Question

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 $n$).