Mapping the Poincaré disk to hyperbolic surfaces in $\mathbb{R}^3$.

Question

Take any hyperbolic surface with constant curvature in $\mathbb{R}^3$, such as Dini's surface, or a hyperboloid of constant curvature.

If I understood things correctly, for any such surface, we should be able to find a conformal mapping from the Poincaré disk to the surface, in such a way that arcs with the same Poincaré metric map to arcs on the surface with the same "usual" arc length.

Three basic questions come to mind:

• Is such a mapping guaranteed to exist?
• Is there a known method with which to find such a mapping?
• Is this mapping unique?

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Image credit to David Dumas.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$If $S$ is a surface in $\Reals^{3}$ of constant curvature $-1$, then sufficiently small pieces of $S$ are indeed locally isometric to small pieces of the hyperbolic plane, and the universal cover of $S$ is globally isometric to a proper open subset of the hyperbolic plane. (There's no local isometry from the entire hyperbolic plane to $S$ by Hilbert's theorem. Equivalently, $S$ is not geodesically complete.)

Methods for finding a "hyperbolic parametrization" of $S$ depend on $S$. For the Dini-type surface in the question, there are formulas depending on definite integrals of explicit elementary functions, similar to the standard descriptions of a surface of rotation of constant Gaussian curvature.

If you'll excuse the self-promotion, in "action-angle" coordinates, the Gaussian curvature formally becomes $-\frac{1}{2}$ times the second derivative of a "profile" function defining the metric. These coordinates are straightforwardly adapted to surfaces invariant under a helical group of Euclidean isometries. For a metric of curvature $-1$, the profile is a quadratic polynomial $\varphi(u) = a + u^{2}$. If $a = 0$, the profile is a tractrix; if $a > 0$ (diagram below), one gets hyperbolic surfaces with helical symmetry, and with "U"-shaped profile curves resembling the profiles of constant-curvature surfaces of rotation.

Regarding uniqueness, if $U_{1}$ and $U_{2}$ are regions in the hyperbolic plane, and if $\Phi_{1}:U_{1} \to S$ and $\Phi_{2}:U_{2} \to S$ are hyperbolic parametrizations, then $\Phi_{2}^{-1} \circ \Phi_{1}$ is a hyperbolic isometry everywhere it's defined.

Answer 2

The Dini surface is just a "rolled up" Tractricoid

The Tractricoid is just an hyperbolic figure where the "pointy end at infinity " is an ideal point and the "meridians" are hyperbolic parallel lines (converging to the "pointy end at infinity ", left in your diagram).

(imagine only one half of the Tractricoid and cut open along an meridian.

The parallels on the tractrix are parts of horocycles.(which are also centered around the "pointy end at infinity ")

The dini surface is just a multiple wider that the normal Tractricoid.

ADDED later

I am still a bit puzzeling with it.

First of all the Dini surface is always only a part of the full hyperbolic plane. (there is always the cusp).

Secondly the Dini surface is isotropic (as long as you stay away from the cusp) so you are free to choose any (non -ideal) point as starting point.

But then what is its metric, the Dini surface has a constant metric $m$, but then I got struggeling to calculate it, or are you even free in this choice? I am not sure.

After this it is relativly easy.

To plot a point R given a starting point O and an ideal point X

• Calculate the the directed length $y$ along the main horocycle $M$ (with centre X) going trough starting point $O$ to the line $l$ going trough X and R. The length along the parallel of the Dini surface is $my $ (ps this can be to far and be outside the dini surface in that case you have to start again with another starting point. (and plot all points again)

• Calculate the the directed length $x$ along line $l$ from the intersection of $M$ and $l$ to $R$. The length along the meridian of the Dini surface is $mx$