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 $\neq-1$.

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

1 Answer 1


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 $\frac{\sqrt\beta}{x_1+\sqrt\beta}(x_2,\dots,x_{n+1})$.)


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