Centralizers of $\mathbb{F}_q$-rational semisimple elements of a finite group of Lie type

Question

Let $\mathbb{G}$ be a connected reductive $\mathbb{F}_q$ algebraic group over its algebraic closure $\bar{\mathbb{F}_q}$, and $\mathbb{T}$ be an $\mathbb{F}_q$-defined maximal torus. Let $\Phi$ be the root system of $\mathbb{G}$ wrt $\mathbb{T}$, and, given $g\in\mathbb{T}$, put $$\Phi(g)=\lbrace\alpha\in \Phi:\alpha(g)=1\rbrace.$$

There is a very clean criterion by Deriziotis for which closed subsystems $\Sigma\le \Phi$ can occur as $\Phi(g)$ for some $g\in \mathbb{T}$; namely, these are precisely the subsystems $\Sigma\le \Phi$ which admit a basis which is a subset of the set of affine simple roots of $\Phi$, and all such subsystems occur for some $g\in \mathbb{T}$. These are often called pseudo-Levi subsystems. Furthermore, if $g\in\mathbb{T}(\mathbb{F}_q)$, then $\Sigma$ is stable under the action of the Frobenius map associated with the $\mathbb{F}_q$-structure on $\mathbb{G}$.

Question Given a subsystem $\Sigma\le \Phi$ as in the last paragraph, stable under the Frobenius map, does there exist $g\in \mathbb{T}(\mathbb{F}_q)$ for which $\Sigma=\Phi(g)$? Do there exist counterexamples for this?

What I know by now: If $\Sigma=\Phi(g)$ is a Levi subsystem, meaning that it has a basis of simple elements of $\Phi$ then one can always take $g$ to be $\mathbb{F}_q$-rational. To show this, one can compute the dimension of the subgroups $$\mathbb{T}_{\Sigma'}=\bigcap_{\alpha\in \Sigma'}\ker(\alpha)\le \mathbb{T}$$ for all $\Sigma\le \Sigma'\le \Phi$, and verify that $$\dim\mathbb{T}_{\Sigma'}\le \mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma),$$ with equality iff $\Sigma'=\Sigma$, and, consequently, deduce that $\mathbb{T}_{\Sigma}^\circ\setminus(\bigcup_{\Sigma<\Sigma'}\mathbb{T}_{\Sigma'})$ is irreducible of dimension $\mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma)$, and therefore admits an $\mathbb{F}_q$-rational point.

In the more general case, where $\Sigma$ is merely pseudo-Levi, this argument fails more-or-less completely. However, in all cases I have computed thus far it seems that one can find elements $g\in\mathbb{T}$ with $\Phi(g)=\Sigma$ whose representing matrices only have the entries $0,1$ and $-1$... I wonder if maybe there is a simpler argument that my dimension computation above overshoots.

Answer 1

As @LSpice already pointed out, you need $q$ to be sufficiently large even in the case of a Levi subgroup. Just take $G = \operatorname{GL}_n(\overline{\mathbb{F}}_q)$ and $G^F = \operatorname{GL}_n(\mathbb{F}_q)$ under the usual Frobenius endomorphism. If $T \leqslant G$ is the maximal torus of diagonal matrices then $(\mathsf{C}_{q-1})^n \cong T^F = C_G(s)$ for some semisimple element $s \in G^F$ if and only if $q-1 \geqslant n$. That is because $s$ needs $n$ distinct eigenvalues to be regular.

Deciding the exact conditions for your subgroup to be realisable as the centraliser of a rational semisimple element involves a detailed case by case analysis. For the exceptional groups Frank Lübeck's GAP calculations Centralizers and numbers of semisimple classes in exceptional groups of Lie type are invaluable here. But if you're just interested in a "$q$ sufficiently large" statement then this was obtained by Deriziotis's advisor R. W. Carter in Corollary 20 of the following paper:

• "Centralizers of semisimple elements in finite groups of Lie type", Proc. London Math. Soc. (3), vol. 37, (1978), 491–507.

It is essentially a counting argument. It's pointed out explicitly in Theorem 21 that a Levi subgroup is always the centraliser of a rational semisimple element assuming $q$ is sufficiently large.

Another paper that streamlines things here is the following paper of Bonnafé:

• "Quasi-isolated elements in reductive groups", Comm. Algebra (7), vol. 33, (2005), 2315–2337.

This paper makes parts of Carter's and Deriziotis' work much clearer. In particular, Bonnafé gives a clean construction for elements whose centraliser is not contained in any proper parabolic subgroup.