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Question 1: Let $G$ be a connected reductive group defined over a number field $K$, and $\pi$ is an irreducible cuspidal automorphic representation of $G(\mathbb{A}_K)$. Then by a theorem of Flath, we have $\pi=\otimes^{'}\pi_v$. All but finitely many $\pi_v$ are unramified and they all generic by Shalike.

I want to know whether there exists a finite place $\mu$ such that $\pi_{\mu}$ is supercuspidal for $G(K_{\mu})$ or there is no relation with cuspidal automorphic representation for $G(\mathbb{A}_K)$ and supercuspidal representation of $G(K_{\mu})$.

Question 2: Let $S$ be a finite set containing infinite places, and assume $\pi_v$ is unramified for $v \notin S$ and $\pi_{\mu}$ is generic for $\mu \in S$, then can I say $\otimes_{v}^{'}\pi_v$ is a cuspidal automorphic representation of $G(\mathbb{A}_K)$? If not, what additional condition should I add?

Question 3: I know there is a proposition below proving by Bernstein-Zelevinsky classification.

$\pi_v$ is an unramified and supercuspidal for $GL_n(K_v)$, where $v$ is a finite place if and only if $n=1$ and $\pi_v$ is an unramified quasi-character.

I want to know whether the proposition holds for general connected reductive p-adic group.

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  • $\begingroup$ I think you'll get a better understanding of Question 2 if you proceed in the reverse direction, that is if you ask yourself what conditions the $\pi_{v}$ must satisfy if they are the local components of an automorphic $\pi$. Then you should realize how rare global automorphic $\pi$ are compared to local ones. Then you'll see that no list of purely local conditions on the $\pi_v$ can ensure that they are the local components of a global $\pi$ (even for $\operatorname{GL}_{2}$, how could such a list guess that some weights are globally forbidden?). $\endgroup$
    – Olivier
    Commented Mar 2, 2018 at 9:01
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    $\begingroup$ It is not true that all of the $\pi_v$ are generic. (You claim that "they all generic by Shalike"; presumably you mean the theorem of Piatetski-Shapiro and Shalika which states that cuspidal automorphic representations of $GL(n)$ are generic, but this does not generalise to groups other than $GL(n)$.) E.g. there are well-known examples of cuspidal automorphic reps of $GSp(4)$ coming from Saito--Kurokawa lifts which are not locally generic at any finite place. $\endgroup$
    – David Loeffler
    Commented Mar 2, 2018 at 14:29

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It is not necessary that a cuspidal $\pi$ have a supercuspidal local component: for example, let $\Delta(\tau)$ be Ramanujan's delta function and let $\pi = \otimes' \pi_p$ denote the associated cupsidal automorphic representation of $\textrm{GL}_2(\mathbb{A}_{\mathbb{Q}})$ (see Kudla's article here). Since $\Delta$ has level 1, $\pi_p$ will be unramified for every prime $p$.

What are your assumptions for question 2? Given a random collection $\{\pi_v\}_v$ of irreducible local representations, almost all of which are unramified, the representation $\otimes' \pi_v$ will be neither cupsidal nor automorphic.

There are many ways to prove question 3. For simplicity, assume $G$ is unramified over $K_v$ (that is, quasi-split and split over an unramified extension), let $K$ denote a hyperspecial maximal compact subgroup, $I\subset K$ an Iwahori subgroup, and let $P = MN$ be a minimal parabolic subgroup. If $\pi$ were both ($K$-)unramified and supercuspidal, then Theorem 3.7 in Cartier's article in the Corvallis proceedings would give $$0\neq \pi^K \subset \pi^I \stackrel{\sim}{\rightarrow}(\pi_N)^{K\cap M}.$$ If $N\neq 1$, then the right-hand side would be zero by supercuspidality. Therefore $G$ must be a torus.

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    $\begingroup$ And/but it is worth emphasizing/recalling that if $\bigotimes \pi_v$ is automorphic and there is a supercuspidal $\pi_{v_o}$ (at some finite $v_o$) then the whole is cuspidal. And that this follows easily by direct computation. $\endgroup$
    – paul garrett
    Commented Mar 1, 2018 at 18:35

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