1
$\begingroup$

I am reading this note on super-Riemann surfaces. In the second paragraph of section 7.4.1 (page 87), there is a statement that I am trying to understand:

The compactified moduli space of closed oriented Riemann surfaces has no boundary.

However, as it is mentioned, the compactified moduli space of surfaces with boundary is an orbifold with boundary.

Can someone please elaborate these points and that why the compactified moduli space of closed oriented Riemann surfaces does not have boundaries and is just compact? A good pedagogical reference on the Deligne-Mumford compactification is highly appreciated.

Improve this question
$\endgroup$
9
  • 4
    $\begingroup$ In some sense Deligne and Mumford's paper is a reference for this, but I guess probably not in the sense you want... $\endgroup$
    – Will Sawin
    Commented Nov 14, 2018 at 2:26
  • 1
    $\begingroup$ why the compactified moduli space of closed oriented Riemann surfaces does not have boundaries and is just compact As written, this is a tautology --- it is compact because it is the compactification of something. Maybe a better way to ask the question is why the Deligne--Mumford compactification is modular, that is, why the points added in actually correspond to something geometric. A reference that I like is Harris--Morrison, Moduli of Curves. $\endgroup$
    – Bort
    Commented Nov 14, 2018 at 10:42
  • 1
    $\begingroup$ An other reference might be the paper by John Hubbard and Sarah Koch called An analytic construction of the Deligne-Mumford compactification of the moduli space of curves projecteuclid.org/download/pdf_1/euclid.jdg/1406552251 $\endgroup$
    – Maxime Scott
    Commented Nov 14, 2018 at 15:31
  • 3
    $\begingroup$ Your intuition of taking an open disk and adding a (real codimension one) boundary is not the right picture here. Adding boundary strata to the moduli space of curves is adding a complex codimension 1 strata. A better analogy is adding a point to $\mathbb{C}$ to get the Riemann sphere $\mathbb{CP}^1\cong S^2$ which is a compact manifold without boundary. $\endgroup$
    – Jim Bryan
    Commented Nov 15, 2018 at 0:55
  • 1
    $\begingroup$ @QGravity. Maybe you can think about it this way. There is a real parameter associated to a boundary component of a Riemann surface with boundary (I believe given by the length of the loop using the (unique) hyperbolic metric on the Riemann surface with boundary). This leads to a real co-dimension 1 boundary in the compactification of the moduli space. Whereas the complex structure on a Riemann surface without boundary is described solely by complex parameters (3g-3 of them) this leads to a complex co-dimension 1 boundary in the compactification of the moduli space. $\endgroup$
    – Jim Bryan
    Commented Nov 16, 2018 at 23:13

0

Reset to default

You must log in to answer this question.