Classifying Riemann surfaces

Wednesday, October 21st, 2009 | Author:

In this post, I will sketch a classification of Riemann surfaces.

For those who haven't heard about the subject before, there is an introduction. For the impatient, look at the bottom of the post, where I have written a very short summary.

Table of contents:


At first, a definition: A Riemann surface is a connected complex manifold of (complex) dimension 1 (so real dimension 2, which qualifies it as a surface). A compact Riemann surface is a Riemann surface which is compact as a topological space.

At first, let's think only about orientable surfaces. Orientability means that the orientation double cover is trivial (not simply-connected), or equivalently, the tangent bundle (or sheaf, if you prefer) admits a global section. In simple terms, orientable means you can take an orientation with you, walk around the surface, return where you were first and the orientation is still the same.
It turns out that every complex manifold is orientable as real manifold, so a Riemann surface is always orientable as a real surface (and this is what we care about for the moment).

Some simple examples of Riemann surfaces:

  • the complex plane \mathbb{C} itself
  • the upper half plane \mathbb{H} (also called the hyperbolic disc; it's biholomorphic to the open unit disc)
  • Complex projective space of one dimension \mathbb{C}P^1 which is topologically a sphere S^2, called Riemann sphere
  • A complex torus E_\tau, that is the quotient of the complex plane \mathbb{C} by a lattice \Lambda_\tau = \mathbb{Z} + \mathbb{Z}\tau with some \tau \in \mathbb{C}, so we have E_\tau = \mathbb{C}/\Lambda_\tau

Some simple counter-examples:

  • The moebius strip is not orientable (so it admits no complex structure).
  • Complex projective space modulo action of z \mapsto -z^{-1} is a real manifold, but no longer a complex one.

So we know what we're talking about. Now what kind of classification do we search? The "best" classification would be one up to biholomorphism, but let's see how far we can get. First, we can seperate different surfaces by their topology. This is a good approximation because each biholomorphism is a homeomorphism.

Topological classification

Note that, topologically, all tori E_\tau are homeomorphic. Their complex structure is different and depends on \tau which can always be chosen to lie in \mathbb{H}, the complex upper half plane. This variable \tau is called the modulus of E_\tau.

However, \mathbb{C}P^1 and E_\tau are not homeomorphic, their genera are different. The genus is a topological invariant which completely classifies the compact (orientable real) surfaces up to homeomorphism. This can be shown by proving that each such surface admits a triangulation and then showing each triangulation is homeomorphic to a standard genus g triangulation. Roughly speaking, the genus is "the number of holes" inside the surface. One can "mathematise" this definition by speaking about loops non-homotopic to the constant loop, which leads to a discussion of the fundamental group \pi_1. The study of fundamental groups is also necessary to understand holomorphic (branched) coverings, which are used in the biholomorphic classification theory.

Biholomorphic classification

There are other surfaces than \mathbb{C}P^1 and tori E_\tau. The famous Riemann Mapping Theorem tells us, that the only simply connected Riemann surfaces are \mathbb{C},\ \mathbb{H} and \mathbb{C}P^1.

Digression about algebraic field extensions

(You don't have to read this)

So the only compact surface without holes is \mathbb{C}P^1. This is interesting, because by definition a meromorphic function on a surface X is a holomorphic map X \rightarrow \mathbb{C}P^1. The collection of all meromorphic functions on a surface X is written \mathcal{M}(X), it's always a field, called the function field of X. If there is an element \eta in \mathcal{M}(X), it is always a branched covering. There is a theorem, that shows each Riemann surface admits a non-constant meromorphic function. This leads to the conclusion, that the map \eta^\ast : \mathcal{M}(\mathbb{C}P^1) \rightarrow \mathcal{M}(X) which sends each f to f \circ \eta is an algebraic field extension of degree n, where n is the degree of the covering \eta. Since the meromorphic functions on \mathbb{C}P^1 are just the rational functions \mathbb{C}(z) (a transcendence degree 1 extension of the complex numbers), each compact Riemann surface has a function field which is an algebraic field extension of \mathbb{C}(z).

The other way works, too: To each algebraic field extension of \mathbb{C}(z) we can define a corresponding compact Riemann surface.

Every compact Riemann surface can be embedded in complex projective space of at most dimension 3. With Chow's Theorem, we get the surprising statement that every compact Riemann surface is algebraic.

Universal coverings

We can understand a Riemann surface by it's coverings, since it will be a quotient of the covering. To be more specific: for each covering p : Y \rightarrow X, the space X is biholomorphic to the quotient of Y with respect to the transformation subgroup which commutes with the projection p. This transformation subgroup is called the group of deck transformations (it's similar to the Galois group of a field extension). There are some theorems about that:
A surface X which is covered by \mathbb{C}P^1 is already biholomorphic to \mathbb{C}P^1.
A surface covered by \mathbb{C} is biholomorphic either to \mathbb{C},\ \mathbb{C}^\times or a torus. On the other hand, every torus is covered by \mathbb{C}. For surfaces with universal covering space \mathbb{H}, the situation is more complicated. For example, one can show that every hyper-elliptic surface is covered by \mathbb{H}.

Automorphisms of simply connected surfaces

Now we can realise every Riemann surface as quotient of the simply connected spaces, since every surface admits a universal covering, which is by definition a simply connected Riemann surface. In order to do the classification, we have to study the automorphism groups of the simply connected surfaces and their subgroups.

  • Aut(\mathbb{C}) = \{z \mapsto az+b \mid a,b \in \mathbb{C}\} (Affine translations)
  • Aut(\mathbb{C}P^1) = PSL_2(\mathbb{C}) (Möbius transformations)
  • Aut(\mathbb{H}) = PSU_{1,1}(\mathbb{C}) (Möbius transformations preserving the unit disc)


One can show that each Riemann surface admits a Riemannian metric because they are all paracompact. Luckily, this metric is unique up to conformal equivalence for a given complex structure. Every such metric is equivalent to one with a constant Gaussian curvature either -1,\ 0 or +1.

Short summary

Every Riemann surface is the quotient of a free, proper holomorphic group action on its universal covering, which is one of the simply connected surfaces \mathbb{C},\ \mathbb{C}P^1,\ \mathbb{H}. The complete list:

  • Elliptic surfaces - constant Gauss curvature +1
    • Genus 0: only the Riemann sphere \mathbb{C}P^1
    • Genus 1 or higher: None.
  • Parabolic surfaces - constant Gauss curvature 0
    • Genus 0: only the complex plane \mathbb{C}
    • Genus 1: either \mathbb{C}^\times or a torus E_\tau, where the tori biholomorphism classes are in one-to-one correspondence to the elements of their moduli space \mathbb{H}.
    • Genus 2 or higher: None.
  • Hyperbolic surfaces - constant Gauss curvature -1
    • Genus 0: only the hyperbolic disc \mathbb{H}
    • Genus 1 and higher: quotients of \mathbb{H} by a Fuchsian group. Sadly, further classification is difficult (but at least for compact surfaces, we can apply the theory of field extensions sketched above)
    • All genus > 1 and especially the so-called hyperelliptic surfaces belong to this class.

I wrote this because I couldn't find a quick overview of the classification, with enough but not too much detail, on the web. If you find mistakes or have any questions, don't hesitate to contact me and/or write a comment :-)

UPDATE: this is the conformal classification. There are various other classifications, see Solomentsevs explanations here and Krushkal's explanations here.

Category: English

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