Transvections and Hecke algebras

0 Introduction

1 Persi’s 26 July note

(1) Let $G=GL_{n}(\mathbb{F}_{q})$ . Let $a,v\in\mathbb{F}^{n}_{q}$ (I think of them as column vectors). Have $a^{t}v=0$ . These determine a transvection

T=T_{av}\in SL_{n}(q)\qquad\hbox{as}\qquad T=I+va^{t}\quad\hbox{or}\quad T(x)=x+v(a^{t}x).

Thus you add a multiple of $v$ to $x$ . If you multiply $v$ by $\theta$ and divide $a$ by $\theta$ that doesn’t change $T$ so lets agree that $v\neq 0$ and the last nonzero coordinate of $v$ is a $1$ . Since the case $a=0$ is the identity lets exclude that too.

(2) A flag $C=(C_{1},\dots,C_{n})$ is a chain of subspaces with $\mathrm{dim}(C_{i})=i$ . Of course $C_{1}\subseteq C_{2}\subseteq\cdots\subseteq C_{n}=\mathbb{F}^{n}_{q}$ . The distance between two flags $C$ and $D$ is a permutation $\pi=\pi_{CD}\in S_{n}$ defined as follows

\pi(i)=j,\qquad\hbox{if $j$ is smallest with $D_{i-1}+C_{j}\supseteq D_{i}$.}

Its not obvious but $\pi$ is a permutation and the map

\begin{matrix}GL_{n}&\to&S_{n}\\ g&\mapsto&\pi_{C_{0}gC_{0}}\end{matrix}\qquad\hbox{is a bijection from $B\backslash G/B\to S_{n}$}

with inverse given by $\sigma=B\sigma B$ . All of this is in Abel’s paper on the Jordan-Holder permutation.

(3) Example. Lets fix $v$ with $v_{n}\neq 0$ and $a$ with $a_{i}\neq 0$ for any $i$ . Let $C$ be the standard flag ( $C_{i}=\langle e_{1},e_{2},\ldots,e_{i}\rangle$ ). Let

D=T_{av}(C).

I claim $\pi_{CD}=(1,n)$ .

Proof.

What is $\pi(i)$ ? We need the smallest $j$ so that $D_{0}+C_{j}\supseteq D$ . Well $D_{0}=0$ and $D_{1}=T(e_{1})=e_{1}+va_{1}$ . The smallest $j$ with $C_{j}\supseteq e_{1}+va_{1}$ is $j=n$ so $\pi(1)=n$ .

What is $\pi(2)$ ? We need $D_{1}+C_{j}\supseteq D_{2}$ . That is

\langle e_{1}+va_{1},C_{j}\rangle\supseteq\langle C_{1}+va_{1},C_{2}+va_{2}\rangle.

Clearly $j=2$ does the job (from $e_{1}+va_{1}$ and $C_{1}$ get $v$ and so $\langle e_{1}+va_{1},e_{2}+va_{2}]\rangle$ .

Similarly $\pi(i)=i$ for $2\leq i\leq n-1$ .

Finally, what about $\pi(n)$ ? Need $D_{n-1}+C_{j}\supseteq D_{n}$ . But $D_{n-1}=\langle e_{1}+va_{1},\ldots,e_{n-1}+va_{n-1}\rangle$ . With $e_{1}$ get $v$ and so all of $D_{n}$ . ∎

(4)

Proposition 1.1.

Let $a,v$ have $v_{j}=1$ and $v_{k}=0$ for $k\neq j$ and $a_{1}=a_{2}=\cdots a_{i-1}=0$ and $a_{i}\neq 0$ . Then, with $C$ the standard flag and $D=T_{av}(C)$ ,

\pi_{CD}=(i,j).

Proof.

For $k<i$ , $a_{k}=0$ , so $T(e_{k})=e_{k}$ so $\pi(k)=k$ . What is $\pi(i)$ ? We need $\langle e_{1},\ldots,e_{i-1}+C_{k}\rangle\supseteq D_{i}=\langle e_{1},\ldots,e_{i-1},e_{i}+ve_{i}\rangle$ . Since $v\in C_{j}$ (and no smaller $j$ will do) then $\pi(i)=j$ .

What is $\pi(i+1)$ ? Need $D_{i}+C_{k}\supseteq D_{i+1}$ . The LHS is

\langle e_{1},e_{2},\ldots,e_{i-1},e_{i}+va_{i},C_{k}\rangle.

The RHS is

\langle e_{1},e_{2},\ldots,e_{i-1},e_{i}+va_{i},e_{i+1}+va_{i+1}\rangle.

We have $v\in LHS$ and so $k=i+1$ .

Similarly $\pi(k)=k$ for $k<j$ .

Lets do $\pi(j)$ . Need

\langle e_{1},\ldots,e_{i-1},e_{i}+va_{i},\ldots,e_{j-1}+va_{j},C_{k}\rangle\supseteq D_{j}.

Clearly $k=i$ does the job ( $v\in C_{j}$ ).

For $k>j$ , $\pi(k)=k$ . ∎

(5) It’s important to note that $T_{av}$ can give the identity. This will happen if say $v_{j}=1$ has all $v_{k}=0$ for $k>j$ if and only if $a_{i}=0$ for $1\leq i\leq j-1$ .

Proof.

$\pi(1)=1$ needs $C_{1}\supseteq e_{1}+va_{1}$ , eg $a_{1}=0$ , $\pi(2)=2$ needs $C_{2}\supseteq\langle e_{1},e_{2}+va_{2}$ , eg $a_{2}=0$ , (for $j>2$ ) $\vdots$ $\pi(j)=j$ needs $C_{j-1}+C_{j}\supseteq\langle e_{j-1}+va_{j}$ . This holds for all $a$ . Similarly, $\pi(k)=k$ for all $k$ . ∎

(6) I think that the same thing holds for any flag $F$ and $T(F)$ , $\pi_{F,T(F)}$ can only be the identity or a transposition!

2 parsing Persi’s note

Let $u=u_{(21^{n})}=x_{12}(1)$ . For $y=(y_{1},\ldots,y_{n})\in\mathbb{C}^{n}$ let $y_{2}$ denote the second entry of $y$ . Then

uy=x_{12}(1)y=y+y_{2}e_{1}

since $x_{12}(1)$ adds the second entry of $y$ to the first entry and leaves all other entries of $y$ the same.

Let $g\in GL_{n}(\mathbb{C})$ so that $gug^{-1}$ is in the same conjugacy class as $u$ . Then

gug^{-1}x=g(g^{-1}x+(g^{-1}x)_{2}e_{1})=x+(g^{-1}x)_{2}ge_{1}.

So let $v=ge_{1}$ . We want to say $a^{t}x=((g^{-1})^{t}x)_{2}$ . This is ok since

(g^{-1}x)_{2}=\langle e_{2},g^{-1}x\rangle=\langle(g^{-1})^{t}e_{2},x\rangle=\langle a,x\rangle=a^{t}x,

provided $a=(g^{-1})^{t}e_{2}$ . In other words, $a$ is the second column of $(g^{-1})^{t}$ . So letting $v$ be the first column of $g$ and $a$ the second column of $(g^{-1})^{t}$ then $at$ is the second row of $g^{-1}$ and so

a^{t}v=0.

Probably there are no other restrictions on $a$ and $v$ other than they be nonzero.

3 Character computation for the coefficient of $T_{\gamma_{\mu}}$ in $D$

3.1 Symmetric functions

For $\lambda=(\lambda_{1},\ldots,\lambda_{n})\in\mathbb{Z}_{\geq 0}^{n}$ with $\lambda_{1}\geq\cdots\geq\lambda_{n}\geq 0$ define the Hall-Littlewood polynomial by

P_{\lambda}=P_{\lambda}(x;0,t)=\frac{1}{v_{\lambda}(t)}\sum_{w\in S_{n}}w\Big{(}x_{1}^{\lambda_{1}}\cdots x_{n}^{\lambda_{n}}\prod_{1\leq i<j\leq n}\frac{x_{i}-tx_{j}}{x_{i}-x_{j}}\Big{)}

where $v_{\lambda}(t)$ is the normalization that make the coefficient of $x^{\lambda}$ in $P_{\lambda}$ equal to $1$ . For $\lambda=(1^{m_{1}}2^{m_{2}}\ldots)$ define

Q_{\lambda}=b_{\lambda}(t)P_{\lambda}\qquad\hbox{where}\qquad b_{\lambda}(t)=\prod_{i=1}^{\ell}\varphi_{m_{i}}(t)\quad\hbox{with}\quad\varphi_{m}(t)=(1-t)(1-t^{2})\cdots(1-t^{m}).

The monomial symmetric functions $m_{\lambda}$ and the Schur functions $s_{\lambda}$ are given by

m_{\lambda}=P_{\lambda}(x;0,1)\qquad\hbox{and}\qquad s_{\lambda}=P_{\lambda}(x;0,0),

the evaluations of $P_{\lambda}$ at $t=1$ and $t=0$ , respectively. For $r\in\mathbb{Z}_{\geq 0}$ define $q_{r}$ by the generating function

\sum_{r\in\mathbb{Z}_{\geq 0}}q_{r}z^{r}=\prod_{i=1}^{n}\frac{1-tx_{i}z}{1-x_{i}z}\qquad\hbox{and define}\qquad q_{\nu}=q_{\mu_{1}}\cdots q_{\mu_{n}},\qquad\hbox{for $\nu=(\nu_{1},\ldots,\nu_{n})\in\mathbb{Z}_{\geq 0}^{n}$.}

The Big Schurs are defined by the $n\times n$ determinant (see [Mac, Ch. III (4.5)])

S_{\lambda}=S_{\lambda}(x;t)=\det(q_{\lambda_{i}-i+j}).

Define $a_{\mu\nu}(t)$ , $K_{\lambda\mu}(0,t)$ and $L_{\nu\lambda}(t)$ by

Q_{\mu}(x;0,t)=\sum_{\nu}a_{\mu\nu}(t)m_{\nu}\qquad Q_{\mu}(x;0,t)=\sum_{\lambda}K_{\lambda\mu}(0,t)S_{\lambda}\qquad\hbox{and}\qquad q_{\nu}(x;t)=\sum_{\lambda}L_{\nu\lambda}(t)s_{\lambda}.

By [Mac, (4,8) and (4.10)], $\langle q_{\nu},m_{\mu}\rangle_{0,t}=\delta_{\nu\mu}$ and $\langle s_{\lambda},S_{\mu}\rangle=\delta_{\lambda\mu}$ in the inner product $\langle,\rangle_{0,t}$ of [Mac, Ch. III] so that

L_{\nu\lambda}(t)=\langle q_{\nu},S_{\lambda}\rangle\qquad\hbox{which gives}\qquad S_{\lambda}=\sum_{\nu}L_{\nu\lambda}(t)m_{\nu}.

Thus

Q_{\mu}=\sum_{\nu}\Big{(}\sum_{\lambda}K_{\lambda\mu}(0,t)L_{\nu\lambda}(t)\Big{)}m_{\nu}.

We have

	$\displaystyle s_{(1^{n})}$	$\displaystyle=m_{(1^{n})}=P_{(1^{n})}\qquad\hbox{and}$
	$\displaystyle s_{(21^{n})}$	$\displaystyle=P_{(21^{n})}+(t+t^{2}+\cdots+t^{n-1})P_{(1^{n})}=m_{(21^{n})}+(n-1)m_{(1^{n})}$

giving

P_{(1^{n})}=m_{(1^{n})}\qquad\hbox{and}\qquad P_{(21^{n-2})}=m_{(21^{n-2})}+((n-1)-(t+t^{2}+\cdots+t^{n-1}))m_{(1^{n})}.

Thus

	$\displaystyle Q_{(1^{n})}$	$\displaystyle=b_{(1^{n})}(t)m_{(1^{n})}\qquad\hbox{and}$
	$\displaystyle Q_{(21^{n-2})}$	$\displaystyle=b_{(21^{n-2})}(t)\big{(}m_{(21^{n-2}])}+((n-1)-(t+t^{2}+\cdots+t^{n-1}))m_{(1^{n})}\big{)},$

and so $a_{(21^{n-2})\nu}(t)=b_{(21^{n-2})}(t)\tilde{a}_{(21^{n-2})\nu}$ , where

\tilde{a}_{(21^{n-2})\nu}(t)=\sum_{\lambda}K_{\lambda\mu}(0,t)L_{\nu\lambda}(t)=\begin{cases}1,&\hbox{if $\nu=(21^{n-2})$,}\\ (n-1)-(t+t^{2}+\cdots+t^{n-1}),&\hbox{if $\nu=(1^{n})$,}\\ 0,&\hbox{otherwise.}\end{cases}

(3.1)

3.2 A character computation

Let $\mathbb{F}_{q}$ be the finite field with $q$ elements, $G=GL_{n}(\mathbb{F}_{q})$ the group of $n\times n$ invertible matrices with entries in $\mathbb{F}$ and let $B$ be the subgroup of upper triangular matrices. Then

	$\displaystyle\mathrm{Card}(B)$	$\displaystyle=(q-1)^{n}q^{\frac{1}{2}n(n-1)}\quad\hbox{and}$
	$\displaystyle\mathrm{Card}(G)$	$\displaystyle=(q^{n}-1)(q^{n}-q)\cdots(q^{n}-q^{n-1})=q^{\frac{1}{2}n(n-1)}(q-1)(q^{2}-1)\cdots(q^{n}-1)$

By [Mac, Ch. II (1.6) and Ch. IV (2.7)],

	$\displaystyle\mathrm{Card}(Z_{G}(u_{(21^{n-2})}))$	$\displaystyle=q^{\|(21^{n-2})\|+2n(21^{n-2})}(1-q^{-1})\cdot(1-q^{-1})(1-q^{-2})\cdots(1-q^{-(n-2)})$
		$\displaystyle=q^{n+(n-1)(n-2)}q^{-1-\frac{1}{2}(n-1)(n-2)}(q-1)^{2}(q^{2}-1)(q^{3}-1)\cdots(q^{n-2}-1)$
		$\displaystyle=q^{n-1+\frac{1}{2}(n-1)(n-2)}(q-1)^{2}(q^{2}-1)(q^{3}-1)\cdots(q^{n-2}-1)$
		$\displaystyle=q^{\frac{1}{2}n(n-1)}(q-1)^{2}(q^{2}-1)(q^{3}-1)\cdots(q^{n-2}-1),$

so that

\mathrm{Card}(\mathcal{C})=\frac{\mathrm{Card}(G)}{|Z_{G}(u_{(21^{n-2})})|}=\frac{(q^{n-1}-1)(q^{n}-1)}{(q-1)}=\frac{(q^{n-1}-1)(q^{n}-1)}{(q-1)}.

and

\frac{|B|}{|Z_{G}(u_{(21^{n-1})})|}=\frac{(q-1)}{[n-2]!},\qquad\hbox{where}\quad[n-2]!=\frac{(1-q)\cdots(1-q^{n-2})}{(1-q)^{n-2}}.

Let $\mathbf{1}_{B}^{G}=\mathrm{Ind}_{B}^{G}(triv)$ be the representation of $G$ obtained by inducing the trivial representation from $B$ to $G$ . The Hecke algebra is $H=\mathrm{End}_{G}(\mathbf{1}_{B}^{G})$ . The algebra $H$ has basis $\{T_{w}\ |\ w\in S_{n}\}$ with multiplication determined by

T_{s_{k}}T_{w}=\begin{cases}(q-1)T_{w}+qT_{s_{k}w},&\hbox{if $\ell(s_{k}w)<\ell(w)$,}\\ T_{s_{k}w},&\hbox{if $\ell(s_{k}w)>\ell(w)$.}\end{cases}

As a $(G,H)$ -bimodule,

\mathbf{1}_{B}^{G}\cong\bigoplus_{\lambda}G^{\lambda}\otimes H^{\lambda},

where the sum is over partitions of $n$ , $G^{\lambda}$ is the irreducible $G$ -module indexed by $\lambda$ and $H^{\lambda}$ is the irreducible $H$ -module indexed by $\lambda$ .

Let $\mathcal{C}$ be the conjugacy class of transvections in $G=GL_{n}(\mathbb{F}_{q})$ . A favorite representative of $\mathcal{C}$ is $u_{(21^{n-2})}=I+E_{12}$ , where $I$ denotes the identity matrix and $E_{12}$ is the matrix with $1$ in the $(1,2)$ entry and $0$ elsewhere. Let

C=\sum_{g\in\mathcal{C}}g,\qquad\quad\hbox{an element of $Z(\mathbb{C}[G])$,}

where $Z(\mathbb{C}G)$ denotes the center of the group algebra of $G$ . The element $C$ acts on $\mathbf{1}_{B}^{G}$ by

\sum_{\lambda}\frac{\chi^{\lambda}_{G}(C)}{\chi^{\lambda}_{G}(1)}\cdot\mathrm{id}_{G^{\lambda}\otimes H^{\lambda}}\ ,

where $\mathrm{id}_{G^{\lambda}\otimes H^{\lambda}}$ is the operator that acts by the identity on the component $G^{\lambda}\otimes H^{\lambda}$ and by $0$ on $G^{\mu}\otimes H^{\mu}$ for $\mu\neq\lambda$ .

Let $z_{\lambda}$ be the element of $H$ that acts on $H^{\lambda}$ by the identity and by $0$ on $H^{\mu}$ for $\mu\neq\lambda$ . Let $D\in H$ be given by

D=\sum_{\lambda}\frac{\chi^{\lambda}_{G}(C)}{\chi^{\lambda}_{G}(1)}z_{\lambda},\qquad\hbox{so that $D$ acts on $\mathbf{1}_{B}^{G}$ the same way that $C$ does.}

Use cycle notation for permutations in $S_{n}$ so that $(i,j)$ denotes the transposition in $S_{n}$ that switches $i$ and $j$ .

D_{(21^{n-2})}=\sum_{i<j}q^{(n-1)-(j-i)}T_{(i,j)}.

Note that at $q=1$ this specializes to the sum of the transpositions in the group algebra of the symmetric group. Use cycle notation for permutations. Then

T_{s_{k}}(T_{(k,j)}+qT_{(k+1,j)})=qT_{(k,k+1,j)}+q(q-1)T_{(k,j)}+qT_{(k+1,k,j)}=(T_{(k,j)}+qT_{(k+1,j)})T_{s_{k}},\qquad\hbox{if $k+1<j$,}

T_{s_{k}}(qT_{(i,k)}+T_{(i,k+1)})=qT_{(i,k,k+1)}+q(q-1)T_{(i,k+1)}+qT_{(i,k+1,k)}=(qT_{(i,k)}+T_{(i,k+1)})T_{s_{k}},\qquad\hbox{if $i<k$,}

and $T_{s_{k}}T_{(k,k+1)}=T_{(k,k+1)}T_{s_{k}}$ so that

if

k\in\{1,\ldots,n-1\}

then

T_{s_{k}}D_{(21^{n-2})}=D_{(21^{n-2})}T_{s_{k}}

.

So $D_{(21^{n-2})}\in Z(H)$ .

Theorem 3.1.

Assume $n\geq 2$ . Then

D=(n-1)q^{n-1}-[n-1]+(q-1)D_{(21^{n-2})}.

Proof.

In terms of the favorite basis $\{T_{w}\ |\ w\in S_{n}\}$ of $H$ , the element $z_{\lambda}$ is given by (see [CR81, (68.29)] or [HLR, (1.6)])

z_{\lambda}=\frac{1}{|G/B|}\sum_{w\in S_{n}}\chi^{\lambda}_{G}(1)q^{-\ell(w)}\chi^{\lambda}_{H}(T_{w^{-1}})T_{w}.

Thus, the expansion of $D$ in terms of the basis $\{T_{w}\ |\ w\in S_{n}\}$ of $H$ is

$\displaystyle D$	$\displaystyle=\frac{1}{\|G/B\|}\sum_{\lambda}\sum_{w\in S_{n}}\frac{\chi^{\lambda}_{G}(C)}{\chi^{\lambda}_{G}(1)}\chi^{\lambda}_{G}(1)q^{-\ell(w)}\chi^{\lambda}_{H}(T_{w^{-1}})T_{w}$
	$\displaystyle=\frac{\|B\|}{\|G\|}\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(C)\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}$
	$\displaystyle=\frac{\|B\|}{\|G\|}\mathrm{Card}(\mathcal{C})\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(u_{21^{n-2}})\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}$
	$\displaystyle=\frac{\|B\|}{\|Z_{G}(u_{(21^{n-1})})\|}\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(u_{21^{n-2}})\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}.$	(3.2)

For a partition $\nu$ let $\gamma_{\nu}$ be the favorite permutation of cycle type $\nu$ (a minimal length permutation which has cycle type $\nu$ ). By [Ra91, Th. 4.14],

\frac{q^{n}}{(q-1)^{\ell(\nu)}}q_{\nu}(0;q^{-1})=\sum_{\lambda}\chi^{\lambda}_{H}(T_{\gamma_{\nu}^{-1}})s_{\lambda}\qquad\hbox{so that}\qquad\chi^{\lambda}_{H}(T_{\gamma_{\nu}^{-1}})=\frac{q^{n}}{(q-1)^{\ell(\nu)}}L_{\nu\lambda}(q^{-1}),

since $\nu$ is a partition of $n$ . By [HR99, Theorem 4.9(c)],

\chi^{\lambda}_{G}(u_{\mu})=q^{n(\mu)}K_{\lambda\mu}(0,q^{-1})

Thus

	$\displaystyle\sum_{\lambda}$	$\displaystyle\chi^{\lambda}_{G}(u_{21^{n-2}})\chi^{\lambda}_{H}(T_{\gamma_{\nu}^{-1}})=\sum_{\lambda}q^{n(21^{n-2})}K_{\lambda(21^{n-2})}(0,q^{-1})\frac{q^{n}}{(q-1)^{\ell(\nu)}}L_{\nu\lambda}(q^{-1})$
		$\displaystyle=\sum_{\lambda}\frac{q^{\frac{1}{2}(n-1)(n-2)+n}}{(q-1)^{\ell(\nu)}}K_{\lambda(21^{n-2})}(0,q^{-1})L_{\nu\lambda}(q^{-1})=\frac{q^{\frac{1}{2}(n-1)(n-2)+n}}{(q-1)^{\ell(\nu)}}a_{(21^{n-2})\nu}(q^{-1}).$

Plugging this into (3.2), the coefficient of $T_{\gamma_{\nu}}$ in $D$ is

	$\displaystyle\frac{\|B\|}{\|Z_{G}(u_{(21^{n-1})})\|}\frac{q^{\frac{1}{2}(n-1)(n-2)+n}}{(q-1)^{\ell(\nu)}}$	$\displaystyle a_{(21^{n-2})\nu}(q^{-1})q^{-\ell(\gamma_{\nu})}=\frac{(q-1)}{[n-2]!}\frac{q^{\frac{1}{2}(n-1)(n-2)+n}}{(q-1)^{\ell(\nu)}}a_{(21^{n-2})\nu}(q^{-1})q^{-(\|\nu\|-\ell(\nu))}$
		$\displaystyle=\frac{q^{\frac{1}{2}(n-1)(n-2)}(q-1)}{[n-2]!}\frac{q^{\ell(\nu)}}{(q-1)^{\ell(\nu)}}a_{(21^{n-2})\nu}(q^{-1})$

Now use (3.1),

b_{(21^{n-2})}(q^{-1})=(1-q^{-1})(1-q^{-1})\cdots(1-q^{-(n-2)})=q^{-1-\frac{1}{2}(n-1)(n-2)}(q-1)^{n-1}[n-2]!,

and

\tilde{a}_{(21^{n-2})\nu}(q^{-1})=\begin{cases}1,&\hbox{if $\nu=(21^{n-2})$,}\\ \displaystyle{(n-1)-q^{1-n}\frac{(q^{n-1}-1)}{q-1},}&\hbox{if $\nu=(1^{n})$},\\ 0,&\hbox{otherwise},\end{cases}

to get that the coefficient of $T_{\gamma_{\nu}}$ in $D$ is

	$\displaystyle\frac{(q-1)}{[n-2]!}$	$\displaystyle q^{\frac{1}{2}(n-1)(n-2)}\frac{q^{\ell(\nu)}}{(q-1)^{\ell(\nu)}}q^{-1-\frac{1}{2}(n-1)(n-2)}(q-1)^{n-1}[n-2]!\cdot\tilde{a}_{(21^{n-2})\nu}(q^{-1})$
		$\displaystyle=(q-1)^{n-\ell(\nu)}q^{\ell(\nu)-1}\tilde{a}_{(21^{n-2})\nu}(q^{-1})=\begin{cases}(q-1)q^{n-2},&\hbox{if $\nu=(21^{n-2})$,}\\ \displaystyle{(n-1)q^{n-1}-[n-1],}&\hbox{if $\nu=(1^{n})$},\\ 0,&\hbox{otherwise},\end{cases}$

Since $\tilde{a}_{(21^{n-2})\nu}(q^{-1})$ is 0 unless $\nu=(1^{n})$ or $\nu=(21^{n})$ and since the coefficient of $T_{\gamma_{(21^{n-2})}}=T_{s_{1}}$ in $D_{(21^{n-2})}$ is $q^{n-2}$ then it follows from [Fr99, (2.2) Main Theorem] that

D=(n-1)q^{n-1}-[n-1]+(q-1)D_{(21^{n-2})}.

∎

The following corollary provides the connection to [Ra97, (3.16),(3.18),(3.20)] and [DR00, Proposition 4.9].

Corollary 3.2.

Let $M_{1}=1$ and let $M_{k}=T_{s_{k-1}}\cdots T_{s_{2}}T_{s_{1}}T_{s_{1}}T_{s_{2}}\cdots T_{s_{k-1}}$ , for $k\in\{2,\ldots,n\}$ . Then

D=\sum_{k=1}^{n}q^{n-k}(M_{k}-1).

Proof.

Using that $M_{k}=T_{s_{k-1}}M_{k-1}T_{s_{k-1}}$ , check, by induction, that

M_{k}=q^{k-1}+(q-1)\sum_{j=1}^{k-1}q^{j-1}T_{(j,k)}.

Thus

	$\displaystyle\sum_{k=1}^{n}q^{n-k}(M_{k}-1)$	$\displaystyle=0+\sum_{k=2}^{n}\big{(}q^{n-k+(k-1)}-q^{n-k}+(q-1)\sum_{j=1}^{k-1}q^{n-k+j-1}T_{(j,k)}\big{)}$
		$\displaystyle=(n-1)q^{n-1}-[n-1]+(q-1)D_{(21^{n-2})}=D.$

∎

3.3 The example for $n=2$

If $n=2$ then $H=\hbox{span}\{1,T_{s_{1}}\}$ with $(1+T_{s_{1}})(q-T_{s_{1}})=0$ and

z_{(2)}=\frac{1}{1+q}(1+T_{s_{1}})\qquad\hbox{and}\qquad z_{(1^{2})}=\frac{q}{1+q}(1-q^{-1}T_{s_{1}}).

In this case $Z(H)=H$ and $D_{(2)}=T_{(1,2)}=T_{s_{1}}=qz_{(2)}-z_{(1^{2})}.$

So $D_{(2)}=T_{s_{1}}$ and $H$ has basis $\{1,T_{s_{1}}\}$ and

\displaystyle\begin{array}[]{rl}1\cdot D_{(2)}&=1\cdot T_{s_{1}}=T_{s_{1}},\\ T_{s_{1}}\cdot D_{(2)}&=T_{s_{1}}\cdot T_{s_{1}}=(q-1)T_{s_{1}}+q,\end{array}\qquad\hbox{so that}\qquad\begin{pmatrix}0&q\\ 1&q-1\end{pmatrix}

is the matrix for multiplying by $D_{(2)}$ . The character values are

	$\displaystyle\chi^{(1^{2})}_{G}(u_{(1^{2})})=q^{n(1^{2})}K_{(1^{2})(1^{2})}(0,q^{-1})=q^{1}\cdot 1=q,$
	$\displaystyle\chi^{(1^{2})}_{G}(u_{(2)})=q^{n(2)}K_{(1^{2})(2)}(0,q^{-1})=q^{0}\cdot 0=0,$
	$\displaystyle\chi^{(2)}_{G}(u_{(1^{2})})=q^{n(1^{2})}K_{(2)(1^{2})}(0,q^{-1})=q^{1}\cdot q^{-1}=1,$
	$\displaystyle\chi^{(2)}_{G}(u_{(2)})=q^{n(2)}K_{(2)(2)}(0,q^{-1})=q^{0}\cdot 1=1,$

giving that $C$ acts on $\mathbf{1}_{B}^{G}$ the same way as

	$\displaystyle D$	$\displaystyle=\mathrm{Card}(\mathcal{C})\Big{(}\frac{\chi^{(1^{2})}(u_{(2)})}{\chi^{(1^{2})}(u_{(1^{2})})}z_{(1^{2})}+\frac{\chi^{(2)}(u_{(2)})}{\chi^{(2)}(u_{(1^{2})})}z_{(2)}\Big{)}=\frac{\mathrm{Card}(G)}{\mathrm{Card}(Z_{G}(u_{(2)}))}\Big{(}\frac{0}{q}z_{(1^{2})}+\frac{1}{1}z_{(2)}\Big{)}$
		$\displaystyle=\frac{(q^{2}-1)(q^{2}-q)}{q^{2+0}(1-q^{-1})}z_{(2)}=\frac{(q^{2}-1)(q-1)}{q-1}z_{(2)}=(q^{2}-1)\frac{1}{1+q}(1+T_{s_{1}})$
		$\displaystyle=(q-1)(1+D_{(2)}).$

and the matrix for multiplying by $D$ in the basis $\{1,T_{s_{1}}\}$ is

(q-1)\left(\begin{pmatrix}1&0\\ 0&1\end{pmatrix}+\begin{pmatrix}0&q\\ 1&q-1\end{pmatrix}\right)=(q-1)\begin{pmatrix}1&q\\ 1&q\end{pmatrix}.

In this case $|B|=(q-1)^{2}q$ and $|Z_{G}(u_{2})|=q(q-1)$ so that

\frac{|B|}{|Z_{G}(u_{2})|}=(q-1).

Then

\begin{array}[]{ll}\chi^{(1^{2})}(T_{\gamma_{(1^{2})}})=\chi^{(1^{2})}(1)=1,&\chi^{(1^{2})}(T_{\gamma_{(2)}})=\chi^{(1^{2})}(T_{s_{1}})=-1,\\ \chi^{(2)}(T_{\gamma_{(1^{2})}})=\chi^{(2)}(1)=1,&\chi^{(2)}(T_{\gamma_{(2)}})=\chi^{(2)}(T_{s_{1}})=q,\end{array}

so that

	$\displaystyle\sum_{\lambda}\chi^{\lambda}_{G}(u_{2})\chi^{\lambda}_{H}(1)$	$\displaystyle=\chi^{(2)}_{G}(u_{2})\chi^{(2)}_{H}(T_{\gamma_{(1^{2})}})+\chi^{(1^{2})}_{G}(u_{2})\chi^{(1^{2})}_{H}(T_{\gamma_{(1^{2})}})=1\cdot 1+0\cdot 1=1,$
	$\displaystyle\sum_{\lambda}\chi^{\lambda}_{G}(u_{2})\chi^{\lambda}_{H}(T_{s_{1}})$	$\displaystyle=\chi^{(2)}_{G}(u_{2})\chi^{(2)}_{H}(T_{\gamma_{(2)}})+\chi^{(1^{2})}_{G}(u_{2})\chi^{(1^{2})}_{H}(T_{\gamma_{(2)}})=1\cdot q+0\cdot(-1)=q,$

so that the coefficient of $1$ in $D$ is $(q-1)\cdot 1\cdot q^{\ell(1)}=(q-1)\cdot 1\cdot 1=(q-1)$ and the coefficient of $T_{s_{1}}$ is $(q-1)\cdot q\cdot q^{-\ell(s_{1})}=(q-1)$ . Thus

D=(q-1)(1+T_{s_{1}}).

For $n=2$ , Theorem 3.1 says that

\displaystyle D

\displaystyle=(q-1)\big{(}(q-1)^{-2}q^{2}a_{(2)(1^{2})}(q^{-1})+(q-1)^{-1}qD_{(2)}\big{)}=q^{-2}\big{(}(q-1)^{2}(1-q^{-1})+(q-1)D_{(2)}\big{)}

3.4 The example of $n=3$

If $n=3$ then $H=\mathrm{span}\{1,T_{s_{1}},T_{s_{2}},T_{s_{1}s_{2}},T_{s_{2}s_{1}},T_{s_{1}s_{2}s_{1}}\}$ with

(1+T_{s_{1}})(q-T_{s_{1}})=0,\quad(1+T_{s_{2}})(q-T_{s_{2}})=0\quad\hbox{and}\quad T_{s_{1}}T_{s_{2}}T_{s_{1}}=T_{s_{2}}T_{s_{1}}T_{s_{2}}.

Then

	$\displaystyle z_{(3)}$	$\displaystyle=\frac{1}{[3]!}(1+T_{s_{1}}+T_{s_{2}}+T_{s_{2}s_{1}}+T_{s_{1}s_{2}}+T_{s_{1}s_{2}s_{1}}),$
	$\displaystyle z_{(1^{3})}$	$\displaystyle=\frac{q^{3}}{[3]!}(1-q^{-1}T_{s_{1}}-q^{-1}T_{s_{2}}+q^{-2}T_{s_{2}s_{1}}+q^{-2}T_{s_{1}s_{2}}-q^{-3}T_{s_{1}s_{2}s_{1}})$
	$\displaystyle z_{(21)}$	$\displaystyle=1-z_{(3)}-z_{(1^{3})}.$

Then

	$\displaystyle(q+1)D_{(21)}$	$\displaystyle=(q+q^{2})T_{s_{1}}+(q+q^{2})T_{s_{2}}+(q+1)T_{s_{1}s_{2}s_{1}}=[3]!(qz_{(3)}-z_{(1^{2})})-(q-q^{3})\cdot 1$
		$\displaystyle=\frac{[3]!}{[3]!}((q-q^{3})+(q+q^{2})T_{s_{1}}+(q+q^{2})T_{s_{2}}+0+0+(q+1)T_{s_{1}s_{2}s_{1}})-(q-q^{3})$
		$\displaystyle=(q+1)(qT_{s_{1}}+qT_{s_{2}}+T_{s_{1}s_{2}s_{1}})$

and the brute force computation showing that $D_{(21)}=qT_{s_{1}}+qT_{s_{2}}+T_{s_{1}s_{2}s_{1}}$ commutes with $T_{s_{1}}$ is

	$\displaystyle T_{s_{1}}D_{(21)}$	$\displaystyle=q(q-1)T_{s_{1}}+q^{2}+qT_{s_{1}s_{2}}+(q-1)T_{s_{1}s_{2}s_{1}}+qT_{s_{2}s_{1}}$
	$\displaystyle D_{(21)}T_{s_{1}}$	$\displaystyle=q(q-1)T_{s_{1}}+q^{2}+qT_{s_{2}s_{1}}+(q-1)T_{s_{1}s_{2}s_{1}}+qT_{s_{1}s_{2}}.$

In this computation already, some flags in $Bs_{1}B$ are ending up in $Bs_{1}s_{2}s_{1}B$ after the application of $D_{(21)}$ .

In the basis $\{1,T_{s_{1}},T_{s_{2}},T_{s_{1}s_{2}},T_{s_{2}s_{1}},T_{s_{1}s_{2}s_{1}}\}$ , the matrices of multiplication by $T_{s_{1}}$ and $T_{s_{2}}$ are

\begin{pmatrix}0&q&0&0&0&0\\ 1&q-1&0&0&0&0\\ 0&0&0&q&0&0\\ 0&0&1&q-1&0&0\\ 0&0&0&0&0&q\\ 0&0&0&0&1&q-1\end{pmatrix}\qquad\hbox{and}\qquad\begin{pmatrix}0&0&q&0&0&0\\ 0&0&0&0&q&0\\ 1&0&q-1&0&0&0\\ 0&0&0&0&0&q\\ 0&1&0&0&q-1&0\\ 0&0&0&1&0&q-1\end{pmatrix}

The matrix of multiplication by $T_{s_{1}s_{2}s_{1}}$ is

\begin{pmatrix}0&0&0&0&0&q^{3}\\ 0&0&0&0&q^{2}&q^{2}(q-1)\\ 0&0&0&q^{2}&0&q^{2}(q-1)\\ 0&0&q&q(q-1)&q(q-1)&q(q-1)^{2}\\ 0&q&0&q(q-1)&q(q-1)&q(q-1)^{2}\\ 1&q-1&q-1&(q-1)^{2}&(q-1)^{2}&(q-1)^{3}+q(q-1)\end{pmatrix}

where the last column comes from the computation

	$\displaystyle T_{s_{1}s_{2}s_{1}}T_{s_{1}}T_{s_{2}}T_{s_{1}}$	$\displaystyle=\big{(}(q-1)T_{s_{1}s_{2}s_{1}}+qT_{s_{1}s_{2}}\big{)}T_{s_{2}s_{1}}$
		$\displaystyle=\big{(}((q-1)^{2}T_{s_{1}s_{2}s_{1}}+q(q-1)T_{s_{2}s_{1}})+(q(q-1)T_{s_{1}s_{2}}+q^{2}T_{s_{1}})\big{)}T_{s_{1}}$
		$\displaystyle=((q-1)^{3}T_{s_{1}s_{2}s_{1}}+q(q-1)^{2}T_{s_{1}s_{2}})+(q(q-1)^{2}T_{s_{2}s_{1}}+q^{2}(q-1)T_{s_{2}})$
		$\displaystyle\qquad+q(q-1)T_{s_{1}s_{2}s_{1}}+(q^{2}(q-1)T_{s_{1}}+q^{3})$

Thus the matrix of multiplication by $D_{21}=qT_{s_{1}}+qT_{s_{2}}+T_{s_{1}s_{2}s_{1}}$ is

\begin{pmatrix}0&q^{2}&q^{2}&0&0&q^{3}\\ q&q(q-1)&0&0&2q^{2}&q^{2}(q-1)\\ q&0&q(q-1)&2q^{2}&0&q^{2}(q-1)\\ 0&0&2q&2q(q-1)&q(q-1)&q^{2}+q(q-1)^{2}\\ 0&2q&0&q(q-1)&2q(q-1)&q^{2}+q(q-1)^{2}\\ 1&q-1&q-1&q+(q-1)^{2}&q+(q-1)^{2}&(q-1)^{3}+3q(q-1)\end{pmatrix}

with $1+2(q-1)+2q+2(q-1)^{2}=2q^{2}-4q+2+4q-2+1=2q^{2}+1$ and $(q-1)^{3}+3q(q-1)=q^{3}-3q^{2}+3q-1+3q^{2}-3q=q^{3}-1$ so that the bottom row sums to $q^{3}-1+2q^{2}+1=q^{3}+2q^{2}$ Thus the matrix of $D=(n-1)q^{n-1}-[n-1]+(q-1)D_{(21)}=2q^{2}-(q+1)+(q-1)D=(q-1)(2q+1)+(q-1)D$ which has row sums

(q-1)\begin{pmatrix}(2q+1)+(2q^{2}+q^{3})\\ (2q+1)+(2q^{2}+q^{3})\\ (2q+1)+(2q^{2}+q^{3})\\ (2q+1)+(2q+3q(q-1)+q^{2}+q^{3}-2q^{2}+q)\\ (2q+1)+(2q+3q(q-1)+q^{2}+q^{3}-2q^{2}+q)\\ (2q+1)+(q^{3}+2q^{2}))\end{pmatrix}

and

\mathrm{Card}(\mathcal{C})=\frac{(q^{n-1}-1)(q^{n}-1)}{q-1}=\frac{(q^{2}-1)(q^{3}-1)}{q-1}=(q-1)(q+1)(q^{2}+q+1)=(q-1)(q^{3}+2q^{2}+2q+1).

Let

M_{2}=T_{s_{1}}T_{s_{1}}\qquad\hbox{and}\qquad M_{3}=T_{s_{2}}T_{s_{1}}T_{s_{1}}T_{s_{2}}

so that

M_{2}=(q-1)T_{s_{1}}+q\qquad\hbox{and}\qquad M_{3}=T_{s_{2}}((q-1)T_{s_{1}}+q)T_{s_{2}}=(q-1)T_{s_{2}s_{1}s_{2}}+q(q-1)T_{s_{2}}+q^{2},

giving

	$\displaystyle q(M_{2}-1)+(M_{3}-1)$	$\displaystyle=(q-1)(T_{s_{1}s_{2}s_{1}}+qT_{s_{2}}+qT_{s_{1}})+2q^{2}-q-1$
		$\displaystyle=(3-1)q^{3-1}-[3-1]+(q-1)D_{(21)}=D.$

Let’s work out the $q_{\mu}$ and the matrix $L$ . By definition

	$\displaystyle\sum_{r}q_{r}(x;t)z^{r}$	$\displaystyle=\prod_{i=1}^{n}\frac{1-tx_{i}z}{1-x_{i}z}\Big{(}\sum_{k}(-1)^{k}e_{k}t^{k}z_{k}\Big{)}\big{(}\sum_{\ell}h_{\ell}z^{\ell}\big{)}$
		$\displaystyle=(1-zte_{1}+z^{2}t^{2}e_{2}-t^{3}z^{3}e_{3}+\cdots)(1+h_{1}z+h_{2}z^{2}+\cdots)$

giving $q_{0}=1$ , $q_{1}=h_{1}-te_{1}=(1-t)s_{(1)}$ ,

q_{2}=t^{2}e_{2}-tzh_{1}e_{1}+h_{2}=t^{2}(s_{(1^{2})}-t(s_{(1^{2})}+s_{(2)})+s_{(2)}=(1-t)\big{(}(-t)s_{(1^{2})}+s_{(2)}\big{)}

and

	$\displaystyle q_{3}$	$\displaystyle=(-t)^{3}e_{3}+(-t)^{2}h_{1}e_{2}+(-t)h_{2}e_{1}+h_{3}$
		$\displaystyle=(-t)^{3}s_{(1^{3})}+(-t)^{2}(s_{(21)}+s^{(1^{3})})+(-t)(s_{(3)}+s_{(21)}+s_{(3)})$
		$\displaystyle=(1-t)\big{(}(-t)^{2}s_{(1^{3})}+(-t)s_{(21)}+s_{(3)}\big{)}.$

Thus

	$\displaystyle q_{(1^{2})}$	$\displaystyle=q_{1}^{2}=(1-t)^{2}(s_{(1^{2})}+s_{(2)}),\qquad\qquad q_{(2)}=(1-t)\big{(}(-t)s_{(1^{2})}+s_{(2)}\big{)}$
	$\displaystyle q_{(1^{3})}$	$\displaystyle=q_{1}^{3}=(1-t)^{3}(s_{(1^{3})}+2s_{(21)}+s_{(3)}\big{)},$
	$\displaystyle q_{(21)}$	$\displaystyle=q_{2}q_{1}=(1-t)^{2}((-t)(s_{(1^{3})}+s_{(21)})+(s_{(3)}+s_{(21)}))=(1-t)^{2}((-t)s_{(1^{3})}+(1-t)s_{(21)}+s_{(3)}),$
	$\displaystyle q_{3}$	$\displaystyle=(1-t)\big{(}(-t)^{2}s_{(1^{3})}+(-t)s_{(21)}+s_{(3)}\big{)}.$

Then

	$\displaystyle\chi^{(1^{2})}_{H}(T_{\gamma_{(1^{2})}})$	$\displaystyle=1=\frac{1}{(1-q^{-1})^{2}}(1-q^{-1})^{2}=\frac{q^{2}}{(q-1)^{2}}L_{(1^{2})(1^{2})}(q^{-1}),$
	$\displaystyle\chi^{(2)}_{H}(T_{\gamma_{(1^{2})}})$	$\displaystyle=1=\frac{1}{(1-q^{-1})^{2}}(1-q^{-1})^{2}=\frac{q^{2}}{(q-1)^{2}}L_{(1^{2})(2)}(q^{-1}),$
	$\displaystyle\chi^{(1^{2})}_{H}(T_{\gamma_{(2)}})$	$\displaystyle=-1=\frac{q}{(1-q^{-1})}(1-q^{-1})(-q^{-1})=\frac{q^{2}}{(q-1)}L_{(2)(1^{2})}(q^{-1}),$
	$\displaystyle\chi^{(2)}_{H}(T_{\gamma_{(2)}})$	$\displaystyle=q=\frac{q}{(1-q^{-1})}(1-q^{-1})=\frac{q^{2}}{(q-1)}L_{(2)(2)}(q^{-1}),$

	$\displaystyle\chi^{(1^{3})}_{H}(T_{\gamma_{(1^{3})}})$	$\displaystyle=1=\frac{1}{(1-q^{-1})^{3}}(1-q^{-1})^{3}=\frac{1}{(1-q^{-1})^{3}}L_{(1^{3})(3)}(q^{-1}),$
	$\displaystyle\chi^{(21)}_{H}(T_{\gamma_{(1^{3})}})$	$\displaystyle=2=\frac{1}{(1-q^{-1})^{3}}(1-q^{-1})^{3}\cdot 2=\frac{1}{(1-q^{-1})^{3}}L_{(1^{3})(21)}(q^{-1}),$
	$\displaystyle\chi^{(3)}_{H}(T_{\gamma_{(1^{3})}})$	$\displaystyle=1=\frac{1}{(1-q^{-1})^{3}}(1-q^{-1})^{3}=\frac{1}{(1-q^{-1})^{3}}L_{(1^{3})(3)}(q^{-1}),$
	$\displaystyle\chi^{(1^{3})}_{H}(T_{\gamma_{(21)}})$	$\displaystyle=-1=\frac{q}{(1-q^{-1})^{2}}(1-q^{-1})^{2}(-q^{-1})=\frac{q^{3}}{(q-1)^{2}}L_{(21)(1^{3})}(q^{-1}),$
	$\displaystyle\chi^{(21)}_{H}(T_{\gamma_{(21)}})$	$\displaystyle=1-q=\frac{q}{(1-q^{-1})^{2}}(1-q^{-1})^{2}(1-q^{-1})=\frac{q^{3}}{(q-1)^{2}}L_{(21)(21)}(q^{-1}),$
	$\displaystyle\chi^{(3)}_{H}(T_{\gamma_{(21)}})$	$\displaystyle=q=\frac{q}{(1-q^{-1})^{2}}(1-q^{-1})^{2}\cdot 1=\frac{q^{3}}{(q-1)^{2}}L_{(21)(3)}(q^{-1}),$
	$\displaystyle\chi^{(1^{3})}_{H}(T_{\gamma_{(3)}})$	$\displaystyle=1=\frac{q^{2}}{(1-q^{-1})}(1-q^{-1})(-q^{-1})^{2}=\frac{q^{3}}{(q-1)}L_{(3)(1^{3})}(q^{-1}),$
	$\displaystyle\chi^{(21)}_{H}(T_{\gamma_{(3)}})$	$\displaystyle=-q=\frac{q^{2}}{(1-q^{-1})}(1-q^{-1})(-q^{-1})=\frac{q^{3}}{(q-1)}L_{(3)(21)}(q^{-1}),$
	$\displaystyle\chi^{(3)}_{H}(T_{\gamma_{(3)}})$	$\displaystyle=q^{2}=\frac{q^{2}}{(1-q^{-1})}(1-q^{-1})\cdot 1=\frac{q^{3}}{(q-1)}L_{(3)(21)}(q^{-1}),$

Since

	$\displaystyle\chi^{(1^{3})}_{G}(u_{(21)})$	$\displaystyle=q^{n(21)}K_{(1^{3})(21)}(0,q^{-1})=q\cdot 0=0,$
	$\displaystyle\chi^{(21)}_{G}(u_{(21)})$	$\displaystyle=q^{n(21)}K_{(21)(21)}(0,q^{-1})=q\cdot 1=q,$
	$\displaystyle\chi^{(3)}_{G}(u_{(21)})$	$\displaystyle=q^{n(21)}K_{(3)(21)}(0,q^{-1})=q\cdot q^{-1}=1,$

and

	$\displaystyle\chi^{(1^{3})}_{G}(u_{(1^{3})})$	$\displaystyle=q^{n(1^{3})}K_{(1^{3})(1^{3})}(0,q^{-1})=q^{3}\cdot 1=q^{3},$
	$\displaystyle\chi^{(21)}_{G}(u_{(1^{3})})$	$\displaystyle=q^{n(1^{3})}K_{(21)(1^{3})}(0,q^{-1})=q^{3}\cdot(q^{-1}+q^{-2})=q^{2}+q,$
	$\displaystyle\chi^{(3)}_{G}(u_{(1^{3})})$	$\displaystyle=q^{n(1^{3})}K_{(3)(1^{3})}(0,q^{-1})=q^{3}\cdot q^{-3}=1,$

then $C$ acts the same way as

	$\displaystyle D$	$\displaystyle=\frac{(q^{n-1}-1)(q^{n}-1)}{(q-1)}\big{(}\frac{0}{q^{3}}z_{(1^{3})}+\frac{q}{q^{2}+q}z_{(21)}+z_{(3)}\big{)}=\frac{(q^{2}-1)(q^{3}-1)}{(q-1)}\big{(}\frac{1}{q+1}z_{(21)}+z_{(3)}\big{)}$
		$\displaystyle=\frac{(q^{2}-1)(q^{3}-1)(q-1)}{(q-1)(q^{2}-1)}\big{(}1-z_{(3)}-z_{(1^{3})}+(q+1)z_{(3)}\big{)}=(q^{3}-1)\big{(}1-z_{(1^{3})}+qz_{(3)}\big{)}$
		$\displaystyle=\frac{(q^{3}-1)}{[3]!}\left(\begin{array}[]{l}[3]!-(q^{3}-q^{2}T_{s_{1}}-q^{2}T_{s_{2}}+qT_{s_{1}s_{2}}+qT_{s_{2}s_{1}}-T_{s_{1}s_{2}s_{1}})\\ +q(1+T_{s_{1}}+T_{s_{2}}+T_{s_{1}s_{2}}+T_{s_{2}s_{1}}+T_{s_{1}s_{2}s_{1}})\end{array}\right)$
		$\displaystyle=\frac{(q^{3}-1)}{[3]!}\big{(}([3]!-q^{3}+q)+(q^{2}+q)T_{s_{1}}+(q^{2}+q)T_{s_{2}}+(1+q)T_{s_{1}s_{2}s_{1}}\big{)}$
		$\displaystyle=\frac{(q^{3}-1)}{[3]!}([3]!-q^{3}+q)+\frac{(q^{3}-1)}{[3]!}(1+q)D_{(21)}=(q^{3}-1)+\frac{(q-1)}{(1+q)}(q-q^{3})+\frac{(q^{3}-1)}{[3]!}\frac{(q^{2}-1)}{(q-1)}D_{(21)}$
		$\displaystyle=(q^{3}-1)-q(q-1)^{2}+(q-1)D_{(21)}=2q^{2}-q-1+(q-1)D_{(21)}$
		$\displaystyle=(3-1)q^{3-1}-(q+1)+(q-1)D_{(21)}.$

Then

	$\displaystyle\sum_{\lambda}$	$\displaystyle\chi^{\lambda}(u_{(21)})\chi^{\lambda}_{H}(T_{\gamma_{(1^{3})}})=\chi^{(1^{3})}(u_{(21)})\chi^{(1^{3})}_{H}(T_{\gamma_{(1^{3})}})+\chi^{(21)}(u_{(21)})\chi^{(21)}_{H}(T_{\gamma_{(1^{3})}})+\chi^{(3)}(u_{(21)})\chi^{(3)}_{H}(T_{\gamma_{(1^{3})}})$
		$\displaystyle=0\cdot 1+q\cdot 2+1\cdot 1=2q+1,$
	$\displaystyle\sum_{\lambda}$	$\displaystyle\chi^{\lambda}(u_{(21)})\chi^{\lambda}_{H}(T_{\gamma_{(21)}})=\chi^{(1^{3})}(u_{(21)})\chi^{(1^{3})}_{H}(T_{\gamma_{(21)}})+\chi^{(21)}(u_{(21)})\chi^{(21)}_{H}(T_{\gamma_{(21)}})+\chi^{(3)}(u_{(21)})\chi^{(3)}_{H}(T_{\gamma_{(21)}})$
		$\displaystyle=0\cdot(-1)+q\cdot(1-q)+1\cdot q=2q-q^{2},$
	$\displaystyle\sum_{\lambda}$	$\displaystyle\chi^{\lambda}(u_{(21)})\chi^{\lambda}_{H}(T_{\gamma_{(3)}})=\chi^{(1^{3})}(u_{(21)})\chi^{(1^{3})}_{H}(T_{\gamma_{(3)}})+\chi^{(21)}(u_{(21)})\chi^{(21)}_{H}(T_{\gamma_{(3)}})+\chi^{(3)}(u_{(21)})\chi^{(3)}_{H}(T_{\gamma_{(3)}})$
		$\displaystyle=0\cdot 1+q\cdot(-q)+1\cdot q^{2}=0,$

4 Some Hecke algebra multiplications

The Hecke algebra is $H=\mathrm{End}_{G}(\mathbf{1}_{B}^{G})$ . The algebra $H$ has basis $\{T_{w}\ |\ w\in S_{n}\}$ with multiplication determined by

T_{s_{k}}T_{w}=\begin{cases}qT_{s_{k}w}+(q-1)T_{w},&\hbox{if $\ell(s_{k}w)<\ell(w)$,}\\ T_{s_{k}w},&\hbox{if $\ell(s_{k}w)>\ell(w)$.}\end{cases}

Take $n=3$ . We have $T_{s_{1}s_{2}s_{1}}=T_{s_{2}s_{1}s_{2}}$ .

$\displaystyle T_{s_{1}}\cdot 1$	$\displaystyle=T_{s_{1}}$	$\displaystyle=0\cdot 1+1\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{1}}\cdot T_{s_{1}}$	$\displaystyle=q+(q-1)T_{s_{1}}$	$\displaystyle=q\cdot 1+(q-1)\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{1}}\cdot T_{s_{2}}$	$\displaystyle=T_{s_{1}s_{2}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+1\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{1}}\cdot T_{s_{1}s_{2}}$	$\displaystyle=qT_{s_{2}}+(q-1)T_{s_{1}s_{2}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+q\cdot T_{s_{2}}+(q-1)\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{1}}\cdot T_{s_{2}s_{1}}$	$\displaystyle=T_{s_{1}s_{2}s_{1}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+1\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{1}}\cdot T_{s_{1}s_{2}s_{1}}$	$\displaystyle=qT_{s_{2}s_{1}}+(q-1)T_{s_{1}s_{2}s_{1}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+q\cdot T_{s_{2}s_{1}}+(q-1)\cdot T_{s_{1}s_{2}s_{1}},$

This gives the matrrix for multiplication by $T_{s_{1}}$ :

\begin{pmatrix}0&q&0&0&0&0\\ 1&q-1&0&0&0&0\\ 0&0&0&q&0&0\\ 0&0&1&q-1&0&0\\ 0&0&0&0&0&q\\ 0&0&0&0&1&q-1\end{pmatrix}

Then, prodcuts by $T_{s_{2}}$ are

$\displaystyle T_{s_{2}}\cdot 1$	$\displaystyle=T_{s_{2}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+1\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{2}}\cdot T_{s_{1}}$	$\displaystyle=T_{s_{2}s_{1}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+1\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{2}}\cdot T_{s_{2}}$	$\displaystyle=q+(q-1)T_{s_{2}}$	$\displaystyle=q\cdot 1+0\cdot T_{s_{1}}+(q-1)\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{2}}\cdot T_{s_{1}s_{2}}$	$\displaystyle=T_{s_{2}s_{1}s_{2}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+1\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{2}}\cdot T_{s_{2}s_{1}}$	$\displaystyle=qT_{s_{1}}+(q-1)T_{s_{2}s_{1}}$	$\displaystyle=0\cdot 1+q\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+(q-1)\cdot T_{s_{2}s_{1}}+0\cdot T_{s_{1}s_{2}s_{1}},$
$\displaystyle T_{s_{2}}\cdot T_{s_{2}s_{1}s_{2}}$	$\displaystyle=qT_{s_{1}s_{2}}+(q-1)T_{s_{1}s_{2}s_{1}}$	$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+q\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+(q-1)\cdot T_{s_{1}s_{2}s_{1}},$

This gives the matrrix for multiplication by $T_{s_{2}}$

\begin{pmatrix}0&0&q&0&0&0\\ 0&0&0&0&q&0\\ 1&0&q-1&0&0&0\\ 0&0&0&0&0&q\\ 0&1&0&0&q-1&0\\ 0&0&0&1&0&q-1\end{pmatrix}

Then products with $T_{s_{1}s_{2}s_{1}}$

	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot 1$	$\displaystyle=T_{s_{1}s_{2}s_{1}}$
		$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+1\cdot T_{s_{1}s_{2}s_{1}},$
	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot T_{s_{1}}$	$\displaystyle=qT_{s_{1}s_{2}}+(q-1)T_{s_{1}s_{2}s_{1}}$
		$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+q\cdot T_{s_{1}s_{2}}+0\cdot T_{s_{2}s_{1}}+(q-1)\cdot T_{s_{1}s_{2}s_{1}},$
	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot T_{s_{2}}$	$\displaystyle=qT_{s_{2}s_{1}}+(q-1)T_{s_{1}s_{2}s_{1}}$
		$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+0\cdot T_{s_{2}}+0\cdot T_{s_{1}s_{2}}+q\cdot T_{s_{2}s_{1}}+(q-1)\cdot T_{s_{1}s_{2}s_{1}},$
	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot T_{s_{1}s_{2}}$	$\displaystyle=(qT_{s_{1}s_{2}}+(q-1)T_{s_{1}s_{2}s_{1}})T_{s_{2}}$
		$\displaystyle=q^{2}T_{s_{1}}+q(q-1)T_{s_{1}s_{2}}+q(q-1)T_{s_{2}s_{1}}+(q-1)^{2}T_{s_{1}s_{2}s_{1}}$
		$\displaystyle=0\cdot 1+q^{2}\cdot T_{s_{1}}+0\cdot T_{s_{2}}+q(q-1)\cdot T_{s_{1}s_{2}}+q(q-1)\cdot T_{s_{2}s_{1}}+(q-1)^{2}\cdot T_{s_{1}s_{2}s_{1}},$
	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot T_{s_{2}s_{1}}$	$\displaystyle=(qT_{s_{2}s_{1}}+(q-1)T_{s_{1}s_{2}s_{1}})T_{s_{1}}$
		$\displaystyle=q^{2}T_{s_{2}}+q(q-1)T_{s_{2}s_{1}}+q(q-1)T_{s_{1}s_{2}}+(q-1)^{2}T_{s_{1}s_{2}s_{1}}$
		$\displaystyle=0\cdot 1+0\cdot T_{s_{1}}+q^{2}\cdot T_{s_{2}}+q(q-1)\cdot T_{s_{1}s_{2}}+q(q-1)\cdot T_{s_{2}s_{1}}+(q-1)^{2}\cdot T_{s_{1}s_{2}s_{1}},$
	$\displaystyle T_{s_{1}s_{2}s_{1}}\cdot T_{s_{2}s_{1}s_{2}}$	$\displaystyle=\big{(}(q-1)T_{s_{1}s_{2}s_{1}}+qT_{s_{1}s_{2}}\big{)}T_{s_{2}s_{1}}$
		$\displaystyle=\big{(}((q-1)^{2}T_{s_{1}s_{2}s_{1}}+q(q-1)T_{s_{2}s_{1}})+(q(q-1)T_{s_{1}s_{2}}+q^{2}T_{s_{1}})\big{)}T_{s_{1}}$
		$\displaystyle=((q-1)^{3}T_{s_{1}s_{2}s_{1}}+q(q-1)^{2}T_{s_{1}s_{2}})+(q(q-1)^{2}T_{s_{2}s_{1}}+q^{2}(q-1)T_{s_{2}})$
		$\displaystyle\qquad+q(q-1)T_{s_{1}s_{2}s_{1}}+(q^{2}(q-1)T_{s_{1}}+q^{3})$
		$\displaystyle=q^{3}\cdot 1+q^{2}(q-1)\cdot T_{s_{1}}+q^{2}(q-1)\cdot T_{s_{2}}+q(q-1)^{2}\cdot T_{s_{1}s_{2}}+q(q-1)^{2}\cdot T_{s_{2}s_{1}}$
		$\displaystyle\qquad+((q-1)^{3}+q(q-1))\cdot T_{s_{1}s_{2}s_{1}},$

The matrix of multiplication by $T_{s_{1}s_{2}s_{1}}$ is

\begin{pmatrix}0&0&0&0&0&q^{3}\\ 0&0&0&0&q^{2}&q^{2}(q-1)\\ 0&0&0&q^{2}&0&q^{2}(q-1)\\ 0&0&q&q(q-1)&q(q-1)&q(q-1)^{2}\\ 0&q&0&q(q-1)&q(q-1)&q(q-1)^{2}\\ 1&q-1&q-1&(q-1)^{2}&(q-1)^{2}&(q-1)^{3}+q(q-1)\end{pmatrix}

Thus the matrix of mutliplication by $qT_{s_{1}}+qT_{s_{2}}$ is

\begin{pmatrix}0&q^{2}&0&0&0&0\\ q&q(q-1)&0&0&0&0\\ 0&0&0&q^{2}&0&0\\ 0&0&q&q(q-1)&0&0\\ 0&0&0&0&0&q^{2}\\ 0&0&0&0&q&q(q-1)\end{pmatrix}+\begin{pmatrix}0&0&q^{2}&0&0&0\\ 0&0&0&0&q^{2}&0\\ q&0&q(q-1)&0&0&0\\ 0&0&0&0&0&q^{2}\\ 0&q&0&0&q(q-1)&0\\ 0&0&0&q&0&q(q-1)\end{pmatrix}

=\begin{pmatrix}0&q^{2}&q^{2}&0&0&0\\ q&q(q-1)&0&0&q^{2}&0\\ q&0&q(q-1)&q^{2}&0&0\\ 0&0&q&q(q-1)&0&q^{2}\\ 0&q&0&0&q(q-1)&q^{2}\\ 0&0&0&q&q&2q(q-1)\end{pmatrix}

and the matrix of multiplication by $D_{21}=qT_{s_{1}}+qT_{s_{2}}+T_{s_{1}s_{2}s_{1}}$ is

\begin{pmatrix}0&q^{2}&q^{2}&0&0&q^{3}\\ q&q(q-1)&0&0&2q^{2}&q^{2}(q-1)\\ q&0&q(q-1)&2q^{2}&0&q^{2}(q-1)\\ 0&0&2q&2q(q-1)&q(q-1)&q^{2}+q(q-1)^{2}\\ 0&2q&0&q(q-1)&2q(q-1)&q^{2}+q(q-1)^{2}\\ 1&q-1&q-1&q+(q-1)^{2}&q+(q-1)^{2}&(q-1)^{3}+3q(q-1)\end{pmatrix}

5 Row reduction for $GL_{3}(\mathbb{F}_{q})$

For $c\in\mathbb{F}_{q}$ define

y_{1}(c)=\begin{pmatrix}c&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix},\qquad\qquad y_{2}(c)=\begin{pmatrix}1&0&0\\ 0&c&1\\ 0&1&0\end{pmatrix},

s_{1}=y_{1}(0)=\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix},\qquad\qquad s_{2}=y_{2}(0)=\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix},

x_{12}(c)=\begin{pmatrix}1&c&0\\ 0&1&0\\ 0&0&1\end{pmatrix},\qquad\qquad x_{23}(c)=\begin{pmatrix}1&0&0\\ 0&1&c\\ 0&0&1\end{pmatrix},\qquad\qquad x_{13}(c)=\begin{pmatrix}1&0&c\\ 0&1&0\\ 0&0&1\end{pmatrix},

and for $d\in\mathbb{F}_{q}^{\times}$ define

h_{1}(d)=\begin{pmatrix}d&0&0\\ 0&0&1\\ 0&0&1\end{pmatrix},\qquad\qquad h_{2}(d)=\begin{pmatrix}1&0&0\\ 0&d&0\\ 0&0&1\end{pmatrix},\qquad\qquad h_{3}(d)=\begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&d\end{pmatrix}.

Then, coset representatives of cosets of $B$ in the double cosets in $B\backslash G/B$

	$\displaystyle B1B$	$\displaystyle=\{B\},$
	$\displaystyle Bs_{1}B$	$\displaystyle=\{y_{1}(c)B\ \|\ c\in\mathbb{F}^{\times}_{q}\},$
	$\displaystyle Bs_{2}B$	$\displaystyle=\{y_{2}(c)B\ \|\ c\in\mathbb{F}^{\times}_{q}\},$
	$\displaystyle Bs_{1}s_{2}B$	$\displaystyle=\{y_{1}(c_{1})y_{2}(c_{2})B\ \|\ c_{1},c_{2}\in\mathbb{F}^{\times}_{q}\},$
	$\displaystyle Bs_{2}s_{1}B$	$\displaystyle=\{y_{2}(c_{1})y_{1}(c_{2})B\ \|\ c_{1},c_{2}\in\mathbb{F}^{\times}_{q}\},$
	$\displaystyle Bs_{1}s_{2}s_{1}B$	$\displaystyle=\{y_{1}(c_{1})y_{2}(c_{2})y_{1}(c_{1})B\ \|\ c_{1},c_{2},c_{3}\in\mathbb{F}^{\times}_{q}\},$

and

	$\displaystyle B$	$\displaystyle=\{h_{1}(d_{1})h_{2}(d_{2})h_{3}(d_{3})x_{23}(c_{3})x_{13}(c_{2})x_{12}(c_{1})\ \|\ c_{1}c_{2}c_{3}\in\mathbb{F}^{q},d_{1},d_{2},d_{3}\in\mathbb{F}^{\times}_{q}\}$
		$\displaystyle=\left\{\begin{pmatrix}d_{1}&d_{1}c_{1}&d_{1}c_{2}\\ 0&d_{2}&d_{2}c_{3}\\ 0&0&d_{3}\end{pmatrix}\ \Big{\|}\ c_{1}c_{2}c_{3}\in\mathbb{F}^{q},d_{1},d_{2},d_{3}\in\mathbb{F}^{\times}_{q}\right\},$

so that $\mathrm{Card}(B)=(q-1)^{3}q^{3}$ (or, in general, $\mathrm{Card}(B)=(q-1)^{n}q^{\frac{1}{2}n(n-1)}$ ).

If $c_{1}\neq 0$ and $c_{2}\neq 0$ then

\begin{pmatrix}c_{1}&1\\ 1&0\end{pmatrix}\begin{pmatrix}c_{2}&1\\ 1&0\end{pmatrix}=\begin{pmatrix}c_{1}c_{2}+1&1\\ 1&0\end{pmatrix}=\begin{pmatrix}c_{1}&0\\ 0&c_{2}\end{pmatrix}\begin{pmatrix}c_{2}+c^{-1}_{1}&1\\ 1&0\end{pmatrix}

If $c_{2}\neq 0$ then

\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\begin{pmatrix}c_{2}&1\\ 1&0\end{pmatrix}=\begin{pmatrix}1&1\\ c_{2}&0\end{pmatrix}=\begin{pmatrix}c_{2}^{-1}&1\\ 1&0\end{pmatrix}\begin{pmatrix}c_{2}&0\\ 0&-1\end{pmatrix}\begin{pmatrix}1&1\\ 0&1\end{pmatrix}

If $c_{2}=0$ then

\begin{pmatrix}c_{1}&1\\ 1&0\end{pmatrix}\begin{pmatrix}0&1\\ 1&0\end{pmatrix}=\begin{pmatrix}1&c_{1}\\ 0&1\end{pmatrix}

Thus the reflection relation is

y_{i}(c_{1})y_{i}(c_{2})=\begin{cases}y_{i}(c_{1}^{-1}+c_{2})h_{i}(c_{2})h_{i+1}(c_{1}),&\hbox{if $c_{1}\neq 0$ and $c_{2}\neq 0$},\\ y_{i}(c_{2}^{-1})h_{i}(c_{2})h_{i+1}(-1)x_{i,i+1}(1),&\hbox{if $c_{1}=0$ and $c_{2}\neq 0$},\\ x_{i,i+1}(c_{1}),&\hbox{if $c_{2}=0$.}\end{cases}

(5.1)

Since

	$\displaystyle\begin{pmatrix}c_{1}&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}\begin{pmatrix}c_{3}&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}$	$\displaystyle=\begin{pmatrix}c_{1}c_{3}+c_{2}&1&0\\ c_{3}&1&0\\ 1&0&0\end{pmatrix}$
		$\displaystyle=\begin{pmatrix}1&0&0\\ 0&c_{3}&1\\ 0&1&0\end{pmatrix}\begin{pmatrix}c_{1}c_{3}+c_{2}&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&c_{1}&1\\ 0&1&0\end{pmatrix}$

then the building relation is

y_{i}(c_{1})y_{i+1}(c_{2})y_{i}(c_{3})=y_{i+1}(c_{3})y_{i}(c_{1}c_{3}+c_{2})y_{i+1}(c_{1}).

(5.2)

The reflection relations and the building relations are the relations for rearranging $y$ s. Since

\begin{pmatrix}1&c_{1}\\ 0&1\end{pmatrix}\begin{pmatrix}1&c_{2}\\ 0&1\end{pmatrix}=\begin{pmatrix}1&c_{1}+c_{2}\\ 0&1\end{pmatrix}

the x-interchange relations are

$\displaystyle x_{ij}(c_{1})x_{ij}(c_{2})$	$\displaystyle=x_{ij}(c_{1}+c_{2}),$
$\displaystyle x_{ij}(c_{1})x_{ik}(c_{2})$	$\displaystyle=x_{ik}(c_{2})x_{ij}(c_{1}),\qquad\qquad$	$\displaystyle x_{ik}(c_{1})x_{jk}(c_{2})$	$\displaystyle=x_{jk}(c_{2})x_{ik}(c_{1}),$
$\displaystyle x_{ij}(c_{1})x_{jk}(c_{2})$	$\displaystyle=x_{jk}(c_{2})x_{ij}(c_{1})x_{ik}(c_{1}c_{2}),\qquad\qquad$	$\displaystyle x_{jk}(c_{1})x_{ij}(c_{2})$	$\displaystyle=x_{ij}(c_{2})x_{jk}(c_{1})x_{ik}(-c_{1}c_{2}),$

where we assume that $i<j<k$ . Since

\begin{pmatrix}d_{1}&0\\ 0&d_{2}\end{pmatrix}\begin{pmatrix}c&1\\ 1&0\end{pmatrix}=\begin{pmatrix}cd_{1}&d_{1}\\ d_{2}&0\end{pmatrix}=\begin{pmatrix}cd_{1}d_{2}^{-1}&1\\ 1&0\end{pmatrix}\begin{pmatrix}d_{2}&0\\ 0&d_{1}\end{pmatrix}

then the h-past-y relation is (letting $h(d_{1},\ldots,d_{n})=h_{1}(d_{1})\cdots h_{n}(d_{n})$ )

h(d_{1},\ldots d_{n})y_{i}(c)=y_{i}(cd_{i}d_{i+1}^{-1})h(d_{1},\ldots,d_{i-1},d_{i+1},d_{i},d_{i+2},\ldots,d_{n}).

(5.3)

The h-past-x relation is

h(d_{1},\ldots,d_{n})x_{ij}(c)=x_{ij}(cd_{i}d^{-1}_{j})h(d_{1},\ldots,d_{n}).

(5.4)

The x-past-y relations are

	$\displaystyle x_{i,i+1}(c_{1})y_{i}(c_{2})=y_{i}(c_{1}+c_{2})x_{i,i+1}(0),$
$\displaystyle x_{ik}(c_{1})y_{k}(c_{2})$	$\displaystyle=y_{k}(c_{2})x_{ik}(c_{1}c_{2})x_{i,k+1}(c_{1}),\qquad x_{i,k+1}(c_{1})y_{k}(c_{2})=y_{k}(c_{2})x_{ik}(c_{1}),$	(5.5)
$\displaystyle x_{ij}(c_{1})y_{i}(c_{2})$	$\displaystyle=y_{i}(c_{2})x_{i+1,j}(c_{1}),\qquad x_{i+1,j}(c_{1})y_{i}(c_{2})=y_{i}(c_{2})x_{ij}(c_{1})x_{i+1,j}(-c_{1}c_{2}),$

where $i<k$ and $i+1<j$ . These follow from

\begin{pmatrix}1&c_{1}\\ 0&1\end{pmatrix}\begin{pmatrix}c_{2}&1\\ 1&0\end{pmatrix}=\begin{pmatrix}c_{1}+c_{2}&1\\ 1&0\end{pmatrix},

\begin{pmatrix}1&c_{1}&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}=\begin{pmatrix}1&c_{1}c_{2}&c_{1}\\ 0&1&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}\begin{pmatrix}1&c_{1}c_{2}&0\\ 0&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&c_{1}\\ 0&1&0\\ 0&0&1\end{pmatrix},

\begin{pmatrix}1&0&c_{1}\\ 0&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}=\begin{pmatrix}1&c_{1}&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}=\begin{pmatrix}1&0&0\\ 0&c_{2}&1\\ 0&1&0\end{pmatrix}\begin{pmatrix}1&c_{1}&0\\ 0&1&0\\ 0&0&1\end{pmatrix},

\begin{pmatrix}1&0&c_{1}\\ 0&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}c_{2}&1&0\\ 1&0&1\\ 0&0&1\end{pmatrix}=\begin{pmatrix}c_{2}&1&c_{1}\\ 1&0&0\\ 0&0&1\end{pmatrix}=\begin{pmatrix}c_{2}&1&0\\ 1&0&1\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&1&c_{1}\\ 0&0&1\end{pmatrix},

\begin{pmatrix}1&0&0\\ 0&1&c_{1}\\ 0&0&1\end{pmatrix}\begin{pmatrix}c_{2}&1&0\\ 1&0&1\\ 0&0&1\end{pmatrix}=\begin{pmatrix}c_{2}&1&0\\ 1&0&c_{1}\\ 0&0&1\end{pmatrix}=\begin{pmatrix}c_{2}&1&0\\ 1&0&1\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&c_{1}\\ 0&1&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&1&-c_{1}c_{2}\\ 0&0&1\end{pmatrix}.

5.1 Transvections in $GL_{3}(\mathbb{F}_{2})$

The size of $G$ is

\mathrm{Card}(G)=(q^{3}-1)(q^{3}-q)(q^{3}-q^{2})=7\cdot 6\cdot 4=42\cdot 4=168.

The number of transvections is

\mathrm{Card}(\mathcal{C}_{u})=\frac{(q^{n}-1)(q^{n-1}-1)}{q-1}=7\cdot 3.

and these should distribute among the flags as either $(5,4,4,0,0,8)$ or $(5,2,2,0,0,1)$ (which differ just by the sizes of the corresponding double cosets). Consider

T_{a,v}(x)=x+v(a^{t}x)\qquad\hbox{so that}\qquad T_{a,v}(e_{i})=e_{i}+a_{i}v,

and the $i$ th column of $T_{a,v}$ is $a_{i}v$ except with an extra $1$ added to the $i$ th entry. So

T_{a,v}=\begin{pmatrix}1+a_{1}v_{1}&a_{2}v_{1}&a_{3}v_{1}&\cdots&a_{n}v_{1}\\ a_{1}v_{2}&1+a_{2}v_{2}&a_{3}v_{2}&\cdots&a_{n}v_{2}\\ \vdots&&&&\vdots\\ a_{1}v_{n}&a_{2}v_{n}&&\cdots&1+a_{n}v_{n}\end{pmatrix}=1+\big{(}a_{i}v_{j}\big{)}.

Case 1: $a=e_{1}+e_{2}$ and $v=e_{1}+e_{2}$ . Then

T_{a,v}(e_{1})=e_{1}+(e_{1}+e_{2})=e_{2},\quad T_{a,v}(e_{2})=e_{2}+(e_{1}+e_{2})=e_{1},\quad T_{a,v}(e_{3})=e_{3}+0(e_{1}+e_{2})=e_{3}

which has matrix

\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}=y_{1}(0).

The flag

\mathbf{F}_{231}=\begin{pmatrix}0&0&1\\ 1&0&0\\ 0&1&0\end{pmatrix}B=\begin{pmatrix}0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}B=y_{1}(0)y_{2}(0)B,

and

T_{a,v}\mathbf{F}_{231}=y_{1}(0)y_{1}(0)y_{2}(0)B=y_{2}(0)B.

Case 2: $a=e_{1}+e_{2}+e_{3}$ and $v=e_{1}+e_{2}$ . Then

T_{a,v}(e_{1})=e_{1}+(e_{1}+e_{2})=e_{2},\quad T_{a,v}(e_{2})=e_{2}+(e_{1}+e_{2})=e_{1},\quad T_{a,v}(e_{3})=e_{3}+(e_{1}+e_{2})=e_{1}+e_{2}+e_{3}

which has matrix

\begin{pmatrix}0&1&1\\ 1&0&1\\ 0&0&1\end{pmatrix}=y_{1}(0)x_{13}(1)x_{23}(1).

The group $GL_{2}(\mathbb{F}_{2})$ consists of $6=(4-1)(4-2)$ matrices which, grouped in conjugacy classes are

\left\{\begin{pmatrix}1&0\\ 0&1\end{pmatrix}\right\}\sqcup\left\{\begin{pmatrix}1&1\\ 0&1\end{pmatrix},\begin{pmatrix}1&0\\ 1&1\end{pmatrix},\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\right\}\sqcup\left\{\begin{pmatrix}1&1\\ 1&0\end{pmatrix},\begin{pmatrix}0&1\\ 1&1\end{pmatrix}\right\}

It is generated by $s_{1}$ and $x_{12}(1)$ which both have order 2, and there are two elements of order 3.

The group $GL_{3}(\mathbb{F}_{2})$ consists of $168=(8-1)(8-2)(8-4)=7\cdot 6\cdot 4$ matrices. It is generated by $s_{1},s_{2},x_{12}(1),x_{23}(1)$ which all have order 2. Let’s write the 8 elements of $B$ in the form

B=\{x_{12}(a)x_{23}(b)x_{13}(c)\ |\ a,b,c\in\{0,1\}\}.

For sanity of computation, record

\begin{array}[]{ll}x_{12}(1)y_{1}(0)=y_{1}(1),&x_{12}(1)y_{1}(1)=y_{1}(0),\\ x_{12}(1)y_{2}(0)=y_{2}(0)x_{13}(1),&x_{12}(1)y_{2}(1)=y_{2}(1)x_{12}(1)x_{13}(1),\\ \\ x_{23}(1)y_{1}(0)=y_{1}(0)x_{13}(1),&x_{23}(1)y_{1}(1)=y_{1}(1)x_{13}(1)x_{23}(1),\\ x_{23}(1)y_{2}(0)=y_{2}(1),&x_{23}(1)y_{2}(1)=y_{2}(0),\\ \\ x_{13}(1)y_{1}(0)=y_{1}(0)x_{23}(1),&x_{13}(1)y_{1}(1)=y_{1}(1)x_{23}(1),\\ x_{13}(1)y_{2}(0)=y_{2}(0)x_{12}(1),&x_{13}(1)y_{2}(1)=y_{2}(1)x_{12}(1),\end{array}

Letting $\stackrel{{\scriptstyle a}}{{\to}}$ denote conjugation by $a$ , Then

\displaystyle\begin{array}[]{l}y_{1}(0)\stackrel{{\scriptstyle s_{1}}}{{\to}}y_{1}(0),\\ y_{1}(1)\stackrel{{\scriptstyle s_{1}}}{{\to}}y_{1}(0)y_{1}(1)y_{1}(0)=y_{1}(0)x_{12}(1)=y_{1}(1),\\ y_{2}(0)\stackrel{{\scriptstyle s_{1}}}{{\to}}y_{1}(0)y_{2}(0)y_{1}(0),\\ y_{2}(1)\stackrel{{\scriptstyle s_{1}}}{{\to}}y_{1}(0)y_{2}(1)y_{1}(0),\end{array}\qquad\qquad\begin{array}[]{l}y_{1}(0)\stackrel{{\scriptstyle s_{2}}}{{\to}}y_{1}(0)y_{2}(0)y_{1}(0),\\ y_{1}(1)\stackrel{{\scriptstyle s_{2}}}{{\to}}y_{1}(0)y_{2}(1)y_{1}(0)\\ y_{2}(0)\stackrel{{\scriptstyle s_{2}}}{{\to}}y_{2}(0),\\ y_{2}(1)\stackrel{{\scriptstyle s_{2}}}{{\to}}y_{2}(0)x_{23}(1),\end{array}

\displaystyle\begin{array}[]{l}y_{1}(0)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)y_{1}(0)x_{12}(1)=y_{1}(1)x_{12}(1),\\ y_{1}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)y_{1}(1)x_{12}(1)=y_{1}(0)x_{12}(1),\\ y_{2}(0)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)y_{2}(0)x_{12}(1)=y_{2}(0)x_{12}(1)x_{13}(1),\\ y_{2}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)y_{2}(1)x_{12}(1)=y_{2}(1)x_{13}(1),\end{array}\qquad\begin{array}[]{l}y_{1}(0)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)y_{1}(0)x_{23}(1)=y_{1}(0)x_{23}(1)x_{13}(1),\\ y_{1}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)y_{1}(1)x_{23}(1)=y_{1}(1)x_{13}(1),\\ y_{2}(0)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)y_{2}(0)x_{23}(1)=y_{2}(1)x_{23}(1),\\ y_{2}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)y_{2}(1)x_{23}(1)=y_{2}(0)x_{23}(1),\end{array}

\displaystyle\begin{array}[]{l}x_{12}(1)\stackrel{{\scriptstyle s_{1}}}{{\to}}y_{1}(1)x_{12}(1),\\ x_{23}(1)\stackrel{{\scriptstyle s_{1}}}{{\to}}x_{13}(1),\\ x_{13}(1)\stackrel{{\scriptstyle s_{1}}}{{\to}}x_{23}(1),\end{array}\qquad\qquad\begin{array}[]{l}x_{12}(1)\stackrel{{\scriptstyle s_{2}}}{{\to}}x_{13}(1),\\ x_{23}(1)\stackrel{{\scriptstyle s_{2}}}{{\to}}y_{2}(1)x_{23}(1),\\ x_{13}(1)\stackrel{{\scriptstyle s_{2}}}{{\to}}x_{12}(1),\end{array}

\displaystyle\begin{array}[]{l}x_{12}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1),\\ x_{23}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)x_{23}(1)x_{12}(1)=x_{23}(1)x_{13}(1)\\ x_{13}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}x_{12}(1)x_{13}(1)x_{12}(1)=x_{13}(1),\end{array}\qquad\begin{array}[]{l}x_{12}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)x_{12}(1)x_{23}(1)=x_{12}(1)x_{13}(1),\\ x_{23}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1),\\ x_{13}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}x_{23}(1)x_{13}(1)x_{23}(1)=x_{13}(1),\end{array}

Let $\mathcal{C}$ be the conjugacy class of $x_{12}(1)$ . Then $\mathcal{C}\cap B$ is

\begin{array}[]{ccccccccc}x_{12}(1)x_{13}(1)\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&x_{12}(1)&\stackrel{{\scriptstyle s_{2}}}{{\to}}&x_{23}(1)&\stackrel{{\scriptstyle s_{1}}}{{\to}}&x_{23}(1)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&x_{23}(1)x_{13}(1)\end{array}

Then $\mathcal{C}\cap Bs_{1}B$ is

\begin{array}[]{ccccccccc}y_{1}(1)x_{12}(1)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{1}(0)&\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&y_{1}(0)x_{23}(1)x_{13}(1)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{1}(1)x_{12}(1)x_{23}(1)\end{array}

Next, $\mathcal{C}\cap Bs_{2}B$ is

\begin{array}[]{ccccccccc}y_{2}(1)x_{23}(1)&\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&y_{2}(0)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{2}(0)x_{12}(1)x_{13}(1)&\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&y_{2}(1)x_{12}(1)x_{23}(1)\end{array}

Finally $\mathcal{C}\cap Bs_{1}s_{2}s_{1}B$ is

\begin{array}[]{ccccccccc}y_{1}(0)y_{2}(0)y_{1}(0)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{1}(1)y_{2}(0)y_{1}(0)x_{12}(1)&\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&y_{1}(1)y_{2}(1)y_{1}(1)x_{12}(1)x_{23}(1)\end{array}

\begin{array}[]{ccccccccc}\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{1}(0)y_{2}(1)y_{1}(1)x_{13}(1)x_{13}(1)&\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}&y_{1}(0)y_{2}(1)y_{1}(0)x_{13}(1)&\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}&y_{1}(1)y_{2}(1)y_{1}(0)x_{12}(1)x_{13}(1)\end{array}

\begin{array}[]{ccccccccc}\stackrel{{\scriptstyle x_{23}(1)}}{{\to}}y_{1}(1)y_{2}(0)y_{1}(1)x_{12}(1)x_{23}(1)x_{13}(1)\stackrel{{\scriptstyle x_{12}(1)}}{{\to}}y_{1}(0)y_{2}(0)y_{1}(1)x_{23}(1)\end{array}

$\displaystyle D$	$\displaystyle=\frac{1}{\|G/B\|}\sum_{\lambda}\sum_{w\in S_{n}}\frac{\chi^{\lambda}_{G}(C)}{\chi^{\lambda}_{G}(1)}\chi^{\lambda}_{G}(1)q^{-\ell(w)}\chi^{\lambda}_{H}(T_{w^{-1}})T_{w}$
	$\displaystyle=\frac{\|B\|}{\|G\|}\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(C)\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}$
	$\displaystyle=\frac{\|B\|}{\|G\|}\mathrm{Card}(\mathcal{C})\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(u_{21^{n-2}})\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}$
	$\displaystyle=\frac{\|B\|}{\|Z_{G}(u_{(21^{n-1})})\|}\sum_{w\in S_{n}}\Big{(}\sum_{\lambda}\chi^{\lambda}_{G}(u_{21^{n-2}})\chi^{\lambda}_{H}(T_{w^{-1}})\Big{)}q^{-\ell(w)}T_{w}.$	(3.2)

Transvections and Hecke algebras

Abstract

0 Introduction

1 Persi’s 26 July note

Proof.

Proposition 1.1.

Proof.

Proof.

2 parsing Persi’s note

3 Character computation for the coefficient of $T_{\gamma_{\mu}}$ in $D$

3.1 Symmetric functions

3.2 A character computation

Theorem 3.1.

Proof.

Corollary 3.2.

Proof.

3.3 The example for $n=2$

3.4 The example of $n=3$

References

4 Some Hecke algebra multiplications

5 Row reduction for $GL_{3}(\mathbb{F}_{q})$

5.1 Transvections in $GL_{3}(\mathbb{F}_{2})$

Transvections and Hecke algebras

Abstract

0 Introduction

1 Persi’s 26 July note

Proof.

Proposition 1.1.

Proof.

Proof.

2 parsing Persi’s note

3 Character computation for the coefficient of TγμT_{\gamma_{\mu}} in DD

3.1 Symmetric functions

3.2 A character computation

Theorem 3.1.

Proof.

Corollary 3.2.

Proof.

3.3 The example for n=2n=2

3.4 The example of n=3n=3

References

4 Some Hecke algebra multiplications

5 Row reduction for G​L3​(𝔽q)GL_{3}(\mathbb{F}_{q})

5.1 Transvections in G​L3​(𝔽2)GL_{3}(\mathbb{F}_{2})

3 Character computation for the coefficient of $T_{\gamma_{\mu}}$ in $D$

3.3 The example for $n=2$

3.4 The example of $n=3$

5 Row reduction for $GL_{3}(\mathbb{F}_{q})$

5.1 Transvections in $GL_{3}(\mathbb{F}_{2})$