A non-continuous p-adic representation

Question

I am looking for an example of a non-continuous homomorphism $$G \to GL_r(\mathbb C_p)$$ from a profinite (topologically finitely generated) group $G$, where $\mathbb C_p$ is the completion of an algebraic closure $\overline {\mathbb Q}_p$ of the field of $p$-adic numbers $\mathbb Q_p$.

Answer 1

Added: I overlooked that in the question $G$ can be any "profinite (topologically finitely generated) group". Then reuns' answer, using an essentially pro-$\ell$ group with $\ell \neq p$, is probably the shortest. Below, I exhibit examples even in the case that $G$ is the pro-$p$ group $(\mathbb Z_p, +)$ (with "the same $p$ as in $\mathbb C_p$").

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According to Quotient of $\mathbb{Z}_p$ by the rational integers, the quotient of additive groups $\mathbb Z_p/ \mathbb Z_{(p)} =: V$ has a natural structure of a $\mathbb Q$-vector space. If we choose a basis of that and project it down to one basis vector, we get a nonzero map of abelian groups:

$$\mathbb Z_p \twoheadrightarrow \mathbb Z_p/\mathbb Z_{(p)} \twoheadrightarrow \mathbb Q$$

Now $\mathbb C_p^\times$, by virtue of $\mathbb C_p$ being algebraically closed, should contain many subgroups which are isomorphic to $\mathbb Q$ or quotients of $\mathbb Q$; for starters, the group of all roots of unity $\mu (\mathbb C_p)$ is isomorphic to $\mathbb Q/\mathbb Z$.

[I am wondering if each element $x \in \mathbb C_p^\times \setminus \mu(\mathbb C_p)$ is contained in a group $\simeq \mathbb Q$ which I would like to call $x^\mathbb Q$; for this, one has to choose $n$-th roots of $x$ but in a compatible way, i.e. $(x^{\frac{1}{n}})^d =x^{1/m}$ for $n=md$. In fact, I wonder if there might even be a big subgroup inside $\mathbb C_p^\times$ which is isomorphic to (the additive group of) a $\mathbb Q$-vector space of cardinality/dimension of the continuum, so we could skip the projection step above. Anyway:]

Prolonging the above via $\mathbb Q \rightarrow \mathbb Q/\mathbb Z \simeq \mu(\mathbb C_p) \subset \mathbb C_p^\times$ gives us non-trivial (albeit not terribly explicit) group homomorphisms

$$\mathbb Z_p \rightarrow \mathbb C_p^\times =GL_1(\mathbb C_p)$$

which are certainly not continuous, as they are trivial on the dense subset $\mathbb Z_{(p)}$.

For general $GL_n$, one can do even more, as now there are obvious subgroups $\simeq \mathbb Q$, like $\pmatrix{1 & \mathbb Q\\0&1} \subset GL_2(\mathbb C_p)$.

Answer 2

Take $G=\Bbb{Z}_\ell^\times$ with $\ell \ne p$,

using the axiom of choice embed $\Bbb{Q}_\ell$ into $\Bbb{C}_p$, so $\Bbb{Z}_\ell^\times$ embeds into $GL_1(\Bbb{C}_p)$.

This map can't be continuous: for all $a\in \Bbb{Z}_\ell^\times$, $\lim_{n \to \infty} a^{(\ell-1)\ell^n}=1$ whereas for $b\in \Bbb{C}_p^\times$, $\lim_{n \to \infty} b^{(\ell-1)\ell^n}=1$ iff $b^{(\ell-1)\ell^n}=1$ for some $n$.