5
$\begingroup$

How can we prove that the moduli space,$M_{g}(n)$, of genus $g$ Riemann surface with $n$ boundary components is homotopy equivalent to $M_{g,n}$, that is ,the moduli space of genus $g$ Riemann surface with $n$ punctures? Thanks! (It is very intuitive, but it seems that I can't make it)

$\endgroup$
1
  • $\begingroup$ It would help to change your title to something more descriptive. $\endgroup$ Jul 6, 2010 at 19:17

1 Answer 1

8
$\begingroup$

A compact Riemann surface of genus $g$ with $n$ boundary components has a unique realization as a hyperbolic surface with geodesic boundary. One may see this by reflecting through the boundary and uniformizing. The uniqueness of the uniformization implies it is invariant under reflection, and therefore the fixed point set is geodesic.

Thus, the moduli space of genus $g$ Riemann surfaces with $n$ boundary components is equivalent to the space of hyperbolic surfaces with totally geodesic boundary. One may now insert a punctured disk into each boundary component, to obtain a Riemann surface with punctures. I don't know of a canonical way to do this, but for example for a boundary component of length $l$, one may attach isometrically the boundary of a punctured Euclidean disk of circumference $l$. The important thing is that this gluing only depends on $l$, and that it induces a conformal structure on the punctured surface. This gives a map between the spaces. Since the mapping class groups are the same, it induces a homotopy equivalence (in the category of orbifolds). Of course, there are some technical details one must carry out to make this argument rigorous. There are several other ways to fill in a punctured disk.

Another possible approach is to use the Weil-Petersson metric on moduli space. One can take the WP nearest point in the Deligne-Mumford compactification of moduli space, which is finite distance away since the WP metric is incomplete. Because the metric is CAT(0), a unique nearest point exists.

$\endgroup$
4
  • $\begingroup$ Hi,Agol,thanks! I have a little question, why do we choose inserting a disk WITH puncture, for example, if we just insert a disk, then can we get something? $\endgroup$
    – HYYY
    Jul 6, 2010 at 3:23
  • 2
    $\begingroup$ If you insert disks, then you get the closed surface of genus g. This factors through the canonical map from $M_{g,n}$ which fills in punctures. The fiber here is more complicated though, this is related to the Birman exact sequence. $\endgroup$
    – Ian Agol
    Jul 6, 2010 at 3:44
  • $\begingroup$ Very small question:Is the moduli space of Riemann surface with genus $g$ and $n$ punctures the same as moduli space of Riemann surface with genus $g$ and $n$ marked points?Thanks! $\endgroup$
    – HYYY
    Jul 6, 2010 at 17:15
  • 1
    $\begingroup$ Yes, puncture at the marked points to go one direction, fill in to go the other. The point is that there is a canonical way to fill in a puncture, since there is a neighborhood conformally equivalent to a punctured surface. $\endgroup$
    – Ian Agol
    Jul 6, 2010 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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