Theorems

Uniformization vs. Classification of Space Forms

Or: one illustration of the relationship between the notions of (Riemannian) isometry and  conformal equivalence.

Uniformisation Theorem

Any connected, simply-connected Riemann surface X is conformally equivalent to $S^2$, the complex plane, or the unit disk.

General scheme of proof (for details see Donaldson’s book, or hyperlinks in the text of the scheme.)

• Any compact (“elliptic”) connected surface X is conformally equivalent to a Riemann sphere: we can show this by demonstrating [the existence of] a meromorphic map from X with a single simple pole.
• One way of doing this is to study the $H^{0,1}_X$ Dolbeault cohomology of X using the following result (the “Main Theorem” in Donaldson’s book): Let $\rho$ be a 2-form on X. There is a solution f to the equation $\Delta f = \rho$ iff $\int_X \rho = 0$, and the solution is unique up to additive constant.
• More explicit constructions using Dirichlet integrals  (similar to those that appeared in the original proofs, due independently to Poincaré and Koebe) may be found in this senior thesis or these notes.
• For noncompact (“parabolic” or “hyperbolic”)  X, we can use similar, but more careful, arguments to exhibit [the existence of] a meromorphic function from X with a single simple pole, which injectively maps X into an open subset of the Riemann sphere which turns out to be $S^2 - I$, where I is a single closed interval (possibly degenerate, i.e. a single point, but not empty.)
• The version of the “Main Theorem” Donaldson uses here (with extra care taken regarding points at infinity) reads: Let $\rho$ be a 2-form of compact support on X with $\int_X \rho = 0$. There is a function $\phi: X \to \mathbb{R}$ with $\Delta\rho = \phi$ s.t. $\rho \to 0$ at infinity in X.
• As above, one can furnish more explicit constructions, using Dirichlet integrals and Green’s functions. See the same references as above for details.

This may be viewed as a generalisation of the Riemann Mapping Theorem, which states that any (connected, simply-connected, proper) open subset of the complex plane is conformally equivalent to the unit disk.

Note that a Riemann surface is an orientable 2-manifold, and conversely any orientable 2-manifold can be given a Riemann surface structure. This suggests (or, at least, causes me to endlessly confuse uniformization with) a similar result, formulated purely in terms of Riemannian geometry.

Classification of space forms

Any complete Riemannian manifold with constant sectional curvature has universal cover (isometric to) one of $S^n$ (spherical), $\mathbb{R}^n$ (Euclidean / flat), or $\mathbb{H}^n$ (hyperbolic).

Outline of Proof (for details see e.g. Chapters 7 and 8 of do Carmo’s Riemannian Geometry)

• First we need a Lemma (Cartan) (determination of metric by means of the curvature): Let $M, \tilde{M}$ be Riemannian manifolds of dimension n. Choose $p \in M$ and $\tilde{p} \in \tilde{M}$ and a linear isometry $i: T_pM \to T_{\tilde{p}} \tilde{M}$.

Proof: Let V be a normal neighborhood of p in M (such that the exponential maps required below are defined) and define a map $f: V \to \tilde{M}$ using the exponential maps at p and $\tilde{p}$.

For any $q \in V$ there exists a unique normalized geodesic $\gamma: [0,t] \to M$; define $P_t$ to be parallel transport along $\gamma$, and $\phi_t$ to be transport along this geodesic push forward to $\tilde{M}$.

Now if $(x,y,u,v) = (\phi_t(x), \phi_t(y), \phi_t(u), \phi_t(v))$ for all x, y, u, v and t (i.e. if the curvature along R and $\tilde{R}$ is equal and compatible with our notion/s of parallel transport) then $f: V \to f(V)$ is a local isometry and $df_p = \mathrm{id}$ (then our manifolds are in fact isometric.)

• Now consider the map from $\tilde{M}$ to model space as in the Cartan lemma above, which is globally defined here by completeness. Invoke the Cartan lemma to show it is a local isometry; then show it is a covering map, hence a diffeomorphism and so a global isometry.
• The concrete details differ slightly between the case of non-positive curvature and the case of positive curvature.
• For NPC: map from $\tilde{M}$ to model space ($\mathbb{H}^n$ or $\mathbb{R}^n$) is well-defined due to NPC (by a Theorem of Hadamard, 7.3.1 in do Carmo), and is a covering map from the following Lemma (here, the inequality is satisfied a fortiori because the exponential maps are diffeomorphisms)

Lemma NPC: If f is a local diffeomorphism from a complete Riemannian manifold M onto a Riemannian manifold N which is expanding in the sense that $|df_p(v)| \geq |v|$ for all $p \in M, v \in T_pM$, then f is a covering map

Sketch of prooff has path lifting property for curves in N: since f is a local diffeomorphism, we can lift small initial segments of paths. By the non-contracting property of the map, paths upstairs are no shorter, and so entire path must remain within compact neighborhood by completeness—but then we can argue that we can lift the entire path.

• In positive curvature: map from $\tilde{M} to S^n$ as in Cartan Theorem (with more careful choice of q) is a local isometry; pick another such map, and observe from the Lemma below that the maps are equal where they agree (here we need two charts, not just one.)
• Combine the two to obtain a map g on all of $S^n$; then argue that g is a covering map by compactness of codomain (proper local diffeomorphisms are covering maps.)

Lemma PC: If f and g are local isometries from a connected Riemmanian manifold M to a Riemannian manifold N s.t. $f(p) = g(p)$ and $df_p = dg_p$ for some $p \in M$, then $f = g$

Sketch of proof: Given hypotheses imply f = g in some normal neighborhood V of p; now use connectedness to propagate this identity outwards along paths in M

This suggests an alternative proof of uniformisation, via the classification of (smooth orientable) surfaces and Riemannian geometry: any connected and simply-connected Riemann surface, being a connected and simply-connected orientable topological manifold of real dimension 2, is [can be given a Riemannian metric] isometric to the 2-sphere, Euclidean 2-space, or hyperbolic 2-space; we may put conformal structures on these conformally equivalent to the sphere, the complex plane, and the unit disk.

Unfortunately, the remaining gap is the hardest bit of the argument. Conformal equivalence is a coarser equivalence than Riemannian isometry—a Riemannian metric yields a conformal structure; equivalent metrics yield equivalent conformal structures, but a conformal equivalence need not be an isometry. e.g. $\mathbb{R}^2_+$ (with a flat metric) and $\mathbb{H}^2$ are conformally equivalent. It might be possible to show that any connected, simply-connected Riemann surface is conformally equivalent to a surface of constant sectional curvature, but a priori this does not seem any easier than the other proofs of uniformisation.

Coarser though conformal equivalence may be as an equivalence relation, it is also surprisingly (?) more subtle. Coarsely speaking, the additional information which the Riemannian metric records beyond the conformal structure seems to provide a nice organizing principle which makes things easier. Slightly more precisely: the set of all Riemann surfaces of a fixed genus g, modulo conformal equivalences, is exactly the moduli space $\mathcal{M}_g$, while the same set, modulo [something only slightly stronger than] metric equivalence, is Teichmüller space Teich(g). The former is a quotient of the latter, and is much more difficult to study.

(Offline) References

• Simon Donaldson, Riemann Surfaces
• Manfredo P. do Carmo, Riemannian Geometry
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