$\DeclareMathOperator\Sl{Sl}\DeclareMathOperator\PSl{PSl}\DeclareMathOperator\Isom{Isom}$Let $\widetilde{\Sl\_2}$ be the Thurson geometry that can either be described as the universal cover of $\PSl(2,\mathbb{R})$, or as the twisted line bundle over the hyperbolic plane $\mathbb{H}^2\mathbin{\tilde{\times}}\mathbb{E}$.


Its isometry group $\Isom(\widetilde{\Sl\_2})$ is a $4$-dimensional Lie group with $2$ connected components, whose identity component $\Gamma$ fits into a short exact sequence $$0\to \Isom(\mathbb{E})\overset{i}{\to} \Gamma\overset{\pi}{\to}\Isom^+(\mathbb{H}^2)\to 0.$$
In fact this is even a central extension since $\Isom(\mathbb{E})\cong\mathbb{R}\triangleleft\Gamma$ is a central normal subgroup. Notice that this s.e.s. does not split, otherwise the group $\Gamma$ would be a direct product and we would have the non-twisted geometry $\mathbb{H}^2\times\mathbb{R}$. (See for example
[[Scott - The geometries of 3-manifolds]](http://homepages.warwick.ac.uk/%7Emasgar/Teach/2012_MA4J2/geometry.pdf) for more details on Thurston geometries.)


Hence, by the general theory of central extensions, there exist a map $\Phi:\Isom^+(\mathbb{H}^2)^2\to\mathbb{R}$ such that $\Gamma$ can be described in the following way: $\Gamma\cong(\mathbb{R}\times\Isom^+(\mathbb{H})^2,\circ\_\Phi)$ where the composition law is the following: $$(a,f)\circ\_\Phi(b,g)=(a+b+\Phi(f,g),fg).$$
To give a little more details we can say that the map $\Phi$ measures how much the s.e.s. fails to split: if $c:\Isom^+(\mathbb{H}^2)\to\Gamma$ is a section (of sets!) of the projection $\pi$ then for two general elements $f,g\in\Isom^+(\mathbb{H}^2)$ the elements $c(f)c(g)$ and $c(fg)$ differ by a unique element $i(a)$, i.e. $c(f)c(g)=i(a)c(fg)$. Then $\Phi$ is defined to be $\Phi(f,g)=a$. (See for example [[Brown - Cohomology of groups](https://link.springer.com/book/10.1007/978-1-4684-9327-6)] IV.3 for more details about extensions with abelian kernel.)


**I am trying to answer the following question**: what is the function $\Phi$ that gives rise to $\Gamma$?


Here is my attempt. First of all let us be aware of the nice series of equivalences of Riemannian manifolds $$\Isom^+(\mathbb{H}^2)\cong \PSl(2,\mathbb{R})\cong U\mathbb{H}^2$$
where $U\mathbb{H}^2\subset T\mathbb{H}^2$ is the unitary bundle on the hyperbolic plane that inherits is metric as a Riemaniann submanifold of the tangent bundle endowed with the Sasaki metric. Note that this metric comes with a couple of nice features:


1. An isometry $f$ of $\mathbb{H}^2$ acts as an isometry of $T\mathbb{H}^2$ via the differential $df$, hence also on $U\mathbb{H}^2$ when restricted on it.
2. The fibers $S^1$ over each point are totally geodesic.


It follows in particular that $df$ sends fibers to fibers and acts on each of them as a rotation of an angle $\theta\_x$ (is this angle constant w.r.t $x$? I think so, but wouldn't know how to prove it). Hence $df$ acts also on the universal cover $\widetilde{\Sl\_2}$ sending each fiber to the corresponding one, and translating it by a length $\theta\_x$. In this way we have a section $\Isom^+(\mathbb{H}^2)\to\Gamma$ defined by $f\to df$.
Now let us take to isometries $f,g\in\Isom^+(\mathbb{H}^2)$ and look at the elements $df,dg,d(fg)^{-1}\in\Gamma$. Of course the projection of their composition acts as the identity on $\mathbb{H}$, so their composition acts as a translation on each fiber. If we choose a point $x\in\mathbb{H}^2$ their composition will act as a rotation of $S^1\cong U\_x\mathbb{H}^2$ of angle $\theta\_x(f,g)$, and it should be pretty straightforward to check that $\Phi(f,g)=\theta\_x(f,g)$ is the function I'm looking for.


**Here are some open points**: first of all I strongly believe that that angle $\theta\_x$ should *not* depend on $x$. This might simplify the calculation a little bit. Nevertheless the calculations still looks to me fairly annoying, as they involve a lot of nasty differentials in coordinates or parallel transports. So I wonder **is there any other more straightforward way to find this (or another) explicit description?** Maybe it could involve some Lie-algebra work that I am not aware of.



An explicit description of $\operatorname{Isom}(\widetilde{\operatorname{Sl}_2})$
Surely the group $\tilde {SL}(2,\mathbb R)$ maps into this isometry group of the manifold $\tilde {SL}(2,\mathbb R)$, and in such a way that the composed map $\tilde {SL}(2,\mathbb R)\to {PSL}(2,\mathbb R)\cong Isom^+(\mathbb H^2)$ is the usual projection. So I imagine that your central extension by $\mathbb R$ comes from the central extension by $\mathbb Z$
$$
0\to \mathbb Z\to \tilde {SL}(2,\mathbb R)\to {PSL}(2,\mathbb R)\to 1.
$$
This means that your cocycle $\Phi$ really wants to be integer-valued (and discontinuous).


