3 Added ruminations on compact support

Addendum 1: To be more precise about the action on $[-1,1]$ we may use$\tanh$ as diffeomorphism from $\mathbb R$ to $(-1,1)$. The rotation group of$\mathrm{SL}_2(\mathbb R)$ lifts to the group of translations of $\mathbb R$ andthey correspond under this diffeomorphism to $\varphi_\lambda(t)=(\lambda t+1)/(t+\lambda)$ of$(-1,1)$ for $\lambda>1$ or $\lambda<-1$ (we miss the identity map which corresponds to$\lambda=\pm\infty$). Now, one possibility of getting an action of $G$ on $\mathbb R$ whichis the identity outside of $(-1,1)$ is by trying to conjugate the one we have bya diffeomorphism $\gamma$ of $(-1,1)$. This requires every conjugatediffeomorphism to be flat at $\pm1$ where a diffeomorphism $\psi$ of$(-1,1)$ is flat at $1$ if $\psi(x)=x+\mathcal{O}(|x-1|^n)$ for all $n$ and $x$close to $1$ (and similarly for $-1$). Now suppose that we have a $\gamma$ such thatconjugating the given action of $G$ by it gives an action all of whose elementsare flat. In particular we have $\gamma(\varphi_\lambda(\gamma^{-1}(x)))=x+\mathcal{O}(|x-1|^n)$for all $n$. In particular, putting $n=2$ we get thatand putting $s=\gamma^{-1}(t)$, using that $\lim_{t\to1}\gamma^{-1}(t)=1$ and that$\lim_{s\to1}\varphi_\lambda'(s)=\varphi_\lambda'(1)=(\lambda-1)/(\lambda+1)$ this gives$\lim_{s\to1}\gamma'(\varphi_\lambda(s))/\gamma'(s)=(\lambda+1)/(\lambda-1)$. This puts a very stringentcondition on $\gamma$ and then also the same condition must be fulfilled for$\varphi_\lambda$ replaced by any element of $G$ (as well as needing the flatnesscondition for all orders $n$).

2 Added comment on non-compact actions

Addendum: Tom's example gives that the dimension of a non-compact groupacting (faithfully) on $S^n$ is unbounded. One question seems to remain namelyif the dimension of a semi-simple group action is bounded or not. By the abovewe get a bound on the dimension of a maximal compact subgroup. In many casesthis seems to bound the dimension of the group itself. For instance, is thedimension bounded if the center is finite? In that case the situation iscompletely described by a Cartan decomposition of the Lie algebra and thequestion is whether the $-1$-part (usually denoted $\mathfrak p$) of it hasdimension bounded by the dimension of the $+1$-part (the Lie algebra of themaximal compact). I myself don't know enough of the real Lie group theory todecide that.

An example beyond that is the universal cover $G$ of $\mathrm{SL}_2(\mathbb R)$.It is contractible and has no non-trivial compact connected subgroup. It alsoacts on the universal cover of $S^1$ (compatibly with the projective action of$\mathrm{SL}_2(\mathbb R)$ on $S^1$) and hence acts continuously on $[-1,1]$(say) fixing the endpoints. However, the action at the endpoints is not flat sothis action does not extend to a smooth action on $\mathbb R$ acting as theidentity outside if $[-1,1]$. If it could be modified to do so one could useTom's argument to get an action of any finite product of copies of $G$ (at leaston $S^1$). One could try to find three vector fields with support in $[-1,1]$fulfilling the defining relations of $\mathfrak{sl}_2(\mathbb R)$ but it lookstricky to me.

1

For 2) I think the following is an answer: Suppose $K$ is a compact Lie subgroup of $\mathrm{Diff}(S^n)$ of dimension $\geq{n+1\choose 2}$. Being compact it is the group of isometries of some Riemannian metric of $S^n$ and we fix one such metric. The stabiliser of a point therefore has dimension at most $n\choose 2$ (the dimension of the orthogonal group of $\mathbb R^n$) and as an orbit has at most dimension $n$, $K$ has at most dimension $n+{n\choose 2}={n+1\choose 2}$ and hence we have equality. This means that the stabiliser contains $\mathrm{SO}_n$ and $K$ acts transitively. In particular the metric fixed by $K$ has constant curvature and then the curvature is positive and thus up to a constant conformal factor is conjugate to the standard metric. This gives a conjugation of $K$ into the isometry group of the standard sphere.