Does the Hairy Ball Theorem prove there is no group structure on the sphere?

Question

It is known that there is no topological group structure on the sphere, but all of the methods of proving it seem quite advanced and not immediately intuitive. It seems like there's a much easier way:

The left action of any non-identity element would be a continuous map, $f$, from $S^2$ to itself with no fixed points. Then you could define a continuous vector field on the sphere by taking the vector pointing from $x$ to $f(x)$.

This has a wrinkle when $f(x)=-x$, but it seems like that can be avoided by taking some $x$ sufficiently close to the identity so that $f(x)$ will be near $x$.

Presumably, there's a flaw in this argument, or it would be used. So what is it?

Answer 1

Yes, that will work. As you note you will have to make your argument that for $y$ near enough to $e$ you have $yx\ne-x$ for all $x$ exact. This will probably be a compactness argument.

Once you have done that you will have another answer to: Can $S^2$ be turned into a topological group?. It will be very similar in spirit to the first argument in the accepted answer.

Answer 2

There's a very nice proof which uses Poincare-Hopf which you can think of as a generalization of hairy ball. It says that the Euler characteristic is equal to the number of zeros (counted with "index") of a tangent vector field with isolated singularities. If $X$ is a topological group, take a nonzero vector at the identity and generate a vector field on $X$ by translating it around (pushforward) by the (transitive) left group action. Thus, any topological group has a nonvanishing tangent vector field, and must have Euler characteristic 0 i.e. a torus. Actually, if you only care to show that the sphere isn't a group, you just need that you can find a nonvanishing vector field and use hairy ball instead of Poincare-Hopf, but this actually gives something much stronger. But note, this proof works in arbitrary dimension (to show a topological group has 0 Euler characteristic)! In fact, the same translation trick shows any topological group has trivial tangent bundle.

To prove Poincare-Hopf, there's a neat argument with "charges" that I first saw in Thurston's book on 3d geometry and topology. There's also a nice argument that a section of the tangent bundle of $X$ is a section of the normal bundle of the diagonal in $X\times X$, and hence corresponds to an "infinitesimal deformation". The index of a vanishing point is the multiplicity of the intersection, and it now becomes an intersection theory problem.