If $L$ is a standard tableau of shape $\lambda /\mu$ let ${\stackrel{\sim }{v}}_L$ be as given in (5.1). It follows from the formulas defining the module ${\stackrel{\sim }{H}}^{(c,\lambda /\mu )}$ that the basis $\{{\stackrel{\sim }{v}}_L\hspace{0.17em}\mid \hspace{0.17em}L\in {ℱ}^{\lambda /\mu }\}$ is simply a renormalized version of the basis $\{v_L\hspace{0.17em}\mid \hspace{0.17em}L\in {ℱ}^{\lambda /\mu }\},$ i.e there are constants ${\kappa }_L\in {ℂ}^{*}$ such that ${\stackrel{\sim }{v}}_L={\kappa }_L{v}_L\text{.}$

Let $s_{i_1}\dots s_{i_p}=w$ be a reduced word for $w\text{.}$ Then, with notations as in Theorem 4.1,

where $L_j=s_{i_{j+1}}\dots s_{i_p}C\text{.}$ Expanding this expression yields

for some constants $b_u\in ℂ\text{.}$ The second equality is a consequence of the fact that, by Proposition 2.1, the tableaux $uC,u<w$ are always standard. Since $\{{\stackrel{\sim }{v}}_L\hspace{0.17em}\mid \hspace{0.17em}L\in {ℱ}^{\lambda /\mu }\}$ is a basis it follows from the triangular relation above that $\{n_L\hspace{0.17em}\mid \hspace{0.17em}L\in {ℱ}^{\lambda /\mu }\}$ is also a basis of ${\stackrel{\sim }{H}}^{(c,\lambda /\mu )}\text{.}$

$\square$

Part (a) follows immediately from the definition of ${\stackrel{\sim }{H}}^{(c,\lambda /\mu )}$ in Theorem 4.1.

(b) The subgroups $S_A,S_B$ and $S_{A\cup B}$ have longest elements

with lengths $\ell \left(w_A\right)=\left(\frac{j-i+1}{2}\right),$ $\ell \left(w_B\right)=\left(\frac{\ell -j}{2}\right),$ and $\ell \left(w_{A\cup B}\right)=\left(\frac{\ell -i+1}{2}\right),$ respectively, and $w_A{w}_B$ is the longest element of $S_A\times S_B\subseteq S_{A\cup B}\text{.}$ Let $t=(t_1,\dots ,t_n)=(a,a{q}^2,a{q}^4,\dots ,a{q}^{2(n-1)})$ where $a=q^{2(c\left(C\left(j\right)\right)-(i-1))}\in {ℂ}^{*}\text{.}$ The positions of $C\left(i\right),\dots ,C\left(\ell \right)$ in $\lambda /\mu$ are such that

By Proposition 3.6 and the fact that $T_s{n}_C=-q^{-1}{n}_C$ for all $i\le s\le \ell ,s\ne j,$

where $\left[r\right]!=\left[r\right][r-1]\dots \left[2\right]\left[1\right]$ and $\left[p\right]=(q^p-q^{-p})/(q-q^{-1})\text{.}$

Let $\pi$ be the permutation such that $P=\pi C$ and let ${\stackrel{\sim }{v}}_P={\tau }_{\pi }{v}_C\text{.}$ Then $s_i\pi w_A{w}_B=w_{A\cup B}$ and, by (3.5),

Each element $w\in S_{A\cup B}$ has a unique expression $w=uv$ such that $v\in S_A\times S_B$ and $\ell \left(w\right)=\ell \left(u\right)+\ell \left(v\right)\text{.}$ The left factor $u$ is the minimal length representative of the coset $w(S_A\times S_B)$ in $S_{A\cup B}\text{.}$ Then

It follows from the formulas for the action on ${\stackrel{\sim }{H}}^{(c,\lambda /\mu )}$ that the element ${\stackrel{\sim }{v}}_P={\tau }_{\pi }{v}_C$ is a nonzero multiple of the basis element $v_P\text{.}$ Furthermore, by (5.4), ${\tau }_k{\stackrel{\sim }{v}}_P=0\text{.}$ So

For each $u\in S_{A\cup B}/(S_A\times S_B)$ except $u=w_{A\cup B}{w}_A{w}_B,$ the tableau $uC$ is standard. In fact these are exactly the standard tableaux $Q$ which are obtained by permuting entries of $C$ which are in the snake. The tableau $w_{A\cup B}{w}_A{w}_{B}C=s_{k}P\text{.}$ Note that $\ell \left(w_{A\cup B}{w}_A{w}_B\right)=\ell \left(w_{A\cup B}\right)-\ell \left(w_A{w}_B\right)\text{.}$ Thus we have

where the sum over all standard tableaux $uC$ which are obtained from $C$ by permuting entries which are in the snake.

$\square$

(a) It is sufficient to show that for all $1\le d\le n$ and all standard tableaux $L$ of shape $\lambda /\mu$ we have

with coefficients $b_Q\in \mathbb{Z}[q,q^{-1}]\text{.}$ For notational convenience let us identify each standard tableau $L$ of shape $\lambda /\mu$ with the permutation $w\in S_n$ such that $wC=L,$ where $C$ is the column reading tableau of shape $\lambda /\mu \text{.}$

From the definitions

Let $L$ be a standard tableau such that $\ell \left(s_{d}L\right)=\ell \left(L\right)+1$ and $s_{d}L$ is not standard. The pair of boxes where the nonstandardness in $s_{d}L$ occurs in either a row or a column:

Case (1): The two boxes in $s_{d}L$ where the nonstandardness occurs determine a snake in the shape $\lambda /\mu$ consisting of all the boxes above the entry $d$ and all the boxes below the entry $d+1\text{.}$ Use the same notation as in Theorem 5.5 so that $P$ is the standard tableau which is the same as the column reading tableau $C$ except that the entries in the snake are in row reading order. Then $s_{k}P$ is nonstandard and there exists a (unique) $v\in S_n$ such that $s_{d}L=v{s}_{k}P$ and $\ell \left(s_{d}L\right)=\ell \left(v\right)+\ell \left(s_{k}P\right)\text{.}$ The permutation $v$ is given by

and $v\left(x\right)=L\left(x\right)$ for all $x\notin \{i,i+1,\dots ,\ell \}\text{.}$ Thus

Let $\ell \left(w_{A\cup B}\right)=\left(\genfrac{}{}{0ex}{}{\ell -i+1}{2}\right)\text{.}$ By Theorem 5.5

and so

The permutations $v$ and $u\le P$ are such that $\ell \left(vu\right)=\ell \left(v\right)+\ell \left(u\right)$ and so $T_v{T}_u=T_{vu}$ for all $u\le P\text{.}$ Furthermore, the tableaux $vuP,$ for $u\le P,$ are exactly the standard tableaux which are obtained by permutations of the entries in $L$ which are in the snake. Thus

Case (2): Suppose that the boxes containing $d$ and $d+1$ in $L$ are the boxes containing k$$ and $k+1$ in the column reading tableau $C\text{.}$

Then $L{s}_k=s_{d}L$ and

by Theorem 5.5 (a).

(b) Let $L$ be a standard tableau of shape $\lambda /\mu$ and let $w$ be the permutation such that $L=wC,$ where $C$ is the column reading tableau of shape $\lambda /\mu \text{.}$ By repeatedly using the relations (1.4) we obtain

where the $p_v(x_1,\dots ,x_n)$ are polynomials in $x_1,\dots ,x_n$ with coefficients in $\mathbb{Z}[q,q^{-1}]\text{.}$ If $x_i$ acts on $v_C$ by integral powers of $q$ then we have

with coefficients $b_v^{\prime }$ in $\mathbb{Z}[q,q^{-1}]\text{.}$

$\square$