# How to compute $t_0$ and $r^0$ in Belavin-Drinfeld's classification of solutions of classical Yang-Baxter equations?

I tried to understand Belavin-Drinfeld's classification of solutions of classical Yang-Baxter equations.

In the book a guide to quantum groups, on page 83, there is an example of solutions of the classical Yang-Baxter equation in the case of $\mathfrak{g} = \mathfrak{sl}_3$.

My questions are

(1) how to compute $t_0$ and $r^0$?

(2) In the case of (b), suppose that \begin{align} r^0 = \frac{1}{3} H_{\alpha} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\beta}. \end{align} I tried to verify that \begin{align} r_{12}^0 + r_{21}^0 = t_0, \\ (\alpha \otimes 1)(r^0) + (1 \otimes \beta)(r^0) = 0. \end{align} We have \begin{align} r_{12}^0 = r^0 = \frac{1}{3} H_{\alpha} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\beta}. \end{align} I think that \begin{align} r_{21}^0 = \tau_{12} r_{12}^0 \tau_{12}. \end{align} How to express $r_{21}^0$ using $H_{\alpha}$, $H_{\beta}$?

We have \begin{align} & (\alpha \otimes 1)(r^0) \\ & = (\alpha \otimes 1)(\frac{1}{3} H_{\alpha} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\beta}) \\ & = \frac{1}{3} \alpha(H_{\alpha}) \otimes H_{\alpha} + \frac{1}{3} \alpha(H_{\beta}) \otimes H_{\alpha} + \frac{1}{3} \alpha(H_{\beta}) \otimes H_{\beta}. \end{align} I think that $\alpha(H_{\alpha})=1$ and $\alpha(H_{\beta})=0$ (is this correct?). Then we have \begin{align} & (\alpha \otimes 1)(r^0) \\ & = (\alpha \otimes 1)(\frac{1}{3} H_{\alpha} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\alpha} + \frac{1}{3} H_{\beta} \otimes H_{\beta}) \\ & = \frac{1}{3} \alpha(H_{\alpha}) \otimes H_{\alpha} + \frac{1}{3} \alpha(H_{\beta}) \otimes H_{\alpha} + \frac{1}{3} \alpha(H_{\beta}) \otimes H_{\beta} \\ & = .\frac{1}{3} \otimes H_{\alpha}. \end{align} Similarly, \begin{align} & (1 \otimes \beta)(r^0) \\ & = .\frac{1}{3} H_{\beta} \otimes 1. \end{align} But we do not have \begin{align} (\alpha \otimes 1)(r^0) + (1 \otimes \beta)(r^0) = 0. \end{align} I think that I made some mistake. Any help would be greatly appreciated!

• $\alpha$ and $\beta$ are simples not fundamentals so redo that $\alpha (H_\beta)$ step. Jan 20, 2016 at 20:58
• @AHusain, thank you very much. Since $\alpha = 2\omega_1-\omega_2$. $\beta = -\omega_1 + 2 \omega_2$, we have $\alpha(H_{\alpha}) = 2$, $\alpha(H_{\beta}) = -1$, $\beta(H_{\alpha})=-1$, $\beta(H_{\beta})=2$. Therefore $(\alpha \otimes 1)(r^0) = \frac{1}{3} \otimes H_{\alpha} - \frac{1}{3} \otimes H_{\beta}$, $(1 \otimes \beta)(r^0) = -\frac{1}{3} H_{\alpha} \otimes 1 + \frac{1}{3} H_{\beta} \otimes 1$. But it seems that $H_{\alpha} \otimes 1 \neq 1 \otimes H_{\alpha}$? Jan 21, 2016 at 3:08
• @AHusain, I think that $(\alpha \otimes 1)(r^0) = \frac{1}{3} H_{\alpha} - \frac{1}{3} H_{\beta}$, $(1 \otimes \beta)(r^0) = -\frac{1}{3} H_{\alpha} + \frac{1}{3} H_{\beta}$. Therefore $(\alpha \otimes 1)(r^0) + (1 \otimes \beta)(r^0) = 0$. Jan 22, 2016 at 2:21

the $H_{\alpha}$ should be defined $H_{\alpha}=E_{11}-E_{22}$.
The Casimir element $t_{0}=\sum_{i}\frac{n}{n-1} E_{ii}\otimes E_{ii}-\sum_{i\neq j}\frac{1}{n}E_{ii}\otimes E_{jj}$. You can see the following paper for more detail. http://arxiv.org/pdf/math/9901079v3.pdf .