Problem in understanding a fact about Belavin-Drinfeld triple

Question

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?

Thanks for your time.

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.

Answer 1

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 $\tau (\pi) \in \Gamma_2 \subseteq \Gamma \subseteq \Phi.$