I have a kind of open-ended question on the definitions regarding automorphic forms.
I need to recall some background to explain myself a little bit better. Let's consider $G$ a semi-simple group over $Q$, and let $A$ be the adeles of $Q$. Let's focus on the discrete automorphic representations. One way to proceed is to look at the space $L$ of $L^2$-functions on $G(A)/G(Q)$, taking values in $C$, which is a Hilbert space. The group $G(A)$ then acts by translations on $L$. We can now consider the irreducible subrepresentations $\pi$ of $L^2$. These irreducible subrepresentations are not exactly the discrete automorphic representations, but it is very close. Namely, in the literature (see e.g. Borel-Casselman), one defines a certain dense subspace $X$ of functions in $L^2$, called "automorphic forms". Then, instead of working with $\pi$, one works with $X \cap \pi$. My question: What is the point of taking this intersection?
(If I understand correctly), by density, $\pi$ can be recovered from $X \cap \pi$ by taking the closure in the $L^2$-space. So $X \cap \pi$ and $\pi$ carry the same amount of information?
The $G(A)$-representation $\pi$ should have some level, which is a compact open subgroup $K \subset G(A)$ such that $\pi^K$ is non-zero. Could it be that, unlike in the case where we would look at $(X \cap \pi)^K$, the space $\pi^K$ could be infinite dimensional? Is this one of the reasons to consider $\pi \cap X$ instead of $\pi$?
One important property of automorphic representations is Flath's tensor product decomposition. Maybe this decomposition fails if you work with $\pi$ instead of $X \cap \pi$? Although I am not really sure, because my feeling is that you can take the tensor product decomposition of $X \cap \pi$, and then complete all the local representations that appear there, and then get a decomposition of $\pi$ ?
Yes, this question is very much on the mark, a sort of "emperor's new clothes" question/issue. Indeed, to meta-explain to my own research students the folly of "defining" automorphic forms in any one of the usual (and orthodox) ways (e.g., smooth, K-finite, $\mathfrak z$-finite, moderate growth), I ask them, "so not everything in $L^2(\Gamma\backslash\mathfrak H)$ is an automorphic form?" And about distributions not being any kind of automorphic form?
Well, sure, we like smooth functions better than not-so-smooth, etc., but/and this "definition" is best when used only to refer to the individual, special automorphic forms/functions (that do span, one way or another, the $L^2$ space and others).
That is, just as the exponentials (or sines and cosines) are very nice functions on the circle, certainly not every $L^2$ function on the circle is as nice. But we still do have reasons to care about $L^2$.
In the case of automorphic things, we might care about spectral decompositions of Poincar\'e series of various sorts, especially (!) some made from not-so-smooth functions, to achieve better outcomes (by the typical irony). Similarly, we might care about genuine proofs of trace formulas and relative trace formulas, not only for "smooth, compactly-supported" data-inputs, but more generally, and the most interesting cases are analytically the least trivial. So there is considerable reason to care about analytically "awkward" situations.
As in the simplest-possible case of Fourier series, the individual exponential functions are very nicely behaved. No doubt. But the convergence of Fourier series is a highly non-trivial topic.