The Stacks project

Lemma 63.31.7. Switching $l$. Let $E$ be a number field. Start with

\[ \rho : \pi _1(X)\to SL_2(E_\lambda ) \]

absolutely irreducible continuous, where $\lambda $ is a place of $E$ not lying above $p$. Then for any second place $\lambda '$ of $E$ not lying above $p$ there exists a finite extension $E'_{\lambda '}$ and a absolutely irreducible continuous representation

\[ \rho ': \pi _1(X)\to SL_2(E'_{\lambda '}) \]

which is compatible with $\rho $ in the sense that the characteristic polynomials of all Frobenii are the same.

Proof. To prove the switching lemma use Theorem 63.31.3 to obtain $f\in C(\overline{\mathbf{Q}}_ l)$ eigenform ass. to $\rho $. Next, use Proposition 63.31.5 to see that we may choose $f\in C(E')$ with $E \subset E'$ finite. Next we may complete $E'$ to see that we get $f\in C(E'_{\lambda '})$ eigenform with $E'_{\lambda '}$ a finite extension of $E_{\lambda '}$. And finally we use Theorem 63.31.2 to obtain $\rho ': \pi _1(X) \to SL_2(E_{\lambda '}')$ abs. irred. and continuous after perhaps enlarging $E'_{\lambda '}$ a bit again. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 63.31: Automorphic forms and sheaves

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03VY. Beware of the difference between the letter 'O' and the digit '0'.