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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.

So I first define $A_\alpha: U_\alpha \to T_x(X)$ by $A_\alpha(t,x) = (t,x,1,0)$ using the chart $\phi_\alpha$. Let $\psi_\alpha$ be a partition of unity subordinate to $U_\alpha$ and define the vector field $$A = \sum_{\alpha \in \mathcal{A}} \psi_\alpha A_\alpha.$$

Here $A_\alpha$ is just constant function on the vector field, hence smooth. And $\psi_\alpha$ guarantees the existance of smooth function over the whole vector field $A$.

Then I show $A$ has no zeros. First by partition of unity, $A$ can be covered by finitely many subcovers. Since the folliation is orientation preserving, $A_\alpha$s will not flip back and change the orientation to $-1$, hence every point is $+1$ and will not cancel out. If $A_\alpha$s have no zero, $A$ is nonvanishing as well.

Is this good?

WishingFish
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  • what is $(t,x) \in U_{\alpha}$ and do you intend the $(p,v)$ notation for tangent vectors, so basically, you're attaching $e_1$ to each point, or perhaps $\partial/\partial x_{\alpha}^1|_{p}$ where $p=(t,x)$... also, silly question, what are you trying to prove? It reads like a fragment to me as it's written. – James S. Cook Aug 09 '13 at 21:45
  • Hello James, it is this guy: http://math.stackexchange.com/questions/460665/folliation-and-non-vanishing-vector-field – WishingFish Aug 09 '13 at 22:59

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