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Which free abelian groups can be realized as the fundamental group of a closed 3-manifold? The only one I can come up with is $\mathbb{Z}$, which is the fundamental group of $S^1 \times S^2$. For the application I have in mind, the key case is $\mathbb{Z}^2$. Here it is easy if you allow boundary (just take $T^2 \times [0,1]$), but I don't see how to do it without the boundary.

flag	asked Jul 14 2010 at 20:54 EL 73●2

Lennart Meier · Accepted Answer · 2010-07-14 21:23:46Z UTC

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(I assume all occuring 3-manifolds to be orientable and closed)

A manifold with a free abelian fundamental group cannot be a connected sum of non-trivial 3-manifolds since its fundamental group is not a free product. A prime manifold is either $S^1\times S^2$ or irreducible (Hatcher's notes on 3-manifolds, 1.4). By 3.9 in the same source, an irreducible 3-manifold $M$ with infinite fundamental group is a $K(\pi,1)$. If $\pi = \mathbb{Z}^n$, then $M$ needs therefore to be homotopy equivalent to $(S^1)^n$. This can only be if $n=3$.

answered Jul 14 2010 at 21:23

Lennart Meier
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I suppose one could point out that if $M$ is non-orientable, one can apply the argument to the 2-fold orientable cover (to rule out a non-orientable 3-manifold with $\pi_1=\mathbb{Z}^2$). In the case of $\mathbb{Z}$, there is also the non-orientable $S^2$-bundle over $S^1$. One also needs the Poincare conjecture to conclude that $M$ is irreducible. Of course, one can make the argument independent of the Poincare conjecture, since a connect sum with a homotopy 3-sphere does not affect the homotopy groups $\pi_k$. – Agol Jul 14 2010 at 21:39

Your first sentence depends on the Poincare conjecture. – Richard Kent Jul 14 2010 at 21:42

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Surely Lennart avoided the Poincare conjecture by defining a 3-manifold to be "trivial" if it is simply connected. – Ben Wieland Jul 15 2010 at 2:26

Agol · Answer 2 · 2010-07-14 21:30:57Z UTC

Only $\mathbb Z$ and $\mathbb Z^3$ (for $T^3$) are free abelian groups that appear as fundamental groups of $3$-manifolds. Hopefully the following is an approximative proof.

The manifold must be prime (otherwise the $\pi_1$ is not ableian), hence it is $K(\pi,1)$. Hence its cohomology are just cohomology of the group $\mathbb Z^n$. So the can not get $\mathbb Z^2$ since $H^3(\mathbb Z^2)=0$, and we can not get $\mathbb Z^n$ with $n>3$ since $H^n(\mathbb Z^n)=\mathbb Z$.

Mariano Suárez-Alvarez · Answer 3 · 2010-07-14 21:33:37Z UTC

John Hempel, in his book $3$-manifolds, shows that if $G$ is a finitely generated abelian group which is a subgroup of the fundamental group of a closed $3$-manifold, then $G$ is one of $\mathbb Z$, $\mathbb Z\oplus\mathbb Z$, $\mathbb Z\oplus\mathbb Z \oplus\mathbb Z$, $\mathbb Z_p$ or $\mathbb Z\oplus\mathbb Z_2$. This is theorem 9.13 in the book.

He also proves, in theorem 9.14, that an abelian group which is not finitely generated and a subgroup of the fundamental group of a $3$-manifold, then it is isomorphic to a subgroup of $\mathbb Q$ (and proposes, as an exercise, to show that all such groups in fact occur)

Closed 3-manifolds with free abelian fundamental groups

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