# How to convert between the hyperboloid model and the Poincare ball when the curvature $\neq-1$?

The Poincare ball model with curvature $$-1$$ is defined as:

$$B^n=\{x\in\mathbb{R}^{n}\,|\, ||x||< 1\}.$$

The hyperboloid model with curvature $$-1/\beta$$ is defined as:

$$H^{n,\beta}=\{x\in\mathbb{R}^{d+1}\,|\, \langle x,x\rangle_L=-\beta\}.$$

For $$\beta=1$$, refer to Projection from Poincaré Ball to Hyperboloid,

we have $$\Pi: H^{n}\rightarrow B^{n},$$ $$\Pi(x_{1}, \cdots, x_{n+1}) = \frac{(x_{2}, \ldots, x_{n+1})}{1 + x_{1}}$$

and $$\Pi^{-1}: B^{n}\rightarrow H^{n},$$ $$\Pi^{-1}(x_{1}, \cdots, x_{n}) = \frac{(1+\|x\|^2,2x_{1}, \ldots, 2x_{n})}{1 - \|x\|_2^2}$$

I want to know how to convert the two models when the curvature $$\neq-1$$.

• With $\beta\ne 1$, do you naturally map to a ball of radius $\sqrt\beta$ instead? Commented Dec 25, 2019 at 22:02
• Yes, I want to do this, but I don't know how to achieve it. Commented Dec 26, 2019 at 3:02

For any $$\beta>0$$, the map $$\Pi$$ will now map to the ball of radius $$\sqrt\beta$$, with formula $$\Pi(x_1,\dots,x_{n+1}) = \frac{\sqrt\beta(x_2,\dots,x_{n+1})}{\sqrt\beta+x_1}.$$ You can work out the inverse.
(I understand the map $$\Pi$$ as a composition of mappings. Thinking of $$x_1$$ as the vertical axis, first project down to the horizontal disk of radius $$\beta$$ at the vertex $$(\sqrt\beta,0)$$ inside the cone $$x_1=\sqrt{\sum x_i^2}$$. Then go to the sphere of radius $$\beta$$ by taking $$\sqrt\beta\big(\frac{\sqrt\beta}{x_1},\frac{x_2}{x_1},\dots,\frac{x_{n+1}}{x_1}\big)$$, and then stereographically project from the south pole to the disk at the equator, getting $$\frac{\sqrt\beta}{x_1+\sqrt\beta}(x_2,\dots,x_{n+1})$$.)