Hyperbolic vs Euclidean balls

Question

I'm trying to prove that, in the Poincaré half-space of dimension 2, a hyperbolic ball with center $P:=(x,y)$ and radius $r$ is exactly, as a set of points, a euclidean ball with center $P_1:=(x,y\cosh(r))$ and radius $r1:=y\sinh(r)$. The distance function I'm using in the Poincaré half-plane is, written in complex coordinates, $$d_h(z,w):=2\ln\frac{|z-w|+|z-\bar{w}|}{2\sqrt{Im(z)Im(w)}}.$$ My idea is to consider a generic point on the euclidean ball boundary cited above, in the form $$A:=(r_1\cos(\theta),r_1\sin(\theta)+y\cosh(r))$$ and then to explicitly compute its distance to the center of the hyperbolic ball with which we started, to prove it is exactly $r$, i.e. I'm trying to prove that $$d_h(A,P)=r.$$ This leads me towards extremely hard computations in which I have to admit I've been losing myself several times. I found the statement of the result stated on the wikipedia page https://en.wikipedia.org/wiki/Poincar%C3%A9_half-plane_model, but I can't find a reference for a proof. Could anyone please share some ideas or references on how to prove this result?

Answer 1

As comments have pointed out, you might be able to solve this for very specific situations, then show that these distances are invariant under isometries and thus hold under rotation around the center and translation of the center. Since knowing about isometries is closely ties with properties of circles, I'll not thake that path here for now. Instead I'll try to tackle the formula “the hard way”.

One thing that makes your formula hard to deal with is the abundance of trigonometric and hyperbolic functions, which come with a huge number of rules for how to cancel them or convert them to one another. Compared to that, polynomials or rational functions are easier to deal with. So I'm a huge fan of the tangent half-angle formula.

\begin{align*} u &:= \tan\tfrac\theta2 & t &:= \tanh\tfrac r2 \\ \cos\theta &= \frac{1-u^2}{1+u^2} & \cosh r &= \frac{1+t^2}{1-t^2} \\ \sin\theta &= \frac{2u}{1+u^2} & \sinh r &= \frac{2t}{1-t^2} \end{align*}

With that you get your points as

$$ P = \begin{pmatrix}x\\y\end{pmatrix} \qquad A = \begin{pmatrix}x+y\sinh r\cos\theta\\ y\cosh r+y\sinh r\sin\theta\end{pmatrix} = \begin{pmatrix}x+y\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}\\ y\frac{1+t^2}{1-t^2}+y\frac{2t}{1-t^2}\frac{2u}{1+u^2}\end{pmatrix} $$

Your distance formula has a logarithm in there, but Wikipedia also gives a different formula for the distance in terms of $\operatorname{artanh}$ which allows us to deal with that factor $2$ more easily by tying back to the substitution above.

\begin{align*} r\overset?=d_h(A,P)&= 2\operatorname{artanh}\frac{\lVert A-P\rVert}{\lVert A-\bar P\rVert} \\ \frac r2&\overset?= \operatorname{artanh}\frac{\lVert A-P\rVert}{\lVert A-\bar P\rVert} \\ \tanh\frac r2=t&\overset?= \frac{\lVert A-P\rVert}{\lVert A-\bar P\rVert} \\ t^2&\overset?= \frac{\left(A_x-P_x\right)^2+\left(A_y-P_y\right)^2} {\left(A_x-P_x\right)^2+\left(A_y+P_y\right)^2} \end{align*}

Substitute the expressions for the coordinates as given above, and you end up with just polynomials and basic arithmetic operations (including divisions but excluding square roots). Now standard tools for dealing with rational functions (e.g. bringing things to a common denominator in order to add or subtract them) will allow you to combine fractions until you end up with a single fraction on the right hand side. At that point you will find that the denominator is one, that the indeterminates $x, y, u$ have already disappeared, and that the numerator is $t^2$ as expected.

Mind you, the manipulations of these polynomials will still be annoying to do by hand. A computer algebra system might help you do this or at least check your intermediate results. But there should be no more points where you need to make a clever decision in order to proceed; the approach should be clear from the start and not depend on any specific combinations of symbols along the way. After the above formulation it's all purely mechanical.

$$ \frac{\left(A_x-P_x\right)^2+\left(A_y-P_y\right)^2} {\left(A_x-P_x\right)^2+\left(A_y+P_y\right)^2} = \\ = \frac{\left(x+y\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}-x\right)^2+\left(y\frac{1+t^2}{1-t^2}+y\frac{2t}{1-t^2}\frac{2u}{1+u^2}-y\right)^2} {\left(x+y\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}-x\right)^2+\left(y\frac{1+t^2}{1-t^2}+y\frac{2t}{1-t^2}\frac{2u}{1+u^2}+y\right)^2} = \\ = \frac{\left(y\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}\right)^2+\left(y\frac{1+t^2}{1-t^2}+y\frac{2t}{1-t^2}\frac{2u}{1+u^2}-y\right)^2} {\left(y\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}\right)^2+\left(y\frac{1+t^2}{1-t^2}+y\frac{2t}{1-t^2}\frac{2u}{1+u^2}+y\right)^2} = \\ = \frac{\left(\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}\frac{2u}{1+u^2}-1\right)^2} {\left(\frac{2t}{1-t^2}\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}\frac{2u}{1+u^2}+1\right)^2} = \\ = \frac{\left(2t\frac{1-u^2}{1+u^2}\right)^2+\left(\left(1+t^2\right)+2t\frac{2u}{1+u^2}-\left(1-t^2\right)\right)^2} {\left(2t\frac{1-u^2}{1+u^2}\right)^2+\left(\left(1+t^2\right)+2t\frac{2u}{1+u^2}+\left(1-t^2\right)\right)^2} = \\ = \frac{\left(2t\frac{1-u^2}{1+u^2}\right)^2+\left(2t\frac{2u}{1+u^2}+2t^2\right)^2} {\left(2t\frac{1-u^2}{1+u^2}\right)^2+\left(2t\frac{2u}{1+u^2}+2\right)^2} = \\ = \frac{\left(\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{2u}{1+u^2}+\frac t1\right)^2} {\left(\frac{1-u^2}{1+u^2}\right)^2+\left(\frac{2u}{1+u^2}+\frac 1t\right)^2} = \\ = \frac{\left(1-u^2\right)^2+\left(2u+\frac t1\left(1+u^2\right)\right)^2} {\left(1-u^2\right)^2+\left(2u+\frac 1t\left(1+u^2\right)\right)^2} = \\ = \frac{1-2u^2+u^4+4u^2+4u\frac t1\left(1+u^2\right)+\frac{t^2}1\left(1+u^2\right)^2} {1-2u^2+u^4+4u^2+4u\frac 1t\left(1+u^2\right)+\frac1{t^2}\left(1+u^2\right)^2} = \\ = \frac{1+2u^2+u^4+4u\frac t1\left(1+u^2\right)+\frac{t^2}1\left(1+u^2\right)^2} {1+2u^2+u^4+4u\frac 1t\left(1+u^2\right)+\frac1{t^2}\left(1+u^2\right)^2} = \\ = \frac{\left(1+u^2\right)^2+4u\frac t1\left(1+u^2\right)+\frac{t^2}1\left(1+u^2\right)^2} {\left(1+u^2\right)^2+4u\frac 1t\left(1+u^2\right)+\frac1{t^2}\left(1+u^2\right)^2} = \\ = \frac{\left(1+u^2\right)+4u\frac t1+\frac{t^2}1\left(1+u^2\right)} {\left(1+u^2\right)+4u\frac 1t+\frac1{t^2}\left(1+u^2\right)} = \\ = \frac{t\bigl(\frac 1t\left(1+u^2\right)+4u+\frac t1\left(1+u^2\right)\bigr)} {\frac1t\bigl(\frac t1\left(1+u^2\right)+4u+\frac1t\left(1+u^2\right)\bigr)} = t^2 $$