Spectral subalgebras

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

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 18 March 2012

Spectral subalgebras

Then $$C_0=\left\{\mu \in A*|\mu \left(xy\right)=\mu \left(yx\right)\right\}\text{ is a commutative algebra,}$$ since, if $l_1,l_2\in C_0$ and $a\in A$ then $$\begin{array}{rcl}\left(l_2{l}_1\right)\left(a\right)& =& \left(l_1\otimes l_2\right){\Delta }^{\mathrm{op}}\left(a\right)=\left(l_1\otimes l_2\right)ℛ\Delta \left(a\right){ℛ}^{-1}& & =& \left(l_1\otimes l_2\right)\Delta \left(a\right){ℛ}^{-1}ℛ=\left(l_1\otimes l_2\right)\Delta \left(a\right)=\left(l_1{l}_2\right)\left(a\right),\end{array}$$ where the third equality uses the definition of $C_0.$

If $\left(A,ℛ\right)$ is a quasitriangular Hopf algebra the $ℛ$ satisfies the quantum Yang-Baxter equation, (QYBE),$${ℛ}^{12}{ℛ}^{13}{ℛ}^{23}={ℛ}^{12}\left(\Delta \otimes \mathrm{id}\right)\left(ℛ\right)=\left({\Delta }^{\mathrm{op}}\otimes \mathrm{id}\right)\left(ℛ\right){ℛ}^{12}={ℛ}^{23}{ℛ}^{13}{ℛ}^{12}.$$

Since $$ℛ=\left(\epsilon \otimes \mathrm{id}\otimes \mathrm{id}\right)\left(\Delta \otimes \mathrm{id}\right)\left(ℛ\right)=\left(\epsilon \otimes \mathrm{id}\otimes \mathrm{id}\right){ℛ}^{13}{ℛ}^{23}=\left(\epsilon \otimes \mathrm{id}\right)\left(ℛ\right).ℛ,$$ and $$ℛ=\left(\mathrm{id}\otimes \mathrm{id}\otimes \epsilon \right)\left(\mathrm{id}\otimes \Delta \right)\left(ℛ\right)=\left(\mathrm{id}\otimes \mathrm{id}\otimes \epsilon \right){ℛ}^{13}{ℛ}^{23}=\left(\mathrm{id}\otimes \epsilon \right)\left(ℛ\right).ℛ,$$ and so $$\left(\epsilon \otimes \mathrm{id}\right)\left(ℛ\right)=1\text{and}\left(\mathrm{id}\otimes \epsilon \right)\left(ℛ\right)=1.$$

Then, since $$ℛ\left(S,\otimes ,\mathrm{id}\right)\left(ℛ\right)=\left(m\otimes \mathrm{id}\right)\left(\mathrm{id}\otimes S\otimes \mathrm{id}\right)\left({ℛ}^{13}{ℛ}^{23}\right)=\left(m\otimes \mathrm{id}\right)\left(\mathrm{id}\otimes S\otimes \mathrm{id}\right)\left(\Delta \otimes \mathrm{id}\right)\left(ℛ\right)=\left(\epsilon \otimes \mathrm{id}\right)\left(ℛ\right)=1,$$ it follows that $$\left(S\otimes \mathrm{id}\right)\left(ℛ\right)={ℛ}^{-1}.$$ Applying this to the pair $\left(A^{\mathrm{op}},{ℛ}^{21}\right)$ gives $\left(S^{-1}\otimes \mathrm{id}\right)\left({ℛ}^{21}\right)={\left({ℛ}^{21}\right)}^{\mathrm{op}},$ and so $$\left(\mathrm{id}\otimes S^{-1}\right)\left(ℛ\right)={ℛ}^{-1}.$$ Then $$\left(S\otimes S\right)\left(ℛ\right)=\left(\mathrm{id}\otimes S\right)\left(S\otimes \mathrm{id}\right)\left(ℛ\right)=\left(\mathrm{id}\otimes S\right)\left({ℛ}^{-1}\right)=\left(\mathrm{id}\otimes S\right)\left(\mathrm{id}\otimes S^{-1}\right)\left(ℛ\right)=ℛ.$$

The map $\phi :C\to Z\left(A\right)$ in the following proposition is ananalogue of the Harish-Chandra homomorphism.

Let $\left(A,ℛ\right)$ be a quasitrtiangular Hopf algebra. Then $$C=\left\{\lambda \in A*|\lambda \left(xy\right)=\lambda \left(y{S}^2\left(x\right)\right)\right\}\text{is a commutative algebra}$$ and the map $$\begin{array}{rcll}\phi :& C& \to & Z\left(A\right)\\ & l& \mapsto & \left(\mathrm{id}\otimes l\right)\left({ℛ}_{21},ℛ\right)\end{array}is\; a\; well\; defined\; algebra\; homomorphism.$$

		Proof.
	If $l_1,l_2\in A*$ and $a\in A$ then $$\begin{array}{rcl}\left(l_1{l}_2\right)\left(a\right)& =& \left(l_1\otimes l_2\right){\Delta }^{\mathrm{op}}\left(a\right)=\left(l_1\otimes l_2\right)\left(ℛ\Delta \left(a\right){ℛ}^{-1}\right)\\ & =& \left(l_1\otimes l_2\right)\left(\Delta \left(a\right){ℛ}^{-1}\left(S^2\otimes S^2\right)\left(ℛ\right)\right),\text{by definiton of C,}\\ & =& \left(l_1\otimes l_2\right)\left(\Delta \left(a\right){ℛ}^{-1},ℛ\right)\\ & =& \left(l_1\otimes l_2\right)\left(\Delta \left(a\right)\right)\\ & =& \left(l_1{l}_2\right)\left(a\right),\end{array}$$ and hence $C$ is a commutative algebra. Let $a\in A.$ First note that $$\begin{array}{rcl}a\otimes 1& =& \left(\mathrm{id}\otimes \epsilon \right)\Delta \left(a\right)=\left(\mathrm{id}\otimes m\right)\left(\mathrm{id}\otimes S^{-1}\otimes \mathrm{id}\right)\left(\mathrm{id}\otimes {\Delta }^{\mathrm{op}}\right)\Delta \left(a\right)\\ & =& \sum _a{a}_{\left(1\right)}\otimes S^{-1}\left(a_{\left(3\right)}\right)a_{\left(2\right)}=\sum _a\left(1\otimes S^{-1}\left(a_{\left(2\right)}\right)\left(a_{\left(11\right)}\otimes a_{\left(12\right)}\right)\right)\\ & =& \sum _a\left(1\otimes S^{-1}\left(a_{\left(2\right)}\right)\right)\Delta \left(a\right),\end{array}$$ since $S^{-1}$ is the antipode of $A^{\mathrm{op}}$, and $$\begin{array}{rcl}a\otimes 1& =& \left(\mathrm{id}\otimes \epsilon \right)\Delta \left(a\right)=\left(\mathrm{id}\otimes m\right)\left(\mathrm{id}\otimes \mathrm{id}\otimes S\right)\left(\mathrm{id}\otimes \Delta \right)\Delta \left(a\right)\\ & =& \sum _a{a}_{\left(1\right)}\otimes a_{\left(2\right)}{S}^{\left(a_{\left(3\right)}\right)}=\sum _a\left(a_{\left(11\right)}\otimes a_{\left(12\right)}\right)\left(1\otimes S\left(a_{\left(2\right)}\right)\right)\\ & =& \sum _a\Delta \left(a_{\left(1\right)}\right)\left(1\otimes S\left(a_{\left(2\right)}\right)\right),\end{array}.$$ Then, since $${ℛ}^{21}ℛ\Delta \left(a\right)={ℛ}^{21}{\Delta }^{\mathrm{op}}\left(a\right)ℛ=\Delta \left(a\right){ℛ}^{21}ℛ,$$$$\begin{array}{rcl}a\phi \left(l\right)& =& a\left(\mathrm{id}\otimes l\right)\left({ℛ}^{21}ℛ\right)=\left(\mathrm{id}\otimes l\right)\left(\left(a\otimes 1\right){ℛ}^{21}{ℛ}^{12}\right)\\ & =& \left(\mathrm{id}\otimes l\right)\left(\sum _a\left(1\otimes S^{-1}\left(a_{\left(2\right)}\right)\right)\Delta \left(a_{\left(1\right)}\right){ℛ}^{21}ℛ\right)\\ & =& \left(\mathrm{id}\otimes l\right)\left(\sum _a\Delta \left(a_{\left(1\right)}\right){ℛ}^{21}ℛ\left(1\otimes S\left(a_{\left(2\right)}\right)\right)\right),\text{by definition of }C,\\ & =& \left(\mathrm{id}\otimes l\right)\left({ℛ}^{21}ℛ\sum _a\Delta \left(a_{\left(1\right)}\right)\left(1\otimes S\left(a_{\left(2\right)}\right)\right)\right)\\ & =& \left(\mathrm{id}\otimes l\right)\left({ℛ}^{21}ℛ\left(a\otimes 1\right)\right)=\left(\mathrm{id}\otimes l\right)\left({ℛ}^{21}ℛ\right)a=\phi \left(l\right)a,\end{array}$$ and so $\phi \left(l\right)\in Z\left(A\right).$ Since $$\begin{array}{rcl}\phi \left(l_1{l}_2\right)& =& \left(\mathrm{id}\otimes l_1{l}_2\right)\left({ℛ}^{21}ℛ\right)=\left(\mathrm{id}\otimes l_1\otimes l_2\right)\left(\left(\mathrm{id}\otimes \Delta \right)\left({ℛ}^{21}ℛ\right)\right)\\ & =& \left(\mathrm{id}\otimes l_1\otimes l_2\right)\left({ℛ}^{21}{ℛ}^{31}{ℛ}^{13}{ℛ}^{12}\right)=\left(\mathrm{id}\otimes l_1\right)\left({ℛ}^{21}\left(\phi \left(l_2\right)\otimes 1\right){ℛ}^{12}\right)\\ & =& \left(\mathrm{id}\otimes l_1\right)\left({ℛ}^{21}ℛ\left(\phi \left(l_2\right)\otimes 1\right)\right),\text{ since }\phi \left(l_2\right)\in Z\left(A\right),\\ & =& \phi \left(l_1\right)\phi \left(l_2\right),\end{array}$$ and so $\phi$ is a homomorphism. $\square$

Central elements $(\mathrm{id}\otimes {\mathrm{qtr}}_V)\left(a\right)$

The following proposition provides Drinfeld's "second central element construction" [Dr, Prop. 1.2 and Prop 3.3]. The model example is the case that $U=U_h𝔤, x_0=e^{h\rho }$ and $a={ℛ}_{21}ℛ.$

Let $(U,ℛ)$ be a quasitriangular Hopf algebra with antipode $S.$ Let $x_0\in U$ and $a\in U\otimes U$ be invertible elements such that $$\begin{array}{ll}{x}_{0}x{x}_0^{-1}=S^2\left(x\right)anda\Delta \left(x\right)=\Delta \left(x\right)a, \; for\; x\in U,& (Ss\; 2.1)\end{array}$$ respectively. Let $$\begin{array}{ll}{C}_0=\{l\in U^{*} | l\left(xy\right)=l\left(yx\right)\}and{C}_1=\{\ell \in U^{*} | \ell \left(xy\right)=\ell \left(y{S}^2\left(x\right)\right)\}.& (Ss\; 2.2)\end{array}$$ Let $Z\left(U\right)$ be the center of $U$ and let $\mathrm{Rep}$ be the Grothendieck group of finite dimensional representations of $U.$ Define $$\begin{array}{ll}\begin{array}{ccccccc}\mathrm{Rep}& \stackrel{\phantom{0}\mathrm{tr}\phantom{0}}{\to }& C_0& \stackrel{\phantom{0}{x}_0^{*}\phantom{0}}{\to }& C_1& \stackrel{\phantom{0}{f}_a\phantom{0}}{\to }& Z\left(U\right)\\ & & l& \mapsto & \ell & \mapsto & (\mathrm{id}\otimes \ell )\left(a\right)\\ V& \mapsto & {\mathrm{tr}}_V& \mapsto & {\mathrm{qtr}}_V& \mapsto & (\mathrm{id}\otimes {\mathrm{qtr}}_V)\left(a\right)\end{array}where\; \ell \left(u\right)=l\left(u{x}_0\right)& (Ss\; 2.3)\end{array}$$ for $u\in U.$ Then

1. $C_k=\{\ell \in U^{*} | \ell \left(xy\right)=\ell \left(y{S}^{2k}\left(x\right)\right)\}$ is a commutative subalgebra of $U^{*};$
2. the maps $x_0^{*}$ and $f_a$ are well defined;
3. $\mathrm{tr}:\mathrm{Rep}\to C_0$ is an algebra homomorphism;
4. if $\Delta \left(x_0\right)=x_0\otimes x_0$ then $x_0^{*}:C_0\to C_1$ is an algebra homomorphism; and
5. if $a={ℛ}_{21}ℛ$ then $f_a:C_1\to Z\left(U\right)$ is an algebra homomorphism.

		Proof.
	Let ${\ell }_1,{\ell }_2\in C_k.$ Then $$\begin{array}{rcl}\left({\ell }_1{\ell }_2\right)\left(xy\right)& =& ({\ell }_1\otimes {\ell }_2)\left(\Delta \left(xy\right)\right)\\ & =& ({\ell }_1\otimes {\ell }_2)\left(\Delta \left(x\right)\Delta \left(y\right)\right)\\ & =& ({\ell }_1\otimes {\ell }_2)\left(\Delta \left(y\right)(S^{2k}\otimes S^{2k})\Delta \left(x\right)\right)\\ & =& ({\ell }_1\otimes {\ell }_2)\left(\Delta \left(y\right)\Delta \left(S^{2k}\left(x\right)\right)\right)\\ & =& ({\ell }_1\otimes {\ell }_2)\left(\Delta \left(y{S}^{2k}\left(x\right)\right)\right)\\ & =& \left({\ell }_1{\ell }_2\right)\left(y{S}^{2k}\left(x\right)\right),\end{array}$$ where the fourth equality follows from the fact that $S^2$ is a coalgebra automorphism. Thus $C_k$ is a subalgebra of $U^{*}.$ Since $(S\otimes S)\left(ℛ\right)=ℛ,$ $$\begin{array}{rcl}\left({\ell }_1{\ell }_2\right)\left(x\right)& =& ({\ell }_1\otimes {\ell }_2)\Delta \left(x\right)\\ & =& ({\ell }_2\otimes {\ell }_1){\Delta }^{\mathrm{op}}\left(x\right)\\ & =& ({\ell }_2\otimes {\ell }_1)\left(ℛ\Delta \left(x\right){ℛ}^{-1}\right)\\ & =& ({\ell }_2\otimes {\ell }_1)\left(\Delta \left(x\right){ℛ}^{-1}(S^{2k}\otimes S^{2k})\left(ℛ\right)\right)\\ & =& ({\ell }_2\otimes {\ell }_1)(\Delta \left(x\right){ℛ}^{-1}ℛ\\ & =& ({\ell }_2\otimes {\ell }_1)\Delta \left(x\right)\\ & =& {\ell }_2{\ell }_1\left(x\right).\end{array}$$ So $C_k$ is commutative. Let $\ell =x_0^{*}\left(l\right).$ Then $$\ell \left(xy\right)=l\left(xy{x}_0\right)=l\left(y{x}_{0}x\right)=l\left(y{x}_{0}x{x}_0^{-1}{x}_0\right)=l\left(y{S}^2\left(x\right)x_0\right)=\ell \left(y{S}^2\left(x\right)\right),$$ and thus $x_0^{*}\left(l\right)\in C_1$ and $x_0^{*}$ is well defined. The identities $$\begin{array}{ll}\epsilon \left(x\right)=\sum _x{S}^{-1}\left(x_{\left(2\right)}\right)x_{\left(1\right)}and\sum _x{x}_{\left(1\right)}S\left(x_{\left(2\right)}\right)=\epsilon \left(x\right),& (Ss\; 2.4)\end{array}$$ from the definition of a Hopf algebra are the relations which provide the isomorphisms in [DRV] (A.8). For $x\in U,$ $$\begin{array}{rcl}x(\mathrm{id}\otimes \ell \left(a\right))& =& (\mathrm{id}\otimes \ell )\left((x\otimes 1)a\right)\\ & =& (\mathrm{id}\otimes \ell )\left(\sum _x(x_{\left(1\right)}\otimes \epsilon \left(x_{\left(2\right)}\right))a\right)\\ & =& (\mathrm{id}\otimes \ell )\left(\sum _x(x_{\left(1\right)}\otimes S^{-1}\left(x_{\left(3\right)}\right)x_{\left(2\right)})a\right)\\ & =& (\mathrm{id}\otimes \ell )\left(\sum _x(x_{\left(1\right)}\otimes x_{\left(2\right)})a(1\otimes S\left(x_{\left(3\right)}\right))\right),since\; \ell \left(bc\right)=\ell \left(c{S}^2\left(b\right)\right),\\ & =& (\mathrm{id}\otimes \ell )\left(\sum _{x}a(x_{\left(1\right)}\otimes x_{\left(2\right)})(1\otimes S\left(x_{\left(3\right)}\right))\right),since\; \Delta \left(x\right)a=a\Delta \left(x\right),\\ & =& (\mathrm{id}\otimes \ell )\left(\sum _{x}a(x_{\left(1\right)}\otimes \epsilon \left(x_{\left(2\right)}\right)))\right)\\ & =& (\mathrm{id}\otimes \ell )\left(a(x\otimes 1)\right)\\ & =& \left((\mathrm{id}\otimes \ell )\left(a\right)\right)x.\end{array}$$ So $(\mathrm{id}\otimes \ell )\left(a\right)\in Z\left(U\right).$ Hence $f_a\left(\ell \right)=(\mathrm{id}\otimes \ell )\left(a\right)$ is an element of the center of $U,$ and $f_a$ is well defined. Let $M,N\in \mathrm{Rep}.$ Let $\left\{m_i\right\}$ be a basis of $M,$ $\left\{m^i\right\}$ a dual basis in $M^{*},$ $\left\{n_j\right\}$ a basis of $N$ and $\left\{n^j\right\}$ a dual basis in $N^{*}.$ Then $$\begin{array}{rcl}({\mathrm{tr}}_M\cdot {\mathrm{tr}}_N)\left(x\right)& =& \sum _x{\mathrm{tr}}_M\left(x_{\left(1\right)}\right){\mathrm{tr}}_N\left(x_{\left(2\right)}\right)\\ & =& \sum _{x,i,j}⟨x_{\left(1\right)}{m}_i\otimes x_{\left(2\right)}{n}_j,m^i\otimes n^j⟩\\ & =& \sum _{i,j}⟨x(m_i\otimes n_j),m^i\otimes n^j⟩\\ & =& {\mathrm{tr}}_{M\otimes N}\left(x\right).\end{array}$$ So $\mathrm{tr}:\mathrm{Rep}\to C_0$ is an algebra homomorphism. Assume $\Delta \left(x_0\right)=x_0\otimes x_0.$ Let $l_1,l_2\in C_0.$ Then $$\begin{array}{rcl}{x}_0^{*}\left(l_1{l}_2\right)\left(x\right)& =& \left(l_1{l}_2\right)\left(x{x}_0\right)\\ & =& (l_1\otimes l_2)\left(\Delta \left(x{x}_0\right)\right)\\ & =& (l_1\otimes l_2)\left(\Delta \left(x\right)(x_0\otimes x_0)\right)\\ & =& (x_0^{*}\left(l_1\right)\otimes x_0^{*}\left(l_2\right))\left(\Delta \left(x\right)\right)\\ & =& \left(x_0^{*}\left(l_1\right)x_0^{*}\left(l_2\right)\right)\left(x\right).\end{array}$$ So $x_0^{*}:C_0\to C_1$ is an algebra homomorphism. Assume $a={ℛ}_{21}ℛ.$ Let ${\ell }_1,{\ell }_2\in C_1.$ Then $$\begin{array}{rcl}(\mathrm{id}\otimes {\ell }_1{\ell }_2)\left({ℛ}_{21}ℛ\right)& =& (\mathrm{id}\otimes {\ell }_1\otimes {\ell }_2)(\mathrm{id}\otimes \Delta )\left({ℛ}_{21}ℛ\right)\\ & =& (\mathrm{id}\otimes {\ell }_1\otimes {\ell }_2)\left({ℛ}_{21}\right)\left({ℛ}_{21}{ℛ}_{13}ℛ\right)\\ & =& (\mathrm{id}\otimes {\ell }_1)({ℛ}_{21}\cdot ((\mathrm{id}\otimes {\ell }_2)\left({ℛ}_{21}ℛ\right)\otimes 1)\cdot ℛ).\end{array}$$ Since $(\mathrm{id}\otimes {\ell }_2)\left({ℛ}_{21}ℛ\right)\in Z(U),$ $$\begin{array}{rcl}{f}_a\left({\ell }_1{\ell }_2\right)& =& (\mathrm{id}\otimes {\ell }_1{\ell }_2)\left({ℛ}_{21}ℛ\right)\\ & =& (\mathrm{id}\otimes {\ell }_1)({ℛ}_{21}ℛ\cdot ((\mathrm{id}\otimes {\ell }_2)\left({ℛ}_{21}ℛ\right)\otimes 1))\\ & =& (\mathrm{id}\otimes {\ell }_1)\left({ℛ}_{21}ℛ\right)(\mathrm{id}\otimes {\ell }_2)\left({ℛ}_{21}ℛ\right)\\ & =& f_a\left({\ell }_1\right)f_a\left({\ell }_2\right).\end{array}$$ So $f_a:C_1\to Z\left(U\right)$ is an algebra homomorphism. $\square$

Let $U=U𝔤$ and let $a=\gamma =\sum _{b}b\otimes b^{*}.$ Let ${\ell }_1,{\ell }_2\in C_1.$ Then $$\begin{array}{rcl}{f}_a\left({\ell }_1{\ell }_2\right)& =& (\mathrm{id}\otimes {\ell }_1{\ell }_2)\left(\gamma \right)\\ & =& (\mathrm{id}\otimes {\ell }_1\otimes {\ell }_2)(\mathrm{id}\otimes \Delta )\left(\gamma \right)\\ & =& (\mathrm{id}\otimes {\ell }_1\otimes {\ell }_2)\left(\sum _{b}b\otimes b^{*}\otimes 1+b\otimes 1\otimes b^{*}\right)\\ & =& f_a\left({\ell }_1\right){\ell }_2\left(1\right)+{\ell }_1\left(1\right)f_a\left({\ell }_2\right).\end{array}$$ Thus, since $\left({\ell }_1{\ell }_2\right)\left(1\right)=({\ell }_1\otimes {\ell }_2)\left(\Delta \left(1\right)\right)=({\ell }_1\otimes {\ell }_2)(1\otimes 1)={\ell }_1\left(1\right){\ell }_2\left(1\right),$ $$\begin{array}{ll}\frac{f_a\left({\ell }_1{\ell }_2\right)}{\left({\ell }_1{\ell }_2\right)\left(1\right)}=\frac{f_1\left({\ell }_1\right)}{{\ell }_1\left(1\right)}+\frac{f_a\left({\ell }_2\right)}{{\ell }_2\left(1\right)}.& (Ss\; 2.5)\end{array}$$

$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({ℛ}_{21}ℛ\right)$ when $U=U_h𝔤$

In the case that $U=U_h𝔤$ and $x_0=v^{-1}u=e^{h\rho },$ $M$ and $N$ are $U-$modules, and ${\ell }_1={\mathrm{qtr}}_M$ and ${\ell }_2={\mathrm{qtr}}_N$ then (d) and (e) above give that, as elements of $U^{*},$ $$({\mathrm{qtr}}_M\cdot {\mathrm{qtr}}_N)\left(x\right)={\mathrm{qtr}}_{M\otimes N}\left(x\right),$$ and $$(\mathrm{id}\otimes {\mathrm{qtr}}_{M\otimes N})\left({ℛ}_{21}ℛ\right)=(\mathrm{id}\otimes {\mathrm{qtr}}_M)\left({ℛ}_{21}ℛ\right)(\mathrm{id}\otimes {\mathrm{qtr}}_N)\left({ℛ}_{21}ℛ\right)$$ (as explained in [Dr, Prop. 3.3] and [Bau, Prop. 2]). In this case, the computation in the proof of (d) and (e), pictorially, is $$\begin{array}{c} \end{array}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}=\begin{array}{c} \end{array}$$

Fix a Cartan subalgebra $𝔥$ in $𝔤.$ For $\gamma \in {𝔥}^{*}$ define the Weyl character $$s_{\gamma }=\frac{a_{\gamma +\rho }}{a_{\rho }},where{a}_{\mu }=\sum _{w\in W_0}\det \left(w\right)X^{w\mu }.$$ The expressions $s_{\gamma }$ and $a_{\mu }$ are elements of the group algebra of ${𝔥}^{*},$ $ℂ\left[{𝔥}^{*}\right]=\mathrm{span}\left\{X^{\nu } \right| \nu \in {𝔥}^{*}\},$ and, if $w\in W_0$ then $$a_{w\mu }=\det \left(w\right)a_{\mu },and{s}_{w\circ \mu }=\det \left(w\right)s_{\mu },$$ where the dot action of $W_0$ on ${𝔥}^{*}$ is given by $$\begin{array}{ll}w\circ \mu =w(\lambda +\rho )-\rho ,for\; w\in W_0, \mu \in {𝔥}^{*}.& (Ss\; 3.1)\end{array}$$ The Weyl denominator formula says that $$\begin{array}{ll}{a}_{\rho }=\prod _{\alpha \in R^{+}}(X^{\alpha /2}-X^{-\alpha /2})& (Ss\; 3.2)\end{array}$$ and the Weyl character formula says that if $\gamma$ is a dominant integral weight, $L\left(\lambda \right)$ is the simple $𝔤-$module of highest weight $\gamma ,$ and $v_1,...,v_n$ is a basis of $L\left(\lambda \right)$ consisting of weight vectors then $$\begin{array}{ll}{s}_{\lambda }=\sum _{i=1}^n{X}^{\mathrm{wt}\left(v_i\right)}=\sum _{\nu \in P}{K}_{\lambda \nu }{X}^{\nu },where{K}_{\lambda \nu }=\dim \left(L{\left(\lambda \right)}_{\nu }\right),& (Ss\; 3.3)\end{array}$$ the dimension of the $\nu$ weight space of $L\left(\lambda \right).$ For $\mu \in {𝔥}^{*}$ define an algebra homomorphism $${\mathrm{ev}}_{\mu }:ℂ\left[{𝔥}^{*}\right]\to ℂ\left[{𝔥}^{*}\right]by{\mathrm{ev}}_{\mu }\left(X^{\nu }\right)=q^{⟨\mu ,\nu ⟩}.$$ Then $$\begin{array}{ll}{\dim }_q\left(L\left(\nu \right)\right)={\mathrm{ev}}_{2\rho }\left(s_{\nu }\right),& (Ss\; 3.4)\end{array}$$ since ${\dim }_q\left(L\left(\nu \right)\right)={\mathrm{tr}}_{L\left(\nu \right)}\left(e^{h\rho }\right)$ and $q=e^{\frac{h}{2}}.$

[TW, Lemma 3.5.1] Let $𝔤$ be a finite dimensional complex semisimple Lie algebra and let $U_h𝔤$ be the corresponding Drinfeld-Jimbo quantum group. etc etc ???? Let $\nu$ be a dominant integral weight so that the irreducible module $L\left(\nu \right)$ of highest weight $\nu$ is finite dimensional. Then $$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({ℛ}_{21}ℛ\right)acts\; on\; L\left(\mu \right) \; by\; {\mathrm{ev}}_{2(\mu +\rho )}\left(s_{\nu }\right){\mathrm{id}}_{L\left(\mu \right)}.$$

		Proof.
	Let $h_1,...,h_r$ be an orthonormal basis of $𝔥.$ By [Dr, §4] (see [LR, (2.13)]) there is an expression $$ℛ=e^{\frac{1}{2}h{\gamma }_0}+\sum b_j^{+}\otimes b_j^{-},where{\gamma }_0=\sum _{l=1}^r{h}_l\otimes h_l,$$ and $b_j^{+}\in U^{+}$ and $b_j^{-}\in U^{-}$ are homogeneous elements of degree greater than $0.$ Let $v_1,...,v_n$ be a basis of weight vectors of $L\left(\nu \right)$ and let $v^1,...,v^n$ be the dual basis in $L{\left(\nu \right)}^{*}.$ Let $v_{\mu }^{+}$ be a highest weight vector in $L\left(\mu \right).$ Since $b_j^{+}{v}_{\mu }^{+}=0$ and $q=e^{\frac{h}{2}},$ ${ℛ}_{21}ℛ\otimes 1$ acts on $v_{\mu }^{+}\otimes v_i\otimes v^i\in L\left(\mu \right)\otimes L\left(\nu \right)\otimes L{\left(\nu \right)}^{*}$ by $$\begin{array}{rcl}({ℛ}_{21}ℛ\otimes \mathrm{id})(v_{\mu }^{+}\otimes v_i\otimes v^i)& =& \left(\left(q^{{\gamma }_0}+\sum _j(b_j^{-}\otimes b_j^{+})\right)\left(q^{{\gamma }_0}+\sum _j(b_j^{+}\otimes b_j^{-})\right)(v_{\mu }^{+}\otimes v_i)\right)\otimes v^i\\ & =& \left(\left(q^{{\gamma }_0}{q}^{{\gamma }_0}+q^{{\gamma }_0}\sum _j(b_j^{-}\otimes b_j^{+})\right)(v_{\mu }^{+}\otimes v_i)\right)\otimes v^i\\ & =& \left(\left(q^{{\gamma }_0}{q}^{{\gamma }_0}(v_{\mu }^{+}\otimes v_i\otimes v^i)+q^{{\gamma }_0}\sum _j(b_j^{-}{v}_{\mu }^{+}\otimes b_j^{+}{v}_i\otimes v^i)\right)(v_{\mu }^{+}\otimes v_i)\right)\otimes v^i\end{array}$$ Since $(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({ℛ}_{21}ℛ\right)$ is central in $U,$ it acts on $v_{\mu }^{+}$ by a scalar. Therefore, since $b_j^{-}$ is a lowering operator, $$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({ℛ}_{21}ℛ\right)=(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left(q^{2{\gamma }_0}\right),$$ as proved in [Dr, Prop. 5.3]. Then $$\begin{array}{rcl}(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({ℛ}_{21}ℛ\right)& =& (\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left(q^{2{\gamma }_0}\right)\\ & =& \sum _i{q}^{2{\gamma }_0}(1\otimes e^{h\rho }){(v_{\mu }^{+}\otimes v_i)|}_{v_{\mu }^{+}\otimes v_i}\\ & =& \sum _i{v}_i{q}^{2\sum _{l=1}^r\mu \left(h_l\right)\mathrm{wt}\left(v_i\right)\left(h_l\right)}{q}^{⟨2\rho ,\mathrm{wt}\left(v_i\right)⟩}{(v_{\mu }^{+}\otimes v_i)|}_{v_{\mu }^{+}\otimes v_i}\\ & =& \sum _i{q}^{2⟨\mu +\rho ,\mathrm{wt}\left(v_i\right)⟩}\\ & =& {\mathrm{ev}}_{2(\mu +\rho )}\left(s_{\nu }\right)\\ & =& \frac{\sum _{w\in W_0}\det \left(w\right)q^{2⟨\mu +\rho ,w(\nu +\rho )⟩}}{\sum _{w\in W_0}\det \left(w\right)q^{2⟨\mu +\rho ,w\rho ⟩}}.\end{array}$$ $\square$

The Turaev-Wenzl identity almost provides an inverse to the Harish-Chandra homomorphism. In the case of $U=U_h{\mathrm{𝔰𝔩}}_2,$ $$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left({\omega }_1\right)})\left({ℛ}_{21}ℛ\right)=(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left({\omega }_1\right)})\left(e^{h\left(\frac{1}{2}(H\otimes H)\right)}\right)=e^{\frac{h}{2}(H+1)}+e^{-\frac{h}{2}(H+1)},$$ and this element acts on $L\left(\mu {\omega }_1\right)$ by the constant $${\mathrm{ev}}_{2(\mu +\rho )}(X^{{\omega }_1}+X^{-{\omega }_1})=q^{\mu +1}+q^{-(\mu +1)}={\mathrm{ev}}_{\mu +\rho }(e^{\frac{h}{2}H}+e^{-\frac{h}{2}H})={\mathrm{ev}}_{\mu +\rho }(K+K^{-1}).$$

$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)}\left({\left({ℛ}_{21}ℛ\right)}^l\right))$ when $U=U_h𝔤$

Drinfeld [Dr, last par. of §4] explains the connection between the construction of central elements in (Ss 2.3) and the construction of central elements in [RTF, Theorem 14]. This is expanded by Baumann [Bau, 3rd par. of §3] as follows. Let $U=U_h𝔤$ and let $\pi :U\to M_n\left(ℂ\right)$ be a representation of $U_h𝔤$ and let ${\pi }_{ij}:U\to ℂ$ be the matrix coefficients of $\pi .$ As in [RTF, Theorem 16(1) and Theorem 18] set $$L^{+}=\left(l_{ij}^{+}\right)=(\pi \otimes \mathrm{id})\left({ℛ}_{21}\right)so\; that{l}_{ij}^{+}=(\mathrm{id}\otimes {\pi }_{ij})\left(ℛ\right).$$ Let $$L^{-}=\left(l_{ij}^{-}\right)=(\pi \otimes \mathrm{id})(\mathrm{id}\otimes S^{-1})\left(ℛ\right)=(\pi \otimes \mathrm{id}){ℛ}^{-1})so\; thatS\left(l_{ij}^{-}\right)=({\pi }_{ij}\otimes \mathrm{id})\left(ℛ\right).$$ Then $e^{h\rho }{\left(L^{+}S\left(L^{-}\right)\right)}^k$ is a matrix with entries in $U$ and $$\begin{array}{rcl}\mathrm{tr}\left(e^{h\rho }{\left(L^{+}S\left(L^{-}\right)\right)}^k\right)& =& \sum _{\genfrac{}{}{0ex}{}{i_1,...,i_k}{j_1,...,j_k}}{q}^{⟨2\rho ,{\lambda }_{i_1}⟩}{l}_{i_1{j}_1}^{+}S\left(l_{j_1{i}_2}^{-}\right)l_{i_1{j}_2}^{+}S\left(l_{j_2{i}_3}^{-}\right)\cdots l_{i_k{j}_k}^{+}S\left(l_{j_k{i}_1}^{-}\right)\\ & =& \sum _{\genfrac{}{}{0ex}{}{i_1,...,i_k}{j_1,...,j_k}}{q}^{⟨2\rho ,{\lambda }_{i_1}⟩}(\mathrm{id}\otimes {\pi }_{i_1{j}_1})\left(ℛ\right)({\pi }_{j_1{i}_2}\otimes \mathrm{id})\left(ℛ\right)\cdots (\mathrm{id}\otimes {\pi }_{i_k{j}_k})\left(ℛ\right)({\pi }_{j_k{i}_1}\otimes \mathrm{id})\left(ℛ\right)\\ & =& \sum _{i_1}{q}^{⟨2\rho ,{\lambda }_{i_1}⟩}({\pi }_{i_1{i}_1}\otimes \mathrm{id})\left({\left({ℛ}_{21}ℛ\right)}^k\right)\\ & =& ?????\; (\mathrm{id}\otimes {\mathrm{qtr}}_{\pi })\left({\left({ℛ}_{21}ℛ\right)}^k\right).\end{array}$$

Examples. Since ${\gamma }^0={\left({ℛ}_{21}ℛ\right)}^0=1\otimes 1$ the definition of $\dim \left(L\left(\nu \right)\right)$ and ${\dim }_q\left(L\left(\nu \right)\right)$ give $$(\mathrm{id}\otimes {\mathrm{tr}}_{L\left(\nu \right)})\left({\gamma }^0\right)=\dim \left(L\left(\nu \right)\right)and(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({\left({ℛ}_{21}ℛ\right)}^0\right)={\dim }_q\left(L\left(\nu \right)\right).$$ Since ${\pi }_0\left((\mathrm{id}\otimes {\mathrm{tr}}_{L\left(\nu \right)})\left(\gamma \right)\right)$ is a degree 1 element of $S\left(𝔥\right),$ $S^1{\left(𝔥\right)}^{W_0}=0$ and ${\pi }_0:Z\left(U\right)\to {\sigma }_{\rho }\left(S{\left(𝔥\right)}^{W_0}\right)$ is an isomorphism, it follows that $$(\mathrm{id}\otimes {\mathrm{tr}}_{L\left(\nu \right)})\left(\gamma \right)=0.$$

If $f=\sum _{\nu \in {𝔥}^{*}}{f}_{\nu }{X}^{\nu }$ is an element of $ℂ{\left[{𝔥}^{*}\right]}^{W_0}$ then $$\begin{array}{rcll}f{s}_{\mu }& =& f\frac{a_{\mu +\rho }}{a_{\rho }}& \\ & =& \frac{1}{a_{\rho }}f\sum _{w\in W_0}\det \left(w\right)w{X}^{\mu +\rho }& \\ & =& \frac{1}{a_{\rho }}\sum _{w\in W_0}\det \left(w\right)wf{X}^{\mu +\rho }& \\ & =& \frac{1}{a_{\rho }}\sum _{\nu \in {𝔥}^{*}}\sum _{w\in W_0}\det \left(w\right)w{f}_{\nu }{X}^{\nu }{X}^{\mu +\rho }& \\ & =& \frac{1}{a_{\rho }}\sum _{\nu \in {𝔥}^{*}}{f}_{\nu }{a}_{\nu +\mu +\rho }& \\ & =& \sum _{\nu \in {𝔥}^{*}}{f}_{\nu }{s}_{\nu +\mu }.& (Ss\; 4.1)\end{array}$$ Two special cases of (Ss 4.1) are $$\begin{array}{ll}{s}_{\gamma }{s}_{\mu }=\sum _{\nu \in P}{K}_{\gamma \nu }{s}_{\nu +\mu },& (Ss\; 4.2)\end{array}$$ and $$\begin{array}{ll}\sum _{w\in W_0}{X}^{w\lambda }=s_{\varnothing }\sum _{w\in W_0}{X}^{w\lambda }=\sum _{w\in W_0}{s}_{w\lambda }=\sum _{w\in W_0}\det \left(w^{-1}\right)s_{w^{-1}\circ w\lambda }=\sum _{w\in W_0}\det \left(w^{-1}\right)s_{\lambda +w^{-1}\rho -\rho }.& (Ss\; 4.3)\end{array}$$ Comparing coefficients of $X^{\nu }$ in (Ss 4.3) gives $$\begin{array}{ll}{\delta }_{v\lambda ,\nu }\left|{\left(W_0\right)}_{\lambda }\right|=\sum _{w\in W_0}{K}_{w\lambda ,\nu }=\sum _{w\in W_0}\det \left(w\right)K_{\lambda +w\rho -\rho ,\nu }& (Ss\; 4.4)\end{array}$$ for some $v\in W_0.$

Baumann's identity

Identify Rep with $ℂ{\left[X\right]}^{W_0}=\mathrm{span}\left\{s_{\lambda } \right| \lambda \; dominant\; integral\}.$ For $l\in {\mathbb{Z}}_{\ge 0}$ let $$\begin{array}{rrcl}{\Psi }_l:& ℂ{\left[X\right]}^{W_0}& \to & Z\left(U_h𝔤\right)\\ & s_{\nu }& \mapsto & (\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({\left({ℛ}_{21}ℛ\right)}^l\right)\end{array}$$

[Bau, Thm. 1] For $l\in {\mathbb{Z}}_{\ge 0}$ define $$m_{\lambda }^{\left(l\right)}=\sum _{w\in W_0}{q}^{2l⟨w\lambda ,\rho ⟩}{s}_{w\lambda }.$$ Then $${\Psi }_l\left(m_{\lambda }^{\left(l\right)}\right)={\Psi }_1\left(m_{l\lambda }^{(1/l)}\right).$$

		Proof.
	First note that $$\begin{array}{rcl}{\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_l\left(s_{\gamma }\right)\right)& =& ({\mathrm{qtr}}_{L\left(\mu \right)}\otimes {\mathrm{qtr}}_{L\left(\gamma \right)})\left({\left({ℛ}_{21}ℛ\right)}^l\right)\\ & =& \sum _{\nu \in P}{K}_{\gamma \nu }{\mathrm{qtr}}_{L(\mu +\nu )}\left(q^{l(⟨\mu +\nu ,\mu +\nu +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\gamma ,\gamma +2\rho ⟩)}\right)\\ & =& \sum _{\nu \in P}{K}_{\gamma \nu }{\dim }_q\left(L(\mu +\nu )\right)q^{l(⟨\mu +\nu ,\mu +\nu +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\gamma ,\gamma +2\rho ⟩)}\\ & =& \sum _{\nu \in P}{K}_{\gamma \nu }\frac{{\mathrm{ev}}_{2\rho }\left(a_{\mu +\nu +\rho }\right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}{q}^{l(⟨\mu +\nu ,\mu +\nu +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\gamma ,\gamma +2\rho ⟩)}\end{array}$$ where the second equality uses (Ss 4.2). Taking ${\mathrm{qtr}}_{L\left(\mu \right)}$ of the LHS of the Baumann identity is $$\begin{array}{rcl}{\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_l\left(m_{\lambda }^l\right)\right)& =& {\mathrm{qtr}}_{L\left(\mu \right)}\left(\sum _{w\in W_0}\det \left(w\right)q^{2l⟨\lambda ,w\rho ⟩}{\Psi }_l\left(s_{\lambda +w\rho -\rho }\right)\right)\\ & =& \sum _{w\in W_0}\det \left(w\right)q^{2l⟨\lambda ,w\rho ⟩}\sum _{\nu \in P}{K}_{\lambda +w\rho -\rho ,\nu }\frac{{\mathrm{ev}}_{2\rho }\left(a\mu +\nu +\rho \right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}{q}^{l(⟨\mu +\nu ,\mu +\nu +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\lambda +w\rho -\rho ,\lambda +w\rho -\rho +2\rho ⟩)}\\ & =& \sum _{\nu \in P}\frac{{\mathrm{ev}}_{2\rho }\left(a\mu +\nu +\rho \right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}\sum _{w\in W_0}\det \left(w\right)K_{\lambda +w\rho -\rho ,\nu }{q}^{2l⟨\lambda ,w\rho ⟩}{q}^{l(⟨\mu +\nu ,\mu +\nu +2\rho ⟩-⟨\mu ,\mu +2\rho ⟩-⟨\lambda +w\rho -\rho ,\lambda +w\rho -\rho +2\rho ⟩)}\\ & =& \sum _{\nu \in P}\frac{{\mathrm{ev}}_{2\rho }\left(a\mu +\nu +\rho \right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}{q}^{l(⟨\nu ,\nu ⟩-⟨\lambda ,\lambda ⟩+2⟨\nu ,\mu +\rho ⟩)}\sum _{w\in W_0}\det \left(w\right)K_{\lambda +w\rho -\rho ,\nu }\\ & =& \sum _{w\in W_0}\frac{{\mathrm{ev}}_{2\rho }\left(a\mu +w\lambda +\rho \right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}{q}^{2l⟨w\lambda ,\mu +\rho ⟩},\end{array}$$ where the last equality uses (Ss 4.4). Taking ${\mathrm{qtr}}_{L\left(\mu \right)}$ of the RHS of the Baumann identity is $$\begin{array}{rcl}{\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_1\left(m_{l\lambda }^{(1/l)}\right)\right)& =& {\mathrm{qtr}}_{L\left(\mu \right)}{\Psi }_1\left(\sum _{v\in W_0}\det \left(v\right)q^{2⟨\lambda ,v\rho ⟩}{s}_{l\lambda +v\rho -\rho }\right)\\ & =& \sum _{v\in W_0}\det \left(v\right)q^{2⟨\lambda ,v\rho ⟩}{\dim }_q\left(L\left(\mu \right)\right){\mathrm{ev}}_{2(\mu +\rho )}\left(s_{l\lambda +v\rho -\rho }\right),\end{array}$$ by Turaev-Wenzl. Thus $$\begin{array}{rcl}{\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_l\left(m_{\lambda }^l\right)\right)& =& \sum _{v\in W_0}\det \left(v\right)q^{2⟨\lambda ,v\rho ⟩}\frac{{\mathrm{ev}}_{2\rho }\left(a_{\mu +\rho }\right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}\frac{{\mathrm{ev}}_{2(\mu +\rho )}\left(a_{l\lambda +v\rho }\right)}{{\mathrm{ev}}_{2(\mu +\rho )}\left(a_{\rho }\right)}\\ & =& \frac{1}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}\sum _{v\in W_0}\det \left(v\right)q^{2⟨\lambda ,v\rho ⟩}{\mathrm{ev}}_{2(\mu +\rho )}\left(a_{l\lambda +v\rho }\right)\\ & =& \frac{1}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)}\sum _{v\in W_0}{q}^{2⟨v^{-1}\lambda ,\rho ⟩}{\mathrm{ev}}_{2(\mu +\rho )}\left(a_{l{v}^{-1}\lambda +\rho }\right)\\ & =& \sum _{w\in W_0}{q}^{2⟨wl\lambda ,\mu +\rho ⟩}\frac{{\mathrm{ev}}_{2\rho }\left(a_{w\lambda +\mu +\rho }\right)}{{\mathrm{ev}}_{2\rho }\left(a_{\rho }\right)},\end{array}$$ since $$\begin{array}{rcl}\sum _{y\in W_0}{q}^{2⟨y\lambda ,\rho ⟩}{\mathrm{ev}}_{2(\mu +\rho )}\left(a_{ly\lambda +\rho }\right)& =& {\mathrm{ev}}_{2(\mu +\rho )}\left(\sum _{y,w\in W_0}{q}^{2⟨y\lambda ,\rho ⟩}\det \left(w\right)X^{lwy\lambda +w\rho }\right)\\ & =& \sum _{y,w\in W_0}{q}^{2⟨wy\lambda ,w\rho ⟩}\det \left(w\right)q^{2⟨\mu +\rho ,lwy\lambda ⟩}{q}^{2⟨\mu +\rho ,w\rho ⟩}\\ & =& \sum _{x,w\in W_0}\det \left(w\right)q^{2⟨\mu +\rho ,lx\lambda ⟩}{q}^{2⟨x\lambda +\mu +\rho ,w\rho ⟩}\\ & =& \sum _{x\in W_0}{q}^{2⟨\mu +\rho ,xl\lambda ⟩}{\mathrm{ev}}_{2\rho }\left(a_{x\lambda +\mu +\rho }\right).\end{array}$$ Thus $${\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_l\left(m_{\lambda }^l\right)\right)={\mathrm{qtr}}_{L\left(\mu \right)}\left({\Psi }_1\left(m_{l\lambda }^{(1/l)}\right)\right)$$ which completes the proof of the Baumann identity (THOUGH NOT THE STATEMENT THAT THE $B_{L(\lambda +w\rho -\rho )}^{\left(l\right)}$ ARE CHARACTERIZED BY THIS IDENTITY. $\square$

It follows from (Ss 3.4)??? that the quantum dimension is $$\begin{array}{rcl}{\dim }_q\left(L\left(\nu \right)\right)& =& {\mathrm{ev}}_{2\rho }\left(s_{\nu }\right)\\ & =& {\mathrm{ev}}_{2\rho }\left(\frac{a_{\nu +\rho }}{a_{\rho }}\right)\\ & =& \frac{\sum _{w\in W}\det \left(w\right)q^{⟨w(\nu +\rho ),2\rho ⟩}}{\sum _{w\in W}\det \left(w\right)q^{⟨w\rho ,2\rho ⟩}}\\ & =& \frac{\sum _{w\in W}\det \left(w\right)q^{⟨2(\nu +\rho ),w\rho ⟩}}{\sum _{w\in W}\det \left(w\right)q^{⟨2\rho ,w\rho ⟩}}\\ & =& \frac{{\mathrm{ev}}_{2(\nu +\rho )}\left(a_{\rho }\right)}{{\mathrm{ev}}_{2\rho }\left(a\rho \right)}\\ & =& \prod _{\alpha \in R^{+}}\frac{{\mathrm{ev}}_{2(\nu +\rho )}(X^{\alpha /2}-X^{-\alpha /2})}{{\mathrm{ev}}_{2\rho }(X^{\alpha /2}-X^{-\alpha /2})}\\ & =& \prod _{\alpha \in R^{+}}\frac{q^{⟨\nu +\rho ,\alpha ⟩}-q^{-⟨\nu +\rho ,\alpha ⟩}}{q^{⟨\rho ,\alpha ⟩}-q^{-⟨\rho ,\alpha ⟩}}.\end{array}$$ Hence $$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({\left({ℛ}_{21}ℛ\right)}^0\right)=\prod _{\alpha \in R^{+}}\frac{\left[⟨\nu +\rho ,\alpha ⟩\right]}{\left[⟨\rho ,\alpha ⟩\right]},where\left[k\right]=\frac{q^k-q^{-k}}{q-q^{-1}},$$ and the Weyl dimension formula is $$(\mathrm{id}\otimes {\mathrm{qtr}}_{L\left(\nu \right)})\left({\gamma }^0\right)=\dim \left(L\left(\nu \right)\right)={\mathrm{ev}}_0\left(s_{\nu }\right)=\underset{q\to 1}{\lim }{\mathrm{ev}}_{2\rho }\left(s_{\nu }\right)=\prod _{\alpha \in R^{+}}\frac{⟨\nu +\rho ,\alpha ⟩}{⟨\rho ,\alpha ⟩}.$$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

[DRV] Z. Daugherty, A. Ram, R. Virk, Appendices to Affine and degenerate affine BMW algebras: Actions on tensor space.

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