A version of the uniformization theorem for the 2-manifolds that I have read about says that any connected 2-manifold is diffeomorphic to either of the 3 constant curvature model spaces modulo an action on them by a discrete subgroup of their isometries acting freely and properly discontinuously. This apparently implies that every connected 2-manifold has a complete Riemannian metric with constant Gaussian curvature.

I don't know of a reference which explains how this proof works.

I would like to understand what this really means and how does this connect to the Hopf-Rinow theorem (which I am more familiar with) which also characterizes geodesically complete spaces in general.

Especially I would like to understand why the two caveats on the group action, "freely" and "properly discontinuously" necessary. Whats the intuition behind them?

I would like to know if this is something special about 2-manifolds or are their similar analogues in higher dimensions? If not then what goes wrong?

Some geometers tell me that theorems which guarantee existence of a complete Riemannian metric on some space are in some sense "weak" since it basically says that one can make the metric "bigger" towards the edges of the spaces so that the geodesic never "hits an end".(and apparently this is not interesting) I don't quite understand these comments.

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    $\begingroup$ If the action is not properly discontinuous then the quotient space need not be a manifold! Freeness ensures that the quotient map is a covering map. The fact that a quotient by a free, properly discontinuous group action is a covering map is not special to 2-manifolds, but the existence of a uniformisation theorem is. $\endgroup$
    – HJRW
    Aug 13, 2010 at 17:45
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    $\begingroup$ Once you classify surfaces up to diffeomorphism it is easy to construct metrics of constant curvature by hand, e.g. 2-torus is R2 modulo the group generated by translations along coordinate axes, giving a flat metric. The case of hyperbolic surfaces isn't much harder: just consider a regular polygon in the hyperbolic disk with angle $2 \pi/4g$ and identify its sides in a standard way. The case of round sphere is trivial, and non-orientable surfaces are handled similarly. $\endgroup$ Aug 13, 2010 at 20:12
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    $\begingroup$ As for you other questions, I do not see why Hopf-Rinow is more relevant than Arzela-Ascoli :), and the reason "uniformization" does not work in higher dimensions is that there are many more higher-dimensional manifolds. On the other hand, geometrization does work in dimension 3 as you surely know. Finally, any manifold admits a complete Riemannian metric which is perhaps what your geometers mean, but this can hardly be compared with geometrization by nice metrics, such as of constant curvature or locally homogeneous. $\endgroup$ Aug 13, 2010 at 20:13
  • $\begingroup$ @Igor Thanks for your insights. So what I understand is that given any 2-manifold one can put on it a metric which is complete. Can you give some explicit examples? Like what is the geodesically complete metric on the open-2-disk? (sounds surprising that there exists one!) And given some metric is there any easy to do check for completeness? Could you elaborate on the statement that "any manifold admits a complete Riemannian metric" ? (I thought one way to interprete the Hopf-Rinow theorem is that the manifold needs to be complete as a metric space to be geodesically complete) $\endgroup$
    – Anirbit
    Aug 14, 2010 at 8:41
  • $\begingroup$ I think the topics your question addresses mainly apply to closed manifolds. However, the disk can be given (for example) the hyperbolic metric, which is complete. I'd guess that any complete metric on the disk would need to be similar, since the point is that it must "take infinitely long" to get to the boundary. In practice, I think it is usually pretty clear whether a manifold is complete or not (once you've given it a metric), and you can generally prove it by hand. I haven't heard the statement that any manifold admits a complete metric before, but I'd guess that you can (cont.) $\endgroup$ Aug 14, 2010 at 12:18

3 Answers 3


There is a proof based on elliptic PDE in a paper of mine with Michael Taylor that was published in the Israel Journal of Math in 2002 Vol 130, 323-346. The emphasis is on the general open case of domains with arbitrary boundary, and the issue is to find a solution of the curvature equation which grows sufficiently strongly to produce a complete metric. --


The uniformization theorem is really about complex structures. It says that up to bi-holomorphism, there are only $3$ connected simply connected $1$-dimensional complex manifolds: the plane, the unit disc, and the Riemann sphere. This theorem is quite difficult to prove, and I do not now a reference (although a collective book on it is in preparation, initiated by Étienne Ghys).

The link with Riemannian geometry is that in real dimension $2$, a complex structure is the same than a conformal structure. So, if you consider an orientable surface, you can endow it with any riemannian metric, whose conformal class gives you a complex structure on it (rotation by an angle of $\pi/2$ plays the role of multiplication by $i$). Then its universal covering is one of the three simply connected complex surfaces above. Since each of them can be endowed with a Riemannian metric of constant curvature whose conformal class corresponds to their complex structure and the groupe of deck transformation acts by isometries, you can push this metric to your original surface.

The main reason you cannot do this in higher dimension is that you do not have complex numbers to help you. A great number of theories have been studied that are higher-dimensional analoguous, though. A few of them are: Yamabe problem (finding a constant scalar curvature metric in a conformal class), geometrization (decompose a $3$-manifold into pieces having nice, homogeneous metrics), Einstein manifolds (constant Ricci curvature metrics). If you want to know more about these topics, there are plenty of references but A panoramic view of geometry by Marcel Berger is my favorite.

  • $\begingroup$ Surely there are references to the uniformization theorem, e.g. see references at the end of math.harvard.edu/theses/senior/chan/fulldraft7.pdf, an expository thesis written under direction of Siu. One such reference is [Abikoff, William, The uniformization theorem. Amer. Math. Monthly 88 (1981), no. 8, 574--592]. $\endgroup$ Aug 14, 2010 at 13:44
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    $\begingroup$ Donaldson's notes on Riemann surfaces cover uniformisation: www2.imperial.ac.uk/~skdona/RSPREF.PDF I find counting degrees of freedom quite instructive. In dim $n$ the curvature tensor has $O(n^4)$ components, the metric $O(n^2)$, and a conformal factor only $1$. Trying to find a metric of prescribed curvature will give a very overdetermined system except in dim $\leq 3$. And it's only in 2 dimensions, where the curvature reduces to a scalar, that one has a hope of obtaining constant curvature by a conformal change. $\endgroup$
    – Tim Perutz
    Aug 14, 2010 at 14:06
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    $\begingroup$ My favorite exposition of the classical proof of the uniformization theorem is in the first chapter of John Hubbard's book on Teichmuller theory. $\endgroup$ Aug 14, 2010 at 15:23
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    $\begingroup$ Notice that this argument proves a much stronger conclusion than the "easy" gluing-of-polygons argument that Igor mentioned. The gluing-of-polygons argument proves that there exists a constant-curvature metric. The complex-analysis argument proves that there exists a constant-curvature metric <i>in the same conformal class as your favorite Riemannian metric</i>. $\endgroup$
    – Dave Futer
    Aug 21, 2010 at 12:26

There is an excellent expository article by Peter Scott titled "The Geometries of 3-Manifolds." The first section is spent talking about uniformization of surfaces (which is what you are interested in), and he also discusses 2-Oribfolds, which are what one gets when the group action is properly discontinuous but not free. The rest of the article is certainly good reading, but not really relevant to what you asked. Most importantly though, he sketches a proof of uniformazation for surfaces. The basic idea is that the only group which acts nicely on $S^2$ is $\mathbb{Z}/2\mathbb{Z}$, with quotient $P^2$. Furthermore, discrete groups of isometries of $\mathbb{R}^2$ with compact quotient are isomorphic to a group with a finite index subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$, and hence the quotient is either a torus or a Klein bottle.Then one can classify the isometries of the hyperbolic plane, $\mathbb{H}^2$, and show that $\mathbb{Z} \oplus \mathbb{Z}$ cannot be a discrete, orientation preserving subgroup of $Isom(\mathbb{H}^2)$, and hence the torus and Klein bottle cannot admit a hyperbolic structures. Furthermore, it is easy to show (with the classification of hyperbolic isometries) that other surfaces do admit hyperbolic structures.

Also, to give you an idea of why free and properly discontinuous are important, it helps to see a few examples. First, consider $\mathbb{R}^2$ and let $G$ be the group generated by rotation around the origin through the angle $2\pi/n$. This action has a fixed point at the origin. Topologically, the quotient is just $\mathbb{R}^2$, but is has a cone point with angle $2\pi/n$ at the origin, so geometrically it is different. This type of space is what is known as an orbifold.

Without proper discontinuity, things get even worse. Consider $S^1$ and the group $G$ generated by rotation through an irrational multiple of $\pi$. The resulting quotient is not even Hausdorff (the orbit of a point under $G$ is dense in $S^1$), and we generally want to avoid that sort of thing.

In dimension 3, things get a little harder, but geometrization still works. A really good reference for this dimension is William Thurston's "3 Dimensional Geometry and Topology," which is easily one of my favorite math books ever written.

In dimension 4, the same sort of approach is a lot harder. The reason is that in dimensions 2 and 3, we have a fair deal of control as to what sorts of groups show up as fundamental groups of (closed) manifolds. It can be shown (I don't recall a reference for this off the top of my head, perhaps someone else has it) that for any finitely presented group $G$, there exists a closed 4-manifold $M$ with $\pi_1(M) = G$. So, we don't have the kind of set-up like in dimension 2 where we know which fundamental groups arise, and then show that they are isometries of one of three model spaces.

Hope this helps.


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