# What closed 3-manifolds have fundamental group $\Bbb Z$?

For certain small groups, it is easy (and desirable) to classify closed (and orientable if necessary) 3-manifolds with that group as their fundamental group. (Essentially due to Waldhausen is that for "large" 3-manifold groups, indecomposable under free product, the 3-manifold is determined up to homeomorphism by its fundamental group, though this was only proved in full generality after geometrization. Large means, precisely, infinite and not $\Bbb Z$).

For finite groups, as a corollary of elliptization, $\Bbb Z/p$ only appears as the fundamental group of lens spaces $L(p,q)$, and one can classify the other finite fundamental groups (before elliptization this was harder and one could only classify up to connected sum with a homotopy sphere.) as having either at most two manifolds per group (and precisely what they are).

$\Bbb Z$ is the simplest infinite group I know. What's the classification of closed 3-manifolds with fundamental group $\Bbb Z$?

• I'm posting and answering this because I've wanted an easy place to reference this fact for a while now, and I don't know of a source that writes out this proof in full. – user98602 Dec 14 '15 at 3:57

The first argument of your proof sounded a little familiar to me, and I realized I saw it in Hempel. So for a maybe more citeable reference and also some other infinite non-$\mathbb Z$ examples, let me just add this answer, if this is fine:

Chapter 5. of Hempel's 3-manifolds book talks about free fundamental groups. Theorem 5.2. states:

Theorem Let $M^3$ be a prime, compact manifold with free fundamental group. Then $M$ is either a bundle $S^2 \to M \to S^1$ or a cube with handles (not necessarily orientable).

Your question spefically asked about $\mathbb Z$ which the theorem answers, but also in the proof you see where it comes in: when he cuts along a non-disk bounding sphere, he necessarily gets something connected with reduced fundamental group rank.

• Can you clarify what a cube with handles means here? Can I just read this as "solid genus $g$ surface"? (that is, a thickening of a wedge of $g$ circles?) Thanks for the pointer, I didn't remember this. – user98602 Dec 15 '15 at 4:08
• I'll second the Hempel recommendation. – Lee Mosher Dec 17 '15 at 16:28

There are precisely two: $S^2 \times S^1$ and a twisted product $S^2 \widetilde \times S^1$.

Suppose $M$ is closed, connected, with fundamental group $\Bbb Z$. Then it must be prime (by Poincare and because $\Bbb Z$ is indecomposable under free product), but not necessarily irreducible (meaning that every embedded 2-sphere bounds a 3-ball).

Indeed, it can't possibly be irreducible: consider $\tilde M$. It has trivial fundamental group, and because $M$ is irreducible, has $\pi_2 M = 0$ by the sphere theorem (which states that if $M$ is orientable, some nontrivial element of $\pi_2 M$ is represented by an embedded sphere). (If $M$ here is not orientable, pass to the oriented double cover.) Because $\Bbb Z$ is infinite, $\tilde M$ is not compact, so has $\pi_3 = H_3 = 0$; the first equality by Hurewicz and the latter by noncompactness. Inductively by Hurewicz we see that $\pi_n = 0$ for all $n$. So $M$ is aspherical, and hence a $K(\Bbb Z,1)$. But this would imply that $$\Bbb Z/2 = H_3(M;\Bbb Z/2) = H_3(S^1;\Bbb Z/2) = 0!$$ So $M$ is prime but not irreducible. We now follow an argument from Hatcher's 3-manifolds notes.

Because $M$ is prime, any embedded sphere either bounds a ball or does not separate $M$. But because $M$ is not irreducible there is some embedded sphere that does not bound a ball. So there is an embedded, non-separating, 2-sphere in $M$. Because $S^2$ is simply connected, any line bundle (which are in bijection with $H^1(X;\Bbb Z/2)$) over it is trivial. In particular, the normal bundle to any $S^2$ in $M$ is trivial.

So embed this separating 2-sphere, which extends to an embedding of $S^2 \times I$. Pick a curve that goes from one side of the thickened $S^2$ to the other (such a thing exists by the non-separating assumption); thicken it to get an embedded $I \times D^2$, whose ends are glued to each side of the $S^2$. More precisely, we can glue these together by a map $\varphi: \{0,1\} \times D^2 \to \{0,1\} \times S^2$ with $\varphi(0,\cdot) \subset \{0\} \times S^2$ and similarly with $\varphi(1,\cdot)$. In different language, I'm attaching a 1-handle to $S^2 \times I$. The results depend only on whether or not $\varphi$ preserves orientation on both components. (That is, if it preserves orientation on both components, or reverses orientation on both components, you get one manifold; if it preserves orientation on one component but reverses orientation on the other, you get a second, distinct manifold). Call the results $X_0$ and $X_1$, where $X_0$ is the orientable one; $X_0$ is what you get if $M$ is orientable, and $X_1$ if it's not. The boundary of $X_i$ is $S^2$; because this $S^2$ does separate $M$, it bounds a ball on the other side. There is only one way up to homeomorphism to glue a ball onto a manifold with boundary $S^2$. So we get that $M$ is one of either $M_0$ or $M_1$, where these are the $X_i$ with that ball capped off.

To finish, we just need to find two manifolds with such a decomposition. $S^2 \times S^1$ is one, where here the embedded $S^2$ is just $S^2 \times \{*\}$; and $S^2 \widetilde \times I = (S^2 \times I)/(x,0) \sim (-x,1)$ is another (again, pick, say, $S^2 \times \{1/2\}$.) The former is orientable; the latter is non-orientable. So $M_0 = S^2 \times S^1$ and $M_1 = S^2 \widetilde \times S^1$ are the only closed 3-manifolds with fundamental group $\Bbb Z$.

In fact something significantly stronger is true. If $Y$ is closed orientable and $\Bbb Z$ injects into the abelianization $\pi_1(Y)$, either $Y$ has an $S^1 \times S^2$ factor or $Y$ has a $\pi_1$-injective surface of genus at least one (An injection $\pi_1(\Sigma_g) \to \pi_1(Y^3)$ for $\Sigma_g$ a genus $g$ surface with $g\geq 1$ induced from an embedding $\Sigma_g \to Y$).

Consider a prime factor of $Y$, $Y'$ with $\Bbb Z$ injecting into the abelianization of $\pi_1(Y')$. By Poincare duality $H_2(Y')\ne 0$, and by a very famous result of Gabai there exists a taut foliation on $Y'$ with a closed leaf $\Sigma$. By a theorem of Novikov, the induced map $\pi_1(\Sigma) \to \pi_1(Y')$ is injective and by another theorem of Novikov if $\Sigma$ is $S^2$ if and only if $Y'$ is $S^1\times S^2$.