I am struggling with the following subgroups of GL$_2(\mathbb{A})$ where $\mathbb{A}$ is (the topological ring of) Adeles over $\mathbb{Q}$:
$$G_\mathbb{Q} := \iota(\text{GL}_2(\mathbb{Q})) $$ where $$\iota : \text{GL}_2(\mathbb{Q}) \to \text{GL}_2(\mathbb{A}) , ~~~\iota(y) = (y,y,...|y)$$ and $$Z_{\mathbb{R}} = \{(id, id, ...| r ~id) : r \in \mathbb{R}^\times \}$$ the notation being to be read as follows: I write $g = (\gamma, g)$ for an element $g$ of GL$_2(\mathbb{A})$ for $\gamma \in \text{GL}_2(\mathbb{A}_{\text{fin}})$, the finite part of the adeles, i.e. $$\text{GL}_2(\mathbb{A}_{\text{fin}}) = \widehat{\prod}_{p \in \mathbb{P}}^{\text{GL}_2(\mathbf{Z}_p)} \text{GL}_2(\mathbf{Q}_p)$$ carries the topology of restricted rectangles (see below), $\mathbb{P} = \{2,3,5,...\}$ is the set of (finite) primes, and $g \in \text{GL}_2(\mathbb{R})$.
The question is the following:
Is $G_\mathbb{Q} Z_\mathbb{R}$ closed in GL$_2(\mathbb{A})$?
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Ok, one possible answer is that $$\omega : G_\mathbf{Q} Z_{\mathbf{R}}\backslash \text{GL}_2(\mathbf{A}) \to \tilde{G_\mathbb{Q}} \backslash \text{GL}_2^1(\mathbb{A}), ~~~ \omega(G_\mathbf{Q} Z_{\mathbf{R}} g) = \tilde{G_\mathbb{Q}} g$$
is a homemorphism. Now $G_\mathbb{Q}$ is discrete in GL$_2(\mathbb{A})$, so it is closed. Hence, so is $\tilde{G_\mathbb{Q}}$. As generally, for every LCH group $G$ and a subgroup $H$, $H\backslash G$ is an LCH space iff. H is closed, $G_\mathbf{Q} Z_{\mathbf{R}}\backslash \text{GL}_2(\mathbf{A})$ is hausdroffian and thus, $G_\mathbf{Q} Z_{\mathbf{R}}$ is closed... but there must be a better way to see this :(
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- Additional information:
GL$_n2(\mathbb{A})$ carries the topology of restricted rectangles meaning that a set is open iff. it it can be written as a union over so-called restricted rectangles, that is sets of the form
$$U = \prod_{p \in E} U_p \times \prod_{p \notin E} \text{GL}_2(\mathbf{Z}_p) \times U_\infty$$
where $E$ is a finite set of prime numbers and $U_p \subset \text{GL}_2(\mathbf{Q}_p)$ is open and $U_\infty \subset \text{GL}_2(\mathbb{R})$ is open.
- That much I know
[*] $G_\mathbb{Q}$ is discrete (hence closed) in GL$_2(\mathbb{A})$
[*] $Z_{\mathbb{R}}$ is closed as it it the preimage of the projection to the $\infty$ part of the closed set $\begin{pmatrix} * & 0 \\ 0 & *\end{pmatrix}$ in GL$_2(\mathbb{R})$.
[*] In general, the product of two closed subgroups is not closed, not even when it makes sense to consider their product (i.e. when one normalizes the other so that their product becomes a subgroup), so it must be something related to the special structure...
Thanks in advance,
FW