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

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?

• Note that 'algebraic-groups' would be a better tag than 'reductive-groups', since you are only concerned about Chevalley groups (split and simple over an arbitrary field, originally of adjoint type but generalized by Steinberg to other isogeny types). Look for example at the later part of Steinberg's $\S2$: math.depaul.edu/cdrupies/research/papers/chevalleygroups.pdf, where he works with representatives of $W$ in $N_G(T)$, etc. Jun 26, 2016 at 14:59
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 $Sp(2n-2)$.