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Let $\pi_p$ be a ramified representation of $GL(n,\mathbb{Q}_p)$. Let $\chi_p$ be a ramified representation of $GL(1,\mathbb{Q}_p)$.

Is it generally known that $L(s,\pi_p\times \chi_p)=1$ if $\chi_p$ is ramified enough (higher conductor)?

I believe this is known for $n=2$ or even general $n$. Can you give some reference?

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    – GH from MO
    Commented Nov 10, 2019 at 14:20

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Yes, we have $L(s,\pi_p\otimes\chi_p)=1$ if the conductor of $\chi_p$ exceeds the conductor of $\pi_p$. By the inductive nature of the local $L$-function (see Jacquet: Principal $L$-functions of the linear group), it suffices to show this when $\pi_p$ is supercuspidal.

So let $\pi_p$ be supercuspidal. If $n\geq 2$, then $L(s,\pi_p\otimes\chi_p)=1$, because $\pi_p\otimes\chi_p$ is supercuspidal. If $n=1$, then $\pi_p\otimes\chi_p$ is a ramified quasicharacter of $\mathbb{Q}_p^\times$ by the assumption on the conductors, hence again $L(s,\pi_p\otimes\chi_p)=1$.

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