# 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.

• Why couldn't an integral curve spiral in towards one pole or the other without intersecting itself? Commented May 10, 2011 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? Commented May 10, 2011 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. Commented May 10, 2011 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... Commented May 10, 2011 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 Commented May 10, 2011 at 6:46

To me 1. looks not intuitive at all. Why should all integral curves be periodic? I think it is really hard to prove this using some kind of real theorem. To have even one periodic orbit you have to put in some effort.

Of cause for logical reasons 1. is obviously true, since Hairy Ball theorem is true. But even one periodic orbit would prove Hairy ball theorem, so I don't see how you could proof that all orbits are periodic without exploiting that you have at least one.

I suggest an other approach:

• There exists at least one periodic orbit. This is intuitive, you can assume the vector field to be smooth and normalized, then by compactness the curvature of the curve has a lower bound. Assume you have an orbit which is not periodic, then it is quite intuitive that the $\omega$-limit is a period orbit, since the orbit is not self intersecting and the curvature has a lower bound.
• If you have a periodic orbit, then there is a fixed point. The intuitive case the equator is an integral curve and then try to push it down to south pole. So at some point all vectors have to point into the same direction. So you need to have a discontinuity or a zero.

Why is this true?

A stupid thing you could do now is to assume you have a non vanishing vector field with a non periodic orbit, then use what I said before, you get a contradiction and can conclude all orbits are periodic.

Using this you could prove 2.: By Jordan Curve theorem you can recursively define curves. Start with $c_{i-1}$ then, this curve divides $\Bbb S^2$ into two areas pick a point $x_n$ in the area which has trivial intersection with $c_{i-2}$. The points $x_i$ determine the integral curves $c_i$. Further you can choose the $x_i$ carefully such that $\text{diam}(c_i)$ approaches a minimal diameter. The sequence $x_i$ has a limit point $x$ which defines an integral curve $c$ with minimal diameter.

To sum it up. My feeling is there is no reasonable way to prove 1. using a constructive method. Also it would be kind of strange to prove something like this since a single period orbit would imply a fixed point. But since Hairy Ball theorem is true you can prove it by contradiction, which to me is kind of odd since you don't get anything from this.

I think using Poincaré-Bendixson is a bit of cracking a nut with a sledgehammer, but kind of intuitive. Recently I posted a more elementary proof which has a similar idea behind it. I also sketched there what I want to do here.

One method that i use for the hairy ball in even dimensions, is to use 'complex euclidean space'. CEn space is algebraically n-dimensional space, using complex numbers, which maps onto 2n real space.

The trick here is to imagine a point $x,y,z...$, such that the ratios of any pair remain constant. Now multiply through by $w$ to get $wx, wy, wz, ...$. Now we put $w=\operatorname{cis}(\omega t)$, where $\omega$ is the speed of rotation. For every point in space, forms a solid rotation around the centre.

This means, that every point has 'somewhere to go', and the hair is complete.

For odd dimensions, you have already shown that the plane $h=0$ is a hairy ball, and that it contains hairy balls inside until you reach the pole. The other proof is that if the even spaces are hairy, then the odd ones can not, because they are sections of hairy balls, and must contain points where the hair is perpendicular to the section, and appear as vortices.