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

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

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

• $S^2\times S^1$ for the first. May 18 '19 at 3:12

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 $$S^2 \times S^1$$.

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).
• If this approach worked, why wouldn't it work using $\mathbb{Z}^3$, which is the fundamental group of a closed orientable 3-manifold: $T^3$. May 19 '19 at 4:15
• 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. May 19 '19 at 15:11
• 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. May 19 '19 at 16:30