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So the consequence of the geometrization (according to 3-manifold group note) is that any finite-volumed hyperbolic 3-manifold is residually finite. So the question is:

Q1. If $M$ is an infinite-volumed hyperbolic 3-manifold, in particular, a compact 3-manifold whose interior admits a hyperbolic structure such that it contains at least one boundary component with genus $\geq 2$, then $\pi_1(M)$ is residually finite?

Since any subgroup of a residually finite group is again residually finite, I can also ask:

Q2. Any infinite-volumed hyperbolic 3-manifold is a covering of a finite-volumed hyperbolic 3-manifold? i.e., if $M = \Bbb H^3/\Gamma$ is infinite-volumed, then I can find $\tilde{\Gamma}>\Gamma$ such that $\Bbb H^3/\tilde{\Gamma}$ is finite-volumed?

I think the first question is positive but couldn't find a reason.

Edit: I think this answer answers Q1 but I'm not sure if it's dealing with my case (infinite volumed case) because I think the main reference of that is 3-manifold groups note and I think that note mainly deals with finite volumed ones.

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    $\begingroup$ The answer you link to does indeed answer Question 1, but is massive overkill. In fact, all tame hyperbolic manifolds have residually finite fundamental group for very easy reasons. $\endgroup$
    – HJRW
    Commented Mar 21 at 9:01
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    $\begingroup$ The first sentence is wrong as written, this result is not a consequence of geometrization, it is due to Maltsev who proved his theorem before Thurston was born. $\endgroup$ Commented Mar 21 at 15:41

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Here's another negative answer for Q2. I'm assuming (as in Sam Nead's answer) that the covering should be locally isometric. By Ahlfors-Bers, a tame infinite volume hyperbolic manifold with ends of the form $\Sigma \times [0,\infty)$ where $\Sigma$ is a higher-genus surface will have uncountably many hyperbolic structures (coming from the Teichmüller space of $\Sigma$). On the other hand, there are only countably many finite-volume hyperbolic manifolds. So most of those uncountably many hyperbolic manifolds don't cover a finite volume one.

The proof of the latter statement is that there are only finitely many homeomorphism types, since each finite volume manifold is the interior of a compact manifold with boundary. But now Mostow rigidity says that the hyperbolic structure is unique.

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Sam Nead's answer does it, but perhaps I can offer a slightly different perspective on Question 1. No complicated hyperbolic gluing results are needed.

I assume we are satisfied with the characterisation of a hyperbolic 3-manifold as a quotient $M=\Gamma\backslash\mathbb{H}^3$, where $\Gamma$ is a torsion-free discrete subgroup of $SL_2(\mathbb{C})$. If $M$ is tame then, by definition, $M$ is homotopy equivalent to a compact $3$-manifold with boundary, and in particular $\Gamma=\pi_1M$ is finitely generated. Therefore, $\Gamma$ is residually finite, by Malcev's theorem that finitely generated linear groups are residually finite [1]. (Note that this is also the theorem one uses in the finite-volume case, so there's no avoiding it.)

By the tameness theorem, $M$ being tame is the same as $\pi_1M$ being finitely generated. But I have no idea what happens outside the finitely generated case. Which is to say, I don't know if every infinitely generated torsion-free Kleinian group is residually finite. Presumably this is well known to someone... [In comments, Misha links to a nice construction of a non-residually finite Kleinian group, but it has torsion.]

[1] A.I. Malcev: On isomorphic matrix representations of infinite groups of matrices (Russian), Mat. Sb. 8 (1940), 405–422 & Amer. Math. Soc. Transl. (2) 45 (1965), 1–18

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    $\begingroup$ A small note on your last question (which I’m sure you know already, but maybe not well known?) is that there exist infinitely generated Kleinian groups which are not residually finite. However, these have torsion, and are not virtually torsion-free AFAIK. $\endgroup$ Commented Mar 21 at 10:10
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    $\begingroup$ @Carl-FredrikNybergBrodda: It's easy to see that there are infinitely generated Kleinian (even Fuchsian) groups that aren't virtually torsion-free, simply because the torsion is unbounded. I would be curious to know a reference for the construction of examples that aren't residually finite. (And, as you say, for hyperbolic manifolds we need torsion-free examples.) $\endgroup$
    – HJRW
    Commented Mar 21 at 10:34
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    $\begingroup$ @HJRW: mathoverflow.net/questions/89439/… $\endgroup$
    – Misha
    Commented Mar 21 at 10:44
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    $\begingroup$ @Misha: Thanks, that's very nice! (And rings a bell, now that I see it again.) $\endgroup$
    – HJRW
    Commented Mar 21 at 10:49
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The answer to the first question is "yes" and the answer to the second is "no", assuming that you are looking for a covering which is a locally isometric. If you do not require a locally isometric covering then the answer to the second is "yes" and it follows from the first.

Claim: Tame hyperbolic three-manifolds have residually finite fundamental groups.

Proof: In fact, such manifolds embed nicely into finite volume ones. Here are the details.

Suppose that $M$ is a compact three-manifold whose interior admits a hyperbolic structure. Let $\partial_0 M$ be the boundary components of $M$ which are tori. If $\partial_0 M = \partial M$ then we are done. Suppose that $\partial_+ M = \partial M - \partial_0 M$ is non-empty - so $M$ has boundary components of higher genus. For each component $S$ of $\partial_+ M$ we choose a "sufficiently complicated" mapping class $f_S$. Let $F_+$ be the union of the $f_S$. We glue two copies of $M$ using $F_+$ to obtain $D(M, F_+)$. The manifold $D(M, F_+)$ is a "double" of $M$; it is hyperbolic and of finite volume. Also, the inclusion of $M$ into $D(M, F_+)$ induces an injection of fundamental groups. So $\pi_1(M)$ is residually finite. $\Box$

Claim: There are infinite-volume hyperbolic three-manifolds which do not cover (locally isometrically) any finite volume hyperbolic three-manifold.

Proof: Suppose that $N$ is hyperbolic, of finite volume, and without cusps. Then there is a constant $\epsilon_N > 0$ so that, at all points of $x$ of $N$ the injectivity radius of $N$ is bounded below by $\epsilon_N$. Thus the same is true for all covers (finite or infinite), using the same constant $\epsilon_N$.

On the other hand, there are infinite volume three-manifolds $M$, homeomorphic to $S \times \mathbb{R}$ ($S$ closed of genus two), which have a sequence of geodesics $\gamma_n$ (exiting one end of $M$) so that the length of $\gamma_n$ tends to zero as $n$ tends to infinity. $\Box$

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  • $\begingroup$ For the proof of Q1, could you explain what theorem you're using to produce $D(M,F_+)$? I think $f_S$ is something like pseudo-Anosov but why that gluing preserves hyperbolicity? $\endgroup$ Commented Mar 21 at 8:34
  • $\begingroup$ For the proof of Q2, I think your $M$ is something like a cyclic cover of a hyperbolic mapping torus. But still, it covers a hyperbolic mapping torus. Why that property shows it does not cover a finite volume hyperbolic 3-manifold? $\endgroup$ Commented Mar 21 at 8:37
  • $\begingroup$ Yes, $f_S$ is a carefully chosen pseudo-Anosov map. We arrange matters so that disks and annuli in $M$ are not glued to disks and annuli in $M'$ (the other copy of $M$). Since the doubled manifold has no obstructions to being hyperbolic (spheres, tori) it will be hyperbolic. There are various versions of this construction in the literature. I'll find you a reference. $\endgroup$
    – Sam Nead
    Commented Mar 21 at 8:38
  • $\begingroup$ In Q2, I assumed that you wanted the covering map to be locally isometric. I'll add that to the answer. $\endgroup$
    – Sam Nead
    Commented Mar 21 at 8:39
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    $\begingroup$ Here are two papers relevant to the "sufficiently complicated" construction: Soma - Volume of hyperbolic 3-manifolds with iterated pseudo-Anosov amalgamations and Lackenby - The Heegaard genus of amalgamated 3-manifolds. $\endgroup$
    – Sam Nead
    Commented Mar 21 at 8:45
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The answer to Q1 is negative in general (allowing infinitely generated fundamental group). See Example 2 which is a discrete torsion-free subgroup $G< PSL_2(\mathbb{C})$, hence $\mathbb{H}^3/G$ is a hyperbolic manifold with non-residually finite fundamental group:

Cremaschi, Tommaso; Souto, Juan, Discrete groups without finite quotients, Topology Appl. 248, 138-142 (2018). ZBL1404.20023.

Of course this also gives a negative answer to Q2, even as a covering space in the topological category.

As observed in some other answers, if $M$ is a hyperbolic 3-manifold and $\pi_1(M)$ is finitely generated, then $M$ admits a Scott-Shalen core and hence has residually-finite fundamental group by the compact case.

Q2 is almost true in the finitely generated case, in the sense that a finitely generated discrete torsion-free group $ G <PSL_2(\mathbb{C})$ is an algebraic limit of geometrically finite groups. In turn, these can be perturbed to be covers of finite-volume 3-manifolds.

Brooks, Robert, Circle packings and co-compact extensions of Kleinian groups, Invent. Math. 86, 461-469 (1986). ZBL0578.30037. MR0860677

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  • $\begingroup$ Very nice!${}{}$ $\endgroup$
    – HJRW
    Commented Mar 25 at 6:57

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