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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?

Zoe Allen
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2 Answers2

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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.

Carsten S
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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.

Toyesh Jayaswal
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