3
$\begingroup$

I have a question about comparing two $\ell$-adic Galois representations. Suppose we have two irreducibible Galois representaions $$ \rho_1,\rho_2: Gal(\overline{\mathbb{Q}}/\mathbb{Q})\to GL_N(\mathbb{Z}_{\ell}). $$ If $\rho_1$ is not isomorphic to $\rho_2$, then there must be a largest integer $a$ such that $$tr(\rho_1)\equiv tr(\rho_2)\ (mod\ \ell^a).$$ Similarly, there is a largest integer $b$ such that up to a conjugation by an element in $GL_N(\mathbb{Z}_{\ell})$, we have $$ \rho_1 =\rho_2 \ (mod\ \ell^b). $$ It is obvious that $b\leq a$. I am wondering is there any known conditions to ensure that $b=a$?

$\endgroup$
1
  • $\begingroup$ $N=1$ for instance. You want all other coefficients of the characterstic polynomials, like $\det(\rho_i)$ to be congruent , too. $\endgroup$ Oct 12, 2018 at 8:43

1 Answer 1

1
$\begingroup$

This is not true if the representations are residually reducible. Suppose that $N=2$, then $b$ can be strictly less than $a$. Let $\bar{\rho}_i$ for $i=1,2$ denote the mod $l$ representations. Then if $\bar{\rho}_1$ is diagonal and $\bar{\rho}_2$ is upper-triangular with the same diagonal entries as $\bar{\rho}_1$ but not isomorphic to $\bar{\rho}_1$ it would follow that $a\geq 1$ and $b=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.