# “Almost-ideals” in the (simple) Lie algebra of an algebraic group?

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.

• 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 $p$. – Tom De Medts Nov 22 '18 at 8:44
• Every (connected) simple linear algebraic group $G$ over a finite field $K$ is quasi split and hence has non-trivial unipotent elements in $G(K)$. – Venkataramana Nov 23 '18 at 3:41
• Venkataramana: and so? – H A Helfgott Nov 23 '18 at 7:05
• 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). – Peter McNamara Nov 23 '18 at 19:55
• 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 $G(K)$. – Venkataramana Nov 25 '18 at 15:26

Let q=|K|. An irreducible algebraic representation of $$G(\overline{K})$$ of highest weight λ remains irreducible when restricted to G(K) if $$\langle \lambda,a^\vee\rangle for all simple coroots $$\alpha^\vee$$.
In the situation at hand, we need to apply Steinberg's theorem to the adjoint representation. There are only a small number of cases where the condition $$\langle \lambda,a^\vee\rangle 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.