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

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Apr 17, 2016 at 16:04	history	edited	user21574		edited tags
Apr 17, 2016 at 4:35	answer	added	Daisy		timeline score: 1
Jan 22, 2016 at 2:21	comment	added	Jianrong Li		@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 21, 2016 at 3:08	comment	added	Jianrong Li		@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 20, 2016 at 20:58	comment	added	AHusain		$\alpha$ and $\beta$ are simples not fundamentals so redo that $\alpha (H_\beta)$ step.
Jan 20, 2016 at 12:19	history	asked	Jianrong Li	CC BY-SA 3.0