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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    $\begingroup$ Sullivan proved a general version of the quasi conformal rigidity using Ahlfors-Bers’ measurable Riemann mapping theorem (this generalized Bers’ simultaneous uniformization theorem). math.stonybrook.edu/~dennis/publications/PDF/DS-pub-0053.pdf As for Q2, you can take a compression body obtained by taking a genus 2 surface times an interval and adding a 1-handle to one side. en.wikipedia.org/wiki/Compression_body $\endgroup$
    – Ian Agol
    Dec 19, 2023 at 21:07
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    $\begingroup$ Once the existence of one compact hyperbolic manifold with boundary components of the given genera is known (probably due to Maskit in 1960s, but definitely follows from Thurston's theorem, using a compression body as in Agol's comment), the rest is due to Bers, who generalized his simultaneous uniformization theorem in L.Bers, Spaces of Kleinian groups. Several Complex Variables, I (Proc. Conf., Univ. of Maryland, College Park, Md., 1970), pp. 9–34 Lecture Notes in Math., Vol. 155. Springer-Verlag, Berlin-New York, 1970 $\endgroup$ Dec 20, 2023 at 0:34
  • $\begingroup$ @IanAgol, Moishe - I also thought of the compression body answer (using Klein combination). However, I am unclear on how the hyperbolic structures are parametrised. Is it (Teichmuller space cross Teichmuller space) modulo a mapping class group? $\endgroup$
    – Sam Nead
    Dec 21, 2023 at 8:54
  • $\begingroup$ One more comment - you are correct that $S^3 - n(X)$ is cylindrical. (There is a two-sphere meeting $X$ in exactly two points; the points are the vertices of the left and right eyeglasses.) However, being cylindrical does not absolutely rule out hyperbolicity. $\endgroup$
    – Sam Nead
    Dec 21, 2023 at 9:28
  • $\begingroup$ @SamNead: More precisely, it is the product of Teichmuller space of the boundary components modulo product of suitable subgroups of mapping class groups. It is all explained in the paper by Bers. $\endgroup$ Dec 22, 2023 at 14:15

1 Answer 1


The answer to question one and question two are both "yes". Here is a very special case:

Suppose that $M$ is a connected oriented compact three-manifold with boundary. Suppose further that $M$ is:

  • irreducible (all two-spheres embedded in $M$ bound balls embedded in $M$),
  • boundary irreducible (the boundary of $M$ is incompressible in $M$),
  • strongly atoroidal (the fundamental group does not contain a copy of $\mathbb{Z}^2$), and
  • acylindrical (if two loops in $\partial M$ are freely homotopic in $M$ then they are freely homotopic in $\partial M$).

It follows that the boundary components all have genus at least two. Let $(\Sigma_i)$ be the list of boundary components.

Then the space of hyperbolic structures on $M$ (with conformal boundaries at infinity) is homeomorphic to the product of the Teichmuller spaces $\mathcal{T}(\Sigma_i)$.

Note that this is both an existence and a uniqueness result. Namely hyperbolic structures are plentiful, and are parameterised by "data at infinity".

In this very special case (since $M$ has many side hypotheses) to obtain just one hyperbolic metric, we "simply" double $M$ across its boundary, and apply either Thurston's hyperbolisation theorem for Haken manifolds (see the first pages of the book Hyperbolic manifolds and discrete groups by Kapovich) or apply Perelman's work.

Here is a somewhat more concrete answer to question one. Suppose that $M$ is a compact, connected, oriented manifold with the desired boundary components. (You can obtain $M$ by taking a connect sum of handlebdies.) Note that we assume that no boundary component of $M$ is a two-sphere or a two-torus. The paper Simple knots in compact, orientable 3-manifolds, by Myers, implies that there is a knot $K$ in $M$ so that (the interior of) $M - K$ is simple - that is, satisfies the four hypotheses above. In particular, $M$ admits a hyperbolic structure. (This again uses Thurston's hyperbolisation theorem.) We now apply Thurston's Dehn filling theorem; we choose a sufficiently complicated slope $r$ on the two-torus cusp and Dehn fill $M - K$ to get the manifold $M_K(r)$. This again satisfies the four hypotheses above and has the same boundary components as $M$. So, by the theorem, we can realise any desired conformal structures on the boundaries of $M_K(r)$.

  • $\begingroup$ Thank you for the answer. I just skimmed Marden's paper but I can't find any statement answering my question. I'm afraid you're misunderstanding the question. The question is the existence of $M$. What you're saying is first assuming $M$ then consider its conformal boundaries. $\endgroup$ Dec 18, 2023 at 10:47
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    $\begingroup$ The point of the result is it gives a purely topological criterion for hyperbolicity. So you can use it to build your own examples. $\endgroup$
    – Sam Nead
    Dec 18, 2023 at 20:25
  • $\begingroup$ I have two questions: 1. I'm not sure about the existence of the stated connected oriented compact topological 3-manifold with boundary with prescribed genus. 2. Even if it exists, how do we know that such a manifold can be hyperbolized? $\endgroup$ Dec 19, 2023 at 9:14
  • $\begingroup$ 1. Such manifolds are plentiful - I'll add a construction to my answer. 2. This is the content of Thurston's "hyperbolisation theorem for Haken manifolds". See the first page of the preface of Kapovich's book. $\endgroup$
    – Sam Nead
    Dec 19, 2023 at 10:13
  • $\begingroup$ I made a mistake in citing Marden - I think his work is more in the direction of a converse to Thurston's hyperbolisation theorem. Namely, "if $M$ is hyperbolic then what can we know about its topology?". $\endgroup$
    – Sam Nead
    Dec 19, 2023 at 10:28

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