# Problem in understanding a fact about Belavin-Drinfeld triple

A Belavin-Drinfeld triple associated to a simple Lie algebra $$L$$ is a triple $$(\Gamma_1, \Gamma_2, \tau)$$ where $$\Gamma_1, \Gamma_2 \subseteq \Gamma$$ ($$\Gamma$$ is a set of simple roots or fundamental roots of $$L$$) and $$\tau : \Gamma_1 \longrightarrow \Gamma_2$$ is a bijection such that

$$(1)$$ $$\tau$$ preserves the dual Killing form $$(\cdot, \cdot),$$

$$(2)$$ for any $$\gamma_1 \in \Gamma_1,$$ there exists $$n \gt 0$$ such that $$\tau^n (\gamma_1) \in \Gamma_2 \setminus \Gamma_1.$$

This definition has been taken from the lecture notes on compact quantum groups by Pavel Etingof and Oliver Schiffmann. In this connection the author mentioned that we can linearly extend $$\tau$$ as a map from $$\mathbb Z \Gamma_1 \to \mathbb Z \Gamma_2.$$ Let $$L_{\Gamma_i}$$ be the subalgebra generated by the $$L_{\pm \alpha}, \alpha \in \Gamma_i$$ for $$i=1,2$$ (i.e. $$L_{\Gamma_i}$$ is the direct sum of the root subspaces corresponding to the roots generated by the roots of $$\Gamma_i$$ for $$i=1,2$$). In other words, if $$\Phi$$ is the set of all roots of $$L$$ then $$L_{\Gamma_i} = \bigoplus_{\beta \in \mathbb Z \Gamma_i \cap \Phi} L_{\beta}$$ for $$i=1,2.$$ At this point the authors mentioned that the extended linear map $$\tau$$ from $$\mathbb Z \Gamma_1 \longrightarrow \mathbb Z \Gamma_2$$ would then induce an isomorphism from $$L_{\Gamma_1} \longrightarrow L_{\Gamma_2}$$ given by $$e_{\beta} \mapsto e_{\tau(\beta)}$$ and $$f_{\beta} \mapsto f_{\tau(\beta)}$$ where $$L_{\beta} = \text {span}\ \{e_{\beta} \}$$ and $$L_{-\beta} = \text {span}\ \{f_{\beta}\},$$ for $$\beta \in \mathbb Z \Gamma_1 \cap \Phi$$ which are known as Chevalley-Serre generators for the root subspaces of $$L.$$ This is where I am confused. I am unable to understand why is it even injective.

Question $$:$$ If $$\beta \in \mathbb Z \Gamma_1 \cap \Phi$$ is it always true that $$\tau (\beta) \in \mathbb Z \Gamma_2 \cap \Phi\$$?

For otherwise injectivity of this induced map would be violated. Could anyone please shed some light on it?

Source $$:$$ Lecture $$5$$ on Belavin-Drinfeld classification of Lie bialgebra structures on a complex simple Lie algebra from the Lecture Notes on Compact Quantum Groups written by Pavel Etingof and Oliver Schiffmann.

You may try to show the following $$:$$

Let $$W$$ be the Weyl group associated to $$L,$$ $$\Phi$$ be the set of all roots of $$L$$ and $$\Gamma$$ be the set of simple roots of $$L.$$ Then

$$(1)$$ Any element $$w \in W$$ can be generated by the simple roots i.e. there exists $$\pi_1, \pi_2, \cdots, \pi_l \in \Gamma$$ such that $$w = s_{\pi_1} \circ \cdots \circ s_{\pi_l}.$$

$$(2)$$ Any root of $$L$$ can be produced by the action of the Weyl group on the simple roots i.e. given $$\lambda \in \Phi$$ there exists $$w \in W$$ and $$\pi \in \Gamma$$ such that $$\lambda = w (\pi).$$

$$(3)$$ The Weyl group merely permutes the roots i.e. given $$w \in W$$ and $$\lambda \in \Phi$$ we have $$w (\lambda) \in \Phi.$$

Combining $$(1)$$ and $$(2)$$ it follows that any root $$\lambda \in \Phi$$ can be generated by the simple roots i.e. given $$\lambda \in \Phi$$ there exists $$\pi_1, \cdots, \pi_l, \pi \in \Gamma$$ such that

$$\tag{*}\lambda = \left ( s_{\pi_1} \circ \cdots \circ s_{\pi_n} \right ) (\pi)$$

Then by using the fact that $$\tau$$ preserves the dual Killing form try to show the following $$:$$

Let $$\lambda \in \mathbb Z \Gamma_1 \cap \Phi$$ be as in $$(*).$$ Then $$\pi_{1}, \cdots, \pi_l, \pi \in \Gamma_1$$ and $$\tag{**}\tau(\lambda) = \left ( s_{\tau(\pi_1)} \circ \cdots \circ s_{\tau (\pi_l)} \right ) (\tau (\pi))$$

The result then follows by combining $$(3)$$ and $$(**)$$ as $$\tau (\pi) \in \Gamma_2 \subseteq \Gamma \subseteq \Phi.$$