## Geodesics on the Hyperbolic Plane

Continuing from the last post with the upper half plane model as our model of the hyperbolic plane. First, it is to be noted that the fractional linear transformations or more commonly called the Moebius transformations maps the upper half plane $H$ to itself.

Consider the transformed point under $T_g$ on $z\in H$ where $g=\begin{pmatrix} a&b\\c&d \end{pmatrix}$ :

$w\equiv T_g(z)=\cfrac{az+b}{cz+d}=\cfrac{ac\vert z\vert^2 + adz+bc\bar{z}+bd}{\vert cz+d\vert^2}$ .

The imaginary part of the transformed point is

$\textrm{Im}(w)=\cfrac{w-\bar{w}}{2i}=\cfrac{(ad-bc)(z-\bar{z})}{2i\vert cz+d\vert^2}=\cfrac{\textrm{Im}(z)}{\vert cz+d\vert^2}$ .

Note that if $\textrm{Im}(z)>0$ then the above shows that $\textrm{Im}(w)>0$ also, and hence the earlier statetment of the map onto $H$ itself.

To continue studying the geometry of the hyperbolic plane, one must have a notion of hyperbolic distance and this is obtained from the previous hyperbolic metric on the upper half plane. One thus defined the infinitesimal element of the hyperbolic distance to be

$ds=\cfrac{\sqrt{dx^2+dy^2}}{y}$ .

Given a curve $\gamma(t)=\{v(t)=x(t)+iy(t)\vert t\in H\}$, its length is thus given by

$h(\gamma)=\displaystyle\int\limits_0^1 \cfrac{\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}}{y(t)}\ dt$ .

For the hyperbolic distance between $z,w\in H$, one takes the minimum value of the length of curves connectig the two points i.e.

$\rho(z,w)=\inf_\gamma h(\gamma)$ .

Later, it will be convenient for us to know exactly what are the shortest hyperbolic curves connecting two point on $H$ i.e. the geodesics. For now, we note that the Moebius transformations themselves are isometries (preserve lengths). Consider the transformed points on a curve $\gamma(t)$ :

$w(t) = T(z) = \cfrac{az+b}{cz+d} = u(t) + iv(t)$ .

Its derivative is

$\cfrac{dw}{dz}=\cfrac{a(cz+d)-c(az+b)}{(cz+d)^2}=\cfrac{1}{(cz+d)^2}$ .

Noting that $v=y/\vert cz+d\vert^2=\textrm{Im}(w)$, we can write the derivative as $\vert\frac{dw}{dz}\vert = v/y$. Thus under the transformation, the hyperbolic distance stays invariant as shown below:

$\begin{matrix} h(T(\gamma))&=\displaystyle\int\limits_0^1\cfrac{\vert\frac{dw}{dt}\vert\ dt}{v(t)}\\ &=\displaystyle\int\limits_0^1 \cfrac{\vert\frac{dv}{dz}\vert\vert\frac{dz}{dt}\vert\ dt}{v(t)} \\ &=\displaystyle\int\limits_0^1\cfrac{\vert\frac{dz}{dt}\vert dt}{y(t)}= h(\gamma) \end{matrix}$ .

Next, we claim that geodesics in $H$ are either (arcs of) semicircles and vertical lines orthogonal to the real axis. We calculate first, the formula of the hyperbolic length of two points on a semicircle with centre $(c,0)$ (see diagram below).

The two points can be coordinatized by the angles $\alpha,\beta$ or parametrized by the following:

$x=c+r\cos t\ ;\quad y=r\sin t$ .

Their differentials are

$dx=-r\sin t dt\ ;\quad dy= r\cos t dt$ .

Plugging into the distance formula on a hyperbolic curve gives

$\rho(\alpha,\beta)=\displaystyle\int\limits_\alpha^\beta \cfrac{\sqrt{(-r\sin t dt)^2+(r\cos t dt)^2}}{r\sin t}= \displaystyle\int\limits_\alpha^\beta \cfrac{r dt}{r\sin t}$

Note the denominator of the final integrand can be rexpressed as $2\sin\frac{t}{2}\cos\frac{t}{2}$ and observing that

$d(\ln\tan\frac{t}{2}) = \cfrac{1}{\tan\frac{t}{2}}\ \cfrac{\sec^2\frac{t}{2}}{2}\ dt$ ,

one obtains

$\rho(\alpha,\beta)=\displaystyle\int\limits_\alpha^\beta d\left(\ln\tan\frac{t}{2}\right)=\ln\left(\cfrac{\tan\beta/2}{\tan\alpha/2}\right)$

Using trigonometric identity,

$\csc\theta-\cot\theta=\cfrac{1}{\sin\theta}-\cfrac{\cos\theta}{\sin\theta}=\cfrac{1-\cos^2\theta/2+\sin^2\theta/2}{2\sin\theta/2\cos\theta/2}=\cfrac{\sin\theta/2}{\cos\theta/2}$ ,

we can reexpress the distance (following Stahl) as

$\rho(\alpha,\beta)=\ln\left(\cfrac{\csc\beta-\cot\beta}{\csc\alpha-\cot\alpha}\right)$ .

For the case of two points on the vertical line

the distance formula (with $dx=0$) gives

$\rho(y_1,y_2)=\displaystyle\int\limits_{y_1}^{y_1} \cfrac{dy}{y}=\ln\left(\cfrac{y_2}{y_1}\right)$ .

Next, we show the distance formula given by the (claimed) geodesics indeed give the infimum distance. Consider an arbitrary curve between $P$ and $Q$ (not collinear vertically):

Using the same coordinatization $x=d+r\cos\theta,\ y=r\sin\theta$ but with $r$ depend on $\theta$, we obtain

$\cfrac{dx}{d\theta}=r'\cos\theta-r\sin\theta\ ;\quad\cfrac{dy}{d\theta}=r'\sin\theta+r\cos\theta$ .

Hence,

$dx^2+dy^2=\left(\cfrac{dx}{d\theta}\right)^2d\theta^2+\left(\cfrac{dy}{d\theta}\right)^2d\theta^2=(r'^2+r^2)d\theta^2$ .

Thus, the distance measured through arbitrary curve connecting $P$ and $Q$ obeys

$\displaystyle\int\limits_\alpha^\beta \cfrac{\sqrt{r'^2+r^2}}{r\sin\theta}\ d\theta\geq\displaystyle\int\limits_\alpha^\beta \cfrac{\sqrt{r^2}}{r\sin\theta} d\theta=\int_\alpha^\beta\csc\theta d\theta=\rho(\alpha,\beta)$ .

This shows the arc of the semicircle connecting $P$ and $Q$ gives the infimum hyperbolic distance. Next we do the same for arbitrary curves joining two vertically collinear points.

Let $x$ be a function of $y$ i.e. $x=x(y)$. Then

$\displaystyle\int\limits_{y_1}^{y_2} \cfrac{\sqrt{dx^2+dy^2}}{y}=\displaystyle\int\limits_{y_1}^{y_2} \cfrac{\sqrt{x'^2dy^2+dy^2}}{y}\geq\displaystyle\int\limits_{y_1}^{y_2} \cfrac{dy}{y}=\ln\cfrac{y_2}{y_1}$ .

Again, the vertical line joining the two points give the least distance in comparison to other curves joining them, hence the earlier-claimed geodesics.

Finally we would like to give the solution to Exercise 4 of Katok’s article where one could essentially transform any semicircle (geodesic) to the vertical line and hence making any other proofs concerning geodesics easier. Let the semicircle be centred at the origin with radius $\alpha$ and the vertical line be the $y$-axis. Both can be parametrized as $z=\alpha e^{i\theta}\ ,\ z=i\tan\theta/2$ respectively. Consider then the transformation $T(z)=-(z-\alpha)^{-1}+\beta$. Plugging in the parametrization gives

$T(z)=\cfrac{1}{\alpha(1-e^{i\theta})}+\beta=\cfrac{1+\beta\alpha(1-e^{i\theta})}{\alpha e^{i\theta/2}(e^{-i\theta/2}-e^{i\theta/2})}$ .

Putting $\beta\alpha = -1/2$ and multiplying top and bottom by $e^{-i\theta/2}$ gives

$T(z)=-\cfrac{1}{2\alpha}\ \cfrac{e^{-i\theta/2}+e^{i\theta/2}}{e^{i\theta/2}-e^{-i\theta/2}}$ .

Note that the second factor is already the reciprocal of $i\tan\theta/2$. Thus by putting $\alpha=1/2$ and composing the transformation with an inversion $S(z)=-1/z$ gives the necessary transformation i.e.

$(S\circ T)(z)=\left(\cfrac{1}{z-\frac{1}{2}} + 1\right)^{-1}$ .

References

1. S. Katok, “Fuchsian Groups, Geodesic Flows on Surfaces of Constant Negative Curvature and Symbolic Coding of Geodesics”, (http://www.math.psu.edu/katok_s/cmi.pdf)
2. R. Hayter, “The Hyperbolic Plane – A Strange New Universe” , (http://www.maths.dur.ac.uk/Ug/projects/highlights/CM3/Hayter_Hyperbolic_report.pdf)
3. S. Stahl, A Gateway to Modern Geometry: The Poincare Half-Plane, (Jones & Bartlett, 2007).
4. J. Hilgert, “Maass Cusp Forms on $\textrm{SL}(2,\mathbb{R})$“, (http://www.math.uni-paderborn.de/~hilgert/Metz05web.pdf)