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Let $F$ be a number field. For an irreducible cuspidal automorphic representation $\pi$ of $\operatorname{GL}_n(\mathbb{A}_F)$, we say that $\pi$ is symplectic (or orthogonal) if $L(s,\pi,\bigwedge^{2})$ (or $L(s,\pi,\operatorname{Sym}^2)$) has a pole at $s=1$.

I am wondering whether if $\pi=\bigotimes \pi_v$ is symplectic (or orthogonal), then $\pi_v$’s are also symplectic (or orthogonal) for all places $v$?

(Here, $\pi_v$ is symplectic (or orthogonal) means that its corresponding Weil–Deligne group representation by local Langlands correspondence is of such type.)

Any comments are welcome!

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  • $\begingroup$ It's easy to see that ($\pi$ is either symplectic or orthogonal) $\Leftrightarrow$ ($\pi$ is selfdual) $\Leftrightarrow$ ($\pi_v$ is selfdual for all $v$), but it's not so obvious if the local factors are symplectic / orthogonal if $\pi$ is. $\endgroup$
    – David Loeffler
    Commented Jan 2, 2022 at 8:03
  • $\begingroup$ @DavidLoeffler, thank you very much for the helpful comment! The definition of orthogonal (or symplectic) of $\pi_v$ I mean is that if $\phi: WD_F \to \operatorname{GL}(n)$ is the Weil-Deligne representation corresponding to $\pi_v$, then we say $\pi_v$ is orthogonal (or symplectic) if the image of $\phi$ can be replaced to $\operatorname{SO}(n)$ (or $\operatorname{Sp}(n)$.) So if $\pi$ is symplectic (orthogonal), then it seems $\pi_v$’s are symplectic (or orthogonal) for all places $v$. $\endgroup$
    – Andrew
    Commented Jan 2, 2022 at 13:15

1 Answer 1

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Here is a proof of the claim using results from Arthur's monograph The Endoscopic Classification of Representations: Orthogonal and Symplectic Groups.

Let $N = 2n$ be an even integer, and $\pi$ a cuspidal automorphic representation satisfying $\pi^\vee \cong \pi$.

  • Since $\pi \cong \pi^\vee$, the Rankin-Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. We have the factorisation $L(\pi \times \pi, s) = L(\pi, S^2, s) L(\pi, \wedge^2, s)$ and both factors are non-vanishing at $s = 1$, so exactly one of them must have a pole; thus $\pi$ is orthogonal or symplectic, in the sense of the question, but never both.
  • The discussion preceding Theorem 1.5.3 of Arthur shows that $\pi$ defines an element of his set $\tilde{\Phi}_{\mathrm{sim}}(N)$ of global parameters, and this lies in the subset $\tilde{\Phi}_{\mathrm{sim}}(G)$ for a uniquely determined quasi-split group $G$ whose Langlands dual $\widehat{G}$ is either $SO_{2n}$ or $Sp_{2n}$.
  • Theorem 1.5.3 of Arthur shows that $\widehat{G}$ is symplectic if $\pi$ is of symplectic type, and $\widehat{G}$ is orthogonal if $\pi$ is of orthogonal type. (This is a very deep theorem, despite sounding like a tautology!)
  • Theorem 1.4.2 of op.cit. now shows that there is a cuspidal automorphic representation $\sigma$ of $G$ such that, for every $v$, the Weil–Deligne representation associated to $\pi_v$ is the image in $GL_N$ of the ${}^L G$-valued parameter associated to $\sigma_v$. So, in particular, it is symplectic (resp. orthogonal) if $\pi$ is.
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  • $\begingroup$ thank you so much for sharing your knowledge and wisdom. It helped me a lot! In the meanwhile, may I ask one more question? If $\phi$ is the Weil-Deligne representation associated to $\pi_v$, we can decompose $\phi$ as $\phi_0 + \phi’ + \phi’^{\vee}$, where $\phi_0$ is the sum of irreducible subrepresentation of $\phi$ of the same type with $\phi$ and $\phi’$ is the sum of irreducible subrepresentation of $\phi$ which is not of the same type with $\phi$. Then I am wondering if it is possible that $\phi_0$ could be zero? In that case, the component group of $\pi_v$ is trivial. $\endgroup$
    – Andrew
    Commented Jan 4, 2022 at 2:30
  • $\begingroup$ I am wondering that $\phi_0$ could be zero at places v where $\pi_v$ is unramified. I think it is possible if $\pi$ is of symplectic type cuspidal automorphic representation of $GL(2)$. $\endgroup$
    – Andrew
    Commented Jan 4, 2022 at 6:36
  • $\begingroup$ It's not clear to me if such a decomposition is always well-defined, but for a symplectic-type $GL(2)$ rep, at almost all finite places the representation will look like $\phi_v \cong \chi_v \oplus \chi_v^\vee$ for a 1-dimensional unramified character $\chi_v$. $\endgroup$
    – David Loeffler
    Commented Jan 4, 2022 at 7:07
  • $\begingroup$ Thank you very much for the comment! In the $GL(2)$ case, I also think as you because square-integrable representation cannot be unramified. So the component group $S_{\phi_v}$ of $\phi_v$ should be trivial group at unramified places $v$. Thank you very much again! $\endgroup$
    – Andrew
    Commented Jan 4, 2022 at 7:58

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