Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

2 Answers 2

up vote 4 down vote accepted

As Akhil writes in his comment, to see that some given Riemann surface is an algebraic curve, you usually need Riemann-Roch. A prototypical example of its use is the usual proof of existence of a Weierstrass equation for an elliptic curves (see e.g. Silverman).

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.

A more powerful approach to modular curves, which will also give you more information about fields of definition, is by interpreting the modular curves are moduli varieties. Milne also sketches some of this in section 8, and much more can be found in his notes on Shimura varieties.

share|improve this answer
Thank you for the helpful resources. –  William D. Oct 11 '11 at 22:56

I think you can check Hartshrone's section on Riemann-Roch. There are some relevant discussions on the 19th century work on this (without using fancy machinery listed in above comments). In fact, nowadays we are so used to modern mathematical language that we often forgot the roots which inspires so many masters in the past.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.