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?
Why is there no foliations of the 2-sphere, or a genus two surface?
2 Answers
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 :
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.
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).
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0I see - I was thinking of passing to the orientation covering - that makes perfect sense! – 2012-08-09
A manifold admits a 1-dimensional foliation iff its Euler characteristic is zero.
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0Do you mean any manifold, or just surfaces? – 2012-08-09