If I have a cusp form $f$, I can consider the associated Galois representation $\rho_l(f)$ for any prime $l$. For two distinct primes $p$ and $q$, what is the relationship between $\rho_p(f)$ and $\rho_q(f)$? Are they basically the same? How can I make the relationship precise?

  • 1
    $\begingroup$ They have the same $L$-function (locally and globally), which is both pretty impressive and quite far from being "basically the same". $\endgroup$
    – Olivier
    Sep 28, 2021 at 9:48

1 Answer 1


At the most basic level, $\rho_p$ and $\rho_q$ are "nothing to do with each other". E.g. the kernels of $\rho_p \bmod p$ and of $\rho_q \bmod q$ cut out two finite Galois extensions of $\mathbf{Q}$ which are essentially unrelated.

On the other hand, there is quite an industry devoted to formulating properties of $\rho_p$ which are independent of $p$ (with the independence sometimes being easy, and sometimes very hard indeed); these results go under the name of "local-global compatibility in the Langlands programme".

Here is an example at a rather deeper level of how the different $\rho_p$'s complement each other. If $f$ corresponds to an elliptic curve of analytic rank 0, then $L(f, 1) / \Omega_E$ is a nonzero rational number; and the BSD conjecture gives a formula for what that number should be. There is a rather substantial body of theory aiming to prove, for a given prime $p$, that the power of $p$ dividing $L(f, 1)/\Omega_E$ is the "correct" one for the BSD formula (i.e. BSD holds up to a $p$-adic unit); and this strategy involves a deep study of the arithmetic of $\rho_p$ -- but you really need $\rho_p$ here, there's no way that $\rho_q$ for $q \ne p$ will tell you anything about the powers of $p$ dividing $L(f, 1)$. If you can do this for all $p$, then you can nail down $L(f, 1)$ exactly, but you need all the $\rho_p$'s available; each one is giving you different information.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .