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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.

THW
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    Are there any examples of such groups that are not isomorphic to a subgroup of $O(3)$? – user7530 Oct 10 '13 at 00:18
  • I am not aware of any other groups. I can see, for example, how one might alter the standard metric so that rotations about the z-axis are isometries, but rotations about x-axis and y-axis fail to be isometries. – THW Oct 10 '13 at 00:55

1 Answers1

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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 $O(3)$.

Anthony Carapetis
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    I will take some time to review this, but it is certainly enough to get me started. Thanks. – THW Oct 10 '13 at 19:29
  • Is there a reference for the isometry group of two-sphere endowed with arbitrary non-round metric? – QGravity Oct 31 '23 at 03:16