A question about the standard maximal compact subgroup of $GL(2,\mathbb{A}))$

Question

Let $F$ be a number field and let $\mathbb{A}$ be the adele ring of $F$. I kow that if $K_v$ is the standard maximal compact subgroup of $GL(2,F_v)$, $K= \prod_v{K_v}$ is the standard maximal compact subgroup of $GL(2,\mathbb{A}))$.My question is: How is the product here defined? Is it restricted direct product? I need to know some little details since i'm trying to learn the subject in question.

Answer 1

Actually, it's both a restricted and regular old product! Set $G = GL(2,F)$ and $G_v = GL(2,F_v)$ for each place $v$. If $S_v \subseteq G_v$ for each place $v$, then, by definition, the restricted product $S = {\prod_v}^{\!\!\prime} S_v$ is defined as the set of all tuples $(s_v)_v$ in the unrestricted product $\prod_v S_v$ such that $s_v \in K_v$ for almost all (all but finitely many) places $v$. So if we begin with $S_v = K_v$, then there is no distinction between the groups ${\prod_v}^{\!\!\prime}K_v$ and $\prod_v K_v$. Moreover, there's no distinction topologically between these spaces either, including (unlike with G) if we assign $\prod_v K_v$ the product topology. One way to see this is that $\prod_v K_v$ belongs to (in fact, is the intersection) of all the topological groups

$$ G_S = \prod_{v \in S}G_v \times \prod_{v \not\in} $$

for $S$ a finite set of places containing all archimedean $v$, the induced topology on $K$ as a subgroup of $G_S$ is the product topology, and a formal definition of $G$ is the topological direct limit

$$ \varinjlim_{S}G_S. $$

So you can see directly from the definition of $\varinjlim$ that the topology on $K$ induced from $G$ is the product topology, as well.

Hopefully you found an answer elsewhere, but I accidentally stumbled on your question and could provide a quick response. Better late than never I suppose! :)