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I am working on path motion planning on different topological spaces. In order to prove the existence of some motion planning algorithms I would like to use that given an even dimensional sphere, we can always find a continuous vector field tangent to it which only vanishes at one point.

My attempts:

First of all I think that that statement should be true since we could use the stereographic projection (which is a diffeomorphism) in order to "bring" or "take back" a smooth vector field defined on $\mathbb{R}^{2n}$ (assuming we are dealing with $\mathbb{S}^{2n}$) which only vanishes at one point.

But I am doing a project about other topic and if I use that, then I should introduce and define charts, diffeomorphisms...etc and I am restricted on the number of pages. That is the reason why I am looking for other argument, moreover, I don't need smoothness so that would be like killing a mosquito with a cannon ball.

What about using that the stereographic projection is an homeomorphism and using the same idea as before? Continuity is preserved by homeomorphisms so I would have finished. But again if could be possible I would like to avoid using stereographic projection.

What I have consulted:

this question on stackexcange

Marcel Berger, Bernard Gostiaux book on differential geometry where they draw a picture in the case of dimension 2

My question:

What other idea do you suggest? Is there an intuitive vector field that works and I am missing it?

Thanks in advance!

Community
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D1811994
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1 Answers1

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It shouldn't be too hard to construct such a vector field. The hard trick is proving that all vectors fields on $S^{2n}$ vanish at some point.

Start with the tangent vectors to a rotation about some axis. This provides a vector field with exactly two $0$ points. Now morph the sphere below to bring these two points together. Everwhere off the joined point, the vector field has undergone a continuous transformation, so it remains a vector field. At the joined point, the behavior is singular, but since the vector field vanishes there, the singularity is smoothed out.

Paul Sinclair
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  • I already have proved that all continuous vectors fields on $\mathbb{S}^{2n}$ vanish at some point. I think I intuitively understand what you mean despite I don't get an analytic expression, which is what I want. I started by picturing such a vector field, for example: – D1811994 Mar 04 '16 at 21:08
  • https://drive.google.com/file/d/0B5QLY6KP0JyvNjlxb256NThteUU/view?usp=sharing – D1811994 Mar 04 '16 at 21:09
  • Now by morphing I think in an homotopy or something like that but I think I won't work. Or at least I don't see how to write down some analytic expression. – D1811994 Mar 04 '16 at 21:12