# Hyperbolic Geometry Note #1: Strange behaviour of length calculations in the Poincaré half-plane model of hyperbolic space

In the Poincaré half-plane model of hyperbolic space, the upper half-plane $\mathbb{H}$ is the set of complex numbers $z$ with positive imaginary part:

$\mathbb{H} = \{z \in \mathbb{C}: \text{Im}(z) > 0 \}$

The boundary of $\mathbb{H}$, $\delta \mathbb{H}$, is the real axis plus the point $\infty$. Formally,

$\delta \mathbb{H} = \{z \in \mathbb{C}: \text{Im}(z) = 0 \} \cup \{ \infty \} \equiv \mathbb{R} \cup \{ \infty \}$.

$\delta \mathbb{H}$ is often referred to as the circle at infinity because it is topologically equivalent to a circle. This is the two-dimensional analogue of the topological equivalence of the 2-sphere and the extended complex plane (the Riemann sphere’) which I wrote about in a previous note. As in that note, the argument is based on stereographic projection. We represent the unit circle in the complex plane $\mathbb{C}$ as the set

$K = \{z \in \mathbb{C}: |z| = 1 \}$

and we construct a homeomorphism

$\pi: K \rightarrow \delta \mathbb{H}$

by imagining a line passing through $i$ at the north pole’ of the circle and another point $z \in K-\{i\}$.

This line meets the real axis at a unique point $\pi(z)$. We define $\pi(i) = \infty$. The map $\pi$ is a homeomorphism from $K$ to $\delta \mathbb{H} = \mathbb{R} \cup \{ \infty \}$, which is a topological way of saying that $K$ and $\delta \mathbb{H}$ are the same’.

What I want to focus on in this note is really the at infinity’ part of the circle at infinity. It is said to be at infinity’ because the upper half-plane model of hyperbolic space has the curious property that points on $\delta \mathbb{H}$ are at an infinite distance from any point in $\mathbb{H}$. This arises from the fact that the hyperbolic length of a path in the upper half-plane $\mathbb{H}$ is obtained by integrating the functon $f(z) = 1/\text{Im}(z)$ along that path.

To clarify, let $\sigma$ be a (piecewise differentiable) path in $\mathbb{H}$ with parametrisation

$\sigma(t): [a, b] \rightarrow \mathbb{H}$

As usual in complex analysis, if $f: \mathbb{H} \rightarrow \mathbb{R}$ is a continuous function, then the integral of $f$ along the path $\sigma$ is defined to be

$\int_{\sigma}f = \int_a^b f(\sigma(t)) |\sigma^{\prime}(t)|dt$

where $|\cdot|$ denotes the usual modulus of a complex number, i.e.,

$|\sigma^{\prime}(t)| = \sqrt{(\text{Re}\sigma^{\prime}(t))^2 + (\text{Im}\sigma^{\prime}(t))^2}$

(Note that no matter what parametrisation is chosen for the path, the value of this integral will be the same so one usually tries to find a parametrisation that is as easy to integrate as possible).

As stated earlier, the hyperbolic length of the path $\sigma$ is obtained by integrating the function $f(z) = 1/\text{Im}(z)$ along $\sigma$, so

$\text{Length}_{\mathbb{H}}(\sigma) = \int_{\sigma}\frac{1}{\text{Im}(z)}dt = \int_a^b \frac{|\sigma^{\prime}(t)|}{\text{Im}(\sigma(t))}dt$

I want to highlight two curious results that follow from this setup. First, consider the points $u + iv$ and $u + kiv$ in $\mathbb{H}$ where $0 < k < 1$ and $v > 0$.

A parametrisation of the vertical’ path between them in the diagram could be

$\sigma(t) = u + kiv + (1 - k)ivt$

$t \in [0, 1]$

We then have

$\int_{\sigma}\frac{1}{\text{Im}(z)}dt = \int_0^1 \frac{|\sigma^{\prime}(t)|}{\text{Im}(\sigma(t))}dt = \int_0^1 \frac{|(1 - k)iv|}{kv + (1 - k)vt}dt = \int_0^1 \frac{(1 - k)v}{kv + (1 - k)vt}dt$

$= [\log(kv + (1 - k)vt)]_0^1 = \log v - \log kv = \log \big(\frac{v}{kv}\big) = \log \big(\frac{1}{k}\big)$

Since the length of the path is $\log(1/k)$ we see that as the lower point gets closer and closer to the real axis (i.e., as $k \rightarrow 0$), the length of the path goes to $\infty$. This is the motivation for the at infinity’ nomenclature for the boundary $\delta \mathbb{H}$.

The second curious result involves the lengths of paths between two points at the same height’ in $\mathbb{H}$, say $-2 + i$ and $2 + i$.

A parametrisation of the `horizontal’ path between them in the leftmost diagram above could be

$\sigma_1(t) = 2t + i$

$t \in [-1, 1]$

We then have

$\int_{\sigma_1}\frac{1}{\text{Im}(z)}dt = \int_{-1}^1 \frac{|\sigma_1^{\prime}(t)|}{\text{Im}(\sigma_1(t))}dt = \int_{-1}^1 \frac{2}{1}dt = 2 \cdot 2 = 4$

so the length of the horizontal path between the two points is $4$. Now consider the situation shown in the remaining two pictures above. These involve piecewise linear paths between the two points that go diagonally from $-2 + i$ to a point $ki$ and then diagonally from $ki$ to $2 + i$. I am leaving $k$ unspecified so that we can study the lengths of the piecewise linear paths as a function of $k$.

A parametrisation of the piecewise linear paths is given by

$\sigma_2(t) = \big[ (2 - i)t + ki(1 + t) \big]I_{[-1, 0]}(t) + \big[ (2 + i)t + ki(1 - t) \big]I_{[0, 1]}(t)$

where $I_A(t)$ denotes the indicator function for set $A$. We then have

$\int_{\sigma_2}\frac{1}{\text{Im}(z)}dt = \int_{-1}^1 \frac{|\sigma_2^{\prime}(t)|}{\text{Im}(\sigma_2(t))}dt = \int_{-1}^0\frac{|2 + (k - 1)i|}{k(1 + t) - t}dt + \int_0^1\frac{|2 + (1 - k)i|}{k(1 - t) + t}dt$

$= \int_{-1}^0 \frac{\sqrt{4 + (k-1)^2}}{k + t(k-1)}dt + \int_0^1 \frac{\sqrt{4 + (k-1)^2}}{k - t(k-1)}dt$

$= \sqrt{4 + (k-1)^2} \bigg\{ \int_{-1}^0 \frac{1}{k + t(k-1)}dt + \int_0^1 \frac{1}{k - t(k-1)}dt \bigg\}$

$= \sqrt{4 + (k-1)^2} \frac{1}{k-1}\bigg\{ \int_{-1}^0 \frac{k-1}{k + t(k-1)}dt - \int_0^1 \frac{-(k-1)}{k - t(k-1)}dt \bigg\}$

$= \frac{\sqrt{4 + (k-1)^2}}{k-1} \bigg\{ \big[\log(k + t(k-1))\big]_{-1}^0 - \big[\log(k - t(k-1))\big]_0^1 \bigg\}$

$= \frac{\sqrt{4 + (k-1)^2}}{k-1} \bigg\{ \log k - (- \log k) \bigg\}$

$= 2\frac{\sqrt{4 + (k-1)^2}}{k-1} \log k$

This final expression gives the lengths of the piecewise linear paths as a function of $k$. If we plot this function of $k$ we see something completely at odds with Euclidean geometry:

We see that for values of $k$ between $1$ and $6.5$ the lengths of the piecewise linear paths are actually less than the length of the straight horizontal line between the two points $-2 + i$ and $2 + i$. The minimum possible path length between the two points is achieved by one of the piecewise linear paths (with a value of $k$ somewhere between $2.25$ and $2.5$) rather than by the straight horizontal path between them. This shows that hyperbolic lengths behave very differently from the lengths we are used to in Euclidean geometry.

Clearly, this approach to calculating distances between points is at odds with the triangle inequality, so it cannot be used directly as the metric for measuring distances between points in hyperbolic space. Instead, given $z, z^{\prime} \in \mathbb{H}$, the hyperbolic distance $d_{\mathbb{H}}(z, z^{\prime})$ between $z$ and $z^{\prime}$ is defined as

$d_{\mathbb{H}}(z, z^{\prime}) = \text{inf}\{\text{Length}_{\mathbb{H}}(\sigma) : \sigma \text{ is a piecewise differentiable path with endpoints } z \text{ and } z^{\prime} \}$

So what we do is consider all piecewise differentiable paths between $z$ and $z^{\prime}$, calculate the hyperbolic length of each such path, then take the shortest. It turns out that this infimum is always actually achieved by some path (a geodesic) and this path is unique. The graph I plotted above is an illustration of this.

The metric $d_{\mathbb{H}}$ clearly does satisfy the triangle inequality, i.e.,

$d_{\mathbb{H}}(x, z) \leq d_{\mathbb{H}}(x, y) + d_{\mathbb{H}}(y, z)$

for all $x, y, z \in \mathbb{H}$. The distance between two points according to this metric must be increased if one goes via a third point, because it is by definition the infimum of all hyperbolic lengths between the points, so those involving any third point must have already been discounted as being longer.