10
$\begingroup$

Let $F$ be a global field, let $\mathbf{G}$ be a reductive algebraic group over $F$, and let $\pi$ be an irreducible discrete automorphic representation of $\mathbf{G}$. Write $\pi$ as a restricted product of local factors $\pi=\bigotimes_\nu\pi_{\nu}$ (here, $\nu$ ranges over the places of $F$), and suppose that for almost all $\nu$, the $\mathbf{G}(F_\nu)$-representation $\pi_\nu$ is tempered. Is it true that $\pi_\nu$ is tempered for all $\nu$?

I am particularly interested in the following setting: Suppose $F$ has positive characteristic, let $D$ be a central division $F$-algebra and take $\mathbf{G}=\mathbf{GL}_1(D)$ or $\mathbf{G}=\mathbf{PGL}_1(D)$.

Notice that the answer to the above question is yes in case $F$ has positive characteristic and $\mathbf{G}=\mathbf{GL}_n$. In fact, in this case, if just one of the local factors is tempered, then all local factors are tempered. This follows from Lafforgue's proof of the Ramanujan conjecture for $\mathbf{GL}_n$ in positive characteristic, asserting that local factors of cuspidal automorphic representations are tempered, and the description of the residual spectrum by M\oeglin and Waldspurger (1989), which implies (after some work) that the local factors of residual representations of $\mathbf{GL}_n$ are never tempered.

Returning to the the case of $\mathbf{G}=\mathbf{GL}_1(D)$ with $\mathrm{char}(F)>0$, let $\pi=\bigotimes_\nu\pi_\nu$ be an automorphic representation of $\mathbf{G}$ such that $\pi_n$ is tempered for almost all $\nu$. The global Jacquet-Langlads correspondence gives an automorphic representation $\rho=\bigotimes_\nu\rho_\nu$ of $\mathbf{GL}_n$ such that $\rho_\nu=\pi_\nu$ for all $\nu\notin S$. By the previous paragraph, this means $\rho_\nu$ is tempered for all $\nu$. The question therefore reduces to a question about the behavior of the global Jacquet-Langlands correspondence at ramified places, namely, does the fact that $\rho_\nu$ is tempered for all $\nu\in S$ implies that $\pi_\nu$ is tempered for all $\nu\in S$. Of course, the answer is yes if $D_\nu:=D\otimes_FF_\nu$ is a division algebra for all $\nu\in S$, so assume this is not the case.

$\endgroup$
1
  • $\begingroup$ Note for general $G$, being tempered at one place is not enough: mathoverflow.net/q/184937 $\endgroup$
    – Kimball
    Commented Feb 2, 2016 at 22:13

1 Answer 1

5
$\begingroup$

It turns out that the answer is "yes" in the case $\mathbf{G}=\mathbf{GL}_1(D)$, where $D$ is a central division $F$-algebra of degree $n$ and $\mathrm{char}(F)>0$. More precisely, the local Jacquet-Langlands at a place $\nu$ takes (when defined) tempered representations of $\mathbf{GL}_n(F_\nu)$ to tempered representations of $\mathbf{GL}_1(D_\nu)$. Since the global correspondence behaves like the local correspondence on each local factors, this gives the result as explained above.

The proof of the statement about the local Jacquet-Langlands correspondence is done by noting the following (non-trivial) facts:

  • The local Jacquet-Langlands correspondence is compatible with the $\times$-product.

  • The tempered representations of $\mathbf{GL}_r(D_\nu)$ are precisely those of the form $\pi_1\times\dots\times \pi_t$ where $\pi_i$ is an irreducible (unitary) square-integrable representation of $\mathbf{GL}_{r_i}(D_\nu)$ and $\sum r_i=r$.

  • The local Jacquet-Langlands (which is defined for a certain class of unitary representations of $\mathbf{GL}_{nr}(F_\nu)$) restricts to a bijection between the square integrable representations of $\mathbf{GL}_r(D_\nu)$ and $\mathbf{GL}_{nr}(F_\nu)$.

This argument is not mine and was communicated to me independently by I.A. Badulescu and A.-M. Aubert.

I wrote a full proof giving references to the previous three facts in Theorem 7.13 in https://arxiv.org/abs/1605.02664 .

It should be noted that this proof uses a lot of heavy machinery such as the Ramanujan conjecture for $\mathbf{GL}_n$ in positive characteristic and the global Jacquet-Langlands correspondence. I was kind of hoping that there would be a less involved proof.

$\endgroup$

You must log in to answer this question.

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