21
$\begingroup$

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 $L^2(G(\mathbb Q)\backslash G(\mathbb A))$?

Improve this question
$\endgroup$
6
  • 2
    $\begingroup$ 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 $\endgroup$
    – Emerton
    Commented Dec 13, 2012 at 6:50
  • 1
    $\begingroup$ 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 $G(F_v)$. $\endgroup$
    – GH from MO
    Commented Dec 13, 2012 at 9:08
  • 3
    $\begingroup$ 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 zentralblatt-math.org/zmath/en/search/… and ams.org/mathscinet-getitem?mr=713736 $\endgroup$
    – Yemon Choi
    Commented Dec 13, 2012 at 10:08
  • 3
    $\begingroup$ 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. $\endgroup$
    – Valerie
    Commented Dec 13, 2012 at 14:31
  • 2
    $\begingroup$ @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. $\endgroup$
    – GH from MO
    Commented Dec 13, 2012 at 14:34

3 Answers 3

Reset to default
26
$\begingroup$

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).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Very good. .... $\endgroup$
    – paul garrett
    Commented Dec 13, 2012 at 23:13
  • $\begingroup$ Perfectly clear answer. This is also the difference to the nilpotent setting, where there are no large compact subgroups. $\endgroup$
    – Marc Palm
    Commented Dec 14, 2012 at 11:35
  • $\begingroup$ Yes, Clozel's appendix is exactly what one needs. Thanks. $\endgroup$
    – Valerie
    Commented Dec 14, 2012 at 14:58
4
$\begingroup$

Freitag and van Dijk proved that the adelic points of a reductive group over a global field is trace class (Theorem 2.3), while every trace class group is of type I (Theorem 1.7). So the answer is yes (again).

Improve this answer
$\endgroup$
1
$\begingroup$

The question is fully answered. I just add some useful older references.

Moore, Calvin C. Decomposition of unitary representations defined by discrete subgroups of nilpotent groups. Ann. of Math. (2) 82 (1965), 146–182.

  • The following paper seems to be relevant, but I only had access to its mathscinet review. It would be nice if anyone could provide a link.

Tadić, M. (YU-ZAGR) Dual spaces of adelic groups. (Serbo-Croatian summary) Glas. Mat. Ser. III 19(39) (1984), no. 1, 39–48.

Bekka, M. E. B. (F-METZ-LM); Cowling, M. (5-NSW-SM) Some irreducible unitary representations of G(K) for a simple algebraic group G over an algebraic number field K. (English summary) Math. Z. 241 (2002), no. 4, 731–741.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .