# Which submanifolds are leaves of a foliation?

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.

• What obstructions are there for extending $\Sigma^n$ to a foliation on all of $M$ (without removing any points)? Oct 20, 2022 at 15:44
• @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.
– mme
Oct 20, 2022 at 16:02
• @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. Oct 20, 2022 at 17:39

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 $$\Sigma$$.