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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?

Taras Banakh
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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 https://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 '18 at 07:08
  • @YCor Yes, I have learned this and have rewritten the question correspondingly as it reduces (via the formula for universal coefficients to the problem of classification of closed 3-manifolds with finite first homology group). – Taras Banakh Sep 21 '18 at 07:24
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    @TarasBanakh the answer of your current question is already in my previous comment. – YCor Sep 21 '18 at 07:25
  • @YCor, But $SL_2(R)$ has fundamental group isomorphic to $\mathbb Z$. Why then this infinite group turns into a finite group in the quotient space $SL_2(R)/\Gamma$? – Taras Banakh Sep 21 '18 at 07:33
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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 '18 at 07:36
  • @YCor For connected sums of Poincare sphere it seems that you are right: thye Mayer-Vietoris ensures that the finiteness of the homology group is preserved by the connected sum. Thank you. – Taras Banakh Sep 21 '18 at 07:36
  • I used Van Kampen, which implies that $\pi_1$ of a connected sum (in dimension $d\ge 3$, using the simple connectedness of the $(d-1)$-sphere) is the free product of the $\pi_1$. – YCor Sep 21 '18 at 07:37
  • The link YCor provided is very good. The Brieskorn spheres are simple examples of homology spheres that come in an infinite family, and are irreducible under connected sum. These have infinite fundamental group except in one case (the Poincare sphere). There is a nice little article of Milnor on these spaces titled something like 'the 3-dimensional Brieskorn manifolds''. – mme Sep 21 '18 at 07:38
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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 '18 at 07:39
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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 '18 at 13:54

2 Answers2

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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:

Jean Raimbault
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  • 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 '18 at 07:44
  • @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 '18 at 07:59
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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 '18 at 08:06
  • @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 '18 at 08:32
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    @YCor On the other hand, I have found this SE-post (https://math.stackexchange.com/questions/421303/non-orientable-3-manifold-has-infinite-fundamental-group) implying that a closed 3-manifold with trivial rational homology group $H_1$ necessarily is orientable. Is it indeed true? – Taras Banakh Sep 21 '18 at 08:40
  • @YCor According to this SE-post (https://math.stackexchange.com/questions/356966/orientability-of-projective-space) the projective space $P^n$ is non-orientable if and only if $n$ is even. So, $P^3$ is orientable. – Taras Banakh Sep 21 '18 at 08:47
  • Yes (I was stupid about $P^3$). So: yes, I forgot to assume orientable (in the light of what I knew), but indeed by this fact, a closed 3-manifold is a rational homology 3-sphere iff its $H_1$ is finite (orientability follows automatically). – YCor Sep 21 '18 at 08:49
  • @YCor Then everything is simply perfect! Thank you very much for your help. It allowed me to improve a bit my (partial) answer to this MO-problem (https://mathoverflow.net/questions/309197) – Taras Banakh Sep 21 '18 at 08:58
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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.

Bruno Martelli
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