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Section 5.5 The Upper Half-Plane Model

The PoincarΓ© disk model is one way to represent hyperbolic geometry, and for most purposes it serves us very well. However, another model, called the upper half-plane model, makes some computations easier, including the calculation of the area of a triangle.

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.

The PoincarΓ© disk model of hyperbolic geometry may be transferred to the upper half-plane model via a MΓΆbius transformation built from two inversions as follows:

  1. Invert about the circle C centered at i passing through -1 and 1 as in Figure 5.5.2.

  2. Reflect about the real axis.

Figure 5.5.2. Inversion in C maps the unit disk to the upper-half plane.

Notice that inversion about the circle C fixes -1 and 1, and it takes i to ∞. Since reflection across the real axis leaves these image points fixed, the composition of the two inversions is a MΓΆbius transformation that takes the unit circle to the real axis. The map also sends the interior of the disk into the upper half plane. Notice further that the MΓΆbius transformation takes ∞ to βˆ’i; therefore, by Theorem 3.5.8, the map can be written as

V(z)=βˆ’iz+1zβˆ’i.

This MΓΆbius transformation is the key to transferring the disk model of the hyperbolic plane to the upper half-plane model. In fact, when treading back and forth between these models it is convenient to adopt the following convention for this section: Let z denote a point in D, and w denote a point in the upper half-plane U, as in Figure 5.5.3. We record the transformations linking the spaces below.

Going between (D,H) and (U,U).

The MΓΆbius transformation V mapping D to U, and its inverse Vβˆ’1, are given by:

w=V(z)=βˆ’iz+1zβˆ’i    and    z=Vβˆ’1(w)=iw+1w+i.

Some features of the upper half-plane model immediately come to light. Since V is a Mâbius transformation, it preserves clines and angles. This means that the ideal points in the disk model, namely the points on the circle at infinity, S1∞, have moved to the real axis and that hyperbolic lines in the disk model have become clines that intersect the real axis at right angles.

Figure 5.5.3. Mapping the disk to the upper half-plane.

Define the hyperbolic distance between two points w1,w2 in the upper half-plane model, denoted dU(w1,w2), to be the hyperbolic distance between their pre-images in the disk model.

Suppose w1 and w2 are two points in V whose pre-images in the unit disk are z1 and z2, respectively. Then,

dU(w1,w2)=dH(z1,z2)=ln((z1,z2;u,v)),

where u and v are the ideal points of the hyperbolic line through z1 and z2. But, since the cross ratio is preserved under MΓΆbius transformations,

dU(w1,w2)=ln((w1,w2;p,q)),

where p,q are the ideal points of the hyperbolic line in the upper half-plane through w1 and w2. In particular, going from w1 to w2 we're heading toward ideal point p.

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 ∞. Thus,

dU(ri,si)=ln((ri,si;0,∞))=riβˆ’0riβˆ’βˆžβ‹…siβˆ’βˆžsiβˆ’0=ln(rs).
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 Vβˆ’1 we send w1 and w2 back to D. We let z1=Vβˆ’1(w1) and z2=Vβˆ’1(w2). Then, let S(z)=eiΞΈzβˆ’z11βˆ’Β―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(Vβˆ’1(w2))| (and 0<k<1). Then, applying V to the situation, 0 gets sent to i and ki gets sent to 1+k1βˆ’ki. Thus, V∘S∘Vβˆ’1 sends w1 to i and w2 to 1+k1βˆ’ki, where by the previous example the distance between the points is known:

dU(w1,w2)=ln(1+k)βˆ’ln(1βˆ’k).

Describing k in terms of w1 and w2 is left for the adventurous reader. We do not need to pursue that here.

We now derive the hyperbolic arc-length differential for the upper half-plane model working once again through the disk model. Recall the arc-length differential in the disk model is

ds=2|dz|1βˆ’|z|2.

Since z=Vβˆ’1(w)=iw+1w+i we may work out the arc-length differential in terms of dw. We will need to take the derivative of a complex expression, which can be done just as if it were a real valued expression. Here we go:

ds=2|dz|1βˆ’|z|2=2|d(iw+1w+i)|1βˆ’|iw+1w+i|2=2|i(w+i)dwβˆ’(iw+1)dw||w+i|2/[1βˆ’|iw+1|2|w+i|2]=4|dw||w+i|2βˆ’|iw+1|2=4|dw|(w+i)(Β―wβˆ’i)βˆ’(iw+1)(βˆ’iΒ―w+1)=4|dw|2i(Β―wβˆ’w)=|dw|Im(w).

This leads us to the following definition:

Definition 5.5.6.

The length of a smooth curve r(t) for a≀t≀b in the upper half-plane model (U,U), denoted L(r), is given by

L(r)=∫ba|rβ€²(t)|Im(r(t)) dt.
Example 5.5.7. The length of a curve.

To find the length of the horizontal curve r(t)=t+ki for a≀t≀b, note that rβ€²(t)=1 and  Im(r(t))=k. Thus,

L(r)=∫ba1k dt=bβˆ’ak.

From the arc-length differential ds=dwIm(w) comes the area differential:

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

A(R)=∬R1y2 dxdy.
Example 5.5.9. The area of a 23-ideal triangle.

Suppose w∈U is on the unit circle, and consider the 23-ideal triangle 1w∞ as pictured.

In particular, suppose the interior angle at w is Ξ±, so that w=ei(Ο€βˆ’Ξ±) where 0<Ξ±<Ο€.

The area of this 23-ideal triangle is thus

A=∫1cos(Ο€βˆ’Ξ±)∫∞√1βˆ’x21y2 dydx=∫1cos(Ο€βˆ’Ξ±)1√1βˆ’x2 dx.

With the trig substituion cos(ΞΈ)=x, so that √1βˆ’x2=sin(ΞΈ) and βˆ’sin(ΞΈ)dΞΈ=dx, the integral becomes

=∫0Ο€βˆ’Ξ±βˆ’sin(ΞΈ)sin(ΞΈ) dΞΈ=Ο€βˆ’Ξ±.

It turns out that any 23-ideal triangle is congruent to one of the form 1w∞ where w is on the upper half of the unit circle (Exercise 5.5.3), and since our transformations preserve angles and area, we have proved the area formula for a 23-ideal triangle.

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 ∞ and an interior vertex on the upper half of the unit circle.

4.

Determine the area of the β€œtriangular” region pictured below. What is the image of this region under Vβˆ’1 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 90∘.

  1. 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.
  2. Prove that the hyperbolic lengths of sides pq and st are equal.
  3. 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).
  4. Prove that the area of the four-sided figure is cβˆ’d.