[Bou, VIII §2 no. 3 Prop. 6 and VIII §6 no. 4 Cor. to Prop. 7] Let $\kappa \in U𝔤$ be the Casimir and let $e^{-h\rho }u\in U_h𝔤$ be the quantum Casimir as defined in (??) and (???), respectively. Then $${\pi }_0\left(\kappa \right)={\sigma }_{\rho }(-⟨\rho ,\rho ⟩+\sum _{i=1}^r{h}_i{h}_i^{*})\text{and}{\mathrm{ev}}_{\nu }\left({\pi }_0\right(\kappa \left)\right)=⟨\nu ,\nu +\rho ⟩,$$ ,

Proof.

Let $h_1,\dots ,h_r$ be a basis of $𝔥$ and let $h_1^{*},\dots ,h_r^{*}$ be the dual basis of $𝔥$ with respect to the restriction of $⟨,⟩$ to $𝔥$. The nondegenerate bilinear form $⟨,⟩:𝔥\times 𝔥\to ℂ$ is equivalent to a vector space isomorphism $$\nu :𝔥\stackrel{\sim }{⟶}{𝔥}^{*}\text{given by}\nu \left(h\right)=⟨h,\cdot ⟩\text{and}{\nu }^{-1}\left(\lambda \right)=\sum _{i=1}^r\lambda \left(h_i\right)h_i^{*},$$ for $h\in 𝔥$ and $\lambda \in {𝔥}^{*}$. The isomorphism $\nu :𝔥\stackrel{\sim }{⟶}{𝔥}^{*}$ and the form $⟨,⟩:𝔥\times 𝔥\to ℂ$ provide a form $⟨,⟩:{𝔥}^{*}\times {𝔥}^{*}\to ℂ$ given by $$⟨\lambda ,\mu ⟩=⟨{\nu }^{-1}\left(\lambda \right),{\nu }^{-1}\left(\mu \right)⟩=⟨\sum _{i=1}^r\lambda \left(h_i\right)h_i^{*},\sum _{j=1}^r\mu \left(h_j^{*}\right)h_j⟩=\sum _{i=1}^r\lambda \left(h_i\right)\mu \left(h_i^{*}\right).$$

Let $x_{\alpha }\in {𝔤}_{\alpha }$. There is a unique $h_{\alpha }\in 𝔥$ such that $$\text{if}h\in 𝔥\text{then}⟨h,h_{\alpha }⟩=\alpha \left(h\right).\text{Thus}\nu \left(h_{\alpha }\right)=\alpha .$$ A different element $h_{{\alpha }^{\vee }}$ is obtained by choosing $y_{\alpha }\in {𝔤}_{-\alpha }$ and $h_{{\alpha }^{\vee }}\in 𝔥$ such that $x_{\alpha },y_{\alpha },h_{{\alpha }^{\vee }}$ span an ${𝔰𝔩}_2$-subalgebra of $𝔤$, $$[x_{\alpha },y_{\alpha }]=h_{{\alpha }^{\vee }},[h_{{\alpha }^{\vee }},x_{\alpha }]=2x_{\alpha },[h_{{\alpha }^{\vee }},y_{\alpha }]=-2y_{\alpha }.$$ Then $2=\alpha \left(h_{{\alpha }^{\vee }}\right)$ and $$2=\alpha \left(h_{{\alpha }^{\vee }}\right)=⟨h_{\alpha },h_{{\alpha }^{\vee }}⟩=⟨h_{\alpha },[x_{\alpha },y_{\alpha }]⟩=-⟨[x_{\alpha },h_{\alpha }],y_{\alpha }⟩=\alpha \left(h_{\alpha }\right)⟨x_{\alpha },y_{\alpha }⟩=⟨h_{\alpha },h_{\alpha }⟩⟨x_{\alpha },y_{\alpha }⟩$$ gives $$⟨x_{\alpha },y_{\alpha }⟩=\frac{2}{⟨h_{\alpha },h_{\alpha }⟩}\text{and}{h}_{{\alpha }^{\vee }}=\frac{2}{⟨h_{\alpha },h_{\alpha }⟩}{h}_{\alpha }.$$ Thus $$\begin{array}{ll}\{h_1,\dots ,h_r,x_{\alpha },\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }\mid \alpha \in R^{+}\}& \text{is a basis of} 𝔤, \text{and}\\ \{h_1^{*},\dots ,h_r^{*},\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha },x_{\alpha }\mid \alpha \in R^{+}\}& \text{is the dual basis of} 𝔤,\end{array}$$ with respect to $⟨,⟩$.

Then $$\begin{array}{rcl}\kappa & =& \sum _{i=1}^r{h}_i{h}_i^{*}+\sum _{\alpha \in R^{+}}{x}_{\alpha }\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }+\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }{x}_{\alpha }\\ & =& \sum _{i=1}^r{h}_i{h}_i^{*}+\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}[x_{\alpha },y_{\alpha }]+2\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }{x}_{\alpha }\\ & =& \sum _{i=1}^r{h}_i{h}_i^{*}+\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{h}_{{\alpha }^{\vee }}+2\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }{x}_{\alpha }\\ & =& \sum _{i=1}^r{h}_i{h}_i^{*}+\sum _{\alpha \in R^{+}}{h}_{\alpha }+2\sum _{\alpha \in R^{+}}\frac{⟨h_{\alpha },h_{\alpha }⟩}{2}{y}_{\alpha }{x}_{\alpha }\end{array}$$ and so $$\text{if}\rho =\nu \left(\frac{1}{2}\sum _{\alpha \in R^{+}}{h}_{\alpha }\right)=\frac{1}{2}\sum _{\alpha \in R^{+}}\alpha \text{then}\frac{1}{2}\sum _{\alpha \in R^{+}}{h}_{\alpha }=\sum _{i=1}^r\rho \left(h_i\right)h_i^{*}.$$ Then $$\begin{array}{rcl}{\pi }_0\left(\kappa \right)& =& \sum _{i=1}^r{h}_i{h}_i^{*}+\sum _{\alpha \in R^{+}}{h}_{\alpha }=\sum _{i=1}^r{h}_i{h}_i^{*}+\rho \left(h_i\right)h_i^{*}+h_i\rho \left(h_i^{*}\right)\\ & =& -⟨\rho ,\rho ⟩+\sum _{i=1}^r\left(h_i+\rho \left(h_i\right)\right)\left(h_i^{*}+\rho \left(h_i^{*}\right)\right)={\sigma }_{\rho }(-⟨\rho ,\rho ⟩+\sum _{i=1}^r{h}_i{h}_i^{*})\end{array}$$ since $$⟨\rho ,\rho ⟩=\sum _{i=1}^r\rho \left(h_i\right)\rho \left(h_i^{*}\right).$$

WE NEED TO SHOW THAT $$⟨\rho ,{\alpha }_i^{\vee }⟩=1,\text{for}i=1,2,\dots ,r.$$ I THINK THIS IS THE IDENTITY WE WANT.

In the case where $𝔤=𝔤{𝔩}_n$, $$\begin{array}{rcl}\kappa & =& \sum _{1\le i,j\le n}{E}_{ij}{E}_{ji}=\sum _{i=1}^n{E}_{ii}{E}_{ii}+\sum _{1\le i<j\le n}{E}_{ij}{E}_{ji}+\sum _{1\le i<j\le n}[E_{ij},E_{ji}]+E_{ji}{E}_{ij}\\ & =& \sum _{i=1}^n{E}_{ii}{E}_{ii}+2\sum _{1\le i<j\le n}{E}_{ij}{E}_{ji}+\sum _{1\le i<j\le n}(E_{ii}-E_{jj}).\end{array}$$ Since $E_{ii}{v}_{\lambda }^{+}={\lambda }_i{v}_{\lambda }^{+}$ and $E_{ij}{v}_{\lambda }^{+}=0$ for $i<j$, $\kappa v_{\lambda }^{+}=c_{\lambda }{v}_{\lambda }^{+}$ where $$\begin{array}{rcl}{c}_{\lambda }& =& \sum _{i=1}^n{\lambda }_i^2+0+\sum _{1\le i<j\le n}({\lambda }_i-{\lambda }_j)=\sum _{i=1}^n{\lambda }_i^2+(n-i){\lambda }_i-(i-1){\lambda }_i\\ & =& \sum _{i=1}^n\left({\lambda }_i+n-2i+1\right){\lambda }_i=\sum _{i=1}^n({\lambda }_i+2n-2i){\lambda }_i-(n-1){\lambda }_i=⟨\lambda ,\lambda +2\delta ⟩-(n-1)\left|\lambda \right|\\ & =& ⟨\lambda +\delta ,\lambda +\delta ⟩-⟨\delta ,\delta ⟩-(n-1)\left|\lambda \right|,\text{where}\delta =(n-1){\epsilon }_1+(n-2){\epsilon }_2+\dots +{\epsilon }_{n-1}.\end{array}$$

The Drinfeld-Jimbo quantum group is a ribbon Hopf algebra

In the definition of the quantum group: If $K_i=K^{{\alpha }_i}$ then

Let ${\stackrel{\sim }{H}}_1,\dots ,{\stackrel{\sim }{H}}_r$ be an orthonormal basis of $𝔥$. The algebra $U_h𝔤$ is a quasitriangular Hopf algebra and the element $ℛ$ can be written in the form, see [Dr, Sect. 4],

[Dr, Prop. ???], [LR, Prop. (2.14)] Let $U_h𝔤$ be a Drinfeld-Jimbo quantum group and let $\rho$ be an element of $𝔥$ such that $⟨{\alpha }_i,\rho ⟩=1$ for all simple roots ${\alpha }_i$. Let $u$ be as given in (???). Then

1. if $a\in U_h𝔤$ then $e^{h\rho }a{e}^{-h\rho }=S^2\left(a\right)$,
2. $e^{-h\rho }u=ua{e}^{-h\rho }$ is a central element in $U_h𝔤$,
3. ${\left(e^{-h\rho }\right)}^2-uS\left(u\right)=S\left(u\right)u$,
4. $e^{-h\rho }u$ acts in an irreducible representation $L\left(\lambda \right)$ of $U_h𝔤$ of highest weight $\lambda$ by the constant $e^{-\frac{h}{2}⟨\lambda ,\lambda +2\rho ⟩}=q^{-⟨\lambda ,\lambda +2\rho ⟩}$,
5. $\Delta \left(e^{-h\rho }u\right)={\left({ℛ}_{21}ℛ\right)}^{-1}(e^{-h\rho }u\otimes e^{-h\rho }u)$,
6. $S\left(e^{-h\rho }u\right)=e^{-h\rho }u$, and
7. $\epsilon \left(e^{-h\rho }u\right)=1$,

Proof.

1. Since both $S^2$ and conjugation by $e^{h\rho }$ are algebra homomorphisms it is sufficient to check this on generators. We shall show how this is done for the generator $X_j$. It follows from the fact that $[\rho ,X_j]=\rho X_j-X_j\rho =⟨{\alpha }_j,\rho ⟩X_j$, that $$e^{h\rho }{X}_j{e}^{-h\rho }=e^{h\rho }{e}^{-h(\rho -⟨{\alpha }_j,\rho ⟩)}{X}_j=e^{h⟨{\alpha }_j,\rho ⟩)}{X}_j=e^h{X}_j=q^2{X}_j=S^2\left(X_j\right).$$
2. This follows from (1) and the relation $ua{u}^{-1}=S^2\left(a\right)$, since $e^{-h\rho }ua{u}^{-1}{e}^{h\rho }=S^{-2}\left(S^2\right(a\left)\right)=a$.
4. Let ${\stackrel{\sim }{H}}_1,\dots ,{\stackrel{\sim }{H}}_r$ be an orthonormal basis of $𝔥$. Let $L\left(\lambda \right)$ be an irreducible $U_h𝔤$-module of highest weight $\lambda$ and let $v_{\lambda }$ be a highest weight vector in $L\left(\lambda \right)$. Since element of ${U_h𝔤}^{\ge 0}$ which are of degree $\ge 1$ annihilate $v_{\lambda }$ it follows that $$u{v}_{\lambda }=\exp \left(\frac{h}{2}\sum _{i=1}^{r}S\left({\stackrel{\sim }{H}}_i\right){\stackrel{\sim }{H}}_i\right)v_{\lambda }=\exp (-\frac{h}{2}\sum _{i=1}^r{\stackrel{\sim }{H}}_i{\stackrel{\sim }{H}}_i)v_{\lambda }=\exp (-\frac{h}{2}\sum _{i=1}^r\lambda \left({\stackrel{\sim }{H}}_i\right)\lambda \left({\stackrel{\sim }{H}}_i\right))v_{\lambda }=\exp (-\frac{h}{2}⟨\lambda ,\lambda ⟩)v_{\lambda }.$$ The result follows since $e^{-h\rho }{v}_{\lambda }=e^{-h⟨\lambda ,\rho ⟩}{v}_{\lambda }$.
5. This follows from $\Delta \left(u\right)={\left({ℛ}_{21}ℛ\right)}^{-1}(u\otimes u)$, since $$\Delta \left(e^{-h\rho }u\right)=\Delta \left(u{e}^{-h\rho }\right)=\Delta \left(u\right)\Delta \left(e^{-h\rho }\right)={\left({ℛ}_{21}ℛ\right)}^{-1}(u\otimes u)(e^{-h\rho }\otimes e^{-h\rho })={\left({ℛ}_{21}ℛ\right)}^{-1}(e^{-h\rho }u\otimes e^{-h\rho }u).$$
3. and 6. and 7. follow from the equality $e^{h\rho }S\left(u\right)=e^{-h\rho }u$ which is proved as follows. Since $e^{h\rho }S\left(u\right)=S\left(e^{-h\rho }u\right)$ is a central element of $U_h𝔤$, it is sufficient to check that both $e^{h\rho }S\left(u\right)$ and $e^{-h\rho }u$ act by the same constant on an irreducible representation $L\left(\lambda \right)$ of $U_h𝔤$. The element $e^{h\rho }S\left(u\right)=S\left(e^{-h\rho }u\right)$ acts on the module $L\left(\lambda \right)$ in the same way that $u{e}^{-h\rho }$ acts on the irreducible module ${L\left(\lambda \right)}^{*}$ which has highest weight $-w_0\lambda$ where $w_0$ is the longest element of the Weyl group. Thus, $u{e}^{-h\rho }$ acts on the irreducible module ${L\left(\lambda \right)}^{*}$ by the constant $$e^{-\frac{h}{2}⟨-w_0\lambda ,-w_0\lambda ⟩}{e}^{-h⟨-w_0\lambda ,\rho ⟩}=e^{-\frac{h}{2}⟨\lambda ,\lambda +2\rho ⟩}=q^{-⟨\lambda ,\lambda +2\rho ⟩}$$ since $w_0\rho =-\rho$ and the inner product is invariant under the action of $w_0$.

Notes and References

See [Bou, Ch. I §3 Prop. 11] for the fact that the Casimir element is in the center of $U𝔤$.

[Dr, §5 remark (1)] explains that the source of the element $\rho$ is coming from the rewriting of the Casimir in a preferred form so that higher degree terms act by 0 on a highest weight vector:

Bibliography