## Moduli Spaces and Base Change

Last time we spoke of representable functors and talked about how to check if a functor is representable. The whole idea being that if we can first construct a functor that SHOULD be the functor of points for the scheme we want, and then we check representability and we’ve got it. Well, it doesn’t always happen like that. In fact, a lot of the time, the functors we get from moduli problems aren’t quite representable as schemes. That’s why people bother with algebraic spaces and with stacks.

So first up, what’s a moduli problem? We take some collection of objects, we then define what it means to have a family of them and finally we need to know when two families are the same. Then we take $S(B)$ to be the collection of families over $B$. This defines a functor! But how? Well, let’s worry about families of schemes first. Then that’s just a flat morphism $X\to B$. Let’s take another potential base $B'$ and a map $B'\to B$. Then we can take $X\times_B B'\to B'$, and this in fact gives a family, in fact, the morphism is flat. So this gives a map $S(B)\to S(B')$, which means we have a contraviarant functor, which is what we need. This process is called base change. In general if we have this situation, with a map between spaces and then a map into the codomain, we can use the fiber product to “change the base.” The terminology comes from the theory of moduli, where we think of these as bases of families.

But now really what we want is for $S(B)$ to be families up to equivalence. We still get a functor, but now we’ll call it $F$. So now the first real definition of a moduli space is just that if $F$ is represented by a scheme $\mathcal{M}$, we call $\mathcal{M}$ a fine moduli space for the moduli problem $F$.

Now, this almost never works out. Let’s look at elliptic curves, for instance. A moduli space must have a point for each isomorphism class, and so for elliptic curves, there’s the j-invariant which classifies them. So then the potential moduli space is $\mathbb{A}^1$ with coordinate $j$. However, this can’t work, because you can have a nontrivial family of elliptic curves such that all the fibers are isomorphic. Just take two copies $\mathbb{A}^1\times E$ and glue them together to get a family over $\mathbb{P}^1$ using the automorphism which is multiplication by -1 on the curve. this is not a trivial family, but it gives the trivial map.

More generally, objects with automorphisms cause problems, and make it so that we can’t solve the moduli problem with schemes.

So we have a few choices. One is to say “Well, schemes weren’t good enough, lets try stacks.” That’s a good one, I can’t say that I really understand the technical details of it though. So we won’t pursue that just now. Another solution is to say “Well, I guess there just isn’t a moduli space to work with here.” That’s just defeatest, and I’ll have none of that on my blog! Instead, we’ll take a third option: relax our definition of moduli space a bit.

So now we take a scheme $\mathcal{M}$ and a natural transformation $\Psi_{\mathcal{M}}:F\to h_{\mathcal{M}}$ and call the pair a coarse moduli space for $F$ if the map $\Psi_{\mathcal{M}}(\mathrm{Spec}(\mathbb{C}):F(\mathrm{Spec}(\mathbb{C}))\to h_{\mathcal{M}}(\mathrm{Spec}(\mathbb{C})$ is an bijection of sets and if given any other scheme $\mathcal{M}'$ and natural transformation $\Psi_{\mathcal{M}'}:F\to h_{\mathcal{M}'}$, there is a unique morphism $\pi:\mathcal{M}\to\mathcal{M}'$ such that the associated natural transformation (we get it from Yoneda) satisfies $\Psi_{\mathcal{M}'}=\pi\circ\Psi_{\mathcal{M}}$.

This says that, although the value on the base isn’t all the families, there’s a good way of getting from the set of all families to the set of maps into the space, and it varies reasonably with changing the base.

So as it happens, the affine line is a coarse moduli space for elliptic curves, and usually coarse moduli spaces are good enough. In the future, we’ll just call coarse moduli spaces “moduli spaces” and we’ll use the modifier fine for the better ones.

Now, I’ve been saying “a coarse moduli space.” Can there be more than one? Well, no, it’s defined by a universal property, and some formal manipulations give us uniqueness up to unique isomorphism, which is good to know. There is one fine moduli space that we should be familiar with before going forward: the Grassmannian. The problem it solves is just the collection of $k$-planes in $\mathbb{A}^n$ which pass through the origin. Any family of these really is just the same as a map into the Grassmannian, and we’ll be using the Grassmannian to construct more fine moduli spaces later.

Before finishing up, a remark: it is a theorem that there is a family of coarse moduli spaces $\mathcal{M}_{g,n}$ of curves of genus $g$ with $n$ distinct points marked on them. That these exist is nontrivial itself, and in fact when $g\geq 2$ we have that the dimension is $3g-3+n$. For $g=1$ we need to also have $n=1$, and we get a 1-dimensional space, the moduli space of elliptic curves (the marked point is the origin for the multiplication, but it doesn’t really matter, because the automorphism group of an elliptic curve is transitive). For $g=0$ we need $n=3$ or more, and in the case $\mathcal{M}_{0,3}$ we get a single object. Why do we do this? Well, we want to really be looking at curves with distinct points up to automorphism, that is, a map $C\to C$ which takes the first set of points to the second.

Advertisements

## About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in AG From the Beginning, Algebraic Geometry, Deformation Theory. Bookmark the permalink.