# Action of $\text{Gal}(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ on $X(T)$ permutes root system

I got these questions while reading Chapter 3 of "Representations of Finite Groups of Lie Type" by Digne-Michel.

Let $T$ be a torus defined over $\mathbb{F}_q$. Then $\text{Gal}(\overline{\mathbb{F}_q}/{\mathbb{F}_q})$ acts on $X(T)$ as it acts on the algebras of $\mathbb{G}_m$ and on $\overline{\mathbb{F}_q}[T]$ via the arithmetic Frobenius. Let $\tau$ be the action of the element $x \mapsto x^q$ on $X(T)$.

If I consider $X(T) \subseteq \overline{\mathbb{F}_q}[T]$, then is this $\tau$ the restriction of the arithmetic Frobenius of $\overline{\mathbb{F}_q}[T]$ to $X(T)$? I mean the isomorphism

$$\text{Hom}(\overline{\mathbb{F}_q}[X,X^{-1}], \overline{\mathbb{F}_q}[T]) \cong \text{Hom}(T,\mathbb{G}_m)$$

is given by $\varphi \mapsto \varphi(X)$, right? And the action of $x \mapsto x^q$ of the Galois group on the left is given by $\varphi \mapsto F_0 \circ \varphi \circ F_0^{-1}$ where $F_0$ denotes the arithmetic Frobenius of both algebras. Then under this isomorphism, the action of $\tau$ on $\chi = \varphi(X)$ would be $\tau(\chi) = F_0 (\varphi (F_0^{-1}(X))) = F_0(\varphi(X)) = F_0(\chi)$, because $F_0(X) = X$ in $\overline{\mathbb{F}_q}[X,X^{-1}]$, right?

Second Question (what I really would like to understand):

The following can be found on p. 40 of Digne-Michel:

"If $T$ is a rational maximal torus of a reductive connected algebraic group $G$, it is clear that $\tau$ permutes the root system of $G$ relative to $T$. Similarly the transpose automorphism $\tau^*$ of $Y (T)$ permutes the coroots."

Unfortunately, this is not clear to me at all.

My Thoughts:

What I know is that if $F: G \to G$ is the Frobenius morphism associated with the $\mathbb{F}_q$-structure of $G$, being an isogeny, we get an induced $p$-morphism (in the sense of Springer), that is, a bijection $b : \Phi \to \Phi$ of the roots and a $p$-power $q(\alpha)$ for every $\alpha \in \Phi$ such that $F^*(b(\alpha)) = q(\alpha) \alpha$ holds for all $\alpha \in \Phi$ (as well as a similar equation for coroots). Here $F^*$ denotes the comorphism of $F$ on $\overline{\mathbb{F}_q}[G]$, $\overline{\mathbb{F}_q}[T]$ or $X(T)$.

From this and $\tau (F^*(\alpha)) = q \alpha$ we get $q b(\alpha) = q(\alpha) \tau(\alpha)$ for all $\alpha \in \Phi$ and if $\tau(\alpha)$ were a root this would imply $b(\alpha) = \tau(\alpha)$ and $q = q(\alpha)$ for all $\alpha \in \Phi$. On the other hand, if I could prove that $q(\alpha) = q$ for all $\alpha \in \Phi$, I would be done, but I do not know if this is the correct approach.

The important thing to note is, that the comorphism $F^* : \mathbb{F}[G] \to \mathbb{F}[G]$ has the property $F^*(\mathbb{F}[G]) = \mathbb{F}[G]^q$ and this implies $F^*(\mathbb{F}[U_{b(\alpha)}]) = \mathbb{F}[U_\alpha]^q$ for all $\alpha \in \Phi$. But from $F(u_\alpha(x)) = u_{b(\alpha)}(c_\alpha x^{q(\alpha)})$ we obtain $F^*(\mathbb{F}[U_{b(\alpha)}]) = \mathbb{F}[U_\alpha]^{q(\alpha)}$. As $\mathbb{F}[U_\alpha]$ is isomorphic to the polynomial ring in one variable, it follows that $q(\alpha) = q$ for all $\alpha$. Then $\tau(\alpha) = b(\alpha)$ for all $\alpha \in \Phi$, so $\tau$ maps $\Phi$ to itself.
• Could you please say that how the relation $F(u_{\alpha}(x))=u_{b(\alpha)}(c_{\alpha} x^{q_{\alpha}})$ will imply that $F^*(\mathbb{F}[U_{b(\alpha)}])={\mathbb{F}[U_{\alpha}]}^{q_{\alpha}}$? – user97635 May 16 '17 at 12:53
• @user97635 : Up to the isomorphisms $u_\alpha$ and $u_{b(\alpha)}$ (or, more precisely, their comorphisms), the comorphism of $F$ is an endomorphism of the polynomial ring in one variable given by $\mathbb{F}[T] \to \mathbb{F}[T], T \mapsto c_\alpha T^{q(\alpha)}$. The image of this map is precisely $\mathbb{F}[T]^{q(\alpha)}$. – Matthias Klupsch May 17 '17 at 6:10