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Let $L$ be a regular closed curve on the sphere $S^{2}.$ Let $\vec{V}$ be a differential vector field in $S^{2}$ such that $\vec{V}$ is not tangent to $L.$ Show that each of the two regions determined by $L$ have at least one point where $\vec{V}$ vanish.

I was thinking about the curve $L$ that separate the upper semicircle from the lower semicircle, and I want to define the vector field $\vec{V}$ as $$V : S^{2} \longrightarrow \mathbb{R}^{2}$$ $$x \mapsto V(x)$$ with $x \in S^{2}$ and $V(x) \in \mathbb{R}^{2}$ and $\mathbb{R}^{2}$ is the plane that divides de sphere in upper and lower semicircles (the way I chose $L$ makes it live in that plane).

This is the way I'm taking, but I don't know how to carry on, I don't know if that vector field that I chose is indeed a differential vector field and that is not tangent to $L,$ and if it is and it is well defined then how I'm suppose to continue with this idea? Thank you very much for your help.

HelaHelheim
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1 Answers1

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Consider the vector field in one of those regions and suppose it is nowhere vanishing there. Since the vector field is never tangent to the boundary, it will either always point inwards or always point outwards (on the boundary). Without loss of generality we can assume it always points inwards. If you take its flow for a small amount of time, you get a self-map of the region which does not have fixed points. Since that region is topologically a disk, this cannot happen.

Pedro
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  • With regards to my 'without loss of generality': as you can see comented here https://en.wikipedia.org/wiki/Schoenflies_problem#Original_formulation , this region is diffeomorphic to a disk. Hence (assuming we are dealing with $D^2 \subset \mathbb{R}^2$), if the vector field $X$ points outwards, just look at $-X$ instead. – Pedro Nov 17 '16 at 04:28