# Expression of the Hyperbolic Distance in the Upper Half Plane

While looking for an expression of the hyperbolic distance in the Upper Half Plane $\mathbb{H}=\{z=x +iy \in \mathbb{C}| y>0\},$ I came across two different expressions. Both of them in Wikipedia.

In the page Poincaré Half Plane Model it is explicitly stated that the distance of $z,w \in \mathbb{H}$ is: $$d_{hyp}(z,w)= Arccosh(1+ \frac{|z-w|^2}{2 Im(z) Im(w)}).$$ While in the page Poincaré Metric it is stated that the metric on the Upper Half Plane is : $$\rho(z,w)=2 Arctangh(\frac{|z-w|}{|z-\bar w|}).$$

At the beginnning I thought it would have been an easy exercise to prove the equivalence of the two expressions. But first I failed in doing that, and then I found, using Mathematica, a counterexample (i.e. $z=2i$ and $w=i$).

Question: If they are not equal, which of the two expressions is the right one? Then, how is the metric related to the induced distance?

The question is probably silly, but I'm often confused about the relationships between "metric" objects.

Thank you very much for your time!

• Very nice question. it was a reference for me today. +1 May 13, 2013 at 5:28
• Just for reference, the equivalence between different expressions for the hyperbolic distance in the upper half plane can be found in Theorem 1.2.6 of Katok, Svetlana. Fuchsian groups. Chicago Lectures in Mathematics. University of Chicago Press, Chicago, IL, 1992.
– F.G.
Jul 8, 2019 at 14:22

To verify the consistency of the formulas, recall these identities:

$$1 - \tanh^2 d = \mathrm{sech}^2 d \qquad \qquad \cosh^2\frac{d}{2}=\frac{1+\cosh d}{2}$$

The "Poincare Metric" formula is equivalent to this form:

$$\tanh\frac{d}{2} = \frac{p}{q}$$

so ...

\begin{align} \mathrm{sech}^2\frac{d}{2} &= \frac{q^2-p^2}{q^2} \\ \implies \qquad \frac{1 + \cosh d}{2} &= \frac{q^2}{q^2-p^2} \\ \implies \qquad \cosh d &= \frac{q^2+p^2}{q^2-p^2} \end{align}

Now, with $w := u + i v$ and $z := x + i y$, and $p := |z-w|$ and $q := |z-\overline{w}|$, we have

\begin{align} q^2+p^2 = |z-\overline{w}|^2+|z-w|^2 &=\left( (x-u)^2+(y+v)^2 \right)+|z-w|^2 \\ &=\left( (x-u)^2+(y-v)^2+4yv \right)+|z-w|^2 \\ &=4yv+2|z-w|^2 \\ q^2-p^2 = |z-\overline{w}|^2-|z-w|^2 &=(x-u)^2+(y+v)^2-(x-u)^2-(y-v)^2 \\ &=4yv \end{align}

Thus,

$$\cosh d = 1 + \frac{\;|z-w|^2}{2yv}$$

• Thank you very much! This is exactly what I was looking for! Jun 26, 2012 at 9:30
• Squared in numerator $\cosh d = 1 + \dfrac{|z-w|^2}{2yv}$ to be added. Aug 14, 2015 at 8:30
• @Narasimham: Corrected, thanks. You have an extremely good eye for detail to spot that error in a three-year-old answer! :)
– Blue
Aug 14, 2015 at 9:55
• :)) looking at it only now! Aug 14, 2015 at 10:15
• I have a doubt. The points $(z,w)$can be located anywhere; both inside, both outside and one on either side of the boundary circle. Right? Aug 15, 2015 at 21:27

The distance between two points is the absolute value of the parameter difference along a unit-speed geodesic that joins your two points. After that, you may derive closed-form expressions as you like.

Given real constants $A$ which can be anything, and then $B > 0,$ the two types of geodesics, parametrized by $t,$ are $$A + \; i \, e^t$$ which is vertical, and $$A + B \, \tanh t \; + \; i \, B \; \mbox{sech} \; t$$ which is a semicircle with center at $A$ on the real axis. Note that, if your two points do not have the same real part, you need to draw the perpendicular bisector of the line segment between them and see the location where that crosses the real axis, that becomes $A.$

Note that $$1 + \sinh^2 t = \cosh^2 t,$$ divide through by $\cosh^2 t$ to get $$\mbox{sech}^2 t + \tanh^2 t = 1.$$ Furthermore, $\cosh t \geq 1,$ so $0 < \mbox{sech} \; t \leq 1.$ We have given a fairly natural way to parametrize a semicircle without its endpoints.

It is a worthwhile exercise to show that these really are geodesics. Or, at least, to confirm they are unit speed.

I have obtained the same result $\ln 2$ for the two expressions. Be carefull with using $arc ~cosh (x)= ln( x+ \sqrt{x^2 -1})$, no $ln( x- \sqrt{x^2 -1})$

Why? This is because you have $$\displaystyle\frac{e^d+e^{-d}}{2}=1+\frac{|z-w|}{2\cdot Im ~z\cdot Im ~w}$$ and to obtain the bigger value $e^d$ in the resultant second degree equation, you need the +.