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Why is the Moduli of smooth curves of a fixed genus not compact/proper?

I know that there is a compactification using stable curves. But is it easy to see that the Moduli of smooth curves is not compact?

Following Sam Need's comment, I should write that I am comfortable with the Moduli of smooth projective curves i.e., as an algebraic variety.

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    $\begingroup$ If one assumes the classical result that, for genus $g\geq2$, the moduli space of smooth curves is "the same" as the space of hyperbolic Riemann surfaces of genus $g$, then the noncompactness is easy to understand. A classical paper by Milnor shows that when one tends to infinity in the moduli space, the systole of the surface tends to zero (and vice versa). Of course this still leaves us with the task of actually constructing families of surfaces with systole tending to zero; this is a bit more complicated than the case of elliptic curves. $\endgroup$
    – Mikhail Katz
    Commented Oct 14, 2023 at 19:00
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    $\begingroup$ You need to improve the question by giving us a few hints about the background you want to assume. Which version of moduli space are you working with? What are its points, its topology? What background references are you comfortable with? And so on. $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 12:58
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    $\begingroup$ There are already plenty of answers/comments addressing your question. But for general intuition, if you have some moduli space which parameterizes objects which can degenerate to an object outside the class you're dealing with, then this indicates the moduli space is not compact. $\endgroup$
    – Donu Arapura
    Commented Oct 15, 2023 at 15:02
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    $\begingroup$ The easiest and most classical way to detect degeneration is by looking at a particular nonzero period and observe that it converges to zero as you move in a family of Riemann surfaces. Examples are easy to arrange by looking at hyperelliptic curves or at conformal connected sums of tori. $\endgroup$
    – Moishe Kohan
    Commented Oct 15, 2023 at 21:14
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    $\begingroup$ The closing votes on this question are unreasonable. It is easy to say that the result is true because the bigshots proved it long ago, but it does appear to be a nontrivial task to give an accessible explanation of this. Giving accessible explanations is of interest to research mathematicians and therefore legitimate on MO. $\endgroup$
    – Mikhail Katz
    Commented Oct 16, 2023 at 12:07

4 Answers 4

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Here is yet another approach. If $M_g$ were proper, its image in $A_g$ under the Torelli map would be closed. But products of lower dimensional Jacobians are in the closure. This last result is attributed to Matsusaka and Hoyt by Mumford in the article entitled "Further comments on boundary points" here: https://www.jmilne.org/math/Documents/woodshole.pdf

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Seeing that there are already two answers that are not entirely convincing, I decided to copy my comment to an answer. If one assumes the classical result that, for genus g≥2, the moduli space of smooth curves is "the same" as the space of hyperbolic Riemann surfaces of genus g, then the noncompactness is easy to understand. A classical paper by Milnor shows that when one tends to infinity in the moduli space, the systole of the surface tends to zero (and vice versa). Of course this still leaves us with the task of actually constructing families of surfaces with systole tending to zero; this is a bit more complicated than the case of an elliptic curve (i.e., a 2-torus).

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    $\begingroup$ The problem with this answer is that it requires two more pieces of work from the original asker: why is this "new" moduli space homeomorphic to the old one? And (as you mention): why are there hyperbolic surfaces with systoles tending to zero? $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 12:45
  • $\begingroup$ One way of doing it is by looking at the particular case of the space of tori with one (geodesic) circle component, and showing that the boundary circle can be arbitarily short. Then one can double this torus with boundary to get a genus 2 surface with small systole. Any work with hyperbolic surfaces is going to involve more calculation than the case of elliptic curves, but it is easy enough to find papers dealing with the case of torus with boundary. @SamNead $\endgroup$
    – Mikhail Katz
    Commented Oct 15, 2023 at 12:51
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    $\begingroup$ Yes, I know that we can metrically pinch a curve, and you know that too.... but it is not going to help the original poster. (And, if we are going to pinch a curve then I prefer the family I gave in my answer. You can prove, using the euclidean metric, that there is a large modulus annulus about the points $z_{2g + 1}$ and $z_{2g + 2}$. Since moduli of annuli are conformal invariants, my family exits). $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 12:56
  • $\begingroup$ @SamNead, do you mean "exists"? $\endgroup$
    – Mikhail Katz
    Commented Oct 15, 2023 at 13:00
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    $\begingroup$ It exists and it exits (moduli space). The latter is the hard part. $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 13:02
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Is Arbarello's 74 paper (specifically, theorem 3.27 there) Weierstrass points and moduli of curves an easy enough argument ?

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Fix $g \geq 1$, a genus. Fix $Z = (z_i)_{i = 1}^{2g + 2}$, a collection of distinct point in the plane. Define $C(Z)$ to be the curve in the complex projective plane obtained from the solutions of $y^2 = \prod (x - z_i)$. So $C(Z)$ is hyperelliptic. Fixing the first $2g + 1$ points of $Z$, we move $z_{2g + 2}$ towards $z_{2g + 1}$. This gives a non-constant family of curves. The limiting curve is stable but not smooth, thus the moduli space is not compact.

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    $\begingroup$ "thus the moduli space is not compact": that's exactly the question — see @Will Sawin's comment. $\endgroup$
    – abx
    Commented Oct 15, 2023 at 8:24
  • $\begingroup$ @abx - I am giving a simple (well, simpler) example. I assume that you agree that this family exits moduli space. So, help me help the questioner. How would you prove that the family exits? (Without talking about metrics, or conformal invariants, which they presumably don't know about...) $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 12:57
  • $\begingroup$ I gave a simple construction in my comment. Anyway, I think the problem is not the construction of such a family, but the proof that its existence implies that the moduli space is not compact. $\endgroup$
    – abx
    Commented Oct 15, 2023 at 13:44
  • $\begingroup$ I wrote "exits" not "exists". The phrase "exits" is apparently not standard, but I meant it to mean "leaves every compact set" which in turn means "is not contained in any compact set". $\endgroup$
    – Sam Nead
    Commented Oct 15, 2023 at 16:01
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    $\begingroup$ Sorry, indeed I misread. I don't see an elementary way to prove that the family exits. As I wrote in a comment, I think one can give an argument based on the Picard-Lefschetz formula, but it requires some care, and is certainly not elementary. $\endgroup$
    – abx
    Commented Oct 15, 2023 at 17:18

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