Pretty much everything in this post is in Mumford’s “Curves and their Jacobians,” but I do a couple of things slightly differently, and I intend to supply a bit more detail in some places.  The goal here is to construct the moduli of curves of genus g for small g.

Let’s start with some definitions.  The set \mathcal{M}_{g,n} is the set of (C,x_1,\ldots,x_n) where C is a genus g curve and x_1,\ldots,x_n\in C are distinct.  Now, sadly, \mathcal{M}_{g,n} is not a variety.  It can be realized as a smooth stack, however, but that’s a bit more than we are going to even try to handle.  Instead, we’ll quote a very good, powerful and classical theorem, which is true (and clearly so, because I am using a boldface for the word theorem):

Theorem: There exists a unique normal quasiprojective variety M_{g,n} such that it is in bijection with \mathcal{M}_{g,n} and if \pi:X\to S is a proper smooth map whose fibers are genus g curves and \sigma_1,\ldots,\sigma_n:S\to X are n disjoint sections, then the induced map of sets \phi:S\to M_{g,n} is a morphism of varieties.

Rather, the above is true when 3g-3+n\geq 0.  When it’s smaller, then the space has expected dimension negative (this only happens for genus 0 with at most 2 points, and genus 1 with no points) and we really need the stacky interpretation to make sense of things.

So why is this the dimension? The computation is fairly quick: first off, each point is a degree of freedom, so \dim M_{g,n}=n+\dim M_{g,0}, and M_{g,0} is just deformations of a curve.  It can be shown that the tangent space to M_{g,0} at a curve C is just H^1(C,T_C), so we must compute this dimension, which by Riemann-Roch is 3g-3 for g\geq 2.  For g=0,1, we need to check things carefully because of the aforementioned stacky problems (it amounts to positive dimensional families of automorphisms for these curves), but the result works out.

We’re really only going to have one example, and then the others will follow from it.  We’ll directly compute M_{0,n} for n\geq 3.  There’s only one genus zero curve, and that’s \mathbb{P}^1, so we’re looking at sets of n ordered points in \mathbb{P}^1.  The automorphisms let us take the first three to 0,1,\infty, so we have n-3 points to pick, and there are no automorphisms left once we’ve done this, so M_{0,n}\cong (\mathbb{P}^1)^{n-3}\setminus\mbox{diagonals}, because the points are distinct.

Next, we’ll look at M_{1,1}, as this is the next simplest that has positive dimension.  There are a few ways we can attack this.  The simplest is by using Weierstrass form and the j-invariant, and this works for characteristic not 2,3, and we’re going to imagine ourselves over k=\mathbb{C}, so it’s fine.  Every elliptic curve can be written as y^2=x(x-1)(x-\lambda) for some \lambda\in k, with \lambda\neq 0,1.  However, this representation isn’t unique.  So whatever our curve is, it’s covered by \mathbb{P}^1\setminus\{0,1,\infty\}.  There’s a function j(\lambda)=256\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}, and it takes the same values at \lambda, 1-\lambda, \frac{1}{\lambda}, \frac{\lambda-1}{\lambda}, \frac{\lambda}{\lambda-1}, and \frac{1}{\lambda-1}, which all give the same elliptic curve! Even better, no other \lambda will give the same curve, so j:\mathbb{P}^1\setminus\{0,1,\infty\}\to M_{1,1} is a surjective map.  And because j can take any value, it shows that M_{1,1}\cong \mathbb{A}^1.

A slightly different presentation is that every elliptic curve can be written as a double cover of \mathbb{P}^1 ramified at four points.  These points are unordered, and so we can look at quartic polynomials on \mathbb{P}^1 with no repeated roots f(x,y).  However, automorphisms of \mathbb{P}^1 let us take the roots to 0,1,\infty,\lambda, so we have f(x,y)=xy(x-y)(x-\lambda y)=0 and its discriminant is nonzero, and we have to identify the same \lambda‘s as above.

This approach is fruitful for identifying a certain subvariety of M_g(=M_{g,0}) for g\geq 2.  This is the locus of hyperelliptic curves.  These are the curve which double cover \mathbb{P}^1 and thus are uniquely determined by their branch points.  For a genus g hyperelliptic curve, we have 2g+2 ramification points.  The second approach above is predicated on the fact that for g=1, every curve is hyperelliptic.  We’ll denote the hyperelliptic locus by H_g.

Now, it’s a fairly easy theorem that H_2=M_2, so we should be able to handle finding a description of this family of curves.  By the above, they branch at six points, so we have to look at six distinct points in \mathbb{P}^1, which is an open subset U\subset\mathbb{P}^6 of the sextic polynomials.  Then, we need to take U/\mathrm{Aut}(\mathbb{P}^1).  We can achieve this by looking at ordered sextuples of points on \mathbb{P}^1, and having an extension of \mathrm{Aut}(\mathbb{P}^1) by S_6 (or perhaps extension in the other direction? I never quite got that bit of language) act by first permuting the points, and then transforming \mathbb{P}^1 so that the first three are 0,1,\infty.  This can actually be calculated directly, and it’s a quotient of \mathbb{A}^3.  The points that need to be identified are the ones where, if \zeta is a primitive fifth root of unity, we have (x,y,z) gives the same curve as (\zeta x, \zeta^2 y, \zeta^3 z).  It’s very straightforward to compute the invariants, and they are x^5,x^3y,xy^2,y^5,x^2z,xz^3, z^5, yz, and give a map \mathbb{A}^3\to \mathbb{A}^8 which factors through the quotient, thus giving us M_2 inside of \mathbb{A}^8.

In general, we have that H_g can be written as a quotient of M_{0,2g+2}, though computing this quotient is not always easy.  This is also less useful in higher genus, because starting in genus 3, there are nonhyperelliptic curves.  In fact, the moduli space M_3 can be described in two pieces: the hyperelliptic locus H_3 can be worked out as above.  For the nonhyperelliptic curves, the canonical map gives them as quartic plane curves, so we have U\subset |4H| the open subset of the linear system of quartics consisting of smooth curves, and we have to take U/PGL(3), the automorphism group of the plane.  This is plainly unirational (for those who know the term), and I vaguely recall seeing a paper proving that it’s actually rational, but I can’t find it at the moment.  So starting at g, these moduli spaces can be very hard to construct directly, and that’s a big part of why this subject is interesting, and why it is so difficult.

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