# An explicit description of $\operatorname{Isom}(\widetilde{\operatorname{Sl}_2})$

$$\DeclareMathOperator\Sl{Sl}\DeclareMathOperator\PSl{PSl}\DeclareMathOperator\Isom{Isom}$$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] 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] 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.

• @LSpice It's SL, rather than Sl. PSl is a bit absurd, since these are 3 initials (Projective Special Linear)
– YCor
Jan 5 at 18:26
• @LSpice Thanks for all the fixing! $\Gamma$ is indeed the indentity componend of $\operatorname{Isom}(\widetilde{SL_2})$ Jan 5 at 19:42
• @LSpice Even if reasonably common, I think Sl, Gl are wrong analogues of Sp (which is correct since p is not an initial).
– YCor
Jan 5 at 20:19
• One can easily describe $\mathrm{Isom}^+(\widetilde{\mathrm{SL}_2})$ as central product of $\widetilde{\mathrm{SL}_2}$ and $\mathbf{R}$, and there's a analogous hardly more complicated description for the whole isometry group. Is this what you're looking for?
– YCor
Jan 5 at 20:21
• @YCor could you explain this better? as far as I know all isometries of $\widetilde{\operatorname{SL}_2}$ are orientation preserving, the identity component of which comes from $\operatorname{Isom}^+(\mathbb{H}^2)$ and the other component from $\operatorname{Isom}^-(\mathbb{H}^2)$ as a central product with $\mathbb{R}$. Jan 5 at 20:29

## 1 Answer

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).