Transformation between different hyperbolic plane coordinate systems

Question

On the hyperbolic plane, we can use 'polar' coordinates $(\rho,\tau)$ giving the metric $$ds^2 =d\rho^2 + \sinh^2(\rho) d\tau^2,$$ or, in the upper-half plane model, we can use the 'Cartesian' coordinates $(x,y)$, giving the metric: $$ ds^2 = \frac{dx^2 + dy^2}{y^2}. $$ I'm wondering if there's an explicit transformation between these coordinates. What seems to work is: $$ x=i\coth(\rho) e^{i\tau}, \quad y = \frac{1}{\sinh(\rho)}e^{i\tau}, $$ however, then $x$ and $y$ are complex numbers, whereas I would think that they should both be real to interpret them as the real and imaginary parts of $z = x+iy$ on the complex upper-half plane. Is there a simpler transformation that you are aware of such that $x$ and $y$ are real?

Answer 1

Thanks to the answer above, I've worked out a solution, for who is interested.

The relation between disk polar coordinates and plane polar coordinates can be found by requiring: $$ds^2 = \frac{4}{(1-r^2)^2}(dr^2+r^2d\theta^2) = d\rho^2 + \sinh^2\rho d\tau^2$$ This is solved by: $$ \theta = \tau, \quad r = \tanh\frac{\rho}{2} $$ Then, we map the disk coordinate $w = re^{i\theta}$ to the upper-half plane via the Cayley transformation: $$ z= i\frac{1-w}{1+w} = i \frac{1-r^2-2i \Im(w)}{1+r^2 +2\Re(w)} $$ Writing $z=x+iy$, we then see that: $$ x = \frac{2\Im(w)}{1+r^2 +2\Re(w)}, \quad y = \frac{1-r^2}{1+r^2 +2\Re(w)} $$ It's clear that the unit circle ($r=1$) is mapped to the real axis $(y=0)$. Now plugging in $\Re(w) = r\cos\theta$ and $\Im(w) = r\sin\theta$, as well as our expressions for $\theta$ and $r$ in terms of $\tau$ and $\rho$, we find: $$ x = \frac{2\tanh\frac{\rho}{2} \sin\tau}{1+2\tanh\frac{\rho}{2}\cos\tau + \tanh^2\frac{\rho}{2}}, \quad y = \frac{1}{\cosh\rho+\cos\tau\sinh\rho} $$ Since the upper-half plane has a PSL$(2,\mathbb{R})$ isometry, we can probably find a nicer expression for $x$ and $y$ after a suitable PSL$(2,\mathbb{R})$ transformation $z\to \frac{az+b}{cz+d}$, $ad-bc=1$, but I haven't tried that yet. To check that this transformation has the right behaviour at the boundary, let's take the limit $\rho \to \infty$: $$ \lim_{\rho\to\infty}y(\rho,\tau) = 0, $$ which makes sense since the boundary should be at $y=0$. Furthermore, the expression for $x$ simplifies a lot at the boundary: $$ \lim_{\rho\to\infty}x(\rho,\tau) = \tan\frac{\tau}{2}. $$

Answer 2

Let me suggest an outline for deriving a transformation, breaking the problem into steps, where the Poincare disc model sits intermediate between the polar coordinate plane model and the upper half plane model. The metric on the Poincare disc model is given by $$ds^2 = \frac{4}{\left( 1 - (x^2+y^2)\right)^2} (dx^2 + dy^2) $$ You can easily convert this formula into polar coordinates on the unit disc, using the ordinary Euclidean--polar coordinate transformations. Noticing that the Poincare disc formula is conformal with respect to the Euclidean metric (i.e. of the form $g(x,y) (dx^2 + dy^2)$), and noticing the rotational symmetry of the Poincare disc metric, you should expect to get a metric of the form $$ds^2 = f(r)^2 (dr^2 + r^2 d\theta^2) $$ where that bit in parentheses is the usual formula for the Euclidean metric expressed in polar coordinates. Of course you'll probably want to work out an explicit formula for $f(r)$; by inspection it ought to be $f(r) = \frac{4}{(1-r^2)^2}$.

There is a well known formula that relates the Poincare disc metric (expressed in a complex $w=x+iy$ coordinate) to the upper half plane model (also expressed in a complex $z=x+iy$ coordinate), and it is easy to derive for yourself: $$w = \frac{iz+1}{z+i} $$ Perhaps you want to rewrite this in $x,y$ coordinates.

So what remains is to convert the Poincare disc polar coordinate model into the planar polar coordinate model. The circular symmetry of both models suggest that you first set $\theta=\tau$, and then work out the transformation of the radial coordinate $\rho = g(r)$, $g : [0,1) \to [0,\infty)$.