6
$\begingroup$

Let $G<\mathrm{GL_n}$ be a simple linear algebraic group defined over a finite field $K$. Let $\mathfrak{g}$ be its Lie algebra. Assume $\mathfrak{g}$ is simple.

Is it necessarily the case that there is no subspace $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}_g$ for every $g\in G(K)$?

Note: it is clear that there is no $\mathfrak{v}\subset \mathfrak{g}$ with $0<\dim(\mathfrak{v})<\dim(\mathfrak{g})$ such that $\mathfrak{v}$ is invariant under $\mathrm{Ad}$ for every $g\in G(\overline{K})$. It is also clear that the answer to the question above is "yes" when the number of elements of $K$ is larger than a constant depending only on $n$: since $\mathfrak{v}$ is not an ideal, there is a $v\in\mathfrak{v}$ such that all $g\in G(\overline{K})$ such that $\mathrm{Ad}_g(v)\in \mathfrak{v}$ lie in a proper subvariety of $G$.

Note 2: A friend has just proposed over the breakfast table that there are linear algebraic groups with no non-trivial rational points over $K$. That would obviously imply an answer of "no" to my question. I am not convinced that such a thing is really possible, at least not when we are talking about the group $G(K)$, $G$ simple (as opposed to more exotic groups of Lie type). EDIT: as per Venkataramana's comment below, this situation cannot, in fact, occur for $G$ simple.

$\endgroup$
  • 2
    $\begingroup$ About your Note 2: I would guess that your friend was thinking in terms of affine group schemes, in which case he might have been thinking e.g. about the group $\mu_p$ of $p$-th roots of unity in characteristic $p$. $\endgroup$ – Tom De Medts Nov 22 '18 at 8:44
  • 1
    $\begingroup$ Every (connected) simple linear algebraic group $G$ over a finite field $K$ is quasi split and hence has non-trivial unipotent elements in $G(K)$. $\endgroup$ – Venkataramana Nov 23 '18 at 3:41
  • $\begingroup$ Venkataramana: and so? $\endgroup$ – H A Helfgott Nov 23 '18 at 7:05
  • 2
    $\begingroup$ Venkatarama's comment implies that in your situation G has non-trivial points over K (c.f. your note 2). The only groups G that can have no non-trivial rational points are tori (and I can only think of examples when K has two elements). $\endgroup$ – Peter McNamara Nov 23 '18 at 19:55
  • $\begingroup$ Sorry: I saw the question just now. It is as @Peter McNamara says. A simple algebraic group over a finite field $K$ has nontrivial points in $G(K)$ since it has non-trivial unipotent elements in $G(K)$. $\endgroup$ – Venkataramana Nov 25 '18 at 15:26
2
$\begingroup$

You are looking for the following theorem of Steinberg:

Let q=|K|. An irreducible algebraic representation of $G(\overline{K})$ of highest weight λ remains irreducible when restricted to G(K) if $\langle \lambda,a^\vee\rangle <q$ for all simple coroots $\alpha^\vee$.

The reference is here. When G is simply connected, Steinberg proves more, namely that all irreducible modules for G(K) arise from this construction.

In the situation at hand, we need to apply Steinberg's theorem to the adjoint representation. There are only a small number of cases where the condition $\langle \lambda,a^\vee\rangle <q$ does not apply, easily classified on a case by case basis. I haven't checked, but I would suspect that in each of these cases, $\mathfrak{g}$ is not simple.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How would one go about enumerating such cases? $\endgroup$ – H A Helfgott Nov 29 '18 at 22:14
  • $\begingroup$ The highest weight of the adjoint representation is the highest root. You have to go through each root system (A-G) and pair this root with each simple coroot. Only the numbers 0,1 and 2 can appear. This is the same computation as what is needed to compute the Cartan matrix for the loop group, so if you know the translation, you can read it off from there. $\endgroup$ – Peter McNamara Nov 30 '18 at 7:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.