Section 5.5 The Upper Half-Plane Model
Definition 5.5.1.
The upper half-plane model of hyperbolic geometry has space U consisting of all complex numbers z such that Im(z)>0, and transformation group U consisting of all MΓΆbius transformations that send U to itself. The space U is called the upper half-plane of C.
Invert about the circle C centered at i passing through -1 and 1 as in Figure 5.5.2.
Reflect about the real axis.
Going between (D,H) and (U,U).
The MΓΆbius transformation V mapping D to U, and its inverse Vβ1, are given by:
Example 5.5.4. The distance between ri and si.
For r>s>0 we compute the distance between ri and si in the upper half-plane model.
The hyperbolic line through ri and si is the positive imaginary axis, having ideal points 0 and β. Thus,
Example 5.5.5. The distance between any two points.
To find the distance between any two points w1 and w2 in U, we first build a map in the upper half-plane model that moves these two points to the positive imaginary axis. To build this map, we work through the PoincarΓ© disk model.
By the transformation Vβ1 we send w1 and w2 back to D. We let z1=Vβ1(w1) and z2=Vβ1(w2). Then, let S(z)=eiΞΈzβz11βΒ―z1z be the transformation in (D,H) that sends z1 to 0 with ΞΈ chosen carefully so that z2 gets sent to the positive imaginary axis. In fact, z2 gets sent to the point ki where k=|S(z2)|=|S(Vβ1(w2))| (and 0<k<1). Then, applying V to the situation, 0 gets sent to i and ki gets sent to 1+k1βki. Thus, VβSβVβ1 sends w1 to i and w2 to 1+k1βki, where by the previous example the distance between the points is known:
Describing k in terms of w1 and w2 is left for the adventurous reader. We do not need to pursue that here.
Definition 5.5.6.
The length of a smooth curve r(t) for aβ€tβ€b in the upper half-plane model (U,U), denoted L(r), is given by
Example 5.5.7. The length of a curve.
To find the length of the horizontal curve r(t)=t+ki for aβ€tβ€b, note that rβ²(t)=1 and Im(r(t))=k. Thus,
Definition 5.5.8.
In the upper half-plane model (U,U) of hyperbolic geometry, the area of a region R described in cartesian coordinates, denoted A(R), is given by
Example 5.5.9. The area of a 23-ideal triangle.
Suppose wβU is on the unit circle, and consider the 23-ideal triangle 1wβ as pictured.
In particular, suppose the interior angle at w is Ξ±, so that w=ei(ΟβΞ±) where 0<Ξ±<Ο.
The area of this 23-ideal triangle is thus
With the trig substituion cos(ΞΈ)=x, so that β1βx2=sin(ΞΈ) and βsin(ΞΈ)dΞΈ=dx, the integral becomes
Theorem 5.5.10.
The area of a 23-ideal triangle having interior angle Ξ± is equal to ΟβΞ±.
Exercises Exercises
1.
What becomes of horocycles when we transfer the disk model of hyperbolic geometry to the upper half-plane model?
2.
What do hyperbolic rotations in the disk model look like over in the upper half-plane model? What about hyperbolic translations?
3.
Give an explicit description of a transformation that takes an arbitrary 23-ideal triangle in the upper half-plane to one with ideal points 1 and β and an interior vertex on the upper half of the unit circle.
4.
Determine the area of the βtriangularβ region pictured below. What is the image of this region under Vβ1 in the disk model of hyperbolic geometry? Why doesn't this result contradict Theorem 5.4.9?
5.
Another type of block. Consider the four-sided figure pqst in (D,H) shown in the following diagram. This figure is determined by two horocycles C1 and C2, and two hyperbolic lines L1 and L2 all sharing the same ideal point. Note that the lines are orthogonal to the horocycles, so that each angle in the four-sided figure is 90β.
- By rotation about the origin if necessary, assume the common ideal point is i and use the map V to transfer the figure to the upper half-plane. What does the transferred figure look like in U? Answer parts (b)-(d) by using this transferred version of the figure.
- Prove that the hyperbolic lengths of sides pq and st are equal.
- Let c equal the hyperbolic length of the leg pt along the larger radius horocycle C1, and let d equal the hyperbolic length of the leg sq on C2. Show that c=exd where x is the common length found in part (b).
- Prove that the area of the four-sided figure is cβd.