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Introduction to Groups and Geometries

David W. Lyons

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Section 3.3 Hyperbolic geometry

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Before the discovery of hyperbolic geometry, it was believed that Euclidean geometry was the only possible geometry of the plane. In fact, hyperbolic geometry arose as a byproduct of efforts to prove that there was no alternative to Euclidean geometry. In this section, we present a Kleinian version of hyperbolic geometry.
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Definition 3.3.1.

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Let D={z:|z|<1} denote the open unit disk in the complex plane. The hyperbolic group, denoted H, is the subgroup of the Möbius group M of transformations that map D onto itself. The pair (D,H) is the (Poincaré) disk model of hyperbolic geometry.
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Comments on terminology: Beware of the two different meanings of the adjective "hyperbolic". To say that a Möbius transformation is hyperbolic means that it is conjugate to a homothety (see Subsection 3.2.6). That is not the same thing as an element of the group of hyperbolic transformations.
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Subsection 3.3.1 The hyperbolic transformation group

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Our first task is to characterize transformations in the group H. We begin with an observation about Möbius transformations that map one "side" of a cline to itself. This is pertinent because the disk D is the "inside" of the cline which is the unit circle. It will be useful to start with a general case.
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Any cline C divides the extended plane into two regions. If C is a Euclidean circle, we might called these regions the "inside" and the "outside" of C. If C is a Euclidean straight line, we simply have one side and the other of C.
Sketch: Suppose that T maps D onto itself. The "other side" of C is the set of points that are symmetric, with respect to C, to the points in D. By Proposition 3.2.27, T maps symmetric points to symmetric points, so T maps E into itself. It is easy to check that, in fact, T maps E onto itself. By elimination, it must be that T maps C onto itself.
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Instructor’s solution for Checkpoint 3.3.3.

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Let C be the unit circle, and let E be C^∖(C∪D) be the set of points on the outside of the unit circle. We claim that that E={z∗C:z∈D}. Indeed, it is clear that if w=z∗=1/z∗, then w lies in E. On the other hand, if w∈E, then z=w∗C=1/w∗ is in D, and w=z∗C. Now suppose that T maps D onto itself, and let w∈E. Then w∗C∈D. By Proposition 3.2.27, we have Tw=(Tz)∗C, which must lie in E. This shows that T maps E into itself. To see that T maps E onto itself, again let w∈E and let z=w∗C. Again applying Proposition 3.2.27, we have T((T−1(w∗C))∗C)=w. Finally, because T is one-to-one and onto when restricted to regions D and E, it must be that T maps C onto itself.
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Given T∈H, let z0∈D be the point that T maps to 0. It must be that T maps the symmetric point 1/z0∗ to ∞. Let z1 be the point that T maps to 1. Then T has the form (see (3.2.3))
Tz=z−z0z−1/z0∗z1−1/z0∗z1−z0.
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Multiplying top and bottom by −z0∗, and setting α=−z0∗z1−1/z0∗z1−z0, we have
Tz=αz−z01−z0∗z.
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A straightforward derivation shows that |α|=1, so that we have (3.3.1) below. Another computation establishes an alternative formula (3.3.2) for T∈H. See Exercise 3.3.6.1.
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Subsection 3.3.2 Classification of clines in hyperbolic geometry

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The clines of Möbius geometry are classified into several types in hyperbolic geometry, as summarized in Table 3.3.6.
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Table 3.3.6. Clines in hyperbolic geometry
hyperbolic curve type cline type
hyperbolic straight line a cline that intersects the unit circle at right angles
hyperbolic circle a circle completely contained in D
horocycle a circle with all but one point in D, tangent to the unit circle
hypercycle a cline that intersects the unit circle at a non-right angle
Show that each of the four categories of clines in Table 3.3.6 is preserved by transformations in the hyperbolic group. That is, show that any transformation in the hyperbolic group takes hyperbolic straight lines to hyperbolic straight lines, takes hyperbolic circles to hyperbolic circles, takes horocycles to horocycles, and takes hypercycles to hypercycles.
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Instructor’s solution for Checkpoint 3.3.7.

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Let T∈H. Because T is a Möbius transformation, T is conformal and preserves the intersection properties that define the four categories of clines in Table 3.3.6.
Show that a hyperbolic straight line that contains 0 must be a diameter of the unit circle.
Hint.
Prove the contrapositive: assume C is a hyperbolic straight line that is also a Euclidean circle, and intersects the unit circle orthogonally at p. Give an argument why C can not contain 0.
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Instructor’s solution for Checkpoint 3.3.8.

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(Proof of the contrapositive) Suppose C is a hyperbolic straight line that is also a Euclidean circle, and suppose that C intersects the unit circle at point p. Let L be the diameter of the unit circle at p. It must be that C is tangent to L at p, and therefore C lies on one side of L. Therefore C can not contain 0.
Show that all hyperbolic straight lines are congruent.
Hint.
Start by showing that any hyperbolic straight line is congruent to R^.
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Instructor’s solution for Checkpoint 3.3.9.

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First, we show that any hyperbolic straight line is congruent to R^. Let C be a hyperbolic straight line, and let p∈C∩D. The transformation Tz=z−p1−p∗z takes p to 0. By Checkpoint 3.3.8, T(C) is a diameter, which can now be rotated to R^. Next, given two hyperbolic straight lines C,D, choose hyperbolic transformations U,V that take C,D (respectively) to R^. Now V−1U is our desired congruence that takes C to D.
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Subsection 3.3.3 Normal forms for the hyperbolic group

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In this subsection, we follow the development of normal forms for general Möbius transformations given in Subsection 3.2.6 to derive normal forms and graphical interpretations for transformations in the hyperbolic group. We begin with an observation about fixed points of a Möbius transformation that maps a cline to itself.
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Now let T be a non-identity element of H. The fact that T maps the unit circle to itself implies that there are exactly three possible cases for fixed points of T.
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    There is a pair of fixed points p,q with |p|<1, |q|>1, and q=1p∗, that is, p,q are a pair of symmetric points (with respect to the unit circle) that do not lie on the unit circle.
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    There is a pair of fixed points that lie on the unit circle.
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    There is a single fixed point that lies on the unit circle.
Give an argument to justify the three cases above.
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Instructor’s solution for Checkpoint 3.3.11.

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By Lemma 3.3.10, fixed points are symmetric with respect to the unit circle C. By Exercise 3.2.8.5, we have z∗C=1z∗. From this we have |z||z∗C|=1, so the norms of |z|,|z∗| are either both 1, or one of the norms is less than 1 and the other is greater. Finally, a single (degenerate) fixed point must have norm 1.
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Figure 3.3.12. Three types of hyperbolic transformations
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For cases 1 and 2 above, the map T acting on the z-plane is conjugate to the map U=S∘T∘S−1 acting on the w-plane by Uw=λw, for some nonzero λ∈C, via the map w=Sz=z−pz−q. In case 1, the map S takes the unit circle to some polar circle, say C, so U must map C to itself. It follows that |λ|=1, so the Möbius normal form type for T is elliptic. The action of T is a rotation about Steiner circles of the second kind (hyperbolic circles) with respect to the fixed points p,q. A transformation T∈H of this type is called a hyperbolic rotation. See Figure 3.3.12.
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For case 2, the map w=Sz=z−pz−q takes the unit circle to a straight line, say L, through the origin, so U=S∘T∘S−1 must map L to itself. It follows that λ is real. Since S maps D to one of the two half planes on either side of L, the map U must take this half plane to itself. If follows that λ must be a positive real number, so the Möbius normal form type for T is hyperbolic. The action of T is a flow about Steiner circles of the first kind (hypercycles and one hyperbolic straight line) with respect to the fixed points p,q. A transformation T∈H of this type is called a hyperbolic translation. See Figure 3.3.12.
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For case 3, the conjugating map w=Sz=1z−p takes T to U=S∘T∘S−1 of the form Uw=w+β for some β≠0. The Möbius normal form type for T is parabolic. The action of T is a flow along degenerate Steiner circles (horocycles) tangent to the unit circle at p. A transformation T∈H of this type is called a parallel displacement. See Figure 3.3.12.
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This completes the list of transformation types for the hyperbolic group. See Table 3.3.13 for a summary.
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Table 3.3.13. Normal forms for the hyperbolic group
hyperbolic transformation type Möbius normal form graphical dynamic
hyperbolic rotation elliptic flow around hyperbolic circles
parallel displacement parabolic flow around horocycles
hyperbolic translation hyperbolic flow along hypercycles
(none) loxodromic
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Subsection 3.3.4 Hyperbolic length and area

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Figure 3.3.14. Constructing the hyperbolic straight line containing two points z1,z2
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Let z1,z2 be distinct points in D. Let T∈H be the transformation that sends z1→0 and z2→u>0. Then T−1(R) is a hyperbolic straight line that contains z1,z2. Let q1=T−1(−1) and q2=T−1(1). See Figure 3.3.14.
Use Proposition 3.3.5 to write a formula for the transformation T in the previous paragraph.
Solution.
Let Sz=z−z11−z1∗z, let t=−arg⁡(Sz2), and let Tz=eitSz, so that we have Tz1=0 and Tz2=u>0. Because T∈H, T is determined by the two parameters z1,t.
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A simple calculation verifies that (0,u,1,−1)=1+u1−u. By invariance of the cross ratio, we have (z1,z2,q2,q1)=1+u1−u. For 0≤u<1, we have
1≤1+u1−u<∞
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with equality on the left if and only if u=0.
Do the simple calculation mentioned above.
Hint.
Use (3.2.4).
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Instructor’s solution for Checkpoint 3.3.16.

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(0,u,1,−1)=0−10+1u+1u−1=1+u1−u
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Now, given any points z1,z2 in D, not necessarily distinct, define the quantity d(z1,z2) by
(3.3.3)d(z1,z2)={ln⁡((z1,z2,q2,q1))z1≠z20z1=z2
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where q1,q2 are the ideal points on the hyperbolic straight line connecting z1,z2 (with each qi at the zi end of the line) as described above, in the case z1≠z2. From the discussion above we have
(3.3.4)d(z1,z2)=ln⁡(1+u1−u)
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where u=|z2−z11−z1∗z2|.
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Instructor’s solution for Checkpoint 3.3.17.

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It is clear that u=|Tz2|, where T is the transformation found in Checkpoint 3.3.15.
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Instructor’s solution for Proposition 3.3.18.

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Let z1,z2∈D, and let q1,q2 be points on the hyperbolic line through z1,z2 that lie on the unit circle, with qi "on the zi side" for i=1,2, that is, qi=T−1zi, where T is the transformation found in Checkpoint 3.3.15. Be invariance of the cross ratio under Möbius transformations, we have
d(z1,z2)=ln⁡(z1,z2,q2,q1)=ln⁡(Tz1,Tz2,Tq2,Tq1)=d(Tz1,Tz2).
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The following Proposition shows that d is a metric on hyperbolic space, and justifies referring to d(z1,z2) as the (hyperbolic) distance between the points z1,z2.
Property 1 follows immediately from (3.3.4). Property 2 is a simple calculation: just write down the cross ratio expressions for d(z1,z2) and d(z2,z1) and compare. The proof of Property 3 is outlined in exercise Exercise 3.3.6.4.
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Now let γ be a curve parameterized by t→z(t)=x(t)+iy(t), where x(t),y(t) are differentiable real-valued functions of the real parameter t on an interval a<t<b. Consider a short segment of γ, say, on an interval t0≤t≤t1. Let z0=z(t0) and z1=z(t1). Then we have d(z(t0),z(t1))=ln⁡(1+u1−u) where u=|z1−z01−z0∗(z1)|. The quantity |z1−z0| is well-approximated by |z′(t0)|dt, where z′(t)=x′(t)+iy′(t) and dt=t1−t0. Thus, u is well-approximated by |z′(t0)|1−|z(t0)|2dt. The first order Taylor approximation for ln⁡((1+u)/(1−u)) is 2u. Putting this all together, we have the following.
(3.3.6)Length(γ)=2∫ab|z′(t)|1−|z(t)|2dt
Show that the first order Taylor approximation of ln⁡((1+u)/(1−u)) is 2u.
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Instructor’s solution for Checkpoint 3.3.21.

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Using the Taylor series
ln⁡(1+u)=u−u22+u33−⋯
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and the fact ln⁡[(1+u)/(1−u)]=ln⁡(1+u)−ln⁡(1−u) we get
ln⁡(1+u1−u)=2(u+u33+u55+⋯).
Find the length of the hyperbolic circle parameterized by z(t)=αeit for 0≤t≤2π, where 0<α<1 .
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Instructor’s solution for Checkpoint 3.3.22.

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For z(t)=αeit, we have z′(t)=iαeit, and |z(t)|=|z′(t)|=α. The circumference of the circle is
(3.3.7)circumference =2∫02π|z′(t)|1−|z(t)|2dt(3.3.8)=2∫02πα1−α2dt(3.3.9)=4πα1−α2.
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While this answer is correct, it is more useful to have the circumference expressed in terms of the hyperbolic radius r=ln⁡(1+α1−α). Solving for α in terms of r, we get α=er−1er+1. Then simplifying (3.3.9), we get
circumference =π(er−e−r)=2πsinh⁡r.
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We conclude this subsection on hyperbolic length and area with an integral formula for the area of a region R in D, following the development in [4]. As a function of the two real variables r and θ, the polar form expression z=reiθ gives rise to the two parameterized curves r→z1(r)=reiθ (where θ is constant) and θ→z2(θ)=reiθ (where r is constant). Using z1′(r)=eiθ and z2′(θ)=ireiθ, the arc length differential ds=2|z′(t)|dt1−|z(t)|2 for the two curves are the following.
2|eiθ|dr1−r2=2dr1−r2(for curve z1)
2|ireiθ|dr1−r2=2rdr1−r2(for curve z2)
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Thus we have dA=4rdrdθ(1−r2)2, so that the area of a region R is
(3.3.10)Area(R)=∬RdA=∬R4rdrdθ(1−r2)2.
Find the area of the hyperbolic disk {|z|≤α}, for 0<α<1.
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Instructor’s solution for Checkpoint 3.3.23.

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We have
Area =∫02π∫0α4rdrdθ(1−r2)2=−2∫02π[∫11−α2duu2]dθ(for u=1−r2)=2∫02π[1u|11−α2]dθ=2∫02πα21−α2dθ=4πα21−α2.
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As was the case for Checkpoint 3.3.22, this solution is correct, but we want an expression in terms of the hyperbolic radius r=ln⁡(1+α1−α). Again, solve for α in terms of r, then simplify. We get
Area =π(er+e−r−2)=4πsinh2⁡(r2).
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Subsection 3.3.5 The upper-half plane model

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Definition 3.3.24. The upper half-plane model of hyperbolic geometry.

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Let U={z:Im(z)>0} denote the upper half of complex plane above the real axis, and let HU denote the subgroup of the Möbius group M of transformations that map U onto itself. The pair (U,HU) is the upper half-plane model of hyperbolic geometry.
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Hyperbolic straight lines in the upper half-plane model are clines that intersect the real line at right angles. The hyperbolic distance between two points z1,z2 in the upper half-plane is
(3.3.12)d(z1,z2)=ln⁡((z1,z2,q2,q1))
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where q1,q2 are the points on the (extended) real line at the end of the hyperbolic straight line that contains z1,z2, with each qi on the same "side" as the corresponding zi. The hyperbolic length of a curve γ parameterized by t→z(t)=x(t)+iy(t) on the interval a≤t≤b is
(3.3.13)Length(γ)=∫ab|z′(t)|y(t)dt.
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The hyperbolic area of a region R in U is
(3.3.14)Area(R)=∬RdA=∬Rdxdyy2.
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Exercises 3.3.6 Exercises

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1.

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Prove Proposition 3.3.5 using the following outline.
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    Complete the proof of (3.3.1) using this outline: Let |z|=1 and apply Corollary 3.3.4. We have
    1=|Tz|=|α||z−z01−z0∗z|.
    Continue this derivation to show that |α|=1.
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    Prove (3.3.2) by verifying the following. Given z0∈D and t∈R, show that the assignments a=eit/21−|z0|2,b=−eit/2z01−|z0|2 satisfy |a|2−|b|2=1 and that
    (3.3.15)az+bb∗z+a∗=eitz−z01−z0∗z.
    Conversely, given a,b∈C with |a|2−|b|2=1, show that the assignments t=2arg⁡a,z0=−ba satisfy z0∈D, and that (3.3.15) holds.
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Instructor’s solution for Exercise 3.3.6.1 (proof of Proposition 3.3.5).

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    We have
    |z−z01−z0∗z|2=(z−z0)(z∗−z0∗)(1−z0∗z)(1−z0z∗)=|z|2+|z0|2−z0z∗−z0∗z1+|z0|2−z0z∗−z0∗z=1
    (because |z|=1 by assumption). It follows that |α|=1.
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    Let |z0|∈D and t∈R be given and let a=eit/21−|z0|2,b=−eit/2z01−|z0|2. We have the following.
    |a|2−|b|2=1−|z0|21−|z0|2=1az+bb∗z+a∗=eit/2z−eit/2z0−e−it/2z0∗z+e−it/2=eitz−z01−z0∗z
    Conversely, let a,b∈C be given with |a|2−|b|2=1 and let t=2arg⁡a,z0=−ba. The assumption |a|2−|b|2=1 implies that |a|>|b|, so |z0|=|b|/|a|<1, as required. We have
    az+bb∗z+a∗=az−az0−a∗z0∗z+a∗=aa∗z−z01−z0∗z=eitz−z01−z0∗z,
    as desired.
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2. Two points determine a line.

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Let p,q be distinct points in D. Show that there is a unique hyperbolic straight line that contains p and q.
Hint.
Start by choosing a transformation that sends p→0. For uniqueness, use Checkpoint 3.3.8.
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Instructor’s solution for Exercise 3.3.6.2.

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Let Tz=z−p1−p∗z, so that Tp=0. Let L be the diameter that passes through Tp,Tq. Then T−1(L) is a hyperbolic straight line that contains p,q. Uniqueness is a direct consequence of Checkpoint 3.3.8.
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3. Dropping a perpendicular from a point to a line.

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Let L be a hyperbolic straight line and let p∈D be a point not on L. Show that there is a unique hyperbolic straight line M that contains p and is orthogonal to L.
Hint.
Start by choosing a transformation that sends p→0. For uniqueness, use Checkpoint 3.3.8.
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Instructor’s solution for Exercise 3.3.6.3.

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Let Tz=z−p1−p∗z, so that Tp=0. Let a,b be the points where T(L) intersects the unit circle. Let K be the diameter that bisects the Euclidean straight line segment connecting a and b. It is clear that K is orthogonal to T(L). It follows that T−1(K) is a hyperbolic straight line that passes through p and is orthogonal to L. Uniqueness is a direct consequence of Checkpoint 3.3.8.
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4. The triangle inequality for the hyperbolic metric.

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Show that d(a,b)≤d(a,c)+d(c,b) for all a,b,c in D using the outline below.
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    Show that the triangle inequality holds with strict equality when a,b,c are collinear and c is between a and b. Suggestion: This is a straightforward computation using the cross ratio expressions for the values of d.
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    Show that the triangle inequality holds with strict inequality when a,b,c are collinear and c is not between a and b.
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    Let p∈D lie on a hyperbolic line L, let q∈D, let M be a line through q perpendicular to L (this line M exists by Exercise 3.3.6.3), and let q′ be the point of intersection of L,M. Show that d(p,q′)≤d(p,q). Suggestion: apply T∈H that takes p→0 and takes L→R. Let t=−arg⁡(Tq) if Re(Tq)≥0 and let t=π−arg⁡(Tq) if Re(Tq)<0. Let r=eitTq. See Figure 3.3.26.
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    Given arbitrary a,b,c, apply a transformation T to send a→0 and b to a nonnegative real point. Drop a perpendicular from Tc to the real line, say, to c′. Apply results from the previous steps of this outline.
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Figure 3.3.26.
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Instructor’s solution for Exercise 3.3.6.4.

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    First use a hyperbolic transformation to take a→0 and b,c real with 0=a<c<b<1. Let p=−1 and q=1. Assuming a,b,c are distinct, we have the following.
    d(a,b)=ln⁡(a,b,q,p)=ln⁡(a−qa−pb−pb−q)=(q−aa−pb−pq−b)=ln⁡(q−a)−ln⁡(a−p)+ln⁡(b−p)−ln⁡(q−b)d(a,c)=ln⁡(a,c,q,p)=ln⁡(q−aa−pc−pq−c)=ln⁡(q−a)−ln⁡(a−p)+ln⁡(c−p)−ln⁡(q−c)d(c,b)=ln⁡(b,c,q,p)=ln⁡(q−cc−pb−pq−b)=ln⁡(q−c)−ln⁡(c−p)+ln⁡(b−p)−ln⁡(q−b)
    From this we have d(a,b)=d(a,c)+d(c,b). Degenerate cases where two or more of a,b,c coincide are trivial.
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    Observe that, for 0<u<1,
    ddud(0,u)=dduln⁡(1+u1−u)=ddu(ln⁡(1+u)−ln⁡(1−u))=11+u+11−u=21−u2>0,
    so d(0,u) is an increasing function of u. This means that if 0=a<b<c (we may assume this without loss of generality), then d(a,c)>d(a,b), so that we have
    d(a,b)<d(a,c)+d(c,b)
    as claimed.
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    Following the suggestion, it is clear (draw a sketch!) that Tq′ is between Tp=0 and r. Thus, by part (a), we have d(0,Tq′)<d(0,r)=d(0,Tq), where the last equality is justified because rotation about 0 is distance preserving.
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    We have
    d(a,b)≤d(a,c′)+d(c′,b)(by parts (a),(b))≤d(a,c)+d(c,b)(by part (c)).
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5.

Hint.
For the "only if" direction, apply Proposition 3.3.2 to the cline R^, and conclude that T must send the real line to itself. Set z1,z2,z3 to be the preimages under T of 0,1,∞, and then use (3.2.3). For the "if" direction, suppose T has the given form. Let y>0 and show that Im(T(x+iy))>0.
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Instructor’s solution for Exercise 3.3.6.5 (proof of Proposition 3.3.25).

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First we prove the "only if" direction. Suppose T∈HU. Applying Proposition 3.3.2 to the cline R^, we conclude that T must send the real line to itself. So the preimages z1=T−1(1),z2=T−1(0),z3=T−1(∞) must be real. Thus we have
Tz=(z,z1,z2,z3)=z−z2z−z3z1−z3z2−z3=az+bcz+d
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for some real a,b,c,d. We have
Ti=ai+bci+d=ac+bd+i(ad−bc)c2+d2
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so ad−bc must be positive because T(U)=U. For the "if" direction, suppose Tz=az+bcz+d with a,b,c,d real, and ad−bc>0. Let z=x+iy with y>0. It is straightforward to check that the imaginary part of Tz is y(ad−bc)(cx+d)2+c2y2>0.
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6. Length integral in the upper half-plane model.

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This exercise is to establish (3.3.13). The strategy is to obtain the differential expression
d(z(t0),z(t1))≈|z′(t)|dty(t)
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for a curve z(t)=x(t)+iy(t) with z(t0)=z0, z(t1)=z1, and dt=t1=t0 using the following sequence of steps.
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    First, map z0,z1 in U to z0′,z1′ in D using a transformation μ that preserves distance.
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    Using the analysis we used to get the disk model length integral formula (3.3.6), we have
    d(z0′,z1′)=ln⁡(1+u1−u)
    where u=|z1′−z0′1−(z0′)∗z1′|.
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    Translate the above expression in terms of z0,z1, and show that the differential approximation is |z′(t)|dty(t).
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Complete the exercise parts below to carry out the strategy just outlined.
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    Show that μz=z−iz+i takes U to D.
  2. 🔗
    Let z0′=μz0 and z1′=μz1. Show that
    z1′−z0′1−(z0′)∗z1′=eitz1−z0z0∗−z1
    for some real t.
  3. 🔗
    Let u=|z1−z0z0∗−z1|. Show that
    ln⁡(1+u1−u)≈|z(t)|dty(t).
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Instructor’s solution for Exercise 3.3.6.6.

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  1. 🔗
    The map μ sends 1→−i, 0→−1, and ∞→1, so μ takes R to the unit circle. Checking point interior to U, we have μ(i)=0, so μ takes U to D.
  2. 🔗
    We have
    z1′−z0′1−(z0′)∗z1′=z1−iz1+i−z0−iz0+i1−(z0−iz0+i)∗z1−iz1+i=((z0+i)(z1−i)−(z0−i)(z1+i)(z0+i)∗)(z1+i)−(z0−i)∗(z1−i))((z0+i)∗(z1+i)(z0+i)(z1+i))=eitz1−z0z0∗−z1
    where t=−2arg⁡(z0+i).
  3. 🔗
    For dt small, we have z1−z0≈dz=|z′(t)|dt, and z0∗−z1≈2iIm(z(t))=2iy(t). Taking norms yields the desired expression for the length differential.
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7. Area integral in the upper half-plane model.

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Adapt the argument in the paragraph preceding the disk model area integral (3.3.10) to establish the upper half-plane area integral (3.3.14).
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Instructor’s solution for Exercise 3.3.6.7.

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Consider parameterized curves x=x(t)+y0, y=x0+y(s) with x(0)=x0,y(0)=y0. The arc length differential ds for the two curves are x′(t)dty=dxy (for x(t)) and y′(t)dty=dyy (for y(t)) so the area differential is dA=dxdyy2.
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Figure 3.3.27. Hyperbolic triangle △ABC
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Area of a hyperbolic triangle..

The following sequence of exercises establishes the area formula for hyperbolic triangles.
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8. Area of a doubly-asymptotic triangle.
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A triangle with one vertex in D and two vertices on the unit circle, connected by arcs of circles that are orthogonal to the unit circle, is called a doubly-asymptotic hyperbolic triangle. Examples are △AA1A2, △BB1B2, and △CC1C2 in Figure 3.3.27.
  1. 🔗
    Explain why any doubly-asymptotic triangle in the upper half-plane is congruent to the one shown in Figure 3.3.28for some angle α.
  2. 🔗
    Now use the integration formula for the upper half-plane model to show that the area of the doubly-asymptotic triangle with angle α (at the vertex interior to U) is π−α.
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Figure 3.3.28. Doubly-asymptotic hyperbolic triangle in the upper half-plane with vertices 1,p,∞ with p on the upper half of the unit circle
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9. Area of an asymptotic n-gon.
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A polygon with n≥3 vertices on the unit circle, connected by arcs of circles that are orthogonal to the unit circle, is called an asymptotic n-gon. An example of an asymptotic hexagon is the figure with vertices A1,A2,B1,B2,C2,C2 connected by the colored hyperbolic lines in Figure 3.3.27. Show that the area of an asymptotic n-gon is π(n−2).
Hint.
Partition the asymptotic n-gon into n doubly-asymptotic triangles.
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10. Area of a hyperbolic triangle.
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Let △ABC be a hyperbolic triangle. Extend the three sides AB, BC, AC to six points on the unit circle. See Figure 3.3.27. Use a partition of the asymptotic hexagon whose vertices are these six points to show that the area of △ABC is
(3.3.16)Area(△ABC)=π−(∠A+∠B+∠C).
Hint.
Partition the asymptotic hexagon with vertices A1,A2,B1,B2,C1,C2. Start with the six overlapping doubly-asymptotic triangles whose bases are colored arcs and whose vertex in D is whichever of A,B,C matches the color of the base. For example, the two red doubly-asymptotic triangles are △AA1A2 and △AC1B2.
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Instructor’s solution for Exercise 3.3.6.8.

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  1. 🔗
    Let △ABC be a doubly-asymptotic triangle in the upper half-plane, and suppose B,C∈R^. Any hyperbolic line intersects R^ in two points (this is easy to see in the disk model, so must also be true in the upper half-plane). Let B′∈R^ be the intersection of side AB with R^ that is not B, and let C′ be the intersection of side AC with R^ that is not C. Let T be the Möbius transformation that takes B→1,B′→−1,C→∞, and let T′ be the Möbius transformation that takes C→1,C′→−1,B→∞. One of T,T′ takes A to the upper half-plane, and hence is an element of HU that make the doubly-asymptotic triangle given in the Figure.
  2. 🔗
    The area of the triangle with vertices at 1,p,∞ is
    ∫cos⁡(π−α)1∫1−x2∞1y2dydx=∫cos⁡(π−α)1[−1y]1−x2∞=∫cos⁡(π−α)111−x2dx=[−arccos⁡x]cos⁡(π−α)1=π−α.
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Instructor’s solution for Exercise 3.3.6.9.

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Making the hint more explicit, draw a radius from 0 to each of the vertices of the asymptotic n-gon. This partitions the asymptotic n-gon into n doubly-asymptotic triangles. The n angles αi made be consecutive radii at 0 sum to 2π, so the areas of the doubly-asymptotic triangles sum to ∑i(π−αi)=nπ−2π=(n−2)π.
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Instructor’s solution for Exercise 3.3.6.10.

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Using Exercise 3.3.6.8, the sum of the areas of the six doubly asymptotic triangles in the hint is
2((π−∠A)+(π−∠B)+(π−∠C)).
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Subtracting the overcounted area 2 Area(△ABC) gives the area of the asymptotic hexagon with vertices A1,A2,B1,B2,C1,C2. We know this quantity is 4π by Exercise 3.3.6.9. Rearrange terms in
2((π−∠A)+(π−∠B)+(π−∠C))−2 Area(△ABC)=4π
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to get the desired area formula.
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