# Conformal factor between Euclidean metric and metric on Poincaré Ball of arbitrary radius

Usually, a Poincaré Ball is given as the set $\mathbb{D}^n = \{ x \in \mathbb{R}^n : \|x\|^2 < 1\}$

Let $g_{E,x}$ be the Euclidean Riemannian metric induced at $x \in \mathbb{R}^n$ -- in that case, $g_{E,x} = I_n$ for all $x$(i.e. it can be represented as the identity matrix), so that the inner product in Euclidean space is just the normal Euclidean inner product we are used to (i.e. if $T_x\mathbb{R}^N$ is the tangent space of $\mathbb{R}^n$ at $x$, and $u,v$ are vectors in that tangent space, then our inner-product is just $g_{E,x}(u,v) = u^TI_nv = u^Tv = \langle u,v \rangle$).

The metric used in the Poincaré ball, $g_{\mathbb{D}^n}$ is conformal to the Euclidean metric because it preserves angles. In other words, for some $x \in \mathbb{D}^n$, and some $u,v \in T_x\mathbb{D}^n$, we have that

$$g_{\mathbb{D}^n,x}(u,v) = \lambda_x^2g_{E,x}(u,v)$$ And that $\lambda_x$ is called the conformal factor; for the Poincaré ball metric, it is $$\lambda_x = \frac{2}{1-\|x\|^2}$$

In my understanding, this conformal factor shows that in the Poincaré ball looks locally like $\mathbb{R}^n$ near the origin and has similar distances, but that near the border, the distances become really huge for even small deviations in where $x$ is located.

This works for the unit Poincaré disk. What I am looking for is the conformal factor for the Poincaré disk of some radius $r > 0$ where $r$ is not necessarily equal to 1.

This particular Poincaré disk can be represented as $$\mathbb{D}_r^n = \{ x \in \mathbb{R}^n : r\|x\|^2 < 1 \}$$ and in this case,

$$g_{\mathbb{D}_r^n,x}(u,v) = (\lambda_x^r)^2g_{E,x}(u,v)$$

What is $\lambda_x^r$? My guess would be $$\lambda_x^r = \frac{2}{1-r\|x\|^2}$$

Is this correct? And even if it is, can anybody explain why? An explanation of this would go a long way toward helping me to understand the Poincaré disk a lot better.

An open ball of radius $r$ corresponds to $$\lVert x\rVert<r$$ or equivalently (for $r>0$) $$\frac{\lVert x\rVert^2}{r^2}<1$$ not $$r\,\lVert x\rVert^2<1$$ as you wrote it. In other words, wherever you use $\lVert x\rVert$ in the unit ball setup, you'd use $\lVert x\rVert/r$ in the arbitrary radius setup. Thus you'd have $$\lambda_x^r=\frac2{1-\frac{\lVert x\rVert^2}{r^2}}=\frac{2r^2}{r^2-\lVert x\rVert^2}\;.$$
Note that as long as you still multiply this by $g_{E,x}$, you still get lengths which are scaled by a factor of $r$ compared to the unit ball. Which in turn means that the curvature of your hyperbolic plane will have a Gaussian curvature not of $-1$ but of $-\frac{1}{r^2}$. If you want to have the standard $-1$ curvature, better use $$\lambda_x^r=\frac{2r}{r^2-\lVert x\rVert^2}$$ to correct for the scale inside $g_{E,x}$. But in that case you essentially have a unit ball model drawn at a larger scale, and the radius you used to draw it has no impact any more on the geometry within the model. Which might be desirable but might also make this consideration boring.
• Aha I see. But I could also cast it as $\|x\| < \frac{1}{r}$ for $r > 0$ -- This is probably unnecessarily convoluted but, believe it or not, this is exactly how I was approaching the problem in my head. Thanks for the answer -- I need some time to grok it first, see if I have any more comments and then I will mark it as accepted if I have no more questions and it seems correct to me! :) – Marko Aug 17 '18 at 2:33
• @Marko: On a certain level, all variable names are just conventions, and you are free to change them however you like. On the other hand, certain symbols carry certain connotations, and $r$ is usually seen as the radius. If you want a symbol for the inverse radius, $k$ or $\kappa$ tend to crop up, particularly when discussing curvatures. So I'd suggest you use on of these to avoid confusion. – MvG Aug 17 '18 at 10:13