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Does there exist a compact hyperbolic 3-manifold $M$ that is not diffeomorphic to a geometrically finite hyperbolic manifold? If yes, can such $M$ have incompressible boundary?

I think the answer should be yes to both questions but I cannot find this in the literature.

Remarks: as usual, a compact hyperbolic manifold is a compact manifold whose interior carries a complete hyperbolic structure. The structure is geometrically finite if it is obtained as the quotient of the hyperbolic 3-space by a geometrically finite group. Thurston's hyperbolization theorem implies:

  1. A compact 3-manifold with non-empty boundary is hyperbolizable if and only if it is irreducible and atoroidal.

  2. Any compact, atoroidal, pared 3-manifold is diffeomorphic to a geometrically finite one.

  3. Any compact hyperbolic 3-manifold is homotopy equivalent to a geometrically finite one.

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I think you've answered your own question negatively, essentially from 1. and 2. You can see this stated in a survey paper of Dick Canary: – Ian Agol Feb 7 '13 at 4:07
Igor, it is also in my book, where I include a proof for orbifolds as well. – Misha Feb 7 '13 at 4:10
Can someone write up an answer, which can then be accepted? – HJRW Feb 7 '13 at 10:12
Ian and Misha, are you saying that every irreducible, atoroidal 3-manifold with non-empty boundary is pared? Canary's survey is not in our library. Where is this statement in Misha's book? Thanks! – Igor Belegradek Feb 7 '13 at 12:58
@Igor: let an annulus A connect two tori T and T'. Now take a regular neighborhood N of A \cup T \cup T': this is a 3-manifold with boundary (contained in your bigger manifold M), and its boundary is necessarily made of three tori T, T', and a new one T''. This new torus T'' is contained in M: either it is incompressible (and hence M is toroidal, against your hypothesis) or is compressible. In the latter case it bounds a solid torus S: hence you get M = N \cup S which is a Seifert manifold. – Bruno Martelli Feb 7 '13 at 14:49
up vote 8 down vote accepted

[Edited several times] As the comments say, the answer to the first and hence to the second question is "no". Suppose that $M$ is the compact manifold and $N$ is its interior. Let $\rho$ be the given hyperbolic structure on $N$. If $M$ is without boundary then the volume of $\rho$ is finite and we are done.

Suppose instead that $M$ has boundary. Since $N$ is hyperbolic, via $\rho$, deduce $M$ is atoroidal (which includes aspherical). Thus $M$ is Haken. Place all tori in the boundary of $M$ into the paring locus $P$. By Thurston's hyperbolization theorem, the interior $N$ admits a hyperbolic metric, $\rho_0$, which is geometrically finite. (The convex core has finite volume and contains all torus boundary components.) See Theorem 1.43 in Kapovich's book.

[A brief note - your hypotheses can be weakened. You assumed (a) $N$ is the interior of a compact manifold and (b) N is hyperbolizable. This can be replaced by (a') $\pi_1(N)$ is finitely generated and the same (b). This is called the "tameness theorem", due to Agol and also Calegari-Gabai.]

In the comments below (above?) Igor asks why an atoroidal manifold with torus boundary, and admitting an essential annulus, is Seifert fibered. This can be found as Lemma 1.16 on page 25 of Hatcher's three-manifold notes.

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Sam, this is not an answer because it misses the most interesting case when $M$ has a boundary component of genus $>1$. In this case $N$ does not admit a finite volume hyperbolic metric. Of course, it might still admit a complete infinite volume hyperbolic metric. – Igor Belegradek Feb 7 '13 at 12:49
You are correct - I was referring to the wrong thing. I have added a temporary fix, and will look for a precise reference. – Sam Nead Feb 7 '13 at 14:11
If what you say is true, why would anyone bother to prove the weaker statement 3 in my list (which is theorem 19.6 in Misha's book)? – Igor Belegradek Feb 7 '13 at 14:22
There is a purely geometric proof, as well. Suppose $N$ is hyperbolic, and $B_S$ and $B_T$ are disjoint horo-tori about $S$ and $T$. Suppose that $A$ is a compact annulus connecting $B_S$ to $B_T$. Lift everything to the universal cover. A component of the lift of $A$ is a strip (quasi-isometric to a line!) that fellow-travels lines in two distinct horospheres, a contradiction. – Sam Nead Feb 7 '13 at 17:13
I will add a reference to my answer. – Sam Nead Feb 7 '13 at 17:53

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