I'm asking this not because I have an idea how one might approach it, but because it seems natural and inherently interesting.

Let $K$ be a number field, $G_K$ its absolute Galois group, and $\ell\neq\ell'$ prime numbers. An $\ell$-adic representation of $K$, i.e. a continuous homomorphism $\rho_\ell$ from $G_K$ to $\mathrm{GL}(n,\mathbb Q_{\ell})$ is said to be compatible with an $\ell'$-adic representation $\rho_{\ell'}$ of $K$ if for almost all (i.e. all but finitely many) places $v$ of $K$, the representations $\rho_\ell,\rho_{\ell'}$ are unramified, and the characteristic polynomials of $\rho_\ell(F_v)$ and $\rho_{\ell'}(F_v)$ are equal and have rational coefficients, where $F_v \in D_v/I_v \subseteq G_K$ denotes the Frobenius at $v$.

Note that by the Chebotarev density, this uniquely determines the trace of $\rho_{\ell'}$, which, in turn, uniquely determines it up to semi-simplification.

If $\mathcal{L}$ is a collection of prime numbers, then a strictly compatible system of $\ell$-adic representations consists of an $\ell$-adic representation $\rho_\ell$ of $K$ for each $\ell \in \mathcal{L}$ such that there is a finite set $S$ of places of $K$ (called the *exceptional set*) such that for $\ell\neq\ell'$, $v \notin S$, and $v \nmid \ell\ell'$, the representations $\rho_\ell$ and $\rho_{\ell'}$ are unramified at $v$, and the characteristic polynomials of $\rho_\ell(F_v)$ and $\rho_{\ell'}(F_v)$ are equal and have rational coefficients.

Can we find an $\ell$-adic representation unramified almost everywhere with integral traces of Frobenius and a prime $\ell'$ such that there is no $\ell'$-adic representation compatible with it? The Fontaine-Mazur conjecture would predict that the representation cannot be geometric.

But what's more, I'm wondering the following: suppose we have a strictly compatible system of $\ell$-adic representations for almost all $\ell$. Then can we necessarily extend this to a strictly compatible system for all $\ell$?