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

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 $\lambda =\left({\lambda }_{1}\ge {\lambda }_{2}\ge \cdots \ge {\lambda }_{t}\ge 0\right)$ 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, $(5,3,3,1)=$ 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 $\lambda \in {\stackrel{ˆ}{Y}}_{m}$ of $Y$ which are on level $m$ are the partitions of $m\text{;}$ $Yˆm= {λ⊢m}.$ A partition $\lambda \in {\stackrel{ˆ}{Y}}_{m}$ is connected by an edge to a partition $\mu \in {\stackrel{ˆ}{Y}}_{m+1}$ 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 $T=\left({\tau }^{\left(0\right)},{\tau }^{\left(1\right)},\dots ,{\tau }^{\left(m\right)}\right)\in {𝒯}^{\lambda }$ 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 $Yˆm(r)= {λ⊢m | l(λ)≤r}.$ A partition $\lambda \in {\stackrel{ˆ}{Y}}_{m}\left(r\right)$ is connected by an edge to a partition $\mu \in {\stackrel{ˆ}{Y}}_{m+1}\left(r\right)$ 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 $\lambda \in \stackrel{ˆ}{Y}\left(r\right),$ 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 $S=\left({\sigma }^{\left(m-2\right)},{\sigma }^{\left(m-1\right)},{\sigma }^{\left(m\right)}\right)\in {𝒯}_{m-2}^{m}\text{.}$ Suppose that ${\sigma }^{\left(m-1\right)}$ is obtained by adding a box to the $k\text{th}$ row of ${\sigma }^{\left(m-2\right)}$ and that ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{\left(m-1\right)}\text{.}$ Now suppose that $T=\left({\sigma }^{\left(m-2\right)},{\tau }^{\left(m-1\right)},{\sigma }^{\left(m\right)}\right)$ is such that $\left(S,T\right)\in {\Omega }_{m-2}^{m}\text{.}$ If $k=l$ then we must have that $∅ Figure 1$ ${\tau }^{\left(m-1\right)}={\sigma }^{\left(m-1\right)}\text{.}$ If $k\ne l$ then either ${\tau }^{\left(m-1\right)}={\sigma }^{\left(m-1\right)},$ or ${\tau }^{\left(m-1\right)}$ is the shape obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{\left(m-2\right)}\text{.}$ Thus, $there is at most one T≠S such that (S,T)∈ Ωm-2m. (4.1)$

The Centralizer Algebras ${𝒵}_{m}$

For the remainder of this section fix $𝔘={𝔘}_{h}\left(𝔰𝔩\left(r+1\right)\right)\text{.}$ 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 $𝔥* = { λ1ε1+⋯+ λr+1εr+1 | ∑i=1r+1 λi=0 } , αi = εi-εi+1, 1≤i≤r, ωi = ε1+⋯+εi- ir+1 ∑i=1r+1 εj,1≤i≤r, 2ρ = ∑i2ρiεi=r ε1+(r-2) ε2+⋯-(r-2) εr-rεr+1.$ The finite dimensional irreducible modules ${\Lambda }_{\lambda }$ of ${𝔘}_{h}\left(𝔰𝔩\left(r+1\right)\right)$ are indexed by the dominant integral weights, $𝔘ˆ= { λ=λ1ε1+⋯+ λrεr- |λ|r+1 ∑j=1r+1 εj | λi ∈ℤ,λ1≥⋯≥ λr≥0 } ,$ where $|\lambda |={\lambda }_{1}+\cdots +{\lambda }_{r}\text{.}$ It is sometimes helpful to identify each dominant integral weight $\lambda$ with the partition $\lambda =\left({\lambda }_{1},\dots ,{\lambda }_{r}\right)\text{.}$ Note that all partitions in $\stackrel{ˆ}{𝔘}$ have at most $r$ rows. It will also be helpful to note that, if $\lambda ={\lambda }_{1}{\epsilon }_{1}+\cdots +{\lambda }_{r}{\epsilon }_{r}-\left(|\lambda |/\left(r+1\right)\right)\sum _{j}{\epsilon }_{j}\in \stackrel{ˆ}{𝔘}$ then $⟨λ,λ+2ρ⟩= ∑i=1rλi2- |λ|2r+1+ ∑i=1r2ρiλi. (4.2)$

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 $Λλ⊗V≅ ⨁μ∈λ+ Λμ, (4.3)$ where the sum is over all partitions $\mu \in \stackrel{ˆ}{𝔘}$ 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 $𝔘={𝔘}_{h}\left(𝔰𝔩\left(r+1\right)\right)\text{-module}$ indexed by the fundamental weight ${\omega }_{1}\text{.}$ The matrices ${Ř}_{i}\in {\text{End}}_{𝔘}\left({V}^{\otimes m}\right)$ satisfy the relations $ŘiŘj = ŘjŘi, |i-j|>1, ŘiŘi+1Ři = Ři+1Ři Ři+1, 1≤i≤m-2, (q(1/r+1)Ři-q) (q(1/r+1)Ři-q-1) = 0, 1≤i≤m-1.$

 Proof. The first two relations follow from Proposition (2.18). From (4.3), we have $V⊗V≅Λ(12) ⊕Λ(2)= Λ2ω1⊕Λω2.$ Use (4.2) to show that $⟨ω2,ω2+2ρ⟩ = 2-(4/(r+1))+ 2ρ1+2ρ2=2r- 4/(r+1), ⟨2ω1,2ω1+2ρ⟩ = 4-(4/(r+1))+ 4ρ1=4+2r-4/ (r+1), ⟨ω1,ω1+2ρ⟩ = 1-(1/(r+1))+ 2ρ1=r+1-1/ (r+1).$ It follows that $q(1/2)⟨ω2,ω2+2ρ⟩-⟨ω1,ω1+2ρ⟩ = q-1-(1/(r+1)), q(1/2)⟨2ω1,2ω1+2ρ⟩-⟨ω1,ω1+2ρ⟩ = q1-(1/(r+1)).$ 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 ${E}_{ST},$ $\left(S,T\right)\in {\Omega }^{m}$ of matrix units.

For each tableau $S=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)}\right)\in {𝒯}^{m}$ define $∇i(S)= ⟨σ(i),σ(i)+2ρ⟩- ⟨σ(i-1),σ(i-1)+2ρ⟩- ⟨ω1,ω1+2ρ⟩. (4.5)$ Let $\left(S,T\right)$ be a pair of tableaux $S= ( σ(0),…, σ(i-1), σ(i), σ(i+1),…, σ(m) ) ,$ and $T= ( σ(0),…, σ(i-1), τ(i), σ(i+1),…, σ(m) ) ,$ 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 $\left(S,T\right)\in {\Omega }_{i-1}^{i+1}\text{.}$ Define $⋄i(S,T)=12 (∇i+1(S)-∇i(T)) +1r+1. (4.6)$ 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 $Dm=∑S∈𝒯m (Dm)SSESS, where (Dm)SS= q∇m(S),$ and $q-2⋄m-1(S,T) =q-(2/r+1) (Dm-1)SS (Dm-1)TT.$ The first of these formulas is a consequence of Corollary (2.25).

Let $S=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)}\right)$ be a tableau in the Bratteli diagram $Y\left(r\right)\text{.}$ Then $∇m(S) = 2(σk(m-1)-k+1) +-2m+2r+1, ⋄m-1(S,S) = σk(m)- σl(m-1)-k+l,$ where ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $k\text{th}$ row of ${\sigma }^{\left(m-1\right)}$ and ${\sigma }^{\left(m-1\right)}$ is obtained by adding a box to the $l\text{th}$ row of ${\sigma }^{\left(m-2\right)}\text{.}$

 Proof. Let $S=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)}\right)$ be a tableau in the Bratteli diagram $Y\left(r\right)\text{.}$ Then, since ${\sigma }^{\left(m\right)}$ differs from ${\sigma }^{\left(m-1\right)}$ by adding a box in the $k\text{th}$ row we have $σ(m-1)= ∑i=1r σi(m-1)εi -m-1r+1 ∑j=1r+1 εj,$ and $σ(m)= (σk(m-1)+1) εk+∑1≤i≤ri≠k σi(m)εi- mr+1 ∑j=1r+1εj.$ Using (4.2) to compute ${\nabla }_{m}\left(S\right)$ we get $∇m(S) = ⟨σ(m),σ(m)+2ρ⟩- ⟨σ(m-1),σ(m-1)+2ρ⟩- ⟨ε1,ε1+2ρ⟩ = (∑i≠k(σi(m-1))2) +(σk(m-1)+1)2 -m2r+1 +(∑i≠k2σi(m-1)ρi) +2(σk(m-1)+1) ρk -(∑i≠k(σi(m-1))2) -(σk(m-1))2+ (m-1)2r+1 -(∑i≠k2σi(m-1)ρi) -2σk(m-1)ρk -1+1r+1-2ρ1 = 2σk(m-1)+2 (ρk-ρ1)+ -2m+2r+1.$ The formula for ${\nabla }_{m}\left(S\right)$ follows since $2\left({\rho }_{k}-{\rho }_{1}\right)=\left(r-\left(2k-1\right)\right)-\left(r-1\right)=-2k-2=2\left(-k+1\right)\text{.}$ The formula for ${\diamond }_{m-1}\left(S,S\right)$ now follows easily since $⋄m-1(S,S) = 12 (∇m(S)-∇m-1(S)) +1r+1 = (σk(m-1)-k+1)+ -m+1r+1- (σl(m-2)-l+1) --(m-1)+1r+1 +1r+1 = σk(m)-k- σl(m-1)+l- 1r+1+1r+1.$ $\square$

Remark. In terms of standard tableaux, the value of ${\diamond }_{i}\left(S,S\right)$ is the "axial distance" between the box containing $i$ and the box containing $i+1\text{.}$ $i i+1$

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 $Ři=∑(S,T)=Ωi-1i+1 (Ři)ST EST,$ where for each $S\in {𝒯}^{m}$ we have $q1/(r+1) (Ři)SS= q⋄i(S,S) [⋄i(S,S)]$ and for each pair $\left(S,T\right)\in {\Omega }_{i-1}^{i+1}$ such that $S\ne T$ we have $q1/(r+1) (Ři)ST= [⋄i(S,S)-1] [⋄i(S,S)+1] [|⋄i(S,S)|]$

 Proof. Since ${Ř}_{i}\in {𝒵}_{i}={\text{End}}_{𝔘}\left({V}^{\otimes \left(i+1\right)}\right)$ commutes with all elements of ${𝒵}_{i-1}={\text{End}}_{𝔘}\left({V}^{\otimes \left(i-1\right)}\right)$ it follows from Corollary (1.5) that $Ři= ∑(S,T)∈Ωm-2m (Ři)STEST,$ 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 $Řm-1-1= Rm-1 Řm-1 Dm-1.$ The relation $\left({q}^{1/\left(r+1\right)}{Ř}_{i}-q\right)\left({q}^{1/\left(r+1\right)}{Ř}_{i}+{q}^{-1}\right)=0$ from Proposition (4.4) can be written in the form ${q}^{1/\left(r+1\right)}{Ř}_{m-1}-{q}^{-1/\left(r+1\right)}{Ř}_{m-1}^{-1}=\left(q-{q}^{-1}\right)$ or, equivalently, in the form $q1/(r+1) Řm-1- q-2/(r+1) Dm-1 q1/(r+1) Řm-1 Dm-1= (q-q-1). (4.9)$ 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 $q1/(r+1) (Řm-1)SS- q-2/(r+1) (Dm-1)SS q1/(r+1) (Řm-1)SS (Dm-1)SS= (q-q-1) δSS,$ or, equivalently, $( 1-q-2/(r+1) (Dm-1)SS (Dm-1)SS ) q1/(r+1) (Řm-1)SS =(q-q-1). (4.10)$ Since the right hand side of this equation is nonzero, the left hand side is also nonzero and we may write $q1/(r+1) (Řm-1)SS= (q-q-1)δSS q-q-2/(r+1) (Dm-1)SS (Dm-1)SS .$ Plugging in the following $q-q-1 q-q-2/(r+1) (Dm-1)SS (Dm-1)SS = q-q-1 1-q-2⋄m-1(S,S) = q⋄m-1(S,S) (q-q-1) q⋄m-1(S,S)- q-⋄m-1(S,S) = q⋄m-1(S,S) [⋄m-1(S,S)]$ 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 $\left(S,T\right)\in {\Omega }_{m-2}^{m}$ and $T\ne S\text{.}$ By the remark in (4.1), $T$ is unique. It follows that $q2/(r+1) (Řm-12)SS =q2/(r+1) ((Řm-1)SS)2 +q2/(r+1) (Řm-1)ST (Řm-1)TS. (4.11)$ On the other hand, the relation $\left({q}^{1/\left(r+1\right)}{Ř}_{m-1}-q\right)\left({q}^{1/\left(r+1\right)}{Ř}_{m-1}+{q}^{-1}\right)=0$ from Proposition (4.4) can be written in the form ${q}^{2/\left(r+1\right)}{Ř}_{m-1}^{2}-1={q}^{1/\left(r+1\right)}{Ř}_{m-1}\left(q-{q}^{-1}\right),$ giving that $q2/(r+1) (Řm-12)SS =(q-q-1) q1/(r+1) (Řm-1)SS+1 (4.12)$ Equating (4.11) and (4.12) and using the formula for ${q}^{1/\left(r+1\right)}{\left({Ř}_{m-1}\right)}_{SS}$ gives $(q1/(r+1)Řm-1)ST (q1/(r+1)Řm-1)TS = 1+(q-q-1) (q1/(r+1)Řm-1)SS- (q1/(r+1)Řm-1)SS2 = (q-(q1/(r+1)Řm-1)SS) (q-1+(q1/(r+1)Řm-1)SS) = (q-q⋄i(S,S)[⋄i(S,S)]) (q-1+q⋄i(S,S)[⋄i(S,S)]) = q ( q⋄i(S,S)- q-⋄i(S,S) ) -q⋄i(S,S) (q-q-1) q⋄i(S,S)- q-⋄i(S,S) · q-1 ( q⋄i(S,S)- q-⋄i(S,S) ) +q⋄i(S,S) (q-q-1) q⋄i(S,S)- q-⋄i(S,S) = ( q⋄i(S,S)+1- q-⋄i(S,S)-1 ) ( q-⋄i(S,S)+1 -q⋄i(S,S)-1 ) q⋄i(S,S)- q-⋄i(S,S) = [⋄i(S,S)+1] [⋄i(S,S)-1] [⋄i(S,S)]2$ 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 $\left(S,T\right)\in {\Omega }_{i-1}^{i+1}$ such that $S\ne T$ then ${\diamond }_{i}\left(S,S\right)=-{\diamond }_{i}\left(T,T\right)\text{.}$ Thus, the formula for ${q}^{1/\left(r+1\right)}{\left({Ř}_{m-1}\right)}_{ST}$ given in Theorem (4.8) is actually symmetric in $S$ and $T\text{.}$

Matrix Units

Given a tableau $T=\left({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m\right)}\right)\in {𝒯}^{m}$ let $T\prime$ denote the tableau $T\prime =\left({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m-1\right)}\right)\in {𝒯}^{m-1}\text{.}$ Let ${\left(T\prime \right)}^{+}$ denote the set of all extensions of $T\prime \text{;}$ $(T′)+= { S∈𝒯m | S′=T′ } .$ Given tableaux $S=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)}\right)$ and $T=\left({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m\right)}\right)$ in ${𝒯}^{m}$ let ${\left({Ř}_{m-1}\right)}_{ST}$ be the constant given by Theorem (4.8) in the case that $\left(\left({\sigma }^{\left(m-2\right)},{\sigma }^{\left(m-1\right)},{\sigma }^{\left(m\right)}\right),\left({\tau }^{\left(m-2\right)},{\tau }^{\left(m-1\right)},{\tau }^{\left(m\right)}\right)\right)\in {\Omega }_{m-2}^{m}$ and let ${\left({Ř}_{m-1}\right)}_{ST}=0$ otherwise.

Let $T\prime =\left({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m-1\right)}\right)\in {𝒯}^{m-1}$ 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 $S=\left({\tau }^{\left(0\right)},\dots ,{\tau }^{\left(m-1\right)},{\sigma }^{\left(m\right)}\right)\in {\left(T\prime \right)}^{+}\text{.}$ By the previous lemma, ${\nabla }_{m}\left(S\right)=2\left({\tau }_{k}^{\left(m-1\right)}-k+1-\left(m-1\right)/\left(r+1\right)\right)$ if ${\sigma }^{\left(m\right)}$ is obtained by adding a box to the $k\text{th}$ row of ${\tau }^{\left(m-1\right)}\text{.}$ Since $τ1(m-1)≥⋯≥ τk(m-1)≥⋯≥ τr(m-1)$ it follows that $τ1(m-1)- m-1r+1 >⋯>τk(m-1) -k+1-m-1r+1 >⋯>τr(m-1) -r+1-m-1r+1.$ $\square$

[RWe1992]. The matrix units ${E}_{ST}\in {𝒵}_{m},$ $\left(S,T\right)\in {\Omega }^{m}$ 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 ${E}_{TT}=\prod _{S\in {𝒯}^{m},S\ne T,S\prime =T\prime }\left({E}_{T\prime T\prime }{Ř}_{m-1}{E}_{T\prime T\prime }-{\left({Ř}_{m-1}\right)}_{SS}{E}_{T\prime T\prime }\right)/\left({\left({Ř}_{m-1}\right)}_{TT}-{\left({Ř}_{m-1}\right)}_{SS}\right)$ (2) Let $\left(S,T\right)\in {\Omega }^{m}\text{.}$ 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 $\left(S,T\right)\in {\Omega }^{m}\text{.}$ If $\text{shp}\left(S\prime \right)\ne \text{shp}\left(T\prime \right)$ then $EST= 1(Řm-1)MN ES′M′ Řm-1 EN′T′ ETT,$ where $M,N\in {𝒯}^{m}$ are of the form M $M=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-2\right)},\text{shp}\left(S\prime \right),\text{shp}\left(S\right)\right)$ and $N=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-2\right)},\text{shp}\left(T\prime \right),\text{shp}\left(S\right)\right)\text{.}$

 Proof. (1) Let $T\prime \in {𝒯}^{m-1}\text{.}$ It follows from the formula for ${Ř}_{m-1}$ that $ET′T′ Řm-1 ET′T′ =∑S∈(T′)+ (Řm-1)SS ESS.$ 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 ${\diamond }_{m-1}\left(S,S\right)=\frac{1}{2}\left({\nabla }_{m}\left(S\right)-{\nabla }_{m-1}\left(S\right)\right)+1/\left(r+1\right)=\frac{1}{2}\left({\nabla }_{m}\left(S\right)-{\nabla }_{m-1}\left(T\prime \right)\right)+1/\left(r+1\right)$ 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 $M=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-2\right)},\text{shp}\left(S\prime \right),\text{shp}\left(S\right)\right)$ and $N=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-2\right)},\text{shp}\left(T\prime \right),\text{shp}\left(S\right)\right)\text{.}$ 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 $1(Řm-1)MN ES′M′Řm-1 EN′T′ = 1(Řm-1)MN ES′M′ ∑(U,V)∈Ωm (Řm-1)UV EUVEN′T′ = 1(Řm-1)MN ES′M′ (Řm-1)MN ENM EN′T′=EST,$ 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 ${\mu }^{\left(m-2\right)}$ 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 ${\mu }^{\left(m-2\right)}$ is uniquely determined by $S$ and $T,$ and $M$ and $N$ can be determined by fixing some tableau $\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-2\right)}\right)\in {𝒯}^{m-2}$ of shape ${\mu }^{\left(m-2\right)}\text{.}$ $\square$

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 ${E}_{ST},$ $\left(S,T\right)\in {\Omega }^{m}$ span the centralizer algebras ${𝒵}_{m}\text{.}$ In view of Theorem (4.14), the matrix units ${E}_{ST},$ $\left(S,T\right)\in {\Omega }^{m}$ 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.