# Coarse Moduli space of plane cubics

I am studying Joe Harris' Algebraic Geometry: A First Course, the section on Moduli Spaces, pg 278. I am stuck in a subtle point. Harris gives on p 279 an argument why there is no coarse moduli space of plane cubics and I not understand this argument:

Example 21.12. Plane Cubics
The fundamental example of a moduli space is one we encountered before in Example 10.16, that of plane cubics. Based on our previous discussion, we see that even a coarse moduli space does not exist for plane cubics. This is due to the various inclusions among closures of orbits of the action of $$PGL_3 \ K$$ ($$K$$ field with $$char(K) \neq 2,3$$)n on the space $$\mathbb{P}^9$$ of plane cubics. For example, if $$\mathcal{M}$$ is the set of isomorphism classes of plane cubics, $$\mathcal{M}$$ will have one point $$p$$ corresponding to irreducible plane cubics with a node, and another point $$q$$ corresponding to cuspidal cubics. But by what we saw in Example 10.16, the point $$q$$ would have to lie in the closure of the point $$p$$!

Explantions on terminology & references:

1. the coarse moduli space is informally introduced at pges 278/279 as: Let $$\{X_{\alpha}\}$$ a collection of certain varieties (e.g. like genus $$g$$ curves, etc.). Then a variety $$\mathcal{M}$$ is called a coarse moduli space with respect this collection $$\{X_{\alpha}\}$$ if
• as underlying set $$\mathcal{M}$$ is bijective to the set of isomorphism classes of $$X_{\alpha}$$.

• for any reduced family $$\pi: \mathcal{V} \to B$$ (ie a flat surjection such that the general fiber is reduced) such that every fiber $$\pi^{-1}(b) = X_b$$ is a member of the collection $$\{X_{\alpha}\}$$ the canonical set-theoretic map

$$\phi_{\pi}: B \to \mathcal{M}$$

given by sending each point $$b \in B$$ to the point of $$\mathcal{M}$$ representing the isomorphism class $$[X_b]$$ of the fiber $$X_b$$ over $$b$$ is a regular map.

1. Above is also refered to Example 10.16. ( How $$PGL_3 \ K$$ Acts on $$\mathbb{P}^9$$) (page 121). It states that for base field $$K$$ with $$char(K) \neq 2,3$$ there exist a natural action of $$PGL_3 \ K$$ on the space of cubic polynomials on $$\mathbb{P}^2$$.

The interesting result was that this action has some interesting closure relationships among diverse orbits under this action. For example the orbit consisting of smooth cubics with $$j$$-invariant $$j$$ contains in its closure the locus of cuspidal cubics (i.e., the orbit of cubics projectively equivalent to $$Y^2Z - X^3$$).

This also implies that as stated above that the closure of the orbit of point $$p$$ corresponding to irreducible plane cubics with a node contains point $$q$$ corresponding to cuspidal cubic.

Question: Why this observation heuristically indicates that if $$q \in \overline{ \{p \} } \subset \mathcal{M}$$ then $$\mathcal{M}$$ cannot be a coarse moduli space of plane cubics? (and therefore a coarse moduli space of plane cubics not exist)

• why the downvote? May 24, 2021 at 22:05
• I have upvoted this stupid downvote. It happens sometimes on this site, without any reason in your case. May 24, 2021 at 23:03

The idea is that the $$PGL_3 (K)$$ action preserves isomorphism classes, but since $$PGL$$ is non-proper, the orbit closures contain limiting cases that are not of the same isomorphism class as the general fiber. So e.g. $$y^2 = x^3 - tx$$ is a family of nodal cubics over $$\mathbb A^1$$ with central fiber a cusp. Since all nodal cubics are abstractly isomorphic to $$\mathbb P^1$$ with $$0$$ and $$\infty$$ identified, the image of $$\mathbb A^1$$ induced by this family in $$\mathcal M$$ is a pair of isolated points ($$0\mapsto [cusp]$$ and $$x \mapsto [node]$$ for $$x \in \mathbb A^1 \setminus 0$$), showing that the map is not even continuous.
• I think there is a subtle point here which should be clarified. We consider the space $\mathcal{M}$ as a variety, therefore it is endowed with Zariski topology which is not Hausdorff. Therefore we cannot separate arbitrary two points therefore and I'm not sure why it is immediately that your map $f: \mathbb{A}^1 \to \mathcal{M}, 0 \mapsto [cusp], x \neq 0 \mapsto [node]$ can't be continuous with resp underlying topologies. Clearly in analytic topology where continuity can be defined by sequences as $f: X \to Y$ is continuous May 25, 2021 at 23:21
• iff for every converging sequence $x_n \to x$ the image sequence converge as well $f(x_n) \to f(x)$ your map is discontinuous. May 25, 2021 at 23:21
• But here our topologies are coarser and we cannot separate $p=[node]$ and $q=[cusp]$ since we observed above that if $\mathcal{M}$ would be coarse moduli space then $q \in \overline{ \{p\} }$. Question: what are open sets in the image $f(\mathbb{A}^1)= \{ [node], [cusp] \} \subset \mathcal{M}$? In order to show that $f$ is discontinuous we have to show that $[cusp]$ is open in $f(\mathbb{A}^1)$, that is that there exist open $U \subset \mathcal{M}$ which contains $[cusp]$ but not $[node]$, but that's not obvious to me. May 25, 2021 at 23:21
• Suppose $f$ is continuous. We know from point-set topology that the continuous image of a connected set is connected. But since the nodal cubic and cuspidal cubic are each represented by closed points of $\mathcal M$, the set $\left\{[node],[cusp]\right\}$ is disconnected. Since $\mathbb A^1$ is connected, this is a contradiction. May 26, 2021 at 1:57
• If $\mathcal{M}$ would be our moduli space why $[node], [cusp]$ should be closed points of $\mathcal{M}$? That's exactly main reason for my confusion: a variety in general can of course have points (=primes) which are not closed. Why in this situation $[node], [cusp]$ you assume to be closed? May 26, 2021 at 23:19