Projective Geometry

Axiomatic Projective Geometry
Emiel Haakma
April 2019
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Preface
Projective geometry is, in essence, a geometry in which parallel lines do not
exist. In this way, it contrasts itself from Euclidean and hyperbolic geometry,
and this difference causes many interesting results. In this paper, projective
geometry will first be extensively introduced, before several definitions are in-
troduced, often unique to this geometry, and many different theorems will be
shown and proven.
To do so, it is important that the reader has some experience with common
strategies for creating mathematical and geometrical proofs. This experience
is assumed in this paper; strategies are not explained any further and instead
simply applied.
It is also expected that the reader has some experience in axiomatic mathemat-
ics, preferably geometry. In a way, we will be going into this paper with no
prior knowledge on the projective plane other than the axioms, so having an
understanding as to how such a system works is crucial.
Lasty, experience in linear algebra is assumed. While it may not be obvious
at first glance, the projective plane has deep links to the Euclidean three-
dimensional space, so understanding the math and theorems that are applied is
important.
This paper it is split up into four chapters. In the first, projective geometry will
be introduced and defined, the second will describe and prove many theorems,
the third will introduce projective maps and show many of their properties, and
finally the fourth chapter will apply all we have learned to a new notion called
harmonic additions.
This research was done as a final project in the bachelor Applied Mathematics
at the Delft University of Technology, in collaboration with Dr. J. Vermeer. My
thanks goes out to him.
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Contents
1 Introduction to projective geometry 4
1.1 Definitions and axioms . . . . . . . . . . . . . . . . . . . . . . . . 4
1.2 Principle of duality . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.3 Finite Projective Plane . . . . . . . . . . . . . . . . . . . . . . . . 8
2 Theorems in the projective plane 10

2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 Desargue’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.3 Pappus’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 17
2.4 Fano’s Axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.5 Theorems of Menelaus and Ceva . . . . . . . . . . . . . . . . . . 26
3 Perspectives and projective maps 29

3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.2 Double Perspective Theorem . . . . . . . . . . . . . . . . . . . . 30
3.3 Fundamental Property . . . . . . . . . . . . . . . . . . . . . . . . 38
4 Harmonic Addition 46
4.1 Definition and Properties . . . . . . . . . . . . . . . . . . . . . . 46
4.2 Harmonic Pairs and Projective Maps . . . . . . . . . . . . . . . . 50
4.3 Dual Harmonic Addition . . . . . . . . . . . . . . . . . . . . . . . 52
3
1 Introduction to projective geometry
This chapter will be focused on introducing the basic concepts behind the projec-
tive plane. This is done through first defining it in the first section, introducing
a core principle in the second, and finally showing what finite projective planes
look like and how they function in the third.
1.1 Definitions and axioms

The very first thing we must do is to define a projective geometry. After all, as
is always the case in math, we cannot talk about something we have not clearly
defined. However, before we can do that, we’ll have to define an axiomatic
theory first:
Definition 1.1. An axiomatic theory is described by:

1. a system of fundamental notions (P1 , P2 , ...),
2. a set of axioms about the fundamental notions.
Now, this may look very confusing, so we will illustrate it with a well-known
example: our very own real number system. This is, after all, nothing more
than an axiomatic theory. In this case, our system of fundamental notions are
(R, +, ∗), in other words, they consist of the set of real numbers, the operation
of addition, and the operation of multiplication. Meanwhile. our set of axioms
consists of the well-known addition axioms, multiplication axioms, the order
axiom, and the completeness axiom.
So now that we have a sense for what an axiomatic theory is, we can define our
most important term:
Definition 1.2. A projective geometry is an axiomatic theory with the triple
(Π, Λ, I) as its fundamental notions and axioms 1.1, 1.2, and 1.3 as the axioms.
Here, Π and Λ are disjoint sets and I is a symmetric relation between Π and
Λ (in other words, a I b ⇔ b I a where a ∈ Π and b ∈ Λ). The elements of Π
are called ’points’, the elements of Λ are called ’lines’, and a I b is read as ’a is
incident with b’.
Now, again, this looks confusing, but effectively all it says is that in other
to define a projective geometry, we need a set of points, a set of lines, and an
incidence relation, as well as the following three axioms:
Axiom 1.1. Given two distinct points, there is exactly one line incident with
both points.
Axiom 1.2. Given two distinct lines, there is exactly one point incident with
both lines.
Axiom 1.3. Π contains at least four points such that no three of them are
incident with one and the same line.
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Before we move on, it should be noted that this axiomatic system is not
unique; there is a wide variety of equivalent sets of axioms for projective geom-
etry, but these 3 are the ones we will be working with.
Now, so far, we have talked about abstract sets for points and lines as well as
an abstract relation for incidence.. The reason for this is that, often, we do
not need to define these exactly to prove theorems. However, if we do, and the
found sets and relation satisfy the axioms, we call it a projective plane. From
here on out, we will often denote a projective plane as P. In the case where
we are talking about multiple different projective planes, we will use a subscript
such as P1 or P2 to differentiate between them.
Next, we will provide a different way of defining lines. Note that every line l is
determined uniquely by the set of points incident with it. If we call this set P,
then we can see there is no problem in saying l = P. As such, from here on out,
we will see lines as nothing more than sets of points. With this, we can also
write ’P ∈ l’ for ’P I l’ and ’P ∈/ l’ for ’not P I l’, where P is a point and l is
a line. In addition to this, we can now denote the point of intersection for two
different lines l and m as l ∩ m; by axiom 1.2, this is exactly one point.
To continue, we will introduce a few more definitions:
Definition 1.3. Three (or more) points P1 , P2 , and P3 are called collinear if
and only if there is a line l such that P1 , P2 , P3 ∈ l. In other words, there is a
line such that all three points are incident with this line.
Definition 1.4. Three (or more) lines l1 , l2 , l3 are called concurrent if and
only if there is a point P such that P ∈ l1 , P ∈ l2 , and P ∈ l3 . In other words,
there is a point such that all three lines are incident with this point.
Definition 1.5. Two projective planes P0 = (Π0 , Λ0 , I0 ) and P1 = (Π1 , Λ1 , I1 )
are called isomorphic if and only if there are one-to-one mappings π : Π0 → Π1
and λ : Λ0 → Λ1 such that P ∈ l ⇔ π(P ) ∈ λ(l). In other words, π and λ
preserve incidence relations.
Finally, we have established almost every important definition. The final
one will be discussed now.
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1.2 Principle of duality
In the context of projective planes, duality refers to switching the words ’point’
and ’line’ in theorems, or interchanging the sets Π and Λ (recall that these are
the sets of points and lines respectively). Before discussing what exactly this
means, we will first show a proof: namely the proof of the dual theorem of axiom
1.3, or the theorem found by interchanging the words ’point’ and ’line’ in that
axiom:
Theorem 1.1 (Dual theorem of axiom 1.3). Λ contains at least four lines such
that no three of them are concurrent.
Proof. By axiom 1.3, we can find four points such that no three of them are
collinear. Clearly, these four points are different, so we will call them P1 , P2 ,
P3 , and P4 . By axiom 1.1, the lines P1 P2 , P1 P3 , P2 P4 , and P3 P4 are unique.
We will show no three of these are concurrent.
By way of contradiction, assume that P1 P2 , P1 P3 , and P3 P4 are all incident
with a point Q. Then, P1 P2 and P1 P3 are both incident with Q and with P1 .
Since these lines are not the same (recall that P1 , P2 , and P3 are not collinear),
we must find Q = P1 by axiom 1.2. But then P3 P4 is incident with P1 , which
means P1 , P3 , and P4 are collinear, which is a contradiction. Thus, P1 P2 , P1 P3 ,
and P3 P4 are not concurrent. This can be proven analogously for the other
triples.
So now, we have proven the dual theorem of axiom 1.3, and clearly, axioms
1.1 and 1.2 are dual to each other. This is very important, as what we have
now shown is that for all of the axioms, their dual theorem is true. Thus now,
for any theorem we proof using the axioms, we can also prove its dual theorem.
This is known as the principle of duality:
Theorem 1.2 (Principle of duality). If, in a theorem that can be proven using
the axioms of projective geometry, we interchange the words ’point’ and ’line’,
we obtain another theorem that is true in projective geometry.
Proof. Let P be a theorem that is true in a system with axioms 1.1, 1.2, and 1.3.
Note that, in that system, theorem 1.1 is true. Therefore, we can use axioms
1.2 and 1.1 as well as theorem 1.1 to prove dP , the theorem dual to P .
This is a very useful result, and we will illustrate using a simple example:
Theorem 1.3. For each line l there are at least three points incident with l.
Proof. We begin with a line l. By axiom 1.3, there exist 4 points P1 , P2 , P3 ,
and P4 such that no three of them are collinear. We will consider three cases:
1. l is incident with two of these points. Without loss of generality, say

P1 , P2 ∈ l. Then, by axiom 1.2, P3 P4 ∩ l exists. This point cannot be P1
or P2 , because that would imply collinearity between that point and P3
and P4 , which is a contradiction. Thus, we have found three points on l.
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2. l is incident with one of these points. Without loss of generality, assume
P1 ∈ l. Then, l intersects P2 P3 , P3 P4 , and P2 P4 by axiom 1.2. None of
these points of intersection can be P1 , as otherwise we have a contradictory
collinearity. In addition, no two of these lines can have the same point
of intersection, as that would imply they are either the same line or have
multiple points of intersection, both of which are contradictions. Thus,
we have found four points on l.
3. l is incident with none of these points. In that case, by the fact that these
points cannot be collinear and P1 ∈ / l, the points P1 P2 ∩l, P1 P3 ∩l, P1 P4 ∩l
are all different (after all, the lines that are intersecting l are all different,
and they all intersect in P1 , which must be the only point of intersection
by axiom 1.2). Thus, we have found three points on l.
This is a useful theorem to have, of course, but more importantly is how

easily we can apply the principle of duality to it:
Theorem 1.4 (Dual theorem of theorem 1.3). For each point P there are at
least three lines incident with P .
Thanks to the principle of duality, we do not have to provide a proof to this.
Simply having the dual theorem is enough.
To close out this paragraph, I would like to make note of another result of the
principle of duality:
Theorem 1.5. For any projective plane P = (Π, Λ, I), its dual plane dP =
(Λ, Π, I) is also a projective plane.
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1.3 Finite Projective Plane
In this paragraph, we will showcase a special case of projective planes, namely
finite ones. A well-known one is Fano’s Plane, which contains 7 points and 7
lines:
Definition 1.6. Fano’s Plane is a projective plane with:

• Π = {A, B, C, D, E, F, G}
• Λ = {ADB, AGE, AF C, BEC, BGF, CGD, F DE}
The incidence relation is as expected.
See figure 1 for an image of what this plane looks like. Now, all of the axioms
are easily verified: pick any two points and there will be a line through both,
any two lines intersect in some point, and of the points A, C, and F , and G, no
three are collinear.
Now, let us define the order of a finite projective plane:
Definition 1.7. A finite projective plane has order n if and only if there is at
least one point such that there are exactly n + 1 lines incident with that point.
Clearly, Fano’s Plane is of order 2. After all, every point has 3 lines incident
with it. Interestingly, this is not a coincidence, as we will now prove in two
steps:
Figure 1: Fano’s Plane
Theorem 1.6. In a projective plane P of order n, every point has exactly n+1
lines incident with it.
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Proof. Let P be an arbitrary point in P and let l be a line with n + 1 points
on it. Label these points P1 , ..., Pn+1 . We recognize two cases:
1. P is not on l. Then for every Pi , there is a line through P and Pi and
since P is not on l, each of these are both distinct from l and from each
other. After all, if there was a line that goes through P , Pi , and Pj , then
that line would be distinct from l and still go through Pi and Pj , which
violates uniqueness in axiom 1.1. Now, we have n + 1 distinct lines going
through P .
2. P is on l. Now, by axiom 1.3, there exist points Q and R not on l. Bt
axiom 1.1, lines RP1 , RP2 , and RP3 exist. Since R is not on l, at least
two of these lines do not contain P , and similarly, at least two of these
lines do not contain Q. Then, there is at least one line m that does not
contain P or Q.
Since Q is not on l, the first case gives us that there are exactly n + 1
lines m1 , ..., mn+1 through Q. And since Q is not on m, m intersects each
of these lines in exactly one point by axiom 1.2, which means there are
n + 1 points S1 , ..., Sn+1 . There cannot be another point on m either;
after all, if there was another point T on m, then the line T Q would be
distinct from the mi . But then there are n + 2 lines through Q, which is
a contradiction. Thus, we now have a line m with exactly n + 1 points on
it and that is not incident with P . We can now apply case 1 to see there
are exactly n + 1 lines through P .
Now, we can trivially say there is a point with n+1 lines through it, which is
the dual statement to the definition of a projective plane of order n. Therefore,
the principle of duality still holds. Thanks to that, we can easily say
Theorem 1.7. In a projective plane P of order n, every line has n + 1 points
incident with it.
Proof. This theorem follows directly from theorem 1.6 and the principle of du-
ality.
Finally, we will showcase a proof of how many points and lines there are
exactly:
Theorem 1.8. In a projective plane of order n, there are exactly n2 + n + 1
points and n2 + n + 1 lines.
Proof. By axiom 1.3, there exists at least one point P , as well as some number of
points distinct from P . For every point, there must be exactly one line through
P for each point distinct from P . By theorem 1.6, there are exactly n + 1 lines
through P . Note that every point in the plane must be on one of these lines.
By theorem 1.7, each of these lines has exactly n points on it other than P .
Thus, the total amount of points is n(n + 1) + 1 = n2 + n + 1. By the principle
of duality, the total amount of lines is the same.
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2 Theorems in the projective plane
This chapter will be focused on introducing many famous theorems of the pro-
jective plane, several of which will see use in later chapters too. To do so, we
must first introduce certain notions in the first section before moving onto the
theorems in later sections.
2.1 Definitions
In this section, we will repeat a few definitions from algebra and soon relate
them to the projective planes that we have established before. To begin, recall
the definitions of fields and division rings:
Definition 2.1. A field is an axiomative theory with (L, +, ∗) as the fundamen-
tal notions, where L is a set, + is an addition operation, and ∗ is a multiplication
operation, and the following axioms:
• a + (b + c) = (a + b) + c and a ∗ (b ∗ c) = (a ∗ b) ∗ c
• a + b = b + a and a ∗ b = b ∗ a
• There exist elements 0L and 1L of L such that a + 0L = a and b ∗ 1L = b
• For any a, there exists an element −a of L such that a + (−a) = 0L
• For any a 6= 0, there exists an element a−1 such that a ∗ a−1 = 1L
• a ∗ (b + c) = (a ∗ b) + (a ∗ c)
Here, a, b, and c are arbitrary elements of L. Such a field is often written as
simply L if the addition and multiplication operations are clear.
Definition 2.2. A division ring is an axiomative theory with the same funda-
mental notions and axioms as a field, with one exception: a ∗ b = b ∗ a does not
have to be true. In other words, multiplication is not commutative.
A common example given for a division ring that is not a field are the
quaternions. As it is just an example, we will not dive too deep into it, but in
short, the set of quaternions is a number system where each number is written
as a + bi + cj + dk, where a, b, c, d ∈ R and i, j, and k are the fundamental
quaternion units, defined by
i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, and ki = −ik = j
It is not hard to show the quaternions form a division ring, but from how multi-
plication is defined, it is clearly not commutative. Thus, the set of quaternions
is not a field.
However, of course, there are quite a few division rings that are, in fact, fields.
It should be clear that the set of fields is a subset of the set of division rings,
but we can say slightly more about it too:
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Theorem 2.1. Every finite division ring is a field.
We present this theorem without a proof, as the proof, as it was originally
given by MacLagan Wedderburn in 1905, calls upon concepts that we do not
wish to introduce within this paper. However, it is certainly useful, as it means
we know more about the context of section 1.3.
Now that we have this theorem and an example of when division rings aren’t
fields, we know that division rings is a relevant definition. Next, we recall the
notions of vector spaces and modules:
Definition 2.3. An vector space over a field L is an axiomative theory with
(V , +, ∗) as the fundamental notions, where this time, V is a set of so-called
vectors, + is vector addition and ∗ is scalar multiplication. The axioms are as
follows:
• X +Y =Y +X
• (X + Y ) + Z = X + (Y + Z)
• There is an element 0V of V such that X + 0Ln = X
• For every X, there is an element −X of V such that X + −X = 0V
• r ∗ (s ∗ X) = (rs) ∗ X
• (r + s) ∗ X = r ∗ X + s ∗ X
• r ∗ (X + Y ) = (r ∗ Y ) + (r ∗ X)
• 1L ∗ X = X
Here, X, Y , and Z are arbitrary elements of V , r and s are arbitrary elements
of L, and 1L is as described in the field axioms.
Definition 2.4. An module over a division ring M is the generalization of the
notion of a vector field to division rings. Their fundamental notions and axioms
are equivalent.
Quickly, we will present two simple theorems without proof, as they are both
easily shown:
Theorem 2.2. For any field L, (Ln , +, ∗) is a vector field.
Theorem 2.3. For any division ring M , (M n , +, ∗) is a module.
These will both be useful, and in general, when we are talking about a vector
field over L or a module over M , these will be the ones we are considering.
It is assumed common definitions from linear algebra, such as subspaces, depen-
dence, linear combinations, spans, and bases, are known and understood. Note
that each of these definitions is equivalent when used in the context of modules.
Now, before we move on, we will show and prove an incredibly important the-
orem in vector spaces as well as some of its corollaries, though we first begin
with a lemma:
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Lemma 2.1. If u1 , u2 , ...uk ∈ V are linearly independent, then so are u1 , u2 −
c2 ∗ u1 , ...uk − ck ∗ u1 , where c2 , ...ck are scalars.
The proof of this lemma is very simple and will not be presented, as it is
more important to move on towards the following theorem:
Theorem 2.4. If u1 , ..., uk+1 are k + 1 vectors such that they are all included
in the span of k vectors v1 , ..., vk , then u1 , ..., uk+1 are linearly dependent.
Proof. Since ui is contained in the span of v1 , ..., vk , we can write each of them as
ui = ai1 v1 + ... + aik vk . Now, by way of contradiction, assume that u1 , ..., uk+1
are linearly independent. Then, clearly, none of the ui are multiples of u1 .
Because of that, we can subtract a multiple of u1 from each of them. If we
do this appropriately, we can eliminate the term with v1 in the expression of
all ui with i ≥ 2. So, to recap, we have now written the ui in the form ui =
bi2 v2 + ... + bik vk for i ≥ 2. However, we can continue on like this, constantly
eliminate a term from the sums. Then, eventually, we will reach a sequence
with just two sums: uk = λkk vk and uk+1 = λk+1k vk . But then, clearly, uk and
uk+1 are linearly dependent. This is a contradiction. Therefore, the ui must be
linearly dependent.
Corollary 2.4.1. If a basis of a module V contains k vectors, then every basis
of V contains k vectors. We call k the dimension of V .
Corollary 2.4.2. Let V be a module with dimension k. Then, a set of k vectors
{u1 , ..., uk } is linearly independent if and only if Span{u1 , ..., uk } = V
The proofs of both corollaries are trivial.
Now that we have these theorems, we can finally link this theory to projective
planes:
Definition 2.5. Let L be a division ring and V be a module over L with

dimension 3. A projective plane over V, written as P(V ), is the projective
plane where:
• The set of points Π is the set of one-dimensional subspaces of V ;
• The set of lines Λ is the set of two-dimensional subspaces of V ;
• For π ∈ Π and λ ∈ Λ, π ∈ λ if and only if π is a subspace of λ.

It can easily be shown that P(V ) satisfies the three axioms of projective geom-
etry, meaning it is indeed a projective plane.
Having this is nice, because it lets us think about projective planes as some-
thing other than abstract bodies, but we can actually make models. After all,
it’s important to realize that all fields are division rings and thus all vector
spaces are modules. Therefore, this definition works just fine for vector spaces
such as, for instance R3 , where our points take the form of lines through the
origin and our lines are planes through the origin.
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2.2 Desargue’s Theorem
It is great that we have managed to define all of this, but now, we have to move
on and try to find theorems that are true in projective planes over modules. Two
of the most important ones are Desargue’s Theorem and Pappus’s Theorem:
Desargue’s Theorem. Let A1 , A2 , A3 , B1 , B2 , and B3 be points with the
following properties:
• The lines A1 B1 , A2 B2 , and A3 B3 are concurrent. Name their point of
intersection C.
• No three of the points C, A1 , A2 , and A3 and no three of the points C,
B1 , B2 , and B3 are collinear
Let P12 = A1 A2 ∩ B1 B2 , P23 = A2 A3 ∩ B2 B3 , and P31 = A3 A1 ∩ B3 B1 . Then,
P12 , P23 , and P31 are collinear.
Figure 2: Desargue’s Theorem
Now, as can be seen, we have not yet provided a proof for this theorem.
The reason for that is that, unfortunately, Desargue’s theorem does not hold in
every projective plane. However, it does hold in the most common ones:
Theorem 2.5. Let V be a module of dimension 3 over a division ring M . Then,
Desargue’s Theorem holds in P(V ).
Proof. Let V and M be as described and assume points A1 , A2 , A3 , B1 , B2 ,
and B3 are as in the hypothesis for Desargue’s Theorem. Now, remember that
these points are one-dimensional vector spaces. This means there are vectors
v1 , v2 , and v3 such that A1 = hv1 i, A2 = hv2 i, and A3 = hv3 i. In addition, since
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A1 , A2 , and A3 are not collinear, v1 , v2 , and v3 are linearly independent. Thus,
these three vectors span V and thus, C lies in the span of v1 , v2 , and v3 . Then,
by definition, there are a1 , a2 , a3 ∈ M such that C = a1 v1 + a2 v2 + a3 v3 . Since
we have that no three of A1 , A2 , A3 , and C are collinear, we have a1 , a2 , a3 6= 0,
as otherwise, C would be a linear combination of two of the vectors, thus making
it collinear with two of the points. Now, remember that Ai = hvi i. Because this
is a span we’re talking about, we can easily define wi = ai vi and say Ai = hwi i,
meaning we get C = hw1 + w2 + w3 i.
We know that for i ∈ {1, 2, 3}, C, Ai , and Bi are collinear. Thus, there are
b1 , b2 , b3 ∈ M such that
B1 = hw1 + w2 + w3 + b1 w1 i = h(b1 + 1)w1 + w2 + w3 i
B2 = hw1 + (b2 + 1)w2 + w3 i

B3 = hw1 + w2 + (b3 + 1)w3 i
Now, we want to find the points Pij . We begin with P12 :
P12 = A1 A2 ∩ B1 B2 = hw1 , w2 i ∩ h(b1 + 1)w1 + w2 + w3 , w1 + (b2 + 1)w2 + w3 i
Clearly, hb1 w1 − b2 w2 i is on both A1 A2 and B1 B2 . Then, by axiom 1.2, this is

the only point on both lines (as the lines are distinct). Therefore
P12 = hb1 w1 − b2 w2 i
and similarly
P23 = hb2 w2 − b3 w3 i
P31 = hb3 w3 − b1 w1 i
Clearly, P31 = −P12 − P23 , meaning these three points are collinear.
Interestingly, this theorem has another side to it, which is quite fascinating:
Theorem 2.6. If Desargue’s Theorem holds in a projective plane P, then there
is a module V such that P = P(V ).
We will not provide a proof for this statement, as it is quite technical and is
not fit for the scope of this paper.
Next, we will discuss the dual of Desargue’s theorem:
Dual of Desargue’s Theorem. Let l1 , l2 , l3 , m1 , m2 , and m3 be lines with
the following properties:
• The points P1 = l1 ∩ m1 , P2 = l2 ∩ m2 , and P3 = l3 ∩ m3 are collinear.
Call their common line n.
• No three of the lines n, l1 , l2 , and l3 and no three of the lines n, m1 , m2 ,
m3 are concurrent.
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Let c12 = (l1 ∩l2 )(m1 ∩m2 ), c23 = (l2 ∩l3 )(m2 ∩m3 ), and c31 = (l3 ∩l1 )(m3 ∩m1 ).
Then c12 , c23 , and c31 are concurrent.
Now, this may sound exceptionally confusing, and that’s understandable.
However, in fact, it is no more than the converse of Desargue’s Theorem! You
can check this for yourself too, but in the end, the point of intersection between
the line c12 , c23 , and c31 is equivalent to the point C from Desargue’s Theorem.
Now, interestingly, we will show the following:
Theorem 2.7. If Desargue’s Theorem holds in a projective plane, then so does
the dual of Desargue’s Theorem.
Proof. Let l1 , l2 , l3 , m1 , m2 , m3 , P1 , P2 , P3 , n, c12 , c23 , and c31 be as described.
Let C be the point of intersection between c23 and c31 , which exists by axiom
1.2. Call Aij = li ∩ lj and Bij = mi ∩ mj for i, j ∈ {1, 2, 3} and i 6= j. Note that,
by definition, Aij , Bij ∈ cij . See also figure 3. Now, we will apply Desargue’s
theorem, but we have to determine which points we use for this. First, P3 will
be the point of intersection between the concurrent lines (called C in Desargue’s
Theorem). Next, P2 , A23 , and B23 will play the roles of the Ai from Desargue’s
Theorem, while P1 , A31 , and B31 are the Bi . It is easily checked that these
points fulfill the conditions from Desargue’s theorem.
Now, clearly, A23 B23 ∩ A31 B31 = c23 ∩ c31 = C. Furthermore, it is not hard
to see that P2 A23 ∩ P1 A31 = A12 and P2 B23 ∩ P1 B31 = B12 . By Desargue’s
Theorem, these are collinear. But then c12 = A12 B12 goes through C, which
was defined as c23 ∩ c31 ! Therefore, c12 , c23 , and c31 all go through C and are
thus concurrent.
Figure 3: Dual of Desargue’s Theorem
It is great to see that the Dual of Desargue’s Theorem holds if Desargue’s

Theorem does, because that means that the Principle of Duality also holds
15
in that case. This is important, as otherwise, we’d already have ruined that
important principle. Luckily, we have not, and so we can continue working with
it.
Now, a question may arise now as to in what sort of plane Desargue’s Theorem
does not hold. After all, we know it holds in the most regular ones we know,
vector spaces, so what would a plane look like where it does not? For that, we
turn towards the so-called Moulton Plane:
Definition 2.6. The Moulton Plane is a projective plane with:
• Π = R2
• Λ = (R ∪ {∞} × R
• Let π = (x, y) ∈ Π and λ = (m, b) ∈ Λ. Then

x = b
 if m = ∞
πIλ ⇔ y = 21 mx + b if m ≤ 0, x ≤ 0

y = mx + b if m > 0 or x > 0

Of course, at first glance, this means nothing. The incidence relation is very
complicated and thus very unintuitive. Thus, in figure 4, an image is shown.
Here, you can see that as lines cross the y-axis, the bend somewhat. It is this
bend that causes Desargue’s Theorem not to hold in the Moulton Plane: the
line through two of the Pij would bend away before reaching the third, thus
causing a lack of collinearity.
Figure 4: Moulton’s Plane

Source: By Kmhkmh - Own work, CC BY 4.0,
https://commons.wikimedia.org/w/index.php?curid=53588728
Clearly, Moulton’s Plane is an infinite plane. But what about finite planes, like
the ones in section 1.3? Well, as it turns out, it is possible to construct finite
projective planes where Desargue’s Theorem doesn’t hold. The catch is that
this is only possible for planes of order 9 or higher. However, we do not provide
a proof of this, as it goes beyond the scope of this paper.
16
2.3 Pappus’s Theorem
In the last section, we mentioned one of two important and well-known theorems
in projective planes. In this section, we will discuss the other:
Pappus’s Theorem. Let l and m be distinct lines. Let A1 , A2 , and A3 be
distinct points on l, while B1 , B2 , and B3 are distinct points on m. These
6 points are also all different from l ∩ m. Let P12 = A1 B2 ∩ B1 A2 , P23 =
A2 B3 ∩ B2 A3 , and P31 = A3 B1 ∩ B3 A1 . Then, P12 , P23 , and P31 are collinear.
Figure 5: Pappus’s Theorem
Theorem 2.8. Let V be a module of dimension 3 over a division ring M . Then

Pappus’s Theorem holds on P(V ) if and only if M is a field.
Proof. Let V and M be as in the theorem and let the points A1 , A2 , A3 , B1 , B2 ,
and B3 as well as lines l and m be as in the hypothesis for Pappus’s Theorem.
We begin by only looking at C, A1 , A2 , B1 , and B2 , where C = l ∩ m.
Let u, v, w ∈ V such that C = hui, A1 = hvi, and B1 = hwi. By definition,
A2 = hu + avi and B2 = hu + bwi with a, b ∈ M \ {0}. Through appropriate
rescaling, we can set A1 = hv 0 i, A2 = hu + v 0 i, B1 = hw0 i, and B2 = hu + w0 i.
Let p, q ∈ M . Since we are aiming to show that M is a field, we must show
commutativity of multiplication. Since 0 and 1 commute with any element of
M, we can safely assume p, q 6= 0, 1. Define A3 = u + pv 0 and B3 = u + qw0 .
Since p, q 6= 0, 1, A3 6= A1 , A2 , C and B3 6= B1 , B2 , C.
Claim: The points P12 , P23 , and P31 are collinear if and only if pq = qp.
To do so, we must first compute expressions for the points Pij .
P12 = A1 B2 ∩ B1 A2 = hv 0 , u + w0 i ∩ hw0 , u + v 0 i = hu + v 0 + w0 i
P31 = A3 B1 ∩ B3 A1 = hu + pv 0 , w0 i ∩ hu + qw0 , w0 i = hu + pv 0 + qw0 i
17
P23 = A2 B3 ∩ B2 A3 = hu + v 0 , u + qw0 i ∩ hu + w0 , u + pv 0 i
= h(p + (p − 1)(q − 1)−1 )u + pv 0 + (p − 1)(q − 1)−1 qw0
This last one seems to come out of nowhere, so we will do a quick calculation
to show it is correct:
(p + (p − 1)(q − 1)−1 )u + pv 0 + (p − 1)(q − 1)−1 qw0
= p(u + v) + (p − 1)(q − 1)−1 (u + qw0 ) ∈ hu + v 0 , u + qw0 i

and
(p + (p − 1)(q − 1)−1 )u + pv 0 + (p − 1)(q − 1)−1 qw0
= (p + (p − 1)(q − 1)−1 + (p − 1)(q − 1)−1 q − (p − 1)(q − 1)−1 q)u
+pv 0 + (p − 1)(q − 1)−1 qw0
= (p + (p − 1)(q − 1)−1 (1 − q) + (p − 1)(q − 1)−1 q)u + pv 0 + (p − 1)(q − 1)−1 qw0
= (p − (p − 1) + (p − 1)(q − 1)−1 q)u + pv 0 + (p − 1)(q − 1)−1 qw0
= (1 + (p − 1)(q − 1)−1 q)u + pv 0 + (p − 1)(q − 1)−1 qw0
= (u + pv 0 ) + ((p − 1)(q − 1)−1 q)(u + w0 ) ∈ hu + w0 , u + pv 0 i
By axiom 1.2, this is the only point on both lines and therefore equal to P23 .
Now, we first show what collinearity of these three points would mean:
P23 ∈ P12 P31
⇔
(p + (p − 1)(q − 1)−1 )u + pv 0 + (p − 1)(q − 1)−1 qw0 ⊆ hu + v 0 + w0 , u + pv 0 + qw0 i
⇔
There exist x, y ∈ M such that
(p+(p−1)(q−1)−1 )u+pv 0 +(p−1)(q−1)−1 qw0 = xu+xv 0 +xw0 +yu+ypv 0 +yqw0
⇔
The following equations hold:
p + (p − 1)(q − 1)−1 = x + y
p = x + yp
(p − 1)(q − 1)−1 q = x + yq
Now, we must solve these equations. From the first one we get x = p + (p −
1)(q − 1)−1 − y, which allows the second one to give us
y = (p − 1)(q − 1)−1 (1 − p)−1
and thus
x = p + (p − 1)(q − 1)−1 (1 − (1 − p)−1 )
18
Using these results and the third equation, we can finally find our equivalence
after a long manipulation of formulae:
P23 ∈ P12 P31
⇔
(p − 1)(q − 1)−1 q =
= p + (p − 1)(q − 1)−1 (1 − (1 − p))−1 + (p − 1)(q − 1)−1 (1 − p)−1 q
⇔
−1
q = (q − 1)(p − 1) p + (1 − (1 − p)−1 ) + (1 − p)−1 q
⇔
(1 − p)q = (1 − p)(q − 1)(p − 1)−1 p + (1 − p) − 1 + q
⇔
(1 − p)q =
−1
= (1 − p)(q − 1)(p − 1) p + (1 − p)(q − 1)(p − 1)−1 − (1 − p)(q − 1)(p − 1)−1 − p + q
⇔
−1
(1 − p)q = (1 − p)(q − 1)(p − 1) (p − 1) + (1 − p)(q − 1)(p − 1)−1 − p + q
⇔
(1 − p)q = (1 − p)(q − 1) − p + q + (1 − p)(q − 1)(p − 1)−1
⇔
0 = (p − 1) − p + q + (1 − p)(q − 1)(p − 1)−1
⇔
1 − q = (1 − p)(q − 1)(p − 1)−1
⇔
(1 − q)(p − 1) = (1 − p)(q − 1) ⇔ p − 1 − qp + q = q − 1 − pq + p
⇔
qp = pq
Since p and q are arbitrary, this is true for all p, q ∈ M . Therefore, Pappus’s
theorem holds if and only if M is a field.
Now, it is important to point out that Pappus’s Theorem and Desargue’s
Theorem are not independent. In fact:
Theorem 2.9. If Pappus’s Theorem holds in a projective plane, then so does
Desargue’s Theorem.
19
As of right now, we do not have everything we need to prove this theorem.
As such, the proof will come in a future section. For now, based on what we
already know, we can actually say something else:
Theorem 2.10. If P is a finite projective plane and Desargue’s Theorem holds
in P, then so does Pappus’s Theorem.
Proof. From theorem 2.6, we know P = P(V ) for some module V and division
ring M . But then M is a finite division ring, and thus a field by theorem 2.1.
Finally, by theorem 2.8, Pappus’s Theorem holds.
Interestingly, this proof relies entirely on the algebra we’ve used earlier,
which may feel weird for a paper on geometry. The reason for this, however, is
that there is currently no geometrical proof known for theorem 2.10.
To conclude this section, let us take a look at the theorem dual to Pappus and
show their relation, in a similar manner to what we did for Desargue’s Theorem.
Dual to Pappus’s Theorem. Let P and Q be distinct points. Let l1 , l2 , and
l3 be distinct lines through P , while m1 , m2 , and m3 are distinct lines through
Q. Let these six lines also be different from P Q. Let n12 = (l1 ∩ m2 )(l2 ∩ m1 ),
n23 = (l2 ∩ m3 )(l3 ∩ m2 ), and n31 = (l3 ∩ m1 )(l1 ∩ m3 ). Then n12 , n23 , and n31
are concurrent.
Figure 6: Dual to Pappus’s Theorem

the Dual to Pappus’s Theorem.
20
Proof. Let A2 = l1 ∩ m2 , A3 = l1 ∩ m3 , B2 = l3 ∩ m1 , and B3 = l2 ∩ m1 ,
as in figure 6. In addition, let C = n12 ∩ n31 = A2 B3 ∩ A3 B2 . Now, we
apply Pappus’s Theorem to the lines l1 and m1 , with the points P , A2 , and
A3 on l1 and Q, B2 , and B3 on m1 . We find C2 = P B2 ∩ A2 Q = l3 ∩ m2 and
C3 = P B3 ∩ A3 Q = l2 ∩ m3 . By Pappus’s Theorem, C, C2 , and C3 are collinear.
But C2 C3 = n23 . Since this line is unique by axiom 1.1, C ∈ n23 . Since C was
defined as n12 ∩ n31 , we have that n12 , n31 , and n23 are concurrent.
Finally, we will present a different way to state the Dual to Pappus’s Theo-
rem:
Pappus’s Hexagon Theorem. Let ABCDEF be a hexagon. If the lines AB,
CF , and DE are concurrent, and the lines BC, AD, and EF are concurrent,
then the lines AF , BE, and CD are concurrent.
Figure 7: Pappus’s Hexagon Theorem
It is very easily shown that this theorem is equivalent with the dual to
Pappus’s Theorem. See also figure 7 for an aid in visualizing what it means.
2.4 Fano’s Axiom

Next, we will present another theorem, this time about quadrilaterals. While
we all have an intuition regarding what a quadrilateral is, it is important to
21
define it clearly:
Definition 2.7. A quadrilateral ABCD consists of four points A, B, C, and
D, no three of which are collinear, and the four lines AB, BC, CD, and DA.
It is important to note that the order of writing matters: the quadrilateral
ABCD is different from the quadrilateral BADC.
With that out of the way, we will need another definition:
Definition 2.8. The diagonal points of a quadrilateral ABCD are the three
points X1 = AB ∩ CD, X2 = AC ∩ BD, and X3 = AD ∩ BC.
Intuitively, the diagonal points as described here are simply the points of
intersection of the lines of the quadrilateral, excluding A, B, C, and D.
Now, with these few definitions written, we can move on to presenting Fano’s
Axiom:
Fano’s Weak Axiom. In a projective plane, there exists a quadrilateral ABCD
such that its diagonal points are non-collinear.
Fano’s Strong Axiom. For any quadrilateral ABCD in a projective plane,
its diagonal points are non-collinear.
Figure 8: Fano’s Axiom
We introduce the theorem in two steps, as that is how it was originally

presented. It should be clear that Fano’s Strong Axiom implies Fano’s Weak
Axiom, though we will return to the reverse implication later.
You may notice the discrepancy between the name of the theorem, which implies
it is an axiom, and its status as a theorem. The reason for this is that in certain
geometries, Fano’s Axiom is used as an axiom. In this paper, however, Fano’s
22
Axiom will be seen as a theorem, as it is not always true, instead requiring a
certain condition:
Theorem 2.12. Let V be a module over a division ring M . Then Fano’s Strong
Axiom holds in P(V ) if and only if 1 + 1 6= 0 in M .
Proof. Let V and M be as described. To start off, notice that the theorem is
equivalent to
Fano’s Axiom does not hold in P(V ) if and only if 1 + 1 = 0 in M .
This is easier to prove, and therefore will be what we go for instead.
First, let ABCD be a quadrilateral ABCD with diagonal points X1 , X2 , and
X3 . Let v1 , v2 , v3 , and v4 be vectors in V so that A = hv1 i, B = hv2 i, C = hv3 i,
and D = hv4 i. Then, through appropriate rescaling, we can get
X1 = hv1 + v2 i = hv3 + v4 i
X2 = hv1 − v3 i = hv4 − v2 i
X3 = hv1 − v4 i = hv3 − v2 i
Now that we have these coordinates, we will prove the theorem in both direc-
tions. Assume 1 + 1 = 0. Then 1 = −1 and
X1 = hv1 + v2 i
X2 = hv1 + v3 i
And
X2 − X1 = hv3 − v2 i = X3
Thus, the three diagonal points are collinear.
Next, assume the three points are on a line. Then, for some a1 , a2 ∈ M
X3 = a1 X1 + a2 X2
v3 − v2 = a1 (v1 + v2 ) + a2 (v1 − v3 )
(a1 + a2 )v1 + (a1 + 1)v2 + (−a2 − 1)v3 = 0
Since A, B, and C are, by definition of a quadrilateral, non-collinear, we have
v1 , v2 , and v3 linearly independent. Thus, by definition
a1 + 1 = 0 ⇒ a1 = −1
−a2 − 1 = 0 ⇒ a2 = −1
a1 + a2 = 0 ⇒ −1 − 1 = 0 ⇒ 1 + 1 = 0
23
Note that the first part of this proof implies that if 1 + 1 = 0, then for any
quadrilateral, its diagonal points must be on a line. However, the second part
of this proof implies that if there exists a quadrilateral with the diagonal points
on a line, then 1 + 1 = 0. From that, we can conclude
Theorem 2.13. Let V be a module over a division ring M . Then, in P(V ),
Fano’s Weak Axiom implies Fano’s Strong Axiom. In other words, if there is a
quadrilateral ABCD with non-collinear diagonal points, all quadrilaterals have
non-collinear diagonal points.
The proof for this requires no more work than we have already done.
Now, we have a nice equivalence relation between Fano’s Weak and Strong
Axioms in projective planes over modules, but we cannot say anything about
other projective planes. In fact, it is currently still an open question whether or
not Fano’s Weak Axiom implies Fano’s Strong Axiom in all projective planes.
Before moving on, let us, just as with the theorems of Desargue and Pappus,
take a look at the dual theorem to Fano’s Strong Axiom:
Dual to Fano’s Strong Axiom. Let ABCD be a quadrilateral in a projective
plane. Let us rename the lines of the quadrilateral as l1 = AB, l2 = BC,
l3 = CD, and l4 = AD. Now, we define the diagonal lines as x1 = (l1 ∩ l2 )(l3 ∩
l4 ) = BD, x2 = (l1 ∩ l3 )(l2 ∩ l4 ), and x3 = (l1 ∩ l4 )(l2 ∩ l3 ) = AC. Then, x1 ,
x2 , and x3 are non-concurrent.
Figure 9: Dual to Fano’s Strong Axiom
24
Theorem 2.14. If Fano’s Strong Axiom holds in a projective plane, then so
does the dual theorem to Fano’s Strong Axiom.
Proof. Let ABCD and its diagonal lines x1 , x2 , and x3 be as defined in the
dual theorem to Fano’s Strong Axiom. Also, let X1 , X2 , and X3 be the diagonal
points of ABCD. By definition, X2 = AC ∩BD = x1 ∩x3 . Thus, if the diagonal
lines are concurrent, then x2 must go through X2 . On the other hand, if x2
does not go through X2 , then the diagonal lines are non-concurrent.
Now, take note that X1 = AB ∩ CD = l1 ∩ l3 and X3 = AD ∩ BC = l2 ∩ l4 .
Thus, by definition, x2 = X1 X3 . By Fano’s Strong Axiom, the diagonal points
of ABCD are non-collinear. Thus, X2 ∈ / x2 and the diagonal lines are non-
concurrent.
25
2.5 Theorems of Menelaus and Ceva
In this section, we will showcase two more theorems, as well as proofs for both.
Throughout this section, V is a module of dimension 3 over a division ring M ,
and we are discussing the theorems within the context of P(V ).
Theorem 2.15 (Theorem of Menelaus). Let P = hui, Q = hvi, and R = hwi be

three non-collinear points. Let P 0 = hv + awi be a point on QR, Q0 = hw + bui
be a point on RP , and R0 = hu + cvi be a point on P Q. Then, P 0 , Q0 , and R0
are collinear if and only if abc = −1.
Figure 10: Theorem of Menelaus
Proof. Note that, if P 0 , Q0 , and R0 are collinear, we have v + aw = p(w +

bu) + q(u + cv). Because P , Q, and R are non-collinear, we have u, v, and w
independent. Thus, we find
P 0 , Q0 , and R0 are collinear
⇔ v = qcv, aw = pw, 0 = (pb + q)u

⇔ q = c−1 , p = a, pb + q = 0
⇔ ab + c−1 = 0
⇔ ab = −c−1
⇔ abc = −c−1 c = −1
26
Figure 11: Theorem of Ceva
Theorem 2.16 (Theorem of Ceva). Let P = hui, Q = hvi, and R = hwi be

three non-collinear points. Let P 0 = hv + awi be a point on QR, Q0 = hw + bui
be a point on RP , and R0 = hu + cvi be a point on P Q. Then, the lines P P 0 ,
QQ0 , and RR0 are concurrent if and only if abc = 1.
Proof. Let X = hxi be the point of intersection between P P 0 and QQ0 , which
exists and is unique by axiom 1.2. Then, for some scalars p1 , p2 , q1 , q2 , we have
p1 u + p2 (v + aw) = x = q1 v + q2 (w + bu). Now, note that RR0 goes through X
if and only if, for some scalars r1 , r2 , x = r1 w + r2 (u + cv). Thus, we get
P P 0 , QQ0 , and RR0 are concurrent
⇔ p1 u = q2 bu = r2 u, p2 v = q1 v = r2 cv, p2 aw = q2 = r1
⇔ p2 = p1 c, q2 = p2 a, p1 = q2 b
⇔ p1 = p1 cab
⇔ cab = 1
⇔ ab = c−1
⇔ abc = c−1 c = 1
What makes both of these theorems interesting is that they have a way of
being described in Euclidean Geometry as well; this is not true for the theorems
of Pappus and Desargue, as they rely on axiom 1.2, but Ceva and Menelaus do
not, at least no heavily, meaning we get:
27
Remark 2.1 (Theorem of Menelaus in Euclidean Geometry). Let ∆ABC be a
triangle and let D ∈ BC, E ∈ AC, and F ∈ AB, with all three distinct from A,
B, and C. Then, D, E, and F are collinear if and only if
AF BD CE
∗ ∗ = −1
F B DC EA
Here, we use signed lengths of segments.
Remark 2.2. Let ∆ABC be a triangle and let D ∈ BC, E ∈ AC, and F ∈
AB, with all three distinct from A, B, and C. Then, AD, BE, and CF are
concurrent if and only if
AF BD CE
∗ ∗ =1
F B DC EA
Here, we use signed lengths of segments.
To conclude, it should be noted that the theorems of Menelaus and Ceva are
very close to being each other’s dual theorems, with only the difference between
1 and -1 being in the way, as well as some of the wording. This wording is
much more convenient, however, as writing the dual of the theorem of Menelaus
results in unintuitive use of spans to indicate lines, while being equivalent with
the Theorem of Ceva.
28
3 Perspectives and projective maps
This chapter will consider the notions of perspectives and projective maps, which
will be defined in the first section. Afterwards, the next two sections will focus
on stating and proving several incredibly important theorems.
3.1 Definitions
As stated, this section will define perspectives and projective maps. However,
each of these take into account one very important fact that we will restate here
again before moving on: lines can be seen as sets of points. This is crucial,
as you will see very quickly. If this weren’t the case, our definitions would not
make sense.
Without further ado, let us introduce the two most important terms for this
chapter:
Definition 3.1. Let l and m be lines and let A be a point that not incident
with l or m. The perspective from l to m through A is the function σA : l −→ m
such that for every B ∈ l, σA (B) = AB ∩ m.
In figure 12, an example of a perspective is given. Clearly, σA (B1 ) = C1

and σA (B3 ) = C3 . However, note what is happening at B2 = l ∩ m. Clearly,
σA (B2 ) = AB2 ∩ m = B2 , meaning it is a fixed point. This is not a coincidence:
Theorem 3.1. For any perspective σA : l −→ m, l ∩ m is a fixed point.
The proof of this is trivial.

This theorem will prove important later on, but for now, let us simply move on
to the next important definition for the time being:
Definition 3.2. A projective map of order n is a composition of n perspectives.
This seems simply and not particularly useful, but as we will see, it is quite
nice to have a name for this.
29
Figure 12: The perspective σA : l −→ m
3.2 Double Perspective Theorem

In this section, we will introduce a certain theorem that will immediately show
why we defined projective maps, but more importantly, we will show its relation
to the theorems of Desargue and Pappus and, finally, prove theorem 2.9, which
states that Pappus’s Theorem implies Desargue’s Theorem.
Of course, we will need several steps to get there, so first, let’s look at the
Double Perspective Theorem:
Double Perspective Theorem. We will split this theorem into two cases:
1. Let l, m, and n be three non-concurrent lines and let σ : l −→ n be a

projective map of order 2 with σ = σA ◦ σB for some perspectives σA :
l −→ m and σB : m −→ n, such that l ∩ n is a fixed point for σ. Then
there is a point C such that σ = σC , where σC : l −→ n is the perspective
from l to n through C.
2. Let l, m, and n be three concurrent lines and let σ : l −→ n be a projective
30
map of order 2 with σ = σA ◦ σB for some perspectives σA : l −→ m and
σB : m −→ n, such that l ∩ n is a fixed point for σ. Then there is a point
C such that σ = σC , where σC : l −→ n is the perspective from l to n
through C.
In other words, any projective map σ : l −→ n of order 2 with l ∩ n a fixed point
is a perspective.
Do not let the separate cases confuse you: in the end, this theorem says little
more than the last line does. The reason for splitting it up is for use in the proof
that Pappus’s Theorem implies Desargue’s Theorem. For notational purposes,
from here in out, we will refer to the Double Perspective Theorem as ’DPT’.
In addition, if we require one of the two cases, we will write it as ’DPT.1’ or
’DPT.2’ respectively.
So, why is this important at all? At first glance, this theorem seems completely
separate from the ones we’ve shown before, focusing entirely on projective maps
and perspectives instead of lines, points, and collinearity. However, as we will
see, this is not the case at all:
Theorem 3.2. In a projective plane, DPT.1 holds if and only if Pappus’s The-
orem does.
Proof. As this is an ’if and only if’-statement, we will need to prove it both
ways. As such, we’ll split it in two:
1. First, assume DPT.1 holds. We will prove Pappus’s Theorem does as well.
To start off, choose l, m, A1 , A2 , A3 , B1 , B2 , B3 , P12 , P23 , and P31 such
that the hypothesis of Pappus’s Theorem is satisfied. We will consider the
perspectives σA1 : B1 A2 −→ m and σA3 : m −→ A2 B3 , as well as the
projective map σ = σA1 ◦ σA3 of order 2. It should be clear that the 3
lines B1 A2 , m, and A2 B3 are non-concurrent, since A2 ∈ / m. In addition,
σA1 (A2 ) = l ∩ m and σA3 (l ∩ m) = A2 , meaning A2 = B1 A2 ∩ A2 B3
is a fixed point for σ. Thus, by DPT.1, there is a point S such that
σ = σS : B1 A2 −→ A2 B3 .
Next, our goal will be to find where S is. To accomplish that, let’s take a
look at what happens to P12 when we apply σ to it. Clearly, σA1 (P12 ) = B2
and σA3 (B2 ) = P23 . Thus, by definition of a perspective, S ∈ P12 P23 .
Now, to continue on, we will define two more points: Q1 = A1 B3 ∩ B1 A2
and Q2 = B1 A3 ∩A2 B3 . It is easily checked that we now have σS (Q1 ) = B3
and σS (B1 ) = Q2 . But then S is on Q1 B3 = A1 B3 and on B1 Q2 = B1 A3 .
By axiom 1.2 and the definition of P31 , the only point that satisfies both
of these requirements is P31 . And now, as we have found earlier, P31 =
S ∈ P12 P23 . Thus, P12 , P23 , and P31 are collinear.
2. Now, we assume Pappus’s Theorem holds and will use this to prove DPT.1.
Let there be two perspectives σA1 : l1 −→ l2 and σA3 : l2 −→ l3 such that
for A2 = l1 ∩ l3 and σ = σA3 ◦ σA1 , we have σ(A2 ) = A2 . In addition, let
l1 , l2 , and l3 be non-concurrent. Then, clearly, both A1 and A3 are on the
31
Figure 13: DPT.1 implies Pappus’s Theorem
line between A2 and σA (A2 ), so A1 , A2 and A3 are collinear.

Next, let us define some points: B1 = l1 ∩l2 , B3 = l2 ∩l3 , and P31 = A1 B3 ∩
A3 B1 . Then choose an arbitrary point P12 on l1 and let B2 = A1 P12 ∩ l2 ,
such that σA1 (P12 ) = B2 , and let P23 = A2 B3 ∩ A3 B2 = l3 ∩ A3 B2 , such
that σA3 (B2 ) = P23 . Now we have σ(P12 ) = P23 .
Now, we can use Pappus’s Theorem to say P12 , P23 , and P31 are on one
line, so if we define σP23 : l1 −→ l3 , we get σP31 (P12 ) = P23 . But since
P12 was chosen arbitrarily, we can conclude that for every point C ∈ l1 ,
we get σP31 (C) = σ(C). Thus, σ = σP31 .
Figure 14: Pappus’s Theorem implies DPT.1
32
This alone is already special, but of course, we aren’t done. Because in order
to work with Pappus’s and Desargue’s Theorem, we need to have a link to the
latter:
Theorem 3.3. In a projective plane, DPT.2 holds if and only if Desargue’s
Theorem does.
Proof. As before, we have an ’if and only if’-statement, so we will split this in
two.
1. First, assume DPT.2 holds. We will show Desargue’s Theorem. Let l1 ,
l2 , l3 , A1 , A2 , A3 , B1 , B2 , B3 , C, P12 , P23 , and P31 be such that the
conditions for Desargue’s Theorem are fulfilled. Consider σP12 : l1 −→ l2
and σP23 : l2 −→ l3 . By DPT.2, there is a point S such that σS =
σP23 ◦σP12 , since clearly σP23 ◦σP12 (C) = C and l1 , l2 , and l3 are concurrent.
Note that σS (A1 ) = A3 and σS (B1 ) = B3 . Thus S = A1 A3 ∩ B1 B3 = P31 .
Now, let Q1 = P12 P23 ∩ l1 and Q3 = P12 P23 ∩ l3 . Then, clearly, σS (Q1 ) =
Q3 , so P31 = S ∈ Q1 Q3 = P12 P23 . Thus, P12 , P23 , and P31 are collinear.
Figure 15: DPT.2 implies Desargue’s Theorem
2. Next, assume Desargue’s Theorem holds. We will show DPT.2. To start,

we look at two perspectives, σF : l1 −→ l2 and σE : l2 −→ l3 , where l1 ,
l2 , and l3 are concurrent. If DPT.2 holds, then there is some S such that
σS = σE ◦ σF .
To show this, we fix A1 ∈ l1 and call A2 = σF (A1 ) and A3 = σE (A2 ).
We also set S = EF ∩ A1 A3 . Now, we pick an arbitrary B1 ∈ l1 , with
33
B2 = σF (B1 ) and B3 = σE (B2 ). By applying Desargue’s Theorem to the
triangles ∆A1 A2 A3 and ∆B1 B2 B3 , we see that regardless of our choice
of B1 , EF goes through A1 A3 ∩ B1 B3 . This also means that B1 B3 goes
through EF ∩ A1 A3 = S. Thus, σS (B1 ) = B3 = (σE ◦ σF )(B1 ) for any
B1 ∈ l1 . Thus, σS = σE ◦ σF .
Figure 16: Desargue’s Theorem implies DPT.2
Now, we finally have links to Pappus’s Theorem and to Desargue’s Theorem.

With that information, we can finally complete the proof we’ve been working
towards:
Desargue’s Theorem.
Proof. Let A1 , A2 , A3 , B1 , B2 , B3 , C, P12 , P23 , and P31 be so that the con-
ditions for Desargue’s Theorem are met. Name l1 = A1 B1 , l2 = A2 B2 , and
l3 = A3 B3 . We will split this up into a few cases:
1. Assume C ∈ / P12 P23 and P23 ∈
/ l1 . We will look at P12 P23 and G =
l1 ∩ P12 P23 . Consider σP12 : l1 −→ l2 and σP23 : l2 −→ l3 . Now, by
theorem 3.2, we know DPT.1 holds. Unfortunately, l1 , l2 , and l3 are
concurrent, so we cannot apply it to σP23 ◦ σP12 yet.
34
Now, let m be an arbitrary line through G with m 6= l1 and m 6= P12 P23 .
Consider the perspectives µP23 : l2 −→ m and νP23 : m −→ l3 . Note that
here, we use that P23 ∈ / l1 . Clearly, σP23 = νP23 ◦ µP23 , so σP23 ◦ σP12 =
νP23 ◦ µP23 ◦ σP12 .
Let’s first take a closer look at µP23 ◦ σP12 . The three lines that are being
considered here are l1 , l2 , and m. It is easily checked that these are non-
concurrent, and in addition, G = l1 ∩ m is a fixed point. This means that
by DPT.1, there is a point S1 such that σS1 = µP23 ◦ σP12 .
So now we have σP23 ◦ σP12 = νP23 ◦ σS1 . Considering the right portion,
we are looking at the lines l1 , m, and l3 , which are clearly non-concurrent.
In addition, we know (νP23 ◦ σS1 )(C) = (σP23 ◦ σP12 )(C) = C. Since
C = l1 ∩ l3 , we can apply DPT.1 to find that there is a point S2 such that
σS2 = σP23 ◦ σP12 .
So all we now have to do is find where S2 is. First off, it’s very easily seen
that σS2 (A1 ) = (σP23 ◦ σP12 )(A1 ) = A3 and σS2 (B1 ) = (σP23 ◦ σP12 )(B1 ) =
B3 . Thus, S2 = A1 A3 ∩ B1 B3 = P31 by axiom 1.2 and the definition of
P31 .
Let G1 = P12 P23 ∩ l2 and G2 = P12 P23 ∩ l3 . Then σP23 (G) = G1 and
σP23 (G1 ) = G2 . Thus, σS2 (G) = G2 , meaning P31 = S2 ∈ GG2 = P12 P23 .
Now we see P12 , P23 , and P31 are collinear, which means we have shown
Desargue’s Theorem!
Figure 17: The first case
35
2. Next, we assume C ∈ / P12 P23 and P23 ∈ l1 . Now, we define G = P12 P23 ∩l3 .
Consider σP12 : l1 −→ l2 and σP23 : l2 −→ l3 . As before, we cannot yet
apply DPT.2, since l1 , l2 , and l3 all go through C. Thus, instead, we
consider a line m through G that’s not equal to l1 or P12 P23 .
Similar to before, we take a look at perspectives µP12 : l1 −→ m and
νP12 : m −→ l2 . To be allowed to define this, we need to argue P12 is on
none of these lines. Now, it should be clear that P12 ∈ / m. Next, note that
since C ∈ / P12 P23 , l1 6= P12 P23 . With P23 ∈ l1 , we have P12 ∈/ l1 . Last,
A1 ∈ / l2 and A1 ∈ P12 A2 , so P12 ∈ / l2 .
Finally, we can say σP12 = νP12 ◦ µP12 . Define σ = σP23 ◦ σP12 . We now
also have σ = σP23 ◦ νP12 ◦ µP12 .
Now, since m, l2 , and l3 aren’t concurrent and since G is clearly a fixed
point, we can apply DPT.1 to find that there is a point S1 such that
σS1 = σP23 ◦ νP12 . Similarly, we can also see that there is a point S2 such
that σS2 = σS1 ◦ µP12 = σP23 ◦ νP12 ◦ µP12 = σ.
Our next goal is to find out where S2 is. It’s very easy to check that
σS2 (A1 ) = A3 and σS2 (B1 ) = B3 . Thus, we can immediately see S2 =
A1 A3 ∩ B1 B3 = P31 . Finally, we set G1 = P12 P23 ∩ l2 . Once again,
it’s easy to check σS2 (P23 ) = (σP23 ◦ σP12 )(P23 ) = σP23 (G1 ) = G. Thus,
P31 = S2 ∈ GP23 = P12 P23 , meaning P12 , P23 , and P31 are collinear.
Figure 18: The second case
3. We have checked two cases for C ∈ / P12 P23 and have proven both. Using
similar methods, we can prove Desargue’s Theorem for C ∈ / P23 P31 or C ∈
/
P12 P31 . Thus, the last case we must check is C ∈ P12 23 and C ∈ P23 P31
and C ∈ P12 P31 . But then P23 ∈ P12 C and P31 ∈ P12 C, meaning they
are still collinear.
36
And with that, we have shown a major result in projective geometry. Pap-
pus’s Theorem implying Desargue’s Theorem is not only a very interesting re-
sult, it is also majorly useful. Interestingly, however, it is not the only thing
Pappus’s Theorem is incredibly useful for, as we will see in the next section.
37
3.3 Fundamental Property
In the previous two sections, we have introduced and then used the concepts
of perspectives and projective maps, but of course, we are not quite done with
them yet. In this section, we will introduce a theorem known as the Fundamental
Property of projective geometry. As may be expected, there is not just a single
formulation for this theorem. In this section, we will showcase three (equivalent)
ways of formulating it and then show they are all equivalent:
Fundamental Property. The following are three equivalent formulations of
this theorem:
1. If p : l1 −→ l2 is any projective map with l1 6= l2 and p(l1 ∩ l2 ) = l1 ∩ l2 ,
then p is a perspective.
2. If A1 , A2 , and A3 are distinct points on l and for some projective p : l −→ l
we have p(Ai ) = Ai for i ∈ {1, 2, 3}, then p = idl .
3. If A1 , A2 , and A3 are distinct points on l1 and B1 , B2 , and B3 are
distinct points on l2 (with l1 6= l2 ), then there exists a unique projective
map p : l1 −→ l2 such that p(Ai ) = Bi for i ∈ {1, 2, 3}.
4. If A1 , A2 , A3 , B1 , B2 , and B3 are distinct points on a line l, then there is
a unique projective map p : l −→ l such that p(Ai ) = Bi for i ∈ {1, 2, 3}.
First, we will proof that these 4 statements are equivalent:
Proof. This proof will be presented in separate steps, each only looking at a
single implication. The order may seem strange at first, but this is because we
wish to begin with the simple proofs.
• 2. ⇒ 3. Let l1 and l2 be lines (l1 6= l2 ) with A1 , A2 , A3 , B1 , B2 , and
B3 as described in 3. Let p1 and p2 be two projective maps such that
p1 (Ai ) = p2 (Ai ) = Bi for i ∈ {1, 2, 3}. We will show p1 = p2 .
Consider p : l1 −→ l1 with p = p−1 2 ◦ p1 . Then we have p(Ai ) =
p−1
2 (p 1 (A i )) = p −1
2 (B i ) = A i for i ∈ {1, 2, 3}. By 2., we have p = idl1 .
Thus, p−1 1 = p −1
2 and p 1 = p2 .
• 3. ⇒ 2. Let l1 be any line with points A1 , A2 , and A3 on l1 . Define
any line l2 and points B1 , B2 , and B3 and let p : l1 −→ l2 be such that
p(Ai ) = Bi for i ∈ {1, 2, 3}. By 3., p is unique. Now, let q : l1 −→ l1 be
such that q(Ai ) = Ai for i ∈ {1, 2, 3}. We will show q = idl1 .
Define r = p ◦ q. Clearly, r(Ai ) = Bi for i ∈ {1, 2, 3}. Thus, by 3., r = p.
But then p ◦ q = p ⇒ p−1 ◦ p ◦ q = p−1 ◦ p ⇒ q = idl1 .
• 3. ⇒ 4. Let l1 be a line with 6 distinct points Ai and Bi for i ∈ {1, 2, 3}.
Also let p1 : l −→ l and p2 : l −→ l such that p1 (Ai ) = p2 (Ai ) = Bi for
i ∈ {1, 2, 3}. We will show p1 = p2 .
Let l2 be another line with three distinct points C1 , C2 , and C3 . By 3.,
there is a unique projective map q : l1 −→ l2 such that q(Ai ) = Ci for
38
i ∈ {1, 2, 3}, as well as a unique projective map r : l1 −→ l2 such that
r(Bi ) = Ci . Now, if we consider r ◦ p1 and r ◦ p2 , we notice that for both
of them, we have that they map Ai to Bi for i ∈ {1, 2, 3}. Thus, they
must both equal q. But then r ◦ p1 = r ◦ p2 ⇒ p1 = p2 .
• 4. ⇒ 2. Let l be a line with 6 distinct points Ai and Bi for i ∈ {1, 2, 3}.
In addition, let p : l −→ l be a projective map such that p(Ai ) = Ai for
i ∈ {1, 2, 3}. We will show p = idl .
By 4., there is a unique projective map q : l −→ l such that q(Ai ) = Bi for
i ∈ {1, 2, 3}. Now consider r = q ◦ p. Clearly, r(Ai ) = Bi for i ∈ {1, 2, 3}.
But then, since q is unique, r = q. From this, we clearly see p ◦ q = q ⇒
p = idl .
• 3. ⇒ 1. Let l1 and l2 be lines and let p : l1 −→ l2 be a projective map

with p(l1 ∩ l2 ) = l1 ∩ l2 . Now, let A1 and A2 be distinct points on l1 and
define B1 = p(A1 ) and B2 = p(A2 ). Finally, let A3 = B3 = l1 ∩ l2 .
We consider the perspective σS : l1 −→ l2 with S = A1 B1 ∩A2 B2 . Clearly,
σS is a projective map (of order 1) such that σS (Ai ) = Bi for i ∈ {1, 2, 3}.
By 3., it is unique. However, note that p is also such a projective map.
Thus, we can conclude that p = σS . Since p was chosen arbitrarily, we
can conclude that we can find a point S for any projective map p.
• 1. ⇒ 3. As this proof is more complicated, we will split it up into 4 cases.
However, regardless of that, we will always consider two lines l1 and l2 ,
with points A1 , A2 , and A3 on l1 and B1 , B2 , and B3 on l2 .
39
– Case 1: Ai 6= l1 ∩ l2 and Bi 6= l1 ∩ l2 for i ∈ {1, 2, 3}.
Let m = A3 B1 and let R be any point such that R ∈ A1 B1 . Consider
σR : l1 −→ m. Define P = σR (A2 ) and S = A3 B3 ∩ P B2 . We also
consider σS : m −→ l2 and σ = σS ◦ σR . It is easily checked that
σ(Ai ) = Bi for i ∈ {1, 2, 3}.
Now, let p : l1 −→ l2 be a projective map such that p(Ai ) = Bi for
i ∈ {1, 2, 3}. Let µ = σS−1 ◦ p : l1 −→ m. Then we have µ(A1 ) = B1 ,
µ(A2 ) = P , and µ(A3 ) = A3 = l1 ∩m. Then, by 1., µ is a perspective,
and it’s easily checked that this must be σR . Thus σS−1 ◦ p = µ =
σR ⇒ p = σS ◦ σR = σ. Since p was chosen arbitrarily, we must have
that σ is the only projective map with σ(Ai ) = Bi for i ∈ {1, 2, 3}.
– Case 2: A3 = l1 ∩ l2 and Bi 6= l1 ∩ l2 for i ∈ {1, 2, 3}
Define m = A1 B3 and R ∈ l2 with R 6= A3 , B3 . Consider σR : l1 −→
m. Clearly, we have σR (A1 ) = A1 and σR (A3 ) = B3 . We also define
P = σR (A2 ) and S = A1 B1 ∩ P B2 . Now, consider σS : m −→ l2 .
Let σ = σS ◦ σR : l1 −→ l2 and see that σ(Ai ) = Bi for i ∈ {1, 2, 3}.
As in the previous case, we let p : l1 −→ l2 be a projective map such
that p(Ai ) = Bi for i ∈ {1, 2, 3} and we look at µ = σS−1 ◦p : l1 −→ m.
We find µ(A1 ) = A1 = l1 ∩ m, µ(A2 ) = P , and µ(A3 ) = B3 . By 1.,
we know µ is a perspective, and so we quickly find µ = σR . Using
the same reasoning as before, we see σS−1 ◦ p = σR ⇒ p = σ, meaning
σ is unique.
– Case 3: Ai 6= l1 ∩ l2 for i ∈ {1, 2, 3} and B3 = l1 ∩ l2

This case is simple and analogous to the previous one; simply replace
p with p−1 . Once you prove the inverse is unique, p itself must be
unique too.
40
– Case 4: A3 = B1 = l1 ∩ l2 .
Define m = A1 B3 and pick R ∈ l2 such that R 6= A3 , B3 . Once again,
we look at σR : l1 −→ m and P = σR (A2 ). Now let S = l1 ∩ P B2
and consider σS : m → l2 and σ = σS ◦ σR . It is easily checked that
σ(Ai ) = Bi for i ∈ {1, 2, 3}.
Now let p : l1 −→ l2 be a projective map with p(Ai ) = Bi for
i ∈ {1, 2, 3}. Just as before, we look at µ = σS−1 ◦ p and conclude, by
1., that µ = σR and thus p = σ, meaning σ is unique. Also see figure
on the next page.
Figure 21: The fourth case
– Case 5: A3 = B3 = l1 ∩ l2
This case is simple. Let p : l1 −→ l2 be such that p(Ai ) = Bi for
i ∈ {1, 2, 3}. Then, clearly p(l1 ∩ l2 ) = p(A3 ) = B3 = l1 ∩ l2 . Thus, p
is a perspective σS : l1 −→ l2 for some S. We immediately see that
there is only one S for which this holds: S = A1 B1 ∩ A2 B2 . Thus,
p = σS is unique.
By simply renumbering points, every other case can be proven now as
well. Thus, 1. ⇒ 3..
Clearly, the four statements are now equivalent.
From here on out, we will refer to the four statements as FP.1, FP.2, FP.3,
and FP.4 respectively.
At the end of the previous section, it was stated that Pappus’s Theorem would
41
prove important once more in this section. So far, however, we have seen nothing
of the sort. As you may have guessed, we will know be moving on to a perhaps
stunning result:
Theorem 3.5. In a projective plane, the Fundamental Property holds if and
only if Pappus’s Theorem holds.
Now, before we prove this, we will first prove two lemmas and a theorem
that will greatly help us in our proof. Before we do so, however, we will be
O P
introducing a new notation for projective maps: p = l − →m− → n means that
for the lines l, m, and n and points O and P , we have σO : l −→ m and
σP : m −→ n, and p = σP ◦ σO . Now, with that in mind, we introduce our
lemmas:
Lemma 3.1. Assume Pappus’s Theorem holds. Let l, m, and n be non-
concurrent lines and let O and P be points, with O ∈/ l, m and P ∈/ m, n. Con-
O P
sider p : l −
→m− → n. Then there exist lines m0 and m00 with l ∩ m = l ∩ m0 and
O0 P
m00 ∩ n = m ∩ n, as well as points O and P such that we have p = l −→ m0 −
→n
0
O 00 P
and p = l − → m −→ n. In other words, we can replace m as well as either of
the points.
O
Proof. Let m0 be a line through l ∩ m not equal to l or m. Clearly, p = l −
→
P O P 0 P
m −→ n = l − → m − → m − → n. Since Pappus’s Theorem holds, so too does
O O P
DPT.2, so we see that there is an O0 such that l −
→ m0 = l −
→m− → m0 . Thus,
O0 P
we get p = l −→ m0 − → n.
Finding P 0 is done in similar fashion.
Lemma 3.2. Assume Pappus’s Theorem holds. Let l, m, and n be non-
O P
concurrent lines and let p = l −
→m−
→ n be a projective map with p(l∩n) 6= l∩n.
O0 P0
Then there exist points O0 ∈ n and P 0 ∈ l such that p = l −→ m0 −→ n for
some line m0 . We will see O0 = OP ∩ n and P 0 = OP ∩ l.
O P
Proof. See figure 22 on the next page. Fix A and A0 on l. Let A − →B− → C and
0 O 0 P 0 0
A − →B − → C . Additionally, set O = OP ∩ n. Note that since p(l ∩ n) 6= l ∩ n,
l∩n∈ / OP . Thus, O0 6= l ∩ n ⇒ O0 ∈ / l. Let D = O0 A ∩ P C, D0 = O0 A ∩ P C 0 ,
and m = DD . Consider the triangles ∆ABD and ∆A0 B 0 D0 and realize that
0
since Pappus’s Theorem holds, so too does Desargue’s Theorem and its dual
theorem. Now, since O = AB ∩ A0 B 0 , O0 = AD ∩ A0 D0 , and P = BD ∩ B 0 D0
are collinear, we have that l = AA0 , m = BB 0 , and m0 = DD0 are concurrent.
So now, we have that m0 goes through l ∩ m. However, this means that if we
find D via A, we can then see m0 = D(l ∩ m). Thus, m0 is only dependent on A.
So, let us choose A00 on l and choose B 00 and C 00 in the same manner as before.
Then if we choose D00 = O0 A00 ∩ m0 , we will find D00 P ∩ n = C 00 , as was also the
O0 P
case for A0 . Thus, we get l −→ m0 −
→ n = p.
We can use a similar argument to find P 0 ∈ n.
42
Figure 22: Lemma 3.2
Theorem 3.6. If Pappus’s Theorem holds, then any projective map can be
written as a projective map of order 2. In other words, for any projective map
O P
p : l1 −→ l2 with l1 6= l2 , there exist points O and P such that p = l1 −
→m−
→ l2
for some line m.
Proof. We state that we only need to proof this for projective maps of order 3, as
the rest follows from induction. In addition, we will show that every projective
map of order 3 can be reduced to the general case p : l −→ m −→ n −→ o, with
l, m, n, and o all distinct lines:
• Case 1: p : l −→ m −→ n −→ m with l, m, and n distinct
By lemma 3.1, we can find an m0 distinct from l, m, and n such that
p = p0 : l −→ m0 −→ n −→ m, which is the general case.
• Case 2: p : l −→ m −→ l −→ o with l, m, and o distinct
We use lemma 3.1 again. We find l0 such that p = p0 : l −→ m −→ l0 −→ o,
which is the general case.
• Case 3: p : l −→ m −→ l −→ m with l and m distinct.
O P Q
Let O, P , and Q be the points such that p = l − → m −→ l −→ m.
Let n be any line not equal to l or m such that l ∩ m ∈ n. Clearly
O P P Q
p=l− →m− →n− →l− → n.
Now, since Pappus’s Theorem holds, so does DPT.2. Thus we can con-
O0 O P
clude that there is a point O0 such that l −→ n = l −
→ m −
→ n and a
0
P P Q O P P
point P 0 such that n −→ m = n −
→l−
→ n. Then p = l −
→m−
→n−
→
0 0
Q O P
l−
→ n = l −→ n −→ m. This is a projective map of order 2.
43
Now we can move on to the general case.
Let p : l −→ m −→ n −→ o with l, m, n, and o distinct lines. If l ∩ m = n ∩ o,
the four lines are concurrent and we can use DPT.2 to quickly find that this is
a perspective. Thus, assume l ∩ m 6= n ∩ o. Now, if l ∩ m ∈ o, we can use lemma
3.1 to replace m with m0 . In this way, we can now also assume l ∩ m ∈ / o.
P Q R
Finally, we use lemma 2 to get the following situation: p = l − →m− →n− →o
with Q ∈ O, R ∈ m, and l ∩ m ∈ / o.
Let h = (m ∩ l)(n ∩ o). Note that since Q ∈ / n, Q 6= n ∩ o and thus Q ∈/ h. Let
P Q R
A, A0 ∈ l with A, A0 −→ B, B 0 − → C, C 0 −
→ D, D0 . In addition, we will consider
Q Q
q=m− → h with points H, H ∈ h such that B, B 0 −
0
→ H, H 0 .
Since Pappus’s Theorem holds, so too does Desargue’s Theorem, and we will
first apply it to the triangles ∆ABH and ∆A0 B 0 H 0 . Clearly, AA0 , BB 0 , and
HH 0 are all collinear, so we find that AB ∩ A0 B 0 = P , BH ∩ B 0 H 0 = Q, and
AH ∩ A0 H 0 are all collinear. If we call this last point M , we notice that M is
equal to P H ∩ AH. In other words, it only depends on our choice of A.
Next, we apply Desargue’s Theorem to ∆CDH and ∆C 0 D0 H 0 . Again, we
easily find the concurrent lines we need, so we get that CD ∩ C 0 D0 = R,
CH ∩ C 0 H 0 = Q, and DH ∩ D0 H 0 are collinear. Just like before, we call this
last point N and note that N only depends on our choice of A, as it is equal to
DH ∩ QR.
Figure 23: Theorem 3.6
44
Now that we have found that M and N only depend on A, we consider the
M N M N
projective map p0 = l −→ h −→ o. Now we see p0 (A) = A −→ H −→ D and,
M N
for any choice of A0 , p0 (A0 ) = A0 −→ H 0 −→ D0 (where D0 = p(A0 )). In other
words, p = p0 . But clearly, p0 is of order 2. Thus, if Pappus’s Theorem holds,
any projective map of order 3 is equal to some projective map of order 2.
A simple induction argument can be used to prove it for projective maps of

greater orders:
Assume the statement holds for projective maps of order n. Let p = l1 −→
l2 −→ · −→ ln −→ ln+1 −→ ln+2 be a projective map of order n + 1. First
assume ln+1 6= l1 . Then we can reduce p to p = l1 −→ m −→ ln+1 −→ ln+2
by the assumption, and then use the same proof as before to reduce this to
p = l1 −→ m0 −→ ln+2 .
Now we assume ln+1 = l1 and ln 6= ln+2 . Then we use lemma 1 to exchange
ln+1 for some other line that is not equal to l1 , allowing us to reduce it once
more.
Next, what if ln+1 = l1 and ln = ln+2 . Now if ln−1 6= ln+1 , we can simply
use lemma 3.1 to exchange ln for some other line to achieve our result. And
if ln−1 = ln+1 , we look at q = ln+1 −→ ln −→ ln+1 −→ ln . By the proof we
used in Case 3 above, this is a perspective, meaning we can reduce p to order
n − 1, which we already know can be reduced to order 2. Thus, we are done
and have proven any projective map p : l1 −→ l2 (with l1 6= l2 ) can be written
as a projective map of order 2.
With that, we can finally move on to the proof of theorem 3.5, which has
now been made simple:
Proof. As expected, we will split this up into two separate proves, as we are
working with an ’if and only if’-statement.
• Assume the Fundamental Property holds. We will show Pappus’s Theorem
also holds.
This is largely trivial. After all, clearly, FP.1 implies DPT.1, which is
equivalent with Pappus’s Theorem by theorem 3.2.
• Assume Pappus’s Theorem holds. We will show the Fundamental Property
also holds.
We will prove FP.1. Let p : l1 −→ l2 (with l1 6= l2 ) be a projective map
such that p(l1 ∩ l2 ) = l1 ∩ l2 . Now, since Pappus’s Theorem holds, so too
does theorem 3.6 and DPT. With the former, we can conclude that p is a
projective map of order 2. But since p(l1 ∩ l2 ) = l1 ∩ l2 , DPT tells us that
p is now a perspective, which is what FP.1 states.
With that, we conclude this section, as we have now shown that Pappus’s
Theorem is equivalent to the Fundamental Property.
45
4 Harmonic Addition
In this shorter chapter, we will introduce a new notion, which we will then
apply many different things we have learned to. It will function as a showcase
of how we can use the different notions and theorems we have found to prove a
variety of statements. To do so, we will first showcase its definition and some
crucial properties that rely on theorems from chapter 2, then we move on to its
relationship to the projective maps of chapter 3, and finally we consider how it
functions under the Principle of Duality from chapter 1.
4.1 Definition and Properties

So first, let us introduce the harmonic additions as we will use them:
Definition 4.1. Let l be a line and let P1 , P2 , and Q1 be distinct points on
l. Let A1 be any point not on l and let A2 be a point on A1 Q1 , distinct from
both A1 and Q1 . Define A3 = P1 A1 ∩ P2 A2 and A4 = P1 A2 ∩ P2 A1 . Finally, let
Q2 = A3 A4 ∩ l. We call Q2 the harmonic addition to Q1 relative to P1 and P2 .
Figure 24: The harmonic addition
Now, at first glance, this definition doesn’t seem very useful. After all, A1
and A2 are both arbitrarily chosen, meaning that we’d be able to get many
different harmonic additions for the same Q1 , P1 , and P2 , right? Well, as it
turns out, that is often not the case. Until we determine that, however, we will
refer to Q2 as determined via A1 and A2 as QA1 A2 . Note QA1 A2 = QA2 A1 . Now
then, let us move on to a proof:
Theorem 4.1. In any projective plane where Desargue’s Theorem and Fano’s
Strong Axiom hold, the harmonic addition to Q1 relative to P1 and P2 is inde-
pendent of our choice of A1 and A2 . In other words, it’s unique.
46
Proof. For this, we will consider two different constructions for Q2 , where we
find it ones using A1 and A2 , and once using B1 and B2 (which are chosen in
the same way as A1 and A2 before) We will consider several different cases:
• Case 1: A1 = B1 .
See figure 25 on the next page. We simply apply Desargue’s Theorem to
the triangles ∆A2 A3 A4 and ∆B2 B3 B4 , as the relevant lines all meet in A1 .
Now, we see that P1 = A2 A4 ∩B2 B4 , P2 = A2 A3 ∩B2 B3 , and A3 A4 ∩B3 B4
are collinear. But P1 P2 = l, so A3 A4 ∩ B3 B4 = A3 A4 ∩ l = QA1 A2 . Thus
QB1 B2 = B3 B4 ∩ l = QA1 A2 .
47
• Case 2: A1 A2 = B1 B2 and A1 6= B1 .
Note that the roles of A1 and A2 are symmetrical. Thus, from the first
case, we get QA1 A2 = QA2 A1 = QA2 B1 = QB1 A2 = QB1 B2 .
• Case 3: m = A1 A2 6= B1 B2 = n.
For notational purposes, we state Qm = QA1 A2 and Qn = QB1 B2 . We can
do this as in the first two cases, we determined that Qm is independent
of where on m A1 and A2 are chosen. Thanks to this, we can also choose
these points in ways that are convenient for us. So we pick A1 ∈ m
arbitrarily and let B1 = P1 A1 ∩ n. Next, choose B2 ∈ n arbitrarily and
let A2 = P2 B2 ∩ m.
Now we will apply Desargue’s Theorem to ∆P1 A2 B2 and ∆P2 A1 B1 , as
the relevant lines meet in Q1 . We then conclude P1 A2 ∩ P2 A1 = A4 ,
A2 B2 ∩ A1 B1 = A3 = B3 , and P1 B2 ∩ P2 B1 = B4 are collinear. Thus
A3 A4 = B3 B4 , giving us Qm = A3 A4 ∩ l = B3 B4 ∩ l = Qn .
Here, we use Fano’s Strong Axiom to ensure Q1 6= Qm . To prove this, we
assume Q1 = Qm by way of contradiction and consider the quadrilateral
P1 P2 A3 A4 . Then by definition, the diagonal points are P1 P2 ∩ A3 A4 =
Qm = Q1 , P1 A3 ∩ P2 A4 = A1 and P1 A4 ∩ P2 A3 = A2 . However, we have
Q1 , A1 , A2 ∈ m, which is a contradiction to Fano’s Strong Axiom. Thus,
Q1 6= Qm .
Figure 26: The third case
So, at least we now know we can speak of the harmonic addition to Q1

relative to P1 and P2 , as it is unique. However, there’s two more properties that
will make it even easier to talk about this definition:
Theorem 4.2. If Q2 is the harmonic addition to Q1 relative to P1 and P2 , then
also Q1 is the harmonic addition to Q2 relative to P1 and P2 . In other words,
the relationship of ’harmonic addition’ is symmetric.
48
Proof. This is largely trivial. After all, if Q2 = QA1 A2 as in figure 24, then
Q1 = QA3 A4 , or the harmonic addition to Q2 relative to P1 and P2 , constructed
via A3 and A4 .
Theorem 4.3. In any projective plane where Desargue’s Theorem and Fano’s
Strong Axiom hold, if Q2 is the harmonic addition to Q1 relative to P1 and P2 ,
then P2 is the harmonic addition to P1 relative to Q1 and Q2 .
Proof. Let A1 , A2 , A3 , and A4 be the points used to construct Q2 as before. By
Fano’s Strong Axiom, Q1 6= Q2 . We will be constructing the harmonic addition
to P1 relative to Q1 and Q2 via A1 and A3 , and show it must be equal to P2 .
Consider the points B3 = Q1 A1 ∩ Q2 A3 and B4 = Q1 A3 ∩ Q2 A1 . Clearly,
B3 B4 ∩ l (where l is the line on which P1 , P2 , Q1 , and Q2 are) is the harmonic
addition we are looking for.
Now, we apply Desargue’s Theorem to the triangles ∆Q2 A1 A4 and ∆Q1 A3 A2 ,
which we can do since the relevant lines are concurrent in P1 . Then we see that
B3 , B4 , and P2 = A1 A4 ∩ A3 A2 are collinear. But then, since P2 ∈ l, P2 =
B3 B4 ∩ l. As such, P2 is the harmonic addition to P1 relative to Q1 and Q2 .
Now that we have accomplished this, from here on out, as long as Desargue’s
Theorem and Fano’s Strong Axiom hold, we will speak of the harmonic pairs
P1 , P2 and Q1 , Q2 , or we can say Q1 and Q2 are harmonic to P1 and P2 .
49
4.2 Harmonic Pairs and Projective Maps
In this section, we will consider exactly how harmonic pairs act when we apply
projective maps to them. As it turns out, they act quite nicely:
Theorem 4.4. Assume Desargue’s Theorem and Fano’s Strong Axiom hold
and let p : l1 −→ l2 (maybe l1 = l2 ) be a projective map. If P1 , P2 , Q1 , Q2 ∈ l1
and P1 and P2 are harmonic to Q1 and Q2 , then p(P1 ) and p(P2 ) are harmonic
to p(Q1 ) and p(Q2 ).
Proof. Since projective maps are simply compositions of perspectives by defini-
tion, a trivial argument allows us to state that if we prove this for a perspective
σS : l −→ m, we have proven it for all projective maps. We will split this up
into two different cases:
• Case 1: Q1 = l ∩ m.
Consider P10 = σS (P1 ) and P20 = σS (P2 ). We will first construct Q2 via
P10 and P20 , so let A3 = P1 P10 ∩ P2 P20 = S and A4 = P1 P20 ∩ P2 P10 . Then,
clearly Q2 = SA4 ∩ l.
Next, we will construct Q02 ∈ m such it’s the harmonic addition to Q1
relative to P10 and P20 . To do so, we use P1 and P2 to construct it. Thus
B3 = P10 P1 ∩ P20 P2 = S and B4 = P10 P2 ∩ P20 P1 = A4 . Thus, Q02 =
B3 B4 ∩ m = SA4 ∩ m.
But now we have SQ2 = SA4 = SQ02 , so σS (Q2 ) = Q02 . Since σS was an
arbitrarily chosen perspective, we have now shown it for any perspective.
Also note that since all 4 points Q1 , Q2 , P1 , and P2 play a similar role
in their relationship as harmonic pairs, this case covers the situation that
any one of them is equal to l ∩ m.
• Case 2: Q1 , Q2 , P1 , P2 6= l ∩ m.
Let P1 and P2 be harmonic to Q1 and Q2 and let P10 = σS (P1 ), P20 =
50
σS (P2 ), Q01 = σS (Q1 ), and Q02 = σS (Q2 ). We will show that Q02 is the
harmonic addition to Q01 relative to P10 and P20 .
Let n = Q1 Q02 and consider the perspectives µS = l −→ n and νS :
n −→ m. Note that σS = νS ◦ µS . Now, by case 1, Q02 = µS is the
harmonic addition to Q1 relative to µS (P1 ) and µS (P2 ). And similarly
by the first case, Q01 = νS (µS (Q1 )) = σS (Q1 ) is the harmonic addition
to Q02 = νS (µS (Q2 )) = σS (Q2 ) relative to P10 νS (µS (P1 )) = σS (P1 ) and
P20 = νS (µS (P2 )) = σS (P2 ). Thus, Q01 , Q02 and P10 , P20 are harmonic pairs.
While this section may have been short, it hopefully showed off well enough
how projective maps interact with other concepts in projective planes.
51
4.3 Dual Harmonic Addition
Finally, we will look at one more concept, which will take us back to the Principle
of Duality. After all, we have not yet considered what harmonic pairs of lines
would be like, so let us do so now:
Definition 4.2. Let A be a point and let l1 , l2 , and m1 be lines through A.
Let n1 be any line not through A and let n2 be a line through m1 ∩ n1 , distinct
from both m1 and n1 . Define n3 = (l1 ∩ n1 )(l2 ∩ n2 ) and n4 = (l1 ∩ n2 )(l2 ∩ n1 ).
We call m2 = (n3 ∩ n4 )A the dual harmonic addition to m1 relative to l1 and
l2 .
Figure 30: The dual harmonic addition
Up until this point, we have seen that theorems tend to imply their dual
theorems, as is the case with Pappus’s Theorem and Desargue’s Theorem. As
it turns out, a similar relationship holds when it comes to harmonic additions:
Theorem 4.5. Assume Desargue’s Theorem and Fano’s Strong Axiom hold.
Let l1 , l2 , m1 , and m2 be concurrent in a point A and let o be a line not through
A. Call P1 = l1 ∩ o, P2 = l2 ∩ o, Q1 = m1 ∩ o, and Q2 = m2 ∩ o. Then P1 and
P2 are harmonic to Q1 and Q2 if and only if m2 is the dual harmonic addition
to m1 relative to l1 and l2 .
Proof. We begin by recalling theorems 2.7 and 2.14, which allow us to use the
dual theorems to Desargue’s Theorem and Fano’s Strong Axiom. Using this
and by applying the principle of duality to theorem 3.7, we find that m2 as the
52
dual harmonic addition to m1 relative to l1 and l2 is unique.
Now then, we first assume we have all lines and points as described and assume
P1 and P2 are harmonic to Q1 and Q2 . We make a construction for Q2 with
A1 = A and A2 any point on AQ1 = m1 . Of course, A3 and A4 are found
as before. We will show that m2 = AQ2 is the dual harmonic addition to m1
relative to l1 and l2 .
To do this, we will construct m2 using n1 = P1 A4 and n2 = P2 A3 . Then
n3 = (l1 ∩ n1 )(l2 ∩ n2 ) = P1 P2 = o and n4 = (l1 ∩ n2 )(l2 ∩ n1 ) = A3 A4 . Then
we easily see n3 ∩ n4 = o ∩ A3 A4 = Q2 . By definition, we then see m2 = AQ2 ,
which is what we were aiming to show.
Now, we assume m2 is as required. By way of contradiction, we assume Q2 isn’t
the harmonic addition to Q1 relative to P1 and P2 . Then we find this harmonic
addition and name it Q02 . By the proof we just gave, m02 AQ02 is the harmonic
addition to m1 relative to l1 and l2 . But we already had m2 as this harmonic
addition, and as we showed at the beginning of this proof, it is unique. This is
a contradiction, so we must have Q1 and Q2 harmonic to P1 and P2 .
53
Conclusion
Throughout this paper, we have seen a variety of interesting results, and practi-
cally all of them directly resulted from the lack of parallel lines in the projective
plane. In all of our proofs, we got to simply assume that the point of intersec-
tion existed thanks to axiom 1.2, which resulted in theorems such as Pappus’s
Theorem, Desargue’s Theorem, and even the existence of perspectives at all.
In addition, we saw the Principle of Duality arise, which showed us that to any
projective plane there is what can be described as a parallel plane in which all
the same rules hold, which is a fascinating result to think about.
Overall, it’s important to consider that what we explored in this paper is but
a small portion of the field of projective geometry. Many more advanced re-
search includes Pappus’s Theorem as a fourth axiom and then considers the
implications that it has. Based on what we have found here, we know that the
Fundamental Property then holds, but as of right now other results go outside
of the scope of this paper. Rest assured, many interesting results are still to be
discussed.
54
References
[1] A. Beutelspacher & U. Rosenbaum. (1998). Projective Geometry: From
Foundations to Applications. Cambridge, Press Syndicate of the University
of Cambridge.
[2] A. Heyting. (1963). Edited by N.G. de Bruijn & J. de Groot & A.G. Zaanen.
Axiomatic Projective Geometry. Groningen, P. Noordhoff N.V. & Amsterdam,
North-Holland Publishing Company.
[3] R. Hartshorne. (1967). Foundations of Projective Geometry.
55

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Axiomatic Projective Geometry

2 Theorems in the projective plane 10

3 Perspectives and projective maps 29

1.1 Definitions and axioms

Definition 1.1. An axiomatic theory is described by:

1. l is incident with two of these points. Without loss of generality, say

This is a useful theorem to have, of course, but more importantly is how

Definition 1.6. Fano’s Plane is a projective plane with:

Figure 1: Fano’s Plane

i2 = j 2 = k 2 = −1, ij = −ji = k, jk = −kj = i, and ki = −ik = j

Definition 2.5. Let L be a division ring and V be a module over L with

• For π ∈ Π and λ ∈ Λ, π ∈ λ if and only if π is a subspace of λ.

Figure 2: Desargue’s Theorem

B1 = hw1 + w2 + w3 + b1 w1 i = h(b1 + 1)w1 + w2 + w3 i

B2 = hw1 + (b2 + 1)w2 + w3 i

P12 = A1 A2 ∩ B1 B2 = hw1 , w2 i ∩ h(b1 + 1)w1 + w2 + w3 , w1 + (b2 + 1)w2 + w3 i

Clearly, hb1 w1 − b2 w2 i is on both A1 A2 and B1 B2 . Then, by axiom 1.2, this is

Figure 3: Dual of Desargue’s Theorem

It is great to see that the Dual of Desargue’s Theorem holds if Desargue’s

Figure 4: Moulton’s Plane

Figure 5: Pappus’s Theorem

Theorem 2.8. Let V be a module of dimension 3 over a division ring M . Then

(p + (p − 1)(q − 1)−1 )u + pv 0 + (p − 1)(q − 1)−1 qw0

= p(u + v) + (p − 1)(q − 1)−1 (u + qw0 ) ∈ hu + v 0 , u + qw0 i

P23 ∈ P12 P31

y = (p − 1)(q − 1)−1 (1 − p)−1

P23 ∈ P12 P31

Figure 6: Dual to Pappus’s Theorem

Theorem 2.11. If Pappus’s Theorem holds in a projective plane, then so does

Figure 7: Pappus’s Hexagon Theorem

2.4 Fano’s Axiom

Figure 8: Fano’s Axiom

We introduce the theorem in two steps, as that is how it was originally

Figure 9: Dual to Fano’s Strong Axiom

Theorem 2.15 (Theorem of Menelaus). Let P = hui, Q = hvi, and R = hwi be

Figure 10: Theorem of Menelaus

Proof. Note that, if P 0 , Q0 , and R0 are collinear, we have v + aw = p(w +

P 0 , Q0 , and R0 are collinear

⇔ v = qcv, aw = pw, 0 = (pb + q)u

Theorem 2.16 (Theorem of Ceva). Let P = hui, Q = hvi, and R = hwi be

P P 0 , QQ0 , and RR0 are concurrent

In figure 12, an example of a perspective is given. Clearly, σA (B1 ) = C1

The proof of this is trivial.

3.2 Double Perspective Theorem

1. Let l, m, and n be three non-concurrent lines and let σ : l −→ n be a

2. Let l, m, and n be three concurrent lines and let σ : l −→ n be a projective

line between A2 and σA (A2 ), so A1 , A2 and A3 are collinear.

Figure 14: Pappus’s Theorem implies DPT.1

Figure 15: DPT.2 implies Desargue’s Theorem

2. Next, assume Desargue’s Theorem holds. We will show DPT.2. To start,

Figure 16: Desargue’s Theorem implies DPT.2

Now, we finally have links to Pappus’s Theorem and to Desargue’s Theorem.

Figure 17: The first case

Figure 18: The second case

• 3. ⇒ 1. Let l1 and l2 be lines and let p : l1 −→ l2 be a projective map

Figure 19: The first case

Figure 20: The second case

– Case 3: Ai 6= l1 ∩ l2 for i ∈ {1, 2, 3} and B3 = l1 ∩ l2

Figure 21: The fourth case