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 $b=a$?

$\begingroup$ $N=1$ for instance. You want all other coefficients of the characterstic polynomials, like $\det(\rho_i)$ to be congruent , too. $\endgroup$ – Chris Wuthrich Oct 12 '18 at 8:43
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 uppertriangular with the same diagonal entries as $\bar{\rho}_1$ but not isomorphic to $\bar{\rho}_1$ it would follow that $a\geq 1$ and $b=0$.