# A Way to make the following “proof” of the Hairy Ball Theorem rigorous?

I plan on giving a talk soon to undergraduates and I'd like to talk about the hairy ball theorem during the talk. I was trying to think of some sort of visually intuitive proof of this fact. (I already know several homotopical proofs), and this is roughly what I came up with:

1. Suppose you have a vector field on a sphere. It's seems reasonable that if it is nonvanishing, all of the integral curves will be circles. (I think that's true? I know either an integral curve is periodic, a point, or a line... and we're assuming it's not a point, and I just feel like you couldn't fit a line in there for some silly reason.)
2. If this is true then there is probably an integral curve with the smallest 'diameter,': a circle divides the sphere into two pieces (not easy to prove, but visually an audience could be convinced), and the 'diameter' is defined to be the smaller of the two different obvious ways one could define the diameter.
3. This curve can't be a point, so it has nonempty 'interior' (again using the Jordan curve theorem, and we pick the 'smaller' interior).
4. Pick a point in the interior and follow it's integral curve.
5. This integral curve has to be contained inside the 'smallest'.
6. Contradiction, since clearly the diameter of this one is less.

Obviously there are lots of things that are not at all obvious. But I would be happy if this argument COULD be made rigorous, even with lots of technical details, because then I wouldn't feel bad giving it without the details since it's very visual.

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Why couldn't an integral curve spiral in towards one pole or the other without intersecting itself? –  Qiaochu Yuan May 10 '11 at 1:31
Right, I was thinking about this. But compactness of the sphere makes me think that the limit point would be a vanishing point of the vector field? –  Dylan Wilson May 10 '11 at 1:34
Maybe. If it's (continuous and) nonvanishing I guess you know that its velocity is bounded from below, so maybe this can't happen. But I am no differential geometer... as for Amy's edit, that was approved by me. –  Qiaochu Yuan May 10 '11 at 1:38
I like to explain it as just that you flow all points along great circles at the same speed, and then at time $\pi$ you end up at the antipodal map. If you hold your hands around an (imaginary) basketball -- one on top and the other below -- you'll see that the antipodal map on $S^2$ reverses handedness! Perhaps your audience would believe that a homotopy of self-maps can't change its degree, if you say it in friendlier words... –  Aaron Mazel-Gee May 10 '11 at 1:46
Step (1) in your argument is contradicted by the Poincare-Bendixon theorem. en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Bendixson_theorem –  Ryan Budney May 10 '11 at 6:46