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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?

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  • $\begingroup$ If your quotient $[G_n]$ is compact, there is no issue with convergence as both are functions on a compact space. Even if $[G_n]$ is not compact, it has finite volume and cuspidality ensures that the functions have rapid decrease on a fundamental domain and so you do not have convergence issues. You may look up Harish-Chandra's lectures on automorphic forms Spriner lecture notes for a resolution of these issues. $\endgroup$ Jan 17, 2013 at 13:49
  • $\begingroup$ Aakumadula, why did you not put this as answer? $\endgroup$
    – Marc Palm
    Jan 17, 2013 at 16:05
  • $\begingroup$ Dear Aakumadula, Thanks for yor kind reply. I will also refer to Harish-Chandra's book. $\endgroup$
    – Jude
    Jan 18, 2013 at 4:01
  • $\begingroup$ Marc, I thought this was very well known to experts (I am not one) so did not put this up as an answer. $\endgroup$ Jan 20, 2013 at 19:28
  • $\begingroup$ Dear Anonymous, no problem. $\endgroup$ Jan 21, 2013 at 10:26

1 Answer 1

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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.

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