By Bers' simultaneous uniformization theorem, if $\Gamma$ is a Fuchsian group, then $\operatorname{QC}(\Gamma)\cong \mathcal{T}(S)\times\mathcal{T}(\overline{S})$ where $S = \Bbb H^2/\Gamma$. In particular, one can explicitly construct a hyperbolic 3-manifold $M$ whose conformal boundary has two components, each is homeomorphic to $S$ with prescribed conformal structures. The general form of the Ahlfors–Bers theory implies that if $M$ has only geometrically finite ends, then its hyperbolic structure is uniquely determined by the corresponding points in the Teichmüller spaces of each conformal boundary of $M$.

I think my question is answered but I can't find:

Q1. Let $\Sigma_1,\dotsc,\Sigma_n$ be closed surfaces of genus $g_i\geq 2$ with prescribed conformal structures. Then there exists a hyperbolic 3-manifold $M$ whose conformal boundaries are precisely $\Sigma_1,\dotsc,\Sigma_n$.

If $n=2$ with $g_1 = g_2$ then it's Bers' simultaneous uniformization theorem. The origin of this question is by questioning:

Q2. Is there a hyperbolic 3-manifold whose conformal boundary has two components which are precisely genus $2$ and genus $3$ surfaces?

I think the answer is NO but I can't come up with an explanation.

A partial answer I found: Let $M = \Bbb H^3/\Gamma$ be such manifold. *Suppose* its regular set $\Omega(\Gamma)$ has two $\Gamma$-invariant components $\Omega_1$ and $\Omega_2$ then each is simply connected. Each of the three types of hyperbolic isometries on $\Omega_1$ corresponds to the same type of isometries on $\Omega_2$ so that we can induce an identity isomorphism $\pi_1(R_1)\to\pi_1(R_2)$ where $R_i = \Omega_i/\Gamma$ and we're done. $\square$

(I don't know if it's possible or not but) I don't know how to exclude the case when $\Omega(\Gamma)$ has infinitely many components so that $\Gamma$ somehow identifies those $\Omega_i$s to produce hyperbolic 3-manifolds with two conformal boundaries I stated above.

Topologically, existence is easy to see but the main point is the existence of hyperbolic structure.

Edit: Consider the space $M = \Bbb S^3-\text{thicken}(X)$, i.e., we thicken $X$ a bit and see its complement of $\Bbb S^3$. This gives topologically a compact manifold with three $\Sigma_2$ boundaries. I think this space satisfies four conditions in Sam's answer. The general case can be constructed by attaching 1-handles to one of the boundary components to increase the genus of the boundary surface and attaching copies of $M$ along a single boundary surface to increase the number of boundary components.

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