A permutation representation

A permutation representation

Let $T\stackrel{\text{def}}{=}\{ws{w}^{-1}:s\in S,w\in W\}$. The elements of $T$ (i.e., the elements conjugate to some Coxeter generator) are called reflections. The definition shows that $S\subseteq T$ and that $t^2=e,\text{for all }t\in T.$ The elements of $S$ are sometimes called simple reflections. Given a word $s_1{s}_2\dots s_k\in S^{\ast }$, define $t_i\stackrel{\text{def}}{=}{s}_1{s}_2\dots s_{i-1}{s}_i{s}_{i-1}\dots s_2{s}_1,\text{for }1\le i\le k,$ and the ordered $k$-tuple $\hat{T}(s_1{s}_2\dots s_k)\stackrel{\text{def}}{=}(t_1,t_2,\dots ,t_k).$ Note that $t_i=(s_1\dots s_{i-1})s_i(s_1\dots s_{i-1}{)}^{-1}\in T,$ $t_i{s}_1{s}_2\dots s_k=s_1\dots {\hat{s}}_i\dots s_k(s_i\text{ omitted}),$ and $s_1{s}_2\dots s_i=t_i{t}_{i-1}\dots t_1.$

Lemma

If $w=s_1{s}_2\dots s_k$, with $k$ minimal, then $t_i\ne t_j$ for all $1\le i<j\le k$.

$n(s_1{s}_2\dots s_k;t)\stackrel{\text{def}}{=}\text{the number of times }t\text{ appears in }\hat{T}(s_1{s}_2\dots s_k).$ Furthermore, for $s\in S$ and $t\in T$, let $\eta (s;t)\stackrel{\text{def}}{=}\{\begin{array}{ll}-1,& \text{if }s=t,\\ +1,& \text{if }s\ne t.\end{array}$ Note that $(-1{)}^{n(s_1{s}_2\dots s_k;t)}={\prod }_{i=1}^k\eta (s_i;s_{i-1}\dots s_{1}t{s}_1\dots s_{i-1}).$ We will consider the group $S(R)$ of all permutations of the set $R=T\times \{+1,-1\}.$ For $s\in S$, define a mapping ${\pi }_s$ of $R$ to itself by ${\pi }_s(t,\epsilon )\stackrel{\text{def}}{=}(sts,\epsilon \eta (s;t)),$and ${\pi }_s\in S(R)$.

Theorem

(i) The mapping $s\mapsto {\pi }_s$ extends uniquely to an injective homomorphism $w\mapsto {\pi }_w$ from $W$ to $S(R)$. (ii) ${\pi }_t(t,\epsilon )=(t,-\epsilon )$, for all $t\in T$.

Proof.

(1) ${\pi }_s^2={\text{id}}_R$. ${\pi }_s^2(t,\epsilon )={\pi }_s(sts,\epsilon \eta (s;t))=(sstss,\epsilon \eta (s;t)\eta (s;sts))=(t,\epsilon )$ (2) Let $s,s^{\prime }\in S$ and $m(s,s^{\prime })=p\ne \infty$. We claim that $({\pi }_s{\pi }_{s^{\prime }}{)}^p={\text{id}}_R.$ To prove this, let $s_i=\{\begin{array}{ll}{s}^{\prime },& \text{if }i\text{ is odd},\\ s,& \text{if }i\text{ is even}\end{array}$ and let $\mathbf{s}$ denote the word $s_1{s}_2\dots s_{2p}=s^{\prime }s{s}^{\prime }s{s}^{\prime }\dots s^{\prime }s$. Let $\hat{T}(\mathbf{s})=(t_1,t_2,\dots ,t_{2p})$; that is, $t_i=s_1{s}_2\dots s_i\dots s_2{s}_1=(s^{\prime }s{)}^{i-1}{s}^{\prime },1\le i\le 2p.$ Since $(s^{\prime }s{)}^p=e$, we have that $t_{p+i}=t_i,1\le i\le p,$ which implies that $n(\mathbf{s};t)$ is even for all $t\in T$. Let $(t^{\prime },{\epsilon }^{\prime })=({\pi }_s{\pi }_{s^{\prime }}{)}^p(t,\epsilon )={\pi }_{s_{2p}}{\pi }_{s_{2p-1}}\dots {\pi }_{s_1}(t,\epsilon ).$ Then, $t^{\prime }=s_{2p}\dots s_{1}t{s}_1\dots s_{2p}=t$, since $s_1{s}_2\dots s_{2p}=(s^{\prime }s{)}^p=e$. Furthermore, using (1.15), we get ${\epsilon }^{\prime }=\epsilon {\prod }_{i=1}^{2p}\eta (s_i;s_{i-1}\dots s_{1}t{s}_1\dots s_{i-1})=\epsilon (-1{)}^{n(\mathbf{s};t)}=\epsilon .$ (3) By the universality property and what has just been shown, the mapping $s\mapsto {\pi }_s$ extends to a homomorphism $w\mapsto {\pi }_w$ of $W$. If $w=s_k{s}_{k-1}\dots s_1$, we compute $\begin{array}{rl}{\pi }_w(t,\epsilon )& ={\pi }_{s_k}{\pi }_{s_{k-1}}\dots {\pi }_{s_1}(t,\epsilon )\\ & =\left(s_k\dots s_{1}t{s}_1\dots s_k,\epsilon \prod _{i=1}^k\eta (s_i;s_{i-1}\dots s_{1}t{s}_1\dots s_{i-1})\right)\\ & =(wt{w}^{-1},\epsilon (-1{)}^{n(s_1{s}_2\dots s_k;t)}).\end{array}$ In particular, the parity of $n(s_1{s}_2\dots s_k;t)$ only depends on $w$ and $t$. (4) Suppose that $w\ne e$. Choose an expression $w=s_k{s}_{k-1}\dots s_1$ with $k$ minimal, and let $\hat{T}(s_1{s}_2\dots s_k)=(t_1,t_2,\dots ,t_k)$. By Lemma 1.3.1, all $t_i$‘s are distinct, so $n(s_1{s}_2\dots s_k;t_i)=1$. Therefore, ${\pi }_w(t_i,\epsilon )=(w{t}_i{w}^{-1},-\epsilon )$ for $1\le i\le k$ by equation (1.16), which shows that ${\pi }_w\ne {\text{id}}_R$. Hence, the homomorphism is injective. (5) We show part (ii) of the theorem by induction on the size of a symmetric expression for $t$. Let $t=s_1{s}_2\dots s_p\dots s_2{s}_1,s_i\in S.$ The case $p=1$ is clear by definition. Then, by induction, $\begin{array}{rl}{\pi }_{s_1\dots s_p\dots s_1}(s_1\dots s_p\dots s_1,\epsilon )& ={\pi }_{s_1}{\pi }_{s_2\dots s_p\dots s_2}(s_2\dots s_p\dots s_2,\epsilon \eta (s_1;s_1\dots s_p\dots s_1))\\ & ={\pi }_{s_1}(s_2\dots s_p\dots s_2,-\epsilon \eta (s_1;s_2\dots s_p\dots s_2))\\ & =(s_1\dots s_p\dots s_1,-\epsilon {\eta }^2(s_1;s_2\dots s_p\dots s_2))\\ & =(t,-\epsilon ).\end{array}$

\tag*{$\square$}