# 1-dimensional foliation on a surface

Is it possible to find a 1-dimensional nonsingular foliation on an orientable surface with one boundary component such that lines of the foliation are transverse to the boundary?

• I take it you want compact surfaces otherwise a half infinite cylinder works. – Dan Rust Aug 25 '14 at 22:18

No such foliation exists. As the surface can only have $1$ boundary component, the component must be a circle (assuming compactness), and so let $S$ be the surface with one boundary component. Let's suppose that $S$ has genus $g$. Let $S+S$ be the surface you get by gluing $S$ to another copy of $S$ along its boundary component. $S+S$ has genus $2g$. If $S$ admits a $1$-dimensional non-singular foliation, transverse at the boundary, then so does $S+S$. But note that the only closed manifolds which admit a $1$-dimensional non-singular foliation are those with Euler characteristic $0$. For surfaces this means $S+S$ must be a torus by the classification of closed surfaces. But the genus of the torus is $1$ which is odd.

• Very nice! The argument is classical. – Bombyx mori Aug 26 '14 at 3:46

I originally (and mistakenly) wrote: Consider the $2 \times 2$ square from $-1$ to $1$ in each of $x$ and $y$, and delete a small disk around the origin. If you identify the two vertical edges of the square and the two horizontal edges, you get a punctured torus.

Now consider the set of straight lines through the origin; these (after deleting the disk around the origin) constitute a foliation that's orthogonal to the boundary.

Let me clean up and give a better answer: This construction fails, because the lines don't match up at the identified edges. If the edges are identified with a flip (yielding a projective plane) then you DO get a foliation, but it's easier to see if you draw the punctured projective plane as a Mobius band; then the foliation lines are exactly the lines perpendicular to the centerline of the band. That's not an example of what's asked for, though, because it's nonorientable.

Indeed, suppose that there were a foliation of the kind requested. Then at each point, there's a tangent line to the foliation, i.e., the foliation yields an everywhere nonzero line field. If you pick an orientation of any one of the lines, it extends to an orientation along the whole leaf of the foliation, and thence to an orientation on the adjacent leaves, and hence to a nonzero vector field on the surface $M$ (I'm being glib here; the fact that $M$'s orientable lets you do this; clearly it doesn't work for the Mobius band case! I'm also assuming that $M$'s path-connected). By filling in the boundary component of $M$ with a disk to get a compact surface-without-boundary $N$, and extending the vector field by vectors pointing towards the disk-center, we find that the manifold $N$ has a vector field with a single singularity of index 1, hence the Euler characteristic of $N$ is 1. But there's no such orientable compact surface-without-boundary; their Euler characteristics are all even numbers.

Conclusion: you can't find such a foliation.

• In this case, don't only the vertical, horizontal and diagonal lines through the origin actually 'match up' when they pass over the edge of the square? The other lines intersect non-trivially. – Dan Rust Aug 25 '14 at 22:22
• If you translate the fundamental domain by (1,1) so that the removed disk is at the corners, you'll notice much more clearly that the diagonal lines here will be singular (they must intersect). – Dan Rust Aug 25 '14 at 22:38
• Completely correct; it does work for the projective plane, but that's nonorientable, so it's not an example. I'll edit what I wrote above. – John Hughes Aug 26 '14 at 3:14
• I Just noticed you fixed this answer not long after my comment. Your new answer is very instructive! – Dan Rust Dec 20 '14 at 0:36
• Thanks. Approbation from Sir Hubert is praise indeed. :) – John Hughes Dec 20 '14 at 0:52