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

  • 2
    $\begingroup$ @LSpice It's SL, rather than Sl. PSl is a bit absurd, since these are 3 initials (Projective Special Linear) $\endgroup$
    – YCor
    Jan 5 at 18:26
  • $\begingroup$ @LSpice Thanks for all the fixing! $\Gamma$ is indeed the indentity componend of $\operatorname{Isom}(\widetilde{SL_2})$ $\endgroup$
    – Dinisaur
    Jan 5 at 19:42
  • $\begingroup$ @LSpice Even if reasonably common, I think Sl, Gl are wrong analogues of Sp (which is correct since p is not an initial). $\endgroup$
    – YCor
    Jan 5 at 20:19
  • $\begingroup$ 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? $\endgroup$
    – YCor
    Jan 5 at 20:21
  • $\begingroup$ @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 $\mathbb{R}$. $\endgroup$
    – Dinisaur
    Jan 5 at 20:29

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.