Let $K$ be a global field, $\mathbb{A}_K$ the ring of adeles, and $U$ a unipotent algebraic group over $K$. Why is $U(\mathbb{A}_K)/U(K)$, when endowed with the quotient topology, compact?

1$\begingroup$ This reduces to the compactness of $\mathbb{A}_K/K$. $\endgroup$– S. Carnahan ♦Commented Sep 2, 2014 at 23:21

$\begingroup$ Can you give details, esp. in the case when $K$ is not perfect? $\endgroup$– Question MarkCommented Sep 3, 2014 at 0:17
1 Answer
You don't mention if $U$ is assumed to be smooth or connected, but it doesn't matter. In general, if $H$ is any affine group scheme of finite type over a global field $K$ and if $H$ does not contain ${\rm{GL}}_1$ as a $K$subgroup scheme (as is the case when $H$ is a unipotent $K$group scheme) then $H(K)\backslash H(\mathbf{A}_K)$ is compact. In effect, the noncompactness of the idele class group is the source of "all" noncompactness for such adelic coset spaces. This is Theorem A.5.5 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148.
But that generality is quite hard in positive characteristic, since imperfectness forces the use of the fullblown structure theory of pseudoreductive groups (with its extra wrinkles in characteristics 2 and 3). For just the unipotent smooth connected case the idea is way easier: directly show that a closed immersion of $K$group schemes induces a closed embedding on such adelic coset spaces (see Lemma 4.2.5 in loc. cit.), and then apply it to $U \hookrightarrow {\rm{R}}_{K'/K}(U_{K'})$ and use that Weil restriction interacts as expected with adelic topologies, so in that way it suffices to solve the problem after a finite extension on $K$. But $U_{\overline{K}}$ has a composition series whose successive quotients are vector groups, and that descends to some finite extension, so you thereby reduce to the case when $U$ is $K$split. That case is then readily reduced to the case of a vector group, which is clear.
QED