If $S$ is a closed connected surface and $\varphi \in \mathrm{Diff}(M)$, then we can build the mapping torus $M_\varphi = \dfrac{S \times [0,1]}{(x,0)\sim (\varphi(x),1)}$. Then we have that $ M_\varphi \to S^1$ is a fibration (it's actually a fiber bundle with fiber $S$).

Now let $M$ be a closed connected $3$-manifold and $f:M \to S^1$ be a fibration (in the homotopy theoretic sense). Under which hypothesis can I say that

  1. $M$ is a fiber bundle with fiber a surface $S$?

  2. $M$ is a mapping torus for some $\varphi \in \mathrm{Diff}(S)$?

Some of my thoughts: $f$ is a fibration, so we know that the fibers over any point are all homotopy equivalent. But $M$ is a closed $3$-manifold, so the fiber over a generic point will be a closed surface. Since two homotopy equivalent closed surfaces are actually diffeomorphic, fibers are "generically" the same. My problem is that in general $f$ can have some critical values where $f^{-1}(p)$ fails to be a surface (it is still a subspace with the same homotopy type of the generic fiber surface).

If $f$ happens to be a submersion then this problem will not occur and all the fibers will be diffeomorphic. In this case, it's clear that all these surfaces can be "packed together" in the structure of a fiber bundle (please correct me if I'm wrong). So here the question is: is this a mapping torus for some $\varphi \in \mathrm{Diff}(S)$? My guess is this is true, because each fiber is non-separating in $M$ and $S^1 \setminus {p}$ is just an open interval $I$ whose preimage in $M$ is of the form $S \times I$, but I don't see any obvious way to recover the glueing $\varphi$.


1 Answer 1


Answer is 'always', provided that you treat your map $f$ up to homotopy and assume homotopy finiteness of the fiber. There are several cases to consider, I will deal with the case when $M$ has contractible universal cover (if you are interested in the case when it is not the case, I will write the details, the problem is solved by appealing to Kneser's theorem). Let $F$ be a homotopy fiber of your fibration. One has to assume that it is a finite cell complex, otherwise one can easily find a counter example by taking an aspherical 3-manifold whose fundamental group maps onto $Z$ with the infinitely generated kernel. (First, find a 2-dimensional example of this situation and then multiply by the circle.)

Now, by the long exact sequence of fibration you obtain that the finitely generated group $N =\pi_1(F)$ is normal in $G=\pi_1(M)$ and the quotient is the infinite cyclic group $Z$. Since Perelman proved the Poincare conjecture for our benefit, the manifold $M$ is also irreducible and, hence, you can apply a theorem which you can find somewhere in Hempel's book "3-manifolds" which states that in this case $N$ is a surface group and $M$ indeed is a fiber bundle over the circle so that the projection map of the bundle is homotopic to $f$.

For the second part. Suppose that you have a fiber bundle $f: M\to S^1$. Take $S$ to be the preimage of a point. Then, by local triviality of $f$, the complement of an open bicolor of $S$ is the product of $S$ and the unit interval. The manifold $M$ is obtained by identifying its boundary surfaces by a homeomorphism $h$, which we can regard as an automorphism of $S$. Now, it is clear that $M$ is the mapping torus of $h$. QED


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