Let $G$ be the absolute Galois group of some number field. Can there be a semisimple continuous representation $G\to GL\_n(\overline{\mathbb{F}\_p})$ (the latter has Zariski topology) with infinite image?



$\mathrm{mod}\:p$ Galois representation with respect to Zariski topology
No. A quick proof uses the existence of Haar measure on compact topological groups like the Galois group.


The kernel would be a closed subgroup of the Galois group with infinite index, and thus would have Haa measure $0$. However, because $GL\_n (\overline{\mathbb F\_p})$ is countable, countably many translates cover the Galois group, so the Galois group would have measure $0$, contradicting the fact that it has measure $1$.


I'm sure a more direct proof that avoids Haar measure can be made to work as well.


