Number of points on a linear algebraic group over a finite field

Question

Let $G$ be a linear algebraic group defined over a finite field $\mathbb{F}_q$ as a variety of dimension $d$. What would be a good, simple lower bound for $G(F_q)$?

One can get something fairly nice from a general lower bound on the number of points on a variety over $\mathbb{F}_q$, in the style of Lang-Weil (Cafure-Matera 2006, Thm 7.5, is the best result I know). However - perhaps one can do something better for a group?

Example: for a Chevalley group, we have $|G(\mathbb{F}_q)|\geq q^d - d q^{d-1}$. Perhaps something like that is also true in general?

Answer 1

Since we reduce points without changing dimension if we pass to identity components, we may, and do, assume that $G$ is connected. I will say just "rational points" in place of "$\mathbb F_q$-rational points".

If $G$ is smooth and unipotent, then it is split unipotent, and hence has $q^d$ rational points.

If $G$ is connected and reductive, then let $B^\pm$ be opposite Borel subgroups of $G$, with unipotent radicals $U^\pm$ and common maximal torus $T$. Write $r$ for the dimension of $T$. As @WillSawin observes, $T$ has at least $(q - 1)^r$ rational points.

@DanielLoughran suggested giving more detail on this bound. I can do no better than to reproduce an argument by @DanielLitt and @DavidESpeyer at my question Pointless groups III, in the spirit of @PeterMcNamara's observation above. Namely, if $F$ is a topological generator of the absolute Galois group of $\mathbb F_q$, then $\lvert T(\mathbb F_q)\rvert$ is the determinant of the action of $q - F$ on the absolute cocharacter lattice $Y \mathrel{:=} X_*(T_{\overline{\mathbb F_q}})$, which has dimension $r$. Since some power of $F$ acts trivially on $Y$, all of its eigenvalues are roots of unity, so $\det(q - F)$ is at least $(q - 1)^r$.

Now the multiplication map $U^- \times T \times U^+ \to G$ is an isomorphism of varieties onto an open subvariety of $G$, we have that $G$ has at least $q^{\dim(U^-)}(q - 1)^r q^{\dim(U^+)} = q^{d - r}(q - 1)^r$ rational points.

We have that $G$ is an extension of a reductive group $G^\text{red}$ by the unipotent radical $\operatorname R_u(G)$ of $G$. If we continue to write $r$ for the dimension of some (hence every) maximal torus in $G$, and put $d_\text u = \dim(\operatorname R_u(G))$ and $d_\text s = \dim(G^\text{red}) - r$, then the special cases that we have already handled show that $G$ has at least $q^{d_\text s}(q - 1)^r q^{d_\text u} = q^{d - r}(q - 1)^r$ rational points. In particular, as @WillSawin conjectured, $G$ always has at least $(q - 1)^d$ rational points. This is obviously sharp if our bound is to depend only on dimension, since equality holds for $G = \operatorname{GL}_1^d$.