I am reading the paper "On fibering certain 3-manifolds" by John Stallings and I was hoping someone could help me through a certain detail. In particular, I am confused at the very end of the proof of Theorem 1 which is as follows:

Theorem (Stallings): Let $M^3$ be a compact 3-manifold such that there is a finitely generated normal subgroup $G$ of $\pi_1(M)$ so that $\pi_1(M)/G \cong \mathbb{Z}$. Then $G$ is in fact the fundamental group of a 2-manifold embedded in $M$.

Stallings breaks the proof up into sections 2,3,4, and 5. I will briefly summarize how each of these sections goes, but my confusion is that I do not understand the alluded to "geometric construction" in section 5 at the end of the proof.

In section 2, Stallings tells us to look at a map $f: M \to S^1$ that topologically realizes the surjective homomorphism $\pi_1(M) \to \pi_1(M)/G \cong \mathbb{Z}$. We are then instructed to pick a point $p \in S^1$ and look at the surface $f^{-1}(p)$. Section 2 proceeds to give a method to ensure that $f^{-1}(p)$ is in fact connected. This uses the finitely generated property of $G$.

In section 3, Stallings look a the now connected surface $f^{-1}(p)$ and, if the inclusion map for this surface to the 3-manifold is not $\pi_1$-injective, he uses the loop theorem to find a disk to do surgery on $f^{-1}(p)$ and then he performs this surgery by homotoping $f$ and looking again at the preimage of $p$.

In section 4, Stallings repeats step 2 on the result of step 3 which has the effect of removing one of the two components of the surgered surface. Then by an Euler characteristic count, we say that this process of repeating steps 2 and 3 must terminate with a map $f : M \to S^1$ with $f^{-1}(p)$ connected and $\pi_1$-injective. (Here Stallings writes that $\pi_1(f^{-1}(p)) \to H_1(M)$ is injective - I imagine that is a typo?).

Now we have reached the section I do not understand. We know that the image of $\pi_1(f^{-1}(p)$ is in the kernel of the map $\pi_1(M) \to \mathbb{Z}$, and we want to argue that it is the entire kernel. Here, Stallings says, "If there is anything else in the kernel of $f_*$, then a geometric construction shows that the kernel of $f_*$ is the union of strictly increasing sequence of groups, and so could not be finitely generated." (Also he mentions in the previous sentence that this is due to Neuwirth and is in his thesis which I have not tracked down.)

What is this geometric construction?

  • 1
    $\begingroup$ Have you transcribed that final sentence of Stallings correctly? It doesn't even seem to be grammatically correct, which is highly uncharacteristic of Stallings, who was a wonderful writer. Also, it doesn't seem to mention any geometric construction. $\endgroup$
    – HJRW
    Nov 25, 2020 at 14:40
  • 1
    $\begingroup$ @HJRW Whoops yes my mistake, oh course - not Stallings. I edited it - thank you! $\endgroup$
    – user101010
    Nov 25, 2020 at 16:06

1 Answer 1


I think about it like this. For convenience, I'll assume $M$ is closed.

Given a homomorphism $\phi:\pi_1M\to\mathbb{Z}$, Stallings explains how to find an essential surface $S\subset M$ with $\pi_1S\leq\ker\phi$, as you outline nicely in the question.

Now we construct the cyclic cover $M_\phi$ of $M$ corresponding to $\ker\phi$ as follows. Let $M_0$ be the result of cutting $M$ along $S$, so $M_0$ has two boundary components, $S_+$ and $S_-$, corresponding to $S$. Then $M_\phi$ can be thought of as an infinite gluing:

$M_\phi=\left(\bigcup_{n\in \mathbb{Z}} M_0\right)/\sim$ ,

where $\sim$ glues $S_-$ in the $n$th copy of $M_0$ to $S_+$ in the $(n-1)$th copy of $M_0$.

Therefore, $\ker\phi$ decomposes as an infinite amalgamated product:

$\ker\phi= \ldots *_{\pi_1 S} \pi_1 M_0 *_{\pi_1 S} \pi_1 M_0 *_{\pi_1 S} \pi_1 M_0 *_{\pi_1S} \ldots $ .

In particular, it is infinitely generated unless the inclusion $S\to M_0$ induces an isomorphism of fundamental groups. But this only occurs if $M_0$ is a product $S\times [0,1]$, which is exactly the case in which $S$ is a fibre.

  • $\begingroup$ Thanks for the response - does the implication that the inclusion induces an isomorphism implies that $M_0 = S \times I$ resting on the Poincare conjecture? $\endgroup$
    – user101010
    Nov 25, 2020 at 20:52
  • $\begingroup$ No, but we should certainly assume that our manifold is irreducible, to make sure it doesn’’t have homotopy-sphere summands. (Without the Poincare Conjecture, that is a necessary hypothesis for Stallings’ theorem.) I don’t have a reference for that step to hand, I’m afraid. $\endgroup$
    – HJRW
    Nov 25, 2020 at 22:31
  • $\begingroup$ Ahh ok - I guess that is some fact about 3-manifolds with boundary where the boundary inclusion induces an isomorphism on $\pi_1$. Do you happen to know a reference for that? Also, I believe you answer does not in any way use that $G$ is not $\mathbb{Z}/2\mathbb{Z}$ and thus relates to this question (at least in the closed case). $\endgroup$
    – user101010
    Nov 26, 2020 at 0:00
  • $\begingroup$ An irreducible 3-manifold is aspherical by the Sphere Theorem, and hence $\pi_1$ is torsion-free. In particular, $G$ cannot be $\mathbb{Z}/2\mathbb{Z}$. $\endgroup$
    – HJRW
    Nov 26, 2020 at 13:17
  • $\begingroup$ If I had to give a geometrisation-free proof of the assertion that $\pi_1S\to \pi_1M_0$ is an isomorphism implies $M_0$ is a product (in the irreducible case), I would appeal to JSJ theory. For instance, I think the techniques of this paper by Neumann--Swarup -- arxiv.org/abs/math/9712227 -- which are quite elementary, should be sufficient. Of course, Stallings' paper predates JSJ theory by over a decade, so he must have had something different in mind. $\endgroup$
    – HJRW
    Nov 26, 2020 at 13:21

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