left(\begin{array}{cc}0& {\xi }^k\\ {\xi }^{-k}& 0\end{array}\right)\hspace{0.17em}|\hspace{0.17em}k=0,1,\dots ,r-1\},\text{where}\hspace{0.17em}\xi =e^{2\pi i/r}\text{.}$$ Conjugation by the matrix $P=\text{??}$ shows that $G(r,r,2)$ is isomorphic to the group of matrices $$G(r,r,2)=\{\left(\begin{array}{cc}{\xi }^k& 0\\ 0& {\xi }^{-k}\end{array}\right),\left(\begin{array}{cc}0& {\xi }^k\\ {\xi }^{-k}& 0\end{array}\right)\hspace{0.17em}|\hspace{0.17em}k=0,1,\dots ,r-1\},\text{where}\hspace{0.17em}\xi =e^{2\pi i/r}\text{.}$$ In this form $G(r,r,2)$ is the group of symmetries of a regular $r\text{-gon}$ (embedded in ${ℝ}^2$ with its center at the origin), $$\begin{array}{c} id s1 s2 s1s2 s2s1 s1s2s1 s2s1s2 s1s2s1s2 \end{array}$$ with $s_1$ being the reflection in $H_{{\alpha }_1}$ and $s_2$ the reflection in $H_{{\alpha }_2}\text{.}$

(a)	The generators $t=\left(\begin{array}{cc}\xi & 0\\ 0& {\xi }^{-1}\end{array}\right)$ and $s=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and the relations $$t^r=1,s^2=1,st=t^{-1}s,$$ form a presentation of $G(r,r,2)\text{.}$
(b)	The generators $s_1=\left(\begin{array}{cc}0& \xi \\ {\xi }^{-1}& 0\end{array}\right)$ and $s_2=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and the relations $$\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{s_1{s}_2{s}_1{s}_2\cdots }=\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{s_2{s}_1{s}_2{s}_1\cdots },s_1^2=1,s_2^2=1,$$ form a presentation of $G(r,r,2)\text{.}$

The conjugacy classes of $G(r,r,2)$ are $$\begin{array}{c}{𝒞}_1=\left\{1\right\}{𝒞}_{w_0}=\{\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{s_1{s}_2{s}_1{s}_2\cdots }\}\\ \begin{array}{ccc}{𝒞}_{s_1}& =& \{\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{\cdots s_1{s}_2}{s}_1\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{s_2{s}_1\cdots }\hspace{0.17em}|\hspace{0.17em}0\le k<r/2\},\\ {𝒞}_{s_2}& =& \{\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{\cdots s_1{s}_2}{s}_2\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{s_2{s}_1\cdots }\hspace{0.17em}|\hspace{0.17em}0\le k<r/2\},\text{and}\\ {𝒞}_k& =& \{{\left(s_1{s}_2\right)}^k,{\left(s_2{s}_1\right)}^k\},1\le k<r/2\end{array}\end{array}$$

The irreducible representations of the dihedral group $G(r,r,2)$ are given as follows. The one dimensional representations ${\rho }^{++}$ and ${\rho }^{--}$ are given by $${\rho }^{++}\left(s_1\right)=1,{\rho }^{++}\left(s_2\right)=1\text{and}{\rho }^{--}\left(s_1\right)=-1,{\rho }^{--}\left(s_2\right)=-1,$$ and, if $r$ is even, there are additional one dimensional representations ${\rho }^{+-}$ and ${\rho }^{-+}$ given by $${\rho }^{+-}\left(s_1\right)=1,{\rho }^{+-}\left(s_2\right)=-1,\text{and}{\rho }^{-+}\left(s_1\right)=-1,{\rho }^{-+}\left(s_2\right)=1\text{.}$$ The two dimensional irreducible representations $S^{\lambda },$ $0<\lambda <r/2,$ are given by $$S^{\lambda }\left(s_2\right)=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)\text{and}{S}^{\lambda }\left(s_1\right)=\left(\begin{array}{cc}0& {\xi }^{\lambda }\\ {\xi }^{-\lambda }& 0\end{array}\right),\text{where}\hspace{0.17em}\xi =e^{2\pi i/r}\text{.}$$

		Proof.
	There are four things to show: (a) The ${\rho }^{\lambda }$ are irreducible, (b) The ${\rho }^{\lambda }$ are representations of $G(r,r,2),$ (c) The ${\rho }^{\lambda }$ are all nonisomorphic, and (d) The ${\rho }^{\lambda }$ are complete set of irreducible representations. (a) It is straightforward to check that ${\rho }^{\lambda }{\left(s_1\right)}^2={\rho }^{\lambda }{\left(s_2\right)}^2=\text{Id}$ for all the given ${\rho }^{\lambda }\text{.}$ It remains to check that ${\rho }^{\lambda }{\left(s_1{s}_2\right)}^r=\text{Id.}$ This follows since $$\begin{array}{c}{\rho }^{\lambda }{\left(s_1{s}_2\right)}^r=1\text{, if}\hspace{0.17em}r\hspace{0.17em}\text{is odd and}\hspace{0.17em}{\rho }^{\lambda }\hspace{0.17em}\text{is one dimensional,}\\ {\rho }^{\lambda }{\left(s_1{s}_2\right)}^r=\pm 1\text{, if}\hspace{0.17em}r\hspace{0.17em}\text{is even and}\hspace{0.17em}{\rho }^{\lambda }\hspace{0.17em}\text{is one dimensional,}\\ {\rho }^{\lambda }\left(s_1{s}_2\right)=\left(\begin{array}{cc}{\xi }^{\lambda }& 0\\ 0& {\chi }^{-\lambda }\end{array}\right)\text{, when}\hspace{0.17em}{\rho }^{\lambda }\hspace{0.17em}\text{is two dimensional.}\end{array}$$ (b) All one dimensional representations are irreducible. Let ${\rho }^{\lambda }$ be a given two dimensional representaitons and let $M_1,$ with basis $\{m_1,m_2\}$ be the corresponding $G(r,r,2)$ module. Let $N$ be a nonzero submodule of $M$ and let $n=c_1{m}_1+c_2{m}_2$ be a nonzero vector in $N\text{.}$ Suppose $c_1\ne 0\text{.}$ Then $$\left(\frac{s_1{s}_2-{\xi }^{-\lambda }}{{\xi }^{\lambda }-{\xi }^{-\lambda }}\right)n=c_1{m}_1\in N,$$ and so $m_1\in N$ and $m_2=s_2{m}_1\in N\text{.}$ So $N=M\text{.}$ So $M$ is irreducible. (c) Since the values $${\chi }^{\lambda }\left(s_1{s}_2\right)={\xi }^{\lambda }+{\xi }^{-\lambda }=2\text{cos}(2\pi \lambda /r),1\le \lambda <r/2,$$ are all distinct, the characters of all the two dimensional representations ${\rho }^{\lambda },$ $1\le \lambda <r/2,$ are distinct. Thus these representations are nonisomorphic. (d) If $m$ is odd $$\sum _{\lambda }{d}_{\lambda }^2=1^2+1^2+\sum _{\lambda =1}^{r/2-1/2}{2}^2=2+4(r/2-1/2)=2r,$$ and, if $m$ is even, $$\sum _{\lambda }{d}_{\lambda }^2=1^2+1^2+1^2+1^2+\sum _{\lambda =1}^{r/2-1}{2}^2=4+4(r/2-1)=2r\text{.}$$ In either case ${\sum }_{\lambda }{d}_{\lambda }^2=\left|G(r,r,2)\right|,$ and so the ${\rho }^{\lambda }$ are a complete set of inequivalent irreducible representations of $G(r,r,2)\text{.}$ $\square$

The Iwahori-Hecke algebra of $G(r,r,2)$

Let $p,q\in {ℂ}^{*}\text{.}$ The Iwahori-Hecke algebra of the dihedral group $H_{r,r,2}=H_{r,r,2}(p,q)$ is the algebra given by generators $T_1,T_2$ and relations $$\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2{T}_1{T}_2\cdots }=\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{T_2{T}_1{T}_2{T}_1\cdots }{T}_1^2=(p-p^{-1})T_1+1,T_2^2=(q-q^{-1})T_2+1\text{.}$$ The last two relations guarantee that $T_1$ and $T_2$ are invertible, $$T_1^{-1}=T_1-(p-p^{-1})\text{and}{T}_2^{-1}=T_2-(q-q^{-1})\text{.}$$ If $r$ is odd then $p=q$ is forced since $${(\underset{\underset{r-1\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots })}^{-1}{T}_2(\underset{\underset{r-1\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots })=T_1\text{.}$$ The elements $$\{1,\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2{T}_1{T}_2\cdots },\underset{\underset{k\hspace{0.17em}\text{factors}}{⏟}}{T_2{T}_1{T}_2{T}_1\cdots },\underset{\underset{r\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2{T}_1{T}_2\cdots }\hspace{0.17em}|\hspace{0.17em}1\le k<r\}$$ form a basis of $H_{r,r,2}\text{.}$ If $p=q=1$ then $H_{r,r,2}=ℂG(r,r,2)$ is the group algebra of the dihedral group of order $2r\text{.}$

Assume that $p,q$ satisfy ????. The irreducible representations of $H_{r,r,2}(p,q)$ are given as follows. The one dimensional representations ${\rho }^{++}$ and ${\rho }^{--}$ are given by $$\begin{array}{ccc}{\rho }^{++}\left(T_1\right)& =& p,\\ {\rho }^{++}\left(T_2\right)& =& q,\end{array}\text{and}\begin{array}{ccc}{\rho }^{--}\left(T_1\right)& =& -p^{-1},\\ {\rho }^{--}\left(T_2\right)& =& -q^{-1},\end{array}$$ and, if $r$ is even, there are additional one dimensional representations ${\rho }^{+-}$ and ${\rho }^{-+}$ given by $$\begin{array}{ccc}{\rho }^{+-}\left(T_1\right)& =& p,\\ {\rho }^{+-}\left(T_2\right)& =& -q^{-1},\end{array}\text{and}\begin{array}{ccc}{\rho }^{-+}\left(T_1\right)& =& -p^{-1},\\ {\rho }^{-+}\left(T_2\right)& =& q\text{.}\end{array}$$ The two dimensional irreducible representations ${\rho }^{\lambda },$ $0<\lambda <r/2,$ are given by $$\begin{array}{c}{\rho }^{\lambda }\left(T_2\right)=\left(\begin{array}{cc}a& (1+ad)\\ 1& d\end{array}\right)\text{and}{\rho }^{\lambda }\left(T_1\right)=\left(\begin{array}{cc}-d\xi & (1+ad)\xi \\ {\xi }^{-1}& -a{\xi }^{-1}\end{array}\right),\text{where}\hspace{0.17em}\xi =e^{2\pi i/r},\\ a=\frac{(q-q^{-1})\xi +(p-p^{-1})}{\xi -{\xi }^{-1}},\text{and}d-\frac{(q-q^{-1}){\xi }^{-1}+(p-p^{-1})}{{\xi }^{-1}-\xi }\text{.}\end{array}$$

		Proof.
	Instead of repeating the proof of Theorem ??? and trying to mind our $p\text{’s}$ and $q\text{’s}$ let us give a proof which shows how this result can be derived. The quadratic relations in ??? are equivalent to $$(T_1-p)(T_1+p^{-1})=0\text{and}(T_2-q)(T_2+q^{-1})=0,$$ and these imply that any one dimensional representation $\chi :H_{r,r,2}\to ℂ$ must have $\chi \left(T_1\right)$ equal to $p$ or $-p^{-1}$ and $\chi \left(T_2\right)$ equal to $q$ or $-q^{-1}\text{.}$ If $r$ is odd then $T_1$ is conjugate to $T_2$ and so $\chi \left(T_1\right)=\chi \left(T_2\right)\text{.}$ This determines the one dimensional representations of $H_{r,r,2}\text{.}$ Let $\rho :H_{r,r,2}\to M_2\left(ℂ\right)$ be an irreducible two dimensional representation of $H_{r,r,2}\text{.}$ By the relation $(T_1-\rho )(T_1+p^{-1})=0$ the eigenvalues of $\rho \left(T_1\right)$ are in the set $\{p,-p^{-1}\}\text{.}$ If $\rho \left(T_1\right)$ has two equal eigenvalues then it is a multiple of the identity. If $v$ is an eigenvector of $\rho \left(T_2\right),$ and $\rho \left(T_1\right)$ is a constant multiple of the identity then $ℂv$ is a submodule of $\rho$ and this contradicts the irreducibility of $\rho \text{.}$ So $\rho \left(T_1\right)$ must have eigenvalues $p$ and $-p^{-1}\text{.}$ Similarly $\rho \left(T_2\right)$ must have eigenvalues $q$ and $q^{-1}\text{.}$ By the defining relations $$T_1\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }=T_1\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_2{T}_1\cdots }=\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }{T}_i,$$ where $i=1$ if $r$ is odd and $i=2$ if $r$ is even. So $$T_2{\left(T_1{T}_2\right)}^m=T_1\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }=\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }{T}_i\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }=\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }\underset{\underset{m\hspace{0.17em}\text{factors}}{⏟}}{T_1{T}_2\cdots }{T}_1={\left(T_1{T}_2\right)}^m{T}_1,$$ and similarly one shows that $T_2{\left(T_1{T}_2\right)}^m={\left(T_1{T}_2\right)}^m{T}_2\text{.}$ So ${\left(T_1{T}_2\right)}^m$ is in the center of $H_{r,r,2}\text{.}$ By Schur’s lemma $$\rho {\left(T_1{T}_2\right)}^m=c\text{Id,}\text{for some}\hspace{0.17em}c\in ℂ\text{.}$$ Since $\rho \left(T_1\right)$ has eigenvalues $p$ and $p^{-1}$ and $\rho \left(T_2\right)$ has eigenvalues $q$ and $q^{-1},$ $\text{det}\left(\rho \left(T_1\right)\right)=\text{det}\left(\rho \left(T_2\right)\right)=-1$ and so $$\text{det}\left(\rho {\left(T_1{T}_2\right)}^m\right)={(-1)}^{2m}=1\text{.}$$ On the other hand, $\text{det}\left(\rho {\left(T_1{T}_2\right)}^m\right)=c^2$ and so $c=\pm 1\text{.}$ Since $c$ does not depend on $p$ and $q$ and ${\left(s_1{s}_2\right)}^m=1$ in the group $G(r,r,2)$ it follows that $c=1\text{.}$ So $$\rho {\left(T_1{T}_2\right)}^m=\text{Id.}$$ Thus the eigenvalues of $\rho \left(T_1{T}_2\right)$ are $r\text{th}$ roots of unity. Since $\text{det}\left(\rho \left(T_1{T}_2\right)\right)={(-1)}^2=1$ we can write (possibly after some change of basis in the module $\rho \text{)}$ $$\rho \left(T_1{T}_2\right)=\left(\begin{array}{cc}{\xi }^k& 0\\ 0& {\xi }^{-k}\end{array}\right)\text{where}\hspace{0.17em}\xi =e^{2\pi i/r}\hspace{0.17em}\text{and}\hspace{0.17em}k\in \mathbb{Z}\text{.}$$ If $k=0$ or (if $r$ is even and) $k=r/2$ then $\rho \left(T_1{T}_2\right)=\pm 1$ and $\rho$ is not irreducible. So without loss of generality we can assume that $0<k<r/2\text{.}$ Suppose that $\rho \left(T_2\right)=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\text{.}$ The relation $T_2^2=(q-q^{-1})T_2+1$ implies that $a+d=q-q^{-1}$ and $ad=bc=1\text{.}$ So we can assume $$\rho \left(T_2\right)=\left(\begin{array}{cc}a& ad+1\\ 1& d\end{array}\right),\text{with}\hspace{0.17em}a+d=q-q^{-1}\text{.}$$ Then $$\rho \left(T_1\right)=\rho \left(T_1{T}_2{T}_2^{-1}\right)=\rho \left(T_1{T}_2\right)\rho {\left(T_2\right)}^{-1}=\left(\begin{array}{cc}{\xi }^k& 0\\ 0& {\xi }^{-k}\end{array}\right)\left(\begin{array}{cc}-d& (ad+1)\\ 1& -a\end{array}\right)=\left(\begin{array}{cc}-d{\xi }^k& (1+ad){\xi }^k\\ {\xi }^{-k}& -a{\xi }^{-k}\end{array}\right)\text{.}$$ Now $$\text{Tr}\left(\rho \left(T_1\right)\right)=p-p^{-1}=-d{\xi }^k-a{\xi }^{-k},\text{and}\text{Tr}\left(\rho \left(T_2\right)\right)=q-q^{-1}=a+d,$$ and these equations determine $a$ and $d$ as given in the statement of the theorem. $\square$

The symmetric group

The symmetric group is the group of permutations of $1,2,\dots ,n,$ i.e. the group of bijections $w:\{1,2,\dots ,n\}\to \{1,2,\dots ,n\}$ under the operation of composition. There are multiple notations for permutations:

(1)	Two line notation: where $w\left(i\right)$ in the second line is below $i$ in the first line. $$w=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 4& 1& 6& 2& 5& 3\end{array}\right)$$
(2)	Cycle notation: Where $(i_1,i_2,\dots ,i_r)$ indicates $w\left(i_1\right)=i_2,$ $w\left(i_2\right)=i_3,$ $\dots ,$ $w\left(i_{r-1}\right)=i_r,$ $w\left(i_r\right)=i_1\text{.}$ Cycles of length $1$ are usually dropped from the notation. $$w=\left(142\right)\left(36\right)\left(5\right)=\left(142\right)\left(36\right)\text{.}$$
(3)	Matrix notation: where the $w\left(i\right)\text{th}$ entry of the $i\text{th}$ column is $1$ and all other entries are $0\text{.}$ $$w=\left(\begin{array}{cccccc}0& 1& 0& 0& 0& 0\\ 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 1\\ 1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 0& 0& 1& 0& 0& 0\end{array}\right)$$
(4)	Diagram notation: where the $i\text{th}$ dot in the top row is connected to the $w\left(i\right)\text{th}$ dot in the bottom row. $$w=\begin{array}{c} \end{array}$$

The symmetric group $s_n$ is the group of $n\times n$ permutation matrices and $\text{Card}\left(S_n\right)=n!\text{.}$

The transpositions in $S_n$ are the permutations $(i,j),$ $1\le i<j\le n$ (in cycle notation). The simple transpositions are $$s_i=(i,i+1)=\begin{array}{c} i i+1 ⋯ ⋯ \end{array},1\le i\le n-1\text{.}$$

The symmetric group $S_n$ can relations be presented by generators $s_1,s_2,\dots ,s_{n-1}$ and relations $$\begin{array}{cccc}{s}_i{s}_j& =& s_j{s}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ s_i{s}_{i+1}{s}_i& =& s_{i+1}{s}_i{s}_{i+1},& 1\le i\le n-2,\\ s_i^2& =& 1,& 1\le i\le n-1\text{.}\end{array}$$

		Proof.
	There are three things to show: (a) The simple transpositions in $S_n$ satisfy the given relations. (b) The simple transpositions generate $S_n\text{.}$ (c) The group given by generators $s_1,\dots ,s_{n-1}$ and relations as in the statement of the theorem has order $\le n!\text{.}$ (a) The following pictures show that the simple tranpositions $s_i=(i,i+1),$ $1\le i\le n-2$ satisfy the given relations. $$\begin{array}{c}\begin{array}{c} \end{array}=\begin{array}{c} \end{array}\\ \begin{array}{c} \end{array}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}\\ \begin{array}{c} \end{array}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}\end{array}$$ (b) The diagram of any permutation can be “stretched” to guarantee that no more than two edges cross at any given point. Then w can be written as a product of simple transpositions corresponding to these crossings. $$w=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}=s_1{s}_3{s}_5{s}_4{s}_2{s}_5{s}_3\text{.}$$ (c) Let $G_n$ be the free group generated by sysmbols $s_1,s_2,\dots ,s_{n-1}$ modulo the relations in the statement of the theorem. Let $G_{n-1}$ be the subgroup generated by $s_1,s_2,\dots ,s_{n-2}\text{.}$ We will show that (1) Every element $w\in G_n$ can be written in the form $w=w_1{s}_{n-1}{w}_2,$ with $w_1,w_2\in G_{n-1}\text{.}$ (2) Every element $w\in G_n$ can be written in the form $w=a{s}_{n-1}{s}_{n-2}\dots s_k,$ $1\le k\le n,$ and $a\in G_{n-1}\text{.}$ (1) Since $s_{n-1}^2=1$ every element $w\in G_n$ can be written in the form $$w=w_1{s}_{n-1}{w}_2{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell },\text{where}\hspace{0.17em}{w}_j\in G_{n-1}\text{.}$$ By induction, either $w_2\in G_{n-2}$ or $w_2=a{s}_{n-2}b$ where $a,b\in G_{n-2}\text{.}$ In the first case $$\begin{array}{ccc}w& =& w_1{s}_{n-1}{w}_2{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_1{w}_2{s}_{n-1}{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_1{w}_2{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell },\end{array}$$ and in the second case $$\begin{array}{ccc}w& =& w_1{s}_{n-1}{w}_2{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_1{s}_{n-1}a{s}_{n-2}b{s}_{n-1}{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_{1}a{s}_{n-1}{s}_{n-2}{s}_{n-1}b{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_{1}a{s}_{n-2}{s}_{n-1}{s}_{n-2}b{w}_3{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\\ & =& w_1^{\prime }{s}_{n-1}{w}_3^{\prime }{s}_{n-1}\cdots w_{\ell -1}{s}_{n-1}{w}_{\ell }\end{array}$$ where $w_1^{\prime }=w_{1}a{s}_{n-2}\in G_{n-1}$ and $w_3^{\prime }=s_{n-2}b{w}_3\in G_{n-1}\text{.}$ In either case the number of $s_{n-1}$ factors in the expression of $w$ has been reduced. The statement in (1) follows by induction. (2) By (1) any element $w\in G_n$ is either an element of $G_{n-1}$ or can be written in the form $w=w_1{s}_{n-1}{w}_2,$ with $w_1,w_2\in G_{n-1}\text{.}$ In the first case $w=w{s}_{n-1}\cdots s_{\ell }$ with $\ell =n$ and the statement is satisfied. If $w=w_1{s}_{n-1}{w}_2$ then, by induction, $w_2=a{s}_{n-2}{s}_{n-3}\dots s_{\ell },$ for some $1\le \ell \le n-2$ and $a\in G_{n-2}\text{.}$ So $$w=w_1{s}_{n-1}a{s}_{n-2}\cdots s_{\ell }=w_{1}a{s}_{n-1}\cdots s_{\ell }=w_1^{\prime }{s}_{n-1}\cdots s_{\ell },$$ where $w_1^{\prime }=w_{1}a\in G_{n-1}\text{.}$ Statement (2) follows. From (2) it follows that $\text{Card}\left(G_n\right)\le \text{Card}\left(G_{n-1}\right)·n,$ and, by induction, that $\text{Card}\left(G_n\right)\le n!\text{.}$ Since $S_n$ is a quotient of $G_n$ and $\text{Card}\left(S_n\right)=n!$ we have $G_n\cong S_n\text{.}$ $\square$

A partition of $n$ is a collection of $n$ boxes in a corner (gravity goes up and to the left and pushes the boxes tightly into the corner). $$\begin{array}{cc}\lambda =\begin{array}{c} \end{array}& \text{(3.2)}\end{array}$$ If ${\lambda }_i$ is the number of boxes in row $i$ then $\lambda =({\lambda }_1\ge {\lambda }_2\ge \dots )$ is a sequence of nonnegative integers and we write $\lambda ⊢n\text{.}$ For example, the partition in (??) is $\lambda =\left(55311\right)$ and $\lambda ⊢15\text{.}$ A standard tableau of shape $\lambda$ is a filling $T$ of the boxes of $\lambda$ with $1,2,\dots ,n$ such that

(a)	the rows of $T$ are increasing left to right,
(b)	the columns of $T$ are increasing top to bottom.

For example $$T=\begin{array}{c} 1 2 3 4 5 3 4 7 12 15 8 9 14 11 16 18 13 17 \end{array}$$ is a standard tableau of shape $\lambda =\left(55311\right)\text{.}$ The rows and columns of the partition $\lambda$ are numbered as for matrices, $$\begin{array}{c}T\left(i\right)\hspace{0.17em}\text{is the box containing}\hspace{0.17em}i\hspace{0.17em}\text{in}\hspace{0.17em}T\text{, and}\\ c\left(b\right)=j-i\text{, if}\hspace{0.17em}b\hspace{0.17em}\text{is in position}\hspace{0.17em}(i,j)\hspace{0.17em}\text{in}\hspace{0.17em}\lambda \text{.}\end{array}$$ The number $c\left(b\right)$ is the content of the box $b\text{.}$ $$\begin{array}{c} 0 1 2 3 4 -1 0 1 2 3 -2 -1 0 -3 -2 -1 -4 -5 \\ \text{Contents of boxes}\end{array}$$

(a)	The irreducible representations $S^{\lambda }$ of the symmetric group $S_n$ are indexed by partitions $\lambda$ with $n$ boxes.
(b)	$\text{dim}\hspace{0.17em}{S}^{\lambda }=$ # of standard tableaux of shape $\lambda \text{.}$
(c)	The irreducible $S_n\text{-module}$ $S^{\lambda }$ is given by $$S^{\lambda }=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}\lambda \}$$ with basis $\left\{v_T\right\}$ and with $S_n$ action given by $$s_i{v}_T=\left(\frac{1}{c\left(T(i+1)\right)-c\left(T\left(i\right)\right)}\right)v_T+(1+\frac{1}{c\left(T(i+1)\right)-c\left(T\left(i\right)\right)})v_{s_{i}T},$$ where $s_i=(i,i+1),$ $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$ siT is the same as $T$ except that $i$ and $i+1$ are switched, $v_{s_{i}T}=0$ if $s_{i}T$ is not standard.

		Proof.
	There are four things to show: (a) The $S^{\lambda }$ are $S_n\text{-modules,}$ (b) The $S^{\lambda }$ are nonisomorphic, (c) The $S^{\lambda }$ are all irreducible, (d) The $S^{\lambda },$ $\lambda ⊢n,$ are all the irreducible $S_n\text{-modules.}$ (a) In order to show that $S^{\lambda }$ is an $S_n\text{-module}$ there are three relations to check (a1) $s_i^2=1,$ (a2) $s_i{s}_j=s_j{s}_i,$ if $|i-j|>1,$ (a3) $s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}\text{.}$ (a1) The action of $s_i$ preserves the subspace $S$ spanned by the vectors $v_Q$ indexed by the standard tableaux in the set $\{Q,s_{i}Q\}\text{.}$ Depending on the relative positions of the boxes containing $i$ and $i+1$ this space is either one or two dimensional. $$\begin{array}{ccc}\begin{array}{c} \\ \text{Case (A)}\end{array}& \begin{array}{c} \\ \text{Case (B)}\end{array}& \begin{array}{c} \\ \text{Case (C)}\end{array}\end{array}$$ In case (A) the space $S$ is one dimensional and $s_i$ acts by the matrix $M\left(s_i\right)=1\text{.}$ In Case (B), $S$ is one dimensional and $s_i$ acts by the matrix $M\left(s_i\right)=-1\text{.}$ In case (C), $S$ is two dimensional and $s_i$ acts on the subspace $S$ by a matrix of the form $$M\left(s_i\right)=\left(\begin{array}{cc}\frac{1}{a-b}& 1+\frac{1}{b-a}\\ 1+\frac{1}{a-b}& \frac{1}{b-a}\end{array}\right)$$ In cases (A) and (B) it is immediate that $M{\left(s_i\right)}^2=1$ and in case (C), since $\text{Tr}\left(M\left(s_i\right)\right)=0$ and $\text{det}\left(M\left(s_i\right)\right)=-1,$ it follows from the Cayley-Hamilton theorem that $M{\left(s_i\right)}^2=\text{Id.}$ The action of $s_i$ and $s_j$ commute if $|i-j|>1$ because $s_i$ only moves $i$ and $i+1$ in $T$ and $s_j$ only moves $j$ and $j+1\text{.}$ The condition $|i-j|>1$ guarantees that $\{i,i+1\}$ and $\{j,j+1\}$ do not intersect. (c) According to the formulas for the action, $s_i$ and $s_{i+1}$ preserve the subspace $S$ spanned by the vectors $v_Q$ indexed by the standard tableaux $Q$ in the set $\{L,s_{i}L,s_{i+1}L,s_i{s}_{i+1}L,s_{i+1}{s}_{i}L,s_i{s}_{i+1}{s}_{i}L\}\text{.}$ Depending on the relative positions of the boxes containing $i,$ $i+1,$ $i+2$ in $L,$ this space is either $1,$ $2,$ $3$ or $6$ dimensional. Representative cases are when these boxes are positioned in the following ways. $$\begin{array}{cccc}\begin{array}{c} \\ \text{Case (1)}\end{array}& \begin{array}{c} \\ \text{Case (2)}\end{array}& \begin{array}{c} \\ \text{Case (3)}\end{array}& \begin{array}{c} \\ \text{Case (4)}\end{array}\end{array}$$ In Case (1) the space $S$ is one dimensional and spanned by the vector $v_Q$ corresponding to the standard tableau $$\begin{array}{c} a b c \end{array}$$ where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of $T_i$ and $T_{i+1}$ on $S$ is given by the matrices $$M\left(s_i\right)=\left(1\right),\text{and}M\left(s_{i+1}\right)=\left(1\right),$$ respectively. In case (2) the space $S$ is two dimensional and spanned by the vectors $v_Q$ corresponding to the standard tableaux $$\begin{array}{ccc}\begin{array}{c} a b c \end{array}& & \begin{array}{c} a c b \end{array}\end{array}$$ where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of $T_i$ and $T_{i+1}$ on $S$ is given by the matrices $$M\left(s_i\right)=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\text{and}M\left(s_{i+1}\right)=\left(\begin{array}{cc}1/2& 3/2\\ 1/2& -1/2\end{array}\right)$$ In case (3) the space $S$ is three dimensional and spanned by the vectors $v_Q$ corresponding to the standard tableaux $$\begin{array}{ccccc}\begin{array}{c} c a b \end{array}& & \begin{array}{c} b a c \end{array}& & \begin{array}{c} a b c \end{array}\end{array}$$ where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of $T_i$ and $T_{i+1}$ on $S$ is given by the matrices $$\begin{array}{ccc}M\left(T_i\right)& =& \left(\begin{array}{ccc}1& 0& 0\\ 0& \frac{1}{x_1-x_3}& 1+\frac{1}{x_3-x_1}\\ 0& 1+\frac{1}{x_1-x_3}& \frac{1}{x_3-x_1}\end{array}\right)\text{and}\\ {\varphi }_S\left(T_{i+1}\right)& =& \left(\begin{array}{ccc}\frac{1}{x_2-x_3}& 1+\frac{1}{x_3-x_2}& 0\\ 1+\frac{1}{x_2-x_3}& \frac{1}{x_3-x_2}& 0\\ 0& 0& 1\end{array}\right)\end{array}$$ respectively, where $x_1=c\left(L\left(i\right)\right),$ $x_2=c\left(L(i+1)\right)$ and $x_3=c\left(L(i+2)\right)\text{.}$ In case (4) the space $S$ is six dimensional and spanned by the vectors $v_Q$ corresponding to the standard tableaux $$\begin{array}{ccccccccccc}\begin{array}{c} c b a \end{array}& & \begin{array}{c} c a b \end{array}& & \begin{array}{c} b c a \end{array}& & \begin{array}{c} a c b \end{array}& & \begin{array}{c} b a c \end{array}& & \begin{array}{c} a b c \end{array}\end{array}$$ where $a=i,$ $b=i+1,$ and $c=i+2\text{.}$ The action of $T_i$ and $T_{i+1}$ on $S$ is given by the matrices $$M\left(s_i\right)=\left(\begin{array}{cccccc}\frac{1}{x_1-x_2}& 1+\frac{1}{x_2-x_1}& 0& 0& 0& 0\\ 1+\frac{1}{x_1-x_2}& \frac{1}{x_2-x_1}& 0& 0& 0& 0\\ 0& 0& \frac{1}{x_1-x_3}& 1+\frac{1}{x_3-x_1}& 0& 0\\ 0& 0& 1+\frac{1}{x_1-x_3}& \frac{1}{x_3-x_1}& 0& 0\\ 0& 0& 0& 0& \frac{1}{x_2-x_3}& 1+\frac{1}{x_3-x_2}\\ 0& 0& 0& 0& 1+\frac{1}{x_2-x_3}& \frac{1}{x_3-x_2}\end{array}\right)$$ and $$M\left(s_{i+1}\right)=\left(\begin{array}{cccccc}\frac{1}{x_2-x_3}& 0& 1+\frac{1}{x_3-x_2}& 0& 0& 0\\ 0& \frac{1}{x_1-x_3}& 0& 0& 1+\frac{1}{x_3-x_1}& 0\\ 1+\frac{1}{x_2-x_3}& 0& \frac{1}{x_3-x_2}& 0& 0& 0\\ 0& 0& 0& \frac{1}{x_1-x_2}& 0& 1+\frac{1}{x_2-x_1}\\ 0& 1+\frac{1}{x_1-x_3}& 0& 0& \frac{1}{x_3-x_1}& 0\\ 0& 0& 0& 1+\frac{1}{x_1-x_2}& 0& \frac{1}{x_2-x_1}\end{array}\right)$$ where $x_1=c\left(L\left(i\right)\right),$ $x_2=c\left(L(i+1)\right)$ and $x_3=c\left(L(i+2)\right)\text{.}$ In each case we compute directly the products $M\left(s_i\right)M\left(s_{i+1}\right)M\left(s_i\right)$ and $M\left(s_{i+1}\right)M\left(s_i\right)M\left(s_{i+1}\right)$ and verify that they are equal. Define elements $x_k,$ $1\le k\le n,$ in the group algebra $ℂS_n$ of the symmetric group $$x_1=0,\text{and}{x}_k=\sum _{i<k}(i,k),2\le k\le n,$$ where $(i,k)\in S_n$ is the transposition which switches $i$ and $k\text{.}$ The action of $x_k$ on the $S_n\text{-module}$ $S^{\lambda }$ is given by $$x_k{v}_T=c\left(T\left(k\right)\right)v_T,\text{for all standard tableaux}\hspace{0.17em}T,$$ where $c\left(T\left(k\right)\right)$ is the content of the box containing $k$ in $T\text{.}$ $\square$

		Proof.
	The proof is by induction on $k$ using the relation $$x_k=s_{k-1}{x}_{k-1}{s}_{k-1}+s_{k-1}\text{.}$$ Clearly, $x_1{v}_T=0=c\left(T\left(1\right)\right)v_T$ for all standard tableaux $T\text{.}$ By the induction assumption and the definition of the action of $s_{k-1}$ on $S^{\lambda },$ $$\begin{array}{ccc}{x}_k{v}_T& =& (s_{k-1}{x}_{k-1}{s}_{k-1}+s_{k-1})v_T\\ & =& s_{k-1}(x_{k-1}{s}_{k-1}+1)v_T\\ & =& s_{k-1}(\frac{c\left(T(k-1)\right)}{c\left(T\left(k\right)\right)-c\left(T(k-1)\right)}{v}_T+c\left(T\left(k\right)\right)(1+\frac{c\left(T(k-1)\right)}{c\left(T\left(k\right)\right)-c\left(T(k-1)\right)})v_{s_{k-1}T}+v_T)\\ & =& s_{k-1}(\frac{c\left(T\left(k\right)\right)}{c\left(T\left(k\right)\right)-c\left(T(k-1)\right)}{v}_T+c\left(T\left(k\right)\right)(1+\frac{c\left(T(k-1)\right)}{c\left(T\left(k\right)\right)-c\left(T(k-1)\right)})v_{s_{k-1}T}+v_T)\\ & =& c\left(T\left(k\right)\right)s_{k-1}{s}_{k-1}{v}_T=c\left(T\left(k\right)\right)v_T\text{.}\text{qed}\end{array}$$ $\square$

Let $T$ be a standard tableau with $n$ boxes and let $T^{-}$ be the standard tableau $T$ except with the box containing $n$ removed. Then $T$ is the unique standard tableau given by adding a box (containing $n\text{)}$ to $T^{-}$ in the diagonal of boxes with content $c\left(T\left(n\right)\right)\text{.}$ Thus, by induction on the number of boxes in $T,$ $$T\hspace{0.17em}\text{is uniquely determined by the sequence}\hspace{0.17em}(c\left(T\left(1\right)\right),c\left(T\left(2\right)\right),\dots ,c\left(T\left(n\right)\right))\text{.}$$

Example. If $(c\left(T\left(1\right)\right),\dots ,c\left(T\left(n\right)\right))=(0,1,-1,0,2,-2,-1,-3,1,3,2,-4,-3,-5,-6)$ then $$T=\begin{array}{c} 1 2 5 10 3 4 9 11 6 7 8 13 12 14 15 \end{array}$$ It follows that the vector $v_T$ is, up to constant multiples, the unique vector in all the $S^{\lambda }$ such that $x_k{v}_T=c\left(T\left(k\right)\right)v_T,$ for all $1\le k\le n\text{.}$ This proves that the $S^{\lambda }$ are nonisomorphic.

(c) Let us show that $S^{\lambda }$ is irreducible. Let $N$ be a nonzero submodule of $S^{\lambda }$ and let $n={\sum }_T{a}_T{v}_T$ be a nonzero vector in $N\text{.}$ Fix $T$ such that $a_T\ne 0\text{.}$ It follows from the lemma and the fact that $T$ is determined by the sequence $(c\left(T\left(1\right)\right),\dots ,c\left(T\left(n\right)\right))$ that the element in $ℂS_n$ given by $$p_T=\prod _{k=1}^n\prod _{\underset{-n\le j\le n}{j\ne c\left(T\left(k\right)\right)}}\frac{x_k-j}{c\left(T\left(k\right)\right)-j},\text{acts on}\hspace{0.17em}{S}^{\lambda }\hspace{0.17em}\text{by}{p}_T{v}_S={\delta }_{ST}{v}_T,$$ for all standard tableaux $S$ of shape $\lambda \text{.}$ So $(1/a_T)p_{T}n=v_T\in N\text{.}$

Let $Q$ be a standard tableau of shape $\lambda$ and let $k$ be minimal such that $k+1$ is southwest (strictly south and strictly west) of $k$ in $Q\text{.}$ $$\begin{array}{c} k k+1 \end{array}$$ If $k$ exists then $s_{k}Q$ is a standard tableau and if $k$ doesn’t exist then $Q$ is the column reading tableau, i.e. $Q$ is given by filling the boxes of $\lambda$ sequentially down columns from left to right. $$\begin{array}{c}C=\begin{array}{c} 1 6 9 12 14 2 7 10 13 15 3 8 11 4 5 \end{array}\\ \text{Column reading tableau}\end{array}$$ By repeating the procedure with $s_{k}Q$ in place of $Q$ repeatedly we construct a sequence $$Q,s_{k_1}Q,s_{k_2}{s}_{k_1}Q,\dots ,s_{k_r}\cdots s_{k_2}{s}_{k_1}Q=C,$$ which ends at the column reading tableau $C\text{.}$ By concatenating this sequence with a similar sequence for $T,$ $$T,s_{{\ell }_1}T,\cdots ,s_{{\ell }_m}\cdots s_{{\ell }_2}{s}_{{\ell }_1}T=C$$ we obtain a sequence $$T,s_{{\ell }_1}T,\cdots ,C,\dots ,s_{k_1}Q,Q=s_{k_1}\cdots s_{k_r}{s}_{{\ell }_m}\dots s_{{\ell }_1}T$$ of standard tableaux of shape $\lambda$ which begins at $T$ and ends at $Q\text{.}$

If $P$ is a standard tableau in this sequence and $v_P\in N$ and $s_{k}P$ is the next standard tableau in the sequence, then $$\begin{array}{ccc}(s_k-\frac{1}{c\left(P(k+1)\right)-c\left(P\left(k\right)\right)})v_P& =& (1+\frac{1}{c\left(P(k+1)\right)-c\left(P\left(k\right)\right)})v_P\\ & =& \left(\frac{c\left(P(k+1)\right)-c\left(P\left(k\right)\right)+1}{c\left(P(k+1)\right)-c\left(P\left(k\right)\right)}\right)v_{s_{k}P}\text{.}\end{array}$$ Since $s_{k}P$ is standard, $c\left(P(k+1)\right)-c\left(P\left(k\right)\right)+1\ne 0$ and so it follows that $v_{s_{k}P}\in N\text{.}$ Thus since $v_T\in N$ it follows that $v_Q\in N\text{.}$

So $v_Q\in N$ for all standard tableaux of shape $\lambda \text{.}$ So $N$ contains a basis of $S^{\lambda }\text{.}$ So $N=S^{\lambda }\text{.}$ So $N$ is irreducible.

Step 4: Let $f^{\lambda }$ be the number of standard tableaux of shape $\lambda \text{.}$ Then $$n!=\sum _{\lambda ⊢n}{\left(f^{\lambda }\right)}^2\text{.}$$

		Proof.
	The proof is accomplished by giving a bijection between $$\text{permutations}\hspace{0.17em}w\in S_n\text{and}\begin{array}{c}\text{pairs}\hspace{0.17em}(P,Q)\hspace{0.17em}\text{of standard tableaux,}\\ \text{of the same shape, with}\hspace{0.17em}n\hspace{0.17em}\text{boxes.}\end{array}$$ The matching of $w\in S_n$ with a pair $(P,Q)$ is accomplished by a recursive algorithm which begins with $(P_0,Q_0)=(\varnothing ,\varnothing )$ and, at step $i,$ “inserts” $w\left(i\right)$ into $(P_{i-1},Q_{i-1})$ to obtain $(P_i,Q_i)\text{.}$ The output of the algorithm is the pair of tableaux $(P,Q)=(P_n,Q_n)\text{.}$ The insertion step for inserting $w\left(i\right)$ into $(P_{i-1},Q_{i-1})\text{:}$ Place $w\left(i\right)$ into the first column of $P_{i-1}$ as follows. If $w\left(i\right)$ is greater than all entries of column 1 of $P_{i-1}$ then place $w\left(i\right)$ at the end of column 1. Otherwise there is a unique entry in column 1 which can be replaced by $w\left(i\right)$ and preserve the standardness condition. (This is the smallest entry larger that $w\left(i\right)\text{.)}$ If $w\left(i\right)$ displaces an entry from column 1 the the displaced entry is inserted into column 2 (in exactly the same way that $w\left(i\right)$ was inserted into column 1). The displaced entries are successively inserted into the next column, and the process continues until there is no displaced entry. The resulting standard tableau is $P_i,$ which contains one more box than $P_{i-1}\text{.}$ The tableau $Q_i$ is obtained by adding a box filled with $i$ to $Q_{i-1}$ so that $P_i$ and $Q_i$ are the same shape. To show that the algorithm produces a bijection it is necessary to confirm that the process can be reversed. To see this note that, given $(P_i,Q_i),$ the position of the box containing $i$ in $Q_i$ tells which box of $P_i$ must be “uninserted” to obtain $P_{i-1}\text{.}$ The column by column insertion of this box will bump out an entry from column 1 of $P_i$ and this entry is $w\left(i\right)$ for the permutation w which corresponds to the pair $(P,Q)\text{.}$ $\square$

Example. Let ${\epsilon }_1,\dots ,{\epsilon }_n$ be a basis of ${ℝ}^n\text{.}$ The inversion set of a permutation $w\in S_n$ is $$R\left(w\right)=\{{\epsilon }_j-{\epsilon }_i\hspace{0.17em}|\hspace{0.17em}i<j\hspace{0.17em}\text{and}\hspace{0.17em}w\left(i\right)>w\left(j\right)\}\text{.}$$ The Bruhat-Chevalley order on $S_n$ is the partial order on $S_n$ given by $$w\le v\text{if}R\left(w\right)\subseteq R\left(v\right)\text{.}$$

Let $\lambda$ be a partition and number the boxes of $\lambda$ along diagonals (northwest to southeast within a diagonal) beginning with the southwest most diagonal and proceeding northeastward. $$\begin{array}{c} 6 9 11 13 15 4 7 10 12 14 3 5 8 2 1 \\ \text{Numbering of boxes}\end{array}$$ Using this numbering identify each standard tableau of shape $\lambda$ with a permutation $w\in S_n$ given by $$w\left(i\right)=\text{entry in box}\hspace{0.17em}i,or,\; equivalently{w}^{-1}\left(j\right)=\text{box containing}\hspace{0.17em}j\text{.}$$ For example the standard tableau in ??? corresponds to the permutation $$w=\left(\begin{array}{ccccccccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14& 15\\ 13& 11& 8& 3& 9& 1& 4& 14& 2& 7& 5& 12& 6& 15& 10\end{array}\right)\in S_{15}\text{.}$$ Let $w$ be a standard tableau with $n$ boxes and let $w^{-}$ be the standard tableau with $n-1$ boxes obtained by removing the box containing $n$ from $w\text{.}$ Then $w$ is the unique standard tableau given by adding a box containing $n$ to $w^{-}$ in the diagonal of boxes with content $c\left(w^{-}\left(n\right)\right)\text{.}$ Thus, by induction on the number of boxes in $w,$ $$\begin{array}{c}w\hspace{0.17em}\text{is uniquely determined by the sequence}\\ (c\left(w^{-1}\left(1\right)\right),c\left(w^{-1}\left(2\right)\right),\dots ,c\left(w^{-1}\left(n\right)\right))\text{.}\end{array}$$ Define $$\begin{array}{ccc}{Z}_{\lambda }& =& \{{\epsilon }_j-{\epsilon }_i\hspace{0.17em}|\hspace{0.17em}i<j\hspace{0.17em}\text{and}\hspace{0.17em}{\text{box}}_i\hspace{0.17em}\text{and}\hspace{0.17em}{\text{box}}_j\hspace{0.17em}\text{are in the same diagonal}\}\\ P_{\lambda }& =& \{{\epsilon }_j-{\epsilon }_i\hspace{0.17em}|\hspace{0.17em}i<j\hspace{0.17em}\text{and}\hspace{0.17em}{\text{box}}_i\hspace{0.17em}\text{and}\hspace{0.17em}{\text{box}}_j\hspace{0.17em}\text{are in adjacent diagonal}\}\\ J_{\lambda }& =& \{{\epsilon }_j-{\epsilon }_i\in P_{\lambda }\hspace{0.17em}|\hspace{0.17em}{\text{box}}_j\hspace{0.17em}\text{is northwest of}\hspace{0.17em}{\text{box}}_i\}\end{array}$$ where, in the definition of $J_{\lambda },$ northwest means strictly north and weakly west. The sets $Z_{\lambda },$ $P_{\lambda }$ and $J_{\lambda }$ completely determine the partition $\lambda \text{.}$

Let ${ℱ}^{\lambda }=\{w\in S_n\hspace{0.17em}|\hspace{0.17em}w\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}\lambda \}\text{.}$ Then

(a)	${ℱ}^{\lambda }=\{w\in S_n\hspace{0.17em}|\hspace{0.17em}R\left(w\right)\cap Z_{\lambda }=\varnothing ,R\left(w\right)\cap P_{\lambda }=J_{\lambda }\}\text{.}$
(b)	${ℱ}^{\lambda }$ is an interval in the Bruhat-Chevalley order on $S_n,$ $${ℱ}^{\lambda }=[w_{\text{min}},w_{\text{max}}]=\{v\in S_n\hspace{0.17em}|\hspace{0.17em}{w}_{\text{min}}\le v\le w_{\text{max}}\},$$ where $w_{\text{min}}$ is the column reading tableau of shape $\lambda$ and $w_{\text{max}}$ is the row reading tableau of shape $\lambda \text{.}$

		Proof.
	(a) The condition $R\left(w\right)\cap Z_{\lambda }=\varnothing$ is equivalent to (A) $w\left(i\right)<w\left(j\right)$ if $i<j$ and ${\text{box}}_i$ and ${\text{box}}_j$ are in the same diagonal. The condition $R\left(w\right)\cap P_{\lambda }=J_{\lambda }$ is equivalent to (B) $w\left(i\right)<w\left(j\right)$ if $i<j,$ ${\text{box}}_i$ and ${\text{box}}_j$ are in adjacent diagonals, and ${\text{box}}_j$ is southeast of ${\text{box}}_i,$ and (C) $w\left(i\right)>w\left(j\right)$ if $i<j,$ ${\text{box}}_i$ and ${\text{box}}_j$ are in adjacent diagonals, and ${\text{box}}_j$ is borhteast of ${\text{box}}_i\text{.}$ $$\begin{array}{ccccc}\begin{array}{c} w(i) w(j) > \\ \text{Case (A)}\end{array}& & \begin{array}{c} w(i) w(j) > \\ \text{Case (B)}\end{array}& & \begin{array}{c} w(i) w(j) < \\ \text{Case (C)}\end{array}\end{array}$$ Thus the condition $R\left(w\right)\cap P_{\lambda }=J_{\lambda }$ is equivalent to the condition for filling a standard tableau and the condition $R\left(w\right)\cap Z_{\lambda }=\varnothing$ is redundant. $\square$

The Iwahori-Hecke algebra of type A

Fix $q\in {ℂ}^{*}\text{.}$ The Iwahori-Hecke algebra $H_n=H_n\left(q\right)$ of type $A_{n-1}$ is the associative algebra with $1$ given by generators $T_1,\dots ,T_{n-1}$ and relations $$\begin{array}{cccc}{T}_i{T}_j& =& T_j{T}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ T_i{T}_{i+1}{T}_i& =& T_{i+1}{T}_i{T}_{i+1},& 1\le i\le n-2,\\ T_i^2& =& 1,& 1\le i\le n-1\text{.}\end{array}$$ If $q=1$ then $H_n=ℂS_n,$ the group algebra of the symmetric group.

The Iwahori-Hecke algebra $H_n$ of type A has basis $T_w,$ $w\in S_n\text{.}$

		Proof.
	$\square$

$\text{dim}\left(H_n\right)\le n!\text{.}$

		Proof.
	We will show two things. (a) Every element of $H_n$ can be written as a linear combination of elements of the form $w_1{T}_{n-1}{w}_2,$ where $w_1,w_2\in H_{n-1}\text{.}$ (b) Every element of $H_n$ can be written as a linear combination of elements of the form $a{T}_{n-1}{T}_{n-2}\cdots T_{\ell },$ where $1\le \ell \le n$ and $a\in H_{n-1}\text{.}$ (a) Every element of $H_n$ can be written as a linear combination of elements of the form $$w_1{T}_{n-1}{w}_2{T}_{n-2}{w}_3{T}_{n-1}\cdots w_{\ell -1}{T}_{n-1}{w}_{\ell },\text{where}\hspace{0.17em}{w}_j\in H_{n-1}\text{.}$$ Then, by induction, $w_2$ is a linear combination of elements of the form $a{T}_{k-2}b$ where $a,b\in H_{n-2}\text{.}$ So $$T_{n-1}{w}_2{T}_{n-1}=T_{n-1}a{T}_{n-2}b{T}_{n-1}=a{T}_{n-1}{T}_{n-2}{T}_{n-1}b=a{T}_{n-2}{T}_{n-1}{T}_{n-2}b,$$ since all elements of $H_{n-2}$ commute with $T_{n-1}\text{.}$ So $w_1{T}_{n-1}{w}_2{T}_{n-2}{w}_3{T}_{n-1}\cdots w_{\ell -1}{T}_{n-1}{w}_{\ell }$ is a linear combination of elements $$w_1^{\prime }{T}_{n-1}{w}_2^{\prime }{T}_{n-2}{w}_4{T}_{n-1}{w}_5\cdots w_{\ell -1}{T}_{n-1}{w}_{\ell },$$ where $w_1^{\prime }=w_{1}a{T}_{n-1}$ and $w_2^{\prime }=T_{n-2}b{w}_3$ are in $H_{n-1}\text{.}$ In this way the number of $T_{n-1}$ factors has been reduced by one. Thus, by induction, $h$ is a linear combination of elements $h\prime \in H_{n-1}$ and elements $h_1{T}_{n-1}{h}_2$ with $h_1,h_2\in H_{n-1}\text{.}$ (b) By (a) any element $h\in H_{n-1}$ can be written as a linear combination of elements of the $h_1{T}_{n-1}{h}_2$ with $h_1,h_2\in H_{k-2}\text{.}$ By induction, $h_2$ can be written as a linear combination of elements of the form $$a{T}_{n-2}{T}_{n-3}\cdots T_{\ell },\text{for some}\hspace{0.17em}1\le \ell \le k-2\hspace{0.17em}\text{and}\hspace{0.17em}a\in H_{n-2}\text{.}$$ So $h_1{T}_{n-1}{h}_2$ is a linear combination of elements of the form $$h_1{T}_{n-1}a{T}_{n-2}{T}_{n-3}\cdots T_{\ell }=h_{1}a{T}_{n-1}{T}_{n-2}\cdots T_{\ell },\text{where}\hspace{0.17em}{h}_{1}a\in H_{n-1}\text{.}$$ So every element $h\in H_{n-1}$ is a linear combination of elements $h\prime \in H_{n-1}$ and elements of the form $$h_1{T}_{n-1}\cdots T_{\ell },\text{with}\hspace{0.17em}{h}_1\in H_{n-1}\text{.}$$ So $\text{dim}\left(H_n\right)\le \text{dim}\left(H_{n-1}\right)·n\text{.}$ Thus, by induction, $\text{dim}\left(H_n\right)\le n!\text{.}$ $\square$

Assume that $q^k\ne 1$ for all $k=2,\dots ,n\text{.}$

(a)	The irreducible representations $H^{\lambda }$ of the Iwahori-Hecke algebra $H_n$ of type A are indexed by partitions $\lambda$ with $n$ boxes.
(b)	$\text{dim}\hspace{0.17em}{H}^{\lambda }=$ # of standard tableaux of shape $\lambda \text{.}$
(c)	The irreducible $H_n\text{-module}$ $H^{\lambda }$ is given by the vector space $$H^{\lambda }=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}\lambda \}$$ with basis $\left\{v_T\right\}$ and with $H_n$ action given by $$T_i{v}_T=\left(\frac{q-q^{-1}}{1-q^{2(c\left(T\left(i\right)\right)-c\left(T(i+1)\right))}}\right)v_T+(q^{-1}+\frac{q-q^{-1}}{1-q^{2(c\left(T\left(i\right)\right)-c\left(T(i+1)\right))}})v_{s_{i}T},$$ where $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$ $s_{i}T$ is the same as $T$ except that $i$ and $i+1$ are switched, $v_{s_{i}T}=0$ if $s_{i}T$ is not standard.

The proof of this theorem is exactly analogous to the proof of Theorem ??? once one knows what the analogues of the Murphy elements are for this setting.

Define elements $X^{{\epsilon }_k},$ $1\le k\le n,$ in the Iwahori-Hecke algebra $H_n$ of type A by $$X^{{\epsilon }_1}=1,\text{and}{X}^{{\epsilon }_k}=T_{k-1}{T}_{k-2}\cdots T_2{T}_1^2{T}_2\cdots T_{k-2}{T}_{k-1},2\le k\le n,$$ The action of $X^{{\epsilon }_k}$ on the $H_n\text{-module}$ $H^{\lambda }$ is given by $$X^{{\epsilon }_k}{v}_T=q^{2c\left(T\left(k\right)\right)}{v}_T,\text{for all standard tableaux}\hspace{0.17em}T,$$ where $c\left(T\left(k\right)\right)$ is the content of the box containing $k$ in $T\text{.}$

		Proof.
	The proof is by induction on $k$ using the relation $$X^{{\epsilon }_k}=T_{k-1}{X}^{{\epsilon }_{k-1}}{T}_{k-1}\text{.}$$ Clearly, $X^{{\epsilon }_1}{v}_T=v_T=q^{2c\left(T\left(1\right)\right)}{v}_T,$ since $c\left(T\left(1\right)\right)=0$ for all standard tableaux $T\text{.}$ By the induction assumption and the definition of the action of $T_{k-1}$ on $H^{\lambda },$ $$\begin{array}{ccc}{X}^{{\epsilon }_k}{v}_T& =& T_{k-1}{X}^{{\epsilon }_{k-1}}{T}_{k-1}{v}_T\\ & =& T_{k-1}(\frac{q^{2c\left(T(k-1)\right)}(q-q^{-1})}{1-q^{2(c\left(T\left(k\right)\right)-c\left(T(k-1)\right))}}{v}_T+q^{2c\left(T\left(k\right)\right)}(q^{-1}+\frac{(q-q^{-1})}{1-q^{2(c\left(T\left(k\right)\right)-c\left(T(k-1)\right))}})v_{s_{k-1}}T)\\ & =& q^{2c\left(T\left(k\right)\right)}{T}_{k-1}(T_{k-1}-(q-q^{-1}))v_T=q^{2c\left(T\left(k\right)\right)}{v}_T\text{.}\end{array}$$ $\square$

The hyperoctahedral group

The hyperoctahedral group $W{B}_n$ is the group of signed permutations of $1,2,\dots ,n,$ i.e. bijections $w:\{-n,\dots ,-2,-1,1,2,\dots ,n\}\to \{-n,\dots ,-2,-1,1,2,\dots ,n\}$ such that $w(-i)=-w\left(i\right)\text{.}$ There are multiple notations for signed permutations

(1)	Two line notation: where $w\left(i\right)$ in the second line is below $i$ in the first line, for $1\le i\le n\text{.}$ $$w=\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& -1& 5& -6& 2& -4\end{array}\right)$$
(2)	In cycle notation as permutations in the symmetric group $S_{2n}\text{.}$ $$w=(1,3,5,2,-1,-3,-5,-2)(4,-6)\text{.}$$
(3)	Matrix notation: where the $(\left|w\left(i\right)\right|,i)$ entry is $1$ if $w\left(i\right)$ is positive and $-1$ if $w\left(i\right)$ is negative, and all other entries are $0\text{.}$ $$w=\left(\begin{array}{cccccc}0& -1& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0\\ 1& 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0& -1\\ 0& 0& 1& 0& 0& 0\\ 0& 0& 0& -1& 0& 0\end{array}\right)$$
(4)	Diagram notation: where the $i\text{th}$ dot in the top row is connected to the $\left|w\left(i\right)\right|\text{th}$ dot in the bottom row and the edge is labeled by $-1$ if $w\left(i\right)$ is negative. $$w=\begin{array}{c} -1 -1 -1 \end{array}$$

The hyperoctahedral group is also called the Weyl group of type $B_n$ and the Weyl group of type $C_n$ and is the same as the group $$O_n\left(\mathbb{Z}\right)=\{A\in M_n\left(\mathbb{Z}\right)\hspace{0.17em}|\hspace{0.17em}A{A}^t=\text{Id}\}\text{.}$$ It is the group of $n\times n$ matrices such that

(a)	there is exactly one nonzero entry in each row and each column,
(b)	each nonzero entry is $\pm 1\text{.}$

The group $W{B}_n$ is isomorphic to the wreath product $(\mathbb{Z}/2\mathbb{Z})\wr S_n$ and has order $2^{n}n!\text{.}$

The reflections in $W{B}_n$ are the elements $$s_{{\epsilon }_i}=(i,-i)s_{{\epsilon }_i-{\epsilon }_j}=(i,j)(-i,-j),s_{{\epsilon }_i+{\epsilon }_j}=(i,-j)(-i,j)$$ The simple reflections are $$s_1=(1,-1)=\begin{array}{c} -1 ⋯ \end{array}\text{and}{s}_i=(i-1,i)=\begin{array}{c} -1 ⋯ ⋯ \end{array}$$ for $2\le i\le n\text{.}$

The hyperoctahedral group $W{B}_n$ can be presented by generators $s_1,s_2,\dots ,s_{n-1}$ and relations $$\begin{array}{cccc}{s}_i{s}_j& =& s_j{s}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ s_i{s}_{i+1}{s}_i& =& s_{i+1}{s}_i{s}_{i+1},& 2\le i\le n-1,\\ s_1{s}_2{s}_1{s}_2& =& s_2{s}_1{s}_2{s}_1,\\ s_i^2& =& 1,& 1\le i\le n\text{.}\end{array}$$

		Proof.
	There are three things to show: (1) The simple reflections in $W{B}_n$ satisfy the given relations. (2) The simple reflections in $W{B}_n$ generate $W{B}_n\text{.}$ (3) The group given by generators $s_1,\dots ,s_n$ and the relations in the statement has order $2^{n}n!\text{.}$ (1) The following pictures show that $s_1,\dots ,s_n\in W{B}_n$ satisfy the relations in the statement of the theorem. $$\begin{array}{ccccc}\begin{array}{c} -1 -1 \end{array}=\begin{array}{c} \end{array}& & \begin{array}{c} -1 \end{array}=\begin{array}{c} -1 \end{array}=\begin{array}{c} -1 \end{array}& & \begin{array}{c} -1 -1 \end{array}=\begin{array}{c} -1 -1 \end{array}=\begin{array}{c} -1 -1 \end{array}\end{array}$$ (2) For each $1\le i\le n$ let $$t_i=(i,-i)=\begin{array}{c} -1 ith ⋯ ⋯ \end{array}=s_i{s}_{i-1}\cdots s_3{s}_2{s}_1{s}_2{s}_3\cdots s_{i-1}{s}_i\text{.}$$ Every $w\in W{B}_n$ can be written as $w=\pi t_{i_1}\cdots t_{i_k}$ where $\pi \in S_n$ is the permutation given by $\pi \left(i\right)=\left|w\left(i\right)\right|$ and $\{i_1,\dots ,i_k\}=\left\{i\hspace{0.17em}\right|\hspace{0.17em}w\left(i\right)\hspace{0.17em}\text{is negative}\}\text{.}$ Pictorially, $$w=\begin{array}{c} -1 -1 -1 \end{array}=\begin{array}{c} -1 -1 -1 \end{array}=\pi t_1{t}_3{t}_6$$ Since $s_2,\dots ,s_n$ generate $S_n$ and $t_k=s_k\cdots s_2{s}_1{s}_2\cdots s_k$ for $1\le k\le n,$ it follows that $s_1,\dots ,s_n$ generate $W{B}_n\text{.}$ Let $G_n$ be the free group generated by $s_1,s_2,\dots ,s_n$ modulo the relations in the statement of the theorem. We will show that every element $w\in G_n$ is either (a) $w\in G_{n-1},$ (b) $w=w_1{s}_n{s}_{n-1}\cdots s_k,$ with $w_1\in G_{n-1},$ $2\le k\le n,$ (c) $w=w_1{s}_n{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_k,$ with $w_1\in G_{n-1},$ $1\le k\le n\text{.}$ Let $w\in G_n$ and assume that $$w=w_1{s}_n{w}_2{s}_n{w}_3{s}_n\cdots w_{\ell -1}{s}_n{w}_{\ell },\text{where}\hspace{0.17em}{w}_j\in G_{n-1}\text{.}$$ First we will show that every element of $G_n$ can be written in the form $$w=w_1{s}_n{w}_2{s}_n,\text{with}\hspace{0.17em}{w}_1{w}_2\in G_{n-1}\text{.}$$ Suppose $w=w_1{s}_n{s}_2{s}_n{w}_3,$ with $w_1,w_2,w_3\in G_{n-2}\text{.}$ Then, by the induction assumption, $$\begin{array}{ccc}w& =& w_1{s}_{n}a{s}_{n-1}b{s}_{n-1}{s}_{n}c{s}_{n-1}d{s}_{n-1}\\ & =& w_{1}a{s}_n\underbrace{s_{n-1}b{s}_{n-1}c}{s}_n{s}_{n-1}d{s}_{n-1}\\ & =& w_{1}a{s}_{n}x{s}_{n-1}y{s}_{n-1}{s}_n{s}_{n-1}d{s}_{n-1}\\ & =& w_{1}ax{s}_n{s}_{n-1}y{s}_n{s}_{n-1}{s}_{n}d{s}_{n-1}\\ & =& w_{1}ax{s}_n{s}_{n-1}{s}_{n}y{s}_{n-1}{s}_{n}d{s}_{n-1}\\ & =& \underbrace{w_{1}ax{s}_{n-1}}{s}_n\underbrace{s_{n-1}y{s}_{n}d}{s}_n{s}_{n-1}\\ & =& A{s}_n\beta s_{n-1}\gamma s_{n-1}{s}_n{s}_{n-1}\\ & =& A\beta s_n{s}_{n-1}\gamma s_n{s}_{n-1}{s}_n\\ & =& A\beta s_n{s}_{n-1}{s}_n\gamma s_{n-1}{s}_n\\ & =& \underbrace{A\beta s_{n-1}}{s}_n\underbrace{s_{n-1}\gamma s_{n-1}}{s}_n\\ & =& w_1^{\prime }{s}_n{w}_2^{\prime }{s}_n\text{.}\end{array}$$ It follows that if $w=w_1{s}_n{w}_2{s}_n{w}_3{s}_n$ with $w_1,w_2,w_3\in G_{n-2}$ then $w=w_1^{\prime }{s}_n{w}_2^{\prime }{s}_n{s}_n=w_1^{\prime }{s}_n{w}_2^{\prime }\text{.}$ Then, by the induction assumption $$w=w_1{s}_n{w}_2{s}_n,$$ with (1) $w_2\in G_{n-2},$ or (2) $w_2=a{s}_{n-1}\cdots s_k$ and $a\in G_{n-2},$ or (3) $w_2=a{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_k$ and $a\in G_{n-2}\text{.}$ Case 1. $w=s_1{s}_n{s}_2{s}_n=w_1{w}_2{s}_n{s}_n=w_1{w}_2\in G_{n-1}\text{.}$ Case 2. $$\begin{array}{ccc}w& =& w_1{s}_{n}a{s}_{n-1}\cdots s_k{s}_n\\ & =& w_{1}a{s}_n{s}_{n-1}{s}_n{s}_{n-2}\cdots s_k\\ & =& w_{1}a{s}_{n-1}{s}_n{s}_{n-1}{s}_{n-2}\cdots s_k\\ & =& w_1^{\prime }{s}_n{s}_{n-1}\cdots s_k,\text{with}\hspace{0.17em}{w}_1^{\prime }\in G_{n-1}\text{.}\end{array}$$ Case 3. $$\begin{array}{ccc}w& =& w_1{s}_{n}a{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_k{s}_n\\ & =& \{\begin{array}{cc}{w}_{1}a{s}_n{s}_{n-1}{s}_n{s}_{n-2}\cdots s_2{s}_1{s}_2\cdots s_k,& \text{if}\hspace{0.17em}k<n-1,\\ w_{1}a{s}_n{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_{n-1}{s}_n,& \text{if}\hspace{0.17em}k=n-1,\end{array}\\ & =& \{\begin{array}{cc}{w}_{1}a{s}_{n-1}{s}_n{s}_{n-1}{s}_{n-2}\cdots s_2{s}_1{s}_2\cdots s_k,& \text{if}\hspace{0.17em}k<n-1,\\ w_{1}a{s}_n{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_{n-1}{s}_n,& \text{if}\hspace{0.17em}k=n-1,\end{array}\\ & =& w\prime a{s}_n{s}_{n-1}\cdots s_2{s}_1{s}_2\cdots s_k,with\hspace{0.17em}{w}_1^{\prime }\in G_{n-1}\text{.}\end{array}$$ So $\text{Card}\left(G_n\right)\le \text{Card}\left(G_{n-1}\right)·2n\text{.}$ So $\text{Card}\left(G_n\right)\le 2^{n}n!\text{.}$ $\square$

Let $(\alpha ,\beta )$ be a pair of partitions such that the total number of boxes in $\alpha$ and $\beta$ is $n\text{.}$ A standard tableau of shape $(\alpha ,\beta )$ is a filling $T$ of the boxes of $\alpha$ and $\beta$ with $1,2,\dots ,n$ such that, in each partition,

(a)	the rows of $T$ are increasing left to right,
(b)	the columns of $T$ are increasing top to bottom.

$$T=\begin{array}{c} 1 5 7 4 10 12 8 13 2 3 9 6 11 α β \end{array}$$ The rows and columns of each partition are indexed as for matrices, $\begin{array}{c}T\left(i\right)=\text{box containing}\hspace{0.17em}i\hspace{0.17em}\text{in}\hspace{0.17em}T,\\ c\left(b\right)=j-i\text{, if the box}\hspace{0.17em}b\hspace{0.17em}\text{is in position}\hspace{0.17em}(i,j),\\ \text{sgn}\left(b\right)=\{\begin{array}{cc}1,& \text{if}\hspace{0.17em}b\hspace{0.17em}\text{is in}\hspace{0.17em}\alpha ,\\ -1,& \text{if}\hspace{0.17em}b\hspace{0.17em}\text{is in}\hspace{0.17em}\beta ,\end{array}\end{array}$ The numbers $c\left(b\right)$ and $\text{sgn}\left(b\right)$ are the content and the sign of the the box $b,$ respectively. $$\begin{array}{cc}\begin{array}{c} 0 1 2 -1 0 1 -2 -1 0 1 2 -1 0 \\ \text{Contents of boxes}\end{array}& \begin{array}{c} 1 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 \\ \text{Signs of boxes}\end{array}\end{array}$$

(a)	The irreducible representations $S^{(\alpha ,\beta )}$ of the hyperoctahedral group $W{B}_n$ are indexed by pairs of partitions $(\alpha ,\beta )$ with $n$ boxes total.
(b)	$\text{dim}\hspace{0.17em}{S}^{(\alpha ,\beta )}=$ # of standard tableaux of shape $(\alpha ,\beta )\text{.}$
(c)	The irreducible $W{B}_n\text{-module}$ $S^{(\alpha ,\beta )}$ is given by $$S^{(\alpha ,\beta )}=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}(\alpha ,\beta )\}$$ with basis $\left\{v_T\right\}$ and with $W{B}_n$ action given by $$\begin{array}{ccc}{s}_1{v}_T& =& \text{sgn}\left(T\left(1\right)\right)v_T,\\ s_i{v}_T& =& {\left(s_i\right)}_{TT}{v}_T+(1+{\left(s_i\right)}_{TT})v_{s_{i}T},& i=2,3,\dots ,n,\end{array}$$ where $s_1=(1,-1)$ and $s_i=(i,i-1)(-i,-(i-1)),$ ${\left(s_i\right)}_{TT}=\{\begin{array}{cc}\frac{1}{c\left(T\left(i\right)\right)-c\left(T(i-1)\right)},& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)=\text{sgn}\left(T(i-1)\right),\\ 0,& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)\ne \text{sgn}\left(T(i-1)\right),\end{array}$ $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$ $\text{sgn}\left(T\left(i\right)\right)$ is the sign of the box containing $i$ in $T,$ $s_{i}T$ is the same as $T$ except that $i$ and $i-1$ are switched and $v_{s_{i}T}=0,$ if $s_{i}T$ is not a standard tableau.

		Proof.
	We must show that (1) The $S^{(\alpha ,\beta )}$ are $W{B}_n$ modules, (2) The $S^{(\alpha ,\beta )}$ are irreducible $W{B}_n\text{-modules,}$ (3) The $S^{(\alpha ,\beta )}$ are inequivalent $W{B}_n\text{-modules,}$ (4) These are all the simple $W{B}_n$ modules. The proofs are similar to the proofs of the analogous statements in the symmetric group case. $\square$

Define elements $x_k,$ $1\le k\le n,$ in the group algebra of the hyperoctahedral group $W{B}_n$ by $$\begin{array}{c}{y}_k=(k,-k),\text{for}1\le k\le n,\\ x_1=0,\text{and}{x}_k=\sum _{i<k}(i,k)(-i,-k)+(i,-k)(-i,k),2\le k\le n\text{.}\end{array}$$ The action of these elements on the $W{B}_n\text{-module}$ $S^{(\alpha ,\beta )}$ is given by $$y_k{v}_T=\text{sgn}\left(T\left(k\right)\right)v_T\text{and}{x}_k{v}_T=c\left(T\left(k\right)\right)v_T,$$ where $\text{sgn}\left(T\left(k\right)\right)$ and $c\left(T\left(k\right)\right)$ are the sign and the content of the box containing $k$ in $T,$ respectively.

		Proof.
	By induction, $$\begin{array}{ccc}{y}_k{v}_T& =& s_k{y}_{k-1}{s}_k{v}_T\\ & =& s_k{y}_{k-1}({\left(s_k\right)}_{TT}{v}_t+(1+{\left(s_k\right)}_{TT})v_{s_{k}T})\\ & =& s_k(\text{sgn}\left(T(k-1)\right){\left(s_k\right)}_{TT}{v}_T+\text{sgn}\left(T\left(k\right)\right)(1+{\left(s_k\right)}_{TT})v_{s_{k}T})\\ & =& \text{sgn}\left(T\left(k\right)\right)s_k(s_k{v}_T+\frac{(\text{sgn}\left(T(k-1)\right)-\text{sgn}\left(T\left(k\right)\right))(1+\text{sgn}\left(T\left(k\right)\right)\text{sgn}\left(T(k-1)\right))}{c\left(T\left(k\right)\right)-c\left(T(k-1)\right)})v_T\\ & =& \text{sgn}\left(T\left(k\right)\right)v_T\text{.}\end{array}$$ $\square$

(d) Note that $$\text{dim}\left(S^{(\alpha ,\beta )}\right)=\left(\genfrac{}{}{0ex}{}{n}{\left|\alpha \right|}\right)f^{\alpha }{f}^{\beta },$$ where $f^{\alpha }$ is the number of standard tableaux of shape $\alpha \text{.}$ Thus $f^{\alpha }$ is the dimension of the irreducible module $S^{\alpha }$ for the symmetric group $S_{\left|\alpha \right|}\text{.}$ Then $$\begin{array}{ccc}\sum _{\alpha ,\beta }\text{dim}{\left(S^{(\alpha ,\beta )}\right)}^2& =& \sum _{\alpha ,\beta }{\left(\genfrac{}{}{0ex}{}{n}{\left|\alpha \right|}\right)}^2{\left(f^{\alpha }\right)}^2{\left(f^{\beta }\right)}^2\\ & =& \sum _{k=0}^n{\left(\genfrac{}{}{0ex}{}{n}{k}\right)}^2\left(\sum _{\alpha ⊢k}{\left(f^{\alpha }\right)}^2\right)\left(\sum _{\beta ⊢n-k}{\left(f^{\beta }\right)}^2\right)\\ & =& \sum _{k=0}^n\frac{n!}{k!(n-k)!}k!(n-k)!\left(\genfrac{}{}{0ex}{}{n}{k}\right)\\ & =& \sum _{k=0}^{n}n!\left(\genfrac{}{}{0ex}{}{n}{k}\right)=2^{n}n!\text{.}\end{array}$$

The Iwahori-Hecke algebra of type $B_n$

Fix $p,q\in {ℂ}^{*}\text{.}$ The Iwahori-Hecke algebra $H$ of type $B_n$ is the algebra given by generators $T_1,\dots ,T_n$ and relations $$\begin{array}{c}{T}_i{T}_j=t_j{T}_i,\text{if}\hspace{0.17em}|i-j|>1,\\ T_1{T}_2{T}_1{T}_2=T_2{T}_1{T}_2{T}_1\text{and}{T}_i{T}_{i+1}{T}_i=T_{i+1}{T}_i{T}_{i+1}\\ T_1^2=(p-p^{-1})T_1+1\text{and}{T}_i^2=(q-q^{-1})T_i+1,2\le i\le n\text{.}\end{array}$$

The Iwahori-Hecke algebra $H{B}_n$ has dimension $\le 2^{n}n!\text{.}$

(a)	The irreducible representations $H^{(\alpha ,\beta )}$ of $H{B}_n$ are indexed by pairs of partitions $(\alpha ,\beta )$ with $n$ boxes total.
(b)	$\text{dim}\hspace{0.17em}{H}^{(\alpha ,\beta )}=$ # of standard tableaux of shape $(\alpha ,\beta )\text{.}$
(c)	The irreducible $H{B}_n\text{-module}$ $H^{(\alpha ,\beta )}$ is given by $$S^{(\alpha ,\beta )}=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\hspace{0.17em}(\alpha ,\beta )\}$$ with basis $\left\{v_T\right\}$ and with $H{B}_n$ action given by $$\begin{array}{ccc}{T}_1{v}_T& =& \text{sgn}\left(T\left(1\right)\right)p^{\text{sgn}\left(T\left(1\right)\right)}{q}^{c\left(T\left(1\right)\right)}{v}_T,\\ T_i{v}_T& =& {\left(T_i\right)}_{TT}{v}_T+(1+{\left(T_i\right)}_{TT})v_{s_{i}T},& i=2,3,\dots ,n,\end{array}$$ where $${\left(T_i\right)}_{TT}=\frac{q-q^{-1}}{1-(CT\left(T(k-1)\right)/CT\left(T\left(k\right)\right))},\text{with}CT\left(T\left(k\right)\right)=\text{sgn}\left(T\left(k\right)\right)p^{\text{sgn}\left(T\left(k\right)\right)}{q}^{c\left(T\left(k\right)\right)},$$ $c\left(T\left(i\right)\right)$ is the content of the box containing $i$ in $T,$ $\text{sgn}\left(T\left(i\right)\right)$ is the sign of the box containing $i$ in $T,$ $s_{i}T$ is the same as $T$ except that $i$ and $i-1$ are switched, and $v_{s_{i}T}=0,$ if $s_{i}T$ is not a standard tableau.

		Proof.
	Let us show that the action satisfies the given relations. The relations $T_i{T}_j=T_j{T}_i,$ $|i-j|>1,$ $i,j\ge 2,$ and $T_i{T}_{i+1}{T}_i=T_{i+1}{T}_i{T}_{i+1},$ $2\le i\le n$ are taken care of by the case of the Iwahori-Hecke algebra of type $A\text{.}$ $\square$

Define $$X^{{\epsilon }_1}=T_1,V^{{\epsilon }_i}=T_i{T}_{i-1}\cdots T_2{T}_1{T}_2\cdots T_{i-1}{T}_i,$$ in $H{B}_n\text{.}$ Then The action of $X^{{\epsilon }_i},$ $1\le i\le n,$ on the $H{B}_n\text{-module}$ $H^{(\alpha ,\beta )}$ is given by $$X^{{\epsilon }_i}{v}_T=CT\left(T\left(i\right)\right)v_T,$$

		Proof.
	By the definition of the action $T_1{v}_T=CT\left(T\left(1\right)\right)v_T=X^{{\epsilon }_1}{v}_T$ and, by induction, $$\begin{array}{ccc}{X}^{{\epsilon }_i}{v}_T& =& T_i{X}^{{\epsilon }_{i-1}}{T}_i{v}_T\\ & =& T_i(CT\left(T(i-1)\right){\left(T_k\right)}_{TT}{v}_T+CT\left(T\left(i\right)\right)(q^{-1}+{\left(T_k\right)}_{TT})v_{s_{k}T})\\ & =& CT\left(T\left(i\right)\right)T_i(T_i-(q-q^{-1}))v_T\\ & =& CT\left(T\left(i\right)\right)v_T\text{.}\end{array}$$ $\square$

Note that $$\begin{array}{ccc}1-\frac{CT\left(T(i-1)\right)}{CT\left(T\left(i\right)\right)}& =& 1-\frac{\text{sgn}\left(T(i-1)\right)}{\text{sgn}\left(T\left(i\right)\right)}·\frac{p^{\text{sgn}\left(T(i-1)\right)}{q}^{c\left(T(i-1)\right)}}{p^{\text{sgn}\left(T\left(i\right)\right)}{q}^{c\left(T\left(i\right)\right)}}\\ & =& \text{sgn}\left(T\left(i\right)\right)p^{\text{sgn}\left(T\left(i\right)\right)}{q}^{c\left(T\left(i\right)\right)}-\text{sgn}\left(T(i-1)\right)p^{\text{sgn}\left(T(i-1)\right)}{q}^{c\left(T(i-1)\right)}\end{array}$$ When $p=q$ $$\begin{array}{ccc}\frac{q-q^{-1}}{1-\frac{CT\left(T(i-1)\right)}{CT\left(T\left(i\right)\right)}}& =& \frac{q-q^{-1}}{\text{sgn}\left(T\left(i\right)\right)q^{\text{sgn}\left(T\left(i\right)\right)+c\left(T\left(i\right)\right)}-\text{sgn}\left(T(i-1)\right)q^{\text{sgn}\left(T(i-1)\right)+c\left(T(i-1)\right)}}\\ & =& \frac{-\text{sgn}\left(T(i-1)\right)q^{-c\left(T(i-1)\right)-\text{sgn}\left(T(i-1)\right)}(q-q^{-1})}{1-\text{sgn}\left(T(i-1)\right)\text{sgn}\left(T\left(i\right)\right)q^{\text{sgn}\left(T\left(i\right)\right)+c\left(T\left(i\right)\right)-\text{sgn}\left(T(i-1)\right)-c\left(T(i-1)\right)}}\\ & =& \{\begin{array}{cc}1/[c\left(T\left(i\right)\right)-c\left(T(i-1)\right)],& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)=\text{sgn}\left(T(i-1)\right),\\ (q-q^{-1})/(1+q^{c\left(T\left(i\right)\right)-c\left(T(i-1)\right)\pm 2}),& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)=-\text{sgn}\left(T(i-1)\right),\end{array}\end{array}$$ and at $q=1$ this is $$\{\begin{array}{cc}1/(c\left(T\left(i\right)\right)-c\left(T(i-1)\right)),& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)=\text{sgn}\left(T(i-1)\right),\\ 0,& \text{if}\hspace{0.17em}\text{sgn}\left(T\left(i\right)\right)=-\text{sgn}\left(T(i-1)\right)\text{.}\end{array}$$

Let $u_0,u_1,\dots ,u_{r-1}\in ℂ$ and let $q\in {ℂ}^{*}\text{.}$ The cyclotomic algebra $H_{r,1,n}$ is the algebra given by generators $X^{{\epsilon }_1},T_2,T_3,\dots ,T_n$ and relations $$\begin{array}{cccc}{T}_i{T}_j& =& T_j{T}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ T_i{T}_{i+1}{T}_i& =& T_{i+1}{T}_i{T}_{i+1},& 2\le i\le n-1,\\ X^{{\epsilon }_1}{T}_j& =& T_j{X}^{{\epsilon }_1},& j>1,\\ X^{{\epsilon }_1}{T}_2{X}^{{\epsilon }_1}{T}_2& =& T_2{X}^{{\epsilon }_1}{T}_2{X}^{{\epsilon }_1},\\ \multicolumn{3}{c}{(X^{{\epsilon }_1}-u_0)(X^{{\epsilon }_1}-u_1)\cdots (X^{{\epsilon }_1}-u_{r-1})=0,}\\ T_i^2& =& (q-q^{-1})T_i+1,& 2\le i\le n\text{.}\end{array}$$ If $p_j={\xi }^j$ where $\xi =e^{2\pi i/r}$ and $q=1$ then $H_{r,1,n}=ℂG(r,1,n)\text{.}$ Define $$\begin{array}{c}{X}^{{\epsilon }_k}=T_i{T}_{i-1}\cdots T_2{X}^{{\epsilon }_1}{T}_2\cdots T_{i-1}{T}_i,\text{for}\hspace{0.17em}1\le i\le n\text{, and}\\ X^{\lambda }={\left(X^{{\epsilon }_1}\right)}^{{\lambda }_1}\cdots {\left(X^{{\epsilon }_n}\right)}^{{\lambda }_n},\text{for}\hspace{0.17em}\lambda ={\lambda }_1{\epsilon }_1+\cdots {\lambda }_n{\epsilon }_n,{\lambda }_i\in \mathbb{Z}\text{.}\end{array}$$ Then $$ℂ\left[X\right]=ℂ\text{-span}\left\{X^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda ={\lambda }_1{\epsilon }_1+\cdots {\lambda }_n{\epsilon }_n,{\lambda }_i\in \mathbb{Z}\}$$ is a commutative subalgebra of $H_{r,1,n}\text{.}$

If $w\in S_n$ define $T_w=T_{i_1}\cdots T_{i_p}$ if $w=s_{i_1}\cdots s_{i_p}$ and $p$ is as small as possible. The cyclotomic Hecke algebra $H_{r,1,n}$ has basis $$X^{\lambda }{T}_w,\text{where}w\in S_n,\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_n{\epsilon }_n,0\le {\lambda }_j\le r-1,$$ and $\text{dim}\left(H_{r,1,n}\right)=r^{n}n!\text{.}$

(a)	The irreducible representations $H^{\lambda }$ of $H_{r,1,n}$ are indexed by $r\text{-tuples}$ $\lambda =({\lambda }^{\left(0\right)},\dots ,{\lambda }^{(r-1)})$ of partitions with $n$ boxes total.
(b)	$\text{dim}\hspace{0.17em}{H}^{\lambda }=$ # of standard tableaux of shape $\lambda \text{.}$
(c)	The irreducible $H_{r,1,n}\text{-module}$ $H^{\lambda }$ is given by $$H^{\lambda }=ℂ\text{-span}\left\{v_T\hspace{0.17em}\right|\hspace{0.17em}T\hspace{0.17em}\text{is a standard tableau of shape}\lambda =({\lambda }^{\left(0\right)},\dots ,{\lambda }^{(r-1)})\}$$ with basis $\left\{v_T\right\}$ and with $H_{r,1,n}$ action given by $$\begin{array}{ccc}{X}^{{\epsilon }_1}{v}_T& =& CT\left(T\left(1\right)\right)v_T,\\ T_i{v}_T& =& {\left(T_i\right)}_{TT}{v}_T+(q^{-1}+{\left(T_i\right)}_{TT})v_{s_{i}T},& i=2,3,\dots ,n,\end{array}$$ where $${\left(T_i\right)}_{TT}=\frac{q-q^{-1}}{1-(CT\left(T(k-1)\right)/CT\left(T\left(k\right)\right))},\text{with}CT\left(b\right)=u_{s\left(b\right)}{q}^{2c\left(b\right)},$$ $c\left(T\left(i\right)\right)$ is the content of the box $b,$ $T\left(i\right)$ is the box containing $i$ in $T,$ $s_{i}T$ is the same as $T$ except that $i$ and $i-1$ are switched, and $v_{s_{i}T}=0,$ if $s_{i}T$ is not a standard tableau.

The action of the elements $X^{{\epsilon }_i},$ $1\le i\le n,$ on $H^{\lambda }$ is given by $$X^{{\epsilon }_i}{v}_T=CT\left(T\left(i\right)\right)v_T,$$ for all standard tableaux $T$ of shape $\lambda \text{.}$

The Weyl group of type $D_n$

The Weyl group of type $D_n$ is the group of $n\times n$ matrices such that

(a)	there is exactly one nonzero entry in each row and each column,
(b)	the nonzero entries are $\pm 1,$
(c)	the product of the nonzero entries is $1\text{.}$

Condition (c) says that there are an even number of $-1$ in the matrix. The Weyl group $W{D}_n$ is a normal subgroup of index $2$ in $W{B}_n$ and has order $2^{n-1}n!\text{.}$ The reflections in $W{D}_n$ are $$s_{{\epsilon }_i-{\epsilon }_j}=(i,j)(-i,-j),\text{and}{s}_{{\epsilon }_i+{\epsilon }_j}=(i,-j)(-i,j),1\le i<j\le n\text{.}$$ The simple reflections in $W{D}_n$ are $$\begin{array}{ccc}{s}_1& =& (-1,2)(1,-2)=\begin{array}{c} -1 -1 ⋯ \end{array},\\ s_i& =& (i-1,i)=\begin{array}{c} ⋯ ⋯ \end{array},2\le i\le n\text{.}\end{array}$$

The Weyl group of type $D_n$ has a presentation given by generators $s_1,\dots ,s_n$ and relations corresponding to the Dynkin diagram $D_n$ and $s_i^2=1,$ for $1\le i\le n\text{.}$ $$\begin{array}{ccc}\begin{array}{cc}{A}_n& \begin{array}{c} 2 3 n-1 n \end{array}\end{array}& & \begin{array}{cc}{B}_n& \begin{array}{c} 1 2 3 n-1 n \end{array}\end{array}\\ \begin{array}{cc}{D}_n& \begin{array}{c} 1 2 3 4 n-1 n \end{array}\end{array}& & \begin{array}{cc}{E}_6& \begin{array}{c} 2 3 4 5 6 1 \end{array}\end{array}\\ \begin{array}{cc}{E}_7& \begin{array}{c} 2 3 4 5 6 7 1 \end{array}\end{array}& & \begin{array}{cc}{E}_8& \begin{array}{c} 2 3 4 5 6 7 8 1 \end{array}\end{array}\\ \begin{array}{cc}{F}_4& \begin{array}{c} 1 2 3 4 \end{array}\end{array}& & \begin{array}{cc}{H}_3& \begin{array}{c} 1 2 3 5 \end{array}\end{array}\\ \begin{array}{cc}{H}_4& \begin{array}{c} 1 2 3 4 5 \end{array}\end{array}& & \begin{array}{cc}{I}_2\left(m\right)& \begin{array}{c} 1 2 m \end{array}\end{array}\end{array}$$

The affine symmetric group

The affine symmetric group ${\stackrel{\sim }{S}}_n$ is the group of permutations with edges labeled by elements of $\mathbb{Z}\text{.}$ For each $\lambda =({\lambda }_1,\dots ,{\lambda }_n)\in {\mathbb{Z}}^n$ let $$t_{\lambda }=t_{({\lambda }_1,\dots ,{\lambda }_n)}=\begin{array}{c} λ1 λ2 ⋯ λn \end{array}$$ Then every element $\stackrel{\sim }{w}\in {\stackrel{\sim }{S}}_n$ can be written uniquely in the form $$\begin{array}{c}\stackrel{\sim }{w}=t_{\lambda }w,\text{with}\hspace{0.17em}\lambda \in {\mathbb{Z}}^n\hspace{0.17em}\text{and}\hspace{0.17em}w\in S_n\text{.}\\ w=\begin{array}{c} 5 32 0 -7 -61 1 \end{array}=\begin{array}{c} 5 32 0 -7 -61 1 \end{array}=t_{(5,32,0,-7,-61,1)}\left(\begin{array}{cccccc}1& 2& 3& 4& 5& 6\\ 3& 6& 2& 4& 1& 5\end{array}\right)\text{.}\end{array}$$ The multiplication in $S_n$ is determined by $$t_{\lambda }{t}_{\mu }=t_{\lambda +\mu }\text{and}w{t}_{\lambda }=t_{w\lambda }w,w\in S_n,\lambda ,\mu \in {\mathbb{Z}}^n,$$ where $\lambda +\mu =({\lambda }_1+{\mu }_1,\dots ,{\lambda }_n+{\mu }_n)$ and $S_n$ is acting on ${\mathbb{Z}}^n$ by $w\lambda =w({\lambda }_1,\dots ,{\lambda }_n)=({\lambda }_{w\left(1\right)},\dots ,{\lambda }_{w\left(n\right)})\text{.}$ So $${\stackrel{\sim }{S}}_n={\mathbb{Z}}^n⊢S_n\text{.}$$

The affine symmetric group ${\stackrel{\sim }{S}}_n$ has a presentation by generators $$\begin{array}{c}{s}_0=\begin{array}{c} -1 0 0 0 0 1 ⋯ ⋯ \end{array},\omega =\begin{array}{c} -1 0 0 0 0 0 ⋯ ⋯ \end{array},\\ s_i=\begin{array}{c} 0 0 0 0 0 0 0 ⋯ ⋯ \end{array},1\le i\le n-1,\end{array}$$ and relations $$\begin{array}{cccc}{s}_i{s}_j& =& s_j{s}_i,& |i-j|>1,\\ s_i{s}_{i+1}{s}_i& =& s_{i+1}{s}_i{s}_{i+1},\\ s_i^2& =& 1,\\ {\omega }^{-1}{s}_i\omega & =& s_{i+1},\end{array}$$ where the indices on the $s_i$ are always interpreted mod $n\text{.}$

		Proof.
	The theorem is proved by using the relations $$\begin{array}{ccc}{s}_0& =& t_{(-1,0,0,\dots ,0,1)}{s}_{n-1}{s}_{n-2}\cdots s_2{s}_1{s}_2\cdots s_{n-2}{s}_{n-1},\\ \omega & =& t_{(-1,0,0,\dots ,0)}{s}_1{s}_2\cdots s_{n-2}{s}_{n-1},\end{array}$$ Using these expressions one can write $t_{\lambda }$ in terms of $s_0,\dots ,s_{n-1}$ and $\omega$ and prove that the equation $t_{\lambda }{t}_{\mu }=t_{\lambda +\mu },$ for $\lambda ,\mu \in {\mathbb{Z}}^n,$ is a consequence of the relations in the statement of the theorem. $\square$

For each $r$ there is a surjective homomorphism $$\begin{array}{ccc}{\stackrel{\sim }{S}}_n& ⟶& G(r,1,n)\\ t_{\lambda }w& ⟼& t_{\bar{\lambda }}w,\end{array}$$ where $\bar{\lambda }=({\bar{\lambda }}_1,\dots ,{\bar{\lambda }}_n),$ and $\bar{i}=i$ $\left(\text{mod}\hspace{0.17em}r\right)\text{.}$ Thus, if $M$ is a $G(r,1,n)\text{-module}$ there is an action of ${\stackrel{\sim }{S}}_n$ on $M$ given by $$\left(t_{\lambda }w\right)m=\left(t_{\bar{\lambda }}w\right)m,\text{for all}\hspace{0.17em}{t}_{\lambda }w\in {\stackrel{\sim }{S}}_n\text{.}$$ So every irreducible $G(r,1,n)$ module is an irreducible ${\stackrel{\sim }{S}}_n\text{-module.}$ Are these all irreducible ${\stackrel{\sim }{S}}_n\text{-modules?}$ The answer is ??? as we shall see in ???, by using Clifford theory.

The affine Hecke algebra

Let $q\in {ℂ}^{*}\text{.}$ The affine Hecke algebra ${\stackrel{\sim }{H}}_n$ is the algebra given by generators $X^{{\epsilon }_1},T_2,T_3,\dots ,T_n$ and relations $$\begin{array}{cccc}{T}_i{T}_j& =& T_j{T}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ T_i{T}_{i+1}{T}_i& =& T_{i+1}{T}_i{T}_{i+1},& 2\le i\le n-1,\\ X^{{\epsilon }_1}{T}_j& =& T_j{X}^{{\epsilon }_1},& j>1,\\ X^{{\epsilon }_1}{T}_2{X}^{{\epsilon }_1}{T}_2& =& T_2{X}^{{\epsilon }_1}{T}_2{X}^{{\epsilon }_1},\\ T_i^2& =& (q-q^{-1})T_i+1,& 2\le i\le n\text{.}\end{array}$$ If $q=1$ then ${\stackrel{\sim }{H}}_n=ℂ{\stackrel{\sim }{S}}_n,$ the group algebra of the affine symmetric group.

Define $$\begin{array}{c}{X}_{{\epsilon }_k}=T_i{T}_{i-1}\cdots T_2{X}^{{\epsilon }_1}{T}_2\cdots T_{i-1}{T}_i,\text{for}\hspace{0.17em}1\le i\le n\text{, and}\\ X^{\lambda }={\left(X^{{\epsilon }_1}\right)}^{{\lambda }_1}\cdots {\left(X^{{\epsilon }_n}\right)}^{{\lambda }_n},\text{for}\hspace{0.17em}\lambda ={\lambda }_1{\epsilon }_1+\cdots {\lambda }_n{\epsilon }_n,{\lambda }_i\in \mathbb{Z}\text{.}\end{array}$$ Then $$X^{\lambda }{X}^{\mu }=X^{\lambda +\mu },\text{for all}\hspace{0.17em}\lambda ,\mu \in {\mathbb{Z}}^n\text{.}$$ Then $$ℂ\left[X\right]=ℂ\text{-span}\left\{X^{\lambda }\hspace{0.17em}\right|\hspace{0.17em}\lambda ={\lambda }_1{\epsilon }_1+\cdots {\lambda }_n{\epsilon }_n,{\lambda }_i\in \mathbb{Z}\}$$ is a commutative subalgebra of ${\stackrel{\sim }{H}}_n$ isomorphic to the group algebra of ${\mathbb{Z}}^n\text{.}$

The affine Hecke algebra is the unique algebra structure on $$ℂ\left[X\right]\otimes H$$ such that $ℂ\left[X\right]=ℂ\left[X\right]\otimes 1$ and $H=1\otimes H$ are subalgebras and $$X^{\lambda }{T}_i=T_i{X}^{s_i\lambda }+(q-q^{-1})\frac{X^{\lambda }-X^{s_i\lambda }}{1-X^{{\epsilon }_i-{\epsilon }_{i-1}}},$$ for all $\lambda \in {\mathbb{Z}}^n$ and $1\le i\le n\text{.}$

Define $$T_0=X^{{\epsilon }_1-{\epsilon }_n}{T}_n{T}_{n-1}\cdots T_3{T}_2{T}_3\cdots T_n\text{and}\omega =X^{-{\epsilon }_1}{T}_2{T}_3\cdots T_n\text{.}$$

The affine Hecke algbera ${\stackrel{\sim }{H}}_n$ is presented by generators $T_0,T_1,\dots ,T_n,\omega$ and relations $$\begin{array}{cccc}{T}_i{T}_j& =& T_j{T}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ T_i{T}_{i+1}{T}_i& =& T_{i+1}{T}_i{T}_{i+1},\\ \omega T_i{\omega }^{-1}& =& T_{i+1},\\ T_i^2& =& (q-q^{-1})T_i+1,& 0\le i\le n-1\text{.}\end{array}$$ where the indices are always taken modulo $n\text{.}$

If $w\in S_n$ let $T_w=T_{i_1}\cdots T_{i_p}$ if $w=s_{i_1}\cdots s_{i_p}$ and $p$ is minimal. The affine Hecke algebra ${\stackrel{\sim }{H}}_n$ has basis $$X^{\lambda }{T}_w,\text{where}w\in S_n,\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_n{\epsilon }_n,{\lambda }_j\in \mathbb{Z}\text{.}$$

The evaluation homomorphism is the surjective algebra homomorphism defined by $$\begin{array}{ccc}{\stackrel{\sim }{H}}_n& ⟶& H_{r,1,n}\\ T_i& ⟼& T_i,\\ X^{{\epsilon }_k}& ⟼& T_k\cdots T_2{X}^{{\epsilon }_1}{T}_2\cdots T_k,\end{array}$$ Via this homomorphism every irreducible $H_{r,1,n}$ module $M$ is an irreducible ${\stackrel{\sim }{H}}_n$ module. Conversely, Let $M$ be an ${\stackrel{\sim }{H}}_n\text{-module.}$ Let $u_0,\dots ,u_{r-1}\in ℂ$ be such that the minimal polynomial $p\left(t\right)$ of the linear transformation of $M$ determined by the action of $X^{{\epsilon }_1}$ divides the polynomial $(t-u_0)(t-u_1)\cdots (t-u_{r-1})\text{.}$ The the ${\stackrel{\sim }{H}}_n$ action on $M$ is an $H_{r,1,n}(u_0,\dots ,u_{r-1};q)$ action on $M\text{.}$

So any $H_{r,1,n}(u_0,\dots ,u_{r-1};q)\text{-module}$ $M$ is an ${\stackrel{\sim }{H}}_n$ and conversely, any ${\stackrel{\sim }{H}}_n$ module $M$ is an $H_{r,1,n}(u_0,\dots ,u_{r-1};q)$ module for some appropriate choice of $r$ and $u_0,\dots ,u_{r-1}\in ℂ\text{.}$ So ${\stackrel{\sim }{H}}_n$ the representation theory “contains the representation theory” of all the algebras $H_{r,1,n}(u_0,\dots ,u_{r-1};q)\text{.}$

The degenerate affine Hecke algebra

Let $x_1,\dots ,x_n$ be commuting variables and let $ℂ[x_1,\dots ,x_n]$ be the polynomials in $x_1,\dots ,x_n\text{.}$ Let $ℂS_n$ be the group algebra of the symmetric group. The graded Hecke algebra is the vector space $ℂ[x_1,\dots ,x_n]\otimes S_n$ with multiplication such that

(a)	$ℂ[x_1,\dots ,x_n]=ℂ[x_1,\dots ,x_n]\otimes 1$ is a subalgebra,
(b)	$ℂS_n=1\otimes ℂS_n$ is a subalgebra,
(c)	$s_i{x}_{i-1}{s}_i=x_i-s_i,$ for $2\le i\le n\text{.}$

If $\gamma =({\gamma }_1,\dots ,{\gamma }_n)$ is an $n\text{-tuple}$ of nonnegative integers let $x^{\gamma }=x_1^{{\gamma }_1}\cdots x_n^{{\gamma }_n}\text{.}$ The elements $$\left\{x^{\gamma }w\hspace{0.17em}\right|\hspace{0.17em}\gamma =({\gamma }_1,\dots ,{\gamma }_n),{\gamma }_i\in {\mathbb{Z}}_{\ge 0},w\in S_n\}$$ form a basis of $H_n\text{.}$ The map $$\begin{array}{ccc}{H}_n& ⟶& ℂS_n\\ s_i& ⟼& s_i\\ x_1& ⟼& 0\\ x_k& ⟼& \sum _{i<k}\left(ik\right)& 2\le k\le n,\end{array}$$ is a surjective algebra homomorphism.

Lusztig’s approah to the passage from the affine Hecke algebra to the graded Hecke algebra is as follows. Let $h:ℂ\left[X\right]\to {ℂ}^{*}$ be $W\text{-invariant}$ $h\left(X^{\alpha }\right)=1\text{.}$ Then $I=\text{ker}\hspace{0.17em}h$ is a maximal ideal of $ℂ\left[X\right]\text{.}$ Then the associated graded of the filtration $$ℂ\left[X\right]\supseteq I\supseteq I^2\supseteq \cdots \text{is}S\left({𝔥}^{*}\right)=\text{gr}ℂ\left[X\right]\text{.}$$ The “derivative” of $f$ is the image $d\left(f\right)$ of $f-h\left(f\right)$ in $I/I^2\text{.}$ Then $$d\left(ff\prime \right)=h\left(f\right)d\left(f\prime \right)+h\left(f\prime \right)d\left(f\right),\text{for}\hspace{0.17em}f,f\prime \in ℂ\left[X\right]\text{.}$$ Then $\stackrel{\sim }{H}$ is a $ℂ[X]$ module and we have a filtration $$\stackrel{\sim }{H}\supseteq I\stackrel{\sim }{H}\supseteq I^2\stackrel{\sim }{H}\supseteq \cdots$$ such that $\left(I^{k}H\right)\left(I^{\ell }H\right)\subseteq I^{k+\ell }H\text{.}$ So consider $$\text{gr}\stackrel{\sim }{H}=\bar{H}=\underset{k\ge 0}{⨁}{\bar{H}}^k,\text{where}{\bar{H}}^k=\left(I^k\stackrel{\sim }{H}\right)/\left(I^{k+1}\stackrel{\sim }{H}\right)\text{.}$$ Let $w$ be the image of $T_w$ in $\stackrel{\sim }{H}/I\stackrel{\sim }{H}={\bar{H}}^0\text{.}$ $$r=d\left(q\right)\in I/I^2\text{.}$$ Then $\bar{H}$ is the graded Hecke algebra (Prop. 4.4 in [Lus1989]). Prop. 4.5 in [Lus1989] says that $$Z\left(\bar{H}\right)=S{\left({𝔥}^{*}\right)}^W\text{.}$$ There should be an analogue of the Pittie-Ram theorem for the graded Hecke algebra.

The degenerate affine Hecke algebra ${\stackrel{\sim }{G}}_n$ is the algebra generated by $ℂ[x_1,\dots ,x_n]$ and $ℂS_n$ with the relations $$\begin{array}{cccc}{x}_i{T}_j& =& T_j{x}_i,& \text{if}\hspace{0.17em}|i-j|>1,\\ s_i{x}_{}& =& x_{i+1}{s}_i-1,\end{array}$$

There is a surjective “evaluation” homomrphism $$\begin{array}{ccc}{\stackrel{\sim }{G}}_n& ⟶& ℂS_n\\ s_i& ⟼& s_i,x_k& ⟼& {\sum }_{i<k}(i,k)\end{array}$$ The degenerate affine Hecke algebra is obtained from the affine Hecke algebra by setting $x_k$ to be the “derivative” of $X^{{\epsilon }_k}$ at $q=1,$ $$x_k=\frac{X^{{\epsilon }_k}-1}{q-1}{|}_{q=1}\text{.}$$

Notes and references

This is a typed exert of Representation theory Lecture notes: Chapter 2 by Arun Ram. Research supported in part by National Science Foundation grant DMS-9622985.

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