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Stallings' celebrated Fibration Theorem states that if a closed irreducible $3$-manifold $M$ admits a short exact sequence \begin{equation} 1 \to N \to \pi_1(M) \to \mathbb{Z} \to 1, \end{equation} where $N$ is finitely generated, then $M$ fibers over $S^1$.

My question is that whether it is possible to give the exact form of $M$ if we already know that $N$ is a surface group. More precisely, assume there is a short exact sequence \begin{equation} 1 \to \pi_1(S) \to \pi_1(M) \to \mathbb{Z} \to 1, \end{equation} where $S$ is a closed surface. Let $t \in \pi_1(M)$ be a preimage of $1 \in \mathbb{Z}$. Conjugation with $t$ induces an automorphism $\varphi$ of $\pi_1(S)$ whose projection $[\varphi] \in Out(\pi_1(S))$ in the outer automorphism group does not depend on the choice of $t$. By the Dehn-Nielsen-Baer Theorem the extended mapping class group $MCG^{\pm}(S)$ is isomorphic to $Out(\pi_1(S))$. Let $f:S \to S$ be a homeomorphism so that $[f] \in MCG^{\pm}(S)$ corresponds to $[\varphi] \in Out(\pi_1(S))$ under the Dehn-Nielsen-Baer isomorphism. My question is whether $M$ is diffeomorphic to the mapping torus $T_f$ of $f:S \to S$. If this is true, does it just follow from Stallings' proof? Or, in case it is known but requires more work, is there a good reference for it?

More concretely, I am interested in the following situation (which might simplify things). If $M$ is a closed hyperbolic $3$-manifold so that $\pi_1(M)$ is (abstractly) isomorphic to the fundamental group $\pi_1(T_f)$ of a mapping torus, then $M$ is diffeomorphic to $T_f$. This is, of course, an easy consequence of the Geometrization Conjecture. However, I would like to know whether there is a proof of this fact that does not depend on the solution of the Geometrization Conjecture nor on Thurston's mapping torus theorem.

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    $\begingroup$ You certainly don't need to invoke Stallings here, since you already assumed the kernel is a surface group. Neither do you need to invoke the full strength of Perelman's geometrisation: Thurston proved that any mapping torus is geometrisable, and in particular in your case $T_f$ has to be hyperbolic. Mostow rigidity then implies that $M$ and $T_f$ are isometric. Alternatively, I think you can avoid geometry entirely using Waldhausen's theorem (see here: mathoverflow.net/questions/242276/…). $\endgroup$
    – HJRW
    Commented Oct 31, 2022 at 15:15
  • $\begingroup$ I explicitely stated that I do not want to make use of Thurston's mapping torus result. But I will have a look at Waldhausen's theorem. Also, I would still be interested in the answer to my first question. $\endgroup$
    – Frieder Jäckel
    Commented Oct 31, 2022 at 16:14
  • $\begingroup$ Apologies, I missed that you wanted to avoid Thurston. In that case, I think Waldhausen is probably what you want. If your "first question" is the case of a general surface-by-cyclic group, as opposed to the hyperbolic case, then I think this should also follow from Waldhausen. $\endgroup$
    – HJRW
    Commented Oct 31, 2022 at 16:27

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Your question ("is $M$ homeomorphic to $T_f$?") is answered in the affirmative by Theorem 2 of Stallings' paper *On fibering certain 3-manifolds". You will also need his Theorem 1. Here are the statements (slightly simplified).

Theorem 1: Suppose that $M$ is a compact connected three-manifold. Suppose that $\Gamma$ is a finitely generated normal subgroup of $\pi_1(M)$ whose quotient group is $\mathbb{Z}$. Then there is a surface $F$ properly embedded in $M$ so that $\Gamma = \pi_1(F)$.

Theorem 2: With hypotheses as in Theorem 1. Suppose that $M$ is irreducible. Suppose that $\Gamma$ is not $\mathbb{Z}/2\mathbb{Z}$. Then $M$ is a surface bundle over the circle, with $F$ isotopic to a fibre.

That is, your hypothesis on the short exact sequence (plus Theorem 1) gives the surface $F$. Your hypothesis that the manifold $M$ is hyperbolic then gives the additional hypotheses of Theorem 2.

Note that Stallings does not cite Waldhausen. I suppose that this is because his situation is a very very simple case of a Haken hierarchy. Once you have $F$ in your hands (and all the group theory hypotheses), it is "easy" to show that $M - F$ is homeomorphic to a product.

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  • $\begingroup$ Thank you for your answer. I'm not sure if I am obsessing over tiny details, but I have the impression that what is stated is not yet enough to conclude what I want. Theorem 2 gives me that $M$ is some mapping torus $T_{f^\prime}$. However, I would like to conclude that $M$ is $T_f$ with the exact $f$ I described in my question. But I guess this should be clear if one of the following two statements is true (which I state in the next comment): $\endgroup$
    – Frieder Jäckel
    Commented Oct 31, 2022 at 19:31
  • $\begingroup$ 1) Let $p:M \to S^1$ be the bundle projection given by Theorem 2. Then the induced map $\pi_1(p):\pi_1(M) \to \pi_1(S^1)$ coincides (up to possibly precomposing with an automorphism of $\pi_1(M)$) with the given surjection $\pi_1(M) \to \mathbb{Z}$. I hope that this is an artefact of Stallings' proof. Is that correct? $\endgroup$
    – Frieder Jäckel
    Commented Oct 31, 2022 at 19:38
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    $\begingroup$ Stallings proves (1) and Waldhausen proves (2). $\endgroup$
    – Sam Nead
    Commented Oct 31, 2022 at 19:53
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    $\begingroup$ When I say "Stallings proves" I mean "You can find this in the second sentence of section two of his paper (on page 93)" A copy of the paper is available here: maths.ed.ac.uk/~v1ranick/papers/stallfib.pdf $\endgroup$
    – Sam Nead
    Commented Oct 31, 2022 at 21:14
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    $\begingroup$ When I say "Waldhausen proves" I mean "I only vaguely remember this". Looking at the reference you suggest, in fact Corollary 6.5 is exactly what you want. Yes, boundary irreducible allows empty boundary. $\endgroup$
    – Sam Nead
    Commented Oct 31, 2022 at 21:24

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