Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 5 down vote favorite
2

I'm trying to see why there is no (one-dimensional) foliation of $S^2$ or an orientable surface of genus two. Originally I was thinking that such a foliation could give me a non-vanishing vector field, which would be a contradiction, but now I have learned that line fields don't necessary lift to vector fields. Is it still something that depends on Euler characteristic $0$ so that the torus is the only orientable surface with a foliation?

share|improve this question

2 Answers

up vote 6 down vote accepted

A line field $L$ on a manifold $M$ need not come from a vector field. The trouble is that you might not be able to orient the lines in the various tangent spaces in a coherent way. However, if $L$ is not orientable, then there exists a $2$-fold cover $\tilde{M}$ of $M$ and a lift $\tilde{L}$ of $L$ to to $\tilde{M}$ which is orientable. This implies that $\tilde{M}$ has a nonvanishing vector field, and thus has Euler characteristic $0$. Since Euler characteristic is multiplicative in covers, this implies that $M$ has Euler characteristic $0$.

In response to your question on Chris Gerig's answer :

  1. A compact manifold of any dimension supports a $1$-dimensional foliation if and only if its Euler characteristic is $0$. In fact, it has a nonvanishing vector field. Indeed, choose a vector field with isolated zeros. Poincare-Hopf tells you that the signs of the zeros add to $0$. It is then a fun exercise to see that you can "move the zeros together and make them collide and cancel" to get a nonvanishing vector field.

  2. Something even more amazing is true. It is true that an $n$-manifold that supports an $(n-1)$-foliation has Euler characteristic $0$ (choose a metric, and after passing to a double cover if necessary, you can find a unit vector field which is orthogonal to the foliation). Shockingly, the converse is also true! This is a very deep theorem of Thurston from the 1970's (before he got interested in $3$-manifolds).

share|improve this answer
On the orientable manifold $\tilde{M}$, why is it clear the the lift $\tilde{L}$ is coherently orientable? – Jason DeVito Aug 9 '12 at 1:32
@JasonDeVito : You define the cover so orientability is trivial. The cover consists of pairs $(p,d)$, where $p$ is a point in $M$ and $d$ is one of the two directions in the foliation. The topology is the obvious one. This has nothing to do with the orientation double cover of the manifold, which is made up of pairs $(p,o)$ where $p$ is a point on the manifold and $o$ is one of the two local orientations at $p$. – Adam Smith Aug 9 '12 at 1:39
I see - I was thinking of passing to the orientation covering - that makes perfect sense! – Jason DeVito Aug 9 '12 at 1:45
up vote 1 down vote

A manifold admits a 1-dimensional foliation iff its Euler characteristic is zero.

share|improve this answer
Could you elaborate? – Aaron Mazel-Gee Aug 9 '12 at 0:26
Do you mean any manifold, or just surfaces? – paragon Aug 9 '12 at 0:30

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.