Computation of Central Elements

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

Last updates: 15 October 2010

Computation of $Z_V^{\left(l\right)}$ when $V=L\left({\epsilon }_1\right)$

Let $𝔤={𝔰𝔬}_{2r+1}$ or $𝔤={𝔰𝔭}_{2r}$ or $𝔤={𝔰𝔬}_{2r}$, and let $V=L\left({\epsilon }_1\right)$, $$y=⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩\epsilon =\left\{\begin{array}{ll}1,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ -1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ 1,& \text{if} 𝔤={𝔰𝔬}_{2r},\end{array}\right.\delta =\left\{\begin{array}{ll}0,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ 1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ -1,& \text{if} 𝔤={𝔰𝔬}_{2r}.\end{array}\right.$$ Let $z_V^{\left(l\right)}$ be the central elements in $U_q𝔤$ defined by $$z_V^{\left(l\right)}=\epsilon \left(id\otimes {tr}_V\right)\left({\left(t+\frac{1}{2}y\right)}^l\right)\text{and let}{z}_V\left(u\right)=\sum _{i\in {\mathbb{Z}}_{\ge 0}}{z}_V^{\left(l\right)}{u}^{-l}.$$ Then $${\pi }_0\left(z_V\left(u\right)+\epsilon u-\frac{1}{2}\right)=\frac{\left(\epsilon u+\frac{1}{2}\right)\left(u+\frac{1}{2}y\right)\left(u-\frac{1}{2}y-1\right)\left(u-\frac{1}{2}\delta \right)}{\left(u-\frac{1}{2}y\right)\left(u+\frac{1}{2}y+1\right)\left(u+\frac{1}{2}\delta \right)}{\sigma }_{\rho }\left(\prod _{i=1}^r\frac{\left(u+h_i+\frac{1}{2}\right)\left(u-h_i+\frac{1}{2}\right)}{\left(u+h_i-\frac{1}{2}\right)\left(u-h_i-\frac{1}{2}\right)}\right).$$

		Proof.
	By (5.20), as operators on $L\left(0\right)\otimes V$, $$t=\frac{1}{2}\left(⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟨-⟩{\epsilon }_1,{\epsilon }_1+2\rho +⟨0,0+2\rho \right)=0\text{and}{\left(\frac{1}{2}y+t\right)}^l={\left(\frac{1}{2}y\right)}^l.$$ So, as operators on $L\left(0\right)\otimes V$, $$z_V^{\left(l\right)}=\epsilon \left(id\otimes {tr}_V\right)\left({\left(\frac{1}{2}y+t\right)}^l\right)=\epsilon \left(id\otimes {tr}_V\right)\left({\left(\frac{1}{2}y\right)}^l\right)=\epsilon dim\left(V\right){\left(\frac{1}{2}y\right)}^l$$ and, since $dim\left(V\right)=\epsilon +y$, $$z_V\left(u\right)=\sum _{l\in {\mathbb{Z}}_{\ge 0}}{z}_V^{\left(l\right)}{u}^{-l}=\sum _{l\in {\mathbb{Z}}_{\ge 0}}\epsilon dim\left(V\right){\left(\frac{1}{2}y\right)}^l{u}^{-l}=\epsilon dim\left(V\right)\frac{1}{1-\frac{1}{2}y{u}^{-1}}.$$ Therefore $$\begin{array}{rcl}{z}_V\left(u\right)+\epsilon u-\frac{1}{2}& =& \frac{1+\epsilon y}{1-\frac{1}{2}y{u}^{-1}}+\epsilon u-\frac{1}{2}=\frac{u+\epsilon yu+\left(\epsilon u-\frac{1}{2}\right)\left(u-\frac{1}{2}y\right)}{u-\frac{1}{2}y}\\ & =& \frac{\epsilon u^2+\frac{1}{2}u+\frac{1}{2}\epsilon yu+\frac{1}{4}y}{u-\frac{1}{2}y}=\frac{\left(\epsilon u+\frac{1}{2}\right)\left(u+\frac{1}{2}y\right)}{u-\frac{1}{2}y}.\end{array}$$ 6.13 By the second identity in (2.6) and the definition of $\Phi$ in Theorem 5.2(a), $$y_k\in {𝒲}_{k-1}\text{acts on}L\left(0\right)\otimes V^{\otimes k}=\left(L\left(0\right)\otimes V^{\otimes k-1}\right)\otimes V\text{as}\frac{1}{2}y+t.$$ If $L\left(\mu \right)$ is and irreducible $U_q𝔤$-module in $L\left(0\right)\otimes V^{\otimes k-1}$, then (5.20) gives that $y_k$ acts on the $L\left(\lambda \right)$ component of $L\left(\mu \right)\otimes V$ by the constant $$\begin{array}{rcl}c\left(\lambda ,\mu \right)& =& \frac{1}{2}y+\frac{1}{2}\left(⟨\mu \pm {\epsilon }_k,\mu \pm {\epsilon }_k+2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩\right)\\ & =& \frac{1}{2}y+\frac{1}{2}\left(\pm 2{\mu }_i\pm 2{\rho }_i+1-y\right)=\pm {\mu }_i\pm \frac{1}{2}\left(y-2i+1\right)+\frac{1}{2}=\pm \left(\frac{1}{2}+\frac{1}{2}y+{\mu }_i-i\right)+\frac{1}{2}\\ & =& \left\{\begin{array}{ll}\left(-\frac{1}{2}+\frac{1}{2}y+{\lambda }_i\right)+\frac{1}{2}& \text{if} \mu \subseteq \lambda ,\\ \left(-\frac{1}{2}-\frac{1}{2}y-\left({\mu }_i-i\right)\right)+\frac{1}{2},& \text{if} \mu \supseteq \lambda ,\end{array}\right.\\ & =& \left\{\begin{array}{ll}\frac{1}{2}y+{\lambda }_i-i,& \text{if} \mu \subseteq \lambda ,\\ -\frac{1}{2}-\left({\mu }_i-i\right),& \text{if} \mu \supseteq \lambda ,\end{array}\right.\\ & =& \left\{\begin{array}{ll}\frac{1}{2}y+c\left(\lambda /\mu \right),& \text{if} \mu \subseteq \lambda ,\\ -\frac{1}{2}y-c\left(\mu /\lambda \right)& \text{if} \mu \supseteq \lambda .\end{array}\right.\end{array}$$ 6.14 As in [Naz, Theorem 2.6], the irreducible ${𝒲}_k$-module ${𝒲}_k^{\mu /0}={𝒲}_k^{\mu }$ has a basis $\left\{v_T\right\}$ indexed by up-down tableaux $T=\left(T^{\left(0\right)},T^{\left(1\right)},\dots ,T^{\left(k\right)}\right)$, where $T^{\left(0\right)}=\varnothing$, $T^{\left(k\right)}=\mu$, and $T^{\left(i\right)}$ is a partition obtained from $T^{(i-1)}$ by adding or removing a box, and $$y_i{v}_T=\left\{\begin{array}{ll}\left(\frac{1}{2}y+c\left(b\right)\right)v_T,& \text{if} b=T^{\left(i\right)}/T^{(i-1)},\\ \left(\frac{1}{2}y-c\left(b\right)\right)v_T,& \text{if} b=T^{(i-1)}/T^{\left(i\right)}.\end{array}\right.$$ Thus $$\prod _{i=1}^{k-1}\frac{\left(u-y_i-1\right)\left(u+y_i+1\right){\left(u-y_i\right)}^2}{{\left(u+y_i\right)}^2\left(u-y_i+1\right)\left(u-y_i-1\right)}$$ acts on the $L\left(\mu \right)\otimes {𝒲}_{k-1}^{\mu }$ isotypic component in the $U_q𝔤\otimes {𝒲}_{k-1}$-module decomposition $$L\left(0\right)\otimes V^{\otimes k-1}\cong \underset{\mu }{\oplus }L\left(\mu \right)\otimes {𝒲}_{k-1}^{\mu },$$ 6.15 by $$\prod _{i=1}^{k-1}\frac{\left(u+c\left(T^{\left(i\right)},T^{(i-1)}\right)-1\right)\left(u+c\left(T^{\left(i\right)},T^{(i-1)}\right)+1\right){\left(u-c\left(T^{\left(i\right)},T^{(i-1)}\right)\right)}^2}{{\left(u+c\left(T^{\left(i\right)},T^{(i-1)}\right)\right)}^2\left(u-c\left(T^{\left(i\right)},T^{(i-1)}\right)+1\right)\left(u-c\left(T^{\left(i\right)},T^{(i-1)}\right)-1\right)}$$ 6.16 for any up-down tableaux $T$ of length $k$ and shape $\mu$. If a box is added (or removed) at step $i$ and then removed (or added) at step $j$, then the $i$ and $j$ factora of this product cancel. Therefore, (6.16) is equal to $$\prod _{b\in \mu }\frac{\left(u+\frac{1}{2}y+c\left(b\right)-1\right)\left(u+\frac{1}{2}y+c\left(b\right)+1\right){\left(u-\frac{1}{2}y-c\left(b\right)\right)}^2}{{\left(u+\frac{1}{2}y+c\left(b\right)\right)}^2\left(u-\frac{1}{2}y+c\left(b\right)+1\right)\left(u-\frac{1}{2}y-c\left(b\right)-1\right)}$$ 6.17 (see [Naz, Lemma 3.8]). Simplifying one row at a time, $$\begin{array}{rcl}\prod _{b\in \mu }\frac{\left(u+\frac{1}{2}y+c\left(b\right)-1\right)\left(u+\frac{1}{2}y+c\left(b\right)+1\right)}{{\left(u+\frac{1}{2}y+c\left(b\right)\right)}^2}& =& \prod _{i=1}^r\frac{\left(u+\frac{1}{2}y-i\right)\left(u+\frac{1}{2}y+{\mu }_i-i+1\right)}{\left(u+\frac{1}{2}y+1-i\right)\left(u+\frac{1}{2}y+{\mu }_i-i\right)}\\ & =& \frac{u+\frac{1}{2}y-r}{u+\frac{1}{2}y+1}\prod _{i=1}^r\frac{\left(u+\frac{1}{2}y+{\mu }_i-i+1\right)}{\left(u+\frac{1}{2}y+{\mu }_i-i\right)},\end{array}$$ 6.18 if $\mu =\left({\mu }_1,\dots ,{\mu }_r\right)$. It follows that (6.17) is equal to $$\begin{array}{r}\frac{\left(u+\frac{1}{2}y-r\right)}{\left(u+\frac{1}{2}y+1\right)}\frac{\left(u-\frac{1}{2}y-1\right)}{\left(u-\frac{1}{2}y+r\right)}\prod _{i=1}^r\frac{\left(u+\frac{1}{2}y+{\mu }_i-i+1\right)}{\left(u+\frac{1}{2}y+{\mu }_i-i\right)}\frac{\left(u-\frac{1}{2}y-\left({\mu }_i-i\right)\right)}{\left(u+\frac{1}{2}y-\left({\mu }_i-i+1\right)\right)}\\ =\frac{\left(u-\frac{1}{2}\delta \right)}{\left(u+\frac{1}{2}y+1\right)}\frac{\left(u-\frac{1}{2}y-1\right)}{\left(u+\frac{1}{2}\delta \right)}{ev}_{\mu +\rho }\left(\prod _{i=1}^r\frac{\left(u+h_i+\frac{1}{2}\right)}{\left(u+h_i-\frac{1}{2}\right)}\frac{u-h_i+\frac{1}{2}}{\left(u-h_i-\frac{1}{2}\right)}\right)\end{array}$$ 6.19 where $$\delta =\left\{\begin{array}{ll}0,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ 1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ -1,& \text{if} 𝔤={𝔰𝔬}_{2r},\end{array}\right.$$ since $${ev}_{\mu }\left({\sigma }_{\rho }\left(h_{{\epsilon }_i^{\vee }}\right)\right)={ev}_{\mu +\rho }\left(h_{{\epsilon }_i^{\vee }}\right)=⟨\mu +\rho ,{\epsilon }_i^{\vee }⟩={\mu }_i+\frac{1}{2}\left(y-2i+1\right)=\frac{1}{2}y+\frac{1}{2}-i$$ and $$\frac{1}{2}y-r=\frac{1}{2}\left(y-2r\right)=\frac{1}{2}\left(y-2r+\delta \right)-\frac{1}{2}\delta =-\frac{1}{2}\delta$$ Combining (6.13) and (6.19), the identity (4.14) gives that $$z_V\left(u\right)+\epsilon u-\frac{1}{2}=\left(z_V\left(1\right)+\epsilon u-\frac{1}{2}\right)\prod _{i=1}^{k-1}\frac{\left(u+y_i-1\right)\left(u+y_i+1\right){\left(u-y_i\right)}^2}{{\left(u+y_i\right)}^2\left(u-y_i+1\right)\left(u-y_i-1\right)}$$ acts on the $L\left(\mu \right)\otimes {𝒲}_{k-1}^{\mu }$-component in the $U_q𝔤\otimes {𝒲}_{k-1}$-module decomposition in (6.15) by $$\frac{\left(\epsilon u+\frac{1}{2}\right)\left(u+\frac{1}{2}y\right)}{u-\frac{1}{2}y}\frac{\left(u-\frac{1}{2}\delta \right)}{\left(u+\frac{1}{2}y+1\right)}\frac{\left(u-\frac{1}{2}y-1\right)}{\left(u+\frac{1}{2}\delta \right)}{ev}_{\mu +\rho }\left(\prod _{i=1}^r\frac{\left(u+h_i+\frac{1}{2}\right)}{\left(u+h_i-\frac{1}{2}\right)}\frac{u-h_i+\frac{1}{2}}{\left(u-h_i-\frac{1}{2}\right)}\right).$$ $\square$

To help illuminate the cancellation done in (6.18), we walk through the example where $\mu =\left(5,5,3,3,1,1\right)$, and so the contents of the boxes are $$\text{TABLEAU GOES HERE.}$$ In this example, the product over the boxes in the first row of the diagram is $$\begin{array}{l}\prod _{b \text{in first row of} \mu }\frac{\left(x+c\left(b\right)-1\right)\left(x+c\left(b\right)+1\right)}{\left(x+c\left(b\right)\right)\left(x+c\left(b\right)\right)}\\ =\frac{\left(x-1\right)\left(x+1\right)}{\left(x+0\right)\left(x-0\right)}\frac{\left(x+0\right)\left(x+2\right)}{\left(x+1\right)\left(x+1\right)}\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+2\right)}\frac{\left(x+2\right)\left(x+4\right)}{\left(x+3\right)\left(x+3\right)}\frac{\left(x+3\right)\left(x+5\right)}{\left(x+4\right)\left(x+4\right)}\\ =\frac{\left(x-1\right)}{\left(x+0\right)}\frac{\left(x+5\right)}{\left(x+4\right)},\text{where} x=y+\frac{1}{2}y.\end{array}$$ Thus, simplifying the product one row at a time, $$\begin{array}{l}\prod _{b\in \mu }\frac{\left(x+c\left(b\right)-1\right)\left(x+c\left(b\right)+1\right)}{\left(x+c\left(b\right)\right)\left(x+c\left(b\right)\right)}\\ =\frac{\left(x-1\right)}{\left(x+0\right)}\frac{\left(x+5\right)}{\left(x+4\right)}\cdot \frac{\left(x-2\right)}{\left(x-1\right)}\frac{\left(x+4\right)}{\left(x+3\right)}\cdot \frac{\left(x-3\right)}{\left(x-2\right)}\frac{\left(x+1\right)}{\left(x+0\right)}\cdot \frac{\left(x-4\right)}{\left(x-3\right)}\frac{\left(x+0\right)}{\left(x-1\right)}\cdot \frac{\left(x-5\right)}{\left(x-4\right)}\frac{\left(x-3\right)}{\left(x-4\right)}\cdot \frac{\left(x-6\right)}{\left(x-5\right)}\frac{\left(x-4\right)}{\left(x-5\right)}\\ =\frac{\left(x-6\right)}{\left(x+0\right)}\frac{\left(x+5\right)}{\left(x+4\right)}\cdot \frac{\left(x+4\right)}{\left(x+3\right)}\cdot \frac{\left(x+1\right)}{\left(x+0\right)}\cdot \frac{\left(x+0\right)}{\left(x-1\right)}\cdot \frac{\left(x-3\right)}{\left(x-4\right)}\cdot \frac{\left(x-4\right)}{\left(x-5\right)}\end{array}$$ leading us to the identity $$\prod _{b\in \mu }\frac{\left(x+c\left(b\right)-1\right)\left(x+c\left(b\right)+1\right)}{\left(x+c\left(b\right)\right)\left(x+c\left(b\right)\right)}=\frac{x-r}{x+0}\prod _{i=0}^r\frac{x-{\mu }_i-i+1}{x+{\mu }_i-i},\text{if} \mu =\left({\mu }_1,\dots ,{\mu }_r\right).$$

Let $𝔤={𝔰𝔬}_{2r+1}$ or $𝔤={𝔰𝔭}_{2r}$ or $𝔤={𝔰𝔬}_{2r}$, and let $V=L\left({\epsilon }_1\right)$ and $z=\epsilon q^y$, where $$y=⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟩\epsilon =\left\{\begin{array}{ll}1,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ -1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ 1,& \text{if} 𝔤={𝔰𝔬}_{2r},\end{array}\right.\delta =\left\{\begin{array}{ll}0,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ 1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ -1,& \text{if} 𝔤={𝔰𝔬}_{2r}.\end{array}\right.$$ Let $Z_V^{\left(l\right)}$ be the central elements in $U_q𝔤$ defined by $$Z_V^{\left(l\right)}=\epsilon \left(id\otimes {qtr}_V\left({\left(z{ℛ}_{12}ℛ\right)}^l\right)\right)$$ and let $$Z_V^{\left(l\right)}=\sum _{l\in {\mathbb{Z}}_{\ge 0}}{Z}_V^{\left(l\right)}{u}^{-l}\text{and}{Z}_V^{-}=\sum _{l\in {\mathbb{Z}}_{\ge 0}}{Z}_V^{(-l)}{u}^{-l}.$$ Then $$\begin{array}{l}{\pi }_0\left(Z_V^{+}\left(u\right)-\frac{z}{q-q^{-1}}-\frac{u^2}{u^2-1}\right)\\ =\left(\frac{z}{q-q^{-1}}\right)\frac{\left(u+q\right)\left(u-q^{-1}\right)\left(u-\epsilon q^{\delta }\right)}{\left(u+1\right)\left(u-1\right)\left(u-\epsilon q^{-\delta }\right)}{\sigma }_{\rho }\left(\prod _{i=1}^r\frac{\left(u-\epsilon L_i^{-2}{q}^{-1}\right)\left(u-\epsilon L_i^2{q}^{-1}\right)}{\left(u-\epsilon L_i^{-2}q\right)\left(u-\epsilon L_i^{2}q\right)}\right).\end{array}$$

		Proof.
	As operators on $L\left(0\right)\otimes V$, $$z\left({ℛ}_{12}ℛ\right)=z{q}^{⟨{\epsilon }_1,{\epsilon }_1+2\rho ⟨-⟩{\epsilon }_1,{\epsilon }_1+2\rho +⟨0,0+2\rho ⟩}=z\text{and}{\left(z{ℛ}_{12}ℛ\right)}^l=z^l.$$ So, as operators on $L\left(0\right)$, $$Z_V^{\left(l\right)}=\epsilon \left(id\otimes {qtr}_V\right)\left({\left(z{ℛ}_{12}ℛ\right)}^l\right)=z^l\epsilon {dim}_q\left(V\right)$$ and, since $\epsilon {dim}_q\left(V\right)=\frac{z-z^{-1}}{q-q^{-1}}+1$, $$Z_V^{+}=\sum _{l\in {\mathbb{Z}}_{\ge 0}}\epsilon {dim}_q\left(V\right)z^l{u}^{-l}=\epsilon {dim}_q\left(V\right)\frac{1}{1-z{u}^{-1}}=\frac{z-z^{-1}+\left(q-q^{-1}\right)}{\left(q-q^{-1}\right)\left(1-z{u}^{-1}\right)}.$$ A similar calculation yields $$Z_V^{+}\left(u\right)=\frac{z-z^{-1}+q-q^{-1}}{\left(q-q^{-1}\right)\left(1-z{u}^{-1}\right)}\text{and}\frac{z-z^{-1}+q+q^{-1}}{\left(q-q^{-1}\right)\left(1-z^{-1}{u}^{-1}\right)}$$ (see [BB], the last displayed equation in the proof of Lemma 7.4). Amazingly, as operators on $L\left(0\right)$, $$Z_V^{+}-Z_V^{\left(0\right)}+\frac{z}{q-q^{-1}}-\frac{1}{u^2-1}=\frac{z}{\left(q-q^{-1}\right)}\frac{\left(1-z^{-1}{u}^{-1}\right)}{\left(1-z{u}^{-1}\right)}\frac{\left(u+q\right)\left(u-q^{-1}\right)}{\left(u+1\right)\left(u-1\right)}$$ 6.21 and $$Z_V^{-}-Z_V^{\left(0\right)}+\frac{-z^{-1}}{q-q^{-1}}-\frac{1}{u^2-1}=\frac{-z^{-1}}{\left(q-q^{-1}\right)}\frac{\left(1-z{u}^{-1}\right)}{\left(1-z^{-1}{u}^{-1}\right)}\frac{\left(u-q\right)\left(u+q^{-1}\right)}{\left(u+1\right)\left(u-1\right)}.$$ 6.22 By (5.39), $$Y_k\text{acts on}L\left(0\right)\otimes V^{\otimes k}\text{as}z{ℛ}_{12}ℛ.$$ If $L\left(\mu \right)$ is an irreducible $U_q𝔤$-module in $L\left(0\right)\otimes V^{\otimes k-1}$, then (5.23) and (5.26) give that $Y_k$ acts on the $L\left(\lambda \right)$ component of $L\left(\mu \right)\otimes V$ by the constant $$\epsilon q^{2c\left(\lambda ,\mu \right)}=\left\{\begin{array}{ll}\epsilon q^{y+2c\left(\lambda /\mu \right)},& \text{if} \mu \subseteq \lambda ,\\ \epsilon q^{-y-2c\left(\mu /\lambda \right)},& \text{if} \mu \supseteq \lambda ,\end{array}\right.=\left\{\begin{array}{ll}z{q}^{2c\left(\lambda /\mu \right)},& \text{if} \mu \subseteq \lambda ,\\ z^{-1}{q}^{-2c\left(\mu /\lambda \right)},& \text{if} \mu \supseteq \lambda ,\end{array}\right.$$ where $c\left(\lambda /\mu \right)$ is computed in (6.14) (see [OR]). By induction, as in [OR, Theorem 6.3(b)], the irreducible $W_k$-module $W_k^{\mu /0}=W_k^{\mu }$ has a basis $\left\{v_T\right\}$ indexed by up-down tableaux $T=\left(T^{\left(0\right)},T^{\left(1\right)},\dots ,T^{\left(k\right)}\right)$, where $T^{\left(0\right)}=\varnothing$, $T^{\left(k\right)}=\mu$, and $T^{\left(i\right)}$ is a partition obtained from $T^{(i-1)}$ by adding or removing a box, and $$Y_i{v}_T=\left\{\begin{array}{ll}z{q}^{2c\left(b\right)},& \text{if} b=T^{\left(i\right)}/T^{(i-1)},\\ z^{-1}{q}^{-2c\left(b\right)},& \text{if} b=T^{(i-1)}/T^{\left(i\right)}\end{array}\right..$$ Thus $$\prod _{i=1}^{k-1}\frac{{\left(u-Y_i\right)}^2\left(u-q^{-2}{Y}_i^{-1}\right)\left(u-q^2{Y}_i^{-1}\right)}{{\left(u-Y_i^{-1}\right)}^2\left(u-q^2{Y}_i\right)\left(u-q^{-2}{Y}_i\right)}$$ acts on the $L\left(\mu \right)\otimes W_{k-1}^{\mu }$ isotypic component in the $U_q𝔤$-module decomposition $$L\left(0\right)\otimes V^{\otimes k-1}\cong \underset{\mu }{\oplus }L\left(\mu \right)\otimes W_{k-1}^{\mu },$$ 6.23 by $$\prod _{i=1}^{k-1}\frac{{\left(u-\epsilon q^{2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)}^2\left(u-q^{-2}{q}^{-2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)\left(u-\epsilon q^2{q}^{-2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)}{{\left(u-\epsilon q^{-2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)}^2\left(u-\epsilon q^2{q}^{-2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)\left(u-\epsilon q^{-2}{q}^{2c\left(T^{\left(i\right)},T^{(i-1)}\right)}\right)}$$ 6.24 for any up-down tableaux $T$ of length $k$ and shape $\mu$. If a box is added (or removed) at step $i$ and then removed (or added) at step $j$, then the $i$ and $j$ factora of this product cancel. Therefore, (6.24) is equal to $$\prod _{b\in \mu }^{}\frac{{\left(u-z{q}^{2c\left(b\right)}\right)}^2\left(u-z^{-1}{q}^{-2\left(c\left(b\right)+1\right)}\right)\left(u-z^{-1}{q}^{-2\left(c\left(b\right)-1\right)}\right)}{{\left(u-z^{-1}{q}^{-2c\left(b\right)}\right)}^2\left(u-z{q}^{2\left(c\left(b\right)+1\right)}\right)\left(u-z{q}^{2\left(c\left(b\right)-1\right)}\right)}$$ 6.25 (see [BB], the next to last displayed equation in the proof of Lemma 7.4). Simplifying one row at a time, $$\begin{array}{rcl}\prod _{b\in \mu }\frac{\left(u-z^{-1}{q}^{-2\left(c\left(b\right)-1\right)}\right)\left(u-z^{-1}{q}^{-2\left(c\left(b\right)+1\right)}\right)}{\left(u-z^{-1}{q}^{-2c\left(b\right)}\right)\left(u-z^{-1}{q}^{-2c\left(b\right)}\right)}& =& \prod _{i=1}^r\frac{\left(u-z^{-1}{q}^{-2\left(-i\right)}\right)\left(u-z^{-1}{q}^{-2\left({\mu }_i-i+1\right)}\right)}{\left(u-z^{-1}{q}^{-2\left(-\left(i-1\right)\right)}\right)\left(u-z^{-1}{q}^{-2\left({\mu }_i-i\right)}\right)}\\ & =& \frac{u-z^{-1}{q}^{2r}}{u-z^{-1}{q}^{2\cdot 0}}\prod _{i=1}^r\frac{u-z^{-1}{q}^{-2\left({\mu }_i-i+1\right)}}{u-z^{-1}{q}^{-2\left({\mu }_i-i\right)}}\end{array}$$ if $\mu =\left({\mu }_1,\dots ,{\mu }_r\right)$. It follows that (6.25) is equal to $$\begin{array}{l}\frac{\left(u-z^{-1}{q}^{2r}\right)}{\left(u-z^{-1}\right)}\frac{\left(u-z\right)}{\left(u-z{q}^{-2r}\right)}\prod _{i=1}^r\frac{\left(u-z^{-1}{q}^{-2\left({\mu }_i-i+1\right)}\right)}{\left(u-z^{-1}{q}^{-2\left({\mu }_i-i\right)}\right)}\cdot \frac{\left(u-z{q}^{2\left({\mu }_i-i\right)}\right)}{\left(u-z{q}^{2\left({\mu }_i-i+1\right)}\right)}\\ =\frac{\left(u-\epsilon q^{\delta }\right)}{\left(u-z^{-1}\right)}\frac{\left(u-z\right)}{\left(u-\epsilon q^{\delta }\right)}{ev}_{\mu +\rho }\left(\prod _{i=1}^r\frac{\left(u-\epsilon L_i^{-2}{q}^{-1}\right)\left(u-\epsilon L_i^2{q}^{-1}\right)}{\left(u-\epsilon L_i^{-2}q\right)\left(u-\epsilon L_i^{2}q\right)}\right)\end{array}$$ 6.26 where $$\delta =\left\{\begin{array}{ll}0,& \text{if} 𝔤={𝔰𝔬}_{2r+1},\\ 1,& \text{if} 𝔤={𝔰𝔭}_{2r},\\ -1,& \text{if} 𝔤={𝔰𝔬}_{2r},\end{array}\right.$$ since $${ev}_{\mu +\rho }\left(L_i^2\right)={ev}_{\mu +\rho }\left(K_{2{\epsilon }_i^{\vee }}\right)=q^{⟨\mu +\rho ,2{\epsilon }_i^{\vee }⟩}=q^{2{\mu }_i+\left(y-2i+1\right)}=q^{y+1+2\left({\mu }_i-i\right)}=\epsilon z{q}^{2\left({\mu }_i-i\right)+1}$$ and $$z^{-1}{q}^{2r}=\epsilon q^{-2r}{q}^{\delta }{q}^{2r}=\epsilon q^{\delta }.$$ Combining (6.21) and (6.26), the identity (4.15) gives that $$\begin{array}{l}{Z}_k^{+}-Z_k^{\left(0\right)}+\frac{z}{q-q^{-1}}-\frac{1}{u^2-1}\\ =\left(Z_1^{+}-Z_1^{\left(0\right)}+\frac{z}{q-q^{-1}}-\frac{1}{u^2-1}\right)\prod _{i=1}^{k-1}\frac{{\left(u-Y_i\right)}^2\left(u-q^{-2}{Y}_i^{-1}\right)\left(u-q^2{Y}_i^{-1}\right)}{{\left(u-Y_i^{-1}\right)}^2\left(u-q^2{Y}_i\right)\left(u-q^{-2}{Y}_i\right)},\end{array}$$ acts on the $L\left(\mu \right)\otimes W_{k-1}^{\mu }$ isotypic component in the $U_q𝔤\otimes W_{k-1}$-module decomposition in (6.23) by $$\begin{array}{l}\frac{z}{\left(q-q^{-1}\right)}\frac{\left(1-z^{-1}{u}^{-1}\right)}{\left(1-z{u}^{-1}\right)}\frac{\left(u+q\right)\left(u-q^{-1}\right)}{\left(u+1\right)\left(u-1\right)}\\ \frac{\left(u-\epsilon q^{\delta }\right)}{\left(u-z^{-1}\right)}\frac{\left(u-z\right)}{\left(u-\epsilon q^{\delta }\right)}{ev}_{\mu +\rho }\left(\prod _{i=1}^r\frac{\left(u-\epsilon L_i^{-2}{q}^{-1}\right)\left(u-\epsilon L_i^2{q}^{-1}\right)}{\left(u-\epsilon L_i^{-2}q\right)\left(u-\epsilon L_i^{2}q\right)}\right)\\ =\frac{z}{\left(q-q^{-1}\right)}\frac{\left(u+q\right)\left(u-q^{-1}\right)}{\left(u+1\right)\left(u-1\right)}\frac{u-\epsilon q^{\delta }}{u-\epsilon q^{-\delta }}{ev}_{\mu +\rho }\left(\prod _{i=1}^r\frac{\left(u-\epsilon L_i^{-2}{q}^{-1}\right)\left(u-\epsilon L_i^2{q}^{-1}\right)}{\left(u-\epsilon L_i^{-2}q\right)\left(u-\epsilon L_i^{2}q\right)}\right).\end{array}$$ $\square$

References

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[Bou] N. Bourbaki, Groupes et Algèbres de Lie, Masson, Paris, 1990.

[DRV] Z. Daugherty, A. Ram, and R. Virk, Affine and graded BMW algebras, in preparation.

[GH1] F. Goodman and H. Hauschild Mosley, Cyclotomic Birman-Wenzl-Murakami algebras. I. Freeness and realization as tangle algebras, J. Knot Theory Ramifications 18 (2009), 1089-1127. MR2554337 (2010j:57014)

[Naz] M. Nazarov, Young's orthogonal form for Brauer's centralizer algebra, J. Algebra 182 (1996), no. 3, 664-693. MR1398116 (97m:20057)

[OR] R. Orellana and A. Ram, Affine braids, Markov traces and the category $𝒪$, Algebraic groups and homogeneous spaces, 423-473, Tata Inst. Fund. Res. Stud. Math., Tata Inst. Fund. Res., Mumbai, 2007. MR2348913 (2008m:17034)

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