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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 $z$...

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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:

$$ \begin{pmatrix} y_0 \\ y_1 \\ y_2 \\ \vdots \\ y_{n-1} \\ y_n \end{pmatrix} \mapsto \frac{2}{1+y_0-2y_n+\frac{\sum_{i=1}^n y_i^2}{1+y_0}} \begin{pmatrix}y_1\\y_2\\\vdots\\y_{n-1}\\1+y_0-y_n\end{pmatrix} - \begin{pmatrix}0\\0\\\vdots\\0\\1\end{pmatrix} =: \begin{pmatrix}z_1\\z_2\\\vdots\\z_{n-1}\\z_n\end{pmatrix} $$

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  • $\begingroup$ Thanks! Let me read that other answer. But what is the projection map that you use initially? $\endgroup$ – user6818 Sep 20 '13 at 20:21
  • $\begingroup$ @user6818: Included an explicit formula, hope this makes things clearer. $\endgroup$ – MvG Sep 20 '13 at 20:41
  • $\begingroup$ When you are projecting down into the disk in $\mathbb{R}^n$ are you still getting a hyperbolic metric there? Is it obvious that the coordinate transformations that you have done actually shift between the two metrics that I wrote down? What is worrying is that your $z_n$ seems to depend on the values of all the $y$s - whereas I would think that there would be a function $f$ such that $e^y = f(z)$ which would convert between my two metrics... $\endgroup$ – user6818 Sep 20 '13 at 23:20
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    $\begingroup$ I think there is a typo in your equation - I think it should be $1 + y_0 - y_n$ in the last entry instead of $1-y_n$. $\endgroup$ – user6818 Sep 21 '13 at 0:39
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    $\begingroup$ Thanks so much for writing this out. In my experience it is impossible to find this explicit formula in the literature. Also, the inverse of this map in dimension $3$ can be written very simply using real and complex quaternions. I did this here: arxiv.org/abs/1701.06709 $\endgroup$ – j0equ1nn Dec 8 '17 at 19:39
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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 $x = \frac{\cos\phi \sinh \theta}{\cosh \theta - \sinh \theta \sin \phi }$, $y = \frac{1}{\cosh \theta - \sinh \theta \sin \phi}$.

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