The Cayley-Klein model of hyperbolic geometry

Let ${\cal C} \subset {\Bbb R}P^2$ be a nondegenerate conic and let $G_{\cal C} \subset PGL(3,{\Bbb R})$ be the subgroup of projective transformations that leave the conic invariant. The aim of the next four exercises is to prove the following result:

Theorem 5.1   The group $G_{\cal C}$ acts transitively on the interior and the exterior of the conic ${\cal C}$.

Because of the results in the previous section,s it is enough to consider the case where the conic is given in homogeneous coordinates by the equation $x^2 + y^2 - z^2 = 0$.

The strategy of the proof is to find a large class of explicit linear transformations on ${\Bbb R}^3$ that preserve the cone $x^2 + y^2 - z^2 = 0$ and to show that by using transformations in this class it is possible to take a line that passes through the origin and lies in the interior (resp. exterior) of the cone to any other line that passes through the origin and lies in the interior (resp. exterior) of the cone.

Exercise 5.1 (-05)   Define the hyperbolic functions

$$\cosh(t) := \frac{e^t + e^{-t}}{2} , \quad \sinh(t) := \frac{e^t - e^{-t}}{2} .$$

Show that

1. $\cosh^2(t) - \sinh^{2}(t) = 1$,
2. $\cosh(t+s) = \cosh(t)\cosh(s) + \sinh(t)\sinh(s)$,
3. $\sinh(t+s) = \cosh(t)\sinh(s) + \sinh(t)\cosh(s)$,
4. the function $\tanh(t) := \sinh(t)/\cosh(t)$ is a bijection between the real line and the interval $(-1,1)$.

If $\theta$ and $t$ are real numbers, we define

$$R_{\theta} := \begin{pmatrix} \cos(\theta) & \sin(\theta) & 0 \\... ... & 0 \\ 0 & \cosh(t) & \sinh(t) \\ 0 & \sinh(t) & \cosh(t) \end{pmatrix} .$$

Exercise 5.2 (05)   Show that

1. $R_{\theta}, H_{t} \in G_{\cal C}$.
2. $R_{\theta + \phi} = R_{\theta} R_{\phi}$ and $H_{t+s} = H_{t} H_{s}$.

Exercise 5.3 (05)   Show that if $y$ belongs to the interval $(-1,1)$, there exists a number $t$ such that $H_{t}$ applied to the vector $(0,0,1)$ is a nonzero multiple of the vector $(0,y,1)$. Conclude that if $y, y' \in (-1,1)$, there exists a number $s$ such that $H_{s}$ applied to the vector $(0,y,1)$ is a nonzero multiple of the vector $(0,y',1)$.

The previous exercise easily implies that $G_{\cal C}$ acts transitively on the interior of ${\cal C}$. Indeed, if $L_{1}$ and $L_{2}$ are two lines in ${\Bbb R}^3$ passing through the origin and inside the cone $x^2 + y^2 - z^2 = 0$, there exist rotations $R_{\theta}$ and $R_{\phi}$ such that $R_{\theta}(L_{1})$ and $R_{\phi}(L_{2})$ lie on the plane $x = 0$. By the previous exercise, there exists a transformation $H_{t}$ such that $H_{t}(R_{\theta}(L_{1})) = R_{\phi}(L_2)$. It follows that $R_{-\phi}(H_{t}(R_{\theta}(L_{1}))) = L_2$, and we are done.

Exercise 5.4 (10)   Use the same techniques as above to show that $G_{\cal C}$ acts transitively on the exterior of ${\cal C}$.

We now pass to the study of the action of $G_{\cal C}$ on the interior of ${\cal C}$, which we will henceforth denote as ${\cal D}$. The first thing we remark is that the action is not doubly transtive: it is in general impossible to transform a pair of distinct points $X,Y \in {\cal D}$ into a pair of prescribed points $X',Y'$. As before, it suffices to consider the case where the conic is a circle.

If $X$ and $Y$ are two distinct points on ${\cal D}$, draw the line that joins them and intersect it with the circle to obtain two points $A$ and $B$. Choose the notation such that $X$ belongs to the segment joining $A$ and $Y$, and $Y$ belonds to the segment joining $B$ and $X$ (Fig. 9). Now define $\rho(X,Y) := [A,B,Y,X]$. Notice that we have chosen this precise ordering (and the definition of $A$ and $B$) so that this cross-ratio be greater than one. If $X = Y$, we define $\rho(X,Y)$ to be one.

Figure 9.

Exercise 5.5 (15)   Let $X,Y,X',Y'$ be four points on ${\cal D}$. Show that there exists a map $T \in G_{\cal C}$ that takes $X$ to $X'$ and $Y$ to $Y'$ if and only if $\rho(X,Y) = \rho(X',Y')$.

If $X$ and $Y$ are two points in ${\cal D}$, define $d(X,Y) := \ln(\rho(X,Y))$. The rest of this section is devoted to proving that $d$ is a distance function and that its geodesics are straight lines.

Exercise 5.6   Show that

1. $d(X,Y) \geq 0$ with equality if and only if $X = Y$.
2. $d(X,Y) = d(Y,X)$.
3. If $X,Y,Z$ are three collinear points, with $Y$ belonging to the segment that joins $X$ and $Z$, then $d(X,Z) = d(X,Y) + d(Y,Z)$.

To show that $d$ is a distance function, it remains to prove the triangle inequality for three noncollinear points $X, Y, Z \in {\cal D}$ (Fig. 10). Since the logarithm is an increasing function, all we must do is show that $\rho(X,Z) < \rho(X,Y) \rho(Y,Z)$, where these quantities are

$$\rho(X,Y) = \frac{\vert X-B\vert}{\vert X-A\vert} \frac{\vert Y-A\vert}{\vert... ... = \frac{\vert Y-D\vert}{\vert Y-C\vert} \frac{\vert Z-C\vert}{\vert Z-D\vert},$$

$$\rho(X,Z) = \frac{\vert X-F\vert}{\vert X-E\vert} \frac{\vert Z-E\vert}{\vert Z-F\vert}.$$

Figures 10 and 11.

Using figure 11 and the invariance of cross-ratios under perspectivities, we have that

$$\rho(X,Y) = \frac{\vert X-R\vert}{\vert X-P\vert} \frac{\vert Q-P\vert}{\vert Q-R\vert},$$    and $$\rho(Y,Z) = \frac{\vert Q-R\vert}{\vert Q-P\vert} \frac{\vert Z-P\vert}{\vert Z-R\vert}.$$

Therefore,

$$\rho(X,Y) \rho(Y,Z) = \frac{\vert X-R\vert}{\vert X-P\vert} \frac{\vert Z-P\vert}{\vert Z-R\vert}.$$

Exercise 5.7   Finish the proof of the triangle inequality by showing that

$$\frac{\vert X-R\vert}{\vert X-P\vert} \frac{\vert Z-P\vert}{\vert... ... \frac{\vert X-F\vert}{\vert X-E\vert} \frac{\vert Z-E\vert}{\vert Z-F\vert}.$$

The disc ${\cal D}$ together with distance $d$ is called the hyperbolic plane.

Definition 5.1   Let $(M,d)$ be a metric space and let $\gamma: [a,b] \rightarrow M$ be a continuous curve. Define the length of $\gamma$ as

$$\sup \{ \sum_{i = 0}^{n-1} d(\gamma(t_{i}),\gamma(t_{i+1})) : a = t_{0} < \cdots < t_{n} = b$$    is a partition of $$[a,b] \}$$

The curve $\gamma$ is called a geodesic segment if its length equals the distance between its endpoints.

Exercise 5.8 (10)   Show that any line segment on ${\cal D}$ is a geodesic segment for the distance $d$.