## 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

## Ribbon Hopf Algebras, Conditional Expectations, and Markov Traces on Centralizer Algebras

Let $\left(𝔘,ℛ,v\right)$ be a ribbon Hopf algebra. Let $W$ be a finite dimensional $𝔘\text{-module.}$ Let $\left\{{w}_{i}\right\}$ be a basis of $W$ and let $\left\{{w}^{i}\right\}$ be the dual basis in ${W}^{*}\text{.}$ Let $⟨ , ⟩$ be the ordinary pairing between $W$ and ${W}^{*}$ so that $⟨\varphi ,w⟩=⟨w,\varphi ⟩=\varphi \left(w\right)$ for elements $\varphi \in {W}^{*}$ and $w\in W\text{.}$ Using this notation, the action of an element $b\in \text{End}\left(W\right)$ can be given in the form $bwi=∑j ⟨bwi,wj⟩ wj.$ The Hopf algebra $𝔘$ acts on ${W}^{*}$ via the antipode $S$ in the standard way, $aϕ=∑j ⟨aϕ,wj⟩ wj=∑j ⟨ϕ,S(a)wj⟩ wj,$ for all $a\in 𝔘$ and $\varphi \in {W}^{*}\text{.}$ We shall often use the relation $⟨a\varphi ,{w}_{j}⟩=⟨\varphi ,S\left(a\right){w}_{j}⟩,$ which follows from this definition. The material in this section is very much motivated by [Wen1988-2, Section 1] and [Wen1990].

Quantum Trace and Quantum Dimension

Define the quantum trace of an element $b\in {\text{End}}_{𝔘}\left(W\right)$ by $trq(b)=Tr (v-1ub)= ∑i ⟨v-1ubwi,wi⟩,$ where the sum is over the basis ${W}_{i}$ of $W\text{.}$ If $a,b\in {\text{End}}_{𝔘}\left(W\right)$ then both $a$ and $b$ commute with ${v}^{-1}u\text{;}$ thus, ${\text{tr}}_{q}\left(ab\right)={\text{tr}}_{q}\left(ba\right)$ for all $a,b\in {\text{End}}_{𝔘}\left(W\right)\text{.}$ Define the quantum dimension of the $𝔘\text{-module}$ $W$ to be $dimq(W)= trq(id),$ where id denotes the identity operator on $W\text{.}$

Let $\stackrel{ˆ}{W}$ be the subset of $\stackrel{ˆ}{𝔘}$ that indexes the irreducible modules ${\Lambda }_{\mu }$ appearing in the decomposition of $W\text{.}$ As a trace on ${\text{End}}_{𝔘}\left(W\right),$ the quantum trace ${\text{tr}}_{q}$ has weights given by $wt(μ)=dimq (Λμ), μ∈Wˆ,$ where ${\Lambda }_{\mu }$ are the irreducible $U\text{-modules}$ appearing in the decomposition of $W\text{.}$

 Proof. By the double centralizer theory we know that as ${\text{End}}_{𝔘}\left(W\right)\otimes 𝔘$ modules, $W\cong \underset{\lambda }{⨁}{𝒵}^{\lambda }\otimes {\Lambda }_{\lambda },$ where ${𝒵}^{\lambda }$ are irreducible modules for ${\text{End}}_{𝔘}\left(W\right)$ and ${\Lambda }_{\lambda }$ are irreducible modules for $𝔘\text{.}$ By taking traces on both sides of this isomorphism we have $trq(b)=Tr (v-1ub)= ∑λ∈Wˆ ηλ(b)χλ (v-1u)= ∑λ∈Wˆ ηλ(b) dimq(Λλ),$ where ${\eta }_{\lambda }$ is the irreducible character of ${\text{End}}_{𝔘}\left(W\right)$ on the module ${𝒵}^{\lambda }$ and ${\chi }^{\lambda }$ is the irreducible character of the irreducible $𝔘\text{-module}$ ${\Lambda }_{\lambda }\text{.}$ Thus the trace of a minimal idempotent ${p}_{\mu }$ in the minimal ideal corresponding to $\mu$ is $wt(μ)=trq (pμ)= ∑λ∈Wˆ ηλ(pμ) dimq(Λλ) =∑λ∈Wˆ δλμdimq (Λλ)= dimq(Λμ).$ $\square$

The Projection onto the Invariants

Let $V$ be a $𝔘\text{-module}$ and let ${V}^{*}$ be the dual module to $V\text{.}$ Let $\left\{{e}_{i}\right\}$ be a basis of $V$ and let $\left\{{e}^{i}\right\}$ be the dual basis in ${V}^{*}\text{.}$ Define $ě: V⊗V* ⟶ V⊗V* x⊗ϕ ⟼ ⟨ (dimq(V))-1 ϕ,v-1ux ⟩ ∑iei⊗ei (3.2)$ Where $⟨\varphi ,{v}^{-1}ux⟩=\varphi \left({v}^{-1}ux\right)$ denotes the evaluation of the functional $\varphi \in {V}^{*}$ at the element ${v}^{-1}ux\in V\text{.}$ It follows from (a) and (b) of the following proposition that

 (1) $ě\in {\text{End}}_{𝔘}\left(V\otimes {V}^{*}\right),$ and (2) $ě$ is the $𝔘\text{-invariant}$ projection onto the invariants in $V\otimes {V}^{*}\text{.}$

 (a) For every $g\in 𝔘$ we have $gě=ěg=\epsilon \left(g\right)ě,$ (b) ${ě}^{2}=ě\text{.}$

 Proof. (a) Let $g\in 𝔘,$ $x\in V,$ $\varphi \in {V}^{*}\text{.}$ Then, by direct computation, $gě(x⊗ϕ) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ g(∑iei⊗ei) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ Δ(g) (∑iei⊗ei) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ ∑g,ig(1) ei⊗g(2)ei = (dimq(V))-1 ⟨ϕ,v-1ux⟩ ∑i,j,k∑g ⟨g(1)ei,ej⟩ ×⟨g(2)ei,ek⟩ (ej⊗ek) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ ∑i,j,k∑g ⟨g(1)ei,ej⟩ ×⟨ei,S(g(2))ek⟩ (ej⊗ek) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ ∑j,k∑g ⟨g(1)S(g(2))ek,ej⟩ (ej⊗ek) = (dimq(V))-1 ⟨ϕ,v-1ux⟩ ∑j,k ⟨ε(g)ek,ej⟩ ej⊗ek = (dimq(V))-1 ε(g)⟨ϕ,v-1ux⟩ ∑jej⊗ej = ε(g)ě(x⊗ϕ),$ where we are using the identity $\sum _{g}{g}_{\left(1\right)}S\left({g}_{\left(2\right)}\right)=\epsilon \left(g\right)$ which follows from the definition of the antipode in a Hopf algebra. On the other hand, since $v$ is in the center of $𝔘$ and ${u}^{-1}xu={S}^{-2}\left(x\right)$ for all $x\in 𝔘,$ $ěg(x⊗ϕ) = ě(∑gg(1)x⊗g(2)ϕ) = (dimq(V))-1 ∑g,i ⟨g(2)ϕ,v-1ug(1)x⟩ ei⊗ei = (dimq(V))-1 ∑g,i ⟨ ϕ,uu-1S(g(2)) v-1ug(1)x ⟩ ei⊗ei = (dimq(V))-1 ∑g,i ⟨ ϕ,v-1uS-1 (g(2)) g(1)x ⟩ ei⊗ei = (dimq(V))-1 ∑g,i ⟨ ϕ,v-1u ε(g)x ⟩ ei⊗ei = ε(g)ě(x⊗ϕ),$ where we are using the identity $\sum _{g}{S}^{-1}\left({g}_{\left(2\right)}\right){g}_{\left(1\right)}=\epsilon \left(g\right)$ which follows from the definition of the skew antipode in a Hopf algebra. (b) This follows from the following easy computation. $ě2(x⊗ϕ) = (dimq(V))-1 ě ( ⟨ϕ,v-1ux⟩ ∑iei⊗ei ) = (dimq(V))-2 ⟨ϕ,v-1ux⟩ ∑i⟨ei,v-1uei⟩ ∑jej⊗ej = (dimq(V))-2 ⟨ϕ,v-1ux⟩ dimq(V) ∑jej⊗ej = ě(x⊗ϕ).$ $\square$

The Conditional Expectation

Let $V$ be a $𝔘\text{-module}$ and let ${V}^{*}$ be the dual $𝔘\text{-module}$ to $V\text{.}$ For each $m,$ $let 𝒵m=End𝔘 (V⊗m) and define 𝒞m+1= End𝔘 (V⊗(m-1)⊗V*) . (3.4)$ Let $\left\{{w}_{s}\right\}$ be a basis of ${V}^{\otimes \left(m-1\right)}$ and let $\left\{{w}^{s}\right\}$ be a dual basis in ${\left({V}^{\otimes \left(m-1\right)}\right)}^{*}\text{.}$ Let $\left\{{e}_{i}\right\}$ be a basis of $V$ and let $\left\{{e}^{i}\right\}$ be a dual basis in ${V}^{*}\text{.}$ Then define an operator ${\epsilon }_{m-1}:{𝒵}_{m}\to \text{End}\left({V}^{\otimes \left(m-1\right)}\right)$ by $εm-1(b) wj= (dimq(V))-1 ∑k,p ⟨ (id⊗v-1u)b (wj⊗ek), wp⊗ek ⟩ wp. (3.5)$ for each $b\in {𝒵}_{m}\text{.}$ The map ${\epsilon }_{m-1}$ is called the conditional expectation. Let $ěm=id⊗id⊗⋯⊗ id⊗ě∈𝒞m+1. (3.6)$

 (a) ${ě}_{m}b{ě}_{m}={\epsilon }_{m-1}\left(b\right){ě}_{m}={ě}_{m}{\epsilon }_{m-1}\left(b\right)$ for all $b\in {𝒵}_{m}\text{.}$ (b) ${\epsilon }_{m-1}\left({a}_{1}b{a}_{2}\right)={a}_{1}{\epsilon }_{m-1}\left(b\right){a}_{2},$ for all ${a}_{1},{a}_{2}\in {𝒵}_{m-1}$ and $b\in {𝒵}_{m}\text{.}$ In particular, ${\epsilon }_{m-1}\left(a\right)=a$ for all $a\in {𝒵}_{m-1}\text{.}$ (c) ${\epsilon }_{m-1}\left(b\right)\in {𝒵}_{m-1}$ for all $b\in {𝒵}_{m}\text{.}$

 Proof. Let $W={V}^{\otimes \left(m-1\right)}\text{.}$ Let $\left\{{w}_{t}\right\}$ and $\left\{{e}_{i}\right\}$ be bases of $W$ and $V$ respectively and let $\left\{{w}^{s}\right\}$ and $\left\{{e}^{i}\right\}$ be dual bases in ${W}^{*}$ and ${V}^{*}$ respectively. (a) Then $ěmběm (ws⊗ei⊗ej) = (dimq(V))-1 ⟨ej,v-1uei⟩ ∑kěmb (ws⊗ek⊗ek) = (dimq(V))-1 ⟨ej,v-1uei⟩ ∑k,t,l ⟨b(ws⊗ek),wt⊗el⟩ ěm(wt⊗el⊗ek) = (dimq(V))-2 ⟨ej,v-1uei⟩ ∑k,t,l,p ⟨b(ws⊗ek),wt⊗el⟩ ⟨ek,v-1uel⟩ ×(wt⊗ep⊗ep) = (dimq(V))-2 ⟨ej,v-1uei⟩ ∑k,t,p ⟨ (id⊗v-1u) b(ws⊗ek), wt⊗ek ⟩ ×(wt⊗ep⊗ep) = εm-1(b) ěm(ws⊗ei⊗ej).$ The remaining assertion follows since ${ě}_{m}$ commutes with elements of $\text{End}\left(W\right)\subseteq \text{End}\left(W\otimes V\otimes {V}^{*}\right)\text{.}$ (b) The action of ${\epsilon }_{m-1}\left({a}_{1}b{a}_{2}\right)$ on a basis element ${w}_{j}$ of $W$ satisfies $εm-1 (a1ba2) wj = (dimq(V))-1 ∑k,p ⟨ (id⊗v-1u) (a1⊗id)b (a2⊗id) (wj⊗ek), wp⊗ek ⟩ wp = (dimq(V))-1 a1∑k,p ⟨ (id⊗v-1u) b(a2wj⊗ek), wp⊗ek ⟩ wp = a1εm-1(b) a2wj.$ (c) Let $x\in 𝔘$ and let $\stackrel{‾}{x}\in \text{End}\left(W\right)$ be the endomorphism of $W$ determined by the action of $x$ on $W\text{.}$ Then, since $b\in {\text{End}}_{𝔘}\left(W\otimes V\right),$ $\stackrel{‾}{x}{ě}_{m}b{ě}_{m}={ě}_{m}\stackrel{‾}{x}b{ě}_{m}={ě}_{m}b\stackrel{‾}{x}{ě}_{m}={ě}_{m}b{ě}_{m}\stackrel{‾}{x}\text{.}$ This implies that $\stackrel{‾}{x}{\epsilon }_{m-1}\left(b\right){ě}_{m}={\epsilon }_{m-1}\left(b\right)\stackrel{‾}{x}{ě}_{m}\text{.}$ Since the map $\text{End}\left(W\right)\to \text{End}\left(W\otimes V\otimes {V}^{*}\right)$ given by $a↦a{ě}_{m}$ is injective, it follows that $\stackrel{‾}{x}{\epsilon }_{m-1}\left(b\right)={\epsilon }_{m-1}\left(b\right)\stackrel{‾}{x}\text{.}$ $\square$

Markov Traces and Framing Anomalies

Assume that $V$ is an irreducible $𝔘\text{-module}$ and let ${𝒵}_{m}={\text{End}}_{𝔘}\left({V}^{\otimes m}\right)\text{.}$ Define traces ${\text{mt}}_{m}:{𝒵}_{m}\to k$ by $mtm(b)= trq(b) dimq(V)m . (3.8)$ The traces ${\text{mt}}_{m}$ are called Markov traces.

Let $Ř$ be the element of ${𝒵}_{2}$ given in (2.16). Since $V$ is irreducible it follows from Schur's lemma that ${𝒵}_{1}\cong k\text{.}$ Thus, ${\epsilon }_{1}:{𝒵}_{2}\to k$ and $ε1(Ř)= αdimq(V), (3.9)$ for some constant $\alpha \in k\text{.}$ The constant $\alpha$ is called the framing anomaly of $Ř\text{.}$

 (a) If $a\in {𝒵}_{m-1}$ then ${\text{mt}}_{m-1}\left(a\right)={\text{mt}}_{m}\left(a\right)\text{.}$ In particular ${\text{mt}}_{m}\left(1\right)=1$ for all $m\text{.}$ (b) For each $b\in {𝒵}_{m},$ ${\text{mt}}_{m}\left(b\right)={\text{mt}}_{m-1}\left({\epsilon }_{m-1}\left(b\right)\right)\text{.}$ (c) For each $a\in {𝒵}_{m-1},$ ${\text{mt}}_{m}\left(a{Ř}_{m-1}\right)={\text{dim}}_{q}{\left(V\right)}^{-1}\alpha {\text{mt}}_{m-1}\left(a\right),$ where $\alpha$ is the framing anomaly of $Ř\text{.}$ (d) The Markov traces ${\text{mt}}_{m}$ have weights given by $wtm(λ)= dimq(Λλ) dimq(V)m ,λ∈𝒵ˆm,$ where ${\Lambda }_{\lambda }$ denotes the irreducible $𝔘\text{-module}$ corresponding to $\lambda \text{.}$

 Proof. (a) By the definition of the Markov trace and the fact that ${v}^{-1}u$ is a grouplike element of $𝔘,$ $mtm(a)= Trq((v-1u⊗v-1u)(a⊗id)) dimq(V)m ,$ by the definition of the quantum trace on ${V}^{\otimes m}\text{.}$ Since traces on tensor products of modules are the products of the individual traces we may write $mtm(a)= Tr(v-1ua)Tr(v-1uid) dimq(V)m ,$ where the first Tr in the numerator is on ${V}^{\otimes \left(m-1\right)}$ and the second is on $V\text{.}$ Then, by the definition of quantum dimension, we get $mtm(a)= Tr(v-1ua)dimq(V) dimq(V)m =mtm-1(a).$ In particular, ${\text{mt}}_{m}\left(1\right)={\text{tr}}_{q}\left({\text{id}}^{\otimes m}\right)/{\text{dim}}_{q}{\left(V\right)}^{m}=1\text{.}$ (b) Let $W={V}^{\otimes \left(m-1\right)}\text{.}$ Let $\left\{{w}_{s}\right\}$ be a basis of $W$ and let $\left\{{w}^{s}\right\}$ be a dual basis in ${W}^{*}\text{.}$ Let $\left\{{e}_{i}\right\}$ be a basis of $V$ and let $\left\{{e}^{i}\right\}$ be a dual basis of ${V}^{*}\text{.}$ Let $b\in {𝒵}_{m}\text{.}$ Since the element ${v}^{-1}u$ is a grouplike element of $𝔘$ we have $dimq(V)trq (εm-1(b)) = dimq(V)∑s ⟨v-1uεm-1(b)ws,ws⟩ = ∑s,k ⟨ (v-1u⊗id) (id⊗v-1u) b(ws⊗ek), ws⊗ek ⟩ = ∑s,k ⟨ (v-1u⊗v-1u) b(ws⊗ek), ws⊗ek ⟩ = ∑s,k ⟨ v-1ub (ws⊗ek), ws⊗ek ⟩ = trq(b),$ where the quantum trace on the left hand side of equation is the quantum trace on ${V}^{\otimes \left(m-1\right)}$ and the quantum trace on the right side of the equation is the quantum trace on ${V}^{\otimes m}\text{.}$ The statement follows by converting to Markov traces. (c) Let ${ě}_{m}$ be the element of ${\text{End}}_{𝔘}\left({V}^{\otimes m}\otimes {V}^{*}\right)$ given by ${ě}_{m}={\text{id}}^{\otimes \left(m-1\right)}\otimes ě,$ where $ě$ is as in (3.2). Then, since $a\in {\text{End}}_{𝔘}\left({V}^{\otimes \left(m-1\right)}\right),$ $a$ commutes with ${ě}_{m}$ and $ěmaŘm-1ěm = aěmŘm-1ěm = a(id⊗(m-1)⊗(ěŘě)) = a(id⊗(m-1)⊗ε1(Ř)ě) = dimq(V)-1 αaěm.$ It follows that ${\epsilon }_{m-1}\left(a{Ř}_{m-1}\right)=\alpha a$ and thus that $mtm(aŘm-1)= mtm-1(εm-1(aŘm-1)) =dimq(V)-1α mtm-1(a).$ (d) This follows immediately from Lemma (3.1) and the definition of the Markov traces. $\square$

 (1) Let $𝔘=\left(𝔘,ℛ,v\right)$ be a ribbon Hopf algebra and let $V={\Lambda }_{\lambda }$ be a irreducible $𝔘\text{-module.}$ Since $v$ is a central element of $𝔘,$ the element $v$ acts by a constant $v\left(\lambda \right)$ on $V={\Lambda }_{\lambda }\text{.}$ Then the framing anomaly $\alpha$ of $Ř$ is given by $\alpha =v{\left(\lambda \right)}^{-1}\text{.}$ (2) Let $𝔤$ be a finite dimensional complex simple Lie algebra and let $𝔘={𝔘}_{h}\left(𝔤\right)$ be the corresponding Drinfel'd-Jimbo quantum group. Suppose that $V={\Lambda }_{\lambda }$ is an irreducible representation of highest weight $\lambda \text{.}$ Then the framing anomaly $\alpha$ of $Ř$ is given by $\alpha ={q}^{⟨\lambda ,\lambda +2\rho ⟩}\text{.}$

 Proof. (1) By Proposition (3.7)(a) it is enough to show that ${ě}_{2}Ř{ě}_{2}={\left({\text{dim}}_{q}\left(V\right)\right)}^{-1}v{\left(\lambda \right)}^{-1}{ě}_{2}$ as elements of ${\text{End}}_{𝔘}\left(V\otimes V\otimes {V}^{*}\right)\text{.}$ Let $\left\{{e}_{i}\right\}$ be a basis of $V$ and let $\left\{{e}^{i}\right\}$ be a dual basis in ${V}^{*}\text{.}$ It follows from the identities (2.5), (2.6) and (2.7) that if $ℛ=\sum _{i}{a}_{i}\otimes {b}_{i}$ and $\left(S\otimes \text{id}\right)\left(ℛ\right)={ℛ}^{-1}=\sum _{j}{c}_{j}\otimes {d}_{j},$ then $∑ibiS2(ai)= ∑jdjS(cj)= ∑jS-1(dj)cj= u-1.$ Let $x,y\in V$ and let $\varphi \in {V}^{*}\text{.}$ Then, $ě2Řě2 (x⊗y⊗ϕ) = (dimq(V))-1 ⟨ϕ,v-1uy⟩ ě2Ř∑kx⊗ek⊗ek = (dimq(V))-1 ⟨ϕ,v-1uy⟩ ě2∑k,ibiek ⊗aix⊗ek = (dimq(V))-2 ⟨ϕ,v-1uy⟩ ∑k,i,l ⟨ek,v-1uaix⟩ biek⊗el⊗el = (dimq(V))-2 ⟨ϕ,v-1uy⟩ ∑i,l (biv-1uaix)⊗el⊗el = (dimq(V))-2 ⟨ϕ,v-1uy⟩ ∑i,l biS2(ai)v-1ux ⊗el⊗el = (dimq(V))-2 ⟨ϕ,v-1uy⟩ ∑l u-1v-1ux ⊗el⊗el = (dimq(V))-1 ě2(v-1x⊗y⊗ϕ) = (dimq(V))-1 v(λ)-1ě2(x⊗y⊗ϕ).$ (2) follows immediately, since, by Proposition (2.14), the element $v={e}^{-h\rho }u$ acts on an irreducible module ${\Lambda }_{\lambda }$ of highest weight $\lambda$ by the constant ${q}^{-⟨\lambda ,\lambda +2\rho ⟩}\text{.}$ $\square$

A Path Algebra Formula for ${ě}_{m}$

Assume that $V$ is an irreducible $𝔘$ module and that the branching rule for tensoring by $V$ is multiplicity free. Let ${𝒵}_{m}={\text{End}}_{𝔘}\left({V}^{\otimes m}\right)$ and ${𝒞}_{m+1}={\text{End}}_{𝔘}\left({V}^{\otimes m}\otimes {V}^{*}\right)$ as in (3.4). Identify the centralizer algebras ${𝒵}_{k},$ $1\le k\le m,$ with path algebras as in Section 1. It can be shown that if the branching rule for tensoring by $V$ is multiplicity free, then the branching rule for tensoring by ${V}^{*}$ is also multiplicity free. It follows that the sequence of centralizer algebras ${𝒵}_{0}\subseteq \cdots \subseteq {𝒵}_{m-1}\subseteq {𝒵}_{m}\subseteq {𝒞}_{m+1}$ can be identified with a sequence of path algebras corresponding to a multiplicity free Bratteli diagram. Let us review the notation.

 (1) $\stackrel{ˆ}{𝔘}$ is a set indexing the irreducible representations of $𝔘\text{.}$ (2) ${\stackrel{ˆ}{𝒵}}_{k}$ is a set indexing the irreducible representations of the algebra ${𝒵}_{k}\text{.}$ (3) By the double centralizer theory ${\stackrel{ˆ}{𝒵}}_{k}$ is naturally identified with the subset of $\stackrel{ˆ}{𝔘}$ containing the indexes of representations that appear in the decomposition of ${V}^{\otimes k}$ into irreducible $𝔘\text{-modules.}$ (4) Let ${\stackrel{ˆ}{𝒞}}_{m+1}$ be an index set for the irreducible representations of ${𝒞}_{m+1}$ which is naturally identified with the subset of $\stackrel{ˆ}{𝔘}$ containing indexes of representations that appear in the decomposition of ${V}^{\otimes m}\otimes {V}^{*}$ into irreducible $𝔘\text{-modules.}$

The notation for paths and tableaux will be as in Section 1. Let ${\text{mt}}_{m}$ denote the Markov trace on ${𝒵}_{m}$ and let ${\text{wt}}_{m}$ denote the weights of the Markov trace.

 (a) Viewing ${\stackrel{ˆ}{𝒵}}_{m-1}$ and ${\stackrel{ˆ}{𝒞}}_{m+1}$ as sets with elements in $\stackrel{ˆ}{𝔘},$ we have ${\stackrel{ˆ}{𝒵}}_{m-1}\subseteq {\stackrel{ˆ}{𝒞}}_{m+1}\text{.}$ (b) One can identify the centralizer algebra ${𝒞}_{m+1}$ with a path algebra in such way that ${ě}_{m}$ is given by the formula $ěm= ∑(S,T)∈Ωm-1m+1 (ěm)STEST,$ where, if $S=\left({\sigma }^{\left(m-1\right)},{\sigma }^{\left(m\right)},{\sigma }^{\left(m+1\right)}\right),$ and $T=\left({\sigma }^{\left(m-1\right)},{\tau }^{\left(m\right)},{\sigma }^{\left(m+1\right)}\right),$ then $(ěm)ST= { wtm(τ(m))wtm(σ(m)) wtm-1(σ(m-1)) , if σ(m+1)= σ(m-1) as elements of 𝔘ˆ, 0, otherwise.$

Proof.

Step 1. Let $M\prime =\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)}\right)\in {𝒯}^{m}$ and let $M″=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-1\right)}\right)\text{.}$ Then $εm-1(EM′M′)= wtm(μ(m)) wtm-1(μ(m-1)) EM″M″≠0.$

 Proof. Suppose that $εm-1(EM′M′) =∑(U″,R″)∈Ωm-1 aU″R″EU″R″ ∈𝒵m-1,$ for some constants ${a}_{U″R″}\in k\text{.}$ Suppose that $\left(S″,T″\right)\in {\Omega }^{m-1}$ and that $S″=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m-1\right)}\right)\text{.}$ Then $mtm(ES″T″EM′M′) = mtm (ES″S″ES″T″EM′M′) = mtm (ES″T″EM′M′ES″S″) = δS″M″ δT″M″ wtm(μ(m)).$ On the other hand, by Proposition (3.7b) $mtm-1 (εm-1(ES″T″EM′M′)) = mtm-1 (ES″T″εm-1(EM′M′)) = mtm-1 (ES″T″εm-1(EM′M′)ES″S″) = mtm-1 (ES″T″aS″T″ET″S″ES″S″) = aS″T″wtm-1 (σ(m-1)).$ By Proposition (3.7a), these two expressions are equal. Since the weights of the Markov trace are nonzero, it follows that $aS″T″= { wtm(μ(m)) wtm-1(μ(m-1)) , if S″=T″=M″, 0, otherwise,$ and, if $S″=T″=M″$ then ${a}_{S″T″}\ne 0\text{.}$ The formula for ${\epsilon }_{m-1}\left({E}_{M\prime M\prime }\right)$ follows. $\square$

Step 2. It follows from Proposition (1.4) that ${ě}_{m}$ has the form $ěm= ∑(S,T)∈Ωm-1m+1 (ěm)STEST, (3.13)$ since, by its definition, ${ě}_{m}$ commutes with all elements of ${𝒵}_{m-1}\text{.}$

Step 3. Let $\left(S,T\right)\in {\Omega }^{m+1}\text{.}$ Suppose $S=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m+1\right)}\right)$ and define $S\prime =\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m\right)}\right)$ and $S″=\left({\sigma }^{\left(0\right)},\dots ,{\sigma }^{\left(m-1\right)}\right)\text{.}$ Define $T\prime$ and $T″$ analogously. Let $M\prime =\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)}\right)\in {𝒯}^{m}$ and let $M″=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m-1\right)}\right)\text{.}$ Then, if ${\left({ě}_{m}\right)}_{SS}\ne 0,$ then $(ěm)SS= wtm(σ(m)) wtm-1(σ(m-1)) , (3.14)$ and $(ěm)SM (ěm)MS= wtm(μ(m)) wtm(σ(m)) wtm-1(σ(m-1))2 . (3.15)$

 Proof. It follows from the path algebra definitions and (3.13) that $ESSěm EM′M′ěm ETT=δS″M″ δM″T″∑M (ěm)SM (ěm)MT EST,$ where the sum is over all tableaux $M$ such that $M=\left({\mu }^{\left(0\right)},\dots ,{\mu }^{\left(m\right)},{\sigma }^{\left(m+1\right)}\right)\text{.}$ Since the Bratteli diagram is multiplicity free there is at most one such $M\text{.}$ Thus $ESSěm EM′M′ěm ETT= δS″M″ δM″T″ (ěm)SM (ěm)MT EST. (3.16)$ Let $S$ and $M″$ be as above. Then $ESSεm-1 (EM′M′) ěmETT = wtm(μ(m)) wtm-1(μ(m-1)) ESSEM″M″ ěmETT = wtm(μ(m)) wtm-1(μ(m-1)) δS″M″ δM″T″ (ěm)ST EST. (3.17)$ Since ${ě}_{m}{E}_{M\prime M\prime }{ě}_{m}={\epsilon }_{m-1}\left({E}_{M\prime M\prime }\right){ě}_{m},$ it follows that (3.16) and (3.17) are equal. Assuming that $S=T$ and that $S″=M″,$ i.e. ${\sigma }^{\left(i\right)}={\mu }^{\left(i\right)},$ for all $i\le m-1,$ this gives the following equation. $(ěm)SM (ěm)MS = wtm(μ(m)) wtm-1(μ(m-1)) (ěm)SS. (3.18)$ The formula in (3.14) follows by setting $M=S\text{.}$ The formula in (3.15) now follows from (3.14) and (3.18) (recall that ${\mu }^{\left(m-1\right)}={\sigma }^{\left(m-1\right)}\text{).}$ $\square$

Step 4. For each $\lambda \in {\stackrel{ˆ}{𝒵}}_{m-1}$ there exist $S$ such that $\left(S,S\right)\in {\Omega }_{\lambda }^{m+1}$ and ${\left({ě}_{m}\right)}_{SS}\ne 0\text{.}$

 Proof. Fix $\lambda \in {\stackrel{ˆ}{𝒵}}_{m-1}$ and let $M$ be such that ${\mu }^{\left(m-1\right)}=\lambda \text{.}$ Assume that ${\left({ě}_{m}\right)}_{SS}=0$ for all $S$ such that $\left(S,S\right)\in {\Omega }_{\lambda }^{m+1}\text{.}$ Then ${\left({ě}_{m}\right)}_{SM}{\left({ě}_{m}\right)}_{MS}=0$ for all $S\text{.}$ So, by (3.16), ${E}_{SS}{ě}_{m}{E}_{M\prime M\prime }{ě}_{m}{E}_{SS}=0$ for all $S\text{.}$ This implies that ${\epsilon }_{m-1}\left({E}_{M\prime M\prime }\right)=0$ which is a contradiction to Step 1. $\square$

Step 5. If $S$ is such that ${\left({ě}_{m}\right)}_{SS}\ne 0$ then ${\sigma }^{\left(m-1\right)}={\sigma }^{\left(m+1\right)}\text{.}$

 Proof. Let $S\in {𝒯}^{m+1}$ be a tableau such that ${\left({ě}_{m}\right)}_{SS}\ne 0\text{.}$ Then, as $𝔘\text{-modules,}$ ${\Lambda }_{{\sigma }^{\left(m+1\right)}}\cong {E}_{SS}b\left({V}^{\otimes m}\otimes {V}^{*}\right)$ for all $b\in {𝒞}_{m+1}$ such that ${E}_{SS}b\ne 0\text{.}$ In particular, since ${\left({ě}_{m}\right)}_{SS}\ne 0,$ $ESSěm ES′S′ ěm=cS′ ESSES″S″ ěm≠0$ and we have that $Λσ(m+1) ≅ ESSěm ES′S′ ěm (V⊗m⊗V*) = cS′ESS ES″S″ ěm (V⊗m⊗V*) ≅ ESS ( ES″S″ V⊗(m-1) ⊗ěm (V⊗V*) ) .$ Since ${E}_{S″S″}{V}^{\otimes \left(m-1\right)}\cong {\Lambda }_{{\sigma }^{\left(m-1\right)}}$ and ${e}_{m}\left(V\otimes {V}^{*}\right)\cong {\Lambda }_{\varnothing }$ it follows that ${\Lambda }_{{\sigma }^{\left(m+1\right)}}$ is isomorphic to an irreducible component in the tensor product ${\Lambda }_{{\sigma }^{\left(m-1\right)}}\otimes {\Lambda }_{\varnothing }\text{.}$ Thus ${\Lambda }_{{\sigma }^{\left(m+1\right)}}\cong {\Lambda }_{{\sigma }^{\left(m-1\right)}},$ and so ${\sigma }^{\left(m+1\right)}={\sigma }^{\left(m-1\right)}$ as elements of $\stackrel{ˆ}{𝔘}\text{.}$ $\square$

Let us complete the proof of the theorem. Part (a) follows from step 5. Recall from Section 1 that there is some freedom in the choice of the matrix units ${E}_{SM}$ and ${E}_{MS}$ when $M\ne S\text{.}$ This freedom allows us to normalize the matrix units ${E}_{SM}$ and ${E}_{MS}$ in any way such that (3.15) holds. In particular, we can choose that normalization so that the formula is as in the theorem. The fact that ${\left({ě}_{m}\right)}_{SM}=0$ if ${\sigma }^{\left(m-1\right)}\ne {\sigma }^{\left(m+1\right)}$ follows from steps 3, 4, and 5.

$\square$

## 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.