Representations of Hecke Algebras of Finite Groups with BN-Pairs of Classical Type

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

Last update: 8 June 2014

Notes and References

This is a typed copy of Peter Norbert Hoefsmit's thesis Representations of Hecke Algebras of Finite Groups with BN-Pairs of Classical Type published in August, 1974.

Chapter 3.$$Degrees of the Irreducible Constituents of $1_B^G$

Definitions and Characters of Parabolic Type

In this chapter we give some results on the irreducible constituents of the induced representation $1_B^G$ of the Borel subgroup $B$ of a finite group $G$ with BN-pair of classical type. The following theorem is basic to the study of these representations.

Theorem (3.1.1) ([CFo1968]) Let $\bar{\mathbb{Q}}$ denote the algebraic closure of $\mathbb{Q}\text{.}$ Each irreducible $\bar{\mathbb{Q}}\text{-character}$ $\chi$ of $H_{\bar{\mathbb{Q}}}(G,B)$ is the restriction to $H_{\bar{\mathbb{Q}}}(G,B)$ of a unique irreducible $\bar{\mathbb{Q}}\text{-character}$ ${\zeta }_{\chi }$ of $G\text{,}$ such that $({\zeta }_{\chi },1_B^G)>0\text{.}$ Moreover every irreducible constituent of $1_B^G$ is obtained in this way. The degree of ${\zeta }_{\chi }$ is given by $$\begin{array}{cc}\text{(3.1.2)}& \text{deg}\hspace{0.17em}{\zeta }_{\chi }=|G:B|\text{deg}\hspace{0.17em}\chi {\left(\sum _{w\in W}{\left(\text{ind}\hspace{0.17em}w\right)}^{-1}\chi \left(S_w\right)\chi \left({\hat{S}}_w\right)\right)}^{-1}\end{array}$$ where ${\hat{S}}_w$ is the basis element of $H_{\bar{\mathbb{Q}}}(G,B)$ corresponding to $w^{-1}$ and $$\text{ind}\hspace{0.17em}w=|B:B\cap B^w|,w\in W,B^w=w^{-1}Bw\text{.}$$

Let $𝒜$ be the generic Ring of a Coxeter system $(W,R)$ defined over $D=\mathbb{Q}[{\mu }_r,r\in R]$ as in (2) and let $k$ be the quotient field of $D\text{,}$ $\bar{K}$ the algebraic closure of $K\text{.}$ It is clear from the relations (2.1.8) (see e.g. [CIK1971]), lemma 2.7) that there exists a unique homomorphism $\nu :𝒜\to D$ such that $\nu \left(a_r\right)={\mu }_r,$ $r\in R\text{.}$

Definition (3.1.3) Let $\chi$ be an irreducible $\bar{K}\text{-character}$ of ${𝒜}^{\bar{K}}\text{.}$ Set $$d_{\chi }=\left(\sum _{w\in W}\nu \left(a_w\right)\right)\text{deg}\hspace{0.17em}\chi {\left(\sum _{w\in W}\nu {\left(a_w\right)}^{-1}\chi \left(a_w\right)\chi \left({â}_w\right)\right)}^{-1}$$ where ${â}_w=a_{w^{-1}}\text{.}$ We call $d_{\chi }$ the generic degree associated with $\chi \text{.}$

Let $M$ be an irreducible $\bar{K}\text{-matrix}$ representation of ${𝒜}^{\bar{K}}\text{.}$ Let $M_{ij}\left(a\right)$ denote the $i,j\text{-th}$ entry of $M\left(a\right),$ $a\in {𝒜}^{\bar{K}}\text{.}$ Thus $M_{ij}$ is a function from ${𝒜}^{\bar{K}}$ to $k\text{.}$ The ring ${𝒜}^{\bar{K}}$ is a symmetric algebra with dual basis $\left\{a_w\right\}$ and $\left\{\nu {\left(a_u\right)}^{-1}{â}_w\right\}$ (see e.g. [Gre1970], Lemma 5.1). Then from ([CRe1961], Lemma 62.8) and Schur's lemma, we have $$\begin{array}{cc}\text{(3.1.4)}& {\displaystyle (M_{ij},M_{rs})=\sum _{w\in W}\nu {\left(a_w\right)}^{-1}{M}_{ij}\left(a_w\right)M_{rs}\left({â}_w\right)=C_M{\delta }_{is}{\delta }_{jr},}\end{array}$$ where $\delta$ is the Kronecker delta and if $\chi$ is the character of $M,$ $$\begin{array}{cc}\text{(3.1.5)}& {\displaystyle C_M{\left(\text{deg}\hspace{0.17em}\chi \right)}^{-1}\sum _{w\in W}\nu {\left(a_w\right)}^{-1}\chi \left(a_w\right)\chi \left({â}_w\right)\text{.}}\end{array}$$ Now let $𝒜$ be the generic ring of a Coxeter system $(W,R)$ of classical type and let $G$ be a finite group with BN-pair of type $(W,r)\text{.}$ Let $\varphi :D\to \bar{\mathbb{Q}}$ be the homomorphism defined by $\varphi \left({\mu }_r\right)=q_r,$ $r\in R,$ $q_r$ the index parameters (see 2.1.5). Let $P=\text{ker}\hspace{0.17em}\varphi$ and let ${\varphi }^{*}:D_P\to \bar{\mathbb{Q}}$ be the extension of $\varphi$ to the ring of fractions $D_P,$ regarded as a subring of $k\text{.}$ Let $\chi$ be an irreducible character of ${𝒜}^{\bar{K}}\text{.}$ The results of chapter 2 (see also [CIK1971], Proposition 7.1) show $\chi \left(a_w\right)\in D_P$ for all $w\in W$ and the $\bar{\mathbb{Q}}\text{-linear}$ map ${\chi }_{\varphi }:{𝒜}_{\varphi ,\bar{\mathbb{Q}}}\to \bar{\mathbb{Q}}$ defined by $$\begin{array}{cc}\text{(3.1.6)}& {\chi }_{\varphi }\left(a_{w\varphi }\right)={\varphi }^{*}\left(\chi \left(a_w\right)\right)\end{array}$$ is an irreducible character of $A_{\varphi ,\bar{\mathbb{Q}}}\text{.}$ The map $\chi \mapsto {\chi }_{\varphi }$ is a bijection between the irreducible characters of ${𝒜}^{\bar{K}}$ and those of $A_{\varphi ,\bar{\mathbb{Q}}}\text{.}$

As $A_{\varphi ,\bar{\mathbb{Q}}}=H_{\bar{\mathbb{Q}}}(G,B),$ we regard the specialized character ${\chi }_{\varphi }$ as an irreducible character of $H_{\bar{\mathbb{Q}}}(G,B)$ and denote the corresponding irreducible constituent of $1_B^G$ in the sense of Theorem (3.1.1) by ${\zeta }_{\chi ,\varphi }\text{.}$

Proposition (3.1.7) With the notations as above we have $${\varphi }^{*}\left(d_{\chi }\right)=\text{deg}\hspace{0.17em}{\zeta }_{\chi ,\varphi }\text{.}$$

		Proof.
	From ([CIK1971], lemma 5.9), $\varphi \left(\sum _{w\in W}\nu \left(a_w\right)\right)=|G:B|$ and $\varphi \left(\nu \left(a_w\right)\right)=\text{ind}\hspace{0.17em}w\text{.}$ The statement now follows from (3.1.2) and (3.1.6) and the definition of $d_{\chi }\text{.}$ $\square$

In particular if ${\varphi }_0:D\to \bar{\mathbb{Q}}$ is defined by ${\varphi }_0\left({\mu }_r\right)=1$ for all $r\in R,$ ${𝒜}_{{\varphi }_0,\bar{\mathbb{Q}}}=\bar{\mathbb{Q}}W$ and (3.1.7) becomes $\text{deg}\hspace{0.17em}\chi =\text{deg}\hspace{0.17em}{\zeta }_{\chi ,{\varphi }_0}\text{.}$

We will evaluate $d_{\chi }$ for the irreducible character $\chi$ of the generic ring corresponding to a Coxeter system of classical type in the next section. We conclude this section with the following.

Let $J\subset R$ and let $W_J=⟨J⟩\text{.}$ $J$ determines a parabolic subgroup $G_J=B{W}_{J}B\text{.}$

Definition (3.1.8) Let $\zeta$ be an irreducible character of $G$ such that $(\zeta ,1_B^G)>0\text{.}$ $\zeta$ is said to be of parabolic type if $(\zeta ,1_{G_J}^G)=1$ for some $J\subset R\text{.}$

From the above there is a natural bijective correspondence ${\zeta }_{\chi ,\varphi }\mapsto {\zeta }_{\chi ,{\varphi }_0}$ between the irreducible $\bar{\mathbb{Q}}\text{-characters}$ ${\zeta }_{\chi ,\varphi }$ of $G$ and the irreducible $\bar{\mathbb{Q}}\text{-characters}$ of $W\text{.}$ In ([CIK1971], Theorem 7.2) it is shown that $$\begin{array}{cc}\text{(3.1.9)}& ({\zeta }_{\chi ,\varphi },1_{G_J}^G)=({\zeta }_{\chi ,{\varphi }_0},1_{W_J}^W)\end{array}$$ for all $J\subset R\text{.}$ Thus to show the irreducible constituents of $1_B^G$ are of parabolic type it is enough to show it for the irreducible characters of the Weyl groups.

Proposition (3.1.10) Every irreducible character $\chi$ of $W\left(A_n\right),$ $W\left(B_n\right),$ $n\ge 2,$ and $W\left(D_n\right),$ $n\ge 4,$ is of parabolic type.

		Proof.
	Let $\left(\alpha \right)$ be a partition of $n\text{.}$ Let $R\left(\alpha \right)$ denote the group of row permutations of the canonical tableau of shape $\left(\alpha \right)\text{.}$ Then $R\left(\alpha \right)$ coincides with $W_J$ for some $J\subset R,$ $R$ the set of distinguished generators for $W\left(A_{n-1}\right)\cong S_n$ given in (3). Order the partitions of $n$ lexicographically and let ${\chi }^{\alpha }$ denote the character of the irreducible representation of $W\left(A_{n-1}\right)$ corresponding to $\left(\alpha \right)\text{.}$ From ([Rob1961], pp. 40-41) $$\begin{array}{cc}\text{(3.1.11)}& {\displaystyle 1_{R\left(\alpha \right)}^{S_n}={\chi }^{\alpha }+\sum _{\beta >\alpha }{m}_{\alpha ,\beta }{\chi }^{\beta },m_{\alpha ,\beta }\ge 0\text{.}}\end{array}$$ Thus $({\chi }^{\alpha },1_{R\left(\alpha \right)}^{S_n})=1,$ which is well known. For a double partition $(\alpha ,\beta )$ of $n\text{,}$ $\left(\alpha \right)=({\alpha }_1,\dots ,{\alpha }_r),$ $\left(\beta \right)=({\beta }_1,\dots ,{\beta }_s)$ let $(\alpha +\beta )$ denote the partition of $n$ defined by $(\alpha +\beta )=({\alpha }_1+{\beta }_1,\dots ,{\alpha }_t+{\beta }_t),$ $t=\text{max}\{r,s\}\text{.}$ Let ${\chi }^{\left[\alpha \right]·\left[\beta \right]}$ denote the character of the outer product representation $\left[\alpha \right]·\left[\beta \right]$ of $S_n$ (see (1.2.2)). From ([Rob1961], Theorem 3.13), $${\chi }^{\left[\alpha \right]·\left[\beta \right]}={\chi }^{\alpha +\beta }+\sum _{\mu >\alpha +\beta }{m}_{\mu ,\alpha +\beta }{\chi }^{\mu },m_{\mu ,\alpha +\beta }\ge 0\text{.}$$ Then by (3.1.11) $$\begin{array}{cc}\text{(3.1.12)}& ({\chi }^{\left[\alpha \right]·\left[\beta \right]},1_{R(\alpha +\beta )}^{S_n})=(X^{\alpha +\beta },{\chi }^{\alpha +\beta })=1\text{.}\end{array}$$ But by Theorem (1.2.2) the restriction of the irreducible representation ${\pi }_{{\varphi }_0}^{\alpha ,\beta }$ of $W\left(B_n\right)$ to $S_n$ is the outer product representation $\left[\alpha \right]·\left[\beta \right]\text{.}$ Thus, letting ${\chi }^{\alpha ,\beta }$ denote the character of ${\pi }_{{\varphi }_0}^{\alpha ,\beta },$ we have $$({\chi }^{\alpha ,\beta },1_{R(\alpha ,\beta )}^{W\left(B_n\right)})=({\chi }^{\left[\alpha \right]·\left[\beta \right]},1_{R(\alpha +\beta )}^{S_n})=1$$ by (3.1.12) and Frobenius reciprocity. As $R(\alpha +\beta )$ is a parabolic subgroup of $W\left(B_n\right),$ the ${\chi }^{\alpha ,\beta }$ are of parabolic type. The representations of $W\left(D_n\right),$ $n\ge 4,$ are handled similarly. If $\left(\alpha \right)\beta \beta ,$ ${\chi }^{\alpha ,\beta }$ remains irreducible when restricted to $W\left(D_n\right)$ by ([You1929]). As $S_n$ is a subgroup of $W\left(D_n\right),$ $$({\chi }^{\alpha ,\beta },1_{R(\alpha ,\beta )}^{W\left(D_n\right)})=({\chi }^{\left[\alpha \right]·\left[\beta \right]},1_{R(\alpha +\beta )}^{S_n})=1$$ If $\left(\alpha \right)=\beta$ the situation is only slightly more complicated. Set $J\prime =\{w_2,\dots ,w_{n-1}\}\subset R,$ the distinguished generators of $W\left(D_n\right)\text{.}$ Then $W_{J\prime }\cong Sn-1\text{.}$ Let ${}_i{\chi }^{\alpha ,\alpha }$ denote the character of ${}_i{\pi }_{{\varphi }_0}^{\alpha ,\alpha },$ $i=1,2\text{.}$ From (2.3.7) $$\begin{array}{cc}\text{(3.1.13)}& {}_i{\chi }^{\alpha ,\alpha }{|}_{W_{J\prime }}={\chi }^{\left[\alpha \right]·[{\alpha }_r-]}+\dots +{\chi }^{\left[\alpha \right]·[{\alpha }_1-]},i=1,2\end{array}$$ where $({\alpha }_j-)$ are the partitions of $n-1$ contained in $\left(\alpha \right)=({\alpha }_1,\dots ,{\alpha }_r)\text{.}$ Now $$\left(\alpha \right)+({\alpha }_r-)>\left(\alpha \right)+({\alpha }_{r-1}-)>\dots >\left(\alpha \right)+({\alpha }_1-)$$ in the lexicographic order so $$\begin{array}{ccc}({}_i{\chi }^{\alpha ,\alpha },1_{R(\left(\alpha \right)+({\alpha }_r-))}^{W\left(D_n\right)})& =& ({}_i{\chi }^{\alpha ,\alpha }{|}_{S_{n-1}},1_{R(\left(\alpha \right)+({\alpha }_r-))}^{S_{n-1}})\\ & =& ({\chi }^{\left(\alpha \right)+({\alpha }_r-)},{\chi }^{\left(\alpha \right)+({\alpha }_r-)}),i=1,2\end{array}$$ by (3.1.12), (3.1.13) and Frobenius reciprocity. Hence ${}_i{\chi }^{\alpha ,\alpha },$ $i=1,2\text{,}$ are of parabolic type. This completes the proof. $\square$

An Induction Formula

Let $\mu =({\mu }_1,\dots ,{\mu }_s)$ be a double partition of $n$ and let ${\chi }^{\mu }$ denote the character of the representation ${\pi }^{\mu }$ of ${𝒜}^k\left(B_n\right)\text{.}$ Set $$C^{\mu }{\left(f^{\mu }\right)}^{-1}\sum _{w\in W\left(B_n\right)}\nu {\left(a_w\right)}^{-1}{\chi }^{\mu }\left(a_w\right){\chi }^{\mu }\left({â}_w\right)$$ We show that the inductive construction of the representations ${\pi }^{\mu }$ yields an inductive formula for $C^{\mu },$ which will provide the means to determine the generic degree associated with ${\chi }^{\mu }\text{.}$

Let $(W\left(B_n\right),R)$ be as in Section 2. Take $J_{n-1}\subset R$ to be the the subset $J_{n-1}=\{w_1,\dots ,w_{n-1}\}$ and let $W_{J_{n-1}}=⟨J_{n-1}⟩\text{.}$ Then $W_{J_{n-1}}\cong W\left(B_{n-1}\right)\text{.}$ Let $M^{\mu }\left(a_w\right)$ denote the matrix of ${\pi }^{\mu }\left(a_w\right)\text{.}$ From the proof of Theorem (2.2.7) we have the decomposition $$\begin{array}{cc}\text{(3.2.2)}& M^{\mu }\left(a_w\right)=M^{({\mu }_s-)}\left(a_w\right)\dotplus \dots \dotplus M^{({\mu }_1-)}\left(a_w\right),w\in W_{J_{n-1}}\end{array}$$ where the sum is taken over those $({\mu }_t-)$ which are non-zero. Let $g_t$ and $f_t$ denote the position, in the arrangement according to the last letter sequence, of the first and last tableau of shape $\left(\mu \right)$ respectively, which upon deletion of $n$ yield standard tableaux of shape $({\mu }_t-)\text{.}$ Then for $a_t,$ $b_r,$ $g_t\le a_t\le f_t,$ $g_r\le b_r\le f_r,$ (3.2.2) implies $$\begin{array}{cc}\text{(3.2.3)}& M_{a_t,b_r}^{\mu }\left(a_w\right)=0\text{for}\hspace{0.17em}t\ne r,w\in W_{J_{n-1}}\end{array}$$ and $$\begin{array}{cc}\text{(3.2.4)}& M_{a_t,b_r}^{\mu }\left(a_w\right)=M_{a_t,b_r}^{({\mu }_t-)}\left(a_w\right),w\in W_{J_{n-1}}\text{.}\end{array}$$ Since $\left|W\left(B_n\right)\right|=2^{n}n!,$ $|W\left(B_n\right):W_{J_{n-1}}|=2n\text{.}$ From ([Bou1968], p. 37), there exists a set $\left\{x_k\hspace{0.17em}\right|\hspace{0.17em}k=1,\dots ,2n\}$ of coset representatives of the left $W_{J_{n-1}}\text{-cosets}$ of $W\left(B_n\right)$ such that $l\left(x_{k}w\right)=l\left(x_k\right)+l\left(w\right),$ $w\in W_{J_{n-1}},$ $k=1,\dots ,2n\text{.}$ The $\left\{x_k\right\}$ are the unique elements of minimal length in the left cosets $x_k{W}_{J_{n-1}}\text{.}$ We will determine these elements explicitly for our choice of $R$ and $J_{n-1}$ in the next section.

Definition (3.2.5) With the notation as above, let $$E^{({\mu }_t-)}=\sum _{k=1}^{2n}\sum _{j=g_t}^{f_t}\nu {\left(a_{x_k}\right)}^{-1}{M}_{1,j}^{\mu }\left(a_{x_k}\right)M_{j,1}^{\mu }\left({â}_{x_k}\right)\text{.}$$

We now prove

Theorem (3.2.6) $C^{\mu }=C^{({\mu }_t-)}{E}^{({\mu }_t-)}\text{.}$

		Proof.
	Choose $p$ such that $g_t\le p\le f_t\text{.}$ Let $\{x_k:k=1,\dots ,2n\}$ be the set of left $W_{J_{n-1}}\text{-coset}$ representatives of minimal length as above. By (3.1.4) and (3.1.5), $$\begin{array}{cc}\text{(3.2.7)}& \begin{array}{ccc}{C}^{\mu }& =& {\displaystyle (M_{1,p}^{\mu },M_{p,1}^{\mu })=\sum _{w\in W\left(B_n\right)}\nu {\left(a_w\right)}^{-1}{M}_{1,p}^{\mu }\left(a_w\right)M_{p,1}^{\mu }\left({â}_w\right)}\\ & =& {\displaystyle \sum _{k=1}^{2n}\sum _{w\in W_{J_{n-1}}}\nu {\left(a_{x_k}\right)}^{-1}\nu {\left(a_w\right)}^{-1}{M}_{1,p}^{\mu }\left(a_{x_k}{a}_w\right)M_{p,1}^{\mu }\left({â}_w{â}_{x_k}\right)}\\ & =& {\displaystyle \sum _{k=1}^{2n}\sum _{w\in W_{J_{n-1}}}\nu {\left(a_{x_k}\right)}^{-1}\nu {\left(a_w\right)}^{-1}\left(\sum _{i=1}^f{M}_{1,i}^{\mu }\left(a_{x_k}\right)M_{i,p}^{\mu }\left(a_w\right)\right)\left(\sum _{i=1}^f{M}_{p,j}^{\mu }\left({â}_w\right)M_{j,1}^{\mu }\left({â}_{x_k}\right)\right)}\\ & =& {\displaystyle \sum _{k=1}^{2n}\sum _{i,j=1}^f\nu {\left(a_{x_k}\right)}^{-1}{M}_{1,i}^{\mu }\left(a_{x_k}\right)M_{j,1}^{\mu }\left({â}_{x_k}\right)\left(\sum _{w\in W_{J_{n-1}}}\nu {\left(a_w\right)}^{-1}{M}_{i,p}^{\mu }\left(a_w\right)M_{p,j}^{\mu }\left({â}_w\right)\right)\text{.}}\end{array}\end{array}$$ The second step follows from the fact that by the choice of the $\left\{x_k\right\},$ $l\left(x_{k}w\right)=l\left(x_k\right)+l\left(w\right)\text{.}$ By (3.2.3) $$\sum _{w\in W_{J_{n-1}}}\nu \left(a_w\right)M_{i,p}^{\mu }\left(a_w\right)M_{p,j}^{\mu }\left({â}_w\right)=0$$ for either $i$ or $j$ not lying between $g_t$ and $f_t,$ while by (3.2.4) and (3.1.4) $$\sum _{w\in W_{J_{n-1}}}\nu {\left(a_w\right)}^{-1}{M}_{i,p}^{\mu }\left(a_w\right)M_{p,j}^{\mu }\left({â}_w\right)=(M_{i,p}^{({\mu }_t-)},M_{p,j}^{({\mu }_t-)})={\delta }_{ij}{C}^{({\mu }_t-)}$$ for $g_t\le i,j\le f_t,$ as ${\pi }^{({\mu }_t-)}$ is an absolutely irreducible representation of ${𝒜}^K\left(B_{n-1}\right)$ by theorem (2.2.14). Combining the above formulae with (3.2.7) gives $$C^{\mu }=C^{({\mu }_t-)}\sum _{k=1}^{2n}\sum _{j=g_t}^{f_t}\nu {\left(a_{x_k}\right)}^{-1}{M}_{1,j}^{\mu }\left(a_{x_k}\right)M_{j,1}^{\mu }\left({â}_{x_k}\right)=C^{({\mu }_t-)}{E}^{({\mu }_t-)}$$ which is the required result. $\square$

While the above theorem provides the induction step for a variety of factorizations of $C^{\mu },$ no such factorization lends itself to an explicit formula for $C^{\mu }$ without tedious calculations. To obtain a formula for $C^{\mu }$ by induction using Theorem (3.2.6) requires the evaluation of $E^{({\mu }_t-)}$ for some row $t\text{.}$ The most convenient choice is the first allowable row from which the last square can be deleted. Let $c$ denote the index of this row. Then in the Young diagram of shape $\left(\mu \right)=(\alpha ,\beta ),$ the row $c$ is a row of the Young diagram of shape $\left(\alpha \right)$ if $\left(\alpha \right)\ne \left(0\right)\text{.}$ Moreover, the standard tableaux of shape $\left(\mu \right)$ which upon deletion of $n$ yield standard tableaux of shape $({\mu }_c-)$ occur last in the ordering according to the last letter sequence. We defer the evaluation of $E^{({\mu }_c-)}$ until the next section. We close this section with a deduction which will prove valuable in the calculations to come.

Proposition (3.2.8) Let $(W\left(B_n\right),R)$ be as before and let $W$ be any element of $W\left(B_n\right)$ which can be expressed as a product of distinct generators chosen from $R$ in increasing order, i.e. $w=w_{i_1}\cdots w_{i_k},$ $w_{i_j}\in R,$ $j=1,\dots ,k$ and $i_1<i_2<\dots <i_k\text{.}$ For the matrix representation $M^{\mu }$ of ${𝒜}^K\left(B_n\right)\text{,}$ $\mu =(\alpha ,\beta )$ a double partition of $n\text{,}$ $$\begin{array}{cc}\text{(3.2.9)}& {\displaystyle M_{c,d}^{\mu }\left(a_w\right)=\sum _{s,t,\dots ,u}{M}_{c,s}^{\mu }\left(a_{i_1}\right)M_{s,t}^{\mu }\left(a_{i_2}\right)\cdots M_{u,d}^{\mu }\left(a_{i_k}\right)}\end{array}$$ and there is at most one non-zero term in the above summation. In particular $M_{c,d}^{\mu }\left(a_w\right)\ne 0$ if and only if there exists a $\bar{w}\in W\left(B_n\right),$ $\bar{w}=w_{j_1}\cdots w_{j_s},$ where $j_1<\dots <j_s$ and $\{j_1,\dots ,j_s\}\subset i_1,\dots ,i_k,$ or $\bar{w}=1,$ such that ${\bar{w}}^{-1}{T}_c^{\mu }=T_d^{\mu }\text{.}$

		Proof.
	As the $\left\{w_{i_j}\right\},$ $j=1,\dots ,k$ are distinct, $a_w=a_{i_1}\cdots a_{i_k}$ and (3.2.9) is just the definition of matrix multiplication. Furthermore the second statement in the proposition follows immediately from the first since by definition (2.2.6) and theorem (2.2.7), a matrix entry $M_{i,j}^{\mu }\left(a_k\right)\ne 0,$ for $k=2,\dots ,n,$ if and only if $i=j$ or $w_k{T}_i^{\mu }=T_j^{\mu }$ while $M_{i,j}^{\mu }\left(a_1\right)\ne 0$ if and only if $i=j\text{.}$ Thus a product of matrix entries of the form $$M_{c,c_1}^{\mu }\left(a_{i_1}\right)M_{c_1,c_2}^{\mu }\left(a_{i_2}\right)\cdots M_{c_{k-1},d}^{\mu }\left(a_{i_k}\right),i_1<i_2<\dots <i_k$$ is non-zero if and only if there exists a $w\in W,$ $w=w_{j_1}\cdots w_{j_s}$ with $j_1<\dots <j_s,$ $\{j_1,\dots ,j_s\}\subset \{i_1,\dots ,i_k\}$ or $w=1$ such that $w^{-1}{T}_c^{\mu }=T_d^{\mu }\text{.}$ Hence to complete the proof of the proposition we must show there is at most one non-zero term in the summation (3.2.9). We do this by induction on $k\text{.}$ It is certainly true for $k=1\text{.}$ Assume true for $k-1\text{.}$ By the rules of matrix multiplication, $$\begin{array}{cc}\text{(3.2.10)}& {\displaystyle M_{c,d}^{\mu }\left(a_w\right)=\sum _s{M}_{c,s}^{\mu }\left(a_{w\prime }\right)M_{s,d}^{\mu }\left(a_{i_k}\right),w\prime =w_{i_1}\cdots w_{i_{k-1}}\text{.}}\end{array}$$ By the induction hypothesis it is enough to show at most one term in (3.2.10) is non-zero. Consider the position of the letters $i_k$ and $i_{k-1}$ in $T_d^{\mu }\text{.}$ (i) If they belong either to the same row or column of the tableau of shape $\left(\alpha \right)$ or the tableau of shape $\left(\beta \right)$ of $T_d^{\mu }\text{,}$ $M_{s,d}^{\mu }\left(a_{i_k}\right)\ne 0$ if and only if $s=d\text{.}$ Thus $$M_{c,d}^{\mu }\left(a_w\right)=M_{c,d}^{\mu }\left(a_{w\prime }\right)M_{d,d}^{\mu }\left(a_{i_k}\right)$$ and the proposition is proved for this case. (ii) If the letters $i_k$ and $i_{k-1}$ do not belong to the same row or column of either the tableau $T^{\alpha }$ or the tableau $T^{\beta }$ of $T_d^{\mu }=(T^{\alpha },T^{\beta }),$ set $T_e^{\mu }=w_{i_k}{T}_d^{\mu }\text{.}$ Then $e\ne d,$ $M_{s,d}^{\mu }\left(a_{i_k}\right)\ne 0$ if and only if $s=d$ or $e$ and $$M_{c,d}^{\mu }\left(a_w\right)=M_{c,d}^{\mu }\left(a_{w\prime }\right)M_{d,d}^{\mu }\left(a_{i_k}\right)+M_{c,e}^{\mu }\left(a_{w\prime }\right)M_{e,d}^{\mu }\left(a_{i_k}\right)\text{.}$$ Suppose $M_{c,d}^{\mu }\left(a_{w\prime }\right)\ne 0\text{.}$ By the inductive hypothesis and the first part of the proposition, there exists a $\bar{w}\in W$ expressible as a product of distinct generators chosen from the set $\{w_{i_1},\dots ,w_{i_{k-1}}\}$ such that $\bar{w}{T}_c^{\mu }=T_d^{\mu }\text{.}$ As $i_1<\dots <i_k,$ the letter $i_k$ is left fixed under the action of $\bar{w}$ on $T_c^{\mu },$ i.e. $i_k$ occupies the same position in the standard tableaux $T_c^{\mu }$ and $T_d^{\mu }\text{.}$ Therefore by the choice of $e\text{,}$ the letter $i_k$ occupies different positions in the standard tableaux $T_c^{\mu }$ and $T_c^{\mu }\text{.}$ But then there cannot exist a $\bar{w}\in W$ expressible as a product of generators taken from $\{w_{i_1},\dots ,w_{i_{k-1}}\}$ such that $\bar{w}{T}_c^{\mu }=T_e^{\mu },$ i.e. $M_{c,e}^{\mu }\left(a_{w\prime }\right)=0\text{.}$ Thus $$M_{c,d}^{\mu }\left(a_{i_1}\cdots a_{i_k}\right)=M_{c,s}^{\mu }\left(a_{i_1}\cdots a_{i_{k-1}}\right)M_{s,d}^{\mu }\left(a_{i_k}\right)$$ where $s=e$ or $d,$ depending on the position of the letter $i_k$ in $T_d^{\mu }\text{.}$ This completes the proof. $\square$

The above proposition is a distinctive property of the shape of the matrices $M^{\mu }\left(a_i\right)$ in that it clearly depends only on the position of the zero entries. Furthermore if $w\in W\left(B_n\right)$ is as in the statement of the proposition, the proposition is clearly valid for $w^{-1}$ as well. Finally, for the diagonal entries $M_{c,c}^{\mu }\left(a_w\right),$ the proposition is valid for any $W$ expressible as a product of distinct generators, not necessarily an increasing (or decreasing) product (see [Rut1948], pp. 43-44).

The Evaluation of $E^{({\mu }_c-)}$

Our aim in this section is an expression for $E^{({\mu }_c-)}$ in terms of the polynomials $\mathrm{\Delta }(m,y)$ and $\mathrm{\Delta }(m,-1)\text{.}$ Throughout, $\mu =(\alpha ,\beta )$ will denote a double partition of $n\text{,}$ with the partition $\left(\alpha \right)$ having $s_{\alpha }$ parts $\left(\alpha \right)=({\alpha }_1,\dots ,{\alpha }_{s_{\alpha }})$ and the partition $\left(\beta \right)$ having $s_{\beta }$ parts $\left(\beta \right)=({\beta }_1,\dots ,{\beta }_{s_{\beta }})\text{.}$ As before $({\mu }_c-)$ denotes the non-zero double partition of $n-1$ obtained from $\left(\mu \right)$ by deletion of a square from the end of row $c$ of the Young diagram $D\left(\mu \right)\text{,}$ i.e. the first allowable row.

We first determine explicitly the elements $\left\{x_k\right\}$ of minimal length in the left $W_{J_{n-1}}\text{-cosets}$ of $(W\left(B_n\right),R),$ $R$ and $J_{n-1}$ defined as in the previous section. We introduce some more notation. For any set of consecutive integers $1\le k,k+1,\dots ,l\le n,$ set $w(k,l)=w_k{w}_{k+1}\cdots w_l$ so that $w(k,k)=w_k\text{.}$ For ease of notation we also define $w(k+1,k)=1\text{.}$ Furthermore, set $w\left(k\right)=w{(2,k)}^{-1}{w}_{1}w(2,k),$ $k=1,\dots ,n\text{.}$ Thus $w\left(1\right)=w_1\text{.}$ $w\left(k\right)$ is the $k\text{-th}$ sign change $-\left(k\right)$ of $W\left(B_n\right)\text{.}$

Lemma (3.3.1) The set $$S=\left\{1\right\}\cup \{w(k,n):k=2,\dots ,n\}\cup \{w\left(k\right)w(k+1,n):k=1,\dots ,n\}$$ is the unique set of elements of minimal length in the left $W_{J_{n-1}}\text{-cosets}$ of $W\left(B_n\right)\text{.}$

		Proof.
	$W\left(B_n\right)$ acts on a fixed orthonormal basis $\{{ϵ}_1,\dots ,{ϵ}_n\}$ of ${ℝ}^n$ as all permutations and sign changes. For the given choice of $R,$ the set of fundamental roots of $W\left(B_n\right)$ is $$\{{ϵ}_1,{ϵ}_2-{ϵ}_1,{ϵ}_3-{ϵ}_2,\dots ,{ϵ}_n-{ϵ}_{n-1}\}$$ and the set of positive roots is $$\{{ϵ}_i,{ϵ}_j-{ϵ}_i,{ϵ}_j+{ϵ}_i\hspace{0.17em}|\hspace{0.17em}1\le i,j\le n,j>i\}$$ see ([Bou1968]). For an element $x=w(k,n),$ $x^{-1}$ is the $n-k+2$ cycle $(n\hspace{0.17em}n-1\dots k-1)$ working from right to left. Hence $x^{-1}$ sends the positive roots ${ϵ}_j-{ϵ}_{k-1},$ $j>k-1,$ to the negative roots ${ϵ}_{j-1}-{ϵ}_n,$ $j=k,\dots ,n\text{.}$ By the choice of $J_{n-1},$ these roots remain negative under the action of $W_{J_{n-1}}\cong W\left(B_{n-1}\right)\text{.}$ Thus $l\left(w\prime x^{-1}\right)\ge l\left(x^{-1}\right)$ for all $w\prime \in W_{J_{n-1}},$ as $l\left(w\right)$ equals the number of positive roots sent to negative roots under $W$ (see [Bou1968] or [Ste1967], appendix). Therefore $x$ is of minimal length in the left $W_{J_{n-1}}\text{-coset}$ $x{W}_{J_{n-1}}\text{.}$ A similar argument shows $l\left(w\prime x^{-1}\right)\ge l\left(x^{-1}\right)$ for $x=w\left(k\right)w(k+1,n),$ $k=1,\dots ,n$ and for all $w\prime \in W_{J_{n-1}}\text{.}$ As $l\left(x\right)\ne l\left(x\prime \right)$ for any $x,x\prime \in S,$ the elements of $S$ must belong to distinct cosets. As $|W\left(B_n\right):W_{J_{n-1}}|=2n$ and $\left|S\right|=2n,$ $S$ must be a set of coset representatives for the left $W_{J_{n-1}}\text{-cosets}$ of $W\left(B_n\right)\text{.}$ This completes the proof. $\square$

Let $a_{w(2,k)}=a(2,k)$ and $a_{w\left(k\right)}=a\left(k\right)$ in $𝒜\left(B_n\right)\text{.}$ As the expression for $w(2,k)$ and $w\left(k\right)$ as a product of generators from $R$ is reduced $$\begin{array}{cc}\text{(3.3.2)}& \begin{array}{ccc}a(2,k)& =& a_2\cdots a_k\\ a\left(k\right)& =& \left(a_k\cdots a_2\right)a_1\left(a_2\cdots a_k\right)\text{.}\end{array}\end{array}$$ In order to state the next proposition concerning the matrices $M^{\mu }\left(a\left(k\right)\right)$ corresponding to the $k\text{-th}$ sign change of $W\left(B_n\right)$ we need some notations. Let ${\rho }_{i,p},$ $i=2,\dots ,n$ denote the axial distance from $i-1$ to $i$ in the standard tableau $T_p^{\mu }\text{.}$ Let $${\rho }_p\left(i\right)=\sum _{j=2}^i({\rho }_{j,p}+1),i=2,\dots ,n$$ and define ${\rho }_p\left(1\right)=0$ for all indices $p=1,\dots ,f^{\mu }\text{.}$

Proposition (3.3.3) The matrix $M^{\mu }\left(a\left(k\right)\right),$ $k=1,\dots ,n$ is a diagonal matrix with the $p,p\text{-th}$ entry equal to $z{x}^{{\rho }_p\left(k\right)},$ where $z=y$ if the letter $k$ appears in the tableau $T^{\alpha }$ of $T_p^{\mu }=(T^{\alpha },T^{\beta })$ and $z=-1$if $k$ appears in the tableau $T^{\beta }$ of $T_p^{\mu }\text{.}$

		Proof.
	The proof is by induction on $k\text{.}$ For $k=1$ the statement of the proposition is just the definition of $M^{\mu }\left(a_1\right)\text{.}$ Now assume $M^{\mu }\left(a(k-1)\right)$ is diagonal. By (3.3.2) $$M^{\mu }\left(a\left(k\right)\right)=M^{\mu }\left(a_k\right)M^{\mu }\left(a(k-1)\right)M^{\mu }\left(a_k\right)\text{.}$$ Write $V_{\mu }^K$ as the direct sum $V_{\mu }^K=⨁V_{p,q}$ of $a_k$ invariant subspaces, where $V_{p,q}$ has basis $\{t_p,t_q\}$ if $(k-1,k)T_p=T_q,$ $p<q,$ and $V_{p,p}$ has basis $\left\{t_p\right\}$ if the letters $k-1$ and $k$ appear either in the same row or column of the same tableau. With this ordering of the basis, $M^{\mu }\left(a\left(k\right)\right)$ has the corresponding block form $M^{\mu }\left(a\left(k\right)\right)=\dotplus M_{p,q}^{\mu }\left(a\left(k\right)\right)\text{.}$ Thus the usual case by case argument on the configuration of the letters $k-1$ and $k$ will suffice. 1. $k-1$ and $k$ in the same row or column of the same $T_p\text{.}$ If $k-1$ and $k$ are in the same row, ${\rho }_{k,p}=1\text{.}$ Therefore ${\rho }_p\left(k\right)={\rho }_p(k-1)+2\text{.}$ As $M_{p,p}^{\mu }\left(a_k\right)=x$ by theorem (2.2.7) $$M_{p,p}^{\mu }\left(a\left(k\right)\right)=x\left(z{x}^{{\rho }_p(k-1)}\right)x=z{x}^{{\rho }_p(k-1)+2}=z{x}^{{\rho }_p\left(k\right)}\text{.}$$ If $k-1$ and $k$ are in the same column, ${\rho }_{k,p}=-1\text{.}$ Therefore ${\rho }_p\left(k\right)={\rho }_p(k-1)\text{.}$ As $M_{p,p}^{\mu }\left(a_k\right)=-1$ by theorem (2.2.7) $$M_{p,p}^{\mu }\left(a\left(k\right)\right)=(-1)\left(z{x}^{{\rho }_p(k-1)}\right)(-1)=z{x}^{{\rho }_p\left(k\right)}\text{.}$$ 2. $k-1$ and $k$ in distinct rows and columns of $T_p^{\mu }\text{.}$ Set $T_q^{\mu }=(k-1,k)T_p^{\mu }$ and take $p<q\text{.}$ The deletion of all letters $\ge k-1$ in $T_p^{\mu }$ and $T_q^{\mu }$ yield the same tableaux of $k-2$ letters. Therefore ${\rho }_p(k-2)={\rho }_q(k-2)\text{.}$ Let $$\begin{array}{c} k-2 k-1 k ϵ1 ϵ2 \end{array}$$ denote the configuration of the letters $k-2,$ $k-1,$ and $k$ in $T_p^{\mu },$ with ${ϵ}_1,$ ${ϵ}_2$ the respective axial distances. With this notation we have, $$\begin{array}{ccc}{\rho }_p(k-1)& =& {\rho }_p(k-2)+{ϵ}_1+1\\ {\rho }_p\left(k\right)& =& {\rho }_p(k-2)+{ϵ}_1+{ϵ}_2+2\\ {\rho }_q(k-1)& =& {\rho }_q(k-2)+{ϵ}_1+{ϵ}_2+1\\ {\rho }_q\left(k\right)& =& {\rho }_q(k-2)+{ϵ}_2+2\text{.}\end{array}$$ Therefore if $k-1$ and $k$ belong to the same tableau of $T_p^{\mu },$ $M_{p,q}^{\mu }\left(a_k\right)=M({ϵ}_2,-1)$ and $M_{p,q}^{\mu }\left(a(k-1)\right)=z{x}^{{\rho }_p(k-2)+{ϵ}_1+1}D(1,x^{{ϵ}_2})$ by theorem (2.2.7) and the induction hypothesis. Direct computation verifies the relation $$M({ϵ}_2,-1)D(1,x^{{ϵ}_2})M({ϵ}_2,-1)=xD(x^{{ϵ}_2},1)\text{.}$$ Therefore $$M_{p,q}^{\mu }\left(a\left(k\right)\right)=z{x}^{{\rho }_p(k-2)+{ϵ}_1+2}D(x^{{ϵ}_2},1)=D(z{x}^{{\rho }_p\left(k\right)},z{x}^{{\rho }_q\left(k\right)})$$ If $k-1$ and $k$ belong to distinct tableaux of $T_p^{\mu },$ $M_{p,q}^{\mu }\left(a_k\right)=M(-{ϵ}_2,y)$ by theorem (2.2.7). As we have taken $p<q$ with respect to the last letter sequence, $k-1$ appears in the tableau corresponding to $\left(\alpha \right)$ of $T_p^{\mu }\text{.}$ Thus by the induction hypothesis $$M_{p,q}^{\mu }\left(a(k-1)\right)=x^{{\rho }_p(k-2)+{ϵ}_1+1}D(y,-x^{{ϵ}_2})\text{.}$$ Direct computation verifies the relation $$M(-{ϵ}_2,y)D(y,-x^{{ϵ}_2})M(-{ϵ}_2,y)=xD(y,-x^{{ϵ}_2})\text{.}$$ Therefore $$M_{p,q}^{\mu }\left(a\left(k\right)\right)=x^{{\rho }_p(k-2)+{ϵ}_1+2}D(-x^{{ϵ}_2},y)=D(-x^{{\rho }_p\left(k\right)},y{x}^{{\rho }_q\left(k\right)})\text{.}$$ This completes the proof of the proposition. $\square$

We use this proposition to affect a reduction of $E^{({\mu }_c-)}\text{.}$ Pairing the $W_{J_{n-1}}\text{-coset}$ representatives $a\left(k\right)a(k+1,n)$ and $a(k+1,n)$ of $W\left(B_n\right)$ (lemma (3.3.1)) we have $$\begin{array}{cc}\text{(3.3.4)}& M_{1,i}^{\mu }\left(a(k+1,n)\right)+M_{1,i}^{\mu }\left(a\left(k\right)a(k+1,n)\right)=(1+M_{1,l}^{\mu }\left(a\left(k\right)\right))M_{1,i}^{\mu }\left(a(k+1,n)\right),i=1,\dots ,f,\end{array}$$ as $M^{\mu }\left(a\left(k\right)\right)$ is diagonal by proposition (3.3.3).

Furthermore for letters $k-1$ and $k$ both appearing in the same row of the canonical tableau $T_1^{\mu }$ of shape $\left(\mu \right)$ (see section 1), $w_k=(k-1,k),$ $k=1,\dots ,n,$ is a row permutation of the canonical tableau. As any row permutation $W$ can be written as a product of such transpositions $$\begin{array}{cc}\text{(3.3.5)}& M_{1,i}^{\mu }\left(a_w\right)=M_{i,1}^{\mu }\left(a_w\right)=0,i\ne 1,\end{array}$$ and $$\begin{array}{cc}\text{(3.3.6)}& M_{1,1}^{\mu }\left(a_w\right)=M_{1,1}^{\mu }\left({â}_w\right)=x\end{array}$$ by theorem (2.2.7), $W$ a row permutation of $T_1^{\mu }\text{.}$ Let $r_i,$ $i=1,\dots ,s,$ $s=s_{\alpha }+s_{\beta },$ denote the last letter in the $i\text{-th}$ row of the canonical tableau $T_1^{\mu }\text{,}$ i.e. $$r_i=\{\begin{array}{cc}{\displaystyle \sum _{j=1}^i{\alpha }_j}& \text{if}\hspace{0.17em}i\le s_{\alpha },\\ {\displaystyle \sum _{j=1}^{s_{\alpha }}{\alpha }_j+\sum _{j=1}^k{\beta }_j}& \text{if}\hspace{0.17em}i=s_{\alpha }+k\text{.}\end{array}$$ Set $$\begin{array}{cc}\text{(3.3.7)}& R_i={\displaystyle \sum _{k=r_{i-1}+1}^{r_i}(1+\nu {\left(a\left(k\right)\right)}^{-1}{M}_{1,1}^{\mu }{\left(a\left(k\right)\right)}^2)\nu {\left(a(k+1,r_i)\right)}^{-1}{M}_{1,1}^{\mu }{\left(a(k+1,r_i)\right)}^2\text{.}}\end{array}$$ Combining (3.3.5) and (3.3.6) with the definitions of $R_i$ and $E^{({\mu }_c-)}$ and using the explicit coset representatives given by lemma (3.3.1) we have $$\begin{array}{cc}\text{(3.3.8)}& {\displaystyle E^{({\mu }_c-)}=\sum _{i=1}^s{R}_i\nu {\left(a(r_i+1,n)\right)}^{-1}\sum _{j=g_c}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(r_i+1,n)\right)M_{j,1}^{\mu }\left(â(r_i+1,n)\right)\text{.}}\end{array}$$ Set $\mathrm{\Delta }(m,y^{-1})=1+x^m{y}^{-1}\text{.}$ Then

Proposition (3.3.9) For $1\le i\le s_{\alpha }$ denote $R_i$ by $R_{{\alpha }_i}\text{.}$ For $s_{\alpha }+1\le i\le s_{\alpha }+s_{\beta },$ denote $R_i$ by $R_{{\beta }_j}$ where $i=s_{\alpha }+j\text{.}$ Then $$\begin{array}{ccc}{R}_{{\alpha }_i}& =& \mathrm{\Delta }({\alpha }_i-2i+1,y)\mathrm{\Delta }({\alpha }_i,-1),\\ R_{{\beta }_i}& =& \mathrm{\Delta }({\beta }_i-2i+1,y-1)\mathrm{\Delta }({\beta }_i,-1)\text{.}\end{array}$$

		Proof.
	First let $1\le i\le s_{\alpha }\text{.}$ The axial distance from $r_i$ to $r_i+1$ is $-{\alpha }_i$ while for $k$ and $k+1$ in the same row of $T_1^{\mu }\text{,}$ the axial distance from $k$ to $k+1$ is $1\text{.}$ Therefore $${\rho }_1(r_{i-1}+1)=\sum _{k=2}^{r_{i-1}+1}({\rho }_{k,1}+1)=\sum _{j=1}^{i-1}(2({\alpha }_j-1)+(1-{\alpha }_j))=\sum _{j=1}^{i-1}({\alpha }_j-1)\text{.}$$ Also $\nu \left(a(r_{i-1}+1+k)\right)=y{x}^{2(m+k)}$ where $m=\sum _{j=1}^{i-1}{\alpha }_j$ and $k=0,\dots ,{\alpha }_i\text{.}$ Therefore $$\nu {\left(a(r_{i-1}+1+k)\right)}^{-1}{M}_{1,1}^{\mu }{\left(a(r_{i-1}+1+k)\right)}^2={\left(y{x}^{2(m+k)}\right)}^{-1}{y}^2{x}^{2(m-(i-1)+2k)}=y{x}^{-2(i-1)+2k}$$ for $k=0,\dots ,{\alpha }_i\text{.}$ Furthermore $\nu \left(a(r_{i-1}+1+k,r_i)\right)=x^{{\alpha }_i-k}$ and $$M_{1,1}^{\mu }{\left(a(r_{i-1}+1+k,r_i)\right)}^2=x^{2(r_i-r_{i-1}-k)}=x^{2({\alpha }_i-k)}$$ for $k=1,\dots ,{\alpha }_i$ by (3.3.6), as we defined $a(m+1,m)=1\text{.}$ Thus (3.3.7) implies $$\begin{array}{ccc}{R}_{{\alpha }_i}& =& {\displaystyle \sum _{k=1}^{{\alpha }_i}(1+y{x}^{-2(i-1)+2(k-1)})x^{{\alpha }_i-k}}\\ & =& {\displaystyle \sum _{k=1}^{{\alpha }_i}(x^{{\alpha }_i-k}+y{x}^{{\alpha }_i+k-2i})}\\ & =& (1+y{x}^{{\alpha }_i-2i+1})(1+\dots +x^{{\alpha }_i-1})\\ & =& \mathrm{\Delta }({\alpha }_i-2i+1,y)\mathrm{\Delta }({\alpha }_i,-1)\text{.}\end{array}$$ We now turn to the second part of the proposition. Observe first that the axial distance from $r_{s_{\alpha }}$ to $r_{s_{\alpha }}+1$ in the canonical tableau $T_1^{\mu }$ is by definition (1.1.6) the axial distance from the last square in the diagram $\left(\alpha \right)$ to the first square in the first row of $\left(\alpha \right)\text{.}$ Therefore the path traversed from the letter $1$ to the letter $r_{s_{\alpha }}+1$ is a closed path in terms of axial distance. As a result $$\sum _{j=2}^{r_{s_{\alpha }}+1}{\rho }_{j,1}=0\text{and}{\rho }_1(r_{s_{\alpha }}+1)=\sum _{j=2}^{r_{s_{\alpha }}+1}{\rho }_{j,1}+1=\left|\alpha \right|\text{.}$$ For $i>s_{\alpha }$ set $l=i-s_{\alpha }\text{.}$ Then $$\begin{array}{ccc}{\rho }_1(r_{i-1}+1)& =& {\displaystyle \left|\alpha \right|+\sum _{k=r_{s_{\alpha }}+2}^{r_{i-1}+1}{\rho }_{k,1}+1}\\ & =& {\displaystyle \left|\alpha \right|+\sum _{j=1}^{l-1}{\beta }_j-1\text{.}}\end{array}$$ Also $\nu \left(a(r_{i-1}+1+k)\right)=-x^{2(m+k)}$ where $m=\left|\alpha \right|+\sum _{j=1}^{l-1}{\beta }_j$ and $k=0,\dots ,{\beta }_l\text{.}$ Therefore $$\begin{array}{ccc}\nu \left(a(r_{i-1}+1+k)\right)M_{1,1}^{\mu }\left(a(r_{i-1}+1+k)\right)& =& {\left(y{x}^{2(m+k)}\right)}^{-1}{x}^{2(m-(l-1)+2k)}\\ & =& y^{-1}{x}^{-2(l-1)+2k}\text{.}\end{array}$$ The argument used in the first part of the proposition can now be used to give $$R_{{\beta }_i}=\mathrm{\Delta }({\beta }_i-2i+1,y^{-1})\mathrm{\Delta }({\beta }_i,-1)\text{.}$$ This completes the proof. $\square$

We now start the evaluation of the sum $$\sum _{j=g_c}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(r_i+1,n)\right)M_{j,1}^{\mu }\left(â(r_i+1,n)\right)$$ in the expression for $E^{({\mu }_c-)}$ given by (3.3.8).

Proposition (3.3.10) Let $c$ be as above. Then

(i)	$M_{i,j}^{\mu }\left(a(k,n)\right)=0$ for $k>r_c+1$ and for all $j\ge g_c\text{.}$
(ii)	$M_{i,j}^{\mu }\left(a(r_c+1,n)\right)=0$ for all $j>g_c$ while $$M_{1,g_c}^{\mu }\left(a(r_c+1)\right)=\prod _{i=1}^{n-r_c}{M}_{e_{i-1},e_i}^{\mu }\left(a_{r_c+i}\right)$$ where $T_{e_0}^{\mu }=T_1^{\mu },$ the canonical tableau of shape $\left(\mu \right)\text{,}$ and $T_{e_i}^{\mu }=w_{r_c+i}{T}_{e_{i-1}}^{\mu },$ $i\ge 1\text{.}$
(iii)	$M_{i,j}^{\mu }\left(a(k,n)\right)={\displaystyle \left(\sum _{i=1}^{f^{\mu }}{M}_{1,i}^{\mu }\left(a(k,r_c)\right)\right)M_{1,g_c}^{\mu }\left(a(r_c+1,n)\right)}$ for $j\ge f_1$ and $k\le r_c\text{.}$
(iv)	${\displaystyle \sum _{i=1}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(r_k+1,r_c)\right)}={\displaystyle \prod _{i=k}^{c-1}\left(\sum _{j=1}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(r_i+1,r_{i+1})\right)\right)}$ for $k<r_c\text{.}$

		Proof.
	(i) and (ii). For $j\ge g_c,$ the letter $n$ occupies the last square in row $c$ by definition of $({\mu }_c-)\text{.}$ Suppose $M_{1,j}^{\mu }\left(a(k,n)\right)\ne 0,$ $j\ge g_c\text{.}$ By proposition (3.2.8) there exists a $w\in W$ such that $w{T}_1^{\mu }=T_j^{\mu },$ $W$ expressible as a product of distinct generators chosen from $\{w_k,\dots ,w_n\}\text{.}$ It follows that $W$ fixes all letters $\le k-1$ in $T_1^{\mu }\text{.}$ Hence if $k>r_c+1,$ the letter $r_c$ occupies the same position in the tableaux $T_1^{\mu }$ and $T_j^{\mu },$ i.e. the last square in row $c\text{.}$ This is impossible. Therefore $M_{1,j}^{\mu }\left(a(k,n)\right)=0$ for $k>r_c+1$ which proves (i). On the other hand, if $k=r_c+1,$ $w{T}_1^{\mu }=T_j^{\mu }$ implies $w=w\prime w_{r_c+1},$ since the letter $r_c$ must be moved under the action of $W\text{.}$ Therefore $r_c+1$ is in the last square of row $c$ in $w_{r_c+1}{T}_1^{\mu }=T_{e_1}^{\mu }\text{.}$ Similarly, as $w_{r_c+1}$ does not occur in the expression of $w\prime$ as a product of distinguished generators, $w\prime =w″w_{r_c+2}$ because the letter $r_c+1$ must be moved under the action of $w\prime \text{.}$ Therefore $r_c+2$ is in the last square of row $c$ in $w_{r_c+2}{T}_{e_1}^{\mu }=T_{e_2}^{\mu }\text{.}$ Continuation of this argument allows us to conclude $j=g_c$ and $$M_{1,g_c}^{\mu }\left(a_{r_c+1}\cdots a_n\right)=M_{1,e_1}^{\mu }\left(a_{r_c+1}\right)M_{e_1,e_2}^{\mu }\left(a_{r_c+2}\right)\cdots M_{e_{n-k},g_c}^{\mu }\left(a_n\right)$$ where $w_{r_c+i}{T}_{e_{i-1}}^{\mu }=T_{e_i}^{\mu }\text{.}$ This proves (ii). (iii) and (iv). Let $T_e^{\mu }$ denote a standard tableau obtained from $T_1^{\mu }$ by any rearrangement of the letters $1,\dots ,r_j,$ for $j\le c\text{.}$ This amounts to a rearrangement of the above letters among the first $j$ rows of $\left(\mu \right)\text{.}$ Because the first $j$ rows, $\text{(}j\le c\text{)}$ are of equal length, any such arrangement of $1,\dots ,r_j$ in the first $j$ rows must have $r_j$ in the last box of row $j\text{.}$ Thus the position of the letters $r_j,\dots ,n$ is the same in $w{T}_e^{\mu }$ and $w{T}_1^{\mu },$ for any $W$ whose reduced expression as a product of elements from $R$ is made up entirely of the generators $w_i,$ $i=r_j+1,\dots ,n\text{.}$ It follows that $$M_{e,i}^{\mu }\left(a(r_j+1,k)\right)=M_{1,i}^{\mu }\left(a(r_j+1,k)\right)$$ for $j\le c$ and $k\ge r_j+1\text{.}$ Therefore $$\begin{array}{ccc}{M}_{1,i}^{\mu }\left(a(r_j+1,r_c)\right)& =& {\displaystyle \sum _{s,t,\dots ,u}{M}_{1,s}^{\mu }\left(a(r_j+1,r_{j+1})\right)M_{s,t}^{\mu }\left(a(r_{j+1}+1,r_{j+2})\right)\cdots M_{u,i}^{\mu }\left(a(r_{c-1}+1,r_c)\right)}\\ & =& {\displaystyle \sum _{s,t,\dots }{M}_{1,s}^{\mu }\left(a(r_j+1,r_{j+1})\right)M_{1,t}^{\mu }\left(a(r_{j+1}+1,r_{j+2})\right)\cdots M_{1,i}^{\mu }\left(a(r_{c-1}+1,r_c)\right)}\\ & =& {\displaystyle \prod _{i=j}^{c-1}\sum _{k=1}^{f^{\mu }}{M}_{1,k}^{\mu }\left(a(r_i+1,r_{i+1})\right)}\end{array}$$ by the above argument and the ordering of the last letter sequence. For the same reason $$\begin{array}{ccc}{M}_{1,i}^{\mu }\left(a(k,n)\right)& =& {\displaystyle \sum _{j=1}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(k,r_c)\right)M_{j,i}^{\mu }\left(a(r_c+1,n)\right)}\\ & =& {\displaystyle \left(\sum _{j=1}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(k,r_c)\right)\right)M_{1,g_c}^{\mu }\left(a(r_c+1,n)\right)\text{.}}\end{array}$$ This proves (iii) and (iv). $\square$

Using this proposition and proposition (3.2.8) it is straight forward from (3.3.8) that $E^{({\mu }_c-)}$ can be written as $$\begin{array}{cc}\text{(3.3.11)}& \begin{array}{ccc}{E}^{({\mu }_c-)}& =& {\displaystyle \sum _{i=1}^c{R}_i\nu {\left(a(r_i+1,n)\right)}^{-1}\sum _{j=g_c}^{f^{\mu }}{M}_{1,j}^{\mu }\left(a(r_i+1,n)\right)M_{j,1}^{\mu }\left(â(r_i+1,n)\right)}\\ & =& D_1{D}_2\end{array}\end{array}$$ where $$\begin{array}{ccc}{D}_1& =& {\displaystyle \sum _{i=1}^c{R}_i\nu {\left(a(r_i+1,r_c)\right)}^{-1}\prod _{j=1}^{c-1}\left(\sum _{k=1}^{f^{\mu }}{M}_{1,l}^{\mu }\left(a(r_j+1,r_{j+1})\right)M_{k,1}^{\mu }\left(â(r_j+1,r_{j+1})\right)\right),}\\ D_2& =& \nu {\left(a(r_c+1,n)\right)}^{-1}{M}_{1,g_c}^{\mu }\left(a(r_c+1,n)\right)M_{g_c,1}^{\mu }\left(â(r_c+1,n)\right)\text{.}\end{array}$$ Consider a part of the Young digram of shape $(\alpha ,\beta )$ consisting of the last box in the $p\text{-th}$ row and the entire $q\text{-th}$ row, $p<q\text{.}$ $$\begin{array}{c} l l+1 ⋯ l+t t boxes row p row q \end{array}$$ Let $T_{e_0}^{\mu }$ be a standard tableau with $t+1$ letters $l,l+1,\dots ,l+t$ distributed in increasing order in this part. Set $T_{e_i}^{\mu }=w_{l+i}\cdots w_{l+1}{T}_{e_0}^{\mu },$ $i=1,\dots ,t\text{.}$ Then $w_{l+i}$ is a row permutation of $T_{e_j}^{\mu }$ for $j\ne i-1$ or $i\text{,}$ as the letters $l+i-1$ and $l+i$ have either not been moved from row $q$ or have been returned to row $q\text{.}$ Hence $$M_{e_j,k}^{\mu }\left(a_{l+i}\right)=M_{k,e_j}^{\mu }\left(a_{l+i}\right)=0$$ for $k\ne e_j$ and $j\ne i-1$ or $i\text{,}$ by theorem (2.2.4). Therefore $$\begin{array}{cc}\text{(3.3.12)}& \begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle \sum _{j=1}^{f^{\mu }}{M}_{e_0,j}^{\mu }\left(a(l+1,l+t)\right)M_{j,e_0}^{\mu }\left(â(l+1,l+t)\right)}}\\ & =& {\displaystyle \prod _{i=1}^t{\left[M_{e_0,e_0}^{\mu }\left(a_{l+i}\right)\right]}^2}\\ & & {\displaystyle +\sum _{k=1}^{t-1}\prod _{j=1}^k{M}_{e_{j-1},e_j}^{\mu }\left(a_{l+j}\right)M_{e_j,e_{j-1}}^{\mu }\left(a_{l+j}\right)\prod _{j=k+1}^t{\left[M_{e_k,e_k}^{\mu }\left(a_{l+j}\right)\right]}^2}\\ & & {\displaystyle +\sum _{i=1}^t{M}_{e_{i-1},e_i}^{\mu }\left(a_{l+i}\right)M_{e_i,e_{i-1}}^{\mu }\left(a_{l+i}\right),}\end{array}\end{array}$$ by proposition (3.2.8). Label these three terms $A\text{,}$ $B,$ and $C$ respectively. As the entries of $M^{\mu }\left(a_i\right)$ depend only on the position of the letters $i-1$ and $i$ in a standard tableau, the above computation is independent of the letter $l$ and depends only on the rows $P$ and $q\text{.}$ Hence set $$\begin{array}{cc}\text{(3.3.13)}& \begin{array}{ccc}{F}_{p,q}& =& \nu {\left(a(l+1,l+t)\right)}^{-1}(A+B)=x^t(A+B)\\ f_{p,q}& =& \nu {\left(a(l+1,l+t)\right)}^{-1}C=x^{t}C\end{array}\end{array}$$ We now rewrite (3.3.11) as follows.

Let $d_j$ be such that $w{(r_{c+1},r_j)}^{-1}{T}_1=T_{d_j}\text{.}$ Then $d_j=e_{r_j-r_c}$ where the $e_i\text{'s}$ are defined as in (ii) of proposition (3.3.10). The proof of (3.3.10 (ii)) shows the letters $r_j,r_j+1,\dots ,r_{i+1}$ occur in the last square of row $c$ and in the $j+1\text{-st}$ row of $T_{d_j}$ in increasing order, i.e. in a configuration as described above. Hence (3.3.10 (ii)) and proposition (3.2.8) show $$\begin{array}{cc}\text{(3.3.14)}& \begin{array}{ccc}\multicolumn{3}{c}{\nu {\left(a(r_c+1,n)\right)}^{-1}{M}_{1,g_c}^{\mu }\left(a(r_c+1,n)\right)M_{g_c,1}^{\mu }\left(â(r_c+1,n)\right)}\\ & =& {\displaystyle \prod _{j=c}^{s-1}\nu {\left(a(r_j+1,r_{j+1})\right)}^{-1}\prod _{i=r_j-r_c+1}^{r_{j+1}-r_c}{M}_{e_{i-1},e_i}^{\mu }\left(a_{r_c+i}\right)M_{e_i,e_{i-1}}^{\mu }\left(a_{r_c+i}\right)}\\ & =& {\displaystyle \prod _{j=c+1}^s{f}_{c,j}\text{.}}\end{array}\end{array}$$ Similarly, for $j<c$ the letters $r_j,r_j+1,\dots ,r_{i+1}$ appear in the last square of row $j$ and in the $j+1\text{-st}$ row of $T_1^{\mu }\text{.}$ Noting that $f_{p,q}=0$ for rows $P$ and $q$ of equal length and both belonging to the same tableau of $T_1^{\mu }$ the computation (3.3.12) shows $$\begin{array}{cc}\text{(3.3.15)}& {\displaystyle \nu {\left(a(r_j+1,r_{j+1})\right)}^{-1}\sum _{i-1}^{f^{\mu }}{M}_{1,i}^{\mu }\left(a(r_j+1,r_{j+1})\right)M_{i,1}^{\mu }\left(â(r_j+1,r_{j+1})\right)=F_{j,j+1}}\end{array}$$ for $j<c\text{.}$ Substitution of (3.3.14) and (3.3.15) into (3.3.11) now gives $$\begin{array}{cc}\text{(3.3.16)}& {\displaystyle E^{({\mu }_c-)}(\sum _{i=1}^{c-1}{R}_i\prod _{j=i}^{c-1}{F}_{j,j+1}+R_c)\prod _{i=c+1}^s{f}_{c,i}\text{.}}\end{array}$$

Proposition (3.3.17) Let $m$ be the axial distance from the first square in the $q\text{-th}$ row to the last square in the $p\text{-th}$ row and assume row $p$ is a row of the tableau $T^{\alpha }$ and row $q$ is a row of the tableau $T^{\beta }$ of $T_{e_0}=(T^{\alpha },T^{\beta })\text{.}$ Let $t$ be the length of the $q\text{-th}$ row. Then

(i)	$f_{p,q}={\displaystyle \frac{\mathrm{\Delta }(m+1,y)\mathrm{\Delta }(m-t,y)}{\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-t+1,y)}},$
(ii)	$F_{p,q}={\displaystyle \frac{{(x-1)}^2\mathrm{\Delta }(t,-1)}{x\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-t+1,y)}}\text{.}$

		Proof.
	We use the notations of (3.3.13). (i) Because the letter $l$ appears in the last box in row $P$ and $l+1$ in the first box of row $q\text{,}$ the axial distance from $l+1$ to $l$ in $T_{e_0}^{\mu }$ is $m$ and the axial distance from $l+i$ to $l+i-1$ in $T_{e_0}^{\mu }$ is $m-i+1,$ $i=1,\dots ,t\text{.}$ Therefore $$M_{e_{i-1},e_i}^{\mu }\left(a_{l+i}\right)M_{e_i,e_{i-1}}^{\mu }\left(a_{l+i}\right)=\frac{x\mathrm{\Delta }(m-i,y)\mathrm{\Delta }(m-i+2,y)}{{\left(\mathrm{\Delta }(m-i+1,y)\right)}^2}$$ by theorem (2.2.4). Thus (3.3.11) implies $$C=\prod _{i=1}^t\frac{x\mathrm{\Delta }(m-i,y)\mathrm{\Delta }(m-i+2,y)}{{\left(\mathrm{\Delta }(m-i+1,y)\right)}^2}=\frac{x^t\mathrm{\Delta }(m+1,y)\mathrm{\Delta }(m-t,y)}{\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-y+1,y)}\text{.}$$ As $f_{p,q}=x^{-t}C,$ we have the required result. (ii) We have $$M_{e_{i-1},e_{i-1}}^{\mu }\left(a_{l+i}\right)=\frac{x-1}{\mathrm{\Delta }(m-i+1,y)},i=1,\dots ,t,$$ by theorem (2.2.7) and as $w_{l+i}$ is a row permutation of $T_{e_j}^{\mu }$ for $i>j+1,$ $M_{e_j,e_j}^{\mu }\left(a_{l+1}\right)=x$ for $i>j+1,$ by theorem (2.2.4). Hence (3.3.11) implies $$A=\frac{x^{2(t-1)}{(x-1)}^2}{{\left(\mathrm{\Delta }(m,y)\right)}^2}\text{.}$$ Furthermore, using (i) and (3.3.11) $$\begin{array}{ccc}B& =& {\displaystyle \sum _{k=1}^{t-1}\left(\prod _{j=1}^k\frac{x\mathrm{\Delta }(m-i,y)\mathrm{\Delta }(m-i+2,y)}{{\left(\mathrm{\Delta }(m-i+1,y)\right)}^2}\right)\frac{{(x-1)}^2{x}^{2(t-1-k)}}{{\left(\mathrm{\Delta }(m-k,y)\right)}^2}}\\ & =& {\displaystyle \sum _{k=1}^{t-1}\frac{x^k\mathrm{\Delta }(m+1,y)\mathrm{\Delta }(m-k,y)}{\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-k+1,y)}·\frac{{(x-1)}^2{x}^{2(t-1-k)}}{{\left(\mathrm{\Delta }(m-k,y)\right)}^2}}\\ & =& {\displaystyle \left(\frac{x^{t-1}{(x-1)}^2\mathrm{\Delta }(m+1,y)}{\mathrm{\Delta }(m,y)}\right)\sum _{k=1}^{t-1}\frac{x^{t-1-k}}{\mathrm{\Delta }(m-k+1,y)\mathrm{\Delta }(m-k,y)}\text{.}}\end{array}$$ An easy induction argument, using the fact that $$x\mathrm{\Delta }(k,-1)\mathrm{\Delta }(m-k-1,y)+\mathrm{\Delta }(m,y)=\mathrm{\Delta }(k+1,-1)\mathrm{\Delta }(m-k,y)$$ shows $$\sum _{k=1}^{t-1}\frac{x^{t-1-k}}{\mathrm{\Delta }(m-k+1,y)\mathrm{\Delta }(m-k,y)}=\frac{\mathrm{\Delta }(t-1,-1)}{\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-t+1,y)}\text{.}$$ Therefore $$B=\frac{x^{t-1}{(x-1)}^2\mathrm{\Delta }(t-1,-1)\mathrm{\Delta }(m+1,y)}{{\left(\mathrm{\Delta }(m,y)\right)}^2\mathrm{\Delta }(m-t+1,y)}\text{.}$$ Finally, the above computations give $$\begin{array}{ccc}A+B& =& {\displaystyle \frac{x^{t-1}{(x-1)}^2}{{\left(\mathrm{\Delta }(m,y)\right)}^2}(x^{t-1}+\frac{\mathrm{\Delta }(t-1,-1)\mathrm{\Delta }(m+1,y)}{\mathrm{\Delta }(m-t+1,y)})}\\ & =& {\displaystyle \frac{x^{t-1}{(x-1)}^2}{{\left(\mathrm{\Delta }(m,y)\right)}^2}\frac{\mathrm{\Delta }(m,y)\mathrm{\Delta }(t,-1)}{\mathrm{\Delta }(m-t+1,y)}}\\ & =& {\displaystyle \frac{x^{t-1}{(x-1)}^2\mathrm{\Delta }(t,-1)}{\mathrm{\Delta }(m,y)\mathrm{\Delta }(m-t+1,y)}\text{.}}\end{array}$$ As $F_{p,q}=x^{-t}(A+B),$ we have the required result. $\square$

Corollary (3.3.18) Let $P$ and $q$ be rows belonging to the same tableau of $T_{e_0}^{\mu }\text{.}$ With the same notations as in proposition (3.3.17) we have

(i)	$f_{p,q}={\displaystyle \frac{\mathrm{\Delta }(m+1,-1)\mathrm{\Delta }(m-t,-1)}{\mathrm{\Delta }(m,-1)\mathrm{\Delta }(m-t+1,-1)}},$
(ii)	$F_{p,q}={\displaystyle \frac{{(x-1)}^2\mathrm{\Delta }(t,-1)}{x\mathrm{\Delta }(m,-1)\mathrm{\Delta }(m-t+1,-1)}}\text{.}$

If furthermore, $p=q-1$ and the rows $q$ and $q-1$ have the same length, then $F_{p,q}=x^{-1}\text{.}$

		Proof.
	If the rows $p$ and $q$ belong to the same tableau of $T_{e_0}^{\mu },$ the matrices $M(k,y)$ are replaced by the matrices $M(k,-1)$ in proposition (3.3.17) by theorem (2.2.7). The matrices $M(k,-1)$ are obtained from $M(k,y)$ by setting $y=-1,$ whence the first statement. For the second statement we have $m=t$ and $\mathrm{\Delta }(1,-1)=1\text{.}$ $\square$

Generic Degrees

Recall from (1) that $\left(\alpha \right)\prime =(a_1^{\prime },\dots ,{\alpha }_{s_{\alpha \prime }}^{\prime })$ denotes the partition conjugate to the partition $\left(\alpha \right)=({\alpha }_1,\dots ,{\alpha }_{s_{\alpha }})\text{.}$

Definition (3.4.1) Let $(\alpha ,\beta )$ be a double partition with corresponding ordered pair of Young diagrams $(D\left(\alpha \right),D\left(\beta \right))\text{.}$ For the $(i,j)\text{-th}$ square of $D\left(\alpha \right)\text{,}$ set $$\begin{array}{ccc}{h}_{i,j}^{\alpha }& =& ({\alpha }_i-j)+({\alpha }_j^{\prime }-i)+1,\\ g_{i,j}^{\alpha }& =& ({\alpha }_i-j)+({\beta }_j^{\prime }-i)+1,\end{array}$$ where ${\beta }_j^{\prime }=0$ for $j>s_{\beta \prime }\text{.}$ For the $(i,j)\text{-th}$ square of $D\left(\beta \right)\text{,}$ set $$\begin{array}{ccc}{h}_{i,j}^{\beta }& =& ({\beta }_i-j)+({\beta }_j^{\prime }-i)+1,\\ g_{i,j}^{\beta }& =& ({\beta }_i-j)+({\alpha }_j^{\prime }-i)+1,\end{array}$$ where ${\alpha }_j^{\prime }=0$ for $j>s_{\alpha \prime }\text{.}$ We call $h_{i,j}^{\alpha }$ [resp. $h_{i,j}^{\beta }\text{]}$ the hook length of the $(i,j)\text{-th}$ square of $D\left(\alpha \right)$ [resp. $D\left(\beta \right)\text{].}$ We call $g_{i,j}^{\alpha }$ [resp. $g_{i,j}^{\beta }\text{]}$ the split hook length of the $(i,j)\text{-th}$ square of $D\left(\alpha \right)$ [resp. $D\left(\beta \right)\text{].}$

From (1.1.1), $h_{i,j}^{\alpha }$ $\left[h_{i,j}^{\beta }\right]$ is the length of the $(i,j)\text{-hook}$ of $D\left(\alpha \right)$ $\left[D\left(\beta \right)\right]\text{.}$ As the last square in the $i\text{-th}$ row of $D\left(\alpha \right)$ has coordinates $(i,{\alpha }_i)$ while the last square in the $j\text{-th}$ column of $D\left(\alpha \right)$ has coordinates $({\alpha }_j^{\prime },j),$ (1.1.5) shows $h_{i,j}^{\alpha }$ equals the axial distance from the $({\alpha }_j^{\prime },j)\text{-square}$ to the $(i,{\alpha }_i)\text{-square}$ plus one in $D\left(\alpha \right)\text{.}$ $h_{i,j}^{\beta }$ has the same interpretation for the diagram $D\left(\beta \right)\text{.}$ The split hook lengths have a corresponding interpretation. Namely, $g_{i,j}^{\alpha }$ is the axial distance from the last square in the $j\text{-th}$ column of $D\left(\beta \right)\text{,}$ the $({\beta }_j^{\prime },j)\text{-square,}$ to last square in the $i\text{-th}$ row of $D\left(\alpha \right)\text{,}$ the $(i,{\alpha }_i)\text{-square,}$ plus one. Similarly $g_{i,j}^{\beta }$ equals the axial distance from the end of the $j\text{-th}$ column of $D\left(\alpha \right)$ to the end of the $i\text{-th}$ row of $D\left(\beta \right)\text{,}$ plus one.

Proposition (3.4.2)

(i)	For the double partition $(\alpha ,\left(0\right))$ $$E^{(({\alpha }_c-),\left(0\right))}=\mathrm{\Delta }({\alpha }_c-c,y)H^{\alpha }$$ where $$H^{\alpha }=\prod _{j=1}^{c-1}\frac{\mathrm{\Delta }(h_{j,{\alpha }_c}^{\alpha },-1)}{x\mathrm{\Delta }(h_{j,{\alpha }_c}^{\alpha }-1,-1)}\prod _{j=1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(h_{c,j}^{\alpha },-1)}{\mathrm{\Delta }(h_{c,j}^{\alpha }-1,-1)}\text{.}$$ Furthermore $$\mathrm{\Delta }({\alpha }_c-c,y)=\mathrm{\Delta }(g_{c,{\alpha }_c}^{\alpha },y)\prod _{j=1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(g_{c,j}^{\alpha },y)}{\mathrm{\Delta }(g_{c,j}^{\alpha }-1,y)}\text{.}$$
(ii)	For the double partition $(\left(0\right),\beta )$ $$E^{(\left(0\right),({\beta }_c-))}=\mathrm{\Delta }({\beta }_c-c,y^{-1})H^{\beta }$$ where $H^{\beta }$ is defined as in (i) the partition $\left(\beta \right)$ replacing the partition $\left(\alpha \right)\text{.}$ Furthermore $$\mathrm{\Delta }({\beta }_c-c,y^{-1})={\left(y{x}^{1-c}\right)}^{-1}\mathrm{\Delta }(-g_{c,{\beta }_c}^{\beta },y)\prod _{j=1}^{{\beta }_c-1}\frac{\mathrm{\Delta }(-g_{c,j}^{\beta },y)}{x\mathrm{\Delta }(-g_{c,j}^{\beta }+1,y)}\text{.}$$
(iii)	For the double partition $(\alpha ,\beta ),$ $\left(\alpha \right)\ne \left(0\right),$ $\left(\beta \right)\ne \left(0\right),$ let $d$ be the row of $D\left(\beta \right)$ such that the $a_c\text{-th}$ column of $D\left(\beta \right)$ ends in row $d$ if ${\alpha }_c\le {\beta }_1\text{.}$ If ${\beta }_1<{\alpha }_c,$ set $d=0\text{.}$ Then $$E^{(({\alpha }_c-),\left(\beta \right))}=E^{(({\alpha }_c-),\left(0\right))}G$$ where $$G=\frac{\mathrm{\Delta }(g_{c,{\alpha }_c}^{\alpha },y)}{\mathrm{\Delta }({\alpha }_c-c,y)}\prod _{j=1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(g_{c,j}^{\alpha },y)}{\mathrm{\Delta }(g_{c,j}^{\alpha }-1,y)}\prod _{j=1}^d\frac{\mathrm{\Delta }(-g_{j,{\alpha }_c}^{\beta },y)}{\mathrm{\Delta }(-g_{j,{\alpha }_c}^{\beta }+1,y)}$$ It is understood that the last product in the definition of $G$ is taken to be equal to $1$ if $d=0\text{.}$

		Proof.
	We will show the expression for $E^{({\mu }_c-)}$ given by (3.3.16) has the desired form for each of the cases mentioned. Direct computations are all that is required. (i) For the double partition $(\alpha ,\left(0\right)),$ the row $c$ is a row of $D\left(\alpha \right)$ and ${\alpha }_i={\alpha }_c$ for $i=1,\dots ,c\text{.}$ Then by Proposition (3.3.9) and (3.3.18) $$\begin{array}{cc}\text{(3.4.3)}& \begin{array}{ccc}{\displaystyle \sum _{i=1}^{c-1}{R}_i\prod _{j=i}^{c-1}{F}_{j,j+1}{R}_c}& =& {\displaystyle \mathrm{\Delta }({\alpha }_c,-1)\sum _{i=1}^c\mathrm{\Delta }({\alpha }_c-2i+1,y)x^{i-c}}\\ & =& x^{1-c}\mathrm{\Delta }({\alpha }_c,-1)\mathrm{\Delta }(c,-1)\mathrm{\Delta }({\alpha }_c-c,y)\text{.}\end{array}\end{array}$$ Let $m_i$ denote the axial distance from the first square in row $i$ to the last square in row $c,$ $i>c\text{.}$ Then $m_{i+1}=m_i+1$ and using Corollary (3.3.18) $$\begin{array}{ccc}{\displaystyle \prod _{i=c+1}^{s_{\alpha }}{f}_{c,i}}& =& {\displaystyle \prod _{i=c+1}^{s_{\alpha }}\frac{\mathrm{\Delta }(m_i+1,-1)\mathrm{\Delta }(m_i-{\alpha }_i,-1)}{\mathrm{\Delta }(m_i,-1)\mathrm{\Delta }(m_i-{\alpha }_i+1,-1)}}\\ & =& {\displaystyle \frac{\mathrm{\Delta }(m_{c+1}-{\alpha }_{c+1},-1)}{\mathrm{\Delta }(m_{c+1},-1)}\prod _{i=c+1}^{s_{\alpha }}\frac{\mathrm{\Delta }(m_{i+1}-{\alpha }_{i+1},-1)}{\mathrm{\Delta }(m_{i+1}-{\alpha }_i,-1)}\text{.}}\end{array}$$ Set ${\alpha }_j=0$ for $j>s_{\alpha }$ and rewrite the above as $$\prod _{i=c+1}^{s_{\alpha }}{f}_{c,i}=\frac{\mathrm{\Delta }(m_{c+1}-{\alpha }_{c+1},-1)}{\mathrm{\Delta }(m_{c+1},-1)}\prod _{i=c+1}^{s_{\alpha }}\prod _{k={\alpha }_{i+1}+1}^{{\alpha }_i}\frac{\mathrm{\Delta }(m_{i+1}-k+1,-1)}{\mathrm{\Delta }(m_{i+1}-k,-1)}$$ where the second product is taken equal to $1$ for all $i$ such that ${\alpha }_{i+1}={\alpha }_i\text{.}$ If $a_{i+1}<{\alpha }_i,$ ${\alpha }_i-{\alpha }_{i+1}$ equals the number of columns of $D\left(\alpha \right)$ which end in row $i$ while if ${\alpha }_{i+1}={\alpha }_i,$ no column ends in row $i\text{.}$ Then for $i\ge c$ such that ${\alpha }_{i+1}<{\alpha }_i$ we have by (3.4.1) and (1.1.5) $$m_{i+1}-k+1={\alpha }_c+(i-c)-k+1=h_{c,k}^{\alpha },{\alpha }_{i+1}+1\le k\le {\alpha }_i\text{.}$$ Thus $$\prod _{k={\alpha }_{i+1}+1}^{{\alpha }_i}\frac{\mathrm{\Delta }(m_{i+1}-k+1,-1)}{\mathrm{\Delta }(m_{i+1}-k,-1)}=\prod _{k={\alpha }_{i+1}+1}^{{\alpha }_i}\frac{\mathrm{\Delta }(h_{c,k}^{\alpha },-1)}{\mathrm{\Delta }(h_{c,k}^{\alpha }-1,-1)}\text{.}$$ Furthermore, by the choice of the row $c,$ $m_{c+1}={\alpha }_c>{\alpha }_{c+1}$ and $h_{c,{\alpha }_c}^{\alpha }=1,$ so that $$\frac{\mathrm{\Delta }(m_{c+1}-{\alpha }_{c+1},-1)}{\mathrm{\Delta }(m_{c+1},-1)}=\frac{1}{\mathrm{\Delta }({\alpha }_c,-1)}\prod _{k={\alpha }_{c+1}+1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(h_{c,k}^{\alpha },-1)}{\mathrm{\Delta }(h_{c,k}^{\alpha }-1,-1)}\text{.}$$ The above computations show $$\begin{array}{cc}\text{(3.4.4)}& {\displaystyle \prod _{i=c+1}^{s_{\alpha }}{f}_{c,i}=\frac{1}{\mathrm{\Delta }({\alpha }_c,-1)}\prod _{i=1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(h_{c,i}^{\alpha },-1)}{\mathrm{\Delta }(h_{c,i}^{\alpha }-1,-1)}\text{.}}\end{array}$$ Finally, $c=h_{1,{\alpha }_c}^{\alpha },$ so that $$\begin{array}{cc}\text{(3.4.5)}& {\displaystyle x^{1-c}\mathrm{\Delta }(c,-1)=x^{1-c}\mathrm{\Delta }(h_{1,{\alpha }_c}^{\alpha },-1)=\prod _{j=1}^{c-1}\frac{\mathrm{\Delta }(h_{j,{\alpha }_c}^{\alpha },-1)}{\mathrm{\Delta }(h_{j,{\alpha }_c}^{\alpha }-1,-1)}\text{.}}\end{array}$$ (3.4.3) - (3.4.5) combine to give the desired expression for $E^{(({\alpha }_c-),\left(0\right))}\text{.}$ As the second partition is the empty partition $\left(0\right),$ ${\alpha }_c-c=g_{c,1}^{\alpha }$ and $g_{c,j}^{\alpha }=g_{c,j+1}^{\alpha }+1,$ $j=1,\dots ,c-1,$ so that the expression for $\mathrm{\Delta }({\alpha }_c-c,u)$ given in the statement of (i) is just a telescopic product. (ii) The proof of (ii) is entirely similar to (i) and will be omitted. Simply replace $\left(\alpha \right)$ with $\left(\beta \right)$ and note that by Proposition (3.3.9) $y$ must be replaced with $y^{-1}$ in (3.4.3). Also ${\beta }_c-c=g_{c,1}^{\beta }$ so that $$\mathrm{\Delta }({\beta }_c-c,y^{-1})={\left(y{x}^m\right)}^{-1}\mathrm{\Delta }(m,y),m=-g_{c,1}^{\beta }\text{.}$$ (iii) In this case the row $c$ is a row of $D\left(\alpha \right)\text{.}$ Thus employing (3.3.16) $$\begin{array}{ccc}{E}^{(({\alpha }_c-),\left(\beta \right))}& =& {\displaystyle (\sum _{i=1}^{c-1}{R}_i\prod _{j=1}^i{F}_{j,j+1}+R_c)\prod _{i=c+1}^{s_{\alpha }}{f}_{c,i}\prod _{i=s_{\alpha }+1}^{s_{\alpha }+s_{\beta }}{f}_{c,i}}\\ & =& {\displaystyle E^{(({\alpha }_c-),\left(0\right))}\prod _{i=s_{\alpha }+1}^{s_{\alpha }+s_{\beta }}{f}_{c,i}}\end{array}$$ It suffices to show $$\begin{array}{cc}\text{(3.4.6)}& {\displaystyle \prod _{i=s_{\alpha }+1}^{s_{\alpha }+s_{\beta }}{f}_{c,i}=G\text{.}}\end{array}$$ Let $m_i,$ $i=1,\dots ,s,$ denote the axial distance from the first box in the $i\text{-th}$ row of $D\left(\beta \right)$ to the last box in row $c$ of $D\left(\alpha \right)\text{.}$ Then $m_{i+1}=m_i+1$ and as in (i) $$\prod _{i=s_{\alpha }+1}^{s_{\alpha }+s_{\beta }}{f}_{c,i}=\frac{\mathrm{\Delta }(m_{s_{\beta }}+1,y)}{\mathrm{\Delta }(m,y)}\prod _{i=1}^{s_{\beta }}\frac{\mathrm{\Delta }(m_i-{\beta }_i,y)}{\mathrm{\Delta }(m_{i+1}-{\beta }_i,y)}\text{.}$$ Let ${\beta }_j=0$ for $j>s_{\beta }$ and let $d$ be as in the statement of the theorem. Rewrite the above as $$\begin{array}{cc}\text{(3.4.7)}& {\displaystyle \left(\frac{\mathrm{\Delta }(m_{d+1}-d_c+1,y)}{\mathrm{\Delta }(m,y)}\prod _{i=1}^d\frac{\mathrm{\Delta }(m_i-{\beta }_i,y)}{\mathrm{\Delta }(m_{i+1}-{\beta }_i,y)}\right)\left(\frac{\mathrm{\Delta }(m_{d+1}-{\beta }_{d+1},y)}{\mathrm{\Delta }(m_{d+1}-{\alpha }_c+1,y)}\prod _{i=d+1}^{s_{\beta }}\frac{\mathrm{\Delta }(m_{i+1}-{\beta }_{i+1},y)}{\mathrm{\Delta }(m_{i+1}-{\beta }_i,y)}\right)\text{.}}\end{array}$$ Label the two bracketed expressions in (3.4.7) $A$ and $B$ respectively. By (3.4.1) and (1.1.5) $$m_{d+1}-{\alpha }_c+1=({\alpha }_c-{\alpha }_c)+(d-c)+1=g_{c,{\alpha }_c}^{\alpha }$$ and $$m_i-{\beta }_i={\alpha }_c+(i-c)-1-{\beta }_i=-(({\beta }_i-{\alpha }_c)+(c-i)+1)=-g_{i,{\alpha }_c}^{\beta }\text{.}$$ Therefore $$A=\frac{\mathrm{\Delta }(g_{c,{\alpha }_c}^{\alpha },y)}{\mathrm{\Delta }({\alpha }_c-c,y)}\prod _{i=1}^d\frac{\mathrm{\Delta }(-g_{i,{\alpha }_c}^{\beta },y)}{\mathrm{\Delta }(-g_{i,{\alpha }_c}^{\beta }+1,y)}\text{.}$$ As ${\beta }_{d+1}<{\alpha }_c,$ and as for ${\beta }_{i+1}\le k\le {\beta }_i-1,$ $$m_{i+1}-k={\alpha }_c+(i-c)-k=g_{c,k+1}^{\alpha },$$ computations identical to those employed in (i) show $$B=\prod _{i=1}^{{\alpha }_c-1}\frac{\mathrm{\Delta }(g_{c,i}^{\alpha },y)}{\mathrm{\Delta }(g_{c,i}^{\alpha }-1,y)}\text{.}$$ The above expressions for $A$ and $B$ and (3.4.7) show (3.4.6) as required. This completes the proof of (iii) and the proposition. $\square$

We can now obtain an elegant formula for $$C^{\mu }={\left(\text{deg}\hspace{0.17em}{\chi }^{\mu }\right)}^{-1}\sum _{w\in W\left(B_n\right)}\nu {\left(a_w\right)}^{-1}{\chi }^{\mu }\left(a_w\right){\chi }^{\mu }\left({â}_w\right)\text{.}$$ For the double partition $\left(\mu \right)=(\alpha ,\beta )$ set $$\begin{array}{cc}\text{(3.4.8)}& \begin{array}{ccc}{H}_{i,j}^{\alpha }& =& x^{1-{\bar{\alpha }}_j}\mathrm{\Delta }(h_{i,j}^{\alpha },-1),\\ H_{i,j}^{\beta }& =& x^{1-{\bar{\beta }}_j}\mathrm{\Delta }(h_{i,j}^{\beta },-1),\\ G_{i,j}^{\alpha }& =& \mathrm{\Delta }(g_{i,j}^{\alpha },y),\\ G_{i,j}^{\beta }& =& {\left(y{x}^m\right)}^{-1}\mathrm{\Delta }(m,y),m=-g_{i,j}^{\beta }\text{.}\end{array}\end{array}$$

Theorem (3.4.9) For the irreducible character ${\chi }^{\mu }$ of ${𝒜}^K\left(B_n\right)\text{,}$ $\left(\mu \right)=(\alpha ,\beta )$ a double partition of $n\ge 2,$ $$C^{\mu }=\prod _{(i,j)\in \left(\alpha \right)}{H}_{i,j}^{\alpha }{G}_{i,j}^{\alpha }\prod _{(i,j)\in \left(\beta \right)}{H}_{i,j}^{\beta }{G}_{i,j}^{\beta }\text{.}$$

		Proof.
	By induction on $n\ge 2\text{.}$ The statement of the theorem can readily be checked for the representations of ${𝒜}^K\left(B_2\right)$ given by Theorem (2.2.7). Let $n\ge 2\text{.}$ The Young diagram $D({\mu }_c-)$ is obtained from the Young diagram $D\left(\mu \right)$ by deleting the last square from row $c\text{.}$ Thus the hook lengths of the squares of $D({\mu }_c-)$ differ from the hook lengths of the squares of $D\left(\mu \right)$ only for squares in the $c\text{-th}$ row and the ${\bar{\alpha }}_c$ or ${\bar{\beta }}_c$ column depending on whether the row $c$ is a row of the diagram $D\left(\alpha \right)$ or the diagram $D\left(\beta \right)$ of $D\left(\mu \right)=(D\left(\alpha \right),D\left(\beta \right))\text{.}$ Indeed, the hook lengths of these squares of $D({\mu }_c-)$ are one less than the corresponding squares of $D\left(\mu \right)\text{.}$ Similarly the split hook lengths of the squares of $D({\mu }_c-)$ differ from the split hook lengths of the squares of $D\left(\mu \right)$ only in the $c\text{-th}$ row and, if $c$ is a row of $D\left(\alpha \right)$ and ${\bar{\beta }}_c\ne 0,$ in the ${\bar{\beta }}_c$ column of $D\left(\beta \right)\text{.}$ Again the split hook lengths of these squares of $D({\mu }_c-)$ are one less than the corresponding squares of $D\left(\mu \right)\text{.}$ Thus, from the computations of Proposition (3.4.2) we have that $E^{({\mu }_c-)}$ is of the form $$E^{({\mu }_c-)}=G_{c,{\lambda }_c}\prod _{(i,j)}\frac{G_{i,j}^{\mu }}{G_{i,j}^{({\mu }_c-)}}\prod _{(s,y)}\frac{H_{s,t}^{\mu }}{H_{s,t}^{({\mu }_c-)}}$$ where $\lambda =\alpha$ or $\beta ,$ depending on whether the row $c$ is a row of $D\left(\alpha \right)$ or $D\left(\beta \right)\text{,}$ and where $(i,j)$ and $(s,t)$ run over the appropriate squares of $D({\mu }_c-),$ mentioned above, for which the hook lengths and the split hook lengths differ from those of $D\left(\mu \right)\text{.}$ As $C^{\mu }=C^{({\mu }_c-)}{E}^{({\mu }_c-)}$ by Theorem (3.2.6), this completes the induction and the proof of the theorem. $\square$

We can now give an explicit expression for the generic degree $d_{\chi }$ (3.1.3), of the irreducible character $\chi$ of ${𝒜}^K\left(B_n\right)\text{.}$ Let $P_{B_n}(x,y)$ be the Poincaré polynomial of ${𝒜}^K\left(B_n\right)\text{,}$ $$\begin{array}{cc}\text{(3.4.10)}& {\displaystyle P_{B_n}(x,y)=\sum _{w\in W}\nu \left(a_w\right)\text{.}}\end{array}$$ From ([Mac1972]), $$P_{B_n}(x,y)=\prod _{i=0}^{n-1}(1+x^{i}y)(1+\dots +x^i)\text{.}$$

Corollary (3.4.11) For the irreducible character ${\chi }^{\mu }$ of ${𝒜}^K\left(B_n\right)\text{,}$ $\left(\mu \right)=(\alpha ,\beta ),$ $$d_{{\chi }^{\mu }}=\frac{P_{B_n}(x,y)}{\prod _{(i,j)\in \left(\alpha \right)}{H}_{i,j}^{\alpha }{G}_{i,j}^{\alpha }\prod _{(i,j)\in \left(\beta \right)}{H}_{i,j}^{\beta }{G}_{i,j}^{\beta }}\text{.}$$

		Proof.
	This follows from the definition of $d_{{\chi }^{\mu }}$ (3.1.3), (3.4.10) and (3.4.9). $\square$

The generic degrees of the representations of ${𝒜}^K\left(A_n\right)$ and ${𝒜}^K\left(D_n\right)$ are readily obtained from Theorem (3.4.9) as well. In particular,

Corollary (3.4.12) Let $\varphi :D=\mathbb{Q}[x,y]\to \mathbb{Q}\left(x\right)$ be the homomorphism defined by $\varphi \left(y\right)=0\text{.}$ Let $\left(\alpha \right)$ be a partition of $n$ and let ${\chi }^{\alpha }$ and ${\chi }^{(\alpha ,\left(0\right))}$ be the irreducible characters of ${𝒜}^K\left(A_{n-1}\right)$ and ${𝒜}^K\left(B_n\right)$ corresponding to $\left(\alpha \right)$ and $(\alpha ,\left(0\right))\text{.}$ Let ${\varphi }^{*}:D_P\to \mathbb{Q}\left(x\right)$ be the extension of $\varphi$ to the ring of fractions, $P=\text{ker}\hspace{0.17em}\varphi \text{.}$ Then $d_{{\chi }^{(\alpha ,\left(0\right))}}\in D_P$ and ${\varphi }^{*}\left(d_{{\chi }^{(\alpha ,\left(0\right))}}\right)=d_{{\chi }^{\alpha }}\text{.}$

		Proof.
	For $w\in W\left(B_n\right),$ define $l_1\left(w\right)$ to be the number of times $w_1$ occurs in a reduced expression for $W$ in $R,$ and set $l_2\left(w\right)=l\left(w\right)-l_1\left(w\right)\text{.}$ Then $$\nu \left(a_w\right)=y^{l_1\left(w\right)}{x}^{l_2\left(w\right)},a_w\in 𝒜\left(B_n\right)\text{.}$$ We first show $$\begin{array}{cc}\text{(3.4.13)}& {\chi }^{(\alpha ,\left(0\right))}\left(a_w\right)=y^{l_1\left(w\right)}{\chi }^{(\alpha ,\left(0\right))}\left({\bar{a}}_w\right)\in D_P,w\in W\left(B_n\right)\text{.}\end{array}$$ Let $M\left(a\right)$ denote the matrix of ${\pi }^{(\alpha ,\left(0\right))}\left(a\right)$ with respect to the basis $\{t_1,\dots ,t_f\}$ of $V_{(\alpha ,\left(0\right))}^K,$ $a\in {𝒜}^K\left(B_n\right)\text{.}$ By (2.2.6) $M\left(a_1\right)=yI$ and thus commutes with $M\left(a_i\right),$ $i=1,\dots ,n\text{.}$ Hence for $w\in W\left(B_n\right)$ $$M\left(a_w\right)={\left(M\left(a_1\right)\right)}^{l_1\left(w\right)}M\left({\bar{a}}_w\right)$$ where $${\bar{a}}_w=\sum _{g\in W\left(A_{n-1}\right)}{c}_g^w{a}_g,c_g^w\in \mathbb{Q}\left[x\right]\text{.}$$ From (2.2.6), ${\chi }^{(\alpha ,\left(0\right))}\left({\bar{a}}_w\right)\in \mathbb{Q}\left(x\right)\subset D_P,$ $D_P$ considered as a subring of $K=\mathbb{Z}(x,y)\text{.}$ Thus we have shown (3.4.13). The rest of the proof is now clear. As $l_1\left(w\right)=l_1\left(w^{-1}\right),$ we have $${\varphi }^{*}\left(C^{(\alpha ,\left(0\right))}\right)=\sum _{w\in W\left(A_{n-1}\right)}\nu {\left(a_w\right)}^{-1}{\chi }^{\alpha }\left(a_w\right){\chi }^{\alpha }\left({â}_w\right)=C^{\alpha }\text{.}$$ Similarly $$\varphi \left(\sum _{w\in W\left(B_n\right)}\nu \left(a_w\right)\right)=\sum _{w\in W\left(A_{n-1}\right)}\nu \left(a_w\right)$$ and the statement of the corollary follows. $\square$

Thus by the above corollary and Theorem (\ref{l349}) we have, for $\left(\alpha \right)$ a partition of $n\text{,}$ $$\begin{array}{cc}\text{(3.4.14)}& {\displaystyle d_{{\chi }^{\alpha }}=\frac{P_{A_{n-1}}\left(x\right)}{\prod _{i,j}{H}_{i,j}^{\alpha }}}\end{array}$$ where $$P_{A_{n-1}}\left(x\right)=\prod _{i=0}^{n-1}(1+\dots +x^i)\text{.}$$

Corollary (3.4.15) Let $\varphi :D=\mathbb{Q}[x,y]\to \mathbb{Q}\left(x\right)$ be the homomorphism defined by $\varphi \left(y\right)=1\text{.}$ Let $\left(\mu \right)=(\alpha ,\beta )$ be a double partition of $n$ with $\left(\alpha \right)\ne \left(\beta \right)$ and let ${\chi }^{\mu }$ and ${\psi }^{\mu }$ be the irreducible characters of ${𝒜}^K\left(B_n\right)$ and ${𝒜}^K\left(D_n\right)$ corresponding to $\left(\mu \right)\text{.}$ Then ${\varphi }^{*}\left(d_{{\chi }^{\mu }}\right)=d_{{\psi }^{\mu }},$ where ${\varphi }^{*}:D_P\to \mathbb{Q}\left(x\right)$ is the extension of $\varphi$ to the ring of fractions $D_P,$ $P=\text{ker}\hspace{0.17em}\varphi \text{.}$

		Proof.
	From Theorem (2.3.9), ${\psi }^{\mu }$ is the restriction to ${𝒜}^K\left(D_n\right)$ of the irreducible character ${\chi }_{\varphi }^{\mu }$ of ${𝒜}_{\varphi ,K}\left(B_n\right),$ $K=\mathbb{Q}\left(x\right),$ where ${\chi }_{\varphi }^{\mu }\left(a_w\right)={\varphi }^{*}\left({\chi }^{\mu }\left(a_w\right)\right)\text{.}$ Thus by the definition of generic degree, to prove the corollary it is sufficient to prove $$\begin{array}{cc}\text{(3.4.16)}& {\displaystyle \sum _{w\in W\left(B_n\right)}\nu {\left(a_{w\varphi }\right)}^{-1}{\chi }_{\varphi }^{\mu }\left(a_{w\varphi }\right){\chi }_{\varphi }^{\mu }\left({â}_{w\varphi }\right)=2\sum _{w\in W\left(D_n\right)}\nu {\left(a_{w\varphi }\right)}^{-1}{\chi }_{\varphi }^{\mu }\left(a_{w\varphi }\right){\chi }_{\varphi }^{\mu }\left({â}_{w\varphi }\right)}\end{array}$$ for $\left(\mu \right)$ as in the statement of the corollary. For any $w\in W\left(B_n\right)$ and corresponding basis element $a_{w\varphi }\in {𝒜}_{\varphi ,\mathbb{Q}\left(x\right)}\left(B_n\right),$ we have $a_{w\varphi }{a}_{w_1\varphi }=a_{w{w}_1\varphi },$ as ${\left(a_{w_1\varphi }\right)}^2=1\text{.}$ Thus, using the orthogonality relations (3.1.4) and the coset decomposition $W\left(B_n\right)=W\left(D_n\right)\cup W\left(D_n\right)w_1,$ we have $$\begin{array}{cc}\text{(3.4.17)}& \begin{array}{ccc}{\varphi }^{*}\left(C^{\mu }\right)& =& {\displaystyle \sum _{w\in W\left(B_n\right)}\nu {\left(a_{w\varphi }\right)}^{-1}{\chi }_{\varphi }^{\mu }\left(a_{w\varphi }\right){\chi }_{\varphi }^{\mu }\left({â}_{w\varphi }\right)}\\ & =& {\displaystyle \sum _{w\in W\left(B_n\right)}\sum _{i=1}^{f^{\mu }}\nu {\left(a_{w\varphi }\right)}^{-1}{M}_{ii}^{\mu }\left(a_{w\varphi }\right)M_{ii}^{\mu }\left({â}_{w\varphi }\right)}\\ & =& {\displaystyle \sum _{w\in W\left(D_n\right)}\sum _{i=1}^{f^{\mu }}\nu {\left(a_{w\varphi }\right)}^{-1}(M_{ii}^{\mu }\left(a_{w\varphi }\right)M_{ii}^{\mu }\left({â}_{w\varphi }\right)+M_{ii}^{\mu }\left(a_{w\varphi }{a}_{w_1\varphi }\right)M_{ii}^{\mu }\left(a_{w_1\varphi }{â}_{w\varphi }\right))}\end{array}\end{array}$$ where $M^{\mu }\left(a_{w\varphi }\right)$ is the matrix of ${\varphi }_{\varphi }^{\mu }\left(a_w\right)$ with respect to the last letter sequence arrangement of the basis $\{t_1,\dots ,t_f\}$ of $V_{\mu }^K\text{.}$ Then $M^{\mu }\left(a_{w_1\varphi }\right)$ is a diagonal matrix with entries $\pm 1$ by (2.3.9) so $$M_{ii}^{\mu }\left(a_{w\varphi }{a}_{w_1\varphi }\right)M_{ii}^{\mu }\left(a_{w_1\varphi }{â}_{w\varphi }\right)=M_{ii}^{\mu }\left(a_{w\varphi }\right)M_{ii}^{\mu }\left({â}_{w\varphi }\right)\text{.}$$ Then (3.4.17) becomes $$\begin{array}{ccc}{\varphi }^{*}\left(C^{\mu }\right)& =& {\displaystyle 2\sum _{w\in W\left(D_n\right)}\sum _{i=1}^{f^{\mu }}\nu {\left(a_{w\varphi }\right)}^{-1}{M}_{ii}^{\mu }\left(a_{w\varphi }\right)M_{ii}^{\mu }\left({â}_{w\varphi }\right)}\\ & =& {\displaystyle 2\sum _{w\in W\left(D_n\right)}\nu {\left(a_{w\varphi }\right)}^{-1}{\chi }_{\varphi }^{\mu }\left(a_{w\varphi }\right){\chi }_{\varphi }^{\mu }\left({â}_{w\varphi }\right)}\end{array}$$ as the restriction of ${\pi }_{\varphi }^{\mu }$ to ${𝒜}^K\left(D_n\right)$ is an absolutely irreducible representation of ${𝒜}^K\left(D_n\right)\text{.}$ This proves (3.4.16) and completes the proof. $\square$

We conclude with an example. We calculate the generic degree of the reflection representation of the generic algebra of classical type, a computation also given in ([CIK1971]) (as a polynomial in one variable). The double partition $\left(\mu \right)=((n-1),\left(1\right))$ yields the reflection representation of ${𝒜}^K\left(B_n\right)$ $$\begin{array}{c} ⋯ n-1 squares (D(n-1),D(1)) \end{array}$$ We have $$C^{((n-1),\left(1\right))}=y^{-1}(1+x^{n-1}y)\prod _{i=-1}^{n-3}(1+x^{i}y)\prod _{i=1}^{n-2}(1+\dots +x^i)\text{.}$$ Thus $$d_{{\chi }^{\mu }}=\frac{P_{B_n}(x,y)}{C^{\mu }}=\frac{y(1+\dots +x^{n-1})(1+x^{n-2}y)}{(1+x^{-1}y)}$$ Setting $y=1$ in the above we have by (3.4.15) the generic degree of the reflection representation of ${𝒜}^K\left(D_n\right)\text{.}$

The partition $\left(\alpha \right)=(n-1,1)$ of $n$ yields the reflection representation of ${𝒜}^K\left(A_{n-1}\right)\text{.}$ $$\begin{array}{c}\begin{array}{c} ⋯ n-1 squares (D(n-1),1) \end{array}\\ D(n-1,1)\end{array}$$ The generic degree of the representation of ${𝒜}^K\left(B_n\right)$ corresponding to the double partition $(\left(\alpha \right),\left(0\right))=((n-1,1),\left(0\right))$ is $$d_{{\chi }^{(\alpha ,\left(0\right))}}=\frac{x(1+x^{n-1}y)(1+\dots x^{n-2})}{(1+x^{-1}y)}$$ Setting $y=0$ in the above we have by (3.4.12) the generic degree of the reflection representation of ${𝒜}^K\left(A_{n-1}\right)\text{.}$

Finally we remark that from (3.1.7) the above corollaries give the degrees of the irreducible constituents of $1_B^G$ for $G$ a finite group with BN-pair with Coxeter system of classical type by substitution of the index parameters in the formula for the generic degree. In particular, these computations apply to the families of Chevalley groups $A_l\left(q\right),$ $B_l\left(q\right),$ $D_l\left(q\right),$ $A_{2l}^1\left(q^2\right),$ $A_{2l-1}^1\left(q^2\right),$ $D_l^{\prime }\left(q^2\right)\text{.}$ The degrees of these characters for the families of type $A_l\left(q\right)$ were already known (see [Ste1951]).

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