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

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

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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 $S^2 \times S^1$.

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    $\begingroup$ +1, en.wikipedia.org/wiki/Prime_manifold was a helpful resource for reading this answer $\endgroup$ – hunter May 18 at 4:39
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    $\begingroup$ Is there no proof without higher homotopy groups? I am just learning the material of that chapter; the hint in the book is to look at torsion in the cup product. $\endgroup$ – topology master May 18 at 16:34
  • $\begingroup$ @topologymaster There might be, I don't know one off the top of my head. Let me think about that for some time. $\endgroup$ – Balarka Sen May 18 at 16:58
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    $\begingroup$ You said you are learning the material of "that chapter"... Which chapter in which book? This would be important information to edit into the text of your question, so that we have a proper understanding of the context of your question. $\endgroup$ – Lee Mosher May 18 at 17:23
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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).

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  • $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. $\endgroup$ – dantopa May 19 at 2:06
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    $\begingroup$ If this approach worked, why wouldn't it work using $\mathbb{Z}^3$, which is the fundamental group of a closed orientable 3-manifold: $T^3$. $\endgroup$ – Jason DeVito May 19 at 4:15
  • $\begingroup$ Precisely because the fundamental group is $\mathbb{Z}^2$- you can send each generator of $T^2$ to one generator of the fundamental group of $M$ and use this to show that cup product in $M$ is not torsion. $\endgroup$ – Barbara Bogumila May 19 at 15:11
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    $\begingroup$ Would you please add the details? I can’t really see how it works/how it is relevant, because in this case $H^2(M)=H_1(M)=\mathbb{Z}^2$ so the image of the cup porduct is clearly non-torsion. $\endgroup$ – topology master May 19 at 16:30

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