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Let $G$ be a reductive group defined over a field $F$. Let $\mathbf{A}$ denote the ring of adeles of $F$. My question is:

Assuming the automorphic quotient $[G]=G(F) \backslash G(\mathbf{A})$ is compact, can we say that all the (non-character) automorphic representations of $G$ are tempered? generic?

I do not precisely understand the relations between the notions of tempered and generic representations beyond the very specific $\mathrm{GL}(n)$ case, so that any reference about these matters is also welcome.

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    $\begingroup$ If the quotient is compact, then so is each $G(F_v)$ for $v$ finite, and this is a very strong condition. An absolutely simple, adjoint group over $F_v$ with this property must be the adjoint quotient of a central division algebra over $F_v$. I don't know enough about the global situation to know whether this answers your question. $\endgroup$ – LSpice May 15 at 1:46
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    $\begingroup$ @LSpice Any reductive group over a global field is quasi-split at all but finitely many primes. (The moduli space of Borel subgroups of $G$ is geometrically $G/B$, which is a smooth variety, so it is itself smooth, hence it has a model smooth over all but finitely many primes. The fibers are flag varieties, hence have rational points, and because the variety is smooth must have rational lifts). One can probably see this more concretely using the fact that any division algebra is ramified at only finitely many primes. So this can't happen. $\endgroup$ – Will Sawin May 15 at 2:11
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    $\begingroup$ @LSpice I think the claim that $G(F_v)$ is compact for all $F_v$ is wrong. Instead, isn't $G(F_v)$ compact for one $F_v$ equivalent to compactness of the automorphic quotient? $\endgroup$ – Will Sawin May 15 at 2:32
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    $\begingroup$ @WillSawin Thanks you both for the discussion. Indeed, $G(F_v)$ compact at every places is too strong. The requirement of $G(F_v)$ compact at only one place is what happens for quaternion algebras, and this is also the case for inner forms of $\mathrm{GSp}(4)$. But is this a more general statement? $\endgroup$ – Desiderius Severus May 15 at 3:10
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    $\begingroup$ @DesideriusSeverus It can happen that the automorphic quotient is compact without any $G(F_v)$ being compact. For instance, take $G=SL(D)$ for $D$ a division algebra of degree $6$ over $\mathbb{Q}$ with invariants $(1/2,1/2,1/3,2/3)$ at four different places and $0$ elsewhere. $\endgroup$ – Aurel May 15 at 12:37
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I'm not sure what you mean by tempered or generic, or if you even know what you mean (defined locally or globally? in terms of representations or parameters?). But basically the answer is no. For instance, if you look at a compact form of SO(5), you can see representations which are "Saito-Kurokawa" lifts, and even "non-holomorphic Saito-Kurokawa" lifts, i.e. they are near equivalent to Saito-Kurokawa packets on the split SO(5) = PGSp(4). The Saito-Kurokawa forms are neither locally generic nor tempered. Alternatively, we can say the endoscopic parameters defined by Arthur for these representations are neither generic nor tempered. Basically representations that are lifts from smaller groups should not be generic or tempered.

For what you might mean by generic or tempered, I don't remember exact references off the top of my head, but presumably you can look in Corvallis and some survey papers by Arthur. For the Saito-Kurokawa case, this is explained in a paper of Gan.

However, for compact U($p$) which are division at a finite place, and $p$ prime, the endoscopic classification announced by Kaletha-Minguez-Shin-White will tell you that the non-abelian representations have generic parameters. These will also be tempered everywhere locally under a cohomological condition by work of Shin. The point is these conditions mean you don't have any endoscopic representations. I'm not sure if this is explicitly recorded in the literature, though it is known to experts. I'm writing something down about this in a preprint that is almost done.

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  • $\begingroup$ I confess this is not a field where I have deep roots nor a good understanding, unfortunately I don't know personally many experts on that. I will look into Corvallis, and KMSW article seems to be in some sense readable and has explicit introductory sections. However, I am exactly working on compact unitary groups on which I am free to assume some local conditions, so that your last statements and your incoming preprint is absolutely of interest to me. $\endgroup$ – Desiderius Severus May 16 at 0:49
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About the hypothesis that $G_k\backslash G_\mathbb A$ is compact: In addition to Will Sawin's good comments, @Aurel's example can easily be expanded a little for clarity, and some other classical-group examples given.

To fill in some details to Aurel's example: Fujisaki's lemma (as in Weil's `Adeles and algebraic groups') shows that $D^\times\backslash D_\mathbb A^1$ is compact for any (central, $n^2$-dimensional) division algebra $D$ over a global field $k$. At the same time, such $D$ splits locally almost everywhere, so the group $SL_1(D)$ is locally $SL_n(k_v)$ almost everywhere. Further, the sum of the local (Brauer-group) invariants can be any [edit: not necessarily even] number of fractions with denominator $n$ summing to an integer [edit: not necessarily $1$]. The local $D_v=D\otimes_k k_v$ is a division algebra (hence $SL_1(D_v)$ is compact) if and only if the least common multiple of the denominators in lowest terms is $n$. For $n$ prime, there are qualitatively only two things that can arise: $SL_n(k_v)$ and $SL_1(\mathrm{division\;algebra})$, but for $n$ composite, as in Aurel's example, it can be arranged so that no local group is compact, but the arithmetic quotient is compact.

A family of examples somewhat more distant from $GL_n$'s is the family of orthogonal groups $G=O(Q)$ of a (non-degenerate, $k$-valued) quadratic form $Q$ on a $k$-vectorspace $V$ of dimension $n$ over $k$. By Mahler's criterion from reduction theory (e.g., see Godement's treatment of reduction theory in Sem. Bourb.), if $Q$ is $k$-anisotropic, then $G_k\backslash G_\mathbb A$ is compact.

By Hasse-Minkowski, $Q$ is $k$-anisotropic if and only if it is $k_v$-anisotropic at at least one place $v$. It is straightforward that $G_v$ is compact if and only if $Q$ is $k_v$-anisotropic. For complex $k_v$, this cannot happen. For real $k_v$, there is a signature $p,q$ at $v$, and $G_v$ is compact if and only if $p=0$ or $q=0$. At finite places, in dimensions $5$ and larger, $Q$ is always isotropic, and $G_v$ is non-compact.

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    $\begingroup$ Thanks for expanding my comment! One tiny remark: the nontrivial local invariants don't have to be an even number, and they must sum to an integer, not necessarily 1. The even number occurs only for quaternion algebras, where is is a consequence of the fact that they are 0 or 1/2 and sum to an integer. $\endgroup$ – Aurel May 16 at 8:45
  • $\begingroup$ @Aurel, oop, yes, you're right. I'll correct that. $\endgroup$ – paul garrett May 16 at 12:31

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