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?
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Suppose we consider the case $n=1$. Then such a representation is automatically semisimple, and it is continuous if and only if its kernel is a closed subgroup of $G$. So I think one can construct such an example if one can construct an infinite Galois extension of number fields $E/F$ such that $\Gal(E/F)$ is isomorphic to a subgroup of $\bar{\mathbb{F}_p}^{\times}. Although I'm not sure how to construct such an extension or whether such an extension exists. $\endgroup$– JefCommented Aug 21, 2020 at 16:30
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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.
answered Aug 21, 2020 at 19:08
-
$\begingroup$ I have a little trouble justifying the first statement of the second paragraph except essentially by referring to the second statement. Am I right? If so, would it be correct to summarise both statements by saying that the point is that countability of the image of, say, $\rho$ implies that $\infty\cdot\operatorname{meas}(\operatorname{ker}(\rho)) = \#\operatorname{im}(\rho)\cdot\operatorname{meas}(\operatorname{ker}(\rho)) = \operatorname{meas}(G) \in (0, \infty)$, which is impossible? $\endgroup$– LSpiceCommented Aug 21, 2020 at 23:47
-
1$\begingroup$ @LSpice Yes, that's a fair point. I was thinking of the measure of a subgroup being $1$ over the index is a separate fact, but indeed the proof of this is the same argument I give after, so your argument is more elegant. $\endgroup$ Commented Aug 22, 2020 at 0:20