How to convert between the hyperboloid model and the Poincare patch for $\mathbb{H}_n$? This question is motivated by the results in this paper, http://calvino.polito.it/~camporesi/JMP94.pdf  In this paper some of its most important results about the asymptotics of symmetric traceless transverse harmonic rank-s tensors on $\mathbb{H}_n = EAdS_n$ in equations 2.27, 2.28, 2.88, 2.89 are all given in hyperbolic coordinates.  
But for reasons of physics one wants to write $\mathbb{H}_n$ in the Poincare patch!
How does one convert between the two? 
Is there a known transformation? 



*

*In the hyperbolic model of $\mathbb{H}_n$ the space is thought of a zero-set in $\mathbb{R}^{n+1}$ of the equation, $x_0^2 - \sum_{i=1}^n x_i ^2 = a^2$ and then one uses the coordinates $y \in [0,\infty)$ and and $\vec{n} \in S^{n-1}$ to write, $x_0 = a cosh y$ and $\vec{x} = a \vec{n} sinh y$ and then the metric is, $ds^2 = a^2 [ dy^2 + sinh ^ 2 yd\Omega_{n-1}^2]$


Here $d\Omega_{n-1}^2$ is the standard metric on $S^{n-1}$. 
(..and this is the metric in equation 2.15 in the linked paper..) 


*

*In the Poincare patch model of $\mathbb{H}_n$ it is thought of as the half-space $x_n > 0$ in $\mathbb{R}^n$ with the metric,  $ds^2 = \frac{a^2}{z^2}(dz^2 + \sum_{i=1}^{n-1}dx_i^2 )$



I would like to know the transformation and the relation between these two models. 
I believe that there is some function connecting the $y$ and the $z$ and then I can substitute that into 2.28 and 2.89 of the paper to see the asymptotics in the Poincare patch. 
I am hoping the relationship is such that it relates large $y$ to small $z$...
 A: I'm thinking in 2D, but the results should generalize to arbitrary dimensions.
You can use the point $(-1, 0, 0, \dots, 0)$ (which is the vertex of the unused sheet of the hyperboloid) as a projection center, and project onto the symmetry plane $x_0=0$. The image of the used sheet of the hyperboloid will be the disc $\left\{x\;\middle\vert\;x_0=0,\lVert x\rVert<1\right\}$ which then forms the Poincaré ball model of hyperbolic space.
From there you can go to the half-space model using the higher-dimensional analogon of a Möbius transformation, as discussed in this answer.
If you use $y$ for the coordinates in the Hyperboloid model, $x$ for coordinates in the ball model and $z$ for coordinates in the half-space model, you can write the two steps like this:
\begin{align*}
\begin{pmatrix}
y_0 \\
y_1 \\
y_2 \\
\vdots \\
y_{n-1} \\
y_n
\end{pmatrix}
&\mapsto
\frac1{y_0+1}
\begin{pmatrix}
y_1 \\
y_2 \\
\vdots \\
y_{n-1} \\
y_n
\end{pmatrix}
=:
\begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_{n-1} \\
x_n
\end{pmatrix}
\\
\begin{pmatrix}
x_1 \\
x_2 \\
\vdots \\
x_{n-1} \\
x_n
\end{pmatrix}
&\mapsto
\frac{2}{1-2x_n+\lVert x\rVert^2}
\begin{pmatrix}x_1\\x_2\\\vdots\\x_{n-1}\\1-x_n\end{pmatrix} -
\begin{pmatrix}0\\0\\\vdots\\0\\1\end{pmatrix}
=:
\begin{pmatrix}z_1\\z_2\\\vdots\\z_{n-1}\\z_n\end{pmatrix}
\end{align*}
Now you can combine them into a single formula:
$$
\begin{pmatrix}
y_0 \\
y_1 \\
y_2 \\
\vdots \\
y_{n-1} \\
y_n
\end{pmatrix}
\mapsto
\frac{2}{1+y_0-2y_n+\frac{\sum_{i=1}^n y_i^2}{1+y_0}}
\begin{pmatrix}y_1\\y_2\\\vdots\\y_{n-1}\\1+y_0-y_n\end{pmatrix} -
\begin{pmatrix}0\\0\\\vdots\\0\\1\end{pmatrix}
=:
\begin{pmatrix}z_1\\z_2\\\vdots\\z_{n-1}\\z_n\end{pmatrix}
$$
A: Thanks for this helpful result, which I could not find elsewhere.  For the common case of $n=2$, it simplifies as follows:  To map from $ds^2 = a^2 \left(d\theta^2 + \sinh^2 \theta d\phi^2\right)$ to
$ds^2 = \frac{a^2}{y^2}\left(dx^2 + dy^2 \right)$,
use the transformation
$x = \frac{\cos\phi \sinh \theta}{\cosh \theta - \sinh \theta \sin \phi }$,
$y = \frac{1}{\cosh \theta - \sinh \theta \sin \phi}$.
