Is a compact, connected, orientable 3-manifold with $\mathbb{Z}^K$ fundemental group uniquely determined?

Question

According to the Kneser-Milnor prime decomposition theorem for 3-manifolds, any compact, connected, orientable 3-manifold $M$ is diffeomorphic to $S^3 / \Gamma_1$ # $\cdots$ # $S^3/ \Gamma_n$ # $(S^2 \times S^1)_1$ # $\cdots$ # $(S^2 \times S^1)_r$ # $K( \pi_1,1)$ # $\cdots$ # $K( \pi_m,1)$, where # is the connect sum, and $\Gamma_i$ is a non-trivial finite subgroup of $SO(4)$ acting orthogonally to $S^3$ (so that the result is a spherical space form).

I suppose my question boils down to if $\pi_1(M)=\mathbb{Z} \times \cdots \times \mathbb{Z}$, k times, does that uniquely identify a manifold as a $(S^2 \times S^1)_1$ # $\cdots$ # $(S^2 \times S^1)_k$, or can the various quotient manifolds or aspherical factors create a more complicated topology without changing the fundamental group?

Answer 1

NO, since the three-torus $T^3$ does not have this form.

EDIT if the OP really means a free product of $\mathbb{Z}$s, so the free group $F_k,$ then the answer is YES. It is a fact (see Hempel's book, chapter 7) that every splitting of the fundamental group of $M^3$ as a free product comes from a connected sum decomposition. On the other hand, a prime three manifold is either a $K(\pi, 1)$ or $S^2 \times S^1.$ In the former case, its fundamental group cannot be $\mathbb{Z}$

Answer 2

Let me address a more general question:

To what extent is a closed, connected $M^3$ determined by its fundamental group?

Following the Geometrization Theorem, we have a complete answer. There are only two ways in which a closed $3$--manifold $M$ can fail to be determined by its fundamental group:

1. $M$ is a lens space, or a connected sum of something with a lens space. It is well-known that lens spaces are not determined up to homeomorphism by their fundamental groups. Note that in this case, you would see a ${Z}/p$ free factor in your group $G$, so it can't arise in the context of your question.

2. $M = N_1 \sharp N_2$, where each $N_i$ is orientable, and each is chiral (fails to have an orientation-reversing symmetry). In this case, reversing the orientation on one factor would produce $M' = N_1 \sharp \overline{N_2}$, which is not homeomorphic to $M$ but has the same fundamental group. This also cannot arise in your context, for exactly the reasons that Igor outlined, and because $S^1 \times S^2$ does have an orientation-reversing symmetry.

Finally, let me point out that although Geometrization may seem like an overly big hammer for this question, in fact one needs the Poincare conjecture. For, if there existed a fake $3$--sphere, one could take the connect sum with that manifold without altering the group.

Answer 3

If you did in fact mean the free abelian group, you can still do it (up to punctures). All references in brackets are to Hempel.

If $\pi_1(M)\cong \mathbb{Z}$, then $M$ is one of $S^2\times S^1$, $S^2\tilde{\times} S^1$, the solid torus, or the solid Klein bottle. [5.3]

If $\pi_1(M)\cong \mathbb{Z}^2$, then $M$ is an I-bundle over the torus. [10.6]

If $\pi_1(M)\cong \mathbb{Z}^3$, then $M$ is the 3-torus. [11.11]

The case $k>3$ doesn't happen. [9.13]