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Let $\mathbb{F}_p$ be a finite filed with $p$ elements, and $G=\mathop{\mathrm{Gal}(\mathbb{F}_p^s/\mathbb{F}_p)}$ be its absolute Galois group. $G$ is a pro-finite group, with the Krull topology, see the 3rd example. Then, for an integer $\ell\neq p$, an $\ell$-adic Galois representation is defined to be a finite dimensional $\mathbb{Q}_\ell$-vector space with a continuous linear action of $G$, see def.1.4 for the general definition.

Does there exist a finite dimensional $\mathbb{Q}_\ell$-vector space with a linear action of $G$ that is not continuous? If yes, could you please provide an example or a reference.

Thank you in advance!

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    $\begingroup$ If you require that $G$ takes values in $GL_n(\mathbf{Z}_\ell)$ rather than $GL_n(\mathbf{Q}_\ell)$, then the answer is "no". This is a baby case of a very powerful theorem of Nikolov and Segal: if G is a topologically finitely generated profinite group, and H is any profinite group, then any homomorphism of abstract groups G --> H is automatically continuous. $\endgroup$ Commented May 8, 2016 at 13:03
  • $\begingroup$ @user337830 If you are reading Fontaine's book may I discuss with you some things that are unclear to me? $\endgroup$ Commented May 10, 2016 at 14:34

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Of course there are -- you really need topological groups rather than simply abstract groups (i.e. discrete topological groups) in order to study representation theory beyond the finite case. All that you're asking is whether there exists a homomorphism $Gal(\Bbb{F}_p^s/\Bbb{F}_p)\rightarrow GL_n(\Bbb{Q}_\ell)$ for some $n$ which isn't continuous wrt the $\ell$-adic topology.

Your Galois group is naturally isomorphic to the profinite completion $\hat{\Bbb{Z}}$ of $\Bbb{Z}$; it is topologically generated by a Frobenius element. In particular, a continuous representation is uniquely determined by the action of a fixed Frobenius. So the problem becomes choosing two homomorphisms $G\rightarrow GL_n(\Bbb{Q}_\ell)$ (viewed as discrete groups) which let Frobenius act in the same way, but are not conjugate. This is easy: as a discrete group, $G$ is no longer toplogically generated by Frobenius, so pick an element which isn't a power of Frobenius and let it act in two non-conjugate ways.

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    $\begingroup$ But it's not clear a priori whether one can just freely set the value of an element which isn't a power of Frobenius. $\endgroup$ Commented May 8, 2016 at 0:21
  • $\begingroup$ Thank you @PL.. I am sorry for my stupidity, but I do not see what guarantees the existence of such group homomorphism $$\rho:\mathop{\mathrm{Gal}}(\mathbb{F}^s_p/\mathbb{F}_p)\to\mathop{\mathrm{Gl}_n}(\mathbb{Q}_\ell)$$ that allows us to send a non-power of Frobenius to some linear operator of our chosen. I will be grateful if you could explain why such homomorphism exists, or provide some reference about it. Thank you again! $\endgroup$
    – user337830
    Commented May 8, 2016 at 5:59
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Often you can construct these kinds of examples with ultra-filters. Here is a sketch of a construction that gives an example that is close to what you are asking (but not exactly, plus there are some steps that I believe are true, but did not check in detail):

Consider first the surjection $\varphi$ of $\widehat {\mathbb{Z}}$ onto $\prod_p \mathbb F_p =R$. Instead of a group, we may view $R$ as a ring under multiplication. Choose a maximal ideal $m$ of $R$ that is not closed (this is what I meant with ``ultra-filter''). The quotient ring $Q = R/m$ is then a field of characteristic $0$ and admits by cardinality considerations an embedding into $\overline {\mathbb{Q}}_p$. Composing these maps we get a mapping $$ \varphi \colon \widehat {\mathbb Z} \rightarrow \overline {\mathbb{Q}}_p $$ and hence a representation $$ \psi \colon \widehat {\mathbb Z} \rightarrow GL_2( \overline {\mathbb{Q}}_p), \quad x \mapsto \left( \begin{matrix} 1 & \varphi(x) \cr 0 & 1 \end{matrix} \right). $$

Note that this is different than what you asked for, because $\psi$ takes values in $GL_2(\overline {\mathbb{Q}}_p)$ as opposed to $GL_2({\mathbb Q}_p)$.

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