# Shortest proof for 'hairy ball' theorem

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.

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

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

• @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
• @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
• @EricWilson Everyone is goint to download at scihub anyway. – Billy Rubina Dec 19 '16 at 15:04

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.

• this also works for general dimensions, with the same argument. define $H(x,t) := \cos(t) \, x + \sin(t) \, \nu_x$, as in hirsch's differential topology, which gives you a homotopy between $id_{\mathbb S^m}$ and $-id_{\mathbb S^m}$. we have $\deg (-id_{\mathbb S^m}) = (-1)^{m+1}$. – cesare borgia Jan 30 '17 at 23:49
• Note that reflection in one coordinate has degree -1, so that the map from x to -x has degree (-1)^(n+1) where n is the dimension of the sphere. Bonus fact that any fixed point free map from the sphere to itself is homotopic to the antipodal map via parameterized normalized line segments between -x and f(x), necessarily not crossing the origin. – Mars Mar 23 '17 at 4:54

The proof can be found on this site:

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

• Right, except this is not "the proof" but is "Milnor's proof". – Moishe Kohan Sep 18 '19 at 23:41