I'm taking introductory courses in both Riemann surfaces and algebraic geometry this term. I was surprised to hear that any compact Riemann surface is a projective variety. Apparently deeper links exist.

What is, in basic terms, the relationship between Riemann surfaces and algebraic geometry?

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    $\begingroup$ If you would like a book on Riemann surfaces with a more algebro-geometric point of view, try Algebraic Curves and Riemann Surfaces by Rick Miranda. $\endgroup$ Aug 31, 2010 at 4:32

7 Answers 7


For simplicity, I'll just talk about varieties that are sitting in projective space or affine space. In algebraic geometry, you study varieties over a base field k. For our purposes, "over" just means that the variety is cut out by polynomials (affine) or homogeneous polynomials (projective) whose coefficients are in k.

Suppose that k is the complex numbers, C. Then affine spaces and projective spaces come with the complex topology, in addition to the Zariski topology that you'd normally give one. Then one can naturally give the points of a variety over C a topology inherited from the subspace topology. A little extra work (with the inverse function theorem and other analytic arguments) shows you that, if the variety is nonsingular, you have a nonsingular complex manifold. This shouldn't be too surprising. Morally, "algebraic varieties" are cut out of affine and projective spaces by polynomials, "manifolds" are cut out of other manifolds by smooth functions, and polynomials over C are smooth, and that's all that's going on.

In general, the converse is false: there are many complex manifolds that don't come from nonsingular algebraic varieties in this manner.

But in dimension 1, a miracle happens, and the converse is true: all compact dimension 1 complex manifolds are analytically isomorphic to the complex points of a nonsingular projective dimension-1 variety, endowed with the complex topology instead of the Zariski topology. "Riemann surfaces" are just another name for compact dimension 1 (dimension 2 over R) complex manifolds, and "curves" are just another name for projective dimension 1 varieties over any field, hence the theorem you described.

As for why Riemann surfaces are algebraic, Narasimhan's book explicitly constructs the polynomial that cuts out a Riemann surface, if you are curious.

  • $\begingroup$ is an algebraic curve over the complex numbers (so complex dimension 1) any different from a compact Riemann surface in any context? From the above I conclude that the answer is no, algebraic curve over complex numbers is the same thing as comp. Riemann surface. $\endgroup$
    – Marion
    Dec 21, 2016 at 13:48
  • $\begingroup$ there is a functor from irreducible algebraic curves to compact riemann surfaces, but birationally equivalent curves correspond to isomorphic riemann surfaces. E.g. the curves y^2 = x^3 and y=x are birational but not isomorphic plane curves, both of whose corresponding riemann surfaces are isomorphic to P^1. So plane curves modulo birational equivalence (isomorphism except on a finite set) are essentially the same as compact riemann surfaces modulo isomorphism. Thus in 1) of my answer below, I should say "non singular" projective plane curves. $\endgroup$
    – roy smith
    Apr 12, 2019 at 0:53

This relationship is a very beautiful one.

Imagine a Riemann surface. There are different ways to introduce it, but since you gave kind of a reference point, let's just define it as a projective variety in the complex projective plane. Now the people call it a surface because it looks two-dimensional from a real point of view. You can also draw a picture of Riemann surface covering a sphere by projection onto a coordinate.

Now what could be an object of study of algebraic geometry? Why, certainly it should be some geometric object defined by algebraic means. Among the different ways to start learning algebraic geometry let's say we selected the abstract definition of an algebraic curve. To recap, this a a geometry locally defined by algebraic equations in some space so that the resulting manifold is one-dimensional.

These algebraic curves can be studied purely abstractly. You can, e.g., define algebraic forms on these, and prove various theorems relating to their geometry.

But the beautiful fact is that those are two sides of the same medal. That's right:

Every Riemann surface is a complex algebraic curve and every compact complex algebraic curve can be embedded into a projective plane and drawn as the Riemann surface.

There are lots of gems in this short statement. For example, as I said there is a way to count the algebraic forms in terms of inner geometry of algebraic curve. This gives some number, which could be 0, 1, 2, etc. On the other hand, if you draw a Riemann surface, you notice that it can be studied in topology and then it has the invariant called the number of handles which could also be 0 (sphere), 1 (torus), 2, etc. It turns out this is exactly the same thing though defined in a completely different way by a completely different branch of mathematics.

The whole algebraic geometry is, so to say, our attempt to make ourselves comfortable about this amazing connection between things we calculate (algebra) and things we draw (geometry).

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    $\begingroup$ Most higher-genus curves cannot be smoothly embedded in the plane, but they fit nicely in three-space. $\endgroup$
    – S. Carnahan
    Oct 27, 2009 at 0:10
  • $\begingroup$ Yes, I corrected it to refer to non-smooth Riemann surfaces. $\endgroup$ Oct 27, 2009 at 0:18
  • $\begingroup$ Is the result you alluded to called Riemann-Roch? $\endgroup$
    – timur
    Jan 20, 2011 at 2:35
  • $\begingroup$ Reemphasizing Carnahan's comment, a plane curve must have genus of the form $(d-1)(d-2)/2$, by the genus-degree formula. Hence, in particular, an algebraic curve of genus $2$ cannot be embedded into the plane. $\endgroup$
    – user94803
    Jul 24, 2016 at 12:52

From the analysis point of view, that every compact Riemann surface X is biholomorphic to a variety in complex projective space follows from an existence theorem for nonconstant meromorphic functions on X that Riemann proved by means of Dirichlet's principle (his proof was not rigorous, because Dirichlet's principle had not yet been made rigorous in his day). Whenever you have enough independent meromorphic functions on a compact complex manifold, you can put it in complex projective space. And a nonsingular variety in complex projective space has nonconstant globally meromorphic functions on it; just use ratios of homogenous coordinates. There are complex tori (compact quotients of C^n by lattices) that carry no nonconstant meromorphic functions at all; they obviously cannot sit in complex projective space. This stuff is nicely explained by Shafarevich in the second volume of his introduction to algebraic geometry.

Dirichlet's principle is an existence theorem for harmonic functions; this is relevant because harmonic functions on Riemann surfaces can be locally completed to holomorphic functions, and thus to meromorphic functions globally if topology allows (a question of monodromy).


Mumford's great short book "Curves and their Jacobians" is about that "amazing synthesis of algebra, geometry and analysis", as Mumford expresses it. The book's goal is to provide readers an overview what the zoo of curves looks like. Arithmetic issues are not discussed.

Ouch, I forgot to recommend this very beautifull and readable book by Clemens. The chapter "Manin and the unity of mathematics" is esp. fascinating as it is about a startling connection with the arithmetics.


To me, excellent as the others are, engelbrekt's is the most direct answer to your question. I.e.

1) Every projective plane curve is a compact Riemann surface, essentially because of the implicit function theorem.

2) Conversely, every compact Riemann surface [immerses as] a projective plane curve because it has [enough] non constant meromorphic functions which almost embed it in the plane. Also every meromorphic function is the pullback of a rational function in the plane. So all the analytic structure is induced from the algebraic structure.

In higher dimensions, complex projective algebraic varieties are a special subcategory of compact complex spaces, namely those that admit [holomorphic is sufficient] embeddings in projective space.

More precisely, an n dimensional compact complex variety has a field of meromorphic functions that has transcendence degree ≤ n, and projective algebraic varieties are a subcategory of those (Moishezon spaces) for which the transcendence degree is n. Indeed I believe Moishezon proved the latter are all birational modifications of projective varieties.

I would also add something about the impact of Riemann surfaces on algebraic geometry. Namely it was Riemann's introduction of the topological and analytic points of view, showing that path integrals and differential forms could be profitably used to study projective algebraic curves, that deepened and revolutionized algebraic geometry forever.

  • $\begingroup$ Riemann's melding of the topological and analytic ran parallel with his interests in and experiments on electrostatics. $\endgroup$ Jan 7, 2022 at 20:07

There is a canonical way to embed a Riemann Surface into projective space.

A Riemann Surface X is called hyperlliptic if it admits a 2 to 1 holomorphic map φ:X → P1. Given a (compact) Riemann Surface X of genus g ≥3 which is not hyperelliptic one can define an embedding φK:X → Pg-1 called the canonical embedding (the construction of this map can found in Rick Miranda's book "Algebraic Curves and Riemann Surfaces").

From this embedding one can find equations for the (image of the) curve X. For example, a nonhyperelliptic curve of genus 3 is given by the vanishing of a quartic polynomial in P2, a nonhyperelliptic curve of genus 4 is defined by the vanishing of a quadratic and a cubic polynomial in P3.

The hyperelliptic case is not hard to understand. A hyperellitic curve of genus g is defined by an equation of the form y2 = h(x), where h is a polynomial of degree 2g + 1 or 2g + 2.

These results can be found in Miranda's book or in Griffiths-Harris "Principles of Algebraic Geometry"

  • $\begingroup$ Actually not in Miranda, who only proves that a compact Riemann surfaces can be embedded in P^n, but not that the image is a projective algebraic curve. $\endgroup$
    – Joël
    Nov 9, 2021 at 2:03

Algebraic curves (one-dimensional projective varieties) over the complex numbers are exactly Riemann surfaces. It confuses everyone at first when one is told "curves are surfaces." Almost everyone else calls $\mathbb{C}$ the complex plane, but algebraic geometers call it the complex line.

One can work in any algebraically closed field, say $\mathbb{A}$, the field of algebraic numbers. But analysis only works in $\mathbb{R}$ or $\mathbb{C}$, which are complete.


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