MINISTRY OF EDUCATION AND TRAINING
HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

HOANG MAI PHUONG

MODELS OF HYPERBOLIC GEOMETRY

BACHELOR THESIS
Major: Geometry

HANOI - 2019


MINISTRY OF EDUCATION AND TRAINING
HANOI PEDAGOGICAL UNIVERSITY N2

DEPARTMENT OF MATHEMATICS

HOANG MAI PHUONG

MODELS OF HYPERBOLIC GEOMETRY

BACHELOR THESIS
Major: Geometry

Supervisor:
Assoc. Prof. Dr. NGUYEN THAC DUNG

HANOI - 2019

Bachelor thesis

HOANG MAI PHUONG

Thesis Acknowledgement
I would like to express my gratitude to the teachers in the geometry group as well as all the teachers of the Department of Mathematics,
Hanoi Pedagogical University No.2. They have facilitated, supported me
during the time I have done the thesis.
In particular, I would like to express my deep respect and gratitude
to Assoc. Prof. Dr. Nguyen Thac Dung, who has directed guidance,
conscientiously counseled, helped me complete this thesis.
Due to limitation of time, capacity and conditions, so the thesis can
not avoid any mistakes. Thus, I am looking forward to receiving valuable
comments from teachers and friends.
Thank you so much.

Hanoi, May 5, 2019.
Student

Hoang Mai Phuong

i


Bachelor thesis

HOANG MAI PHUONG


Thesis Assurance
The thesis is completed after self-study and synthetic process with
the guidance of Assoc. Prof. Dr. Nguyen Thac Dung. It is written by
following the textbook entitled “Elementary Differential Geometry” by
Andrew Pressley. Most of the contents of this thesis are taken from the
chapter 11: “Hyperbolic geometry” in the mentioned textbook.
While the thesis is completed, I consulted some documents which have
been mentioned in the bibliography section.
I assure that this thesis is not copied from any other theses. I certify
that these statements are true and I will be responsible for their correctness.

Hanoi, May 5, 2019.
Student

Hoang Mai Phuong

ii


Contents

Preface

1

1 Preliminaries

2

1.1


Smooth surfaces and the first fundamental form . . . . .

2

1.2

The M¨obius transformation and the inversion . . . . . .

6

2 Some models of hyperbolic geometry

11

2.1

Upper half-plane model . . . . . . . . . . . . . . . . . . .

11

2.2

Isometries of H . . . . . . . . . . . . . . . . . . . . . . .

20

2.3

Poincar´e disc model . . . . . . . . . . . . . . . . . . . . .


27

2.4

Hyperbolic parallels . . . . . . . . . . . . . . . . . . . . .

35

2.5

Beltrami - Klein model . . . . . . . . . . . . . . . . . . .

40

Conclusion

50

Bibliography

51

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HOANG MAI PHUONG


Preface
1. Rationale
Hyperbolic Geometry as known as non - Euclidean Geometry, is an
important mathematical object. It is a geometry that discards one of
Euclid’s axioms and indicates that the geometry of space in which Euclid’s parallel axiom fails. Hyperbolic geometry is closely connected to
many other parts mathematics such as: differntial geometry, complex
analysis, topology, etc. . . So, It has attracted many international mathematical researchers and gained many significant achievements. They
have focused on not only researching some axioms, postulates but also
exploiting the special models of Hyperbolic Geometry and using its important properties to solve the problems. Desiring to study more about
the properties, some related problems about models in Hyperbolic Geometry and its importance, I chosen the topic:“Models of Hyperbolic
geometry” as the subject of undergraduate thesis.
2. Aims of the study and Research questions
Starting acquainted with the scientific research and learn more about
the models of hyperbolic geometry.
3. Research methods
- Researching on textbook and materials related to problem’s reseach
- Analysis and synthesis of knowledge.
4. Main contents
Chapter 1: “Preliminaries”
Chapter 2: “Some models of hyperbolic geometry”

1


Chapter 1
Preliminaries
1.1

Smooth surfaces and the first fundamental form


Definition 1.1.1. A subset S of R3 is a surface if, for every point p ∈ S,
there is an open set U in R2 and an open set W in R3 containing p such
that S ∩ W is homeomorphic to U . A subset of a surface S of the form
S ∩ W , where W is an open subset of R3 , is called an open subset of
S. A homeomorphism σ : U → S ∩ W as in this definition is called a
surface patch or parametrization of the open subset S ∩ W of S.
Definition 1.1.2. A surface patch σ : U → R3 is called regular if it
smooth and the vectors σu and σv are linearly independent at all points
(u, v) ∈ U . If S is a surface, an allowable patch for S is a regular surface
patch σ : U → R3 such that σ is a homeomorphism from U to an open
subset of S. A smooth surface is a surface S such that, for any point
p ∈ S, there is an allowable surface patch σ as above such that p ∈ σ(U ).
Example 1.1.2.1. The plane is given by formula:
σ (u, v) = a + up + vq
is a smooth surface. It’s clearly smooth and σu = p and σv = q are
linearly independent because p and q were chosen to be perpendicular
unit vectors.
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Definition 1.1.3. A tangent vector to a smooth surface S at a point
p ∈ S is the tangent vector at p of a smooth curve in S passing through
p. The tangent space Tp S of S at p is the set of all tangent vectors to S
at p.
Definition 1.1.4. Let p be a point of a smooth surface S. The first
fundamental form of S at p associates to tangent vectors v, w ∈ Tp S the

scalar
v, w

p,S

= v · w.

Example 1.1.4.1. Suppose that σ(u, v) is a surface patch of S. Then,
any tangent vector to S at a point p in the image of σ can be expressed
uniquely as a linear combination of σu and σv . Define maps du : Tp S → R
and dv : Tp S → R by
du(v) = λ

dv(v) = µ

if v = λσu + µσv ,

for some λ, µ ∈ R. We can see that du and dv are linear maps. Then,
using the fact that , is a symmetric bilinear form, we have
v, v = λ2 σu , σv + 2λµ σu , σv + µ2 σu , σv .
Writing
E = σu 2 ,

F = σu · σ v ,

G = σv 2 ,

this becomes
v, v = Eλ2 + 2F λµ + Gµ2 = Edu(v)2 + 2F du(v)dv(v) + Gdv(v)2
Traditionally, the expression

Edu2 + 2F dudv + Gdv 2
is called the first fundamental form of the surface patch σ(u, v).
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Example 1.1.4.2. For the plane
σ (u, v) = a + up + vq
with p and q being perpendicular unit vectors, we have σu = p, σv = q,
so E = σu

2

= p

2

= 1, F = σu · σv = p · q = 0, G = σv

2

= q

2

= 1.


Hence, the first fundamental form is du2 + dv 2 .
Definition 1.1.5. If S1 and S2 are surfaces, a conformal map f : S1 →
S2 is a local diffeomorphism such that, if γ1 and γ˜1 are any two curves
on S1 that intersect, say at a point p ∈ S1 , and if γ2 and γ˜2 are their
images under f , the angle of intersection of γ1 and γ˜1 at p is equal to
the angle of intersection of γ2 and γ˜2 at f (p).
Proposition 1.1.6. A local diffeomorphism f : S1 → S2 is conformal if
and only if, for any surface patch σ of S1 , the first fundamental forms
of the patches σ of S1 and f ◦ σ of S2 are proportional.
In particular, a surface patch σ(u, v) is conformal if and only if its
first fundamental form is λ(du2 + dv 2 ) for some smooth function λ(u, v).
Example 1.1.6.1. The catenoid can be parametrized by
σ(u, v) = ((cosh u) cos v, (cosh u) sin v, u)
is conformal.
Indeed, now we will find the first fundamental form of σ(u, v).
We have:
σu = (sinh u cos v, sinh u sin v, 1)
σv = (− cosh u sin v, cosh u cos v, 0) .
These imply,
E = σu

2

= cosh2 u, F = 0, G = σv
4

2

= cosh2 u.



Bachelor thesis

HOANG MAI PHUONG

So, the first fundamental form is
cosh2 u(du2 + dv 2 ).
By Proposition 1.1.6, σ(u, v) is conformal.
Example 1.1.6.2. We consider the unit sphere S 2 . If q is any point of
S 2 other than the north pole n = (0, 0, 1), the straight line joining n and
q intersects the xy-plane at some point p. The map that takes q to p is
called stereographic projection from S 2 to the plane, and we denote it by
Π. We are going to show that Π is conformal.
Let p = (u, v, 0), q = (x, y, z). Since, p, q, n lie on a straight line, there
is a scalar ρ such that
q − n = ρ(p − n),
and hence
(x, y, z) = (0, 0, 1) + ρ((u, v, 0) − (0, 0, 1)) = (ρu, ρv, 1 − ρ).
Hence, ρ = 1 − z, u =

(1.1)

x
y
,v =
and we have
1−z
1−z

Π(x, y, z) =


x
y
,
,0 .
1−z 1−z

On the other hand, from Eq. 1.1 and x2 + y 2 + z 2 = 1 we get ρ =
2
and hence
u2 + v 2 + 1
q=

2u
2v
u2 + v 2 − 1
,
,
.
u2 + v 2 + 1 u2 + v 2 + 1 u2 + v 2 + 1

If we denote the right-hand side by σ1 (u, v), then σ1 is a parametrization of S 2 with the north pole removed. Parametrizing the xy-plane by
σ2 (u, v) = (u, v, 0), we then have
Π(σ1 (u, v)) = σ2 (u, v).
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The first fundamental form of σ2 is du2 + dv 2 . As to σ1 , we get
(σ1 )u =
(σ1 )v =

2 v 2 − u2 + 1
(u2 + v 2 + 1)

2

,

−4uv
4u
,
2
(u2 + v 2 + 1) (u2 + v 2 + 1)2

2 u2 − v 2 + 1
−4uv
4v
,
,
(u2 + v 2 + 1)2 (u2 + v 2 + 1)2 (u2 + v 2 + 1)2

This gives
E1 = (σ1 )u · (σ1 )v =

2


4 v 2 − u2 + 1

+ 16u2 v 2 + 16u2

(u2 + v 2 + 1)

4

=

4
.
(u2 + v 2 + 1)2

Similarly,
F1 = 0,

G1 =

4
.
(u2 + v 2 + 1)2

We can see that, the first fundamental form of σ2 is λ times that of σ1
1 2
2
where λ =
u + v 2 + 1 . Therefore, Π is conformal.
4
Theorem 1.1.7. (Green’s theorem) Let C be a positively oriented simple

closed curve in a plane and let D be the region bounded by C. Let f (x, y)
and g(x, y) be defined on an open region containing D and be smooth
functions (i.e., functions with continous partial derivatives of all orders).
Then,
D

∂g ∂f
dxdy =
−
∂x ∂y

f (x, y)dx + g(x, y)dy.
C

Definition 1.1.8. A curve ϕ on a surface S is called a geodesic if ϕ(t)
¨
is zero or perpendicular to the tangent plane of the surface at the point
ϕ(t)
¨ i.e., parallel to its unit normal, for all values of the parameter t.

1.2

The M¨
obius transformation and the inversion

Definition 1.2.1. A M¨obius transformation is a map of the form:
M (z) =
6

az + b

cz + d


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HOANG MAI PHUONG

where a, b, c, d ∈ C and ad − bc = 0
The simplest examples of M¨obius transformation are the following:
Translation Ta (z) = z + a, a ∈ C
Complex dilations Da(z) = az, a ∈ C, a = 0
1
z
Proposition 1.2.2. Every M¨obius transformation is a composite of M¨obius
The map K(z) =

transformation of the above three types.
Proof. If c = 0 then a = 0 and
M (z) =

az + b a
b
z+
=
d
d
a

= Da Kd T b (z).
a


If c = 0,
a (ad − bc)/c
−
= Ta/c D−(ad−bc)/c K(cz + d)
c
cz + d
= Ta/c D−(ad−bc)/c KTd Dc (z).

M (z) =

The proof is complete.
It is clear geometrically that translations and complex dilations take
straight lines to straight lines and circles to circles.
Proposition 1.2.3. Every Circle in C∞ can be described by an equation
of the form
az z¯ + ¯bz + b¯
z+c=0

(1.2)

where a, c ∈ R, b ∈ C. In fact, that equation represents a straight line if
a = 0 and b = 0, and a circle if a = 0 and |b|2 > ac.
Proof. Any straight line in the xy-plane has equation of the form
px + qy + r = 0,
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where p, q, r ∈ R and p, q are not both zero. Writing z = x + iy, this
is of the form (1.2) with a = 0, b = p + iq, c = 2r. On the other hand,
every circle has equation of the form
(x + p)2 + (y + q)2 = r2 ,
where p, q ∈ R and r > 0. This is of the form (1.2) with a = 1, b =
p + iq, c = p2 + q 2 − r2 .
Proposition 1.2.4. Every M¨obius transformation takes Circles to Circles.
Proof. This is clear geometrically for translations and complex dilations.
By Proposition 1.2.2, it is sufficient to prove for the map K. If w = K(z)
1
then z = and so if z lies on the circle (1.2) then
w
¯b
a
b
+
+
+ c = 0,
|w|2 w w
c|w|2 + ¯bw¯ + bw + a = 0.
If c = 0, this is the equation of a line; if c = 0, it is the equation of a
2
circle since ¯b > ca.
Note that this proof actually shows that K takes circles passing
through the origin to lines and all other circles to circles.
Definition 1.2.5. A conjugate M¨obius transformation is a map of the
form:
z −→


a¯
z+b
c¯
z+d

where a, b, c, d ∈ C and ad − bc = 0
Proposition 1.2.6. Conjugate-M¨obius transformations take Circles to
Circles.
Proof. This follows from Proposition 1.11 and the fact that C, being the
reflection on the real axis and takes lines to lines and circles to circles.
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Definition 1.2.7. The inversion Ia,r in the circle Ca,r with centre a ∈ C
and radius r > 0 is
Ia,r (z) = a +

r2
z¯ − a
¯

Proposition 1.2.8. The inversion Ia,r takes a circle C to itself if and
only if C intersects Ca,r perpendicularly. In that case, Ia,r takes each of
the two regions into which C divides the plane to themselves.
Proof. Since Ia,r interchanges the interior and exterior of Ca,r , it is clear

that if Ia,r fixes C then C and Ca,r must intersect. Suppose that C is
a circle that intersects Ca,r at points P , Q. Then Ia,r interchanges the
segments of C that are inside and outside Ca,r , respectively. Since Ia,r is
conformal, the angles made by the interior and exterior segments of C
with Ca,r at P must be equal. Since the sum of these angles is π, hence
Ca,r and C must intersect perpendicularly at P .

Conversely, assume that C is a circle that intersects Ca,r at right angles,
at P , Q. Then Ia,r takes C to a circle that intersects Ca,r at right angles at
P , Q. But it is clear that there is a unique circle C with these properties
(for the centre of the circle must be the intersection of the tangent lines
to Ca,r at P ,Q)
Assume that Ia,r takes C to itself, and let C has centre b and radius
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s. Since Ca,r and C intersect perpendicularly, |b − a|2 = r2 + s2 . Hence,
s2
s2
Ia,r (b) = b − ¯
⇔ |Ia,r (b) − b| =
|b − a|
b−a
¯
because s < |b − a|. Thus, Ia,r ∈ D, the interior of C. It follows that

Ia,r takes every point c ∈ D to a point of D.
Suppose that Ia,r (c) is outside C. Now, Ia,r takes the line segment
joining b and c to a (smooth) curve γ joining the point Ia,r (c) inside C
to the point Ia,r (c) outside C, so γ must intersect C at a point d. Since
Ia,r takes C to itself, e = Ia,r (d) ∈ C. But since d is on γ, e is a point of
the line segment with endpoints b, c. This is a contradiction. It follows
that Ia,r takes D to itself, and since Ia,r is a bijection (indeed, it is equal
to its own inverse), it must take the region exterior C to itself.

10


Chapter 2
Some models of hyperbolic
geometry
2.1

Upper half-plane model

The upper half-plane is the set of complex numbers with positive imaginary parts. It is denoted by:
H = (v, w) ∈ R2 |w > 0 = {z ∈ C|Im(z) > 0} .
The upper half-plane is equipped with the first fundamental form:
dv 2 + dw2
.
w2
The half-plane along with that first fundamental form composed a model
which has name “Upper half-plane model”.
Proposition 2.1.1. Hyperbolic angles in H are the same as Euclidean
angles.
Proof. As we know, a surface patch σ(v, w) is conformal if and only if its

first fundamental form is λ(dv 2 + dw2 ) for some smooth function λ(v, w).
dv 2 + dw2
Moreover, the first fundamental form of H is
. Then, σ(u, v)
w2
is conformal. The proof is complete.
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The hyperbolic lines are the geodesics in H. By a result in chapter
9 in [1], we can determine all geodesics in H. Indeed, we now have
proposition:
Proposition 2.1.2. The geodesics in H are the half-lines parallel to
imaginary axis and the semicircles with centres on the real axis.
We have some simple properties of hyperbolic lines:
Proposition 2.1.3.
(i). There is a unique hyperbolic line passing through any two distinct
points of H.
(ii). The parallel axiom does not hold in H.
Proof.
(i). Let a, b be two points in H. There are two cases to consider.
Case 1: Suppose Re(a) = Re(b). Then the Euclide line L =
{z ∈ C|Re(z) = Re(a)} passes through a and b and is parallel to
imaginary axis. By the uniqueness of the Euclidean line, we see
that l = H ∩ L is the unique hyperbolic line passing through a, b
is the half-line containg a, b.

Case 2: Suppose Re(a) = Re(b). As the Euclidean line through
a and b is no longer perpendicular to the real axis, we need to
construct a Euclidean circle centered on the real axis and passing
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HOANG MAI PHUONG

through the two points. Let Lab be the Euclidean line segment
joining a and b and let K be the perpendicular bisector of Lab .
Every Euclidean circle that passes through a, b has its center on
K. As a, b have non-equal real parts, the Euclidean line K is not
parallel to the real axis, so K intersects the real axis at a unique
point c and |a − c| = |b − c|. Then let C be a Euclidean circle
centered at c with radius |a − c| = |b − c|. So l = H ∩ C is the
unique hyperbolic line passing through a, b is the semicircle with
centre c and radius |a − c| = |b − c|.

(ii). There are two cases to consider.
Case 1: Suppose that l is contained in a Euclidean line L and a
be any point not on l in H. As a is not in L, a is on Euclidean K
that is parallel to L. Since L is perpendicular to real axis, then K
is perpendicular to real axis and K ∩H is a hyperbolic line through
a and parallel to l.

13

Bachelor thesis

HOANG MAI PHUONG

Now, we construct another hyperbolic line by taking a point x ∈
real axis and let n be the hyperbolic line through a and x. Because
Re(x) = Re(a), so K and n are distinct. And we can construct
infinitely hyperbolic lines because there are many infinitely points
on real axis between L and K.
Case 2: Suppose that l is contained in a Euclidean circle M . As
a is in H and not on l, there is a Euclidean circle N through a
and concentric to M . As concentric circle are disjoint, N ∩ H is a
hyperbolic line through a and parallel to l. To construct another
hyperbolic line, take any point b on real axis between M and N .
Let p be the hyperbolic line through a and b. Then p is disjoint
from l. As there are infinitely many points between M and N , so
we can construct infinitely many hyperbolic lines through a and
parallel to l.

The proof is complete.
There is a unique hyperbolic line passing through any two points
a, b ∈ R. It defines the hyperbolic distance between a and b to be the
length of hyperbolic line segment joining a, b.
Definition 2.1.4. The hyperbolic distance, d(z1 , z2 ), where zj = xj +iyj ,
t2

x (t)2 + y (t)2

is given by the infimum of

t1

y(t)

with ϕ(tj ) = zj = x(tj ) + iy(tj ).

14

dt taken over all paths ϕ(t)


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Proposition 2.1.5. The hyperbolic distance between two points a, b ∈ H
is
dH (a, b) = 2tanh−1

|b − a|
|b − a
¯|

where a
¯ is the complex conjugate of the complex number a.
Proof. There are two cases. It depends on whether the hyperbolic line
passing through a and b is a half-line or a semicircle.
Case 1: The hyperbolic line joing a, b is a half-line.
Suppose that a, b lie on a half-line geodesic. Then,
a = r + is, b = r + it (r, s, t ∈ R, t > s)

By Definition 2.4, we have:
t

dw
t
= ln
w
s

dH (a, b) =

= d. So,

t
= ed
s

s

In the other hand,
−1 |t

2tanh

− s|
= 2tanh−1
|t + s|

ed − 1
ed + 1

= 2tanh−1 tanh

d
2

= d.

Case 2: The hyperbolic line passing through a, b is a semicircle.
Suppose that a, b lie on semicircle with the centre c (c ∈ real axis)
The semicircle can be parametrixed by v = c + r cos θ; w = r sin θ. Then,
ψ

dH (a, b) =

dv
dθ

2

+

2

dw
dθ

ψ

dθ =

w2
ϕ

ϕ

ψ

dθ
θ
= ln tan
sin θ
2

=
ϕ

r2 sin2 θ + r2 cos2 θ
dθ
r2 sin2 θ

ψ
ϕ

15

ψ
2
= ln
ϕ.

tan
2
tan


Bachelor thesis

HOANG MAI PHUONG

where ϕ = arg(a − c), ψ = arg(b − c), (d is independent of the radius)

Now, we have:
ψ
2 −1
ϕ
ψ
ψ
tan
tan − tan
d ed − 1
2
2
2
tanh = d
=
=
ψ
ψ
ϕ
2

e +1
tan
tan + tan
2 +1
2
2
ψ
tan
2
ϕ
ϕ
ψ
ψ
ψ−ϕ
sin cos − sin cos
sin
2
2
2
2 =
2 .
=
ψ
ϕ
ϕ
ψ
ψ+ϕ
sin cos + sin cos
sin
2

2
2
2
2
tan

(2.1)

Moreover, b (r cos ψ, r sin ψ), a (r cos ϕ, r sin ϕ). Then,
|b − a|2 = r2 ((cos ψ − cos ϕ)2 + (sinψ − sin ϕ)2 )
ψ−ϕ
.
= 2r2 (1 − cos(ψ − ϕ)) = 4r2 sin2
2

(2.2)

Similarly,
|b − a
¯|2 = 4r2 sin2

16

ψ+ϕ
.
2

(2.3)

Bachelor thesis

HOANG MAI PHUONG

From (2.1), (2.2), (2.3), we get:
tanh

d |b − a|
=
.
2 |b − a
¯|

This implies
d = 2tanh−1

|b − a|
.
|b − a
¯|

We are done.
The last is the area of a hyperbolic polygon (i.e., polygon whose sides
are hyperbolic lines).
Theorem 2.1.6. Let P be a n-sides hyperbolic polygon in H with internal angles α1 , α2 , . . . , αn . Then, the area of hyperbolic polygon is
A(P) = (n − 2)π − α1 − α2 − ... − αn .

Proof. Let α1 , α2 , . . . , αn be the vertices of P and C is its boundary,
consisting of n hyperbolic line segments a1 a2 , a2 a3 , . . . , an a1 (α1 is the
internal angle of P at the vertex a1 ).

Since, the first fundamental form is
dv 2 + dw2
,
w2
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and the area of P is
P

dvdw
.
w2

Using Green’s theorem, we have
pdv + qdw =
C

P

∂p
∂q
−
dvdw,
∂v ∂w

where p, q are smooth function of (v, w)
1
Taking p = and q = 0 gives
w

P

dvdw
=
w2

C

dv
.
w

(2.4)

To evaluate integral (2.4), we will prove that the following lemma:
Lemma 2.1.7. Let a, b be two endpoints of a line segment of a hyperbolic
line in H such that the forms of part of a semicircle with centre p on the
real axis and suppose that radius vector joining p to a and p to b makes
angles ϕ and ψ, respectively, with the positive real axis. Then,
dv
= ϕ − ψ.
l w
Proof. We parametrize the hyperbolic line by v = r cos θ, w = r sin θ,
where r is the radius of the semicircle. Then,
ψ

ψ

−r sin θdθ
=−
r sin θ

dv
=
w
ϕ

ψ

ϕ

dθ = ϕ − ψ.
ϕ

We are done.
Returning to proof Theorem 2.1.6:
Let ϕi and ψi be the angles are defined in Lemma 2.1.7 that corresponding to the side ai ai+1 , (i = 1, n).

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Bachelor thesis

HOANG MAI PHUONG

From (2.4) and Lemma 2.1.7, we have

P

n

dvdw
=
w2

(ϕi − ψi ).

(2.5)

i=1

Now, we consider the change in direction of outward-pointing normal
of P as we traverse its boundary in a anticlockwise direction. As we
traverse the side with end points ai , ai+1 , the outward normal rotates
anticlockwise through an angle ψi − ϕi . In the other hand, at the vertex
ai , it rotates by π − αi . Thus, if we traverse the boundary of P, the
outward normal rotates through an angle
n

(ψi − ϕi − ai ).

nπ +
i=1

But this angle of rotation is 2π, so we have:

n

(ψi − ϕi − ai ).

2π = nπ +
i=1

And then,
n

n

(ϕi − ψi ) = (n − 2)π −
i=1

ai
i=1

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(2.6)


Bachelor thesis

HOANG MAI PHUONG

From (2.4), (2.5), (2.6), we obtained:

P

dvdw
= (n − 2)π −
w2

n

ai .
i=1

The proof is complete.
In particular, for a triangle with angles α, β, ϕ, the area is π−α−β−ϕ.
Otherwise, the sum of the angles of a Euclidean triangle with straight
line sides: π = α + β + ϕ. The area of a triangle on the unit sphere with
geodesic sides: α + β + ϕ − π.

2.2

Isometries of H

We now turn to the geometry description of isometries of H. There are
four simplest types as follows:
(i). Translations parallel to the real axis Ta (z), given by:
Ta (z) = z + a, a ∈ R.
(ii). Reflextion through a line parallel to imaginary axis, given by:
Ra (z) = 2a − z, a ∈ R.
Ra (z) is the reflextion of z in the line Re(z) = a, thought of as a
mirror; each point of this line is fixed by R + a.
(iii). Dilations by a factor a > 0, given by:
Da (z) = az.

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