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Work over an algebraically closed field ($\mathbb C$, if you prefer) and fix $g\geq 2$. By $M_g$ I mean, of course, the moduli space of smooth projective curves of genus $g$. I know it is in general not proper.

Question 1. Is it non proper for every $g$? And if not, what is an example of a proper $M_g$?

The motivation for the above question is an argument I read on some lecture notes. This is how I understood it: let us consider the (injective) Torelli morphism $M_g\to A_g$ (defined by $[C]\mapsto [J(C)]$), where $A_g$ is the moduli space of PPAVs. Then, if $M_g$ were proper, its image would coincide with the Torelli locus $T_g\subset A_g$. But $T_g$ contains products of PPAVs, and no such product can be the Jacobian of a curve. Contradiction.

Now I'm lost.

Question 2,3. How to see that $T_g$ contains products of PPAVs? And why a product is not the Jacobian of any curve?

Also, in passing, where can I find a clean definition of the theta divisor on a PPAV? I am confused about this point.

Thanks in advance.

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See page 3 and page 4 in Dan Edidin's amazing notes on the moduli space of curves: – Harry Nov 5 '12 at 23:02
Another related reference is Matthieu Romagny's notes on models of curves:… – Harry Nov 5 '12 at 23:03
@Harry: Thanks for the references! The first one is very nice. There is an example showing that $M_3$ is not complete, but I still wonder if the argument in my question says that $M_g$ is never complete... – Brenin Nov 6 '12 at 21:49
up vote 4 down vote accepted

Question 1: $M_g$ ($g\ge 2$) is never proper. A quick explanation is that $M_g$ is irreducible and has a compactification whose boundary is given by the singular stable curves of genus $g$. The boundary is non-empty because for any $g\ge 2$, there exists a singular stable curve of genus $g$ (attach a smooth projective curve of genus $1$ transversally to a smooth projective of genus $g-1$), so $M_g$ is different from its compactification, hence is not proper.

A more elementary way to see the non-properness is to use the valuative criterion. Consider the hyperelliptic curve of genus $g$ defined by $$ y^2=(x^2-t)(x^{2g}-1)$$ over $\mathbb C((t))$. Then over any finite extension of $\mathbb C[[t]]$ (necessarily of the form $\mathbb C[[t^{1/n}]$), the morphism $$\mathrm{Spec}( \mathbb C((t^{1/n})))\to M_g $$ can't extend to $\mathrm{Spec}(\mathbb C[[t^{1/n}]])$, because the curve $$ y^2=(x^2-t_n)(x^{2g}-1), \quad t_n=t^{1/n}$$ has bad reduction. We have to prove this for all finite extension of $\mathbb C[[t]]$ because $M_g$ is a coarse moduli space.

Question 2: If $C_1, ..., C_n$ are projective smooth curves of positive genus, form a chain with $C_i$ intersecting transversally $C_{i+1}$ at exactly one point. The stable curve $C$ obtained this way can be deformed to a smooth projective $X$. The Jacobian $C$ (equal to the product of the Jacobians of the $C_i$'s) deformes to the Jacobian of $X$. This shows that $\prod_i \mathrm{Jac}(C_i)$ is a limit of Jacobians, hence belongs to $T_g$.

For the rest of the questions, I don't have a satisfactory answer to offer.

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Thank you for your super concrete answer! When you say that the curve over $\mathbb C((t_n))$ has bad reduction do you mean it has a singular point over that field? I am sorry I know this term only for elliptic curves. – Brenin Sep 8 '13 at 20:06
@atricolf: sorry for not being precise. In fact the curve is defined over the DVR $\mathbb C[[t]]$ and the closed fiber is singular. If $M_g$ were proper, then the initial curve would have good reduction over some $\mathbb C[[t_n]]$ by the coarse moduli space interpretation. In fact, the equation $y^2=(x^2-t)(x^{2g}-1)$ defines a singular stable curve, this imples that it can't have good reduction over any extension. – Cantlog Sep 8 '13 at 21:59
Today I have been thinking about your answer and I almost get it. You have been very clear but I need a final help. For me "coarse" means that $\mathbb C$-points in $M_g$ correspond to isomorphism classes of smooth curves over $\mathbb C$, and there is a universal natural transformation $\mathfrak M_g\to \textrm{Hom}(-,M_g)$ where $\mathfrak M_g$ is the moduli functor. How do you use this to prove that $f:M_g\to\textrm{Spec}\,\mathbb C$ is not proper? Summing up, I cannot visualize the usual "square" in the valuative criterion. – Brenin Sep 9 '13 at 16:27
The usual square doesn't work for a <b>coarse</b> moduli space, because the latter is ... not fine. This means that a morphism $S\to M_g$ does not define a smooth curve over $S$, but only over some finite extension of $S$. You can search for a survey text of Matthieu Romagny on models of algebraic curves. Ah I see Harry already gave the link. – Cantlog Sep 9 '13 at 19:10

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