ma243 geometry revision guide...2 euclidean geometry we start with our classical euclidean geometry....
TRANSCRIPT
2
MA243
Geometry
Revision Guide
Written by Itamar Aharoni
Contents
1 Preface 3
2 Euclidean Geometry 4
Definition: Line Segment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Definition: Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Definition: Collinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Definition: Collinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Definition: Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
Example: Discrete Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Definition: Metric Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Definition: Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Example: Dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Definition: Metric Vector Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Example: Metric Vector Space and Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . 5
Definition: Induced Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Example: Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
Lemma: Cauchy Schwartz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Definition: Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Distance Preserving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Isometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
Definition: Affine Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
Theorem: Normal Form of an orthogonal matrix . . . . . . . . . . . . . . . . . . . . . . . . . 16
Definition: Hyperplane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Definition: Reflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Theorem: Sum of Angles in A Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Proposition: Pons Asinorum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
Proposition: Parallel Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Definition: Group Homomorphism and Isomorphisms . . . . . . . . . . . . . . . . . . . . . . . 26
Definition: Semi-direct Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
Example: Semi-direct Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Spherical Geometry 28
Definition: Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Definition: Great Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Definition: Tangent Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Definition: Spherical Angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Definition: Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Definition: Spherical Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
1
2
Theorem: Sum of Angles of a Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4 Hyperbolic Geometry 34
Definition: Lorentzian Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Definition: Lorentzian Norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Definition: Hyperbolic Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Definition: Hyperbolic sine and cosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
Definition: The likes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
Proposition: Cauchy Schwartz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Definition: Light-cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Definition: Lorentzian Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Definition: Hyperbolic Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Definition: Lorentzian cross . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Definition: Hyperbolic distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
Definition: Lorentz transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Definition: Lorentz orthonormal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Example: Lorentz orthonormal basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
Definition: Lorentz group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
Definition: Positive Lorentz group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Theorem: Characterization of Isom(H1, dHn
). . . . . . . . . . . . . . . . . . . . . . . . . . 46
Theorem: Intersection of Hyperbolic Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
Example: A family of lines in H2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Definition: Lorentz normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Definition: Hyperbolic Angle V1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Definition: Hyperbolic Angle V2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
Definition: Orthogonal Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Theorem: Investigation of Line Intersection . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Lemma: Norm of cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
Theorem: Sum of angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
5 Projective Geometry 55
Definition: Projective Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Definition: Projective Space Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Definition: Projective subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Definition: Projective line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
Definition: Projective Span . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Definition: Linear independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Definition: Projective transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Definition: Group of Projective transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Definition: Projective frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
Example: standard projective frame . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
Proposition: Bijection PGLn+1 (R) to action on basis . . . . . . . . . . . . . . . . . . . . . . . 58
3
Lemma: Classification of Projective Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Definition: Projective Triangle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
Definition: Introspective Triangles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Theorem: Desargues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
Theorem: Pappo’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61
6 Axiomatic Projection Geometry 63
Definition: Axiomatic Projection Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
Definition: Division Ring (Skew Field) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
Definition: Projective plane of a Division ring . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
Theorem: Hilbert . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
1 Preface
In the module, we are introduced to 5 different types of Geometries - with the goal for you to understand the
differences between them. We start with our beloved Euclidean Geometry, and from it expand to Spherical,
Hyperbolic and finally Projective and Axiomatic Projection Geometry. In each of these, we first explore the
basic definitions - what are lines, angles, triangles etc. Followed by that, we prove simple theorems in each.
The underlying goal in the module is for you to appreciate the differences between each geometry. While
reading, note how in each our classical results such as the sum of angles in a triangle or the cosine rule
suddenly differ. The following guide forms a condensed summary of the definitions and theorems in the
module. While I tried to make is as complete as possible, it is by no means an official document and
it is to be used at your own risk.
This revision guide for ST111 Probability A has been designed as an aid to revision, not a substitute for it.
This guide is useful for revising through definitions, theorems and proofs found in the course.
Disclaimer: Use at your own risk. No guarantee is made that this revision guide is accurate or complete,
or that it will improve your exam performance.
Authors
Written by Itamar E. Aharoni.
Any corrections or improvements should be reported by email to [email protected].
4
2 Euclidean Geometry
We start with our classical Euclidean Geometry. First, we begin with establishing the notions of a line and
line segment.
Definition 2.1: Line Segment.
The line segment between two vectors x, y ∈ Rn is defined as
L = {x+ λ (y − x) : λ ∈ [0, 1]} (1)
Definition 2.2: Line.
Let u, v ∈ Rn. The line starting at u in the direction of v is
L = {u+ λv : λ ∈ R} ⊆ Rn
With that in mind, we continue to define the notion of collinearity. I.e. what does it mean for three points
to be linearly dependent, or to lie on the same line.
Definition 2.3: Collinearity.
We say that the vectors x, y, z ∈ Rn are collinear if there exits a line L s.t. x, y, z ∈ L
Trivially, if three points are collinear then we can place one of them in the line segment between the other
two. Hence, if {x+ λ (y − x)} is the segment between the two, the third must lie in it and so
∃λ s.t. z = x+ λ (y − x)
and so, we arrive at the alternative definition
Definition 2.4: Collinearity.
We say that the vectors x, y, z ∈ Rn are collinear if there exits a real λ ∈ R s.t.
z = x+ λ (y − x) (2)
We now continue to define distance, or in our new language - a Metric. This will be a generalization of our
notion of distance in Euclidean geometry.
Definition 2.5: Metric.
A metric on a set X is a function d (x, y)→[0,∞) s.t.
1. d (x, y) = 0⇔ x = y for all x, y ∈ X
2. d (x, y) = d (y, x) for all x, y ∈ X
3. d (x, y) ≤ d (x, z) + d (z, y) for all x, y, z ∈ X (triangle inequality)
5
Example 2.1: Discrete Metric.
Let X be a set, the function d (x, y) =
0 x = y
1 x 6= yIs a metric
Definition 2.6: Metric Space.
A Metric Space is a pair of the form (X, d) where X is a set and d is a metric on it
Recall how in Euclidean geometry the distance between points x and y can be calculated using their dot
product
d (x, y) = ‖x− y‖ =√
(x− y) · (x− y)
Hence, before we define our euclidean distance using the Metric terminology - we first generalize the notion
of a dot product.
Definition 2.7: Scalar Product.
A Scalar Product on a vector space V is a bilinear form 〈 , 〉 : V → R s.t.
1. 〈v, w〉 = 〈w, v〉 ∀v, w ∈ V
2. 〈v, v〉 ≥ 0 ∀v ∈ V
3. If 〈v, w〉 = 0 for all w ∈ V then v = 0
Example 2.2: Dot product.
Our familiar dot product ”·” is a scalar product on Rn.
Definition 2.8: Metric Vector Space.
A pair of the form (V, 〈 , 〉) where V is a vector space and 〈 , 〉 is a scalar product is called a Metric Vector
Space
Example 2.3: Metric Vector Space and Scalar Product.
Taking V = Rn, we have that the standard dot product is a scalar product, and V together with the dot
product forms a metric space.
Definition 2.9: Induced Norm.
Given a metric vector space, the norm | | : V → R is defined as
‖v‖ =√〈v, v〉
Remark. . We can also use norms to define metrics, as seen in the following examples
Example 2.4.
Let X = Rn , the function d (x, y) = ‖x− y‖ is a metric.
Example 2.5: Metric.
Let X = R , the function d (x, y) = |x− y| is a metric.
6
Proof. Parts 1 and 2 of the definition follow directly. Let us prove part 3.
Let x, y, z ∈ R
d (x, y) = |x− y| (3)
= |x− z + z − y| (4)
≤ |x− z|+ |z − y| (5)
= d (x, z) + d (z, y) (6)
And so, it fulfills all three conditions of our definition and is a metric.
The following property, called the Cauchy Schwartz Inequality will be used repeatedly throughout the module.
It is incredibly useful in future proofs and assignments, and so you should make yourself familiar with it. Its
first and most important use for us follows as a corollary after the Lemma.
Lemma 2.1: Cauchy Schwartz Inequality.
For all vectors x, y ∈ Rn holds
〈x, y〉2 ≤ ‖x‖2 ‖y‖2
Proof. Let x, y ∈ Rn and assume y 6= 01. By non-negativity for all z holds
〈x, y〉2
‖y‖2≤ 〈x, y〉
2
‖y‖2+ ‖z‖2 (7)
Choose z = x− 〈x,y〉y‖y‖2 , then we have that
〈x, y〉2
‖y‖2≤ 〈x, y〉
2
‖y‖2+
∥∥∥∥∥x− 〈x, y〉y‖y‖2
∥∥∥∥∥2
(8)
Let us now use the linearity of our inner product. Using it, and the definition of the euclidean norm, we get∥∥∥∥∥x− 〈x, y〉y‖y‖2
∥∥∥∥∥2
= 〈x− 〈x, y〉y‖y‖2
, x− 〈x, y〉y‖y‖2
〉 (9)
= 〈x− 〈x, y〉y‖y‖2
, x〉 − 〈x− 〈x, y〉y‖y‖2
,〈x, y〉y‖y‖2
〉 (10)
= 〈x, x〉 − 〈x, 〈x, y〉y‖y‖2
〉 − 〈x, 〈x, y〉y‖y‖2
〉+ 〈 〈x, y〉y‖y‖2
,〈x, y〉y‖y‖2
〉 (11)
= ‖x‖2 +
∥∥∥∥∥ 〈x, y〉y‖y‖2
∥∥∥∥∥2
− 2〈x, 〈x, y〉y‖y‖2
〉 (12)
1Otherwise 0 ≤ 0 and we’re done
7
We can now input the result above into equation (8) to yield
〈x, y〉2
‖y‖2≤ 〈x, y〉
2
‖y‖2+ ‖x‖2 +
‖〈x, y〉y‖2
‖y‖4− 2〈x, 〈x, y〉y〉‖y‖2
(13)
≤ 〈x, y〉2
‖y‖2+ ‖x‖2 + 〈x, y〉2 ‖y‖
�2
‖y‖�4− 2�
��〈x, y〉���〈x, y〉‖y‖2
(14)
≤ 〈x, y〉2
‖y‖2+ ‖x‖2 +
〈x, y〉2
‖y‖2− 2〈x, y〉2
‖y‖2(15)
≤ ‖x‖2 (16)
Meaning we received that
〈x, y〉2
‖y‖2≤ ‖x‖2 (17)
Or
〈x, y〉2 ≤ ‖x‖2 ‖y‖2 (18)
Corollary 2.2.
The Euclidean Metric on Rn also fulfills the triangle inequality
Proof. let x, y, z ∈ Rn, by definition
d (x, y) = ‖x− y‖2 = ‖x− z + z − y‖2 (19)
= 〈x− z + z − y, x− z + z − y〉 (20)
= 〈x− z + z − y, x− z〉+ 〈x− z + z − y, z − y〉 (21)
= 〈x− z, x− z〉+ 〈z − y, x− z〉+ 〈x− z, z − y〉+ 〈z − y, z − y〉 (22)
= ‖x− z‖2 + ‖z − y‖2 + 2〈x− z, z − y〉 (23)
By Cauchy Schwartz
‖x− z‖2 + ‖z − y‖2 + 2〈x− z, z − y〉 ≤ ‖x− z‖2 + ‖z − y‖2 + 2 ‖x− z‖ ‖z − y‖ (24)
≤ (‖x− z‖+ ‖z − y‖)2(25)
≤ (d (x, z) + d (y, z))2
(26)
And so,
d (x, y) ≤ d (x, z) + d (y, z) (27)
With the metric defined, we can define angles and triangles.
8
Definition 2.10: Angle.
Let x, y, z ∈ Rn, the angle between x and y and at the z, denoted θ = 〈x, y, z〉, is defined as
cos (θ) =〈x− y, z − y〉‖x− y‖ ‖z − y‖
, θ ∈ [0, π]
Definition 2.11: Triangle.
Let A,B,C ∈ Rn be three distinct points. The Triangle between them, denoted 4ABC, is defined as
4ABC = [A,B] ∪ [B,C] ∪ [C,A] (28)
Now let us develop our tools to examine the space. Our goal will be to develop sufficient tools to prove
basic theorems such as the sum of angles in triangles. We begin with the notion of an isometry. That is, a
function that simply shifts or rotates the space - but keeps its structure unchanged.
Definition 2.12: Distance Preserving.
Let (X, d) , (X ′, d′) be two metric spaces. A map f : X → X ′ is said to be distance preserving if
d (x, y) = d (f(x), f(y)) ∀x, y ∈ X
Recall the definition of isometry in Introduction to Geometry
Definition 2.13: Isometry.
A map is an isometry if it is both distance preserving and bijective
And so, we can finally define what it means for a space to be a Euclidean Space like the one we know and
love.
Definition 2.14: Euclidean Space.
A Euclidean Space is a metric space which admits an isometry (X, d) ∼= (Rn, d)
Definition 2.15: Motion.
A motion is an isometry from Rn to itself. I.e. an isometry T : (Rn, d)→ (Rn, d′)
We now show that motions take lines to lines. To this avail, we start with the following proposition
Proposition 2.3.
Three vectors x, y, z ∈ Rn are collinear if and if and only if
d (x, y) = d (x, z) + d (z, y)
up to permutation of x, y, z
Proof. We divide the proof into two parts, for each of the implications.
9
Suppose x, y, z are collinear, we wish to prove that d (x, y) = d (x, z) + d (z, y).
As they are collinear, all three lie on the same line segment. One of them must be between the latter two,
without loss of generality suppose z lies between x and y. We can write
z = x+ λ (y − x) , λ ∈ [0, 1] (29)
Hence
d (x, z) + d (z, y) = ‖x− z‖+ ‖z − y‖ (30)
= ‖x− [x+ λ (y − x)]‖+ ‖[x+ λ (y − x)]− y‖ (31)
= ‖�x− �x− λ (y − x)‖+ ‖x+ λ (y − x)−y‖ (32)
= ‖λ (y − x)‖+ ‖λ (y − x) + (x− y)‖ (33)
= ‖λ (y − x)‖+ ‖(1− λ) (y − x)‖ (34)
= �λ ‖y − x‖+����(1− λ) ‖y − x‖ (35)
= ‖y − x‖ = d (x, y) (36)
In summary
d (x, z) + d (z, y) = ‖y − x‖ = d (x, y) (37)
Now Suppose that d (x, z) + d (z, y) = d (x, y), we wish to show the three are collinear. Let us square both
sides
d (x, y)2
= (d (x, z) + d (z, y))2
(38)
‖x− y‖2 = (‖x− z‖+ ‖z − y‖)2(39)
‖x− y‖2 = ‖x− z‖2 + ‖z − y‖2 + 2 ‖x− z‖ ‖z − y‖ (40)
‖x− z + z − y‖2 = ‖x− z‖2 + ‖z − y‖2 + 2 ‖x− z‖ ‖z − y‖ (41)
By linearity of the inner product we have that
‖x− z + z − y‖2 = 〈x− z + z − y, x− z + z − y〉 (42)
= 〈x− z, x− z + z − y〉+ 〈z − y, x− z + z − y〉 (43)
= 〈x− z, x− z〉+ 〈x− z, z − y〉+ 〈z − y, x− z〉+ 〈z − y, z − y〉 (44)
= ‖x− z‖2 + 2〈z − y, x− z〉+ ‖z − y‖2 (45)
And so, inputting the above result into equation (41), we get that
����‖x− z‖2 +���
�‖z − y‖2 + 2〈x− z, y − z〉 =����‖x− z‖2 +���
�‖z − y‖2 + 2 ‖x− z‖ ‖z − y‖ (46)
〈x− z, y − z〉 = ‖x− z‖ ‖z − y‖ (47)
Thus, here the Cauchy Schwartz inequality is an equality. Let us denote
u = x− z , v = y − z (48)
10
Under the new names, we have
〈u, v〉2 = ‖u‖2 ‖v‖2 (49)
Let us now calculate both sides of our equality. The right hand sides gives us
‖u‖2 ‖v‖2 =
(n∑i=1
u2i
)(n∑i=1
v2i
)=
n∑i,j=1
v2i u
2i (50)
and the left hand side
〈u, v〉2 =
(n∑k=1
uivi
)2
=
(n∑k=1
ukvk
)(n∑l=1
ulvl
)=
n∑k,l=1
ukvkulvl (51)
and so, we get that
n∑k,l=1
ukvkulvl =
n∑i,j=1
v2i u
2i (52)
We can now rewrite both sumsn∑
i,j=1
v2i u
2i =
n∑i=j=1
v2i u
2i +
n∑i<jj=1
v2i u
2j + u2
jv2i (53)
n∑k,l=1
ukvkulvl =
n∑k=l=1
ukvkulvl +
n∑k<ll=1
ukvkulvl + ulvlukvk =
n∑k=l=1
u2ku
2l +
n∑k<ll=1
2ukvkulvl (54)
And so, combining equations (51), (53) and (54) we have
�����
n∑i=j=1
v2i u
2i +
n∑i<jj=1
v2i u
2j + u2
jv2i =
�����n∑
k=l=1
u2ku
2l +
n∑k<ll=1
2ukvkulvl (55)
n∑i<jj=1
v2i u
2j + u2
jv2i =
n∑k<ll=1
2ukvkulvl (56)
0 =
n∑i<jj=1
v2i u
2j + u2
jv2i −
n∑k<ll=1
2ukvkulvl (57)
0 =
n∑i<jj=1
v2i u
2j + u2
jv2i − 2uiviujvj (58)
0 =
n∑i<jj=1
[(uivj)− (ujvi)]2
(59)
And as each term is non-negative, and the whole sum equals zero, we have that
(uivj)− (ujvi) = 0 ∀i, j (60)
(uivj) = (ujvi) ∀i, j (61)
11
Lastly, as u 6= 02 there must exists i s.t. ui 6= 0. Then
vj = ujviui
(62)
And denoting λ = viui
we have
vj = λuj (63)
Or, as our equality runs for all j
v = λu (64)
Lastly, returning to our definitions for v, u in (48)
z − y = λ (x− z) (65)
y = z + λ (z − x) (66)
and the three are collinear. We can now prove our desired result about motions.
Proposition 2.4.
A motion T preserves collinearity, meaning it sends lines to lines.
More precisely,
T [(1− λ)x+ λy] = (1− λ)T (x) + λT (y)
For all x, y ∈ Rn and λ ∈ R. Meaning it also leaves the segment ratio between the points unchanged.
Proof. First, let us show that it takes collinear points to collinear points.
Let x, y, z ∈ Rn be collinear. By proposition 2.3 we have that
d (x, y) = d (x, z) + d (z, y) (67)
As T is an isometry
d (T (x), T (y)) = d (T (x), T (z)) + d (T (z), T (y)) (68)
And so, again by proposition 2.3, we have that T (x), T (y), T (z) are collinear.
Let us now prove the stronger claim in our theorem, the precise equation.
Assume z is a point between x and y
z = (1− λ)x+ λy , z ∈ [0, 1] (69)
by the argument above T (x) , T (y) and T (z) are collinear and ∃λ′ ∈ [0, 1] s.t.
T (z) = (1− λ′)T (x) + λ′T (y) (70)
2otherwise x = z and the points are trivially collinear
12
As T preserves the midpoint. Using this, we want to show that λ = λ′.
As T is an isometry
d (x, z) = ‖x− z‖ = ‖x− (1− λ)x− λy‖ = ‖x− x− λx− λy‖ = ‖λ (x− y)‖ = λ ‖x− y‖ = λd (x, y) (71)
|| (72)
d (T (x), T (z)) = ‖T (x)− T (z)‖ = ‖T (x)− (1− λ′)T (x)− λ′T (y)‖ = λ′d (T (x), T (y)) = λ′d (x, y) (73)
Meaning
λd (x, y) = λ′d (x, y) (74)
λ = λ′ (75)
In conclusion, we have that
T [(1− λ)x+ λy] = (1− λ)T (x) + λT (y) (76)
Remark. . In particular T (u+ v) = T (u) + T (v). You can see this by taking u = 2x, v = 2y, λ = 1/2.
Splendid! Let us explore several more properties of isometries. Our first goal will be to find the general form
of distance preserving maps and isometries. Both are closely related to Linear Transformations from Linear
Algebra.
Definition 2.16: Affine Map.
A map f : Rn → Rk is affine if there exists some vector b ∈ Rk and a linear map L : Rn → Rk s.t.
f = L+ b
Remark. . Trivially we have that f(0) = b and L = f − b
We now show that Distance Preserving Maps are affine. To this avail, we prove the following theorem.
Proposition 2.5.
The following are equivalent for maps f : Rn → Rk:
1. f is affine
2. f (λx+ µy)− f(0) = λ [f(x)− f(0)] + µ [f(y)− f(0)] for all x, y ∈ Rn and λ, µ ∈ R
3. f ((1− λ)x+ λy) = (1− λ) f(x) + λf(y) for all x, y ∈ Rn and λ ∈ R
Remark. . Condition two says exactly that L := f(x)− f(y) is linear
Proof. We want to show all three parts are equivalent. To this end, we show pairwise equivalence.
• Two → One:
Set L = f(x)− f(0). It is linear and so the condition holds
13
• Two → Three:
Clear, as three is a special case of two where we set λ = 1− µ
• Three → Two:
To prove this, first let us first note that
λx+ µy =
(1− 1
2
)(2λx) +
1
2(2µy) (77)
2λx = (1− 2λ) · 0 + 2λ (x) (78)
2µy = (1− 2µ) · 0 + 2µ (y) (79)
(80)
With these at hand, let us evaluate f (λx+ µy). By property three and using (77)
f (λx+ µy) =
(1− 1
2
)f (2λx) +
1
2f (2µy) (81)
f (λx+ µy) =1
2f (2λx) +
1
2f (2µy) (82)
Again by property 3, now using (78) and (79)
f (λx+ µy) =1
2[(1− 2λ) f (0) + 2λf (x)] +
1
2[(1− 2µ) f (0) + 2µf (y)] (83)
=1
2[f(0) + 2λ (f(x)− f(0))] +
1
2[f(0) + 2µ (f(y)− f(0))] (84)
= f(0) + λ (f(x)− f(0)) + µ (f(y)− f(0)) (85)
Which is exactly condition two.
Corollary 2.6.
Every distance preserving map is affine.
Proof. In prop. 2.4 we showed that distance preserving maps satisfy part (3) of our proposition. Hence they
are affine.
Splendid, we now continue to describing the general form of isometries. First recall a result from Algebra II
Theorem 2.7.
A matrix A ∈Mn×n (R) orthogonal3 if and only if
‖Av‖ = ‖v‖ , ∀v ∈ Rn
Proof. Proved in the module Linear Algebra II: Advanced Linear Algebra.
3Meaning AT = A
14
Proposition 2.8.
Let L : Rn → Rn be linear with L(x) = Ax, A ∈Mn×n(R). The following are equivalent
1. L is a motion
2. ‖L(x)‖ = ‖x‖ for all x ∈ Rn
3. A is an orthogonal matrix
Proof.
• One → Two:
By definition of norm we have that for all x ∈ Rn
‖x‖ = ‖x− 0‖ = d (x, 0) (86)
And in particular
‖L(x)‖ = d (L(x), 0) (87)
As L is a linear map L (0) = 0 and so
d (L(x), 0) = d (L(x), L(0)) (88)
Further, we assumed L is a motion and so
d (L(x), L(0)) = d (x, 0) = ‖x‖ (89)
Meaning
‖L(x)‖ = ‖x‖ (90)
• Two → One:
let x, y ∈ Rn, by definition of distance in Rn
d (L(x), L(y)) = ‖L(x)− L(y)‖ (91)
By linearity of L
‖L(x)− L(y)‖ = ‖L (x− y)‖ (92)
By property two
‖L (x− y)‖ = ‖x− y‖ = d (x, y) (93)
Meaning, we found that
d (L(x), L(y)) = d (x, y) (94)
15
• Three → Two:
First note that if A is the representative matrix of L under the standard basis then
AAT = I ⇔ 〈L(ei), L(ej)〉 =
0 i 6= j
1 i = j(95)
By definition of the standard basis
〈ei, ej〉 =
0 i 6= j
1 i = j(96)
and so, combining the two we have that A is orthogonal if and only if
〈L(ei), L(ej)〉 = 〈ei, ej〉 ∀i, j (97)
And by linearity of the inner product, this means that A is orthogonal if and only if
〈x, y〉 = 〈L(x), L(y)〉 (98)
And so, assuming three holds - we can plug x = y and two will holds as well
• Two → Three:
Note that
〈x, y〉 =1
4‖x+ y‖2 − 1
4‖x− y‖2 (99)
And so, assuming two holds then by linearity
〈L(x), L(y)〉 =1
4‖L(x+ y)‖2 − 1
4‖L(x− y)‖2 =
1
4‖x+ y‖2 − 1
4‖x− y‖2 = 〈x, y〉 (100)
and by the note in the previous bullet point A must be orthogonal.
With this at hand, we can finally describe the general form of an isometry.
Corollary 2.9.
Every motion T : Rn → Rn is of the form T (x) = Ax+ b where b ∈ Rn and A ∈Mn×n (R) is an orthogonal
matrix
Proof. By prop. 2.6 a motion is of the form T (x) = Ax + b, and by the proposition above the matrix A
must be orthogonal.
Remark. . We can strengthen our description by providing more detail on the structure of our matrix A, as
seen in the proof below.
16
Theorem 2.10: Normal Form of an orthogonal matrix.
Let L : Rn → Rn be a motion. Then there is an orthonormal basis of Rn under which the representative
matrix of L is of the form
Ik | 0
− | − 0
0 | −IhB1 0 0
0 0. . . 0
0 0 Bl
, Where Bi =
(cos (θi) − sin (θi)
sin (θi) cos (θi)
)(101)
with 0 ≤ k, l, h , k + h+ 2l = n
Proof. Let us extend L to J : Cn → Cn, where now J(x+ iy) = J(x) + iJ(y). From MA106 we know that
the eigenvalues of J are the roots of the polynomial
cJ (x) = det [J − xI] (102)
Since cJ(x) has only real coefficients, if λ is a root, so is λ̄. let λ be an eigenvalue of J . By proposition 2.8
|λ| ‖x‖ = ‖λx‖ = ‖L(x)‖ = ‖x‖ (103)
|λ| ‖x‖ = ‖x‖ (104)
and so, |λ| = 1. By linear algebra, there exists a basis of eigenvectors in Cn
B = {v1, . . . , vk, w1, . . . , wh, u1, u′1, . . . , µl, µ
′l} (105)
where
• v1, . . . , vk are eigenvectors for the eigenvalue 1
• w1, . . . , wh are eigenvectors for the eigenvalue −1
• u, u′ are eigenvectors of complex conjugate pairs
Under this basis, the representative matrix of J is
[J ] =
Ik | 0
− | − 0
0 | −Ihλ1 0 0 0 0
0 0 λ̄1 0 0 0
0 0. . . 0
0 0 0 λl 0
0 0 0 0 λ̄l
(106)
17
Moreover, we can assume the basis is orthonormal as we can always apply the Gram Schmidt algorithm
orthonormalize it. Let v, w be two eigenvectors of J . By proposition 2.8
〈v, w〉 = 〈J(v), J(w)〉 = 〈λv, µw〉 (107)
By the definition of the inner product in Cn
〈λv, µw〉 = λµ̄〈v, w〉 (108)
Hence if 〈v, w〉 6= 0 then
〈v, w〉 = λµ̄〈v, w〉 (109)
λµ̄ = 1 (110)
µ̄ = λ−1 (111)
But as |λ| = 1 we have that λ−1 = λ̄. Therefore
µ̄ = λ̄ (112)
µ = λ (113)
Meaning if λ 6= µ then 〈v, w〉 = 0 and the two vectors must be orthogonal. Hence the eigen-spaces of each
eigenvalue are orthogonal.
Let us now evaluate the matrix
A =
(λ 0
0 λ̄
)(114)
where λ = cos (θ) + i sin (θ). Let z = (x+ iy) ∈ Cn be an eigenvector of this matrix where x, y ∈ Rn. By
definition of eigenvectors
Az = λz (115)
A (x+ iy) = λ (x+ iy) (116)
Ax+ i [Ay] = [cos (θ) + i sin (θ)] (x+ iy) (117)
Ax+ i [Ay] = x cos (θ) + iy cos (θ) + ix sin (θ)− y sin (θ) (118)
Ax+ i [Ay] = [x cos (θ)− y sin (θ)] + i [x sin (θ) + y cos (θ)] (119)
Therefore, as x, y ∈ Rn we must have
Ax = x cos (θ)− y sin (θ) (120)
Ay = x sin (θ) + y cos (θ) (121)
Further, note that A is an orthogonal matrix. Hence by proposition 2.8
‖x‖2 = ‖Ax‖2 (122)
= ‖x cos (θ)− y sin (θ)‖2 (123)
= cos2 (θ) ‖x‖2 + sin2 (θ) ‖y‖2 − 2 cos (θ) sin (θ) 〈x, y〉 (124)
18
and through re-arranging we get that[1− cos2 (θ)
]‖x‖2 = sin2 (θ) ‖y‖2 − 2 cos (θ) sin (θ) 〈x, y〉 (125)
sin2 (θ) ‖x‖2 = sin2 (θ) ‖y‖2 − 2 cos (θ) sin (θ) 〈x, y〉 (126)
sin (θ) ‖x‖2 = sin (θ) ‖y‖2 − 2 cos (θ) 〈x, y〉 (127)
Further, again using proposition 2.8 and linearity of the inner product
〈x, y〉 = 〈Ax,Ay〉 = 〈x cos (θ)− y sin (θ) , x sin (θ) + y cos (θ)〉 (128)
= cos (θ) sin (θ) ‖x‖2 + cos2 (θ) 〈x, y〉 − sin2 (θ) 〈x, y〉 − sin (θ) cos (θ) ‖y‖2 (129)
= cos (θ)[sin (θ) ‖x‖2
]+ cos2 (θ) 〈x, y〉 − sin2 (θ) 〈x, y〉 − sin (θ) cos (θ) ‖y‖2 (130)
We can now replace sin (θ) ‖x‖2 with its value from (127)
〈x, y〉 = cos (θ)[sin (θ) ‖y‖2 − 2 cos (θ) 〈x, y〉
]+ cos2 (θ) 〈x, y〉 − sin2 (θ) 〈x, y〉 − sin (θ) cos (θ) ‖y‖2 (131)
=(((((((
(cos (θ) sin (θ) ‖y‖2(((((
(((−2 cos2 (θ) 〈x, y〉(((((((
+ cos2 (θ) 〈x, y〉 − sin2 (θ) 〈x, y〉 −((((((((
sin (θ) cos (θ) ‖y‖2 (132)
= − cos2 (θ) 〈x, y〉 − sin2 (θ) 〈x, y〉 (133)
= −(cos2 (θ) + sin2 (θ)
)〈x, y〉 (134)
= −〈x, y〉 (135)
In conclusion, we have that
〈x, y〉 = −〈x, y〉 (136)
〈x, y〉 = 0 (137)
And so, x is orthogonal to y. Therefore, the two form an orthogonal basis for Cλ ⊕Cλ̄ and{
x‖x‖ ,
y‖y‖
}is an
orthonormal basis of it. Under this basis, A becomes
A′ =
(cos (θ) sin (θ)
− sin (θ) cos (θ)
)(138)
Therefore, we can now divide the representative matrix we found in (106) into sub-matrices - grouping λi
and λ̄i together.
[J ] =
Ik | 0
− | −0 | −Ih
λ1 0
0 λ̄1
λ2 0
0 λ̄2
. . .
λl 0
0 λ̄l
(139)
19
And for each of the small matrices we can replace the two eigenvectors representing it by the corresponding
x, y from above to get
Ik | 0
− | − 0
0 | −IhB1 0 0
0 0. . . 0
0 0 Bl
, Where Bi =
(cos (θi) − sin (θi)
− sin (θi) cos (θi)
)(140)
Remark. . Observe the following
• Note that Bi = I if θi = 0 and Bi = −I if θ = π
• we do not allow θi = 0, instead we choose h maximal
• We always choose θi = π where possible, so we have k minimal
And so, we have proved that every isometry is of the form Ax + b, where A is of the form above. Terrific!
In particular, we now focus our attention to a special type of isometries - reflection by a plane. Essentially,
we are mirroring the entire space, with our plane being the mirror. We do so as they will provide us with
another way to describe our motions.
Definition 2.17: Hyperplane.
A hyperplane of Rn is an affine subspace of dimension n− 1 of the form
Π = V + b
where V is a vector subspace of Rn of dimension n− 1, and b ∈ Rn.
Definition 2.18: Reflection.
Let Π be a hyperplane, its corresponding reflection ρΠ : Rn → Rn is defined as the unique isometry s.t.
ρΠ (x) = x , ∀x ∈ Π
ρπ 6= id
meaning it keeps the hyper-plane the same, and does not alter any points on it.
Remark. . In assignment 2 we find the general form of ρΠ, which is
ρΠ (x) =
x if x ∈ Π
−x if x /∈ Π
Armed with our new definition, we can actually use it to provide an alternative description of motions.
Essentially, the following theorem states that every motion can be viewed as repetitive mirroring by different
planes.
20
Theorem 2.11.
Every isometry of Rn is the composition of at most n+ 1 reflections.
In particular reflections generate all isometries on Rn.
Proof. Let T : Rn → Rn be an isometry, written as T (x) = Ax+ b.
First let us examine its fixed space
Fix (T ) := {x ∈ Rn : T (x) = x} (141)
and assume it is not empty. Take x0 ∈ Fix (T ), we wish to show that Fix (T ) = (Fix (T )− x0) + x0 is an
affine subspace. In other words, we wish to show that (Fix (T )− x0) is a vector space.
let x − x0, y − x0 ∈ (Fix (T )− x0) and λ, µ ∈ R. As T is an isometry it can be written as T (x) = Ax + b.
Hence by direct computation we get
T [λ (x− x0) + µ (y − x0) + x0] = A [λ (x− x0) + µ (y − x0) + x0] + b (142)
= λA (x− x0) + µA (y − x0) +Ax0 + b (143)
= λ (Ax−Ax0) + µ (Ay0 −Ax0) +Ax0 + b (144)
Note that T (x) = x as x ∈ Fix (T ) . Ergo Ax+ b = x or equivalently Ax = x− b, with the same holding for
y and x0. We can input these in equation (144)
T [λ (x− x0) + µ (y − x0) + x0] = λ[(x��−b
)−(x0��−b
)]+ µ
[(y0��−b
)−(x0��−b
)]+(x0��−b
)��+b (145)
= λ (x− x0) + µ (y − x0) + x0 (146)
Therefore
λ (x− x0) + µ (y − x0) + x0 ∈ Fix (T ) (147)
λ (x− x0) + µ (y − x0) ∈ (Fix (T )− x0) (148)
Meaning (Fix (T )− x0) is a vector subspace, and Fix (T ) is an affine subspace.
In conclusion, we have that either Fix (T ) is empty, or it is an affine subspace. We now move to prove a
stronger claim - from which this theorem will follow.
Remark. . We now continue to an even stronger version of the theorem. From which this particular case
follows.
Theorem 2.12.
If T is an isometry on (Rn, d) with dim [Fix (T )] = n− l, then T is the composition of at most l+1 reflections
for 0 ≤ l ≤ n.
Remark. . The theorem above follows, using the fact that l is at most n.
Proof. by induction on l
21
• l = 0:
If l = 0 then dim [Fix (T )] = n and so Fix (T ) = Rn. Meaning T is the identity map which is a
composition of 0 reflections.
• Inductive Step:
As dimension of Fix (T ) is n− l ≤ n− 1 there exists p ∈ Rn s.t. p /∈ Fix (T ). Let Π be the hyperplane
orthogonal to the segment [p, T (p)] through its midpoint, and ρ the reflection by Π. We now consider
ρ ◦ T .
First note that by construction ρ (T (p)) = p, meaning p ∈ Fix (ρ ◦ T ). We now wish to show that
Fix (T ) ⊆ Fix (ρ ◦ T ).
Let x ∈ Fix (T ), as T is an isometry
d (p, x) = d (T (p), T (x)) = d (T (p) , x) (149)
By definition, Π is the set of points equidistant from p and T (p) and so x ∈ Π. Hence ρ (x) = x and
ρ (T (x)) = ρ (x) = x =⇒ x ∈ Fix (ρ ◦ T ) (150)
=⇒ Fix (T ) ⊆ Fix (ρ ◦ T ) (151)
Combining Fix (T ) ⊆ Fix (ρ ◦ T ) and p ∈ Fix (ρ ◦ T ) we have
Fix (T ) t {p} ⊆ Fix (ρ ◦ T ) (152)
Therefore
dim (Fix (ρ ◦ T )) > dim (Fix (T )) = n− l (153)
dim (Fix (ρ ◦ T )) ≥ n− (l − 1) (154)
And by the induction assumption ρ ◦ T is a composition of k reflection with 0 ≤ k ≤ (l − 1) + 1
ρ ◦ T = ρ1 ◦ ρ2 ◦ . . . ◦ ρk (155)
T = ρ ◦ ρ1 ◦ ρ2 ◦ . . . ◦ ρk (156)
Yay ! we have found that T is a composition of at most l + 1 reflections and by induction our proof
holds for all l ∈ N.
In the particular case of R2, we can be even more specific in our description. This will be our last theorem
- before starting our journey in proving the geometric results.
Theorem 2.13.
Every isometry of R2 is either
• A composition of a translation and rotation by angle theta, where the rotation is given by the matrix(cos (θ) − sin (θ)
sin (θ) cos (θ)
)(157)
22
• A composition of a translation and reflection
Proof. Let T = Ax + b be a motion on R2, we need to show it is either a reflection or rotation. First note
that if {v1, v2} is an orthonormal basis of R2 then there is a rotation Rθ s.t. either
Rθ
(1
0
)= v1 , Rθ
(0
1
)= v2 (158)
Or
Rθ
(1
0
)= v2 , Rθ
(0
1
)= v1 (159)
By the normal form theorem, our isometry T can be expressed as T = Rθ−1BRθ where
Rθ =
(cos (θ) − sin (θ)
sin (θ) cos (θ)
)(160)
B = Rα Or I Or − I (161)
In the first case
T = Rθ−1BRθ = Rθ−1RαRθ = RαRθ−1Rθ = Rα (162)
Therefore in the first case T is a rotation.
We can now gleefully move to the second case
T = Rθ−1BRθ = Rθ−1
(1 0
0 −1
)Rθ =
(1− 2 sin2 (θ) −2 cos (θ) sin (θ)
−2 cos (θ) sin (θ) 1− 2 cos2 (θ)
)(163)
To verify it is indeed a reflection, let us look at the line L passing through the origin with unit normal vector
n. for all x ∈ R the orthogonal projection4 of x onto n is
Pn (x) = 〈x, n〉n (164)
hence the reflection by the line L can be written as
ρL (x) = x− 2〈x, n〉n (165)
We can now calculate the representative matrix of ρL under the standard basis
[ρL]E =
(1− 2n2
1 −2n2n1
−− 2n1n2 1− 2n2n1
)(166)
which is identical to the matrix we arrived at (163), taking
n1 = sin (θ) (167)
n2 = cos (θ) (168)
We can now rejoice, as we have found that in the second case we just have a reflection by a line. The third
case B =
(−1 0
0 1
)is analogous and so we have proved the theorem and are happy.
4The part of x in n
23
We now begin with our geometric results.
Theorem 2.14: Sum of Angles in A Triangle.
The sum of the interior angles of a triangle is π.
Proof. Let us translate 4ABC by CA to the right - denote this isometry T1. Further, rotate it at B by π
and translate it by CB - denote this isometry by T2. Under these isometries we have
α = α′ = α′′ (169)
β = β′ = β′′ (170)
γ = γ′ = γ′′ (171)
And so α+ β + γ = α′ + β′′ + γ = π.
In the following theorem, we essentially provide three equivalent conditions for triangles to be equilateral.
The first is trivial, merely stating that two sides are equal. Afterwards, we prove it is equivalent to stating
that the base angels are equal - a result you hopefully recall from Introduction to Geometry. The last,
and most interesting, describes the triangle in terms of an isometry. The isometry sends it to a congruent
triangle5, but with the sides mirrored.
Proposition 2.15: Pons Asinorum.
The following conditions for 4ABC are equivalent
• d (A,B) = d (A,C)
• ∠ABC = ∠ACB
• There is an isometry T of R2 s.t.
T (4ABC) = 4ABC , T (A) = A , T (B) = C , T (C) = B
Proof. We want to show all three parts are equivalent. To this end, we show pairwise equivalence.
• One ⇔ Two:
Define O to be the intersection of [B,C] with its orthogonal line through A. By definition of angles
d (O,A) = d (A,B) sin (θ) (172)
= d (A,C) sin (θ′) (173)
Where θ = ∠ABC , θ′ = ∠ACB. Hence
d (A,C) sin (θ′) = d (A,B) sin (θ) (174)
Meaning d (A,C) = d (A,B)⇔ θ = θ′.
5Meaning to an essentially equal triangle
24
• Three ⇒ One:
By definition of isometries
d (A,B) = d (T (A), T (B)) = d (A,C) (175)
• One ⇒ Three:
Let L be the reflection along the line defined by [O,A] Where O is the same point as the one defined
above. The reflection satisfies our conditions and again we are happy.
The following lemma will aid us in proving the parallel postulate. Essentially, it means that the number of
intersections between two lines does not change when you translate one or the other6. If they are parallel
they will still be parallel, and if they only intersect once, they still only intersect once.
Lemma 2.16.
Let L′, L′′ be two lines which intersect in exactly one point. Then for all x ∈ Rn the line L′+x also intersects
L′′ in exactly one point.
Proof. Up to rotation and translation we we can assume that the intersection is at p = 0 and that L′′ = Re1.
Denote L′ = λv, we now want to show that L′′ and L′ + x intersect at exactly one point. This is equivalent
to stating that there are unique µ, λ ∈ R s.t.
x+ λv = µe1 (176)
In components x1 + λv1 = µ ∗ 1
x2 + λv2 = µ ∗ 0=⇒
x1 + λv1 = µ
x2 + λv2 = 0(177)
(178)
As L′ and L′′ intersect in a unique point, we have that Rv 6= Re1. This gives us that v2 6= 0 and the only
unique solution is
λ = −x2
v2, µ = x1 6=
x2
v2v1 (179)
Proposition 2.17: Parallel Postulate.
Let L be a line, p ∈ R2 a point not on the line. Then there is a unique line through P which does not intersect
L.
Proof. We divide the proof into two parts, for uniqueness and existence.
6Careful, my language is rather sloppy here. What could go wrong if they were parallel?
25
• Uniqueness:
Assume that there exists a line L′ not intersecting L and passing through P , we wish to show it is
unique. Let Q ∈ L, then trivially Q ∈ L ∩ (L′ +Q) and in particular Q ∈ L ∩ (L′ +Q− P ). Assume
by contradiction that Q− P /∈ L′.By the lemma above, if there is a point x ∈ [L′ + (Q− P )] ∩ L then
x ∈ [L′ + (Q− P ) + (P −Q)] ∩ L = L′ ∩ L (180)
meaning L and L′ intersect at x, which is a contradiction.
Therefore our assumption is wrong and
Q− P ∈ L′ (181)
and the line L′ must be the unique line containing both P and Q− P .
• Existence:
Choose arbitrary Q ∈ L and define L′ = L + (P −Q), we need to show the two do not intersect and
that P ∈ L′.
– P ∈ L′:We know that Q ∈ L, therefore
Q+ (P −Q) = P ∈ L′ (182)
– L ∩ L′ = Φ:
Choose v s.t. L = Q+ Rv, then L′ = P + Rv. Suppose L ∩ L′ 6= Φ, then ∃λ, µ ∈ R s.t.
Q+ λv = P + µv (183)
P = Q+ v (λ− µ) (184)
Which means P ∈ L, contradicting our assumption. Therefore our assumption is wrong and
L ∩ L′ = Φ
Lastly, we finish the chapter with an aside in Algebra. We draw a similarity between isometries and isomor-
phisms. First, just like the group of bijections on a set in Algebra - we can define a group of isometries.
Definition 2.19.
Let (X, d) be a metric space. The group of isometries of (X, d) is the set
Isom (X, d) := {f : X → X : f is an isometry} (185)
with the operation fg := f ◦ g
Remark. . We leave it as an exercise for you to verify that the group axioms hold.
26
Lemma 2.18.
The set Isom (X, d) is bijective to O (n)× Rn where
O(n) ={A ∈Mn×n (R) : AAT = I
}(186)
Proof. As they are all isometries, they are all distance preserving and can be written in the form
ρ = A+ b where A ∈Mn×n (R) , b ∈ Rn (187)
We now recall the notions of a group Homomorphism and Isomorphisms from Introduction to Abstract
Algebra
Definition 2.20: Group Homomorphism and Isomorphisms.
Let H,G be groups. A function φ : H → G is said to be a homomorphism if
φ (h1h2) = φ (h1)φ (h2)
for all h1, h2 ∈ H.
If φ is also a bijection, it is an isomorphism between G and H and we say that G ∼= H.
Remark. . The idea is that φ ”preserves” the group operation.
Proposition 2.19.
A homomorphism φ is injective if and only if ker (φ) = id
Proof. left as exercise
In this terminology, we can describe our group of isometries using orthogonal matrices and vectors. Recall
that a motion was always of the form Lx+ b where L was an orthogonal map. Intuitively, it gives rise to the
notion that motions are generally a combination of such an orthogonal matrix and a vector. Let us formalize
this notion.
Proposition 2.20.
The map φ : O (n)×L Rn → Isom (Rn, d) defined by
φ (A, b) = (x 7→ Ax+ b)
is a group isomorphism
Proof. Let (A, b) , (A′, b′) ∈ O (n)×L Rn, by direct calculation we get that
φ [(A, b) (A′, b′)] = φ [(AA′, b+Ab′)] (188)
= (x 7→ AA′x+ b+Ab′) (189)
= φ (A, b)φ (A′, b′) (190)
And so φ is a homomorphism. Let us now prove that it is a bijection.
27
• Injective:
First note that Id ∈ Isom (Rn, d) is the function s.t.
Ax+ b = x , ∀x ∈ Rn (191)
Meaning b = 0, A = I. Hence if
φ (A, b) = Id (192)
Then A = I, b = 0, but this is just the identity in O (n)×L Rn. Ergo
ker (φ) = {Id} (193)
And by the above proposition φ is injective.
• Surjective:
Every isometry in (Rn, d) is of the form Ax+ b = φ (A, b)
Lastly, we define our last object of the chapter. You will explore it further in assignment 2.
Definition 2.21: Semi-direct Product.
The semi-direct product of groups H,G relative to the homomorphism φ : G→ Aut (H) is the group defined
by:
• the elements are the members of the set G×H
• for all (g, h) , (g′, h′) ∈ G×H the group operation is
(g, h) (g′, h′) = (g · g′, h · φ (g) (h′))
and is denoted by G×L H
Remark. . Note that φ (g) (h′) is the function that the group homomorphism assigns to g, taken at h′
Example 2.6: Semi-direct Product.
Let G be a group and φ : G → Aut (G) be the identity homomorphism. Then G ×L G = G ×G is just the
product group
Example 2.7.
Let G = O (m) , H = Rm, and φ : G→ H given by A 7→ (x 7→ Ax)
28
3 Spherical Geometry
We now move to our first ”non-orthodox” geometry - Spherical Geometry. Compared with our Euclidean
plane, now our points, lines, and triangle all lie on a sphere. First we begin with the definitions
Definition 3.1: Sphere.
Define the nth dimensional sphere of radius r as the set of points
Snr ={
(x1, x2, . . . xn+1) : x21 + . . . x2
n+1 = r2}
(1)
Also, as a matter of notation denote Sn := Sn1 .
Definition 3.2: Great Circle.
A spherical line, or Great Circle, is the intersection of Snr with a plane through the origin.
Remark. . Any two non antipodal points7 P and Q determine a unique great circle - determined by the
plane containing O,P,Q. On the other hand, if two points are antipodal then they are aligned with O and
the plane cannot be unique.
To define the notion of angle, we first must define tangent lines.
Definition 3.3: Tangent Line.
The tangent line at P of an arc of the great circle PQ is the intersection of the plane defining the great
circle and the orthogonal complement of OP .
Definition 3.4: Spherical Angle.
The spherical angle between two arcs of the great circle PQ and PR is the angle between their tangent lines
Definition 3.5: Distance.
The spherical distance between two points P,Q ∈ Snr is
d (P,Q) = rθ = r arccos
(〈P,Q〉r2
)I.e. you take the great circle and calculate the length of the arc PQ.
Remark. . The distance between two antipodal points is pi times the radius, i.e. d (P,−P ) = rπ
We have defined the spherical distance, now let us prove it satisfies our general version of distance - metric.
To this end, we first prove a crucial proposition.
Proposition 3.1.
Let P,Q,R ∈ S2, and denote
α = d (Q,R)
β = d (P,Q)
γ = d (P,R)
a = angle between PQ and PR
7Two points x, y are antipodal if and only if y = −x
29
Then
cos (α) = cos (β) cos (γ) + sin (β) sin (γ) cos (a)
Proof. Up to rotation we can assume that P = (0, 0, 1) and Q belongs to the plane O×Z8. We now denote
Q′ = (1, 0, 0) (2)
R′ = (cos (a) , sin (a) , 0) (3)
Q = (sin (β) , 0, cos (β)) (4)
R = (cos (γ) · P + sin (y) ·R′) = (sin (γ) cos (a) , sin (γ) sin (a) , cos (γ)) (5)
Therefore
cos (α) = 〈Q,R〉 = sin (β) sin (γ) cos (a) + cos (β) cos (γ) (6)
Proposition 3.2.
Spherical Distance is a metric
Proof. To prove it is a metric, we need only show the triangle inequality as the other two conditions are
trivial. Meaning we wish to show that given α, β, γ from the prior proof, we have
α ≤ β + γ (7)
Where α = β + γ if and only if P,Q,R lie on the same great circle with P on the shortest arc between P,R.
First note that by definition α, β, γ ∈ [0, π]. Moreover, by the prior theorem
cos (α) = cos (β) cos (γ) + sin (β) sin (γ)︸ ︷︷ ︸∈[0,1]
cos (a)︸ ︷︷ ︸∈[−1,1]
(8)
≥ cos (β) cos (γ)− sin (β) sin (γ) (9)
≥ cos (β + γ) (10)
Meaning
cos (α) ≥ cos (β + γ) (11)
And by definition of cosine we have that α ≤ β + γ, with equality holding if and only if cos (α) = −1 or
α = π. But both cases imply that P,Q,R lie on the same great circle.
Last but not least, we have one last definition under our belt before we can continue to the theorems for this
section.
8The proof of this is exercise 1 of worksheet 3
30
Definition 3.6: Spherical Triangle.
A spherical triangle consists of three distinct points P,Q,R ∈ Snr and the arcs of the great circles joining
them.
To start with our theorem part of the section, we examine how isometries on the sphere affect great circles.
Proposition 3.3.
An isometry T : S2 → S2 preserves antipodal points and great circles.
Remark. . Recall that great circles are the spherical analogue to lines. Hence this theorem forms an analogue
to our proof that isometries of Rn take preserve lines.
Proof. We shall divide the proof to two parts, one for antipodal points and one for great circles.
• Antipodal Points:
Note that if T is an isometry then d (T (P ), T (Q)) = d (P,Q). Further, by definition P,Q ∈ S2 are
antipodal if and only if d (P,Q) = π.
Therefore
d (T (P ) , T (Q)) = π ⇔ d (P,Q) = π (12)
T (P ) , T (Q) are antipodal⇔ P,Q are antipodal (13)
• Great Circles:
Let é be a great circle and P,Q,∈ é two points s.t. d (P,Q) = π2 r, We wish to show that for all R ∈
é holds T (R) ∈ é . We distinguish between four different cases, depending on the location of R.
– R belongs to the shortest arc between Q and P :
Then d (P,R) + d (R,Q) = d (P,Q), and as T is an isometry
d (T (P ) , T (R)) + d (T (R) , T (Q)) = d (T (P ) , T (Q)) (14)
Meaning, by the proposition 3.2, that T (P ) , T (Q) , T (R) lie on the same great circle.
– R belongs to the shortest arc between Q and −P :
Then d (R,Q) + d (Q,P ) = d (R,P ), and as T is an isometry
d (T (R) , T (Q)) + d (T (Q) , T (P )) = d (T (R) , T (P )) (15)
And again T (P ) , T (Q) , T (R) lie on the same great circle.
– R belongs to the shortest arc between −Q and P : this case is analogous to the previous one
– R belongs to the shortest arc between −Q and −P :
We now have
d (R,−P ) + d (−P,Q) = d (R,Q) (16)
d (T (R), T (−P )) + d (T (−P ), T (Q)) = d (T (R), T (Q)) (17)
(18)
31
and so T (R), T (Q), T (−P ) lie on the same great circle. Further, as the circles defined by
T (−P ), T (Q) and T (P ), T (Q) are identical - we have that T (R), T (Q), T (P ) lie on the same
great circle.
Akin to our previous chapter, we now examine the general form of isometries of the sphere.
Corollary 3.4.
Let T : S2 → S2 be an isometry, then T (x) = Ax for some A ∈ O (3)
Proof. We start by extending T to the function T̂ : R3 → R3 by defining
T̂ (x) =
‖x‖r T
(x‖x‖ · r
)x 6= 0
0 x = 0(19)
First note that it acts identically to T on S2. Let x ∈ S2, then by definition ‖x‖ = r and
T̂ (x) =‖x‖rT
(x
‖x‖ · r
)=r
rT(xr· r)
= T (x) (20)
Now let x /∈ S2, we wish to show it is a linear isometry. By proposition 2.8 it suffices to show it is linear and
preserves the norm.
• T̂ preserves norm:
let x 6= 0, then x‖x‖ · r ∈ S2 and
∥∥∥T̂ (x)∥∥∥ =
∥∥∥∥‖x‖r T
(x
‖x‖· r)∥∥∥∥ =
‖x‖r
∥∥∥∥T ( x
‖x‖· r)∥∥∥∥︸ ︷︷ ︸
T is an isometry on S2
=‖x‖r
∥∥∥∥ x
‖x‖r
∥∥∥∥ =‖x‖r· r = ‖x‖ (21)
• T̂ is a linear map:
We need to show that T is both homogeneous and linear.
– homogeneity: Let λ, x 6= 09, then
T̂ (λx) =‖λx‖r
T
(λx
‖λx‖· r)
(22)
= |λ| ‖x‖rT
(λ
|λ|x
‖x‖· r)
(23)
9The case where either is zero is trivial
32
Now assume λ > 0, then
T̂ (λx) = |λ| ‖x‖rT
(λ
|λ|x
‖x‖· r)
(24)
= λ‖x‖rT
(λ
λ
x
‖x‖· r)
(25)
= λ‖x‖rT
(x
‖x‖· r)
(26)
= λT̂ (x) (27)
Meaning if λ > 0 then T̂ is homogeneous. Now assume λ < 0 and return to equation (23)
T̂ (λx) = |λ| ‖x‖rT
(λ
|λ|x
‖x‖· r)
(28)
= −λ‖x‖rT
(λ
−λx
‖x‖· r)
(29)
= −λ‖x‖rT
(− x
‖x‖· r)
(30)
(31)
By proposition 3.3 we have that T preserves antipodal-points and so
T̂ (λx) =�−λ‖x‖r
[�−T
(x
‖x‖· r)]
(32)
= λ‖x‖rT
(x
‖x‖· r)
(33)
= λT̂ (x) (34)
And so, T is homogeneous.
– Linearity: let x, y ∈ R3. If the two are collinear then x + y = (1 + λ)x for some λ ∈ R. In this
case, as T is homogeneous, we are done. Let us now assume the two are not collinear.
Then x, y and O determine a plane, with x + y contained in it as well. Let us denote the great
circle containing x‖x‖ · r and y
‖y‖ · r by é .
We now denote the angles x and y make with x+ y by β and α respectively. Similarly we denote
the angles T(
x‖x‖ · r
)and T
(y‖y‖ · r
)make with T
(x+y‖x+y‖ · r
)by β′ and α′. As T preserves
spherical distances we have that α = α′, β = β′.
We can now write
T̂ (x+ y) =‖x+ y‖
r
[cos (α)T
(x
‖x‖· r)
+ sin (α)T
(y
‖y‖· r)]
(35)
but observe that this is the same expression we get for T̂ (x) + T̂ (y), and so T is linear.
We have thus found that T̂ is a linear map, and as it preserves the norm it is a motion. Hence there
exists a matrix A ∈ O (3) s.t. T̂ (x) = Ax for all x ∈ R3, and in particular
T (x) = Ax ∀x ∈ S2 (36)
33
Lastly, we finish the chapter by proving our spherical analogue to the sum of angles in a triangle.
Theorem 3.5: Sum of Angles of a Triangle.
Let 4ABC be a triangle in S2 with interior angles a, b, c, then
S (4ABC) = a+ b+ c− π
In particular, as long as 4ABC is non degenerate then a+ b+ c > π
Proof. Let Σa be the intersection of S2 with the region of R2 between the plane and OPQ and OPR , with
Σb,Σc defined similarly.
We then have that these areas span our sphere, and intersect at the triangle
Σa ∩ Σb ∩ Σc = 4ABC (37)
Σa ∪ Σb ∪ Σc = S2 (38)
Note that S (Σa) = 2a2πS
(S2), and so
S (Σa) + S (Σa) + S (Σa) =
(a+ b+ c
π
)S(S2)
= S(S2)
+ 4S (4ABC) (39)(a+ b+ c
π
)S(S2)
= S(S2)
+ 4S (4ABC) (40)
S (4ABC) =
(a+ b+ c
π− 1
)S(S2)
4(41)
Lastly, as S(S2)
= 4π, we are left with
S (4ABC) = a+ b+ c− π (42)
34
4 Hyperbolic Geometry
In our next geometry, we wish to examine the analogue to Spherical Geometry - but under the following
inner product.
Definition 4.1: Lorentzian Inner Product.
Let x, y ∈ Rn, we define their Lorentz inner product as
〈x, y〉L = −x1y1 +
n∑i=2
xiyi
Definition 4.2: Lorentzian Norm.
For all x ∈ Rn define
‖x‖L =√〈x, x〉L
Remark. . Note that only one of three occurs
‖x‖L = 0 Or ‖x‖L ∈ R+ Or ‖x‖L ∈ iR+
And so, we define the Hyperbolic Space to be
Definition 4.3: Hyperbolic Space.
The Hyperbolic Space of Hn of(Rn+1, ‖·‖L
)is defined as
Hn ={x ∈ Rn+1 : ‖x‖L = −1 ∧ x1 > 0
}Remark. . Note that the definition is identical to that of the Unit-Sphere, with the euclidean norm replaced
with the lorentz norm and the added property that x1 > 0.
Before we proceed, let us define the hyperbolic trigonometric functions. These will be used as a replacement
for our normal trigonometric functions, and will be essential for us to define distance
Definition 4.4: Hyperbolic sine and cosine.
For all θ ∈ R define
sinh (θ) =eθ + e−θ
2
cosh (θ) =eθ − e−θ
2
Remark. . Note that similarly to our usual identity we have that
cosh2 (θ) + sinh2 (θ) = −1
Lemma 4.1.
The function cosh (x) is injective on the interval [0,∞) with image [1,∞)
35
Proof. Given point y ≥ 1, we wish to show the equation
coshx = y (1)
has a unique solution in x ∈ R≥1. The above equation is equivalent to
ex + e−x
2= y (2)
e2x + 1 = 2exy (3)
We know that ex is injective on R≥0, and so our condition is equivalent to showing that
u2 + 1 = 2uy (4)
has a unique solution for u = ex. But this is just a quadratic and so
u =2y ±
√4y2 − 4
2(5)
= y ±√y2 − 1 (6)
Note that this is always strictly positive as√y2 − 1 < y for all y. Ergo we have that
x = log(y ±
√y2 − 1
)(7)
And we always have a solution, meaning our function is surjective. Finally, to show that it is injective we
must rule out one of our two solutions. Note that only one of them is positive, as
y −√y2 − 1 < y < 1 =⇒ x = log
(y −
√y2 − 1
)< 0 (8)
And so, it is ruled out as our proof was only for positive x. Therefore we are left with only one solution.
In conclusion, for all y ≥ 1 there is a unique solution to the equation
cosh (x) = y (9)
with x ≥ 0 and we are done.
Definition 4.5: The likes.
Let x ∈ Rn, we call it
1. Space-like if
‖x‖L ∈ R+ Or Equivalently x21 <
n∑i=2
x2i
2. Time-like if
‖x‖L ∈ i× R+ Or Equivalently x21 >
n∑i=2
x2i
36
3. Light-like if
‖x‖L = 0 Or Equivalently x21 =
n∑i=2
x2i
Remark. . We say x is positive or negative depending on the value of x1
Proposition 4.2: Cauchy Schwartz Inequality.
Let x, y ∈ Rn be time-like and positive vectors, then
〈x, y〉L ≤ ‖x‖L ‖y‖L (10)
Proof. By careful inspection of our prior proof of the Cauchy Schwartz inequality. Soon we shall prove a
stronger result, which will ease the pain of this proof being omitted.
Definition 4.6: Light-cone.
The Light-cone Cn−1 of Rn is the set of points
Cn−1 = {x ∈ Rn : x is light-like}
Definition 4.7: Lorentzian Distance.
The Lorentzian distance between any two vectors x, y ∈ Rn is defined as
d (x, y)H = ‖x− y‖L
Definition 4.8: Hyperbolic Line.
A Hyperbolic Line L is defined to be the intersection of Hn with a plane in Rn+1 through 0 with L 6= φ
Remark. . Two points x, y ∈ H define a unique hyperbolic line passing through them
Definition 4.9.
The representative matrix of the bilinear form defined by the Lorentzian inner product is denoted by J .
Note that by inspection it is equal to
J = Diag {−1, 1, 1, . . . , 1} (11)
Definition 4.10: Lorentzian cross.
Lorentzian cross product of two vectors x, y ∈ R3 is defined as
x×L y =
−x2y3 + x3y2
x3y1 − x1y3
x1y2 − x2y1
(12)
Note that
x×L y = J (x×L y) = Jx×L Jy (13)
The following lemma will be used repeatedly throughout the chapter, and proves important properties of
the Lorentz inner product and Lorentz cross product.
37
Lemma 4.3.
Note that for all x, y, z, w ∈ R3 the Lorentz cross product holds
1. 〈x×L y, x〉L = 〈x×L y, y〉L = 0
2. It is bilinear
3. It is anti-symmetric
4. it is not associative
5. 〈x×L y, z ×L w〉L = 〈x,w〉L〈y, z〉L − 〈x, z〉L〈y, w〉L
(5′) ‖x×L y‖L = 〈x, y〉2L − ‖x‖2L ‖y‖
2L
6. if x, y are both time-like and positive then
‖x×L y‖L = −‖x‖L ‖y‖L sinh [η (x, y)] (14)
where η (x, y) = arccosh(〈x,y〉L‖x‖L‖y‖L
). In particular, we have that if x, y are positive time-like vectors
their Lorentzian cross product is space-like.
7. x×L (y ×L z) = 〈x, y〉L · z − 〈z, x〉L · y.
Proof. We prove this part by part
1. By definition
〈x×L y, x〉L = 〈J (x×L y) , x〉 = 〈J [J (x× y)] , x〉 = 〈x× y, y〉 = 0 (15)
2. Directly from definition
3. Directly from definition
4. Directly from definition
5. By definitions of Lorentzian inner and cross product
〈x×L y, z ×L w〉 = − (x2y3 + x3y2) (−z2w3 + z3w2) + (x3y1 − x1y3) (z3w1 − z1w3) + (x1y2 − x2y1) (z1w2 − z2w1)
(16)
Similarly
〈x,w〉L〈y, z〉L = (−x1w1 + x2w2 + x3w3) (−y1z1 + y2z2 + y3z3) (17)
−〈x, z〉L〈y, w〉L = − (−z1z1 + x2z2 + x3z3) (−y1w1 + y2w2 + y3w3) (18)
And by comparing the terms we get our desired result.
(5′) By plugging in z = x,w = y into (5).
38
6. By (5′)
‖x×L y‖2L = ‖x×L y‖2L − ‖x‖2L ‖y‖
2L (19)
By definition of of η (x, y) we have that
‖x×L y‖2L = cosh [η (x, y)] ‖x‖2L ‖y‖2L − ‖x‖
2L ‖y‖
2L (20)
= ‖x‖2L ‖y‖2L {cosh [η (x, y)]− 1} (21)
= ‖x‖2L ‖y‖2L sinh2 [η (x, y)] (22)
Note now that this implies that x×L y is space-like. Further, taking square root we now have that
‖x×L y‖L = ‖x‖L ‖y‖L sinh [η (x, y)] (23)
7. By direct calculation
Corollary 4.4.
If x, y ∈ R3 are both positive and time-like then
〈x, y〉L ≤ ‖x‖L ‖y‖L < 0 (24)
Proof. By property (5′) of previous proposition we have that
〈x, y〉2L ≥ ‖x‖2L ‖y‖
2L (25)
as they are both time-like. Taking square roots we get that
|〈x, y〉L| ≥ −‖x‖L ‖y‖L (26)
− |〈x, y〉L| ≤ ‖x‖L ‖y‖L (27)
Lastly, note that
〈x, y〉L = −x1y1 + x2y2 + x3y3 = −x1y1 +
(x2
x3
)·
(y2
y3
)(28)
we can now apply the normal Cauchy Schwartz inequality on (28) to get
〈x, y〉L ≤ −x1y1 +√x2
2 + x23
√y2
2 + y23 (29)
As both are time-like we have that x21 > x2
2 + x23 with the same holding for y. Ergo
〈x, y〉L < −x1y1 + |x1| |y1| (30)
Lastly, as both are positive
〈x, y〉L < −x1y1 + x1y1 = 0 (31)
Hence, returning to (27) we have that
|〈x, y〉L| ≤ ‖x‖L ‖y‖L (32)
39
Corollary 4.5.
If v, w ∈ R3 are both space-like then
|〈v, w〉L| ≤ ‖v‖L ‖w‖L ⇐⇒ v ×L w is time-like (33)
|〈v, w〉L| = ‖v‖L ‖w‖L ⇐⇒ v ×L w is light-like (34)
|〈v, w〉L| ≥ ‖v‖L ‖w‖L ⇐⇒ v ×L w is space-like (35)
Proof. Let v, w ∈ R3 be two space-like vectors, by property (5′) of lemma 4.3
〈v, w〉2L = ‖v‖2L︸︷︷︸>0
‖w‖2L︸ ︷︷ ︸>0
+ ‖v ×L w‖2L (36)
and the corollary follows directly from the condition on (v ×L w).
Definition 4.11: Hyperbolic distance.
We define the distance between any two points x, y ∈ Hn as
d (x, y) = arccosh (−〈x, y〉L)
Remark. . Observe that
1. This is well defined by lemma 4.1
2. Notice that it is analogous to our definition of spherical distance
3. It turns out10 that this is the length of the segment of the hyperbolic line between x and y
Proposition 4.6.
The hyperbolic distance we defined constitutes a metric.
Proof. We prove this in the case n = 3. To achieve this, three different conditions must be proven. Take
x, y, z ∈ H2 and let’s start
• First, we must show that d (x, y)H = 0 if and only if x = y. Assume that x = y, then as x ∈ H2 we
trivially have that 〈x, x〉L = 〈x, y〉L = −1. Hence
d (x, y)H = arccosh [−d (x, y)H] = arccosh [− (−1)] = 0 (37)
Conversely, if d (x, y)H = 0 then
0 = arccosh [−d (x, y)H] (38)
d (x, y)H = −1 (39)
But also, as both belong to the hyperbolic space
‖x‖L ‖y‖L = −1 (40)
10But we do not prove
40
Hence we have that d (x, y)H = ‖x‖L ‖y‖L. By lemma 4.3 part (5′) we have that11
‖x×L y‖2L = 0 (41)∥∥∥∥∥∥∥−1
0
0
×L
y1
y2
y3
∥∥∥∥∥∥∥
2
L
= 0 (42)
∥∥∥∥∥∥∥ 0
−y3
−y2
∥∥∥∥∥∥∥
2
L
= 0 (43)
y22 + y2
3 = 0 (44)
And as this is just a sum of squares y2 = y3 = 0 and y is linearly dependent on x = −e1. Furthermore,
as both of them belong to H3
i = ‖y‖L = ‖λx‖L = |λ| ‖x‖L = |λ| i (45)
|λ| = 1 (46)
Meaning x = ±y. Finally, by the definition of the hyperbolic plane both x and y are positive. Therefore
x = y and we are done.
• Symmetry is clear as the inner product as symmetric
• We now wish to prove the triangle inequality. Using our known trigonometric identities on cosh we
have that
cosh (d (x, y)H + d (y, z)H) = cosh (d (x, y)H) cosh (d (y, z)H) + sinh (d (x, y)H) sinh (d (y, z)H) (47)
by definition of hyperbolic distances we have that
cosh (d (x, y)H + d (y, z)H) = [−〈x, y〉L] [−〈y, z〉L] + sinh (d (x, y)H) sinh (d (y, z)H) (48)
Replacing the sinh with the values from lemma 4.3 part (6) we have12
cosh (d (x, y)H + d (y, z)H) = [−〈x, y〉L] [−〈y, z〉L] + ‖x×L y‖L ‖y × z‖L (49)
We now wish to deal with the second term on the RHS. To do this, we wish to use the Cauchy Schwartz
inequality on x×L y and y ×L z by evaluating [(x×L y)×L (y ×L z)]. By property (7) of lemma 4.3
(x×L y)×L (y ×L z) = 〈x×L y, z〉L︸ ︷︷ ︸=0 by (1) of lemma 4.3
−〈z, x×L y〉L︸ ︷︷ ︸:=λ
y = λy (50)
Therefore, as y is time-like so is (x×L y)×L (y ×L z) and we can apply the inequality
|〈x×L y, y ×L z〉L| ≤ ‖x×L y‖L ‖y ×L z‖L (51)
11We’ll see in assignment 6 that we may assume that x = −e112Note that the norms are all equal to (−1) as we are in H3
41
With this at hand, we can set our eyes towards (49)
cosh (d (x, y)H + d (y, z)H) ≥ [−〈x, y〉L] [−〈y, z〉L] + 〈x×L y, y ×L z〉L (52)
≥ 〈x, y〉L〈y, z〉L + 〈x×L y, y ×L z〉L (53)
Applying lemma (4.3) part (5) we get
cosh (d (x, y)H + d (y, z)H) ≥((((((〈x, y〉L〈y, z〉L + 〈x, z〉L 〈y, y〉L︸ ︷︷ ︸
=−1
−((((((〈x, y〉L〈y, z〉L (54)
cosh (d (x, y)H + d (y, z)H) ≥ −〈x, z〉L (55)
(56)
Further, as arccosh(θ) is monotonic on R+
d (x, y)H + d (y, z)H ≥ arccosh (−〈x, z〉L)︸ ︷︷ ︸=d(x,z)H
(57)
d (x, y)H + d (y, z)H ≥ d (x, z)H (58)
And we have proved the triangle inequality.
Remark. . This holds for Hm as well by reducing to the case m = 3 with the proof in our book
We now continue to explore the isometries on the Hyperbolic plane. These differ from our isometries on the
sphere or euclidean space due to our use of the Lorentz inner product. Analogously to before, we introduce
the following two definitions - towards theorem 4.7.
Definition 4.12: Lorentz transformation.
A map φ : Rn → Rn is a Lorentz transformation if for all x, y ∈ Rn
〈φ (x) , φ (y)〉L = 〈x, y〉L
Definition 4.13: Lorentz orthonormal.
A basis v1, v2 . . . , vn is called Lorentz orthonormal if
〈vi, vj〉L =
0 i 6= j
−1 i = j = 1
1 i = j and i 6= 1
Example 4.1: Lorentz orthonormal basis.
The trivial basis E is a Lorentz orthonormal basis
Theorem 4.7.
A map φ : Rn → Rn is a Lorentz transformation if and only if it
• it is linear
42
• φ (e1) , . . . , φ (en) is a Lorentz orthonormal basis
Remark. . Analogous to our theorem for euclidean isometries
Proof. We need to prove both implications of the if and only if, first let us assume φ is a Lorentz transfor-
mation. We start with the second condition first
〈φ (ei) , φ (ej)〉L = 〈ei, ej〉L =
0 i 6= j
−1 i = j = 1
1 i = j and i 6= 1
(59)
and so φ (e1) , . . . , φ (en) is a Lorentz orthonormal set. We now show φ (e1) , . . . , φ (en) is a basis. It is of size
n = dimRn, let us show it is a linearly independent set by considering a linear combination of 0
n∑i=1
λiφ (ei) = 0 (60)
Fix i ∈ [n] and let us take the inner product of both sides with φ (ei)
〈n∑i=1
λiφ (ei) , φ (ei)〉L = 〈0, φ (ei)〉L (61)
By linearity
n∑i=1
λi〈φ (ei) , φ (ei)〉L = 0 (62)
(63)
From (59) we then deduce that −λi = 0 or λi = 0. In any case we have shown this must be the trivial linear
combination of 0 and φ (e1) , . . . , φ (en) is a Lorentz orthonormal basis.
We now just need to show φ (x) is linear. Let x ∈ Rn, we know there exists a unique representation
φ (x) =
n∑i=1
biφ (ei) (64)
As it is Lorentz orthonormal
−x1 = 〈x, e1〉L = 〈φ (x) , φ (ei)〉L = 〈n∑i=1
bi (ei) , φ (e1)〉L =
n∑i=1
bi〈φ (ei) , φ (e1)〉L = −b1 (65)
Meaning x1 = b1. The same can be done for all bi to get that
bi = xi ∀i ∈ [n] (66)
φ (x) =
n∑i=1
xiφ (ei) (67)
43
Lastly, to see that φ is linear just note that this means that
φ
(n∑i=1
xiei
)=
n∑i=1
xiφ (ei) (68)
We now need to prove the other implication. Assume that φ is linear and that φ (e1) , . . . , φ (en) forms a
Lorentz orthonormal basis - we wish to show it is a Lorentz transformation.
Let x, y ∈ Rn, by linearity of φ
〈φ (x) , φ (y)〉L = 〈n∑i=1
xiφ (ei) ,
n∑j=1
yjφ (ej)〉L =
n∑i=1
n∑j=1
xiyj〈φ (xi) , φ (ej)〉L = −x1y1 +
n∑i=2
xiyj = 〈x, y〉L
(69)
With this at hand, we define our hyperbolic analogue to orthogonal matrices and the orthogonal group.
Definition 4.14: Lorentz group.
The Lorentz group O (1,m− 1) is the group formed from the set{A : ATJA = J
}endowed with the matrix multiplication operation
Theorem 4.8.
Let A be an n× n matrix, the following conditions are equivalent
1. The map defined by φ (x) = Ax is a Lorentz transformation
2. ATJA = J
3. ‖Ax‖L = ‖x‖L for all x
Proof. First let us prove an identity needed in the proof, using nothing but matrix multiplication
(ATJA
)ij
=[(ATJ
)A]ij
=
n∑k=1
(ATJ
)ikAkj =
n∑k=1
(n∑l=1
ATilJlk
)Akj =
n∑k=1
n∑l=1
AliJlkAkj (70)
Note that Jlk = 0 whenever l 6= k and so
(ATJA
)ij
=
n∑k=1
AkiJkkAkj = −A1iA1j +
n∑k=2
AkiAkj = 〈Ai↓, Aj↓〉L (71)
For all 1 ≤ i, j ≤ n. With this identity in hand we can go ahead and prove our theorem.
44
• (1) =⇒ (2):
Assume that φ (x) = Ax is a Lorentz transformation. By definition, the ith column of i is φ (ei). Hence
by our identity (ATJA
)ij
= 〈Ai↓, Aj↓〉L = 〈φ (ei) , φ (ej)〉L = 〈ei, ej〉L (72)
Hence ATJA = J by definition of J , and we reached (2).
• (2) =⇒ (1):
Assume that ATJA = J and define φ (x) = Ax, by our identity
Jij =(ATJA
)ij
= 〈Ai↓, Aj↓〉L = 〈φ (ei) , φ (ej)〉L (73)
Jij = 〈φ (ei) , φ (ej)〉L (74)
And so φ (e1) , . . . , φ (en) is a Lorentz orthonormal basis. Further, φ is clearly linear, and so by theorem
4.7 we have that φ is a Lorentz transformation.
• (1) =⇒ (3):
Assume that φ (x) = Ax is a Lorentz transformation. Trivially
‖Ax‖2L = 〈Ax,Ax〉L = 〈φ (x) , φ (x)〉L = 〈x, x〉L = ‖x‖2L (75)
And taking square roots yields
‖Ax‖L = ‖x‖L (76)
• (3) =⇒ (1):
Assume that for all x ∈ Rn holds ‖Ax‖L = ‖x‖L. To prove this, note that for all x, y ∈ Rn holds
〈x, y〉L =1
4‖x+ y‖L −
1
4‖x− y‖L (77)
by the linearity of the Lorentz inner product. Hence
〈Ax,Ay〉L =1
4‖Ax+Ay‖L −
1
4‖Ax−Ay‖L (78)
=1
4‖A (x+ y)‖L −
1
4‖A (x− y)‖L = 〈x, y〉L (79)
Definition 4.15: Positive Lorentz group.
The positive Lorentz group, denoted PO (1,m− 1), is the subgroup of O (1,m− 1) of matrices sending
positive time-like vectors to positive time-like vectors
With the PO group defined, we can prove our hyperbolic geometry analogue to theorem 2.8 of euclidean
geometry. That is, we describe the general form of isometries of the hyperbolic space.
Theorem 4.9.
A map φ : Hn → Hn is an isometry for the hyperbolic metric if and only if φ (x) = Ax for some matrix
A ∈ PO (1,m)
45
Proof. Note that we must prove both implications of this theorem. First suppose that φ (x) = Ax for some
A ∈ PO (1,m). We begin by showing that φ is well defined, i.e. that its range is in fact Hn.
Let x ∈ Hn, then by theorem 4.8
〈Ax,Ax〉L = 〈x, x〉L = −1 (80)
furthermore, as A ∈ PO (1,m− 1) we also have that Ax is positive. Hence Ax ∈ Hn.
We now show it is an isometry. Let x, y ∈ Rn, by theorem 4.8
d (φ (x) , φ (y))H = d (Ax,Ay)H = arccosh [−〈Ax,Ay〉L] = arccosh [−〈x, y〉L] = d (x, y)H (81)
Thus we proved the first implication of our if and only if statement.
Let us now assume that φ = (φ1, φ1, . . . , φn+1) is an isometry, we wish to prove the existence of such matrix
A. First, assume φ (e1) = e1, then for all x ∈ Hn we have
φ1 (x) =〈φ (x) , e1〉 (82)
=− 〈φ (x) , φ (e1)〉L (83)
= cosh d (φ (x) , φ (e1))H (84)
= cosh d (x, e1)H (φ is an isometry)
=− 〈x, e1〉L = x1 (85)
Meaning that for all x ∈ Hnholds
φ (x) = (x1, φ2 (x) , . . . , φn+1 (x)) (86)
Not note that we can impose an isometry Rn → Hnby
(x2, x3, . . . , xn) 7→(√
1 + x22 + x2
3 + . . .+ x2n, x2, . . . , xn
)(87)
using the condition that ‖x‖L = −1 for all x ∈ Hn. Therefore, we define the map φ̄ : Rn ∼= Hn → Hn ∼= Rn
by
φ̄ (x2, x3, . . . , xn) =
(φ2
(√1 + x2
2 + x23 + . . .+ x2
n, x2, . . . , xn
), . . . , φn+1
(√1 + x2
2 + x23 + . . .+ x2
n, x2, . . . , xn
))(88)
Therefore , ∀x, y ∈ Hn holds
〈φ (x) , φ(y)〉L =− x1y1 + 〈 ¯φ (x), ¯φ (y)〉 (89)
=− x1y1 + 〈x, y〉 (90)
Ergo
〈φ̄ (x) , φ̄ (y)〉 = 〈x, y〉 (91)
46
Meaning φ is a motion. Hence ∃A ∈ O (n) s.t.
φ̄ (x) = Ax (92)
And so
φ (x) =
1 0 . . . 0
0
A
0
︸ ︷︷ ︸
=B
x (93)
where B ∈ PO (n).
Finally, assume φ (e1) 6= e1. We complete φ (e1) to a Lorentz orthonormal basis for Rn+1, and denote B the
transformation taking that basis to En+1. Our prior proof holds for φ ◦B, and so ∃A ∈ PO (n) s.t.
φ ◦B = Aφ = AB−1 (94)
and as B is the motion taking one orthonormal base to another, it is also in PO (n) meaning we found
AB−1 = C ∈ PO (n) s.t.
φ (x) = Cx (95)
Theorem 4.10: Characterization of Isom(H1, dHn
).
The group PO (1) consists of matrices of the form(cosh θ sinh θ
sinh θ cosh θ
),
(cosh θ − sinh θ
sinh θ − cosh θ
)(96)
Proof. We know that all isometries of the sphere are of the form(x
y
)7→
(a c
b d
)︸ ︷︷ ︸=A∈PO
(x
y
)(97)
Therefore, A must satisfy
1. −a2 + b2 = −1 As the first column is a unit vector
2. −c2 + d2 = −1 As the second column is a unit vector
3. −ac+ bd = 0 As the columns are Lorentz orthogonal
4. a > 0 As Ae1 must be positive
47
Since sinh is bijective ∃θ s.t. sinh (θ) = b. By (1)
a2 = 1 + sinh2 θ = cosh2 θ (98)
Using the condition a > 0 we have a = cosh (θ). By (3)
−c cosh θ + d sinh θ = 0 (99)
c =sinh θ
cosh θd (100)
As cosh θ 6= 0 for all θ. By (2)
−(
sinh θ
cosh θd
)2
+ d2 = 1 (101)
d2
(1− sinh2 θ
cosh2 θ
)= 1 (102)
d2 1
cosh2 θ= 1 (103)
d = ± cosh θ (104)
c = ∓ sinh θ (105)
Theorem 4.11: Intersection of Hyperbolic Lines.
Let L1 = H2 ∩Π1, L2 = H2 ∩Π2. Then
L1 ∩ L2 = H2 ∩ (Π1 ∩Π2) = H2 ∩ Rv
for some non-zero v ∈ R3, and
L1 ∩ L2 = H2 ∩ Rv =
one point If v is time-like
φ Otherwise(106)
Proof. For the first equation note that Π1,Π2 are planes in R3 and so their intersection is a line.
Now assume their intersection is non-empty, and so it of the form λv for all λ ∈ R. A point λv on both the
line and the hyperbolic space satisfies
λv1 > 0 (107)
‖λv‖2L = −1 =⇒ λ2 ‖v‖2L = −1 =⇒ λ2 =1
‖v‖2L(108)
the latter of the two equations has a non-negative solution if and only if ‖v‖L < 0, meaning if and only if
v is time-like. Further, whenever a solution exists λ = ±‖v‖−1L . This gives us two options for λ, only one
of which satisfying our first condition. Hence, whenever the intersection is non-empty it contains only one
point.
48
Example 4.2: A family of lines in H2.
For each c ∈ R let us consider the plane Πc = {cx1 + x3 = 0}, this has euclidean normal(c01
). Define the
line
Lc = H2 ∩Πc (109)
We wish to find the intersection point. First we have the two conditions that
cx1 + x3 = 0 =⇒ x1 = −x3
c(110)
−x1 + x22 + x2
3 = −1 =⇒ −x23
c2+ x2
2 + x23 = −1 (111)
And so, by rearranging we get
x23
(1− 1
c2
)= −
(1 + x2
2
)(112)
For a solution to exist we demand (1− 1
c2
)< 0 (113)
c2 − 1
c2︸ ︷︷ ︸=d
< 0 (114)
c ∈ (−1, 1) (115)
Further, if c ∈ (−1, 1) \ {0} then
x3 = ±√− (1 + x2
2) d−1 (116)
And for all c ∈ (−1, 1) \ {0} we found the intersection.
Definition 4.16: Lorentz normal.
The Lorentz normal or Lorentz orthogonal to a plane P = {ax1 + bx2 + cx3 = 0} is n =(abc
)Definition 4.17: Hyperbolic Angle V1.
Let L1 = H2 ∩ Π1, L2 = H2 ∩ Π2 be two lines in the hyperbolic space intersecting at a point P . The
hyperbolic angle between L1 and L2 at P is defined as
arccos
(〈n1, n2〉L‖n1‖L ‖n2‖L
)∈ (0, π]
Where ni is the Lorentz normal of plane Πi
Remark. . Note that this is well defined as∣∣∣ 〈n1,n2〉L‖n1‖L‖n2‖L
∣∣∣ < 1. To see this observe that n1 is perpendicular
to Π1 which intersects H2 and so contains time-like vectors. Hence n1 is perpendicular to a time-like vector
and by assignment 3 it is space-like. The same holds for n2 and by Cauchy Schwartz our inequality holds.
Definition 4.18: Hyperbolic Angle V2.
Let x, y ∈ Rn be space-like vectors with x×L y being time-like. Then the hyperbolic angle between x and y
is defined as
θH (x, y) = arccos
(〈x, y〉L‖x‖L ‖y‖L
)∈ (0, π]
49
Let L1 = H2∩Π1, L2 = H2∩Π2 be two lines in the hyperbolic space intersecting at a point x. The hyperbolic
angle between L1 and L2 at x is defined as
θH (x×L y, x×L z)
where y ∈ L1, z ∈ L2.
Remark. . This will be the definition we use most often
Definition 4.19: Orthogonal Lines.
Let L1 = H2 ∩ Π1, L2 = H2 ∩ Π2 be two lines in the hyperbolic space intersecting at a point x. They are
said to be (Lorentz) orthogonal if
θH (x×L y, x×L z) =π
2
Or alternatively if
〈x×L y, x×L z〉L = 0
Theorem 4.12: Investigation of Line Intersection.
Let L1 = H2 ∩Π1, L2 = H2 ∩Π2 be two lines in the hyperbolic space, with Π1 ∩Π2 = Rv. Then
L1 ∩ L2 =
φ If v is space-like
φ If v is light-like
Single Point If v is time-like
Proof. We separate by cases
1. Let v be space like:
By theorem 4.11 we have the general form of L1 ∩ L2. In this case there is a unique line orthogonal
to both L1 and L2. Let n3 be Lorentz orthogonal to n1, n2, then the space Π3 defined by it is Lorentz
orthogonal to Π1,Π2.
By definition v ∈ Π1,Π2 and so v is Lorentz orthogonal to n1, n2 as well - hence v and n3 are linearly
dependent. And so, v is normal to Π3 - and Π3 is the unique plane Lorentz orthogonal to v. Lastly,
let us verify L3 = Π3 ∩H2 does in fact intersect L1, L2.
By definition of normal
Π1 ∩Π3 ={w ∈ R3 : 〈w, n1〉L = 0 ∧ 〈w, n3〉L = 0
}= R (n1 ×L n3) (117)
We know n1, n3, n1 ×L n3 are Lorentz orthogonal and so they are linearly independent. Let t be a
time-like vector, the three form a basis and so
〈t, t〉L = λ21 〈n1, n1〉L︸ ︷︷ ︸
>0
+λ22 〈n2, n2〉L︸ ︷︷ ︸
>0
+λ23〈n1 ×L n3, n1 ×L n3〉L (118)
Meaning we must have 〈n1×Ln3, n1×Ln3〉L < 0, and so n1×Ln3 is time like and by the above theorem
the intersection Π1 ∩Π3 is non empty. The same reasoning applies to Π2.
50
2. Let v be light like:
Again by the previous theorem their intersection is empty, we claim there is no hyperbolic line orthog-
onal to both L1 and L2. Let Π3 be orthogonal to both, hence by the same reasoning n3 and v are
linearly dependent, meaning n3 light like. Ergo Π3 ∩ H2 = φ as a normal to a time-like vector must
be space-like.
3. Let v be time like:
By the theorem4.11 their intersection contains a single point. Moreover, by the same reasoning as
before there is no common orthogonal line.
Example 4.3.
Let Πc = {cx1 − x2 = 0} , then any Πc1 ,Πc2never intersect
Let us now prove the hyperbolic equivalent to our Euclidean Parallel Postulate.
Theorem 4.13.
Let L = Π ∩H2 be a hyperbolic line and P ∈ H2 \ L then
1. There exists a unique hyperbolic line W orthogonal to L passing through P . Denote Q = Q ∩ L.
2. There is a unique hyperbolic line Morthogonal to PQ passing through P and M ∩ L = φ
3. There are infinitely many lines through P that do not intersect L
Proof. Let L = Π ∩ H2 be a hyperbolic line. As Π intersects the hyperbolic space, it contains time like
vectors and so Π⊥ = Rw where w is space like. Therefore
R3 = Π⊕Π⊥ = Π⊕ Rw (119)
let us write P as a decomposition of the two
P = q + v , Where q ∈ Π, v ∈ Rw (120)
Further, we know v is space-like and want to show q is time-like. We compute
〈P, P 〉L = 〈q + v, q + v〉L = 〈q, q〉L + 〈v, v〉L︸ ︷︷ ︸<0
+2 〈q, v〉L︸ ︷︷ ︸=0 As v∈Π⊥
(121)
And as P is in the hyperbolic space it must be time-like, meaning 〈q, q〉L < 0. Or equivalently q is time-like.
We now define a Lorentz orthonormal basis of R3
q
|‖q‖L|,
q ×L v
‖q ×L v‖L,
v
‖v‖L(122)
51
And so there exists a Lorentz transformation that maps
e1 7→q
|‖q‖L|(123)
e2 7→q ×L v
‖q ×L v‖L(124)
e3 7→v
‖v‖L(125)
Therefore we may assume that
e1 =q
|‖q‖L|, e2 =
q ×L v
‖q ×L v‖L, e3 =
v
‖v‖L(126)
Under this new basis we have that
Π = (Rw)⊥
= (Rv)⊥
= (Re3)⊥
={x ∈ R3 : x3 = 0
}(127)
P = q + v ∼= te1 + ye3 =⇒ q = (t, 0, y) (128)
As P ∈ H2 we have t =√
1 + y2. The line W orthogonal to L is determined by the plane Π1 ={x ∈ R3 : x2 = 0
}, and this line clearly contains P . Further, we have that
L ∩W = Π ∩{x ∈ R3 : x2 = 0
}∩H2 =
{x ∈ R3 : x3 = x2 = 0 ∧ ‖x‖L = −1
}= {(1, 0, 0)} (129)
The unique line M orthogonal to W through P is determined by M = Π1∩H2 where Π2 is Lorentz orthogonal
to Π1. Note that if(n1n2n3
)is the normal to Π1, we must have n2 = 0 from our choice Π1 =
{x ∈ R3 : x2 = 0
}.
Further as P ∈ Π1 we have
−n1
(1 + y2
)+ n3y = 0 (130)
n1 =1√
1 + y2n3 (131)
and therefore Π1 has normal
(1√
1+y2
01
)and as Π2 is perpendicular to Π1 all we have that
Π2 =
{x ∈ R3 : − 1√
1 + y2x1 + x3 = 0
}(132)
and the line M = Π2 ∩H2 was determined uniquely.
Lastly, hyperbolic lines through P are of the form M ′ = Π3 ∩H2
Π3 =
{x ∈ Rn : − 1√
1 + y2x1 +mx2 + x3 = 0
}(133)
where
(− 1√
1+y2
m1
)is a Lorentz orthogonal to the plane, as P must satisfy the equation to the plane. Among
these lines the ones that do not intersect L are those s.t. Π ∩Π3 is not time-like.
52
Let us now write down their intersection
Π ∩Π3 =
{x ∈ Rn : − 1√
1 + y2x1 +mx2 + x3 = 0 ∧ x3 = 0
}=
{x ∈ Rn : − 1√
1 + y2x1 +mx2 = 0
}(134)
Pick x1 =√
1 + y2, then x2 = ym and so
Π ∩Π3 = R
√
1 + y2
ym
0
(135)
and this vector is not time-like for all m s.t.
−(1 + y2
)+y2
m2≥ 0 (136)
m2 ≤ y2
1 + y2(137)
Meaning that for all values of m ∈[− |y|√
1+y2, |y|√
1+y2
]the intersection of M ′ and L is empty.
We now wish to calculate the sum of angles in a hyperbolic triangle. To this end, we first prove the following
lemma.
Lemma 4.14: Norm of cross product.
Let x, y ∈ R3 be space-like and x×L y be time-like, then
|‖x×L y‖L| = ‖x‖L ‖y‖L sin θH (x, y)
Proof. By Lemma 4.3 part (5′)
‖x×L y‖2L =〈x, y〉2L − ‖x‖2L ‖y‖
2L (138)
= ‖x‖2L ‖y‖2L cos2 θH (x, y)− ‖x‖2L ‖y‖
2L = ‖x‖2L ‖y‖
2L
(cos2 θH (x, y)− 1
)(139)
= ‖x‖2L ‖y‖2L sin2 θH (x, y) = ‖x‖L ‖y‖L sin θH (x, y) (140)
Theorem 4.15: Sum of angles.
Let T be a hyperbolic triangle with vertices x, y, zand respective angles α, β, γ. Then
α+ β + γ < π
Proof. Let
u =x×L y
‖x×L y‖L, v =
z ×L y
‖z ×L y‖L, w =
z ×L x
‖z ×L x‖L(141)
53
These are three space like unit vectors, for which
u×L v =(x×L y)×L (z ×L y)
‖x×L y‖L ‖z ×L x‖L=〈x×L y, z〉Ly −
=0︷ ︸︸ ︷〈y, x×L y〉L z
‖x×L y‖L ‖z ×L x‖L=
〈x×L y, z〉L‖x×L y‖L ‖z ×L x‖L
y (142)
by lemma 4.3 part (5). Because y is time-like we have that u ×L v is time-like, with an analogous proof
holding for v ×L w. Now let us calculate
cos [θH (u, v) + θH (v, w)] = cos θH (u, v) cos θH (v, w)− sin θH (u, v) sin θH (v, w) (143)
=〈u, v〉L〈v, w〉L
‖u‖L ‖v‖L ‖v ‖w‖L‖L− sin θH (u, v) sin θH (v, w) (144)
But by choice of u, v and w they are unit vectors and so
cos [θH (u, v) + θH (v, w)] = 〈u, v〉L〈v, w〉L − sin θH (u, v) sin θH (v, w) (145)
And by the preceding lemma
cos [θH (u, v) + θH (v, w)] = 〈u, v〉L〈v, w〉L − ‖u× v‖L ‖v × w‖L (146)
By Cauchy Schwartz for space like vectors
cos [θH (u, v) + θH (v, w)] > 〈u, v〉L〈v, w〉L + 〈u×L v, v ×L w〉L (147)
By lemma 4.3 again
cos [θH (u, v) + θH (v, w)] >(((((((〈u, v〉L〈v, w〉L + 〈u,w〉L〈v, v〉L −(((((
((〈v, w〉L〈u, v〉L (148)
cos [θH (u, v) + θH (v, w)] > 〈u,w〉L 〈v, v〉L︸ ︷︷ ︸=1
(149)
cos [θH (u, v) + θH (v, w)] > 〈u,w〉L = cos θH (u,w) (150)
Lastly, using simple trigonometry we get that
θH (u,w) > θH (v, w) + θH (u, v) (151)
Or (152)
2π − θH (u,w) < θH (v, w) + θH (u, v) (153)
We now separate to cases depending on the two inequalities. If the first inequality holds then
θH (u,w) = arccos〈x×L y, z ×L x〉L‖x×L y‖L ‖z ×L x‖L
= arccos−〈x×L y, x×L z〉L‖x×L y‖L ‖x×L z‖L
= π − α (154)
θH (u, v) = arccos〈x×L y, z ×L y〉L‖x×L y‖L ‖z ×L y‖L
= β (155)
θH (v, w) = arccos〈z ×L y, z ×L x〉L‖z ×L y‖L ‖z ×L x‖L
= γ (156)
meaning
π − α > β + γ (157)
π > α+ β + γ (158)
54
and our desired result is true. Let us now assume the latter inequality is true and reach a contradiction.
Assume without loss of generality that α is the largest angle of the triangle, by the inequality
2π − (π − α) < β + γ (159)
π + α < β + γ (160)
which is a contradiction with our choice of α. Therefore our first inequality must hold and we are happy.
Remark. . This result can be strengthened to
Area (T ) = π − (α+ β + γ) (161)
Theorem 4.16.
Given that α, β, γ ∈ (0, π) are three angles s.t. α + β + γ < π then there exists a hyperbolic triangle with
angles α, β, γ which is unique up to isometry.
Proof. In the book
55
5 Projective Geometry
We continue to the last geometry we explore, projective geometry. On every real vector space, we can define
its projective geometry as follows,
Definition 5.1: Projective Space .
Let V be a real vector space, its Projective space is the quotient
P (V ) = ((V \ 0))/
((v ∼= λv))
for v ∈ V and λ ∈ R.
Remark. . We can also think of P (V ) as the set of lines in V through the origin, i.e. where
[v]︸︷︷︸∈P(V )
∼= Rv
We denote the map between the two by π (x)
Definition 5.2: Projective Space Rn.
In particular for the vector space V = Rn+1 we write
Pn = P(Rn+1
)In which case
Pn =((Rn+1 \ 0
))/((v ∼= λv)) ∼=
{Lines through origin in Rn+1
} ∼= (Sn)/
((x ∼= −x))
Where the last isometry follows from the fact that if we look at a line in Rn+1, it is uniquely defined by the
two points it intersects the unit sphere. In Rn we associate the element [x] = [x1, . . . , xn] with the vector
(x1, . . . , xn)
Remark. . There exists an invective map Rn → Pn that takes (x1 . . . , x2) 7→ [1, x1, . . . , xn]. It is injective
as if two vectors lie on the same line, one is a scalar multiple of the other, but as both have 1 in their first
component that multiple has to be 1.
We now define the notion of projective lines.
Definition 5.3: Projective subspace.
A k dimensional Projective subspace of P (V ) is the image in P (V ) of a k + 1 dimensional subspace of V
under π.
Remark. . for W ⊆ V holds π (W ) ∼= P (W )
Remark. . We define dimP (W ) = dimW − 1 for any vector subspace W ⊆ Rn+1
Definition 5.4: Projective line.
A Projective line is the image in π of a subspace of dimension 2.
Terrific, with lines defined we continue to define the spanning set in our projective geometry.
56
Definition 5.5: Projective Span.
Let Σ ⊆ P (V ) a subset and Σ̃ = { Union of lines in Σ} ∼= π−1 (Σ), then we define the span of Σ̃ and Σ by
〈Σ̃〉 = Span Σ̃
〈Σ〉 = π(〈Σ̃〉)
And this is a Projective subspace of V .
Definition 5.6: Linear independence.
We say a set p1, . . . , pk ∈ P (V ) is linearly independent if the span P (p1, . . . , pk) is of dimension k − 1. We
say that it generates V if
P (〈p1, . . . , pk〉) = P (V )
Remark. . A set p1, . . . , pk ∈ P (V ) is linearly independent if and only if there exist representatives pi = [vi]
for all i s.t. v1, . . . , vk are linearly independent in V .
Remark. . Let W,E be two vector subspaces of V , then
dim (P (W ) ∩ P (E))︸ ︷︷ ︸=P(W∩E)
= dimP (W ) + dimP (E)− dim〈P (W ) ∪ P (E)〉
To see this, note that
dim (P (W ) ∩ P (E)) = dimP (W ∩ E) = dim (W ∩ E)− 1 = dimW + dimE − dim〈W ∩ E〉 − 1
= dimW − 1 + dimE − 1− [dim〈W ∩ E〉 − 1] = dimP (W ) + dimP (E)− dim〈P (W ) ∪ P (E)〉
Definition 5.7: Projective transformation.
A linear bijection L : V → V induces a map L̄ : P (v)→ P (V ) mapping [v] 7→ [L (v)].
Remark. . To check that it is in fact a well defined map take [v] = [λv]. As L is a linear map it will take v
and λv to the same line .
Remark. . The two functions L and λL define the same induced map
λ̄L ([v]) = [λLv] = [L (v)] = L̄ ([v])
Definition 5.8: Group of Projective transformations.
The group of Projective transformations of P (v) is
PGLn+1 (R) = (GLn+1 (R))/
(λIn+1 ∀λ ∈ R∗)
where GLn+1 (R) = {A ∈Mn+1 (R) : |A| 6= 0}. A Projective transformation is a member of PGLn+1 (R),
which induces a map Pn → Pn giving [v] 7→ [Av].
Having defined the span, we continue to the projective space’s equivalent of a basis.
Definition 5.9: Projective frame.
A projective frame of Pn consist of n+ 2 elements p0, p1 . . . , pn+1 s.t. any distinct n+ 1 of them form a basis
of Pn.
57
Lemma 5.1.
The points p0, p1 . . . , pn+1 form a projective frame if and only if pi = [fi] for fi ∈ Rn+1 s.t. f1, . . . , fn+1 are
a basis for Rn+1 and
f0 =
n+1∑i=1
λifi Where λ 6= 0∀i
Proof. Suppose by p0, p1 . . . , pn+1 form a projective frame, then p1, . . . , pn+1 span Rn+1. By definition the
representatives p1 = [f1] , . . . , pn+1 = [fn+1] span Rn+1and so ∃λi s.t.
f0 =
n+1∑i=1
λifi (1)
Assume by contradiction that λj = 0 for some j. Then
f0 =
n+1∑i=1i6=j
λifi (2)
I.e. there is nontrivial linear combination 0 in terms of f0, f1, . . . , fj−1, fj+1, . . . , fn+1
0 =
n+1∑i=1i 6=j
λifi − f0 (3)
But this means they are linearly dependent, which is a contradiction that they must span Rn+1.
Now let p0, p1 . . . , pn+1 be points where pi = [fi] for fi ∈ Rn+1 s.t. f1, . . . , fn+1 are a basis for Rn+1 and
f0 =
n+1∑i=1
λifi Where λ 6= 0∀i (4)
we wish to show that ∀j ∈ [n+ 1] the set f0, f1, . . . , fj−1, fj+1, . . . , fn+1 spans Pn. Alternatively, we wish
them to be linearly independent. Assume that there exists coefficients µi s.t.
0 =
n+1∑i=0i 6=j
µifi (5)
By assumption
0 = µf0 +
n+1∑i=1i 6=j
µifi = µ0
n+1∑i=1
λifi +
n+1∑i=1i6=j
µifi = µ0λjfj +
n+1∑i=1i 6=j
(µi + λi) fi (6)
Since by assumption f1, . . . , fn+1 are linearly independent, we have that
µ0λj = 0 (7)
(µi + λi) = 0∀i (8)
by assumption we know that λi 6= 0 for all i and so µi = 0∀i. Hence the vectors are linearly independent,
and the pi are a projective frame.
58
Example 5.1: standard projective frame .
The standard projective frame of Pn is [( 11...1
)], [e1] , . . . , [en+1] (9)
Proposition 5.2: Bijection PGLn+1 (R) to action on basis.
There is a bijection PGLn+1 (R) to the group of projective frames of Pn taking
[A] 7→
{[A
( 11...1
)], [Ae1] , . . . , Aen+1
}
Remark. . This is analogous to our bijection between linear transformations and their representing matrix
by the euclidean basis.
Proof. Firstly note that the map is well defined by our prior lemma as ∀A ∈ PGLn+1 (R) we have that A is
invertible by definition - and so it preserves linear independence.
Let us remark that
A
( 11...1
)=
n+1∑i=1
Aei (10)
Now take A,B ∈ PGLn+1 (R) s.t. the translation maps both to the same set, i.e.[A
( 11...1
)]=
[B
( 11...1
)]and [Aei] = [Bei]∀i (11)
and we need to show that [A] = [B] ∈ PGLn+1 (R). By the second condition above we have that
Aei = λiBei , ∀i, Where λi 6= 0 (12)
For our condition [A] = [B] we need A = λB with λ fixed. But by the first equation above
A
( 11...1
)= µB
( 11...1
)(13)
n+1∑i=1
Aei = µ
n+1∑i=1
Bei (14)
n+1∑i=1
λiBei = µ
n+1∑i=1
Bei (15)
n+1∑i=1
(λi − µ)Bei = 0 (16)
But since ei are linearly independent we have that λi = µ for all i, or equivalently A = µB. Hence [A] = [B].
59
Lastly, we need to show the map is surjective. Take a projective frame represented by[n+1∑i=1
λifi
], [f1] . . . , [fn] (17)
There is a unique matrix A ∈ PGLn+1 (R) s.t.
Aei = fi∀i (18)
We compute
A
( 11...1
)=
n∑i=1
Aei (19)
And take A′ s.t. Aei = λfi. Now
A′
( 11...1
)=
n∑i=1
A′ei =
n∑i=1
λfi = f0 (20)
and so [A′] is our required preimage.
Akin to previous chapters, let us now examine line intersections and triangles in projective space.
Lemma 5.3: Classification of Projective Lines.
Every two distinct projective lines in P2 intersect in a unique point and any two distinct points in P2 belong
to a unique line
Proof. Let L1 = P (V1) , L2 = P (V2) be two distinct lines in P2 for V1, V2 ⊆ R3. Then
L1 ∩ L2 = P (V1) ∩ P (V2) = P
V1 ∩ V2︸ ︷︷ ︸Planes in R3
= P(line in R3
)= One Point (21)
For the other direction let x, y ∈ P2 with representatives p1, p2 ∈ S2. As x, y are distinct p1, p2 cannot be
antipodal, and so define a great circle. The image of that circle is the unique line of P2 containing x, y.
Alternative Proof for the first implication. Recall that we can view P2 as(S2)/
(x ∼= −x) . Let L1, L2 be
the images of two spherical lines ζ1, ζ2. Then L1 ∩ L2 is the image of ζ1 ∩ ζ2, but this is the intersection of
two great circles, which we know intersect only at two antipodal points. Hence, both points are congruent
under our quotient and L1 ∩ L2 intersect at only one point.
Definition 5.10: Projective Triangle.
A projective triangle consists of three non-collinear points x, y, z ∈ P2 and the three segments of projective
lines joining them.
60
Definition 5.11: Introspective Triangles.
Two projective triangles 4PQR,4P ′Q′R′ are introspective from some point O ∈ P2 if
OPP ′ Are Collinear
OQQ′ Are Collinear
ORR′ Are Collinear
Theorem 5.4: Desargues.
Let 4PQR,4P ′Q′R′ be two projective triangles in P2 introspective from a point O, then the points
A = 〈P,Q〉 ∩ 〈P ′, Q′〉
B = 〈P,R〉 ∩ 〈P ′, R′〉
C = 〈R,Q〉 ∩ 〈R′, Q′〉
are collinear
Proof. Let us choose representatives p, q, r, p′, q′, r′, o ∈ R3 for P,Q,R, P ′, Q′, R′, O ∈ P2 respectively. By
our assumption O,P and P ′ are collinear and so their representatives o, p and p′ are collinear in R3 -with
the same holding for the latter two triads.
Therefore there exist λ, λ′, µ, µ′, δ, δ′ ∈ R s.t.
O =λp+ λ′p′ (22)
O =µq + µ′q′ (23)
O =δr + δ′r′ (24)
By the first two equations
λp− µq = −λ′p′ + µ′q2 =⇒ [λp− µq] ∈ 〈P ′, Q′〉 (25)
Further, trivially we have that [λp− µq] ∈ 〈P,Q〉 and so
[λp− µq] ∈ 〈P,Q〉 ∩ 〈P ′, Q′〉 (26)
But by lemma 5.3 that intersection contains only one point A, meaning
[λp− µq] ∈ 〈P,Q〉 ∩ 〈P ′, Q′〉 = A (27)
Similarly, using the second and third equations
µq − δr = −µ′q′ + δ′r′ =⇒ [µq − δr] ∈ 〈Q,R〉 ∩ 〈Q′, R′〉 = C (28)
And using first and third equations
λp− δr = λ′p′ + δ′r′ =⇒ [λp− δr] ∈ 〈P,R〉 ∩ 〈P ′, R′〉 = B (29)
61
And so we found representatives of A,B,C. By definition, A,B,C are collinear if and only if their represen-
tatives admit a linear rotation, which is simple to find
1 ∗ (λp− µq) + 1 ∗ (λp− δr)− 1 ∗ (µq − δr) = 0 (30)
And so A,B,C are collinear.
Remark. . This theorem also holds in R2 if we further assume that the lines intersect one another
Remark. . The converse also holds. If we have two triangles with collinear intersection of edges then there
exists an O s.t. the triangles are introspective with respect to it. The proof is essentially the same in the
backwards direction.
Remark. . The theorem also holds for Pn for n ≥ 2, this can be proved by proving it for n = 3 and reducing
all other cases to that.
Theorem 5.5: Pappo’s Theorem.
Let L,L′ be two projective lines in P2 and P,Q,R ∈ L and P ′Q′, R′ ∈ L′ be distinct points none of which is
in L ∩ L′. Then if we denote
A = 〈Q,R′〉 ∩ 〈Q′, R〉
B = 〈P,R′〉 ∩ 〈P ′, R〉
C = 〈P,Q′〉 ∩ 〈P ′, Q〉
The points A,B,C are collinear
Proof. By assumption any three points among P,Q, P ′ and Q′ are linearly independent, in other words
P,Q, P ′ and Q′ form a projective frame of P2. By our theorem we can choose a projective transformation
from P2 to itself which maps
[e1] 7→ P , [e2] 7→ Q (31)
[e3] 7→ P ′ ,
[( 11...1
)]7→ Q′ (32)
Since projective transformation preserve projective lines we may assume without loss of generality that
[e1] = P , [e2] = Q (33)
[e3] = P ′ ,
[( 11...1
)]= Q′ (34)
Under the new naming we have that
R ∈ 〈P,Q〉 = 〈[e1] , [e2]〉 =[(
x0x10
)](35)
R′ ∈ 〈P ′, Q′〉 = 〈[e3] ,
[( 11...1
)]〉 =
[(x′0x′0x′2
)](36)
62
Let us now compute C. The span 〈P,Q′〉 is defined by the plane of R3 orthogonal to13 e1 ×(
111
)=(
0−11
).
We can do the same for 〈P ′, Q〉 to get the normal(−1
00
), and to get C we find the line orthogonal to both
normals
C =(
0−11
)×(−1
00
)=(
0−1−1
)(37)
Similarly for A and B
B =
(((100
)×(x′0x′0x′2
))×((
001
)×(x0x10
)))=
(0−x′2−x′0
)×(−x1
x00
)=
(−x0x′2
−x1x′0
−x1x′2
)(38)
A =
(((010
)×(x′0x′0x′2
))×((
111
)×(x0x10
)))=
(x′20−x′0
)×( −x1
x0x1−x0
)=
(x0x′0
x1x′0−(x1−x0)′2x0x′2
)(39)
and so, we found representatives for A,B and C in R3. By definition A,B and C lie on the same projective
line if and only if these representatives are linearly dependent. By calculation
−(
x0x′0
x1x′0−(x1−x0)′2x0x′2
)︸ ︷︷ ︸
[B]
+ (x1 − x0)
(x0x′0
x1x′0−(x1−x0)′2x0x′2
)︸ ︷︷ ︸
[C]
=
(x0x′0
x1x′0−(x1−x0)′2x0x′2
)︸ ︷︷ ︸
[A]
(40)
and so they are linearly dependent, A,B,C are collinear, and we’re happy.
13which is P ×Q′
63
6 Axiomatic Projection Geometry
We finish the module with an aside on Axiomatic Projective Geometry, being the most abstract and general
geometry we deal with.
Definition 6.1: Axiomatic Projection Plane.
An axiomatic projection plane is two sets P (the set of points) and L (the set of lines) with the relation
I ⊆ P × L where I denotes which points belong to which lines, s.t.
1. Every line contains at least three points, or alternatively ∀l ∈ L∃x, y, z ∈ P s.t. (x, l) , (y, l) , (z, l) ∈ I
2. Every point is contained in at least three distinct lines, or alternatively ∀x ∈ P∃l, k, h ∈ L s.t.
(x, l) , (x, h) , (x, k) ∈ I
3. There is a unique line between any two distinct points, or alternatively ∀x, y ∈ P∃!l ∈ L s.t. (x, l) , (y, l) ∈I
4. Any two distinct lines intersect in exactly one point , or alternatively ∀l, k ∈ L∃!x ∈ P s.t. (x, l) , (x, k) ∈I
Example 6.1.
P2 is an axiomatic projection plane.
Example 6.2.
Let P,L be two sets of 7 points each, where P is the 3 basis vectors of R3 completed into a cube14 and the
lines are the 2D vector subspaces of F32
Definition 6.2: Division Ring (Skew Field).
A division ring is a ring K s.t. every non-zero element k ∈ K is invertible (also called a Skew Field)
Example 6.3.
The Quaternions are a division ring
Definition 6.3: Projective plane of a Division ring.
The projective plane of a division ring K is defined as the axiomatic plane with the set of points
P2 (K) =(K3 \ 0
)/(v∼=λvv∈K2\0λ∈K\0
)(1)
and set of lines being the set of 2D right subspaces of K3, where a right subspace V is
V ={x ∈ K3 : x1a+ x2b+ x3c = 0
}, For some a, b, c ∈ K (2)
Or more precisely
L =({
(a, b, c) ∈ K3})/
((a, b, c) ∼= (a, b, c)λ) (3)
14or alternatively F32 \ 0
64
With the relation I defined as[(xyz
)]∈[(
abc
)]⇐⇒ 〈
(xyz
),(abc
)〉 = xa+ yb+ zc = 0 (4)
and note that this is well defined by our choice of directions for the multiplication of the congruences.
Remark. . P2 (K) is an axiomatic projective plane, and satisfies Desargues’ theorem.
Remark. . We can extend this definition to define Pn (K) for all n ≥ 2
Example 6.4.
P2 (R) = P2
Example 6.5.
P2 (F2) = example 5.2
Theorem 6.1: Hilbert.
Let I ⊆ P ×L be an axiomatic projective plane which satisfies Desargues’ theorem. Then there is a bijection
P ∼= P2 (K) for some division ring K. Moreover, P2 (K) satisfies Pappo’s theorem if and only if K is
commutative .
Sketch Proof. The crucial idea is if we define P1 (K) =(K2 \ 0
)/(v ∼= λv) without a point is equivalent to
K. A line in P2 (K) is isomorphic to a copy of P1 (K). Thus, we want to choose a line in L and remove a
point.
From the axioms there exists four points x, y, o and i s.t. no three of them belong to the same line. Define
the set K = 〈o, i〉 \ 0 and to endow it with a ring structure denote
A′ = 〈x,A〉 ∩ 〈o, y〉 (5)
B′ = 〈A, z〉 ∩ 〈y,B〉 (6)
And define
A+B = 〈x,B′〉 ∩ 〈o, z〉 (7)
Further, denote the 0-element of the ring by o. To see this note
o′ = 〈x, o〉 ∩ 〈o, y〉 = o (8)
B′ = 〈o, z〉 ∩ 〈x,B〉 = B (9)
o+B = 〈x,B〉 ∩ 〈o, z〉 = B (10)
And we can do the same thing for addition from the right. The definition for multiplication is similar, with
the identity being i. Note that associativity uses Desargues’ theorem.
Lastly, once we defined the ring we need to find the bijection P ∼= P2 (K). First we look at P with the line
xy removed, this satisfies
P \ 〈x, y〉 7→ K2 (11)
A 7→ (〈o, i〉 ∩ 〈y,A〉, 〈o, i〉 ∩ 〈x,A〉) (12)
But this map only get us points where the second component is non-zero, and so we need to extend it to the
inclusion of 〈x, y〉. The idea is that the missing points are of the form (〈x, y〉, 0)which is a line.
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