Show that roots are preserved by $\tau.$

Question

Let $L$ be a finite dimensional semisimple Lie algebra with Cartan subalgebra $H.$ Let $\Gamma$ be the set of all fundamental roots or simple roots of $L.$ Let $\Gamma_1, \Gamma_2 \subset \Gamma$ and $\tau : \Gamma_1 \longrightarrow \Gamma_2$ be a bijection preserving the dual of the Killing form $(\cdot, \cdot)$ on $H^{\ast} \times H^{\ast}.$ Let $\Phi_1$ and $\Phi_2$ be the roots of $L$ generated by $\Gamma_1$ and $\Gamma_2$ respectively. Then can we conclude that $\overline {\tau} (\Phi_1) = \overline {\tau} (\Phi_2)$ where $\overline {\tau}$ is the $\mathbb Z$-linear extension of $\tau$ from $\mathbb Z \Gamma_1 \longrightarrow \mathbb Z \Gamma_2\ $?

I am thinking of some simple arguments say if $\alpha + \beta$ is a root of $L$ generated by $\alpha, \beta \in \Gamma_1$ but $\tau (\alpha) + \tau (\beta)$ is not a root then $\left (\tau (\alpha), \tau (\beta) \right ) \geq 0.$ But that implies $(\alpha, \beta) \geq 0.$ What can we conclude from here?

Any help in this regard would be warmly appreciated. Thanks for your time.

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