Is the fundamental group of a closed orientable hyperbolic $3$-manifold isomorphic to a non-discrete subgroup of $\mathrm{PSL}(2, \mathbb{C})$?

Question

Consider the fundamental group $\pi_1(M)$ of a closed orientable hyperbolic $3$-manifold $M$. Certain identifications $\tilde{M} \approx \mathbb{H}^3 \approx \mathrm{PSL}(2, \mathbb{C}) / \mathrm{Stab}(p_0)$, for some $p_0 \in \mathbb{H}^3$, result in an injective group homomorphism $\rho : \pi_1(M) \to \mathrm{PSL}(2, \mathbb{C})$ which has discrete and co-compact image. Is it always possible to instead realize $\pi_1(M)$ as a non-discrete subgroup $\Gamma'$ of $\mathrm{PSL}(2, \mathbb{C})$?

Answer 1

Here is a construction, but at this moment I do not have a proof of nondiscreteness (in general).

Let $\Gamma< SL(2, {\mathbb C})$ be a cocompact discrete subgroup (a uniform lattice): The fundamental group of an orientable hyperbolic 3-manifold always lifts from $PSL(2, {\mathbb C})$ to $SL(2, {\mathbb C})$. Let $\gamma_1,...,\gamma_n$ denote generators of $\Gamma$. The matrix coefficients of $\gamma_1,...,\gamma_n$ generate a finitely generated subfield $F\subset {\mathbb C}$. It was observed by Selberg that (local) rigidity of lattices implies that $F$ is a number field, i.e. a finite algebraic extension of ${\mathbb Q}$. At the same time, $F$ cannot be contained in ${\mathbb R}$ since $\Gamma$ is not contained in $SL(2, {\mathbb R})$. Thus, we have a nontrivial Galois group $Gal(F/{\mathbb Q})$. For each $\sigma\in Gal(F/{\mathbb Q})$, we obtain a faithful representation $$ \rho_\sigma: \Gamma\to SL(2, {\mathbb C}) $$ sending each $\gamma\in \Gamma$ to $\gamma^\sigma$ (you act on matrix coefficients of $\gamma$ by the field automorphism $\sigma$). I think that with few exceptions, $\rho_\sigma(\Gamma)$ is a nondiscrete subgroup of $SL(2, {\mathbb C})$. The exceptions that I know are (non-uniform) arithmetic lattices commensurable to Bianchi groups $SL(2, {\mathcal O}_d)$. In this case, $F={\mathbb Q}(\sqrt{-d})$ and the only nontrivial element of $Gal(F/{\mathbb Q})$ is the complex conjugation. I think (but I am not sure) that these are the only exceptions and these are ruled out by the cocompactness assumption. The reason for my belief in nondiscreteness is the Mostow Rigidity Theorem which states that if $\Gamma^\sigma:=\rho_\sigma(\Gamma)$ is discrete then the group isomorphism $$ \rho_\sigma: \Gamma\to \Gamma^\sigma $$ is induced by an automorphism of $SL(2, {\mathbb C})$ (and such automorphisms are either inner automorphisms or compositions of inner automorphisms with the complex conjugation).