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I want to make a project at differential geometry about the Hairy Ball theorem and its applications. I was thinking of including a proof of the theorem in the project. Using the Poincare-Hopf Theorem seems easy enough, but I was thinking that this proves the desired result using a stronger theorem (just like proving Liouville's Theorem in complex analysis using Picard's theorem).

Is there a simple proof of the fact that there is no continuous non-zero vector field on the even dimensional sphere? It is good enough if the proof works only for $S^2$, because that is the case I will be focusing on in the applications.

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[insert joke about hairy balls here]. –  zzzzBov Dec 12 '11 at 15:36
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3 Answers

up vote 7 down vote accepted

There is an elegant and self-contained proof by Milnor: Analytic proofs of the "hairy ball theorem", American Math. Monthly, 85 (1978), 521-524. The paper is reprinted in his collected works and can be downloaded for pay here:

http://www.jstor.org/pss/2320860

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Thank you. I'll probably use this :) –  Beni Bogosel Dec 12 '11 at 12:30
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Please consider adding the proof to your answer. Linking to an answer behind a paywall is not nearly as useful to the community. –  Eric Wilson Dec 12 '11 at 13:43
    
@Eric Wilson: I don't have the paper at my fingertips, but I remembered it as a real gem, so I retrieved its coordinates. I regret the paywall as much as anybody else. Fortunately the Monthly is widely available in math libraries. –  Christian Blatter Dec 12 '11 at 15:04
    
@BeniBogosel If you (or anyone else) found this useful, perhaps you can edit the answer to include a summary of the contents. –  Eric Wilson Dec 12 '11 at 15:14
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@EricWilson: The sketch of the proof can be found in the following blog post: topologicalmusings.wordpress.com/2008/07/22/… I thank Christian Blatter for posting the link. In the library of my university I could read it. The blog post summarizes the proof. –  Beni Bogosel Dec 12 '11 at 18:04
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The simplest I can remember off the top of my head is this:


Assume there is such a vector field. Let $v_x$ denote the vector at the point $x$. Now, define the homotopy $H: S^2\times [0, 1] \rightarrow S^2$ by the following: $H(x, t)$ is the point $t\pi$ radians away from $x$ along the great circle defined by $v_x$. This gives a homotopy between the identity and the antipodal map on $S^2$, which is impossible, since the antipodal map has degree $-1$. Hence there can be no such vector field.

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The proof can be found on this site:

http://people.ucsc.edu/~lewis/Math208/hairyball.pdf

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