Hyperbolic Geometry Note #2: On Möbius transformations as isometries of the upper half plane

Möbius transformations are conformal (i.e., angle-preserving), orientation-preserving, area-preserving isometries in the upper half-plane model of hyperbolic space. It piqued my interest to discover that the form of the metric in this model (which I discussed in Hyperbolic Geometry Note #1) is specifically chosen to ensure that Möbius transformations are indeed isometries. It turns out that in metric spaces one can sometimes `fiddle’ with the form of the metric in order to obtain geometries with particular properties, and the upper half-plane model is an example of this. In this note I want to explore in detail how this `fiddling’ with the metric works.

For $z \in \mathbb{H}$ , a Möbius transformation would take the form

$\gamma(z) = \frac{az + b}{cz + d}$

where the coefficients $a$ , $b$ , $c$ , and $d$ are real numbers satisfying the condition $ad - bc > 0$ . This is an isometry in $\mathbb{H}$ (if one uses the metric discussed in my previous note) because it preserves lengths, i.e., for distinct $z, z^{\prime} \in \mathbb{H}$ we have

$d_{\mathbb{H}}(\gamma(z), \gamma(z^{\prime})) = d_{\mathbb{H}}(z, z^{\prime})$

I want to first prove this in detail before going on to show how the form of the metric guarantees that this result will hold. To prove the isometry property we note that if $\sigma$ is a path from $z$ to $z^{\prime}$ then $(\gamma \circ \sigma)(z) \equiv \gamma(\sigma(z))$ is a path from $\gamma(z)$ to $\gamma(z^{\prime})$ . The shortest such path (whose length would by definition be the hyperbolic distance between the two points) must also be of this form, so all we need to prove is that

$\text{Length}_{\mathbb{H}}(\gamma \circ \sigma) = \text{Length}_{\mathbb{H}}(\sigma)$

where (as discussed in my previous note) the hyperbolic length of a path $\sigma$ is obtained by integrating the function $f(z) = 1/\text{Im}(z)$ along $\sigma$ , so

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

We need two auxiliary results concerning Möbius transformations, namely, expressions for $|\gamma^{\prime}(z)|$ and $\text{Im}(\gamma(z))$ . For any $z \in \mathbb{H}$ we have

$|\gamma^{\prime}(z)| = \big| \frac{(cz + d)a - (az + b)c}{(cz + d)^2} \big| = \big| \frac{ad - bc}{(cz + d)^2} \big| = \frac{ad - bc}{|cz + d|^2}$

and if we write $z = u + iv$ we have

$\text{Im}(\gamma(z)) = \frac{\gamma(z) - \overline{\gamma(z)}}{2i}$

$= \frac{1}{2i}\big\{\frac{az + b}{cz + d} - \frac{a\overline{z} + b}{c\overline{z} + d}\big\}$

$= \frac{1}{2i}\frac{(az+b)(c\overline{z}+d) - (a\overline{z}+b)(cz+d)}{(cz+d)(c\overline{z}+d)}$

$= \frac{1}{2i}\frac{azd + bc\overline{z} + bd - a\overline{z}d - bcz -bd}{|cz + d|^2}$

$= \frac{1}{2i}\frac{2iv(ad - bc)}{|cz + d|^2}$

$= \frac{ad - bc}{|cz + d|^2}v$

$= \frac{ad - bc}{|cz + d|^2} \text{Im}(z)$

Using these two results and the chain rule we obtain

$\text{Length}_{\mathbb{H}}(\gamma \circ \sigma) = \int \frac{|(\gamma \circ \sigma)^{\prime}(t)|}{\text{Im}(\gamma \circ \sigma)(t)}dt$

$= \int \frac{|(\gamma^{\prime}(\sigma(t))||\sigma^{\prime}(t)|}{\text{Im}(\gamma \circ \sigma)(t))}dt$

$= \int \frac{ad - bc}{|c\sigma(t) + d|^2} |\sigma^{\prime}(t)|\frac{|c\sigma(t) + d|^2}{ad - bc}\frac{1}{\text{Im}(\sigma(t))}dt$

$= \int \frac{|\sigma^{\prime}(t)|}{\text{Im}(\sigma(t))}dt = \text{Length}_{\mathbb{H}}(\sigma)$

which proves the isometry result.

It turns out that if we want this isometry result to hold, we must define the hyperbolic length using the function $1/\text{Im}(z)$ as above. To see why this is the case, let $\rho: \mathbb{H} \rightarrow \mathbb{R}$ be a continuous positive function. Define the $\rho$ -length of a path $\sigma: [a, b] \rightarrow \mathbb{H}$ in the usual way to be

$\text{Length}_{\rho}(\sigma) = \int_{\sigma} \rho = \int_a^b \rho(\sigma(t))|\sigma^{\prime}(t)|dt$

Now suppose that $\text{Length}_{\rho}$ is invariant under Möbius transformations of $\mathbb{H}$ , i.e., if $\gamma$ is a Möbius transformation we have

$\text{Length}_{\rho}(\gamma \circ \sigma) = \text{Length}_{\rho}(\sigma)$

On the left hand side we have

$\text{Length}_{\rho}(\gamma \ \circ \ \sigma) = \int_a^b \rho((\gamma \ \circ \ \sigma)(t))|(\gamma \ \circ \ \sigma)^{\prime}(t)|dt = \int_a^b \rho((\gamma \ \circ \ \sigma)(t))|\gamma^{\prime}(\sigma(t))||\sigma^{\prime}(t)|dt$

while on the right hand side we have

$\text{Length}_{\rho}(\sigma) = \int_a^b \rho(\sigma(t))|\sigma^{\prime}(t)|dt$

These two expressions can only be (identically) equal if for all $t$ we have

$\rho((\gamma \circ \sigma)(t))|\gamma^{\prime}(\sigma(t))| = \rho(\sigma(t))$

Writing $z = \sigma(t)$ , we can express this condition on the function $\rho$ as

$\rho(\gamma(z)) |\gamma^{\prime}(z)| = \rho(z)$

Now suppose we take $\gamma(z) = z + b$ in this last equation, with $b$ any real number and $z = x + iy$ . We get

$\rho(z + b) |1| = \rho(z)$

$\rho(z + b) = \rho(z)$

This shows that the real part of the argument of $\rho$ does not matter, because (for example) we can set $b = -x$ and get

$\rho(iy) = \rho(z)$

We deduce that $\rho(z)$ depends only on the imaginary part of $z$ , so we can write

$\rho(z) = \rho(y)$

Now suppose we take $\gamma(z) = kz$ with $k > 0$ in the equation

$\rho(\gamma(z)) |\gamma^{\prime}(z)| = \rho(z)$

We get

$\rho(kz) |k| = \rho(z)$

which (since $\rho$ depends only on the imaginary part of its argument) we can write as

$k \rho(ky) = \rho(y)$

This equation can only hold for all $y$ in the upper half plane if on the left hand side we have

$\rho (ky) = \frac{c}{ky}$

and on the right hand side we have

$\rho(y) = \frac{c}{y}$

where $c$ is some positive constant. Therefore we see that if Möbius transformations of $\mathbb{H}$ are to be isometries, it must be the case that the function used in calculations of distances between points in $\mathbb{H}$ must be (up to a normalising constant $c > 0$ ) the reciprocal $1/\text{Im}(z)$ .

It is this result that really characterises the upper half-plane model of hyperbolic space, leading to the result that the geodesics in the upper half plane are either vertical lines or semicircles with endpoints on the real axis. The reasoning (in brief) is as follows.

geodesics

It can easily be shown that the imaginary axis in $\mathbb{H}$ is a geodesic: any path joining two distinct points $ia$ and $ib$ has a hyperbolic length greater than or equal to $\log b/a$ , with equality to $\log b/a$ only when the path is a straight line along the imaginary axis. It can also easily be shown that a path can be found between any two points in $\mathbb{H}$ which is either a vertical line or a semicircle with endpoints on the real axis, and that both of these can be mapped bijectively to the imaginary axis by a suitable Möbius transformation. (This Möbius transformation is either a simple translation if the path between the two points is a vertical line, or a transformation of the form

$\gamma(z) = \frac{z - \zeta_1}{z - \zeta_2}$

if the path is a semicircle with real endpoints $\zeta_1 < \zeta_2$ ). Since Möbius transformations are isometries, this must mean that such vertical lines and semicircles are also geodesics. It can finally be shown that there are no other such paths in $\mathbb{H}$ , so the Poincaré half-plane model of hyperbolic space is characterised by the fact that geodesics in $\mathbb{H}$ are either vertical lines or semicircles with endpoints on the real axis.

drchristiansalas

mathematical ideas blog

Hyperbolic Geometry Note #2: On Möbius transformations as isometries of the upper half plane