# Why should moduli spaces be compact?

This is going to be a vague question. It seems like enumerative geometry people really like their moduli spaces to be compact (/projective). For example as soon as we mention $M_{0,n}$ we start talking about finding out a good choice for $\overline{M_{0,n}}$, etc. Could someone explain what exactly is the point of doing that, other than compact spaces being "nicer"? Maybe with a concrete example of using the compactness of a moduli space to conclude something nice? Long-windedness would be appreciated.

## 2 Answers

One-line answer: Moduli spaces need to be compact so that you have access to a robust intersection theory on them.

The basic idea is that the answers to enumerative problems are given by computing the number of intersection points of a collection of subvarieties of the moduli space that intersect transversely (i.e. are in a general position). In order to actually figure out how these subvarieties intersect, it is often desirable to deform one or more of them so that the collection lies in some special configuration. In order for this to go anywhere, you need an intersection theory which has the property that if a subvariety $V$ can be deformed to a subvariety $V'$, then for any other subvariety $W$, the intersection product $V \cdot W$ (whatever exactly that is) should be equal to the intersection product $V' \cdot W$.

Compactness is needed to insure that when you deform one variety to another, some of the intersection points don't move off "to infinity" and disappear. For example, suppose you want to intersect two lines in the regular affine plane. If they are in "general position", they will meet at a single point, but if you deform one line so that it is parallel to the other, they will no longer intersect at all (the intersection point has moved to the "line at infinity" in the projective plane - which is a natural compactification of the affine plane). So we've failed to produce an intersection theory that is invariant under deformations.

What takes a lot more work is to show that if you have a compact space, then you can construct an intersection theory that is invariant under deformations. A good introductory survey to intersection theory is in Appendix A of Hartshorne's Algebraic Geometry.

• Your answer feels like you want the ambient space that contains what the moduli is collecting to be compact, rather than the moduli space itself. Am I wrong to say that? – user27126 Nov 8 '12 at 18:09
• @Sanchez: there is no notion of "the ambient space." A moduli space does not come a priori with a preferred embedding into some other space. – Qiaochu Yuan Nov 8 '12 at 18:32
• @QiaochuYuan, Yes. But when you do intersection theory, you must work on an ambient space? What Professor Joyce talked about in his answer feels like something about the ambient space. Or maybe I should phrase my question as: How does the compactness of the ambient space (so as to avoid missing intersections at infinity) correspond to the compactness of the moduli space? – user27126 Nov 8 '12 at 18:35
• @Sanchez: The ambient space is the moduli space. The subvarieties correspond to certain conditions. For example, there is a (already compact!) space $Gr(2,4)$ (a Grassmannian), which parameterizes (projective) lines in (projective) 3-space. It is a four-dimensional variety. It contains three-dimensional subvarieties that parameterize the lines which intersect a given line. A question like "how many lines in 3-space meet 4 given lines?" translates to "how many points of intersection are there when I intersect 4 of these three-dimensional subvarieties?" ... – Michael Joyce Nov 8 '12 at 19:24
• Well the 4 three-dimensional subvarieties have to be in suitably general position so that they intersect transversely, which means that the four given lines have to be in general position. But it would be a lot easier if we could assume the given lines are in a simpler position - such as if two of the lines were contained in a plane. Deformation arguments allow you to do so, but only because the moduli space $Gr(2,4)$ is compact. (BTW, the answer to the problem is that there are 2 such lines. This is the simplest non-trivial calculation in Schubert calculus.) – Michael Joyce Nov 8 '12 at 19:27

I get the feeling that compactification adds limiting cases. Take for example $M_g$, the moduli space of smooth algebraic curves of genus $g$. Conpactifying this space allows us to add some curves with nodes which, although not members of $M_g$, can be thought of as limiting cases.