I am interested in a topological classification of connected closed 3-manifold $M$ that have finite homology group $H_1(M)$.

Since $H_1(M)$ is the abelization of the fundamental group $\pi_1(M)$, each closed 3-manifold with finite homotopy group has finite homology group. It is known that each closed 3-manifold with finite homotopy group $\Gamma$ is a spherical 3-manifold (i.e., is the orbit space $S^3/_\sim$ of the 3-sphere, endowed with a free action of the group $\Gamma$).

Question. Is each closed 3-manifold with trivial homology group a spherical 3-manifold? Equivlalently, is the fundamental group $\pi_1(M)$ of a closed 3-manifold finite if its first homology group $H_1(M)$ is finite?

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    You have the connected sum of $n$ Poincare spheres, so this already gives you an infinite list. Also see constructions of homology 3-spheres in en.wikipedia.org/wiki/Homology_sphere, it includes infinitely many of the form $G/\Gamma$ with $G$ the universal covering of $\mathrm{SL}_2(\mathbf{R})$ and $\Gamma$ a cocompact lattice. – YCor Sep 21 at 7:08
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    @TarasBanakh the answer of your current question is already in my previous comment. – YCor Sep 21 at 7:25
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    The fundamental group in my example is $\Gamma$, and, as a cocompact lattice in the universal covering of $SL_2(R)$, it has an infinite central subgroup $Z$ such that $\Gamma/Z$ is isomorphic to a cocompact lattice in $SL_2(R)$. So it's not only infinite, but contains free subgroups. – YCor Sep 21 at 7:36
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    @YCor Please write down your comments as an answer because it is going to be not so trivial (at least for me). As I understand you have a series of examples of closed 3-manifolds with finite first homology group but infinite fundamental groups? – Taras Banakh Sep 21 at 7:39
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    Here is a counterexample which is not hyperbolic: The Hantzsche-Wendt manifold $M$ is a 3-dimensional flat manifold with finite $H_1(M,\mathbb{Z})$. The fundamental group is a Bieberbach group which is an extension of $\mathbb{Z}^3$ by the finite group $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$ and is thus not finite. – Steffen Kionke Sep 21 at 13:54
up vote 18 down vote accepted

The answer is no by Yves' comments. Let me add that there are plenty of explicit constructions of closed hyperbolic 3--manifolds with finite homology, and this is a generic phenomenon (for example random Heegard gluings have zero first Betti number and are hyperbolic and numerical experiments on the census manifolds exhibit an overwhelming proportion of manifolds with zero first Betti number). This hints to there being no hope to get a classification.

For various results about hyperbolic rational homology spheres (probabilistic, numerical, explicit constructions of infinite families) see for example the papers of Nathan Dunfield and coauthors:

  • Thank you very much for this quick answer. By the way, is there any special name for closed 3-manifolds with finite first homology group? – Taras Banakh Sep 21 at 7:44
  • @YCor Aha! Then the finitude of the first homology group with integer coefficients is the same as triviality of the first homology group with rational coefficients (by the formula of universal coefficients)? – Taras Banakh Sep 21 at 7:59
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    @TarasBanakh yes, it's called "rational 3-homology sphere". (Symmetry of the rational Betti numbers implies, if $H_1$ is finite, that $H_2$ is finite as well.) – YCor Sep 21 at 8:06
  • @YCor But there exists also a question of orientability? So, a rational homology 3-sphere = closed oriented 3-manifold with finite integral $H_1$? Or the orientability is automatic by some (unknown to me) reason? – Taras Banakh Sep 21 at 8:32
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    @YCor On the other hand, I have found this SE-post (math.stackexchange.com/questions/421303/…) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true? – Taras Banakh Sep 21 at 8:40

Pick any knot in the three-sphere, and perform any Dehn surgery on it with some coefficient $p/q \neq 0$. This means that you remove the tubular neighborhood of the knot and you glue it back in a different way, parametrized by $p/q$. The manifold you get has $H_1(M,\mathbb Z) = \mathbb Z/_{p\mathbb Z}$. You get plenty of distinct 3-manifolds in this way. For instance, if the knot is hyperbolic, you get plenty of closed hyperbolic manifolds if $p$ or $q$ is sufficiently large. You can also require that $p=1$ and find plenty of homology spheres.

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