Question. Let $M^{n+1}$ be a closed manifold without boundary. Which closed submanifolds $\Sigma^n \subset M^{n+1}$ (of codimension one) are leaves of a foliation of $M$ minus some finite collection of points? Does one know a priori the number of points one is forced to remove from $M$?

I suspect the answer might be well-known, but it's a bit out of my area of expertise, and I haven't been able to find the answer.

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    $\begingroup$ What obstructions are there for extending $\Sigma^n$ to a foliation on all of $M$ (without removing any points)? $\endgroup$ Oct 20, 2022 at 15:44
  • $\begingroup$ @JasonDeVito When $M$ is actually closed and $\Sigma$ is separating, this amounts to a foliation on both sides with $\partial M_i = \Sigma$ a leaf (with some smoothness compatibility conditions on the boundary, which maybe are hard to guarantee? I am not sure.) Because the two sides are compact manifolds with boundary whose boundary is a leaf, this is answered by Bill Thurston's Theorem 2(a) here. When you allow some noncompactness you have a lot more freedom. $\endgroup$
    – mme
    Oct 20, 2022 at 16:02
  • $\begingroup$ @mme: Thanks! I didn't check details, but I think the compability on the boundary shouldn't be too hard using a collar. It seems that using this, one should be able to assume that near the boundary, the foliation is the obvious foliation of $\partial M\times [0,1)$. Then, of course, things glue with no problems. $\endgroup$ Oct 20, 2022 at 17:39

1 Answer 1


If the normal bundle of $\Sigma$ in $M$ is orientable, then there always exists such a foliation. The idea is that one can construct a smooth function $f$ on $M$ such that $\Sigma$ is the set of zeros of $f$ and $\mathrm{d}f$ does not vanish on a tubular neighborhood of $\Sigma$. Then, since, by Theorem 6.2, Chapter II of Golubitsky and Guillemin's Stable Mappings and their Singularities, the Morse functions on $M$ are an open dense subset of $C^\infty(M,\mathbb{R})$, there will be a Morse function $g$ on $M$ that is sufficiently close to $f$ in the $C^\infty$ topology that the locus $g=0$ is ambiently isotopic to $\Sigma$. By composing with a diffeomorphism of $M$, we can assume that $\Sigma$ is the zero locus of $g$. Since $g$ is a Morse function, it has isolated critical points (and therefore a finite number of them). Away from the critical points the level sets of $g$ define a foliation of $M$ whose zero level set is $\Sigma$. As for the minimal number of such critical points, that will depend on the topology of the two 'sides' of $\Sigma$ in $M$.

In the non-orientable case, you can do the following. First, fix a Riemannian metric on $M$ and foliate a small tubular neighborhood of $\Sigma$ by level sets of the distance function. For any sufficiently small positive $\epsilon$, the set of points $\Sigma_\epsilon$ with distance $\epsilon$ from $\Sigma$ will be a smooth hypersurface that is a double cover of $\Sigma$, and its normal bundle will be trivial. We can apply the above argument to the part of $M$ that lies outside of the (open) $\epsilon$-tube about $\Sigma$ to foliate it by smooth hypersurfaces outside of a finite number of critical points of some appropriate Morse function. Again, bounding the number of isolated singularities will depend on the topology of $M$ and $\Sigma$.

  • $\begingroup$ That's very neat; thanks for your answer! $\endgroup$
    – Leo Moos
    Oct 20, 2022 at 21:41

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