Number of rational points of a quotient of connected linear algebraic groups

Question

Let $H$ be a closed connected subgroup of a connected linear algebraic group $G$ over an algebraically closed field of characteristic $p>0$, and let $\sigma=\sigma_q$ be the standard Frobenius endomorphism, where $q = p^e$ for some natural number $e$. Assume that both $G$ and $H$ are $\sigma$-stable.

A classic result by Rosenlicht (see Corollary 16.5 in Borel), says that the canonical map $\pi:G_\sigma \rightarrow (G/H)_\sigma$ is surjective (here $X_\sigma$ denotes the $\sigma$-fixed points of $X$).

When $H$ is a proper subgroup of $G$, is it the case that $\frac{|G_\sigma|}{|H_\sigma|} \geq (q-1)$?

Answer 1

As you have pointed out, [Borel - Linear algebraic groups, Corollary 16.5(ii)], allows us to identify the set of rational points in a quotient (by a connected group) with the quotient of the groups of rational points. We do so freely.

$\newcommand\card[1]{\lvert#1\rvert}$Suppose first that $G$ is solvable, and let $T_G$ be a maximal torus. Then the inclusion of $T_G$ in the reductive quotient of $G$ is an isomorphism, by, say, [Milne - Algebraic groups, Theorem 16.33]. Since the unipotent radical of $G$ is split, by [Milne, Corollary 16.24], we have that $\card{G(\mathbb F_q)}$ equals $\card{T_G(\mathbb F_q)}q^{\dim(G/T_G)}$. An analogous computation holds for $H$, so, with the obvious notation, $\card{G(\mathbb F_q)/H(\mathbb F_q)}$ equals $\card{(T_G/T_H)(\mathbb F_q)}q^{\dim(G/T_G) - \dim(H/T_H)}$. If $T_H$ is a proper subgroup of $T_G$, then $T_G/T_H$ is a torus of dimension at least $1$, hence has at least $q - 1$ points. (See my answer to Number of points on a linear algebraic group over a finite field.) Otherwise, $\dim(G/T_G) - \dim(H/T_H)$ is positive.

Now drop the assumption that $G$ is solvable. There is a Borel subgroup $B_G$ of $G$ such that $(B_G \cap H)_\text{red}$ is a Borel subgroup $B_H$ of $H$. (This is stated in [Borel, Proposition 11.14(2)], with two caveats. First, the statement there requires passing to the identity component; but $(B_G \cap H)_\text{red}$ normalizes the parabolic subgroup $(B_G \cap H)_\text{red}^\circ = B_H$, hence is contained in it, so actually we don't need to pass to identity components. Second, the result there seems only to guarantee that we may choose a Borel subgroup of $G_{\overline{\mathbb F_q}}$ containing $(B_H)_{\overline{\mathbb F_q}}$; but examining the proof shows that we may take $B_G$ to be any Borel subgroup of $G$ containing $B_H$, i.e., any point of the non-empty set of rational fixed points of $B_H$ on the flag variety of $G$.) Thus the natural map from $B_G(\mathbb F_q)/B_H(\mathbb F_q)$ to $G(\mathbb F_q)/H(\mathbb F_q)$ is an injection. If $B_H$ doesn't equal $B_G$, then $\card{B_G(\mathbb F_q)/B_H(\mathbb F_q)}$, hence $\card{G(\mathbb F_q)/H(\mathbb F_q)}$, is at least $q - 1$. Otherwise, $H$ is a proper parabolic subgroup of $G$, and we may, and do, replace $H$ and $G$ by their images in the reductive quotient of $G$. By, for example, [Conrad, Gabber, and Prasad - Pseudo-reductive groups, Proposition 2.1.12(1)], the unipotent radical of an opposite parabolic to $H$, which has at least $q$ rational points, injects into $G/H$.