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How does one prove that every simply connected Riemann surface is conformally equivalent to the open unit disk, the complex plane, or the Riemann sphere, and these are not conformally equivalent to each other?

I would like to know about different ways of proving it, and appropriate references. This is not to know the best way; but to know about various possible approaches. Therefore I wouldn't be choosing a best answer.

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Please skip the proof that these three are not conformally equivalent! :) – Anweshi Jan 2 '10 at 20:41

10 Answers 10

As has been pointed out, the inequivalence of the three is elementary.

The original proofs of Koebe and Poincare were by means of harmonic functions, i.e. the Laplace equation ${\Delta}u = 0$. This approach was later considerably streamlined by means of Perron's method for constructing harmonic functions. Perron's method is very nice, as it is elementary (in complex analysis terms) and requires next to no topological assumptions. A modern proof of the full uniformization theorem along these lines may be found in the book "Conformal Invariants" by Ahlfors.

The second proof of Koebe uses holomorphic functions, i.e. the Cauchy-Riemann equations, and some topology.

There is a proof by Borel that uses the nonlinear PDE that expresses that the Gaussian curvature is constant. This ties in with the differential-geometric version of the Uniformization Theorem: Any surface (smooth, connected 2-manifold without boundary) carries a Riemannian metric with constant Gaussian curvature. (valid also for noncompact surfaces).

There is a proof by Bers using the Beltrami equation (another PDE).

For special cases the proof is easier. The case of a compact simply connected Riemann surface can be done by constructing a nonconstant meromorphic function by means of harmonic functions, and this is less involved than the full case. There is a short paper by Fisher, Hubbard and Wittner where the case of domains on the Riemann sphere is done by means of an idea of Koebe. (Subtle point here: Fisher et al consider non-simply connected domains on the Riemann sphere. The universal covering is a simply connected Riemann surface, but it is not obvious that it is biholomorphic to a domain on the Riemann sphere, so the Riemann Mapping Theorem does not apply).

The Uniformization Theorem lies a good deal deeper than the Riemann Mapping Theorem. The latter is the special case of the former where the Riemann surface is a simply connected domain on the Riemann sphere.

I decided to add a comment to clear up a misunderstanding. The theorem that a simply connected surface (say smooth, connected 2-manifold without boundary) is diffeomorphic to the plane (a.k.a. the disk, diffeomorphically) or the sphere, is a theorem in topology, and is not the Uniformization Theorem. The latter says that any simply connected Riemann surface is biholomorphic (or conformally equivalent; same in complex dimension $1$) to the disk, the complex plane or the Riemann sphere.

But the topology theorem is a corollary to the Uniformization Theorem. To see this, suppose $X$ is a simply connected (smooth etc.) surface. Step (1): Immerse it in $\mathbb{R}^3$ so as to miss the origin. Step (2): Put the Riemann sphere (with its complex structure!) in $\mathbb{R}^3$ in the form of the unit sphere. Step (3): For every tangent space $T_pX$ on $X$, carry the complex structure $J$ from the corresponding tangent space on the Riemann sphere by parallell transport (Gauss map) to $T_pX$. This is well-defined by choosing a basepoint and recalling that $X$ is simply connected. Step (4): Presto! $X$ is now a Riemann surface (it carries a complex structure), so it is biholomorphic to the disk or the plane or the Riemann sphere, thus diffeomorphic to one of the three.

Of course, I have glided over the question of immersing the surface in 3-space, because this is topology. Actually, I vaguely recall that there is a classification of noncompact topological surfaces by Johannsen (sp?), and no doubt the topological theorem would immediately fall out of that.

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I learned the proof from this paper, Uniformization of Riemann Surfaces. They actually provide three methods of proof, and I found the first of these the easiest to follow. Its not too difficult to go through the proof in detail if you know a bit of topology and complex analysis. The idea is to construct a global analytic function by minimizing a Dirichlet integral. Then, by working out the properties of "flow lines" - on which the function has constant imaginary part - you can get a good idea of what the map looks like, and then show that it is either a mapping onto the Riemann sphere, the Riemann sphere with the origin removed (equiv to the complex plane) or the Riemann sphere with a line segment removed (equiv to the open unit disk).

The second method they provide relies on triangulating the surface. The proof constructs the mapping inductively on larger and larger sets of triangles, using the Riemann mapping theorem to construct maps and the Schwarz reflection principle to join them together.

They also provide a third method, based on sheaf cohomology, although I am not so familiar with this method. The idea is to construct geometric realizations of projective structures on the Riemann surface. However, it does not seem to be quite complete, and there are some unresolved problems posed at the end.

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Do you have the link to the paper? – Ilya Nikokoshev Jan 2 '10 at 23:25
The link should be provided in my answer, if you click on it. (initially made a mistake, but I edited and fixed the link). – George Lowther Jan 2 '10 at 23:30

On my point of view the best modern reference is the book by J. Hubbard, Teichmuller theory and its applications.

A very interesting (and little known) proof is due to Lavrentiev, and can be found in an Appendix to the book of Goluzin, Geometric theory of functions of a complex variable.

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I've just taken a course which concluded with a sketch of the uniformisation theorem for Riemann surfaces, following the last chapter of Gamelin's Complex Analysis. The idea is:

-If the Green's function exists for your surface, use it to construct a conformal map from the surface to a bounded region in the complex plane. Now apply the Riemann mapping theorem.

-If the Green's function doesn't exist, construct a meromorphic variant called the bipole Green's function. Similar to the first case, we can use this to construct an injective map from the surface to the Riemann sphere. Now if this map misses more than one point of the Riemann sphere, simply-connectedness of the domain (and hence the image) means that the image is bounded, so we can use the Riemann mapping theorem. Otherwise the map misses one point (so the surface is conformally equivalent to the complex plane) or is surjective (conformally equivalent to the Riemann sphere).

Complex analysis is very far from my field, so I'm afraid I can't explain this any further (and I apologise for any inaccuracies in the above).

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I was going to answer Anweshi's other complex analysis question too (predictably, the fundamental theorem of algebra turned up in our courses too) but the question seemed to have disappeared, it says "page not found" when I click on it?? – Amy Pang Jan 2 '10 at 22:53
Anweshi deleted it in when he feared that it will be closed. Anweshi was harassed too much and even got a ban for asking questions. – Anweshi Jan 3 '10 at 15:24
The moderator Anton revived that question after Anweshi's request. – Anweshi Jan 3 '10 at 18:35

There is a nice book about the uniformization theorem by Henri Paul de Saint Gervais, Uniformisation des surfaces de Riemann. The author is actually a collective of French mathematicians.

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The Riemann sphere isn't conformally equivalent to the others because it is not homeomorphic to them. :)

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And the complex plane and the disc aren't because the map from the plane to the disc would be a bounded non-constant analytic function. – Ryan Budney Jan 2 '10 at 20:36
Yes of course .. :) You know what I had mainly in mind; it is the other part, that every simply connected Riemann surface is conformally equivalent to one of the above three. – Anweshi Jan 2 '10 at 20:37

I believe there is a formal proof of the classification along the following lines.

Being simply connected means that whenever you have curve, you can always get a disk inside. If there is more than one way to glue a disk, you must have a Riemann sphere. If there is always exactly one way, you can take the picture and put it step-by-step on a complex plane. Once there, you have either the whole plane or you're inside the complement of a ray. In the latter case, you are between a disk and a disk, so there's some approximation thing that says you're also a disk.

The three cases can be distinguished by their global symmetry group: there are three different constant curvature metrics, with the curvature resp. 1, 0, -1 for the sphere, plane, and disk case. Therefore you have three slightly different symmetry groups which can be written as the isometries of quadratic forms in 3 dimensions: $SO(x^2 + y^2 + z^2)$, $SO(x^2+y^2)$, $SO(x^2+y^2-z^2)$.

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A good reference? – Anweshi Jan 2 '10 at 20:42
Sorry, don't know any specific reference, that's just how I think about them :) – Ilya Nikokoshev Jan 2 '10 at 20:44
Interesting... How do you go about rigorously proving statements like, "If there is more than one way to glue a disk, you must have a Riemann sphere", and "If there is always exactly one way, you can take the picture and put it step-by-step on a complex plane." ? I suppose, if these two are done, then the rest is the Riemann mapping theorem .. maybe that is what Gerald Edgar above had in mind. – Anweshi Jan 2 '10 at 20:58
The first is very geometric: imagine you have circle and fill it twice by a disk. By definition, you now have a map of a sphere. Now to say that these two ways are inequivalent is exactly the same as saying that this sphere cannot be filled with a ball. So your $\pi_2$ is nontrivial; this is only possible if your whole surface is a sphere. – Ilya Nikokoshev Jan 2 '10 at 21:00
The second proceeds as follows: You take a triangulation of your surface and construct a map from your surface to the plane one triangle at a time. Because every loop can be extended to one and only one disk, you never have a contradiction. I've heard it only as a proof sketch and not actual proof though. – Ilya Nikokoshev Jan 2 '10 at 23:22

One proof not yet cited is by Ricci flow. It is a proof of the differential geometric version of Uniformization : in each conformal class of Riemannian metric, there is a metric of constant curvature. The idea is very natural : by construction, metric of constant curvature are fixed points of Ricci flow so take a general metric, evolve it by Ricci flow and show that it converges. This is essentially a work of Hamilton (reference : "The Ricci flow :an introdution" Chow, Knopf) which was completed by Chen, Lu and Tian.

I don't claim that it is the best proof of unifomization theorem, it is just a way to see in a simple case how the Ricci flow can be used to obtain classification results. (The point is that for surfaces Ricci flow has no singularity in finite time, which is not the case in 3 dimensions).

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There is an entire book dedicated to this theorem:Uniformisation des surfaces de Riemann-Henri Paul de Saint-Gervais= (Aurélien Alvarez, Christophe Bavard, François Béguin, Nicolas Bergeron, Maxime Bourrigan, Bertrand Deroin, Sorin Dumitrescu, Charles Frances, Étienne Ghys, Antonin Guilloux, Frank Loray, Patrick Popescu-Pampu, Pierre Py, Bruno Sévennec, Jean-Claude Sikorav.) A lot of historical insight!

Here you find some video-lectures on the same subject by one of the authors:Etienne Ghys:

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The book is in french,but the lectures are in english. – pi2000 Aug 5 '12 at 6:05
It is the same book as in Ben McKay answer.(sorry),but the link to video is new – pi2000 Aug 5 '12 at 6:08

Isn't this called the "Riemann mapping theorem"? At least the only nontrivial part of it. Ahlfors proves it using Montel's theorem.

For the rest:

These are not equivalent: The sphere is compact, the others not. The disk admits a bounded nonconstant analytic function, the plane does not.

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How does it become a triviality, if Riemann mapping theorem is assumed? – Anweshi Jan 2 '10 at 20:40
The Riemann mapping theorem, as far as I know, only states that a simply connected region in the complex plane is conformally equivalent to the unit disk (unless it is the entire plane). – Harald Hanche-Olsen Jan 2 '10 at 20:42

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