Affine and Projective Geometry

Home , Affine plane, Collinearity, Incidence geometry, Pappus configuration

Aﬃne and Projective Geometry

Adrien Deloro

Şirince ’17 Summer School Contents

Table of Contents ii

List of Lectures iii

Introduction1

Week I: Affine and Projective Planes2 I.1 Affine planes ...... 3 I.1.1 The Definition ...... 3 I.1.2 Membership as Incidence ...... 3 I.1.3 The Affine Plane over a Skew-field...... 4 I.2 Projective planes...... 7 I.2.1 The need for Projective planes ...... 7 I.2.2 The Definition ...... 8 I.2.3 From Affine to Projective, and Back...... 9 I.3 Arguesian Planes...... 12 I.3.1 Desargues’s Projective Theorem ...... 12 I.3.2 Free Planes and Non-Arguesian Planes...... 15 I.3.3 Pappus’ Theorem...... 17 I.3.4 Pappus Implies Desargues...... 19 I.4 Coordinatising Arguesian Planes Explicitly ...... 21 I.4.1 The Frame ...... 22 I.4.2 Addition...... 23 I.4.3 Multiplication...... 26 I.5 Duality ...... 28 I.5.1 One Finitary Aspect...... 28 I.5.2 Duality ...... 30 I.5.3 The Dual Plane...... 31 I.5.4 Dualising Desargues and Pappus...... 32

Week II: Group-Theoretic Aspects 34 II.1 Arguesianity and the Group of Dilations...... 34 II.1.1 Dilations and Translations...... 34 II.1.2 The (p,p)-Desargues Property...... 37 II.1.3 The (c,p)-Desargues Property...... 39 II.2 A Vector Space from an Aﬃne Plane...... 40 II.2.1 A Ring ...... 40 II.2.2 A Skew-Field...... 41 II.2.3 Dimensionality...... 42

i II.3 An Application: SO(3,R) ...... 44 II.3.1 The Geometry and Non-Degeneracy Axioms ...... 44 II.3.2 The First Axiom...... 45 II.3.3 The Second Axiom...... 46 II.3.4 The Third and Fourth Axioms ...... 47 II.4 Projectivities ...... 48 II.4.1 Projectivities...... 48 II.4.2 Group-theoretic Interpretation of Pappus’ Theorem . . . 49

ii List of Lectures

Week I Lecture 1 (Introduction; Aﬃne Planes)...... 2 Lecture 2 (Projective Planes)...... 7 Lecture 3 (Desargues’ Theorem) ...... 11 Lecture 4 (Pappus’ Theorem)...... 17 Lecture 5 (Coordinates)...... 21 Lecture 6 (Duality)...... 28

Week II Lecture 7 (Coordinatising like adult people (1))...... 34 Lecture 8 (Coordinatising like adult people (2))...... 40 Lecture 9 (The Geometry of SO3(R))...... 43 Lecture 10 (Projectivities)...... 48

iii Introduction

If parallel lines meet at inﬁnity, then all trains for the inﬁnite are 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, equations of 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 out of print but on the internet one ﬁnds a modern typeset. [4] Hilbert, David. Foundations of Geometry. Only Chapter 5 on Desargues’ Theorem.

1 Chapter I: Affine and Projective Planes

Lecture 1 (Introduction; Affine Planes)

The Quartet and the Orchestra. We begin with an analogy. Beethoven’s Ninth Symphony (op. 125) uses an extensive orchestra and a choir. After com- pletion Beethoven quickly returned to composing string quartets (op. 130–132 and 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 on the other, is significant. None is “better” music than the other: the symphony targets universality, whereas the quartets aim at spirituality and purity. And mathematics is like music. There are various layers of mathematical structures, from elementary aspects to the whole orchestra of modern mathematics. And none is better for the true mathematician will enjoy both. For instance when studying the real plane R2 we have very different notions and 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 where one 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 to be on some lines or not. But of course for this to be interesting one must start with intuition, or physical observation in our small part of the universe (a sheet of paper can do), in order to ﬁnd axioms which describe the behaviour.

2 I.1. Aﬃne planes

I.1.1. The Deﬁnition

Definition (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 a line or not;

subject to the following axioms AP1, AP2, AP3 listed hereafter (we shall have to pause for auxiliary deﬁnitions).

2 AP1. ∀(a, b) ∈ P a 6= b → (∃!` ∈ L aI` ∧ bI`). (The meaning is clear: given any two distinct points, there is a unique line incident 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 no point incident to both:

` k m iﬀ (` = m) ∨ (∀a ∈ P¬(aI` ∧ aIm))

AP2. ∀(a, `) ∈ P × L ∃!m ∈ L (aIm) ∧ (` k m). (Intuitive meaning: through any point there is a unique line parallel to a given line. It is convenient to call this line `a.)

Deﬁnition (collinear). Points a1, . . . , an are collinear if there is a line ` such that each akI`.

AP3. There exist three non-collinear points.

Axioms AP1, AP2, AP3 deﬁne aﬃne planes. Notice that we speak only about points and lines; words such as angle, distance, or even between are banned from 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 Incidence Notice that we are using an abstract relation I; however we are used to thinking that 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 relation is the usual membership. We need to give a precise meaning to “similar”.

3 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 incident to at least two points. Then there is an affine plane A0 = (P0, L0, ∈) (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 of points. To each ` ∈ L associate S` = {a ∈ P : aI`}, a subset of P. Now 0 consider L = {S` : ` ∈ L}, a family of subsets of P. Finally let f be the 0 identity function on P and g : L → L map ` to S`. We claim that (P, L,I) and (P0, L0, ∈) are isomorphic via f and g. Clearly f is a bijection; g is clearly onto but it also is one-to-one. For if S` = Sm then since there are at least two distinct points a, b ∈ P incident to `, one has a, 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 definition of 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 use the common phrases: “a is on `”, “a belongs to `”, “` contains a”.

I.1.3. The Affine Plane over a Skew-field We wrote the definition of an affine plane by painting from nature. It is time to 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}

2 −→ −→ where p ∈ R 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 = ` −→ and `2 = ` −→ are equal iﬀ v1 and v2 are −−→p1,v1 p2,v2 proportional vectors, and p1p2 is a multiple of those.

Veriﬁcation. First suppose `1 = `2. Since p2 ∈ `1 there is t ∈ R with p2 = −→ −−→ −→ −→ p1 + tv1. So p1p2 = tv1 is a multiple of v1. But also p1 ∈ `2 so there is s ∈ R −→ with p1 = p2 + sv2; now −→ −→ −→ p1 = p2 + sv2 = p1 + tv1 + sv2 −→ which simpliﬁes into t−→v + s−→v = 0 . Since we know that neither −→v nor −→v is −→ 1 2 1 2 0 , we deduce that they are proportional.

4 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 + tv1. Now by −−→ −→ −→ −→ hypothesis there are λ, µ ∈ R with p1p2 = λv2 and v1 = µv2 (we even know that µ 6= 0). Then: −−→ −→ −→ q = p2 − p1p2 + tv1 = p2 + (tµ − λ)v2 ∈ `2

−→ −→ −→ −→ Claim 2. Two lines `1 = `p1,v1 and `2 = `p2,v2 are parallel iﬀ v1 and v2 are proportional vectors. −→ −→ Veriﬁcation. Suppose `1 k `2; we wish to show that v1 and v2 are proportional vectors. If `1 = `2 then we know the conclusion from the previous claim. So suppose `1 6= `2; since they are parallel this means that there is no point on both lines. Hence the equation: −→ −→ p1 + t1v1 = p2 + t2v2

2 has no solution for (t1, t2) ∈ R . −→ ai Work in coordinates, say pi = (xi, yi) and vi = . We just understood bi that the system of equations:

x + t a = x + t a (S): 1 1 1 2 2 2 y1 + t1b1 = y2 + t2b2

2 has no solution for (t1, t2) ∈ R . Notice that multiplicative coeﬃcients are on the 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 −1 there is a unique solution t2 = (x1 − x2) · a2 to the ﬁrst equation; now t2 being known there is a unique solution in t1 to the second. This is a contradiction 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 the −1 right by λ = a1 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 gives rise to a (unique) solution in t1: so (S) has a solution, a contradiction. −1 Therefore b2 = a2λ = a2a1 b1; now clearly −→ a2 −1 a1 −1−→ v2 = −1 = a2a1 = a2a1 v1 a2a1 b1 b1 −→ −→ so v1 and v2 are proportional vectors.

5 2 Claim 3. A (R) satisﬁes axioms AP1, AP2, AP3. Veriﬁcation. −→ −→ −−→ −→ AP1. Let p1 6= p2 be points. Then v = p1p2 6= 0 and ` = `p1, v clearly contains 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 contains p 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 the associativity law (λµ)ν = λ(µν). But we never used the commutativity law λµ = µλ. This is why we insisted on the difference between left and right; check it 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 a structure (F; 0, 1, +, ·) where all axioms of a field are supposed to hold, except that · 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 checked or known that H is a skew-field, but not a field. An extremely important phenomenon is however the following, which we shall 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 and lines, is an affine plane. Example. Here is the smallest affine plane:

a b

c d

One point is removed: diagonals do not meet. 2 We recognize A (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 ourselves that diagonals should not meet indeed, since both are directed by the same vector: 1 0 1 1 0 − = = − 1 0 1 0 1 Diagonals are parallel!

6 Remark. We started the class by promising to use only points and lines, and to even forget about vectors and coordinates. However we have just spent a lot of 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 by Hilbert, is surprinsingly beautiful. Slightly abusing set-theoretic notation, we have so far:

2 {A (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 planes Below is Desargues’s celebrated theorem. We give the full aﬃne statement with numerous cases, and will prove it only later, when we have the nice, uniﬁed, projective statement. The purpose of giving such a long case division being precisely 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 and work in A2(F). Let abc and a0b0c0 be triangles (involving six distinct points). Suppose that lines (aa0), (bb0), (cc0) are three distinct lines, and either that they are parallel or that they meet at the same point. Then: • either (ab) k (a0b0), (ac) k (a0c0), and (bc) k (b0c0);

• or (ab) k (a0b0) but (ac) ∩ (a0c0) = {b00} and (bc) ∩ (b0c0) = {a00} satisfy (a00b00) k (ab); • or two similar cases (involving the intersection c00 of (ab) and (a0b0)); • or a00, b00, c00 are collinear.

We hope that the complexity of the statement strongly suggests to unify cases by making parallels intersect. The power of this idea was acknowledged and fully exploited in the Renaissance. Below is a picture of the most striking

7 case.

a c0 c o

b b0

a00 c00 b00

It is a good exercise, in order to get familiar with the statement, to draw the other cases (here, no parallelism is involved).

I.2.2. The Deﬁnition Deﬁnition (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 shall return in another lecture. We move to the main class of examples. Remember that a vector line is a 1-dimensional vector subspace, which can be viewed as a line 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 relies on linear algebra in dimension 3.

8 PP1. Let α 6= β be two projective points, i.e. two distinct vector lines 3 3 3 Lα,Lβ ⊆ F . Then set H = hLα,Lβi ⊆ F , a vector plane of F . In projective terms, H is a line λ. Now by deﬁnition, Lα ⊆ H translates into αIλ, and βIλ likewise. Uniqueness is clear as well, since there is no other vector plane containing both Lα and Lβ.

PP2. Now start with two distinct projective lines λ 6= µ. By construction they correspond to distinct vector planes Hλ 6= Hµ, and we know that Hλ ∩ 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) deﬁne suitable projective points.

I.2.3. From Affine to Projective, and Back Another way to obtain projective planes is by “completing” affine planes, a procedure we now explain. In the affine world, some intersections are missing: if ` k m are distinct, parallel affine lines, then we should add a point. But if ` k m k 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 working with pairs of parallel lines, but with sets of parallel lines. Fortunately there is a mathematical tool for that. Lemma. Let A be any affine plane. Then k 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 a general proof, a proof which will use only axioms AP1, AP2, AP3. Remember that in this abstract setting ` and m are parallel iff ` = m or ` and m do not meet. We freely replace A by an isomorphic plane where incidence is given by membership. There are three things to check: Reflexivity. Every line ` satifies ` = `, so ` k `.

Symmetry. If ` k m then either ` = m or ` ∩ m = ∅; in either case we have m k `. Transitivity. Suppose ` k m and m k n; we show that ` k n. If ` = n we are done. Otherwise we want to show ` ∩ n = ∅. So suppose not; by AP1 there is a ∈ ` ∩ n. Now by AP2, there is a unique line parallel to m and containing a; however this applies to both ` and n. So actually ` = n, a contradiction showing that ` ∩ n = ∅. Hence ` k n, as desired.

The equivalence class [`] of a line ` can be thought of as its direction. So it 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 too short: we must force a projective line to contain the direction. This explains why we go through `ˆ below.

9 Definition (projectivisation Aˆ of an affine plane). Let A = (P, L, ∈) be an ˆ affine plane. For each ` ∈ L, let ` = ` ∪ {[`]}. Now let `∞ = {[`]: ` ∈ L}, ˆ Pˆ = P ∪ `∞, and Lˆ = {` : ` ∈ L} ∪ {`∞}. We call Aˆ = (Pˆ, Lˆ, ∈) the projectivisation of A. Proposition. For any affine plane A, Aˆ is a projective plane. Proof. There are three axioms to check.

PP1. Let α 6= β be two points in Pˆ. We see several cases.

• If α = a ∈ P and β = b ∈ P, then we know from AP1 that there is a unique line ` ∈ L containing both. Clearly α, β ∈ `ˆ. We also have uniqueness. If another line λ ∈ Lˆ contains α and β, then λ cannot be `∞, so λ =m ˆ for some aﬃne line m ∈ L. Then a, b ∈ m, so by ˆ ˆ uniqueness in AP1 we have m = ` and therefore λ =m ˆ = `. So ` is the only line of Lˆ containing α and β.

• Suppose α = a ∈ P but β ∈ P\Pˆ = `∞ (the other case is similar). Then by deﬁnition, there is ` ∈ L with β = [`]. Now by AP2 there is a unique m ∈ L with a ∈ m and m k `. Then on the one hand α ∈ mˆ , and on the other hand [`] = [m] ∈ mˆ . Now to uniqueness. If a line λ contains a and β, then it cannot be `∞. So it is of the form nˆ for some aﬃne line n. Now α ∈ nˆ implies a ∈ n, and β = [`] ∈ nˆ implies [`] = [n], that is, ` k n. By uniqueness in AP2 we have n = m, and therefore λ =n ˆ =m ˆ .

• Now suppose α, β ∈ `∞. Clearly `∞ is the only line in Lˆ incident to 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 involves choosing the line to be removed.

Definition (affinisation Pˇ of a projective plane). Let P = (P, L, ∈) be a projective plane and fix one λ ∈ L. Let Pˇ = P\ λ; for each µ ∈ L \ {λ}, set µˇ = µ \ λ, and let Lˇ = {µˇ : µ ∈ L \ {λ}}. We call Pˇ = (Pˇ, Lˇ) the affinisation of P with respect to λ. Proposition. For any projective plane P and line λ ∈ L, Pˇ is an affine plane. Moreover Pbˇ ' P. Proof. Exercise.

Be extremely careful though that the isomorphism type of Pˇ depends on the ˇ line λ you chose to remove. In particular, Aˆ depends on the choice of line we remove from Aˆ, and need not be isomorphic to A. Remark. So far the picture looks like this: n o n o A\2(F): F a skew-ﬁeld ⊆ Aˆ : A an aﬃne plane ⊆ {projective planes};

10 2 {P (F): F a skew-ﬁeld} ⊆ {projective planes} Actually applying the projectivisation procedure to A2(F) gives what we expect.

Proposition. Let F be a skew-ﬁeld. Then A\2(F) and P2(F) are isomorphic.

3 Proof. We shall construct an isomorphism. Let H0 ≤ F be a vector plane 3 and x ∈ F be a vector not in H0. Let H1 = x + H0, an affine translate of H0. 2 2 Clearly A (F), H0, and H1 are isomorphic affine planes; in particular A\(F) and Hc1 are isomorphic projective planes. So it suffices to see that Hc1 and P2(F) are isomorphic, and this is easy thanks to elementary geometry.

a x

H1 x + L0

H0 L0

3 Let L ≤ F be a vector line. If L 6≤ H0, then L will intersect H1 in a point say a; map L to a. If L ≤ H0, then there is no intersection; we feel that L should be mapped to a point at infinity. In our previous construction, such were 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 the ˆ affine plane H1; we map H to ` as in the projectivisation construction. If on the other hand H = H0 then we map H to `∞. It is an exercise to finish the proof:

2 • check that we have a bijection between points of P (F) and points of Hc1;

2 • check that we have a bijection between lines of P (F) and lines of Hc1; • check that these bijections preserve incidence.

End of Lecture 2.

Lecture 3 (Desargues’ Theorem)

11 I.3. Arguesian Planes

We have introduced projective geometry in order to unify case divisions. Today we return to Desargues’ Theorem.

I.3.1. Desargues’s Projective Theorem

Theorem (Desargues’ projective theorem). Let F be a skew-field, and work in P2(F). Suppose abc, a0b0c0 are two triangles such that (aa0), (bb0), and (cc0) meet. Then a00 = (bc) ∩ (b0c0), b00 = (ac) ∩ (a0c0), and c00 = (ab) ∩ (a0b0) are collinear. Notice that by suitably choosing the line to be removed, this proves all versions for A2(F). And now, thanks to projective geometry, only one case must be checked. There are proofs in the plane (using coordinates of course) but we shall introduce a powerful method. Imagine for a moment that the configuration is no longer “flat”, but that triangles (abc) and (a0b0c0) are not in the same plane; say (abc) is in some plane π and (a0b0c0) in π0 6= π.

a c0

b b0

a00 c00 b00

Then observe how a00 ∈ (bc) ⊆ π; likewise, b00, c00 ∈ π. But similarly, a00, b00, c00 ∈ π0. Since we expect two planes to intersect over a line, this implies collinearity of a00, b00, c00. 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.

12 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 lines in F4, lines are 2-dimensional vector subspaces of F4, and planes are vector hyperplanes of F4, is a projective space. Of course one can always replace an abstract projective space by one where incidence relations are of the form p ∈ ` ⊆ π, which makes notation and terminology a lot easier.

Proof of Desargues’ Theorem. We embed P2(F) into P3(F) and (des)argue there. We shall prove a bit more, namely that the desired statement holds for conﬁgurations of P3(F). There are two cases.

Case 1. First suppose that π = (abc) and π0 = (a0b0c0) are distinct planes. Then it is the argument we skteched above: a00 ∈ (bc) ⊆ π; also b00, c00 ∈ π and a00, b00, c00 ∈ π0. It must be checked that π ∩π0, which contains a line `, is actually exactly 0 `. For if some point p ∈ π ∩ π \ `, then taking two points q1, q2 ∈ `, we see by PS2 that π is the only plane containing p, q1, q2; the same applies to π0, against π 6= π0. So a00, b00, c00 are on π ∩ π0 = `, and this proves collinearity. Case 2. Now suppose (abc) = (a0b0c0) = π. Notice that π contains all points of the initial conﬁguration, so we are in the genuine planar case. The idea is however to use Case 1 by slightly “lifting” the conﬁguration, and then project back.

13 e 0 a1 a0

a1 c0 a c o

b b0

a00 c00 b00

00 c1

00 b1

Let e∈ / π; there is such a point as P2(F) = π embeds into P3(F). Now take a1 ∈ (ae) be neither a nor e; there is such a thing by PS6. Finally 0 0 let a1 = (oa1) ∩ (ea ). (This can be imagined as rotating triangle abc along axis (bc), and similarly rotating a0b0c0 along (b0c0).) Of course changing triangles, we must also change our intersections; let 00 0 0 00 0 0 00 b1 = (a1c)∩(a1c ) and c1 = (a1b)∩(a1b ); notice that a however remains unchanged.

0 0 0 Claim 1. Triangles a1bc and a1b c satisfy the assumption of Desargues Theorem and are not coplanar.

0 0 0 Veriﬁcation. We ﬁrst prove that (a1a1), (bb ), and (cc ) meet. But this is 0 clear as o ∈ (a1a1) by construction. Now we check that these triangles are

14 not in the same plane. Otherwise that plane would be π; since a, a1 ∈ π, one then has e ∈ (aa1) ⊆ π, a contradiction. 00 00 00 By Case 1, we know that a , b1 , c1 are collinear. We now project back. Let p be the projection function from e onto π, deﬁned as follows: for any point x 6= e, line (ex) meets π in a point p(x). However p(e) is not deﬁned. Notice that: • p is the identity on π; • p preserves incidence, i.e. if x ∈ `, then p(x) ∈ p(`).

00 00 00 00 Claim 2. p(b1 ) = b and p(c1 ) = c .

Veriﬁcation. Since (ea1) = (ea) and a ∈ π, one has p(a1) = a. But also 0 0 0 0 (ea1) = (ea ) so p(a1) = a . In particular,

00 0 0 {p(b1 )} = p ((a1c) ∩ (a1c )) 0 0 = p((a1c)) ∩ p((a1c )) 0 0 = (p(a1)p(c)) ∩ (p(a1)p(c )) = (ac) ∩ (a0c0) = {b00}

00 00 and p(c1 ) = c likewise. 00 00 00 00 00 00 Hence p(a ) = a , p(b1 ) = b , and p(c1 ) = c are collinear, as desired.

Definition (arguesian plane). Call arguesian a projective plane which satisfies the 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 matter of 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 Planes We now give another family of projective planes. Such planes are not always arguesian; in particular, by Desargues’ Theorem for P2(F), we thus see projective planes 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 new lines and points, so it must be done inductively.

Deﬁnition (conﬁguration). Let P0 be any abstract set of points and L0 be any abstract set of lines, with an incidence relation I0.

15 We call it a conﬁguration if there are no two distinct lines sharing more than 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} of distinct 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: enumerate non-meeting distinct lines by Yn; for any such pair {`, m} in Yn add a new point p{`,m} incident to ` and m but to no other line; then set P2n+2 = P2n+1 ∪ {p{`,m} : {`, m} ∈ Yn}. S S • Finally let P = Pn and L = Ln. N N

(P, L) is called the free projective plane over (P0, L0).

This is a nice construction provided there is enough data to start it; PP3 has 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; we may 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 line through 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 of the value of n, at stage n + 1 we have introduced a line ` joining a and b. The argument for uniqueness of ` is similar.

PP2. Entirely similar.

PP3. This is almost by assumption but not entirely. There are four points a, b, c, d ∈ P0 ⊆ P which are not collinear in L0. We must see that they remain non-collinear in the sense of L. But at stage 1, when we add line ` = (ab), it does not contain c. And we never add another line through a and b. So there never is a line containing a, b, c: they remain non-collinear also in L. Likewise for the other triples.

Example. We start with the conﬁguration consisting of four points and no

16 lines.

1 0 1 3 0 2 1 0 0 2 1

3 2 3

We have indicated the stage of appearance of the various objects. The reader should imagine what happens at the following next steps.

Example (a non-arguesian projective plane). Start with a configuration of 0 0 0 00 00 00 ten points P0 = {o, a, b, c, a , b , c , a , b , c } and take L0 to be all lines of the Desargues configuration but not line (a00b00c00). Let P be the free plane over (P0, L0). Then P is a projective plane and does not satisfy Desargues Theorem. Because there is no line joining a00, b00, c00 at first, 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); it does not embed into a projective space. Hilbert’s Grundlagen contain a completely different example (Theorem 33 there), 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’ Theorem We move to another theorem, seemingly unrelated.

Theorem (Pappus’s aﬃne theorem). Let F be a commutative ﬁeld, and work in A2(F). Let a, b, c be collinear points and c0, b0, a0 be collinear points (being six non- collinear points in total). Suppose that (ab0) k (ba0) and (c0b) k (b0c). Then (ac0) k (ca0).

17 c

o c0 b0 a0

Proof. The proof requires working in coordinates, and of course commutativity of the base ﬁeld will play its role. There are two cases. Case 1. Suppose lines (ac) and (c0a0) are parallel.

a τ1 b τ2 c

c0 b0 a0

0 0 0 0 Let τ1 be the translation mapping a to b. Then since (b c ) = (c a ) k 0 0 0 0 (ac) = (ab) which is the axis of the translation, τ1((b c )) = (c a ). 0 0 0 0 Also, τ1((ab )) k (ab ) k (ba ), but τ1((ab )) contains τ1(a) = b. So by 0 0 AP2, one has τ1((ab )) = (ba ) 0 0 0 0 0 0 In particular, τ1(b ) ∈ (c a ) ∩ (ba ) and τ1(b ) = a . 0 One can prove likewise that the translation τ2 taking b to c also takes c 0 0 0 to b . So the translation τ = τ2τ1 = τ1τ2 takes a to c and c to a . As a conclusion (ac0) k τ((ac0)) = (ca0). 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., of multiplication ·) was not required.

Case 2. This is more interesting. Suppose that (ac) and (c0a0) meet at say o. This is the picture we gave earlier, and we shall adapt the translation argument using another tool.

Let h1 be the homothety with centre o mapping a to b, and h2 be the homothety with centre o mapping b to c. Notice that: 0 0 0 b = h1(a) ∈ h1((ab )) k (ab ) k (ba ) 0 0 0 so h1((ab )) = (ba ). Moreover, since h1 is a homothety, h1((ob )) = 0 0 0 0 (ob ) = (oa ). As a conclusion, h1(b ) = a . 0 0 Likewise, one proves h2(c ) = b . As a conclusion,

0 0 0 0 (ac ) k h2 ◦ h1((ac )) = (h2h1(a) h2h1(c )) = (c h2h1(c ))

18 But here appears commutativity of F: two homotheties with same centre 0 0 0 0 0 commute. So actually h2h1(c ) = h1h2(c ) = a , and now (ac ) k (ca ), as desired.

Theorem (Pappus’ projective theorem). Let F be a commutative ﬁeld and work in P2(F). Let a, b, c be collinear points and c0, b0, a0 be collinear points (being six non- collinear points in total). Then c00 = (ab0) ∩ (ba0), a00 = (cb0) ∩ (bc0), and b00 = (ac0) ∩ (ca0) are collinear.

o a b c

c0 b0 a0

c00 b00

a00

Proof. Let λ = (c00a00) and affinise P with respect to λ; we know Pˇ ' A2(F), and in particular is a pappian plane. But now removing λ from the picture has the effect of sending c00 and a00 at infinity, so that in Pˇ one has (ab0) k (ba0) and (cb0) k (bc0). We do not know whether the intersection (ac) ∩ (c0a0) is at infinity or not, but in either case, using the affine version, we know that (ac0) k (ca0). Going back to the projective plane P, this means that b00 ∈ λ. Definition (pappian plane). Call pappian an affine or projective plane satisfying 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 Desargues

Proposition. Let P be a pappian projective plane. Then P is also arguesian. Proof. Let (o, a, b, c, a0, b0, c0, a00, b00, c00) be a Desargues conﬁguration; we aim at showing that a00, b00, c00 are collinear. We shall introduce four new points and apply Pappus’ Theorem three times. Let: • i = (ab) ∩ (a0c0);

• j = (oi) ∩ (bc);

19 • j0 = (oi) ∩ (b0c0); • k = (oa) ∩ (bc0). Claim 1. b00, j, k are collinear. Veriﬁcation. Consider the triples a, i, b and o, c, c0. Notice that they are collinear by construction and the Desargues assumption.

b i a

o c c0 Moreover: • (ac) = (ab00) and (ic0) = (a0c0) = (a0b00) so (ac) ∩ (ic0) = b00; • (oi) = (oj) and (cb) = (jb) so (oi) ∩ (cb) = j; • (ao) = (ak) and (bc0) = (bk) so (ao) ∩ (bc0) = k. By Pappus’ Theorem, we ﬁnd the desired collinearity. Claim 2. c00, j0, k are collinear. Veriﬁcation. Now consider the triples i, a0, c0 and o, b, b0. They are also collinear by construction and the Desargues assumption.

c0 a0 i

o b b0

Now: • (ib) = (ab) = (ac00) and (a0b0) = (a0c00), so (ib) ∩ (a0b0) = c00; • (oa0) = (oa) = (ok) and (bc0) = (bk), so (oa0) ∩ (bc0) = k; • (io) = (ij0) and (c0b0) = (c0j0) so (io) ∩ (c0b0) = j0. Again by Pappus’ Theorem, the desired points are collinear. Claim 3. a00, b00, c00 are collinear. Veriﬁcation. Finally consider the triples i, j, j0 and b, c0, k. By construction, they are collinear. j0 j i

b c0 k

20 But: • (ic0) = (a0c0) = (a0b00) and (jk) = (jb00) by the ﬁrst claim, so (ic0) ∩ (jb00) = b00; • (bj) = (bc) = (ba00) and (c0j0) = (b0c0) = (b0a00), so (bj) ∩ (c0j0) = a00;

• (ib) = (ab) = (ac00) and (j0k) = (j0c00) by the second claim, so (ib) ∩ (j0k) = c00. 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 speaking one 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. Then one may construct a skew-field F with P ' P2(F). Now F is finite, so by Wedderburn’s Theorem, F is actually commutative: hence P is pappian. We repeat that it is an open question (presumably untractable) to give a direct, geometric proof — at some point one would have to do some finite field theory in geometric form.

Remark. Let F be any non-commutative skew-ﬁeld, for instance the quaternions H. Then P2(H) is arguesian but not pappian.

End of Lecture 4.

Lecture 5 (Coordinates)

I.4. Coordinatising Arguesian Planes Explicitly

Theorem (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, we shall give more abstract arguments. For the moment we give a corollary and describe 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.

21 The proof of the corollary uses the Theorem and results we already know. If P ' P2(F) then certainly we may embed P into P3(F). The second implication is what we actually proved for Desargues’ Theorem. And the last is Hilbert’s coordinatisation 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 = Pˇ (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 that these operations equip ` with the structure of a skew-field F is actually the 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 ' Aˆ ' A\2(F) ' P2(F). We shall not cover (2) entirely, and we shall omit (3). Details are to be found in Hilbert’s Foundations of Geometry, chapter V. The proof starts here.

I.4.1. The Frame

m mx

πm(x) ` x x

π`(x) ` 0 n 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 k n is not parallel to `, this makes sense. Deﬁne πm : A → m similarly. Notice that π` is the identity on ` (and similarly for πm).

22 The prime function. Now consider:

` → m 0 a 7→ a := na ∩ m

0 clearly a bijection between ` and m. For any x ∈ A one has (π`(x)) = πm(x). Finally let 0 = ` ∩ m and notice 00 = 0.

I.4.2. Addition

m mb

0 a `a0 pa,b ` 0 a b a + b n

Given a, b in `, deﬁne:

• pa,b = `a0 ∩ 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 k is an equivalence relation, pa,b is uniquely deﬁned. So is a+b ∈ `. Neutrality is obvious as the construction then 0 degenerates: `00 = ` so p0,a = ` ∩ ma = a, and m0 = m so pa,0 = `a0 ∩ m = a , 0 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.

m ma mb 0 b `b0 pb,a qb

0 a `a0 qa pa,b ` 0 a b n

Introduce:

• qa = ma ∩ `a0 ;

23 • qb = mb ∩ `b0 .

Claim 1. Points 0, qa, qb are collinear. 0 0 Veriﬁcation. Let r = (0qa)∩`b0 . Consider the conﬁguration (0, a , b , qa, r, a, b).

b0 a0

qa r 0 a b

0 0 0 0 Then (a a) = na and (b b) = nb are parallel; also, (a qa) = `a0 and (b r) = `b0 are parallel. So by Desargues’s Theorem, (aqa) = ma and (br) are parallel. Therefore also (br) k ma k mb; now the only parallel to mb through b being mb, one has r ∈ (br) = mb. Since also r ∈ `b0 by construction, it must be r = 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 0 (qb, pb,a, b , qa, 0, pa,b, b).

pb,a

0 qa qb

pa,b b

0 Then (b o) = m and (pb,aqa) = ma are parallel, and also (ob) = ` and (qapa,b) = 0 `a0 are parallel, so by Desargues’s Theorem, (b b) = nb and (pb,apa,b) are parallel.

Since (pb,apa,b) k 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 use soon again.

Lemma. In an arguesian aﬃne plane, let x1, x2, y1, y2, r, s be such that (x1x2) k (y1r) k (y2s) and (y1y2) k (x1s) k (x2r).

24 If (x1y1) k (x2y2), then (rs) k (x1y1).

y2 s

y1 r

x1 x2

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 it without 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 ∩ à0 (which makes sense as n and ` are not parallel), then b = ò ∩ mq. We claim that a + b = 0. Indeed, pa,b = à0 ∩ mb = à0 ∩ mq = q ∈ no, so a + b = π`(pa,b) = π`(q) = 0.

Lemma. + is associative. Proof. Let a, b, c ∈ `; we may suppose that all three are distinct, and distinct from 0. Consider the following picture.

(b + c)0 pb+c,a

(a + b)0 pa+b,c b0 pb,c pb,a

a a + b c b + c

0 • We know that (pb,a pb,c) = `b0 is parallel to ((a + b) p(a+b),c)) = `(a+b)0 , 0 which is parallel to ((b + c) pb+c,a) = `b+c.

0 0 • 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. This 0 0 means that ((a + b) pb,a) = na+b is parallel to nb+c = ((b + c)(b + c) ) = 0 (pb,c (b + c) ).

25 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. Multiplication We introduce the new point 1 = `∩n; we need 1 6= 0, and this is why we wanted `, m, n not to concur. We introduce 10 as well. Now for a, b ∈ ` we deﬁne a · b as follows.

ta,b

α a0 b 10 α 0 1 b a · b

0 Let α = (1a ), then ta,b = αb ∩ m (t is for “times”); ﬁnally a · b = π`(ta,b). 0 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 the construction makes sense. Lemma. Any a ∈ ` \{0} has a two-sided inverse.

Proof. Here is another exercise in notation: let b = α10 ∩ `; this makes sense only since α and ` are not parallel, i.e. only since a0 6= 0. We can check without 0 a picture that ta,b = αb ∩ m = α10 ∩ m = 1 . But the other exercise is to check 0 that tb,a = 1 as well.

Lemma. · is associative.

Proof. Let a, b, c be on `; we may suppose that all are distinct, and that neither is 0 nor 1. Let p = a · b. The picture is as follows.

26 (p · c)0

βc

(b · c)0

πc β nb·c p0

0 b q1 a0 π

0 1 b c b · c

α αb

0 Remember that in our notation, p = a · b and π = (1p ). Let q1 = π ∩ nb and q2 = πc ∩ nb·c.

Claim 1. Points 0, q1, q2 are collinear.

Veriﬁcation. Set r = (oq1) ∩ πc. Apply Desargues’ Theorem to the tuple 0 0 (0, 1, c, b , (b · c) , q1, r).

(b · c)0 b0

q1 r o 1 c

Check that:

• all required collinearities hold;

0 0 • (1b ) = β k βc = (c (b · c) );

• (1q1) = π k πc = (cr).

0 0 By Desargues’ Theorem, (b q1) = nb k ((b · c) r), so r ∈ nb·c and r = q2. Claim 2. Lines (bp0) and ((b · c)(p · c)0) are parallel.

27 Veriﬁcation. Now apply Desargues’ Theorem to the conﬁguration (o, p0, (p · 0 c) , q1, q2, b, (b · c)).

(p · c)0 p0

q1 q2 o b b · c

• All required collinearities hold (by the previous claim);

0 0 • (q1p ) = π k πc = (q2 (p · c) );

• (q1b) = nb k nb·c = (q2 (b · c)).

0 0 So by Desargues’s Theorem again, (bp ) = αb k ((b · c)(p · c) ). As a conclusion, ((b · c)(p · c)0) k α 0 so ((b · c)(p · c) ) = αb·c and

0 0 (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 that we skipped the two distributivity laws), one must still coordinatise A, which is not trivial and uses Desargues again. At this stage the student should learn to read not lecture notes, but major books 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 Aspect We 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.

28 We shall prove this counting Proposition in a moment, but before we give an example and a consequence.

Example. Let F be the field with q elements (incidently, q must be a prime power as we know). Then every line P2(F) has as many points as there are vector 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 − 1 = q + 1 q − 1 points on each line. This section will describe a much better idea than doing the other computation; which we however perform for the moment. The number of lines through each point is the number of vector hyperplanes of F3 containing a given vector 3 3 2 line L; projecting F F /L ' F this amounts to counting vector lines in the plane F2, so we find q + 1 again. The Proposition above tells us this is part of a general phenomenon, not only for 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 Aˆ 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 projective plane P; it is not the number of points in P, but is the most relevant datum since 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 plane of order n ≡ 1 or 2[4], then n must be sum of two squares. In particular there are no projective planes of order 6 nor 14. • It follows immediately from Hilbert’s Coordinatisation Theorem and the fact that every finite field has prime power order, that there is no arguesian projective plane of order 10. It can actually be showed by extensive computation that there is no projective plane of order 10 at all. Whether there exists one of order 12 is open (of course it would be non-arguesian).

Proof of the projective counting Proposition. Let ` be a line and p∈ / `. To a ∈ ` 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 `0 is another line not containing p, then composing we ﬁnd a bijection ` → `0.

29 0 0 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 points there 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 duality principle.

• It is also natural to ask about the degree of isotropy of the space, i.e. to wonder 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. Duality We return to one of the above remarks: there is a remarkable symmetry in the axioms when one exchanges points and lines. Deﬁnition (dual statement). Let ϕ be a statement about projective planes. Let ϕ∗ be the statement obtained by exchanging the words “point” and “line” (and reversing ∈ 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 the projective plane, so it is not a suitable ϕ. But the property that “Pappus implies Desargues” is an example of such a statement.

Proof. Since ϕ∗∗ is always the formula ϕ, one implication is enough. We want to argue by symmetry of the axioms for projective planes. We know that ∗ PP1 = PP2, but we should also do something about PP3.

Claim 1. Let P be an incidence geometry satisfying PP1 and PP2. Then in P, 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 which are collinear. So lines (ab), (ac), (bc), (bd) are four in number and no three of ∗ them 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. Now exchange the words “points” and “lines” in the proof: the resulting text is a ∗ ∗ ∗ ∗ proof of ϕ using only axioms PP1 = PP2, PP2 = PP1, and PP3 . Since the latter is a consequence of PP1, PP2, and PP3, it holds in any projective plane. Therefore so does ϕ∗.

30 Exercise. To check whether you have understood the claim, prove that in an incidence 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 Plane Be careful that the Duality Theorem applies only to statements ϕ true in all planes P. If you have a statement ϕ which is true in one plane P and are looking for 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, and take I∗ to be 3. Then P∗ = (P∗, L∗,I∗) is a projective plane, called the dual plane of P. By construction, P satisfies ϕ iff P∗ satisfies ϕ∗. Remark. We may prefer to have membership as the the incidence relation, in which 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 P1 was the first construction and ∗ ∗ ∗ P2 the second. Let f : P1 = L → L = P2 be the identity map. Now let ∗ ∗ g : L1 = P → {Lp : p ∈ P} = L2 take p to Lp. Then of course, letting ∗ ∗ ∗ ∗ p1 = ` ∈ P1 and m1 = q ∈ L1, one has:

∗ ∗ ∗ ∗ ∗ p1I1 m1 ⇔ ` 3 q ⇔ ` ∈ Lq ⇔ f(p1) ∈ g(m1)

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 same underlying 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 the ordinary duality bracket turns left into right, so it exchanges left-vector spaces with right-vector spaces. Since on the other hand P2(Fop) is a construction obtained from left-vector spaces, we need to do something. Let K = Fop for clarity; notice that right K-vector spaces are exactly left F-vector spaces, and prepare for the proof. It will also save notation to write 2 2 PF = P (F) and PK = P (K).

31 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-)vector plane, 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⊥

3 These are clearly bijections. Now let (p, `) ∈ P(PK) × L(PK ), say p = L ≤ K and ` = H ≤ K3. Since orthogonality reverses inclusion: p ∈ ` ⇔ L ≤ H ⇔ H⊥ ≥ L⊥ ⇔ g(`) 3 f(p)

As a conclusion, ' (L( ), P( ), 3) ' ∗, as desired. PK PF PF PF

I.5.4. Dualising Desargues and Pappus We 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 satiﬁes Desargues∗. So let ω, α, α0, β, β0, γ, γ0 be lines such that: • ω, α, α0 are concurrent; • ω, β, β0 are concurrent;

• ω, γ, γ0 are concurrent. To prove Desargues∗ is to prove that lines α00 = (β∩γβ0∩γ0), β00 = (α∩γα0∩γ0), and γ00 = (α ∩ β α0 ∩ β0) are concurrent. We should name these points and a few more; let:

0 0 0 • a0 = α ∩ α , b0 = β ∩ β , and c0 = γ ∩ γ ; • a = β ∩ γ and a0 = β0 ∩ γ0; • b = α ∩ γ and b0 = α0 ∩ γ0;

• c = α ∩ β and c0 = α0 ∩ β0. With this notation we now have α00 = (aa0), and so on.

32 Consider the following Desargues conﬁguration (and be careful about the swap on line ω):

c0 b c a0 b0 c0

The various collinearities occur because of lines ω, α, α0. Let us compute the Desargues points.

• (c0b) = γ and (b0c) = β meet at a;

0 0 0 0 0 • (c0b ) = γ and (b0c ) = β meet at a ; • (b0b) = β00 and (c0c) = γ00 meet at say x.

Since P is arguesian, x ∈ (aa0) = α00. So α00, β00, γ00 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 commutative ﬁeld F; hence P∗ ' P2(F)∗ ' P2(Fop) = P2(F) ' P, and it is pappian.

Remark. There is a geometric proof (without Hilbert coordinatisation) that if P 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 P is pappian, then P ' P∗. Be extremely careful that P ' P∗ and arguesian does not imply that the underlying skew-ﬁeld F need be commutative: one only has Fop ' F, as in the quaternions!

End of Lecture 6.

33 Chapter II: Group-Theoretic Aspects

Lecture 7 (Coordinatising like adult people (1))

II.1. Arguesianity and the Group of Dilations

We wish to coordinatise using the power of group theory. We remember from last week (or feel from experience) that addition is related to translations, and multiplication to homotheties. In the usual case of A2(F) these generate a group of interest.

II.1.1. Dilations and Translations

Deﬁnition (automorphism). Let A be an aﬃne plane. An automorphism of A is a bijection of A preserving collinearity. It will be convenient, when working with one f : A → A, to use classical “high-school” notation x0 = f(x).

Deﬁnition (the group of dilations). A dilation of A is an automorphism which takes any line to a parallel line: ∀a 6= b ∈ A, (a0b0) k (ab). Let D be the group they form under composition.

In projective terms, a dilation is an automorphism of Aˆ fixing the line at infinity pointwise. Of course in A2(F), translations and homotheties are dilations. We capture the difference with the following general definition. Definition (translation, homothety).

• A translation of A is either the identity map IdA, or a ﬁxed point-free dilation.

• A homothety of A is a dilation having at least one ﬁxed point (this encompasses IdA of course). It is not clear at all how many of these maps there are; we shall study the question later.

Proposition. If a dilation f has two distinct ﬁxed points, then f = IdA.

34 Proof. Suppose f ﬁxes a and b. Using our convention, this means a0 = a and b0 = b. The proof will introduce two cases, points not on (ab) being easier to understand. This kind of case division will be prominent in the lecture.

a0 = a d b0 = b

• Let c∈ / (ab). Then a = a0 ∈ (a0c0) k (ac) since f is a dilation; so 0 by uniqueness in AP2 we know that c ∈ (ac) are collinear. Likewise c0 ∈ (bc); these lines meet at c0 = c, which is fixed as well. • Now let d ∈ (ab). Using our previous c which was fixed under f (there exists 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). Clearly Ha is a subgroup of D; but what about translations? We must investigate further.

Lemma. If t is a translation 6= IdA then for any a, b ∈ A, one has (at(a)) k (bt(b)).

Proof. Suppose there is a counterexample: then (aa0) and (bb0) meet at say c.

We prove that c is a ﬁxed point of t, which will contradict t 6= IdA.

a a0 c

b b0

Notice that c0 ∈ (a0c0) k (ac) since t is a dilation, and since a, a0, c are collinear we ﬁnd c0 ∈ (aa0). Likewise, c0 ∈ (bb0) so actually c0 = c.

Exercise. Prove the converse. Therefore the following deﬁnition makes sense.

Deﬁnition (direction). The direction dt of a translation t 6= IdA is the equivalence class modulo parallelism of (at(a)) for any a ∈ A. (The direction of IdA is not deﬁned.)

35 Remember that given a point a and line `, we use `a to denote the unique line parallel to ` through a.

0 0 0 0 Corollary. If t 6= IdA is a translation and b∈ / (aa ) then b = (ab)a ∩ (aa )b.

a a0

b b0

Proof. Since t is a dilation, (a0b0) k (ab). But also, since t is a translation, (bb0) k (aa0). Since b∈ / (aa0), these lines are not parallel: they meet at b0.

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 normal subset of D and there is only one non-trivial thing to prove: that T is stable under composition. So let t1, t2 be translations; we want to show t2t1 ∈ T . We may supppose

that neither of t1, t2 is IdA. To prove that the dilation t2t1 is a translation, we show that it is either ﬁxed point-free or IdA. So suppose t2t1 has a ﬁxed point.

t2 t1(a)

t1(b) a b

Observe how:

((t2(b) t2t1(b)) k (b t1(b)) since t2 ∈ D

k (a t1(a)) since t1 ∈ T

k (t2(a) t2t1(a)) since t2 ∈ D

k (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)) k (ab)

so t2t1(b) ∈ (ab). The two lines meet at b: therefore t2t1 ﬁxes not only a, but

also b. Being a dilation, it must be IdA, a contradiction. Corollary. If two translations coincide at one point, they are equal.

36 −1 Proof. If t1(a) = t2(a), then t2 t1 is a translation with a ﬁxed point, so it is IdA.

II.1.2. The (p,p)-Desargues Property Deﬁnition ((p,p)-Desargues). The (parallel, parallel)-Desargues property (for short: (p, p)-Desargues) is the following statement: Let (aa0), (bb0), (cc0) be three distinct parallel lines. Suppose (ab) k (a0b0) and (ac) k (a0c0). Then (bc) k (b0c0).

a a0

b b0

c c0

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 at most one translation t mapping given a to given b. So transitivity is equivalent to sharp transitivity.

Proof. First suppose that T is transitive on A; we prove (p,p)-Desargues. Take a (p,p)-Desargues configuration like on the picture and let t be the translation 0 0 0 0 mapping a to a . Then we know that t(b) = (ab)a0 ∩ (aa )b = b and t(c) = c likewise. Now since t is a dilation, we find (b0c0) k (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, a0), we define a 0 0 partial map tˇa,a0 : A \ (aa ) → A by tˇa,a0 (b) = (ab)a0 ∩ (aa )b. Notice that it is well-defined (but a partial map). 0 0 Claim 1. If b∈ / (aa ) and b = tˇa,a0 (b), then tˇa,a0 and tˇb,b0 agree wherever both are defined. Verification. This is exactly the (p,p)-Desargues assumption! 0 This defines a global map ta,a0 : A → A, and obviously ta,a0 (a) = a . We 0 must check it is a translation. Notice that if ta,a0 (b) = b , then ta,a0 = tb,b0 .

Claim 2. ta,a0 is a dilation. Veriﬁcation. Bijectivity is obvious from the following picture.

a a0

37 We claim that ta,a0 is an automorphism. Suppose x, y, z are collinear (and we may suppose they are distinct); we show that so are the images x0, y0, z0.

x x0

y y0

z z0

0 0 But we know that ta,a0 = tx,x0 , so actually y = (xx )y ∩ (xy)x0 ∈ (xy)x0 0 and z ∈ (xz)x0 = (xy)x0 by collinearity of x, y, z. This proves collinearity of x0, y0, z0. We claim that t is a dilation. But we already know that (x0y0) k (xy), so this is clear. Finally t is a translation. Otherwise it has a ﬁxed point b.

a a0

b0 = b

Up to changing base point we may suppose b∈ / (aa0). Now (ab) k (a0b0) = (a0b) so b ∈ (aa0), a contradiction.

Corollary. If A has (p,p)-Desargues, then T is abelian and D = T n Ha for any a ∈ A.

Proof. Let t1, t2 be translations. We want to show t2t1 = t1t2; we may assume

that neither is IdA, so both have a direction. Case 1. Suppose the directions are diﬀerent.

a t2(a)

t1(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 t1t0 has not the same direction as t1 and t2. Therefore, by case 1: −1 −1 t2t1 = (t2t1)(t0t0 ) = t2(t1t0)t0 −1 −1 −1 = (t1t0)(t2t0 ) = t1(t0t2)t0 = t1t2t0t0 = t1t2

38 We turn to proving D = T nHa. 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 t −1 with t(a) = d(a). Hence t d(a) = a and d ∈ T · Ha.

Notice that abelianity only requires the existence of translations having different directions (a consequence of (p, p)-Desargues).

II.1.3. The (c,p)-Desargues Property Deﬁnition ((c,p)-Desargues). The (concurrent, parallel)-Desargues property (for short: (c, p)-Desargues) is the following statement: Let (aa0), (bb0), (cc0) be three distinct concurrent lines. Suppose (ab) k (a0b0) and (ac) k (a0c0). Then (bc) k (b0c0). a0 a b b0 o c c0

The property is also called the big Desargues axiom.

Proposition. A has (c,p)-Desargues iﬀ for any line ` and point a ∈ `, Ha acts transitively on ` \{a}.

Remark. Here again, since an element of Ha already ﬁxes a and a non-identity dilation 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 enough translations. Let a, a0 be distinct points of A. Let o ∈ (aa0)\{a, a0} (here we must assume that lines have more than three 0 points). By assumption, there is g1 ∈ Ho doing g1(a) = a . Also let b∈ / (ao), 0 0 00 and let b = (ab)a0 ∩ (aa )b, then b = g1(b). By assumption, there is g2 ∈ Ha0 00 0 doing g2(b ) = b .

o a a0

b b0 b00

We claim that t = g2g1 is a translation taking a to a’. Of course t(a) = 0 0 00 0 g2(a ) = a , but also t(b) = g2(b ) = b . So if t has a ﬁxed point c, then one must have (ac) k (a0c0) = (a0c), and therefore c ∈ (aa0). But c ∈ (bb0) 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 Plane

In this lecture we ﬁnally see how to recover a vector geometry from an aﬃne plane. We start with as few assumptions as we can and add them progressively.

II.2.1. A Ring

Remember that the direction of a translation t 6= IdA is the equivalence class dt of (at(a)) for any a ∈ A; this is well-deﬁned. Suppose that T is abelian. Then End(T ) is an associative, non-commutative ring with distinguished elements:

0 : T → T 1 : T → T ; t 7→ IdA t 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 any

t ∈ T \{IdA}, one has λ(t) 6= IdA and dλ(t) = dt. In particular a direction-preserving endomorphism of T must be injective by construction.

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 that neither λ, µ, nor λ + µ is 0. Let t ∈ T \{IdA}. Then by deﬁnition λ(t) 6= IdA and 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), so λ + µ is direction-preserving.

Claim 2. F is stable under ·. Veriﬁcation. Here again we take λ, µ ∈ F and we may assume that neither λ nor µ is 0. Then both are injective, and therefore λ · µ 6= 0. Now for any

t ∈ T \{IdA}, d(λ·µ(t) = dλ(µ(t)) = dµ(t) = dt, so by deﬁnition, λ · µ ∈ F.

40 II.2.2. A Skew-Field We 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 → T t 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 for all λ ∈ F \{0} there is a unique h ∈ Ha with λh = λ. We shall prove the lemma in a minute. However observe that letting then µ = λh−1 , we have µ ∈ F and λµ = µλ = 1 in F, so our proposition will be proved. 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 the kernel, 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 only a 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 is × an isomorphism Ha ' F .

41 Proof of the Lemma.

t λ(t) a

ta,xx

λ(ta,x)

h(x)

By (p,p)-Desargues, for any x ∈ A there is a translation ta,x taking a to x. 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. But we 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 translations 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)) k dtx,y k (xy)

An injective map with this property is always a dilation (preservation of collinearity and surjectivity being consequences left as an exercise). So h ∈ D and since we already know h(a) = a, we have h ∈ Ha.

Claim 2. λh = λ. −1 Veriﬁcation. Take any t ∈ T ; say t = ta,b for some b. Then λh(t) = hth is −1 the translation taking a to hth (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. Dimensionality At first it is something of a surprise to be able to define a skew-field with as weak a property as (p,p)-Desargues. Now the F-module T is actually a vector space; we may think we are quite close to coordinatising, but as a matter of fact the dimension theory of T is not clear at all.

42 Proposition. Suppose A has (c, p)-Desargues. Then: 0 0 • for t, t ∈ T \{Id}, one has dt = dt0 iﬀ there is λ ∈ F with t = λt; • T is 2-dimensional.

Proof. Let t, t0 be as in the ﬁrst claim. If there is λ ∈ F with t0 = λt then λ 6= 0, so by deﬁnition, λ is direction-preserving; hence dt0 = dt. 0 Now suppose dt0 = dt, and say t = to,a, t = to,a0 . By (c, p)-Desargues, 0 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,a0 = t0

The first claim being proved we can move to the second. It suffices to show that 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 that the direction of t is neither d1 nor d2. Fix a ∈ A and let b = t(a).

d2 d1

a b

Then lines (d1)b and (d2)a, not being parallel, meet at say c, which is clearly distinct from a and b. So tc,b has direction d1, and therefore there is λ ∈ F with tc,b = λt1. Likewise, the direction of ta,c is d2, whence µ ∈ F with ta,c = µt2. Now t = ta,b = tc,b ◦ ta,c = (λ(t1)) ◦ (µ(t2)) as transformations of A, 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 be two points in A. Then for anny q ∈ A one has equivalences:

q ∈ (p1p2) iﬀ q = p1 or (p1q) k (p1p2) iﬀ q = p or d = d 1 tp1q tp1p2

iﬀ ∃λ ∈ F tp1,q = λtp1,p2

So lines are now given in parametric form. We have retrieved the F-vector space structure 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)

t We consider the group SO3(R) = {M ∈ GL3(R): M · M = I3}, with proper subgroup 1 B = : M ∈ SO ( ) < SO ( ) M 2 R 3 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 3 of SO3(R) these elements are also called half-turns of the space R . We let I be the set of involutions of G.

II.3.1. The Geometry and Non-Degeneracy Axioms Let G be any group equipped with a subgroup B ≤ G; we also let I be the set of 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 exactly the translates of the conjugates of B. Moreover, a line contains 1 iff there is b ∈ B with 1 = gbh, so actually g−1h−1 = b and hg = b−1, meaning gBh = gBg−1. Of course such a line contains 1, and as a conclusion, lines through 1 are exactly the conjugates of B. Such lines are parametrised by G/NG(B). −1 −1 Two lines g1Bh1 and g2Bh2 are equal iff g2 g1 and h1h2 lie in NG(B)? −1 Two planes g1I and g2I are equal iff g2 g1 is inverted by all involutions in G.

We shall prove that in the case of SO3(R) with our choice of B, this geometry is 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 us the 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 3 copy of (Z/2Z) . 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 has continuum-many points: so PS6 is clear. Now we take the unit and three involutions, say:

1  −1  −1  i =  −1  , w =  1  , s =  −1  −1 −1 1

For easy practice, we check that these four points satisfy PS6. First, if they are coplanar, then there is g ∈ G with 1, i, w, s ∈ gI: so g must be an involution.

44 But then gi, gw, gs must be involutions as well, and we recall from group theory that the product of two involutions is an involution iﬀ they commute. So g must be an involution commuting to i, w, s, which form a Klein 4-group. Hence G contains an elementary abelian 2-group of order 8. However it is quickly computed 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 assume that 1, i, w are collinear and will derive a contradiction. The deﬁnition says that there are g, h ∈ G with 1, i, w ∈ gBh. Then we may assume −1 −1 gBh = gBg contains i and w. But gBg ' B ' SO2(R) contains a unique involution, a contradiction.

• Suppose we took the three involutions i, w, s. The deﬁnition now says that i, w, s ∈ gBh for some g, h; now there are b, b0 ∈ B such that i = gbh and w = gb0h. Then:

s = iw = i−1w = h−1b−1g−1gb0h = h−1b−1b0h ∈ h−1Bh

and w ∈ h−1Bh likewise. But h−1Bh contains a unique involution: this is a contradiction ﬁnishing the proof.

II.3.2. The First Axiom Lemma.

• In any group where the geometry was deﬁned in a similar fashion, PS1 is equivalent to: G = ` gBg−1. G/NG(B)

• The above property holds in SO3(R).

Proof. Remember that PS1 is the axiom that through any two distinct points x, y ∈ G there is a unique line. Using left-translation this is equivalent to saying −1 that z = x y lies in a unique line also containing 1. So PS1 is equivalent to saying that any z 6= 1 lies in a unique conjugate of B, which is exactly the set equation we wrote. The equivalence holds for general G where lines are of the given form. We now prove that SO3(R) enjoys this property for chosen B.

Existence. Let z ∈ G = SO3(R). Since the dimension is 3, z has as an eigenvalue 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, this means that there is an orthonormal basis B in which x has matrix λ x0 = M

with λ = ±1 and M ∈ O2(R). The base change matrix associated to B 0 is orthogonal, so x and x are conjugate in O3(R). But notice that up to changing the order of the last two vectors of B, we may assume that B is

45 0 a direct orthonormal basis. So x and x are conjugate in SO3(R). There are 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 orthogonal reﬂection, so up to changing basis again x is conjugate to

−1   1  = w −1

which is easily conjugate to i ∈ B. Uniqueness. Now suppose that gBg−1 and hBh−1 meet at some x 6= 1. 3 Notice that B = StabG(e1) in the standard action of SO3(R) on R . So x ∈ 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) −1 preserve norms, actually λ = ±1. If ge1 = he1 then h g ∈ StabG(e1) = −1 −1 B ≤ NG(B), so certainly gBg = hBh , as desired. If on the other hand ge1 = −he1, then we consider our involution w again, which had −1 −1 the eﬀect that we1 = −e1. So now wh ge1 = e1 so wh g ∈ NG(B). −1 But one can see that w inverts B, so w ∈ NG(B), and here again h g ∈ NG(B), as desired.

II.3.3. The Second Axiom Lemma. Let G be a group with involutions, in which the geometry is deﬁned as above. If no element 6= 1 is inverted by all involutions in G, then planes gI, hI are equal only if g = h. PS2 is equivalent to: any two elements which do not lie in a common conjugate 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 otherwise every element is inverted by all involutions in G), there is an involution j 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 for any two points x, y not collinear with 1, there is a unique element (necessarily an 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 conjugate of B. So PS2 is now equivalent to: any two elements not lying in a same conjugate of B are inverted by a unique involution. We prove that this condition holds. Let g, h not be in a common conjugate of 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 hwi where w inverts B (and B is 2-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 space with axes Lj 6= Lk. We claim that j and k commute iﬀ Lj ⊥ Lk.

II.3.4. The Third and Fourth Axioms

Lemma. PS3 is equivalent to: G = B · I. The above holds in SO3(R).

Proof. Literally speaking, PS3 means that any line xBy intersects any plane gI. But translating this is equivalent to saying that any line g−1xBy intersects I, i.e. that any coset of a conjugate of B contains an involution. Now I is invariant under conjugation, so this is equivalent to saying that any translate gB of B contains an involution j, or again: for all g ∈ G there is (b, j) ∈ B × J such that g = jb. We show that this is the case in SO3(R). Let g ∈ SO3(R); if ge1 = −e1 then 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 clearly je1 = 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 this is true. The above holds in SO3(R).

Proof. Left-translating as usual, PS4 is equivalent to: any plane contains a line made of involutions. Let g ∈ G. Then g belongs to a conjugate B0 of B inverted by some involution say j. Then jB0 ⊆ I ∩ g · I since: jB0 consists of involutions, and g−1jB0 = jgB0 = jB0. Note that if PS4 holds then there is a line ` ⊆ I. Conjugating ` is of the form 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 projective space 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 is isomorphic 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 PS1 3 to PS6. This would create a definable skew-field inside G ' P (K); general model theory implies that K is actually a commutative, algebraically closed field. 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 that 2 Aut(P (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. Projectivities We return to the map ` → m through point p. Deﬁnition (perspectivity). Let ` 6= m be two lines and p∈ / ` ∪ m. The perspectivity ` −→ m is the map taking a ∈ ` to a0 = (ap) ∩ m. p

a0 a p

` m

Deﬁnition (projectivity). A projectivity from ` to m is any composition of perspectivities ` −→ `1 ... −→ `n = m. a1 an If m = ` the map is called a self-projectivity of `.

Let G` be the group of self-projectivities of `. Notice that since there is always 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 f1 and 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. Choose any p ∈ m1 \ `. Now let qx = (px) ∩ m2 and qy = (py) ∩ m2. Finally, let r = m1 ∩ m2. The picture is as follows.

48 m2 m1

f1 f2 x y ` qx

We consider the two perspectivities ` −→ m1 and m1 −→ `. One should qx qy follow on the picture that they have the following eﬀect:

` −→ m1 −→ ` qx qy f1 7→ f1 7→ f1 f2 7→ r 7→ f2 x 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’ Theorem

Recall that G` is always 3-transitive on `. Sharpness actually characterises a known 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 Hilbert coordinatisation, 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 ` is equivalent to that of PGL2(F) on F ∪ {∞}, which is known to be sharply 3-transitive. The converse must be of a purely geometric nature. Let (o, a, b, c, a0, b0, c0) be a Pappus configuration. Let a00 = (bc0) ∩ (cb0), b00 = (ac0) ∩ (ca0), and c00 = (ab0)∩(ba0). We want to show that a00, b00, c00 are collinear. Set `00 = (b00c00); we shall prove a00 ∈ `00. We introduce two auxiliary points: • q = (aa0) ∩ `00;

• z = `0 ∩ `00.

49 o a b c `

c0 b0 `00 a0 z q `0 c00 b00

a00

First consider the following projectivity and its eﬀect:

` −→ `00 −→ `0 a0 a a 7→ q 7→ a0 b 7→ c00 7→ b0 c 7→ b00 7→ c0 o 7→ z 7→ z

We need two more auxiliary points, which we shall not show on picture: • α00 = (b0c) ∩ `00; • γ0 = (bα00) ∩ `0.

Next consider this other projectivity (undeﬁned point x plays no real role, since (xb) = (bb0)): ` −→ `00 −→ `0 b0 b a 7→ c00 7→ a0 b 7→ x 7→ b0 c 7→ α00 7→ γ0 o 7→ z 7→ z

Now the assumption that G` is sharply 3-transitive on ` immediately implies that for given lines m, m0, there is a unique projectivity m → m0 taking a distinct triple of m to one of m0. This means that γ0 = c0. Hence α00 ∈ (bγ0) = (bc0) and α00 ∈ (b0c) so α00 = a00 lies on `00 = (b00c00), as desired.

End of Lecture 10.