# Global symplectic (orthogonal) type of automorphic representation compels its type to all its local components?

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.)

• 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. Jan 2, 2022 at 8:03
• @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 $v$. Jan 2, 2022 at 13:15

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.
• 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. Jan 4, 2022 at 2:30
• 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 $GL(2)$. Jan 4, 2022 at 6:36
• 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 $\chi_v$. Jan 4, 2022 at 7:07
• 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! Jan 4, 2022 at 7:58