Consider the moduli of smooth curves $M_{g,n}$ (genus $g$, $n$ marked points) and its Deligne-Mumford compactification $\overline{M}_{g,n}$ of stable nodal curves (genus $g$, $n$ marked points). This is usually only defined for $2g-2+n>0$. It seems the point of this condition is that $M_{g,n}$ has no infinite automorphisms. (For example for genus 0, we need to specify three marked points in order to pin down a unique automorphism of the curve.) Stability of a nodal curve can then be defined to mean that the automorphism groups are finite, which can be characterized in terms of the irreducible components of the normalization.

Here's my question, is it possible to give a useful definition of $\overline{M}_{g,n}$ when $2g-2+n \le 0$? I guess defining "stable" in terms of finite automorphisms is a bad idea in this case because then $M_{g,n}$ will not be contained in side $\overline{M}_{g,n}$. So is there another definition that will still give us a proper algebraic stack (possibly in the sense of Artin, i.e., just allow infinite automorphisms)?

Any references are appreciated, the ones I looked at all have this numerical condition on $g,n$. I guess it is because algebraic stacks in the Artin sense are too complicated to work in practice. But I am still curious about it.

  • $\begingroup$ One big issue when considering e.g. $(g,n)=(1,0)$ is that the corresponding category fibered in groupoids (given by genus $1$ curves and pullback squares) is not even a stack, i.e. descent is not always effective! Examples are due to Raynaud: see e.g. arxiv.org/abs/1501.04304 $\endgroup$
    – Jef
    Commented Feb 3, 2021 at 14:09

1 Answer 1


A keyword to google for is prestable curves. They are defined just as stable curves except the conditions that ensure finite automorphism groups are removed. There is a stack $\mathfrak M_{g,n}$ of $n$-pointed prestable curves of genus $g$. Stabilization defines a map $\mathfrak M_{g,n} \to \overline M_{g,n}$ whenever $2g-2+n>0$, but the stack makes sense for general $g,n$. Example: $\mathfrak M_{0,2}$ has an open substack parametrizing smooth genus zero curves with two marked points, which is isomorphic to $\mathrm B\mathbb G_m$. The stack $\mathfrak M_{g,n}$ is smooth of dimension $3g-3+n$ and locally of finite type; even though it is not separated and not Deligne-Mumford it is not so bad.

As pointed out by Jef in a comment there is an issue of descent when $g=1$, $n=0$. For this case to work out correctly, in the definition of a family $X \to S$ of prestable curves one must allow $X$ to be an algebraic space. With this definition of prestable curves there is a perfectly well behaved stack $\mathfrak M_{1,0}$, containing as an open substack the relative classifying space of the universal elliptic curve over $M_{1,1}$.

  • $\begingroup$ Thanks, this is interesting. So basically you suggest not to impose any stability condition at all and look at all nodal curves (with geometrically connected fibers). I'm worried about properness though since ideally we'd want to have a compactification of $M_{g,n}$. From your answer, I infer that there is no standard solution to this problem in the literature? $\endgroup$ Commented Feb 4, 2021 at 12:56
  • $\begingroup$ Also is there any reference for the second paragraph (for the construction where X is an algebraic space) ? $\endgroup$ Commented Feb 4, 2021 at 13:00
  • $\begingroup$ No, the stack is not proper, since it is not separated. Note that a separated stack has proper stabilizers, so no proper stack can contain $M_{0,2}=\mathrm B \mathbb G_m$, so it's not reasonable to hope for a proper stack here, either. $\endgroup$ Commented Feb 5, 2021 at 8:49
  • $\begingroup$ For the second paragraph I don't have a reference off hand, but it's not something deep. Genus one curves don't necessarily have effective descent because they don't have a natural polarization. In the world of algebraic spaces descent is effective by definition, and the problem goes away. $\endgroup$ Commented Feb 5, 2021 at 8:50
  • $\begingroup$ Thanks @Dan Petersen, that comment answers the question definitively; no compactification of $M_{0,2}$ exists. $\endgroup$ Commented Feb 8, 2021 at 9:04

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