# Why are Riemann surfaces algebraic curves?

I've never fully understood the connection between Riemann surfaces and algebraic varieties. I'm particularly interested in the case of the modular curve of level N--I know how the Riemann surface is constructed by taking a quotient of the upper half-plane by the action of a congruence subgroup of the modular group, but not how the resulting manifold translates into a curve. From what I've read, it appears that the associated curve is defined by equations satisfied by functions defined on the manifold, but I don't understand which functions are involved in these equations. What exactly is the relation between the two types of objects?

• My understanding is that the modular curve $X_0(N)$ is most naturally viewed as a stack, in that it parametrizes elliptic curves with a point of given torsion, rather than a scheme. As to your general question: there is an equivalence of categories between compact Riemann surfaces and smooth proper curves over $\mathbb{C}$. This is given by the analytification functor that associates to a complex variety a complex analytic space. It is nontrivial, though, that this is an equivalence (it can be shown by using Riemann-Roch to embed any Riemann surface in $\mathbb{P}^n$, and then invoking GAGA). – Akhil Mathew Oct 10 '11 at 5:20
• I do not know of any 20th century work on Riemann Surfaces but algebraic curves were the basic motivation for Riemann to consider Riemann Surfaces, here is a classical approach by Klein zentralblatt-math.org/zmath/en/search/…) – Dinesh Oct 10 '11 at 8:26
• @Dinesh But that doesn't address the question. It is trivial that an algebraic curve is a Riemann surface. The other direction is quite a bit more subtle. – Alex B. Oct 10 '11 at 11:53
• Oh, I just realized I forgot to mention why $X_0(N)$ isn't naturally a scheme: it's because such data naturally has automorphisms (multiply the torsion point by $-1$!). – Akhil Mathew Oct 10 '11 at 13:29
• @AlexB.Thanks.I don't know that, I'm just a newbie in this subject :-) – Dinesh Oct 10 '11 at 16:22

But for the modular curves, things are actually simpler. It is easy to see that a modular curve is a covering of $\Gamma(1)\backslash \mathbb{H}^*$. Now, the latter is the simplest Riemann surface there is, the Riemann sphere, and that is clearly an algebraic curve. There are lots of ways of seeing that $\Gamma(1)\backslash \mathbb{H}^*$ is the Riemann sphere, some of which are sketched in Milne's notes, Proposition 2.21. An explicit isomorphism of $\Gamma(1)\backslash \mathbb{H}$ with $\mathbb{C}$ is provided by the $j$-function.
You can now use the covering to obtain an equation for a modular curve. For the minute details of this calculation for the case $\Gamma_0(N)$, see Milne's notes, Theorem 6.1. This does not need Riemann-Roch, and only relies on explicit computations with the $j$-function.