# Folliation and non-vanishing vector field.

The canonical foliation on $\mathbb{R}^k$ is its decomposition into parallel sheets $\{t\} \times \mathbb{R}^{k-1}$ (as oriented submanifolds). In general, a foliation $\mathcal{F}$ on a compact, oriented manifold $X$ is a decomposition into $1-1$ immersed oriented manifolds $Y_\alpha$ (not necessarily compact) that is locally given (preserving all orientations) by the canonical foliation in a suitable chart at each point. For example, the lines in $\mathbb{R}^2$ of any fixed slope (possibly irrational) descend to a foliation on $T^2 = \mathbb{R}^2/\mathbb{Z}^2$.

(a) If $X$ admits a foliation, prove that $\chi(X) = 0$. (Hint: Partition of unity.)

(b) Prove (with suitable justification) that $S^2 \times S^2$ does not admit a foliation as defined above.

Theorem A compact, connected, oriented manifold $X$ possesses a nowhere vanishing vector field if and only if its Euler characteristic is zero.

Question: How could $X$ in this problem satisfy the connectness property in the theorem? Can I just say if it is not connected, treat each connected component individually?

• The result you're trying to prove is not true. For example, $S^2\times S^2$ admits a nontrivial foliation (the leaves are the subsets of the form $S^2\times\{\mbox{constant}\}$), but it does not admit a non-vanishing vector field. – Jack Lee Aug 6 '13 at 0:12
• Dear Professor Lee: I confess I attempted to reduce the problem, and apparently my reduction was wrong. I put the original problem. – WishingFish Aug 6 '13 at 18:47
• OK, part of the confusion is due to the fact that this is a highly nonstandard definition of "foliation." It would be more accurate to call this a "codimension-1 foliation." Anyway, with this definition, the answer by @studiosus is a good way to solve the problem. (Following the hint, you could use a partition of unity to paste together a global vector field when the foliation is orientable; but I think the idea of using a Riemannian metric is a bit cleaner.) – Jack Lee Aug 6 '13 at 19:24
• I probably should figure this out myself, but do you mind directing me which of your books/chapters cover this? – WishingFish Aug 6 '13 at 19:36
• Chapter 19 of Intro to Smooth Manifolds treats foliations, but I don't have any treatment of the relation between codimension-1 foliations and Euler characteristics (or nonvanishing vector fields). – Jack Lee Aug 6 '13 at 19:55

I assume that your manifold and foliation are smooth and foliation is of codimension 1, otherwise see Jack Lee'a comment. Then pick a Riemannian metric on $X$ and at each point $x\in M$ take unit vector $u_x$ orthogonal to the leaf $F_x$ through $x$: There are two choices, but since your foliation is transversally orientable, you can make a consistent choice of $u_x$. Then $u$ is a nonvanishing vector field on $X$.

In fact, orientability is irrelevant: Clearly, it suffices to consider the case when $X$ is connected. Then you can pass to a 2-fold cover $\tilde{X}\to X$ so that the foliation ${\mathcal F}$ on $X$ lifts to a transversally oriented foliation on $\tilde{X}$. See Proposition 3.5.1 of

A. Candel, L. Conlon, "Foliations, I", Springer Verlag, 1999.

Then $\chi(\tilde{X})=0$. Thus, $\chi(X)=0$ too. Now, recall that a smooth compact connected manifold admits a nonvanishing vector field if and only if it has zero Euler characteristic. Thus, $X$ itself also admits a nonvanishing vector field.
• I confess I attempted to reduce the problem, and apparently my reduction was wrong. I put the original problem. Please take a look at it. Perhaps I shall read "The canonical foliation on $\mathbb{R}^k$ is its decomposition into parallel sheets $\{t\} \times \mathbb{R}^{k-1}$ " as "given the foliation in this question has codimension 1"? – WishingFish Aug 6 '13 at 18:49