Is there an orientable closed compact $3$-manifold such that its fundamntal group is $\mathbb{Z}$?

Question

Is there an orientable closed compact $3$-manifold such that its fundamental group is $\mathbb{Z}$?

How about $\mathbb{Z^2}$?

Answer 1

If $M$ is a closed oriented $3$-manifold with $\Bbb Z^2$ fundamental group, then pass to the universal cover $\tilde{M}$ and note that $\pi_2(\tilde{M}) = 0$ as otherwise by sphere theorem there would be an embedded sphere contained in $\tilde{M}$, which by the covering map descends to a homotopically nontrivial embedded sphere in $M$, i.e., $M$ is not irreducible. Unless $M = S^2 \times S^1$, in which case the fundamental groups don't match, it's not prime either, forcing $M$ to be a connected sum. But that forces $\pi_1$ to be a free product which $\Bbb Z^2$ isn't (Note: Actually, we're using Poincare conjecture here: if $M = M_1 \# M_2$ then $\Bbb Z^2$ is $\pi_1(M_1) * \pi_2(M_2)$, which without loss of generality implies $\pi_1(M_1) = 0$, so $M_1$ is a simply connected closed $3$-manifold, i.e., $S^3$, so $M$ cannot be decomposed as a nontrivial connected sum. But I think this can be avoided by arguing by prime decomposition theorem that $M$ is a connected sum of $S^2 \times S^1$ with a bunch of homotopy $3$-spheres, which doesn't have fundamental group $\Bbb Z^2$ anyway)

By Hurewicz theorem $\pi_3(\tilde{M}) = H_3(\tilde{M})$, also zero as $\tilde{M}$ is noncompact $3$-dimensional. All the higher homology groups, hence the higher homotopy groups by Hurewicz, are subsequently zero. This implies $\tilde{M}$ is contractible, hence $M$ is a $K(\pi, 1)$-space, but as $\pi_1 = \Bbb Z^2$ here and $K(\Bbb Z^2, 1)$ is homotopy equivalent to $T^2$, $M$ must be homotopy equivalent to $T^2$. But $\Bbb Z = H_3(M) \neq H_3(T^2) = 0$ so that can't happen. There are no closed oriented $3$-manifolds with $\Bbb Z^2$ fundamental group.

$S^1 \times S^2$ is the unique closed oriented $3$-manifold with fundamental group $\Bbb Z$ by following a same trail of arguments as above: if $\pi_2(\tilde{M}) \neq 0$ then there's a homotopically nontrivial embedded sphere in $M$. If you put aside the case $M = S^2 \times S^1$, that forces $M$ to be non-prime, i.e., $\pi_1$ is a nontrivial free product, which again $\Bbb Z$ isn't. The rest of the argument to show there are no other cases is identical.

Here's a sketch of an argument for why $S^2 \times S^1$ is the unique prime non-irreducible closed oriented 3-manifold. Suppose $M$ is prime non-irreducible, then there's a homotopically nontrivial sphere $S$ in $M$ that doesn't bound a ball. Take an embedded closed $\epsilon$-neighborhood $S \times [-1, 1]$ of $S$ in $M$ and let $\gamma$ be an arc from $S \times \{-1\}$ to $S \times \{1\}$ which doesn't intersect $S$; this exists because $M \setminus S$ is not disconnected ($M$ is prime!). Take union of $S \times \{-1, 1\}$ with an embedded unit normal bundle of $\gamma$ (which has to be diffeomorphic to $[0, 1] \times S^1$ fixing the boundary, i.e., an "orientation-preserving tube", because $M$ is oriented) to obtain another embedded sphere in $M$ whose interior is union of a tubular neighborhood of $S$ and a tubular neighborhood of $\gamma$ which deformation retracts to $S^2 \vee S^1$. This new embedded sphere has to bound a $3$-ball in the exterior, as $M$ is prime. This gives a CW-decomposition of $M$ as a $D^3$ attatched to $S^2 \vee S^1$, and it's not too hard to check this is indeed $S^2 \times S^1$.

Answer 2

You can probably also use the fact that if the fundamental group is $\mathbb{Z}^2$ then you have that the cup product of two elements of dimension $1$ in the cohomology ring have a non-torsion image in dimension $2$ (think: torus- send generators to generators) and then arrive at a contradiction (by, hint hint- using the fact that the manifold is of dimension $3$ so you can exploit duality).