Theorems

# Bers simultaneous uniformisation

quasifuchsian group is a Kleinian group whose limit set is contained in an invariant Jordan curve. A quasifuchsian group of the first kind is one whose limit set is the whole of the invariant Jordan curve. The choice of the name “quasifuchsian” is justified by the following

Proposition: A Klenian group $\Gamma$ is quasifuchsian iff it is quasiconformally conjugate (in $\mathrm{PSL}_2\mathbb{C}$) to a Fuchsian group.

Proof: The reverse direction is easy: boundary extensions of quasiconformal maps send limit sets of Fuchsian groups, which are round circles, to invariant Jordan curves.

The forward direction appears in a paper of Bers, and proceeds by noting that the domain of discontinuity $\Omega(\Gamma)$ has two disjoint components, conformally mapping any one of them (with the boundary) to a round disc, and proving that we can extend this to a quasiconformal mapping of the other component.

Given a closed genus-g surface $\Sigma_g$, let $QF(\Sigma_g)$ denote the space of quasifuchsian deformations of the fundamental group $\pi_1(\Sigma_g)$, i.e. all conjugates of $\pi_1(\Sigma_g)$ by quasiconformal maps.

Theorem (Bers). $QF(\Sigma_g) \cong \mathcal{T}(\Sigma_g) \times \mathcal{T}(\overline{\Sigma_g})$ (for $g \geq 2$.)

In other words, given any two Riemann surfaces of the same genus $g \geq 2$, there is some (hyperbolizable) 3-manifold which simultaneously uniformizes them, in the sense that its conformal boundary consists of two components, which are precisely the given Riemann surfaces.

(John Hubbard in his tome on Teichmüller theory compares the result to the works of Hieronymous Bosch: somehow unnatural and horrifying, but still a work of art. I don’t know if I agree, but it’s an interesting description anyhow … )

Proof: Let $\Gamma := \pi_1(\Sigma_g)$, and define a map $\Theta: QF(\Sigma_g) \to \mathcal{T}(\Sigma_g) \times \mathcal{T}(\overline{\Sigma_g})$ by $\phi\Gamma\phi^{-1} \mapsto \left( (\phi(S), \phi), (\bar\phi(S), \bar\phi) \right)$.

It is fairly straightforward to show that $\Theta$ is surjective: given $((X,g), (Y,h)) \in \mathcal{T}(\Sigma_g) \times \mathcal{T}(\overline{\Sigma_g})$, $g \coprod h$ lifts to a quasiconformal map $f: U \coprod L \to \mathbb{H}^2 \coprod \overline{\mathbb{H}^2}$, where U and L denote the upper and lower half-planes resp. We may then check that the quasifuchsian group corresponding to f is sent to the point $((X,g), (Y,h)) \in \mathcal{T}(\Sigma_g) \times \mathcal{T}(\overline{\Sigma_g})$ we started with.

To show $\Theta$ is injective: suppose $\Theta(\rho_1) = \Theta(\rho_2)$, so that there exists some conformal map j taking the conformal boundary of $N_{\rho_1}$ to that of $N_{\rho_2}$, which lifts to a conformal map $\tilde{j}: \Omega(\rho_1) \to \Omega(\rho_2)$ between the respective domains of discontinuity.

Now recall the limit sets $\Lambda(\rho_1)$ and $\Lambda(\rho_2)$ are both invariant Jordan curves, so we may canonically extend $\tilde{j}$ over the limit sets by sending a point on one limit set (thought of as a direction at infinity) to the the corresponding point (direction at infinity) on the other limit set. This gives us a map $\hat{j}: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$, which we may argue is quasiconformal.

Now an extension theorem of Douady-Earle allows us to naturally extend this quasiconformal map $\hat{j}$, or more specifically its quasisymmetric restriction to the limit sets, to a quasiconformal map of the interior disc (i.e. the quasiconformal copy of the hyperbolic plane) which it bounds. That such a map existed we already knew, but the naturality properties that come with the Douady-Earle extension, plus the fact that the quasiconformal $\hat{j}$ is conformal on a subset of $\hat{\mathbb{C}}$ of full measure and hence conformal, tells us that this induced extension is in fact conformal, i.e. in fact $\rho_1(\Gamma)$ is conformally conjugate to $\rho_2(\Gamma)$, as desired.

It is clear (or, at least, reasonable) that the map is continuous, as is its inverse, and we are done (a more rigorous proof of this last part would involve going through the construction of the map more carefully, and probably invoking the continuous dependence given by the measurable Riemann mapping theorem at some point.)

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# Brouwer’s invariance of domain

I am recording here an elementary* result used in the proof of the Teichmueller Existence Theorem.

Theorem (Brouwer). Any injective continuous map from $\mathbb{R}^n$ to itself is open.

*the statement itself sounds elementary enough, but the proof (all known proofs) seem to require some amount of machinery from algebraic topology. Note that one corollary of the theorem is that Euclidean spaces of different dimensions are not homeomorphic, another statement which appears elementary (and intuitive), but is difficult to prove without additional machinery.

It seems that the category of continuous maps contains enough curious things—Peano and other space-filling curves and the like—to quite thoroughly upset our intuitions, although going to the rather more restrictive category of smooth things does restore much of our intuition.

Proof of Theorem: It suffices to show that any such map sends open balls to open balls, or, more narrowly, that for any injective continuous map $f: B^n \to \mathbb{R}^n$$f(0)$ lies in the interior of $f(B^n)$ (where $B^n$ denotes the closed unit ball.)

Let f be as in the hypothesis of the Theorem. $f: B^n \to f(B^n)$ is a continuous bijection between compact Hausdorff spaces and hence a homeomorphism. $f^{-1}: f(B^n) \to B^n$ is continuous; by the Tietze extension theorem, $f^{-1}$ may be extended to a continuous map $G: \mathbb{R}^n \to \mathbb{R}^n$.

G has a zero at $f(0)$, and moreover we have the following

Lemma: If $\tilde{G}: f(B^n) \to \mathbb{R}^n$ is a continuous map with $\|G - \tilde{G}\|_\infty \leq 1$ (i.e. is a small perturbation of G), then $\tilde{G}$ has a zero in $f(B^n)$.

Proof of Lemma: Applying Brouwer’s fixed-point theorem to the function $x \mapsto x - \tilde{G}(f(x)) = (G - \tilde{G})(f(x))$ (from the closed unit ball to itself) yields a $x \in B^n$ s.t. $x = x - \tilde{G}(f(x))$, i.e. $\tilde{G}(f(x)) = 0$.

If $f(0)$ were not an interior point of $f(B^n)$, we may construct a small perturbation of G that no longer has a zero on $f(B^n)$, contradicting the above lemma.

At this point I refer the interested reader to Terry Tao’s blogpost (where the gist of this proof was also taken from; he was interested in Hilbert’s fifth problem, whose solution also makes use of invariance of domain) for the details of the construction.

Corollary Any proper injective continuous map $f: \mathbb{R}^n \to \mathbb{R}^n$ is a homeomorphism.

Proof of Corollary: By invariance of domain, f is open. Now it suffices to show that f is surjective (and hence bijective) for it to be a homeomorphism; but surjectivity follows from f being a proper map into a metrizable space, and hence closed—given $E \subset X$ closed and a sequence of points $f(x_n) \in f(E)$ accumulating to $y \in Y$ and $\epsilon > 0$, $f(x_n) \in B(y,\epsilon)$ (the closed ball) for all $n \geq N = N(\epsilon) \gg 0$. By properness, we have, for all $n \geq N$, $x_n \in K \subset E$ with K compact. Then, up to subsequence, $x_n \to x \in K$. By continuity, $f(x_n) \to f(x) = y$, so $y = f(x) \in f(E)$.

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# Mostow Rigidity: several proofs

Mostow rigidity (or, technically, the special case thereof, for $\mathrm{PO}(1,3)$) states that if M and N are two closed (i.e. compact and boundary-less) hyperbolic 3-manifolds, and $f: M \to N$ is a homotopy equivalence, then f is homotopic to an isometry $g: M \to N$.

This is a strong rigidity result. If M and N were hyperbolic surfaces, by contrast, there are plenty of homotopy equivalences between them which are not homotopic to isometries; indeed, this is the very subject of Teichmüller theory (technically, we’re looking at moduli space, not Teichmüller space, but eh.)

We can state it in algebraic terms, by identifying the hyperbolic 3-manifolds $M \cong \mathbb{H}^3 / \Gamma_M$ and $N \cong \mathbb{H}^3 / \Gamma_N$ with their respective fundamental groups $\Gamma_M = \pi_1(M)$ and $\Gamma_N = \pi_1(N)$, which are (uniform) lattices in $\mathrm{Isom}^+(\mathbb{H}^3) \cong \mathrm{PO}(1,3)$.

Mostow rigidity then states any isomorphism between the lattices is given by conjugation in $\mathrm{PO}(1,3)$. Again, contrast this with the case of lattices in $\mathrm{Isom}^+(\mathbb{H}^2) \cong \mathrm{PO}(1,2)$ (i.e. fundamental groups of hyperbolic surfaces), which have plenty of outer automorphisms (elements of the [extended] mapping class group.)

Here we describe, at a relatively high level, various proofs of this result.

### Step 1: lift and extend

Let $\tilde{h}: \mathbb{H}^3 \to \mathbb{H}^3$ be a lift of h to the universal cover(s). This map is a quasi-isometry, by a Milnor-Švarc argument:

• The Cayley graphs of $\Gamma_M$ and $\Gamma_N$ are both quasi-isometric to $\mathbb{H}^3$ by Milnor-Švarc, the quasi-isometry from the Cayley graphs into hyperbolic space being the orbit map.
• If we now construct quasi-inverses from hyperbolic space to the Cayley graphs, then $\tilde{h}$ is coarsely equivalent to a map between the Cayley graphs obtained by composing one orbit map with the quasi-inverse of the other. To wit: $\tilde{h}$ sends $\gamma x_0$ to $(h_* \gamma) x_0$, and the induced map $h_*$ is an isomorphism of the fundamental groups

The quasi-isometry $\tilde{h}$ from the hyperbolic space $\mathbb{H}^3$ to itself extends to a self-homeomorphism of the Gromov boundary $\partial \mathbb{H}^3$:

This is a standard construction which holds for any quasi-isometry between hyperbolic spaces in the sense of Gromov: the boundary extension is defined by $\partial\tilde{h}([\gamma]) = [\tilde{h} \circ \gamma]$, and Gromov hyperbolicity is used to show that this map is a well-defined homeomorphism; roughly speaking, the idea is that quasi-isometries are like “biLipschitz maps for far-sighted people”, and at the boundary at infinity, the uniformly bounded distortions become all but invisible.

### Step 2: boundary map is quasiconformal

Now we claim that $\partial\tilde{h}: \hat{\mathbb{C}} \to \hat{\mathbb{C}}$ is quasiconformal.

This follows e.g. from the following geometric lemma: if L is a geodesic and P a totally-geodesic hyperbolic plane in $\mathbb{H}^3$ with $L \perp P$, then $\mathrm{diam}(\pi_{\hat(L)}(\tilde{h}(P)) \leq D$ for some constant D depending only on the quasi-isometry constants of $\tilde{h}$. (Here $\hat{L}$ is the unique geodesic uniformly close to the quasigeodesic $\tilde{h}(L)$.)

The proof of the lemma uses three key ingredients from the geometry of hyperbolic space: thinness of ideal triangles, quasigeodesic stability—which says that the image of any geodesic segment by a quasi-isometry is uniformly (depending only on the quasi-isometry constants) close to the actual geodesic with the same endpoints, which we will call the “straightening” of the quasigeodesic—, and the (1-)Lipschitz property of projection to a geodesic.

Let $x \in L \cap P$ be the unique point of intersection. If $y \in P$, let z be the point in $\partial_\infty P$ corresponding to the ray from x through y. Now consider the ideal triangle with L as one of its sides and as the third vertex. h(x) is uniformly close to the images of the two sides ending at z , and also to straightenings $\bar{M_1},\bar{ M_2}$ thereof; since projection to a geodesic is Lipschitz, the projection of h(x) to the straightening $\hat{L}$ of the quasigeodesic h(L) is uniformly close to the projections of $\bar{M_1}, \bar{M_2}$.

But now the image of the geodesic ray [x, z] lies uniformly close to some geodesic ray, whose projection onto $\bar{L}$ lies between the projections of h(x) and h(z). Hence any point on this geodesic ray has image which projects between h(x) and h(z); again using the Lipschitz property of the projection, we obtain uniform bounds on the diameter we are trying to control, as desired.

Now note that any circle in $\partial\mathbb{H}^3 \cong \hat{\mathbb{C}}$ is the boundary of a totally geodesic hyperbolic plane in $\mathbb{H}^3$, and our lemma then tells us that the image of the circle under $\tilde{h}$ has its outradius/inradius ratio bounded above by $e^D$.

### Step 3: boundary map is conformal

And then we would be done, for this implies $\tilde{h}$ is (up to homotopy) a hyperbolic isometry, which descends to an isometry $h: M \to N$.

#### An argument along the lines of Mostow’s original

Suppose $\tilde{h}$ is not conformal. Consider the $\Gamma_M$-invariant measurable line field $\ell_h$ on $\hat{\mathbb{C}}$ obtained by taking the direction of maximal stretch at each point.

Rotating this line field by any fixed angle $\phi$ yields a new measurable line field $\ell_h^\phi$. But now $\bigcup_{0 \leq \phi \leq \pi/2} \ell_h^\phi$ yields a measurable $\Gamma_M$-invariant proper subset of $T^1M$, which contradicts the ergodicity of the geodesic flow on $T^1 M$.

An argument following Tukia, in the spirit of Sullivan

Suppose $\tilde{h} =: \phi$ is differentiable at z. Let $\gamma$ be a Möbius transformation fixing z and $\infty$, with z an attracting fixed point and with no rotational component. Now, by the north-south dynamics that hyperbolic Möbius transformations exhibit, $\gamma^n \phi \gamma^{-n} \to d\phi_z$ as $n \to\ infty$; on the other hand, $\gamma^n \phi \gamma^{-n}$ conjugates $\gamma^n\Gamma_1\gamma^{-n}$ to $\gamma^n\Gamma_2\gamma^{-n}$.

Again, using convergence dynamics, we have $\gamma^n\Gamma_i\gamma^{-n} \to \Gamma_i^\infty$ as $n \to \infty$, for i = 1, 2, where both $\Gamma_i^\infty$ are cocompact and of equal covolume.

Now $d\phi_z$ conjugates $\Gamma_1^\infty$ to $\Gamma_2^\infty$; thus $\phi$ is conformal a.e., and hence conformal.

### Alternative perspective I: the Gromov norm

Gromov gives a proof which follows Step 1 as above, but then proceeds differently: he uses the Gromov norm to show that $\tilde{h}$ is conformal, without (separately) showing that it is quasiconformal. We sketch an outline of this here; see Calegari’s blog and/or Lücker’s thesis for details.

The Gromov norm is a measure of complexity for elements of the (singular) homology—more precisely, it is the infimum, over all chains representing the homology element, of $L^1$ norm of the chain.

It is a theorem of Gromov that the Gromov norm of (the fundamental class of) an oriented hyperbolic n-manifold M is equivalent, up to a normalization dependent only on n, to the hyperbolic volume of M. The proof proceeds by

• observing that it suffices to take the infimum over rational chains of geodesic simplices; this suffices to give the lower bound, with a constant given in terms of volume of an ideal simplex;
• using an explicit construction involving almost-ideal simplices with almost-equidistributed vertices to obtain a upper bound.

Then, since $\tilde{h}$ is a homotopy equivalence (and by looking at this last construction carefully), it—or, rather, its extension to $\overline{\mathbb{H}^3}$—sends regular simplices close to regular ideal simplices. Simplices in the chains from the upper bound construction above are (almost) equidistributed, and the error can be made arbitrarily small as we use the construction to get arbitrarily close to the bound.

Translating the vertices of our ideal simplices, we get a dense set of equilateral triangles that are sent by $\partial\tilde{h}$ to equilateral triangles on the boundary. This implies $\partial\tilde{h}$ is conformal a.e., and hence conformal.

(In short: Gromov’s theorem and its proof allows us to control volume in terms of explicitly-constructed singular chains and to control what happens to simplices in those chains under our homotopy equivalence.)

### Alternative perspective II: dynamically characterizing locally-symmetric Riemannian metrics

Besson, Courtois and Gallot have an entirely different proof, which uses a characterization of locally-symmetric Riemannian metrics in terms of entropy to conclude that, since f is a degree-1 map between locally-symmetric spaces, M and N are homothetic. It then follows from e.g. an equal-volume assumption that M and N would be isometric.

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# Ergodicity of the Geodesic Flow

### The geodesic flow

Given any Riemannian manifold M, we may define a geodesic flow $\varphi_t$ on the unit tangent bundle $T^1M$ which sends a point (x, v) to the point $(\varphi_t x, \varphi_t^* v)$, where

• $\varphi_t x$ is the point distance from x along the geodesic ray emanating from x in the direction of v, and
• $\varphi_t^* v$ is the parallel transport of v along the same ray

(it’s a mouthful, isn’t it? It’s really simpler than all those words make it seem.) Note, at each point, we remember not just where we are—the point $x \in M$—, but also where we’re going—the direction vector $v \in T_x M$; if we were to forget this second piece of information, we would become a little unmoored: here we are … where should we go next?

### Ergodicity

When M is a closed (i.e. compact, no boundary) hyperbolic surface, or more generally closed with strictly negative curvature, this geodesic flow is ergodic, i.e. any subset of $\Sigma$ or M invariant under the flow has either zero measure, or full measure. Here the measure on our Riemannian manifold is the pushforward of the Lebesgue measure through the coordinate charts.

Since linear combinations of step functions are dense in the space of bounded measurable functions, we may equivalently define ergodicity as: any measurable function invariant under the flow is a.e. constant.

(Side note: with more assumptions on the curvature we may relax the compactness assumption to a finite volume assumption)

### The Hopf argument (for closed hyperbolic manifolds)

This is essentially due to the exponential divergence of geodesics in negative curvature , and the splitting of the tangent spaces $T_ v T^1M = E^s_v \oplus E^0_v \oplus E^u_v$ into stable, tangent (flowline), and unstable distributions; these give rise to three maximally transverse foliations, the stable foliation $W^s$, the unstable foliation $W^u$, and the foliation by flowlines $W^0$.

The flow is exponentially contracting in the forward time direction on the leaves of the stable foliation $W^s$, and on which the flow is exponentially contracting in the reverse time direction the leaves of the unstable foliation $W^u$. In other words, the flow is Anosov.

We may describe these foliations explicitly in the case of constant negative curvature—if we take $\gamma$ to be the geodesic tangent to $v \in T^1M$,

• $W^s$(v) is (the quotient image of) the unit normal bundle to the horosphere through $\pi(v) \in M$ tangent to the forward endpoint of $\gamma$ in $\partial_\infty\mathbb{H}^n \cong \partial_\infty\widetilde{M}$. “forward” here being taken with reference to how v is pointing along $\gamma$;
• $W^u(v)$ is (the quotient image of) the unit normal bundle to the horosphere through $\pi(v)$ tangent to the backward endpoint of $\gamma$ in $\partial_\infty\mathbb{H}^n \cong \partial_\infty\widetilde{M}$;
• $W^0(v) = \gamma$.

#### Step 1

Suppose f is a $\phi$-invariant function; by replacing f with min(f, C) if needed, WMA f is bounded. Since continuous functions are dense in the set of measurable functions on M, we may approximate f in $L^1$ by bounded continuous functions $h_\epsilon$.

By the Birkhoff ergodic theorem, forward time averages [w.r.t. $\phi$] exist for $h_\epsilon$.

By an argument involving the $\phi$-invariance of f and the triangle inequality, f is well-approximated (in $L^1$) by the forward time averages of $h_\epsilon$.

#### Step 2

The forward time averages of $h_\epsilon$ are constant a.e., since by invariance these averages are already constant a.e. on (each of) the leaves of $W^0$, and they are also constant a.e. on (each of the) unstable and stable leaves, by uniform continuity of $h_\epsilon$.

#### Step 3

To conclude that time averages, and hence our original arbitrary integrable function, are constant a.e. on M, we (would like to!) use Fubini’s theorem: locally near each $(x_0,v_0) \in T^1M$, the set of (x, v) along each of the foliation directions at which the time averages are equal to those at $(x_0,v_0)$ has full measure, by the previous Step.

By Fubini’s theorem applied to the three foliation directions, we (would) conclude that the set of nearby (x, v) at which the time averages are equal to those at $(x_0,v_0)$ has full measure. Hence the time averages are locally constant, and since $T^1M$ is connected we are done.

### But! (Also more generally, for K < 0)

The problem is that while our stable and unstable leaves are differentiable, the foliations need not be—i.e. the leaves may not vary smoothly in their parameter space.

To justify the use of a Fubini-type argument one instead shows that that these foliations are absolutely continuous.

The proof then immediately generalizes to all compact manifolds with (not necessarily constant) negative sectional curvature. For more general negatively-curved nanifolds, the stable and unstable foliations $W^s$ and $W^u$ may still be described in terms of unit normal bundles over horospheres, where horospheres are now described, more generally, as level sets of Busemann functions.

The proof of absolute continuity of the foliations proceeds as follows

1. Showing that the stable and unstable distributions $E^s$ and $E^u$ (also the “central un/stable” or “weak un/stable” distributions, i.e. $E^{s0} := E^s \oplus E^0$ and  $E^{u0} := E^u \oplus E^0$) (of any $C^2$ Anosov flow) are Hölder continuous—i.e. given $x, y \in M$, the Hausdorff distance in $TTM$ between the stable subspace $E^s(x)$ and the stable subspace $E^s(y)$ is $\leq A \cdot d(x,y)^\alpha$.
Roughly speaking, this is true because any complementary subspace to $E^s$ will become exponentially close to $E^s$ under the repeated action of the geodesic flow, by the same mechanism that makes power iteration tick; and the distance function on M is Lipschitz. Analyzing the situation more carefully, and applying a bunch of simplifying tricks such as the adjusted metric described in Brin’s section 4.3, yields the desired Hölder continuity.
2. Using this, together with the description of horospheres as limits of sequences of spheres with radii increasing to $+\infty$, to establish that between any pair of transversals for the un/stable foliation, we have a homeomorphism which is $C^1$ with bounded Jacobians, and hence absolutely continuous.
Very slightly less vaguely, Hölder continuity of $E^{u0}$, together with the power iteration argument as above, implies tangents to transversals to the stable foliation $W^s$ become exponentially close; given regularity of the Riemannian metric, this implies the Jacobians of the iterated geodesic flow on these transversals become exponentially close. By a chain rule argument and another application of the power iteration argument, this implies that the Jacobians of the map between transversals are bounded.
This condition on the foliations is known as transversal absolute continuity, and implies, by a general measure theoretic argument (see section 3 of Brin’s article), absolute continuity of the foliations.
3. Note that this last step, at least as presented in Brin, appears to require the use of pinched negative curvature.

### References

Eberhard Hopf, “Ergodic theory and the geodesic flow on surfaces of constant negative curvature.” Bull. Amer. Math. Soc. 77 (1971), no. 6, 863–877.

Yves Coudene, “The Hopf argument.

Misha Brin, “Ergodicity of the Geodesic Flow.” Appendix to Werner Ballman’s Lectures on Spaces of Nonpositive Curvature.

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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
Standard

# Convergence of power series: Cauchy-Hadamard

The domain of convergence of a general holomorphic function can be quite arbitrary—in one (complex) dimension any domain is the domain of convergence for some holomorphic function, and in higher dimensions domains of holomorphy may be characterized using certain notions of convexity: holomorphic convexity, which involves the propagation of maximal-principle-type estimates, or pseudoconvexity, which is similar but uses the larger class of plurisubharmonic functions instead of holomorphic functions

The domain of convergence of functions defined by power series, however, are much more restricted and exhibit more symmetry: in one dimension, the domain of convergence of a power series $f(z) = \sum_{n=0}^\infty c_n z^n$ is a disc of radius R satisfying $\dfrac 1R = \limsup_{n \to \infty} \left( |c_n|^{1/n} \right)$. This we may prove by noting that, for $|z| > R$ with R as above, the terms of the series do not converge to zero, and hence the series cannot converge; for $|z| < R$, the series will converge by comparison with a geometric series. This is the Cauchy-Hadamard theorem. (On the boundary of the disc of convergence the behavior of the power series is more subtle—we can extract some information using results such as Abel’s theorem.)

There is an analogue in higher dimensions, which states that (using multi-index notation) a power series $f(z) = \sum_\alpha c_\alpha z^\alpha$ converges in a polydisc with polyradius $\rho$ iff $\limsup_{|\alpha| \to \infty} \sqrt[|\alpha|]{|c_\alpha|\rho^\alpha} \leq 1$; the proof of this is exactly analogous to the one given above. Note that the union of all such polydiscs may not itself be a polydisc–e.g. the domain of convergence of the power series $\sum_{k=0}^\infty z_1^k z_2^k$ is precisely the set $\{|z_1z_2|| < 1\}$, which is not a polydisc.

Nevertheless these domains of convergence still have a certain sort of radial symmetry—they belong to the more general sort of domains known as a complete (a.k.a. logarithmically-convex) Reinhardt domains. These are domains U for which $(z_1, \dots, z_n) \in U$ implies $(e^{i\theta_1} z_1, \dots, e^{i\theta_n} z_n) \in U$ for all $\theta_1, \dots, \theta_n \in \mathbb{R}$ (this is the “Reinhardt” part) and $(w_1, \dots, w_n) \in U$ whenever $|w_i| \leq |z_i|$ (this is the complete, or log-convex part.) Note that one-dimensional complete Reinhardt domains are precisely discs centered at the origin.

We can think of the Cauchy-Hadamard condition in several variables as giving a constraint on the polyradius $\rho$, so that the domain of convergence $\Omega$ is the union of all polydiscs with polyradius satisfying the given constraint or, equivalently, the logarithmic image $\mathrm{Log}(\Omega)$ of the domain of convergence is the set of all polyradii satisfying the given constraint. $\mathrm{Log}(\Omega)$ is convex (and closed to the right) but $\Omega$ need not be an orthant, as would be the case if $\Omega$ were a polydisc.

Standard
Theorems

# Arzelà–Ascoli

Version 1 (215)

Let K be a compact subset of $\mathbb{R}$ and let $Y \subset C(K)$ be an equicontinuous family of functions. Assume that f(x) is bounded for each fixed $x \in K$ as f varies over Y. Then Y is pre-compact, i.e. every sequence in Y has a subsequence converging to some $f \in C(K)$ (equivalently, the closure of Y in C(K) is compact.)

In other words

If $\mathcal{F}$ is an equicontinuous and uniformly bounded family of continuous functions on $\mathbb{R}$, then any sequence drawn from $\mathcal{F}$ has a subsequence which converges (uniformly on any compact set) to some continuous function.

In complex analytic terms

Any equicontinuous and uniformly bounded collection of continuous functions forms a normal family: this is the second half of Montel’s theorem (the first half states that a uniformly bounded family of continuous functions is equicontinuous.)

Version 2 (Tao)

Let Y be a metric space, be a compact metric space, and let $(f_\alpha)_{\alpha \in A}$ {(f_\alpha)_{\alpha \in A}} be a family of bounded continuous functions from X to Y. The following are equivalent:

• (i) $\{f_\alpha \mid \alpha \in A\}$ is a precompact subset of the bounded continuous functions from X to Y.
• (ii) $(f_\alpha)_{\alpha \in A}$ is pointwise precompact (i.e. for every $x \in X$, the set $\{f_\alpha(x) \mid \alpha \in A\}$ is precompact in Y) and equicontinuous.
• (iii) $(f_\alpha)_{\alpha \in A}$ is pointwise precompact and uniformly equicontinuous (= equicontinuous + uniform bound on size of neighborhoods (radius of balls))

Proof (Arzelà–Ascoli diagonalisation)

Now a standard technique in analysis, as first shown to you by Sarnak

Given $\epsilon > 0$, by equicontinuity of $\mathcal{Y}$ there is a $\delta > 0$ such that $\left| f(x) - f(y) \right| < \epsilon$ whenever $d_X(x,y) < \delta$.

Key observation: since $K \subset \cup_{k \in K} B(k, \delta)$ is compact, there exists a finite subcover $\cup_{j=1}^n B(p_j, \delta) \supset K$.

Let $M_j = \sup_{f \in \mathcal{Y}} \left| f(p_j) \right| < \infty$ and set $M = \max \{M_j \:|\: j=1, \cdots, N\} < \infty$.

Given $x \in K$, $x \in B(p_j, \delta)$ for some $j = 1, \cdots, N$; then $\left| f(x) – f(p_j) \right| < \epsilon$. Then

$\left|f(x)\right| < M_j + \epsilon \leq M + \epsilon$

for all $x \in K$; hence $f$ is uniformly bounded as $f$ varies over $\mathcal{Y}$ and $x$ varies over $K$.

Next, let $E \subset K$ be a countable dense subset of $K$. We claim that any sequence $\{f_n\}\stdseq \subset \mathcal{Y}$ has a subsequence $\{f_{n_j}\}_{j=1}^{\infty}$ for which $\{f_{n_j}(e)\}_{j=1}^{\infty}$ converges for each $e \in E$.

This we prove using a diagonal argument (not quite Cantor’s.) Enumerate $E = \{e_1, e_2, \cdots \}$. Write out a table

$\begin{array}{l l l l l} & f_1 & f_2 & f_3 & \cdots \\ e_1 & f_1(e_1) & f_2(e_1) & f_3(e_1) & \cdots \\ e_2 & f_1(e_2) & f_2(e_2) & f_3(e_2) & \cdots \\ e_3 & f_1(e_3) & f_2(e_3) & f_3(e_3) & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array}$

First construct a family of sequences as follows: let $A_1$ be a convergent subsequence of the elements of the first row.

For $n > 1$, let $A’_n$ be the subsequence of the $n\thc$ row of elements picked out by the indices of $A_{n-1}$, and let $A_n$ be a convergent subsequence of $A’_n$.

At each step a convergent subsequence exists because $f \in \mathcal{Y}$ is uniformly bounded, and hence $f_i(e_j) \in S$ for all $i,j \in \nats$, where $S$ is some closed, bounded (hence compact) interval of reals.

Now define the sequence $\{f_{n_j}\}_{j=1}^{\infty}$ as follows: for each $j$, look at the $j\thc$ element of $A_j$, which is of the form $f_k(x_l)$ for some natural numbers $k,l$. Then $n_j := k$, i.e. $f_{n_j} = f_k$. By recursion, this subsequence converges pointwise for every $e_i \in E$.

Finally, we will show that the subsequence we have just constructed is convergent in $C(K)$, by showing that it is uniformly Cauchy, i.e. given any $\epsilon > 0$, there exists a $N \in \nats$ such that $\left| f_n(x) – f_m(x) \right| < \epsilon$ for all $x\in K$ whenever $n,m \geq N$.

We have the finite open cover $\cup_{j=1}^n B(p_j, \delta) \supset K$. For each $j = 1, \cdots, N$, $B(p_j, \delta)$ contains some element of $E$—call it $e’_j$—since $E$ is dense.

Note $\{f_n(e’_j)\}\stdseq$ is a Cauchy sequence for each fixed $j$, since we have already proven that $\{f_n(e)\}\stdseq$ converges for any $e \in E$. Hence there exists a natural number $R_j$ such that whenever $n,m \geq R_j$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$.

Pick $R = \max\{R_1, \cdots, R_N\}$. Then if $n,m \geq R$, we have $\left| f_n(e_j) – f_m(e_j) \right| < \epsilon$ for all $j=1, \cdots, N$.

Now, given any $x \in K$, $X \in B(p_j, \delta)$ for some natural number $j$. Then

$\left| f_n(x) – f_m(x) \right| \leq \left| f_n(x) – f_n(e’_j) \right| + \left| f_n(e’_j) – f_m(e’_j) \right| + \left| f_m(e’_j) – f_m(x) \right| < \epsilon$

for all $n,m \geq R$.

There are also other proofs (e.g. using functional analysis, which I shall perhaps explore one day.)

In short

A way of obtaining convergence / compactness in function space from stronger continuity / boundedness assumptions.

Standard