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Let $\pi$ be a generic $A$-parameter, that is an isobaric automorphic representation of linear group.

Decompose $\pi= \otimes \pi_v$ as a restricted tensor product. Then by the local Langlands correpondence, we can attach the corresponding Weil-Deligne representation $\phi_v$ and we can think the local component group $S_{\phi_v}$ such that $|S_{\phi_v}|$, the number of elements of $S_{\phi_v}$ is $2^{n_{\phi_v}}$. (Here, $n_{\phi_v}$ is the number of irreducible symplectic type sub-representation of $\phi_v$.)

Then I am wondering that if $\pi_v$ is unramified, then $S_{\phi_v}$ is signleton? Because, irreducible symplectic type sub-representation of $\phi_v$ correspond to essentially square-integrable representation of linear groups, but ess. sqaure-integrable representation cannot be unramified.

Please correct me if there is some mistake in the above reasoning!

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  • $\begingroup$ Is it because of square-integrability vs. $\textit{essentially}$ square-integrability? An essentially square-integrable representation can be unramified. $\endgroup$
    – Subhajit Jana
    Dec 6, 2021 at 8:14
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    $\begingroup$ Thanks for the comment! But I think essentially sq.integrable representaton seems cannot be unramified. Isn't it? $\endgroup$
    – Andrew
    Dec 7, 2021 at 10:40
  • $\begingroup$ The phrase "essentially square-integrable" confused me. I thought it is the same as "tempered". Actually, it means "square-integrable mod center". So my previous comment is not correct. $\endgroup$
    – Subhajit Jana
    Dec 7, 2021 at 11:20
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    $\begingroup$ @Subhajit, More precisely, essentially square-integrable is square integrable up to character twisting. $\endgroup$
    – Andrew
    Dec 7, 2021 at 13:29

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