In this paper, Danny Calegari shows that taut foliations in (let's say closed for simplicity) 3-manifolds are precisely those which admit a map $f: M \to S^2$ which restricts to a branched cover on each leaf. I'm trying to unpack what this means for the topology of $M$.

The TL;DR is my understanding is that this implies that $M$ admits a foliation by circles. But this implies $M$ is Seifert fibered, which is a contradiction because $M$ could be, say, hyperbolic.

I'm not sure where the following argument goes wrong. Calegari's argument implies that we can choose $f$ so that a given total transversal $\gamma$ of our foliation is precisely the set of critical points of $f$. Moreover, we can assume that these branch points are all order 2. Let's say that $\gamma = f^{-1}(0)$.

Away from $0$, $f$ is a submersion and so $\ker f$ is a 1-dimensional distribution. At a point in $\gamma$, our assumptions imply that there are coordinates where $f$ looks like $$(x, y, z) \mapsto (x^2 - y^2, x^2 + y^2).$$ Thus we see that the distribution $D = \ker f$ extends over $\gamma$, by being generated in these coordinates by the vector field $\frac{\partial}{\partial z}$. So we have a global 1-dimensional distribution $D$, which is tangent to each of our fibers of $f$ and moreover tangent to $\gamma$. So it integrates to a foliation which is precisely this union of circles.

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    $\begingroup$ I haven’t read that paper, so this might be totally misguided. But why are the leaves of the 1-dimensional distribution closed? All closed oriented 3-manifolds support 1-dimensional foliations (since their Euler characteristic is 0, you can find a nonvanishing vector field on them). $\endgroup$ Commented Sep 16, 2022 at 3:18
  • $\begingroup$ The leaves here are precisely the preimages of points under $f$, so they should be compact. $\endgroup$ Commented Sep 16, 2022 at 5:44
  • $\begingroup$ I think that you mean "coordinates where $f$ looks like $(x, y, z) \mapsto (x^2 - y^2, 2xy)$". In your version $f$ is not surjective near the origin. Also, I think that $D = \mathrm{ker} f$ should be $D = \mathrm{ker} \, d\!f$. $\endgroup$
    – Sam Nead
    Commented Sep 16, 2022 at 7:21

1 Answer 1


To see what is going on, I think it is helpful to carefully work out an explicit example. In Section 2.1 of Calegari's paper, he explains how to deal with 3-manifolds that fiber over the circle. Here is a brief description of what he does.

I will start by setting up notation (and I won't follow Calegari here since I prefer to write it a little differently). Let $\Sigma_g$ be a compact oriented genus $g$ surface and let $\psi\colon \Sigma_g \rightarrow \Sigma_g$ be an orientation-preserving diffeomorphism. Let $M_{\psi}$ be the mapping torus of $\psi$, so $M_{\psi} = \Sigma_g \times [0,1] / \sim$ where $(p,1) \sim (\psi(x),0)$ for all $x \in \Sigma_g$. The fibers of the map $M_{\psi} \rightarrow S^1$ are all genus-$g$ surfaces, and form a taut foliation.

This points to one place where I think you slightly misunderstand what Calegari does. It's easy to find a transversal to the above taut foliation that hits each fiber exactly once. But you can't ensure that this is the entire branch locus and that the branching is simple since there might not be a branched cover $\Sigma_g \rightarrow S^2$ with a single branch point of order $2$. For what he is doing, you'll have to take a more complicated transversal with several components.

Let $P \subset S^2$ be a set of $4g$ distinct points. We can then easily construct a map $\pi\colon \Sigma_g \rightarrow S^2$ that is a degree $(g+1)$ branched cover, where the branch locus is exactly $P$ and each point of $p$ is a simple branch point (i.e., has degree $2$).

Now comes the key step in Calegari's argument. He argues that we can find an orientation-preserving diffeomorphism $\phi\colon S^2 \rightarrow S^2$ taking $P$ to $P$ such that $\psi\colon \Sigma_g \rightarrow \Sigma_g$ is a lift of $\phi$, i.e., such that $$\pi(\psi(x)) = \phi(\pi(x)) \quad \quad \text{for all $x \in \Sigma_g$}.$$ Letting $M_{\phi}$ be the mapping torus of $\phi$, we thus get a map $F\colon M_{\psi} \rightarrow M_{\phi}$ that is a fiberwise branched cover.

Calegari stops the argument here, and omits the final step of why this gives a map $f\colon M_{\psi} \rightarrow S^2$ that is a branched cover on each fiber. Here is what I think he has in mind. The mapping class group of $S^2$ is trivial, so the map $\phi\colon S^2 \rightarrow S^2$ is isotopic to the identity. Of course, this isotopy moves the points in $P$. This implies that there is a fiber-preserving diffeomorphism $\lambda\colon M_{\phi} \rightarrow M_{\text{id}} = S^2 \times S^1$. The map $f\colon M_{\psi} \rightarrow S^2$ is then $$M_{\psi} \stackrel{F}{\longrightarrow} M_{\phi} \stackrel{\lambda}{\longrightarrow} S^2 \times S^1 \stackrel{\text{proj}}{\longrightarrow} S^2.$$ One feature of this is that it is not the case that the branch points in the fibers all map to the same points in $S^2$; indeed, as you go around the circle of fibers the images of branch points in $S^2$ trace out a braid. I think it would hard to avoid this.

Since the composition $\lambda \circ F$ is fiber-preserving, your $1$-dimensional foliation I think will be exactly the standard foliation of the mapping torus $M_{\psi}$ whose leaves are unions of the images of sets $x \times [0,1]$ in $M_{\psi}$. If $x$ is a periodic point of $\psi$, then this leaf will be a closed circle, but for non-periodic points it will be $\mathbb{R}$.

  • $\begingroup$ Okay I think this makes sense, thank you! What, then, can we say about the structure of the branch point set? Can we make it an embedded graph, for example? $\endgroup$ Commented Sep 16, 2022 at 21:21
  • $\begingroup$ The branch point set is a link in the original three-manifold, everywhere transverse to the leaves of the foliation. The point that Andy is making is that, while transverse, this link is need not be "orthogonal" to the leaves (with coordinates given locally by $f$). $\endgroup$
    – Sam Nead
    Commented Sep 16, 2022 at 21:52
  • $\begingroup$ @AudreyRosevear: I think you can always ensure that the branch point set is a disjoint union of circles, each transverse to the foliation. $\endgroup$ Commented Sep 16, 2022 at 21:58
  • $\begingroup$ Oh I got it that makes sense. Thank you! $\endgroup$ Commented Sep 16, 2022 at 22:34

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