# Why not add cuspidal curves in the moduli space of stable curves?

Let $$\mathcal{M}_{g,n}$$ be the moduli space (stack) of stable smooth curves of genus $$g$$ with $$n$$ marked points over $$\mathbb{C}.$$ It's known that by adding stable nodal curves to $$\mathcal{M}_{g,n}$$, the resulting space $$\overline{\mathcal{M}}_{g,n}$$ is compact. But why is it so? For example, consider the following family of elliptic curves in $$\mathcal{M}_{1,1}$$, $$y^2=x^3+t,$$ where $$t \in \mathbb{C^*}$$. Then this family of elliptic curves degenerates into the cuspidal cubic curve $$y^2 = x^3.$$ So why is $$\overline{\mathcal{M}}_{1,1}$$ compact?

• I believe this has to do with the fact that $\mathscr M_{g,n}$ is stacky and so the valuative criterion doesn't hold on the nose. You might need to take an etale extension of your curve before you can extend the map from the generic point. In this particular case, this is related to the fact that the cuspidal singularity is unstable: After an unramified base change of $\mathbb C[[t]]$, the family will become either a node or good reduction. – Asvin Sep 25 '19 at 18:03
• Also, a cusp would have a $\mathbb{C}^*$-worth of automorphisms, while the usual moduli spaces of stable curves are designed to be as rigid as possible and in fact are Deligne-Mumford. – Qfwfq Sep 25 '19 at 20:57
• @Qfwfq Do you mean the cuspidal curve $y^2=x^3$ has an automorphism group $\mathbb{C^*}$, even with a marked point at the cusp? – Yuhang Chen Sep 26 '19 at 0:10
• @Asvin I don't follow your last sentence. What do you mean by saying "the cuspidal singularity is unstable"? – Yuhang Chen Sep 26 '19 at 0:15
• @yYuhangChen i was basically referring to the last part of David's answer where he talks about how after adding a 6th root of t, you get an isotrivial (hence good redn) curve. – Asvin Sep 26 '19 at 1:09

If you add cuspidal curves, then $$\overline{\mathcal{M}}_{1,1}$$ will no longer be separated, which is the scheme/stack analogue of Hausdorff. Specifically, consider the families $$y_1^2 = x_1^3 + t^6 \ \mbox{and}\ y_2^2 = x_2^3 + 1$$ (so the second family is a constant family with no $$t$$-dependence). For all nonzero $$t$$, they are isomorphic by the change of variables $$y_1 = t^3 y_2$$, $$x_1 = t^2 x_2$$. So they should give the same map from $$\mathbb{C}^{\ast}$$ to moduli space (namely, a constant map). If the cuspidal curve corresponded to a point of moduli space, then this map would have two limits.
The situation is similar with regard to the family $$y_3^2 = x_3^3 + t$$ that you consider. On the level of coarse moduli spaces, this family also corresponds to a constant map $$\mathbb{C}^{\ast} \to \overline{\mathcal{M}}_{1,1}$$. The subtlety is that the families $$y_2^2 = x_2^3 + 1$$ and $$y_3^2 = x_3^3 + t$$ are not isomorphic over $$\mathrm{Spec}\ \mathbb{C}[t^{\pm 1}]$$, but only over the cover where we adjoin a $$6$$-th root of $$t$$. The definition of coarse moduli space is meant exactly to accommodate families which are not isomorphic but become isomorphic after a finite cover.
• Thanks for the enlightening answer (especially the last paragraph)! Only after seeing your example, then I realize that the family of curves I considered have a same $j$-invariant, which is zero. So this "zero map" is the constant map $\mathbb{C}^* \to \overline{\mathcal{M}}_{1,1}$ you mentioned, on the level of coarse moduli spaces. – Yuhang Chen Sep 26 '19 at 2:24
• @Qfwfq If $S$ is reduced and $X \to S$ and $Y \to S$ are two families which become isomorphic after being pulled back along a finite surjective cover $T \to S$, then the two families correspond to the same map from $S$ to the coarse moduli space. – David E Speyer Sep 27 '19 at 3:01
• I was under the impression that something like a converse held as well -- that if $X \to S$ and $Y \to S$ were two families corresponding to the same map from $S$ to the coarse moduli space, that there was a finite surjection $\pi: T \to S$ such that $\pi^{\ast} X \to T$ and $\pi^{\ast} Y \to T$ are isomorphic. But having failed to find either a proof or reference, I am coming to suspect I'm wrong. – David E Speyer Sep 27 '19 at 3:01