Let $M$ be a compact smooth manifold, and $\mathcal{F}$ be a foliation on $M$. Assume that $L$ is a leaf of $\mathcal{F}$ for which there is $x\in L$ with the property that every neighborhood of $x$ in $M$ intersects a compact leaf of $\mathcal{F}$. Must $L$ itself be compact?

Note that the existence of such $x$ implies that all points of $L$ have the same property.

The conclusion that $L$ is compact seems to be true when the codimension of $\mathcal{F}$ equals one, and this goes back to Reeb. I am interested in what happens in general codimension, but particularly in the case where the codimension of $\mathcal{F}$ equals two.



1 Answer 1


Here's a counterexample where the foliation has codimension two. Consider the integral curves of a flow on the 3-torus $\mathbb{R}^3 \, / \,\mathbb{Z}^3$, defined by $x' = 0, y'=\cos 2\pi x, z'=\sin 2 \pi x$.

All leaves have constant $x$, and they are compact exactly when $x$ is rational. A leaf with irrational $x$ will not be compact, but any neighborhood of it will intersect a compact leaf.

  • 1
    $\begingroup$ Thank you! This is a very pretty counter-example. $\endgroup$
    – Ivo Terek
    Commented Apr 18 at 0:37

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