Compactness of the automorphic quotient and genericity

Question

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.

Answer 1

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.

Answer 2

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.