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?

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    $\begingroup$ 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. $\endgroup$
    – Jack Lee
    Commented Aug 6, 2013 at 0:12
  • $\begingroup$ Dear Professor Lee: I confess I attempted to reduce the problem, and apparently my reduction was wrong. I put the original problem. $\endgroup$ Commented Aug 6, 2013 at 18:47
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    $\begingroup$ 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.) $\endgroup$
    – Jack Lee
    Commented Aug 6, 2013 at 19:24
  • $\begingroup$ I probably should figure this out myself, but do you mind directing me which of your books/chapters cover this? $\endgroup$ Commented Aug 6, 2013 at 19:36
  • $\begingroup$ 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). $\endgroup$
    – Jack Lee
    Commented Aug 6, 2013 at 19:55

1 Answer 1


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.

You should read this book (and, maybe, its sequel, "Foliations, II") if you want to learn more about foliations.

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.

Incidentally, Bill Thurston proved in 1976 (Annals of Mathematics) that the converse is also true: Zero Euler characteristic for a compact connected manifold implies existence of a smooth codimension 1 foliation. This converse is much harder.

  • $\begingroup$ is it completely obvious that your vector field is smooth ? I have a feeling if you write it down it will turn out you don't need orientability $\endgroup$
    – Albert
    Commented Aug 6, 2013 at 1:58
  • $\begingroup$ Yes, it is obvious that the field is smooth: The point is that tangent spaces to foliation vary smoothly. To see why orientability is needed look at foliation of Moebius band by circles. The central circle cannot have a normal field. $\endgroup$ Commented Aug 6, 2013 at 3:21
  • $\begingroup$ 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"? $\endgroup$ Commented Aug 6, 2013 at 18:49
  • $\begingroup$ @WishingFish: Yes, this means that your foliation has codimension 1. $\endgroup$ Commented Aug 7, 2013 at 0:05
  • $\begingroup$ Hi studiousus, your answer helped me figure out this problem, thank you. Because I have been trying to figure out why the hint partition of unity, then I realized this problem probably demands partition of unity to assure the smooth assignment of vector field. $\endgroup$ Commented Aug 9, 2013 at 16:49

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