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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. – Ryan Budney May 10 '11 at 6:46

To show that 1) really needs a big argument: Is false for arbitrary spheres. Consider the Reeb flow of any generic ellipsoid of $E\subset \mathbb{C}^n$. There even exist smooth non-vanishing vector fields on $S^3$, which do not have any periodic orbits. Actually this is true for any three-fold. This is a famous result of K. Kuperberg

How about this for a sketch for a proof of the Hairy ball theorem for the two sphere, using your ideas:

Take a point $x_1\in S^2$, its $\omega$ limit set is a periodic orbit by Poincaré-Bendixon. Now remove a point of the sphere outside the periodic orbit. We obtain $\mathbb{R}^2$ with a (non-complete) flow with a periodic orbit. This periodic orbit is an embedded $S^1$, lets call it $\gamma_1$ and divides the plane $\mathbb{R}^2$ in two regions: an unbounded one and a bounded one. Let's denote the latter by $D_1$. Note that on $D_1$ the flow is complete.

Take a point $x_2\in D_1$. The $\alpha$ and $\omega$ limit sets of $x_2$ are periodic orbits. They cannot coincide if $x_2$ is not on a periodic orbit. Hence one of the periodic orbits must be not equal to $\gamma_1$. Call this periodic orbit $\gamma_2$. Now $\gamma_2$ divides the plane again into an unbounded and bounded region, call the bounded one $D_2$. Clearly $D_2\subset D_1$. Repeat the argument to find a nested sequence of compacts $\ldots\subset D_2\subset D_1$. These are all invariant for the flow. Then $\bigcap_i D_i$ is non-empty as it is a decreasing set of compact sets and each one is properly contained in the other it must be a point. Since the intersection of invariant sets is invariant the point must be an invariant set of the flow. Thus it must be a vanishing point of the vector field.

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The last argument fails. Consider discs with radii $1+\frac1n$. They form a properly contained sequence of compact sets and their intersection is disc with radius $1$. One way to deal with this problem is to use transfinite induction. You have to reach a point before you hit $\omega_1$ because between each two discs there is an unique rational point. – tom Apr 7 at 8:33
@tom: Thanks, I will think more about this. – Thomas Rot Apr 7 at 8:48

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