What groups can be realized as the isometry group of the two-sphere?

Question

Regarding $S^2 \subseteq \mathbb{E}^3$ as a Riemannian manifold with the inherited metric from Euclidean three-space, then it is well known that the isometry group is $O(3)$. What I am curious about, however, is the following: Given a Lie group $G$ (of dimension $\le 3$), when can I find a Riemannian metric on $S^2$ for which $G$ is the isometry group? For what groups is this possible? (There must be other candidates for the isometry group of $S^2$ as an arbitrary metric would almost certainly result in the a trivial group of isometries.)

The question seems to be direct enough, but I am unaware of any resources or work on the problem. Any help would be greatly appreciated.

Answer 1

I'm rusty on all of this, so hopefully someone can point out any flaws in my reasoning.

By uniformization, every metric on $S^2$ is conformal to the round metric; so the orientation-preserving isometry group should be a compact subgroup of the Möbius group. The Möbius group has unique maximal compact subgroup $SO(3)$, and thus the conformal isometries should be a closed subgroup of $SO(3)$. Throwing in the anticonformal isometries should move this to $O(3)$.

Now, the closed subgroups of $O(3)$ are Lie subgroups; which I think in this case can be classified as $S^1 \times C_2$, $S^1$ or discrete (which is the same as finite in a compact group). We can realise these all as isometry groups of a metric on the sphere:

• to get $S^1 \times C_2$, take the standard metric and fatten a band around the equator.
  • to reduce this to just $S^1$, add a bump at one pole.
• for any finite subgroup $H\subset O(3)$, take the standard metric and add a bump at each point in a $H$-orbit.

Thus the isometry groups of metrics on $S^2$ are exactly the closed subgroups of $O(3)$.