54
Affine and Projective Geometry Adrien Deloro Şirince ’17 Summer School
Affine and Projective Geometry
Adrien Deloro
Şirince ’17 Summer School
Contents
Table of Contents ii
List of Lectures iii
Introduction 1
Week I: Affine and Projective Planes 2I.1 Affine planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
I.1.1 The Definition . . . . . . . . . . . . . . . . . . . . . . . . 3I.1.2 Membership as Incidence . . . . . . . . . . . . . . . . . . 3I.1.3 The Affine Plane over a Skew-field . . . . . . . . . . . . . 4
I.2 Projective planes . . . . . . . . . . . . . . . . . . . . . . . . . . . 7I.2.1 The need for Projective planes . . . . . . . . . . . . . . . 7I.2.2 The Definition . . . . . . . . . . . . . . . . . . . . . . . . 8I.2.3 From Affine to Projective, and Back . . . . . . . . . . . . 9
I.3 Arguesian Planes . . . . . . . . . . . . . . . . . . . . . . . . . . . 12I.3.1 Desargues’s Projective Theorem . . . . . . . . . . . . . . 12I.3.2 Free Planes and Non-Arguesian Planes . . . . . . . . . . . 15I.3.3 Pappus’ Theorem . . . . . . . . . . . . . . . . . . . . . . . 17I.3.4 Pappus Implies Desargues . . . . . . . . . . . . . . . . . . 19
I.4 Coordinatising Arguesian Planes Explicitly . . . . . . . . . . . . 21I.4.1 The Frame . . . . . . . . . . . . . . . . . . . . . . . . . . 22I.4.2 Addition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23I.4.3 Multiplication . . . . . . . . . . . . . . . . . . . . . . . . . 26
I.5 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28I.5.1 One Finitary Aspect . . . . . . . . . . . . . . . . . . . . . 28I.5.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30I.5.3 The Dual Plane . . . . . . . . . . . . . . . . . . . . . . . . 31I.5.4 Dualising Desargues and Pappus . . . . . . . . . . . . . . 32
Week II:Group-Theoretic Aspects 34II.1 Arguesianity and the Group of Dilations . . . . . . . . . . . . . . 34
II.1.1 Dilations and Translations . . . . . . . . . . . . . . . . . . 34II.1.2 The (p,p)-Desargues Property . . . . . . . . . . . . . . . . 37II.1.3 The (c,p)-Desargues Property . . . . . . . . . . . . . . . . 39
II.2 A Vector Space from an Affine Plane . . . . . . . . . . . . . . . . 40II.2.1 A Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40II.2.2 A Skew-Field . . . . . . . . . . . . . . . . . . . . . . . . . 41II.2.3 Dimensionality . . . . . . . . . . . . . . . . . . . . . . . . 42
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II.3 An Application: SO(3, R) . . . . . . . . . . . . . . . . . . . . . . 44II.3.1 The Geometry and Non-Degeneracy Axioms . . . . . . . . 44II.3.2 The First Axiom . . . . . . . . . . . . . . . . . . . . . . . 45II.3.3 The Second Axiom . . . . . . . . . . . . . . . . . . . . . . 46II.3.4 The Third and Fourth Axioms . . . . . . . . . . . . . . . 47
II.4 Projectivities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48II.4.1 Projectivities . . . . . . . . . . . . . . . . . . . . . . . . . 48II.4.2 Group-theoretic Interpretation of Pappus’ Theorem . . . 49
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List of Lectures
Week ILecture 1 (Introduction; Affine Planes) . . . . . . . . . . . . . . . . . . 2Lecture 2 (Projective Planes) . . . . . . . . . . . . . . . . . . . . . . . 7Lecture 3 (Desargues’ Theorem) . . . . . . . . . . . . . . . . . . . . . 11Lecture 4 (Pappus’ Theorem) . . . . . . . . . . . . . . . . . . . . . . . 17Lecture 5 (Coordinates) . . . . . . . . . . . . . . . . . . . . . . . . . . 21Lecture 6 (Duality) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
Week IILecture 7 (Coordinatising like adult people (1)) . . . . . . . . . . . . . 34Lecture 8 (Coordinatising like adult people (2)) . . . . . . . . . . . . . 40Lecture 9 (The Geometry of SO3(R)) . . . . . . . . . . . . . . . . . . . 43Lecture 10 (Projectivities) . . . . . . . . . . . . . . . . . . . . . . . . . 48
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Introduction
If parallel lines meet at infinity, then all trains for the infiniteare running to a disaster.
Kryszanowski
Prerequisites
In order to follow this class, one needs to know:
• high-school geometry: in the plane, using coordinates, vectors, equationsof lines;
• vector spaces, linear maps, matrices;
• fields: essentially the definition;
• groups (subgroups, normal subgroups, . . . ) and group actions (stabilisers,orbits, . . . ) will play a prominent role in week 2.
Recommended Reading
[1] Artin, Emil. Geometric Algebra. Only Chapter 2.
[2] Cameron, Peter. Projective Planes and Spaces. Mostly Chapters 1, 2, 3.Notes available online on the author’s webpage.
[3] Hartshorne, Robin. Foundations of Projective Geometry. Technically outof print but on the internet one finds a modern typeset.
[4] Hilbert, David. Foundations of Geometry. Only Chapter 5 on Desargues’Theorem.
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Chapter I: Affine andProjective Planes
Lecture 1 (Introduction; Affine Planes)
The Quartet and the Orchestra. We begin with an analogy. Beethoven’sNinth Symphony (op. 125) uses an extensive orchestra and a choir. After com-pletion Beethoven quickly returned to composing string quartets (op. 130–132and 135), which are no less praised by amateurs than the symphony. The dis-crepancy of means and effects between the monumentality of the massive orches-tration on the one hand, and the elegant economy of just four instruments onthe other, is significant. None is “better” music than the other: the symphonytargets universality, whereas the quartets aim at spirituality and purity.
And mathematics is like music. There are various layers of mathematicalstructures, from elementary aspects to the whole orchestra of modern mathe-matics. And none is better for the true mathematician will enjoy both.
For instance when studying the real plane R2 we have very different notionsand tools:
• points and lines;
• coordinates and vectors;
• distances;
• angles; the dot product;
• the underlying complex multiplication.
In this class we shall however restrict ourselves to the most basic layer whereone has only points and lines.
Painting from Nature. This leads to the notion of an abstract affine plane.It is just a collection of points and lines with the possibility for some points tobe on some lines or not. But of course for this to be interesting one must startwith intuition, or physical observation in our small part of the universe (a sheetof paper can do), in order to find axioms which describe the behaviour.
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I.1. Affine planes
I.1.1. The DefinitionDefinition (affine plane). An affine plane A is given by an incidence geometry,i.e.:
• a set of points P;
• a set of lines L;
• a relation I (for “incidence”) on P × L, which says when a point is on aline or not;
subject to the following axioms AP1, AP2, AP3 listed hereafter (we shall haveto pause for auxiliary definitions).
AP1. ∀(a, b) ∈ P2 a 6= b→ (∃!` ∈ L aI` ∧ bI`).(The meaning is clear: given any two distinct points, there is a unique lineincident to both. It is convenient to call this line (ab).)
The second axiom requires a definition.
Definition (parallel). Two lines `,m are parallel if either ` = m or there is nopoint incident to both:
` ‖ m iff (` = m) ∨ (∀a ∈ P¬(aI` ∧ aIm))
AP2. ∀(a, `) ∈ P × L ∃!m ∈ L (aIm) ∧ (` ‖ m).(Intuitive meaning: through any point there is a unique line parallel to agiven line. It is convenient to call this line `a.)
Definition (collinear). Points a1, . . . , an are collinear if there is a line ` suchthat each akI`.
AP3. There exist three non-collinear points.
Axioms AP1,AP2,AP3 define affine planes. Notice that we speak onlyabout points and lines; words such as angle, distance, or even between are bannedfrom our vocabulary.
Exercise. AP3 is added to prevent degenerate configurations, the most ex-treme of which being P = ∅ = L.
List all configurations satisfying AP1 and AP2 but not AP3.
I.1.2. Membership as IncidenceNotice that we are using an abstract relation I; however we are used to thinkingthat a point belongs to a line, in the set-theoretic sense. Indeed, any not-too-small affine plane can be replaced by a similar one where the incidence relationis the usual membership. We need to give a precise meaning to “similar”.
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Definition (isomorphism). Two incidence geometries (P1,L1, I1) and (P2,L2, I2)are isomorphic if there are bijections f : P1 → P2 and g : L1 → L2 such that:
∀(a, `) ∈ P1 × L1 aI1`b⇔ f(a)I2g(`)
Proposition. Let A = (P,L, I) be an affine plane where each line is incidentto at least two points. Then there is an affine plane A′ = (P ′,L′,∈) (where ∈is the usual membership relation) which is isomorphic to A.
Proof. The idea is to keep the same points, and write new lines as sets ofpoints. To each ` ∈ L associate S` = {a ∈ P : aI`}, a subset of P. Nowconsider L′ = {S` : ` ∈ L}, a family of subsets of P. Finally let f be theidentity function on P and g : L → L′ map ` to S`.
We claim that (P,L, I) and (P ′,L′,∈) are isomorphic via f and g. Clearlyf is a bijection; g is clearly onto but it also is one-to-one. For if S` = Smthen since there are at least two distinct points a, b ∈ P incident to `, one hasa, b ∈ S` = Sm, so a and b are also incident to m. But by AP1, one must have` = m. So g is injective as well.
Now for any (a, `) ∈ P × L, one has aI` iff a ∈ S`, which is the definitionof an isomorphism.
From now on we shall leave aside the case where there are lines with only one(or even zero) point; such lines should be removed from L and simply forgotten.So we are free to consider that ∈ will always be the incidence relation.
As a consequence, instead of always saying that a is incident to ` we can usethe common phrases: “a is on `”, “a belongs to `”, “` contains a”.
I.1.3. The Affine Plane over a Skew-fieldWe wrote the definition of an affine plane by painting from nature. It is timeto check that the usual plane R2 is indeed an affine plane in our abstract sense.
Proposition. R2 with the usual points and lines is an affine plane.
Proof. Points are pairs of real numbers (x, y) ∈ R2; lines are sets of the form
`p,−→v = {p+ t−→v : t ∈ R}
where p ∈ R2 and −→v 6= −→0 . Notice that one always has p ∈ `p,−→v .We must check axioms AP1,AP2,AP3; we shall actually give the proof
as it will reveal something of interest.
Claim 1. Two lines `1 = `p1,−→v1 and `2 = `p2,
−→v2 are equal iff −→v1 and −→v2 areproportional vectors, and −−→p1p2 is a multiple of those.
Verification. First suppose `1 = `2. Since p2 ∈ `1 there is t ∈ R with p2 =p1 + t−→v1. So −−→p1p2 = t−→v1 is a multiple of −→v1. But also p1 ∈ `2 so there is s ∈ Rwith p1 = p2 + s−→v2 ; now
p1 = p2 + s−→v2 = p1 + t−→v1 + s−→v2
which simplifies into t−→v1 + s−→v2 = −→0 . Since we know that neither −→v1 nor −→v2 is−→0 , we deduce that they are proportional.
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Conversely we shall prove one inclusion, and the other will follow by sym-metry. So let q ∈ `1. By definition there is t ∈ R with q = p1 + t−→v1. Now byhypothesis there are λ, µ ∈ R with −−→p1p2 = λ−→v2 and −→v1 = µ−→v2 (we even knowthat µ 6= 0). Then:
q = p2 −−−→p1p2 + t−→v1 = p2 + (tµ− λ)−→v2 ∈ `2
Claim 2. Two lines `1 = `p1,−→v1 and `2 = `p2,
−→v2 are parallel iff −→v1 and −→v2 areproportional vectors.
Verification. Suppose `1 ‖ `2; we wish to show that −→v1 and −→v2 are proportionalvectors. If `1 = `2 then we know the conclusion from the previous claim. Sosuppose `1 6= `2; since they are parallel this means that there is no point onboth lines. Hence the equation:
p1 + t1−→v1 = p2 + t2
−→v2
has no solution for (t1, t2) ∈ R2.
Work in coordinates, say pi = (xi, yi) and −→vi =(aibi
). We just understood
that the system of equations:
(S) :{x1 + t1a1 = x2 + t2a2y1 + t1b1 = y2 + t2b2
has no solution for (t1, t2) ∈ R2. Notice that multiplicative coefficients are onthe right and variables on the left; this surprise will turn into a valuable fact.
Remember that we want to prove that −→v1 and −→v2 are proportional vectors,and that the assumption is that (S) has no solution. There are three cases,two of which are equivalent.
Case 1a. Suppose a1 = 0. Then since −→v1 6=−→0 one has b1 = 0. If a2 6= 0 then
there is a unique solution t2 = (x1 − x2) · a−12 to the first equation; now
t2 being known there is a unique solution in t1 to the second. This is acontradiction to (S) having no solution, so actually a2 = 0. Then since−→v2 6=
−→0 one has b2 6= 0, and clearly −→v1 and −→v2 are proportional vectors.
Case 1b. If b1 = 0 we reach the desired conclusion similarly.
Case 2. Suppose a1 6= 0 and b1 6= 0. Then multiply the first equation on theright by λ = a−1
1 b1, and subtract to the second:
y1 − x1λ = y2 − x2λ+ t2(b2 − a2λ)
If b2 − a2λ 6= 0 then there is a (unique) solution in t2, which then givesrise to a (unique) solution in t1: so (S) has a solution, a contradiction.Therefore b2 = a2λ = a2a
−11 b1; now clearly
−→v2 =(
a2a2a−11 b1
)= a2a
−11
(a1b1
)= a2a
−11−→v1
so −→v1 and −→v2 are proportional vectors.
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Claim 3. A2(R) satisfies axioms AP1,AP2,AP3.
Verification.
AP1. Let p1 6= p2 be points. Then −→v = −−→p1p2 6=−→0 and ` = `p1,
−→v clearlycontains p1 and p2. Uniqueness is clear as well.
AP2. Let p be a point and ` = `q,−→v be a line. Then clearly m = `p,−→v containsp and is parallel to `; but also uniqueness is clear.
AP3. Simply take (0, 0), (0, 1), and (1, 0).
The proof shows a bit more. We used many properties of the real numbers,but they were only algebraic properties: the most notable of which was theassociativity law (λµ)ν = λ(µν). But we never used the commutativity lawλµ = µλ. This is why we insisted on the difference between left and right; checkit in the above computation.
So we can generalise not only to fields but also to skew-fields.
Definition (skew-field). A skew-field (also known as: division algebra) is astructure (F; 0, 1,+, ·) where all axioms of a field are supposed to hold, exceptthat · is not required to be commutative.
Example. To give an example of a skew-field which is really not commutative,think about the quaternion algebra H = R⊕Ri⊕Rj⊕Rk, subject to the usual“Hamilton identities”. There one has ij = k 6= −k = ji. It should be checkedor known that H is a skew-field, but not a field.
An extremely important phenomenon is however the following, which weshall not prove.
Theorem (Wedderburn). Any finite skew-field is actually commutative.
Returning to affine planes, here is what we proved when working over R.
Proposition. Let F be any skew-field. Then A2(F), with the usual points andlines, is an affine plane.
Example. Here is the smallest affine plane:
a b
c d
One point is removed: diagonals do not meet.We recognize A2(F2), where F2 is the field with two elements 0 and 1, subject
to all expected identities and 1 + 1 = 0. Then we can easily convince ourselvesthat diagonals should not meet indeed, since both are directed by the samevector: (
11
)−(
00
)=(
11
)=(
10
)−(
01
)Diagonals are parallel!
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Remark. We started the class by promising to use only points and lines, andto even forget about vectors and coordinates. However we have just spent a lotof time on vector geometry.
The entire idea of the class is to investigate the presence of coordinate sys-tems (say, skew-fields) in abstract affine planes. The answer, discovered byHilbert, is surprinsingly beautiful.
Slightly abusing set-theoretic notation, we have so far:
{A2(F) : F a skew-field} ⊆ {A an affine plane}
Does one have equality?
End of Lecture 1.
Lecture 2 (Projective Planes)
I.2. Projective planes
I.2.1. The need for Projective planesBelow is Desargues’s celebrated theorem. We give the full affine statement withnumerous cases, and will prove it only later, when we have the nice, unified,projective statement. The purpose of giving such a long case division beingprecisely to convince ourselves that the dichotomy “intersect” vs. “are parallel”is too heavy and should be made simpler.
Theorem (Desargues’ Theorem: all affine cases). Let F be a skew-field andwork in A2(F).
Let abc and a′b′c′ be triangles (involving six distinct points). Suppose thatlines (aa′), (bb′), (cc′) are three distinct lines, and either that they are parallelor that they meet at the same point.
Then:
• either (ab) ‖ (a′b′), (ac) ‖ (a′c′), and (bc) ‖ (b′c′);
• or (ab) ‖ (a′b′) but (ac) ∩ (a′c′) = {b′′} and (bc) ∩ (b′c′) = {a′′} satisfy(a′′b′′) ‖ (ab);
• or two similar cases (involving the intersection c′′ of (ab) and (a′b′));
• or a′′, b′′, c′′ are collinear.
We hope that the complexity of the statement strongly suggests to unifycases by making parallels intersect. The power of this idea was acknowledgedand fully exploited in the Renaissance. Below is a picture of the most striking
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case.
o
a
a′
bb′
c
c′
a′′ b′′c′′
It is a good exercise, in order to get familiar with the statement, to drawthe other cases (here, no parallelism is involved).
I.2.2. The DefinitionDefinition (projective plane). A projective plane is an incidence geometry(P,L, I) satisfying the following axioms PP1,PP2,PP3.
PP1. Through any two distinct points there is a unique line.
PP2. Any two distinct lines meet in a unique point.
PP3. There are four points no three of which are collinear.
The definition has a lovely symmetry called self-duality, to which we shallreturn in another lecture. We move to the main class of examples. Rememberthat a vector line is a 1-dimensional vector subspace, which can be viewed as aline through the origin; define a vector plane likewise.
Definition (projective plane over a skew-field P2(F)). Let F be a skew-field.The projective plane over F has:
• as points, the vector lines of F3;
• as points, the vector planes of F3;
• as incidence relation, set-theoretic inclusion ⊆.
Proposition. P2(F) is a projective plane.
Proof. We must check three things, but this is easy as our construction relieson linear algebra in dimension 3.
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PP1. Let α 6= β be two projective points, i.e. two distinct vector linesLα, Lβ ⊆ F3. Then set H = 〈Lα, Lβ〉 ⊆ F3, a vector plane of F3.In projective terms, H is a line λ. Now by definition, Lα ⊆ H translatesinto αIλ, and βIλ likewise.Uniqueness is clear as well, since there is no other vector plane containingboth Lα and Lβ .
PP2. Now start with two distinct projective lines λ 6= µ. By constructionthey correspond to distinct vector planes Hλ 6= Hµ, and we know thatHλ ∩ Hµ is a vector line L, which we think as a projective point α.Obviously αIλ and αIµ, but uniqueness is clear as well.
PP3. Clear: vector lines through (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1) definesuitable projective points.
I.2.3. From Affine to Projective, and BackAnother way to obtain projective planes is by “completing” affine planes, aprocedure we now explain. In the affine world, some intersections are missing:if ` ‖ m are distinct, parallel affine lines, then we should add a point. But if` ‖ m ‖ n then we feel that it should be the same point added for the intersection` ∩ m and for the intersection m ∩ n. This suggests that we are not workingwith pairs of parallel lines, but with sets of parallel lines. Fortunately there isa mathematical tool for that.
Lemma. Let A be any affine plane. Then ‖ is an equivalence relation.
Proof. Of course we already know that in the case of A2(F) for F a skew-field,as we gave in that case a detailed description of parallelism. But we want ageneral proof, a proof which will use only axioms AP1,AP2,AP3. Rememberthat in this abstract setting ` and m are parallel iff ` = m or ` and m do notmeet. We freely replace A by an isomorphic plane where incidence is given bymembership.
There are three things to check:
Reflexivity. Every line ` satifies ` = `, so ` ‖ `.
Symmetry. If ` ‖ m then either ` = m or ` ∩m = ∅; in either case we havem ‖ `.
Transitivity. Suppose ` ‖ m and m ‖ n; we show that ` ‖ n. If ` = n we aredone. Otherwise we want to show ` ∩ n = ∅. So suppose not; by AP1there is a ∈ `∩ n. Now by AP2, there is a unique line parallel to m andcontaining a; however this applies to both ` and n. So actually ` = n, acontradiction showing that ` ∩ n = ∅. Hence ` ‖ n, as desired.
The equivalence class [`] of a line ` can be thought of as its direction. Soit is intuitive that we should add new points “at infinity” for these directions;there will also be a line “at infinity”. But ordinary, affine lines ` are now tooshort: we must force a projective line to contain the direction. This explainswhy we go through ˆ below.
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Definition (projectivisation A of an affine plane). Let A = (P,L,∈) be anaffine plane. For each ` ∈ L, let ˆ = ` ∪ {[`]}. Now let `∞ = {[`] : ` ∈ L},P = P ∪ `∞, and L = {ˆ : ` ∈ L} ∪ {`∞}.
We call A = (P, L,∈) the projectivisation of A.
Proposition. For any affine plane A, A is a projective plane.
Proof. There are three axioms to check.
PP1. Let α 6= β be two points in P. We see several cases.
• If α = a ∈ P and β = b ∈ P, then we know from AP1 that there isa unique line ` ∈ L containing both. Clearly α, β ∈ ˆ. We also haveuniqueness. If another line λ ∈ L contains α and β, then λ cannotbe `∞, so λ = m for some affine line m ∈ L. Then a, b ∈ m, so byuniqueness in AP1 we have m = ` and therefore λ = m = ˆ. So ˆis the only line of L containing α and β.
• Suppose α = a ∈ P but β ∈ P \ P = `∞ (the other case is similar).Then by definition, there is ` ∈ L with β = [`]. Now by AP2 thereis a unique m ∈ L with a ∈ m and m ‖ `. Then on the one handα ∈ m, and on the other hand [`] = [m] ∈ m. Now to uniqueness. Ifa line λ contains a and β, then it cannot be `∞. So it is of the formn for some affine line n. Now α ∈ n implies a ∈ n, and β = [`] ∈ nimplies [`] = [n], that is, ` ‖ n. By uniqueness in AP2 we haven = m, and therefore λ = n = m.
• Now suppose α, β ∈ `∞. Clearly `∞ is the only line in L incidentto both α and β.
PP2. Exercise.
PP3. Obvious from AP3 and the definition of projectivisation.
Downgrading from projective to affine is not hard either, but it involveschoosing the line to be removed.
Definition (affinisation P of a projective plane). Let P = (P,L,∈) be a projec-tive plane and fix one λ ∈ L. Let P = P \λ; for each µ ∈ L\{λ}, set µ = µ\λ,and let L = {µ : µ ∈ L \ {λ}}.
We call P = (P, L) the affinisation of P with respect to λ.
Proposition. For any projective plane P and line λ ∈ L, P is an affine plane.
Moreover P ' P.
Proof. Exercise.
Be extremely careful though that the isomorphism type of P depends on theline λ you chose to remove. In particular, ˇA depends on the choice of line weremove from A, and need not be isomorphic to A.
Remark. So far the picture looks like this:{A2(F) : F a skew-field
}⊆{A : A an affine plane
}⊆ {projective planes};
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{P2(F) : F a skew-field} ⊆ {projective planes}
Actually applying the projectivisation procedure to A2(F) gives what weexpect.
Proposition. Let F be a skew-field. Then A2(F) and P2(F) are isomorphic.
Proof. We shall construct an isomorphism. Let H0 ≤ F3 be a vector planeand x ∈ F3 be a vector not in H0. Let H1 = x+H0, an affine translate of H0.Clearly A2(F), H0, and H1 are isomorphic affine planes; in particular A2(F)and H1 are isomorphic projective planes. So it suffices to see that H1 andP2(F) are isomorphic, and this is easy thanks to elementary geometry.
H0
H1
x
O
a
L
L′
x+ L′
Let L ≤ F3 be a vector line. If L 6≤ H0, then L will intersect H1 in a pointsay a; map L to a. If L ≤ H0, then there is no intersection; we feel that Lshould be mapped to a point at infinity. In our previous construction, suchwere equivalence classes (directions) of lines of H1; so we map L to [x + L],the direction of x+ L which is a line of the affine plane H1.
Now let H be a vector plane. If H 6= H0 then H ∩ H1 is a line ` of theaffine plane H1; we map H to ˆ as in the projectivisation construction. If onthe other hand H = H0 then we map H to `∞.
It is an exercise to finish the proof:
• check that we have a bijection between points of P2(F) and points of H1;
• check that we have a bijection between lines of P2(F) and lines of H1;
• check that these bijections preserve incidence.
End of Lecture 2.
Lecture 3 (Desargues’ Theorem)
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I.3. Arguesian PlanesWe have introduced projective geometry in order to unify case divisions. Todaywe return to Desargues’ Theorem.
I.3.1. Desargues’s Projective TheoremTheorem (Desargues’ projective theorem). Let F be a skew-field, and work inP2(F).
Suppose abc, a′b′c′ are two triangles such that (aa′), (bb′), and (cc′) meet.Then a′′ = (bc) ∩ (b′c′), b′′ = (ac) ∩ (a′c′), and c′′ = (ab) ∩ (a′b′) are collinear.
Notice that by suitably choosing the line to be removed, this proves allversions for A2(F). And now, thanks to projective geometry, only one case mustbe checked.
There are proofs in the plane (using coordinates of course) but we shallintroduce a powerful method. Imagine for a moment that the configuration isno longer “flat”, but that triangles (abc) and (a′b′c′) are not in the same plane;say (abc) is in some plane π and (a′b′c′) in π′ 6= π.
o
a
a′
bb′
c
c′
a′′ b′′c′′
Then observe how a′′ ∈ (bc) ⊆ π; likewise, b′′, c′′ ∈ π. But similarly,a′′, b′′, c′′ ∈ π′. Since we expect two planes to intersect over a line, this im-plies collinearity of a′′, b′′, c′′.
Formalising this idea requires some material.
Definition. A projective 3-dimensional space is an incidence structure of points,lines, and planes, satisfying the six axioms below.
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PS1. Any two points are uniquely collinear.
PS2. Any three non-collinear points are uniquely coplanar.
PS3. Any line and plane meet.
PS4. Any two planes share a line.
PS5. There are four non-coplanar points no three of which are collinear.
PS6. Every line has at least three points.
Example. Let F be a skew-field. Then P3(F), where points are vector linesin F4, lines are 2-dimensional vector subspaces of F4, and planes are vectorhyperplanes of F4, is a projective space.
Of course one can always replace an abstract projective space by one whereincidence relations are of the form p ∈ ` ⊆ π, which makes notation and termi-nology a lot easier.
Proof of Desargues’ Theorem. We embed P2(F) into P3(F) and (des)arguethere. We shall prove a bit more, namely that the desired statement holdsfor configurations of P3(F).
There are two cases.
Case 1. First suppose that π = (abc) and π′ = (a′b′c′) are distinct planes.Then it is the argument we skteched above: a′′ ∈ (bc) ⊆ π; also b′′, c′′ ∈ πand a′′, b′′, c′′ ∈ π′.It must be checked that π∩π′, which contains a line `, is actually exactly`. For if some point p ∈ π ∩ π′ \ `, then taking two points q1, q2 ∈ `, wesee by PS2 that π is the only plane containing p, q1, q2; the same appliesto π′, against π 6= π′.So a′′, b′′, c′′ are on π ∩ π′ = `, and this proves collinearity.
Case 2. Now suppose (abc) = (a′b′c′) = π. Notice that π contains all pointsof the initial configuration, so we are in the genuine planar case. Theidea is however to use Case 1 by slightly “lifting” the configuration, andthen project back.
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o
a
a′
bb′
c
c′
a′′ b′′c′′
e
a1
a′1
b′′1
c′′1
Let e /∈ π; there is such a point as P2(F) = π embeds into P3(F). Nowtake a1 ∈ (ae) be neither a nor e; there is such a thing by PS6. Finallylet a′1 = (oa1) ∩ (ea′).(This can be imagined as rotating triangle abc along axis (bc), and simi-larly rotating a′b′c′ along (b′c′).)Of course changing triangles, we must also change our intersections; letb′′1 = (a1c)∩(a′1c′) and c′′1 = (a1b)∩(a′1b′); notice that a′′ however remainsunchanged.
Claim 1. Triangles a1bc and a′1b′c′ satisfy the assumption of DesarguesTheorem and are not coplanar.
Verification. We first prove that (a1a′1), (bb′), and (cc′) meet. But this is
clear as o ∈ (a1a′1) by construction. Now we check that these triangles are
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not in the same plane. Otherwise that plane would be π; since a, a1 ∈ π,one then has e ∈ (aa1) ⊆ π, a contradiction.By Case 1, we know that a′′, b′′1 , c′′1 are collinear. We now project back.Let p be the projection function from e onto π, defined as follows: forany point x 6= e, line (ex) meets π in a point p(x). However p(e) is notdefined. Notice that:
• p is the identity on π;• p preserves incidence, i.e. if x ∈ `, then p(x) ∈ p(`).
Claim 2. p(b′′1) = b′′ and p(c′′1) = c′′.
Verification. Since (ea1) = (ea) and a ∈ π, one has p(a1) = a. But also(ea′1) = (ea′) so p(a′1) = a′. In particular,
{p(b′′1)} = p ((a1c) ∩ (a′1c′))= p((a1c)) ∩ p((a′1c′))= (p(a1)p(c)) ∩ (p(a′1)p(c′))= (ac) ∩ (a′c′)= {b′′}
and p(c′′1) = c′′ likewise.Hence p(a′′) = a′′, p(b′′1) = b′′, and p(c′′1) = c′′ are collinear, as desired.
Definition (arguesian plane). Call arguesian a projective plane which satisfiesthe Desargues Theorem.
(A word on terminology: nobody ever worked in descartesian coordinates;and likewise, the correct adjective coined after Desargues is arguesian.)
We just saw that whenever F is a skew-field, P2(F) is arguesian. As a matterof fact we proved two things:
• P2(F) always embeds into a projective space (for instance, P3(F));
• any projective plane which embeds into a projective space is arguesian.
I.3.2. Free Planes and Non-Arguesian PlanesWe now give another family of projective planes. Such planes are not alwaysarguesian; in particular, by Desargues’ Theorem for P2(F), we thus see projectiveplanes not of the form P2(F).
The general idea behind a free construction is to add everything one needs,with no redundancies. Here it will mean adding lines to link any two points,but also adding points to intersect any two lines. Of course this creates newlines and points, so it must be done inductively.
Definition (configuration). Let P0 be any abstract set of points and L0 be anyabstract set of lines, with an incidence relation I0.
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We call it a configuration if there are no two distinct lines sharing morethan two points, and there are no two points being on more than two lines.
Definition (free projective plane). Let (P0,L0) be a configuration.
• At stage 2n + 1, enumerate the set Xn of all unordered pairs {a, b} ofdistinct points which are not linked by a line of L2n. For each {a, b} ∈ Xn,add a new line `{a,b} incident to a and b, but to no other point.Then set P2n+1 = P2n, L2n+1 = L2n ∪ {`{a,b} : {a, b} ∈ Xn}.
• At stage 2n+ 2 proceed likewise to force intersections of lines: enumeratenon-meeting distinct lines by Yn; for any such pair {`,m} in Yn add a newpoint p{`,m} incident to ` and m but to no other line; then set P2n+2 =P2n+1 ∪ {p{`,m} : {`,m} ∈ Yn}.
• Finally let P =⋃
N Pn and L =⋃
N Ln.
(P,L) is called the free projective plane over (P0,L0).
This is a nice construction provided there is enough data to start it; PP3has to be assumed.
Proposition. Suppose (P0,L0) contains four points no three of which are collinear,then the free projective plane over (P0,L0) is a projective plane.
Proof. As usual, three things must be checked.
PP1. Let a 6= b ∈ P be two distinct points. Say that a ∈ Pn and b ∈ Pm; wemay assume n ≥ m.First suppose that n = 0 and a, b are already linked by a line ` ∈ L0.Then ` is unique as such; moreover, we never introduce another linethrough a and b since {a, b} /∈ Xn. So ` remains unique even in L.Now suppose that there is no ` ∈ L0 containing a and b. Regardless ofthe value of n, at stage n + 1 we have introduced a line ` joining a andb. The argument for uniqueness of ` is similar.
PP2. Entirely similar.
PP3. This is almost by assumption but not entirely. There are four pointsa, b, c, d ∈ P0 ⊆ P which are not collinear in L0. We must see thatthey remain non-collinear in the sense of L. But at stage 1, when weadd line ` = (ab), it does not contain c. And we never add another linethrough a and b. So there never is a line containing a, b, c: they remainnon-collinear also in L. Likewise for the other triples.
Example. We start with the configuration consisting of four points and no
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lines.
00
0
0
1
1
1
1
1
1
2
2
2
3
3
3
We have indicated the stage of appearance of the various objects. The readershould imagine what happens at the following next steps.
Example (a non-arguesian projective plane). Start with a configuration often points P0 = {o, a, b, c, a′, b′, c′, a′′, b′′, c′′} and take L0 to be all lines of theDesargues configuration but not line (a′′b′′c′′).
Let P be the free plane over (P0,L0). Then P is a projective plane and doesnot satisfy Desargues Theorem. Because there is no line joining a′′, b′′, c′′ atfirst, and we never add a line joining three points at once.
In particular the free plane we just constructed is not of the form P2(F); itdoes not embed into a projective space.
Hilbert’s Grundlagen contain a completely different example (Theorem 33there), but it is rather involved. So is the so-called Moulton plane. Finally,finite non-arguesian planes do exist.
End of Lecture 3.
Lecture 4 (Pappus’ Theorem)
I.3.3. Pappus’ TheoremWe move to another theorem, seemingly unrelated.
Theorem (Pappus’s affine theorem). Let F be a commutative field, and workin A2(F).
Let a, b, c be collinear points and c′, b′, a′ be collinear points (being six non-collinear points in total). Suppose that (ab′) ‖ (ba′) and (c′b) ‖ (b′c). Then(ac′) ‖ (ca′).
17
o
a
b
c
c′ b′ a′
Proof. The proof requires working in coordinates, and of course commutativityof the base field will play its role. There are two cases.
Case 1. Suppose lines (ac) and (c′a′) are parallel.
a b c
c′ b′ a′
τ1 τ2
Let τ1 be the translation mapping a to b. Then since (b′c′) = (c′a′) ‖(ac) = (ab) which is the axis of the translation, τ1((b′c′)) = (c′a′).Also, τ1((ab′)) ‖ (ab′) ‖ (ba′), but τ1((ab′)) contains τ1(a) = b. So byAP2, one has τ1((ab′)) = (ba′)In particular, τ1(b′) ∈ (c′a′) ∩ (ba′) and τ1(b′) = a′.One can prove likewise that the translation τ2 taking b to c also takes c′to b′. So the translation τ = τ2τ1 = τ1τ2 takes a to c and c′ to a′. As aconclusion (ac′) ‖ τ((ac′)) = (ca′).We merely used commutativity of composition of translations; i.e., that(F2;−→0 ,+) is an abelian group. So far commutativity of F itself (i.e., ofmultiplication ·) was not required.
Case 2. This is more interesting. Suppose that (ac) and (c′a′) meet at say o.This is the picture we gave earlier, and we shall adapt the translationargument using another tool.Let h1 be the homothety with centre o mapping a to b, and h2 be thehomothety with centre o mapping b to c.Notice that:
b = h1(a) ∈ h1((ab′)) ‖ (ab′) ‖ (ba′)
so h1((ab′)) = (ba′). Moreover, since h1 is a homothety, h1((ob′)) =(ob′) = (oa′). As a conclusion, h1(b′) = a′.Likewise, one proves h2(c′) = b′. As a conclusion,
(ac′) ‖ h2 ◦ h1((ac′)) = (h2h1(a) h2h1(c′)) = (c h2h1(c′))
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But here appears commutativity of F: two homotheties with same centrecommute. So actually h2h1(c′) = h1h2(c′) = a′, and now (ac′) ‖ (ca′),as desired.
Theorem (Pappus’ projective theorem). Let F be a commutative field and workin P2(F).
Let a, b, c be collinear points and c′, b′, a′ be collinear points (being six non-collinear points in total). Then c′′ = (ab′) ∩ (ba′), a′′ = (cb′) ∩ (bc′), andb′′ = (ac′) ∩ (ca′) are collinear.
o a b c
a′b′
c′
a′′
b′′c′′
Proof. Let λ = (c′′a′′) and affinise P with respect to λ; we know P ' A2(F),and in particular is a pappian plane.
But now removing λ from the picture has the effect of sending c′′ and a′′at infinity, so that in P one has (ab′) ‖ (ba′) and (cb′) ‖ (bc′).
We do not know whether the intersection (ac) ∩ (c′a′) is at infinity or not,but in either case, using the affine version, we know that (ac′) ‖ (ca′). Goingback to the projective plane P, this means that b′′ ∈ λ.
Definition (pappian plane). Call pappian an affine or projective plane satis-fying the relevant version of Pappus’ Theorem.
Exercise. Suppose that F is a skew-field such that A2(F) (or equivalently,P2(F)) is pappian. Show that F is commutative.
I.3.4. Pappus Implies DesarguesProposition. Let P be a pappian projective plane. Then P is also arguesian.
Proof. Let (o, a, b, c, a′, b′, c′, a′′, b′′, c′′) be a Desargues configuration; we aimat showing that a′′, b′′, c′′ are collinear. We shall introduce four new pointsand apply Pappus’ Theorem three times. Let:
• i = (ab) ∩ (a′c′);
• j = (oi) ∩ (bc);
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• j′ = (oi) ∩ (b′c′);
• k = (oa) ∩ (bc′).
Claim 1. b′′, j, k are collinear.
Verification. Consider the triples a, i, b and o, c, c′. Notice that they arecollinear by construction and the Desargues assumption.
ai
b
o c c′
Moreover:
• (ac) = (ab′′) and (ic′) = (a′c′) = (a′b′′) so (ac) ∩ (ic′) = b′′;
• (oi) = (oj) and (cb) = (jb) so (oi) ∩ (cb) = j;
• (ao) = (ak) and (bc′) = (bk) so (ao) ∩ (bc′) = k.
By Pappus’ Theorem, we find the desired collinearity.
Claim 2. c′′, j′, k are collinear.
Verification. Now consider the triples i, a′, c′ and o, b, b′. They are also collinearby construction and the Desargues assumption.
ia′
c′
o b b′
Now:
• (ib) = (ab) = (ac′′) and (a′b′) = (a′c′′), so (ib) ∩ (a′b′) = c′′;
• (oa′) = (oa) = (ok) and (bc′) = (bk), so (oa′) ∩ (bc′) = k;
• (io) = (ij′) and (c′b′) = (c′j′) so (io) ∩ (c′b′) = j′.
Again by Pappus’ Theorem, the desired points are collinear.
Claim 3. a′′, b′′, c′′ are collinear.
Verification. Finally consider the triples i, j, j′ and b, c′, k. By construction,they are collinear.
ij
j′
b c′ k
20
But:
• (ic′) = (a′c′) = (a′b′′) and (jk) = (jb′′) by the first claim, so (ic′) ∩(jb′′) = b′′;
• (bj) = (bc) = (ba′′) and (c′j′) = (b′c′) = (b′a′′), so (bj) ∩ (c′j′) = a′′;
• (ib) = (ab) = (ac′′) and (j′k) = (j′c′′) by the second claim, so (ib) ∩(j′k) = c′′.
Pappus’ Theorem gives the conclusion.This is exactly the desired collinearity for the Desargues property. Of course
we treated only the case where all points where distinct. Rigorously speakingone ought to handle the degenerate cases as well, but this is tedious.
There is an affine version but also a striking partial converse.
Proposition. Let P be a finite, arguesian projective plane. Then P is pappian.
Proof. There is no geometric proof known of this apparently geometric prob-lem. We rely on the next day’s major theorem. Let P be arguesian. Thenone may construct a skew-field F with P ' P2(F). Now F is finite, so byWedderburn’s Theorem, F is actually commutative: hence P is pappian.
We repeat that it is an open question (presumably untractable) to give adirect, geometric proof — at some point one would have to do some finite fieldtheory in geometric form.
Remark. Let F be any non-commutative skew-field, for instance the quater-nions H. Then P2(H) is arguesian but not pappian.
End of Lecture 4.
Lecture 5 (Coordinates)
I.4. Coordinatising Arguesian Planes ExplicitlyTheorem (Hilbert coordinatisation). Let P be an arguesian projective plane.Then there is a skew-field F with P ' P2(F).
The proof will be long and quite explicit; next week, using group theory, weshall give more abstract arguments. For the moment we give a corollary anddescribe the strategy.
Corollary. Let P be a projective plane. Then the following are equivalent:
• there is a skew-field F with P ' P2(F);
• P embeds into a projective, 3-dimensional space;
• P is arguesian.
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The proof of the corollary uses the Theorem and results we already know. IfP ' P2(F) then certainly we may embed P into P3(F). The second implicationis what we actually proved for Desargues’ Theorem. And the last is Hilbert’scoordinatisation theorem.
We move to explaining how to prove the theorem.
1. Let P be an arguesian projective plane and fix some line λ. Let A = P(with respect to this line); we know that A is an arguesian affine plane.
2. We fix one line ` of A and define two operations +, · on it. Checking thatthese operations equip ` with the structure of a skew-field F is actuallythe long part of the proof and we shall skip the tedious bits.
3. Once this is done, one may coordinatise A using F, namely A ' A2(F).This is actually not as obvious as it seems, and Desargues is used again.
4. Going up, one finally concludes P ' A ' A2(F) ' P2(F).
We shall not cover (2) entirely, and we shall omit (3). Details are to befound in Hilbert’s Foundations of Geometry, chapter V. The proof starts here.
I.4.1. The Frame
0π`(x)
πm(x)
x
`
m
n
`x
mx
nx
Notation for directions. Let `,m, n be three pairwise non-parallel lines(later we shall require them to be non-concurrent).
For x ∈ A we let:
• `x be the line parallel to ` through x;
• mx be the line parallel to m through x;
• nx be the line parallel to n through x.
Projections. Also construct:
π` : A → `x 7→ ` ∩ nx
Since nx ‖ n is not parallel to `, this makes sense. Define πm : A→ m similarly.Notice that π` is the identity on ` (and similarly for πm).
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The prime function. Now consider:
` → ma 7→ a′ := na ∩m
clearly a bijection between ` and m. For any x ∈ A one has (π`(x))′ = πm(x).Finally let 0 = ` ∩m and notice 0′ = 0.
I.4.2. Addition
0 a b a+ b
a′ pa,b`
m
n
`a′
mb
Given a, b in `, define:
• pa,b = `a′ ∩mb (p is for “plus”);
• a+ b = π`(pa,b).
Proposition. (`; 0,+) is an abelian group.
The proof is a series of lemmas.
Lemma. + is well-defined; for any a ∈ ` one has 0 + a = a+ 0 = a.
Proof. Since ` and m are not parallel and ‖ is an equivalence relation, pa,b isuniquely defined. So is a+b ∈ `. Neutrality is obvious as the construction thendegenerates: `0′ = ` so p0,a = `∩ma = a, and m0 = m so pa,0 = `a′ ∩m = a′,whence a+ 0 = π`(a′) = a.
Lemma. Addition is commutative.
Proof. We may suppose that neither a nor b is 0. We may also suppose a 6= b.Consider the following picture.
0 a b
a′
b′
pa,b
pb,a
qa
qb
`
m
n
`a′
`b′
ma mb
Introduce:
• qa = ma ∩ `a′ ;
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• qb = mb ∩ `b′ .
Claim 1. Points 0, qa, qb are collinear.
Verification. Let r = (0qa)∩`b′ . Consider the configuration (0, a′, b′, qa, r, a, b).
0
a′
b′
qa r
ab
Then (a′a) = na and (b′b) = nb are parallel; also, (a′qa) = `a′ and (b′r) =`b′ are parallel. So by Desargues’s Theorem, (aqa) = ma and (br) are parallel.Therefore also (br) ‖ ma ‖ mb; now the only parallel to mb through b beingmb, one has r ∈ (br) = mb. Since also r ∈ `b′ by construction, it must ber = qb. By construction, 0, qa, qb = r are now collinear.
Claim 2. Lines (pb,a pa,b) and n are parallel.
Verification. Now consider the Desargues configuration(qb, pb,a, b′, qa, 0, pa,b, b).
qb
pb,a
b′
qa0
pa,bb
Then (b′o) = m and (pb,aqa) = ma are parallel, and also (ob) = ` and (qapa,b) =`a′ are parallel, so by Desargues’s Theorem, (b′b) = nb and (pb,apa,b) areparallel.
Since (pb,apa,b) ‖ n, one has npb,a= npa,b
, or in other words, b + a =π`(pb,a) = π`(pa,b) = a+ b.
Actually the proof contains the following abstract bit, which we shall usesoon again.
Lemma. In an arguesian affine plane, let x1, x2, y1, y2, r, s be such that (x1x2) ‖(y1r) ‖ (y2s) and (y1y2) ‖ (x1s) ‖ (x2r).
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If (x1y1) ‖ (x2y2), then (rs) ‖ (x1y1).
x1 x2
y1
y2
r
s
We shall return to this in a minute.
Lemma. + has opposites.
Proof. It is an excellent exercise, in order to understand the notation, to do itwithout a picture. Let a ∈ `; we may assume that a 6= 0. We must find b ∈ `such that a+ b = 0; by commutativity this will suffice. Let q = no∩ `a′ (whichmakes sense as n and ` are not parallel), then b = `o ∩mq.
We claim that a + b = 0. Indeed, pa,b = `a′ ∩mb = `a′ ∩mq = q ∈ no, soa+ b = π`(pa,b) = π`(q) = 0.
Lemma. + is associative.
Proof. Let a, b, c ∈ `; we may suppose that all three are distinct, and distinctfrom 0. Consider the following picture.
a a+ b c b+ c
(a+ b)′
b′
(b+ c)′
pb,apb,c
pb+c,a
pa+b,c
• We know that (pb,a pb,c) = `b′ is parallel to ((a+ b)′ p(a+b),c)) = `(a+b)′ ,which is parallel to ((b+ c)′ pb+c,a) = `b+c.
• We also know that ((a+ b)′ (b+ c)′) = m is parallel to (pb,a pb+c,a) = ma,which is parallel to (pb,c pa+b,c) = mc.
• Finally we know that a + b = b + a, so pb,a ∈ n(b+a) = na+b. Thismeans that ((a+ b)′ pb,a) = na+b is parallel to nb+c = ((b+ c) (b+ c)′) =(pb,c (b+ c)′).
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By our general earlier lemma for arguesian affine planes, we deduce that(pb+c,a pa+b,c) is parallel to n. This means:
(b+ c) + a = π`(pb+c,a) = π`(pa+b,c) = (a+ b) + c
Consequently a+ (b+ c) = (a+ b) + c, and addition is associative.
This completes the study of addition.
I.4.3. MultiplicationWe introduce the new point 1 = `∩n; we need 1 6= 0, and this is why we wanted`,m, n not to concur. We introduce 1′ as well. Now for a, b ∈ ` we define a · bas follows.
0 1 b a · b
1′a′
ta,b
α
αb
Let α = (1a′), then ta,b = αb ∩m (t is for “times”); finally a · b = π`(ta,b).In retrospect we see that in our notation, ta,b = (a · b)′.
Lemma. This is well-defined; for any a ∈ `, 1 ·a = a ·1 = a and 0 ·a = a ·0 = 0.
Proof. Notice that since 1 /∈ m, α is never parallel to m; consequently theconstruction makes sense.
Lemma. Any a ∈ ` \ {0} has a two-sided inverse.
Proof. Here is another exercise in notation: let b = α1′ ∩ `; this makes senseonly since α and ` are not parallel, i.e. only since a′ 6= 0. We can check withouta picture that ta,b = αb ∩m = α1′ ∩m = 1′. But the other exercise is to checkthat tb,a = 1′ as well.
Lemma. · is associative.
Proof. Let a, b, c be on `; we may suppose that all are distinct, and that neitheris 0 nor 1. Let p = a · b. The picture is as follows.
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0 1 b c b · c
a′
b′
p′
(b · c)′
(p · c)′
q1
q2
α αb
πc
nb·cβ
π
βc
Remember that in our notation, p = a · b and π = (1p′). Let q1 = π ∩ nband q2 = πc ∩ nb·c.
Claim 1. Points 0, q1, q2 are collinear.
Verification. Set r = (oq1) ∩ πc. Apply Desargues’ Theorem to the tuple(0, 1, c, b′, (b · c)′, q1, r).
o
b′
(b · c)′
q1 r
1c
Check that:
• all required collinearities hold;
• (1b′) = β ‖ βc = (c (b · c)′);
• (1q1) = π ‖ πc = (cr).
By Desargues’ Theorem, (b′q1) = nb ‖ ((b · c)′r), so r ∈ nb·c and r = q2.
Claim 2. Lines (bp′) and ((b · c) (p · c)′) are parallel.
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Verification. Now apply Desargues’ Theorem to the configuration (o, p′, (p ·c)′, q1, q2, b, (b · c)).
o
p′
(p · c)′
q1 q2
bb · c
• All required collinearities hold (by the previous claim);
• (q1p′) = π ‖ πc = (q2 (p · c)′);
• (q1b) = nb ‖ nb·c = (q2 (b · c)).
So by Desargues’s Theorem again, (bp′) = αb ‖ ((b · c) (p · c)′).As a conclusion,
((b · c) (p · c)′) ‖ α
so ((b · c) (p · c)′) = αb·c and
(p · c)′ ∈ αb·c ∩m = (a · (b · c))′
Projecting, this proves (a · b) · c = a · (b · c).
One should then proceed to proving that · is also distributive on both sides(since it need not be commutative); this is in Hilbert’s book, but fairly tedious.
After it has been checked that (`; 0, 1,+, ·) is a skew-field (and we repeat thatwe skipped the two distributivity laws), one must still coordinatise A, which isnot trivial and uses Desargues again.
At this stage the student should learn to read not lecture notes, but majorbooks such as the celebrated Grundlagen, and we leave the topic.
End of Lecture 5.
Lecture 6 (Duality)
I.5. Duality
I.5.1. One Finitary AspectWe have never been talking about cardinalities; the moment has come.
Proposition. Let P be a finite projective plane. Then there is n such that:
• every line has n+ 1 points;
• every point belongs to n+ 1 lines;
• there are n2 + n+ 1 points in total, and equally many lines.
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We shall prove this counting Proposition in a moment, but before we givean example and a consequence.
Example. Let F be the field with q elements (incidently, q must be a primepower as we know). Then every line P2(F) has as many points as there arevector lines in one vector plane of F3: so we are counting the number of non-zero vectors in say F2 modulo being proportional, finding:
q2 − 1q − 1 = q + 1
points on each line.This section will describe a much better idea than doing the other computa-
tion; which we however perform for the moment. The number of lines througheach point is the number of vector hyperplanes of F3 containing a given vectorline L; projecting F3 � F3/L ' F2 this amounts to counting vector lines in theplane F2, so we find q + 1 again.
The Proposition above tells us this is part of a general phenomenon, not onlyfor P2(F). But we have been using some vector geometry to do the counting.
Corollary. Let A be a finite affine plane. Then there is n such that:
• every line has n ppoints;
• every point is on n+ 1 lines;
• there are n2 points in total;
• there are n2 + n lines, falling in n+ 1 parallel classes of n lines each.
Proof. Count in A using the projective version (still to be proved) then forget.
Remark. The following further remarks are borrowed from Cameron’s notes.
• The integer n as in the proposition is called the order of the projectiveplane P; it is not the number of points in P, but is the most relevant datumsince if P ' P2(F) for some finite field F, then n = |F|.
• There is a Theorem by Bruch and Ryser that if there is a projective planeof order n ≡ 1 or 2[4], then n must be sum of two squares. In particularthere are no projective planes of order 6 nor 14.
• It follows immediately from Hilbert’s Coordinatisation Theorem and thefact that every finite field has prime power order, that there is no arguesianprojective plane of order 10.It can actually be showed by extensive computation that there is no pro-jective plane of order 10 at all. Whether there exists one of order 12 isopen (of course it would be non-arguesian).
Proof of the projective counting Proposition. Let ` be a line and p /∈ `. Toa ∈ ` associate line (ap). This defines a bijection between ` and Lp = {m ∈L : p ∈ m}. In particular, they have the same number of elements. Moreover if`′ is another line not containing p, then composing we find a bijection `→ `′.
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Now given any two distinct lines `, `′, there is a point p /∈ ` ∪ `′, by PP3.As a conclusion any two lines are in bijection. Likewise, given any two pointsthere is a line through neither. We thus have all desired bijections.
Remark. The proof is simple but two ideas could be exploited.
• The idea that one can exchange points and lines will lead us to the dualityprinciple.
• It is also natural to ask about the degree of isotropy of the space, i.e. towonder to which extent we could have just “carried the structure around”from one point to another, and to start studying automorphism groups.
I.5.2. DualityWe return to one of the above remarks: there is a remarkable symmetry in theaxioms when one exchanges points and lines.
Definition (dual statement). Let ϕ be a statement about projective planes. Letϕ∗ be the statement obtained by exchanging the words “point” and “line” (andreversing ∈ into 3 suitably).
We call ϕ∗ the dual statement.
Example. The dual statement of PP1 is PP1∗ = PP2. Likewise, PP2
∗ =PP1, but this is not surprising since ϕ∗∗ = ϕ.
Theorem (duality theorem). Let ϕ be a statement about projective planes.Then ϕ holds in every projective plane iff ϕ∗ does.
One should be careful that a property like being arguesian depends on theprojective plane, so it is not a suitable ϕ. But the property that “Pappus impliesDesargues” is an example of such a statement.
Proof. Since ϕ∗∗ is always the formula ϕ, one implication is enough. We wantto argue by symmetry of the axioms for projective planes. We know thatPP1
∗ = PP2, but we should also do something about PP3.
Claim 1. Let P be an incidence geometry satisfying PP1 and PP2. Then inP, PP3 is equivalent to:
PP3. There are four distinct lines, no three of which are concurrent.
Verification. Suppose PP3. There are four points a, b, c, d no three of whichare collinear. So lines (ab), (ac), (bc), (bd) are four in number and no three ofthem concur. Hence PP3
∗ holds.The converse can be obtained similarly, but we don’t really need it (think
why).Suppose that all projective planes satisfy ϕ. Then by Gödel’s complete-
ness theorem there is a proof of ϕ using only axioms PP1,PP2,PP3. Nowexchange the words “points” and “lines” in the proof: the resulting text is aproof of ϕ∗ using only axioms PP1
∗ = PP2, PP2∗ = PP1, and PP3
∗. Sincethe latter is a consequence of PP1,PP2, and PP3, it holds in any projectiveplane. Therefore so does ϕ∗.
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Exercise. To check whether you have understood the claim, prove that in anincidence structure satisfying PP1 and PP2, the following are equivalent:
• every line has at least three points;
• every point is on at least three lines.
I.5.3. The Dual PlaneBe careful that the Duality Theorem applies only to statements ϕ true in allplanes P. If you have a statement ϕ which is true in one plane P and are lookingfor one plane where P∗ holds, then the following construction is for you.
Definition. Let P = (P,L,∈) be a projective plane. Set P∗ = L, L∗ = P, andtake I∗ to be 3. Then P∗ = (P∗,L∗, I∗) is a projective plane, called the dualplane of P.
By construction, P satisfies ϕ iff P∗ satisfies ϕ∗.
Remark. We may prefer to have membership as the the incidence relation, inwhich case one may try the following variation.
For p ∈ P let Lp = {` ∈ L : p ∈ `}. Now take P∗ = L, L∗ = {Lp : p ∈ P},and I∗ =∈. We claim that this construction is isomorphic to the previous one.
To see it, first use better notation: say P∗1 was the first construction andP∗2 the second. Let f : P∗1 = L → L = P∗2 be the identity map. Now letg : L∗1 = P → {Lp : p ∈ P} = L∗2 take p to Lp. Then of course, lettingp∗1 = ` ∈ P∗1 and m∗1 = q ∈ L∗1, one has:
p∗1I∗1m∗1 ⇔ ` 3 q ⇔ ` ∈ Lq ⇔ f(p∗1) ∈ g(m∗1)
Remark. In general, P 6' P∗. However P ' P∗∗ always and canonically.
Let F be a skew-field. Remember that the opposite skew-field Fop has sameunderlying set, same addition, but multiplication:
a ·op b = b · a
Proposition. Let F be a skew-field. Then (P2(F))∗ ' P2(Fop).
Remark. In particular, whenever F is a field, one has (P2(F))∗ ' P2(F).The converse may fail. For instance, although H is not commutative, i.e. H 6=
Hop, one has H ' Hop (use quaternion conjugation); so (P2(H))∗ ' P2(Hop) 'P2(H).
Proof of the Proposition. Be careful that the proof will look obvious but theordinary duality bracket turns left into right, so it exchanges left-vector spaceswith right-vector spaces. Since on the other hand P2(Fop) is a constructionobtained from left-vector spaces, we need to do something.
Let K = Fop for clarity; notice that right K-vector spaces are exactly leftF-vector spaces, and prepare for the proof. It will also save notation to writePF = P2(F) and PK = P2(K).
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If L ≤ K3 is a (left-K-)vector line, then L⊥ ≤ K3 is a (right-K-)vector plane,so L⊥ ≤ F3 is a (left-F-)vector plane. Likewise, if H ≤ K3 is a (left-K-)vectorplane, then H⊥ ≤ F3 is a (left-F-)vector line. So map:
f : P(PK) → L(PF)L 7→ L⊥
g : L(PK) → P(PF)H 7→ H⊥
These are clearly bijections. Now let (p, `) ∈ P(PK)×L(PK), say p = L ≤ K3
and ` = H ≤ K3. Since orthogonality reverses inclusion:
p ∈ ` ⇔ L ≤ H ⇔ H⊥ ≥ L⊥ ⇔ g(`) 3 f(p)
As a conclusion, PK ' (L(PF),P(PF),3) ' P∗F, as desired.
I.5.4. Dualising Desargues and PappusWe move to self-duality of the configurations we encountered.
Proposition. Let P be a projective plane. Then P is arguesian iff P∗ is.
First proof. One implication suffices since P∗∗ ' P. Suppose P is arguesian.Then using Hilbert coordinatisation, there is a skew-field F with P ' P2(F).Then P∗ ' (P2(F))∗ ' P2(Fop) is arguesian as well.
Second proof. To prove that P∗ satisfies Desargues, we check that P satifiesDesargues∗.
So let ω, α, α′, β, β′, γ, γ′ be lines such that:
• ω, α, α′ are concurrent;
• ω, β, β′ are concurrent;
• ω, γ, γ′ are concurrent.
To prove Desargues∗ is to prove that lines α′′ = (β∩γβ′∩γ′), β′′ = (α∩γα′∩γ′),and γ′′ = (α ∩ β α′ ∩ β′) are concurrent.
We should name these points and a few more; let:
• a0 = α ∩ α′, b0 = β ∩ β′, and c0 = γ ∩ γ′;
• a = β ∩ γ and a′ = β′ ∩ γ′;
• b = α ∩ γ and b′ = α′ ∩ γ′;
• c = α ∩ β and c′ = α′ ∩ β′.
With this notation we now have α′′ = (aa′), and so on.
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Consider the following Desargues configuration (and be careful about theswap on line ω):
a0
c0
b0
b c
b′c′
The various collinearities occur because of lines ω, α, α′. Let us compute theDesargues points.
• (c0b) = γ and (b0c) = β meet at a;
• (c0b′) = γ′ and (b0c
′) = β′ meet at a′;
• (b′b) = β′′ and (c′c) = γ′′ meet at say x.
Since P is arguesian, x ∈ (aa′) = α′′. So α′′, β′′, γ′′ concur at x, as desired.
Theorem. If P is pappian, then P ' P∗ is pappian.
Proof. Coordinatising, since P is pappian one has P ' P2(F) for some com-mutative field F; hence P∗ ' P2(F)∗ ' P2(Fop) = P2(F) ' P, and it is pap-pian.
Remark. There is a geometric proof (without Hilbert coordinatisation) that ifP is pappian then so is P∗: do it as an exercise.
On the other hand I am not entirely sure about a geometric proof that if Pis pappian, then P ' P∗.
Be extremely careful that P ' P∗ and arguesian does not imply that theunderlying skew-field F need be commutative: one only has Fop ' F, as in thequaternions!
End of Lecture 6.
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Chapter II:Group-Theoretic Aspects
Lecture 7 (Coordinatising like adult people (1))
II.1. Arguesianity and the Group of DilationsWe wish to coordinatise using the power of group theory. We remember fromlast week (or feel from experience) that addition is related to translations, andmultiplication to homotheties. In the usual case of A2(F) these generate a groupof interest.
II.1.1. Dilations and TranslationsDefinition (automorphism). Let A be an affine plane. An automorphism of Ais a bijection of A preserving collinearity.
It will be convenient, when working with one f : A → A, to use classical“high-school” notation x′ = f(x).
Definition (the group of dilations). A dilation of A is an automorphism whichtakes any line to a parallel line: ∀a 6= b ∈ A, (a′b′) ‖ (ab). Let D be the groupthey form under composition.
In projective terms, a dilation is an automorphism of A fixing the line atinfinity pointwise.
Of course in A2(F), translations and homotheties are dilations. We capturethe difference with the following general definition.
Definition (translation, homothety).
• A translation of A is either the identity map IdA, or a fixed point-freedilation.
• A homothety of A is a dilation having at least one fixed point (this en-compasses IdA of course).
It is not clear at all how many of these maps there are; we shall study thequestion later.
Proposition. If a dilation f has two distinct fixed points, then f = IdA.
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Proof. Suppose f fixes a and b. Using our convention, this means a′ = a andb′ = b. The proof will introduce two cases, points not on (ab) being easier tounderstand. This kind of case division will be prominent in the lecture.
a′ = a b′ = b
c
d
• Let c /∈ (ab). Then a = a′ ∈ (a′c′) ‖ (ac) since f is a dilation; soby uniqueness in AP2 we know that c′ ∈ (ac) are collinear. Likewisec′ ∈ (bc); these lines meet at c′ = c, which is fixed as well.
• Now let d ∈ (ab). Using our previous c which was fixed under f (thereexists such a point by AP3), since d /∈ (ac) we find that d is fixed too.
So all points are fixed and f = IdA.
Corollary.
• Any dilation is determined by the image of two distinct points.
• A non-identity homothety has a unique fixed point, called its centre.
We let Ha be the set of homotheties fixing a (this encompasses IdA). ClearlyHa is a subgroup of D; but what about translations? We must investigatefurther.
Lemma. If t is a translation 6= IdA then for any a, b ∈ A, one has (at(a)) ‖(bt(b)).
Proof. Suppose there is a counterexample: then (aa′) and (bb′) meet at say c.We prove that c is a fixed point of t, which will contradict t 6= IdA.
a
b
a′
b′
c
Notice that c′ ∈ (a′c′) ‖ (ac) since t is a dilation, and since a, a′, c are collinearwe find c′ ∈ (aa′). Likewise, c′ ∈ (bb′) so actually c′ = c.
Exercise. Prove the converse.
Therefore the following definition makes sense.
Definition (direction). The direction dt of a translation t 6= IdA is the equiv-alence class modulo parallelism of (at(a)) for any a ∈ A. (The direction of IdAis not defined.)
35
Remember that given a point a and line `, we use `a to denote the uniqueline parallel to ` through a.
Corollary. If t 6= IdA is a translation and b /∈ (aa′) then b′ = (ab)a′ ∩ (aa′)b.
a
b
a′
b′
Proof. Since t is a dilation, (a′b′) ‖ (ab). But also, since t is a translation,(bb′) ‖ (aa′). Since b /∈ (aa′), these lines are not parallel: they meet at b′.
Proposition. T E D is a normal subgroup.
Be careful that we claim nothing about abelianity!
Proof. By definition, IdA is a translation; clearly the inverse of one is still one;moreover if (t, h) ∈ T × D then hth−1 has no fixed point. So T is a normalsubset of D and there is only one non-trivial thing to prove: that T is stableunder composition.
So let t1, t2 be translations; we want to show t2t1 ∈ T . We may suppposethat neither of t1, t2 is IdA. To prove that the dilation t2t1 is a translation, weshow that it is either fixed point-free or IdA. So suppose t2t1 has a fixed point.
a
t1(a)t2
b
t1(b)
Observe how:
((t2(b) t2t1(b)) ‖ (b t1(b)) since t2 ∈ D‖ (a t1(a)) since t1 ∈ T‖ (t2(a) t2t1(a)) since t2 ∈ D‖ (t2(b) b) since t2 ∈ T
It follows that t2t1(b) ∈ (b t2(b)).On the other hand, using that t2t1 is a dilation:
(a t2t1(b)) = (t2t1(a) t2t1(b)) ‖ (ab)
so t2t1(b) ∈ (ab). The two lines meet at b: therefore t2t1 fixes not only a, butalso b. Being a dilation, it must be IdA, a contradiction.
Corollary. If two translations coincide at one point, they are equal.
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Proof. If t1(a) = t2(a), then t−12 t1 is a translation with a fixed point, so it is
IdA.
II.1.2. The (p,p)-Desargues PropertyDefinition ((p,p)-Desargues). The (parallel, parallel)-Desargues property (forshort: (p, p)-Desargues) is the following statement: Let (aa′), (bb′), (cc′) be threedistinct parallel lines. Suppose (ab) ‖ (a′b′) and (ac) ‖ (a′c′). Then (bc) ‖ (b′c′).
a a′
b b′
c c′
The property is also called the small Desargues axiom.
Theorem. A has (p,p)-Desargues iff T is transitive on A.
Remark. Since distinct translations never coincide anywhere, there can be atmost one translation t mapping given a to given b. So transitivity is equivalentto sharp transitivity.
Proof. First suppose that T is transitive on A; we prove (p,p)-Desargues. Takea (p,p)-Desargues configuration like on the picture and let t be the translationmapping a to a′. Then we know that t(b) = (ab)a′ ∩ (aa′)b = b′ and t(c) = c′
likewise. Now since t is a dilation, we find (b′c′) ‖ (bc).The converse is more interesting, and will require a few partial maps. Sup-
pose (p,p)-Desargues holds. For any pair of distinct points (a, a′), we define apartial map ta,a′ : A \ (aa′)→ A by ta,a′(b) = (ab)a′ ∩ (aa′)b. Notice that it iswell-defined (but a partial map).
Claim 1. If b /∈ (aa′) and b′ = ta,a′(b), then ta,a′ and tb,b′ agree wherever bothare defined.
Verification. This is exactly the (p,p)-Desargues assumption!This defines a global map ta,a′ : A → A, and obviously ta,a′(a) = a′. We
must check it is a translation. Notice that if ta,a′(b) = b′, then ta,a′ = tb,b′ .
Claim 2. ta,a′ is a dilation.
Verification. Bijectivity is obvious from the following picture.
a a′
x′
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We claim that ta,a′ is an automorphism. Suppose x, y, z are collinear (andwe may suppose they are distinct); we show that so are the images x′, y′, z′.
x x′
y y′
z z′
But we know that ta,a′ = tx,x′ , so actually y′ = (xx′)y ∩ (xy)x′ ∈ (xy)x′and z′ ∈ (xz)x′ = (xy)x′ by collinearity of x, y, z. This proves collinearity ofx′, y′, z′.
We claim that t is a dilation. But we already know that (x′y′) ‖ (xy), sothis is clear.
Finally t is a translation. Otherwise it has a fixed point b.
a a′
b′ = b
Up to changing base point we may suppose b /∈ (aa′). Now (ab) ‖ (a′b′) = (a′b)so b ∈ (aa′), a contradiction.
Corollary. If A has (p,p)-Desargues, then T is abelian and D = T n Ha forany a ∈ A.
Proof. Let t1, t2 be translations. We want to show t2t1 = t1t2; we may assumethat neither is IdA, so both have a direction.Case 1. Suppose the directions are different.
a
t1(a)
t2(a)
Then t2(t1(a)) = (at1(a))t2(a) ∩ (at2(a))t1(a) = t1t2(a).
Case 2. Suppose the directions are the same. The above no longer applies.But there is another direction by AP3; now also using (p,p)-Desargues,there is a translation t0 in this direction. It is clear that translation t1t0has not the same direction as t1 and t2. Therefore, by case 1:
t2t1 = (t2t1)(t0t−10 ) = t2(t1t0)t−1
0
= (t1t0)(t2t−10 ) = t1(t0t2)t−1
0 = t1t2t0t−10
= t1t2
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We turn to proving D = TnHa. We already know T E D and T ∩Ha = {IdA}.Now let d ∈ D be any dilation. By (p, p)-Desargues there is a translation twith t(a) = d(a). Hence t−1d(a) = a and d ∈ T ·Ha.
Notice that abelianity only requires the existence of translations having dif-ferent directions (a consequence of (p, p)-Desargues).
II.1.3. The (c,p)-Desargues PropertyDefinition ((c,p)-Desargues). The (concurrent, parallel)-Desargues property(for short: (c, p)-Desargues) is the following statement: Let (aa′), (bb′), (cc′)be three distinct concurrent lines. Suppose (ab) ‖ (a′b′) and (ac) ‖ (a′c′). Then(bc) ‖ (b′c′).
o
a
a′
b b′
cc′
The property is also called the big Desargues axiom.
Proposition. A has (c,p)-Desargues iff for any line ` and point a ∈ `, Ha actstransitively on ` \ {a}.
Remark. Here again, since an element of Ha already fixes a and a non-identitydilation has at most one fixed point, sharpness is for free.
Proof. Exercise, following the steps of the characterisation of (p,p)-Desargues.
Corollary. If A has (c, p)-Desargues, then it has (p, p)-Desargues.
Proof. The statement is not obvious geometrically. We give an algebraic proof,using the equivalences we know. It suffices to show that there are enoughtranslations. Let a, a′ be distinct points of A.
Let o ∈ (aa′)\{a, a′} (here we must assume that lines have more than threepoints). By assumption, there is g1 ∈ Ho doing g1(a) = a′. Also let b /∈ (ao),and let b′ = (ab)a′ ∩ (aa′)b, then b′′ = g1(b). By assumption, there is g2 ∈ Ha′
doing g2(b′′) = b′.
o a a′
b b′
b′′
We claim that t = g2g1 is a translation taking a to a’. Of course t(a) =g2(a′) = a′, but also t(b) = g2(b′′) = b′. So if t has a fixed point c, then onemust have (ac) ‖ (a′c′) = (a′c), and therefore c ∈ (aa′). But c ∈ (bb′) likewise,and these lines do not meet. Hence t has no fixed point; it is a translation.
39
End of Lecture 7.
Lecture 8 (Coordinatising like adult people (2))
II.2. A Vector Space from an Affine PlaneIn this lecture we finally see how to recover a vector geometry from an affineplane. We start with as few assumptions as we can and add them progressively.
II.2.1. A RingRemember that the direction of a translation t 6= IdA is the equivalence class dtof (at(a)) for any a ∈ A; this is well-defined.
Suppose that T is abelian. Then End(T ) is an associative, non-commutativering with distinguished elements:
0 : T → Tt 7→ IdA
; 1 : T → Tt 7→ t
Addition in End(T ) is given by pointwise composition on A:
(λ+ µ)(t) = λ(t)µ(t)
whereas multiplication is composition in End(T ):
(λ · µ)(t) = λ(µ(t))
Definition. An endomorphism λ : T → T is direction-preserving if: for anyt ∈ T \ {IdA}, one has λ(t) 6= IdA and dλ(t) = dt.
In particular a direction-preserving endomorphism of T must be injective byconstruction.
Definition. Let F = {λ ∈ End(T ) : λ is direction-preserving} ∪ {0}.
Proposition. F is a subring of End(T ).
Proof. Clearly F contains 0 and 1; so it suffices to check stability under +,−, ·.Opposition will be an exercise.
Claim 1. F is stable under +.
Verification. Let λ, µ ∈ F; we show λ+ µ ∈ F. Of course we may assume thatneither λ, µ, nor λ+ µ is 0. Let t ∈ T \ {IdA}. Then by definition λ(t) 6= IdAand µ(t) 6= IdA, but also λ(t)µ(t) 6= IdA as otherwise λ+ µ = 0.
Therefore dλ(t) = dt = dµ(t); clearly this is also dλ(t)µ(t) = d(λ+µ)(t), soλ+ µ is direction-preserving.
Claim 2. F is stable under ·.
Verification. Here again we take λ, µ ∈ F and we may assume that neitherλ nor µ is 0. Then both are injective, and therefore λ · µ 6= 0. Now for anyt ∈ T \ {IdA}, d(λ·µ(t) = dλ(µ(t)) = dµ(t) = dt, so by definition, λ · µ ∈ F.
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II.2.2. A Skew-FieldWe have not used any form of Desargues property so far (only abelianity of T ).
Definition. For h ∈ D any dilation, consider the transformation:
λh : T → Tt 7→ hth−1
Quickly notice that λh is always a group automorphism of T , and direction-preserving (by definition of a dilation): hence λh ∈ F.
Proposition. If A has (p, p)-Desargues, then F is a skew-field.
Proof. We need to prove that any λ ∈ F \ {0} admits a two-sided inverse for ·.The proof entirely relies on a geometric lemma.
Lemma. Suppose that A has (p,p)-Desargues. Let a ∈ A be fixed. Then forall λ ∈ F \ {0} there is a unique h ∈ Ha with λh = λ.
We shall prove the lemma in a minute. However observe that letting thenµ = λh−1 , we have µ ∈ F and λµ = µλ = 1 in F, so our proposition will beproved.
We sketch a further consequence. It then means that we a group homo-morphism Ha � F×; this is actually an isomorphism. For if h ∈ Ha lies in thekernel, then it means that λh = 1, or equivalently, that for any t ∈ T , ht = th.However as soon as t 6= IdA, one gets ht(a) = th(a) = t(a) so h fixes not onlya but also t(a) 6= a; being a dilation it implies h = IdA, as desired.
Therefore the geometric lemma exactly means that for any a ∈ A, there isan isomorphism Ha ' F×.
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Proof of the Lemma.
at λ(t)
ta,x
λ(ta,x)
x
h(x)
By (p,p)-Desargues, for any x ∈ A there is a translation ta,x taking a tox. Define:
h(x) = (λ(ta,x)) (a)
We claim that h ∈ Ha and λh = λ.
Claim 1. h ∈ Ha.
Verification. First notice that h is well-defined since ta,x exists and is unique.Also h(a) = λ(IdA)(a) = IdA(a) = a since λ is a group endomorphism. Butwe have a number of things to check that it is a dilation.
Let y ∈ A be another point. Then ta,y = tx,yta,x since these two transla-tions coincide at a. Therefore, using that λ is a group endomorphism:
h(y) = (λ(tx,y)λ(ta,x)) (a) = λ(tx,y)(h(x))
In particular, if x 6= y, then tx,y 6= IdA and since λ ∈ F \ {0} is direction-preserving, λ(tx,y) 6= IdA, proving h(y) 6= h(x). So h is injective. But also,always since λ is direction-preserving,
(h(x) h(y)) ‖ dtx,y‖ (xy)
An injective map with this property is always a dilation (preservation ofcollinearity and surjectivity being consequences left as an exercise). So h ∈ Dand since we already know h(a) = a, we have h ∈ Ha.
Claim 2. λh = λ.
Verification. Take any t ∈ T ; say t = ta,b for some b. Then λh(t) = hth−1 isthe translation taking a to hth−1(a) = ht(a) = h(b) = λ(ta,b)(a) = λ(t)(a).So λh(t) = λ(t), for any translation; hence λh = λ in F.
So h ∈ Ha and λh = λ. Remember that this proves not only the lemma,but also the proposition.
II.2.3. DimensionalityAt first it is something of a surprise to be able to define a skew-field with asweak a property as (p,p)-Desargues. Now the F-module T is actually a vectorspace; we may think we are quite close to coordinatising, but as a matter of factthe dimension theory of T is not clear at all.
42
Proposition. Suppose A has (c, p)-Desargues. Then:
• for t, t′ ∈ T \ {Id}, one has dt = dt′ iff there is λ ∈ F with t′ = λt;
• T is 2-dimensional.
Proof. Let t, t′ be as in the first claim. If there is λ ∈ F with t′ = λt thenλ 6= 0, so by definition, λ is direction-preserving; hence dt′ = dt.
Now suppose dt′ = dt, and say t = to,a, t′ = to,a′ . By (c, p)-Desargues,there is h ∈ Ho doing h(a) = a′. Then let λ = λh. We know that:
λt = hth−1
= to,hth−1(o)
= to,ht(o)
= to,h(a)
= to,a′
= t′
The first claim being proved we can move to the second. It suffices to showthat two translations with different directions span T . So let t1, t2 be such,with directions say d1, d2. Let t be another translation; we may suppose thatthe direction of t is neither d1 nor d2. Fix a ∈ A and let b = t(a).
a
d1d2
b
c
Then lines (d1)b and (d2)a, not being parallel, meet at say c, which isclearly distinct from a and b. So tc,b has direction d1, and therefore there isλ ∈ F with tc,b = λt1. Likewise, the direction of ta,c is d2, whence µ ∈ F withta,c = µt2. Now t = ta,b = tc,b ◦ ta,c = (λ(t1)) ◦ (µ(t2)) as transformations ofA, which in vector space notation rewrites: t = λ · t1 + µ · t2.
We are done; the isomorphism is easily constructed now. For let p1 6= p2 betwo points in A. Then for anny q ∈ A one has equivalences:
q ∈ (p1p2) iff q = p1 or (p1q) ‖ (p1p2)iff q = p1 or dtp1q = dtp1p2
iff ∃λ ∈ F tp1,q = λtp1,p2
So lines are now given in parametric form. We have retrieved the F-vector spacestructure underlying the affine plane A and coordinatising is easy.
End of Lecture 8.
Lecture 9 (The Geometry of SO3(R))
43
II.3. An Application: SO(3, R)We consider the group SO3(R) = {M ∈ GL3(R) : M · tM = I3}, with propersubgroup
B ={(
1M
): M ∈ SO2(R)
}< SO3(R)
which is isomorphic to SO2(R) (the circle group).Recall that an involution of G is an element x 6= 1 with x2 = 1; in the case
of SO3(R) these elements are also called half-turns of the space R3. We let I bethe set of involutions of G.
II.3.1. The Geometry and Non-Degeneracy AxiomsLet G be any group equipped with a subgroup B ≤ G; we also let I be the setof involutions of G. On G we define an incidence geometry as follows:
• points are elements of G;
• lines are sets of the form gBh for g, h ∈ G;
• planes are sets of the form gI for g ∈ G.
Remark. Notice at once that since gBh = gh · h−1Bh, the lines are exactlythe translates of the conjugates of B.
Moreover, a line contains 1 iff there is b ∈ B with 1 = gbh, so actuallyg−1h−1 = b and hg = b−1, meaning gBh = gBg−1. Of course such a linecontains 1, and as a conclusion, lines through 1 are exactly the conjugates of B.Such lines are parametrised by G/NG(B).
Two lines g1Bh1 and g2Bh2 are equal iff g−12 g1 and h1h
−12 lie in NG(B)?
Two planes g1I and g2I are equal iff g−12 g1 is inverted by all involutions in
G.
We shall prove that in the case of SO3(R) with our choice of B, this geometryis a projective 3-dimensional space. Return to the axioms: six must be proved,and we begin with PS5 and PS6 which are almost obvious. This will give usthe opportunity to understand the definition better.
Lemma. PS6 is equivalent to: B has order at least three.Suppose that B contains at most one involution, and that G contains no
copy of (Z/2Z)3. Then G satisfies PS5.PS5 and PS6 are satisfied in SO3(R).
Proof. First, since every line is in bijection with B ' SO2(R), every line hascontinuum-many points: so PS6 is clear. Now we take the unit and threeinvolutions, say:
i =
1−1
−1
, w =
−11−1
, s =
−1−1
1
For easy practice, we check that these four points satisfy PS6. First, if they arecoplanar, then there is g ∈ G with 1, i, w, s ∈ gI: so g must be an involution.
44
But then gi, gw, gsmust be involutions as well, and we recall from group theorythat the product of two involutions is an involution iff they commute. So gmust be an involution commuting to i, w, s, which form a Klein 4-group. HenceG contains an elementary abelian 2-group of order 8. However it is quicklycomputed in SO3(R) that CG(i, w, s) = 1, a contradiction.
Now suppose that three of them are collinear. There are two cases.
• Suppose we took 1 and two of the three involutions; we may assumethat 1, i, w are collinear and will derive a contradiction. The definitionsays that there are g, h ∈ G with 1, i, w ∈ gBh. Then we may assumegBh = gBg−1 contains i and w. But gBg−1 ' B ' SO2(R) contains aunique involution, a contradiction.
• Suppose we took the three involutions i, w, s. The definition now saysthat i, w, s ∈ gBh for some g, h; now there are b, b′ ∈ B such that i = gbhand w = gb′h. Then:
s = iw = i−1w = h−1b−1g−1gb′h = h−1b−1b′h ∈ h−1Bh
and w ∈ h−1Bh likewise. But h−1Bh contains a unique involution: thisis a contradiction finishing the proof.
II.3.2. The First AxiomLemma.
• In any group where the geometry was defined in a similar fashion, PS1 isequivalent to: G =
∐G/NG(B) gBg
−1.
• The above property holds in SO3(R).
Proof. Remember that PS1 is the axiom that through any two distinct pointsx, y ∈ G there is a unique line. Using left-translation this is equivalent to sayingthat z = x−1y lies in a unique line also containing 1. So PS1 is equivalent tosaying that any z 6= 1 lies in a unique conjugate of B, which is exactly the setequation we wrote.
The equivalence holds for general G where lines are of the given form. Wenow prove that SO3(R) enjoys this property for chosen B.
Existence. Let z ∈ G = SO3(R). Since the dimension is 3, z has as aneigenvalue λ; since z ∈ O3(R), λ = ±1.Let L be an eigenspace of dimension 1 for λ and P = L⊥; since x ∈O3(R), x restricts to P and x|P ∈ O(P ) ' O2(R). In matrix form, thismeans that there is an orthonormal basis B in which x has matrix
x′ =(λ
M
)with λ = ±1 and M ∈ O2(R). The base change matrix associated to Bis orthogonal, so x and x′ are conjugate in O3(R). But notice that up tochanging the order of the last two vectors of B, we may assume that B is
45
a direct orthonormal basis. So x and x′ are conjugate in SO3(R). Thereare two cases.If λ = 1, then using determinants we see M ∈ SO2(R): x is SO3(R)-conjugate to an element in B and we are done. If on the other hand λ =−1, then M ∈ O2(R) \ SO2(R); we know that such M is an orthogonalreflection, so up to changing basis again x is conjugate to−1
1−1
= w
which is easily conjugate to i ∈ B.
Uniqueness. Now suppose that gBg−1 and hBh−1 meet at some x 6= 1.Notice that B = StabG(e1) in the standard action of SO3(R) on R3. Sox ∈ Stab(ge1, he1).If ge1 and he1 span a vector plane P , then x|P = IdP ; also, since x ∈SO3(R), x restricts to the line L = P⊥ and must be in SO(L) = {IdL}.Therefore x = I3, against the definition.Therefore ge1 and he1 are collinear, say ge1 = λhe1; since g, h ∈ SO3(R)preserve norms, actually λ = ±1. If ge1 = he1 then h−1g ∈ StabG(e1) =B ≤ NG(B), so certainly gBg−1 = hBh−1, as desired. If on the otherhand ge1 = −he1, then we consider our involution w again, which hadthe effect that we1 = −e1. So now wh−1ge1 = e1 so wh−1g ∈ NG(B).But one can see that w inverts B, so w ∈ NG(B), and here again h−1g ∈NG(B), as desired.
II.3.3. The Second AxiomLemma. Let G be a group with involutions, in which the geometry is definedas above.
If no element 6= 1 is inverted by all involutions in G, then planes gI, hI areequal only if g = h.
PS2 is equivalent to: any two elements which do not lie in a common con-jugate of B are inverted by a unique involution.
This holds provided: G has PS1 and
Proof. Suppose gI = hI. Then for any involution i (and there is one, since oth-erwise every element is inverted by all involutions in G), there is an involutionj with gi = hj. Now h−1g = ji, but oddly enough:
i · ji · i−1 = ij = i−1j−1 = (ji)−1
so i inverts ji = h−1g, and i was arbitrary. The assumption implies that h = g.Recall that PS2 is the statement that any three non-collinear points lie
in a unique plane. With our definition, this means that for any three non-collinear points a, b, c there is a unique Left-translating, this means that forany two points x, y not collinear with 1, there is a unique element (necessarilyan involution since jI must contain 1) j such that x, y ∈ jI.
46
As we know, being collinear with 1 means belonging to a common conjugateof B. So PS2 is now equivalent to: any two elements not lying in a sameconjugate of B are inverted by a unique involution.
We prove that this condition holds. Let g, h not be in a common conjugateof B. But using PS1 each lies in some (unique) conjugate of B, so conjugating,PS2 is equivalent to:
Notice that the set of involutions inverting B is exactly NG(B) \B.
Commutation distance. If NG(B) = B o 〈w〉 where w inverts B (and B is2-divisible), and commutation distance is 2, then that should work.
Here it is the case: if j 6= k are involutions, they are half-turns of the spacewith axes Lj 6= Lk. We claim that j and k commute iff Lj ⊥ Lk.
II.3.4. The Third and Fourth AxiomsLemma. PS3 is equivalent to: G = B · I.
The above holds in SO3(R).
Proof. Literally speaking, PS3 means that any line xBy intersects any planegI. But translating this is equivalent to saying that any line g−1xBy intersectsI, i.e. that any coset of a conjugate of B contains an involution. Now I isinvariant under conjugation, so this is equivalent to saying that any translategB of B contains an involution j, or again: for all g ∈ G there is (b, j) ∈ B×Jsuch that g = jb.
We show that this is the case in SO3(R). Let g ∈ SO3(R); if ge1 = −e1then we find g ∈ wB and we are done. Otherwise the bisector L of the angle(e1, ge1) is well-defined. Let j be the half-turn with axis L. Then clearlyje1 = ge1, and this means jg ∈ B or equivalently g ∈ jB, as desired.
Lemma. PS4 is equivalent to: [NG(B) : B] = 2 and every element in NG(B)\B inverts B +
If B is inverted by some involution and conjugates of B cover G, then thisis true.
The above holds in SO3(R).
Proof. Left-translating as usual, PS4 is equivalent to: any plane contains aline made of involutions.
Let g ∈ G. Then g belongs to a conjugate B′ of B inverted by someinvolution say j. Then jB′ ⊆ I ∩ g · I since: jB′ consists of involutions, andg−1jB′ = jgB′ = jB′.
Note that if PS4 holds then there is a line ` ⊆ I. Conjugating ` is of theform gB for some g ∈ G. So gB consists of involutions.
Ajouter : B n’est pas formé d’involutions.
Exercise. Axiomatise better than I did the situation in which the projectivespace method will apply.
Corollary (Nesin). The field R is definable in the group SO3(R).
Proof. The field obtained by coordinatisation is present definably. It also isisomorphic to R.
47
Corollary (Nesin). A bad group of Morley rank 3 has no involutions.
Proof. If it did, then one could using the same geometry recover axioms PS1to PS6. This would create a definable skew-field inside G ' P3(K); generalmodel theory implies that K is actually a commutative, algebraically closedfield. Actually in its conjugation action on I, G acts by (projective) automor-phisms, so G acts as an automorphism group of I ' P2(K). It is known thatAut(P2(K)) ' PGL3(K) n Aut(K), so by simplicity G embeds into PGL3(K).There must be something elegant.
See Nesin’s Non-solvable groups of Morley rank 3 in J. Algebra.
End of Lecture 9.
Lecture 10 (Projectivities)
II.4. Projectivities
II.4.1. ProjectivitiesWe return to the map `→ m through point p.
Definition (perspectivity). Let ` 6= m be two lines and p /∈ ` ∪m. The per-spectivity ` −→
pm is the map taking a ∈ ` to a′ = (ap) ∩m.
` m
a
a′
p
Definition (projectivity). A projectivity from ` to m is any composition ofperspectivities ` −→
a1`1 . . . −→
an
`n = m.If m = ` the map is called a self-projectivity of `.
Let G` be the group of self-projectivities of `. Notice that since there isalways a perspectivity between two lines, G` and Gm are always isomorphic.
Proposition. G` acts 3-transitively on `.
Proof. Let f1, f2 be two distinct points on `: we show that the stabiliser of f1and f2 in G` acts transitively on ` \ {f1, f2}. Let x and y be points there.
Let m1 6= ` be a line through f1, and m2 6= ` be a line through f2. Chooseany p ∈ m1 \ `. Now let qx = (px) ∩ m2 and qy = (py) ∩ m2. Finally, letr = m1 ∩m2. The picture is as follows.
48
p
f1 f2 x y
r
qx
qy
m1m2
`
We consider the two perspectivities ` −→qx
m1 and m1 −→qy
`. One shouldfollow on the picture that they have the following effect:
` −→qx
m1 −→qy
`
f1 7→ f1 7→ f1f2 7→ r 7→ f2x 7→ p 7→ y
So the resulting self-projectivity of ` fixes f1 and f2, and takes x to y.
II.4.2. Group-theoretic Interpretation of Pappus’ TheoremRecall that G` is always 3-transitive on `. Sharpness actually characterises aknown configuration.
Proposition. Let P be a projective plane. Then G` is sharply 3-transitive on `iff P is pappian.
Proof. Suppose P is pappian. Then we know it is arguesian; using Hilbertcoordinatisation, there is a skew-field F with P ' P2(F). Since it is pappian,we also know that F is a commutative field. Now the action of G` on ` isequivalent to that of PGL2(F) on F ∪ {∞}, which is known to be sharply3-transitive.
The converse must be of a purely geometric nature. Let (o, a, b, c, a′, b′, c′)be a Pappus configuration. Let a′′ = (bc′) ∩ (cb′), b′′ = (ac′) ∩ (ca′), andc′′ = (ab′)∩(ba′). We want to show that a′′, b′′, c′′ are collinear. Set `′′ = (b′′c′′);we shall prove a′′ ∈ `′′.
We introduce two auxiliary points:
• q = (aa′) ∩ `′′;
• z = `′ ∩ `′′.
49
o a b c
a′b′
c′
`
`′
a′′
b′′c′′
qz
`′′
First consider the following projectivity and its effect:
` −→a′
`′′ −→a
`′
a 7→ q 7→ a′
b 7→ c′′ 7→ b′
c 7→ b′′ 7→ c′
o 7→ z 7→ z
We need two more auxiliary points, which we shall not show on picture:
• α′′ = (b′c) ∩ `′′;
• γ′ = (bα′′) ∩ `′.
Next consider this other projectivity (undefined point x plays no real role,since (xb) = (bb′)):
` −→b′
`′′ −→b
`′
a 7→ c′′ 7→ a′
b 7→ x 7→ b′
c 7→ α′′ 7→ γ′
o 7→ z 7→ z
Now the assumption that G` is sharply 3-transitive on ` immediately im-plies that for given lines m,m′, there is a unique projectivity m → m′ takinga distinct triple of m to one of m′. This means that γ′ = c′.
Hence α′′ ∈ (bγ′) = (bc′) and α′′ ∈ (b′c) so α′′ = a′′ lies on `′′ = (b′′c′′), asdesired.
End of Lecture 10.
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