TITLE: Does the group G(K) have a cocompact solvable closed subgroup?
QUESTION [5 upvotes]: Let $K$ be a (locally compact) local field and $G$ be a linear algebraic $K$-group.
Does the topological group $G(K)$ have a cocompact solvable closed subgroup?
If $\mathrm{char}(K)=0$, it is true that $G(K)$ has a cocompact solvable closed subgroup, see Proposition 9.3 in "A. Borel and J. Tits: Groupes reductifs" (Publ. IHES, 1965: link). So the only case left is when the characteristic is positive.
It is also well-known that for any local field $K$, $\mathrm{GL}(n,K)$ has a cocompact solvable closed subgroup.
What about $G=H_{2n+1}\rtimes \mathrm{Sp}_{2n}$, which is the semidirect product of the Heisenberg group $H_{2n+1}$ with the symplectic group $\mathrm{Sp}_{2n}$?

REPLY [3 votes]: The answer below (is a detailed edit of a previous answer which) provides a (more-or-less) self contained proof of the fact that $\mathbf{G}(K)$ contains a cocompact solvable closed subgroup.
Let $K$ be a local field and $\mathbf{G}$ a linear algebraic group defined over $K$. I denote $G=\mathbf{G}(K)$ and assume as I may that $G$ is Zariski dense in $\mathbf{G}$. Note that by [Borel's "Linear Algebraic Group, AG14.4] the Zariski closure of $G$ in $\mathbf{G}$ is itself a $K$-algebraic group whose $K$-points set is again $G$, thus replacing $\mathbf{G}$ with the Zariski closure of $G$ justifies this assumption.
Below topological notions, unless otherwise said, are taken with respect to the Hausdorff (ie $K$-analytic) topology.
Lemma 1:
If $G$ is not solvable then $\mathbf{G}$ has an irreducible $K$-representation of dimension $>1$.
Proof:
Assuming every $K$-representation is 1-dimensional, fixing an injective $K$-representation $\mathbf{G}\to \text{GL}_n$ the image of $G$ is easily seen to be in a $K$-Borel subgroup of $\text{GL}_n$ (here we use the fact that $G$ is Zariski-dense in $\mathbf{G}$).
Lemma 2:
If $G$ is not solvable then there exists a $K$-algebraic variety $\mathbf{X}$ endowed with a $K$-action of $\mathbf{G}$ and a point $x\in X=\mathbf{X}(K)$ which is not $G$-fixed such that $Gx$ is compact.
Proof:
Using lemma 1, we fix an irreducible $K$-representation $\mathbf{G}\to \text{GL}_n$ with $n>1$
and consider the associated $K$-action of $\mathbf{G}$ on $\mathbf{X}=\mathbb{P}^{n-1}$.
Observe that $X=\mathbf{X}(K)$ is compact.
By Zorn Lemma, using compactness, 
we can find a minimal closed, non-empty, $G$-invariant subset $Y\subset X$.
We fix such $Y$ and a point $x\in Y$.
By the fact that $n>1$ and the representation is $K$-irreducible we get that $x$ is not $G$-fixed.
Note that $\overline{Gx}$ is a non-empty $G$-invariant closed subspace of $Y$, thus by minimality, $Y=\overline{Gx}$.
We recall that the action of $G$ on $X$ has locally closed orbits (this is proved in the appendix of http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf). 
It follows that $\partial(Gx)$ is a proper closed invariant subset of $Y$, hence by minimality $\partial(Gx)=\emptyset$.
We conclude that $Y=Gx$ and indeed $Gx$ is compact.
Corollary:
If $G$ is not solvable then there exists a proper $K$-algebraic group $\mathbf{H}<\mathbf{G}$ such that $H=\mathbf{H}(K)$ is cocompact in $G$.
Proof:
Fix $\mathbf{X}$ and $x\in X$ as in Lemma 2 and let $H$ be the stabilizer of $x$ in $G$ and $\mathbf{H}$ be the Zariski-closure of $H$ in $\mathbf{G}$.
By [Borel's "Linear Algebraic Group, AG14.4], $\mathbf{H}$ is a $K$-group and clearly $H=\mathbf{H}(K)$ (note however that $\mathbf{H}$ might not be the stabilizer of $x$ in $\mathbf{G}$).
We get a continuous bijective $G$-map $G/H\to Gx$. By a standard argument in the theory of locally compact groups (using Baire category) this map is open, thus $H<G$ is cocompact.
By the fact that $x$ is not $G$ fixed, $H<G$ is proper. Thus also $\mathbf{H}<\mathbf{G}$ is proper.
Theorem:
There exists a $K$-subgroup $\mathbf{S}<\mathbf{G}$ such that $S=\mathbf{S}(K)$ is solvable and cocompact in $G$.
Proof:
The collection
$$ \{\mathbf{H}\mid \mathbf{H}<\mathbf{G} \text{ is a $K$-algebraic subgrop such that $\mathbf{G}(K)/\mathbf{H}(K)$ is compact} \} $$
is not empty, as it contains $\mathbf{G}$, hence it contains a minimal element by Noetherianity. Let $\mathbf{S}$ be such a minimal element.
Denote $S=\mathbf{S}(K)$ and note that by minimality $S$ is Zariski dense in $\mathbf{S}$.
If $S$ is not solvable, then by the corollary there exists $\mathbf{H}<\mathbf{S}$ such that $H=\mathbf{H}(K)$ is cocompact in $S$, hence also in $G$,
contradicting the minimality of $\mathbf{S}$.
Thus $S$ is indeed solvable.