4 Hyperbolic Geometry
4.1 History: Saccheri, Lambert and Absolute Geometry
For 2000 years after Euclid, many mathematicians believed that his parallel postulate could not be
an independent axiom. Rigorous work on this problem was undertaken by Giovanni Saccheri (1667–
1733) & Johann Lambert (1728–1777); both attempted to force contradictions by assuming the nega-
tion of the parallel postulate. While this approach ultimately failed, their insights supplied the foun-
dation of a new non-Euclidean geometry. Before considering their work, we define some terms and
recall our earlier discussion of parallels (pages 10–13).
Definition 4.1. Absolute or neutral geometry is the axiomatic system comprising all of Hilbert’s
axioms except Playfair. Euclidean geometry is therefore a special case of neutral geometry.
A non-Euclidean geometry is (typically) a model satisfying most of Hilbert’s axioms but for which
parallels might not exist or are non-unique:
There exists a line ℓ and a point P ∈ ℓ through which there are no parallels or at least two.
For instance, spherical geometry is non-Euclidean since there are no parallel lines—Hilbert’s axioms
I-2 and O-3 are false, as is the exterior angle theorem.
Results in absolute geometry The conclusions of Euclid’s first 28 theorems are valid.
• Basic constructions: bisectors, perpendiculars, etc.
• Triangle congruence theorems: SAS, ASA, SAA, SSS.
• Exterior angle theorem and its consequences:
ℓ
m
P
β
α
– Side/angle comparison and triangle inequality (Exercise 2.3.5).
– Existence of a parallel m to a line ℓ through a point P ∈ ℓ via congruent angles
α
∼
=
β =⇒ ℓ ∥ m
Arguments making use of unique parallels The following results were proved using Playfair’s
axiom or the parallel postulate, whence the arguments are false in absolute geometry:
• A line crossing parallel lines makes congruent angles: in the picture, ℓ ∥ m =⇒ α
∼
=
β. This is
the uniqueness claim in Playfair: the parallel m to ℓ through P is unique.
• Angles in a triangle sum to 180°.
• Constructions of squares/rectangles.
• Pythagoras’ Theorem.
While our arguments for the above are false in absolute geometry, we cannot instantly claim that the
results are false, for there might be alternative proofs! To show that these results truly require unique
parallels, we must exhibit a model in which they are false—such will be described in the next section.
The existence of this model explains why Saccheri and Lambert failed in their endeavors; the parallel
postulate (Playfair) is indeed independent of Euclid’s (Hilbert’s) other axioms.
50
The Saccheri–Legendre Theorem
We work in absolute geometry, starting with an extension of
the exterior angle theorem based on Euclid’s proof.
Suppose △ABC has angle sum Σ
△
and construct M and E fol-
lowing Euclid to the arrangement pictured. Observe:
M
A
B
C
E
1. ∡ACB + ∡CAB = ∡ACB + ∡ACE < 180° is the exterior angle theorem. More generally, the
exterior angle theorem says that the sum of any two angles in a triangle is strictly less than 180°.
2. △ABC and △EBC have the same angle sum
Σ
△
=
•
+
•
+
•
+
•
Just look at the picture—remember that we do not know whether Σ
△
= 180°!
3. △EBC has at least one angle (∡EBC or ∡BEC) measuring ≤
1
2
∡ABC.
Iterate this construction: if ∡EBC ≤
1
2
∡ABC, start by bisecting CE; otherwise bisect BC . . . The result
is an infinite sequence of triangles △
1
= △EBC, △
2
, △
3
, . . . with two crucial properties:
(a) All triangles have same angle sum Σ
△
= Σ
△
1
= Σ
△
2
= ···.
(b) △
n
has at least one angle measuring α
n
≤
1
2
n
∡ABC.
Now suppose Σ
△
= 180° + ϵ is strictly greater than 180°. Since lim
1
2
n
= 0, we may choose n large
enough to guarantee α
n
< ϵ. But then the sum of the other two angles in △
n
would be greater than
180°, contradicting the exterior angle theorem (observation 1)! We have proved a famous result.
Theorem 4.2 (Saccheri–Legendre). In absolute geometry, triangles have angle sum Σ
△
≤ 180°.
Saccheri’s failed hope was to prove equality without invoking the parallel postulate.
Saccheri and Lambert Quadrilaterals
Two families of quadrilaterals in absolute geometry are named in honor of these pioneers.
Definition 4.3. A Saccheri quadrilateral ABCD satisfies
AD
∼
=
BC and ∡DAB = ∡CBA = 90°
AB is the base and CD the summit.
The interior angles at C and D are the summit angles.
A Lambert quadrilateral has three right-angles; for instance AMND
in the picture.
B
A
M
N
C
D
We draw these with curved sides to indicate that the summit angles need not be right-angles, though
we haven’t yet exhibited a model which shows they could be anything else. Regardless of how they
are drawn, AD, BC and CD are all segments!
51
The apparent symmetry of a Saccheri quadrilateral is not an illusion.
Lemma 4.4. 1. If the base and summit of a Saccheri quadrilateral
are bisected, we obtain congruent Lambert quadrilaterals.
2. The summit angles of a Saccheri quadrilateral are congruent.
3. In Euclidean geometry, Saccheri and Lambert quadrilaterals
are rectangles (four right-angles).
B
A
M
N
C
D
Parts 1 and 2 are exercises. We could interpret part 3 as saying that Saccheri and Lambert quadrilat-
erals are as close as we can get to rectangles in absolute geometry.
Proof of 3. By part 1 we need only prove this for a Saccheri quadrilateral. Following the exterior angle
theorem,
←→
AB is a crossing line making congruent right-angles, whence AD ∥ BC.
However
←→
CD also crosses the same parallel lines. By the parallel postulate, the summit angles sum
to a straight edge. Since these are congruent, they are both right-angles.
We now show that drawing acute summit angles is justified by the Saccheri–Legendre Theorem.
Theorem 4.5. The summit angles of a Saccheri quadrilateral measure ≤ 90°.
Proof. Suppose ABCD is a Saccheri quadrilateral with base AB.
Extend CB to E (opposite side of AB to C) such that BE
∼
=
DA.
Let M be the midpoint of AB.
SAS implies ∠DAM
∼
=
△EBM; the vertical angles at M are con-
gruent, whence M lies on DE.
A
B
C
D
E
M
By Saccheri–Legendre, the (congruent) summit angles at C and D sum to
∡ADC + ∡DCB = ∡ADM + ∡EDC + ∡DCE = ∡CED + ∡EDC + ∡DCE ≤ 180°
Exercises 4.1. Work in absolute geometry; you cannot use Playfair’s Axiom or the parallel postulate!
1. Use the first picture to prove parts 1 and 2 of Lemma 4.4.
2. Use the first picture to give an alternative proof of Theorem 4.5.
3. Suppose □ABCD has four right-angles (second picture). Ap-
ply the Saccheri–Legendre Theorem to prove that AC splits
□ABCD into two congruent triangles, and conclude that the
opposite sides are congruent.
Why is this question easier in Euclidean geometry?
4. A pair of Saccheri quadrilaterals have congruent bases (e.g.,
AB) and perpendicular sides (AD, BC). Prove that the quadri-
laterals are congruent.
B
A
M
N
C
D
A
B
C
D
5. (Hard!) Suppose Saccheri quadrilaterals have congruent summits and perpendicular sides.
Prove that the quadrilaterals are congruent.
52
4.2 Models of Hyperbolic Geometry
In the 1820-30s, J
´
anos Bolyai, Carl Friedrich Gauss and Nikolai Lobachevsky independently took the
next step, each describing versions of non-Euclidean geometry.
21
Rather than attempting to establish
the parallel postulate as a theorem within Euclidean geometry, a new geometry was defined based
on an alternative to the parallel postulate.
Axiom 4.6 (Bolyai–Lobachevsky/Hyperbolic Postulate). Given a line ℓ and a point P ∈ ℓ, there
exist at least two parallel lines to ℓ through P.
Hyperbolic Geometry is the resulting axiomatic system: Hilbert with Playfair’s axiom replaced by the
hyperbolic postulate. Consistency was proved in the late 1800s by Beltrami, Klein and Poincar
´
e, each
of whom created models by defining point, line, etc., in novel ways. One of the simplest is named for
Poincar
´
e, though it was first proposed by Beltrami.
22
Definition 4.7. The Poincar´e disk is the interior of the unit circle
(x, y) ∈ R
2
: x
2
+ y
2
< 1
or
z ∈ C :
|
z
|
< 1
A hyperbolic line is a diameter or a circular arc meeting the unit circle at
right-angles.
In the picture we have a hyperbolic line ℓ and a point P: also drawn
are several parallel hyperbolic lines to ℓ passing through P.
Points on the boundary circle are termed omega-points: these are not in
the Poincar
´
e disk and are essentially ‘points at infinity.’
P
ℓ
By the incidence axioms, there exists a unique hyperbolic line joining any two points in the Poincar
´
e
disk. Such may straightforwardly be described using equations in analytic geometry.
Lemma 4.8. Every hyperbolic line in the Poincar
´
e disk model
is one of the following:
• A diameter passing through (c, d) = (0, 0) with Euclidean
equation dx = cy.
• The arc of a (Euclidean) circle with equation
x
2
+ y
2
−2ax −2by + 1 = 0 where a
2
+ b
2
> 1
and (Euclidean) center and radius
C = (a, b) and r =
p
a
2
+ b
2
−1
1
r
O
P
Q
C = (a, b)
21
Bolyai indeed is the source of the term ‘absolute geometry.’
22
The key results of hyperbolic geometry—including almost everything in Sections 4.3, 4.4 & 4.5—can be discussed
synthetically without reference to a model. While efficient, such an approach would be both ahistorical and masochistic
for a first exposure: an explicit model allows us to visualize theorems and to verify examples via calculation.
53
Example 4.9. We compute the hyperbolic line through P = (
1
3
,
1
2
) and Q = (
1
2
, 0) in the Poincar
´
e
disk: this is the picture shown in Lemma 4.8.
Substitute into x
2
+ y
2
−2ax −2by + 1 = 0 to obtain a system of equations for a, b:
(
1
9
+
1
4
−
2
3
a − b + 1 = 0
1
4
− a + 1 = 0
=⇒ (a, b) =
5
4
,
19
36
The required hyperbolic line
←→
PQ therefore has equation
x
2
+ y
2
−
5
2
x −
19
18
y + 1 = 0 or
x −
5
4
2
+
y −
19
36
2
=
545
648
The undefined terms point, line, on and between now make sense. To complete the model, we need to
define congruence of hyperbolic segments and angles.
Definition (4.7 continued). The hyperbolic distance between points P, Q in the Poincar
´
e disk is
23
d(P, Q) := cosh
−1
1 +
2
|
PQ
|
2
(1 −
|
P
|
2
)(1 −
|
Q
|
2
)
!
where
|
PQ
|
is the Euclidean distance and
|
P
|
,
|
Q
|
are the Euclidean
distances of P, Q from the origin.
Hyperbolic segments are congruent if they have the same length.
The angle between hyperbolic rays is that between their tangent lines:
angles are congruent if they have the same measure.
θ
Lemma 4.10. The hyperbolic distance of P from the origin is
d(O, P) = cosh
−1
1 +
|
P
|
2
1 −
|
P
|
2
= ln
1 +
|
P
|
1 −
|
P
|
Example 4.11. We calculate the sides and angles in the isosceles right-triangle
with vertices O = (0, 0), P = (
1
2
, 0) and Q = (0,
1
2
).
|
P
|
=
1
2
=
|
Q
|
,
|
PQ
|
2
=
1
4
+
1
4
=
1
2
d(O, P) = d(O, Q) = ln
1 +
1
2
1 −
1
2
= ln 3 = cosh
−1
5
3
≈ 1.099
d(P, Q) = cosh
−1
1 +
2 ·
1
2
(1 −
1
4
)
2
!
= cosh
−1
25
9
≈ 1.681
θ
P
Q
O
23
It seems reasonable for hyperbolic functions to play some role in hyperbolic geometry! For reference:
cosh x =
e
x
+ e
−x
2
, sinh x =
e
x
−e
−x
2
, cosh
2
x −sinh
2
x = 1, cosh
−1
x = ln(x +
p
x
2
−1)
54
To find the interior angle θ, implicitly differentiate the equation for the hyperbolic line
←→
PQ:
x
2
+ y
2
−
5
2
x −
5
2
y + 1 = 0 =⇒
dy
dx
P
=
4x −5
5 − 4y
P
= −
3
5
=⇒ θ = tan
−1
3
5
≈ 30.96°
By symmetry, we have the same angle at Q. With a right-angle at O, we conclude that the angle sum
is approximately Σ
△
= 151.93°!
As a sanity check, we compare data for △OPQ and the Euclidean triangle with the same vertices
Property Hyperbolic Triangle Euclidean Triangle
Edge lengths 1.099 : 1.099 : 1.681 0.5 : 0.5 : 0.707
Relative edge ratios 1 : 1 : 1.530 1 : 1 : 1.414
Angles 30.06°, 30.96°, 90° 45°, 45°, 90°
The hyperbolic triangle has longer sides and a relatively longer hypotenuse. Moreover, its side lengths
do not satisfy the Pythagorean relation a
2
+ b
2
= c
2
(though cosh a cosh b = cosh c . . .).
The next result is an exercise; it says that distance increases
smoothly as one moves along a hyperbolic line.
Lemma 4.12. Fix P and a hyperbolic line through P. Then the
distance function Q 7→ 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
look like one expects: the circle of hyperbolic radius δ centered at
P is the set of points Q such that d(P, Q) = δ.
In the picture are several hyperbolic circles and their centers; one has several of its radii drawn.
Observe how the centers are closer (in a Euclidean sense) to the boundary circle than one might
expect: this is since hyperbolic distances measure greater the further one is from the origin.
In fact (Exercise 4.2.6) hyperbolic circles in the Poincar
´
e disk model are also Euclidean circles! Their
hyperbolic radii moreover intersect the circles at right-angles, as we’d expect.
Theorem 4.13. The Poincar
´
e disk is a model of hyperbolic geometry.
Sketch Proof. A rigorous proof would require us to check the hyperbolic postulate and all Hilbert’s
axioms except Playfair. Instead we verify Euclid’s postulates 1–4 and the hyperbolic postulate 5.
1. Lemma 4.8 says we can join any given points in the Poincar
´
e disk by a unique segment.
2. A hyperbolic segment joins two points inside the (open) Poincar
´
e disk. The distance formula in-
creases (Lemma 4.12) unboundedly as P moves towards the boundary circle, so we can always
make 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. The first picture on page 53 shows multiple parallels!
55
Other Models of Hyperbolic Space: non-examinable
There are several other models of hyperbolic space. Here are three of the most common.
Klein Disk Model This is similar to the Poincar
´
e disk, though lines are chords of the unit circle
(‘Euclidean’ straight lines!) and the distance function is different:
d
K
(P, Q) =
1
2
ln
|
PΘ
||
QΩ
|
|
PΩ
||
QΘ
|
where Ω, Θ are where the chord
←→
PQ meets the boundary circle.
The cost is that the notion of angle is different. The picture shows
perpendicularity: Given a hyperbolic line find the tangents to where
it meets the boundary circle. Any chord whose extension passes
through the intersection of these tangents is perpendicular to the orig-
inal line. Measuring other angles is difficult!
Ω
Θ
P
Q
Gauss’ famous theorem egregium says that this problem is unavoidable; there is no model in which
lines and angles both have the same meaning as in Euclidean geometry.
Poincar´e Half-plane Model Widely used in complex analysis, the
points comprise the upper half-plane (y > 0) in R
2
, while hyperbolic
lines are verticals or semicircles centered on the x-axis
x = constant or (x − a)
2
+ y
2
= r
2
and angles are the same as in Euclidean space. The expression for
hyperbolic distance remains horrific! The picture shows several hy-
perbolic lines and a hyperbolic triangle.
y
x
Hyperboloid Model Points comprise the upper sheet (z ≥ 1) of the hyperboloid x
2
+ y
2
= z
2
−1.
A hyperbolic line is the intersection of the hyperboloid with a plane through the origin. Isometries
(congruence) can be described using matrix-multiplication and hyperbolic distance is relatively easy:
given P = (x, y, z) and Q = (a, b, c), hyperbolic distance is
d(P, Q) = cosh
−1
(cz − ax − by)
Difficulties include working in three dimensions and the fact
that angles are awkward.
The relationship to the Poincar
´
e disk is via projection. Place
the disk in the x, y-plane centered at the origin and draw a
line through the disk and the point (0, 0, −1). The intersec-
tion of this line with the hyperboloid gives the correspon-
dence.
56
Exercises 4.2. Answer all questions within the Poincar
´
e disk model.
1. (a) Find the equation of the hyperbolic line joining P = (
1
4
, 0) and Q = (0,
1
2
).
(b) Find the side lengths of the hyperbolic triangle △OPQ where O = (0, 0) is the origin.
(c) The triangle in part (b) is right-angled at O. If o, p, q represent the hyperbolic lengths of
the sides opposite O, P, Q respectively, check that the Pythagorean theorem p
2
+ q
2
= o
2
is false. Now compute cosh p cosh q: what do you observe?
2. Find the omega points for the hyperbolic line with equation x
2
+ y
2
−4x + 10y + 1 = 0
3. Let P =
1
2
,
q
5
12
and Q =
1
2
, −
q
5
12
(a) Compute the hyperbolic distances d(O, P), d(O, Q) and d(P, Q), where O is the origin.
(b) Compute the angle ∡POQ.
(c) Show that the hyperbolic line ℓ =
←→
PQ has equation x
2
−
10
3
x + y
2
+ 1 = 0.
(d) Calculate
dy
dx
to show that a tangent vector to ℓ at P is
√
15i + 7j. Hence compute ∡OPQ.
4. We extend Example 4.11. Let c ∈ (0, 1) and label O = (0, 0), P = (c, 0) and Q = (0, c).
(a) Compute the hyperbolic side lengths of △OPQ.
(b) Find the equation of the hyperbolic line joining P = (c, 0) and Q = (0, c).
(c) Use implicit differentiation to prove that the interior angles at P and Q measure tan
−1
1−c
2
1+c
2
.
What happens as c → 0
+
and as c → 1
−
?
5. Let 0 < r < 1 and find the hyperbolic side lengths and interior angles of the equilateral triangle
with vertices (r, 0), (−
r
2
,
√
3r
2
) and (−
r
2
, −
√
3r
2
).What do you observe as r → 0
+
and r → 1
−
?
6. (a) Use the cosh distance formula to prove that the hyperbolic circle of hyperbolic radius
ρ = ln 3 and center C = (
1
2
, 0) in the Poincar
´
e disk has Euclidean equation
x −
2
5
2
+ y
2
=
4
25
(b) Prove that every hyperbolic circle in the Poincar
´
e disk is in fact a Euclidean circle.
7. We sketch a proof of Lemma 4.12.
(a) Prove that f (x) = cosh
−1
x = ln(x +
√
x
2
−1) is strictly increasing on the interval (1, ∞).
(b) By part (a), it is enough to show that
|
PQ
|
2
1−
|
Q
|
2
increases as Q moves away from P along a
hyperbolic line. Appealing to symmetry, let P = (0, c) lie on the hyperbolic line with
equation x
2
+ y
2
−2by + 1 = 0. Prove that
|
PQ
|
2
1 −
|
Q
|
2
=
(b − c)y + bc −1
1 − by
and hence show that this is an increasing function of y when c < y <
1
b
.
57
4.3 Parallels, Perpendiculars & Angle-Sums
From now on, all examples will be illustrated using the Poincar
´
e
disk, though the main results hold in any model. Recall (page 50)
that we may use anything from absolute geometry; as a sanity check,
think through how the picture illustrates the following result.
Lemma 4.14. Through a point P not on a line ℓ there exists a unique
perpendicular to ℓ .
We now consider a major departure from Euclidean geometry.
P
B
A
Q
R
M
ℓ
m
Theorem 4.15 (Fundamental Theorem of Parallels). Given P ∈ ℓ, drop
the perpendicular PQ. Then there exist precisely two parallel lines m, n to ℓ
through P with the following properties:
1. A ray based at P intersects ℓ if and only if it lies between m and n in
the same fashion as
−→
PQ.
2. m and n make congruent acute angles µ with
−→
PQ.
ℓ
R
Q
P
n
m
µ
Definition 4.16. The lines m, n are the limiting, or asymptotic, parallels to ℓ through P. Every other
parallel is an ultraparallel. The angle of parallelism at P relative to ℓ is the acute angle µ.
More generally, parallel lines ℓ, m are limiting if they ‘meet’ at an omega-point.
The proof depends crucially on ideas from analysis, particularly continuity & suprema. As you read
through, consider how everything except the last line is valid in Euclidean geometry!
Proof. Points R ∈ ℓ are in continuous bijective correspondence with the real numbers (Lemma 4.12).
It follows that we have a continuous increasing function
f : R → (−90°, 90°) where f (r) = ∡QPR
By Saccheri–Legendre, ±90° ∈ range f . Since dom f = R is an interval, the intermediate value
theorem forces range f to be a subinterval I ⊆ (−90°, 90°).
Given R ∈ ℓ, transfer QR to the other side of Q to obtain S ∈ ℓ. By SAS,
∡QPS = −∡QPR whence I = range f is symmetric: θ ∈ I ⇐⇒ −θ ∈ I.
Define µ := sup I ∈ (0°, 90°] to be the least upper bound; by symmetry,
inf I = −µ. Let m and n be the lines making angles ±µ respectively.
Plainly every ray making angle θ ∈ (−µ, µ) intersects ℓ.
Suppose m intersected ℓ at M. Let
˜
M ∈ ℓ lie on the other side of M from
Q. Since f is increasing, we see that ∡QP
˜
M > µ, which contradicts
µ = sup I. It follows that m is parallel to ℓ. Similarly n ∥ ℓ and we have
part 1.
ℓ
R
S
Q
P
n
m?
M
˜
M
µ
Finally m = n ⇐⇒ µ = 90°. In such a case there would exist only one parallel to ℓ through P,
contradicting the hyperbolic postulate.
58
The picture suggests a bijective relationship between µ and the perpendicular distance. Here it is; we
postpone a simplified argument to Exercise 4.3.6, and the full result to the next section.
Corollary 4.17. The perpendicular distance δ = d(P, Q) and the angle of parallelism are related via
cosh δ = csc µ or equivalently tan
µ
2
= e
−δ
Examples 4.18. 1. Let ℓ be the hyperbolic line x
2
+ y
2
−4x + 1 = 0.
Intersect with x
2
+ y
2
= 1 to find Ω =
1
2
,
√
3
2
and Θ =
1
2
, −
√
3
2
.
By symmetry, the perpendicular from P = ( 0, 0) to ℓ has equation
y = 0 and results in Q = (2 −
√
3, 0).
The limiting parallels through P have equations y = ±
√
3x, from
which the angle of parallelism is µ = tan
−1
√
3 = 60°.
In accordance with Corollary 4.17, we easily verify that
P
Ω
Θ
Q
δ = d(P, Q) = ln
1 + (2 −
√
3)
1 − (2 −
√
3)
= ln
√
3 ↭ e
−δ
=
1
√
3
= tan
µ
2
2. We find the limiting parallels and the angle of parallelism when
P =
−
3
10
,
4
10
and x
2
+ y
2
+ 2x + 4y + 1 = 0
First find the omega-points by intersecting with x
2
+ y
2
= 1:
Ω = (−1, 0), Θ =
3
5
, −
4
5
Plainly
←→
PΘ is the diameter y = −
4
3
x with slope −
4
3
.
P
Ω
Θ
Q
µ
For
←→
PΩ, substitute into the usual expression x
2
+ y
2
−2ax −2by + 1 = 0 and implicitly differ-
entiate:
x
2
+ y
2
+ 2x −
13
8
y + 1 = 0 =⇒
dy
dx
P
=
16( 1 + x)
13 − 16y
P
=
16 ·
7
10
13 −
64
10
=
56
33
The angle of parallelism is half that between the tangent vectors
−33
−56
and
3
−4
:
µ =
1
2
cos
−1
−33
−56
·
3
−4
−33
−56
3
−4
=
1
2
cos
−1
5
13
≈ 33.69°
Corollary 4.17 can now be used to find the perpendicular distance d(P, Q) = ln
3+
√
13
2
.
Without the development of later machinery, it is very tricky to compute Q. If you want a serious
challenge, see if you can convince yourself that Q =
93(−29+2
√
117)
1865
,
26(−29+2
√
117)
1865
.
59
Angles in Triangles, Rectangles and the AAA Congruence
We finish this section three important differences between hyperbolic and Euclidean geometry.
Theorem 4.19. In hyperbolic geometry:
1. There are no rectangles (quadrilaterals with four right-angles). In particular, the summit angles
of a Saccheri quadrilateral are acute.
2. The angles in a triangle sum to strictly less than 180°.
3. (AAA congruence) If the angles of △ABC and △DEF are congruent in pairs, then the triangles
are congruent (△ABC
∼
=
△DEF).
Note that AAA is a congruence theorem in hyperbolic geometry, not a similarity theorem (compare
with Theorem 2.42). Also revisit the observations on page 50; the Theorem largely shows that Euclid’s
arguments making use of the parallel postulate genuinely require it!
Proof. Given a rectangle □ABCD, reflect across CD (Exercise 4.1.4) and repeat to obtain an infinite
family of congruent rectangles. Let P ∈ CD and drop perpendiculars to R ∈ AB and C
1
as shown.
□PRBC is a rectangle: if not, then one of □ARPD or □PRBC would have angle sum exceeding 360°,
contradicting Saccheri–Legendre (Theorem 4.2). Similarly □DPC
1
D
1
is a rectangle.
By Exercise 4.1.3,
−→
BP splits □PRBC into a pair of congruent triangles. In particular,
−→
BP crosses CD
at the same angle as it leaves B. By vertical angles at P, the ray
−→
BP emanates from the upper-right
vertex of □DPC
1
D
1
at the same angle as it does for □ABCD.
Iterate the process to obtain the picture, each time dropping the perpendicular from P
k
to CD to
produce the equidistant sequence Q
1
, Q
2
, Q
3
, . . . (the fact that all the small rectangles are congruent is
essentially Exercise 4.1.4 again). Since CD is finite, the sequence (Q
k
) eventually
24
passes D: some Q
n
lies on the opposite side of
←→
AD. It follows that P
n
∈
−→
BP does also, whence
−→
BP intersects
←→
AD.
A
B
C
D
R
P
P
1
P
2
P
3
P
4
Q
1
Q
2
Q
3
Q
4
C
1
D
1
C
2
D
2
C
3
D
3
Since P ∈ CD was generic, it follows that any ray based at B on the same side as AD must intersect
←→
AD. Otherwise said,
←→
BC is the only parallel to
←→
AD through B, contradicting the hyperbolic postulate.
Parts 2 and 3 are addressed in Exercises 7 and 8.
24
This is the Archimedean property from analysis: a > b =⇒ ∃n ∈ N such that nb > a.
60
Exercises 4.3. 1. Use Theorem 4.19 to prove the following within hyperbolic geometry.
(a) Two hyperbolic lines cannot have more than one common perpendicular.
(b) Saccheri quadrilaterals with congruent summits and summit angles are congruent.
2. A point P lies a perpendicular distance δ = d(P, Q) = ln
√
3 =
1
2
ln 3 from a hyperbolic line ℓ.
A ray
−→
PR makes angle 45° with the perpendicular
−→
PQ. Determine whether
←→
PQ intersects ℓ, is a
limiting parallel, or an ultraparallel.
3. Suppose ℓ intersects m at a right-angle and that m, n are parallel.
(a) In Euclidean geometry, prove that ℓ intersects n at a right-angle.
(b) What are the possible arrangements in hyperbolic geometry? Draw some pictures.
4. Verify cosh δ = csc µ for the point P = (0, 0) and the hyperbolic line (x −1)
2
+ (y −1)
2
= 1.
δ
P
µ
ℓ
O
Q
Ω
Question 4 Question 5
5. Let ℓ be the line x
2
+ y
2
−4x + 2y + 1 = 0 and drop a perpendicular from O to Q ∈ ℓ.
(a) Explain why Q has co-ordinates (
2
√
5
t, −
1
√
5
t) for some t ∈ (0, 1).
(Hint: where is the ‘center’ of ℓ, viewed as a Euclidean circle?)
(b) Show that the hyperbolic distance δ = d(O, Q) of ℓ from the origin is ln
1+
√
5
2
.
(c) Let Ω = (0, −1). Compute µ = ∡QOΩ explicitly and verify that cosh δ = csc µ.
6. We generalize Example 4.18.1. Suppose P = (0, 0) is the origin, and let Q = (r, 0) where
0 < r < 1. Also let ℓ be the hyperbolic line passing through Q at right-angles to PQ.
(a) Find the equation of ℓ and prove that its limiting parallels through P have equations
±2ry = (1 −r
2
)x
(Hint: what does symmetry tell you about ℓ?)
(b) Let µ be the angle of parallelism of P relative to ℓ and δ = d(P, Q) the hyperbolic distance.
Prove that cosh δ = csc µ.
(Hint: csc
2
µ = 1 + cot
2
µ = 1 +
1
tan
2
µ
= . . .)
(c) By differentiating the expression cosh δ = csc µ, verify the claim that δ and µ are bijectively
related.
61
7. We work in neutral geometry. Suppose △ABC has longest side AB (the other sides are no
larger—the triangle could be equilateral!).
(a) Use side-angle comparison (Exercise 2.3.5) to prove that C
lies strictly between the perpendiculars at A, B.
(b) Drop the perpendicular from C to M ∈ AB. Prove that M
is interior to AB.
(Hint: show that the other possibilities are contradictions)
A
B
C
M?
(c) Suppose there exists a triangle with angle sum 180°. Show that there exists a right-triangle
with angle sum 180° and therefore a rectangle.
(Since rectangles are impossible in hyperbolic geometry, this proves part 2 of Theorem 4.19)
(d) Explain why parts (a) and (b) are needed to prove (c): what might happen if AB isn’t the
longest side?
8. We prove the AAA congruence theorem in hyperbolic geometry (Theorem 4.19, part 3).
Suppose, for contradiction, that non-congruent triangles △ABC and △DEF have angles con-
gruent in pairs (∠A
∼
=
∠D, etc.). Without loss of generality, assume DE < AB. By segment
transfer, there exist unique points:
• G ∈
AB such that DE
∼
=
AG.
• H ∈
−→
AC such that DF
∼
=
AH.
(a) Explain why △DEF
∼
=
△AGH.
(b) The picture shows the three generic locations for H.
i. H is interior to AC.
ii. H = C.
iii. C lies between A and H.
A
B
C = H?
G
H?
H?
By connecting GH, in each case explain why we have a contradiction.
62
4.4 Omega-triangles
Recall that limiting parallels (Definition 4.16) ‘meet’ at an omega-point.
Definition 4.20. An omega-triangle or ideal-triangle is a ‘triangle’ where
at least two ‘sides’ are limiting parallels. Alternatively (in the Poincar
´
e
disk model), one or more of the ‘vertices’ is an omega-point.
The three types of omega-triangle depend on how many omega-points
they have. In the picture, △PQΩ has one omega-point, △PΩΘ has
two and △ΩΘΞ three!
P
Ω
Θ
Ξ
Q
Amazingly, many of the standard results of absolute geometry also apply to omega-triangles! The
first can be thought of as the AAA congruence theorem where one ‘angle’ is zero.
Theorem 4.21 (Angle-Angle Congruence for Omega-triangles). Suppose △ABΩ and △PQΘ are
omega-triangles, each with a single omega-point. If the angles are congruent in pairs
∠ABΩ
∼
=
∠PQΘ ∠BAΩ
∼
=
∠QPΘ
then the finite sides of each triangle are also congruent: AB
∼
=
PQ.
Remember that omega-points are not really part of hyperbolic geometry—their appearance in our
description is an artifact of the Poincar
´
e disk model. It therefore doesn’t make sense to speak of con-
gruent ‘infinite’ sides or of congruent ‘angles at omega-points.’ However, if one defines congruence
in terms of isometries (Section 4.6), then this idea becomes more reasonable.
Proof. Assume, WLOG and for contradiction, that AB > PQ. Transfer ∠QPΘ to A to obtain R ∈ AB
such that AR
∼
=
PQ. Transferring ∠PQΘ creates a ray r based at R on the same side as Ω. Exercise 3
verifies that r =
−→
RΩ. Our hypothesis is therefore that the pictured angles at B and R are congruent.
Let M be the midpoint of BR and drop the perpendicular to C ∈
←→
BΩ.
Let S ∈
←→
RΩ lie on the opposite side of
←→
BR to C such that RS
∼
=
BC.
By Side-Angle-Side we have △MBC
∼
=
△MRS. In particular:
• △MRS is right-angled(!) at S.
• Congruent vertical angles at M force M to lie on the segment CS.
The angle of parallelism of S relative to
←→
BΩ is therefore ∠CSΩ = 90°,
which contradicts the Fundamental Theorem (4.15).
There are two other possible orientations:
Ω
A
B
R
M
C
S
• C could lie on the opposite side of B from Ω. In this case SAS is applied to the same triangles
but with respect to congruent supplementary angles.
• In the special case that C = B, the magenta angles are right-angles and the same contradiction
appears: the angle of parallelism of R with respect to
←→
BΩ is 90°.
63
Theorem 4.22 (Exterior Angle Theorem for Omega-Triangles). Suppose △DEΩ has a single
omega-point and that D ∗ E ∗ F. Then ∠FEΩ > ∠EDΩ.
Proof. We show that the two other cases are impossible.
(∠FEΩ
∼
=
∠EDΩ) This is the contradictory arrangement described in the
previous proof where D = B, E = R, F = A.
(∠FEΩ < ∠EDΩ) Transfer the latter to E to produce
−→
EX interior to
∠DEΩ with ∠FEX
∼
=
∠EDΩ.
Since
←→
EΩ is a limiting parallel to
←→
DΩ, the Fundamental Theorem
says that
−→
EX intersects
←→
DΩ at some point Y.
But now △DEY contradicts the standard exterior angle theorem
(∠FEY
∼
=
∠EDY).
Ω
Y
F
D
E
X
The final congruence theorem is an exercise based on the previous picture.
Corollary 4.23 (Side-Angle Congruence for Omega-triangles). Suppose △DEΩ and △PQΘ both
have a single omega-point. If ∠EDΩ
∼
=
∠QPΘ and DE
∼
=
PQ then ∠DEΩ
∼
=
∠PQΘ.
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 (including
Ω), then it must pass through exactly one of the other sides.
• (Omega Crossbar Thm) If a line passes through an interior point and exactly one vertex (includ-
ing Ω) of an omega-triangle, then it passes through the opposite side. This is partly embedded
in the proof of Theorem 4.22.
Perpendicular Distance and the Angle of Parallelism: Applied to right-angled omega-triangles, the AA
and SA theorems prove that the angle of parallelism is a bijective function of the perpendicular dis-
tance. Moreover, by transferring the right-angle to the positive x-axis and the other vertex to the
origin, we obtain the arrangement in Exercise 4.3.6, thus completing the proof of Corollary 4.17.
Exercises 4.4. 1. Let △PQΩ be an omega-triangle. Prove that ∡PQΩ + ∡QPΩ < 180°.
2. Let ℓ and m be limiting parallels. Explain why they cannot have a common perpendicular.
3. In the proof of the AA congruence, explain why r cannot intersect either
−→
AΩ or
−→
BΩ.
4. Prove the Side-Angle congruence theorem for omega-triangles with one omega-point.
5. What would an ‘omega-triangle’ look like in Euclidean geometry? Comment on the three re-
sults in this section: are they still true?
64
4.5 Area and Angle-defect
In this section we consider one of the triumphs of Johann Lambert: the relationship between the
sum of the angles in a triangle and its area. We start with a loose axiomatization of area as a relative
measure. Until explicitly stated otherwise, we work in absolute geometry.
Axiom I Two geometric figures have the same area if and only if they may be sub-divided into
finitely many pairs of mutually congruent triangles.
25
Axiom II The area of a triangle is positive.
Axiom III The area of a union of disjoint figures is the sum of the areas of the figures.
Definition 4.24 (Angle defect). Let Σ
△
be the sum of the angles in a triangle. Measured in radians,
the angle-defect of △ is π − Σ
△
.
Since triangles in absolute geometry have Σ
△
≤ π (Theorem 4.2), it follows that
0 ≤ π −Σ
△
≤ π
In Euclidean geometry the defect is always zero, while in hyperbolic geometry the defect is strictly
positive (Theorem 4.19). A ‘triangle’ with three omega-points would have defect π.
Lemma 4.25. Angle-defect is additive: If a triangle is split into two sub-
triangles, then the defect of the whole is the sum of the defects of the parts.
This is immediate 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!
C
B
A
Q
γ
β
α
δ
ζ
ϵ
Theorem 4.26 (Area determines angle-sum in absolute geometry). If triangles have the same area,
then their angle-sums are identical.
Of course this trivial in Euclidean geometry where all triangles have the same angle-sum!
Proof. The lemma provides the induction step: if △
1
and △
2
have the same area, then their interiors
are disjoint unions of a finite collection of mutually congruent triangles:
△
1
=
n
[
k=1
△
1,k
and △
2
=
n
[
k=1
△
2,k
where △
1,k
∼
=
△
2,k
Each pair △
1,k
, △
2,k
has the same angle-defect, whence the angle-defects of △
1
and △
2
are equal:
defect(△
1
) =
n
∑
k=1
defect(△
1,k
) =
n
∑
k=1
defect(△
2,k
) = defect(△
2
)
25
To allow infinitely many infinitesimal sub-triangles would require ideas from calculus and complicate our discussion.
65
Angle-sum determines area in hyperbolic geometry
The converse in hyperbolic geometry relies on a beautiful and reversible construction relating tri-
angles and Saccheri quadrilaterals. The construction itself is valid in absolute geometry, though the
ultimate conclusion that angle-sum determines area is not. If the initial discussion seems difficult,
pretend you are in Euclidean geometry and think about rectangles.
Lemma 4.27. 1. Given △ABC, choose a side BC. Bisect the remaining sides at E, F and drop per-
pendiculars from A, B, C to
←→
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
conclusions:
• The triangle and quadrilateral have equal area.
• The sum of the summit angles of the quadrilateral
equals the angle sum of the triangle.
We chose BC to be the longest side of △ABC—this isn’t nec-
essary, though it helpfully forces E, F to lie between H, I.
A
B
C
D
E
F
G
H I
Proof. 1. Two applications of the SAA congruence (follow the arrows!) tell us that
△BEH
∼
=
△AEG and △CFI
∼
=
△AFG
We conclude that BH
∼
=
AG
∼
=
CI whence HICB is a Sac-
cheri quadrilateral. The area and angle-sum correspon-
dence is immediate from the picture.
2. Suppose the midpoint of AC were at J = F. By part 1, we
may create a new Saccheri quadrilateral with summit BC
using the midpoints E, J.
The perpendicular bisector of BC bisects the bases of both
Saccheri quadrilaterals (Lemma 4.4), creating △EUV
with two right-angles: contradiction.
J
U
V
A
B
C
D
E
F
G
H I
We now prove a special case of the main result.
Lemma 4.28. Suppose hyperbolic triangles △ABC and △PQR have congruent sides BC
∼
=
QR and
the same angle-sum. Then the triangles have the same area.
Proof. Construct the quadrilaterals corresponding to △ABC and △PQR with summits BC
∼
=
QR.
These have congruent summits and summit angles: by Exercise 4.3.1b they are congruent.
The final observation is what makes this special to hyperbolic geometry. In the Euclidean case, Saccheri
quadrilaterals are rectangles, and congruent summits do not force congruence of the remaining sides.
66
Theorem 4.29. In hyperbolic geometry, if △ABC and △PQR have the same angle-sum then they
have the same area.
Proof. If the triangles have a pair of congruent edges, the previous result says we are done. Other-
wise, we use Lemma 4.27 to create a new triangle △LBC which matching the same Saccheri quadri-
lateral as △ABC.
WLOG suppose
|
AB
|
<
|
PQ
|
and construct the Saccheri quadrilateral with summit BC. Select K on
←→
EF such that
|
BK
|
=
1
2
|
PQ
|
and extend such that K is the midpoint of BL.
• By Lemma 4.27,
Area(△LBC) = Area(HICB) = Area(△ABC)
• By Theorem 4.26, △LBC has the same angle-sum as
△ABC and thus △PQR.
• △LBC and △PQR share a congruent side (LB
∼
=
PQ)
and have the same angle-sum. Lemma 4.28 says their
areas are equal.
A
B
C
D
E
F
G
H
I
L
K
Since both area and angle-defect are additive, we immediately conclude:
Corollary 4.30. The angle-defect of a hyperbolic triangle is an additive function of its area. By
normalizing the definition of area,
26
we may conclude that
π −Σ
△
= Area △
Note finally how the AAA congruence (Theorem 4.19, part 3) is related to the corollary:
△ABC
∼
=
△DEF
AAA
⇐⇒ angles congruent in pairs
⇓ ⇓
equal area
Cor 4.30
⇐⇒ same angle-defect
26
We have really only proved that π −Σ
△
is proportional to Area △. However, it can be seen that these quantities are
equal if we use the area measure arising naturally from the hyperbolic distance function (see page 80).
Corollary 4.30 is a special case of the famous Gauss–Bonnet theorem from differential geometry: for any triangle on a
surface with Gauss curvature K, we have
Σ
△
−π =
ZZ
△
K dA
We’ve now met all three special constant-curvature examples of this:
Euclidean space is flat (K = 0) so the angle-defect is always zero.
Hyperbolic space has constant negative curvature K = −1, whence
RR
△
dA = −(Σ
△
−π) is the angle-defect.
Spherical geometry A sphere of radius 1 has constant positive curvature K = 1 and
RR
△
dA is the angle-excess Σ
△
−π.
67
Example (4.11, cont). The isosceles right-triangle with vertices O, P = (
1
2
, 0) and Q = (0,
1
2
) has
angle-sum and area
π
2
+ 2 tan
−1
3
5
≈ 151.93° =⇒ area = π −
π
2
+ 2 tan
−1
3
5
=
π
2
−2 tan
−1
3
5
≈ 0.490
A Euclidean triangle with the same vertices has area
1
2
·
1
2
·
1
2
=
1
8
= 0.125.
Generalizing this (Exercise 4.2.4), the triangle with vertices O, P =
(c, 0) and Q = (0, c) has area
π −
π
2
+ 2 tan
−1
1 − c
2
1 + c
2
=
π
2
−2 tan
−1
1 − c
2
1 + c
2
As expected, lim
c→0
+
area(c) = 0. In the other limit, the triangle becomes
an omega-triangle with two omega-points and lim
c→1
−
area(c) =
π
2
: an
infinite ‘triangle’ with finite ‘area’!
O
The limit c → 1
−
Our discussion in fact provides an explicit method for cutting a triangle into sub-triangles and rear-
ranging its pieces to create a triangle with equal area.
A
B
C
L
K
S
1
S
3
△
1
△
3
P
R
Q
S
2
△
2
Suppose △
1
and △
2
have equal area and construct the quadrilaterals S
1
and S
2
. Let L, K be chosen
so that BL
∼
=
QR and K is the midpoint of BL. We now have:
• △
1
, △
2
, △
3
, S
1
, S
2
, S
3
have the same area.
• The summit angles of S
1
, S
2
, S
3
are congruent (half the angle-sum of each triangle).
• S
2
, S
3
are congruent since they have congruent summits and summit angles.
We can now follow the steps in Lemma 4.27 to transform △
1
to △
2
:
△
1
→ S
1
→ △
3
→ S
3
∼
=
S
2
→ △
2
where each arrow represents cutting off two triangles and moving them. Indeed this works even for
triangles in Euclidean geometry!
68
Exercises 4.5. 1. Use Corollary 4.30 to find the area of the hyperbolic triangle with given vertices.
Your answers to exercises from Section 4.2 should supply the angles!
(a) O = (0, 0), P = (
1
2
,
q
5
12
) and Q = (
1
2
, −
q
5
12
).
(b) O = (0, 0), P = (
1
4
, 0), Q = (0,
1
2
).
(c) P = (r, 0), Q =
−
r
2
,
√
3r
2
, R =
−
r
2
, −
√
3r
2
where 0 < r < 1.
2. In the proof of Theorem 4.29, explain why we can find K such that
|
BK
|
=
1
2
|
PQ
|
.
3. Show that there is no finite triangle in hyperbolic geometry that achieves the maximum area
bound π.
(Hard!) For a challenge, try to prove that omega-triangles also satisfy the angle-defect formula:
Area = π −Σ
△
, so that only triangles with three omega-points have maximum area.
4. Let Ω
1
, . . . , Ω
n
be n distinct omega-points arranged counter-clockwise around the boundary
circle of the Poincar
´
e disk. A region is bounded by the n hyperbolic lines
←−→
Ω
1
Ω
2
,
←−→
Ω
2
Ω
3
, . . . ,
←−−→
Ω
n
Ω
1
What is the area of the region? Hence argue that the ‘area’ of hyperbolic space is infinite.
5. An omega-triangle has vertices O = (0, 0), Ω = (1, 0) and P = (0, h) where h > 0.
(a) Prove that the hyperbolic segment PΩ is an arc of a circle with equation
(x −1)
2
+ (y − k)
2
= k
2
for some k > 0.
(b) Prove that the area of △OPΩ is given by
A(h) = sin
−1
2h
1 + h
2
6. Suppose two Saccheri quadrilaterals in hyperbolic geometry have the same area and congruent
summits. Prove that the quadrilaterals are congruent.
69
4.6 Isometries and Calculation
There are (at least!) two major issues in our approach to hyperbolic geometry.
Calculations are difficult In analytic (Euclidean) geometry we typically choose the origin and orient
axes to ease calculation. We’d like to do the same in hyperbolic geometry.
We assumed too much We defined distance, angle and line separately, but these concepts are not inde-
pendent! In Euclidean geometry, the distance function, or metric, defines angle measure via the
dot product,
27
and (with some calculus) the arc-length of any curve. One then proves that the
paths of shortest length (geodesics) are straight lines: the metric defines the notion of line!
Isometries provide a related remedy for these issues. To describe these it is helpful to use an alterna-
tive definition of the Poincar
´
e disk and its distance function.
Definition 4.31. The Poincar´e disk is the set D := {z ∈ C :
|
z
|
< 1}
equipped with the distance function
d(z, w) :=
ln
|
z − Ω
||
w − Θ
|
|
z − Θ
||
w − Ω
|
where Ω, Θ are the omega-points for the hyperbolic line through z, w
(defined as circular arcs intersecting the boundary perpendicularly).
z
w
Ω
Θ
We’ll see shortly (Corollary 4.38) that this is the same as the original cosh formula (page 54); it is
already easy to check that d(z, 0) = ln
1+
|
z
|
1−
|
z
|
as in Lemma 4.10 (if w = 0, then Ω, Θ = ±
z
|
z
|
).
For candidate isometries we need functions f : D → D for which d
f (z), f (w)
= d(z, w) . These
follow from some standard results of complex analysis that we state without proof.
Theorem 4.32 (M¨obius/fractional-linear transformations). If a, b, c, d ∈ C and ad − bc = 0, then the
function f (z) =
az+b
cz+d
has the following properties:
1. (Invertibility) f : C ∪{∞} → C ∪{∞} is bijective, with inverse f
−1
(z) =
dz−b
−cz+a
.
2. (Conformality) If curves intersect, then their images under f intersect at the same angle.
3. (Line/circle preservation) Every line/circle
28
is mapped by f to another line/circle.
4. (Cross-ratio preservation) Given distinct z
1
, z
2
, z
3
, z
4
, we have
f (z
1
) − f (z
2
)
f (z
3
) − f (z
4
)
f (z
2
) − f (z
3
)
f (z
4
) − f (z
1
)
=
(z
1
−z
2
)(z
3
−z
4
)
(z
2
−z
3
)(z
4
−z
1
)
27
Writing
|
u
|
=
|
PQ
|
for the length of a line segment, we see that for any u, v,
u · v =
1
2
|
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.
28
In C ∪ {∞} a line is just a circle containing ∞!
70
The isometries of the Poincar
´
e disk are a subset of the M
¨
obius transformations.
Theorem 4.33. The orientation-preserving
29
isometries of the Poincar
´
e disk have the form
f (z) = e
iθ
α − z
αz −1
where
|
α
|
< 1 and θ ∈ [0, 2π) (∗)
All isometries can be found by composing f with complex conjugation (reflection in the real axis).
Referring to the properties in Theorem 4.32:
1. The isometries are precisely the set of M
¨
obius transformations which map D bijectively to itself;
omega-points are also mapped to omega-points.
2. Isometries preserve angles.
3. 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. If Ω, Θ are the omega-points on
←→
zw, then (by 2 and 3), f (Ω) and f (Θ) are the omega-points for
the hyperbolic line through f (z), f (w). Preservation of the cross-ratio says 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 (∗) moves α to the origin, where calculating distances
and angles is easy!
Example 4.34. Let P =
1
2
and Q =
2
3
+
√
2
3
i. Move P to the origin using
an isometry
30
with α = P:
f (z) =
α − z
αz −1
=
1 − 2z
z −2
=⇒ f (P) = O
f (Q) =
1 −
4
3
−
2
√
2
3
i
2
3
−2 +
√
2
3
i
= −
1 + 2
√
2i
−4 +
√
2i
=
i
√
2
Let us compare distances:
P
Q
O = f (P)
f (Q)
f
d
f (P), f ( Q)
= ln
1 +
|
f (Q)
|
1 −
|
f (Q)
|
= ln
1 +
1
√
2
1 −
1
√
2
= ln
√
2 + 1
√
2 − 1
= ln(3 + 2
√
2) (Definition 4.31)
d(P, Q) = cosh
−1
1 +
2
|
PQ
|
2
(1 −
|
P
|
2
)(1 −
|
Q
|
2
)
!
= cosh
−1
1 +
2
4
(1 −
1
4
)(1 −
2
3
)
!
= cosh
−1
3 = ln(3 +
p
3
2
−1) = ln(3 + 2
√
2) = d
f (P), f ( Q)
If we trust the original cosh-formula (page 54), then the points really are the same distance apart!
Indeed the hyperbolic segment PQ has been transformed by f to a segment f (P) f (Q) of the y-axis.
29
If 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.’
30
We could also include a rotation (e
iθ
= −i) to move f (Q) to the positive x-axis, but there is no real benefit.
71
Recall (e.g., Example 4.11) how we previously computed angles. Isometries make this much easier.
Example 4.35. Given A = −
i
2
, B = −
i
5
and C =
1
5
(3 −i), we find d(A, B), d(A, C) and ∡BAC.
Start by moving A to the origin and consider f (B), f (C):
f (z) =
−
i
2
−z
i
2
z −1
=
2z + i
2 − iz
=⇒ f (B) =
−
2i
5
+ i
2 −
1
5
=
i
3
f (C) =
2
5
(3 −i) + i
2 −
i
5
(3 −i)
=
2( 3 −i) + 5i
10 − i(3 −i)
=
2 + i
3 − i
=
(2 + i)(3 + i)
10
=
1 + i
2
By mapping A to the origin, two sides of the triangle are now Eu-
clidean straight lines and the computations are easy:
d(A, B) = d
O, f (B)
= ln
1 +
1
3
1 −
1
3
= ln 2
d(A, C) = d
O, f (C)
= ln
1 +
1
√
2
1 −
1
√
2
= 2 ln(
√
2 + 1)
∡BAC = ∡ f (B) f (A) f (C) = arg
i
3
−arg
1 + i
2
=
π
2
−
π
4
=
π
4
To compute the final side and angles, isometries moving B and then
C to the origin could be used.
A
B
C
O = f (A)
f (B)
f (C)
Interpretation of Isometries (non-examinable)
As in Euclidean geometry, isometries can be interpreted as rotations, reflections and translations.
Here is the dictionary in hyperbolic space.
Translations Move α to the origin via T
−α
(z) =
α−z
αz−1
The picture shows repeated applications of T
−α
to seven initial points.
Compose these to translate α to β:
T
β
◦ T
−α
(z) =
( αβ −1)z + α − β
( α − β)z + αβ −1
Rotations R
θ
(z) = e
iθ
z rotates counter-clockwise around the origin. To ro-
tate around α, one computes the composition
T
α
◦ R
θ
◦ T
−α
The picture shows repeated rotation by 30° =
π
6
around α.
Reflections P
θ
(z) = e
2iθ
z reflects across the line making angle θ with the real
axis. Composition permits more general reflections, e.g.,
T
α
◦ P
θ
◦ T
−α
α
O
α
72
Hyperbolic Trigonometry
By employing isometries in the abstract, we develop expressions that allow us to solve triangles
directly in terms of the side-lengths and angle measures.
Given a right-triangle, we may suppose an isometry has already moved the right-angle to the origin
and the other sides to the positive axes as in the picture. The non-hypotenuse side-lengths are
a = ln
1 + p
1 − p
= cosh
−1
1 + p
2
1 − p
2
, b = cosh
−1
1 + q
2
1 − q
2
To measure the hypotenuse, translate p to the origin via an isometry
f (z) =
p − z
pz −1
=⇒ f (iq) =
p −iq
ipq −1
=⇒
|
f (iq)
|
2
=
p
2
+ q
2
p
2
q
2
+ 1
a
b
c
B
A
b
a
c
B
A
p
iq
0
f (p)
f (iq)
=⇒ cosh c =
1 +
|
f (iq)
|
2
1 −
|
f (iq)
|
2
=
1 + p
2
q
2
+ p
2
+ q
2
1 + p
2
q
2
− p
2
−q
2
=
1 + p
2
1 − p
2
·
1 + q
2
1 − q
2
= cosh a cosh b
Applying the hyperbolic identity sinh
2
b = cosh
2
b −1, we obtain
sinh b =
2q
1 − q
2
=⇒ tanh b =
sinh b
cosh b
=
2q
1 + q
2
Writing f (iq) in real and imaginary parts allows us to find the slope (that is, tan B):
f (iq) =
p −iq
ipq −1
=
−p(1 + q
2
) + iq(1 − p
2
)
p
2
q
2
+ 1
=⇒ tan B =
q(1 − p
2
)
p(1 + q
2
)
=
tanh b
sinh a
Trigonometric identities such as csc
2
B = 1 + cot
2
B quickly yield the other functions:
Theorem 4.36. In a hyperbolic right-triangle with adjacent a, opposite b, and hypotenuse c,
sin B =
sinh b
sinh c
cos B =
tanh a
tanh c
tan B =
tanh b
sinh a
cosh c = cosh a cosh b
This last is Pythagoras’ Theorem for hyperbolic right-triangles.
Pythagoras is easy to remember, as are the
opp
hyp
,
adj
hyp
,
opp
adj
patterns for the basic trig functions. Other-
wise, these expressions (and the next Corollary) are open-book—they are not worth memorizing.
Examples 4.37. 1. A right-triangle (as above) has a = cosh
−1
3 ≈ 1.76, b = cosh
−1
5 ≈ 2.29. Then,
cosh c = cosh a cosh b = 15 =⇒ c = cosh
−1
15 ≈ 3.40
sin A =
sinh a
sinh c
=
s
cosh
2
a −1
cosh
2
c −1
=
s
3
2
−1
15
2
−1
=
1
2
√
7
=⇒ A ≈ 10.9°
sin B =
sinh b
sinh c
=
s
cosh
2
b −1
cosh
2
c −1
=
s
5
2
−1
15
2
−1
=
r
3
28
=⇒ B ≈ 19.1°
We could use the other trig expressions: e.g., tan A =
tanh a
sinh b
=
sinh a
cosh a sinh b
=
√
8
3
√
24
=
1
3
√
3
. . .
73
2. Consider the pictured Lambert quadrilateral with side-lengths a, b, v, h and diagonal d. By the
sine and tangent formulæ,
sin β =
sinh b
sinh d
, cos β = sin(
π
2
− β) =
sinh h
sinh d
=⇒
tanh b
sinh a
= tan β =
sin β
cos β
=
sinh b
sinh h
=⇒ sinh h = sinh a cosh b
a
b
d
h
v
β
By doubling the quadrilateral, we obtain a Saccheri quadrilateral with base 2a, congruent sides
b, and summit 2h. Since cosh b > 1 and is strictly increasing, observe:
• The summit of a Saccheri quadrilateral is longer than its base.
• If the base 2a is fixed, the summit is a strictly increasing function of the side-length b.
The goal of trigonometry is to ‘solve’ triangles: given minimal numerical data, to compute the re-
maining sides and angles. As in Euclidean geometry, you can attack general problems by dropping
perpendiculars and using the results of Theorem 4.36, though it is helpful to generalize this by de-
veloping the sine and cosine rules.
Corollary 4.38 (Sine/Cosine rules and the Cosh-distance formula). Label a general triangle with
angle-measures A, B, C opposite sides with (hyperbolic) lengths a, b, c.
Sine Rule Drop a perpendicular from C and observe that sin A =
sinh h
sinh b
and
sin B =
sinh h
sinh a
. Eliminate sinh h to obtain the first equality in
sinh a
sin A
=
sinh b
sin B
=
sinh c
sin C
Drop a different altitude for the other equality.
a
b
c
h
B
A
Cosine Rule I Repeat the argument of Theorem 4.36 for a triangle with vertices 0, p and qe
iC
to obtain
cosh c = cosh a cosh b −sinh a sinh b cos C
Expressing the right side in terms of p, q (cosh a =
1+p
2
1−p
2
, etc.) yields the original cosh-formula
for distance (page 54).
Cosine Rule II Hyperbolic geometry admits a second version:
cos C = sin A sin B cosh c −cos A cos B
A proof is in Exercise 16.
In hyperbolic geometry, the triangle congruence theorems (SAS, ASA, SSS, SAA and AAA) pro-
vide suitable minimal data. The second version of the cosine rule has no analogue in Euclidean
geometry—it is particularly helpful for solving triangles given ASA or AAA data.
31
31
Unlike in Euclidean geometry, knowing two angles doesn’t automatically give you the third! For SAS and SSS start
with the cosine rule. SAA data is best solved by dropping a perpendicular and using Theorem 4.36.
74
Examples 4.39. 1. (SAS) An isosceles triangle has C =
π
3
and a = b = cosh
−1
2 ≈ 1.32. We have
sinh a = sinh b =
p
cosh
2
a −1 =
√
3. By the sine and cosine rules,
cosh c = 2 · 2 −
√
3
√
3 ·
1
2
=
5
2
=⇒ c = cosh
−1
5
2
≈ 1.57
sin B = sin A =
sin C sinh a
sinh c
=
√
3
2
·
√
3
√
21/4
=
3
√
21
=
r
3
7
=⇒ A = B ≈ 40.9°
cosh
−1
2
cosh
−1
2
c
A
B
π
3
2. (Equilateral AAA) An equilateral triangle has interior angles 30°. The second cosine rule says
cosh c =
cos A cos B + cos C
sin A sin B
=
√
3
2
·
√
3
2
+
√
3
2
1
2
·
1
2
= 3
√
3
=⇒ a = b = c = cosh
−1
(3
√
3) ≈ 2.33
a
b
c
30°
30°
30°
3. (Right-angled AAA) A triangle has angles
π
6
,
π
4
and
π
2
. Rather than using the second version
of the cosine rule, we indicate part of its proof by employing the tan-formula twice,
1
√
3
= tan
π
6
=
tanh a
sinh b
=
sinh a
cosh a sinh b
1 = tan
π
4
=
tanh b
sinh a
=
sinh b
sinh a cosh b
Multiply these together and apply hyperbolic Pythagoras,
a
c
b
π
4
π
6
1
√
3
=
1
cosh a cosh b
=
1
cosh c
=⇒ c = cosh
−1
√
3 = ln(
√
3 +
√
2) ≈ 1.15
Since sinh c =
p
cosh
2
c −1 =
√
2, the remaining sides are easy to compute:
1
√
2
= sin
π
4
=
sinh b
sinh c
=⇒ b = sinh
−1
1 = cosh
−1
√
2 ≈ 0.88
cosh a =
cosh c
cosh b
=
r
3
2
=⇒ a = cosh
−1
r
3
2
= sinh
−1
1
√
2
≈ 0.66
4. (ASA) Solve a triangle with angles
π
6
,
π
3
and a distance cosh
−1
3 between them.
Apply the second version of the cosine rule:
cos C = sin A sin B cosh c −cos A cos B
=
1
2
·
√
3
2
·3 −
√
3
2
·
1
2
=
√
3
2
=⇒ C =
π
6
a
b
cosh
−1
3
π
3
C
π
6
The triangle is isosceles, whence a = cosh
−1
3 ≈ 1.76 also. By the cosine rule,
cosh b = cosh a cosh c −sinh a sinh c cos B = 9 −
p
3
2
−1
2
1
2
= 5
=⇒ b = cosh
−1
5 = ln(5 +
√
24) ≈ 2.29
75
Hyperbolic Tilings (just for fun!)
Example 4.39.3 can be used to make a regular tiling of hyperbolic space. Take eight congruent copies
of the triangle and arrange them around the origin as in the first picture. Now reflect the quadrilateral
over each of its edges and repeat the process in all directions ad infinitum. The result is a regular tiling
of hyperbolic space comprising four-sided figures with six meeting at every vertex!
π
m
π
n
(m, n) = (4, 6) The fundamental triangle (m, n) = (5, 4)
In hyperbolic space, many different regular tilings are possible. Suppose such is to be made using
regular m-sided polygons, n of which are to meet at each vertex: each polygon comprises 2m copies
of the pictured fundamental right-triangle, whose angles must be
π
2
,
π
m
and
π
n
. Since the angles sum to
less than π radians, we see that there exists a regular tiling of hyperbolic space precisely when
π
2
+
π
m
+
π
n
< π ⇐⇒ (m −2)(n −2) > 4
The first example is m = 4 and n = 6. In the other picture tiling, n = 4 pentagons (m = 5) meet
at each vertex (the interior of each pentagon has been colored congruently for fun). This pentagonal
tiling was produced using the tools found here and here: have a play!
The multitude of possible tilings in hyperbolic geometry is in contrast to Euclidean geometry, where
a regular tiling requires equality
(m −2)(n −2) = 4
The three solutions (m, n) = (3, 6), (4, 4), (6, 3) correspond to the only tilings of Euclidean geome-
try by regular polygons (equilateral triangles, squares and regular hexagons); unlike in hyperbolic
geometry, Euclidean tilings may have arbitrary side-lengths.
For related fun, look up M.C. Escher’s Circle Limit artworks, some of which are based on hyperbolic
tilings. If you want an excuse to play video games while pretending to study geometry, have a look
at Hyper Rogue, which relies on (sometimes irregular) tilings of hyperbolic space.
Exercises 4.6. The questions marked ∗ require abstract calculations with complex algebra. Feel free
to skip these if your previous experience with this is minimal.
1. Use Definition 4.31 to prove that d(z, 0) = ln
1+
|
z
|
1−
|
z
|
.
(Hint: what are the omega-points for the line through 0 and z?)
76
2. (a) Use an isometry to find angle ∡ABC when A = 0, B =
i
2
, and C =
1+i
2
.
(b) Now compute ∡ACB, and thus find the angle sum and area of the triangle.
3. Find the area of each triangle in Examples 4.39.
4. * Identify a M
¨
obius transformation f (z) =
az+b
cz+d
with the matrix
a b
c d
. If g is another M
¨
obius
transformation, prove that the composition f ◦ g corresponds to the product of the matrices
related to f , g. Verify that f
−1
(z) =
dz−b
a−cz
corresponds to the inverse matrix.
32
5. (a) A triangle has vertices A =
1
3
, B =
1
2
and C, where ∡BAC = 45° and b = d(A, C) =
cosh
−1
3. Compute a = d(B, C) using the hyperbolic cosine rule.
(b) The isometry f (z) =
1−3z
z−3
moves A to the origin. What is f (B) and therefore f (C)?
(Hint: remember that f is orientation preserving)
(c) Use the inverse of the isometry f to compute the co-ordinates of C. As a sanity-check, use
the cosh distance formula to recover your answer to part (a).
6. * Suppose f (z) =
α−z
αz−1
for some constant α ∈ C with
|
α
|
= 1. If
|
z
|
= 1, prove that
|
f (z)
|
= 1.
Argue that the functions f in Theorem 4.33 really do map the interior of the unit disk to itself.
7. (a) * Show that the isometry T
β
◦ T
−α
which translates α to β (page 72) is the translation T
−γ
where γ =
β−α
αβ−1
followed by a rotation around the origin.
(b) * In what rare situations is the composition of two translations another (pure) translation?
8. Use the power series cosh x = 1 +
1
2
x
2
+
1
4!
x
4
+ ··· to expand the hyperbolic Pythagorean
theorem cosh c = cosh a cosh b to order 4 (a
4
, a
2
b
2
, etc.). What do you observe?
9. A hyperbolic right-triangle has non-hypotenuse sides a = cosh
−1
2 and b = cosh
−1
3. Find the
hypotenuse, the angles and the area of the triangle.
10. Given ASA data c = cosh
−1
(
√
2 +
√
3) , A =
π
4
, B =
π
6
, find the remaining data for the triangle.
11. An equilateral hyperbolic triangle has side-length a and angle A. Prove that cosh a =
cos A
1−cos A
.
If A = 45°, what is the side-length?
12. Find the interior angles and side-lengths for the quadrilateral and pentagonal tiles on page 76.
13. A railway comprises two rails (lines) which start perpendicular to a common sleeper (cross-
beam). Why would it be difficult to build a railway in hyperbolic geometry?
(Hint: consider Example 4.37.2)
14. * As suggested in Corollary 4.38, prove both the cosine rule and the cosh distance formula.
15. You are given isosceles ASA data: angles A = B and side c. Prove that cosh c ≤ 2 csc
2
A −1.
What happens when this is equality?
16. (a) Prove the second cosine rule when C =
π
2
(see the trick in Example 4.39.3).
(b) (Hard!) Prove the full version by dropping a perpendicular from B = B
1
+ B
2
and observ-
ing that
cos A
sin B
1
=
cos C
sin B
2
=
cos C
sin(B−B
1
)
···
77
The Poincar´e Disk for Differential Geometers (non-examinable)
Most of this last optional section should be accessible to anyone who’s taken basic vector-calculus.
All we really need is the Poincar
´
e disk model with its distance function d(z, w) and a description of
the isometries (Theorems 4.32, 4.33).
Consider the infinitesimally separated points z and z + dz. Map z to
the origin via an isometry
f : ξ 7→
z − ξ
zξ −1
Then z + dz is mapped to
P := f (z + dz) =
−dz
z(z + dz) −1
=
dz
1 −
|
z
|
2
z + dz
z + dw
z
0
P
Q
f
where we deleted z dz since it is infinitesimal compared to 1 −
|
z
|
2
.
Since isometries preserve length and angle, this construction has several consequences.
Infinitesimal distance, arc-length, and geodesics
The hyperbolic distance from z to z + dz is
d(z, z + dz) = d(O, P) = ln
1 +
|
P
|
1 −
|
P
|
= ln(1 +
|
P
|
) −ln(1 −
|
P
|
) = 2
|
P
|
=
2
|
dz
|
1 −
|
z(t)
|
2
(∗)
where the approximation ln(1 ±
|
P
|
) = ±
|
P
|
is used since
|
P
|
is infinitesimal. If z(t) parametrizes a
curve in the disk, then the infinitesimal distance formula allows us to compute its arc-length
Z
t
1
t
0
2
|
z
′
(t)
|
1 −
|
z(t)
|
2
dt
Example 4.40 (Circles and ‘hyperbolic π’). Suppose that a circle has hyperbolic radius δ. By moving
its center to the origin via an isometry, we may parametrize in the usual manner:
z(t) = r
cos θ
sin θ
, θ ∈ [0, 2π) where δ = ln
1 + r
1 − r
equivalently r =
e
δ
−1
e
δ
+ 1
Its circumference (hyperbolic arc-length) is then
Z
2π
0
2r
1 − r
2
dθ =
4πr
1 − r
2
= 2π sinh δ = 2π
δ +
1
3!
δ
3
+
1
4!
δ
5
+ ···
> 2πδ
where we used the Maclaurin series to compare.
A hyperbolic circle has a larger circumference : diameter ratio than for a Euclidean circle (π). More-
over, this ratio is not constant: one might say that the hyperbolic version of π is a function (
π sinh δ
δ
)!
32
Since multiplying a, b, c, d by a non-zero scalar doesn’t change f , the set of M
¨
obius transformations is isomorphic to the
projective special linear group PSL
2
(R). The orientation-preserving isometries of hyperbolic space form a proper subgroup.
78
Our arc-length integral approach also allows us to show that hyperbolic lines are really what we
want them to be: lines of shortest distance between points.
Theorem 4.41. The geodesics—paths of minimal length between two points—in the Poincar
´
e disk
are precisely the hyperbolic lines.
Following the comments on page 70, the distance function really does define the concept of hyper-
bolic line.
Proof. First suppose b lies on the positive x-axis. Parametrize a curve from 0 to b via
z(t) = x(t) + iy(t) where 0 ≤ t ≤ 1, z(0) = 0, z(1) = b
Its arc-length satisfies
Z
1
0
2
|
z
′
(t)
|
1 −
|
z(t)
|
2
dt =
Z
1
0
2
p
x
′2
+ y
′2
1 − x
2
−y
2
dt ≥
Z
1
0
2
|
x
′
|
1 − x
2
dt ≥
Z
1
0
2x
′
(t)
1 − x(t)
2
dt =
Z
b
0
2 dx
1 − x
2
= ln
1 + b
1 − b
= d(0, b)
where we have equality if and only if y(t) ≡ 0 and x(t) is increasing. The length-minimizing path is
therefore along the x-axis.
More generally, given points A, B, apply an isometry f such that f (A) = 0 and f (B) = b lies on the
positive x-axis. The geodesic from A to B is therefore the image of the segment 0b under the inverse
isometry f
−1
. By the properties of M
¨
obius transforms, this is an arc of a Euclidean circle through A, B
intersecting the unit circle at right-angles, our original definition of a hyperbolic line.
Area Computation
If dx and idy are infinitesimal horizontal and vertical changes in z = x + iy, then the area of the
infinitesimal rectangle spanned by z → z + dx and z → z + idy is the area element
dA =
2 dx
1 −
|
z
|
2
2 dy
1 −
|
z
|
2
=
4 dx dy
(1 − x
2
−y
2
)
2
The area of a region R in the Poincar
´
e disk is therefore given by the double integral
ZZ
R
4 dxdy
(1 − x
2
−y
2
)
2
=
ZZ
R
4r dr dθ
(1 −r
2
)
2
=
ZZ
R
sinh δ dδ dθ
where the last expression is written in polar co-ordinates using the hyperbolic distance δ. In this way
the measure of area also depends on the distance function.
Example (4.40, cont). The area of a hyperbolic circle with hyperbolic radius δ is
Z
2π
0
Z
δ
0
sinh δ dδ dθ = 2π(cosh δ −1) = π
δ
2
+
2
4!
δ
4
+
2
6!
δ
6
+ ···
> πδ
2
Again, this is larger than you’d expect in Euclidean geometry.
79
Angle Measure and the First Fundamental Form
If we repeat the distance translation (∗, page 78) for a second infinitesimal segment z → z + dw, it can
be checked that the angle between the original segments is precisely that between the infinitesimal
vectors dz and dw. This is precisely the conformality observation in Theorem 4.32 and moreover
shows how the distance function determines the angle measure.
If you’ve studied differential geometry, then a more formal way to think about this is to use the
first fundamental form or metric: essentially the dot product of infinitesimal tangent vectors. For the
Poincar
´
e disk model, (∗) says that this is
I =
4( dx
2
+ dy
2
)
(1 − x
2
−y
2
)
2
=
4( dr
2
+ r
2
dθ
2
)
(1 −r
2
)
2
Since this is a scalar multiple of the standard Euclidean metric dx
2
+ dy
2
, angle measures are identi-
cal.
33
It also gels with the fact that arc-length is the integral
Z
q
I
z
′
(t), z
′
(t)
dt
Using this language, two of the major theorems of introductory differential geometry quickly put a
couple of remaining issues to bed.
Gauss’ Theorem Egregium The first fundamental form determines the Gaussian curvature K. In this
case K = −1 is constant and negative, as you should easily be able to verify if you’ve studied
differential geometry.
Gauss–Bonnet Theorem The angle-sum Σ
△
of a geodesic triangle in a space with Gaussian curva-
ture K satisfies
Σ
△
−π =
ZZ
△
K
This establishes our earlier assertion that Area △ = π −Σ
△
(Corollary 4.30).
33
Recall that the angle ψ between vectors u, v satisfies u ·v =
|
u
||
v
|
cos ψ. For infinitesimal vectors we use I = λ
2
(dx
2
+
dy
2
) instead of the dot product, where λ =
2
1−r
2
. The resulting angle is the same as if we use the Euclidean metric
dx
2
+ dy
2
, since factors of λ
2
cancel on both sides.
80
