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A hyperbolic 3-manifold has finite volume if and only if it is either closed or has toroidal boundary and it is not homeomorphic to $T^2\times I$.

This statement is from 3-Manifold Groups, page 18 (the link is editted) by Matthias Aschenbrenner, Stefan Friedl and Henry Wilton, it seems that the three references in the book toward this statement only give partial results (when the boundary components are already cusps).

Edits: The precise statement in my opinion should be:

Let $M$ be a compact three dimensional manifold with incompressible toroidal boundary (possibly none). If the interior of $M$ admits a hyperbolic structure, then $M$ either has finite volume, or is homeomorphic to $T^2\times I$.

Thanks for any solutions or hints.

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  • $\begingroup$ You find a complete proof in Chapter D of Benedetti-Petronio: “Lectures on Hyperbolic Geometry”, link.springer.com/content/pdf/10.1007%2F978-3-642-58158-8_4.pdf $\endgroup$
    – ThiKu
    Commented Jan 28, 2022 at 18:39
  • $\begingroup$ Fredy, is your statement a little off? The way you have written it, it sounds like the manifolds are compact. If they're compact, they all have finite volume. Maybe your language indicates you are restricting to the class of the interiors of compact 3-manifolds? $\endgroup$
    – Ryan Budney
    Commented Jan 28, 2022 at 19:13
  • $\begingroup$ Btw it is page 12 (not page 18) in your linked pdf and one actually has to require that the toroidal boundary is incompressible. $\endgroup$
    – ThiKu
    Commented Jan 28, 2022 at 22:13
  • $\begingroup$ @ThiKu. My apologies. My version of the book is linked here $\endgroup$
    – Fredy
    Commented Jan 29, 2022 at 1:15
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    $\begingroup$ You are right, but when the manifold has finite volume there is nothing need to prove, the problem is to show $M$ is $T^2\times I$ when $M$ has infinite volume. $\endgroup$
    – Fredy
    Commented Jan 29, 2022 at 15:22

1 Answer 1

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I think you are trying to ask the following question.

Suppose that $M$ is a compact connected oriented three-manifold. Suppose that $M^\circ$, the interior of $M$, admits a hyperbolic metric. Then when must this hyperbolic metric have finite volume?

As Ryan points out, if $M$ is closed, then $M^\circ = M$ is compact and thus has finite volume. Also, as you note, the interior of $M = T^2 \times I$ admits (many) hyperbolic metrics, but all have infinite volume. There is another such manifold: namely the solid torus $D^2 \times S^1$.

I think that the place where you are confused (please correct me if I am wrong) is the case where $M$ has a boundary of higher genus. Here $M^\circ$ again always has infinite volume.

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    $\begingroup$ Thanks for your clarification! What actually puzzles me is that why $T^2\times I$ is the only exception (if I insist the boundary to be incompressible tori)? $\endgroup$
    – Fredy
    Commented Jan 29, 2022 at 1:47
  • $\begingroup$ Right, suppose that $M$ has only torus boundary components, and $M^\circ$ has a hyperbolic metric of infinite volume. We are assuming that all ends of the interior are homeomorphic to $T^2 \times \mathbb{R}$. An end of this type is either parabolic (cross sections shrink as they exit) or "flaring" (cross sections expand as they exit). Proving this is an exercise in classifying $\mathbb{Z}^2$ subgroups of the isometries of hyperbolic three-space. Parabolic ends have finite volume; flaring ends have infinite volume. So suppose that there is at least one flaring end $E$. $\endgroup$
    – Sam Nead
    Commented Jan 29, 2022 at 17:35
  • $\begingroup$ Lift to the cover $X_E$ of $M^\circ$ corresponding to $\pi_1(E)$. By the classification (of $\mathbb{Z}^2$ actions on hyperbolic space) the cover $X_E$ has one flaring end (where $E$ lifts) and one parabolic end. The parabolic end has finite volume. Thus the index of $\pi_1(E)$ in $\pi_1(M)$ is finite. Thus $X_E$ is a finite cover of $M^\circ$. Since we are not allowing orbifolds, and since the deck group action preserves the cross-sections, in fact $X_E$ equals $M^\circ$. $\endgroup$
    – Sam Nead
    Commented Jan 29, 2022 at 17:41
  • $\begingroup$ Thanks, this solves my problem perfectly! $\endgroup$
    – Fredy
    Commented Jan 30, 2022 at 2:11

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