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.)