CLASS NOTES FOR MATH 3400:
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
ROBERT JUDD
Abstract. We discuss aspects of Euclidean geometry including isometries of the plane, affine
tranformations in the plane and symmetry groups. We then explore similar concepts in the sphere
and projective space, and explore elliptic and hyperbolic geometry.
1. Introduction
This course is intended for graduate and upper level undergraduate students. Only a basic math
background is required: an introduction to proofs and a familiarity with vector and matrix algebra
(scalar product, cross product, inverses of matrices) in R2 and R3 .
Almost every area of mathematics may be viewed from a geometric viewpoint and this viewpoint
often leads to new results or new proofs of existing results. It is the intuition that one can gain
from looking at a problem geometrically that can be very important in finding a solution. However,
in order to best appreciate this approach one must first learn the basic ideas that underpin the
study of geometry.
Geometry as a subject in itself was first studied by listing the axioms that seemed to embody the
fundamental concepts and then using these axioms to prove “known” results, i.e. to put every-day
ideas into a rigorous framework. In Section 2 we briefly discuss this axiomatic approach and show
how it can be used to construct the familar Euclidean geometry. In the following sections we use
a more analytic method to investigate the properties of the Euclidean plane, the 2-Sphere and the
projective plane.
2. Axiomatic Geometry
Each theorem in a deductive system must be derived from previous theorems which must in turn
be derived from previous theorems. Since this process must start somewhere we take a set of agreed
upon notions as fact. These are called axioms or postulates. From these we may deduce theorems.
We shall illustrate the axiomatic method with an example.
Definition 2.1. Blobs and rocks are objects that satisfy the following axioms:
P1
P2
P3
P4
There exist at least two rocks.
For any two distinct rocks there exists a unique blob containing both.
Given any blob there is a rock not in it.
Given any blob B and a rock r not in B there exists a unique blob C containing r disjoint
from B.
1
2
ROBERT JUDD
In the spirit of being unable to define every concept we want to use, we have not defined what
it means for one object to contain another. In fact “contains” is simply a binary relation, but
does not necessarily mean “live inside”. Notice that there is no axiom above preventing rocks from
containing blobs.
We need some more definitions before we can proceed.
Definition 2.2. If two blobs B and C contain the same rock r, then we say that B and C intersect
or meet in r.
Definition 2.3. Two blobs B and C are disjoint if they have no rock in common.
Definition 2.4. The unique blob containing the pair of rocks r and s will be denoted by rs.
Note that if a blob contains rocks p, q and r then pq, qr and pr all denote the same blob. We
are not claiming that blobs may contain 3 rocks; in fact this is one of the questions that we shall
attempt to answer. We would like to know if there is a minimum or maximum for the number of
rocks, the number of blobs, and the number of rocks a blob may contain.
Theorem 2.5. There are at least two disjoint blobs.
Proof.
(P1)
(P2)
(P3)
(P4)
Let
Let
Let
Let
p and q be two different rocks.
B = pq be the blob containing p and q.
r be a rock not in B and, consequently, different from p and q.
C be a blob disjoint from B containing r.
It is now clear that B and C are two disjoint blobs as required.
Theorem 2.6. Every blob contains at least one rock.
Proof. Assume that there exists an empty blob B. Since B is empty it follows that any non-empty
blob is disjoint from B. Let r and s be two distinct rocks (P1). Then A = rs (P2) is disjoint from
B. Let t be a rock not in A (P3) and C = rt (P2). Blob C is also disjoint from B and C 6= A. We
now have two different blobs, A and C, disjoint from B that contain r. This contradicts (P4) and
hence there are no empty blobs.
Theorem 2.7. Two distinct blobs, each disjoint from a third blob are disjoint from each other.
Proof. Let A and B be two distinct blobs, both disjoint from a blob C. Suppose A and B are
not disjoint. Let r be a rock common to both A and B. We have produced two different blobs
that contain the same rock not in C and are disjoint from C. This contradicts (P4) and hence the
assumption that A and B are not disjoint.
Theorem 2.8. Every blob contains at least two rocks.
Proof. Suppose there exists a blob B containing exactly one rock r. Let s be a rock not in B (P3),
and C = rs (P2). Let t be a rock not in C (hence, not in B) (P3). Let A be a blob containing t
that is disjoint from C (P4). A must be disjoint from B as the only rock of B is a rock of C. Blobs
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
3
C and B are different and disjoint from A. But B and C are not disjoint from each other, a fact
that contradicts Theorem 2.7 and hence no blob may contain only one rock.
Definition 2.9. Two sets A and B, finite or infinite, have the same size if there is a one-to-one
correspondence between the elements of set A and the elements of set B.
Theorem 2.10. The number of blobs containing a given rock r is the same as the number of blobs
containing any other rock s.
Proof. Let r and s be two different rocks (P1). We shall construct a bijection from the set of blobs
containing r and the set of blobs containing s. Let B = rs be the unique blob containing r and s
(P2). Let t be a rock not in B (P3) and C a blob containing t disjoint from B (P4). Note that
B is the unique blob containing r disjoint from C. Thus, each blob containing r, except B, will
meet C in a rock (P4) which is unique (P2). (Note that two different blobs cannot have two rocks
in common). Thus there is a one-to-one correspondence, or bijection α, from blobs through r that
meet C to rocks in C. Repeating the above argument with s we obtain a bijection β from blobs
through s that meet C to rocks in C. A one-to-one correspondence between blobs through r and
blobs through s is now easily obtained by defining f (B) = B and f (A) = β −1 (α(A)), for blobs
A 6= B that contain r.
Theorem 2.11. All blobs contain the same number of rocks, i.e., have the same size.
Proof. Let B be a blob (P1,P2) and r a rock not in B (P3). Except for the unique blob containing
r disjoint from B (P4), every blob containing r has a unique rock in common with B (P2, Theorem 2.7). Thus, different blobs containing r determine different rocks of B and each rock of B
determines a distinct blob containing r. Notice that the number of rocks in blob B is one less than
the number of blobs containing the rock r. Similarly, for any other blob C let s be a rock not in C.
The number of rocks in C is one less than the number of blobs containing s. Since the number of
blobs containing r is the same as the number of blobs containing s (Theorem 2.10) it follows that
the number of rocks in B is the same as the number of rocks in C.
If you provide interpretations for undefined terms (such as blob and rock) that convert postulates
into true statements then all of the derived theorems (properly interpreted) are also true. This is
called a model. A popular model, for instance, is obtained by interpreting rocks as points and blobs
as lines in the Euclidean plane.
Homework: Provide a model for the above axioms other than lines and points.
Solution 1. Consider a tetrahedron. Let rocks be the vertices of the tetrahedron and blobs be the
edges. Notice how all the theorems proved above continue to hold.
Solution 2. Consider the same tetrahedron. Let rocks be the faces of the tetrahedron and blobs
be the edges. Again, all the theorems proved above continue to hold.
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ROBERT JUDD
3. Isometries in the Euclidean Plane
In this section we shall consider the Euclidean plane to be the vector space R2 together with
the usual pythagorean distance. We then discuss functions from the plane to itself which preserve
distance. These are called isometries. They are the familiar reflections, rotations, translations
and glide reflections along with combinations of these. The goal of this section and the next is to
provide a complete description of isometries.
3.1. Notation. We shall let Rn denote the vector space {(x1 , . . . , xn ) : xi ∈ R}. For x, y ∈ Rn ,
P
hx, yi = ni=1 xi yi is the inner or dot product of x and y. We shall use this to define the standard
p
distance function on Rn to be d(P, Q) = |P − Q| where |x| = hx, xi, x ∈ Rn . The n-dimensional
Euclidean space En may now be defined as En = (Rn , | · |), i.e., the set Rn together with the
distance function d.
3.2. Examples. This is not the only distance function that may be defined. Other spaces that
are of interest in mathematics are the `np spaces (1 ≤ p ≤ ∞). For 1 ≤ p < ∞ define the function
|·|p : Rn → Rn by
!1/p
n
X
|x|p =
|xi |p
for x = (x1 , . . . , xn ) ∈ Rn ,
1
and for p = ∞ let |x|∞ = max1≤i≤n |xi |. We then define `np as (Rn , |·|p ) for 1 ≤ p ≤ ∞. Notice
that En = `n2 .
We can get an idea of how these different distance functions distort spacial relationships if we
compare the “unit spheres” of these spaces, that is the locus of points {x ∈ R2 , |x|p = 1}. Shown
below are the unit spheres for `2p , with p = 1, 2 and ∞ (working from the inside out).
Definition 3.1. An isometry of E2 is an onto function T : E2 → E2 such that d(T x, T y) = d(x, y).
In other words, isometries preserve distance.
Isometries satisfy certain important properties:
• Isometries are one-to-one functions, since if T x = T y then 0 = d(T x, T y) = d(x, y) which
implies x = y.
• Isometries have inverses which are also isometries. Since an isometry T is a bijection so is
T −1 . Then d(T −1 x, T −1 y) = d(T T −1 x, T T −1 y) = d(x, y).
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
5
• The composition of two isometries is again an isometry. If S and T are isometries, then
d(ST x, ST y) = d(T xT y) = d(x, y).
It is clear that composition of isometries is associative and hence the set of isometries of E2 forms
a group under composition. We denote this group by I(2). (For the definition of a group see
Appendix A.)
3.3. A Concrete Example. We make a small digression here to present a concrete example of
some isometries which fix a given triangle using the familiar ideas of trigonometry.
√
√
Consider the points A(0, 2), B(− 3, −1) and C( 3, −1) in E2 . We are interested in isometries
T : E2 → E2 that map the triangle 4ABC to itself. For simplicity we denote this triangle by 4.
Suppose S and T are two such isometries that also satisfy SA = T A, SB = T B, and SC = T C.
Must S equal T ? In other words, could there be a point x in E2 such that T x 6= Sx? The answer
to this important question is no but we postpone the proof until later (see Theorem 3.2).
Since a vertex of 4 must map to a vertex of 4 and an isometry of our set is fully defined by its
behavior on the vertices of 4, it follows that there are no more than 6 such isometries (as there
are only 3! = 6 permutations of a set of 3 elements). It is, in fact, easy to come up with 6 such
isometries, which must then necessarily constitute all the isometries that map 4 to 4.
Let [xyz] denote the unique isometry that satisfies T A = x, T B = y, T C = z. Our desired set of
isometries includes the following (it would be useful to draw a picture as you are reading this list):
[ABC]
[CAB]
[BCA]
[ACB]
[CBA]
[BAC]
the identity transformation, also denoted by I.
a rotation by 2π/3 around the origin.
a rotation by −2π/3 around the origin.
a reflection about the line through A and the midpoint of BC.
a reflection about the line through B and the midpoint of AC.
a reflection about the line through C and the midpoint of AB.
Since the composition of two isometries results in an isometry we can compute a “multiplication
table” under the composition operator ◦:
◦
I
[CAB]
[BCA]
[ACB]
[CBA]
[BAC]
I
I
[CAB]
[BCA]
[ACB]
[CBA]
[BAC]
[CAB] [BCA] [ACB] [CBA] [BAC]
[CAB]
[BCA]
I
[CBA]
[BAC]
[ACB]
[BCA]
I
[CAB]
[BAC]
[ACB]
[CBA]
[ACB]
[BAC]
[CBA]
I
[BCA]
[CAB]
[CBA]
[ACB]
[BAC]
[CAB]
I
[BCA]
[BAC]
[CBA]
[ACB]
[BCA]
[CAB]
I
As expected, the composition of two isometries yields an isometry and no new isometries are
produced through this process. A quick glance through the above multiplication table for the
isometries of 4 shows that this set, under composition, is a subgroup of I(2).
The various isometries of 4 can be easily computed by studying their effect on the unit vectors
(1, 0) and (0, 1). For example, [CAB] exerts a rotation of 2π/3 around the origin. Accordingly,
6
ROBERT JUDD
√
[CAB](1, 0) = (cos 2π/3, sin 2π/3) = (−1/2, 3/2) and [CAB](0, 1) = (− sin 2π/3, cos 2π/3) =
√
(− 3/2, −1/2). This can be compactly represented in matrix form as:
"
#" #
"
√
√ #
−1/2 − 3/2 x
1 −x − 3y
√
[CAB](x, y) = √
=
2
3/2 −1/2
y
3x − y
Similarly,
"
#" #
√ #
√
−1/2
3/2 x
1 x − 3y
√
[BCA](x, y) =
=− √
2
3x + y
− 3/2 −1/2 y
"
"
#" # " #
x
−x
=
y
y
√
Consider now the line `2 through B and the midpoint M = ( 3/2, 1/2) of AC. Since M =
(cos π/6, sin π/6), we have that M makes an angle π/6 with the x-axis. Thus, when reflected about
√
ell2 , the vector (1, 0) becomes (cos 2π/6, sin 2π/6) = (1/2, 3/2). Similarly, (0, 1) forms an angle
√
π/3 with `2 . To reflect it we rotate it 2π/3 degrees obtaining (cos −π/6, sin π/6) = ( 3/2, −1/2).
Thus
"
#" #
"
√
√ #
1/2
3/2 x
1 x + 3y
√
[CBA](x, y) = √
=
2
3/2 −1/2 y
3x − y
−1 0
[ACB](x, y) =
0 1
Similarly,
#" #
"
√
√ #
1/2
− 3/2 x
1 x − 3y
√
√
[BAC](x, y) =
=
2 − 3x − y
− 3/2 −1/2
y
"
3.4. Reflection about an arbitrary line. We may directly calculate how to reflect a point x
through an arbitrary line ` that passes through a point p. The reflected point is denoted by x0 . Let
n be a unit normal to the line and θ, the angle between n and the vector x − p (see the diagram
below). Thus, hx − p, ni = |x − p| cos θ is the length of the projection of x − p onto a line orthogonal
to `, and hx−p, nin is the vector perpendicular to ` from ` to x. Thus, to project x we must displace
x twice this distance in a direction opposite to hx − p, nin. In other words x0 = x − 2hn, x − pin.
x
n
`
Kθ
p
U
U x0
We now go back to the question of what it takes to uniquely define an isometry.
Theorem 3.2. If T is an isometry that maps a triangle 4ABC to itself, then T is unique, i.e.,
there is no other isometry S such that T A = SA, T B = SB, T C = SC and T x 6= Sx for some
x ∈ R2 .
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
7
Proof. Assume T A = SA = a, T B = SB = b, T C = SC = c, where a, b, c are not collinear. If
x ∈ {A, B, C} we are done. Otherwise, since both T and S are isometries,
d(T A, T x) = d(A, x) = d(SA, Sx) and hence d(a, T x) = d(a, Sx) = ra > 0 ,
d(T B, T x) = d(B, x) = d(SB, Sx) and hence d(b, T x) = d(b, Sx) = rb > 0 ,
d(T C, T x) = d(C, x) = d(SC, Sx) and hence d(c, T x) = d(c, Sx) = rc > 0 .
This means that both T x and Sx must lie on the boundary of each of the circles centered at a,
b, and c with radii ra , rb and rc , respectively. The circles centered at a and b intersect at most
twice. If they intersect once we are done. Otherwise, since a, b, c are not collinear c cannot be
equidistant from both intersections and only one of them lies at distance rc from c. This intersection
is T x = Sx.
Definition 3.3 (Subgroups of I(2)). We have the following interesting subgroups of I(2) that will
occur later:
I(2) all isometries of E2 .
O(2) isometries of E2 that fix the origin, i.e., O(2) = {T ∈ I(2), T (0, 0) = (0, 0)}. This is also
called the orthogonal group of E2 .
SO(2) all rotations about the origin.
Note that SO(2) is a subgroup of O(2) which is a subgroup of I(2). O(2) is, in a sense, “twice
the size” of SO(2) as we can obtain two members of O(2) from each member of T of SO(2) by
either applying a reflection or not before the rotation.
4. Reflections, Translations and Rotations in E2
In this section we discuss reflections: isometries of E2 given by reflecting in a line. The product
of two reflections is either a translation or a rotation, and the product of three reflections is another
reflection. In fact we shall see that any isometry may be written as the product of at most three
reflections.
4.1. Reflections. In order to write down a formula for a reflection we first need to know how to
reflect a point in a line.
Theorem 4.1. If ` and m are perpendicular lines, then they have a unique point in common.
Proof. Let ` = P + [v] and m = Q + [w] with |v| = |w| = 1. Since ` and m are perpendicular it
follows that hv, wi = 0 and we may write
P − Q = hP − Q, viv + hP − Q, wiw .
This gives P − hP − Q, viv = Q + hP − Q, wiw and setting F = P − hP − Q, viv = Q + hP − Q, wiw
we see that F is on both lines.
If there are two points F, G on both ` and m, then F − G ∈ [v] ∩ [w] = {0} and hence F − G = 0,
i.e. F = G. Thus the point F is unique.
8
ROBERT JUDD
The following corollary is an immediate consequence of this theorem.
Corollary 4.2. Let X be a point and ` in E2 , then there exists a unique line m through X
perpendicular to ` and
(i) m = X + [n] where n is a unit normal to ` (that is, if [v] is the direction of ` with
p
p
v = (v1 , v2 ), then n = ±(v2 / v12 + v22 , −v1 / v12 + v22 ));
(ii) ` and m intersect in F = X − hX − P, nin where P is any point on `;
(iii) d(X, F ) = |hX − P, ni|.
We can now use the formula for the intersection of two perpendicular lines to find the equation
for the reflection of a point in the line, and thus define reflections.
Definition 4.3. The reflection of a point X in a line ` = P + [v] is the point X 0 that satisfies
1
(X + X 0 ) = F
2
where F = X − hX − P, nin is the intersection of ` with the line through X perpendicular to `.
X0
F
`
X
Figure 1. The reflection of the point X in the line `
Now,
1
(X + X 0 ) = F , so
2
1
1
X + X 0 = X − hX − P, nin and hence
2
2
1 0 1
X = X − hX − P, nin thus
2
2
0
X = X − 2hX − P, nin .
Definition 4.4 (Reflection). The reflection in line ` (where ` = P + [v]) is the bijection Ω` :
E2 → E2 given by
Ω` X = X − 2hX − P, nin ,
where n is a unit normal vector to `.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
9
4.2. Translations. We next investigate what happens when we compose two reflections. The
result depends on whether or not the lines of reflection are parallel. In the following diagrams X 0
is the reflection of X in `, while X 00 is the reflection of X 0 in m.
X 00
X
X0
X0
X 00
`
`
P
m
m
` parallel to m:
translation perpendicular to `
X
` not parallel to m: rotation about P
Figure 2. Reflection of X in line `, then in line m
If m and n are parallel lines, P is on m and Q is at the foot of the perpendicular from P to n,
and N is the unit normal to m (and hence also to n), then
Ωm Ωn X = Ωn X − 2hΩn X − P, N iN
= X − 2hX − Q, N iN − 2h(X − 2hX − Q, N iN ) − P, N iN
= X − 2hX − Q, N iN − 2hX − P, N iN + 4hX − Q, N iN
= X − 2hX − P, N iN + 2hX − Q, N iN
= X + 2hP − Q, N iN
= X + 2(P − Q)
Definition 4.5 (Translation). Let ` be a line and m, n lines perpendicular to `. The transformation
Ωm Ωn is a translation along `. If m 6= n, then the translation is non-trivial.
Theorem 4.6. If T is a non-trivial translation along a line `, then ` has a direction vector v such
that T x = x + v and conversely.
Proof. Let N be a unit direction vector for `, let P ∈ E2 and let α, β be distinct lines perpendicular
to `. Let a, b be the unique real numbers such that P + aN ∈ α and P + bN ∈ β. Then
Ωα Ωβ X = x + 2((P + aN ) − (P + bN )) = x + 2(a − b)N .
If T is not the identity, then a 6= b and 2(a − b)N is the required direction.
10
ROBERT JUDD
m
n
Ω m Ωn X
Ωn X
X
P
Q
N-
`
Figure 3. Reflection of X in line n, then in line m
Conversely, for any λ ∈ R let Tλ x = x + λN and let a, b satisfy λ = 2(a − b). Setting α =
P + aN + [N ⊥ ] and β = P + bN + [N ⊥ ], where N ⊥ is a unit normal vector to `, gives Tλ = Ωα Ωβ
as required.
Definition 4.7. For v ∈ R let τv be the translation τv x = x + v.
4.3. Rotations. Let ` = P + [v] be a line with |v| = 1. There exists a unique θ ∈ (−π, π] such
that v = (cos θ, sin θ). We shall use the unit normal to v, N = (− sin θ, cos θ). We have shown that
reflection in the line ` is given by
Ω` X = X − 2hX − P, N iN .
Now let `0 = 0 + [v] be the line through the origin parallel to `; we shall show that Ω` = τP Ω`0 τ−P .
We have
Ω` X − P = X − P − 2hX − P, N iN and
Ω`0 X = X − 2hX, N iN , hence
Ω` X − P = Ω`0 (X − P ) .
Thus Ω` X = Ω`0 (X − P ) + P as required.
Let x = (x1 , x2 ) ∈ R2 , then hx, N i = −x1 sin θ + x2 cos θ and writing the vectors in columns we
have
" # " #
"
#
x1
x1
− sin θ
Ω `0
=
− 2(−x1 sin θ + x2 cos θ)
x2
x2
cos θ
"
#
(1 − 2 sin2 θ)x1 + (2 sin θ cos θ)x2
=
(2 sin θ cos θ)x1 + (1 − 2 cos2 θ)x2
"
#" #
cos 2θ sin 2θ
x1
=
.
sin 2θ − cos 2θ x2
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
The matrix
"
cos 2θ sin 2θ
refθ =
sin 2θ − cos 2θ
11
#
is therefore a reflection in the line through the origin with direction (cos θ, sin θ).
We saw in Figure 2 that reflection in two non-parallel lines gives a rotation. If m is a second line
through P with direction (cos φ, sin φ), then
"
#
cos 2(θ − φ) − sin 2(θ − φ)
refθrefφ =
.
sin 2(θ − φ) cos 2(θ − φ)
We write a matrix of this form as
"
cos θ − sin θ
rotθ =
sin θ cos θ
#
.
Since rotθ takes the vector (1, 0) and (0, 1) to (cos θ, sin θ) and (− sin θ, cos θ) respectively, we can
think of rotθ as a rotation.
Definition 4.8 (Rotation). If α and β are lines through a point P , then the isometry Ωα Ωβ is
called a rotation about P.
4.4. Isometries that fix the origin. Let T be an isometry such that T (0) = 0. What is T ?
Let us attempt to express T using matrices and vectors. Since T is affine, we have a matrix A
and vector b with A invertible and T x = Ax + b.
T (0) = A(0) + b
T (0) = 0 + b
T (0) = b
Since T (0) = 0, b = 0. So T x = Ax, A invertible.
"
a
A=
c
Now we have to find A. Let
#
b
d
A is invertible, so ad − bc 6= 0. We can also use the distance preserving property of isometries.
Consider
" # the
" action
# " of# T on
" #(0, 0), (0, 1), (1, 0) and (1, 1).
1
a b 1
a
T
=
=
. So a2 + c2 = 1.
0
c d 0
c
" # "
#" # " #
0
a b 0
b
T
=
=
. So b2 + d2 = 1.
1
c d 1
d
" # "
#" # "
#
1
a b 1
a+b
T
=
=
. So (a + b)2 + (c + d)2 = 2.
1
c d 1
c+d
⇒ a2 + b2 + c2 + d2 + 2ab + 2cd = 2
⇒ ab + cd = 0
⇒ cd = −ab
Given T x = Ax, T −1 x = A−1 x.
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ROBERT JUDD
A−1
=
1
∆
"
#
d −b
, where ∆ = ad − bc.
−c a
T
−1
"
#" # " #
" #
d
1
1 d −b 1
∆
= −c
=
∆ −c a
0
0
∆
d2
c2
+
=1
∆ 2 ∆2
⇒ c2 + d2 = ∆2 .
⇒
T
−1
"
#" # " #
" #
−b
0
1 d −b 0
= ∆
=
a
∆ −c a
1
1
∆
b2
a2
+
=1
∆ 2 ∆2
⇒ a2 + b2 = ∆2 .
⇒
a2 + b2 + c2 + d2 = 2∆2
⇒ 2 = 2∆2
⇒ ∆ = ±1
Let us reconstruct A.
a2 + b2 = a2 + c2 = b2 + d2 = c2 + d2 = 1; cd = −ab.
Let a = cos θ; b = ± sin θ. Since a2 + c2 = 1, c = ± sin θ. Also c2 + d2 = 1 ⇒ d = ± cos θ.
Lets pick each possibility.
"
#
cos θ sin θ
θ
= ref ( )
2
sin θ − cos θ
"
# "
#
cos θ sin θ
cos(−θ) − sin(−θ)
=
= rot(−θ)
− sin θ cos θ
sin(−θ) cos(−θ)
"
#
cos θ − sin θ
= rotθ
sin θ cos θ
"
# "
#
cos θ − sin θ
cos(−θ) sin(−θ)
−θ
=
= ref ( )
2
− sin θ − cos θ
sin(−θ) − cos(−θ)
Thus if T fixes the origin, it must be either a reflection or a rotation.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
13
4.5. Isometries that have a fixed point. We have shown that if T is an isometry with T (0) = 0,
then T is either a reflection or a rotation. If T has a fixed point P , then T is either a rotation
about P or a reflection in a line through P .
Since T fixes P , T P = P . τP x = x + P . Then τ−P T τP (0) = 0.
τ−P T τP (0) = τ−P T (P ) = τ−P (P ) = 0
Therefore τ−P T τP =ref(θ) or rot(θ). So T = τP ref(θ)τ−P or τP rot(θ)τ−P .
4.6. Isometries with no fixed points. Let T be an isometry with no fixed points. Let P = T 0.
Then (τ−P T )(0) = 0.
So τ−P T =refθ or rotθ. Hence T = τP refθ or T = τP rotθ.
We have thus proven the following theorems:
Theorem 4.9 (Classification of isometries of E2 ). An isometry of E2 must be one of the following:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
ref(θ);
rot(θ);
τv ref(θ)τ−v ;
τv rot(θ)τ−v ;
τv ref(θ);
τv rot(θ).
Theorem 4.10. The group I(E2 ) of isometries on E2 is generated by { rot θ, ref θ : θ ∈ (−π, π]}
∪ {τP : P ∈ E2 }.
Theorem 4.11 (Three Reflection Theorem for parallel lines). If α, β, γ are all parallel, then there
exists a line δ parallel to α, β, γ such that Ωα Ωβ Ωγ = Ωδ .
Proof. Let ` be a line perpendicular to α, β, γ. Since Ωα Ωβ is a translation, if follows that if we
can find a line m perpendicular to l such that Ωα Ωβ = Ωm Ωγ , then Ωα Ωβ Ωγ = Ωm as required,
since Ω−1
γ = Ωγ .
Now, if P is a point on `, N is a unit vector perpendicular to `, and m is any line perpendicular
to `, then we may choose a, b, c and d so that
P + aN ∈ α and P + bN ∈ β so Ωα Ωβ (x) = x + 2(a − b)N
P + dN ∈ m and P + cN ∈ γ so Ωm Ωγ (x) = x + 2(d − c)N
To obtain Ωα Ωβ = Ωm Ωγ we must choose d such that 2(a − b) = 2(d − c), i.e. d = a − b + c.
Therefore m = P + (a − b + c)N which completes the proof.
Corollary 4.12. If T = Ωα Ωβ is a translation along a line `, and m is any line perpendicular to
`, then there exists lines n and n0 such that
T = Ωα Ωβ = Ωm Ωn = Ωn0 Ωm .
Proof. Apply the previous theorem to Ωm Ωα Ωβ to find a line n such that Ωm Ωα Ωβ = Ωn . Then
Ωα Ωβ = Ωm Ωn . Similarly for Ωα Ωβ Ωm .
14
ROBERT JUDD
Theorem 4.13 (Three Reflection Theorem for concurrent lines). Let α, β and γ be lines intersecting in a point P . There exists a line δ through P such that Ωα Ωβ Ωγ = Ωδ .
Proof (Version 1). Case 1: P = (0, 0). Let θ, φ, ψ be such that Ωα = ref (θ) and Ωβ = ref (φ) and
Ωγ = ref (ψ). We may calculate directly (by matrix multiplication)
Ωα Ωβ Ωγ = ref(θ − φ + ψ) = Ωθ−φ+ψ ,
a reflection in the line through the origin with direction vector (cos(θ − φ + ψ), sin(θ − φ + ψ)).
Case 2: P 6= (0, 0). Now, τ−P Ωα τP = ref(θ), τ−P Ωβ τP = ref(φ) and τ−P Ωγ τP = ref(ψ) are all
reflections in lines through the origin. Thus
(τ−P Ωα τP )(τ−P Ωβ τP )(τ−P Ωγ τP ) = τ−P Ωα Ωβ Ωγ τP = ref(θ − φ + γ) ,
and so Ωα Ωβ Ωγ = τP (ref(θ − φ + γ))τ−P , a reflection in the line through P with direction (cos(θ −
φ + γ), sin(θ − φ + γ)).
Proof (Version 2). If P is the origin, then the reflections are linear transformations and the determinant of each matrix is −1 so the determinant of the product is −1 and hence a reflection (because
rotations have determinant +1). Of course, we haven’t said anything about determinants, so to
make this into a real proof one would first have to do the work on determinants. For P not the
origin the proof is as for Case 2 above.
Proof (Version 3). Again we start with P the origin. We know that if Ωα = ref (θ) and Ωβ = ref (φ)
and Ωγ = ref (ψ), then Ωα Ωβ = rot(2(θ − φ)) is a rotation. If δ is any line through the origin with
direction (cos κ, sin κ), then Ωδ Ωγ = rot(2(κ − ψ)). Setting δ = θ − φ + κ gives Ωα Ωβ = Ωδ Ωγ .
Multiplying both sides on the right by Ωγ completes the proof. As before the case for P not the
origin is as for Case 2 in Version 1.
Corollary 4.14. Let T = Ωα Ωβ be a rotation about point P and let ` be any line through P . There
exist lines m and m0 through P with T = Ω` Ωm = Ωm0 Ω` .
The proof is essentially the same as that of Corollary 4.12.
Theorem 4.15. Let Ωα , Ωβ , Ωγ be three distinct reflections not all parallel or concurrent. Then
Ωα Ωβ Ωγ is a glide reflection.
Note: A glide refleciton is a reflection followed by a translation along the line of reflection, i.e.
τv Ωn where v is parallel to n.
Proof. Case 1: α and β intersect in P . We shall apply Corollary 4.14 to construct a line n and
lines m, m0 perpendicular to n so that Ωα Ωβ Ωγ = Ωm Ωm0 Ωn , a glide reflection as required.
Let ` be the line through P perpendicular to γ and let F be the intersection of ` and γ. Next,
using Corollary 4.14, choose a line m through P so that Ωα Ωβ = Ωm Ωl and hence
Ωα Ωβ Ωγ = Ωm Ωl Ωγ .
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
15
Let n be the line through F perpendicular to m and let m0 be the line through F perpendicular
to n. Now, n and m0 are perpendicular and intersect in F , ` and γ are perpendicular and intersect
in F , so they are both half turns and hence represent the same isometry (see exercises). Thus,
Ωm0 Ωn = Ωl Ωγ and therefore
Ωα Ωβ Ωγ = Ωm Ωm0 Ωn ,
a glide reflection.
m
`
α
β
m0
P
γ
n
F
Case 2: β and γ intersect. Apply Case 1 to (Ωα Ωβ Ωγ )−1 = Ωγ Ωβ Ωα = τv Ωn to obtain Ωγ Ωβ Ωα =
τv Ωn and hence
Ωα Ωβ Ωγ = (τv Ωn )−1 = Ωn τ−v = τ−v Ωn ,
again a glide reflection.
Homework Due Wednesday, October 17 2001. Please hand this in neatly typed, with proper
proofs, good spelling and decent grammar. If it isn’t typed, then it will not be accepted.
4.1 Recall that if ` is a line, X is any point, P a point on `, F the point at the foot of the
perpendicular from X to ` and N is a unit normal to `, then F = X − hX − P, N iN and
d(X, F ) = |hX − P, N i|.
Also note that F has the property that if Q is any other point on `, then d(X, F ) <
d(Q, X). Therefore we may define d(X, `) = |hX − P, N i|.
Show that if N 0 is another unit normal to ` and P 0 ∈ ` then |hX −P 0 , N 0 i| = |hX −P, N i|.
What is the significance of this?
4.2 The distance between two parallel lines `, m is the unique number d(`, m) such that
d(x, m) = d(y, `) = d(`, m) ∀x ∈ `, y ∈ M .
16
ROBERT JUDD
(i) Prove that if N is a unit normal to `, then d(`, m) = |hx − y, N i| for any x ∈ `, y ∈ m.
(ii) Let n = { 12 (x + y) : x ∈ `, y ∈ m}. Show that n is a line parallel to ` and m and
midway between them. i.e. d(n, m) = d(n, `) and d(n, m) + d(n, `) = d(m, `).
4.3 Recall Ω` x = x − 2hx − P, N iN where N is a unit normal to ` and P is a point on `. Prove
hx − P 0 , N 0 iN 0 = hx − P, N iN for any other unit normal N 0 to ` and point P 0 on `. What
does this show?
4.4 Let ` = P + [v], m = Q + [v] be lines and |v| = 1. Show Ω` Ωm = τw where w =
2hP − Q, v ⊥ iv ⊥ and τ−w = Ωm Ω` .
4.5 A Half Turn is a rotation Ω` Ωm where ` is perpendicular to m.
(i) Show that Ωα and Ωβ commute if and only if α is perpendicular to β.
(ii) If `, m, α, β intersect at point P and ` is perpendicular to m, α is perpendicular to β
then Ω` Ωm = Ωα Ωβ .
(iii) Show that the half turn HP about point P is given by HP X = −X + 2P (X ∈ E2 ).
(iv) Show that the product of 2 half turns is a translation along the line joining the centers
of rotations.
Theorem 4.16. If T is a glide reflection and Ωα is a reflection, then Ωα T is a translation or a
rotation.
Proof. A glide reflection is of the form Ω` τv = τv Ω` where v is parallel to `.
Case 1: α is parallel to `. Now, Ωα T = Ωα Ω` τv = τu τv = τu+v since τu = Ωα Ω` is a translation.
Case 2: α intersects ` at point P . Choose m, n perpendicular to ` so that P ∈ m and τv = Ωm Ωn .
Since α, ` and m intersect in P it follows from the three reflections theorem for concurrent lines
that there exists a line δ such that Ωα Ω` Ωm = Ωδ . Now,
Ωα T = Ωα Ω` τv = Ωα Ω` Ωm Ωn = Ωδ Ωn ,
a rotation or translation.
Theorem 4.17. Every isometry of E2 may be written as the product of at most three reflections.
Proof. Recall that all isometries may be written as one of the following:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
ref(θ);
rot(θ);
τv ref(θ)τ−v ;
τv rot(θ)τ−v ;
τv ref(θ);
τv rot(θ).
Thus it is sufficient to show that the theorem is true for each of these. (i) and (ii) are special
cases of (v) and (vi), and (v) may be written directly as the product of three reflections. This
leaves us with (iii), (iv) and (vi).
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
17
For (iii) let Ωα = ref(θ) and τv = Ωm Ωn where 0 ∈ n then
τv ref(θ)τ−v = Ωm Ωn Ωα Ωn Ωm = Ωm Ωδ Ωm ,
where Ωn Ωα Ωn = Ωδ for some line δ since these three lines intersect at the origin.
Next, let rot(θ) = Ωα Ωβ and τv = Ωm Ωn where α and β intersect in the origin and 0 ∈ n. Again
Ωn Ωα Ωβ = Ωδ for some line δ since these three lines intersect at the origin. Thus
τv rot(θ)τ−v = Ωm Ωn Ωα Ωβ Ωn Ωm = Ωm Ωδ Ωn Ωm .
Now, if m is parallel to δ, then Ωm Ωδ = τw for some w and Ωm Ωδ Ωn Ωm = τw τ−v = τw−v , a
translation which may be written as the product of two reflections.
Otherwise m and δ intersect in P , and let τ−v = Ωn0 Ωm0 where P ∈ n0 . By the three reflections
theorem for concurrent lines there exists a line γ through P with Ωm Ωδ Ωn0 = Ωγ . Thus
Ωm Ωδ Ωn Ωm = Ωm Ωδ τ−v = Ωm Ωδ Ωn0 Ωm0 = Ωγ Ωm0 ,
the product of two reflections. This completes the proof for (iv).
For the final case (vi) let rot(θ) = Ωα Ωβ and τv = Ωm Ωn where 0 ∈ n. There exists a line γ
through the origin with Ωn Ωα Ωβ = Ωγ and hence
τv rot(θ) = Ωm Ωn Ωα Ωβ = Ωm Ωγ ,
the product of two reflections as required.
5. Affine Transformations
Definition 5.1. A collineation is a bijection (i.e., both 1-1 and onto) T : E2 → E2 such that if
points p, q, r are collinear then so are T p, T q, T r.
Definition 5.2. An affine transformation is a map T : E2 → E2 of the form T x = Ax + b where
A is a non-singular 2 × 2 matrix and b is a vector in R2 .
Definition 5.3. A map T : V → W between two vector spaces is linear if
(1) T (u + v) = T u + T v, for all u, v ∈ V .
(2) T (λu) = λT u, for all scalars λ and u ∈ V .
Example: the isometries of the triangle 4ABC are all linear maps.
The set of all 2D invertible linear maps is a group which we denote by GL(2).
Note that affine transformations are not, in general, linear as an affine transformation has both
a linear and a translation component.
Theorem 5.4. All collineations are affine transformations and vice versa. In other words, Definitions 5.1 and 5.2 are equivalent.
How do isometries fit in all of this? All isometries are affine (but the converse is not true). To
see this it suffices to show that reflections are affine as every isometry can be expressed as the
composition of three reflections. Consider a reflection through a line ` through p = (px , py ) with
18
ROBERT JUDD
unit normal n = (nx , ny ). For any point q = (x, y) T q = q − 2h< q − p, nin which has the form
T q = Aq + b,
" # "
#" #
" #
x
(1 − 2n2x ) (−2nx ny ) x
nx
T
=
+ hp, ni
2
y
(−2nx ny ) (1 − 2ny ) y
ny
Furthermore A is non-singular as det A = (1 − 2n2x )(1 − 2n2y ) − 4n2x n2y = 1 − 2(n2x + n2y ) = −1.
Theorem 5.5. Every isometry is an affine transformation.
Theorem 5.6. Let AF(2) denote the set of affine transformations on E2 , then AF(2) is a group
under composition.
Proof. Let T x = Ax + c and Sx = Bx + d be affine transformations. Then T Sx = T (Bx + d) =
A(Bx + d) + c = (AB)x + (Ad + c) = M x + f . Since M −1 = B −1 A−1 , T S is affine as well,
so AF (2) is closed under composition. The identity matrix is the identity of the group, and
T −1 x = A−1 (x − c) = A−1 x − A−1 c
Definition 5.7. A direction [v] is the set of vectors proportional (i.e., scalar multiples) to a nonzero vector v. Thus, [v] = {tv : t ∈ R}, v ∈ R2 \ {0}.
Definition 5.8. For any point P ∈ R2 and direction [v] the line through P with direction [v] is
the set ` = {x ∈ E2 , x − P ∈ [v]}. We write ` = P + [v].
Theorem 5.9. Let T be an affine transformation such that T x = Ax + b and ` = P + [v] a line in
E2 . Then T ` is also a line in E2 .
Proof. A point on ` has the form p + tv. Thus, T (p + tv) = A(p + tv) + b = (Ap + b) + tAv =
T p + t(Av) ∈ T P + [Av]. In other words, T ` is the line through T p with direction [Av].
Corollary 5.10. Let T x = Ax + b be an affine transformation. A line P + [v] is a fixed line of T
if and only if the following two conditions hold,
(i) Av = λv for some λ ∈ R \ {0} (i.e. v is an eigenvector of A).
(ii) (A − I)P + b ∈ [v].
Proof. We saw in Theorem 5.9 that if ` = P + [v], then T ` = T P + [Av]. Thus, T ` = ` if and
only if [Av] = [v] (i.e. Av = λv for some λ ∈ R \ {0}) and T P ∈ `, that is T P = P + tv for some
t ∈ R \ {0}. But T P = AP + b which gives (A − I)P + b = tv ∈ [v].
Definition 5.11. The set of lines perpendicular to a fixed line ` in E 2 is called a pencil of parallels.
Theorem 5.12. Let T be an affine transformation, then
(i) if two fixed lines of T intersect, then they do so in a fixed point of T ;
(ii) if two fixed lines of T are parallel, then every line in a pencil containing these lines is fixed
by T ;
(iii) if two lines are parallel, then their images in T are parallel.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
19
Proof. Let lines ` and m intersect in the point P and be fixed by T . Then T P ∈ T ` = ` and
T P ∈ T m = m, so that T P is on both ` and m and hence in their intersection. Thus T P = P
which proves (i).
Let distinct lines ` and m be parallel with direction [v]. Let T x = Ax + b for x ∈ E2 and let
X ∈ E2 . Since T fixes ` and m, it follows from Corollary 5.10 that Av = λv for some λ ∈ R \ {0}.
Again, by Corollary 5.10 we must show that (A − I)X + b ∈ [v] to prove (ii). From Section 4 it
is clear that we may find points P ∈ ` and Q ∈ m such that Q − P is perpendicular to v and X
lies on the line through P and Q. Since ` 6= m, it follows that P 6= Q and hence we may write
X = tP + (1 − t)Q for some t ∈ R. Now, by Corollary 5.10 (A − I)P + b, (A − I)Q + b ∈ [v], thus
(A − I)X + b = (A − I)(tP + (1 − t)Q) + b = t(A − I)P + tb + (1 − t)(A − I)Q + (1 − t)b
= t((A − I)P + b) + (1 − t)((A − I)Q + b) ∈ [v] ,
which completes the proof of (ii).
For (iii) we observe that if ` and m are parallel lines with direction [v], then both T ` and T m
have direction [Av].
We have defined an affine transformation to be a bijection T : E2 → E2 such that T x =
Ax + b (x ∈ E2 ) where A ∈ GL(2) (i.e. a 2 × 2 invertible matrix) and b ∈ R2 . We may also
represent T as a linear transformation in R3 that fixes the plane z = 1. If A and b are given as
"
#
" #
a11 a12
b1
A=
,
b=
,
a21 a22
b2
then we may write y = Ax + b as
  
 
y1
a11 a12 b1
x1
  
 
y2  = a21 a22 b2  x2  ,
1
0
0
1
1
and composition in AF(2) is equivalent to multiplication of 3 × 3 matrices. Thus AF(2) is a
subgroup of GL(3), the General Linear group on R3 , consisting of 3 × 3 invertible matrices.
Theorem 5.13. Let P, Q be points in E2 and T an affine transformation. Then
(i) T ((1 − t)P + tQ) = (1 − t)T P + tT Q∀t ∈ R, i.e. affine transformations preserve order
along lines, and
(ii) a point X lies between P and Q if and only if T X lies between T P and T Q, furthermore
d(P, X)
d(T P, T X)
=
.
d(Q, X)
d(T Q, T X)
Proof. We can calculate directly
T ((1 − t)P + tQ) = A((1 − t)P + tQ) + b = (1 − t)AP + (1 − t)b + tAQ + tb
= (1 − t)(AP + b) + t(AQ + b) = (1 − t)T P + tT Q ,
as required for (i).
20
ROBERT JUDD
The point X lies between P and Q if, and only if, the exists t ∈ [0, 1] with X = (1 − t)P + tQ.
It is thus clear from (i) that T X lies between T P and T Q. Now,
d(P, X)
|tP − tQ|
t|P − Q|
t
=
=
=
d(Q, X)
|(1 − t)P − (1 − t)Q|
(1 − t)|P − Q|
1−t
and a similar argument gives the same value for d(T P, T X)/d(T Q, T X).
Theorem 5.14. Let T be an affine transformation.
(i) If T fixes two distinct points, then T fixes every point on the line through these two points.
(ii) If T fixes three non-collinear points, then T is the identity.
Proof. (i) is immediate from part (i) of the previous theorem. For (ii) let T be an affine transformation and let P, Q, R be three non-collinear points fixed by T .
Let X ∈ E2 . If X is on a line through any of the pairs of points P, Q, R, then X is fixed by (i).
Otherwise let A be the midpoint of the line from P to Q. The line through X and A cannot be
parallel to both lines P R and QR, thus it meets one of these in B distinct from A.
Q
X
A
R
B
P
Now both A and B are fixed by T and X is on a line through A and B. Thus X is fixed using
part (i).
Theorem 5.15 (Fundamental Theorem of Affine Transformations). Given 2 non-collinear triples
P, Q, R and P 0 , Q0 , R0 , there exists a unique affine transformation T such that T P = P 0 , T Q = Q0
and T R = R0 .
Proof. Since {Q − P, R − P } and {Q0 − P 0 , R0 − P 0 } are bases for R2 there is a 2 × 2 invertible
matrix A such that A(Q − P ) = (Q0 − P 0 ) and A(R − P ) = (R0 − P 0 ) (see Lemma 5.16 below).
Recall that if B ∈ E2 , then τB is the isometry TB X = X + B and let T = τP 0 Aτ−P , then it is easy
to check that T P = P 0 , T Q = Q0 and T R = R0 . Hence such a transformation exists.
To show the T is unique, suppose S is an affine transformation that agrees with T on P, Q, R.
But now S −1 T is affine and fixes three non-collinear points P, Q, R. Hence S −1 T = I and so S = T
as required.
Lemma 5.16. Let {u, v} and {x, y} be two pairs of non-parallel vectors in R2 . Then there exists
an invertible 2 × 2 matrix A such that Au = x and Av = y.
Proof. Let i = (1, 0), j = (0, 1), then it suffices to show that for any pair {u, v} of non-parallel
vectors there exists an invertible 2 × 2 matrix A such that Ai = u and Aj = v.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
21
Let u = (a, b), v = (c, d), then a simple calculation shows that
"
#
a c
A=
b d
is the required matrix. We only need check that A is invertible, i.e. det(A) 6= 0. If det(A) = 0,
then ad − bc = 0. If a 6= 0, then d = bc/a and v = (c/a)u, a contradiction. Otherwise, if a = 0,
then since b 6= 0 it follows that c = 0 and v = (d/b)u, again a contradiction.
6. Geometry of the Sphere S2
6.1. Preliminaries in E3 . We need to include the properties of the cross product in R3 since we
shall use this heavily in S2 .
Definition 6.1 (Cross Product in R3 ). The cross product of two vectors u, v in R is the unique
vector z = u × v ∈ R3 satisfying
hz, xi = det(x, u, v) for every x ∈ R3 .
In fact we may calculate u × v directly from u = (u1 , u2 , u3 ) and v = (v1 , v2 , v3 ) as

ı̄
̄ k̄


u × v = det u1 u2 u3  = (u2 v3 − u3 v2 , −u1 v3 + u3 v1 , u1 v2 − u2 v1 )
v1 v2 v3

One should check that the cross product is well defined; we leave this and the rest of the proof
of the next theorem as an exercise.
Theorem 6.2. The cross product has the folloing properties:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
u × v is well defined;
hu × v, ui = 0 = hu × v, vi;
u × v = −v × u;
hu × v, wi = hu, v × wi;
(u × v) × w = u × (v × w) = hu, wiv − hv, wiu;
u × v = 0 if and only if ukv;
u × v 6= 0 if and only if {u, v, u × v} is a basis of R3 ;
|u × v|2 = |u|2 |v|2 − hu, vi2 .
Definition 6.3. A triple {u, v, w} of a mutually orthogonal unit vectors is called an orthonormal
triple.
Proposition 6.4. If {u, v, w} is a orthonormal triple then
x = hx, uiu + hx, viv + hx, wiw for every x ∈ R3 .
Proposition 6.5. If u is any unit vector then there exist v, w in R3 such that {u, v, w} is an
orthonormal triple.
Q
Definition 6.6. A plane in E3 is a set of points
satisfying:
22
ROBERT JUDD
Q
(i) There is a point of E3 not in .
Q
(ii)
cannot be contained in a line.
Q
Q
(iii) Given any two points in
the line through those points is also in .
Definition 6.7. Let v, w ∈ R3 , then [v, w] = {su + tw : t, s ∈ R} is the span of v, w.
Theorem 6.8. We may represent a plane in the following ways:
(i) If v, w ∈ R3 are non parallel vectors and P ∈ E3 then P + [v, w] is a plane. It is “the
plane through P spanned by v and w”.
(ii) If N ∈ R3 is a unit vector and P ∈ E3 then {x ∈ R3 : hx − P, N i = 0} is a plane. It is
“the plane through P with normal vector N .”
Q
(iii) If P, Q, R are non-colinear points in E3 , then there is a unique plane
containing P, Q, R.
This plane is called “the plane P, Q, R.”
6.2. Lines in S2 . We next want to define what a line is in S2 . If you start at any point of S2 and
move “straight ahead” then you will trace out a “great circle” and end up back where you started.
In terms of the surrounding context, this is the intersection of a plane in R3 through the origin
with S2 . These will be our lines.
Definition 6.9. A line in S2 is the set ` = {x ∈ S2 : hu, xi = 0} for some unit vector u in R3 .
We refer to u as the pole of `.
Notice that the line ` with pole u is the intersection of the plane through the origin in R3 having
normal vector u with S2 .
Definition 6.10. Two points P and Q of S2 are antipodal if P = −Q.
Theorem 6.11. If u is a pole of `, then so is −u. If P lies on `, then so does −P .
Theorem 6.12. If P and Q are distinct points that are not antipodal, then there is a unique line
containing P and Q.
Proof. (i) Existence: let ` be the line with pole P × Q/|P × Q|. Both P and Q lie on ` since P and
Q are both perpendicular to P × Q.
(ii) Uniqueness: Let u be a pole of any line through P and Q, then hP, ui = 0 and hQ, ui = 0.
From Theorem 6.2 (v) (P × Q) × u = hP, ui Q − hQ, ui P = 0, i.e. u is parallel to P × Q. But since
|u| = 1 this means u = ±(P × Q)/|P × Q|, in other words u is a pole of the line ` from part (i). Theorem 6.13. Let `, m be distinct lines in S2 , then they have exactly 2 points of intersection and
these points are antipodal.
Proof. Let u, v be poles of `, m respectively. The two lines are distinct so u, v are not parallel
and hence u 6= ±v, or u × v 6= 0. Points ±(u × v)/|u × v| are on both ` and m since they are
perpendicular to both u and v. Thus `, m have at least 2 points of intersection.
If P is a third point of intersection, then P is not antipodal to either ±(u × v)/|u × v| so by the
previous theorem there is a unique line through (u × v)/|u × v| and P , a contradiction.
Corollary 6.14. No two lines of S2 are parallel.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
23
Note: even lines with a common perpendicular intersect.
Definition 6.15. The distance between two points P, Q ∈ S2 is given by d(P, Q) = cos−1 hP, Qi
Remark: The range of the inverse cosine function is [0, π] so that 0 ≤ d(P, Q) ≤ π.
Theorem 6.16. Let P, Q, R ∈ S2 , then
(i)
(ii)
(iii)
(iv)
d(P, Q) ≥ 0;
d(P, Q) = 0 if and only if P = Q;
d(P, Q) = d(Q, P );
d(P, Q) + d(Q, R) ≥ d(P, R).
Proof. We shall prove (iv). Recall that |u × v|2 = |u|2 |v|2 − hu, vi2 (Theorem 6.2). Thus, hu, vi2 ≤
|u|2 |v|2 . Now, let r = d(P, Q), p = d(Q, R), q = d(P, R), then hP × R, Q × Ri2 ≤ |P × R|2 |Q × R|2 .
Calculating the left and right sides separately gives
hP × R, Q × Ri2 = hP, R × Q × Ri2
= hP, (hR, Ri Q − hQ, Ri R)i2
= (hR, Ri hP, Qi − hQ, Ri hP, Ri)2
= (cos r − cos q cos p)2 ,
and
|P × R|2 |Q × R|2 = (|P |2 |R|2 − hP, Ri2 )(|Q|2 |QR|2 − hQ, Ri2 )
= (1 − cos2 q)(1 − cos2 p)
= sin2 q sin2 p .
Thus (cos r − cos q cos p)2 ≤ sin2 q sin2 p and hence cos r − cos q cos p ≤ sin q sin p since 0 ≤ p, q ≤ π.
This gives cos r ≤ sin q sin p + cos q cos p and then cos r ≤ cos(q − p). Since cos x is decreasing on
[0, π], we obtain r ≥ q − p provided that 0 ≤ q − p ≤ π, i.e. r + p ≥ q. Since 0 ≤ p, q ≤ π, clearly
p − q ≤ π. If q − p ≤ 0 ≤ r then q ≤ p + r as required.
Corollary 6.17. If d(P, Q) + d(Q, R) = d(P, R), then P , Q and R are collinear.
Proof. An inspection of the proof of the theorem shows that if d(P, R) = d(P, Q) + d(Q, R), then
hP × R, Q × Ri2 = |P × R|2 |Q × R|2 . From Theorem 6.2 (vii) hu, vi2 = |u|2 |v|2 − |u × r|2 thus
(P × R) × (Q × R) = 0. If P × R = 0 then P, R are parallel, so they must be equal or antipodal
and Q is on some line between them. Similarly if Q × R = 0, then P is on some line between Q
and R.
Otherwise P ×R and Q×R are parallel. Now, the line containing P and R has a pole P ×R/|P ×R|
and the line containing Q and R has a pole Q × R/|Q × R|. Since these are parallel, the lines P R,
QR have the same poles and are therefore equal.
In fact more is true. We shall shortly prove that if equality holds, then Q lies on a half line
between P and R.
24
ROBERT JUDD
Definition 6.18. Two lines are perpendicular if their poles are perpendicular.
Theorem 6.19. Let ` and m be distinct lines in S2 , then there is a unique line n such that both `
and m are perpendicular to n. Moreover, the intersection points of ` and m are the poles of n.
Proof. Let ` have pole u and m have pole v. As in Theorem 6.13 the point P = u × v/|u × v| is in
the intersection of ` and m. Let n be the line with pole P , then both u and v are perpendicular to
P , and hence ` and m are perpendicular to n.
Theorem 6.20. Let ` be a line in S2 , P ∈ S2 . If P is not a pole of `, then there exists a unique
line m through P perpendicular to `.
Proof. Let ` have pole u, let v = (u × P )/|u × P | and let m be the line with pole v. Since P is
perpendicular to v it follows that P ∈ m and m is perpendicular to ` since u is perpendicular to v.
Now let n be a different line perpendicular to ` containing P . Thus both m and n are perpendicular to ` and hence their points of intersection are the poles of `. Since P is on both m and n,
it follows that P is a pole of `. Finally, then, if P is not a pole of `, the line m must be unique. 6.3. Isometries of S2 . We would like to define reflection in the line ` of S2 with pole u, as
Ω` x = x − 2hx, uiu. We must first check that the range of Ω` is at least contained in S2 :
|Ω` x|2 = hx − 2hx, uiu, x − 2hx, uiui = hx, xi − 4hx, ui2 + 4hx, ui2 = 1 ,
so Ω` : S2 → S2 . Furthermore,
Ω` Ω` x = Ω` x − 2hΩ` x, uiu
= x − 2hx, uiu − 2hx − 2hx, uiu, uiu
= x − 2hx, uiu − 2hx, uiu + 4hx, uiu
=x.
Thus Ω` is a good candidate for reflection. In fact, more is true.
Theorem 6.21. Let n ∈ R3 be a unit vector, let T : R3 → R3 be T x = x − 2hx, nin, then
(i) T is linear;
(ii) hT x, T yi = hx, yi for all x, y ∈ R3 .
Proof. In order to show linearity we must prove T (x + y) = T x + T y and T (ax) = aT x for every
x, y ∈ R3 and a ∈ R.
T (x + y) = (x + y) − 2h(x + y), nin = (x − 2hx, nin) + (y − 2hy, nin) = T x + T y ,
and
T (ax) = ax − 2hax, nin = a(x − 2hx, nin) = aT x ,
as required.
For the second part,
hT x, T yi = (x − 2hx, nin, y − 2hy, nini = hx, yi − 4hx, nihy, ni + 4hx, nihy, ni = hx, yi .
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
25
Corollary 6.22. Let Ω` x = x − 2hx, uiu for x ∈ S2 , where u is a pole of `, then
Ω` : S2 → S2
d(Ω` x, Ω` y) = d(x, y) for all x, y ∈ S2
Ω` Ω` x = x for all x ∈ S2
Ω` is a bijection.
Ω` x = x if and only if x ∈ `
p
p
Proof. For x ∈ S2 , 1 = |x| = hx, xi = hΩ` x, Ω` xi = |Ω` x|, so Ω` x ∈ S2 . Part (ii) is immediate
from the linearity of Ω` and (iii) we have already calculated.
That Ω` is one-to-one is a consequence of part (ii), and for x ∈ S2 let y = Ω` x, then Ω` y = x
and hence Ω` is also onto.
Finally, if Ω` x = x and n is a pole of `, then x = x − 2hx, nin which forces hx, ni = 0 since n 6= 0.
In other words, x is perpendicular to a pole of ` and so x ∈ `.
(i)
(ii)
(iii)
(iv)
(v)
Definition 6.23. A rotation about a point P is the composition of two reflections in lines α and
β that intersect in P .
Theorem 6.24. If α, β, and γ are lines that intersect in a point P , then there exists a line δ
through P such that Ωα Ωβ Ωγ = Ωδ .
Theorem 6.25. If α, β, and γ are lines that intersect in a point P then there exists lines δ and
δ 0 such that Ωα Ωβ = Ωδ Ωγ = Ωγ Ωδ0 .
Definition 6.26. A translation along the line ` is Ωα Ωβ where α and β are lines perpendicular
to `.
Note: If α and β are distinct lines then there is a unique line `, perpendicular to both α and β
which intersect in the poles of `. So, a translation along ` is the same as a rotation about a pole of
`. The same theorems hold for translations as reflections.
Theorem 6.27. Every isometry of S2 may be written as the composition of at most three reflections.
Theorem 6.28. The composition of two rotations of S2 is another rotation.
Theorem 6.29. Every rotation may be written as the composition of two half-turns.
6.4. Segments in S2 . In order to discuss angles and triangles in S2 , we first need some more
information on lines in S2 . We may parameterize a line ` in S2 as follows. Let ` have pole u and
let P and Q be points chosen such that {u, P, Q} is orthonormal (of course, then P and Q are on
`) and set α(t) = (cos t)P + (sin t)Q.
Theorem 6.30. Let `, u, P, Q be as above, then
(i) ` = {α(t) : t ∈ R};
(ii) Each point of ` occurs exactly once as α(t) for t ∈ [0, 2π), i.e. ` = {α(t) : t ∈ [0, 2π)};
(iii) d(α(t1 ), α(t2 )) = |t1 − t2 | if 0 ≤ |t1 − t2 | ≤ π.
26
ROBERT JUDD
Definition 6.31. A subset s of S2 is called a segment if there exist points P and Q with P
perpendicular to Q (ie. hP, Qi = 0) and numbers t1 < t2 with (t2 − t1 ) < 2π such that s =
{(cos t)P + (sin t)Q : t1 ≤ t ≤ t2 } and α(t) = (cos t)P + (sin t)Q.
Note that t1 , t2 , P, Q are not unique. However, if the segment s may also be represented as
s = {(cos t)P 0 + (sin t)Q0 : t01 ≤ t ≤ t02 } and α0 (t) = (cos t)P 0 + (sin t)Q0 , then
• |t1 − t2 | = |t01 − t02 | (the same length),
• {α(t1 ), α(t2 )} = {α0 (t01 ), α0 (t02 )} (the same end points) and
• P × Q = ±P 0 × Q0 (the line containing the segment is unique).
Theorem 6.32. Let A and B be any perpendicular points (ie. hA, Bi = 0) in S2 . Let s be a segment
on line AB, then there exists a unique interval [a, b] with 0 ≤ a < 2π and s = {(cos t)A + (sin t)B :
a ≤ t ≤ b}.
Theorem 6.33. Let A and B be non-antipodal points, then there exist exactly two segments with
A and B as their endpoints. The union of the two segments is the line AB. The intersection is the
set {A, B}
Definition 6.34. The longer segment is the major segment, the shorter segment is the minor
segment. If A and B are antipodal, then each of the (infinitely many) segments, with A and B as
endpoints, is called a half-line.
The length of the minor segment AB is d(A, B), the length of the major segment AB is 2π −
d(A, B) and the length of the half-line AB is π.
Theorem 6.35. Let P, Q, X be distinct points in S2 . If P and Q are not antipodal, then X lies
on minor P Q if and only if d(P, X) + d(X, Q) = d(P, Q).
Proof. Let P 0 be perpendicular to P and write minor P Q = {(cos t)P + (sin t)P 0 : 0 ≤ t ≤ L}
where L = d(P, Q). If X lies on minor P Q, then X = (cos φ)P + (sin φ)P 0 for some 0 < φ < L.
Now, d(P, X) = cos−1 hP, Xi = cos−1 (cos φ) = φ, and d(X, Q) = cos−1 hX, Qi = cos−1 (cos L cos φ +
sin L sin φ) = cos−1 (cos(L − φ)) = L − φ. Thus d(P, X) + d(X, Q) = φ + L − φ = L = d(P, Q) as
required.
To prove the converse let X = (cos φ)P + (sin φ)P 0 for some
−π < φ ≤ π. We shall consider four cases:
P0
[φ between P and Q] 0 < φ < L, d(P, X) = φ, d(X, Q) = L−φ
and d(P, X) + d(X, Q) = φ + L − φ = L = d(P, Q).
Q
[φ between Q and −P] L < φ ≤ π, d(P, X) = φ, d(X, Q) =
φ − L and d(P, X) + d(X, Q) = 2φ − L 6= L since φ 6= L.
−P
P
[φ between P and −Q] L − π ≤ φ < 0, d(P, X) = −φ,
d(X, Q) = L − φ and d(P, X) + d(X, Q) = L − 2φ 6= L since
−Q
φ 6= 0.
[φ between −Q and −P] −π < φ < L − π, d(P, X) = −φ,
d(X, Q) = 2π + φ − L and d(P, X) + d(X, Q) = 2π − L 6= L since L 6= π.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
27
Definition 6.36. A ray is a half-line with one end point removed. The other end point is called
the origin of the ray.
Suppose P and Q are points, and minor P Q has length L and is written minor P Q = {(cos t)P +
(sin t)P 0 : 0 ≤ t ≤ L}. Then {(cos t)P + (sin t)P 0 : 0 ≤ t < π} is the unique ray through Q with
−−→
origin P . We write this ray as P Q (which is unique).
Definition 6.37. The union of two rays r1 and r2 with common origin P is called an angle with
vertex P and arms r1 and r2 . If r1 = r2 it is the zero angle. If r1 and r2 are on the same line
and r1 6= r2 it is the straight angle.
The angle with vertex Q and arms QP and QR is written as ]P QR.
Definition 6.38. Let ]P QR be an angle. A point X is in the interior of ]P QR if minor XP
does not intersect the line QR and minor XR does not intersect the line QP . The interior is called
the lune of ]P QR.
Definition 6.39. The radian measure of ]P QR is
Q×P Q×R
−1
cos
,
.
|Q × P | |Q × R|
Definition 6.40 (Triangle). Let P, Q, R be non-collinear points the triangle M P QR is the unon
of the minor segments P Q, QR, P R these are the sides of the triangle the interior of the triangle
is the set of points X which are in the interior of each of the three angles.
There are other possible definitions for a triangle, but this has the following properties:
• M P QR lies in a “hemisphere”.
• 3 non-collinear points determine a unique triangle.
• π < angle sum < 2π
Spherical trigonometry. Let ABC be a triangle sides of length a = d(B, C), b = d(A, C) and
c = d(A, B), so that a = cos−1 hB, Ci etc.
In the plane law of cosines
b2 + c2 − a2
cos A =
2bc
We shall use the equation hA × B, A × Ci = hB, Ci − hA, CihA, Bi from Theorem 6.2(ix) to obtain
a version of this for the sphere. First note from Theorem 6.2(viii) that | B × C |2 =| B |2 | C |2
−hB, Ci2 = 1 − cos2 a = sin2 a so that B × C = sin a. Now,
hB, Ci − hA, CihA, Bi = cos a − cos b cos c
and from the defintion of angle we have
cos A =
A×B
A×C
,
|A×B | |A×C |
.
Thus,
cos A =
the law of cosines on the sphere.
hA × B, A × Ci
cos a − cos b cos c
=
,
sin b sin c
sin b sin c
28
ROBERT JUDD
We next investigate the law of sines. Rearranging the above we obtain
cos(b − c) − cos a
= 2 sin2 (A/2)
sin b sin c
cos a − cos(b + c)
1 + cos A =
= 2 cos2 (A/2)
sin b sin c
1 − cos A =
and if we multiply these together we get
sin2 A = 4 sin2 (A/2) cos2 (A/2) =
(cos(b − c) − cos a)(cos a − cos(b + c))
K
=
sin2 b sin2 c
sin2 b sin2 c
where
K = 4 sin((b − c + a)/2) sin((b − c − a)/2) sin((a + b + c)/2) sin((a − b − c)/2) .
Setting a + b + c = 2s and dividing both sides by sin2 a we see that
sin2 A
4 sin(s) sin(s − a) sin(s − b) sin(s − c)
=
,
2
sin a
sin2 a sin2 b sin2 c
and since the right hand side is symmetric in a, b, c this gives
sin A
sin B
sin C
=
=
.
sin a
sin b
sin c
We must now justify our choice of K: we must show that K = (cos(b − c) − cos a)(cos a − cos(b + c)).
This follows from the basic trigonometric identities; we give the proof that cos a − cos(b + c) =
−2 sin((a + b + c)/2) sin((a − b − c)/2) and leave the reader to show that cos(b − c) − cos a =
−2 sin((b − c + a)/2) sin((b − c − a)/2).
(b + c)
cos a − cos(b + c) = 2 cos (a/2) − cos
−1
2
2
2 (b + c)
= 2 cos (a/2) − cos
2
(b + c)
(b + c)
(b + c)
= 2 cos2 (a/2) sin2
+ cos2 (a/2) cos2
− cos2
2
2
c
(b + c)
(b + c)
= 2 cos2 (a/2) sin2
− sin2 (a/2) cos2
2
2
(b + c)
(b + c)
= −2 sin
cos(a/2) + sin(a/2) cos
2
2
(b + c)
(b + c)
sin
cos(a/2) − sin(a/2) cos
2
2
(a + b + c)
(a − b − c)
= −2 sin
sin
2
2
2
2
We have the following analogue of Pythagoras’s Theorem for right triangles.
Theorem 6.41. Let ABC be a triangle with sides of lengths a, b, c as above. If AC is perpendicular
to AB, then cos a = cos b cos c.
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
29
Proof. Let u be a pole of AB such that C = (cos b)A + (sin b)u. Then B = (cos c)A ± (sin c)u × A
and hence
cos a = cos(d(B, C)) = cos(cos−1 hB, Ci) = hB, Ci
= h(cos b)A + (sin b)u, (cos c)A ± (sin c)u × Ai = cos b cos c .
Theorem 6.42. Let L be a line X a point not on L nor a pole of L, let m be the line through X
perpendicular to L let F be the point in L ∩ m closest to X and −F its antipode, then for every Y
on L except ±F we have d(X, F ) < d(X, Y ) < d(X, −F ).
Proof. Let X = C, F = A, Y = B in the previous theorem, so that a = d(X, Y ), b = d(X, F ),
c = d(F, Y ), d(X, −F ) = π − b and b < π/2. Thus we have cos a = cos b cos c and since b < π/2,
cos b > 0 so that cos a and cos c have same sign. If both are positive, then cos a = cos b cos c < cos b
since cos c < 1 and cos b > 0. Now, cosine is decreasing on [0, π] which gives b < a < π/2 < π − b.
If both are negative, then b < π/2 < a and cos(π − a) = cos b cos(π − c) < cos b (since cos(π − θ) =
− cos θ) and as before b < π − a so that b < a < π − b. If both zero, then b < π/2 = a < π − b Definition 6.43. In the notation of the previous theorem F is the foot of the perpendicular from
X to L, and d(X, F ) is written as d(X, L), the distance from X to L.
Concurrence theorems. We have the same results as in E2 for the intersection of the perpendicular bisectors of the sides of a triangle, and the intersection of the angle bisectors of the angles
of a triangle.
Theorem 6.44. The perpendicular bisectors of the three sides of a triangle are concurrent.
The proof is the same as in E2 .
Proof. Let ∆P QR be the triangle and let M be the intersection of the perpendicular bisectors of
P Q and QR. Now, d(P, M ) = d(Q, M ) and d(Q, M ) = d(R, M ) so that d(P, M ) = d(R, M ) and
hence M is also on the perpendicular bisector of P R.
Theorem 6.45. Let P , Q and R be non-collinear in S2 , let p = QR, q = P R, r = P Q and let
reflections Ωu interchange p, q, and Ωv interchange q, r. Then there exists a line w concurrent
with u, v such that Ωw interchanges p, r.
Proof. Let M be the intersection of u, v and let ` be the line through M perpendicular to q.
By the three reflection theorem there exists a line w through M with Ωv Ωl Ωu = Ωw . Now,
Ωw p = Ωv Ωl Ωu p = Ωv Ωl q = Ωv q = r, as required.
Corollary 6.46. The lines bisecting the three angles of a triangle are concurrent.
Proof. Exercise. Note that there is something to prove, since (see below) there are at least two
reflections interchanging any pair of lines.
30
ROBERT JUDD
Lemma 6.47. Given any pair of distinct, non-parallel, lines p, q in S2 , there exist exactly two
reflections that interchange them.
Proof. Exercise.
7. The Projective Plane P2
Definition 7.1. The Projective plane P2 can be defined in the following three ways.
(1) P2 is the set of all pairs {x, −x} of antipodal points in S2 .
(2) P2 is the set of lines through the origin O in E3 .
(3) P2 is the set of equivalence classes of triples (x1 , x2 , x3 ) where xi ∈ R, not all zero (i.e.
non-zero vectors in R3 , where vectors are equivalent if they are parallel).
We can represent the same point in P2 as any useful multiple of (x1 , x2 , x3 ). For example, (1, 1, 1)
and (2, 2, 2) represent the same point.
Let π : S2 → P2 be the mapping π : x 7→ {x, −x} (projection). π is a two-to-one map from S2
to P2 .
Definition 7.2. A line of P2 is a set π` where ` is a line in S2 . If u is a pole of `, then πu is the
pole of π`.
Theorem 7.3.
(i) Two distinct lines in P2 intersect in exactly one point.
(ii) Two distinct points in P2 determine a unique line.
Proof. Let ` and m be distinct lines in S2 and let P ∈ `, Q ∈ m be such that πP = πQ. Then
P = ±Q and since πQ = π(−Q) we may assume P = Q. Thus a point of intersection of π` and
πm must be the projection of a point of intersection of ` and m. Let u, v be the poles of ` and m
respectively. Since ` and m are distinct it follows that u, v are not antipodal and ±u × v/|u × v|
are the two points of intersection of ` and m. But π(u × v/|u × v|) = π(−u × v/|u × v|) are the
same point in P2 .
Let πx, πy be distinct in P2 , then x, y are not antipodal in S2 . Thus there exists a unique line
` containing x, y and so πx, πy ∈ π`.
7.1. Homogeneous Coordinates.
Definition 7.4. Let {e1 , e2 , e3 } be a basis for R3 . Then each x ∈ R3 determines a unique triple
(x1 , x2 , x3 ) by x = x1 e1 + x2 e2 + x3 e3 . If πx ∈ P2 , λ ∈ R \ {0} and λx = t1 e1 + t2 e2 + t3 e3 ,
then (t1 , t2 , t3 ) is called a homogeneous coordinate vector of πx and t1 , t2 , t3 are called the
homogeneous coordinates of πx.
Theorem 7.5. Let P , Q, R, S be distinct points in P2 , such that no three are collinear. Then
there exists a basis of R3 with respect to which the four points have homogeneous coordinate vectors
(1, 0, 0), (0, 1, 0), (0, 0, 1), (1, 1, 1).
EUCLIDEAN AND NON-EUCLIDEAN GEOMETRY
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Proof. Let v1 , v2 , v3 in R3 be representatives of P , Q, R. Since P , Q, R are not collinear,
{v1 , v2 , v3 } are linearly independent and hence a basis for R3 . If v4 is any representative of S,
then v4 = k1 v1 + k2 v2 + k3 v3 , for some k1 , k2 , k3 ∈ R. Set e1 = k1 v1 , e2 = k2 v2 , e3 = k3 v3 , then
{e1 , e2 , e3 } is the required basis.
Theorem 7.6. Let x and y be homogeneous coordinate vectors of two distinct points in P2 . Then
λx + µy, for λ, µ ∈ R is a typical point on the line they determine.
Proof. P2 is the set of equivalence classes of the set {(x, y, z) : x, y, z ∈ R not all zero } under the
equivalence relation ∼, where (x, y, z) ∼ (u, v, w) if there exists t ∈ R \ {0} with u = tx, v = ty,
w = tz. A homogeneous coordinate vector is any equivalence class representative.
Let the basis for the homogeneous coordinate vectors be {e1 , e2 , e3 } and let x = (x1 , x2 , x3 ),
y = (y1 , y2 , y3 ). Given vectors u = x1 e1 + x2 e2 + x3 e3 , v = y1 e1 + y2 e2 + y3 e3 in R3 , the plane
in R3 containing u and v is given by {λu + µv : λ, µ ∈ R}. Lines in P2 correspond to planes in
R3 through the origin O. So every homogeneous coordinate vector λx + µy corresponds to a point
λu + µv in the plane containing u, v (execept O) and vice versa.
Note: For the moment we shall use triangle in P2 to mean three non-collinear points.
Theorem 7.7 (Desargues’s Theorem). Let P QR, P 0 Q0 R0 be triangles in P2 such that the lines
P P 0 , QQ0 , RR0 are concurrent. Then P Q ∩ P 0 Q0 , QR ∩ Q0 R0 , P R ∩ P 0 R0 are collinear.
Proof. Let X be the point of concurrence. If X is collinear with any two of P , Q, R, then two lines
(e.g. P Q, P 0 Q0 ) would coincide and the theorem is trivial.
Otherwise no three of P , Q, R, X are collinear and we may represent them as homogeneous
coordinate vectors P = (1, 0, 0), Q = (0, 1, 0), R = (0, 0, 1), X = (1, 1, 1). Now, P 0 is on XP so
P 0 = λ(1, 0, 0) + µ(1, 1, 1) = (λ + µ, µ, µ), so we can write P 0 = (p, 1, 1), for some p ∈ R. Similarly
Q0 = (1, q, 1) and R0 = (1, 1, r).
The equation of the line P Q is x3 = 0. A point on P 0 Q0 looks like λ(p, 1, 1) + µ(1, q, 1) = (λp +
µ, λ+µq, λ+µ). So the equation is (1−q)x1 +(1−p)x2 +(pq−1)x3 = 0. They intersect in the point L.
Now x3 = 0 and hence (1−q)x1 +(1−p)x2 = 0. If x1 = p−1, then x2 = 1−q. So L = (p−1, 1−q, 0).
Similarly QR ∩ Q0 R0 = M , where M = (0, q − 1, 1 − r) and P R ∩ P 0 R0 = N = (1 − p, 0, r − 1).
Since the sum of these homogeneous coordinate vectors is 0, the points L, M , N are collinear. A. Appendix: Groups
We collect here a few useful definitions of groups for completeness.
Definition A.1. A group is a set G together with an operator ◦ : G × G → G, denoted by (G, ◦),
that satisfies:
(i) Associativity: (a ◦ b) ◦ c = a ◦ (b ◦ c), for all a, b, c ∈ G
(ii) Identity: there exists e ∈ G such that a ◦ e = e ◦ a = a, for all a ∈ G.
(iii) Inverses: for all a ∈ G, there exists a−1 ∈ G such that a ◦ a−1 = a−1 ◦ a = e.
Definition A.2. A subset H ⊂ G of a group (G, ◦) is a subgroup of G if
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ROBERT JUDD
(i) a ◦ b ∈ H, ∀ a, b ∈ H
(ii) a−1 ∈ H, ∀ a ∈ H
Equivalently, H is a subgroup of G if and only if a ◦ b−1 ∈ H, ∀a, b ∈ H.
B. Appendix: Equivalence relations
A relation R on a set S is a subset of S × S = {(x, y) : x, y ∈ S}. For example the relation < on
R is given by the set {(x, y) : x, y ∈ R, x < y}. We say xRy to mean X is related to y by R.
Definition B.1. A relation R is
(i) reflexive if aRa for all a ∈ S.
(ii) symmetric if aRb implies bRa for all a, b ∈ S.
(iii) transitive if aRb and bRc implies aRc for all a, b, c ∈ S.
An equivalence relation is one which satisfies all three.
Definition B.2. An equivalence class of an equivalence relation ∼ is the set [a] = {b ∈ S : b ∼
a}. The element a of S is an equivalence class representative.
For example a ∼ b in Z if a = b mod 7. The equivalence classes are [0], [1], . . . , [6].
Theorem B.3. Equivalent classes are disjoint and independent of the representative.
C. Appendix: Linear Independence of Vectors
Definition C.1. Vectors v1 , v2 , . . . , vn in Rn are linearly independent if a1 v1 +a2 v2 +· · ·+an vn =
0 implies ai = 0 for i = 1, . . . , n.
Definition C.2. If v1 , v2 , . . . , vn are linearly independent in Rn , then every vector v ∈ Rn can be
written as v = a1 v1 + · · · + an vn . The set {v1 , v2 , . . . , vn } is a basis for Rn .
Typed by:
Rob Judd, Mario Lopez, Gaurav Ghare, Jennifer Proal, Haifa Al-Nasser, Oscar Hasbun and Aditya
Nagrath.
E-mail address: [email protected]
URL: http://www.math.du.edu/∼rjudd/geometry/