Let $\overline{\mathcal{M}}_g$ be the moduli stack of genus $g$ stable (nodal) curves and let $\overline{M}_g$ denote its coarse moduli space. In 1969, in the paper "The irreducibility of the space of curves of given genus", Deligne and Mumford constructed and showed that $\overline{\mathcal{M}}_g$ is a smooth Deligne-Mumford stack, and proved that $\overline{M}_g$ is an irreducible projective variety. Although $\overline{\mathcal{M}}_g$ is smooth, $\overline{M}_g$ may have finite quotient singularity.

When is $\overline{M}_g$ smooth?

The simplest case is $\overline{M}_0$, which is a point and hence smooth by definition. My question is then equivalent to the following: for which $g>0$ is $\overline{M}_g$ smooth? Is it never smooth?

Edit 1:
I forgot to mention the base scheme is $\text{Spec }\mathbb{C}$.

  • 8
    $\begingroup$ $M_g$ is not smooth for $g>1$, because of the presence of curves with extra automorphisms. $\endgroup$
    – abx
    Nov 22, 2019 at 7:25
  • 2
    $\begingroup$ Yes. This goes back to H. Rauch, The singularities of the modulus space, Bull. Amer. Math. Soc. 68 (1962), 390–394. $\endgroup$
    – abx
    Nov 22, 2019 at 8:23
  • 2
    $\begingroup$ No, I am talking here about $M_g$, not $\overline{M}_g$. But that's fairly trivial: take a smooth curve with an automorphism fixing a point, and add an elliptic tail. $\endgroup$
    – abx
    Nov 22, 2019 at 9:53
  • 9
    $\begingroup$ This is obvious, just take hyperelliptic curves. $\endgroup$
    – abx
    Nov 22, 2019 at 14:21
  • 7
    $\begingroup$ Sorry, this is enough for me. Look at the reference I gave. $\endgroup$
    – abx
    Nov 22, 2019 at 16:05

2 Answers 2


A more detailed description of the singular locus of $\mathrm{M}_g$ is as follows.

Theorem. Let $\mathrm{C}$ be a smooth curve of genus $g$.

If $g=2$, then $[\mathrm{C}]$ is a singular point of $\mathrm{M}_2$ if and only if $\mathrm{C}$ is given by $y^2=x^6-x$.

If $g=3$ and $\mathrm{C}$ is not hyperelliptic (resp. hyperelliptic), then $[\mathrm{C}]$ is a singular point of $\mathrm{M}_3$ if and only if $\mathrm{Aut(C)}$ is nontrivial (resp. $\mathrm{Aut(C)}$ is not $\mathbf{Z}/2\mathbf{Z}$).

If $g\geqslant 4$, then $[\mathrm{C}]$ is a singular point of $\mathrm{M}_g$ if and only if $\mathrm{Aut(C)}$ is nontrivial.

An algebro-geometric reference for this result is this paper by H. Popp.

  • $\begingroup$ This is a better answer than mine. Can you add the case $g=1$ to make it a complete answer? $\endgroup$ Dec 2, 2019 at 17:16
  • $\begingroup$ @YuhangChen: we have $\mathrm{M}_1\simeq\mathbf{A}^1$ (in particular there are no singularities), but I think it is most natural to restrict the discussion to $g\geqslant 2$. $\endgroup$
    – ssx
    Dec 4, 2019 at 19:19
  • $\begingroup$ Yes, and $\overline{M}_1 \cong \mathbb{P}^1$, right? $\endgroup$ Dec 5, 2019 at 11:54
  • $\begingroup$ @YuhangChen: Sure $\endgroup$
    – ssx
    Dec 7, 2019 at 17:28

The question is mostly answered by abx's comments. Here I'm posting a self-contained answer so that it will be helpful for the others.

Case 1: $g=0$

$M_0 = \overline{M}_0$ is just a point and is smooth, as already mentioned in the question.

Case 2: $g=1$
Now there is a subtlety. Actually $\overline{\mathcal{M}}_1$ is not a Deligne-Mumford stack because every elliptic curve has an infinite group of automorphisms, i.e., translations. It's more meaningful to talk about $\mathcal{M}_{1,1}$, the moduli stack of curves with genus $g=1$ and one marked point (we don't want to forget the group structure of elliptic curves, which has the marked point as the identity) and its compactification $\overline{\mathcal{M}}_{1,1}$. We know the coarse moduli space $\overline{M}_{1,1}$ is isomorphic to the weighted projective line $\mathbb{P}(4,6)$ and hence is smooth.

Case 3: $g\geq2$

By Theorem 1 in the paper "The singularities of the modulus space" by Rauch in 1962, we know $M_g$ must have some point of singularity and hence its compactification $\overline{M}_g$ is singular.


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.