You cannot comb a sphere

Cover a ball in hair and try to comb it flat, every hair lying down, with no parting and no tuft sticking up. You will fail. Not because you are clumsy: every possible combing fails, and the reason is a single whole number that cannot make up its mind.

What is being claimed

Take the ordinary sphere — the surface of a ball, every point at distance one from the centre. At each point, put an arrow that lies flat against the surface — that is what a hair combed flat is. A choice of one arrow at every point is a vector field on the sphere. It is continuous if nearby points get nearly the same arrow: no partings, no sudden flips. The theorem says a continuous field of flat arrows must be zero somewhere: there is always a bald spot. (A circle can be combed flat, and so can a doughnut — the arrows wind around its hole. The sphere has no such hole.)

The result goes back to Poincaré and is often called the hairy ball theorem. This page follows Peter McGrath’s two-page proof, which uses one idea: counting turns.

Why it is worth knowing

It settles every possible combing at once, without looking at any of them. Our own reading, not the note's, which contains no applications: treat the horizontal part of the wind over an idealised round planet as flat arrows that vary smoothly, and the theorem says that at every instant the horizontal wind is zero somewhere.

Counting turns

Draw a circle on the sphere with a flat plane. Walk once round it, comparing the direction you are walking with the hair beside you at each step. Back at the start you face the same way against the same hair, so the angle has changed by a whole number of turns. That whole number is the circle's rotation number.

Now suppose some combing has no bald spot anywhere. Then every circle has a rotation number. Slide or shrink a circle a little and its count can only change a little — but it is a whole number, so it cannot change at all. Every circle can be slid into every other, so they all share one number. Call it n.

Two ways to count, two answers

First, take the equator. Walk it one way and you get n. Walk it the other way and the total turning flips sign, so you get minus n. But walking it backwards is itself one of the circles, so its count is also n. Then n equals minus n, and n is zero.

Second, take a very small circle. The hair hardly changes across it, so it looks like a fixed direction, and a small circle in a flat plane turns exactly once against a fixed direction. So n is plus or minus one.

Zero is not plus or minus one. The only assumption was that no hair was missing, so it is false — whichever combing you tried.

Where the argument bites

In the interactive, the tiny circles at the two ends count plus one and minus one. Without a bald spot they would have to agree. So somewhere between them the count jumps, and it can jump only at a circle running through a spot where the hair vanishes.

In short

Count how often a circle's direction turns against the hair. With no bald spot every circle would share one count, forced to zero by the equator walked backwards and to one by a tiny circle. It cannot be both, so every flat combing of a sphere leaves a point bare.