I'm trying to determine the complete set of smooth group actions of $\mathrm{SO}(3)$ on $S^2$. That is, I'm trying to determine all smooth $\sigma: \mathrm{SO}(3) \rightarrow \mathrm{Diff}(S^2)$. The two most obvious actions are the trivial action $\sigma_t(R) = I$ where all elements of $\mathrm{SO}(3)$ act by the identity, and the canonical action $\sigma_c(R) = R$ where $\mathrm{SO}(3)$ acts by rotations on $S^2$. When I asked a related question, it was pointed out that if $f \in \mathrm{Diff}(S^2)$, then $\sigma_f(R) = f \circ \sigma_c(R) \circ f^{-1}$ is generally a different group action. However, I realize that in a way that $\sigma_c$ and $\sigma_f$ are not essentially different. In particular, one can define an equivalence relation on the group actions where two actions $\sigma_1, \sigma_2$ are equivalent if there exists an $F \in \mathrm{Diff}(S^2)$ such that $$ \sigma_2(R)(F(x)) = F(\sigma_1(R)x)$$ where $R \in \mathrm{SO}(3)$ and $x \in S^2$. Under this relation, $\sigma_c$ and $\sigma_f$ are equivalent for all $f$ (by taking $F = f$). My question is this: are the standard rotation action and the trivial action $\sigma_c$ and $\sigma_t$ the only actions up to equivalence? I conjecture the answer is yes, but I'm not sure how to go about proving this. If the answer is no, is there a known classification of all such group actions?
Thanks in advance, any help is very much appreciated.
The group $G=SO(3)$ contains only few closed subgroups: zero-dimensional (finite) subgroups, 1-dimensional subgroups (the group of rotations around a line $L$ and its index 2 extension consisting of transformations preserving the line $L$) and the unique 3-dimensional subgroup, $SO(3)$ itself. Consider first a topological action $SO(3)\times S^2\to S^2$ of $SO(3)$ on the 2-dimensional sphere. Pick a point $x\in S^2$. Then the $G$-stabilizer of $x$ is a closed subgroup $G_x< G$. We have the continuous orbit map $G\to S^2$, $f_x: g\mapsto gx$. Since $f_x(g'g)=f_x(g')$ for all $g\in G_x, g'\in G$, the map $f_x$ descends to a continuous 1-1 map $G/G_x\to S^2$, hence, a homeomorphism to its image. Let $k$ denote the dimension of the subgroup $G_x$. Thus, the quotient space is either 3-dimensional, 2-dimensional or a single point (when $G_x=G$). If $G/G_x$ is 3-dimensional, we obtain a topological embedding of a 3-dimensional manifold $G/G_x$ to the 2-dimensional manifold $S^2$, which is absurd. Suppose that $k=1$. Then $G/G_x$ is 2-dimensional, it is either homeomorphic to $S^2$ (if $G_x$ is connected) or the projective plane $\mathbb R P^2$. The latter, of course, cannot be embedded in $S^2$. Hence, our map $G/G_x\to S^2$ is an embedding $S^2\to S^2$ and, hence, a homeomorphism. In particular, the map $G/G_x\to S^2$ is onto. Thus, in this case, the group $G$ acts transitively on $S^2$: $f_x(G)=S^2$. Lastly, if $G_x=G$ then the entire group $G=SO(3)$ fixes the point $x$. If the latter happens for all $x\in S^2$ then all elements of $G$ act as the identity on $S^2$. Thus, the homomorphism $G\to Homeo(S^2)$ given by our topological action has image $\{1\}$. The only interesting case, hence, is when $G$ acts transitively on $S^2$, this is the case I will consider now.
I will now assume that the action of $G$ on $S^2$ is transitive and smooth, i.e. the map $G\times S^2\to S^2$ is smooth. Pick a randomly chosen Riemannian metric $h$ on $S^2$ and a biinvariant (i.e. invariant under both left and right multiplication) Riemannian metric on the group $G$ itself. (Up to scaling, this metric on the tangent space at the identity element of $G$ equals $h(u,v)=Tr(uv)$, where $u, v$ are skew-symmetric matrices. I will not need this fact.) Let $d\mu(g)$ denote the volume form on $G$ associated with this Riemannian metric. Now, let's average the Riemannian metric $h$ on $S^2$ over the group $G$ using the above volume form: $$ \bar{h}=\int_G g^*(h)d\mu(g). $$ The averaged metric $\bar{h}$ is invariant under the action of the group $G$. Since $G$ acts transitively on $S^2$, the Gaussian curvature of $\bar{h}$ is constant. By rescaling $\bar{h}$, we can assume that the curvature is equal to $1$. Hence, there exists an isometry $$ F: (S^2, \bar{h})\to (S^2, h_0), $$ where $h_0$ is the standard Riemannian metric on $S^2$ induced by its embedding in $\mathbb R^3$ as the unit sphere. Since the metric $\bar h$ is $G$-invariant, the metric $h_0$ is invariant under the action of the subgroup $F G F^{-1}< Diff(S^2)$. Lastly, every self-diffeomorphism of $S^2$ preserving $h_0$ is an element of $O(3)$. Hence, $F G F^{-1}< SO(3)$, which implies the equality since $F G F^{-1}$ is a Lie subgroup of $SO(3)$ and $G\cong SO(3)$.