# $\mathbb{F}_q$-rational elements in unipotent classes of a finite group of Lie-type

I've tried posting this question on MSE, but didn't manage to get an answer there, so I'm trying again here. Sorry in advance if this question is trivial or trivially false. I haven't managed to find a satisfactory proof (or reference of one), or a counterexample for it.

Let $k$ be the algebraic closure of the finite field $\newcommand{\dbF}{\mathbb{F}}\dbF_q$, of characteristic $p$. If necessary, assume $p\gg0$. Let $G$ be a simple algebraic group over $k$ and let $\sigma$ be an endmomorphism of $G$ such that the $\sigma$-fixed $G_\sigma$ subgroup of $G$ is finite.

My Question Let $N\subseteq G$ be a unipotent conjugacy class. Does $N$ contain an element $n\in N$ which is fixed under $\sigma$?

Note that, by [1, I, 3.4], it is enough to show that $N$ is stable under the action of $\sigma$.

This question has a positive answer for $G=GL_n$, with the ordinary Frobenius map $(a_{i,j})\mapsto(a_{i,j}^q)$ since conjuagcy classes of matrices over and algebraically closed field are parametrized by the Jordan normal forms, and the Jordan normal form of a unipotent element is clearly $\sigma$-fixed.

More generally, if $G$ is split over $\dbF_q$, I think the question still has a positive answer, since any unipotent element can be conjugated to an element of the form $\prod_{\alpha\in \Phi^+} x_\alpha(\lambda_\alpha)$, where $\Phi^+$ is the set of positive roots with respect to an $\dbF_q$-split maximal torus within a $\sigma$-stable Borel. I haven't checked this thoroughly, but I think that at this point, by acting by conjugation by elements of the maximal torus, we may assume that the $\lambda_\alpha$'s are in $\dbF_q$, and, since $\sigma$-doesn't permute the roots in $\Phi$, this element is $\sigma$-fixed.

I'm not sure how to extend this type of argument to non-split groups ove $\dbF_q$, but checking small cases of the unitary group, it seems like the question continues to be true for these cases.

In any case, I'm not sure where to go on from here, and I'm not sure that the type of argument I'm trying to pursue here is indeed true, so I would appreciate any type of input anyone would be willing to offer.

Thank you!

1 Springer, T. A.; Steinberg, R., Conjugacy classes, Sem. algebr. Groups related finite Groups Princeton 1968/69, Lect. Notes Math. 131, E1-E100 (1970). ZBL0249.20024.

• As you say, you just need to check whether a class is invariant under the automorphism induced by the Frobenius. In good characteristic it is enough to check that the weighted Dynkin diagram of the class is invariant under the automorphism. If $G$ is simple then this is true unless $G$ is type $\mathsf{D}_{2n}$. In this case there are a family of classes which are interchanged by the automorphism. Sep 17, 2018 at 12:48
• In any case, you should look at the book "Unipotent and nilpotent classes in simple algebraic groups and Lie algebras" by Liebeck and Seitz. It has all the information you will need. Sep 17, 2018 at 12:49
• Excellent, thank you! I was unaware of this reference, I'll check it out. Sep 17, 2018 at 13:02
• See Corollary 6, p. 6 in the book of Liebeck-Seitz mentioned by Jay. Sep 18, 2018 at 8:09
• By the way, it is true that every unipotent class of split $G$ has a representative of the form $\prod_{\alpha \in \Phi^+} x_{\alpha}(c_{\alpha})$, where $c_{\alpha} \in \mathbb{F}_q$, and we can even arrange $c_{\alpha} \in \{1,-1\}$. Consequently $G$ has only finitely many conjugacy classes of unipotent elements, which is a non-trivial theorem due to Lusztig. To prove that such representatives exist I suppose you need to know the unipotent conjugacy classes and proceed case-by-case (at least I am not aware of any other proof). Sep 18, 2018 at 8:57