Is the moduli space of smooth complex curves of genus $g\geq2$ isomorphic to the affine space $\mathbb A_{\mathbb C}^{3g-3}$?

(Note: I am not asking about the compactification of this space, which by definition is not affine!)

I know that the complex dimension is $3g-3$, and so it is locally isomorphic to the vector space $\mathbb C^{3g-3}$. But is it globally $\mathbb A_{\mathbb C}^{3g-3}$?

I can't find a reference that answers this.


Genus $g=1$ is not in the range of my question, but I was trying to use it for intuition nevertheless. One of way of computing the moduli space of smooth elliptic curves gives me $\mathbb A^1_{\mathbb C}$ as the moduli space (i.e. using the $j$-invariant) and another way gives me $\mathbb{CP}^1\backslash\{0,1,\infty\}$ (by studying $2:1$ covers of $\mathbb{CP}^1$ branched over $4$ pairwise-distinct points). Both are one-dimensional and noncompact, but which is the better candidate for the moduli space? I know that the issue of moduli of elliptic curves is quite a subtle one, and the real answer lies in the world of stacks, but if your only two options were $\mathbb A^1_{\mathbb C}$ and $\mathbb{CP}^1\backslash\{0,1,\infty\}$, which would you choose? ("Neither" is not allowed!) Does the $j$-invariant simply give deformations around some particular elliptic curve of one's choice, and hence makes the former space into the tangent space to the moduli space, while the latter space is the "actual" moduli space (in some sense)?


It occurred to me that the $0$ and $1$ that are being omitted from the projective line in the second candidate space could be the two "special" isomorphism classes of elliptic curves (with automorphism groups $\mathbb Z_6$ and $\mathbb Z_4$, respectively). If we were to restore these two points, this would correct the "discrepancy" between the two candidate moduli spaces. The isomorphism of affine varieties from $\mathbb A^1_{\mathbb C}$ to $\mathbb{CP}^1\backslash\{\infty\}$, i.e. from the space of $j$-invariants to double covers of $\mathbb{CP}^1$ branched over $4$ points, would be $a\mapsto1728^{-1}a$.

For this to be correct, it would require that the $j=0$ and $j=1728$ curves are branched over $4$ non-distinct points (perhaps a double branch point for one and a triple for the other)...could any of this hold water?

(Or perhaps $j=0$ and $j=1728$ can't be realized as double covers at all?)

Should any of this experimentation with the $g=1$ case inform the $g>1$ case? What insights should it be giving me about the original question?


3 Answers 3


Take a look at the book "Moduli of curves" by Harris and Morrison. It has answers to your questions. In particular, for all $g>2$ the moduli space contains complete curves, hence cannot be affine. It is a theorem by Diaz that for all $g>22$ the moduli space is not even uniratonal.

Edit: I misremembered who proved what (should have checked the book before writing the answer). It is a theorem of Harris and Mumford (Theorem 6.59 of "Moduli of curves") that $M_g$ has general type for $g\ge 23$.

In any case, (theorem 2.33 of the same book) since for $g\ge 3$, $M_g$ contains complete curves (actually, it contains a complete curve through every point), it cannot be Stein. In particular, it is not isomorphic to the affine space.

  • $\begingroup$ Thank you! By the way, do you have any thoughts on the tangential questions in my Update #1 and #2? $\endgroup$ Commented Jan 15, 2015 at 3:04
  • $\begingroup$ @MathsByTheSea: No, I do not have much to say there; answering those would require some thinking and time. $\endgroup$ Commented Jan 15, 2015 at 3:07
  • $\begingroup$ That's fair enough. I may move them to an independent question at some point. :-) $\endgroup$ Commented Jan 15, 2015 at 3:09
  • 3
    $\begingroup$ I think Harris and Mumford proved the theorem you attribute to Diaz. (Relatively recently Farkas also proved that $\mathcal M_{22}$ is of general type.) $\endgroup$
    – user64687
    Commented Jan 15, 2015 at 8:53
  • $\begingroup$ @Asal is right (as usual). Notice that a variety of general type is as far as can be from a rational variety: the Kodaira dimension of the former is the dimension of the variety whereas the Kodaira dimension of the latter is $-\infty$. These are the extremities of the spectrum of possible values for that birational invariant, the Kodaira dimension. $\endgroup$ Commented Jan 15, 2015 at 18:17

a) The moduli space $\mathcal M_g$ of smooth projective curves is not isomorphic to $\mathbb A^n_\mathbb C$ because its only global holomorphic functions are just the constants: $\mathcal O_\text {hol}(\mathcal M_g)=\mathbb C$.
Hence the holomorphic variety underlying $\mathcal M_g$ is not even Stein (whereas of course $\mathbb A^n_\mathbb C$ is Stein) , a fact confirmed by studiosus's answer that it contains complete curves.

b) Another reason why $\mathcal M_g$ is not isomorphic to $\mathbb A^n_\mathbb C$ is that it is singular.
For $g\geq 4$ we have the nice criterion that a point $[C]\in \mathcal M_g$ is singular iff the corresponding curve $C$ has a non-trivial automorphism.
And there always exist such curves of genus $g$ with non-trivial automorphisms: the hyperelliptic curves for example.

  • $\begingroup$ Very nice! Thank you. $\endgroup$ Commented Jan 15, 2015 at 11:54

This is just a comment on the questions in the updates, but it won't fit nicely in the comment box.

About the question in update 1: the answer is that $\mathbf A^1$ is the moduli space.

The reason the other space isn't is the following: let $C_\lambda$ be the curve branched over $\{0,1,\lambda,\infty\}$. Then $C_\lambda \cong C_\mu$ iff $\mu$ is one of the 6 numbers $\{\lambda,1-\lambda,\frac{1}{\lambda}\,\frac{1}{1-\lambda},\frac{\lambda-1}{\lambda},\frac{\lambda}{\lambda-1} \}$. If you want to check this, note that $C_\lambda$ has affine model $y^2=x(x-1)(x-\lambda)$, and it's fairly straightforward to compute the $j$-invariant from that. (See the formula below.)

In fact the symmetric group $S_3$ acts on $\mathbf P^1 - \{0,1,\infty\}$, with two distinct transpositions acting by the maps $\lambda \mapsto \frac{1}{\lambda}$ and $\lambda \mapsto 1-\lambda$. The quotient of $\mathbf P^1 - \{0,1,\infty\}$ by this action is in fact $\mathbf A^1$, the correct moduli space.

About the question in update 2: no, this is not the correct picture. As mentioned in the previous point, the $j$-line is a quotient of $\mathbf P^1 - \{0,1,\infty\}$. So the latter space contains points corresponding to curves with all possible $j$-invariants.

If you want to see this more explicitly, the elliptic curve $C_\lambda$ defined above with affine model $y^2=x(x-1)(x-\lambda)$ has $j$-invariant

$$j(\lambda) = 256 \frac{(\lambda^2-\lambda+1)^3}{\lambda^2(\lambda-1)^2}.$$

So to get $j$-invariant 0 we just need $\lambda$ to be a root of the quadratic polynomial in the numerator: solving gives $\lambda=\frac{1}{2}(1 \pm \sqrt{-3})$. I leave $j=1728$ as an exercise.

I also mention that whenever $C \rightarrow C'$ is a $d:1$ cover of smooth curves, the ramification index at any point is at most $d$. So for a $2:1$ cover, it isn't possible to have anything other than simple branching.

Finally, let me second @studiosus' recommendation to look at Moduli of curves by Harris and Morrison.

  • $\begingroup$ This is great --- you've certainly cleared up a few of my confusions. Thanks! $\endgroup$ Commented Jan 15, 2015 at 11:57
  • $\begingroup$ Sorry Asal Beag Dubh, I'm sure I'm being silly, but how can the quotient of a subset of $\mathbb C$ (i.e. $\mathbb{P}^1\backslash\{0,1,\infty\}$) be $\mathbb C$? $\endgroup$ Commented Jan 15, 2015 at 14:01
  • $\begingroup$ @MathsByTheSea: I'm not sure what to say, except "it happens!" :) For example, consider my formula for $j(\lambda)$. You want to show that this function is a surjection $\mathbf C - \{0,1\} \rightarrow \mathbf C$ --- i.e. to show that for any $a$, the equation $256(\lambda^2-\lambda+1)^3=a\lambda^2(\lambda-1)^2$ has a solution different from $0$ or $1$. Note that plugging in 0 or 1 will give 0 on the r.h.s. but not the l.h.s. so these are not solutions for any $a$; on the other hand this eqn. has degree 6, so for any $a$ the fundamental thm. of algebra guarantees 6 sols (w. multiplicities). $\endgroup$
    – user64687
    Commented Jan 15, 2015 at 16:42

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