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Suppose $G$ is a semisimple algebraic group over the rational numbers, $\pi$ is a cuspidal automorphic representation of $G$, and $r: \widehat{G}(\mathbf{C}) \rightarrow \mathrm{GL}_N(\mathbf{C})$ is an irreducible representation of the Langlands dual group. Suppose moreover that $G$ is split at every finite place, and $\pi$ is unramified at every finite place. Then the $L$-function $L(\pi,r,s)$ is unambiguously defined. Is it conjectured that this $L$-function is non-vanishing at $s=1$?

Note that if $G = \mathrm{PGL}_n$ and $r$ is the standard representation, then the nonvanishing is known.

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At least it should be true under suitable assumptions. Jacquet-Shalika proved nonvanishing of automorphic $L$-functions of unitary cuspidal representations of GL($m$) at $s=1$. An automorphic $L$-function $L(\pi, r, s)$ as in your question is conjectured to correspond to an isobaric sum $\boxplus \pi_i$ of cuspidal $L$-functions on GL($m_i$)'s such that $L(\pi, r, s) = \prod L(\pi_i, s)$. If this is a unitary isobaric sum, one gets nonvanishing at $s=1$ by Jacquet-Shalika.

I believe the right condition for this to be a unitary isobaric sum (for any $r$) is that $\pi$ is tempered and unitary, but I don't know a precise reference at for general $G$, $r$. At the level of packets, temperedness is conjecturally the same as generic (Shahidi). There is a brief discussion in the case of classical groups of the connection between unitary isobaric sums and genericity at the end of Section 10 in CKPSS's "Functoriality for the classical groups" paper.

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  • $\begingroup$ On a related note, Shahidi has proven that the product of $L(s,\pi,r,s)$, indexed by the representations $r$, is nonvanishing at $s = 1$. Of course, this doesn't preclude the possibility that some of these individual $L$-functions have a pole at $s = 1$. See Theorem 5.1 of this paper: doi.org/10.2307/2374219 $\endgroup$ Oct 23, 2022 at 14:43
  • $\begingroup$ I was under the impression that when transferring to $GL(N)$, we may get something like (for example) $L(\pi,r,s) = L(\pi_1,s-1) L(\pi_2,s+1)$, so evaluating the left-hand side at $s=1$ involves the L-function of a (unitary!) cusp form at $s=0$. In other words, in the isobaric sum you mention, the constituents need not be unitary. Do I have to worry about that? (I don't know the details, which is why I didn't include this caveat in the question statement.) $\endgroup$
    – Joseph
    Oct 23, 2022 at 15:28
  • $\begingroup$ Regarding Peter's comment, to clarify: It looks like Shahidi answers this question, when $\pi$ is assumed globally generic and $L(\pi,r,s)$ is an L-function approachable via the Langlands-Shahidi method (in particular, it appears in the constant term of a cuspidal Eisenstein series.) $\endgroup$
    – Joseph
    Oct 23, 2022 at 15:44
  • $\begingroup$ @Joseph You're right in general there can be shifts---I was just thinking about unitary constituents. So I should probably an assumption like temperedness. Let me get back to this when I have more time. $\endgroup$
    – Kimball
    Oct 23, 2022 at 17:16
  • $\begingroup$ Okay, I revised the answer and I think it's correct now, though I can't find a precise reference for unitary isobaric condition. $\endgroup$
    – Kimball
    Oct 24, 2022 at 1:29

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