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I foresee that to experts of automorphic forms this question will sound unimportant or useless or even not worthy of an answer; but none of these are going to stop me from asking it!

The question is simple: let $G$ be a reductive (or even semisimple) algebraic group over $\mathbb Q$. Is it true that the adelic group $G(\mathbb A)$ is of Type I (i.e., direct integral decomposition of its unitary representations is unique)? And if the answer is negative, then how do people actually get around it in the study of multiplicity of automorphic forms in $L^2(G(\mathbb Q)\backslash G(\mathbb A))$?

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Dear Valerie, Certainly $G(F)$ is Type I for any local field $F$. My first instinct is that this would imply the corresponding statement for $G(\mathbb A)$, although I'm not familiar enough with these things to be very confident. Do you have any sense of whether the local field result implies the adelic result? Regards, Matthew – Emerton Dec 13 '12 at 6:50
I agree with Emerton. The proof that the product of finitely many Type I groups is of Type I should carry over to the restricted direct product of $G(F_v)$, i.e. to $G(\mathbb{A})$. Here it might be useful to know that any irreducible unitary representation of $G(\mathbb{A})$ is the completion of a restricted direct product of irreducible unitary representations of $G(F_v)$. – GH from MO Dec 13 '12 at 9:08
I know nothing about adeles or algebraic groups, but in view of G's last comment would like to mention the (irrelevant?) fact that restricted direct products (wrt some family of compacts) of type I groups need not be Type I, IIRC: cf. some old work of Blackadar… and – Yemon Choi Dec 13 '12 at 10:08
Dear Emerton and GH: I suppose that the general argument does not carry over; for instance in an old paper of Calvin Moore (Annals of Math, 1965) it is shown that nilpotent adelic groups are not type I (whereas nilpotent groups over local fields certainly are). My gut feeling is that $G(\mathbb A)$ not being Type I is not a problem when working with the discrete spectrum, but I am not sure about the continuous spectrum. – Valerie Dec 13 '12 at 14:31
@Yemon: I see, thanks. So I am not so sure any more. At any rate, often a group $G$ is shown to be of Type I by showing that for any irreducible unitary representation $\pi$ of $G$, the operator $\pi(f)$ is compact for any compactly supported continuous function $G\to\mathbb{C}$. In the context of $L^2(G(\mathbb{Q})\backslash G(\mathbb{A})$, this property is known to hold for the cuspidal subspace, and this is the way to see that the cuspidal subspace is a direct sum of irreducible subspaces. – GH from MO Dec 13 '12 at 14:34
up vote 18 down vote accepted

I believe the answer is yes. Let's begin by recalling that if one wants to show that a locally compact group $G$ is of type I, it suffices to show that $G$ contains a "large" compact subgroup $K$, in the sense that for every $\pi \in \hat{G}$ and $\sigma \in \hat{K}$, the multiplicity of $\sigma$ in $\pi|_K$ is finite. This is how Harish-Chandra showed that a real reductive group is of type I (take $K$ to be a maximal compact), and also how Bernstein showed that a $p$-adic reductive group is of type I (take $K$ to be a compact open subgroup).

Now let $G$ be a connected reductive group over $\mathbb Q$. Then, away from a finite set $S$ of places (containing $\infty$), $G$ is unramified and has a model over $\mathbb Z_p$. Let's abuse notation and denote this model by $G$. It suffices to show that $G(\mathbb A^S) = \prod'_{p \not\in S} G(\mathbb Q_p)$ is of type I. The desired large $K$ turns out to be $K = \prod_{p \not\in S} G(\mathbb Z_p)$. This assertion essentially appears (without proof) as Theorem 4 in Flath's article in the Corvallis proceedings. The details are spelled out in the appendix to Clozel's article in the IAS/Park City 2002 lecture notes on automorphic forms (MR2331351; a Google Books preview is available here).

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Very good. .... – paul garrett Dec 13 '12 at 23:13
Perfectly clear answer. This is also the difference to the nilpotent setting, where there are no large compact subgroups. – Marc Palm Dec 14 '12 at 11:35
Yes, Clozel's appendix is exactly what one needs. Thanks. – Valerie Dec 14 '12 at 14:58

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