# Classification of closed 3-manifolds with finite first homology group?

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?

• 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 '18 at 7:08
• @TarasBanakh the answer of your current question is already in my previous comment. – YCor Sep 21 '18 at 7:25
• 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 7:36
• @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 7:39
• 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

• @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 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 '18 at 8:32
• @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 '18 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.