2

It is known from this question that $S^2$ cannot be made into a topological group. Indeed, $S^2$ cannot even be made into an $H$-Space, a much looser requirement than a topological group.

However, an $H$-Space does require an identity element, so I was wondering if different loose requirements could allow some somewhat nice multiplication on $S^2$. In particular, does there exist a continuous product $m:S^2\times S^2 \rightarrow S^2$ such that $(S^2,m)$ is a quasigroup? If this does exist, can we also require that the left and right division operations associated with this quasigroup are also continuous, making $S^2$ a topological quasigroup?

volcanrb
  • 2,796
  • 9
  • 23

1 Answers1

5

In fact, in a very precise sense $S^2$ cannot support any nontrivial algebraic structure! This was proved by Walter Taylor in his paper Spaces and equations. And the situation doesn't change even if we allow equations to be satisfied only "up to error" - see this old question of mine.

Noah Schweber
  • 245,398