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 $V^{{\omega }_1},$ Type $A_r$

We shall use the notations for partitions given in [Mac1979]. In particular, a partition $\lambda$ of the positive integer $m,$ denoted $\lambda ⊢m,$ is a decreasing sequence $\lambda =({\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_t\ge 0)$ of non-negative integers such that ${\lambda }_1+\cdots +{\lambda }_t=m\text{.}$ The length $l\left(\lambda \right)$ is the largest $j$ such that ${\lambda }_j>0\text{.}$ The Ferrers diagram of $\lambda$ is the left-justified array of boxes with ${\lambda }_i$ boxes in the $i\text{th}$ row. For example, $$(5,3,3,1)=\begin{array}{c} \end{array}$$ is a partition of length 4. Given two partitions $\lambda ,\mu$ we write $\lambda \subseteq \mu$ if ${\lambda }_i\le {\mu }_i$ for all $i\text{.}$ We have that $\lambda \subseteq \mu$ if the Ferrers diagram of $\lambda$ is a subset of the Ferrers diagram of $\mu \text{.}$

The Bratteli diagram given in Fig. 1 is called the Young lattice. The shapes $\lambda \in {\hat{Y}}_m$ of $Y$ which are on level $m$ are the partitions of $m\text{;}$ $${\hat{Y}}_m=\{\lambda ⊢m\}\text{.}$$ A partition $\lambda \in {\hat{Y}}_m$ is connected by an edge to a partition $\mu \in {\hat{Y}}_{m+1}$ if $\mu$ can be obtained by adding a box to $\lambda \text{.}$ The Young lattice $Y$ is a multiplicity free Bratteli diagram.

Classically, a standard tableau of shape $\lambda ⊢m$ is a filling of the boxes in the Ferrers diagram of $\lambda$ with the numbers $1,2,\dots ,m$ such that the numbers are increasing left to right in the rows and increasing down the columns. Each tableau $T\in {𝒯}^{\lambda }$ in the Bratteli diagram $Y$ can be identified in a natural way with a standard tableau of shape $\lambda \text{.}$ Let $P$ be a standard tableau of shape $\lambda$ and let $T=({\tau }^{\left(0\right)},{\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)})\in {𝒯}^{\lambda }$ be the tableau such that ${\tau }^{\left(i\right)}$ is the partition given by the set of boxes of $P$ which contain the numbers $1,2,\dots ,i\text{.}$ One easily shows that this identification is a bijection between the standard tableaux $P$ of shape $\lambda$ and the tableaux $T\in T^{\lambda }\text{.}$

The $r\text{-truncated}$ Young lattice is the Bratteli diagram $Y\left(r\right)$ which is given by the sets $${\hat{Y}}_m\left(r\right)=\{\lambda ⊢m\hspace{0.17em}|\hspace{0.17em}l\left(\lambda \right)\le r\}\text{.}$$ A partition $\lambda \in {\hat{Y}}_m\left(r\right)$ is connected by an edge to a partition $\mu \in {\hat{Y}}_{m+1}\left(r\right)$ if $\lambda \subseteq \mu ,$ or equivalently, if $\mu$ can be obtained by adding a box to $\lambda \text{.}$ The $r\text{-truncated}$ Young lattice can be obtained by removing all the partitions with more than $r$ rows (and the edges connected to them) from the full Young lattice $Y\text{.}$ It is easy to see that tableaux in the $r\text{-truncated}$ Young lattice correspond to standard tableaux of shapes $\lambda \in \hat{Y}\left(r\right),$ in exactly the same way as tableaux in $Y$ correspond to standard tableaux. Note also that the full Young lattice can be viewed as the limit of the $r\text{-truncated}$ Young lattices 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 $Y\left(r\right)\text{.}$

Fix $S=({\sigma }^{(m-2)},{\sigma }^{(m-1)},{\sigma }^{\left(m\right)})\in {𝒯}_{m-2}^m\text{.}$ Suppose that ${\sigma }^{(m-1)}$ is obtained by adding a box to the $k\text{th}$ row of ${\sigma }^{(m-2)}$ and that ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{(m-1)}\text{.}$ Now suppose that $T=({\sigma }^{(m-2)},{\tau }^{(m-1)},{\sigma }^{\left(m\right)})$ is such that $(S,T)\in {\Omega }_{m-2}^m\text{.}$ If $k=l$ then we must have that $$\begin{array}{c}\begin{array}{c} ∅ \end{array}\\ \text{Figure 1}\end{array}$$ ${\tau }^{(m-1)}={\sigma }^{(m-1)}\text{.}$ If $k\ne l$ then either ${\tau }^{(m-1)}={\sigma }^{(m-1)},$ or ${\tau }^{(m-1)}$ is the shape obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{(m-2)}\text{.}$ Thus, $$\begin{array}{cc}\text{there is at most one}\hspace{0.17em}T\ne S\hspace{0.17em}\text{such that}\hspace{0.17em}(S,T)\in {\Omega }_{m-2}^m\text{.}& \text{(4.1)}\end{array}$$

The Centralizer Algebras ${𝒵}_m$

For the remainder of this section fix $𝔘={𝔘}_h\left(𝔰𝔩(r+1)\right)\text{.}$ Let ${\epsilon }_1,\dots ,{\epsilon }_{r+1}$ be an orthonormal basis of ${ℝ}^{r+1}\text{.}$ Then ${𝔥}^{*},$ the simple roots, ${\alpha }_i,$ the fundamental weights, ${\omega }_i,$ and the element $2\rho$ are given by $$\begin{array}{ccc}{𝔥}^{*}& =& \{{\lambda }_1{\epsilon }_1+\cdots +{\lambda }_{r+1}{\epsilon }_{r+1}\hspace{0.17em}|\hspace{0.17em}\sum _{i=1}^{r+1}{\lambda }_i=0\},\\ {\alpha }_i& =& {\epsilon }_i-{\epsilon }_{i+1},1\le i\le r,\\ {\omega }_i& =& {\epsilon }_1+\cdots +{\epsilon }_i-\frac{i}{r+1}\sum _{i=1}^{r+1}{\epsilon }_j,1\le i\le r,\\ 2\rho & =& \sum _{i}2{\rho }_i{\epsilon }_i=r{\epsilon }_1+(r-2){\epsilon }_2+\cdots -(r-2){\epsilon }_r-r{\epsilon }_{r+1}\text{.}\end{array}$$ The finite dimensional irreducible modules ${\Lambda }_{\lambda }$ of ${𝔘}_h\left(𝔰𝔩(r+1)\right)$ are indexed by the dominant integral weights, $$\widehat{𝔘}=\{\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_r{\epsilon }_r-\frac{\left|\lambda \right|}{r+1}\sum _{j=1}^{r+1}{\epsilon }_j\hspace{0.17em}|\hspace{0.17em}{\lambda }_i\in \mathbb{Z},{\lambda }_1\ge \cdots \ge {\lambda }_r\ge 0\},$$ where $\left|\lambda \right|={\lambda }_1+\cdots +{\lambda }_r\text{.}$ It is sometimes helpful to identify each dominant integral weight $\lambda$ with the partition $\lambda =({\lambda }_1,\dots ,{\lambda }_r)\text{.}$ Note that all partitions in $\widehat{𝔘}$ have at most $r$ rows. It will also be helpful to note that, if $\lambda ={\lambda }_1{\epsilon }_1+\cdots +{\lambda }_r{\epsilon }_r-(\left|\lambda \right|/(r+1))\sum _j{\epsilon }_j\in \widehat{𝔘}$ then $$\begin{array}{cc}⟨\lambda ,\lambda +2\rho ⟩=\sum _{i=1}^r{\lambda }_i^2-\frac{{\left|\lambda \right|}^2}{r+1}+\sum _{i=1}^{r}2{\rho }_i{\lambda }_i\text{.}& \text{(4.2)}\end{array}$$

Let $V={\Lambda }_{{\omega }_1}$ the irreducible $𝔘\text{-module}$ of highest weight ${\omega }_1\text{.}$ The decomposition rule for tensoring by $V$ is given by $$\begin{array}{cc}{\Lambda }_{\lambda }\otimes V\cong \underset{\mu \in {\lambda }^{+}}{⨁}{\Lambda }_{\mu },& \text{(4.3)}\end{array}$$ where the sum is over all partitions $\mu \in \widehat{𝔘}$ that are gotten by adding a box to the partition $\lambda \text{.}$ It follows that the Bratteli diagram for tensor powers of $V=V^{{\omega }_1}$ is the $r\text{-truncated}$ Young lattice $Y\left(r\right)\text{.}$

Let $V=V^{{\omega }_1}$ be the irreducible $𝔘={𝔘}_h\left(𝔰𝔩(r+1)\right)\text{-module}$ indexed by the fundamental weight ${\omega }_1\text{.}$ The matrices ${Ř}_i\in {\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ satisfy the relations $$\begin{array}{cccc}{Ř}_i{Ř}_j& =& {Ř}_j{Ř}_i,& |i-j|>1,\\ {Ř}_i{Ř}_{i+1}{Ř}_i& =& {Ř}_{i+1}{Ř}_i{Ř}_{i+1},& 1\le i\le m-2,\\ (q^{(1/r+1)}{Ř}_i-q)(q^{(1/r+1)}{Ř}_i-q^{-1})& =& 0,& 1\le i\le m-1\text{.}\end{array}$$

		Proof.
	The first two relations follow from Proposition (2.18). From (4.3), we have $$V\otimes V\cong {\Lambda }_{\left(1^2\right)}\oplus {\Lambda }_{\left(2\right)}={\Lambda }_{2{\omega }_1}\oplus {\Lambda }_{{\omega }_2}\text{.}$$ Use (4.2) to show that $$\begin{array}{ccc}⟨{\omega }_2,{\omega }_2+2\rho ⟩& =& 2-(4/(r+1))+2{\rho }_1+2{\rho }_2=2r-4/(r+1),\\ ⟨2{\omega }_1,2{\omega }_1+2\rho ⟩& =& 4-(4/(r+1))+4{\rho }_1=4+2r-4/(r+1),\\ ⟨{\omega }_1,{\omega }_1+2\rho ⟩& =& 1-(1/(r+1))+2{\rho }_1=r+1-1/(r+1)\text{.}\end{array}$$ It follows that $$\begin{array}{ccc}{q}^{(1/2)⟨{\omega }_2,{\omega }_2+2\rho ⟩-⟨{\omega }_1,{\omega }_1+2\rho ⟩}& =& q^{-1-(1/(r+1))},\\ q^{(1/2)⟨2{\omega }_1,2{\omega }_1+2\rho ⟩-⟨{\omega }_1,{\omega }_1+2\rho ⟩}& =& q^{1-(1/(r+1))}\text{.}\end{array}$$ The result now follows from Proposition (2.22) part (3) and the standard fact that ${\Lambda }_{\left(1^2\right)}$ is antisymmetric part of $V^{\otimes 2}$ and ${\Lambda }_{\left(2\right)}$ is the symmetric part of $V^{\otimes 2}\text{.}$ $\square$

A Path Algebra Formula for ${Ř}_i$

Let $Y=Y\left(r\right)$ be the Bratteli diagram for tensor powers of $V=V^{{\omega }_1}\text{.}$ Identify the centralizer algebras ${𝒵}_m={\text{End}}_{𝔘}\left(V^{\otimes m}\right)$ with the path algebras corresponding to the Bratteli diagram $Y\left(r\right)\text{.}$ 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 $$\begin{array}{cc}{\nabla }_i\left(S\right)=⟨{\sigma }^{\left(i\right)},{\sigma }^{\left(i\right)}+2\rho ⟩-⟨{\sigma }^{(i-1)},{\sigma }^{(i-1)}+2\rho ⟩-⟨{\omega }_1,{\omega }_1+2\rho ⟩\text{.}& \text{(4.5)}\end{array}$$ 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 $i-1\text{.}$ In other words the pair $(S,T)\in {\Omega }_{i-1}^{i+1}\text{.}$ Define $$\begin{array}{cc}{\diamond }_i(S,T)=\frac{1}{2}({\nabla }_{i+1}\left(S\right)-{\nabla }_i\left(T\right))+\frac{1}{r+1}\text{.}& \text{(4.6)}\end{array}$$ 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 $$q^{-2{\diamond }_{m-1}(S,T)}=q^{-(2/r+1)}{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{TT}\text{.}$$ The first of these formulas is a consequence of Corollary (2.25).

Let $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ be a tableau in the Bratteli diagram $Y\left(r\right)\text{.}$ Then $$\begin{array}{ccc}{\nabla }_m\left(S\right)& =& 2({\sigma }_k^{(m-1)}-k+1)+\frac{-2m+2}{r+1},\\ {\diamond }_{m-1}(S,S)& =& {\sigma }_k^{\left(m\right)}-{\sigma }_l^{(m-1)}-k+l,\end{array}$$ where ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $k\text{th}$ row of ${\sigma }^{(m-1)}$ and ${\sigma }^{(m-1)}$ is obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{(m-2)}\text{.}$

		Proof.
	Let $S=({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)})$ be a tableau in the Bratteli diagram $Y\left(r\right)\text{.}$ Then, since ${\sigma }^{\left(m\right)}$ differs from ${\sigma }^{(m-1)}$ by adding a box in the $k\text{th}$ row we have $${\sigma }^{(m-1)}=\sum _{i=1}^r{\sigma }_i^{(m-1)}{\epsilon }_i-\frac{m-1}{r+1}\sum _{j=1}^{r+1}{\epsilon }_j,$$ and $${\sigma }^{\left(m\right)}=({\sigma }_k^{(m-1)}+1){\epsilon }_k+\sum _{\underset{i\ne k}{1\le i\le r}}{\sigma }_i^{\left(m\right)}{\epsilon }_i-\frac{m}{r+1}\sum _{j=1}^{r+1}{\epsilon }_j\text{.}$$ Using (4.2) to compute ${\nabla }_m\left(S\right)$ we get $$\begin{array}{ccc}{\nabla }_m\left(S\right)& =& ⟨{\sigma }^{\left(m\right)},{\sigma }^{\left(m\right)}+2\rho ⟩-⟨{\sigma }^{(m-1)},{\sigma }^{(m-1)}+2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩\\ & =& \left(\sum _{i\ne k}{\left({\sigma }_i^{(m-1)}\right)}^2\right)+{({\sigma }_k^{(m-1)}+1)}^2-\frac{m^2}{r+1}\\ & & +\left(\sum _{i\ne k}2{\sigma }_i^{(m-1)}{\rho }_i\right)+2({\sigma }_k^{(m-1)}+1){\rho }_k\\ & & -\left(\sum _{i\ne k}{\left({\sigma }_i^{(m-1)}\right)}^2\right)-{\left({\sigma }_k^{(m-1)}\right)}^2+\frac{{(m-1)}^2}{r+1}\\ & & -\left(\sum _{i\ne k}2{\sigma }_i^{(m-1)}{\rho }_i\right)-2{\sigma }_k^{(m-1)}{\rho }_k-1+\frac{1}{r+1}-2{\rho }_1\\ & =& 2{\sigma }_k^{(m-1)}+2({\rho }_k-{\rho }_1)+\frac{-2m+2}{r+1}\text{.}\end{array}$$ The formula for ${\nabla }_m\left(S\right)$ follows since $2({\rho }_k-{\rho }_1)=(r-(2k-1))-(r-1)=-2k-2=2(-k+1)\text{.}$ The formula for ${\diamond }_{m-1}(S,S)$ now follows easily since $$\begin{array}{ccc}{\diamond }_{m-1}(S,S)& =& \frac{1}{2}({\nabla }_m\left(S\right)-{\nabla }_{m-1}\left(S\right))+\frac{1}{r+1}\\ & =& ({\sigma }_k^{(m-1)}-k+1)+\frac{-m+1}{r+1}-({\sigma }_l^{(m-2)}-l+1)\\ & & -\frac{-(m-1)+1}{r+1}+\frac{1}{r+1}\\ & =& {\sigma }_k^{\left(m\right)}-k-{\sigma }_l^{(m-1)}+l-\frac{1}{r+1}+\frac{1}{r+1}\text{.}\end{array}$$ $\square$

Remark. In terms of standard tableaux, the value of ${\diamond }_i(S,S)$ is the "axial distance" between the box containing $i$ and the box containing $i+1\text{.}$ $$\begin{array}{c} i i+1 \end{array}$$

One can choose the identification (Section 1) of the centralizer algebras ${𝒵}_m$ with the path algebras corresponding to the Bratteli diagram $Y\left(r\right)$ so that the matrices ${Ř}_i$ are given by the formula $${Ř}_i=\sum _{(S,T)={\Omega }_{i-1}^{i+1}}{\left({Ř}_i\right)}_{ST}{E}_{ST},$$ where for each $S\in {𝒯}^m$ we have $$q^{1/(r+1)}{\left({Ř}_i\right)}_{SS}=\frac{q^{{\diamond }_i(S,S)}}{\left[{\diamond }_i(S,S)\right]}$$ and for each pair $(S,T)\in {\Omega }_{i-1}^{i+1}$ such that $S\ne T$ we have $$q^{1/(r+1)}{\left({Ř}_i\right)}_{ST}=\frac{\sqrt{[{\diamond }_i(S,S)-1][{\diamond }_i(S,S)+1]}}{\left[\left|{\diamond }_i(S,S)\right|\right]}$$

		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 Corollary (1.5) that $${Ř}_i=\sum _{(S,T)\in {\Omega }_{m-2}^m}{\left({Ř}_i\right)}_{ST}{E}_{ST},$$ 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 $${Ř}_{m-1}^{-1}=R_m^{-1}{Ř}_{m-1}{D}_{m-1}\text{.}$$ The relation $(q^{1/(r+1)}{Ř}_i-q)(q^{1/(r+1)}{Ř}_i+q^{-1})=0$ from Proposition (4.4) can be written in the form $q^{1/(r+1)}{Ř}_{m-1}-q^{-1/(r+1)}{Ř}_{m-1}^{-1}=(q-q^{-1})$ or, equivalently, in the form $$\begin{array}{cc}{q}^{1/(r+1)}{Ř}_{m-1}-q^{-2/(r+1)}{D}_m^{-1}{q}^{1/(r+1)}{Ř}_{m-1}{D}_{m-1}=(q-q^{-1})\text{.}& \text{(4.9)}\end{array}$$ Let $S\in {𝒯}^m$ and view (4.9) as an equation in the path algebra. Since the matrices $D_m$ and $D_{m-1}$ are diagonal, taking the $E_{SS}\text{-entry}$ of this equation yields $$q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}-q^{-2/(r+1)}{\left(D_m^{-1}\right)}_{SS}{q}^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}{\left(D_{m-1}\right)}_{SS}=(q-q^{-1}){\delta }_{SS},$$ or, equivalently, $$\begin{array}{cc}(1-q^{-2/(r+1)}{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{SS})q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}=(q-q^{-1})\text{.}& \text{(4.10)}\end{array}$$ Since the right hand side of this equation is nonzero, the left hand side is also nonzero and we may write $$q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}=\frac{(q-q^{-1}){\delta }_{SS}}{q-q^{-2/(r+1)}{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{SS}}\text{.}$$ Plugging in the following $$\begin{array}{ccc}\frac{q-q^{-1}}{q-q^{-2/(r+1)}{\left(D_m^{-1}\right)}_{SS}{\left(D_{m-1}\right)}_{SS}}& =& \frac{q-q^{-1}}{1-q^{-2{\diamond }_{m-1}(S,S)}}\\ & =& \frac{q^{{\diamond }_{m-1}(S,S)}(q-q^{-1})}{q^{{\diamond }_{m-1}(S,S)}-q^{-{\diamond }_{m-1}(S,S)}}\\ & =& \frac{q^{{\diamond }_{m-1}(S,S)}}{\left[{\diamond }_{m-1}(S,S)\right]}\end{array}$$ gives the first formula in Theorem (4.8). Now let us prove the second formula in Theorem (4.8). 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 the remark in (4.1), $T$ is unique. It follows that $$\begin{array}{cc}{q}^{2/(r+1)}{\left({Ř}_{m-1}^2\right)}_{SS}=q^{2/(r+1)}{\left({\left({Ř}_{m-1}\right)}_{SS}\right)}^2+q^{2/(r+1)}{\left({Ř}_{m-1}\right)}_{ST}{\left({Ř}_{m-1}\right)}_{TS}\text{.}& \text{(4.11)}\end{array}$$ On the other hand, the relation $(q^{1/(r+1)}{Ř}_{m-1}-q)(q^{1/(r+1)}{Ř}_{m-1}+q^{-1})=0$ from Proposition (4.4) can be written in the form $q^{2/(r+1)}{Ř}_{m-1}^2-1=q^{1/(r+1)}{Ř}_{m-1}(q-q^{-1}),$ giving that $$\begin{array}{cc}{q}^{2/(r+1)}{\left({Ř}_{m-1}^2\right)}_{SS}=(q-q^{-1})q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}+1& \text{(4.12)}\end{array}$$ Equating (4.11) and (4.12) and using the formula for $q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{SS}$ gives $$\begin{array}{ccc}\multicolumn{3}{c}{{\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{ST}{\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{TS}}\\ & =& 1+(q-q^{-1}){\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{SS}-{\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{SS}^2\\ & =& (q-{\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{SS})(q^{-1}+{\left(q^{1/(r+1)}{Ř}_{m-1}\right)}_{SS})\\ & =& (q-\frac{q^{{\diamond }_i(S,S)}}{\left[{\diamond }_i(S,S)\right]})(q^{-1}+\frac{q^{{\diamond }_i(S,S)}}{\left[{\diamond }_i(S,S)\right]})\\ & =& \frac{q(q^{{\diamond }_i(S,S)}-q^{-{\diamond }_i(S,S)})-q^{{\diamond }_i(S,S)}(q-q^{-1})}{q^{{\diamond }_i(S,S)}-q^{-{\diamond }_i(S,S)}}\\ & & ·\frac{q^{-1}(q^{{\diamond }_i(S,S)}-q^{-{\diamond }_i(S,S)})+q^{{\diamond }_i(S,S)}(q-q^{-1})}{q^{{\diamond }_i(S,S)}-q^{-{\diamond }_i(S,S)}}\\ & =& \frac{(q^{{\diamond }_i(S,S)+1}-q^{-{\diamond }_i(S,S)-1})(q^{-{\diamond }_i(S,S)+1}-q^{{\diamond }_i(S,S)-1})}{q^{{\diamond }_i(S,S)}-q^{-{\diamond }_i(S,S)}}\\ & =& \frac{[{\diamond }_i(S,S)+1][{\diamond }_i(S,S)-1]}{{\left[{\diamond }_i(S,S)\right]}^2}\end{array}$$ 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. $\square$

Remark. If $(S,T)\in {\Omega }_{i-1}^{i+1}$ such that $S\ne T$ then ${\diamond }_i(S,S)=-{\diamond }_i(T,T)\text{.}$ Thus, the formula for $q^{1/(r+1)}{\left({Ř}_{m-1}\right)}_{ST}$ given in Theorem (4.8) is actually symmetric in $S$ and $T\text{.}$

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 (4.8) 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 all 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 the previous lemma, ${\nabla }_m\left(S\right)=2({\tau }_k^{(m-1)}-k+1-(m-1)/(r+1))$ if ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $k\text{th}$ row of ${\tau }^{(m-1)}\text{.}$ Since $${\tau }_1^{(m-1)}\ge \cdots \ge {\tau }_k^{(m-1)}\ge \cdots \ge {\tau }_r^{(m-1)}$$ it follows that $$\begin{array}{cc}{\tau }_1^{(m-1)}-\frac{m-1}{r+1}& >\cdots >{\tau }_k^{(m-1)}-k+1-\frac{m-1}{r+1}\\ & >\cdots >{\tau }_r^{(m-1)}-r+1-\frac{m-1}{r+1}\text{.}\end{array}$$ $\square$

[RWe1992]. 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 $M,N\in {𝒯}^m$ are of the form M $M=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)},\text{shp}\left(S\prime \right),\text{shp}\left(S\right))$ and $N=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)},\text{shp}\left(T\prime \right),\text{shp}\left(S\right))\text{.}$

		Proof.
	(1) Let $T\prime \in {𝒯}^{m-1}\text{.}$ It follows from the formula for ${Ř}_{m-1}$ that $$E_{T\prime T\prime }{Ř}_{m-1}{E}_{T\prime T\prime }=\sum _{S\in {\left(T\prime \right)}^{+}}{\left({Ř}_{m-1}\right)}_{SS}{E}_{SS}\text{.}$$ The identity (1) follows if we show that the values ${\left({Ř}_{m-1}\right)}_{SS}$ are all different as $S$ runs over all tableaux in ${\left(T\prime \right)}^{+}\text{.}$ Since ${\diamond }_{m-1}(S,S)=\frac{1}{2}({\nabla }_m\left(S\right)-{\nabla }_{m-1}\left(S\right))+1/(r+1)=\frac{1}{2}({\nabla }_m\left(S\right)-{\nabla }_{m-1}\left(T\prime \right))+1/(r+1)$ it follows that the values ${\left({Ř}_{m-1}\right)}_{SS}$ are all different as $S$ runs over all tableaux in ${\left(T\prime \right)}^{+}$ if and only if the values ${\nabla }_m\left(S\right)$ are all different as $S$ runs over all tableaux in ${\left(T\prime \right)}^{+}\text{.}$ Statement (1) now follows from Lemma (4.13). (2) follows from the definition (1.3) of the embedding of path algebras. (3) We must show two things: (a) For each possible choice of $M$ and $N$ the formula determines $E_{ST}\text{.}$ (b) There exist tableaux $M$ and $N$ 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 $N=({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)},\text{shp}\left(T\prime \right),\text{shp}\left(S\right))\text{.}$ Suppose that $M$ and $N$ are given. Since $\text{shp}\left(S\prime \right)\ne \text{shp}\left(T\prime \right),$ it follows from (4.1) that $M$ and $N$ are the unique extensions of $M\prime$ and $N\prime$ respectively, such that $\text{shp}\left(M\right)=\text{shp}\left(N\right)=\text{shp}\left(S\right)\text{.}$ By Theorem (4.8) we know that the values ${\left({Ř}_{m-1}\right)}_{MN}$ are nonzero. It follows that $$\begin{array}{ccc}\frac{1}{{\left({Ř}_{m-1}\right)}_{MN}}{E}_{S\prime M\prime }{Ř}_{m-1}{E}_{N\prime T\prime }& =& \frac{1}{{\left({Ř}_{m-1}\right)}_{MN}}{E}_{S\prime M\prime }\sum _{(U,V)\in {\Omega }^m}{\left({Ř}_{m-1}\right)}_{UV}{E}_{UV}{E}_{N\prime T\prime }\\ & =& \frac{1}{{\left({Ř}_{m-1}\right)}_{MN}}{E}_{S\prime M\prime }{\left({Ř}_{m-1}\right)}_{MN}{E}_{NM}{E}_{N\prime T\prime }=E_{ST},\end{array}$$ proving (a). To see that (b) is true we reason as follows. Suppose that $\text{shp}\left(S\prime \right)$ is a partition that is the same as $\text{shp}\left(S\right)$ except that there is a box missing from the $k\text{th}$ row. Suppose that $\text{shp}\left(T\prime \right)$ is a partition that is the same as $\text{shp}\left(S\right)$ except that there is a box missing from the $l\text{th}$ row. Since $\text{shp}\left(S\prime \right)\ne \text{shp}\left(T\prime \right)$ we know that $k\ne l\text{.}$ Then there is a unique partition ${\mu }^{(m-2)}$ that is the same as $\text{shp}\left(S\right)$ except that there is a box missing from the $l\text{th}$ row and a box missing from the $k\text{th}$ row. The partition ${\mu }^{(m-2)}$ is uniquely determined by $S$ and $T,$ and $M$ and $N$ can be determined by fixing some tableau $({\mu }^{\left(0\right)},\dots ,{\mu }^{(m-2)})\in {𝒯}^{m-2}$ of shape ${\mu }^{(m-2)}\text{.}$ $\square$

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

		Proof.
	It follows from the identification of the centralizer algebras ${𝒵}_m$ with the path algebras that the matrix units $E_{ST},$ $(S,T)\in {\Omega }^m$ span the centralizer algebras ${𝒵}_m\text{.}$ In view of Theorem (4.14), the matrix units $E_{ST},$ $(S,T)\in {\Omega }^m$ can be written in terms of the ${Ř}_i$ matrices. The statement follows. $\square$

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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