Question

Let's consider only global case.

Let $G_n$ be classical algebraic group over global # field (eg, $GL(n),SO(n), U(n)$...) and let $\pi_n$ be its irr. cusp. reps of $G_n$.

Then we can define the period of two reps $\pi_{n+1}$ & $\pi_n$ as follows;

For $f_{n+1}\in \pi_{n+1}$ and $f_{n}\in \pi_{n}$ ,

define P($f_{n+1},f_{n}$):=$\int_{[G_n]}f_{n+1}(g)f_{n}(g)dg$

(here, we consider $f_{n+1}$ as a function of $G_n$ by restriction and [$G_n$]:=$G_n(F)$$Z_{G_n}$($\mathbb{A}$)\$G_n$($\mathbb{A}$))

Then several articles say that it gives an element of $Hom_{G_n}$($\pi_{n+1} \otimes \pi_n,\mathbb{C}$).

But I am wondering how one can ensure the above period converges.

And I am also wondering that whenever people uses the global seesaw identity, many people just use the Fubini theorem without proof. However, in contrast to global case, when they working in local field case, they always manifest the absolute convergence issue before using seesaw identity.

Why does they not concern about absolute convergence issue in working global situation? Is there some mechanism I don't know which resolves such issues automatically?

Answer 1

I see now what Marc meant: this will look like an "unanswered" question. So here goes: both the functions $f_1,f_2$ are bounded since they are cuspidal (in fact they are of rapid decrease on Siegel sets). The space $[G_n]$ has finite volume since you have quotiented out by the centre. Therefore, the product is integrable and hence converges.