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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)

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

1 Answer 1

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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.

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  • $\begingroup$ 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 $\endgroup$ May 25, 2021 at 23:21
  • $\begingroup$ 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. $\endgroup$ May 25, 2021 at 23:21
  • $\begingroup$ 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. $\endgroup$ May 25, 2021 at 23:21
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    $\begingroup$ 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. $\endgroup$ May 26, 2021 at 1:57
  • $\begingroup$ 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? $\endgroup$ May 26, 2021 at 23:19

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