2
$\begingroup$

My question is from A primer on mapping class group, p.295:

enter image description here I can see $X=\Delta/\Gamma$ has an induced hyperbolic structure, but why conversely any such hyperbolic structure gives a complex structure on $X$? Also for the bijection, how do we know these induced maps are inverse to each other?

$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

The key fact which answers both of your questions is that the group of orientation preserving hyperbolic isometries of $\Delta$ is identical to the group of biholomorphic automorphisms of $\Delta$.

So, just as any Riemann surface structure on $S_g$ is the quotient of $\Delta$ by a group of bilomorphic automorphisms and hence has an induced hyperbolic structure, similarly any hyperbolic structure on $S_g$ is the quotient of $\Delta$ by a group of orientation preserving hyperbolic isometries and hence has an induced Riemann surface structure.

$\endgroup$
2
  • $\begingroup$ Thanks, is there a reference for the proof the key fact you mentioned? Indeed, I was trying to use the bijection to prove the key fact in the case of any genus case. $\endgroup$
    – Kat
    Commented Jun 30, 2020 at 22:48
  • 2
    $\begingroup$ There are two theorems, saying that the two groups are both identical to the group of Möbius transformations, i.e. to fractional linear transformations of the appropriate form. For the biholomorphic automorphisms, this is done in a complex analysis textbook. For the hyperbolic isometries, I would look in a hyperbolic geometry textbook. $\endgroup$
    – Lee Mosher
    Commented Jul 1, 2020 at 3:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .