Is the circle a homogeneous space for G=SL(2,R)? I think G/B is real analytically isomorphic to the circle where B is Borel subgroup of G. Is there a transitive algebraic action of G on the circle? Can someone give that action with formulas for every entry given as rational functions in the entries of g and the circle (take circle as unit circle in the plane).
I tried projecting onto the K circle in the KAN=KB decomposition and then multiplying by that rotation but it didn’t work.
Maybe something with Möbius transformation acting on the circle bounding the upper half plane viewed as a subset of the Reimannian sphere?
Note: the action cannot be an action by isometries the circle is not a symmetric space for SL(2,R) basically because SO(2,R) is not a group quotient of SL(2,R)
Yes, actually in (essentially) two distinct ways.
$G=\mathrm{SL}_2(\mathbf{R})$ acts transitively on the circle $\mathbb{P}^1_\mathbf{R}$; the stabilizer of $\infty$ is $B$ (group of upper triangular matrices), so $G/B$ is homeomorphic to the circle. In this case, the action of $G$ on $B$ is not faithful: the kernel is $\{\pm \mathrm{id}\}$.
Also, $G$ acts on the circle $(\mathbf{R}^2\smallsetminus\{0\})/\mathbf{R}_{>0}$, and the stabilizer of $(1,0)$ for this action is $B^+$, the subgroup of index 2 of $B$ with positive diagonal. In this case, the action of $G$ on $G/B^+$ is faithful.
Note that there's a continuous $G$-equivariant surjection $G/B^+\to G/B$, which is a connected 2-fold covering.