2. SupposeT is a Möbius transformation of the form T(z) =eiθ z−z0

1−z0z

,

where z0 is inD. Prove that T maps Dto D by showing that if|z|<1 then
|T(z)|<1. Hint: It is easier to prove that|z|2_{<}_{1}_{implies}_{|}_{T}_{(}_{z}_{)}_{|}2_{<}_{1}_{.}

3. Construct the xed points of the hyperbolic translation dened by the
inversion of two nonintersecting clines that intersect S1_{∞} _{at right angles, as}

shown in the following diagram.

S1∞

4. Supposepandqare two points inD. Construct two clines orthogonal

to S1∞ that, when inverted about in composition, send p to 0 and q to the

positive real axis.

5. Prove that any two horocycles in(D,H)are congruent.

6. Prove thatHis a group of transformations.

### 5.2 Figures of Hyperbolic Geometry

The Euclidean transformation group,E, consisting of all (Euclidean) rotations and translations, is generated by reections about Euclidean lines. Similarly, the transformations in H are generated by hyperbolic reections, which are inversions about clines that intersect the unit circle at right angles. This suggests that these clines ought to be the lines of hyperbolic geometry. Denition 5.2.1. A hyperbolic line in(D,H)is the portion of a cline inside Dthat is orthogonal to the circle at innity S1∞. A point onS1∞ is called an

ideal point. Two hyperbolic lines are parallel if they share one ideal point.

Figure 5.2.2: A few hyperbolic lines in the Poincaré disk model. Theorem 5.2.3. There exists a unique hyperbolic line through any two distinct points in the hyperbolic plane.

Proof. Let p and q be arbitrary points in D. Construct the point p∗

symmetric topwith respect to the unit circle, S1∞. Then there exists a cline

through p, q and p∗, and this cline will be orthogonal to S1∞, so it gives a

hyperbolic line throughpandq. Since there is just one cline through p, qand p∗, this hyperbolic line is unique.

Which hyperbolic lines happen to be portions of Euclidean lines (instead of Euclidean circles)? A Euclidean line intersects a circle at right angles if and only if it goes through the center of the circle. Thus, the only hyperbolic lines that also happen to be Euclidean lines are those that go through the origin.

One may also use a symmetric points argument to arrive at this last fact. Any Euclidean line goes through ∞. To be a hyperbolic line (i.e., to be orthogonal to S1∞), the line must also pass through the point symmetric to

∞ with respect to the unit circle. This point is 0. Thus, to be a hyperbolic line in(D,H), a Euclidean line must go through the origin.

Theorem 5.2.4. Any two hyperbolic lines are congruent in hyperbolic geom- etry.

Proof. We rst show that any given hyperbolic line L is congruent to the hyperbolic line on the real axis. Supposepis a point on L, andv is one of its ideal points. By Lemma 5.1.5 there is a transformationT inHthat mapspto 0,v to 1, andp∗ to∞. ThusT(L)is the portion of the real axis insideD, and

L is congruent to the hyperbolic line on the real axis. Since any hyperbolic line is congruent to the hyperbolic line on the real axis, the group nature ofH ensures that any two hyperbolic lines are congruent.

v 1 p 0 L T(L)

Figure 5.2.5: Any hyperbolic line is congruent to the hyperbolic line on the real axis.

Theorem 5.2.6. Given a point z0 and a hyperbolic line L not through z0,

there exist two distinct hyperbolic lines throughz0 that are parallel toL.

Proof. Consider the case wherez0 is at the origin. The lineLhas two ideal

points, call them uand v, as in Figure 5.2.7. Moreover, since Ldoes not go through the origin, Euclidean segmentuv is not a diameter of the unit circle. Construct one Euclidean line through 0 and u, and a second Euclidean line through 0 andv. (These lines will be distinct becauseuvis not a diameter of the unit circle.) Each of these lines is a hyperbolic line through 0, and each shares exactly one ideal point with L. Thus, each is parallel to L. The fact that the result holds for generalz0 is left as an exercise.

5.2. Figures of Hyperbolic Geometry 79

u

v

0

L

Figure 5.2.7: Through a point not on a given hyperbolic line L there exist two hyperbolic lines parallel toL.

Figure 5.2.7 illustrates an unusual feature of parallel lines in hyperbolic geometry: there is no notion of transitivity. In Euclidean geometry we know that if line L is parallel to line M, and line M is parallel to line N, then line L is parallel to line N. This is not the case in hyperbolic geometry.

Example 5.2.8: Hyperbolic triangles.

Three points in the hyperbolic plane D that are not all on a single

hyperbolic line determine a hyperbolic triangle. The hyperbolic triangle∆pqr is pictured below. The sides of the triangle are portions of hyperbolic lines.

Are all hyperbolic triangles congruent? No. Since transformations inHare Möbius transformations they preserve angles, so triangles with dierent angles are not congruent.

p q r

The next section develops a distance function for the hyperbolic plane. As in Euclidean geometry, we want to be able to compute the distance between two points, the length of a path, the area of a region, and so on. Moreover, the distance function should be an invariant; the distance between points should not change under a transformation inH. With this in mind, consider again a hyperbolic rotation about a pointp, as in Figure 5.1.6(a). It xes the point p and moves points around type II clines of p andp∗. If the distance between points is unchanged under transformations in H, then all points on a given type II cline ofpandp∗ will be the same distance away fromp. This leads us to dene a hyperbolic circle as follows.

Denition 5.2.9. Supposepis any point inD, andp∗is the point symmetric

topwith respect to the unit circle. A hyperbolic circle centered at pis a Euclidean circleC insideDthat is a type II circle ofpandp∗.

Figure 5.2.10 shows a typical hyperbolic circle. This circle is centered at point pand contains point q. Construction of such a circle may be achieved with compass and ruler as in Exercise 5.2.5.

p q

p∗ C

Figure 5.2.10: A hyperbolic circle centered atpthroughq.

Theorem 5.2.11. Given any points p and q in D, there exists a hyperbolic

circle centered at pthrough q.

Proof. Givenp, q∈_{D}, constructp∗, the point symmetric topwith respect
to S1∞. Then by Exercise 3.5.15 there exists a type II cline ofp andp∗ that

goes through q. This type II cline lives within _{D} because _{S}1

∞ is also a type

II cline ofpand p∗, and distinct type II clines cannot intersect. This type II cline is the hyperbolic circle centered atpthroughq.

Exercises

1. Suppose C is a hyperbolic circle centered at z0 through point p.

Show that there exists a hyperbolic line L tangent to C at p, and that L is perpendicular to hyperbolic segmentz0p.

2. Constructing a hyperbolic line through two given points.

a. Given a pointpin_{D}, construct the pointp∗symmetric topwith respect
to the unit circle (see Figure 3.2.18).

b. Supposeqis a second point in_{D}. Construct the cline throughp,q, and
p∗. Call this cline C. Explain whyC intersects the unit circle at right angles.
c. Consider the portion of cline C you constructed in part (b) that lies in

D. This is the unique hyperbolic line throughpandq. Mark the ideal points

of this hyperbolic line.

3. Can two distinct hyperbolic lines be tangent at some point inD? Explain.

4. SupposeLis a hyperbolic line that is part of a circleC. Can the origin of the complex plane be in the interior ofC? Explain.

5. Constructing a hyperbolic circle centered at a pointpthrough a pointq. Supposepandqare two points inD, and thatqis not on the line through

pandp∗ _{- the point symmetric to}_{p}_{with respect to the unit circle.}

a. Find the center of the Euclidean circle through p, p∗_{, and} _{q}_{. Call the}

center pointo.

b. Construct the segment oq.

c. Construct the perpendicular to oq at q. This perpendicular intersects the Euclidean line throughpandp∗. Call the intersection point o0.

d. Construct the Euclidean circle centered at o0 throughq.