If a Weyl element preserves a root, then it has a representative which preserves the root space?

Question

Let $G$ be a reductive group defined over a field $F$. Let $\Sigma$ be the set of roots of $G$ with respect to a Borel subgroup $B=TU$ with torus $T$. Let $W=N_G(T)/T$ be the Weyl group of $G$. For $\alpha\in \Sigma$, let $U_\alpha$ be the root space of $\alpha$. Denote $x_\alpha:F\rightarrow U_\alpha$ the fixed isomorphism.

Let $\alpha\in \Sigma$ be a root, and $w\in W$ be a Weyl element such that $w(\alpha)=\alpha$. My question is: is it true that $w$ has a representative $\dot w\in G$ such that $$\dot w x_\alpha(r)\dot w^{-1}=x_\alpha(r),\forall r\in F?$$

If this is false in general, in what cases it is true?

Thanks in advance.

Answer 1

This is true if $G$ is split. First of all $w$ is represented by some $\tilde w\in N_G(T)\cap G(k)$ (see Borel-Tits, for example). Then $\text{Ad}\, \tilde w$ acts on $\mathfrak g_\alpha$ by some scalar $c\in k^*$. If there is $t\in T(k)$ with $\alpha(t)=c$ then $\dot w=\tilde wt^{-1}$ will do the trick. Now if $\alpha$ is primitive in the weight lattice $\Xi(T)$ then $\alpha:T(k)\to k^*$ is even surjective. Otherwise $\alpha$ is the long root of a factor of type $C_n$, $n\ge1$ and $\frac \alpha2$ is a weight. This factor is even direct (a complement in $T$ would be the intersection of the kernels $\ker \bar w\frac \alpha2$, $\bar w\in W$). Since this factor is $\cong Sp(2n)$ one can represent $w$ by an element of $Sp(2n-2)$.