Let $E/F$ be a quadratic extension of number fields and $v$ is a place of $F$.

Let $\chi_1,\chi_2$ be the unramified characters of $F_v^{\times}$.

If $B(\chi_1,\chi_2)$ is the unramified principal series representation of $GL_2(F_v)$, what is the $BC(\pi)$, the base change of $\pi$ to $GL_2(E_v)$?

I suppose that $BC(\pi)=B(\chi_1 \circ \text{Norm}_{E_v/F_v},\chi_2 \circ \text{Norm}_{E_v/F_v})$. Is this right?


The answer to your final question should be yes.

Let me assume that $v$ is inert in $E$, so that $E_v / F_v$ is a honest quadratic extension. Assuming that $B(\chi_1, \chi_2)$ is irreducible, it corresponds via local Langlands (a theorem for $\mathrm{GL}_2(K)$ and any $p$-adic field $K$) to the 2-dimensional representation of the Weil group of $F_v$ given by the sum of the characters $\chi_1$ and $\chi_2$.

The operation of base change on the $p$-adic side corresponds to restriction of Galois representations, thus the base change of $B(\chi_1, \chi_2)$ corresponds via local Langlands to the restriction of $\chi_1 \oplus \chi_2$ to the Weil group of $E_v$.

When we restrict $\chi_1 \oplus \chi_2$ to the Weil group of $E_v$ and then make this homomorphism factor through the abelianization $E_v^*$, local class field theory tells us that this corresponds exactly to pre-composition of $\chi_1$ and $\chi_2$ with the Norm map. Now go back to the $p$-adic side and you have your statement.

  • $\begingroup$ Thanks for your comment. You said only the case where $v$ is inert in $E$. If $v$ splits in $E$, what is $BC(\pi)$? In this case, since $GL_2(E_v)\simeq GL_2(F_v)\times GL_2(F_v)$, I suppose it should be $B(\chi_1,\chi_2) \boxtimes B(\chi_1,\chi_2)$. Am I right? $\endgroup$
    – Monty
    Nov 14 '16 at 13:49
  • 1
    $\begingroup$ Yes, I think that is correct. $\endgroup$
    – user94041
    Nov 14 '16 at 18:30

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