# Projection from Poincaré Ball to Hyperboloid

Let the following set denote the points on the Poincaré Ball with sectional curvature 1. $$B^n=\{x\in\mathbb{R}^{n+1}\,|\, ||x||< 1 \,\land\, x_1>0\}$$ and let the following set denote the points on the hyperboloid $$H^n=\{x\in\mathbb{R}^{n+1}\,|\, \langle x,x\rangle_*=-1\}$$ where $$\langle \cdot,\cdot\rangle_*$$ denotes the Minkowski inner product: $$\langle x,y\rangle = -x_1y_1+x_2y_2+...+x_{n+1}y_{n+1}.$$ What is the formula for the (bijective) projection $$\Pi\colon H^n\to B^n$$?

The following thread illustrates nicely how it's a projection along a line. However i don't know what is the vertical dimension. So I'm unsure on how to derive it.

Here's my approach (inspired by: http://bjlkeng.github.io/posts/hyperbolic-geometry-and-poincare-embeddings/)

Let $$x=(x_1,...,x_{n+1})\in H^n$$, then we get $$\Pi(x)=y=(y_1,...,y_n)\in B^n$$ as follows: $$y_i=\frac{x_i}{1+x_1}.$$ But I'm not sure if this formula is correct.

The formula that I'm getting for $$\Pi: H^{n}\rightarrow B^{n}$$ is $$\Pi(x_{1}, \cdots, x_{n+1}) = \frac{(x_{2}, \ldots, x_{n+1})}{1 + x_{1}}$$ This gives you $$y_{i} = \frac{x_{i+1}}{1 + x_{1}} \qquad\text{ for }\qquad 1\le i\le n.$$
As a side-note, there are variations based on how the Minkowski product is defined. If you had $$\langle x, y \rangle = x_{1}y_{1} + \cdots + x_{n}y_{n} -x_{n+1}y_{n+1}$$ where you treated the $$x_{n+1}$$-coordinate as the special one, you would have instead gotten the formula $$\Pi(x_{1}, \cdots, x_{n+1}) = \frac{(x_{1}, \ldots, x_{n})}{1 + x_{n+1}},$$ which might appear nicer because the indices are aligned better. John Lee's Riemannian Manifolds: An Introduction to Curvature book for example uses the latter convention.
• Thank you very much! What I did is i used the notation $y=(y_1=0,y_2,y_3,...,y_{n+1})$. Then everything worked out with $y_i=x_i/(1+x_1)$ for $i\geq 2$. – ndrizza Jun 29 at 22:23