Let $F$ be a local field (whose residue field is $q$) and $E$ its quadratic extension. Let $\pi$ be a irreducible principal series representation $\pi(\chi_1, \chi_2)$ of $GL_F(2)$ especially where $\chi_1, \chi_2$ are unitary characters. Then I know $L_F(s,\pi)=\frac{1}{1-\chi_1(\varpi)q^s} \cdot \frac{1}{1-\chi_2(\varpi)q^s}$. Then what is $L_E(s,BC(\pi))$ in terms of $\chi_1, \chi_2$?
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$\begingroup$ Is $E$ meant to be the unique unramified extension (in char not $2$? $\endgroup$– paul garrettJan 21, 2014 at 20:45
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$\begingroup$ @Garrett, you are right!, E is unramified extension. I missed it. $\endgroup$– MontyJan 22, 2014 at 3:05
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$\begingroup$ Do you mean to assume that $\chi_i$ are unramified characters? $\endgroup$– David LoefflerJan 22, 2014 at 7:34
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$\begingroup$ @David, you are right. $\chi_i$ are unramified characters. $\endgroup$– MontyJan 22, 2014 at 10:05
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1 Answer
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If $\pi = \pi(\chi_1, \chi_2)$ is an irreducible principal series, then $BC(\pi)$ is the irreducible principal series attached to the unramified characters $\chi_i \circ N_{E/F}$ of $E^\times$. So if the $\chi_i$ are unramified then we have
$$L_E(s, BC(\pi)) = (1 - \chi_1(\varpi)^2 q^{-2s})^{-1}(1 - \chi_2(\varpi)^2 q^{-2s})^{-1}.$$
(The point of base change is that it corresponds to restriction of Weil-Deligne representations from $W_F$ to $W_E$ under the local Langlands correspondence, and this relation is completely obvious on the Galois side, since the Frobenius of $E$ is the square of the Frobenius of $F$.)
answered Jan 22, 2014 at 11:50