A Ribbon Hopf Algebra Approach to the Irreducible Representations of Centralizer Algebras: The Brauer, Birman-Wenzl, and Type A Iwahori-Hecke Algebras

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

Last update: 13 March 2014

Centralizer Algebras of Tensor Powers of $V={\Lambda }_{{\omega }_1},$ Type $B_r$

The Bratteli diagram is given in Fig. 2. The shapes $\lambda \in {\hat{B}}_m$ of $B$ which are on level $m$ are the partitions of $m-2k,$ $0\le k\le \lfloor m/2\rfloor \text{;}$ $${\hat{B}}_m=\{\lambda ⊢m-2k,0\le k\le \lfloor m/2\rfloor \}\text{.}$$ $$\begin{array}{c}\begin{array}{c} ∅ ∅ ∅ \end{array}\\ \text{Figure 2}\end{array}$$ A partition $\lambda \in {\hat{B}}_m$ is connected by an edge to a partition $\mu \in {\hat{B}}_{m+1}$ if $\mu$ can be obtained from $\lambda$ by adding a box to $\lambda$ or by removing a box from $\lambda \text{.}$ The diagram $B$ is a multiplicity free Bratteli diagram. The tableaux $T\in {𝒯}^{\lambda }$ in the Bratteli diagram $B$ are called up-down tableaux since they are sequences of partitions in which each partition differs from the previous one by either adding or removing a box.

The $r\text{-truncated}$ Bratteli diagram $B\left(r\right)$ is given by the sets $${\hat{B}}_m\left(r\right)=\{\lambda ⊢m-2k,0\le k\le \lfloor m/2\rfloor \hspace{0.17em}|\hspace{0.17em}l\left(\lambda \right)\le r\}\text{.}$$ A partition $\lambda \in {\hat{B}}_m\left(r\right)$ is connected by an edge to a partition $\mu \in {\hat{B}}_{m+1}\left(r\right)$ if $\mu$ can be obtained by adding or removing a box from $\lambda \text{.}$ The Bratteli diagram $B\left(r\right)$ is a multiplicity free Bratteli diagram. It can be obtained from the Bratteli diagram $B$ by removing all the partitions with more than $r$ rows (and the edges connected to them). It is easy to see that tableaux in the $r\text{-truncated}$ Bratteli diagram $B\left(r\right)$ are up-down tableaux that never pass through a partition of length greater than $r\text{.}$ The Bratteli diagram $B$ can be viewed as the limit of the Bratteli diagrams $B\left(r\right),$ as $r$ goes to infinity.

For the remainder of this section let us fix $r,$ and, unless otherwise specified, all paths and tableaux shall be from the Bratteli diagram $B\left(r\right)\text{.}$

Fix $S=({\sigma }^{m-2},{\sigma }^{(m-1)},{\sigma }^{\left(m\right)})\in {𝒯}_{m-2}^m$ and assume that ${\sigma }^{(m-2)}\ne {\sigma }^{\left(m\right)}$ as partitions. Then there is at most one $T\ne S$ such that $(S,T)\in {\Omega }_{m-2}^m\text{.}$

		Proof.
	Given a partition $\lambda$ let us write $\mu =\lambda +{\epsilon }_k$ (resp. $\mu =\lambda -{\epsilon }_k\text{)}$ to denote that $\mu$ is obtained by adding (resp. removing) a box to (resp. from) the $k\text{th}$ row of $\lambda \text{.}$ Fix $S=({\sigma }^{(m-2)},{\sigma }^{(m-1)},{\sigma }^{\left(m\right)})\in {𝒯}_{m-2}^m$ and assume that ${\sigma }^{(m-2)}\ne {\sigma }^{\left(m\right)}$ as partitions. Suppose that ${\sigma }^{(m-1)}={\sigma }^{(m-2)}+{\delta }_1{\epsilon }_k$ and that ${\sigma }^{\left(m\right)}={\sigma }^{(m-1)}+{\delta }_2{\epsilon }_l,$ where ${\delta }_1$ and ${\delta }_2$ are either $\pm 1\text{.}$ If $T$ exists then $T=({\sigma }^{(m-2)},{\tau }^{(m-1)},{\sigma }^{\left(m\right)})$ is given by ${\tau }^{(m-1)}={\sigma }^{(m-2)}+{\delta }_2{\epsilon }_l$ and ${\sigma }^{\left(m\right)}={\tau }^{(m-1)}+{\delta }_1{\epsilon }_k\text{.}$ The path $T$ exists when ${\tau }^{(m-1)}={\sigma }^{(m-2)}+{\delta }_2{\epsilon }_l$ is a partition and not equal to ${\sigma }^{(m-1)}\text{.}$ $\square$

The Centralizer Algebras ${𝒵}_m$

For the remainder of this section fix $𝔤$ to be a complex simple Lie algebra of type $B_r,$ $C_r$ or $D_r$ and let $𝔘={𝔘}_h\left(𝔤\right)$ be the corresponding quantum group. We shall use the standard notations ([Bou1981], pp. 252-258) for the root systems of Types $B_r,$ $C_r,$ and $D_r$ so that ${\epsilon }_1,\dots ,{\epsilon }_r$ are an orthonormal basis of ${𝔥}^{*}$ and the element $2\rho$ is given by $$\begin{array}{cc}2\rho =\sum _{k=1}^{r}2{\rho }_k{\epsilon }_j=\sum _{k=1}^r(y-2k+1){\epsilon }_k,& \text{(5.2)}\end{array}$$ where $$\begin{array}{cc}y=\{\begin{array}{cc}2r,& \text{in Type}\hspace{0.17em}{B}_r,\\ 2r+1,& \text{in Type}\hspace{0.17em}{C}_r,\\ 2r-1,& \text{in Type}\hspace{0.17em}{D}_r\text{.}\end{array}& \text{(5.3)}\end{array}$$ The finite dimensional irreducible representations of ${𝔘}_h\left(𝔤\right)$ which appear as irreducible summands in the tensor powers of $V={\Lambda }_{{\omega }_1}$ of ${𝔘}_h\left(𝔤\right)$ are indexed by the dominant integral weights in the set $$\widehat{𝔘}=\{\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_r{\epsilon }_r\hspace{0.17em}|\hspace{0.17em}{\lambda }_i\in \mathbb{Z},{\lambda }_1\ge \cdots \ge {\lambda }_r\ge 0\}\text{.}$$ We shall identify each dominant integral weight $\lambda \in \widehat{𝔘}$ with the partition $\lambda =({\lambda }_1,\dots ,{\lambda }_r)\text{.}$ It will be helpful to note that, if $\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_r{\epsilon }_r\in \widehat{𝔘}$ then $$\begin{array}{cc}⟨\lambda ,\lambda +2\rho ⟩=\sum _{i=1}^r{\lambda }_i^2+\sum _{i=1}^{r}2{\rho }_i{\lambda }_i\text{.}& \text{(5.4)}\end{array}$$

Let $V={\Lambda }_{{\omega }_1}$ the irreducible $𝔘\text{-module}$ of highest weight ${\omega }_1\text{.}$ In type $C_r,$ the decomposition rule for tensoring by $V$ is given by $$\begin{array}{cc}{\Lambda }_{\lambda }\otimes V\cong \underset{\mu \in {\lambda }^{\pm }}{⨁}{\Lambda }_{\mu },& \text{(5.5)}\end{array}$$ where the sum is over all partitions $\mu \in \widehat{𝔘}$ that are gotten by adding or removing a box from the partition $\lambda \text{.}$ It follows that in Type $C_r$ the Bratteli diagram for tensor powers of $V={\Lambda }_{{\omega }_1}$ is given by $B\left(r\right)\text{.}$ In Types $B_r$ and $D_r$ the tensor product rule given in (5.5) holds whenever $\left|\lambda \right|<r-1$ but must be modified slightly when $\left|\lambda \right|\ge r-1\text{.}$ In order to avoid this complication

(5.6) For the remainder of this section, we shall assume that in Types $B_r$ and $D_r$ we have that $r\gg 0\text{;}$ in particular, $m<r$ and $i<r$ whenever the constants $m$ and $i$ are used,

The Elements ${Ě}_i$

The weights of the Markov traces on ${𝒵}_m={\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ are given by $$\begin{array}{cc}{\text{wt}}_m\left(\lambda \right)=\frac{{\text{dim}}_q\left({\Lambda }_{\lambda }\right)}{{\left({\text{dim}}_q\left(V\right)\right)}^m},\lambda \in \widehat{𝔘}& \text{(5.7)}\end{array}$$ where the quantum dimension of $V={\Lambda }_{{\omega }_1}$ is given by $${\text{dim}}_q\left(V\right)=\{\begin{array}{cc}\left[2r\right]+1,& \text{in Type}\hspace{0.17em}{B}_r,\\ [2r+1]-1,& \text{in Type}\hspace{0.17em}{C}_r,\\ [2r-1]+1,& \text{in Type}\hspace{0.17em}{D}_r\text{.}\end{array}$$ Since all automorphisms of the Dynkin diagram corresponding to $𝔤$ fix the node corresponding to the fundamental weight ${\omega }_1$ it follows that $V={\Lambda }_{{\omega }_1}\cong V^{*}={\left({\Lambda }_{{\omega }_1}\right)}^{*}\text{.}$ As in (3.2), define $ě\in {\text{End}}_{𝔘}(V\otimes V)$ to be the $𝔘\text{-invariant}$ projection onto the invariants, ${\Lambda }_{\left(0\right)}\subseteq V\otimes V\text{.}$ Define $${Ě}_i=\delta {\text{dim}}_q\left(V\right)(\text{id}\otimes \cdots \otimes \text{id}\otimes ě\otimes \text{id}\otimes \cdots \otimes \text{id})\in {\text{End}}_{𝔘}\left(V^{\otimes m}\right),$$ where the factor $ě$ appears as a transformation on the $i\text{th}$ and the $(i+1)\text{st}$ tensor factors and $$\begin{array}{cc}\delta =\{\begin{array}{cc}1,& \text{in Types}\hspace{0.17em}{B}_r\hspace{0.17em}\text{and}\hspace{0.17em}{D}_r,\\ -1,& \text{in Type}\hspace{0.17em}{C}_r\text{.}\end{array}& \text{(5.8)}\end{array}$$ By Theorem (3.12), there is a natural identification of the centralizer algebras ${𝒵}_m$ with the path algebras corresponding to the Bratteli diagram $B\left(r\right)$ so that $${Ě}_{m-1}=\sum _{(S,T)\in {\Omega }_{m-1}^m}{\left({Ě}_{m-1}\right)}_{ST}{E}_{ST}$$ where, if $S=({\sigma }^{(m-2)},{\sigma }^{(m-1)},{\sigma }^{\left(m\right)})$ and $T=({\sigma }^{(m-2)},{\tau }^{(m-1)},{\sigma }^{\left(m\right)}),$ then $$\begin{array}{cc}{\left({Ě}_{m-1}\right)}_{ST}=\{\begin{array}{cc}\frac{\delta \sqrt{{\text{dim}}_q\left({\Lambda }_{{\sigma }^{(m-1)}}\right)·{\text{dim}}_q\left({\Lambda }_{{\tau }^{(m-1)}}\right)}}{{\text{dim}}_q\left({\Lambda }_{{\sigma }^{(m-2)}}\right)}& \text{if}\hspace{0.17em}{\sigma }^{(m-2)}={\sigma }^{\left(m\right)},\\ 0& \text{otherwise}\end{array}& \text{(5.9)}\end{array}$$ where we have replaced the weights of the Markov trace by $q\text{-dimensions.}$

Let $V={\Lambda }_{{\omega }_1}$ be the irreducible $𝔘={𝔘}_h\left(𝔤\right)\text{-module}$ indexed by the fundamental weight ${\omega }_1\text{.}$ The matrices ${Ř}_i$ and ${Ě}_i$ in ${\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ satisfy the relations

(a)	${Ř}_i{Ř}_j={Ř}_j{Ř}_i,|i-j|>1,$
(b)	${Ř}_i{Ř}_{i+1}{Ř}_i={Ř}_{i+1}{Ř}_i{Ř}_{i+1},1\le i\le m-2,$
(c)	$({Ř}_i-z^{-1})({Ř}_i-q)({Ř}_i+q^{-1})=0,1\le i\le m-1\text{.}$
(d)	${Ě}_i{Ř}_{i-1}^{\pm 1}{Ě}_i=z^{\pm 1}{Ě}_i$ and ${Ě}_i{Ř}_{i+1}^{\pm 1}{Ě}_i=z^{\pm 1}{Ě}_i,$
(e)	${Ř}_i-{Ř}_i^{-1}=(q-q^{-1})(1-{Ě}_i),$
(f)	${Ě}_i{Ř}_i^{\pm 1}={Ř}_i^{\pm 1}{Ě}_i=z^{\mp 1}{Ě}_i,$

where $$z=\delta q^y=\{\begin{array}{cc}{q}^{2r},& \text{in Type}\hspace{0.17em}{B}_r,\\ -q^{2r+1},& \text{in Type}\hspace{0.17em}{C}_r,\\ q^{2r-1},& \text{in Type}\hspace{0.17em}{D}_r\text{.}\end{array}$$

		Proof.
	(a) and (b) follow from Proposition (2.18). From (5.5), we have $$V\otimes V\cong {\Lambda }_{\varnothing }\oplus {\Lambda }_{\left(1^2\right)}\oplus {\Lambda }_{\left(2\right)}={\Lambda }_0\oplus {\Lambda }_{2{\omega }_1}\oplus {\Lambda }_{{\omega }_1}\text{.}$$ Use (5.4) to show that $$\begin{array}{c}⟨0,0+2\rho ⟩=0,⟨{\epsilon }_1+{\epsilon }_2,{\epsilon }_1+{\epsilon }_2+2\rho ⟩=2y-2,\\ ⟨2{\epsilon }_1,2{\epsilon }_1+2\rho ⟩=2y+2,\text{and}⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩=y\text{.}\end{array}$$ It follows that $$\begin{array}{ccc}{q}^{(1/2)⟨0,0+2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩}& =& q^{-y},\\ q^{(1/2)⟨{\epsilon }_1+{\epsilon }_2,{\epsilon }_1+{\epsilon }_2+2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩}& =& q^{-1},\\ \text{and}{q}^{(1/2)⟨2{\epsilon }_1,2{\epsilon }_1+2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩}& =& q\text{.}\end{array}$$ Relation (c) now follows from Corollary (2.22); the signs of the eigenvalues of ${Ř}_i$ are determined by which summands are in ${\bigwedge }^2\left(V\right),$ $${\bigwedge }^2\left(V\right)=\{\begin{array}{cc}{\Lambda }_{\left(1^2\right)},& \text{in Types}\hspace{0.17em}{B}_r\hspace{0.17em}\text{and}\hspace{0.17em}{D}_r,\\ {\Lambda }_{\left(1^2\right)}\oplus {\Lambda }_0,& \text{in Type}\hspace{0.17em}{C}_r\text{.}\end{array}$$ (d) follows from Proposition (3.11) part (2) and the fact that $q^{⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩}=q^y\text{.}$ Let us prove (e). By Corollary (2.22), ${Ř}_1$ acts by the eigenvalue $z^{-1}$ on the irreducible summand ${\Lambda }_0$ in $V\otimes V\text{.}$ Thus, it follows from relation (c) that $${Ě}_i=\delta {\text{dim}}_q\left(V\right)\frac{({Ř}_i-q)({Ř}_i+q^{-1})}{(z^{-1}-q)(z^{-1}+q^{-1})}\text{.}$$ Using this formula and the relation $$\delta {\text{dim}}_q\left(V\right)=\frac{z-z^{-1}}{q-q^{-1}}+1,$$ it can be easily checked that relation (c) is equivalent to relation (e). The relation ${Ř}_1{Ě}_1=z{Ě}_1,$ follows by noting that, except for the constant $\delta {\text{dim}}_q\left(V\right),$ ${Ě}_i$ is the projection onto the invariants ${\Lambda }_0\subseteq V\otimes V$ and that ${Ř}_i$ acts by constant $z^{-1}$ on ${\Lambda }_0\text{.}$ All of the relations in (f) follow silimlarly. $\square$

A Path Algebra Formula for ${Ř}_i$

Let $B\left(r\right)$ be the Bratteli diagram for tensor powers of $V={\Lambda }_{{\omega }_1}$ (with the assumptions in (5.6)). Identify the centralizer algebras ${𝒵}_m={\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ with the path algebras corresponding to the Bratteli diagram $B\left(r\right)\text{.}$ Recall that the path algebras have a natural basis $E_{ST},$ $(S,T)\in {\Omega }^m$ of matrix units.

For each tableau $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})\in {𝒯}^m$ define $$\begin{array}{cc}{\nabla }_i\left(S\right)=⟨{\sigma }^{\left(i\right)},{\sigma }^{\left(i\right)}+2\rho ⟩-⟨{\sigma }^{(i-1)},{\sigma }^{(i-1)}+2\rho ⟩-⟨{\omega }_1,{\omega }_1+2\rho ⟩\text{.}& \text{(5.11)}\end{array}$$ Let $(S,T)$ be a pair of tableaux $$S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{(i-1)},{\sigma }^{\left(i\right)},{\sigma }^{(i+1)},\dots ,{\sigma }^{\left(m\right)}),$$ and $$T=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{(i-1)},{\tau }^{\left(i\right)},{\sigma }^{(i+1)},\dots ,{\sigma }^{\left(m\right)}),$$ in ${𝒯}^m$ such that $S$ and $T$ are the same except possibly at the shape at level $i\text{.}$ In other words the pair $(S,T)\in {\Omega }_{i-1}^{i+1}\text{.}$ Define $$\begin{array}{cc}{\diamond }_i(S,T)=\frac{1}{2}({\nabla }_{i+1}\left(S\right)-{\nabla }_i\left(T\right))\text{.}& \text{(5.12)}\end{array}$$ These constants are defined so that, if $D_m={Ř}_{m-1}{Ř}_{m-2}\cdots {Ř}_2{Ř}_1{Ř}_1{Ř}_2\cdots {Ř}_{m-2}{Ř}_{m-1},$ then $$D_m=\sum _{S\in {𝒯}^m}{\left(D_m\right)}_{SS}{E}_{SS},\text{where}{\left(D_m\right)}_{SS}=q^{{\nabla }_m\left(S\right)},$$ and $$q^{-2{\diamond }_{m-1}(S,T)}={\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}\text{.}$$ The first of these formulas is a consequence of Corollary (2.25).

Let $y$ be as given in (5.3).

(a)

Let $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ be a tableau in the Bratteli diagram $B\left(r\right)\text{.}$ Then $${\nabla }_m\left(S\right)=\{\begin{array}{cc}2({\sigma }_k^{(m-1)}-k+1),& \text{when}{\sigma }^{\left(m\right)}={\sigma }^{(m-1)}+{\epsilon }_k,\\ 2(-{\sigma }_k^{(m-1)}-y+k),& \text{when}{\sigma }^{\left(m\right)}={\sigma }^{(m-1)}-{\epsilon }_k\text{.}\end{array}$$

(b)

Let $S=({\sigma }^{(m-2)},{\sigma }^{\left(m-1\right)},{\sigma }^{\left(m\right)})$ and $T=({\sigma }^{(m-2)},{\tau }^{(m-1)},{\sigma }^{\left(m\right)})$ be such that $(S,T)\in {\Omega }_{m-2}^m\text{.}$ Then $${\diamond }_{m-1}(S,T)=\{\begin{array}{cc}\pm ({\sigma }_k^{\left(m\right)}-k-{\tau }_l^{(m-1)}+l),& \text{if}{\tau }^{(m-1)}={\tau }^{(m-2)}\pm {\epsilon }_l\hspace{0.17em}\text{and}\hspace{0.17em}{\sigma }^{\left(m\right)}={\sigma }^{(m-1)}\pm {\epsilon }_k,\\ \pm ({\tau }_l^{(m-1)}-l+{\sigma }_k^{\left(m\right)}-k+y+1),& \text{if}{\sigma }^{\left(m\right)}={\sigma }^{(m-1)}\pm {\epsilon }_k\hspace{0.17em}\text{and}\hspace{0.17em}{\tau }^{(m-1)}={\tau }^{(m-2)}\mp {\epsilon }_l\text{.}\end{array}$$

		Proof.
	Let $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ be a tableau in the Bratteli diagram $B\left(r\right)\text{.}$ Then, since ${\sigma }^{\left(m\right)}$ differs from ${\sigma }^{(m-1)}$ by either adding or removing a box in the $k\text{th}$ row, $${\sigma }^{(m-1)}=\sum _{j=1}^r{\sigma }_j^{(m-1)}{\epsilon }_j\text{and}{\sigma }^{\left(m\right)}=({\sigma }_k^{(m-1)}\pm 1){\epsilon }_k+\sum _{\underset{j\ne k}{1\le j\le r}}{\sigma }_j^{(m-1)}{\epsilon }_j\text{.}$$ Using (5.4) to compute ${\nabla }_m\left(S\right)$ we get $$\begin{array}{ccc}{\nabla }_m\left(S\right)& =& ⟨{\sigma }^{\left(m\right)},{\sigma }^{\left(m\right)}+2\rho ⟩-⟨{\sigma }^{(m-1)},{\sigma }^{(m-1)}+2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩\\ & =& \left(\sum _{j\ne k}{\left({\sigma }_j^{(m-1)}\right)}^2\right)+{({\sigma }_k^{(m-1)}\pm 1)}^2+\left(\sum _{j\ne k}2{\sigma }_j^{(m-1)}{\rho }_j\right)\\ & & +2({\sigma }_k^{(m-1)}\pm 1){\rho }_k-\left(\sum _{j\ne k}{\left({\sigma }_j^{(m-1)}\right)}^2\right)-{\left({\sigma }_k^{(m-1)}\right)}^2\\ & & -\left(\sum _{j\ne k}2{\sigma }_j^{(m-1)}{\rho }_j\right)-2{\sigma }_k^{(m-1)}{\rho }_k-1-2{\rho }_1\\ & =& \pm 2{\sigma }_k^{(m-1)}+2(\pm {\rho }_k-{\rho }_1)\text{.}\end{array}$$ The formula for ${\nabla }_m\left(S\right)$ follows since ${\rho }_k-{\rho }_1=(y-2k+1)-(y-1)=2(-k+1)$ and $-{\rho }_k-{\rho }_1=-(y-2k+1)-(y-1)=2(-y+k)\text{.}$ (b) The formulas for ${\diamond }_{m-1}(S,T)$ now follow from the definition of ${\diamond }_{m-1}$ and the formula for ${\nabla }_m$ in (a). $\square$

One can choose the identification (Section 1) of the centralizer algebras ${𝒵}_m$ with the path algebras corresponding to the Bratteli diagram $B\left(r\right)$ (with the assumption in (5.6)) so that the matrices ${Ř}_i$ are given by the formula $${Ř}_i=\sum _{(S,T)\in {\Omega }_{i-1}^{i+1}}{\left({Ř}_i\right)}_{ST}{E}_{ST},$$ where, for each $S=({\sigma }^{(i-1)},{\sigma }^{\left(i\right)},{\sigma }^{(i+1)}),$ $${\left({Ř}_i\right)}_{SS}=\{\begin{array}{cc}\frac{q^{{\diamond }_i(S,S)}}{\left[{\diamond }_i(S,S)\right]},& \text{if}{\sigma }^{(i-1)}\ne {\sigma }^{(i+1)},\\ \frac{q^{{\diamond }_i(S,S)}}{\left[{\diamond }_i(S,S)\right]}(1-\frac{\delta {\text{dim}}_q\left({\Lambda }_{{\sigma }^{\left(i\right)}}\right)}{{\text{dim}}_q\left({\Lambda }_{{\sigma }^{(i-1)}}\right)}),& \text{if}{\sigma }^{(i-1)}={\sigma }^{(i+1)},\end{array}$$ and for each pair $(S,T)=(({\sigma }^{(i-1)},{\sigma }^{\left(i\right)},{\sigma }^{(i+1)}),({\sigma }^{(i-1)},{\tau }^{\left(i\right)},{\sigma }^{(i+1)}))\in {\Omega }_{i-1}^{i+1}$ such that $S\ne T,$ $${\left({Ř}_i\right)}_{ST}=\{\begin{array}{cc}\frac{\sqrt{[{\diamond }_i(S,S)-1][{\diamond }_i(S,S)+1]}}{\left[\left|{\diamond }_i(S,S)\right|\right]},& \text{if}{\sigma }^{(i-1)}\ne {\sigma }^{(i+1)},\\ -\frac{q^{{\diamond }_i(S,T)}}{\left[{\diamond }_i(S,T)\right]}\frac{\delta \sqrt{{\text{dim}}_q\left({\Lambda }_{{\sigma }^{\left(i\right)}}\right)·{\text{dim}}_q\left({\Lambda }_{{\tau }^{\left(i\right)}}\right)}}{{\text{dim}}_q\left({\Lambda }_{{\sigma }^{(i-1)}}\right)},& \text{if}{\sigma }^{(i-1)}={\sigma }^{(i+1)},\end{array}$$ where ${\diamond }_i(S,T)$ and $\delta$ are given by (5.11) and (5.12) respectively.

		Proof.
	Since ${Ř}_i\in {𝒵}_i={\text{End}}_{𝔘}\left(V^{\otimes (i+1)}\right)$ commutes with all elements of ${𝒵}_{i-1}={\text{End}}_{𝔘}\left(V^{\otimes (i-1)}\right)$ it follows from Proposition Corollary (1.5) that $${Ř}_i=\sum _{(S,T)\in {\Omega }_{i-1}^{i+1}}{\left({Ř}_i\right)}_{ST}{E}_{ST},$$ for some constants ${\left({Ř}_i\right)}_{ST}\text{.}$ In view of the imbeddings ${𝒵}_0\subseteq {𝒵}_1\subseteq \cdots \subseteq {𝒵}_m,$ it is sufficient to show that the formulas for ${Ř}_i$ hold for $i=m-1\text{.}$ By definition $D_m={Ř}_{m-1}{Ř}_{m-2}\cdots {Ř}_2{Ř}_1{Ř}_1{Ř}_2\cdots {Ř}_{m-2}{Ř}_{m-1}$ and it follows that ${Ř}_{m-1}^{-1}=D_m^{-1}{Ř}_{m-1}{D}_{m-1}\text{.}$ Thus, we may rewrite the relation (5.10e) in the form $$\begin{array}{cc}{Ř}_{m-1}-D_m^{-1}{Ř}_{m-1}{D}_{m-1}=(q-q^{-1})(1-{Ě}_{m-1})\text{.}& \text{(5.15)}\end{array}$$ Let $(S,T)\in {\Omega }^m$ and view (5.15) as an equation in the path algebra. Since the matrices $D_m$ and $D_{m-1}$ are diagonal, taking the $E_{ST}\text{-entry}$ of this equation yields $${\left({Ř}_{m-1}\right)}_{ST}-{\left(D_m^{-1}\right)}_{SS}{\left({Ř}_{m-1}\right)}_{ST}{\left(D_{m-1}\right)}_{TT}=(q-q^{-1})({\delta }_{ST}-{\left({Ě}_{m-1}\right)}_{ST}),$$ or, equivalently, $$\begin{array}{cc}(1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}){\left({Ř}_{m-1}\right)}_{ST}=(q-q^{-1})({\delta }_{ST}-{\left({Ě}_{m-1}\right)}_{ST}),& \text{(5.16)}\end{array}$$ Hence, $${\left({Ř}_{m-1}\right)}_{ST}=\frac{(q-q^{-1})({\delta }_{ST}-{\left({Ě}_{m-1}\right)}_{ST})}{1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}},\text{if}1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}\ne 0\text{.}$$ Plugging in the following $$\begin{array}{ccc}\frac{q-q^{-1}}{1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}}& =& \frac{q-q^{-1}}{1-q^{-2{\diamond }_{m-1}(S,T)}}\\ & =& \frac{q^{{\diamond }_{m-1}(S,T)}(q-q^{-1})}{q^{{\diamond }_{m-1}(S,T)}-q^{-{\diamond }_{m-1}(S,T)}}\\ & =& \frac{q^{{\diamond }_{m-1}(S,T)}}{\left[{\diamond }_{m-1}(S,T)\right]}\end{array}$$ we get $${\left({Ř}_{m-1}\right)}_{ST}=\frac{q^{{\diamond }_{m-1}(S,T)}}{\left[{\diamond }_{m-1}(S,T)\right]}({\delta }_{ST}-{\left({Ě}_{m-1}\right)}_{ST}),\text{if}1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}\ne 0\text{.}$$ All except the last of the formulas in Theorem (5.14) now follow immediately from (5.9) and the following lemma. $\square$

Let $(S,T)\in {\Omega }_{m-2}^{m}m\text{.}$ If $S=T$ or if ${\sigma }^{(m-2)}\ne {\sigma }^{\left(m\right)}$ then $1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}\ne 0\text{.}$

		Proof.
	Consider the equation (5.16). Case 1. If $S\ne T$ and ${\sigma }^{(m-2)}={\sigma }^{\left(m\right)}$ then ${\delta }_{ST}=0$ and ${\left({Ě}_i\right)}_{ST}\ne 0$ since the weights ${\text{wt}}_k\left(\mu \right)$ are all nonzero. Thus the right hand side of (5.16) is nonzero. This implies that $1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}$ is nonzero. Case 2. If $S=T$ and ${\sigma }^{(m-2)}\ne {\sigma }^{\left(m\right)}$ then ${\left({Ě}_i\right)}_{ST}=0$ and ${\delta }_{ST}\ne 0\text{.}$ Thus the right hand side of (5.16) is nonzero. This implies that $1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}$ is nonzero. Case 3. Suppose $S=T$ and ${\sigma }^{(m-2)}={\sigma }^{\left(m\right)}\text{.}$ Clearly $1-{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{SS}$ is nonzero if and only if ${\diamond }_m(S,S)\ne 0\text{.}$ Then, by Proposition (5.13), there is some $k$ such that ${\diamond }_m(S,S)=\pm (2{\sigma }_k^{(m-1)}-2k+y)\text{.}$ This value is nonzero in Types $C_r$ and $D_r$ since $y$ is odd, and is nonzero in type $B_r$ since, by the assumption in (5.6), $2k<y$ and ${\sigma }_k^{(m-1)}\ge 0\text{.}$ $\square$

Now let us prove the last formula in Theorem (5.14). Let $S\in {𝒯}_{m-2}^m$ and suppose that $T\in {𝒯}_{m-2}^m$ is such that $(S,T)\in {\Omega }_{m-2}^m$ and $T\ne S\text{.}$ By Lemma (5.1), $T$ is unique. It follows that $$\begin{array}{ccc}{\left({Ř}_{m-1}^2\right)}_{SS}& =& \sum _{L\in {𝒯}_m}{\left({Ř}_{m-1}\right)}_{SL}{\left({Ř}_{m-1}\right)}_{LS}\\ & =& {\left({\left({Ř}_{m-1}\right)}_{SS}\right)}^2+{\left({Ř}_{m-1}\right)}_{ST}{\left({Ř}_{m-1}\right)}_{TS}& \text{(5.18)}\end{array}$$ On the other hand $$\begin{array}{ccc}{Ř}_{m-1}^2& =& {Ř}_{m-1}({Ř}_{m-1}-{Ř}_{m-1}^{-1})\\ & =& {Ř}_{m-1}(q-q^{-1})(1-{Ě}_{m-1})\\ & =& (q-q^{-1})({Ř}_{m-1}-z^{-1}{Ě}_{m-1}),\end{array}$$ since ${Ě}_{m-1}{Ř}_{m-1}=z^{-1}{Ě}_{m-1}\text{.}$ Since ${\sigma }^{(m-2)}\ne {\sigma }^{\left(m\right)},$ it follows that ${\left({Ě}_{m-1}\right)}_{SS}=0$ and thus that $$\begin{array}{cc}{\left({Ř}_{m-1}^2\right)}_{SS}=(q-q^{-1}){\left({Ř}_{m-1}\right)}_{SS}+1\text{.}& \text{(5.19)}\end{array}$$ Equating (5.18) and (5.19) and using the formula for ${\left({Ř}_{m-1}\right)}_{SS}$ gives $$\begin{array}{ccc}{\left({Ř}_{m-1}\right)}_{ST}{\left({Ř}_{m-1}\right)}_{TS}& =& (q-q^{-1}){\left({Ř}_{m-1}\right)}_{SS}+1-{\left({Ř}_{m-1}^2\right)}_{SS}\\ & =& \frac{[{\diamond }_{m-1}(S,S)-1][{\diamond }_{m-1}(S,S)+1]}{{\left[{\diamond }_{m-1}(S,S)\right]}^2}\end{array}$$ exactly as in the proof of Theorem (4.8). It follows from the remarks at the end of Section 1 that we can choose the normalization of the elements $E_{ST}$ so that ${\left({Ř}_{m-1}\right)}_{ST}$ and ${\left({Ř}_{m-1}\right)}_{TS}$ are as given in the theorem.

Matrix Units

Given a tableau $T=({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m\right)})\in {𝒯}^m$ let $T\prime$ denote the tableau $T\prime =({\tau }^{\left(0\right)},\dots ,{\tau }^{(m-1)})\in {𝒯}^{m-1}\text{.}$ Let ${\left(T\prime \right)}^{+}$ denote the set of all extensions of $T\prime \text{;}$ $${\left(T\prime \right)}^{+}=\{S\in {𝒯}^m\hspace{0.17em}|\hspace{0.17em}S\prime =T\prime \}\text{.}$$ Given tableaux $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ and $T=({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m\right)})$ in ${𝒯}^m$ let ${\left({Ř}_{m-1}\right)}_{ST}$ be the constant given by Theorem (5.14) in the case that $(({\sigma }^{(m-2)},{\sigma }^{(m-1)},{\sigma }^{\left(m\right)}),({\tau }^{(m-2)},{\tau }^{(m-1)},{\tau }^{\left(m\right)}))\in {\Omega }_{m-2}^m$ and let ${\left({Ř}_{m-1}\right)}_{ST}=0$ otherwise.

Let $T\prime =({\tau }^{\left(0\right)},\dots ,{\tau }^{(m-1)})\in {𝒯}^{m-1}$ and let ${\left(T\prime \right)}^{+}$ be the set of extensions of $T\prime \text{.}$ Then the values ${\nabla }_m\left(S\right)$ are different as $S$ ranges over all elements of ${\left(T\prime \right)}^{+}\text{.}$

		Proof.
	Let $S=({\tau }^{\left(0\right)},\dots ,{\tau }^{(m-1)},{\sigma }^{\left(m\right)})\in {\left(T\prime \right)}^{+}\text{.}$ By Proposition (5.13), $${\nabla }_m\left(S\right)=2({\tau }_k^{(m-1)}-k+1)\text{or}{\nabla }_m\left(S\right)=2(-{\tau }_k^{(m-1)}-y+k),$$ for some positive integer $k\text{.}$ Let $l$ be the largest value of $k$ such that $S\in {\left(T\prime \right)}^{+}\text{.}$ By the assumption in (5.6), $2l-1<y\text{.}$ Since $$\begin{array}{c}{\tau }_1^{(m-1)}\ge \cdots \ge {\tau }_k^{(m-1)}\ge \cdots \ge {\tau }_l^{(m-1)},\text{and}\\ -{\tau }_l^{(m-1)}\ge \cdots \ge -{\tau }_k^{(m-1)}\ge \cdots \ge -{\tau }_1^{(m-1)},\end{array}$$ it follows that $$\begin{array}{c}{\tau }_1^{(m-1)}>\cdots >{\tau }_k^{(m-1)}-k+1>\cdots >{\tau }_l^{(m-1)}-l+1,\text{and}\\ -{\tau }_l^{(m-1)}-y+l>\cdots >-{\tau }_k^{(m-1)}-y+k>\cdots >-{\tau }_1^{(m-1)}-y+1\text{.}\end{array}$$ Since ${\tau }_l^{(m-1)}\ge -{\tau }_l^{(m-1)}$ and $-l+1>-y+l,$ it follows that $${\tau }_l^{(m-1)}-l+1>-{\tau }_l^{(m-1)}-y+l\text{.}$$ The result follows. $\square$

The proofs of the following results are essentially the same as the proofs of Theorem (4.14) and Corollary (4.15).

The matrix units $E_{ST}\in {𝒵}_m,$ $(S,T)\in {\Omega }^m$ are given in terms of the ${Ř}_i,$ $1\le i\le m-1,$ inductively, by the following formulas.

(1)	Let $T\in {𝒯}^m\text{.}$ Then $E_{TT}=\prod _{S\in {𝒯}_m,S\ne T,S\prime =T\prime }(E_{T\prime T\prime }{Ř}_{m-1}{E}_{T\prime T\prime }-{\left({Ř}_{m-1}\right)}_{SS}{E}_{T\prime T\prime })/({\left({Ř}_{m-1}\right)}_{TT}-{\left({Ř}_{m-1}\right)}_{SS})$
(2)	Let $(S,T)\in {\Omega }^m\text{.}$ If $\text{shp}\left(S\prime \right)=\text{shp}\left(T\prime \right)$ then $E_{ST}=E_{S\prime T\prime }{E}_{TT}$ where $E_{TT}$ is given by (1).
(3)	Let $(S,T)\in {\Omega }^m\text{.}$ If $\text{shp}\left(S\prime \right)\ne \text{shp}\left(T\prime \right)$ then $$E_{ST}=\frac{1}{{\left({Ř}_{m-1}\right)}_{MN}}{E}_{S\prime M\prime }{Ř}_{m-1}{E}_{N\prime T\prime }{E}_{TT},$$ where $M$ and $N$ are tableaux in ${𝒯}^m$ of the form $M=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)},\text{shp}\left(S\prime \right),\text{shp}\left(S\right))$ and $N=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)},\text{shp}\left(T\prime \right),\text{shp}\left(S\right))\text{.}$

The centralizer ${𝒵}_m={\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ is generated by the matrices ${Ř}_i,$ $1\le i\le m-1\text{.}$

Notes and References

This is a typed version of the paper A Ribbon Hopf Algebra Approach to the Irreducible Representations of Centralizer Algebras: The Brauer, Birman-Wenzl, and Type A Iwahori-Hecke Algebras by Robert ${\text{Leduc}}^{*}$ and Arun ${\text{Ram}}^{†}\text{.}$

The paper was received June 24, 1994; accepted September 12, 1994.

${}^{*}$Supported in part by National Science Foundation Grant DMS-9300523 to the University of Wisconsin.
${}^{†}$Supported in part by National Science Foundation Postdoctoral Fellowship DMS-9107863.

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