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I want to ask some question on conjugate self dual representation.

Let $E/F$ be a quadratic extension of number fields and $c:GL_n(\mathbb{A}_E) \to GL_n(\mathbb{A}_E)$ be a automorphism induced by conjugation $^{-}:E \to E$. For a representation $\sigma$ of $GL_n(\mathbb{A}_E)$, define $\sigma^{c}=\sigma \circ c$.

If $L(s,\sigma,As^{(-1)^{n-1}})$ has a simple pole at $s=1$, then it is easy to check that $\sigma^{c} \simeq \sigma^{\vee}$. (i.e. $\sigma$ conjugate self dual)

For a conjugate self dual representation $\sigma$, it is known that the central character of $\sigma$ is trivial or the unique non-trivial quadratic character of $\mathbb{A}_E^{\times}$.

Then I am wondering that if $L(s,\sigma,As^{(-1)^{n-1}})$ has a simple pole at $s=1$, then the central character of $\sigma$ should be trivial not quadratic character. I also want to know whether $a$ is even.

Thank you very much for your sharing knowledge.

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  • $\begingroup$ Conjugate self-dual just implies $\omega^c = \omega^{-1}$, where $\omega$ is the central character. So quadratic is possible. $\endgroup$
    – Kimball
    Commented Feb 4, 2020 at 16:14
  • $\begingroup$ @Kimball, Yes! You are right! But the condition that Asai L-function has a simple pole at s=1 is much stronger than just conjugate self dual. $\endgroup$
    – Monty
    Commented Feb 4, 2020 at 16:29

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