5 Hyperbolic Geometry

5.1 History: Saccheri, Lambert and Absolute Geometry

As evidenced by its absence from his first 28 theorems, Euclid clearly found the parallel postulateawkward; indeed many subsequent mathematicians believed it could not be an independent axiom.Two of the earliest to work rigorously on this were Giovanni Saccheri (1667–1733) & Johann Lambert(1728–1777), who attempted to force contradictions by assuming the negation of the parallel postu-late. While they failed at their primary purpose, their insights eventually provided the foundationof a well-defined non-Euclidean geometry. Before considering their work, we define some terms andrecall our earlier discussion regarding parallels.

Definition 5.1. Absolute or neutral geometry refers to the axiomatic system comprising all of Hilbert’saxioms except Playfair’s axiom. Euclidean geometry is therefore a special case of neutral geometry.

A non-Euclidean geometry is typically a model satisfying some (or all) of Euclid’s/Hilbert’s axiomsand for which Playfair’s axiom is false. That said,

There exists a line ` and a point P 6∈ ` through which there exist either no parallels or atleast two.

For example, spherical geometry is non-Euclidean since there are no parallel lines (Hilbert’s axiomsI-2 and O-3 are also false, as is the exterior angle theorem).

Results in absolute geometry Everything in the first 28 theorems of Euclid, including:

• Basic constructions: bisectors, perpendiculars, etc.

• The Exterior Angle Theorem.

• Triangle congruence theorems: SAS, ASA, SAA, SSS.

• Congruent/equal angles imply parallels: i.e.

α ∼= β =⇒ ` ‖ m

`

mP

β

α

This is equivalent to the existence of a parallel m to a given line ` through a point P 6∈ `.

Arguments requiring unique parallels We have previously discussed the following results whoseproofs relied on Playfair’s axiom: the arguments are therefore false in absolute geometry:

• A line crossing parallel lines makes equal angles: in the picture, ` ‖ m =⇒ α ∼= β. This is theuniqueness part of Playfair’s postulate: the parallel m to ` through P is unique.

• Angles in a triangle sum to a straight edge (180°).

• Constructions of squares/rectangles.

• Pythagoras’ Theorem.

While our arguments for the above are certainly false, we cannot immediately claim that the resultsare false in absolute geometry: there might be alternative proofs! To show that the results genuinelyrequire unique parallels, we need to exhibit a model: we shall describe such in the next section. Theexistence of this model demonstrates why Saccheri and Lambert failed in their efforts: the parallelpostulate (Playfair) is indeed independent of Euclid’s (Hilbert’s) other axioms.

1

The Saccheri–Legendre Theorem As an example of the work of Saccheri and Lambert, we consideran extension of the Exterior Angle Theorem based on Euclid’s proof.

Given4ABC, take M to be the midpoint of AC and extendBM to E such that BM ∼= ME. SAS congruence shows that4AMB ∼= 4CME whence the angles/sides marked in thepicture are all congruent. Observe:

1. m∠ACB + m∠CAB = m∠ACB + m∠ACE < 180°:this is essentially Euclid’s Exterior Angle Theorem.

M

A

B C

E

2. The sum of the angles in the triangles4ABC and4EBC are equal:1 to see this, just transfer theorange and green angles. . .

3. At least one of the angles ∠EBC or ∠BEC has at most half the measure of ∠ABC.

Iterating this construction produces a sequence of triangles 40 = 4ABC, 41 = 4BEC, 42,43, . . .each of which has the same angle sum and such that at least one of the angles in4n has measure

αn ≤12n m∠ABC

If the sum Σ of the angles in4ABC were greater than 180°, then

Σ = 180° + ε

for some ε > 0. Since 12n → 0, we may choose n large enough so that αn < ε. But then the sum of the

other two angles in4n would be greater than 180°, contradicting the Exterior Angle Theorem!We have therefore proved:

Theorem 5.2 (Saccheri–Legendre). In absolute geometry, the sum of the angles in4ABC is at most 180°.

Saccheri’s failed hope was to prove equality without having to invoke the parallel postulate. Theresult is remarkable nonetheless.

Saccheri and Lambert Quadrilaterals Saccheri and Lambert both considered quadrilaterals in theabsence of the parallel postulate. Two families of such are named in their honor.

Definition 5.3. A Saccheri quadrilateral ABCD satisfies

AD ∼= BC and m∠DAB = m∠CBA = 90°

We call AB the base and CD the summit. The interior angles at Cand D are the summit angles.

A Lambert quadrilateral has three right-angles; for instanceAMND in the picture.

BA M

N

CD

We draw these with ‘curvy’ sides to indicate that the remaining angles (at C and D) need not be right-angles, though, as yet, we have no model which shows that they could be anything else. Regardlessof how they are drawn, AD, BC and CD are all lines!

1With the parallel postulate, we could use congruence of angles ∠BAC ∼= ∠ECA to conclude that CE ‖ BA, from whichthe sum of the angles in a triangle is 180° and the observation is trivial. We cannot do this in absolute geometry!

2

Lemma 5.4. 1. Every Saccheri quadrilateral is a pair of Lambert quadrilaterals. In particular, if one bisectsthe base and summit of a Saccheri quadrilateral, one obtains two congruent Lambert quadrilaterals.

2. The summit angles of a Saccheri quadrilateral are congruent.

3. In Euclidean Geometry, Saccheri and Lambert quadrilaterals are rect-angles (four right-angles).

We leave parts 1 and 2 as an exercise: you need only the SAS and SSScongruence theorems.

BA

CD

Proof of 3. By part 1 we need only prove this for a Saccheri quadrilateral. Following the exterior angletheorem,

←→AB is a crossing line making equal (right-) angles, whence AD ‖ BC.

However←→CD also crosses the same parallel lines, whence, by the parallel postulate, the angles at C, D

must sum to a straight edge. Since these angles are congruent, they must both be right-angles.

We can now prove another of the conclusions of Saccheri & Lambert.

Theorem 5.5. In absolute geometry, the summit angles of a Saccheri quadrilateral measure ≤ 90°.

Proof. Let M be the midpoint of AB and extend CB to E (on theopposite side of AB to C) such that BE ∼= DA. SAS says that4DAM ∼= 4EBM, whence M lies on the line DE.The summit angles at C and D therefore sum to

m∠ADC + m∠BCD = m∠ADM + m∠EDC + m∠DCE= m∠CED + m∠EDC + m∠DCE≤ 180°

by the Saccheri–Legendre Theorem.

A B

CD

E

M

The Bolyai–Lobashevsky (Hyperbolic) Postulate

In the early 1800’s James Boylai, Carl Friedrich Gauss and Nikolai Lobashevsky independently tookthe next step: rather than attempting to prove the parallel postulate as a theorem of Euclidean geom-etry, they defined an new geometry based on the first four of Euclid’s postulates plus an alternativeto the parallel postulate:

Given a line and a point not on the line, there exist at least two parallel lines through the point.

The resulting axiomatic system2 is known as hyperbolic geometry. Consistency was proved in the late1800’s by Beltrami, Klein and Poincare, each of whom created models of hyperbolic geometry bydefining point, line, etc., in novel ways.

2In the modern approach we assume all of Hilbert’s axioms for Euclidean geometry, replacing Playfair’s axiom with thehyperbolic postulate.

3

5.2 Models of Hyperbolic Geometry

There are many models of hyperbolic geometry. In this section we consider several. The simplest isarguably the Poincare Disk Model, named for Henri Poincare though first proposed by Beltrami.

The fundamental objects of the Poincare model are:

Points The Poincare disk: the interior of the unit circle

(x, y) : x2 + y2 < 1

Lines Any straight line or circular arc meeting the unit circleat right angles. We shall call these hyperbolic lines to avoidconfusion.

In the picture the point P does not lie on the hyperbolic line `.Drawn are several parallel hyperbolic lines passing through P.

P

`

Lemma 5.6. Every hyperbolic line in the Poincare disk model is one of the following:

• A diameter passing through (a, b) 6= (0, 0) with Euclidean equation bx = ay.

• The arc of a (Euclidean) circle with equation

x2 + y2 − 2ax− 2by + 1 = 0 where a2 + b2 > 1 (∗)The center of the circle is C = (a, b) and r =

√a2 + b2 − 1 is its radius.

Moreover, there exists a unique hyperbolic line joining any two points in the Poincare disk.

Proof. The second part should be clear from the picture below: the dashed triangle is tangent to twocircles intersecting at right angles and is therefore right-angled itself. Finally, if two points do not lieon a diameter, then substituting their co-ordinates into (∗) yields a pair of linear equations for a, b:think about why these equations must have a unique solution. . .

Example Find the equation of the hyperbolic line joiningthe points P = (0, 1

2 ) and Q = ( 12 , 1

3 ) in the Poincare disk.

Substituting P and Q into (∗) yields a system of equationsfor a, b:

14 − b + 1 = 014 +

19 − a− 2

3 b + 1 = 0=⇒ (a, b) =

(1936

,54

)The required hyperbolic line has equation

x2 + y2 − 1918

x− 52

y + 1 = 0 or(x− 19

36

)2

+

(y− 5

4

)2

=545648

1

r

O

PQ

C

4

Distance, Segment Congruence and Circles

To complete the model, we need a notion of congruence for hyperbolic line segments and angles.This requires a few more definitions.

Omega-points These are points lying on the unit circle. They are not points in the model, but areuseful for calculations.

Distance Given points P 6= Q in the Poincare disk, there is aunique hyperbolic line on which they lie (Lemma 5.6). Thisintersects the boundary circle at the omega-points Ω, Θ. Thehyperbolic distance between P and Q is defined to be

d(P, Q) : =∣∣∣∣ln |PΘ| |QΩ||PΩ| |QΘ|

∣∣∣∣= cosh−1

(1 +

2 |PQ|2

(1− |P|2)(1− |Q|2)

)P

Q

Ω

Θ

where |PΘ|, etc., are measured in Euclidean fashion and |P| , |Q| are the Euclidean distances ofP, Q from the origin.3 The expressions are simpler if Q = O is the origin:

d(O, P) = ln1 + |P|1− |P| = cosh−1 1 + |P|2

1− |P|2

Congruence of segments Hyperbolic line segments are congruent if they have the same length.

Example: A right triangle Consider the triangle in the Poincare diskwith vertices O = (0, 0), P = ( 1

2 , 0) and Q = (0, 12 ). The hyperbolic

distances are easily computed:

|P| = 12= |Q| , |PQ|2 =

14+

14=

12

d(O, P) = d(O, Q) = ln1 + 1

2

1− 12

= ln 3 = cosh−1 53≈ 1.099

d(P, Q) = cosh−1

(1 +

2 · 12

(1− 14 )

2

)= cosh−1 25

9= ln

25 + 4√

349

≈ 1.681

P

Q

O

For a difficult exercise, try to find the omega-points for the line←→PQ and see that you obtain the same

expression for the distance.4

3Showing the equivalence of these expressions is hard without the machinery of isometries which we’ll introduce later.Note that the first distance formula is independent of the labelling of P, Q, Ω, Θ: for instance, we may swap the order ofthe omega-points and obtain the same measure for distance. It should seem reasonable for hyperbolic functions to playsome role in hyperbolic space! As a primer:

cosh x =ex + e−x

2, sinh x =

ex − e−x

2, tanh x =

sinh xcosh x

=ex − e−x

ex + e−x and cosh−1 x = ln(x +√

x2 − 1)

4Hint: Ω, Θ have x = 4±√

3410 , y = 4∓

√34

10 . . .

5

We leave the next result is an exercise: it says that distance increases smoothly as one moves awayfrom a point along a hyperbolic line.

Lemma 5.7. Fix P and a hyperbolic line through P. Let Q lie on this line. Then the distance function d(P, Q)maps the set of points on one side of P differentiably and bijectively onto the interval (0, ∞).

The Lemma means that hyperbolic circles are well-defined and looklike one expects: the circle of hyperbolic radius δ centered at P is theset of points Q such that d(P, Q) = δ.

In the picture we’ve drawn several hyperbolic circles and their centers.One of the hyperbolic circles has several of its radii (hyperbolic seg-ments) drawn. Notice how the centers are closer (in a Euclidean sense)to the boundary circle than one might expect: this is since hyperbolicdistances measure greater the further one is from the origin.

You might be suspicious (and you’d be correct) that hyperbolic circles in the Poincare disk model arealso Euclidean circles. Moreover, their hyperbolic radii intersect the circles at right-angles. To provethese facts at the moment requires some thoroughly evil algebra!

Angles and Angle Congruence

The final piece missing from our model is angle. This is surprisingly easy to state.

Definition 5.8. The angle between two hyperbolic rays may be mea-sured as the angle between their (Euclidean) tangent lines. Angles inthe Poincare disk model are congruent if and only if they have thesame angle measure.

With a little differential geometry, one can prove that this angle mea-sure arises naturally: the distance function d(P, Q) defines a ‘dot prod-uct’ which defines angle measure. We’ll sketch some of the ideasaround this later, though the full argument is outwith the scope of thiscourse. What’s more important presently is the following:

θ

Theorem 5.9. The Poincare disk is a model of hyperbolic geometry.

Sketch Proof. A rigorous proof would require us to check the hyperbolic postulate and all Hilbert’saxioms except Playfair. Instead we check Euclid’s postulates 1–4 and the hyperbolic postulate 5.

1. Lemma 5.6 says we can join any given points in the Poincare disk by a unique hyperbolic line.

2. A hyperbolic line segment joins two points inside the (open) Poincare disk. The distance for-mula increases (Lemma 5.7) unboundedly as P approaches an omega-point, so we can alwaysmake a hyperbolic line longer.

3. Hyperbolic circles are defined above.

4. All right-angles are equal since the notion of angle is unchanged from Euclidean geometry.

5. See the first picture on page 4.

6

Examples

1. Recall our triangle with vertices O = (0, 0), P = ( 12 , 0) and Q = (0, 1

2 ). The hyperbolic line←→PQ

has equation

x2 − 52

x + y2 − 52

y + 1 = 0

Implicit differentiation yields

2x− 52+

[2y− 5

2

]dydx

= 0 =⇒ dydx

=4x− 55− 4y

=⇒ dydx

∣∣∣∣P= −3

5

Since the side OP is horizontal, we conclude that the interior angle to the triangle at P is

tan−1 35≈ 30.96°

By symmetry, we have the same angle at Q. With a right-angle at O, we conclude that the sumof the angles in the triangle is approximately 151.93° < 180°!

2. Consider the hyperbolic triangle with vertices

O = (0, 0), P =

(12

,

√512

), Q =

(12

,−√

512

)

We can easily compute the side-lengths of this triangle:

|P| = |OP| = |Q| = |OQ| =√

14+

512

=

√23

|PQ| = 2

√512

=

√53

O

P

Q

d(O, P) = d(O, Q) = cosh−1(

1 +4/3

(1− 0)(1− 2/3)

)= cosh−1 5 = ln(5 + 2

√6) ≈ 2.292

d(P, Q) = cosh−1(

1 +10/3

(1− 2/3)(1− 2/3)

)= cosh−1 31 = ln(31 + 8

√15) ≈ 4.127

As a sanity check, we compare some data regarding the hyperbolic triangle 4OPQ and theEuclidean triangle4OPQ.

Property Hyperbolic Triangle Euclidean TriangleEdge lengths 2.292 : 2.292 : 4.127 0.816 : 0.816 : 1.291Relative edge ratios 1 : 1 : 1.801 1 : 1 : 1.581Angles 8.81°, 8.81°, 104.48° 37.76°, 37.76°, 104.48°

Observe that the hyperbolic side lengths are longer, and that the base of the isosceles triangleis relatively longer in the hyperbolic case. Calculating the angles can be done as above (home-work). Again notice that the angles in the hyperbolic triangle sum to ≈ 122.1° < 180°.

7

Other Models of Hyperbolic Space: non-examinable

There are several other models of hyperbolic space. Here are three of the most common.

The Klein Disk Model The approach is similar to the Poincare disk except that lines are taken tobe chords of the unit circle and the distance function is defined differently:

dK(P, Q) =12

∣∣∣∣ln |PΘ| |QΩ||PΩ| |QΘ|

∣∣∣∣where Ω, Θ are the points where the chord meets the circle. It iseasier to compute distances in this model since hyperbolic linesare the same as Euclidean lines.

The cost is that the notion of angle is different. Perpendicularitycomes from the following idea: Take a hyperbolic line and findthe tangents to the unit circle where it meets. Any chord whoseextension passes through the intersection of these tangents is or-thogonal to the original line. Measuring other angles is difficult!

A famous result from differential geometry (Gauss’ Theorem Egregium) says this problem is un-avoidable: hyperbolic geometry is negatively curved, whereas Euclidean geometry is flat. If H is amodel of hyperbolic geometry, it is impossible to find a mapping f : H → R2 which preserves boththe concepts of straight line and angle. Poincare’s model preserves angles but results in ‘bendy’ lines:Klein’s hyperbolic lines are ‘Euclidean straight’ but his notion of angle is ugly. The best we can do isto have one concept or the other: we cannot have both.5

The Poincare Half-plane This model is equivalent to the Poincare disk via a modified stereographicprojection and is widely used in complex analysis. The set of points comprises the upper half-plane(y > 0) in R2. Hyperbolic lines are vertical lines or semicircles centered on the x-axis:

x = constant, or (x− a)2 + y2 = r2

Advantages are the simple expressions for lines and that angles are measured as in Euclidean space.The expression for hyperbolic distance is still horrific!

y

xThe Poincare half-plane: a hyperbolic triangle is shaded

5The same problem arises when trying to make a map of part of the earth (a positively curved geometry). One can havemaps which preserve distance or angle, but not both.

8

The Hyperboloid Model Points comprise the upper sheet (z ≥ 1) of the hyperboloid x2 + y2 =z2 − 1. A hyperbolic line is the intersection of this with a plane through the origin. Isometries can bedescribed using matrix-multiplication and the formula for hyperbolic distance is particularly easy:between two points P = (x, y, z) and Q = (a, b, c) in this model, the hyperbolic distance is

d(P, Q) = cosh−1(cz− ax− by)

Difficulties include working in three dimensions and the factthat angles are awkward.

The relationship between the Hyperboloid and Poincaredisk models is via projection. Place the disk in the x, y-planecentered at the origin and draw a line through a point in thedisk and the point (0, 0,−1). The intersection of this linewith the hyperboloid gives the correspondence.In the picture, the orange and green points on the blue hy-perbolic lines correspond in the two models.

Under this correspondence, it is easy to check that theinverse-cosh formulæ for hyperbolic distance are in accord.

5.3 Parallels and Perpendiculars in Hyperbolic Geometry

From now on, all our pictures and examples will be within Poincare disk model, though the theoremsare model-independent. Recall (page 1) that we may use anything from absolute geometry, includingthe construction of perpendiculars and bisectors, and all the triangle congruence results.In case you need to be convinced, here is a result in absolute geometry proved in the style of Euclidbut illustrated in the Poincare disk model.

Lemma 5.10. Through a point P not on a line ` there exists a unique perpendicular to `.

Proof. Choose a point A ∈ ` and join AP. If AP is perpendicular to `, we are done.

Otherwise, ` is not tangent to the circle centered at P withradius |AP|. It follows that there exists a second intersectionpoint B ∈ `.

Construct the circles with radius |AB| centered at A and Brespectively: these have two intersections Q, R. Let m =

←→QR.

We leave the following as an exercise:

• m intersects ` at right-angles (M in the picture).

• P ∈ m.

• M is the midpoint of AB.

To help, note that the blue and green arcs are radii of theirrespective circles; we have several isosceles triangles. . .

P

B

A

Q

R

M

`

m

For uniqueness, suppose we have two perpendiculars to ` through P intersecting ` at distinct pointsM, N. Then4PMN has two right-angles, contradicting Saccheri–Legendre (Theorem 5.2).

9

The Fundamental Theorem of Parallels in Hyperbolic Geometry

Now we see a primary distinction from Euclidean geometry.

Theorem 5.11. Given a hyperbolic line ` and a point P not on `, thereare exactly two parallel lines m, n to ` through P with the following properties:

1. Every line through P lying within the angle made by one of the paral-lels m, n and the perpendicular from P to ` intersects `; all others areparallel to `.

2. The lines m, n make acute angles with the perpendicular.

`

R

Q

P

n

m

Proof. Drop the perpendicular (Lemma 5.10) from P to ` at Q. The lines through P are in bijectivecorrespondence with angles θ ∈ (−90°, 90°] measured with respect to PQ. Given a point R ∈ `, theangle θ is a continuous6 function of R. The set of angles θ formed by all such intersecting lines istherefore a subinterval I ⊆ (−90°, 90°].

Transfer the segment QR to the other side of Q to produce a point S ∈ `. SAS says4PQR ∼= 4PQS,whence the angle corresponding to the line PS is −θ. It follows that the interval I is symmetric:

θ ∈ I ⇐⇒ −θ ∈ I

Let µ = sup I ≤ 90° be the least upper bound of I; by symmetry, −µ = inf I is the greatest lowerbound. Let m and n be the lines making angles ±µ respectively.

Suppose m intersected ` at M. Let N ∈ ` lie on the other side of M from Q. Then PN makes a largerangle than µ: contradiction. It follows that m is parallel to `.

Finally m = n ⇐⇒ µ = 90°. In this case there exists only one parallel to ` through P, contradictingthe hyperbolic postulate.

Definition 5.12. The parallels m, n in the Fundamental Theorem are limiting parallels or asymptoticparallels to `. All other parallels to ` through P are ultraparallel. The angle of parallelism at P relative to` is the acute angle µ made by a limiting parallel and the perpendicular PQ.

We shall see shortly that the perpendicular distance |PQ| is a function of the angle of parallelism µ.

Example Let ` be the hyperbolic line with equation

x2 + y2 − 4x + 1 = 0

(Euclidean center (2, 0) and radius√

3). Its omega points are

Ω =

(12

,

√3

2

), Θ =

(12

,−√

32

)

P = (0, 0) and Q = (2−√

3, 0) ∈ ` are joined by the hyperbolic linewith equation y = 0 which intersects ` at right-angles. The limitingparallels have equations y = ±

√3x and the angle of parallelism is 60°.

P

Ω

Θ

Q

We can check that the perpendicular hyperbolic distance is d(P, Q) = ln√

3 = 12 ln 3.

6The proof depends crucially on continuity: feel free to omit if you’ve not taken an analysis class.

10

Example Find the limiting parallels and the angle of parallelismwhen

P =

(− 3

10,

410

)and x2 + y2 + 2x + 4y + 1 = 0

First find the omega-points by intersecting the line with the bound-ary circle x2 + y2 = 1:

Ω = (−1, 0), Θ =

(35

,−45

)P lies on the diameter containing Θ, whence

←→PΘ immediately has

equation y = − 43 x.

P

Ω

Θ

Q

For the line←→PΩ, substitute into the usual expression x2 + y2− 2ax− 2by+ 1 = 0 for a hyperbolic line.

We obtain

x2 + y2 + 2x− 138 y + 1 = 0 with Euclidean center (a, b) = (−1, 13

16 )

The angle of parallelism is harder work. Certainly←→PΘ has slope − 4

3 . For←→PΩ, implicit differentiation

yields a slope of

dydx

=a− xy− b

=16(1 + x)13− 16y

=⇒ dydx

∣∣∣∣P=

16 · 710

13− 6410

=5633

The angle of parallelism at P is therefore half the angle between the vectors( −33−56

)and

( 3−4): that is

µ = cos−1

( −33−56

)·( 3−4)∣∣( −33

−56

)∣∣ ∣∣( 3−4)∣∣ = 1

2cos−1 −99 + 224√

332 + 562√

32 + 42=

12

cos−1 22565 · 5

=12

cos−1 513≈ 33.69°

If you want a serious challenge, let PQ be the perpendicular and try to compute7 the co-ordinates ofQ ∈ ←→ΩΘ and the distance |PQ|.

Angles in Triangles, Rectangles and the AAA congruence

We finish this section with three important observations that follow from the Fundamental Theo-rem/Hyperbolic Postulate; we leave the proofs to the homework.

Theorem 5.13. In hyperbolic geometry:

1. There are no rectangles (quadrilaterals with four right-angles).

2. The angles in a triangle always sum to less than 180°.

3. (Angle-Angle-Angle Congruence) If 4ABC and 4DEF have angles congruent in pairs, then theirsides are congruent in pairs and so4ABC ∼= 4DEF.

7This is very hard without some extra hints and machinery. We shall eventually have a much better method for answer-

ing questions like these, so do not despair! The answers are Q =

(93(−29+2

√117)

1865 , 26(−29+2√

117)1865

)and |PQ| = ln 3+

√13

2

11

5.4 Omega-triangles

The concept of limiting parallels through a point relative to a line allows us to extend the definitionof triangle to allow sides to be pairs of limiting parallels.

Definition 5.14. An omega-triangle or ideal-triangle is a ‘triangle’one or more of whose vertices is an omega-point: at least two ofthe sides of an omega-triangle form a pair of limiting parallels.

There are three possible types of omega-triangle depending onhow many omega-points they contain. For example, in the pic-ture, 4PQΩ and4PQΘ have one omega-point, 4PΩΘ has twoand4ΩΘΞ has three!

While the usual exterior angle and congruence theorems applyto (non-omega) hyperbolic triangles, it is perhaps surprising thatmany of the standard results also apply to omega-triangles!

P

Ω

Θ

Ξ

Q

The first of these can be thought of as the extension of the AAA congruence theorem to omega-triangles.

Theorem 5.15 (Angle-Angle Congruence for Omega-triangles). Suppose4PQΩ and4RSΘ are omega-triangles with a single omega-point. If the the angles are congruent in pairs

∠PQΩ ∼= ∠RSΘ ∠QPΩ ∼= ∠SRΘ

then the finite sides8 of each triangle are also congruent: PQ ∼= RS.

Proof. Transfer the angle ∠SRΘ to P and choose the point T on the ray−→PQ such that PT ∼= RS. If T = Q we are done.

Otherwise, WLOG, we have the picture, where our assumptions statethat the marked orange angles at Q and T are congruent. Let M be themidpoint of QT and drop the perpendicular to

←→QΩ at N.

Choose L ∈ ←→ΩT on the opposite side of←→QT to N such that TL ∼= NQ.

The red angles are congruent, as are the pairs of green and blue lines:SAS says 4MQN ∼= 4MTL, whence M lies on the line LN and wehave a right-angle(!) at L. But then the angle of parallelism of L relativeto←→QΩ is 90°: contradiction.

There are several other possible orientations:

• T could lie on the same side of Q as P but the resulting argumentis the same, simply reverse the roles of Q, T.

Ω

P

M

Q N

L

T

Q

• N could lie on the opposite side of Q from Ω. In this case SAS is applied to the same trianglesbut with respect to the congruent orange angles.

• In the special case that N = Q we have that the orange angles are right-angles and the samecontradiction appears.

8It doesn’t really make sense to speak of the ‘infinite’ sides, or of the ‘angles’ at the omega points, being congruent. Ifone defines congruence in terms of the existence of an isometry (later), then the claim is more reasonable.

12

Corollary 5.16 (Exterior Angle Theorem for Omega-Triangles). Suppose 4QTΩ has a single omega-point. Extend TQ to P. Then ∠PQΩ > ∠QTΩ.

Proof. To see that ∠PQΩ and ∠QTΩ cannot be congruent, consider thepicture in the proof of the AA congruence theorem. The orange anglescannot be congruent since the picture is a contradiction!

If ∠PQΩ < ∠QTΩ, then we’re in the case of the picture on the right:there exists a point X inside the angle ∠TQΩ such that

∠PQX ∼= ∠QTΩ

Since←→QΩ is a limiting parallel to

←→TΩ, it follows that

−→QX intersects TΩ

at a point Y. But now we have a triangle 4QTY contradicting thestandard Exterior Angle Theorem.

Ω

Y

P

Q

T

QX

The final congruence theorem is an exercise based on the previous picture.

Corollary 5.17 (Side-Angle Congruence for Omega-triangles). Suppose4QTΩ and4RSΘ are omega-triangles with a single omega-point. If ∠QTΩ ∼= ∠RSΘ and QT ∼= RS then ∠TQΩ ∼= ∠SRΘ.

A triangle with one omega-point only has three pieces of data: two finite angles and one finite edge.The AA and SA congruence theorems say that two of these determine the third.

Other observations

Pasch’s Axiom: Versions of this are theorems for omega-triangles.

• If a line crosses a side of an omega-triangle and does not pass through any vertex (includingΩ), then it must pass through exactly one of the other sides.

• If a line passes through an interior point and exactly one vertex (including Ω) of an omega-triangle, then it passes through the opposite side.

The second statement is partly embedded in the proof of Corollary 5.16.

‘Angle’ Congruence: One could argue that omega-triangles with two omega-points and a single con-gruent angle are congruent. This statement is essentially vacuous as there is nothing to check: allsides have infinite length and the only real angles are assumed to be congruent! As before, once wehave the machinery of isometries it makes more sense to describe such ‘triangles’ as congruent.

Perpendicular Distance and the Angle of Paralllelism: Suppose 4PQΩ has a right-angle at Q, so that∠QPΩ is the angle of parallelism of P relative to

←→QΩ. The AA/SA congruence theorems show that

the perpendicular distance is a bijective function of the angle of parallelism! We leave the verificationof this function as an exercise, but state it here.

Corollary 5.18. Suppose P is a point not on a hyperbolic line `. Let µ be the angle of parallellism at P and δbe the hyperbolic length of the perpendicular from P to `. Then

cosh δ = csc µ or equivalently e−δ = tanµ

2

13

5.5 Area and Angle-defects

Our major result is one of the triumphs of Johann Lambert, the astonishing statement that the sumof the angles in a hyperbolic triangle determines its area.

We start with a loose axiomatization of area as a purely relative measure in neutral geometry.

Axiom I Two geometric figures have the same area if and only if they can be sub-divided intofinitely many pairs of mutually congruent triangles.9

Axiom II The area of triangle is positive.

Axiom III The area of the union of disjoint figures is the sum of the areas of the figures.

Area determines angle-sum in neutral geometry

We start with a useful quantity.

Definition 5.19. The angle-defect (in radians) of a triangle 4 is π − Σ4, where Σ4 is the sum of theangles in the triangle.

Since triangles in neutral geometry have angle-sum ≤ π (Saccheri-Legendre), it follows that

0 ≤ π − Σ4 ≤ π

In Euclidean geometry the defect is always zero, while in hyperbolic geometry (Theorem 5.13) thedefect is strictly positive. The defect is π if and only if we have a ‘triangle’ with three omega-points.

Lemma 5.20. Angle-defect is additive: If a triangle is divided into two sub-triangles, then the defect of the whole is the sum of the defects of the parts.

The proof should be clear from the picture:

[π − (α + γ + ε)] + [π − (β + δ + ζ)] = π − (α + β + γ + δ)

since ε + ζ = π. Notice that angle-sum is not additive!

Theorem 5.21. In neutral geometry, if two triangles have the same area, thentheir angle-sums are identical.

Of course this is utterly trivial in Euclidean geometry!

C

B

AQ

γ

β

α

δ

ζε

Proof. This is induction with the Lemma providing the induction step: if 41 and 42 have the samearea, then their interiors are disjoint unions of a finite collection of mutually congruent triangles:

41 =n⋃

k=1

41,k and 42 =n⋃

k=1

42,k where 41,k∼= 42,k

Each pair41,k,42,k has the same angle-defect, whence the angle-defects of41 and42 are equal:

defect(41) =n

∑k=1

defect(41,k) =n

∑k=1

defect(42,k) = defect(42)

9To allow infinitely many infinitessimal sub-triangles would be to require ideas from calculus and complexify ourinduction argument.

14

Angle-sum determines area in hyperbolic geometry

The converse discussion relies on a reversible construction relating triangles and Saccheri quadrilat-erals. The construction itself is valid in neutral geometry, even though the ultimate conclusion thatangle-sum determines area is not.

Lemma 5.22. To every triangle there corresponds a Saccheri quadrilateral with the same area and whosesummit angles sum to the angle-sum of the triangle. Explicitly:

1. Given a triangle 4ABC, choose a side BC. Bisect the remaining sides at E, F and drop perpendicularsfrom A, B, C to the line

←→EF. Then HICB is a Saccheri quadrilateral with base HI.

2. Conversely, given a Saccheri quadrilateral HICB with summit BC, let A be any point such that←→HI

bisects AB at E. Then the intersection F =←→HI ∩ AC is the midpoint of AC.

Both constructions yield the same picture and the following con-clusions:

• The areas of the triangle and the quadrilateral are equal.

• The sum of the summit angles of the quadrilateral equalsthe angle sum of the triangle (this is trivial in EuclideanGeometry since all Saccheri quadrilaterals are rectangles!).

For clarity, we’ve chosen BC to be the longest side of4ABC; thisforces E, F to lie between H, I.

A

B CD

E FG

H I

Proof. 1. By two applications of the SAA congruence theorem (follow the arrows. . . )

4BEH ∼= 4AEG and 4CFI ∼= 4AFG

We conclude that BH ∼= AG ∼= CI whence HICB is aSaccheri quadrilateral. The area and angle-sum corre-spondences are immediate from the picture.

2. Suppose the midpoint were at J 6= F. We may thencreate a new Saccheri quadrilateral with base BC usingthe midpoints E, J as in part 1. However, the perpen-dicular bisector of BC (at D) also bisects the bases ofboth Saccheri quadrilaterals perpendicularly at U, Vthus creating a triangle 4EUV with two right angles:contradiction.

J

U

V

A

B CD

EFGH I

We first prove a special case of the main result.

Lemma 5.23. Suppose hyperbolic triangles 4ABC and 4PQR have congruent sides BC ∼= QR and thesame angle-sum. Then the triangles have the same area.

Proof. Create the Saccheri quadrilaterals S1,S2 corresponding to 4ABC and 4PQR with summitsBC ∼= QR as in the first part of the Lemma. These have congruent summits and summit angles andare thus congruent (exercise).

15

It is the final observation that makes this result special to hyperbolic geometry: in Euclidean geom-etry all Saccheri quadrilaterals are rectangles and having congruent summits is insufficient to forcecongruence.

Theorem 5.24. In hyperbolic geometry, if 4ABC and 4PQR have the same angle-sum then they have thesame area.

Proof. If the triangles have a congruent pair of edges, we are done by the previous lemma. Otherwise,we create a new triangle4LBC which matches the same Saccheri quadrilateral as4ABC.

Suppose |PQ| 6= |AB|. We can find10 K on the line←→EF such that |BK| = 1

2 |PQ|. Extend to L such thatK is the midpoint of BL.

• By Lemma 5.22,

Area(4LBC) = Area(HICB) = Area(4ABC)

• By Theorem 5.21, 4LBC has the same angle-sum as4ABC and thus4PQR.

• 4LBC and 4PQR share a congruent side and havethe same angle-sum. Lemma 5.23 says their areas areequal.

A

B CD

EF

GH I

L

K

Since both area and angle-defect are additive, we immediately conclude:

Corollary 5.25. The angle-defect of a hyperbolic triangle is an additive function of its area: ∃k > 0 such that

angle-defect(4ABC) = k ·Area(4ABC)

• In fact the constant is k = 1 when we use the area measure arising naturally from the usualhyperbolic distance function and measure angles in radians: we can therefore write11

Area(4) = π − Σ

• Angle-Angle-Angle Congruence (Theorem 5.13, part 3) is related to the corollary:

4ABC ∼= 4DEF AAA⇐⇒ angles congruent in pairs

⇓ ⇓equal area Cor⇐⇒ same angle-defect

10At least one side of one of the triangles4PQR and4ABC must be at least as long as one of the sides of the other! Bysuitably relabelling, we can form the correct configuration.

11The corollary is a special case of the famous Gauss–Bonnet theorem from differential geometry:

Σ− π =∫∫4

KdA where K is the Gauss curvature.

where K is the Gauss curvature. Hyperbolic space has constant negative curvature K = −1 so the area∫∫4 dA is precisely

the angle-defect π − Σ. Euclidean geometry is flat (K = 0) so the defect is always zero. Similarly, the surface of a sphere ofradius 1 has constant positive curvature (K = 1) whence the area of a spherical triangle is the angle-excess Σ− π.

16

Examples Recall the triangles on page 7 within the Poincare disk model.

1. The triangle with vertices (0, 0), ( 12 ,±

√5

12 ) has angles approximately 8.81°, 8.81°and 104.48°,and thus

area ≈ π − π

180(8.81 + 8.81 + 104.48) ≈ 1.011

A Euclidean triangle with the same vertices has area ≈ 0.322.

2. The isosceles right-triangle with vertices O, P = ( 12 , 0) and Q = (0, 1

2 ) has

area = π −(

π

2+ 2 tan−1 3

5

)=

π

2− 2 tan−1 3

5≈ 0.490

3. Generalizing (see homework), the triangle with vertices O, P = (c, 0) and Q = (0, c) has area

π −(

π

2+ 2 tan−1 1− c2

1 + c2

)=

π

2− 2 tan−1 1− c2

1 + c2

Notice that limc→0+

area(c) = 0 and limc→1−

area(c) = π2 .

It is worth noting that our discussion provides an explicit method for subdividing a triangle intosubtriangles and rearranging its pieces to create a triangle with equal area.

A

B C

L

K

S1

S3

41

43

P

R

Q

S2

42

In the picture, suppose 41 and 42 have equal area. Let L, K be chosen as in the proof of Theorem5.24 so that QR ∼= BL, then all three triangles have the same area, as do the Saccheri quadrilateralsS1,S2,S3. Moreover, S2 and S3 are congruent.12 One can follow the steps in Lemma 5.22 to transform41 to42:

41 → S1 → 43 → S3 ∼= S2 → 42

Indeed this works even with triangles in Euclidean geometry: try it!

12This follows from our discussion: S2,S3 have a common summit and the same area, whence their angle-sums are thesame and thus their summit-angles are congruent!

17

5.6 Isometries and Calculation in Hyperbolic Geometry

There are (at least!) two major issues in our approach to hyperbolic geometry.

1: Calculations are difficult In analytic (Euclidean) geometry we typically choose the origin andorientation of axes to ease calculation. We’d like to do the same thing in hyperbolic geometry. Sincehyperbolic lines through the origin in the Poincare disk are genuine (Euclidean) straight lines, thiswill make computing length and angle easier. Since it will become important for us to view thePoincare disk as lying in the complex plane, we state the distance formulæ in this language:

Lemma 5.26. 1. Consider the Poincare disk as the subset

D = z ∈ C : |z| < 1

of the complex plane. If z ∈ D \ 0, then the omega-pointson the line joining 0 and z are Ω = z

|z| and Θ = − z|z| ,

whence

d(0, z) =∣∣∣∣ln |0−Ω| |z−Θ||0−Θ| |z−Ω|

∣∣∣∣ = ln(

1 + |z|1− |z|

)

z

Ω = z|z|

Θ = − z|z|

0

2. Suppose a point lies a Euclidean distance r and hyperbolic distance δ from the origin. Then

δ = ln(

1 + r1− r

)= cosh−1 1 + r2

1− r2 , r2 =cosh δ− 1cosh δ + 1

, r =eδ − 1eδ + 1

= tanhδ

2

One can observe the equivalence of the logarithm and inverse-cosh expressions for hyperbolic dis-tance beginning to emerge.

2: We’ve assumed too much In the Poincare disk we defined distance, angle and line separately, butthese concepts are not independent! In Euclidean geometry, the distance function, or metric,

|PQ| =√(P1 −Q1)2 + (P2 −Q2)2

defines angle measure via the dot product,13 and (with a little calculus) the arc-length of any curve.One can prove that the paths of shortest length between points (geodesics) are straight lines: the metrictherefore also defines the notion of line. Again, the same should be true in hyperbolic space.

There is a related remedy for these issues: isometries, the rigid motions of hyperbolic space. Ratherthan derive these from scratch, we state without proof some facts from complex analysis.

13Writing |u| = |PQ| for the length of a line segment, we see that for any u, v,

u · v =12

(|u + v|2 − |u|2 − |v|2

)so that the metric defines the dot product. Now define angle measure via

u · v = |u| |v| cos θ

18

Definition 5.27. A Mobius or fractioanl-linear transformation is a function of the form f (z) =az + bcz + d

where a, b, c, d ∈ C and ad− bc 6= 0.

Theorem 5.28. Every Mobius transformation f has the following properties:

1. f is a bijection from the extended complex plane C∪ ∞ to itself.

2. f preserves angles (conformality): if two curves in C intersect at a given angle, then the images of thesecurves under f intersect at the same angle.

3. Preservation of the class of circles and lines: every circle or line14 is mapped to another circle or line.

4. Preservation of the cross-ratio of any four distinct points z1, z2, z3, z4:(f (z1)− f (z2)

)(f (z3)− f (z4)

)(f (z2)− f (z3)

)(f (z4)− f (z1)

) =(z1 − z2)(z3 − z4)

(z2 − z3)(z4 − z1)

The isometries of the Poincare disk are built from a subset of the Mobius transformations.

Theorem 5.29. The orientation-preserving15 isometries of the Poincare disk have the form

f (z) = eiθ α− zαz− 1

where |α| < 1 and θ ∈ [0, 2π)

All isometries can be formed by composing f with complex conjugation (reflection in the real axis).

Referring to the properties in Theorem 5.28:

1. The isometries are precisely the set of Mobius transformations which map D bijectively to itself.

2. Preservation of angles is required for a rigid motion.

3. The unit circle is preserved, so that all omega-points are mapped to omega-points. Moreover,the class of hyperbolic lines is preserved: any circle or line intersecting the unit circle at right-angles is mapped to another such (angle-preservation is used here).

4. Let Ω and Θ be the omega-points on the hyperbolic line joining z, w ∈ D. By properties 2 and3, we see that f (Ω) and f (Θ) are the omega-points on the hyperbolic line through f (z), f (w).Preservation of the cross-ratio shows that f is an isometry:

d( f (z), f (w)) =

∣∣∣∣ln | f (z)− f (Ω)| | f (w)− f (Θ)|| f (z)− f (Θ)| | f (w)− f (Ω)|

∣∣∣∣ = ∣∣∣∣ln |z−Ω| |w−Θ||z−Θ| |w−Ω|

∣∣∣∣ = d(z, w)

How does this help us compute? The isometry f moves α to the origin; one can then choose θ toorient whichever direction you like along the positive x-axis.

14In C∪ ∞ a line is just a circle through ∞. . .15If C is to the left of

−→AB, then f (C) is to the left of

−−−−−−→f (A) f (B). This is the usual ‘right-hand rule.’

19

Example 1 Consider the points P = 12 and Q = 2

3 +√

23 i. Move P to the origin using an isometry

with α = P:

f (z) = eiθ α− zαz− 1

= eiθ 1− 2zz− 2

But now

f (Q) = f

(23+ i√

23

)= eiθ 1− 4

3 − 2√

23 i

23 − 2 +

√2

3 i

= − 1 + 2√

2i−4 +

√2i

eiθ =i√2

eiθ

Choosing eiθ = −i places f (Q) = 1√2

on the positive realaxis. Now check:

P

Q

O = f (P) f (Q)

d(P, Q) = cosh−1

(1 +

2 |PQ|2

(1− |P|2)(1− |Q|2)

)= cosh−1

(1 +

24

(1− 14 )(1− 2

3 )

)= cosh−1 3 = ln(3 + 2

√2)

d( f (P), f (Q)) = ln1 + 1√

2

1− 1√2

= ln

√2 + 1√2− 1

= ln(3 + 2√

2)

The points really are the same distance from each other! Indeed the hyperbolic line in red passingthrough the points P, Q (equation x2 + y2 − 5

2 x + 1 = 0) is transformed by f to (a segment of) thex-axis.

Example 2 You are given the points A = − i2 , B = − i

5 and C = − 15 (3 + i). Find:

(a) The distance between A and B.

(b) The distance between A and C.

(c) The angle ∠BAC.

We could use the cosh-formula to answer parts (a) and (b). Finding the angle would first require theequations of the hyperbolic lines

←→AB,←→AC, then their tangent vectors, etc.: a lot of work. Instead we

appeal to isometries. Start by moving A to the origin: our isometry must have the form

f (z) = eiθ− i2 − z

i2 z− 1

=2z + i2− iz

eiθ

for some θ. Now try moving B to the positive x-axis:16

f(− i

5

)=− 2i

5 + i2− 1

5

eiθ =i3

eiθ

16This strictly is unnecessary, though it unambiguously defines the isometry.

20

We choose eiθ = −i to obtain f (B) = 13 . Our isometry is therefore

f (z) =2z + iz + 2i

which maps

f (C) =− 2

5 (3 + i) + i− 1

5 (3 + i) + 2i=−6 + 3i−3 + 9i

=2− i

1− 3i

=(2− i)(1 + 3i)

10=

1 + i2

Two of the hyperbolic lines have been transformed tostraight lines through the origin and we can easily answerthe questions:

(a) d(A, B) = d( f (A), f (B)) = ln 1+ 13

1− 13= ln 2

(b) d(A, C) = d( f (A), f (C)) = ln1+ 1√

2

1− 1√2

= 2 ln(√

2 + 1)

(c) ∠BAC = arg( 1+i2 ) = π

4

A

BC

O = f (A)f (B)

f (C)

The picture hopefully makes clear the notion of an orientation-preserving isometry. The original trian-gle has been rotated and translated but not reflected: the sequences A, B, C and f (A), f (B), f (C) bothtrace their respective triangles the same way (counter-clockwise).

Interpretation of Isometries From our discussion of isometries in Euclidean geometry, we expectto be able to interpret all isometries as combinations of rotations, reflections and translations. Hereis the dictionary in hyperbolic space: note that it is not worth memorizing these expressions; thecomputations are unpleasant, but the idea is identical to how we constructed general rotations andreflections in Euclidean space.

21

Translation of α to the origin is accomplished by the isometry

Tα(z) =α− zαz− 1

Tα then moves the origin to−α, etc. Indeed it maps anythingon the hyperbolic line

←→0α to another such point.

The picture shows repeated applications of Tα to seven col-ored points.To move the origin to α, we need the inverse function

T−α =−α− z−αz− 1

=α + zαz + 1

To translate α to β, do a composition!

T−β Tα(z) =(αβ− 1)z− α + β

(α− β)z + αβ− 1

α

O

Rotations Rθ(z) = eiθz rotates θ radians counter-clockwisearound the origin. To rotate θ around α, one computes thecomposition

T−α Rθ Tα

The picture shows repeated rotation by 30° = π6 around α.

Reflection across the line making angle θ with the real axis isPθ(z) = e2iθz. To reflect across the hyperbolic line through αpassing though α with gradient θ, we’d compute

T−α Pθ Tα

α

22

Hyperbolic Trigonometry

The goal of trigonometry is to ‘solve’ triangles: given one or more pieces of numerical data (say thecombination side-angle-side), we want to compute remaining side lengths and angle measures. Thetriangle congruence theorems (in hyperbolic geometry SAS, ASA, SSS, SAA, AAA) all provide exam-ples of the minimal data required in order to solve a triangle.17

We start with a right-angled triangle in hyperbolic geometry. Applying an isometry, we may assumeour triangle has a right-angle at the origin and that its other vertices lie along the real and imaginaryaxes at points p and iq respectively. The hyperbolic side-lengths of the non-hypotenuse sides are

a = cosh−1(

1 + p2

1− p2

), b = cosh−1

(1 + q2

1− q2

)To measure the third side, apply another isometry,moving p to the origin via

f (z) =p− zpz− 1

But then

a

b c

p

iq

0B

A

b

a

c

f (p)

f (iq)

BA

f (iq) =p− iq

ipq− 1=−p(1 + q2) + iq(1− p2)

p2q2 + 1=⇒ tan B =

q(1− p2)

p(1 + q2)and | f (iq)|2 =

p2 + q2

p2q2 + 1

We therefore see that

cosh c =1 + | f (iq)|2

1− | f (iq)|2=

1 + p2 + q2 + p2q2

1− p2 − q2 + p2q2 =1 + p2

1− p2 ·1 + q2

1− q2 = cosh a cosh b

Moreover, applying the usual hyperbolic identities,

sinh2b = cosh2b− 1 =⇒ sinh b =2q

1− q2 =⇒ tanh b =sinh bcosh b

=2q

1 + q2

as well as standard trig identities such as sec2B = 1 + tan2B, we rapidly conclude the following:

Theorem 5.30. In a right-triangle with hyperbolic adjacent a, opposite b, and hypotenuse c,

cos B =tanh atanh c

sin B =sinh bsinh c

tan B =tanh bsinh a

cosh c = cosh a cosh b

The final expression is Pythagoras’ Theorem18 for hyperbolic triangles.

17For exam purposes, the formulæ in this section will be provided if needed.18This might not look much like Pythagoras’, but if you use the Maclaurin series for cosh, we obtain

1 +12

c2 +14!

c4 + · · · =(

1 +12

a2 +14!

a4 + · · ·)(

1 +12

b2 +14!

b4 + · · ·)

=⇒ c2 +1

12c4 + · · · = a2 + b2 +

112

(12a2b2 + a4 + b4) + · · ·

which, to 2nd order, is Pythagoras’ Theorem in Euclidean geometry. If the triangle is very small, then higher powers ofa, b, c are much smaller than their squares, whence c2 ≈ a2 + b2. The smaller the triangle, the more it looks Euclidean.

23

Corollary 5.31 (Hyperbolic Cosine Rule). Apply the above argument to a triangle with vertices 0, p, qeiθ toobtain

cosh c = cosh a cosh b− sinh a sinh b cos C

where cosh a =1 + p2

1− p2 , sinh a =

√cosh2a− 1 =

2p1− p2 , etc.

Expressing the right-hand side of this in terms of p, q and applying the Euclidean cosine rule yields the cosh-formula for distance, as defined on page 5.

Corollary 5.32 (Hyperbolic Sine Rule). In any hyperbolic triangle,

sinh asin A

=sinh bsin B

=sinh csin C

Proof. For instance, drop the altitude as in the picture and apply Theo-rem 5.30 to see that

sin A =sinh hsinh b

sin B =sinh hsinh a

Now equate the sinh h terms.

a

bh

B

A

Armed with these results, one can solve any hyperbolic triangle numerically given the informationin one of the triangle congruence theorems. Admittedly some (particularly ASA and AAA) are verymessy to compute with, but others are straightforward. The SSS congruence for instance is equivalentto three applications of the hyperbolic cosine rule.

Examples

1. A right-triangle has angles π6 , π

4 and π2 : find its sides

(This is an example of the AAA theorem in the nice caseof a right-triangle).

Using the tan-formula,

1√3= tan

π

6=

tanh bsinh a

=sinh b

sinh a cosh b

1 = tanπ

4=

tanh asinh b

=sinh a

cosh a sinh b

Now multiply together and use Pythagoras’, a

cb

π4

π6

1√3=

1cosh a cosh b

=1

cosh c=⇒ c = cosh−1

√3 = ln(

√3 +√

2) ≈ 1.1462

We quickly see that sinh c =√

cosh2 c− 1 =√

2, whence the sine-rule yields the other sides:

sinh b = sinπ

4· sinh c

sin π= 1 =⇒ b = sinh−1 1 = cosh−1

√2 ≈ 0.8814

=⇒ cosh a =cosh ccosh b

=

√32

=⇒ a ≈ 0.6565

24

2. A hyperbolic triangle has sides with hyperbolic length a = b = cosh−1 2 with angle betweenthem C = π

3 . Find the remaining data points for the triangle.(This is a numerical version of the SAS congruence)

We have sinh a = sinh b =√

cosh2a− 1 =√

3. By the cosine rule,

cosh c = 2 · 2−√

3√

3 · 12=

52

=⇒ c = cosh−1 52

Now use the sine rule:

sin B = sin A =sin C sinh a

sinh c=

√3

2·√

3√21/2

=3√21

=

√37

Hyperbolic Tilings

The first example above can be used to make a regular tilingof hyperbolic space.Take eight congruent copies of the triangle and arrange themaround the origin as in the picture. Now reflect the quadri-lateral over each of its edges and repeat the process in alldirections. We obtain a tiling of hyperbolic space comprisingregular four-sided figures with six meeting at every vertex!

In hyperbolic space, many different tilings are possible. Indeedby considering the fundamental right-triangle as above, if atiling is to be made using regular m-sided polygons, n of whichare to meet at each vertex, then each polygon comprises 2mcopies of the fundamental triangle, whose angles are π

m andπn . Moreover, since the angles in a triangle sum to less than πradians, we see that there exists a regular tiling of hyperbolicspace whenever

π

2+

π

m+

π

n< π ⇐⇒ (m− 2)(n− 2) > 4

The first example is when m = 4 and n = 6. The second ex-ample involves m = 5 and n = 4, where the interiors of thepolygons have been colored. This example was produced us-ing the tools found here and here: have a play!

This situation is in contrast to Euclidean geometry, where a tiling of m-sided polygons with n meet-ing at each vertex requires equality (m − 2)(n − 2) = 4: there are only three distinct regular tilingsin Euclidean geometry, however all can be scaled to arbitrary side-length. In hyperbolic geometry,there are infinitely many distinct tilings, but each has a unique side-length.

For related fun, look up M.C. Escher’s Circle Limit artworks, some of which are based on hyperbolictilings. If you want an excuse to play video games while (pretending) to study hyperbolic geometry,have a look at Hyper Rogue, which relies on (sometimes irregular) tilings.

25

Arc-length, Geodesics and Area: non-examinable

This section should be accessible to anyone who’s taken a vector-calculus course covering line-integrals and surface area. With a little calculus, we can find an expression for the hyperbolic arc-length along a parameterized curve in hyperbolic space. We can then use this to describe hyperbolicarea using integrals.

Theorem 5.33. If z(t) for t ∈ [t0, t1] is a parametrized hyperbolic curve in the Poincare disk, then its arc-length is∫ t1

t0

2 |z′(t)|1− |z(t)|2

dt

Proof. We need to compute the arc-length of a curve z over an infinitessimal segment from w := z(t)to z(t + ∆t). Take an isometry moving w to the origin O:

z 7→ w− zwz− 1

=z− w

1− wz

Then z(t + ∆t) is mapped to

P =z(t + ∆t)− w

1− wz(t + ∆t)=

z(t + ∆t)− z(t)1− z(t)z(t + ∆t)

≈ z′(t)∆t1− z(t)(z(t) + ∆tz′(t))

≈ z′(t)

1− |z(t)|2∆t

Since ln(1 + x) ≈ x for small x, we see that

d(z(t), z(t + ∆t)) ≈ d(O, P) = ln(

1 + |OP|1− |OP|

)= ln(1 + |OP|)− ln(1− |OP|)

≈ 2 |OP| = 2 |z′(t)|1− |z(t)|2

∆t

Since all approximations becomes equality as ∆t→ 0, we have the result.

Definition 5.34. A geodesic is a path of shortest length between two points.

Theorem 5.35. The geodesics in the Poincare disk model are the hyperbolic lines as we’ve defined them.

Proof. Given points A, B, move A to the origin and B to P via an isometry which leaves P on thepositive x-axis. The hyperbolic arc from A to B is then transformed to the straight line segment fromO to P. It is enough therefore for us to show that the shortest hyperbolic path from O to P = a isalong the x-axis.

Parameterize a curve from O to P by z(t) = t + iy(t) where 0 ≤ t ≤ a and y(0) = y(a) = 0. Wecompute its arc-length:

∫ a

0

2 |z′(t)|1− |z(t)|2

dt =∫ a

0

2√

1 + y′2

1− t2 − y2 dt ≥∫ a

0

21− t2 dt = ln

1 + a1− a

= d(0, P)

where we have equality if and only if y(t) ≡ 0. The length-minimizing path is therefore a hyperbolicline.

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Area Suppose we’re at a point z = x + iy and we increase the x, y co-ordinates by small amountsdx, dy respectively. By Theorem 5.33, this produces (approximately) a rectangle with hyperbolic side-lengths19

2 dx1− x2 − y2

2 dy1− x2 − y2

It follows that the infinitessimal area element is the product of these, whence the area of a region Rin the Poincare disk is∫∫

R

4 dxdy(1− x2 − y2)2 =

∫∫R

4r dr dθ

(1− r2)2

where the second integral is written in polar co-ordinates. Changing variables to use the hyperbolicdistance δ = ln

( 1+r1−r

), we have:

Theorem 5.36. The area of a region R in the Poincare disk, described with respect to the hyperbolic distance δand polar angle θ is∫∫

Rsinh δ dδ dθ

Circles Suppose that a circle has hyperbolic radius δ. We may perform an isometry to move itscenter to the origin. It would then be parameterized by

z(t) = r(

cos tsin t

), t ∈ [0, 2π) where r =

eδ − 1eδ + 1

Its circumference (hyperbolic arc-length) is then∫ 2π

0

2r1− r2 dθ =

4πr1− r2 = 2π sinh δ = 2π

(δ +

13!

δ3 +14!

δ5 + · · ·)> 2πδ

where we used Maclaurin series to compare. Its area is∫ 2π

0

∫ δ

0sinh δ dδ dθ = 2π(cosh δ− 1) = π

(δ2 +

24!

δ4 +26!

δ6 + · · ·)> πδ2

A hyperbolic circle therefore has larger ratios of circumference : radius and area : radius2 than for aEuclidean circle. Moreover, these ratios are not constant: one might say20 that the hyperbolic versionof π is a function!

19If you know some differential geometry, this says that the hyperbolic metric is

ds2 =4(dx2 + dy2)

(1− x2 − y2)2 =4 dz2

(1− |z|2)2

The fact that this is a multiple of the standard Euclidean metric dx2 + dy2 means that angle-measure is identical in Eu-clidean geometry and the Poincare disk (this gives a ‘simple’ proof of assertion made after Definition 5.8). Indeed thedifferential geometer’s definition of the Poincare disk model is nothing more than the the unit disk equipped with thismetric. Everything we’ve done can be recovered starting from this.

20It would be more accurate to say that there is no distinct hyperbolic version of π, merely that the ratio π = c2 sinh δ is

constant for all hyperbolic circles with circumference c and radius δ.

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