Let $\widetilde{Sl_2}$ be Thurson geometry that can either me described as the universal cover of $PSl(2,\mathbb{R})$, or as the twisted line bundle over the hyperbolic plane $\mathbb{H}^2\tilde{\times}\mathbb{E}$.

Its isometry group $\mathrm{Isom}(\widetilde{Sl_2})$ is a $4$-dimensional Lie group with $2$ connected components, whose identity component fits into a short exact sequence $$0\to \mathrm{Isom}(\mathbb{E})\overset{i}{\to} \Gamma\overset{\pi}{\to}\mathrm{Isom}^+(\mathbb{H}^2)\to 0.$$
In fact this is even a central extension being $\mathrm{Isom}(\mathbb{E})\cong\mathbb{R}\triangleleft\Gamma$ 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]][1] for more details on Thurston geometries).

Hence, by the general theory of central extentions, there exist a map $\Phi:\mathrm{Isom}^+(\mathbb{H}^2)^2\to\mathbb{R}$ such that $\Gamma$ can be described in the following way $\Gamma\cong(\mathbb{R}\times\mathrm{Isom}^+(\mathbb{H})^2,\circ_\Phi)$ 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 $c:\mathrm{Isom}^+(\mathbb{H}^2)\to\Gamma$ is a section (of sets!) of the projection $\pi$ then for two general elements $f,g\in\mathrm{Isom}^+(\mathbb{H}^2)$ 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][2]] IV.3 for more details about extentions with abelian kernel).

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


Here is my attempt. First of all let us be aware of this nice series of equivalences of Riemannian manifolds $$\mathrm{Isom}^+(\mathbb{H}^2)\cong PSl(2,\mathbb{R})\cong U\mathbb{H}^2$$
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 geodetic.

It follows in particular that $df$ send 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 $\mathrm{Isom}^+(\mathbb{H}^2)\to\Gamma$ defined by $f\to df$.
Now let us take to isometries $f,g\in\mathrm{Isom}^+(\mathbb{H}^2)$ 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 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 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 coodinates or parallel transports. So I wonder **is there any other more straightforward way to find this (or another) explicit descripion?** Maybe it could involve some Lie algebra work that I am not aware of.


  [1]: http://homepages.warwick.ac.uk/~masgar/Teach/2012_MA4J2/geometry.pdf
  [2]: https://link.springer.com/book/10.1007/978-1-4684-9327-6