I have a few (semi-)related questions regarding certain Hilbert space representations of locally compact groups that come up in the theory of automorphic forms.
Let $G$ be a unimodular locally compact Hausdorff group, $\Gamma$ a discrete (hence closed) subgroup of $G$, and $Z$ a closed subgroup of the center of $G$. The example of interest to me is:
(*) $G$ is the group of adelic points of a connected reductive $\mathbf{Q}$-group $\mathscr{G}$, $\Gamma=\mathscr{G}(\mathbf{Q})$, and $Z$ is the group of adelic points of the maximal split torus in the center of $\mathscr{G}$.
I would be happy to know the answer to the following question just in case (*).
Is the group $Z\Gamma$ closed in $G$, and if so, is it unimodular?
The answer to the first part of the question would be yes if one of $Z$, $\Gamma$ were compact, but this isn't so in many cases of interest to me (e.g. $G=\mathrm{GL}_n$). I suspect the answer to both parts of the question are "yes" because of the following definition which appears in the literature (I will assume $Z\Gamma$ is closed and unimodular to make the definition, although I'm hoping, at least in the case of interest mentioned above, that this is unnecessary):
For a unitary character $\omega$ of $Z$, let $L^2(Z\Gamma\setminus G,\omega)$ be the space of equivalence classes (the equivalence being almost every equality relative to the unique-up-to-scaling $G$-invariant measure on $\Gamma\setminus G$) of Borel measurable $f:\Gamma\setminus G\rightarrow\mathbf{C}$ with the following properties: for each $z\in Z$, $f(\Gamma zg)=\omega(z)f(\Gamma g)$ for almost every $\Gamma g\in\Gamma\setminus G$, $\vert f\vert$ is Borel measurable on $Z\Gamma\setminus G$ (this makes sense because of the unitary of $\omega$), and $\int_{Z\Gamma\setminus G}\vert f\vert^2<\infty$ (the integral being taken with respect to the unique-up-to-scaling $G$-invariant measure on $Z\Gamma\setminus G$). Assuming I've got this definition right, it will be a Hilbert space with the natural inner product.
My second question concerns the definition of the so-called cuspidal subspace $L_0^2(Z\Gamma/G,\omega)$, and I'm only interested in the case of (*) with $\mathscr{G}=\mathrm{GL}_n$, so $G=\mathrm{GL}_n(\mathbf{A}_\mathbf{Q})$, $\Gamma=\mathrm{GL}_n(\mathbf{Q})$, and $Z=\mathbf{A}_\mathbf{Q}^\times$ is the center of $G$. The (apparently) standard definition of $L_0^2(Z\Gamma\setminus G,\omega)$ is: the subspace of $L^2(Z\Gamma\setminus G,\omega)$ consisting of equivalence classes containing a function $f$ such that $\int_{\mathbf{A}_\mathbf{Q}/\mathbf{Q}}f\big(\big(\begin{smallmatrix}1&x\\0&1\end{smallmatrix}\big)g\big)dx=0$ for almost every $g\in G$ ($dx$ means Haar measure on $\mathbf{A}_\mathbf{Q}/\mathbf{Q}$). There are some things to check to make precise sense of this definition, but the important thing is that it should be a closed subspace of the full $L^2$-space. This is taken for granted in, e.g., Bump's book on automorphic representations and Godement's notes on Jacquet-Langlands. The closedness doesn't seem completely obvious to me, and I worry about it because some other authors, notably Lang in his book on $\mathrm{SL}_2$, define the cuspidal subspace as the Hilbert space completion of the space of bounded continuous functions on $\Gamma\setminus G$ with the appropriate properties. I would hope these definitions yield the same space but it isn't clear to me.
Assuming the definition I've given of $L_0^2(Z\Gamma\setminus G,\omega)$ is technically correct (i.e. I haven't left out any conditions on the functions under consideration), is it closed in $L^2(Z\Gamma\setminus G,\omega)$?
First, yes, $Z\Gamma$ is closed in $G$. In any particular example this is pretty easy to prove.
Second, for groups bigger than $GL_2$, even the imprecise definition of "cuspform" needs to refer to all conjugacy classes of (unipotent radicals of) rational parabolics. For $GL_n$, these are specified by (ordered) partitions $n_1+\ldots+n_\ell=n$ of $n$, specifying the upper-block-triangular matrices with diagonal blocks of sizes $n_j$, respectively. Their unipotent radicals are upper-triangular with identity matrices in those blocks. Then the imprecise definition is that all the integrals over $N_k\backslash N_\mathbb A$ vanish.
You're right that there is something fishy here, since it doesn't really make sense to integrate $L^2$ functions over smaller subsets, and since these supposed integrals are definitely not continuous functionals on $L^2$, we cannot conclude in any obvious way that the kernels are closed... :)
(Lang's definition dodges the imprecision, but doesn't go far enough...)
For this reason (and also for proving automorphic Plancherel) it is better to say that cuspforms are the orthogonal complement in $L^2$ to all pseudo-Eisenstein series with smooth, compactly-supported data. In effect, this is an integrated form of the naive/imprecise version often seen. The compact support (and continuity) of the data for the pseudo-Eisenstein series produces continuous, compactly-supported (mod the center) functions on $G_k\backslash G_\mathbb A$, so certainly in $L^2$, so giving sensible integrals against $L^2$ functions. Thus, the kernels are closed, and the simultaneous kernel is an intersection of closeds, so is closed.