6
$\begingroup$

I'm reading the book from Jensen's "Surfaces in Classical Geometries". Could anyone help me understand why $\mathbf{H}(3)$ is diffeomorphic to $\mathbf{SL}\left( 2,\mathbf{C}\right) \mathbf{% /SU}\left( 2\right) $?

The following is a print.

enter image description here

$\endgroup$
  • $\begingroup$ @Stephen Hi Stephen, I put another print explaining your question. $\endgroup$ – Mancala Aug 22 '18 at 0:40
3
$\begingroup$

This follows from the general statement that if a lie group $G$ acts transitively on a space $X$, and if given $x\in X$ we define $G_x:=\{g\in G \mid gx=x \}$, then $X\cong G/G_x$. This can be seen as a generalization of the orbit-stabalizer theorem, as when $X$ is a finite set and $G$ is a finite group, $|G/G_x|=|G|/|G_x|$.

We wish to produce a diffeomorphism between $G/G_x$ and $X$, and so we should start by having a map. We note that the map $G/G_x\to X$ sending $[h]$ to $hx$ is well defined, and one can prove that it is smooth and bijective. I'm not immediately seeing why the inverse map is smooth, but I will edit if I can find/think of a simple explanation.

$\endgroup$
  • 3
    $\begingroup$ "I'm not immediately seeing why the inverse map is smooth..." The point is that you can check this by proving that the differential is an isomorphism at each point. This reduces to checking it at $x$, which amounts to showing that the kernel of the differential there is precisely the Lie algebra of the isotropy group. Anyway, +1. $\endgroup$ – Stephen Aug 22 '18 at 1:20

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.