Projective linear special group diffeomorphic to $S^1\times \mathbb{R}^2$ How can I prove that $\mathbb{P}SL_2(\mathbb{R})$ is diffeomorphic 
to $S^1\times \mathbb{R}^2$?
I was thinking about embedding $S^1\subset \mathbb{C}$ as rotations
and $\mathbb{R}^2$ as dilatations (diagonal matrices).
Obviously such matrices commute, then we have a well-defined smooth morphism
$S^1\times \mathbb{R}^2$, but, how I now that this surject on $\mathbb{P}SL_2(\mathbb{R})$?
 A: One way to prove this would be to construct a diffeomorphism between $\operatorname{PSL}_2(\mathbb{R})$ and the unit tangent bundle $\operatorname{UT}(\mathbb{H})$ of the upper-half plane $\mathbb{H} = \mathbb{R} \times \mathbb{R}_{>0} = \{ z \in \mathbb{C} \mid \Im{z} > 0 \}$ with its standard hyperbolic metric. As $\mathbb{H}$ is contractible, $\operatorname{UT}(\mathbb{H}) \cong S^1 \times \mathbb{H}$ is a trivial bundle, and this is of course diffeomorphic to $S^1 \times \mathbb{R}^2$.
The diffeomorphism $\operatorname{PSL}_2(\mathbb{R}) \to \operatorname{UT}(\mathbb{H})$ can be constructed by mapping the identity $I$ to $(i,i)$ (the tangent vector at $i \in \mathbb{H}$ pointing upwards in the vertical direction), and mapping any fractional linear transformation $$f(z) = \frac{az + b}{cz+d} \in \operatorname{PSL}_2(\mathbb{R})$$
to $(f(i), f'(i)i)$ (i.e., rotate the tangent vector $i$ by the argument of the complex derivative of $f$).
In other words, taking the derivative of $f:\mathbb{H} \to \mathbb{H}$ yields a map $df:T(\mathbb{H}) \to T(\mathbb{H})$, which restricts to a map $df:\operatorname{UT}(\mathbb{H}) \to \operatorname{UT}(\mathbb{H})$ (as fractional linear transformations are isometric), and the diffeomorphism $\operatorname{PSL}_2(\mathbb{R}) \to \operatorname{UT}(\mathbb{H})$ sends $f$ to $df(i,i)$. That this is a bijection amounts to the statement that the action of $\operatorname{PSL}_2(\mathbb{R})$ on $\operatorname{UT}(\mathbb{H})$ is simply transitive. This is explained e.g. here.
