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Sorry, I asked this two days ago, but this time I modified it to be easily read and added more specific explanation. I hope to get your illuminating comment on whether my approach is right.

I am computing the order of some $GL(2)$-local $L$-function at $s=0$.

Let $F$ be a local field and $E=F \times F$. Let $\chi$ be the character of $E^\times$ whose restriction to $F^{\times}$ is trivial, $\pi$ be a tempered representation of $GL_2(F)$ and $BC_E(\pi)$ be the base change of $\pi$ to $GL_E(2).$ (since $E=F \times F$, we assume $BC(\pi)=\pi \boxtimes \pi^{\lor}$).

Then I want to calculate the order of $L_E(s,BC(\pi)\otimes \chi)$ at $s=0$.

I suppose it has a pole at $s=0$ whose order is $-2$ or $-4$, but I am not sure.

Let me briefly write my thought on this.

(Since $\chi$ is trivial on $E$, we can write $\chi=\chi_1 \times \chi_1^{-1}$ where $\chi_i$ is a character of $F^{\times}$. Since $\pi$ is tempered, if we let $\pi=B(\lambda_1,\lambda_2)$ for two unitary characters $\lambda_i$ of $F^{\times}$, then $$L_E(s,BC(\pi)\otimes \chi)=L_F(s,B(\lambda_1 \chi_1,\lambda_2 \chi_1) \boxtimes B(\lambda_1^{-1}\chi_1^{-1},\lambda_2^{-1}\chi_1^{-1}))$$

Thus it equals to $$\frac{1}{(1-q^{-s})(1-\lambda_1\lambda_2^{-1}(\varpi)q^{-s})(1-\lambda_1^{-1}\lambda_2(\varpi)q^{-s})(1-q^{-s})}$$ and so if $\lambda_1\lambda_2^{-1}=1$, it has pole at $s=0$ whose order is -4, and if $\lambda_1\lambda_2^{-1}\ne1$, it has pole at $s=0$ of order -2.)

My thought is right?

Since I have been stumbling on this so many times, if you shed a little light on this, I will be very appreciated.

Any comment will be welcome.

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1 Answer 1

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Professor Shunsuke Yamana kindly answered me for this and I want to share his comment with others.

He says, $L_E(s,BC(\pi)\otimes \chi)$ is not the product of L-functions but a product of local L-factors. So I think it should be $L_F(s,\pi \otimes \chi_1) \cdot L_F(s,\pi^{\lor} \otimes \chi_1^{-1})$ if $\chi=(\chi_1,\chi_1^{-1})$.

Thus if $\pi=B(\lambda_1,\lambda_2)$, then I think $L_E(s,BC(\pi)\otimes \chi)=\frac{1}{1-\chi_1\lambda_1(\varpi)q^{-s}}\cdot \frac{1}{1-\chi_2\lambda_1(\varpi)q^{-s}} \cdot \frac{1}{1-\chi_1^{-1}\lambda_1^{-1}(\varpi)q^{-s}}\cdot \frac{1}{1-\chi_2^{-1}\lambda_1^{-1}(\varpi)q^{-s}}$ and so it might not have a pole at $s=0$ depending the choice of $\chi_1$.

(ps: I express deep thanks to prof.Yamana for his illumination)

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