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.




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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.



Order of some $L$-function at $s=0$
$L\_E(s,BC(\pi)\otimes \chi)$ 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 $\chi\_1$.


