7

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 $z$...

user6818
  • 1,365

2 Answers2

5

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:

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

MvG
  • 44,006
  • Thanks! Let me read that other answer. But what is the projection map that you use initially? – user6818 Sep 20 '13 at 20:21
  • @user6818: Included an explicit formula, hope this makes things clearer. – MvG Sep 20 '13 at 20:41
  • 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... – user6818 Sep 20 '13 at 23:20
  • I can map the hyperboloid model of $\mathbb{H}n$ to a ball model in $\mathbb{R}^n$ by declaring, $u = \frac{2a sinh y}{1 + cosh y}$. Then the metric on the ball $B^n$ is, $ds^2 = \frac{du^2 + u^2 d\Omega{n-1}^2 }{(1 - \frac{u^2}{4a^2})^2}$. Is this the disk metric that you would also get? I would believe that there is now a coordinate transformation that goes from this metric to the Poincare patch...I mean when you say $\vert x \vert ^2$ shouldn't you mean the norm of the position vector of the point $\vec{x}$ in this hyperbolic metric? – user6818 Sep 20 '13 at 23:27
  • If the position vector is now at a radial coordinate position $u$ then the length of that vector is, $\int_0 ^u \frac{du}{1- \frac{u^2}{4a^2}} = a ln \vert \frac{2a+u}{2a - u}\vert $. Then by your mobius transformation between the n-ball and the UHP in $\mathbb{R}^n$ shouldn't I get, $z = -1 + \frac{2 ( 1 - u) }{1-2u +aln \vert \frac{2a+u}{2a - u}\vert }$ - but the problem is this, in the first coordinate system when $y \rightarrow \infty$ we had $u \rightarrow 2a$ (the boundary of the ball) and then in these coordinates $z \rightarrow -1$ (...but I would want it to go to $z =0$!..) – user6818 Sep 20 '13 at 23:33
  • It would be helpful if you could state what is the form of the hyperbolic metric in your "y" coordinates to begin with and what does the same metric in the z coordinates look like. – user6818 Sep 20 '13 at 23:48
  • @user6818: I'm not so much thinking in terms of metrics as in terms of models and projections / maps between them. Each model has its own formula for hyperbolic metric, but I trust others have verified that these maps which I learned about will preserve metrics, I haven't verified that myself (yet). So I know how to go from hyperboloid with its hyperbolic metric to ball with its metric to half space with its hyperbolic metric. I haven't checked your metric is the common one, but I simply trust that as well. – MvG Sep 20 '13 at 23:53
  • @user6818: My $\lVert x\rVert^2$ is Euclidean. It comes from an inversion in a sphere. Here I checked that the boundary of the unit ball will map to the boundary of the half plane, so if you agree about infinite points mapping to the unit sphere, that should convince you. But I guess your different result might come from reading $\lVert x\rVert^2$ as hyperbolic length. Perhaps you want to try this for a number of points, see if it works out numerically? – MvG Sep 20 '13 at 23:57
  • Is the equation that I write converting between $z$ and $u$ right? (..this I am doing following your Mobius transformation but interpreting your $\vert x \vert^2$ to mean the length of the vector in the hyperbolic ball metric...) – user6818 Sep 20 '13 at 23:59
  • @MgV Also you seem to pick out two coordinates very specially $y_0$ and $y_n$. I believe what I called $z$ in the Poincare patch is your $z_n$ and my $y$ in the hyperbolic metric is your $y_0$ but what is $y_n$? Given a function in the hyperbolic metric, how do I determine which is $y_n$? (...as you can see in the quoted equations in the linked paper - everything is in terms of that one special radial coordinate "y" in the hyperbolic metric..and I believe that all these results can be converted in terms of $z$...) – user6818 Sep 21 '13 at 00:06
  • @MgV Okay..so then using the defining equation for the hyperboloid as $y_0 ^2 = 1+ \sum _{i=1}^n y_i ^2$ I can rewrite the transformation on the $z$ coordinate as $z = -1 + \frac{1-y_n}{y_0 - y_n}$ - so this gets complicated to interprete - because the asymptotics in the paper say that the large $y_0$ values of the function don't depend on which direction the limit is taken - but in terms of $z$ coordinates you seem to be saying that the value of $y_n$ also matters? - or am I to read this as saying that for large $y_0$ I can approximate, $z \sim -1 + \frac{1}{y_0}$? – user6818 Sep 21 '13 at 00:21
  • As $y_0 \rightarrow \infty$ along any direction I should hit the half-plane boundary $z=0$ - this isn't obvious in your expression... – user6818 Sep 21 '13 at 00:33
  • 1
    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 $1-y_n$. – user6818 Sep 21 '13 at 00:39
  • let us continue this discussion in chat – MvG Sep 21 '13 at 00:50
  • I am actually in a hurry now - so do let me know if the final conclusion is this - that $z$ depends on both $y_0$ and $y_n$ but in the spirit of the asymptotics given in the paper I can always take the limit of $y_0 \rightarrow \infty$ at a any fixed $y_n$ and then I get in that limit $z \sim \frac{1}{y}$ and then I can directly use the equations 2.88 and 2.89 of the linked paper substituting $y = \frac{1}{z}$ and see 2.88 as small $z$ asymptotics in the Poincare patch coordinates? (..and then I wonder if I can Taylor expand the $e^{ \frac {#}{z}}$ about $z=0$...is that possible?...) – user6818 Sep 21 '13 at 00:58
  • 1
    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: https://arxiv.org/abs/1701.06709 – j0equ1nn Dec 08 '17 at 19:39
  • (the choice of coordinates is different there though) – j0equ1nn Dec 08 '17 at 19:47
2

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 $x = \frac{\cos\phi \sinh \theta}{\cosh \theta - \sinh \theta \sin \phi }$, $y = \frac{1}{\cosh \theta - \sinh \theta \sin \phi}$.

user400108
  • 21