Chapter 2.$$Representations of the Generic Ring Corresponding to a Coxeter System of Classical Type

We need to show the relations (B1 - B5) are satisfied with $Z_i^{\mu }$ in place of $a_i\text{.}$ We argue by induction on $n\text{.}$ For $n=2$ it is a case by case verification. The double partitions $(\left(2\right),\left(0\right)),$ $(\left(0\right),\left(2\right)),$ $(\left(1^2\right),\left(0\right))$ and $(\left(0\right),\left(1^2\right))$ are clearly seen to yield the well known one-dimensional representations of ${𝒜}^K\left(B_2\right)$ ([CIK1971], 10). For the double partition $(\left(1\right),\left(1\right))$ there are two standard tableaux, $$\begin{array}{c} 1 2 \end{array}\text{and}\begin{array}{c} 2 1 \end{array}\text{.}$$ From (2.2.6) $M^{(\left(1\right),\left(1\right))}\left(1\right)=D(y,-1)$ and $M^{(\left(1\right),\left(1\right))}\left(2\right)=M(0,y)\text{.}$ Direct computation verifies the relation $$M(0,y)D(y,-1)M(0,y)D(y,-1)=D(y,-1)M(0,y)D(y,-1)M(0,y)\text{.}$$ Thus the relations (B1 - B3) are satisfied with $Z_1^{\mu }$ and $Z_2^{\mu }$ in place of $a_1$ and $a_2$ by the above computation, (2.2.2) and (2.2.5).

Now let $\left(\mu \right)=({\mu }_1,\dots ,{\mu }_s)$ be a double partition of $n\text{.}$ Deletion of the letter $n$ from a standard tableau automatically yields a standard tableau involving $n-1$ letters. In fact deletion of $n$ from all standard tableaux having $n$ at the end of the $i\text{-th}$ column will yield all standard tableaux of shape $({\mu }_1,\dots ,{\mu }_i-1,\dots ,{\mu }_s)\text{.}$ Denoting this partition by $({\mu }_i-)$ and using the fact that all standard tableaux with $n$ in the $i\text{-th}$ row precede all tableaux with $n$ in the $j\text{-th}$ row for $i>j$ when ordered according to the last letter sequence, we have $$\begin{array}{cc}\text{(2.2.8)}& V_{\mu }^K=V_{({\mu }_s-)}^K\oplus \dots \oplus V_{({\mu }_1-)}^K\end{array}$$ and the corresponding matrix block form $$M^{\mu }\left(i\right)=M^{({\mu }_s-)}\left(i\right)\dotplus \dots \dotplus M^{({\mu }_1-)}\left(i\right),i<n$$ as by (2.2.6), $M^{\mu }\left(i\right)$ depends only on the letters $i-1$ and $i\text{.}$ It is understood that $({\mu }_i-)$ is taken to equal zero if $n$ cannot appear in the $i\text{-th}$ row and the above summation, here and elsewhere, will be taken over those $({\mu }_i-)$ which are non-zero. By the induction hypothesis it therefore suffices to check the relations (B1 -B5) as they pertain to $Z_n^{\mu }\text{.}$

The matrix $M^{\mu }\left(n\right)$ from (2.2.6) is composed of the matrices $M(k,y)$ and $M(k,-1)$ centered about the diagonal along with diagonal entries $x$ and $-1$. Thus the relation $${\left(Z_n^{\mu }\right)}^2=xI+(x-1)Z_n^{\mu }$$ follows from (2.2.2), (2.2.4) and (2.2.5).

Let $V_{i,j}$ denote the subspace of $V_{\mu }^K$ with basis $t_1^{i,j},\dots ,t_{s_{i,j}}^{i,j},$ corresponding to the standard tableaux of shape $\left(\mu \right)$ with the letter $n$ appearing in the $i\text{-th}$ row and $n-1$ appearing in the $j\text{-th}$ row, the ordering of the basis taken according to the last letter sequence. Then $$V_{\mu }^K=\underset{i,j}{⨁}{V}_{i,j},$$ the summation taken over all allowable $i,j$ such that $n$ appears in row $i$ and $n-1$ appears in row $j\text{,}$ and this decomposition is consistent with the last letter sequence arrangement of the basis of $V_{\mu }^K\text{.}$ Thus, whenever $n$ and $n-1$ are in distinct rows and columns, we have $V_{i,j}\cong V_{j,i}$ as ${𝒜}^K\left(B_{n-2}\right)\text{-modules}$ for $n$ appearing in row $i\text{,}$ $n-1$ appearing in row $j\text{,}$ where $W\left(B_{n-2}\right)=⟨w_1,\dots ,w_{n-1}⟩\text{.}$

Suppose first that $n$ and $n-1$ appear in distinct rows and columns, in the tableaux corresponding to $V_{i,j}\text{;}$ $n$ in row $i\text{,}$ $n-1$ in row $j\text{.}$ Then $n$ appears in row $j$ and $n-1$ appears in row $i$ in the tableaux corresponding to $V_{j,i}$ and the map $t_p^{i,j}\mapsto t_p^{j,i},$ $p=1,\dots ,s_{i,j}=s_{j,i},$ gives an isomorphism $V_{i,j}\cong V_{j,i}$ as ${𝒜}^K\left(B_{n-2}\right)\text{-modules,}$ as the configuration of the first $n-2$ letters in the tableau corresponding to $t_p^{i,j}$ is the same as the configuration of the first $n-2$ letters in the tableau corresponding to $t_p^{j,i}\text{.}$ In particular the matrix of $Z_k^{\mu },$ $k=1,\dots ,n-2,$ on $V_{i,j}\oplus V_{j,i}$ is $$S_k=\left(\begin{array}{cc}{A}_k& \\ 0& A_k\end{array}\right)$$ where $A_k$ is the matrix of ${\pi }^{\mu }\left(a_k\right)$ on $V_{i,j}\text{.}$ On the other hand, the matrix of $Z_n^{\mu }$ on $K{t}_p^{i,j}\oplus K{t}_p^{j,i}$ is, by (2.2.6), $M(l,y)$ or $M(l,-1),$ $l$ the axial distance from $n$ to $n-1\text{.}$ Thus the matrix of $Z_n^{\mu }$ on $V_{i,j}\oplus V_{j,i}$ is $$S_n=\left(\begin{array}{cc}{a}_{11}I& a_{12}I\\ a_{21}I& a_{22}I\end{array}\right)$$ where $\left(a_{ij}\right)=M(l,y)$ or $M(l,-1),$ $l$ the $s_{i,j}\times s_{i,j}$ identity matrix. Then $S_n{S}_k=S_k{S}_n$ for $k=1,\dots ,n-2$ and (B5) holds. The only other possibility is when $n$ and $n-1$ appear in the same row or column of the tableaux corresponding to $V_{i,j}\text{.}$ But in this case the matrix of $Z_n^{\mu }$ on $V_{i,j}$ is the scalar matrix $xI$ or $-I$ by (2.2.6) and thus commutes with $Z_n^{\mu }$ on $V_{i,j},$ $k=1,\dots ,n-2\text{.}$ This proves (B5) for all cases.

To check the relation (B4), we consider the restriction of $Z_{n-1}^{\mu }$ and $Z_n^{\mu }$ to subspaces with basis $\left\{t_i\right\}$ corresponding to all tableaux $T_i^{\mu }$ having a fixed arrangement of the first $n-3$ letters and all possible rearrangements of the letters $n-2,$ $n-1$ and $n\text{.}$ Let $$V_{\mu }^k=\underset{p}{⨁}{V}_p$$ denote the corresponding decomposition of $V_{\mu }^k,$ the ordering of the basis of each $V_p$ taken with respect to the last letter sequence. Then each $V_p$ is invariant under $Z_{n-1}^{\mu }$ and $Z_n^{\mu }$ and it suffices to check (B4) for the various possible arrangements of the last 3 letters in a case by case basis.

In what follows, $M_p\left(i\right),$ $i=n$ or $n-1,$ will denote the matrix of $Z_i^{\mu }$ on $V_p\text{.}$

Case 1 — the letters $n-2,$ $n-1$ and $n$ in the same row or column. Then $V_p$ is one dimensional and $M_p\left(n\right)=M_p(n-1)=x$ or $-1$ by (2.2.6). Thus $$M_p\left(n\right)M_p(n-1)M_p\left(n\right)=M_p(n-1)M_p\left(n\right)M_p(n-1)$$ and (B4) is satisfied.

Case 2 — the letters $n-2,$ $n-1$ and $n$ in two adjacent rows and two adjacent columns of the same diagram. Then $V_p$ is two-dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is $$\text{(a)}\begin{array}{c} n-2 n-1 n \end{array},\begin{array}{c} n-1 n-2 n \end{array}\text{or (b)}\begin{array}{c} n-2 n-1 n \end{array},\begin{array}{c} n-2 n n-1 \end{array}$$ ordered according to last letter sequence. Then by (2.2.6), $M_p(n-1)=M(2,1)$ and $M_p\left(n\right)=D(x,-1)$ in (a) and $M_p(n-1)=D(x,-1)$ and $M_p\left(n\right)=M(2,-1)$ in (b). Thus (B4) is satisfied in both cases by direct verification of the relation $$M(2,-1)D(x,-1)M(2,-1)=D(x,-1)M(2,-1)D(x,-1)\text{.}$$

Case 3 — the letters $n-2,$ $n-1$ and $n$ in two rows and three columns or three rows and two columns. Then $V_p$ is three dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is one of $$\begin{array}{c}\text{(a)}\begin{array}{cc}1& n-2 n-1 n \end{array}\begin{array}{cc}2& n-2 n n-1 \end{array}\begin{array}{cc}3& n-1 n n-2 \end{array}\\ \text{(b)}\begin{array}{cc}1& n-2 n-1 n \end{array}\begin{array}{cc}2& n-1 n-2 n \end{array}\begin{array}{cc}3& n n-2 n-1 \end{array}\\ \text{(c)}\begin{array}{cc}1& n-2 n-1 n \end{array}\begin{array}{cc}2& n-2 n n-1 \end{array}\begin{array}{cc}3& n-1 n n-2 \end{array}\\ \text{(d)}\begin{array}{cc}1& n-2 n-1 n \end{array}\begin{array}{cc}2& n-1 n-2 n \end{array}\begin{array}{cc}3& n n-2 n-1 \end{array}\end{array}$$ ordered according to the last letter sequence. If we set $$S_1=\left(\begin{array}{ccc}c& ·& ·\\ ·& a_{11}& a_{12}\\ ·& a_{21}& a_{22}\end{array}\right),S_2=\left(\begin{array}{ccc}{b}_{11}& b_{12}& ·\\ b_{21}& b_{22}& ·\\ ·& ·& c\end{array}\right)$$ we have, by (2.2.6), in case (a), $M_p(n-1)=S_1$ and $M_p\left(n\right)=S_2$ where $\left(a_{ij}\right)=M(d_1,ϵ),$ $\left(b_{ij}\right)=M(d_2,ϵ),$ $ϵ=y$ or $-1$, and $c=x\text{.}$ Here $d_1$ is the axial distance form $n-1$ to $n-2$ in 2 and $d_2$ is the axial distance from $n$ to $n-1$ in 1 so that $d_2=d_1+1\text{.}$ In case (b), $M_p(n-1)=S_2$ and $M_p\left(n\right)=S_1$ with the same entries in $S_i$ as in case (a). The analysis of (c) and (d) is similar except that now $c=-1$ in $S_1$ and $S_2\text{.}$ Thus in all cases (B4) is satisfied by

Lemma (2.2.9) Let $S_1,$ $S_2$ be as above and let $\left(a_{ij}\right)=M(a,y),$ $\left(b_{ij}\right)=M(b,y)\text{.}$ Then for

		Proof.
	Observe that $S_1{S}_2{S}_1=S_2{S}_1{S}_2$ iff (1) $c{b}_{11}(c-b_{11})=a_{11}{b}_{12}{b}_{21},$ (2) $c{a}_{22}(c-a_{22})=b_{22}{a}_{12}{a}_{21},$ (3) $b_{ij}(a_{11}{b}_{22}+c(b_{11}-a_{11}))=0,i\ne j,$ (4) $a_{ij}(a_{11}{b}_{22}+c(b_{22}-a_2))=0,i\ne j,$ (5) $a_{11}{b}_{22}(b_{22}-a_{11})=c(a_{12}{a}_{21}-b_{12}{b}_{21})\text{.}$ For (1), $$\begin{array}{cc}\text{(2.2.10)}& c{b}_{11}(c-b_{11})-a_{11}{b}_{12}{b}_{21}=\frac{(x-1)}{\mathrm{\Delta }{(b,y)}^2\mathrm{\Delta }(a,y)}(\mathrm{\Delta }(a,y)(c^2\mathrm{\Delta }(b,y)-c(x-1))-x\mathrm{\Delta }(b-1,y)\mathrm{\Delta }(b+1,y))\text{.}\end{array}$$ Now for $c=x,$ $$c^2\mathrm{\Delta }(b,y)-c(x-1)=x\mathrm{\Delta }(b+1,y)$$ and for $c=-1,$ $$c^2\mathrm{\Delta }(b,y)-c(x-1)=x\mathrm{\Delta }(b-1,y)\text{.}$$ Hence (2.2.10) equals zero for $c=x,$ $b+1=a$ and for $c=-1,$ $b-1=a\text{.}$ The relation (2) is entirely similar. For (3) and (4) we have $$a_{11}{b}_{22}+c(b_{11}-a_{11})=\frac{(x-1)}{\mathrm{\Delta }(a,y)\mathrm{\Delta }(b,y)}(x^{b}y(x-1)+c(\mathrm{\Delta }(a,y)-\mathrm{\Delta }(b,y)))$$ and $$a_{11}{b}_{22}+c(b_{22}-a_{22})=\frac{(x-1)}{\mathrm{\Delta }(a,y)\mathrm{\Delta }(b,y)}(x^{b}y(x-1)+c(x^{a}y\mathrm{\Delta }(b,y)-x^{b}y\mathrm{\Delta }(a,y)))\text{.}$$ But for $c=x,$ $b-1=a$ and $c=-1,$ $b+1=a,$ $$\begin{array}{ccc}\mathrm{\Delta }(a,y)-\mathrm{\Delta }(b,y)& =& x^{a}y\mathrm{\Delta }(b,y)-x^{a}y\mathrm{\Delta }(a,y)\\ & =& x^{b}y(1-x)\text{.}\end{array}$$ For (5), first note the useful factorization $$\begin{array}{cc}\text{(2.2.11)}& \begin{array}{ccc}\multicolumn{3}{c}{{\displaystyle \left(\frac{x\mathrm{\Delta }(a-1,y)\mathrm{\Delta }(a+1,y)}{\mathrm{\Delta }{(a,y)}^2}\right)-\left(\frac{x\mathrm{\Delta }(b-1,z)\mathrm{\Delta }(b+1,z)}{\mathrm{\Delta }{(b,z)}^2}\right)}}\\ & =& {\displaystyle \frac{{(x-1)}^2}{\mathrm{\Delta }{(a,y)}^2\mathrm{\Delta }{(b,z)}^2}(x^{a}y{(x^{b}z+1)}^2-x^{b}z{(x^{a}y+1)}^2)}\\ & =& {\displaystyle \frac{{(x-1)}^2(x^{a+b}yz-1)(x^{b}z-x^{a}y)}{\mathrm{\Delta }{(a,t)}^2\mathrm{\Delta }{(b,z)}^2}\text{.}}\end{array}\end{array}$$ Now $$\begin{array}{cc}\text{(2.2.12)}& {\displaystyle \frac{1}{c}{a}_{11}{b}_{22}(b_{22}-a_{11})=\frac{{(x-1)}^2}{\mathrm{\Delta }{(b,y)}^2\mathrm{\Delta }{(a,y)}^2}\left(\frac{x^{b+1}y-x^{b}y}{c}\right)(x^{a+b}{y}^2-1)\text{.}}\end{array}$$ But $$\frac{1}{c}(x^{b+1}y-x^{b}y)=x^{b}y-x^{a}y$$ for $c=x,$ $b-1=a$ and for $c=-1,$ $b+1=a\text{.}$ So for both cases, (2.2.12) equals $$a_{12}{a}_{21}-b_{12}{b}_{21}$$ using (2.2.11) with $z=y\text{.}$ $\square$

Case 4 — the letters $n-2,$ $n-1$ and $n$ in three distinct rows and three distinct columns. Then $V_p$ is 6-dimensional with basis elements corresponding to tableaux where the configuration of the last 3 letters is $$\begin{array}{c}\begin{array}{cc}1& n-2 n-1 n \end{array}\begin{array}{cc}2& n-1 n-2 n \end{array}\begin{array}{cc}3& n-2 n n-1 \end{array}\\ \begin{array}{cc}4& n-1 n n-2 \end{array}\begin{array}{cc}5& n n-2 n-1 \end{array}\begin{array}{cc}6& n n-1 n-2 \end{array}\end{array}$$ ordered according to the last letter sequence. Let $$S_1=\left(\begin{array}{cccccc}{a}_{11}& a_{12}& ·& ·& ·& ·\\ ·& ·& b_{11}& b_{12}& ·& ·\\ ·& ·& b_{21}& b_{22}& ·& ·\\ ·& ·& ·& ·& c_{11}& c_{12}\\ ·& ·& ·& ·& c_{21}& c_{22}\end{array}\right),S_2=\left(\begin{array}{cccccc}{c}_{11}& ·& c_{12}& ·& ·& ·\\ ·& b_{11}& ·& ·& b_{12}& ·\\ c_{21}& ·& c_{22}& ·& ·& ·\\ ·& ·& ·& a_{11}& ·& a_{12}\\ ·& b_{21}& ·& ·& b_{22}& ·\\ ·& ·& ·& a_{21}& ·& a_{22}\end{array}\right)\text{.}$$ Then if all rows are in the same diagram we have by (2.2.6), $M_p(n-1)=S_1$ and $M_p\left(n\right)=S_2$ where $\left(a_{ij}\right)=M(d_1,-1),$ $\left(b_{ij}\right)=M(d_2,-1)$ and $\left(c_{ij}\right)=M(d_3,-1)\text{.}$ Here $d_1$ is the axial distance from $n-1$ to $n-2$ in 1, $d_2$ is the axial distance from $n-1$ to $n-2$ in 3 and $d_3$ is the axial distance from $n-1$ to $n-2$ in 5 so that $d_1+d_3=d_2$ and all $d_i\ge 2\text{.}$ If two rows are in one diagram and the third in the second diagram, we assume, without loss of generality, the lowest box to be in the second diagram. Superimposing the second diagram upon the first again does not alter the relation $d_1+d_3=d_2$ except that now only $d_1\ge 2\text{.}$ In this case $M_p(n-1)=S_1$ and $M_p\left(n\right)=S_2,$ where now $\left(a_{ij}\right)=M(d_1,-1),$ $\left(b_{ij}\right)=M(d_2,y)$ and $\left(c_{ij}\right)=M(d_3,y)\text{.}$ Thus for both cases (B4) is satisfied by

Lemma (2.2.13) Let $S_1$ and $S_2$ be as above and let $\left(a_{ij}\right)=M(s,w),$ $\left(b_{ij}\right)=M(p,y),$ $\left(c_{ij}\right)=M(t,z)$ with $s+t=p\text{.}$ Then for

		Proof.
	First, either (i) or (ii) clearly imply (iii). Now $S_1{S}_2{S}_1=S_2{S}_1{S}_2$ iff (1) $a_{kl}(c_{ii}{a}_{ii}+b_{ii}(a_{jj}-c_{ii}))=0,$ $k\ne l,$ $i\ne j\text{;}$ (2) $c_{kl}(c_{ii}{a}_{ii}+b_{ii}(c_{jj}-a_{ii}))=0,$ $k\ne l,$ $i\ne j\text{;}$ (3) $b_{kl}(c_{ii}{a}_{ii}-c_{ii}{b}_{jj}-a_{jj}{b}_{ii})=0,$ $k\ne l,$ $i\ne j\text{;}$ (4) $a_{ii}{c}_{ii}(a_{ii}-c_{ii})=b_{ii}(c_{ij}{c}_{ji}-a_{ij}{a}_{ji}),$ $i\ne j\text{;}$ (5) $a_{ii}{b}_{jj}(a_{ii}-b_{jj})=c_{ji}(b_{ij}{b}_{ji}-a_{ij}{a}_{ji}),$ $i\ne j\text{;}$ (6) $c_{ii}{b}_{jj}(c_{ii}-b_{jj})=a_{ji}(b_{ij}{b}_{ji}-c_{ij}{c}_{ji}),$ $i\ne j\text{;}$ and as these relations are symmetric in the $\left(a_{ij}\right)$ and $\left(c_{ij}\right),$ it suffices to prove the lemma for (i). For (1), (2), and (3), we observe that $$\begin{array}{ccc}{E}_1& =& \mathrm{\Delta }(s,-1)+x^s\mathrm{\Delta }(t,y)-\mathrm{\Delta }(p,y)=0,\\ E_2& =& x^{t}y\mathrm{\Delta }(s,-1)-\mathrm{\Delta }(t,y)+\mathrm{\Delta }(p,q)=0\text{.}\end{array}$$ Also set $$D=\frac{{(x-1)}^2}{\mathrm{\Delta }(p,y)\mathrm{\Delta }(t,y)\mathrm{\Delta }(s,-1)}\text{.}$$ We then obtain $$\begin{array}{ccc}{c}_{11}{a}_{11}+b_{11}(a_{22}-c_{11})=-D{E}_1& =& 0,\\ c_{22}{a}_{22}+b_{22}(a_{11}-c_{22})=x^{p}yD{E}_2& =& 0,\\ c_{11}{a}_{11}+b_{11}(c_{22}-a_{11})=D{E}_2& =& 0,\\ c_{22}{a}_{22}+b_{22}(c_{11}-a_{22})=x^{p}yD{E}_1& =& 0,\\ c_{11}{a}_{22}-c_{11}{c}_{22}-a_{22}{b}_{11}=x^{s}D{E}_2& =& 0,\\ c_{22}{a}_{11}-c_{22}{b}_{11}-a_{11}{b}_{22}=-x^{t}yD{E}_1& =& 0\text{.}\end{array}$$ For (4) we have $$\frac{1}{b_{11}}{a}_{11}{c}_{11}(a_{11}-c_{11})=\frac{1}{b_{22}}{a}_{22}{c}_{22}(a_{22}-c_{22})=\frac{{(x-1)}^2(x^{s+t}y+1)(x^{t}y+x^s)}{\mathrm{\Delta }{(s,-1)}^2\mathrm{\Delta }{(t,y)}^2}=c_{12}{c}_{21}-a_{12}{a}_{21}$$ from (2.2.11), setting $a=t,$ $b=s,$ $z=-1\text{.}$ The relations (5) and (6) are handled in an entirely similar manner. $\square$