Vakgroep Zuivere Wiskunde en Computeralgebra

Juni 2001

SOME CONTRIBUTIONS TO THE

(GEOMETRIC) THEORY OF

GENERALIZED POLYGONS

Eline Govaert

Promotor: Prof. Dr. H. Van Maldeghem

Proefschrift voorgelegd

aan de Faculteit Wetenschappen tot het behalen van de graad van Doctor in de Wetenschappen

richting Wiskunde 2

Puzzling (with) Polygons

One of the most puzzling questions of non-mathematicians to a PhD-student in geometry is ‘what exactly are you spending your days with?’. Well, I have been puzzling. My puzzles were - as most puzzles are - made of ordinary polygons, put together in a smart way and forming a puzzling object called a ‘generalized polygon’.

The official birth announcement of the generalized polygons was made in Tits’ paper of 1959, ‘Sur la trialité et certains groupes qui s’en déduisent’. In this paper, Tits discovers the simple group ^{3}D_{4} by classifying certain maps, called trialities of D_{4}-geometries. The related geometries are what we call today ‘generalized hexagons’. In a small appendix to the famous ’59-paper, Tits introduces the notion of a generalized n-gon. Of course, these structures must have been in his head for some time then, and for example the projective planes - which are exactly the generalized 3-gons - had already been studied extensively at that moment. But anyway, from ’59 on, the generalized ngons come out of the shadow of the groups, and become geometries studied on their own. Some years later, Tits introduces - again for group-theoretical reasons - the notion of a ‘building’. The building bricks of buildings are the generalized polygons, which stresses again their importance in the study of incidence geometry.

The original paper of Tits already gives some examples of generalized polygons, arising from ‘classical objects’ such as quadrics and Hermitian varieties in projective space for example, which translates in nice geometric properties for these polygons. But one can do better: a free construction process provides examples of generalized n-gons, for every n. Recently, another method (using model theory) to construct infinitely many new examples of n-gons has been discovered. So at least there are polygons enough to investigate. Now what are the problems that one can look at?

From the point of view of discrete mathematics and combinatorics, one can be interested in finite examples of generalized polygons, i.e. polygons with a finite number of points. A famous theorem of Feit & Higman - proved with purely algebraic methods - states that such finite examples only exist for n ∈ {3, 4, 6, 8}. Surprisingly, these values of n turn up at various places in the theory - and this will also be the case in this thesis. Group-minded people are perhaps more interested in polygons with a very large automorphism group, the so-called Moufang polygons. The classification of all Moufang polygons (which only exist for n ∈ {3, 4, 6, 8}...) was announced by Tits in ’76, is only recently completed by Tits & Weiss and is in the process of being published. This piece of the polygon-research is strongly related with algebraic objects as root systems and specific algebras. Within topology, the polygons are represented by the topological n-gons. The notion of a compact polygon for example has been very useful to prove topological counterparts of theorems about finite polygons. Last but not least, one can take the geometric point of view - which will also be the point of view of this thesis.

One thing that can happen when puzzling, is that some pieces disappear. Some puzzles then become worthless, but not our generalized polygons. Indeed, missing pieces will be the link between the different chapters of this work.

In the first part of this thesis, we are concentrating on our favourite puzzles, being the classical generalized hexagons. These hexagons have nice geometric properties. Now the question is: if we are only given some of the pieces of one of our favourite puzzles, can we recognize from which one they come? So in fact we look for pieces of our favourite puzzles (=geometric properties of the hexagons) that are typical for this puzzle, in this way obtaining characterizations of classical generalized hexagons. The characterizations we obtain are mainly based on regularity properties. In the second part, we leave the hexagons and consider really ‘general’ n-gons. Here we deal with puzzles missing so many pieces, that it is not clear any more that they actually arise from generalized polygons. These structures are called ‘forgetful polygons’, since their definition looks like the definition of a generalized polygon, where some lines seem to have been forgotten. It remains a question however whether a ‘forgetful polygon’ is necessary a forgetful generalized polygon. We investigate this more in detail for the case of the quadrangles (and so we are back at a small value of n). Next, two puzzles come into play, and we want to decide, given only partial information about them, whether they are the same or not. Let us be a bit more precise. A generalized polygon can be seen as a graph, which allows us to define a distance function. If two generalized n-gons are given, and a bijective map between them preserving a specific distance, does this map extend to an isomorphism? The answer to this question is ‘yes’ in a lot of cases but, concerning counterexamples, again the small n-values turn up. In an appendix, we concentrate on the following problem. Suppose we are given some pieces (= some points) of one of our favourite puzzles (being in this case the finite dual split Cayley hexagon), and the following rule of play holds: whenever we have two pieces between which only one piece is missing (= two collinear points), we are allowed to plug in this missing piece (= add all the points on the joining line). With how many pieces do we have to start to end up with the complete puzzle (= to generate the whole point set)? We investigate this problem for some small cases with the help of the computer. So far the rough sketch of the pieces of knowledge added by this thesis to the big puzzle of the theory of generalized polygons. A more extensive introduction to each problem can be found at the beginning of the relevant chapters.

One of the most interesting things about mathematical puzzles is that they are never solved ‘completely’: answering one question, related problems arise.

This is the reason why you will meet the symbol ? △ at various places in this thesis. Some of these questions only turned up when writing this thesis, others kept us puzzling for a while, but the pieces we collected were not enough to complete the whole thing.

At least three aspects distinguished me during these three years of research from a ‘normal’ puzzler. Indeed, many hobby puzzlers would envy the financial support of the FWO, the Fund for Scientific Research-Flanders. Also, puzzling was not at all a lonely business, being surrounded by the interest of the ‘geometric part’ of the maths department here. The puzzles I considered had no key included (since there was no key granted after all), but what I had was much better: a constant help desk, providing me puzzles, puzzle pieces and prepared to listen to all my puzzling attempts of solving. Thanks Hendrik.

Last but not least, I want to thank my parents and my sister for both listening to my puzzling doubts, and remembering me from time to time that puzzling is a game after all. My mother once taught me that, when making a puzzle, the corner pieces are the most important ones. Thanks for being these corners.

Eline Govaert June 2001 iv

Contents

1 Introductory guide 1

1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.1 Incidence geometries . . . . . . . . . . . . . . . . . . . 2 1.1.2 Definition of a generalized n-gon . . . . . . . . . . . . . 3 1.1.3 A bunch of terminology . . . . . . . . . . . . . . . . . 4

1.2 Restrictions on the parameters . . . . . . . . . . . . . . . . . . 5 1.3 Some words on regularity . . . . . . . . . . . . . . . . . . . . . 6 1.4 Generalized quadrangles . . . . . . . . . . . . . . . . . . . . . 7

1.4.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4.2 Classical generalized quadrangles . . . . . . . . . . . . 8 1.4.3 Generalized quadrangles of order s = 2. . . . . . . . . . 8 1.4.4 The construction of Payne . . . . . . . . . . . . . . . . 9

1.5 Classical generalized hexagons and octagons . . . . . . . . . . 9 1.5.1 Generalized hexagons . . . . . . . . . . . . . . . . . . . 9 1.5.2 Generalized octagons . . . . . . . . . . . . . . . . . . . 13

1.6 Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.7 Ovoidal subspaces . . . . . . . . . . . . . . . . . . . . . . . . . 14 1.8 Characterizations of classical hexagons . . . . . . . . . . . . . 15 1.9 Coordinatization . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.9.1 Labelling of the elements . . . . . . . . . . . . . . . . . 18 1.9.2 Coordinates of classical generalized hexagons . . . . . . 20

1.10 More geometries . . . . . . . . . . . . . . . . . . . . . . . . . . 22 2 Characterizations of (dual) classical generalized hexagons 25

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.2 Characterizations of H(q) and T(q^{3}, q), q even . . . . . . . . . 27 2.3 Characterizations of H(q)^{D} and T(q, q^{3}) . . . . . . . . . . . . . 38 2.4 A characterization of Moufang hexagons using ovoidal subspaces 47 2.5 Two characterizations of the Hermitian spread in H(q) …. 51

3 Forgetful Polygons 59 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.2 Definitions and first examples . . . . . . . . . . . . . . . . . . 60 3.3 Classification for n odd . . . . . . . . . . . . . . . . . . . . . . 63 3.4 Square forgetful pentagons . . . . . . . . . . . . . . . . . . . . 70 3.5 Classification results for n even . . . . . . . . . . . . . . . . . 78 3.6 Short forgetful quadrangles . . . . . . . . . . . . . . . . . . . . 93

3.6.1 General properties . . . . . . . . . . . . . . . . . . . . 93 3.6.2 Examples of short forgetful quadrangles . . . . . . . . 97 3.6.3 Characterization results . . . . . . . . . . . . . . . . . 99 3.6.4 Short forgetful quadrangles ‘arising’ from generalized

quadrangles. . . . . . . . . . . . . . . . . . . . . . . . . 109

4 Distance-preserving maps 111 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 4.2 Main theorems and some words about the proof . . . . . . . . 112 4.3 Some examples by pictures . . . . . . . . . . . . . . . . . . . . 115 4.4 Proof of the Point-Line Theorem . . . . . . . . . . . . . . . . 118

4.4.1 Case i < \frac{n}{2} . . . . . . . . . . . . . . . . . . . . . . . . . 118 4.4.2 Case i = \frac{n}{2} + 1, i even . . . . . . . . . . . . . . . . . . 119 4.4.3 Case \frac{n}{2} < i < n − 2 . . . . . . . . . . . . . . . . . . . . 120 4.4.4 Case i = n − 2 . . . . . . . . . . . . . . . . . . . . . . 123 4.4.5 Case i = n − 1 . . . . . . . . . . . . . . . . . . . . . . 133

4.4.6 Case i = n/2 . . . . . . . . . . . . . . . . . . . . . . . 138 4.5 Some exceptions and applications to the Point-Line Theorem . 144 4.6 The exception in the Flag Theorem . . . . . . . . . . . . . . . 152 4.7 Proof of the Flag Theorem . . . . . . . . . . . . . . . . . . . . 156

4.7.1 Case i < n/2 . . . . . . . . . . . . . . . . . . . . . . . 156 4.7.2 Case i = n/2 . . . . . . . . . . . . . . . . . . . . . . . 158 4.7.3 Case n/2 < i < n − 2 . . . . . . . . . . . . . . . . . . . 160 4.7.4 Case i = n − 2 . . . . . . . . . . . . . . . . . . . . . . 163

4.8 Proof of the Special Flag Theorem . . . . . . . . . . . . . . . 186 4.8.1 Case i_{p} ≤ n/2 . . . . . . . . . . . . . . . . . . . . . . . 187 4.8.2 Case n/2 < i_{p} < n − 2 . . . . . . . . . . . . . . . . . . 188 4.8.3 Case i_{p} = n − 2 . . . . . . . . . . . . . . . . . . . . . . 193

A Minimal generating sets in H(q)^{D} 195 A.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 A.2 14 ≤ m ≤ 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 A.3 m = 14 if q ∈ {4, 8, 9, 16} . . . . . . . . . . . . . . . . . . . . . 199 A.4 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205

B Veelhoeken in Veelvoud 207 B.1 Van veelhoek naar veralgemeende veelhoek . . . . . . . . . . . 207 B.2 Ken uw klassiekers . . . . . . . . . . . . . . . . . . . . . . . . 208 B.3 Vergeethoeken . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 B.4 Veelhoeken door een speciale bril . . . . . . . . . . . . . . . . 212 B.5 Strategische verzamelingen in een klassieke zeshoek . . . . . . 213

Bibliography 215

Index of Notations 221 viii

Chapter 1

Introductory guide

The aim of the first chapter is to mention some highlights in the history of generalized polygons, and to introduce the notions that are needed to understand the pieces of knowledge we will add in the further chapters. This introduction is based on the monograph ‘Generalized Polygons’ (Van Maldeghem [57]).

We opt to give three equivalent definitions of generalized polygons. The first one uses a lot of n-gons, and therefore explains why the polygons considered here can really be called ‘generalized’. As Chapter 4 deals with distancepreserving maps, we give a definition in terms of the distance function. The third definition characterizes the incidence graph in a very compact way. The next sections contain the inevitable list of notions and properties that starts every introductory course on generalized polygons. We stress the notion of regularity, and introduce you to two classes of classical generalized hexagons, namely the split Cayley and the twisted triality hexagons. Since Chapter 2 concerns characterizations of these hexagons, we give an overview of the known geometric characterizations of the classical generalized hexagons. Finally, we explain how one can coordinatize generalized hexagons. Indeed, coordinatization is the main tool in appendix A. The calculations there are done by computer, but here we wanted to give the reader the same information about coordinatization as we told the computer.

1.1 Definitions

1.1.1 Incidence geometries

An incidence geometry (or geometry for short) is a triple Γ = (P, L, I), where P and L are two disjoint nonempty sets the elements of which are called points and lines, respectively, and where I ⊆ P × L ∪ L × P is a symmetric relation, called the incidence relation. When a point p is incident with a line L, we also say that p lies on L or that L goes through p. Points and lines are called the elements of the geometry. A flag is a pair {p, L}, with p a point incident with the line L. The set of flags of the geometry is denoted by F. Adjacent flags are distinct flags which have an element in common. An antiflag is a pair {p, L} where p and L are not incident. The dual of the geometry Γ is the geometry Γ^{D} obtained by interchanging the roles of points and lines, i.e. Γ^{D} = (L, P, I). We define the double 2Γ of Γ as the geometry with point set F, line set P ∪L and natural incidence relation (this definition is in fact the dual of the one given in [57]). A geometry is called thick if every element is incident with at least three other elements. A thin element is an element incident with exactly two other elements. If there exist constants s, t such that every line is incident with exactly s + 1 points, and every point is incident with exactly t + 1 lines, the pair (s, t) is called the order of Γ. If s = t, one says that Γ has order s. A geometry is said to be finite if both P and L are finite sets. A subgeometry of Γ is a geometry Γ′ = (P′, L′, I′) with P′ ⊆ P, L′ ⊆ L and I′ the restriction of the relation I to P′ × L′ ∪ L′ × P′.

A path in Γ between the elements x and y is a sequence (x = x_{0}, x_{1}, …, x_{k} = y) of points and lines such that x_{i−1}Ix_{i}, for all i ∈ {1, 2, …, k}. If x_{i−1} ≠ x_{i+1}, for all i ∈ {1, 2, …, k − 1}, the path is said to be non-stammering. The number k is called the length of the path. A non-stammering path of length k is called a k-path. If x = y, we talk about a closed path, and a non-stammering closed path is called a circuit. An ordinary ngon is a closed path (x_{0}, x_{1}, …, x_{2n} = x_{0}) of length 2n > 2 for which all x_{i}, i ∈ {0, 1, …, 2n − 1} are distinct. The distance δ(x, y) between two elements x, y of Γ is the length of a shortest path joining x and y, if such a path exists. If not, then the distance between x and y is by definition ∞. Similarly, one can define a distance function on the set of flags.

The incidence graph of a geometry Γ is the graph with as vertex set V = P ∪ L and as edges the flags of Γ (hence adjacency in the incidence graph coincides with incidence in the geometry). The girth if this graph is the length of a minimal circuit.

1.1.2 Definition of a generalized n-gon

Let n ≥ 1 be a natural number. A weak generalized n-gon is a geometry Γ = (P, L, I) such that the following two axioms are satisfied:

(i) there are no ordinary k-gons in Γ, with k < n,

(ii) every two elements of P ∪ L are contained in an ordinary n-gon.

If a weak generalized n-gon Γ is thick, we call it a generalized n-gon. As shown in [57] (Lemma 1.3.2), thickness is equivalent with the existence of at least one ordinary (n + 1)-gon in Γ.

A generalized 2-gon (digon) is a rather trivial geometry in which every point is incident with every line. The generalized 3-gons are exactly the projective planes. Instead of 4-gons, 5-gons, 6-gons, 8-gons and n-gons, we shall also speak of quadrangles, pentagons, hexagons, octagons and polygons. A generalized n-gon consists of lots of ordinary n-gons, so-called apartments. This terminology is inherited from the theory of buildings. Indeed, generalized polygons are exactly the buildings of rank 2 (i.e. with two types of elements, namely points and lines), see for instance [57], section 1.3.7. Note that the dual of a generalized polygon is again a generalized polygon. Also, the double of a generalized n-gon is a weak generalized 2n-gon with thin points and thick lines.

All weak non-thick generalized polygons arise either from ordinary polygons or from generalized polygons by inserting paths (see Structure Theorem of Tits, [57], section 1.6). So we will mainly be interested in (thick) generalized n-gons. In this case, one can prove that the generalized polygon Γ has an order (s, t) (s and t are also called the parameters of Γ). If n is odd, then necessarily s = t (see [57], Lemma 1.5.3). There exist (thick) generalized n-gons for every n. Examples are provided by a free construction process due to Tits [52] (or see [57], section 1.3.13). We will give explicit examples of generalized quadrangles an hexagons below.

To conclude this section, we give two alternative definitions of generalized n-gons, the first in terms of the distance function, the second one using the incidence graph.

Lemma 1.1.1 ([57], Lemma 1.3.5) A thick geometry Γ = (P, L, I) is a generalized n-gon if and only if the following axioms hold for the distance δ:

(i) If x, y ∈ P ∪ L and δ(x, y) = k < n, then there is a unique path of length k joining x to y.

(ii) For every x ∈ P ∪ L, we have n = max{δ(x, y) : y ∈ P ∪ L}.

Lemma 1.1.2 ([57], Lemma 1.3.6) A geometry Γ = (P, L, I) is a generalized n-gon if and only if the incidence graph of Γ is a connected graph of diameter n and girth 2n, such that each vertex lies on at least three edges.

1.1.3 A bunch of terminology

Let Γ be a generalized n-gon, n ≥ 3. Two points p and q at distance 2 are called collinear, and we denote by pq the unique line joining p and q. Dually, two lines L and M at distance 2 are called concurrent, and the unique point p = L ∩ M incident with both is called the intersection point of L and M. For two elements x and y at distance 2, we also write x ⊥ y. Two elements x and y of Γ lying at maximal distance n are said to be opposite. If two elements x and y of Γ lie at distance k < n, the unique k-path between x and y is denoted by [x, y]. If k is even, x ⋈ y denotes the unique element of [x, y] at distance k/2 from both x and y. The unique element of [x, y] incident with x is called the projection of y onto x, and denoted by proj_{x}y. By definition, we put proj_{x}x = x. When it suits us, we consider a path as a set so that we can take intersections of paths. For instance, if [x, y] = (x = x_{0}, x_{1}, …, x_{i}, x_{i+1}, …, x_{k} = y) and [x, z] = (x = x_{0}, x_{1}, …, x_{i}, x′_{i+1}, x′_{ℓ} = z) with no x_{j} equal to any x′_{j′}, for some 0 < i ≤ k, l and all j, i < j ≤ k, and all j′, i < j′ ≤ l, then we write [x, y] ∩ [x, z] = [x, x_{i}].

For any element x of Γ, and any integer i ≤ n, we denote by Γ_{i}(x) the set of elements of Γ at distance i from x, and by Γ_{≠i}(x) the set of elements of Γ not at distance i from x. If κ is a set of natural numbers, then Γ_{κ}(x) denotes the set of elements z of Γ for which δ(x, z) ∈ κ. If p is a point, the set Γ_{1}(p) of all lines through p is called a line pencil. Dually, if L is a line, the set Γ_{1}(L) of all points on L is called a point row. For a point x, the set x^{⊥} := Γ_{2}(x) is called the perp of x. The perp of a set X of points is the set of points collinear with every element of X. So X^{⊥} = ⋂_{x∈X} x^{⊥}. For a point x of Γ, we denote by x^{⊥ ⊥} the set of points not opposite x.

A (weak) sub-n-gon Γ′ of a generalized n-gon is a subgeometry which is itself a (weak) generalized n-gon. If every line pencil of Γ′ coincides with the corresponding line pencil of Γ, the subpolygon Γ′ is called ideal. Dually, if every point row of Γ′ coincides with the corresponding point row of Γ, the subpolygon Γ′ is called full.

Let Γ = (P, L, I) and Γ′ = (P′, L′, I′) be two generalized n-gons. An isomorphism or collineation of Γ onto Γ′ is a bijection α : P → P′, inducing a bijection of L onto L′ so that incidence and non-incidence is preserved, i.e. pIL ⟺ p^{α}I′L^{α}, for all p ∈ P and L ∈ L. Lemma 1.3.14 of [57] states that, if n ≥ 4, any bijection from P onto P′ preserving collinearity and noncollinearity, extends to an isomorphism from Γ to Γ′. An anti-isomorphism of Γ onto Γ′ is a collineation of Γ onto the dual Γ′^{D} of Γ′. An automorphism (anti-automorphism) of Γ is an isomorphism (anti-isomorphism) of Γ onto itself.

We say that a generalized polygon Γ = (P, L, I) is embedded in the projective space PG(d, 𝕂) if distinct points and lines of Γ are distinct points and lines of PG(d, 𝕂), with the natural incidence, and the point set of P generates PG(d, 𝕂).

1.2 Restrictions on the parameters

Theorem 1.2.1 (Feit & Higman [21]) Finite generalized n-gons, n ≥ 3, exist only for n ∈ {3, 4, 6, 8}.

Lemma 1.2.2 ([57], 1.5.5) Let Γ = (P, L, I) be a finite (weak) generalized n-gon of order (s, t), with n ∈ {3, 4, 6, 8}. Then we have

|P| = \{ \begin{matrix}s^{2} + s + 1 & if n = 3, \\ (1 + s)(1 + st) & if n = 4, \\ (1 + s)(1 + st + s^{2}t^{2}) & if n = 6, \\ (1 + s)(1 + st)(1 + s^{2}t^{2}) & if n = 8. \end{matrix}

Dually,

|L| = \{ \begin{matrix}s^{2} + s + 1 & if n = 3, \\ (1 + t)(1 + st) & if n = 4, \\ (1 + t)(1 + st + s^{2}t^{2}) & if n = 6, \\ (1 + t)(1 + st)(1 + s^{2}t^{2}) & if n = 8. \end{matrix}

The following theorem is a combination of results of Feit & Higman [21], Higman [31] and Haemers & Roos [30].

Theorem 1.2.3 Let Γ be a finite generalized n-gon of order (s, t).

• If n = 4, then s ≤ t^{2} and t ≤ s^{2}.

• If n = 6, then st is a square; s ≤ t^{3} and t ≤ s^{3}.

• If n = 8, then 2st is a square, in particular s ≠ t; s ≤ t^{2} and t ≤ s^{2}.

Theorem 1.2.4 (Thas [38],[40],[41],[42]) Let Γ′ be an ideal weak sub-n-gon of order (s′, t) of a finite generalized n-gon Γ of order (s, t), with s′ ≠ s. Then one of the following cases occurs.

• n = 4 and s ≥ s′t and s ≥ t ≥ s′;

• n = 6 and s ≥ s′^{2}t and s ≥ t ≥ s′;

• n = 8 and s ≥ s′^{2}t.

A finite generalized hexagon of order (s, t) for which s = t^{3} or t = s^{3} is called an extremal hexagon.

The following two properties are used in Chapter 3. A triad of points in a generalized quadrangle is a triple of pairwise non-collinear points.

Proposition 1.2.5 (i) (Bose & Shrikhande [6]) Let Γ be a finite generalized quadrangle of order (s, t) with t = s^{2}. Then for every triad {x, y, z} of points, there are exactly s + 1 points collinear with x, y and z.

(ii) (Thas [40]) Let Γ be a finite generalized quadrangle of order (s, t), and Γ′ an ideal subquadrangle of order (s′, t) satisfying s = s′t. Then every point of Γ not in Γ′ lies on a unique line of Γ′.

1.3 Some words on regularity

Let Γ be a generalized n-gon, and 2 ≤ i ≤ \frac{n}{2}. If two elements x and y are opposite, the set Γ_{i}(x)∩Γ_{n−i}(y) = x_{[i]}^{y} is called the distance-i-trace of y with respect to x. For i = 2, it is convenient to call the distance-2-trace x_{[2]}^{y} simply a trace, and denote it by x^{y}. The element x is distance-i-regular if distinct distance-i-traces with respect to x have at most 1 element in common (i.e. the distance i-traces with respect to x behave as lines, since they intersect in 0,1 or all elements). The element x is regular if it is distance-i-regular, for all 2 ≤ i ≤ \frac{n}{2}. A generalized polygon is said to be point-distance-i-regular respectively line-distance-i-regular if all points respectively all lines are distance-i-regular. Instead of distance-i-regular, we sometimes use i-regular for short (note that no confusion with the notion of 3-regularity in generalized quadrangles as defined in Payne & Thas [34] can occur, since distance-3-regularity is not defined for generalized quadrangles). A regulus is a distance-\frac{n}{2}-trace x_{[ \frac{n}{2} ]}^{y}, which we also denote by ⟨x, y⟩. If Γ is distance- \frac{n}{2}-regular, a regulus is determined by two of its elements u, v. In this case, we denote by R(u, v) the unique regulus containing the elements u and v. If v and v are points (lines), we also talk about the point regulus (line regulus) through u and v.

Now let Γ be a point-distance-2-regular generalized hexagon, and p, q two opposite points of Γ. Define the following geometry Γ(p, q). A point x of Γ belongs to Γ(p, q) if x ∈ y^{z} or x ∈ z^{y}, with y ∈ p^{q}, z ∈ q^{p}, and y, z opposite points. The lines of Γ(p, q) are the lines of Γ containing at least two points of Γ(p, q). Incidence is natural. Then Γ(p, q) is the unique weak non-thick ideal subhexagon of Γ containing p and q (see [57], Lemma 1.9.10). Also, one proves that the point set of Γ(p, q) is the union of the point sets of two projective planes Γ^{+}(p, q) and Γ^{−}(p, q). The points of Γ^{+}(p, q) are the points of Γ(p, q) at distance 0 or 4 from p; the lines of Γ^{+}(p, q) are the traces x^{y}, where x is a point of Γ^{−}(p, q), and y is a point of Γ^{+}(p, q) opposite x. Similarly for Γ^{−}(p, q) = Γ^{+}(q, p).

For later purposes, we mention the following result.

Theorem 1.3.1 (Van Maldeghem [56] A generalized octagon cannot be point-distance-2-, nor point-distance-3-regular.

1.4 Generalized quadrangles

1.4.1 Definition

We give an equivalent (more common) definition of generalized quadrangles.

A generalized quadrangle is a thick incidence geometry Γ = (P, L, I) such that the following axioms are satisfied:

(i) If p is a point of Γ not incident with the line L of Γ, then there exists a unique point incident with L and collinear with p.

(ii) The geometry Γ contains at least one antiflag.

We refer to axiom (i) as ‘the main axiom’ for a generalized quadrangle.

1.4.2 Classical generalized quadrangles

We mention some examples of classical^{1} generalized quadrangles, including all finite ones.

• Let Q be a non-singular quadric in PG(d, 𝕂), with 𝕂 a field, of Witt index 2 (i.e. the quadric contains lines but no subspaces of higher dimension). The points and lines of the quadric form a (weak) generalized quadrangle Q(d, 𝕂). In the finite case (putting 𝕂 = GF(q)) only the dimensions d = 3, 4, 5 occur. We then obtain a (weak) generalized quadrangle of order (q, 1) (sometimes also called a grid), (q, q) and (q, q^{2}), respectively. All lines of these quadrangles are regular.

• Let H be a non-singular Hermitian variety in PG(d, 𝕃), with 𝕃 a skew field, of Witt index 2. The points and lines of the Hermitian variety form a generalized quadrangle H(d, 𝕃). In the finite case (putting 𝕃 = GF(q^{2})) only the dimensions d = 3, 4 occur. We then obtain a generalized quadrangle of order (q^{2}, q) and (q^{2}, q^{3}), respectively.

• Let τ be a symplectic polarity in PG(3, 𝕂), 𝕂 a field. The points of PG(3, 𝕂) together with the absolute lines of τ define a generalized quadrangle W(𝕂), called the symplectic quadrangle. In the finite case (putting 𝕂 = GF(q)) we obtain a quadrangle of order (q, q) (denoted W(q)).

One has the following isomorphisms between the finite classical generalized quadrangles (see Payne & Thas [34]):

• Q(4, q) ≅ W(q)^{D}

• W(q) ≅ W(q)^{D} ⟺ q is even

• Q(5, q) ≅ H(3, q^{2})^{D}.

1.4.3 Generalized quadrangles of order s = 2.

For later purposes, we mention the following results.

• By a result of Cameron [10], a generalized quadrangle of order (2, t) is necessarily finite, and hence has t = 2 or t = 4.

1For a motivation of the word ‘classical’ we refer to [57], Chapter 2.

• There is a unique generalized quadrangle of order 2, namely the symplectic quadrangle W(2). One has the following well-known construction of this generalized quadrangle. Let S = {1, 2, …, 6}. A duad is an unordered pair ij of distinct elements of S. A syntheme is a set {ij, kl, mn} of three duads for which {i, j, k, l, m, n} = S. Now the geometry Γ = (P, L, I) with as points the duads of S, as lines the synthemes and symmetrized containment as the incidence relation, is the unique generalized quadrangle of order 2.

• There is a unique generalized quadrangle of order (2, 4), namely the quadrangle Q(5, 2).

1.4.4 The construction of Payne

Let Γ = (P, L, I) be a finite generalized quadrangle of order q admitting a regular line L. We define the following geometry Γ′ = (P′, L′, I′). The lines of Γ′ are the lines of Γ different from L and not intersecting L. The points are of two types. Points of type (A) are the points of Γ not on the line L. Points of type (B) are the distance-2-traces containing L. Incidence is the incidence of Γ if defined, and symmetrized containment otherwise. Then the geometry Γ′ is a generalized quadrangle of order (q + 1, q − 1).

1.5 Classical generalized hexagons and octagons

1.5.1 Generalized hexagons

The examples of generalized hexagons given below are called ‘classical’, because they live on classical objects (namely quadrics) in projective space. They first appeared in Tits [49]. We sketch this construction, for proofs we refer to [57], section 2.4.

Let Q^{+}(7, 𝕂) be the non-singular hyperbolic quadric in PG(7, 𝕂) (with standard equation X_{0}X_{1} + X_{2}X_{3} + X_{4}X_{5} + X_{6}X_{7} = 0). This quadric has Witt index 4, i.e. it contains 3-spaces (called the ‘generators’), but no subspaces of higher dimension. The generators can be divided into two families: two generators belong to the same family if they are disjoint or they intersect in a line. Any plane of Q^{+}(7, 𝕂) is contained in exactly two generators, one of each family. Now one can define the following geometry Ω(𝕂). There are four types of elements. The 0-points are the points of Q(7, 𝕂), the lines

Figure 1.1: the D_{4}-diagram and a triality

are the lines of Q^{+}(7, 𝕂), the 1-points are the 3-spaces of the first family of generators and the 2-points are the 3-spaces of the second family. Denote by P^{(i)} the set of i-points. Incidence is containment if defined, and a 1-point and a 2-point are incident in Ω(𝕂) if the corresponding 3-spaces intersect in a plane of the quadric. The diagram of this geometry Ω(𝕂) is the so-called D_{4}-diagram (for an introduction to the theory of diagrams, see for instance Buekenhout [9]). One sees that the 0-,1- and 2-points play the same role in this geometry. In [57], section 2.4.6, it is explained how one can label the 1- and the 2-points in the same way as the 0-points, i.e. with an 8-tuple (x_{0}, x_{1}, …, x_{7}).

A triality of Ω(𝕂) is a map

θ : L → L, P^{(0)} → P^{(1)}, P^{(1)} → P^{(2)}, P^{(2)} → P^{(0)},

preserving incidence and such that θ^{3} is the identity.

An i-point p is called absolute if pIp^{θ}, a line is absolute if it is fixed by θ. Now one shows that if θ satisfies a weak additional assumption (basically saying that there are enough absolute points and lines), the geometry with as points the absolute i-points for a fixed i, and as lines the absolute lines, is a generalized hexagon.

Let σ be an automorphism of the field 𝕂 of order 1 or 3, and consider the following map:

τ_{σ} : P^{(i)} → P^{(i+1)} : (x_{j})_{j∈J} → (x_{j}^{σ})_{j∈J}, i = 0, 1, 2 mod 3.

Then τ_{σ} is a triality, and the associated geometry of absolute i-points and lines is a generalized hexagon of order (|𝕂|, |𝕂^{(σ)}|), with 𝕂^{(σ)} the subfield of 𝕂 consisting of those elements of 𝕂 that are fixed by σ. If σ = 1, the associated hexagon is called the split Cayley hexagon H(𝕂). For 𝕂 = GF(q), this hexagon is denoted by H(q). Its order is (q, q). If σ ≠ 1, the associated hexagon is called the twisted triality hexagon T(𝕂, 𝕂^{(σ)}, σ). The dual of this hexagon is denoted by T(𝕂^{(σ)}, 𝕂, σ). In the finite case, putting 𝕂 = GF(q^{3}), σ is necessarily the map x → x^{q}, ∀x ∈ GF(q^{3}). The corresponding hexagon is denote by T(q^{3}, q), and has order (q^{3}, q). Its dual is denoted by T(q, q^{3}) and has order (q, q^{3}).

Note that Tits [49] classifies all trialities of the geometry Ω(𝕂) having at least one absolute point. The only trialities producing thick generalized hexagons are the ones given above (the original definition of Tits of a generalized hexagon was weaker, such that it included also the geometries of absolute points and lines of the other trialities).

There are two other types of generalized hexagons that are also said to be classical, namely the mixed hexagons H(𝕂, 𝕂′) and the hexagons of type ^{6}D_{4} (and their duals). These hexagons are closely related to split Cayley hexagons over a field with characteristic 3 and to twisted triality hexagons respectively (they arise from them by a generalization in the choice of the coordinates, see [57], 3.5.3 and 3.5.9), but they do not exist in a finite version however. In the following, when talking about the finite classical hexagons, we will always mean the hexagons H(q) and T(q^{3}, q). Their duals H(q)^{D} and T(q, q^{3}) are referred to as the finite dual classical hexagons. We now list some properties of the classical hexagons.

• The split Cayley hexagon H(𝕂) is selfdual if and only if 𝕂 is a perfect field of characteristic 3.

• All points of a split Cayley hexagon H(𝕂) are regular. All lines are regular if and only if the field 𝕂 has characteristic 3.

• All points and lines of a mixed hexagon are regular.

• All points of a twisted triality hexagon or a hexagon of type ^{6}D_{4} are regular. No line of such a hexagon is distance-2-regular.

Apart from the examples of generalized hexagons given above, other important examples are the ones related to the exceptional groups of type E_{6} and E_{8}. Also these ones only exist in the infinite case.

Tits’ description of the split Cayley hexagon All the points of the split Cayley hexagon represented on the quadric Q^{+}(7, 𝕂) lie in fact in a certain hyperplane γ of PG(7, 𝕂). Conversely, all the points of Q^{+}(7, 𝕂) lying in this hyperplane γ are points of the hexagon. One now obtains the following description of the split Cayley hexagon H(𝕂). The points of the hexagon are all the points of a non-singular parabolic quadric Q(6, 𝕂) in PG(6, 𝕂) with standard equation

X_{0}X_{4} + X_{1}X_{5} + X_{2}X_{6} = X_{3}^{2}.

The lines of the hexagon are exactly the lines of Q(6, 𝕂) whose Grassmann coordinates satisfy the following six linear equations:

\begin{matrix}p_{12} = p_{34} & p_{54} = p_{32} & p_{20} = p_{35} \\ p_{65} = p_{30} & p_{01} = p_{36} & p_{46} = p_{31}. \end{matrix}

We now list some ‘translations’ of objects existing in the hexagon H(𝕂) to this representation on the quadric Q(6, 𝕂).

• Two points in the hexagon are opposite if and only if they are noncollinear on the quadric (also the twisted triality hexagon has this property).

• A line of the quadric is either a line or a distance-2-trace of the hexagon.

• The points collinear in the hexagon with a point p are exactly the points of a fixed plane through p lying on the quadric. (The twisted triality hexagon has a similar property: the points collinear in the hexagon with a point p are contained in a plane through p lying on the quadric.)

• A line regulus in the hexagon is also a regulus (one set of generators of a hyperbolic quadric Q^{+}(3, 𝕂)) on the quadric. A point regulus in the hexagon consists of the points of a plane intersecting the quadric in a non-singular conic.

• The points and lines of a thin ideal subhexagon Γ(p, q) (see section 1.3) all lie in a hyperplane intersecting the quadric Q(6, q) in a non-singular hyperbolic quadric Q^{+}(5, q). The ‘planes’ Γ^{+}(p, q) and Γ^{−}(p, q) are indeed projective planes lying on the quadric Q(6, q).

The following result is due to Cohen & Tits [11].

Theorem 1.5.1 A finite generalized hexagon of order (s, t) with s = 2 is isomorphic to one of the classical hexagons H(2), H(2)^{D} or T(2, 8).

1.5.2 Generalized octagons

The classical generalized octagons are generally called the Ree-Tits octagons. In the finite case, they only exist over fields GF(q) with q = 2^{2e+1}, and have - up to duality - order (q, q^{2}). There is no elementary description known of these classical octagons. In [57], section 2.5, a construction starting from a building of type F_{4} is given. In Joswig & Van Maldeghem [32], one can find a description with coordinates.

1.6 Groups

Projectivities For two opposite elements x and y of a generalized n-gon Γ, the projection map defines a bijection (denoted by [x; y]) from the set Γ_{1}(x) to the set Γ_{1}(y). For elements x_{0}, …, x_{k} with x_{i} opposite x_{i+1} for 0 ≤ i < k, the composition [x_{0}; x_{1}] … [x_{k−1}; x_{k}] is called a projectivity from x_{0} to x_{k}. If x_{0} = x_{k}, then we obtain a permutation of the set Γ_{1}(x_{0}). The set of all such permutations of Γ_{1}(x_{0}) is a group, called the group of projectivities of x_{0} and denoted by Π(x_{0}).

Elations and homologies Let Γ be a generalized n-gon, and γ a fixed path of length n−2. A γ-elation (or elation for short) is a collineation of Γ fixing all elements incident with at least one element of γ. Let v, w be two opposite elements of Γ. A {v, w}- homology is a collineation fixing every element incident with v or w.

The Moufang property Let γ = (v_{1}, v_{2}, …, v_{n−1}) be a fixed (n − 2)-path in a generalized n-gon Γ. Let v_{0} be a fixed element incident with v_{1}, v_{0} ≠ v_{2}, and denote by V the set of all elements incident with v_{0} different from v_{1}. Let v_{n} be a fixed element incident with v_{n−1}, v_{n} ≠ v_{n−2}, and denote by A the set of all apartments containing v_{0}, γ and v_{n}. Then the set of all γ-elations forms a group G acting semi-regularly on both V and A (see [57], Proposition 4.4.3). If G acts transitively on the set V (or equivalently on the set A), then the path γ is called a Moufang path. If all (n − 2)-paths are Moufang paths, Γ is said to be a Moufang polygon. For a Moufang polygon, the collineation group generated by all elations is often called the little projective group of Γ.

A famous result of Tits [51], [53] and Weiss [58] states that Moufang ngons exist only for n ∈ {3, 4, 6, 8}. All Moufang n-gons are classified, see Tits & Weiss [54]. In particular, all classical generalized polygons are Moufang polygons. The Moufang generalized hexagons are exactly the hexagons H(𝕂), T(𝕂, 𝕂^{(σ)}, σ) and their duals, the mixed hexagons, the hexagons of type ^{6}D_{4} and the hexagons related to the exceptional groups E_{6} and E_{8}.

The split Cayley hexagon For later purposes, we mention that the little projective group of the hexagon H(𝕂) is the group G_{2}(𝕂). It is exactly the group of automorphisms of this generalized hexagon that are induced by the group PGL(7, 𝕂) of linear transformations of PG(6, 𝕂) (the full automorphism group of H(𝕂) is isomorphic to the semi-direct product G_{2}(𝕂) : Aut(𝕂)). In the finite case (putting 𝕂 = GF(q)), we have

|G_{2}(q)| = q^{6}(q^{6} − 1)(q^{2} − 1).

Polygons arising from a BN-pair Let Γ be a generalized polygon, and G a group of automorphisms of Γ. Suppose that G acts transitively on the ordered apartments of Γ. Then one says that Γ is a Tits polygon with respect to G. Fix an apartment Σ and a flag F in Σ. Denote by B the stabilizer in G of the flag F and by N the stabilizer in G of the (unordered) apartment Σ. Then one says that (B, N) is a Tits system in G for Γ. Conversely, from such a Tits system, one can reconstruct the polygon Γ (see Tits [50]). A polygon arising in this way is called a polygon arising from a BN-pair.

Weyl group Consider an apartment Σ of a generalized n-gon Γ. The group of symmetries of Σ is the dihedral group D_{2n}. Let {p, L} be a flag of Σ, and denote by s_{p} (s_{L}) the reflection about p (L). Then D_{2n} can be seen as the Coxeter group with generators s_{p} and s_{L}, i.e. D_{2n} = ⟨s_{p}, s_{L}||(s_{p}s_{L})^{n} = (s_{L}s_{p})^{n} = 1, s_{p}^{2} = s_{L}^{2} = 1⟩. Using the language of buildings, D_{2n} is the Weyl group of the apartment Σ. This allows us to define the Coxeter distance between flags of Γ. Let f_{1} and f_{2} be two flags of Γ, and Σ′ an apartment containing f_{1} and f_{2}. Then the Coxeter distance δ^{∗}(f_{1}, f_{2}) between f_{1} and f_{2} is the unique element of D_{2n} mapping f_{1} to f_{2}.

1.7 Ovoidal subspaces

An ovoid O of a generalized n-gon Γ, n even, is a set of mutually opposite points such that every element of Γ is at distance at most n/2 from at least one element of O. A spread of a generalized polygon is the dual of an ovoid.

Let Γ be a generalized quadrangle. Then equivalently, an ovoid O of Γ is a set of points such that each line of Γ is incident with a unique point of O. A regular ovoid of a generalized quadrangle of order (s, t) is an ovoid O having the property that for any two points o_{1} and o_{2} of O, |{o_{1}, o_{2}}^{⊥⊥}| = t + 1 and {o_{1}, o_{2}}^{⊥⊥} ⊂ O. A geometric hyperplane H of a generalized quadrangle is a proper subset of the point set such that for an arbitrary line L, either all the points of L belong to H, or |L ∩ H| = 1.

Let Γ be a generalized hexagon. Reformulating the general definition, an ovoid O of Γ is a set of mutually opposite points such that each point of Γ not in O is collinear with a unique point of O. The finite hexagon T(q^{3}, q) and its dual T(q, q^{3}) do not admit ovoids (by a counting argument). In fact, A. Offer recently proved that any finite generalized hexagon admitting an ovoid necessarily has order (q, q). An ovoid of a hexagon of order (q, q) contains q^{3} + 1 points. Thas [44] gives the following construction of a spread of the hexagon H(q), which works for all possible values of q.

Let H(q) be defined on the quadric Q = Q(6, q), and let γ be a hyperplane of PG(6, q) intersecting Q in a non-singular elliptic quadric Q^{−}(5, q). Then the lines of H(q) lying in γ constitute a spread of both the hexagon H(q) and the quadrangle Q(5, q). This spread is called the Hermitian or classical spread of the split Cayley hexagon. One has the following characterization of this spread.

Theorem 1.7.1 (Bloemen, Thas & Van Maldeghem [4], Theorem 9) If S is a spread of H(q) for which holds that S is the union of q^{2} line-reguli through L, for each L ∈ S, then S is the Hermitian spread.

For more constructions of ovoids of the classical hexagons, we refer the reader to Bloemen, Thas & Van Maldeghem [4]. An ovoidal subspace O of a generalized hexagon Γ is a proper subset of the point set of Γ such that each point of Γ not in O is collinear with a unique point of O. By Brouns & Van Maldeghem [7], an ovoidal subspace is either an ovoid, (the point set of) a full subhexagon, or the set of points at distance 1 or 3 from a given line M. Dually, one defines a dual ovoidal subspace. A dual ovoidal subspace is either a spread (type S), (the line set of) an ideal subhexagon (type H), or the set of lines at distance 1 or 3 from a given point p (type P).

1.8 Characterizations of classical hexagons

Ronan’s characterizations using regularity

Theorem 1.8.1 (Ronan [35]) If Γ is a point-distance-2-regular generalized hexagon, then all points and lines are distance-3-regular and Γ is a Moufang hexagon. Conversely, up to duality, all points of any Moufang hexagon are regular.

Every finite Moufang hexagon is classical or dual classical. This follows from a group-theoretical result of Fong & Seitz [22], [23], or alternatively from the classification of all Moufang polygons of Tits & Weiss [54]. Hence one has the following:

Theorem 1.8.2 A finite generalized hexagon is point-distance-2-regular if and only if it is classical.

Theorem 1.8.3 Ronan [36] A finite extremal hexagon is classical if and only if it is distance-3-regular.

Let Γ be a generalized hexagon, and x a point of Γ. Let y and z be two points opposite x and at distance 4 from each other, such that the point y ⋈ z lies at distance 4 from x. Then the set x^{y} ∩ x^{z} is called an intersection set. By definition, an intersection set is never empty. (Note that our definition of intersection set is slighty different from the one in Ronan [37].) An intersection set of H(q) or T(q^{3}, q) contains 1 or q + 1 points. An intersection set of H(q)^{D}, q not a power of 3, or T(q, q^{3}) contains 2, respectively q^{2} + 1 points.

Theorem 1.8.4 (Ronan [37]) Let Γ be a distance-3-regular generalized hexagon.

(i) If for every intersection set x^{y} ∩ x^{z}, one has |x^{y} ∩ x^{z}| = 1 or x^{y} = x^{z}, then Γ is point-distance-2-regular and hence a Moufang hexagon.

(ii) If for every intersection set x^{y} ∩ x^{z}, one has |x^{y} ∩ x^{z}| > 1, then Γ is line-distance-2-regular and hence a Moufang hexagon.

The characterizations above restricted to the finite case are summarized in Figure 1.2.

Hyperbolic and imaginary lines Let Γ be a generalized n-gon, and x, y two non-collinear points of Γ at mutual distance 2j. The distance-j hyperbolic line H(x, y) is the set of points not opposite all elements not opposite x and y. If x and y are opposite, one also speaks of an imaginary line, notation I(x, y). In fact, as shown in van Bon, Cuypers & Van Maldeghem [55], a distance-j hyperbolic line is exactly the intersection of all distance-j traces containing x and y. A distance-j

Figure 1.2: The finite Moufang hexagons

hyperbolic line H(x, y) is called long if the projection of H(x, y) onto any element L of Γ at distance n − 1 from all points of H(x, y) is surjective onto Γ_{1}(L) whenever it is injective. One proves that a long hyperbolic line H(x, y) coincides with any distance-j-trace containing any two of its points. For a lot of results concerning hyperbolic lines in generalized n-gons, we refer to [55]. We now restrict to hexagons.

Theorem 1.8.5 (van Bon, Cuypers & Van Maldeghem [55])

(i) All distance-2 hyperbolic lines of a generalized hexagon Γ are long if and only if Γ is isomorphic to H(𝕂).

(ii) All imaginary lines of a generalized hexagon Γ are long if and only if Γ is isomorphic to H(𝕂′), with 𝕂′ a perfect field of characteristic 2.

Characterizations using subpolygons

Theorem 1.8.6 (i) (Van Maldeghem [57], Corollary 6.3.7) A generalized hexagon Γ is point-distance-2-regular (and hence a Moufang hexagon) if and only if every ordinary heptagon in Γ is contained in at least one ideal split Cayley hexagon.

(ii) (De Smet & Van Maldeghem [20]) A finite generalized hexagon is isomorphic to T(q^{3}, q) if and only if every ordinary heptagon is contained in a proper ideal subhexagon.

A characterization using span-regularity A point p of a generalized hexagon is called span-regular if p is distance-2-regular, and if for every point x collinear with p, and every two points a and b opposite x such that p ∈ x^{a} ∩x^{b}, the condition |x^{a} ∩x^{b}| ≥ 2 implies x^{a} = x^{b}. The following characterization weakens the conditions of Theorem 1.8.1 in the finite case.

Theorem 1.8.7 (De Smet & Van Maldeghem [20], Brouns & Van Maldeghem [7]) Let Γ be a finite hexagon containing a dual ovoidal subspace all the points of which are span-regular. If any two opposite points of Γ are contained in a thin ideal subhexagon, then Γ is classical.

A characterization using intersections of traces

Theorem 1.8.8 (Thas [43]) Let Γ be a finite generalized hexagon of order (s, t), s ≥ t. Then Γ is isomorphic to H(q) if and only if any two distance-2-traces with respect to the same point meet in at least one point.

For other characterizations, including some algebraic ones, we refer the reader to [57], Chapter 6.

1.9 Coordinatization

Generalized polygons can be coordinatized in a way similar to the projective planes. In this thesis, we will only need coordinatization of the classical generalized hexagons. The aim of this section is to give a rough idea how the labelling of the elements works (in the case of hexagons), and to show how one works with these coordinates. For a detailed description of the general coordinatization theory, we refer the reader to [57], Chapter 3.

1.9.1 Labelling of the elements

Let Γ be a generalized hexagon of order (s, t), and choose a fixed apartment Σ, called the hat-rack of the coordinatization. The elements of Σ are denoted as in Figure 1.3. Let R_{1} and R_{2} be two sets of cardinality s and t respectively, both containing a zero element, but no symbol ∞. The elements of the hatrack are given coordinates as indicated on Figure 1.3 (coordinates of points will be written in parentheses, those of lines in square brackets). The points incident with L_{0}, different from x_{0} or x_{1} get a label (a), a ∈ R_{1} \ {0} in such a way that there is a bijection between R_{1} and Γ_{1}(L_{0}) \ {x_{0}}. Similarly, the points on the line L_{1} different from x_{0} get a label (0, b), and the points

Figure 1.3: The hat-rack of the coordinatization

incident with L_{2} different from x_{1} are labelled (0, 0, a′) ^{2}. Let y = (a) be a point on L_{0} different from x_{0}. Then the projection y′ of y onto L_{5} gets label (a, 0, 0, 0, 0). Similarly, the points of L_{3} different from x_{4} and the points of L_{4} different from x_{3} are labelled (0, 0, 0, a′) and (0, 0, 0, 0, b) respectively. Dually (now using the set R_{2}) we give coordinates to the lines incident with the points of Σ. In this way, all elements incident with an element of the hat-rack Σ are already given coordinates. Now let y_{0} be an arbitrary point opposite the point x_{0}, and put (y_{0} = z_{5}, M_{4}, z_{3}, M_{2}, z_{1}, L_{0}) the 5-path between y_{0} and L_{0}. Suppose z_{1} = (a_{1}). Let a_{3} (a_{5}) be the last coordinate of the projection of z_{3} (z_{5}) onto the line L_{3} (L_{1}), and l_{2} (l_{4}) the last coordinate of the projection of M_{2} (M_{4}) onto the point x_{4} (x_{2}). Then we label the point z_{i}, i = 3, 5 by (a_{1}, l_{2}, …, a_{i}) and the line M_{i}, i = 2, 4 by [a_{1}, …, l_{i}] (see Figure 1.3).

In this way, all the points opposite x_{0}, and all the elements of Γ for which the projection onto x_{0} is the line L_{0} are given coordinates. Dually, all the lines opposite L_{0} are labelled, hence also all other elements of Γ. For example, consider a point y collinear with x_{0}, not on the line L_{0}. The line x_{0}y has coordinate [k]. The projection of y onto the line L_{4} has coordinate (0, 0, 0, 0, b), and y itself gets the label (k, b).

Now each i-tuple, i ∈ {0, 1, …, 5} (calling (∞) and [∞] 0-tuples) consisting alternately of elements of R_{1} and R_{2} corresponds to exactly one element of P ∪ L and conversely. Also, if the number of coordinates of two different elements differs by at least 2, these elements are not incident; if the number of coordinates differs by 1, the elements are incident if and only the coordinatetuple of one of these elements is obtained from the coordinate-tuple of the other one by deleting the last coordinate. The only incident elements with the same number of coordinates different from 5 are (∞) and [∞]. To obtain a complete description of the incidence in terms of coordinates, we only need a criterion to decide when two elements with 5 coordinates are incident. This is given by the operations Ψ_{i} and Φ_{i}, i = 1, 2, 3, 4. Let p be a point with coordinates (a, ℓ, a′, ℓ′, a′′) and L a line with coordinates [k, b, k′, b′, k′′]. Then Ψ_{i}(k, a, ℓ, a′, ℓ′, a′′) gives the (n−i)-th coordinate of the projection of the line [k] onto the point p. Dually, Φ_{i}(a, k, b, k′, b′, k′′) gives the (n−i)-th coordinate of the projection of the point (a) onto the line L. Using these operations, it can be expressed when the point p and the line L are incident (see Chapter 3 of [57]).

2In [57], it is explained how the labelling of the points on L_{1} and L_{2} can be related to the labelling of the points on L_{0}. We do not assume this ‘normalization’ here. This translates in the fact that we will have to make more choices - compared to [57] - to determine the coordinates of H(𝕂) in the next paragraph.

1.9.2 Coordinates of classical generalized hexagons

Consider the split Cayley hexagon H(𝕂). This hexagon lies on the quadric with equation X_{0}X_{4}+X_{1}X_{5}+X_{2}X_{6} = X_{3}^{2} in PG(6, 𝕂). It is convenient to be able to go from the 6-dimensional space to the hexagon and back. Therefore, we choose coordinates in the following way (putting R_{1} = R_{2} = 𝕂).

\begin{aligned}(1, 0, 0, 0, 0, 0, 0) & → & (∞), \\ (a, 0, 0, 0, 0, 0, 1) & → & (a), \\ (0, 0, 0, 0, 0, 1, 0) & → & (0, 0), \\ (0, 1, 0, 0, 0, 0, 0) & → & (0, 0, 0), \\ (0, 0, 1, 0, 0, 0, 0) & → & (0, 0, 0, 0), \\ (0, 0, 0, 0, 1, 0, 0) & → & (0, 0, 0, 0, 0), \\ (b, 0, 0, 0, 0, 1, 0) & → & (0, b), \\ (0, 1, 0, 0, 0, 0, −a′) & → & (0, 0, a′), \end{aligned}

\begin{aligned}X_{1} = X_{2} = X_{3} = X_{4} = X_{5} = 0 & → & [∞], \\ X_{1} = X_{2} = X_{3} = X_{4} = X_{6} + kX_{5} = 0 & → & [k], \\ X_{0} = X_{2} = X_{3} = X_{4} = X_{5} = 0 & → & [0, 0], \\ X_{1} = X_{3} = X_{4} = X_{6} = X_{0} = 0 & → & [0, 0, 0], \\ X_{0} = X_{2} = X_{3} = X_{5} = X_{6} = 0 & → & [0, 0, 0, 0], \\ X_{0} = X_{1} = X_{3} = X_{5} = X_{6} = 0 & → & [0, 0, 0, 0, 0], \\ X_{0} + lX_{1} = X_{2} = X_{3} = X_{4} = X_{5} = 0 & → & [0, l], \\ X_{0} − k′X_{2} = X_{1} = X_{3} = X_{4} = X_{6} = 0 & → & [0, 0, k′]. \end{aligned}

POINTS

Coordinates in H(𝕂) Coordinates in PG(6, 𝕂)
(∞) (1, 0, 0, 0, 0, 0, 0)
(a) (a, 0, 0, 0, 0, 0, 1)
(k, b) (b, 0, 0, 0, 0, 1, −k)
(a, l, a′) (−l − aa′, 1, 0, −a, 0, a^{2}, −a′)
(k, b, k′, b′) (k′ + bb′, k, 1, b, 0, b′, b^{2} − b′k)
(a, l, a′, l′, a′′) (−al′ + a′^{2} + a′′l + aa′a′′, −a′′, −a, −a′ + aa′′,
1, l + 2aa′ − a^{2}a′′, −l′ + a′a′′)

LINES

Coordinates in H(𝕂) Coordinates in PG(6, 𝕂)
[∞] ⟨(1, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 0, 1)⟩
[k] ⟨(1, 0, 0, 0, 0, 0, 0), (0, 0, 0, 0, 0, 1, −k)⟩
[a, l] ⟨(a, 0, 0, 0, 0, 0, 1), (−l, 1, 0, −a, 0, a^{2}, 0)⟩
[k, b, k′] ⟨(b, 0, 0, 0, 0, 1, −k), (k′, k, 1, b, 0, 0, b^{2})⟩
[a, l, a′, l′] ⟨(−l − aa′, 1, 0, −a, 0, a^{2}, −a′),
(−al′ + a′^{2}, 0, −a, −a′, 1, l + 2aa′, −l′)⟩
[k, b, k′, b′, k′′] ⟨(k′ + bb′, k, 1, b, 0, b′, b^{2} − b′k),
(b′^{2} + k′′b, −b, 0, −b′, 1, k′′, −kk′′ − k′ − 2bb′)⟩

Table 1.1: Coordinatization of H(𝕂).

This determines the coordinates of each point and line of H(𝕂). The complete ‘dictionary’ for translation between the hexagon and the projective space is given in Table 1.1 (where ⟨p_{1}, p_{2}⟩ denotes the line through the points p_{1} and p_{2}). The operations Ψ_{i} and Φ_{i} are the following:

\{ \begin{matrix}Ψ_{1}(k, a, l, a′, l′, a′′) = a^{3}k + l − 3a′′a^{2} + 3aa′, \\ Ψ_{2}(k, a, l, a′, l′, a′′) = a^{2}k + a′ − 2aa′′, \\ Ψ_{3}(k, a, l, a′, l′, a′′) = a^{3}k^{2} + l′ − kl − 3a^{2}a′′k − 3a′a′′ + 3aa′′^{2}, \\ Ψ_{4}(k, a, l, a′, l′, a′′) = −ak + a′′, \end{matrix}

and

\{ \begin{matrix}Φ_{1}(a, k, b, k′, b′, k′′) = ak + b, \\ Φ_{2}(a, k, b, k′, b′, k′′) = a^{3}k^{2} + k′ + kk′′ + 3a^{2}kb + 3bb′ + 3ab^{2}, \\ Φ_{3}(a, k, b, k′, b′, k′′) = a^{2}k + b′ + 2ab, \\ Φ_{4}(a, k, b, k′, b′, k′′) = −a^{3}k + k′′ − 3ba^{2} − 3ab′. \end{matrix}

A similar table can be given for the coordinatization of the twisted triality hexagon. We will only need the coordinatization of T(q^{3}, q) once in this thesis, and refer to [57], Table 3.4 for the ‘translations’ used. Clearly, to coordinatize the duals of these hexagons, one only has to replace round and square brackets in the coordinatization above. Finally we remark that for a point p of the hexagon H(𝕂) or T(𝕂, 𝕂^{(σ)}, σ), the equations of the plane of the quadric containing p^{⊥} can be calculated explicitely, see for instance Thas & Van Maldeghem [46].

Example Let Γ = H(q), q even, q ≠ 2, coordinatized as above. Suppose we are given the points p and p′ with coordinates (a, 0, 0, 0) and (c, 0, d, 0, 0) respectively, a, c, d ∈ GF(q) \ {0}. Then

\begin{aligned}(a, 0, 0, 0) & → & (0, a, 1, 0, 0, 0, 0), \\ (c, 0, d, 0, 0) & → & (d^{2}, 0, c, d, 1, 0, 0). \end{aligned}

The points not opposite the point p′ in Γ (which are the points of the quadric collinear on the quadric with p′) lie in the hyperplane γ with equation X_{0} + d^{2}X_{4} + cX_{6} = 0, hence δ(p, p′) ≤ 4. Since the point (a, 0) (corresponding to the point (0, 0, 0, 0, 0, 1, a) in PG(6, q)) is collinear with p but opposite p′, δ(p, p′) = 4. We now look for the projection L of p′ onto p. The line L has coordinates [a, 0, 0, 0, x], and every point of L lies at distance ≤ 4 from p′. An arbitrary point on L different from (a, 0, 0, 0) has hexagon-coordinates (t, at^{3} + x, at^{2}, a^{2}t^{3} + ax, at) (this is obtained by using the expressions for the Φ_{i} to calculate the coordinates of the projection of (t) onto the line L). Now expressing that such a point is contained in γ for all t ∈ GF(q), we obtain x = \frac{d^{2}}{a}.

1.10 More geometries

A strongly regular graph (notation srg(v, k, λ, μ)) is a graph with v vertices such that

(i) every vertex is adjacent to exactly k vertices;

(ii) for any two adjacent vertices x and y, there are exactly λ vertices adjacent to both x and y;

(iii) for any two non-adjacent vertices x and y, there are exactly μ vertices adjacent to both x and y.

The complement of a strongly regular graph G with parameters (v, k, λ, μ) is denoted by G^{C}, and is again a strongly regular graph with parameters (v, v − k − 1, v − 2k + μ − 2, v − 2k + λ).

An association scheme is a pair (X, R) with X a finite set and R = (R_{0}, R_{1}, …, R_{d}) a partition of X × X such that the following conditions are satisfied.

(i) R_{0} = {(x, x)|x ∈ X}

(ii) The relations R_{i} are symmetric (i.e., (x, y) ∈ R_{i} ⇒ (y, x) ∈ R_{i}).

(iii) There exist integers p_{jk}^{i}, called the intersection numbers, having the following property: for all (x, y) ∈ R_{i}, there exist exactly p_{jk}^{i} elements z ∈ X such that (x, z) ∈ R_{j} and (y, z) ∈ R_{k}.

For an association scheme (X, R), the matrices L_{i} with (L_{i})_{jk} = p_{jk}^{i} are called the intersection matrices. Define the matrix Δ =diag(p_{00}^{0}, …, p_{dd}^{0}). Then in Brouwer, Cohen & Neumaier [8], section 2.2, it is shown that the matrices L_{i} have d + 1 common eigenvectors. We denote these eigenvectors (normalized such that (u_{i})_{0} = 1) with u_{0}, …, u_{d}. Then define v_{i} = Δ(u_{i}), f_{i} = |X|/(u_{i}, v_{i}) and q_{ijk} = ∑_{l} p_{ll}^{0}(u_{i})_{l}(u_{j})_{l}(u_{k})_{l}. The numbers f_{i} are called the multiplicities of the association scheme, and are necessarily integers (since they arise as multiplicities of eigenvalues of the adjacency matrices, see again [8], section 2.2).

Theorem 1.10.1 (Krein conditions [8], Theorem 2.3.2) q_{ijk} ≥ 0, for 0 ≤ i, j, k ≤ d.

24

Characterizations of (dual) classical generalized hexagons

2.1 Introduction

In the first section, we give a characterization of the finite classical hexagons defined over a field of characteristic two. This characterization is based on a property of these hexagons that was proved in Thas & Van Maldeghem [45], saying that no point of the hexagon can lie ‘far away’ from all the points of a point regulus. We formulate this property for an arbitrary finite polygon satisfying s ≥ t, obtaining in this way a generalization of an existing characterization of the generalized quadrangle W(q). The obtained characterization excludes all finite octagons.

We then start from the following observation. Fix a point regulus R in a finite (dual) classical hexagon, and a line L lying at distance 3 from a point x of R and at distance 5 from all the points of R \ {x}. In the case of the split Cayley hexagon over a field of characteristic 2, all the points of R \ {x} project onto the same point of L. By asking the property just mentioned for lines L in a particular position with respect to the regulus R, we obtain a characterization of the split Cayley hexagon over a finite field of even characteristic. In fact, this characterization weakens the condition of ‘having long imaginary lines’, as defined in van Bon, Cuypers & Van Maldeghem [55]. In the case of the dual classical hexagons, and the classical hexagons over a field of odd characteristic, the projection of the points of R onto L determines a bijection. We use this to obtain a characterization of all dual classical hexagons over a finite field of even characteristic, and a characterization of some extremal hexagons.

In a classical hexagon, all intersection sets containing at least two points, contain t + 1 points. In a dual classical hexagon however, the size of an intersection set containing at least two points, is 1 + t/q (with (q, t) the order of the hexagon). One could now ask whether a generalized hexagon for which all intersection sets have size 1+t/q, is necessarily dual classical. This condition seems to be too loose to characterize these hexagons. Instead, we consider intersections of traces that are in a slightly more general position than intersection sets (called ‘(3,4)-position’ later on), also containing 1+t/q points. In this way, we do obtain a characterization of all finite dual classical hexagons.

Payne & Thas [34] define the notion of anti-regular point in a generalized quadrangle. It is a conjecture that all finite anti-regular generalized quadrangles are isomorphic to the dual of W(q), q odd. We generalize this notion for hexagons, and prove that all finite anti-regular generalized hexagons are isomorphic to the dual of H(q), q not a power of three.

In Ronan [35] the point-distance-2-regular hexagons are characterized as follows: if a generalized hexagon is 3-regular, and all intersection sets behave as they should, then the points are 2-regular (see Theorem 1.8.4). We weaken the conditions of this characterization in two ways. First, we keep the assumption on the intersection sets, but ask the 3-regularity only for a certain subset of the points (namely for the points of an ovoidal subspace). Secondly, we start from a 3-regular hexagon, but only ask that all intersection sets with respect to a point in a certain subset (namely the points on lines of a dual ovoidal subspace) have the right size. The characterizations of the first three sections of this chapter are contained in Govaert [24] and Govaert & Van Maldeghem [25] and [26].

2.2 Characterizations of H(q) and T(q^{3}, q), q even 27

In the last section, we turn our attention to the Hermitian spread of the finite split Cayley hexagon. This spread arises by intersecting the underlying quadric Q with a hyperplane intersecting Q in an elliptic quadric Q^{−}(5, q). Since the points and lines of Q^{−}(5, q) form a generalized quadrangle, we obtain a generalized quadrangle Γ_{S} ‘hidden’ inside our hexagon. This quadrangle can be described using the spread (for example, the points of the quadrangle are exactly the points on spread lines). So the geometry Γ_{S} can be defined for an arbitrary spread S of the split Cayley hexagon. The fact that Γ_{S} is a generalized quadrangle is equivalent with a certain configuration of spread lines that is not allowed, and turns out to characterize the Hermitian spread. For an arbitrary spread, one can consider the group of projectivities induced by the spread lines, i.e. considering only those projectivities that use spread lines. For the Hermitian spread, the associated group is a Singer group. We prove that this group characterizes the Hermitian spread as a spread of H(q). The results of this section will appear in Govaert & Van Maldeghem [27].

2.2 Characterizations of H(q) and T(q^{3}, q), q even

Theorem 2.2.1 Let Γ be a finite generalized hexagon. Then Γ is isomorphic to H(q) or to T(q^{3}, q), both with q even, if and only if Γ_{≤4}(x) ∩ Γ_{≤3}(L) ∩ Γ_{≤3}(M) is nonempty for any point x and any pair of lines L, M of Γ.

Proof. Let first Γ be isomorphic to H(q) or to T(q^{3}, q), both with q even, and L, M two lines of Γ. Let for any point x of Γ, S_{x} = Γ_{≤4}(x)∩Γ_{≤3}(L)∩Γ_{≤3}(M). If δ(L, M) ≤ 2, then proj_{L}x ∈ S_{x}. If δ(L, M) = 4, then, with N = L ⋈ M, the point proj_{N}x ∈ S_{x}. If finally L and M are opposite, then it is proved in Thas & Van Maldeghem [45], Lemma 5.2 that S_{x} ≠ ∅ for any point x if and only if q is even.

So from now on we assume that Γ is a finite generalized hexagon with the property that Γ_{≤4}(x) ∩ Γ_{≤3}(L) ∩ Γ_{≤3}(M) is nonempty for any point x, and for any two lines L, M. Let (s, t) be the order of Γ.

We first show that Γ is distance-3-regular. So let L and M be two opposite lines, and let x, y, z ∈ ⟨L, M⟩, x ≠ y ≠ z ≠ x. Let N ∈ ⟨x, y⟩, M ≠ N ≠ L. Let z′ = proj_{M}z, M′ = proj_{z′}N, z′′ = proj_{N}z′ and N′ = proj_{z′′}z′. We project the points of the regulus ⟨L, M⟩ onto the line N′. Since z′′ is the image of at least two points (namely x and y), this projection is not injective. Because the regulus ⟨L, M⟩ and the line N′ both contain s + 1 points, the projection of ⟨L, M⟩ onto N′ cannot be surjective either, so there is a point w on N′ which is not the projection of any point of ⟨L, M⟩. If N′ would lie at distance 5 from every point of ⟨L, M⟩, then the point w on N′ would be opposite every point of ⟨L, M⟩, a contradiction with the assumed property of Γ. Hence N′ is at distance 3 from some point u of ⟨L, M⟩. If u ≠ z, then we obtain an ordinary j-gon, j < 6, through the points proj_{M}u, z′ and proj_{N′}u, a contradiction. Hence u = z, implying that M′ = zz′. We have shown that

Figure 2.1: Proof of Theorem 2.2.1 (size of an intersection set is t + 1).

the line zz′ = proj_{z}M is at distance 4 from N. Similarly (by interchanging the roles of L and M), the line proj_{z}L is at distance 4 from N, implying that z is at distance 3 from N. Hence Γ is distance-3-regular.

Next we show that all intersection sets containing at least two points, have size t + 1. Let x, y, L, M and N be as in the previous paragraph and put a = proj_{L}y. Let L_{a} be any line through a, ay ≠ L_{a} ≠ L (see Figure 2.1). Let c_{N} be the projection of x onto N and define c_{L}, c_{M} similarly. Let b be the projection of c_{M} on L_{a}. Then x^{y} ∩x^{b} is an intersection set already containing the points c_{L} and c_{M}. We prove that x^{y} = x^{b}. Let c′_{N} be the projection of b onto xc_{N}. It suffices to show that c_{N} = c′_{N}. Suppose by way of contradiction these points are different. Let y′ be the projection of y onto proj_{c′_{N}} b. Note that c′_{N} ≠ y′ ≠ c′_{N} ⋈ b. Hence the projection y′′ of y′ onto L is distinct from both a and c_{L}. Let d = y′ ⋈ y′′. We now project the points of the regulus ⟨L, N⟩ onto the line dy′. Since both x and y project onto the point y′, this projection is not injective, so it cannot be surjective either. If the line dy′ would lie at distance 5 from all points of ⟨L, N⟩, this would imply that the line dy′ contains a point opposite all the points of ⟨L, N⟩, a contradiction. So we may assume that there is a point z ∈ ⟨L, N⟩ lying at distance 3 from dy′. If proj_{L}z ≠ y′′, then the points z, proj_{L}z, y′′, d and proj_{dy′}z define a pentagon, a contradiction. Now clearly, the point z lies on the line y′′d, y′′ ≠ z ≠ d. Let M′ be the line through c_{M} and c_{M} ⋈ b. The regulus ⟨x, b⟩ contains the lines L, M′ and y′c′_{N}. Since δ(d, L) = δ(d, y′c′_{N}) = 3, the distance-3-regularity implies that also δ(d, M′) = 3. Note that δ(z, M) = 3. But now the points z, d, proj_{M′}d, c_{M} and proj_{M}z define a pentagon, the final contradiction, showing x^{y} = x^{b}. So all intersection sets containing at least two points, have size t + 1. By Theorem 1.8.4 (i), all points are distance-2-regular, and by Theorem 1.8.2, Γ is a finite Moufang hexagon isomorphic to H(q) or to T(q^{3}, q). By Thas & Van Maldeghem [45], q is even. □

In the proof of Theorem 2.2.1, we used the finiteness assumption in the following way: if one projects the points of a regulus ⟨L, M⟩ onto a line N at distance 5 from every point of ⟨L, M⟩, then the fact that this projection is not injective implies that it cannot be surjective either. This becomes exactly the additional assumption when generalizing Theorem 2.2.1 to the infinite case. We now obtain the following:

Theorem 2.2.2 Let Γ be a generalized hexagon, not necessarily finite. Suppose that for any point x and any two lines L, M the set Γ_{≤4}(x) ∩ Γ_{≤3}(L) ∩ Γ_{≤3}(M) is nonempty (a). Suppose moreover that if L, M are opposite, and N is a line at distance 5 from all elements of the regulus ⟨L, M⟩, then the projection of ⟨L, M⟩ onto N is injective whenever it is surjective (b). Then Γ is a point-distance-2-regular hexagon and hence a Moufang hexagon.

In the infinite case however, the conditions stated in Theorem 2.2.2 do not completely characterize the characteristic 2 case. We investigate when the infinite split Cayley hexagon Γ ≅ H(𝕂) satisfies conditions (a) and (b) in Theorem 2.2.2. (For other types of Moufang hexagons, the condition on the underlying field becomes more involved.) Let L and M be two opposite lines of Γ, x a point of Γ and Q = Q(6, 𝕂) the quadric on which Γ is defined. The regulus ⟨L, M⟩ corresponds with the points of a conic C (lying in a plane π) on Q, and the set Γ_{≤4}(x) corresponds with the set of points of Q lying in the tangent hyperplane ξ_{x} at Q in x. Let x be a point such that ξ_{x} and π intersect in a line R_{x} and put S_{x} = Γ_{≤4}(x) ∩ ⟨L, M⟩.

• char 𝕂 = 2. The line R_{x} is a line through the nucleus of C, hence R_{x} intersects C for all choices of x if and only if the field 𝕂 is perfect. So let 𝕂 be a perfect field of characteristic 2. Then (a) is satisfied. Property (b) follows from the fact that all imaginary lines of Γ ≅ H(𝕂), char 𝕂 = 2, are long (see Theorem 1.8.5). Indeed, this fact implies that an imaginary line coincides with a regulus containing two of its points (see Lemma 2.4 in van Bon, Cuypers & Van Maldeghem [55]). The

Figure 2.2: An (i, j)-intersection set.

projection of an imaginary line onto a line at distance 5 from all its points is either constant or injective (see [55], Corollary 2.3), hence (b).

• char 𝕂 ≠ 2. The intersection of R_{x} with C is nonempty for all choices of x if and only if the field 𝕂 is quadratically closed. So let 𝕂 be a quadratically closed field of odd characteristic. Then (a) is satisfied, but (b) cannot be satisfied. Indeed, let x be a point for which S_{x} contains two points y and z. Let N be a line through x, proj_{x}y ≠ N ≠ proj_{x}z. Since x is opposite every point of ⟨L, M⟩ \ {y, z}, the line N lies at distance 5 from every point of ⟨L, M⟩. The projection of ⟨L, M⟩ onto N is not injective (since both y and z project onto x). If (b) is satisfied, then there would be a point x′ on N opposite every point of ⟨L, M⟩, contradicting (a).

We now want to generalize Theorem 2.2.1 to octagons. For this purpose, we first generalize the notion of an intersection set, and result 1.8.4 (i).

Let Γ be a generalized 2m-gon and 2 ≤ i ≤ j ≤ m. Let x be an element of Γ. An (i, j)-intersection set S with respect to x is a set x_{[i]}^{y} ∩ x_{[i]}^{z}, where y and z are opposite x, where |x_{[j]}^{y} ∩ x_{[j]}^{z}| ≥ 2 and where |y_{[j]}^{x} ∩ z_{[j]}^{x}| = 1 if j < m. Note that, if (s, t) is the order of Γ, then 2 ≤ |S| ≤ t + 1 if x is a point, and 2 ≤ |S| ≤ s + 1 if x is a line.

A (2,2)-intersection set in a generalized hexagon now corresponds more or less with an intersection set as defined in section 1.8. The only difference is that a (2,2)-intersection set by definition already contains two elements. Using this new terminology, Theorem 1.8.4 (i) becomes:

If for every (2, j)-intersection set x_{[2]}^{z}∩x_{[2]}^{y} of a generalized hexagon
Γ, one has x_{[2]}^{z} = x_{[2]}^{y}, then Γ is point-distance-2-regular.

Indeed, we show that, if Γ is a generalized hexagon satisfying the conditions above, then Γ is distance-3-regular. Let x, y and z be three different points of a regulus ⟨L, M⟩, and N a line of ⟨x, y⟩, L ≠ N ≠ M. Since x^{y} ∩ x^{z} is a (2,3)-intersection set, x^{y} = x^{z}, hence δ(z, proj_{N}x) = 4. Interchanging the roles of x and y, we obtain y^{x} = y^{z}, hence δ(z, proj_{N}y) = 4. So δ(z, N) = 3, showing the distance-3-regularity.

The following lemma is a generalization of Theorem 1.8.4 (i).

Lemma 2.2.3 Let Γ be an arbitrary generalized 2m-gon, m ≥ 2, let x be a point of Γ, and let 2 ≤ i ≤ m. Then x is distance-i-regular if and only x_{[i]}^{y} = x_{[i]}^{z}, for every (i, j)-intersection set x_{[i]}^{y} ∩ x_{[i]}^{z} with respect to x, for all j, i ≤ j ≤ m.

Proof. Let x be a distance-i-regular point. Then every distance-i-trace in x is determined by two points. Since an (i, j)-intersection set x_{[i]}^{y} ∩ x_{[i]}^{z} with respect to x contains two elements by definition, x_{[i]}^{y} = x_{[i]}^{z}. So let now x be a point of Γ and suppose that x_{[i]}^{y} = x_{[i]}^{z}, for every (i, j)-intersection set x_{[i]}^{y} ∩x_{[i]}^{z} with respect to x, for all j, i ≤ j ≤ m. Let y and z be two points opposite x such that x_{[i]}^{y}∩x_{[i]}^{z} contains the distinct points u_{0} and w_{0}. Define the elements u_{k} and w_{k}, 0 ≤ k ≤ 2m − i, as u_{k}Iu_{k+1} and w_{k}Iw_{k+1}, 0 ≤ k < 2m − i, with u_{2m−i} = w_{2m−i} = y. Similarly, define the elements v_{k} and r_{k}, 0 ≤ k ≤ 2m−i, as v_{k}Iv_{k+1} and r_{k}Ir_{k+1}, 0 ≤ k < 2m − i, with v_{0} = u_{0}, r_{0} = w_{0} and v_{2m−i} = r_{2m−i} = z. We claim that, in order to prove that x_{[i]}^{y} = x_{[i]}^{z}, we may assume that w_{2m−2i} = r_{2m−2i}. Indeed, suppose that w_{k} = r_{k}, for some k, 0 ≤ k < 2m − 2i. We prove that we can assume w_{k+1} = r_{k+1} . Since w_{k} is opposite u_{2m−2i−k}, one has δ(r_{k+1}, u_{2m−2i−k}) = 2m − 1, so there is a unique chain

r_{k+1}Ir′_{k+2}I … Ir′_{2m−i} = y′ = v′_{2m−i}Iv′_{2m−i−1}I … Iv′_{2m−2i−k} = u_{2m−2i−k}

(y′ is the unique point of this chain opposite x). If i+k ≤ m, then S = x_{[i]}^{y}∩x_{[i]}^{y′} is an (i, i+k)-intersection set with respect to x (indeed, the element u_{2m−2i−k} lies at distance i + k from both y and y′, and the element w_{k} belongs to x_{[i+k]}^{y} ∩ x_{[i+k]}^{y′} and lies opposite u_{2m−2i−k}). If i + k > m, then S = x_{[i]}^{y} ∩ x_{[i]}^{y′} is an (i, 2m − i − k)- intersection set with respect to x (indeed, the element w_{k} lies at distance 2m − i − k from both y and y′, and the element u_{2m−2i−k} belongs to x_{[2m−i−k]}^{y} ∩ x_{[2m−i−k]}^{y′} and lies opposite w_{k}). Thus in both cases, x_{[i]}^{y} = x_{[i]}^{y′}. Since we were interested in S = x_{[i]}^{y} ∩ x_{[i]}^{z}, we can as well consider S′ = x_{[i]}^{y′} ∩ x_{[i]}^{z}. So we may indeed assume w_{k+1} = r_{k+1}. Proceeding like this we can assume that w_{2m−2i} = r_{2m−2i}. But then S′ = x_{[i]}^{y} ∩ x_{[i]}^{z} is an (i, i)-intersection set with respect to x (since S′ contains the element u_{0}, and the element w_{2m−2i} lies at distance i from both y and z and is opposite u_{0}), hence x_{[i]}^{y} = x_{[i]}^{z}. This shows that x is distance-i-regular. □

Let Γ be a generalized 2m-gon and 2 ≤ i ≤ m. Let x be an element of Γ. A half (i, i)-intersection set S with respect to x is a set x_{[i]}^{y} ∩ x_{[i]}^{z}, where y and z are opposite x, and |y_{[i]}^{x} ∩ z_{[i]}^{x}| = 1 if i < m, |y_{[i]}^{x} ∩ z_{[i]}^{x}| ≥ 1 if i = m.

The following lemma is an immediate generalization of Theorem 1.8.4 (ii).

Lemma 2.2.4 Let Γ be a generalized 2m-gon, and 2 ≤ k ≤ m − 1.

(i) Suppose m is even. If Γ is line-distance-m-regular, and all half (m + 1 − j, m + 1 − j)-intersection sets with respect to any line X contain at least two elements, for 2 ≤ j ≤ k, then Γ is line-distance-k-regular.

(ii) Suppose m is odd. If Γ is distance-m-regular, and all half (m + 1 − j, m + 1 − j)-intersection sets with respect to any point x contain at least two elements, for 2 ≤ j ≤ k, then Γ is line-distance-k-regular.

Proof. Suppose Γ is a generalized 2m-gon satisfying the conditions of the lemma, for a fixed k, 2 ≤ k ≤ m − 1. Let L be a line of Γ and L_{[k]}^{M_{1}} ∩ L_{[k]}^{M_{2}} a (k, j)-intersection set with respect to L, k ≤ j. We prove that L_{[k]}^{M_{1}} = L_{[k]}^{M_{2}}. Let v be an element of L_{[j]}^{M_{1}} ∩L_{[j]}^{M_{2}}, and w an element of M_{1}_{[j]}^{L} ∩M_{2}_{[j]}^{L}, v and w opposite. Let X be the element of the path [L, w] at distance m from L. Let p be a point on L, proj_{L}v ≠ p ≠ proj_{L}w. Let [p, M_{1}] = (p = y_{0}, y_{1}, …, y_{2m−1} = M_{1}) and [p, M_{2}] = (p = z_{0}, z_{1}, …, z_{2m−1} = M_{2}). We prove that y_{i} = z_{i}, for i ≤ k. So suppose by way of contradiction that y_{r} = z_{r}, but y_{r+1} ≠ z_{r+1}, for a number r, 0 ≤ r ≤ k − 1. Denote by Y (Z) the unique element of the path [p, M_{1}] ([p, M_{2}]) at distance m from L and Y′ (Z′) the unique element of the path [v, M_{1}] ([v, M_{2}]) at distance m from L. Then S = X_{[m−r]}^{Y} ∩ X_{[m−r]}^{Z} is a half (m−r −1, m−r −1)-intersection set with respect to X. By assumption, S contains an element a, proj_{X}a ≠ proj_{X}L. Since proj_{w}M_{1} ≠ proj_{w}M_{2}, also proj_{X}a ≠ proj_{X}w. Let R_{1} (R_{2}) be the element of the path [Y, a] ([Z, a]) at distance m from Y (Z). The regulus ⟨X, Y ⟩ contains the elements L, M_{1} and R_{1}. Since δ(Y′, M_{1}) = δ(Y′, L) = m, also δ(Y′, R_{1}) = m. Similarly, δ(Z′, R_{2}) = m. But now we obtain a circuit of length at most 4m+2(r−j+1) (determined by the paths [a, R_{1}], [R_{1}, Y′], [Y′, v], [v, Z′], [Z′, R_{2}] and [R_{2}, a]), a contradiction. So L_{[k]}^{M_{1}} = L_{[k]}^{M_{2}}. By applying (the dual of) Lemma 2.2.3, we conclude that the line L is k-regular. □

Theorem 2.2.5 Let Γ be a finite generalized octagon of order (s, t) with s ≥ t ≥ 2 such that Γ_{≤6}(x) ∩ Γ_{≤4}(y) ∩ Γ_{≤4}(z) is nonempty for all points x, y, z of Γ. Then Γ is point-distance-3-regular (and hence does not exist).

Proof. Let Γ be a finite generalized octagon satisfying the conditions of the lemma. Note that s > t by Theorem 1.2.3. Let x be any point of Γ. We first show that x is distance-4-regular. So let y and z be opposite x and u, v ∈ x_{[4]}^{y} ∩ x_{[4]}^{z}, u ≠ v. Let L be any line through x and let y′ = proj_{L}y, z′ = proj_{L}z. We show that y′ = z′ and proj_{y′}y = proj_{z′}z. Suppose by way of contradiction that y′ ≠ z′. Let r be the unique point collinear with z and at distance 4 from z′. Suppose that all elements of ⟨x, y⟩ are at distance ≥ 6 from r. If we project all points of the regulus ⟨x, y⟩ which are at distance 6 from r onto r, then, since |⟨x, y⟩| = t + 1 = |Γ_{1}(r)|, and since v and u are projected onto the same element, we see that there is at least one line N through r which lies at distance 7 from all those points, and hence from every element of ⟨x, y⟩. Since t < s, the projection of ⟨x, y⟩ onto N cannot be surjective, hence there is a point on N opposite every element of ⟨x, y⟩, a contradiction. Consequently there is a point w of ⟨x, y⟩ at distance ≤ 4 from r. If proj_{x}w ≠ xy′, then we obtain an ordinary j-gon, j ≤ 7, through r, w, x, z′, a contradiction. Hence w is the unique point at distance 2 from y′ and at distance 4 from y. Now δ(r, w) ≤ 4 implies that y′ = z′ and proj_{y′}y = proj_{z′}z. Interchanging the roles of x and y, we see that also δ(z, proj_{w}y) = 5, implying that z lies at distance 4 from the point w. This shows that x is distance-4-regular. In particular, all (3,4)-intersection sets with respect to x have size t + 1.

Next we show that all (3, 3)-intersection sets with respect to x have size t+1. Let again y and z be opposite x such that x_{[3]}^{y} ∩ x_{[3]}^{z} is a (3,3)-intersection set. Let L be a line at distance 5 from x and 3 from both y and z and let M be a line of x_{[3]}^{y} ∩ x_{[3]}^{z}, M opposite L. We prove that x_{[3]}^{y} = x_{[3]}^{z}. Let K be any line through x and put y′ = proj_{K}y and z′ = proj_{K}z. Suppose by way of contradiction that y′ ≠ z′. Consider the line N at distance 3 from both z and z′. As before (using t < s), N cannot be at distance 7 from all elements of ⟨x, y⟩, hence N is at distance ≤ 5 from some element w ∈ ⟨x, y⟩.

If proj_{x}w ≠ xy′, then we obtain an ordinary j-gon, j ≤ 7, containing the points w, w ⋈ x, x, z′ and proj_{N}w. So proj_{x}w = xy′, which implies y′ = z′. Put N_{z} = proj_{z′}z and N_{y} = proj_{y′}y. We now show that N_{y} = N_{z}.

Suppose by way of contradiction that N_{y} ≠ N_{z}. Let a be any point incident with N_{z}, a ≠ z′, and let a′ be the projection of a onto the line proj_{y}M (note that a′ ≠ y). Let L_{a} be the line at distance 3 from both a and a′. As before, L_{a} is not at distance 7 from every element of ⟨x, y⟩, so there exists a point w′ ∈ ⟨x, y⟩ at distance ≤ 5 from L_{a}. If w′ ≠ proj_{M}y, then we obtain an ordinary j-gon, j ≤ 7, containing the points w′, w′ ⋈ y, y, a′, proj_{L_{a}}a′ and proj_{L_{a}}w′. So w′ = proj_{M}y and δ(w′, L_{a}) ≤ 5 implies that a′ = w′ ⋈ y. So every point of the line N_{z} different from z′ lies at distance 6 from the point y ⋈w′, a contradiction. We conclude that N_{y} = N_{z}, showing that x_{[3]}^{y} = x_{[3]}^{z}. By Lemma 2.2.3, the point x is distance-3-regular. A point-distance-3-regular octagon does not exist (see Theorem 1.3.1), hence the result. □

Combining the characterization of W(q) in Thas [39], with Theorems 2.2.1 and 2.2.5, we obtain the following.

Theorem 2.2.6 Let Γ be a finite generalized 2m-gon of order (s, t), with s ≥ t ≥ 2. Then Γ is isomorphic to W(q), to H(q′) or to T(q′^{3}, q′), with q′ even, if and only if Γ_{≤2m−2}(x) ∩ Γ_{≤m}(v) ∩ Γ_{≤m}(w) is nonempty for any point x, and for any pair of elements v, w, with v, w points if m is even, and v, w lines if m is odd.

Remarks

1. In the case m = 3, the condition s ≥ t in the previous theorem is not needed.

2. It makes no sense to consider the cases v and w points if m is odd, or v and w lines if m is even. Indeed, let for example Γ be a generalized hexagon, and v, w opposite points in Γ. Let z ∈ v^{w}, and x a point at distance 3 from vz and 4 from v and z. The point x lies at distance 5 from every line of ⟨v, w⟩, hence ⟨v, w⟩ ∩ Γ_{≤3}(x) = ∅.

3. Theorem 2.2.6 characterizes the finite classical hexagons over a field of characteristic 2, but the assumptions kill every finite octagon, despite the fact that the Ree-Tits octagons too are defined over a field of characteristic 2. This shows once again that the Ree-Tits octagons play a special role in the theory of finite generalized polygons.

? △ Generalize Theorem 2.2.6 to the infinite case, i.e. find a ‘nice’ condition that forces 2m ≤ 6.

Another attempt to generalize Theorem 2.2.1 to finite octagons is given in the following theorem.

Theorem 2.2.7 Let Γ be a finite generalized octagon of order (s, t). Suppose that for any point x and any two lines L, M, the set Γ_{≤6}(x)∩Γ_{≤3}(L)∩Γ_{≤5}(M) is nonempty. Then s > t. If moreover, all (2, 2)-intersection sets with respect to any line contain s + 1 lines, then Γ is line-distance-2-regular (and hence does not exist).

Proof. Let Γ be a finite generalized octagon of order (s, t) such that Γ_{≤6}(x)∩ Γ_{≤3}(L) ∩ Γ_{≤5}(M) is nonempty, for any point x and any two lines L, M. We first show that t ≤ s. Suppose by way of contradiction that s < t. Let L and M be two opposite lines, and x, y ∈ L_{[3]}^{M}. Put x′ = proj_{M}x and v = x ⋈ x′. Let N be an arbitrary line concurrent with vx′, not through v or x′, and r = proj_{N}y. Let X be the set of points of L_{[3]}^{M} at distance 6 from r, and Y the set of points of L_{[3]}^{M} at distance 8 from r. Note that X ∪ Y = L_{[3]}^{M} and x, y ∈ X. We now project the points of X onto N; since t > s, at least one line N′ through r is not the projection of any point of X, and hence N′ lies at distance 7 from all points of L_{[3]}^{M}. The points of X lie opposite all points of N′ \{r}. We now project the points of Y onto the line N′. Since |Y | ≤ s−1, at least one point w on N′ lies opposite every point of L_{[3]}^{M}, a contradiction. This shows that t > s.

Now let L be a line of Γ. We show that all (2,3)-intersection sets with respect to L contain s + 1 points. Let L_{[3]}^{M_{1}} ∩ L_{[3]}^{M_{2}} be a (2,3)-intersection set with respect to L. Let r be the point at distance 5 from L and at distance 3 from both M_{1} and M_{2}, and r′ a point in L_{[3]}^{M_{1}} ∩ L_{[3]}^{M_{2}}, r′ opposite r. Let v be an arbitrary point on L. We show that R_{1} := proj_{v}M_{1} = proj_{v}M_{2} =: R_{2}. Suppose by way of contradiction that R_{1} ≠ R_{2}. Let r′′ be the point collinear with r and at distance 3 from L. Let N be the line concurrent with M_{2} at distance 4 from R_{2}. If all points of L_{[3]}^{M_{1}} lie at distance 7 from N, then (since |L_{[3]}^{M_{1}}| = |N| and both r′′ and r′ project onto the same element of N) there is a point on N opposite every point of L_{[3]}^{M_{1}}, a contradiction. Hence there is a point x ∈ L_{[3]}^{M_{1}} at distance ≤ 5 from N. If x is not incident with R_{1}, we obtain an ordinary j-gon, j ≤ 7, through x, v, proj_{N}v and proj_{N}x . Hence xIR_{1}, implying that R_{1} = R_{2}, so L^{M_{1}} = L^{M_{2}}.

Figure 2.3: A weakening of the condition for long imaginary lines

Completely similar, one shows that all (2, 4)-intersection sets with respect to L contain s + 1 points. Together with the assumption about the (2,2)- intersection sets, (the dual of) Lemma 2.2.3 implies that Γ is line-distance-2-regular, and hence cannot exist by Theorem 1.3.1. □

Let x, y be opposite points of the hexagon H(q), q even, and L, M distinct lines in ⟨x, y⟩. Then the imaginary line I(x, y) is long and hence coincides with the point regulus ⟨L, M⟩. It immediately follows from the definition of imaginary line that, if R is a line of Γ at distance 3 from the point x and at distance 5 from all points of ⟨L, M⟩ \ {x}, then all points of ⟨L, M⟩ \ {x} project onto the same point of R. We now ask the above property for lines R in a particular position with respect to the regulus ⟨L, M⟩ and obtain a characterization of H(q), q even.

Consider the following property in a finite generalized hexagon Γ:

(I) Let L and M be two arbitrary opposite lines, x, y different points of ⟨L, M⟩ and x′ = proj_{M}x. Let N be an arbitrary line concurrent with xx′, not through x or x′. Then proj_{N}y = proj_{N}z, for all z ∈ ⟨L, M⟩ \ {x}.

Theorem 2.2.8 A finite generalized hexagon Γ satisfies condition (I) if and only if Γ is isomorphic to H(q), q even.

Proof. Suppose Γ is a finite generalized hexagon in which (I) holds. We first show that Γ is distance-3-regular. So let L and M be two opposite lines, x, y, z different points of ⟨L, M⟩ and N ∈ ⟨x, y⟩. We have to prove that δ(z, N) = 3. Put p = proj_{L}y, p′ = proj_{M}y, x′ = proj_{N}x, y′ = proj_{N}y and z′ = proj_{xx′}z. We show that z′ = x′. Suppose by way of contradiction that z′ ≠ x′ and put z′′ = proj_{py}z′. Suppose first proj_{z′}z ≠ proj_{z′}z′′. But this contradicts (I) since the projections of x and z onto the line through z′′ and z′ ⋈z′′ do not coincide, so proj_{z′}z = proj_{z′}z′′. Note that z′ ⋈z ≠ z′ ⋈z′′ since otherwise, there would be an ordinary pentagon through the points z′′, p, proj_{L}z, z and z ⋈z′. Let u be the projection of z′ onto yp′. But now, noting that proj_{z′}u ≠ proj_{z′}z, the projections of x and z onto the line through u and u⋈z′ do not coincide, again contradicting (I), so x′ = z′. Interchanging the roles of x and y, we see that y′ = proj_{yy′}z. But this creates an ordinary pentagon containing x′, y′ and z, unless δ(z, N) = 3.

We next show that an imaginary line coincides with a regulus containing two of its points. Since Γ is finite, this will imply that all imaginary lines are long, which proves the result in view of Theorem 1.8.5. So let I(x, y) be an imaginary line, and suppose by way of contradiction that there is a point z ∈ R(x, y) not belonging to I(x, y). This implies there exists a point a not opposite the points x and y, with δ(z, a) = 6. If δ(x, a) = 2 or δ(y, a) = 2, or if proj_{a}x = proj_{a}y, then the 3-regularity implies that δ(z, a) = 4, a contradiction. So suppose δ(x, a) = δ(y, a) = 4 and proj_{a}x ≠ proj_{a}y. Put b = x ⋈ a and c = proj_{xb}y. Again by the 3-regularity, the point z lies at distance 3 from the line through c and c ⋈ y. Now by (I), proj_{ab}z = proj_{ab}y = a, the final contradiction. □

Consider the following weaker version of condition (I):

(I′) Let L and M be two arbitrary opposite lines, x, y different points of ⟨L, M⟩ and x′ = proj_{M}x, y′ = proj_{L}y. Let N be an arbitrary line concurrent with xx′, not through x or x′ and at distance 4 from yy′. Then proj_{N}y = proj_{N}z, for all z ∈ ⟨L, M⟩ \ {x}.

Corollary 2.2.9 A finite generalized hexagon Γ of order (s, t), t ≤ s, satisfies condition (I′) if and only if Γ is isomorphic to H(q), q even.

Proof. Suppose Γ satisfies (I′), and let L, M, x, x′, y, y′ be as in (I′). Let z be a point of ⟨L, M⟩, x ≠ z ≠ y, and put z′ = proj_{L}z. Let v be an arbitrary point on xx′, x ≠ v ≠ x′ and v′ = proj_{zz′}v. If proj_{v}yy′ = proj_{v}zz′, then condition (I′) implies that δ(y, v′ ⋈ v) = 4, which creates a pentagon containing z′, y′ and v′ ⋈ v, so proj_{v}yy′ ≠ proj_{v}zz′. This shows that t + 1 = |Γ_{1}(v)| ≥ |⟨x, y⟩| = s+1, so t = s. Now clearly, every line through v different from xx′ lies at distance 4 from a line aa′, with aIL and a′ ∈ ⟨L, M⟩. Hence (I) is satisfied and the result follows. □

2.3 Characterizations of H(q)^{D} and T(q, q^{3})

Consider the following property in a finite generalized hexagon Γ.

(C) If a point x is at distance 4 from an element y of the point regulus R, and if all elements of R\{y} are opposite x, then all elements of R\{y} are at distance 4 from x⋈y.

Lemma 2.3.1 (i) The dual classical hexagons H(q)^{D} and T(q, q^{3}) satisfy property (C).

(ii) The classical hexagons H(q) and T(q^{3}, q) satisfy property (C) if and only if q is odd.

Proof. Let Γ be a finite (dual) classical hexagon. Let x be a point of Γ at distance 4 from an element y of the point regulus R = ⟨M, N⟩, and suppose x is opposite every element of R \ {y}. Put L = proj_{x}y. Note that, by the distance-3-regularity, M and N can be chosen arbitrarily in ⟨y, z⟩, for any z ∈ R \ {y}. In particular, we may choose M at distance ≤ 4 from L. If δ(L, M) = 2, then x ⋈ y is incident with M, hence lies at distance 4 from every element of R \ {y}. So in this case, property (C) is satisfied. Suppose now δ(L, M) = 4. We show that this leads to a contradiction.

Let first Γ be dual classical. Note that the point x ⋈ y does not belong to M, hence x ⋈ y is opposite every element of R \ {y}. We project the points of the regulus R onto the line L. Since |R| = q + 1, and since also x is opposite every element of R \ {y}, there must be some point w incident with L which is at distance 4 from at least two points z_{1}, z_{2} ∈ R \ {y}. Now by the distance-3-regularity we may choose N in such a way that it meets proj_{z_{1}}w (see Figure 2.4). Note that {proj_{y}N, proj_{z_{1}}N, proj_{z_{2}}N} ⊆ N^{M}, and {proj_{y}N, proj_{z_{1}}N} ⊆ N^{L}. By the line-distance-2-regularity, we must have δ(L, proj_{z_{2}}N) = 4. But then, we obtain a pentagon containing z_{2} ⋈ w, z_{2}, w and the line intersecting both L and proj_{z_{2}}N, the final contradiction. Hence (C) is satisfied.

Let now Γ be a classical hexagon. Put y′ = proj_{M}y and v = proj_{yy′}L. We prove (using coordinates) that the projection θ of ⟨M, N⟩\{y} onto L\{v} is

2.3 Characterizations of H(q)^{D} and T(q, q^{3}) 39

Figure 2.4: Proof of Lemma 2.3.1 if Γ is dual classical.

a bijection if and only if q is odd. Let first Γ ≅ T(q^{3}, q). Choose coordinates in the following way: y = (∞), yy′ = [∞], M = [00], N = [000], v = (b), b ∈ GF(q^{3}) \ {0} and L = [b, k], k ∈ GF(q). A point p of ⟨M, N⟩ \ {y} then has coordinates (0, 0, a, 0, 0), a ∈ GF(q^{3}). The projection of p onto L is the intersection of L with the tangent hyperplane ξ_{p} of Q(7, q^{3}) at p. We calculate this intersection in the projective space PG(7, q^{3}) (and therefore, we use Table 3.4 in [57]). Note that

p = (0, 0, a, 0, 0) ↔ (a^{q+q^{2}}, 0, 0, a^{q^{2}}; 1, 0, 0, −a^{q}),

L = [b, k] ↔ (−k, 1, 0, b^{q}; 0, b^{q+q^{2}}, 0, −b^{q^{2}}) + t(b, 0, 0, 0; 0, 0, 1, 0),

with t ∈ (GF(q^{3}) ∪ {∞}), and

ξ_{p} ↔ X_{0} − a^{q}X_{3} + a^{q+q^{2}}X_{4} + a^{q^{2}}X_{7} = 0.

The point ξ_{p} ∩ L is then the point of L for which

t_{p} = \frac{k + a^{q}b^{q} + a^{q^{2}}b^{q^{2}}}{b} .

If θ is not bijective, then there exist two points p(0, 0, a, 0, 0) and p′(0, 0, a′, 0, 0) belonging to R \ {y} such that t_{p}=t_{p′}, or

b^{q}(a^{q} − a′^{q}) = b^{q^{2}}(a′^{q^{2}} − a^{q^{2}}).

Since a ≠ a′ this becomes

b^{q^{2}−q}(a′ − a)^{q^{2}−q} = −1.

If q is even, then one can choose p and p′ such that (a′ − a) = b^{−1}, hence θ is not bijective. Suppose now q is odd. Then −1 has to be a (q^{2} − q)th power, implying

d^{l(q^{2}−q)} = d^{\frac{q^{3}−1}{2}} (= −1),

with d a generating element of the multiplicative group GF(q^{3}) \ {0}. This implies that

2l(q^{2} − q) ≡ 0 mod (q^{3} − 1),

or

2lq ≡ 0 mod (q^{2} + q + 1).

Since q^{2} + q + 1 is odd, and not a multiple of q, l is necessarily a multiple of (q^{2} + q + 1), or l = l′(q^{2} + q + 1). This implies

d^{l(q^{2}−q)} = d^{l′(q^{2}+q+1)(q^{2}−q)} = d^{l′q(q^{3}−1)} = 1,

hence −1 = 1, a contradiction, hence θ is a bijection. If Γ ≅ H(q), we obtain

t_{p} = \frac{k + 2ab}{b}.

Now it is clear that also in this case, θ is a bijection if and only if q is odd. So the line L does not contain a point x opposite every point of R \ {y} if and only if q is odd, showing (ii). □

Lemma 2.3.2 Let Γ be a finite generalized hexagon satisfying property (C). Then Γ is distance-3-regular.

Proof. Let x, y, z be three distinct points of a regulus ⟨L, M⟩ and let N ∈ ⟨x, y⟩, L ≠ N ≠ M. We have to show that δ(z, N) = 3. Let v = proj_{N}x and let w = proj_{N}y.

Suppose first that there exists a line N′ through v, N′ ≠ vx, at distance 5 from every element of ⟨L, M⟩ \ {x}. Consider the projection of ⟨L, M⟩ onto N′. Since v is the image of at least two points of ⟨L, M⟩ (namely x and y), this projection is not injective, hence neither surjective. So there is a point incident with N′ opposite every element of ⟨L, M⟩ \ {x}. Property (C) implies now that every point of ⟨L, M⟩ \ {x} is at distance 4 from v. As a consequence, every point z′ of ⟨L, M⟩ \ {y} is at distance 5 from every line N′′ through w, N′′ ≠ N (indeed, otherwise there arises a circuit of length δ(z′, v)+δ(v, N′′)+δ(N′′, z′) ≤ 4+3+3). Hence, interchanging the roles of v and w, we see that δ(z, v) = δ(z, w) = 4, which is only possible if δ(z, N) = 3.

Now suppose that for every line N′ ≠ vx through v there exists a point of ⟨L, M⟩ \ {x} at distance 3 from N′. By the previous paragraph, we can also assume that for every line N′′ ≠ wy through w, there exists a point of ⟨L, M⟩ \ {y} at distance 3 from N′′. Put z′ = proj_{L}z and z′′ = proj_{M}z. Note that at least one of the lines zz′ and zz′′ is opposite N, otherwise δ(z, N) = 3 and we can go home. Suppose δ(zz′′, N) = 6. Similarly, z is opposite at least one of v, w. Suppose δ(v, z) = 6. Let r be the projection of v onto zz′′ (then z ≠ r ≠ z′′). Put R_{1} = proj_{v}r and R_{2} = proj_{r}v (then R_{1} ≠ N). By assumption, there exists a point y′ ∈ ⟨L, M⟩\{x} at distance 3 from R_{1} (and note that proj_{R_{1}}y′ ≠ r ⋈v because otherwise the points y′, proj_{M}y′, z′′, r and r ⋈v define a pentagon). We consider the projection of ⟨L, M⟩ onto the line R_{2}. Since x and y′ are both mapped onto r ⋈v, this projection is, as above, not surjective, and hence, again using property (C) as before, r must be at distance 4 from every element of ⟨L, M⟩ \ {z}, a final contradiction as, for example, δ(r, x) = 6. This shows the distance-3-regularity. □

Combining Lemma 2.3.2 and Theorem 1.8.3, we now obtain a characterization of some extremal classical hexagons.

Corollary 2.3.3 If Γ is an extremal hexagon satisfying Property (C), then it is isomorphic to T(q, q^{3}) or to T(q′^{3}, q′), q′ odd.

Lemma 2.3.4 Let Γ be a finite generalized hexagon satisfying property (C). Then the following property holds in Γ: if a point x is at distance at most 4 from at least three points y_{1}, y_{2}, y_{3} of a point regulus R, then x is at distance 2 from a unique element of R and at distance 4 from all other elements of R.

Proof. Let Γ, x, R, y_{1}, y_{2} and y_{3} be as in the lemma. Note that Γ is distance-3-regular. If one of the points y_{1}, y_{2}, y_{3} is collinear with x, or if at least two of these points have the same projection onto x, then the property above immediately follows from the distance-3-regularity. So suppose now that δ(x, y_{i}) = 4, i = 1, 2, 3 and that the projections of the points y_{1}, y_{2}, y_{3} onto x are all different. We look for a contradiction. Put N = proj_{x}y_{1}. If N is at distance 3 from an element of R \ {y_{1}} then, by the distance-3-regularity, proj_{x}y_{2} = N, a contradiction with the assumption. So we may assume that every point of R \ {y_{1}} is at distance 5 from N. We then consider the projection of R \ {y_{1}} onto N. Since |R \ {y_{1}}| = |N| − 1 and both y_{2} and y_{3} have the same image, there is some point w incident with N and distinct from x ⋈ y_{1} opposite every element of R \ {y_{1}}. Property (C) implies that δ(y_{2}, x⋈y_{1}) = 4, a contradiction. □

If we want to use property (C) to characterize (some) dual classical hexagons, Lemma 2.3.1 (ii) shows that the condition q even is certainly necessary. It turns out that this condition is also sufficient.

Theorem 2.3.5 Let Γ be a finite generalized hexagon of order (q, t), q even. Then Γ has property (C) if and only if Γ is isomorphic to H(q)^{D} or to T(q, q^{3}).

Proof. Let Γ be a finite generalized hexagon of order (q, t), q even, satisfying property (C). Let x and y be two opposite points of Γ, and L an arbitrary line of Γ. We claim that there exists a line of the line regulus ⟨x, y⟩ at distance at most 4 from L. If L lies at distance ≤ 3 from a point of R(x, y), then the claim follows because of the 3-regularity. So we may assume that L is at distance 5 from every point of R(x, y). Hence, by Lemma 2.3.4, no point incident with L is the projection of at least 3 elements of R(x, y). So every point on L is the projection of 0,1 or 2 points of R(x, y). Since |R(x, y)| = q + 1 is odd, there is a point v on L which is the projection of exactly one point z of R(x, y). So v is opposite every point of R(x, y) \ {z}. Property (C) now implies that v ⋈ z is at distance 4 from every point of R(x, y) \ {z}. By the distance-3-regularity, the projection of any element of R(x, y) \ {z} onto v ⋈ z is a line of ⟨x, y⟩ lying at distance 4 from L. This shows the claim. Now the property just shown is exactly the dual of the condition in Theorem 2.2.1, hence the result follows. □

? △ The property mentioned in Lemma 2.3.4, together with the 3-regularity, is equivalent with property (C). But does this property on its own characterize the finite dual classical hexagons over a field of even characteristic ?

Let x be a point of a generalized hexagon Γ, and y and z two points opposite x. We say that the points y and z are in (3, 4)-position with respect to x if there exist points v, w ∈ x^{y} ∩ x^{z} such that proj_{v}y = proj_{v}z but proj_{w}y ≠ proj_{w}z (∗). So there is a line at distance 3 from x, y and z and a point at distance 4 from both y, z and collinear with x (see the picture in Appendix B).

Lemma 2.3.6 Let Γ be a finite generalized hexagon of order (q, t) isomorphic to H(q)^{D}, q not divisible by 3, or to T(q, q^{3}). Then |x^{y} ∩ x^{z}| = 1 + t/q for all points x, y and z such that y and z are in (3, 4)-position with respect to x.

Proof. Let Γ be as above and v, w, x, y, z as in (∗). Let L = proj_{v}y = proj_{v}z, let M_{y} = proj_{w}y, M_{z} = proj_{w}z and u = v ⋈ y. If z and u are collinear, then x^{y} ∩ x^{z} is an intersection set, hence |x^{y} ∩ x^{z}| = 1 + t/q (see section 1.8). Suppose δ(u, z) = 4. Let a be the unique point of ⟨L, M_{z}⟩ at distance 4 from y. Because of the distance-3-regularity, we have x^{z} = x^{a}. Note that x^{y} ∩ x^{a} is an intersection set. Since we are working with the finite dual classical hexagons, we have |x^{a} ∩ x^{y}| = 1 + t/q. This proves the lemma. □

Theorem 2.3.7 Let Γ be a finite generalized hexagon of order (q, t). Then Γ is isomorphic to H(q)^{D}, q not divisible by 3, or to T(q, q^{3}) if and only if |x^{y} ∩ x^{z}| ≤ 1 + t/q for all points x, y and z such that y and z are in (3, 4)-position with respect to x.

Proof. Let Γ be a finite generalized hexagon of order (q, t) satisfying the condition above on points in (3,4)-position. Note that this condition implies that every intersection set contains at most 1 + t/q points. We claim that each intersection set contains exactly 1 + t/q points. Indeed, let p^{p′} ∩ p^{p′′} be an intersection set, put r = p′ ⋈ p′′ and r′ = r ⋈ p. Now project the points of p^{p′} \ {r′} onto the line p′′r. Since every point of p′′r different from r (and there are q of these points) is the projection of at most t/q points of p^{p′} \ {r′}, every point of p′′r different from r is the projection of exactly t/q points, hence the claim.

Let x and y be opposite points, and {v, w} ⊆ x^{y}. Put u = v ⋈y. Let L be an arbitrary line through x, wx ≠ L ≠ vx and p a point on L, p ≠ x. If there is a point z ∈ u^{w} \ {v} at distance 4 from p, then u^{w} ∩ u^{p} is an intersection set, implying that p is the projection of exactly t/q points of u^{w} \ {v} onto L. Hence each point on L different from x is the projection of either 0 or t/q elements of u^{w} \ {v}. Since |L \ {x}| = q and |u^{w} \ {v}| = t, each point on L different from x is the projection of exactly t/q points of u^{w} \ {v}.

Now we prove that Γ is distance-3-regular. Put w′ = y ⋈ w. Let z be a point of ⟨uv, ww′⟩ different from x or y. Consider the set

S = {(a, b)|a ∈ x^{z} \ {v, w}, b ∈ u^{w} \ {v} with δ(a, b) = 4}.

We count the number of elements in S. Fixing a, we obtain that |S| = (t − 1) · \frac{t}{q} by the previous paragraph. Now fix b. The points a then belong to (x^{b} ∩ x^{z}) \ \{v, w\}. If b ≠ y, then there at most \frac{t}{q} − 1 choices for a, since b and z lie in (3,4)-position with respect to x. Put ℓ = |x^{y} ∩ x^{z} \ {v, w}|. Now we obtain

(t − 1) · \frac{t}{q} ≤ (t − 1)(\frac{t}{q} − 1) + ℓ,

implying ℓ = t − 1, which means that x^{y} = x^{z}. Also, proj_{r}y = proj_{r}z for all r ∈ x^{y} ∩x^{z}. Indeed, if not, then the points y and z would lie in (3,4)-position with respect to x, hence |x^{y} ∩ x^{z}| ≤ \frac{t}{q} + 1 < t + 1. This shows that Γ is distance-3-regular. Since every intersection set contains 1 + t/q points, and hence more than one point, Γ is line-distance-2-regular by Theorem 1.8.4 (ii). By Theorem 1.8.2, Γ is classical. □

Remark. In the previous theorem, we did not assume that t is divisible by q, or that q ≤ t.

Let x, y, z be points of a generalized hexagon Γ. If y and z are opposite x, then we denote

x^{{y,z}} = {r ∈ x^{y} ∩ x^{z}|proj_{r}y ≠ proj_{r}z}.

Let Γ be a finite generalized hexagon of order (q, q). We say that Γ is antiregular if for any three points x, y, z, with y, z lying opposite x, the condition |x^{{y,z}}| ≥ 2 implies that |x^{{y,z}}| = 3 and x^{{y,z}} = x^{y} ∩ x^{z}. We say that Γ is weak anti-regular if for any three points x, y, z, with y, z lying opposite x, the condition |x^{{y,z}}| ≥ 2 implies that |x^{{y,z}}| ≥ 3 and x^{{y,z}} = x^{y} ∩ x^{z}.

Lemma 2.3.8 The generalized hexagon H(q)^{D}, with q not divisible by 3, is anti-regular (and hence also weak anti-regular).

Proof. Let Γ be the hexagon H(q)^{D}, q not a multiple of 3. Let y, z be points opposite a point x, and v, w ∈ x^{{y,z}}. Put L_{y} = proj_{v}y, L_{z} = proj_{v}z, M_{y} = proj_{w}y, M_{z} = proj_{w}z, u′ = w ⋈ y and u′′ = w ⋈ z. We first claim that the pointwise stabilizer of {x, v, w, z, M_{y}} in the automorphism group of Γ acts transitively on the set Γ_{1}(v) \ {vx, L_{z}}. Choose coordinates such that v = (∞), L_{z} = [∞] and the apartment through x, v, w and z is the hat-rack of the coordinatization. Consider the following homologies h_{A}:

\begin{aligned}h_{A} & : & (a, l, a′, l′, a′′) & → & (A^{−1}a, l, Aa′, Al′, A^{2}a′′) \\ & & [k, b, k′, b′, k′′] & → & [Ak, A^{2}b, Ak′, Ab′, k′′], \end{aligned}

with A ∈ GF(q) \ {0} (it suffices to give the action of h_{A} on the elements opposite (∞) and [∞], see 4.5.3 in [57]). Then each h_{A} fixes the apartment through x, v, z and w, and every line through the point w (since a line through w different from xw or M_{z} has coordinates [0, 0, 0, 0, k′′]). Also, these homologies h_{A} act transitively on the lines through u′′ different from M_{z} or u′′z (since these lines have coordinates [k, 0, 0, 0, 0]), and hence on the lines through v different from vx or L_{z}. This shows the claim.

Now consider the following set

T = {x^{u_{L}}|u_{L} = (proj_{L}u′)⋈u′, L ∈ Γ_{1}(v) \ {vx}}.

Since the intersection of two traces belonging to T is an intersection set, two traces contained in T only have the points v and w in common. So, if R is an arbitrary line through x different from xv or xw, the q points u_{L} all project onto a different point of R, implying that the set T meets x^{z} \ {v, w} in exactly q −1 points. Note that the traces x^{z} and x^{u_{Lz}} only have the points v and w in common, since u_{L_{z}} and z are in (3,4)-position with respect to x. We claim that the q − 1 traces of T \ {x^{u_{Lz}} } meet the trace x^{z} in a constant number of points. Indeed, we look for the image under h_{A} of x^{z} ∩ x^{u_{L}}, L ≠ L_{z}. The homology h_{A} leaves the set x^{z} invariant. The image under h_{A} of the trace x^{u_{L}} is the trace x^{r}, with r the unique point collinear with u′′ := h_{A}(u′) (which is a point on M_{y}) and at distance 3 from L′ = h_{A}(L). Hence |x^{z} ∩ x^{u_{L}}| = |x^{z} ∩ x^{r}|. Because of the distance-3-regularity, we have x^{r} = x^{u_{L′}}, implying |x^{z} ∩ x^{u_{L}}| = |x^{z} ∩ x^{u_{L′}}|. Now by the transitivity of the homologies h_{A} mentioned in the first paragraph, it follows that each trace of T \ {x^{u_{Lz}} } meets x^{z} in a constant number (=3) of points. So in particular, |x^{y} ∩ x^{z}| = 3.

Now let a be the unique point of x^{y} ∩ x^{z} different from v and w. If proj_{a}y = proj_{a}z, then the points y and z are in (3,4)-position with respect to x, implying |x^{y} ∩ x^{z}| = 2, a contradiction. Hence x^{{y,z}} = x^{y} ∩ x^{z}, which shows the lemma. □

Remark. Lemma 2.3.8 can also easily be proved using coordinates instead of the more geometric arguments given above.

Theorem 2.3.9 A finite generalized hexagon Γ of order (q, q) is weak antiregular if and only if it is isomorphic to H(q)^{D}, with q not divisible by 3.

Proof. Let Γ be a finite weak anti-regular hexagon of order (q, q). Let y and z be points in (3,4)-position with respect to a point x; v, w ∈ x^{y} ∩ x^{z}, with proj_{v}y = proj_{v}z and proj_{w}y ≠ proj_{w}z. We prove that |x^{y} ∩ x^{z}| = 2. The result will then follow from Theorem 2.3.7. Put u = v ⋈y and u′ = w ⋈y. Suppose first that z and u are collinear, and let by way of contradiction

Figure 2.5: Characterizations obtained in sections 2.2 and 2.3.

r ∈ x^{y} ∩ x^{z} \ {v, w}. Clearly, proj_{r}y ≠ proj_{r}z. But now, the weak antiregularity implies that also proj_{v}y ≠ proj_{v}z, a contradiction.

Suppose now that δ(u, z) = 4. Let T = {x^{r}|r ∈ Γ_{2}(u′) ∩ Γ_{4}(v) ∩ Γ_{6}(x)}. By the previous paragraph, every two elements of T have exactly the points v, w in common. By the weak anti-regularity, every set T ∩x^{z}, with T ∈ T \{x^{y}}, contains at least one element different from v, w. Since this gives rise to at least q−1 elements of x^{z} \{v, w}, there is no room anymore in x^{z} for elements of x^{y} \ {v, w}, implying |x^{y} ∩ x^{z}| = 2. □

Remark. We call a finite generalized 2n-gon Γ of order (q, q) anti-regular if for any three points x, y, z, with z, y both opposite x, such that |Γ_{2}(x) ∩ Γ_{2n−2}(y) ∩ Γ_{2n−2}(z)| ≥ n − 1 and |Γ_{1}(w) ∩ Γ_{2n−3}(y) ∩ Γ_{2n−3}(z)| = 0, for at least n − 1 elements w of Γ_{2}(x) ∩ Γ_{2n−2}(y) ∩ Γ_{2n−2}(z), we have that |Γ_{2}(x) ∩ Γ_{2n−2}(y) ∩ Γ_{2n−2}(z)| = n and |Γ_{1}(w) ∩ Γ_{2n−3}(y) ∩ Γ_{2n−3}(z)| = 0, for all elements w of Γ_{2}(x) ∩ Γ_{2n−2}(y) ∩ Γ_{2n−2}(z). In this way, one can see that the definition given earlier for an anti-regular hexagon generalizes the definition of anti-regularity given in Payne & Thas [34].

? △ Proof Theorem 2.3.9 under the weaker assumption that ‘a certain set’ of points of Γ is anti-regular.

Some of the characterizations obtained in sections 2.2 and 2.3 are summarized in Figure 2.5.

2.4 A characterization of Moufang hexagons using ovoidal subspaces

Recall that for two opposite points x, y in a generalized hexagon, we say that the pair (x, y) is 3-regular if the set ⟨x, y⟩ is determined by any two of its lines.

Theorem 2.4.1 Let Γ be a generalized hexagon in which all intersection sets x^{y} ∩ x^{z} either have size 1 or satisfy x^{y} = x^{z}. If Γ contains an ovoidal subspace O all the points of which are 3-regular, then Γ is 3-regular and hence a Moufang hexagon.

Proof. The condition about the intersection sets is equivalent with the fact that every two opposite points x and y are contained in a (unique) thin ideal subhexagon, which we denote by D(x, y) (this follows for instance from Van Maldeghem [57], Lemma 1.9.10). Let x and y be two opposite points: we prove that the pair (x, y) is 3-regular. It is sufficient to find a point z at distance 3 from two lines of ⟨x, y⟩ such that one of the pairs (x, z) or (y, z) is 3-regular. We may assume that neither x nor y belongs to O.

(a) Suppose there is a point a ∈ D(x, y) opposite x and at distance 4 from y such that the pair (a, x) is 3-regular. Let L_{a} be the line through a ⋈ y at distance 3 from x and let z be an arbitrary point in ⟨L_{a}, M⟩, L_{a} ≠ M ∈ ⟨x, y⟩, z ≠ x. We show that x^{y} = x^{z}. Put u = proj_{L_{a}}z, u′ = proj_{M}a, X = proj_{u′}a and z′ the point of ⟨L_{a}, X⟩ collinear with u. Then x^{y} = x^{a} and x^{z′} = x^{z} because we work in the thin subhexagon D(x, y). Also x^{a} = x^{z′} by the 3-regularity, hence x^{y} = x^{z}.

(b) Suppose in addition to (a) there is a point b ∈ D(x, y) opposite y and at distance 4 from x such that the pair (b, y) is 3-regular. Then the pair (x, y) is 3-regular. Indeed, let L_{b} be the line through b⋈x at distance 3 from y. If L_{a} ≠ L_{b}, put M = L_{a} and N = L_{b}. If L_{a} = L_{b}, put M = L_{a} and N ∈ ⟨x, y⟩, N ≠ L_{a}. Applying (a), we see that for an arbitrary point z ∈ ⟨M, N⟩, x ≠ z ≠ y, x^{y} = x^{z} and y^{x} = y^{z}, so z lies at distance 3 from every line of ⟨x, y⟩, and the pair (x, y) is 3-regular.

Let first O be an ovoid not containing x or y. Then D(x, y) contains 0, 1 or 2 points of O. Suppose first D(x, y) contains two points a and b of O. Up to interchanging x and y, one of the following situations occurs:

• δ(a, x) = 4 = δ(b, y) and δ(a, y) = 6 = δ(b, x). It immediately follows from (b) that the pair (x, y) is 3-regular.

• δ(a, x) = 2, δ(b, x) = 4 and δ(b, y) = 6. Note that b lies at distance 4 from the point a ⋈ y. Let L_{b} be the line of ⟨x, y⟩ at distance 3 from b and L_{a} the line of ⟨x, y⟩ through a. Then (a) shows that

(1) y^{x} = y^{z′}, for all points z′ ∈ ⟨L_{a}, L_{b}⟩, z′ ≠ y.

Consider the point v of y^{x} on L_{b}. Suppose first that the unique point o of O collinear with v does not lie on vy. Put u = proj_{L_{a}}o. Note that u ≠ a since o ∉ D(x, y). Let finally z = u ⋈ (proj_{L_{b}}u). Then applying (a), we obtain

(2) z^{y} = z^{x} = z^{w}, for all w ∈ ⟨L_{a}, L_{b}⟩, w ≠ z.

Combining (1) and (2) as in (b), we see that the pair (x, y) is 3-regular. So we may now assume that o lies on vy. Consider an arbitrary point p of D(x, y) collinear with v, different from y or v ⋈ x. Since the line vp does not contain a point of O, we can apply the previous argument (noting that D(x, y) = D(x, p)) to obtain that the pair (x, p) is 3regular. But now again applying (b) shows that also the pair (x, y) is 3-regular.

• δ(a, x) = 2 = δ(b, y).

Let p be a point of D(x, y) collinear with a⋈y, different from a and y, and p′ a point of D(x, y) collinear with b ⋈ x, different from b and x. Then the previous paragraph shows that both (x, p) and (y, p′) are 3regular pairs, so applying (b) gives that also the pair (x, y) is 3-regular.

Suppose now D(x, y) contains exactly 1 point a of O. Then we have the following cases to consider:

• δ(a, x) = 2.

Let v be a point of x^{y} different from a and put w = v ⋈ y, w′ = a ⋈ y. Denote by o the unique point of O collinear with v. If o lies on vw, then put z = o ⋈ (proj_{aw′}o). If o does not lie on vw, then put u = proj_{aw′}o and z = u ⋈ (proj_{vw}u). Now D(x, z) contains two points of O, so the pair (x, z) (and hence the pair (x, y)) is 3-regular.

• δ(a, x) = 4 and δ(a, y) = 6.

Let again L_{a} be the line of ⟨x, y⟩ at distance 3 from a. By (a), we already know that (3) y^{x} = y^{z}, for all points z, y ≠ z ∈ ⟨L_{a}, M⟩, M ∈ ⟨x, y⟩, M ≠ L_{a}.

Choose a point v ∈ y^{x}, v not on L_{a} such that the unique point o ∈ O collinear with v does not lie on the line vy. Let L_{v} be the line of ⟨x, y⟩ through v. If o lies on L_{v}, then put z = o ⋈ (proj_{L_{a}}o). From the previous case, it is then clear that (x, z) is a 3-regular pair. If o does not lie on L_{v}, put u = proj_{L_{a}}o and z = u⋈proj_{L_{v}}u. Then by (a),

(4) z^{y} = z^{x} = z^{w}, for all points w, z ≠ w ∈ ⟨L_{v}, M⟩, M ∈ ⟨x, y⟩, M ≠ L_{v}.

Combining (3) and (4), we again see that (x, y) is a 3-regular pair.

Suppose finally D(x, y) does not contain any point of O. Similarly as before, we can find a point z at distance 3 from two lines of ⟨x, y⟩ for which the hexagon D(x, z) contains a point of O, from which the result follows.

Suppose now O = Γ_{1}(M) ∪ Γ_{3}(M), M a line of Γ at distance 5 from x and y. Then D(x, y) contains the line M, or D(x, y) intersects O in either 0 points or 2 collinear points. (Indeed, if D(x, y) contains a point of M, then M is a line of D(x, y). If D(x, y) contains a point p ∈ Γ_{3}(M), but no point of M, then the line L = proj_{p}M of D(x, y) contains two points of O. If D(x, y) would contain a point p′ of O, p′ not on L, then also proj_{L}p′ ∈ D(x, y), a contradiction.) As before, the case that D(x, y) contains no point of O can be reduced to one of the other cases. If D(x, y) contains M, then the pair (x, y) is 3-regular because of (b). So we only have to consider the case that D(x, y) contains exactly two collinear points a and b of O. We consider the following situations:

• δ(a, x) = 2 = δ(b, y).

Since M is concurrent with the line ab, we can find a point z ∈ O at distance 3 from ab and another line of ⟨x, y⟩, hence the result follows.

• δ(a, y) = 6 = δ(b, x) (hence δ(a, x) = 4 = δ(b, y)). This is clear because of (b).

• δ(a, x) = 2, δ(b, x) = 4 and δ(b, y) = 6.

Note that the line ab is concurrent with M, and that a and b lie at distance 3 from M. Let L_{a} be the line of ⟨x, y⟩ through a and L′ an arbitrary line of ⟨x, y⟩ different from L_{a}. Let u be the projection onto L′ of the intersection of M and ab. Let finally z be the unique point of ⟨L_{a}, L′⟩ collinear with u. Clearly, D(x, z) contains M, hence the pair (x, z) is 3-regular, and so is (x, y).

Suppose finally that O is a full subhexagon. If D(x, y) contains at least one point p of O, then it has at least an ordinary hexagon through p in common with O. From this, it is easily seen that, if D(x, y) ∩ O is nonempty, either x or y lie in O (and then we are done), or D(x, y) contains points o_{1}, o_{2} ∈ O, δ(o_{1}, x) = δ(o_{2}, y) = 4 and δ(o_{2}, x) = δ(o_{1}, y) = 6. In the latter case, (x, y) is 3-regular because of (b). Again as before, the case that D(x, y) contains no point of O can be reduced to the previous one. □

Let p be a point of a generalized hexagon Γ. We say that the point p is intersection-regular, if for every intersection set p^{x} ∩ p^{y}, x and y opposite p, the condition |p^{x} ∩ p^{y}| > 1 implies p^{x} = p^{y}. ^{1} A subset B of the line set of a generalized hexagon Γ is called a line blocking set if every line of Γ not contained in B intersects at least one line of B.

Theorem 2.4.2 Let Γ be a distance-3-regular hexagon, and B a line blocking set in Γ. Suppose that all the points lying on any line of B are intersectionregular. Then Γ is point-distance-2-regular and hence a Moufang hexagon.

Proof. Let Γ and B be as above, and denote by \overline{B} the points lying on any line of B. Let p be an arbitrary point of Γ, and x, y points opposite p such that p^{x} ∩ p^{y} is an intersection set containing at least two points. We show that p^{x} = p^{y}. This will prove the result, in view of Theorem 1.8.4 (i). Put z = x⋈y, v = p⋈z and w ∈ p^{x} ∩ p^{y}, w ≠ v. If p ∈ \overline{B}, we are done, so assume p ∉ \overline{B}. Let o_{1} be a point of \overline{B} on the line vp, o_{2} a point of \overline{B} on the line wp and R a line of B containing o_{1}. Then we distinguish the following cases.

(i) o_{1} ≠ v Let L be a line through p, vp ≠ L ≠ wp. Suppose by way of contradiction that y′ := proj_{L}y ≠ proj_{L}x. Put x′ = proj_{xz}y′. Let a be the point of the regulus R(z, w) collinear with o_{1}, and b the point of the regulus R(y′, z) collinear with o_{1}. Note that δ(a, proj_{y}w) = δ(a, proj_{x}w) = 3 and δ(b, proj_{x′}y′) = δ(b, proj_{y}y′) = 3. Also, δ(a, b) = 4 (indeed, if not, the points a, b and the distance-4-paths [b, y] and [a, y] would define a pentagon). Now o_{1}^{x′} ∩ o_{1}^{y} is an intersection set (containing the points v and b), hence o_{1}^{x′} = o_{1}^{y}, contradicting a ∈ o_{1}^{y} but δ(a, x′) = 6.

(ii) o_{1} = v and R = vz Let w′ be an arbitrary point of p^{x} \{v, w}. Then z^{w} ∩z^{w′} is an intersection set (containing the points v and x) hence, since z ∈ \overline{B}, z^{w} = z^{w′}. Since y ∈ z^{w}, we have δ(y, w′) = 4, showing p^{x} = p^{y}.

1Using the terminology of Ronan [37], a point is intersection-regular exactly if all intersection sets with respect to this point have size 1.

2.5 Two characterizations of the Hermitian spread in H(q) 51

(iii) o_{1} = v, R ≠ vz and o_{2} ≠ w. Put a = x ⋈ w, b = y ⋈ w and c = proj_{R}a. Since v^{a} = v^{b} (indeed, v^{a} ∩ v^{b} is an intersection set containing the points p and z), δ(b, c) = 4. Put x′ = a ⋈ c and y′ = b ⋈ c. Because p^{x} ∩ p^{x′} is an intersection set (containing the points v and w), and o_{2} ≠ w, case (i) above implies that p^{x} = p^{x′}. Similarly, p^{y} = p^{y′}. Since also p^{x′} = p^{y′} by case (ii), we obtain p^{x} = p^{y}.

(iv) o_{1} = v and o_{2} = w. Let L be an arbitrary line through p, vp ≠ L ≠ wp. Suppose by way of contradiction that y′ := proj_{L}y ≠ proj_{L}x =: x′. We can assume that either both x′ and y′ belong to \overline{B} or, without loss of generality, x′ ∉ \overline{B}. Put x′′ = proj_{yz}x′. Now p^{x} ∩p^{x′′} is an intersection set (containing v and x′) and because of the assumptions, L contains a point of \overline{B} different from x′. Hence we can apply (iii) to obtain p^{x} = p^{x′′}. Since w ∈ p^{x}, this implies δ(w, x′′) = 4, a contradiction.

□

Remark. Let O be a dual ovoidal subspace of a hexagon Γ. Then O is a line blocking set, having the property that every line of Γ not contained in O intersects exactly one line of O. If Γ is a finite generalized hexagon of order (s, t), then the set O has roughly size st^{2}, and \overline{O} (the set of points lying on any line of O) has roughly size s^{2}t^{2}, while the point set of Γ itself has roughly size s^{3}t^{2}.

2.5 Two characterizations of the Hermitian spread in H(q)

Let S be a spread of the generalized hexagon Γ = H(q), and define the following geometry Γ_{S}. The points of Γ_{S} are the points of Γ on lines of the spread. For a point p not on any line of S, we denote by V_{p}^{S} the set of q + 1 points of Γ_{S} collinear with p (if there is only one spread S around, we use the notation V_{p} for short). Now the lines of Γ_{S} are the lines of the spread together with the sets V_{p}^{S}, p ∈ Γ\Γ_{S}. Incidence is symmetrized containment. It is easy to check that Γ_{S} is a generalized quadrangle of order (q, q^{2}) if and only if the spread S satisfies the following property:

(◇) Let L_{1}, L_{2} and L_{3} be three different lines of the spread S, and x_{1} a point on L_{1}. Put x_{2} = proj_{L_{2}}x_{1} and x_{3} = proj_{L_{3}}x_{2}. If δ(x_{1}, x_{3}) = 4, then necessarily x_{1} ⋈x_{2} = x_{1} ⋈x_{3} = x_{2} ⋈x_{3}.

Figure 2.6: A forbidden configuration in H(q).

Property (◇) says that a configuration as in Figure 2.6 (where the bold lines are spread lines) is forbidden. Note that the lines L_{1}, L_{2} and L_{3} in (◇) necessarily belong to the same line regulus.

Theorem 2.5.1 A spread S of the finite generalized hexagon H(q) is Hermitian if and only if the geometry Γ_{S} is a generalized quadrangle, if and only if S satisfies property (◇).

Proof. If S is the Hermitian spread of H(q), then Γ_{S} is indeed a generalized quadrangle (namely the quadrangle Q(5, q)), so assume now we have a spread S of H(q) satisfying property (◇). We prove that for any two lines of S, the regulus defined by these lines is contained in S. The result will then follow from Theorem 1.7.1.

Let p be a point of Γ = H(q) not on any line of the spread. We claim that V_{p} is in fact a distance-2-trace in Γ. Let a and a′ be two different points of V_{p} and suppose by way of contradiction that the trace defined by a and a′ contains a point b, b ∉ V_{p}. Let L be the line of S through a, and let L′ be an arbitrary line of Γ through b, different from bp. Let finally L′′ be the unique spread line that is concurrent with L′ and put y = proj_{L′′}a. Note that proj_{a}y ≠ L (since spread lines are necessarily opposite). Because of the distance-2-regularity, the trace defined by a and b is equal to p^{y}, so δ(y, a′) = 4. If we denote by N the spread line through a′, then proj_{a′}y ≠ N. But now we obtain a configuration forbidden by (◇), by considering the spread lines L, N and L′′, together with the ordinary hexagon through a, p, a′ and y, a contradiction. This shows our claim.

Let L_{0} and L_{1} be two different lines of S, and M ∈ R(L_{0}, L_{1}), L_{0} ≠ M ≠ L_{1}. We show that M ∈ S. Let p and p′ be two different points belonging to the point regulus ⟨L_{0}, L_{1}⟩. By the previous paragraph, we know that V_{p} = p^{p′} and V_{p′} = p′^{p}. But this implies that both proj_{M}p and proj_{M}p′ have to lie on lines of S. Because spread lines are opposite, this is only possible if M ∈ S, which shows that R(L_{0}, L_{1}) is contained in S.

The theorem is proved. □

? △ Suppose Γ is a finite generalized hexagon of order (q, q) having a spread S for which the geometry Γ_{S} is the classical quadrangle Q(5, q). Is Γ itself classical ?

For an arbitrary generalized hexagon Γ, one could ask whether the existence in Γ of ‘a lot of’ spreads having property (◇) is enough to conclude that Γ is classical. We now give a first result in this direction.

Let Γ be a generalized hexagon of order (q, q), and Σ a set of spreads of Γ such that the following conditions hold:

(1) Each element of Σ has property (◇).

(2) For any two points a and b of Γ, with δ(a, b) = 4, there exists a spread S ∈ Σ such that a and b lie on lines of S.

(3) Let S ∈ Σ, and a and b two points of Γ contained in a set V_{p}^{S}, with p = a ⋈ b. For any point p′ at distance 4 from a and opposite b and p, such that p′ lies at distance 5 from the line of S through a, there exists a spread S′ ∈ Σ for which p′ lies on a line of S′ and V_{p}^{S} = V_{p}^{S′}.

Proposition 2.5.2 A finite generalized hexagon of order (q, q) contains a set of spreads Σ satisfying (1), (2), (3) if and only if Γ is isomorphic with H(q).

Proof. Let Γ ≅ H(q). Let Σ be the set of Hermitian spreads of Γ. Then clearly properties (1) and (2) hold. Now let a, b and p′ be as in (3). Then these three points define a plane intersecting the quadric Q(6, q) in two concurrent lines. Through such a plane, there always exists a hyperplane intersecting Q(6, q) in a non-singular elliptic quadric. Hence (3) holds. Now let Γ be a finite generalized hexagon of order (q, q) containing a set of spreads Σ satisfying (1), (2) and (3). By condition (2), it suffices to prove that, for any spread S of Σ, each V_{p}^{S} is a trace determined by two of its points. Fix a set V_{p}^{S}, points a, a′ of V_{p}^{S}, and a point b on pa′, b different from p and a′. Let p′ be a point opposite p, at distance 4 from both a and b. Now we claim that p′ does not lie at distance 4 from a point of V_{p}^{S} \ {a}. Indeed, suppose by way of contradiction there is a point a′′ ∈ V_{p}^{S} \{a} at distance 4 from p′. Suppose first p′ lies at distance 5 from the line of S through a. By condition (3), there exists a spread S′ ∈ Σ such that p′ lies on a line of S′, and V_{p}^{S} = V_{p}^{S′}. But now the forbidden configuration occurs in the ordinary hexagon through a, a′′ and p′, the contradiction. Next, consider the case that p′ lies at distance 3 from the line of S through a. We can assume that the line proj_{a′′}p′ belongs to S (indeed, otherwise we obtain a contradiction with the previous case). Put p_{1} the projection of a′ onto the line proj_{p′}a, and p_{2} the projection of a′ onto the line proj_{p′}a′′. Then as before, the line proj_{a′}p_{1} belongs to S. But now the point p_{2} lies at distance 4 from a′, at distance 5 from the line of S through a′ and opposite a ∈ V_{p}, hence (again by the previous case), p_{2} lies opposite every element of V_{p} \ {a′}, contradicting δ(p_{2}, a′′) = 4.

This shows the claim. Hence an arbitrary point y opposite p and at distance 4 from both a and a′, will be at distance 4 from every point of V_{p}, which shows that V_{p} is a trace determined by two of its points. □

Let Γ be a generalized hexagon admitting a spread S, and let L_{0} be a line of S. Consider all projectivities [L_{0}; L_{1}] … [L_{k}, L_{0}] of the line L_{0} for which the lines L_{i}, 0 ≤ i ≤ k, belong to S. These projectivities form a group, called the group of projectivities of L_{0} with respect to S, denoted by Π_{S}(L_{0}).

The groups Π_{S}(L), L ∈ S are all isomorphic, so we can define Π_{S} = Π_{S}(L), for L an arbitrary line of the spread.

Lemma 2.5.3 If S is the Hermitian spread of H(q), then Π_{S} is a Singer group.

Proof. Let S be the Hermitian spread of H(q). Since S is also a spread of the quadrangle Q(5, q), the result follows from De Bruyn [15] (remembering that collinearity in Q(5, q) corresponds to non-opposition in H(q), so we are really talking about the same projectivities). We give an alternative proof using coordinates in the hexagon Γ = H(q). Let L_{0}, L_{1} be two opposite lines of Γ, and choose coordinates such that L_{0} = [∞] and L_{1} = [0, 0, 0, 0, 0]. Let π be a hyperplane of PG(6, q) containing the lines L_{0}, L_{1}, and intersecting Q(6, q) in an elliptic quadric determining a Hermitian spread of Γ. Then for q odd, π has equation X_{1} = mX_{5}, with m a non square in GF(q), or mX_{1} + X_{3} + nX_{5}, with 1 − 4mn a non square in GF(q), and for q even, π has equation mX_{1}+X_{3}+nX_{5}, with Tr(mn) = 1. We proceed with the case that π has equation X_{1} = mX_{5} (the other cases are completely similar). A line N of S not contained in R(L_{0}, L_{1}) has coordinates [mb′, −mk′′, k′, b′, k′′], k′, b′, k′′ ∈ GF(q) (see for instance Bloemen, Thas & Van Maldeghem [4], section 3.1.2). Now it suffices to show that the projectivity σ = [L_{0}; L_{1}][L_{1}; N][N; L_{0}] has exactly two ‘imaginary’ fixed points independent of the choice of N. Let p = (x) be a point on L_{0}. Put p_{1} = proj_{L_{1}}p, p_{2} = proj_{N}p_{1} and p′ = proj_{L_{0}}p_{2}. Then one easily calculates that the point p′ has coordinate (x′), with x = \frac{Ax′−B}{−mBx′+A}, A = k′ − mb′k′′ and B = b′^{2} − mk′′^{2}. Hence p = p^{σ} if and only if x^{2} = 1/m, from which the result follows. □

Theorem 2.5.4 A spread S of the finite generalized hexagon H(q) is Hermitian if and only if Π_{S} is a Singer group.

Proof. Let Γ = H(q) and Q = Q(6, q) the quadric on which Γ is defined. Let L_{0} and L_{1} be two opposite lines of Γ. We first determine the number of Hermitian spreads containing the regulus R(L_{0}, L_{1}). Note that R(L_{0}, L_{1}) defines a 3-space L intersecting Q in a hyperbolic quadric Q^{+}(3, q). Let α be the polarity associated with the quadric Q. Then the plane L^{α} intersects Q in a nondegenerate conic C which is the point regulus ⟨L_{0}, L_{1}⟩. Let γ′ be a 5-space containing L and N the line in which γ′ and the plane L^{σ} intersect. Then either N ∩ C = {p_{1}, p_{2}} (in this case, γ′ intersects Q in a hyperbolic quadric Q^{+}(5, q) corresponding to the thin ideal subhexagon Γ(p_{1}, p_{2}) of Γ), N ∩ C = {p} (in this case, γ′ intersects Q in a cone pQ(4, q) corresponding to the set p^{⊥ ⊥} in Γ) or N ∩ C = ∅ (then γ′ intersects Q in an elliptic quadric Q^{−}(5, q) determining a Hermitian spread of Γ). So the number of Hermitian spreads through R(L_{0}, L_{1}) is equal to the number of lines of L^{α} not intersecting the conic C, which is (q^{2} − q)/2.

Next, we show that for two opposite lines L_{0} and L_{1} of Γ, there is a bijective correspondence between the set of Hermitian spreads containing the regulus R(L_{0}, L_{1}) and the set of Singer groups in PGL_{2}(q). Put G = G_{2}(q). Let S_{1} be a Hermitian spread containing R(L_{0}, L_{1}), and put H_{1} = Π_{S_{1}}. Let H_{2} be an arbitrary Singer group acting on L_{0}. Since every two Singer groups in PGL_{2}(q) are conjugate, there exists an element σ′ ∈ PGL_{2}(q) for which H_{1}^{σ′} = H_{2}. Now choose σ in G_{L_{0}} such that σ/Γ_{1}(L_{0}) = σ′. Since the pointwise stabilizer of a line in H(q) acts transitively on the lines opposite this line (this follows from the Moufang condition, see [57], Lemma 5.2.4 (ii)) we can choose an element β fixing L_{0} pointwise such that L_{1}^{σβ} = L_{1}. Now σβ maps S_{1} to a Hermitian spread S_{2} containing R(L_{0}, L_{1}) for which Π_{S_{2}} = H_{2}. This shows that every Singer group is the group belonging to a Hermitian spread containing R(L_{0}, L_{1}). Furthermore, this spread is unique, since there are as many Hermitian spreads through a certain regulus as there are different Singer groups in PGL_{2}(q) (namely (q^{2} − q)/2).

Let now S = {L_{0}, L_{1}, …, L_{q^{3}}} be a spread of Γ such that G = Π_{S} is a Singer group. Because of the previous paragraph, we can define S_{H} to be the unique Hermitian spread containing R(L_{0}, L_{1}) for which G = Π_{S_{H}}. Let A be the set of lines M of the hexagon opposite L_{0} and L_{1} such that the projectivity β_{M} := [L_{0}; M][M; L_{1}][L_{1}; L_{0}] belongs to G \ {e}. Let N be an arbitrary line of the hexagon opposite L_{0} and L_{1}. If N ∈ R(L_{0}, L_{1}), then β_{N} = e, so N ∉ A. Suppose N ∉ R(L_{0}, L_{1}). Let γ be a 5-space containing L_{0}, L_{1} and N. Then, as in the first paragraph of the proof, γ intersects the quadric Q(6, q) either in an elliptic quadric (case 1), a hyperbolic quadric (case 3) or in a cone pQ(4, q), with p ∈ ⟨L_{0}, L_{1}⟩ (case 2).

(1) Note that all the lines of S_{H} not belonging to R(L_{0}, L_{1}) are contained in A. In this way, we obtain q^{3} − q elements of A. If N does not belong to S_{H}, then L_{0}, L_{1} and N define a Hermitian spread corresponding to a group G′ ≠ G (see the second paragraph of the proof), so β_{N} ∉ G \ {e} and N ∉ A.

(2) Clearly, N lies at distance 3 from p. Let p′ be the projection of p onto N. Then p′ is a fixed point of β_{N}. Suppose there is a fixed point w on N different from p′. Put w_{0} = proj_{L_{0}}w and w_{1} = proj_{L_{1}}w. Because w is a fixed point, we have δ(w_{0}, w_{1}) = 4. Put w′ = w_{0} ⋈ w_{1}. Note that δ(w, w′) = 6 since we assumed N ∉ R(L_{0}, L_{1}). Because of the 2-regularity, (w′)^{w} = (w′)^{p}. This implies that w lies at distance 4 from the point of (w′)^{p} on the unique line L of R(L_{0}, L_{1}) not opposite N, which is only possible if p′ ∈ L. In this case every point on N is fixed. Indeed, let y be an arbitrary point on N, y ≠ p′. Put z_{0} = proj_{L_{0}}y and z the point of ⟨L_{0}, L_{1}⟩ collinear with z_{0}. Let finally z_{1} = proj_{L_{1}}z and z_{2} = proj_{L}z. Then {z_{0}, z_{1}, z_{2}} ⊆ z^{p}. Since δ(y, z_{2}) = δ(y, z_{0}) = 4, also δ(y, z_{1}) = 4, showing that y is a fixed point of β_{N}. So if p′ lies on a line of R(L_{0}, L_{1}), β_{N} = e; if not, then β_{N} has exactly one fixed point.

(3) Let Γ′ be the ideal subhexagon of order (1, q) defined by the intersection of γ and Q, and let p, p′ be the unique two points belonging to ⟨L_{0}, L_{1}⟩∩ Γ′. Because L_{0} and N are opposite lines, δ(p, N) ≠ 1. If δ(p, N) = 3, then L_{0}, L_{1} and N are contained in a hyperplane corresponding to case (2). Hence we may suppose that δ(p, N) = 5. Put w = proj_{N}p and v = p ⋈ w. Note that v is a point of p^{p′} not on the lines L_{0} or L_{1}. If w lies on a line of R(L_{0}, L_{1}), then w ∈ (p′)^{p} and we are back in case (2), so suppose vw is not a line of R(L_{0}, L_{1}). Put u_{i} = proj_{L_{i}}p and u′_{i} = proj_{L_{i}}p′, i = 0, 1. Then δ(w, u′_{0}) = δ(w, u′_{1}) = 4, so w is a fixed point of β_{N}. Also w′ := proj_{N}u_{0} = proj_{N}u_{1}, so w′ is a second fixed point of β_{N}. Suppose β_{N} has a fixed point f different from w and w′. Put f_{i} = proj_{L_{i}}f, i = 0, 1, and f′ = f_{0} ⋈ f_{1} (δ(f_{0}, f_{1}) = 4 since f is a fixed point). Note that δ(f, f′) = 6. Because of the 2-regularity, we have (f′)^{p} = (f′)^{f}. But this is a contradiction, since the point of (f′)^{p} on the line of R(L_{0}, L_{1}) through v lies opposite f. So in this case, β_{N} has exactly two fixed points.

It follows that A contains exactly q^{3} − q lines (and these lines all belong to the spread S_{H}). Also, β_{N} is the identity if and only if N is a line of R(L_{0}, L_{1}) (N ≠ L_{0}, L_{1}) or N is concurrent with a line of R(L_{0}, L_{1}). Hence, since lines of a spread are mutually opposite, at most q − 1 lines of S \ {L_{0}, L_{1}} can give the identity. Consequently, at least (and hence exactly) q^{3} − q lines of S belong to A. Hence the spreads S and S_{H} can only differ in lines of the regulus R(L_{0}, L_{1}). But now, applying the same argument to a regulus R(L_{0}, M), M a line of A, shows that S = S_{H}. □

58

Forgetful Polygons

3.1 Introduction

For every non-incident point-line pair {p, L} of an affine plane, there is exactly one line through p parallel with L. By replacing exactly one by at most one, the definition of a semi-affine plane is obtained. Since parallelism defines an equivalence relation, we can also say that in a semi-affine plane, any two lines are either concurrent or equivalent. Dually, a dual semi-affine plane is an incidence structure (P, L, I), together with an equivalence relation on the point set P such that every two lines meet, and every two points are either collinear or equivalent. In Dembowski [18], it is shown that every finite dual semi-affine plane Γ arises from a projective plane, in such a way that the points and lines of Γ are projective points and lines, and the equivalence classes are (pieces of) lines of the projective plane. In this chapter, we generalize the notion of a dual semi-affine plane to generalized polygons. The idea is that in the definition of generalized polygon, ‘being collinear’ is replaced by ‘being collinear or equivalent’. We then ask the usual axioms about the distance between points and lines, but we do not ask anything about the distance between a point/line and a class, or between two classes. So the definition of these new structures looks pretty much the same as the one of a generalized polygon, except for the fact that some lines seem to have been ‘forgotten’. This is the reason why we called these structures ‘forgetful polygons’. In a way, the equivalence classes of points can be seen as ‘the holes’ in the memory of the forgetful polygon.

The main question now reads: does every forgetful polygon arise from a generalized polygon, by replacing (pieces of) lines by equivalence classes? Since one can construct infinite forgetful polygons using free constructions, we restrict ourselves to the finite case. For forgetful n-gons with n odd, the answer to the above question is ‘yes’, implying that there are no finite forgetful odd-gons, apart from the dual semi-affine planes. The case n even seems much harder. In this case, a positive answer to the question can be obtained if the forgetful polygon still remembers that at least one line and one equivalence class have the same number of points. If this condition is not satisfied, we prove that the forgetful polygon has ‘nice’ properties, for example the equivalence classes of points all have the same size. These classes can be much shorter than the lines however, and the memory of such a forgetful polygon seems to be too short to prove that they actually arise from generalized polygons. This is the reason why they got the name ‘short forgetful polygons’.

We next turn our attention to the smallest short forgetful polygons, being the forgetful quadrangles. Here one can do a little better: if a forgetful quadrangle still remembers that there was a class and a line for which the number of points incident with it differs by at most one, then again the memory can be freshened up and the forgetful quadrangle indeed arises from a generalized quadrangle. We further investigate the structure of a short forgetful quadrangle, which gives rise to interesting objects such as strongly regular graphs. Furthermore, we give two families of examples of short forgetful quadrangles for which the classes are really short. Also these examples arise from generalized quadrangles, but not always in the expected way (this is, the classes are not always lines of the ambient generalized quadrangle). We then give some characterizations of the known examples of short forgetful quadrangles. We end this chapter with the following question: is it possible to classify all the forgetful quadrangles arising from a generalized quadrangle by ‘forgetting lines’? At the moment, this question does not seem to be much easier than the original classification job...

3.2 Definitions and first examples

Let (P, L, I) be an incidence structure, and ∼ an equivalence relation on the point set P. Denote by C the set of non-trivial equivalence classes of ∼. We say that a point x is incident with a class K of C if x belongs to K (and we also use the notation xIK for this). For two elements x and y of P ∪ L ∪ C, a forgetful path of length j between x and y is a sequence (x = z_{0}, z_{1}, …, z_{j−1}, z_{j} = y) of different elements of P ∪ L ∪ C such that z_{i}Iz_{i+1}, for i = 0, …, j − 1. If j is the length of a shortest forgetful path connecting x and y, we say that x and y are at distance j (notation δ(x, y) = j).

Now Γ = (P, L, I, ∼) is a forgetful n-gon, n ≥ 3, if the following three axioms are satisfied:

(FP1) If x, y ∈ P ∪ L and δ(x, y) = k < n, then there is a unique forgetful path of length k joining x to y.

(FP2) For every x ∈ P ∪ L, we have n =max{δ(x, y) : y ∈ P ∪ L}.

(FP3) Every line contains at least three points, every point is incident with at least three elements of L ∪ C.

If C = ∅, then clearly Γ is a generalized n-gon. So the generalized polygons give the first examples of forgetful polygons.

Let Γ = (P, L, I, ∼) be a forgetful n-gon. The elements of C are called the classes. So except when mentioned differently, when talking about a class, we always mean a non-trivial class. The cardinality of the biggest class of C is denoted by g. A point which is only equivalent with itself is called an isolated point. Let x and y be two elements of P ∪ L ∪ C at distance < n. Then the element proj_{x}y is defined in the obvious way, and we use the notation [x, y] for the shortest forgetful path between x and y. The set of lines through a point p (the set of points incident with a line L) is denoted by L_{p} (P_{L}). The order of L_{p} (P_{L}) is called the degree of p (L) and denoted by |p| (|L|). In the following, if talking about a ‘path’ in Γ, we will always mean a ‘forgetful path’.

Let Δ = (P′, L′, I′) be a finite generalized n-gon of order (s, t), n ∈ {4, 6, 8}. Then the following construction yields a forgetful n-gon Γ = (P, L, I, ∼).

(I) Let D be a set of disjoint lines of Δ. Put P = P′ and L = L′ \ D. Two different points of P are said to be equivalent if and only if they both lie on the same line of D.

If Δ is a finite generalized quadrangle, then we also have the following examples.

(II) Let L be a line of Δ, X_{1} a subset of the points of L, 1 ≤ |X_{1}| ≤ s + 1, and X_{2} = Δ_{1}(L) \ X_{1}. Denote by V the set of lines that intersect L Figure 3.1: A forgetful quadrangle of type (II), and the special case |X_{1}| = 1.

Figure 3.2: A forgetful quadrangle of type (III).

in a point of X_{1}. Then put P = P′ \ X_{1}, L = L′ \ (V ∪ {L}). Two different points of P are said to be equivalent if and only if they both lie on the same line of V ∪ {L} (see Figure 3.1).

(III) Let R and L be two different lines of Δ through a point r. Let V be the set of lines concurrent with L, but not through the point r. Then put P = (P′ \Δ_{1}(L))∪{r}, L = L′ \(V ∪{R, L}). Two different points of P are said to be equivalent if and only if they both lie on the same line of V ∪ {R} (see Figure 3.2).

Note that the above constructions also hold if Δ is not finite. However one can obtain more examples of infinite forgetful polygons by using free constructions, which shows that trying to classify forgetful polygons without the finiteness restriction is impossible.

3.3 Classification for n odd 63

3.3 Classification for n odd

Theorem 3.3.1 There does not exist a finite forgetful generalized n-gon, n odd and n ≥ 5.

Proof.

Let Γ = (P, L, I, ∼) be a finite forgetful n-gon, n odd and n ≥ 5. The aim is to show that the geometry (P, L ∪ C, I) is a finite generalized n-gon, hence Γ does not exist. Therefore it suffices to prove that, if K and K′ are classes, and L an arbitrary line, then δ(K, L) ≤ n − 1 (this is done in Step 1 and 2) and δ(K, K′) ≤ n − 1 (Step 3). In Step 0, we collect some general observations.

Step 0

Note that for two classes K and K′, and an arbitrary line L, we certainly have that δ(K, L) ≤ n + 1 and δ(K, K′) ≤ n + 1. Let K be an arbitrary class, r ∈ K, and N a line for which δ(N, K) = n − 1 and δ(N, r) = n. Then |N| = |r| + 1. Indeed, the map

σ : L_{r} → P_{N} \ {proj_{N}K}
L → p,  with δ(p, L) = n − 2,

is a bijection. Suppose M is a line at distance n + 1 from K. Since for any point p of K, the map

\begin{aligned}σ : & L_{p} & → & P_{M} \\ & L & → & p′, & with δ(p′, L) = n − 2, \end{aligned}

is a bijection, |M| = |p|. Similarly, one shows that if z is an isolated point at distance n from a line N, then |z| = |N|.

Now fix a class K and suppose M is a line at distance n + 1 from K, and mIM. Let x be a point of K of degree k for which N := proj_{m}x is a line (note that such a point certainly exists). Then from the previous observations follows that |M| = k, every point of K has degree k and every line at distance n − 1 from K contains k + 1 points. Note also that |M| = k implies k ≥ 3, because of axiom (FP3). Let A be the element of [x, m] at distance \frac{n+1}{2} from x, a = proj_{A}x and a′ = proj_{A}m. The aim is now to show that the existence of the line M leads to a contradiction. In Step 1, we treat the case that, if n ≡ 1 mod 4, then A is not a class of size 2, and if n ≡ 3 mod 4, then a′ is not a class of size 2. In Step 2, we get rid of the remaining cases.

Step 1

Suppose first that n ≠ 5 and that in the case n = 7, |A| ≥ 3, or a is a class of size at least 3. Let z be a point at distance \frac{n−3}{2} from A such that a ≠ proj_{A}z ≠ a′, and N′ := proj_{z}A is a line. Let v be a point at distance \frac{n−1}{2} from a′ for which A ≠ proj_{a′}v ≠ proj_{a′}m and proj_{v}a′ is a line (note that it is always possible to choose z and v like this because of the assumptions on the degrees of a, a′ and A). (The picture for n = 9 is given in Figure 3.3.) We claim that both z and v are isolated points with degree k + 1.

Let z′ be a point on N′ different from z and from proj_{N′}A. Let x′ be a point of K different from x. Then δ(x′, z′) = n − 1. We define the lines L and R as follows.

• Case n ≡ 1 mod 4. Note that A is a line. Let L be a line at distance \frac{n−3}{2} from l := x′ ⋈z′ for which proj_{l}x′ ≠ proj_{l}L ≠ proj_{l}z′. Then |L| = k + 1 since δ(L, K) = n − 1. Let R be a line at distance \frac{n−3}{2} from a, with proj_{a}x ≠ proj_{a}R ≠ A. Then |R| = k + 1 since δ(R, K) = n − 1.

• Case n ≡ 3 mod 4. If n ≠ 7, let r be the point of [x′, z′] at distance \frac{n+1}{2} from x′, and R a line at distance \frac{n−5}{2} from r for which proj_{r}x′ ≠ proj_{r}R ≠ proj_{r}z′. If n = 7, let R be a line intersecting a, not through A or proj_{a}x. Then in both cases, |R| = k + 1 since δ(R, K) = n − 1. Let L be a line at distance \frac{n−1}{2} from a′′ := proj_{a}x, with proj_{a′′}x ≠ proj_{a′′}L ≠ a. Then |L| = k + 1 since δ(L, K) = n − 1.

Suppose z is not isolated, and let z′′ be a point in the class Z containing z, z′′ ≠ z. Since δ(R, Z) = δ(M, Z) = n − 1 and δ(R, z′′) = δ(M, z′′) = n, we obtain |R| = |z′′| + 1 = |M| (see Step 0), a contradiction with |R| = k + 1 and |M| = k. Hence z is isolated and |z| = |L| = k + 1. The point z was arbitrarily chosen on N′, so every point on N′ different from proj_{N′}A is isolated and has degree k + 1. From this also follows that each line T intersecting K, different from proj_{x}m contains k + 1 points (indeed, one can always find a point of N′ \ {proj_{N′}A} lying at distance n from T).

Suppose v is not isolated, and let v′ be a point in the class V containing v, v′ ≠ v. Then |M| = |v′| + 1 = |N′|, a contradiction with |N′| = k + 1 and |M| = k, hence v is isolated. If n ≡ 1 mod 4, then |v| = |R| = k + 1. If n ≡ 3 mod 4, then consider a line X through x, proj_{x}m ≠ X ≠ proj_{x}v (note that this is possible since |x| = k ≥ 3). Since δ(v, X) = n, we obtain |v| = |X| = k + 1. This shows that both z and v are isolated points of degree k + 1, as claimed.

Figure 3.3: Proof of Step 1 for n = 9

Let R_{z} be a line through z, different from N′. Because δ(v, R_{z}) = n, this line contains k + 1 points, hence δ(R_{z}, K) = n − 1. The map

\begin{aligned}σ : & L_{z} \ \{N′\} & → & K \ \{x\} \\ & R_{z} & → & y, & with δ(y, R_{z}) = n − 2, \end{aligned}

is a bijection, from which follows that |K| = |z| = k + 1. Finally, we consider the point m. If m is isolated, then, with X a line through x different from proj_{x}m, δ(m, X) = n implies that |m| = |X| = k + 1. Every point of K lies at distance n−1 from m, hence we need k +1 lines through m lying at distance n−1 from K, a contradiction since there are at most k such lines (indeed, δ(M, K) = n + 1). If m is not isolated, let m′ be a point of the class K′ containing m, m′ ≠ m. Then δ(m′, x) = n − 1. Put B = proj_{x}m′. If B ≠ K, then since |B| = k + 1, δ(B, K′) = n − 1 and δ(m, B) = n, the point m has degree k. If B = K, put y = proj_{K}m′. Let Y be any line through y different from proj_{y}m′. Then, since |Y | = k + 1, δ(Y, K′) = n − 1 and δ(m, Y) = n, the point m again has degree k. Since |K| = k + 1, we need at least k lines through m lying at distance n − 1 from K, a contradiction since there are at most k − 1 such lines.

Case n = 5. Let z be a point on the line A, a ≠ z ≠ m, and L a line intersecting ax not through a or x. If a would be equivalent with a point a′′, a′′ ≠ a, then |M| − 1 = |a′′| = |L| − 1 (by Step 0), a contradiction with |L| = k + 1 and |M| = k. Hence the point a is isolated. Let N′′ be a line through a, ax ≠ N′′ ≠ am (such a line exists, since the degree of the isolated point a is at least 3). Note that |N′′| = k + 1. Similarly as above for the point a, we obtain that z is isolated (now using N′′ instead of L). From δ(z, L) = 5 and |L| = k + 1 follows |z| = k + 1. By considering a line at distance 5 from both a and z, we conclude that |a| = k + 1. Suppose first that m has degree at least 3. Then let v be a point collinear with m, v not on the lines M or A. Again by using the same argument as above for the point a, we obtain that v is isolated and has degree k + 1. From this follows that every line R_{z} through z, R_{z} ≠ A, contains k + 1 points. Suppose now m has degree 2, and let r be a point equivalent with m, r ≠ m. Then |r| = k since |N′′| = k + 1, implying that also in this case every line R_{z} through z, R_{z} ≠ A, contains k + 1 points. We now proceed similarly as in the case n > 5.

Case n = 7, |A| = 2 and a, a′ are lines. We show that this case cannot occur. Let Z be the class containing A, and z ∈ Z, z ≠ A. Put a′′ = proj_{a}x. Let Y be a line at distance 4 from a′′x, and at distance 5 from both x and a′′. Then |Y | = k + 1. Let y be a point on a, a′′ ≠ y ≠ A and Y′ a line at distance 3 from a′′, and at distance 4 from a′′x and a. Then |Y′| = k + 1. If y ∼ y′, y′ ≠ y, then |M| − 1 = |y′| = |Y′| − 1, a contradiction, hence y is isolated. Since δ(y, Y) = 7, |y| = k + 1. Let finally x′ be a point of K, x′ ≠ x. Then δ(x′, z) = 6. The line proj_{x′}z contains k + 1 points (because it lies at distance 7 from y), implying that |A| = k ≥ 3. This is a contradiction with the assumption.

Case n = 7, a a class of size 2. Define z and v as in the general case. We have to give another argument to conclude that the points z and v are isolated (since the line R defined in the general case cannot be found). Let x′ ∈ K, x′ ≠ x. Then δ(x′, m) = 6. Put R = proj_{m}x′.

• Suppose first R is a line. Then |R| = k + 1. If z ∼ z′, z′ ≠ z, then |M| − 1 = |z′| = |R| − 1, a contradiction, hence the point z is isolated. Also, |z| = |R′| = k + 1, with R′ a line at distance 3 from a′′ = proj_{a}x for which a ≠ proj_{a′′}R′ ≠ a′′x.

• Suppose now R is a class and let w be the point of [x′, R] at distance 2 from x′. Using the same argument as for the point z above (with a′ in the role of R), we obtain that w is isolated. Also, |w| = k + 1 (since |w| = |Az|). Hence an arbitrary line through a′′ = proj_{a}x, different from a′′x, contains k + 1 points. Again using the same argument as above, it now follows easily that also in this case the point z is isolated. By considering a line intersecting a′′x not through x or a′′, we obtain |z| = k + 1.

If v ∼ v′, v ≠ v′, then |Az| − 1 = |v′| = |M| − 1, a contradiction, hence v is isolated. For a line X intersecting M, not through m, holds that |z| = |X| = |v|, hence |v| = k + 1. The rest of the proof is similar as in the general case.

This finishes the case n ≡ 1 mod 4 and A not a class of size 2, or n ≡ 3 mod 4, and a′ not a class of size 2. Note that for n =5 (7), the element A (a′) can be chosen to be a line, so Step 2 does not concern these cases.

Step 2

n ≡ 1 mod 4, n > 5 and A a class of size 2.

In the following, we will construct a point p′ such that δ(p′, A) = n, δ(p′, a) = n − 1 for which both proj_{p′}a and the element of [a, p′] at distance \frac{n+1}{2} from a are lines, and such that |p′| = k if p′ is not isolated, and |p′| = k + 1 if p′ is isolated. Step 1 then implies that every line through p′ lies at distance n − 2 from a point of A. Hence |A| ≥ k ≥ 3, a contradiction.

Case n = 9. Let p′ be a point at distance 5 from the line X = proj_{x}m, with x ≠ proj_{X}p′ ≠ proj_{X}m and such that the path [X, p′] only consists of points and lines. If p′ is isolated, then let y be a point at distance 3 from K and 4 from x, and y′ the point of [y, p′] at distance 2 from y. By considering a line R at distance 3 from y′, proj_{y′}y ≠ proj_{y′}R ≠ proj_{y′}p′, we see that |p′| = |R| = k + 1. If p′ is contained in a non-trivial class Z, then let p′′ ∈ Z, p′′ ≠ p′, and x′ ∈ K, x′ ≠ x. Let L be a line at distance 3 from the point z := x′ ⋈ p′′, proj_{z}p′′ ≠ proj_{z}L ≠ proj_{z}x′. Since |L| = k + 1, p′ has degree k. So we constructed a point p′ as claimed.

Case n ≡ 1 mod 8, n > 9 Let p be the point of [x, a] at distance \frac{δ(x,a)}{2} − 2 = \frac{n−9}{4} from x, and p′ a point at distance \frac{3n−11}{4} from p such that proj_{p}a ≠ proj_{p}p′ ≠ proj_{p}x and such that the path [p, p′] only consists of points and lines, except possibly for the element proj_{p}p′. Suppose first that p′ is isolated. We show that |p′| = k + 1. Consider a point y at distance 6 from x such that proj_{x}p ≠ proj_{x}y ≠ K. Then δ(p′, y) = n − 1. Let γ be the union of the paths [p′, y] and [y, x]. Let z be the element of γ at distance \frac{n+3}{2} from x, and z′ a point at distance \frac{n−9}{2} from z such that proj_{z}p′ ≠ proj_{z}z′ ≠ proj_{z}y. Then any line through z′ different from the projection of K onto z′ contains k+1 points (since it lies at distance n − 1 from K), and lies at distance n from p′, hence |p′| = k + 1. Suppose now that p′ is contained in a class K′, and p′′ is a point of K′ different from p′. We show that |p′| = k. Consider a point y at distance 4 from x such that proj_{x}p ≠ proj_{x}y ≠ K. Then δ(p′′, y) = n − 1. Let c be the point of [y, p′′] at distance \frac{n−5}{2} from y and let c′ be a point at distance \frac{n−9}{2} from c such that proj_{c}p′′ ≠ proj_{c}c′ ≠ proj_{c}y. Then any line R_{c′} through c′ different from the projection of K onto c′ contains k +1 points (because it lies at distance n−1 from K). Since δ(R_{c′}, K′) = n − 1 and δ(R_{c′}, p′) = n, Step 0 implies that |p′| = |R_{c′}| − 1 = k. Now the point p′ satisfies all the conditions above.

Case n ≡ 5 mod 8, n > 5 Let p be the point of [x, a] at distance \frac{n−5}{4} from x, and p′ a point at distance \frac{3n−7}{4} from p such that proj_{p}a ≠ proj_{p}p′ ≠ proj_{p}x and such that the path [p, p′] only consists of points and lines, except possibly for the element proj_{p}p′. Suppose first that p′ is isolated. Let z be the point of [x, p′] at distance \frac{n−5}{2} from x, and z′ a line at distance \frac{n+1}{2} from z such that proj_{z}x ≠ proj_{z}z′ ≠ proj_{z}p′. Then |p′| = |z′| = k + 1. Suppose now that p′ is contained in a class K′, and p′′ is a point of K′ different from p′. Let y be a point of K, y ≠ x. Then δ(p′′, y) = n − 1. Consider a line z′ at distance \frac{n−3}{2} from the point z = y ⋈p′′ for which proj_{z}p′′ ≠ proj_{z}z′ ≠ proj_{z}y. Then |p′| = |z′| − 1 = k.

n ≡ 3 mod 4, n > 7 and a′ a class of size 2.

Similarly as in the case n ≡ 1 mod 4, we construct a point p′ at distance n from a′ and at distance n−1 from A, for which both proj_{p′}A and the element of [A, p′] at distance \frac{n+3}{2} from A are lines, and such that either |p′| = k + 1 if p′ is isolated, or |p′| = k if p′ is not isolated. The result will then follow.

Case n ≡ 3 mod 8 Let p be the point of [x, A] at distance \frac{n−3}{4} from x, and p′ a point at distance \frac{3n−9}{4} from p such that proj_{p}A ≠ proj_{p}p′ ≠ proj_{p}x and such that the path [p, p′] only consists of points and lines except possibly for the element proj_{p}p′. Suppose first that p′ is isolated. Let y be a point at distance 4 from x such that proj_{x}A ≠ proj_{x}y ≠ K. Then δ(p′, y) = n − 1. Let γ be the union of the paths [x, y] and [y, p′], and z the point of γ at distance \frac{n+1}{2} from x. Let finally z′ be a line at distance \frac{n−5}{2} from z such that proj_{z}y ≠ proj_{z}z′ ≠ proj_{z}p′. Then |p′| = |z′| = k + 1. Suppose now that p′ is contained in a class K′, and p′′ is a point of K′ different from p. Consider a point y at distance 2 from x such that proj_{x}p ≠ proj_{x}y ≠ K. Let z be the point of [y, p′′] at distance \frac{n−3}{2} from y, and z′ a line at distance \frac{n−5}{2} from z such that proj_{z}y ≠ proj_{z}z′ ≠ proj_{z}p′′. Then |p′| = |z′| − 1 = k.

Case n ≡ 7 mod 8, n > 7 Let p be the point on [x, A] at distance \frac{n−7}{4} from x, and p′ a point at distance \frac{3n−13}{4} from p such that proj_{p}A ≠ proj_{p}p′ ≠ proj_{p}x and such that the path [p, p′] only consists of points and lines, except possibly for the element proj_{p}p′. Suppose first that p′ is isolated. Let z be the point of [x, p′] at distance \frac{n−7}{2} from x, and z′ a line at distance \frac{n+3}{2} from z such that proj_{z}x ≠ proj_{z}z′ ≠ proj_{z}p′. Then |p′| = |z′| = k + 1. Suppose now that p′ is contained in a class K′, and p′′ is a point of K′ different from p′. Let y be a point at distance 3 from K and at distance 4 from x. Then δ(p′′, y) = n − 1. Let z be the point of [y, p′′] at distance \frac{n−3}{2} from y, and z′ a line at distance \frac{n−5}{2} from z such that proj_{z}y ≠ proj_{z}z′ ≠ proj_{z}p′′. Then |p′| = |z′| − 1 = k.

Hence we have shown that for any class K and an arbitrary line M, δ(M, K) ≤

(n − 1.) / (Step 3)

Suppose there exist two classes K_{1} and K_{2} at distance n+1 from each other. We look for a contradiction. Let x ∈ K_{1} be arbitrary. Since by the results of Step 1 and 2, any line through x lies at distance n − 1 from K_{2}, the map

\begin{aligned}σ : & L_{x} & → & K_{2} \\ & L & → & y, & with δ(L, y) = n − 2, \end{aligned}

is a bijection, hence |K_{2}| = |x| =: k and all points in K_{1} have the same degree k. Fix points x ∈ K_{1} and y ∈ K_{2}. Note that δ(x, y) = n − 1. Let z be the element of [x, y] at distance \frac{n−3}{2} from x. If n ≡ 1 mod 4, then we can assume without loss of generality that z is a line (indeed, if z is a class, then interchange the roles of K_{1} and K_{2}). If n = 5, let w be a point on z, different from x or x⋈y. If n > 5, let w be a point at distance \frac{n−3}{2} from z such that proj_{z}x ≠ proj_{z}w ≠ proj_{z}y (this is possible because z was assumed not to be a class) and proj_{w}z is a line (this is not possible if n = 7, z has degree 2 and x⋈y is a line, see case (1) below). Let finally W be a line through w, W ≠ z if n = 5, W ≠ proj_{w}z if n > 5. Then |W | = k + 1, since δ(K_{1}, W) = n − 1. But also δ(K_{2}, W) = n − 1 by Step 1 and 2, hence there is a point in K_{2} which has degree k. By repeating the argument at the beginning of this step, we obtain that also |K_{1}| = k, and every point of K_{2} has degree k. Note that this implies that every line intersecting K_{1} or K_{2} contains k + 1 points. Let the point w be as above. If w is isolated then, by considering a line intersecting K_{1}, but not through x, we obtain that |w| = k + 1. Since every line through w lies at distance n − 1 from K_{2}, we need at least k + 1 points in K_{2}, a contradiction. Suppose now w is contained in a non-trivial class K, and let w′ be a point of K, w′ ≠ w. Then |w′| = k. Indeed, |w′| = |R| − 1, with R an arbitrary line through x different from proj_{x}y. But then the map

\begin{aligned}σ : & L_{w′} & → & K_{1} \ \{x\} \\ & L & → & v, & with δ(L, v) = n − 2, \end{aligned}

is a bijection, hence |K_{1}| = k + 1, the final contradiction since |K_{1}| = k.

(1) n = 7, x⋈y a line and z non-isolated. Let L be a line intersecting x⋈y, δ(L, K_{1}) = δ(L, K_{2}) = 6. Then |L| = k +1, hence every point in K_{2} \ {y} has degree k. By repeating the argument at the beginning of this step, we obtain that also |K_{1}| = k, and every point of K_{2} has degree k. Let w be a point in the class containing z, w ≠ z. Let Y be a line intersecting K_{1}, Y not through x. Then |w| = |Y | − 1 = k. Since every line through w lies at distance 6 from a point of K_{2} \ {y}, |K_{2}| = k + 1, a contradiction.

This shows that for two classes K_{1} and K_{2}, δ(K_{1}, K_{2}) ≤ n − 1. Now the theorem follows. □

3.4 Square forgetful pentagons

A semi-plane is an incidence structure (P, L, I) together with an equivalence relation on the point set and the line set respectively, such that every two lines (points) are either concurrent (collinear) or equivalent. These structures were introduced in Dembowski [19] (appendix 7.4). The aim of this section is to generalize the notion of a semi-plane to n-gons. So in fact, we want to make the notion of a forgetful polygon selfdual.

Let (P, L, I) be an incidence structure, ∼_{P} an equivalence relation on the point set P and ∼_{L} an equivalence relation on the line set L. Denote by C_{P}, C_{L}, the set of non-trivial classes of ∼_{P} respectively ∼_{L}. Completely similar as in section 3 we define square forgetful paths and distances between elements of P ∪ L ∪ C_{P} ∪ C_{L}, isolated points/lines and the degree of a point/line.

Now Γ = (P, L, I, ∼_{P}, ∼_{L}) is a square forgetful n-gon, n odd, n ≥ 3, if the following axioms are satisfied:

(DFP1) If x, y ∈ P ∪ L and δ(x, y) = k < n, then there is a unique square forgetful path of length k joining x to y.

(DFP2) For every x ∈ P ∪ L, we have n =max{δ(x, y) : y ∈ P ∪ L}.

(DFP3) Every line and every point class is incident with at least three points, every point and every line class is incident with at least three lines.

Note that a square forgetful 3-gon is a semi-plane (but not conversely, since in the definition of semi-plane, nothing is required for the size of the equivalence classes). The classification of all finite semi-planes (see Dembowski [19]) is not completed. In fact, there exists an example of a semi-plane that does not arise from a projective plane by ‘forgetting’ points and lines, see Baker [2]. We now give some partial results for the classification of finite square forgetful

Figure 3.4: possibilities if |p| = k.

pentagons. Moreover, we show that if one adds the assumption that at least one point class and one line class have ‘the right size’ (this is, one more than the degree of a non-isolated element), then this structure arises from a finite generalized pentagon, and hence cannot exist. The generalization of this result for n ≥ 7, as well as the full classification of all finite square forgetful n-gons seems to be out of reach at this moment.

Let Γ be a finite square forgetful pentagon admitting non-isolated points and lines. Throughout this section, we denote by K_{p} the point class containing the non-isolated point p, and by α_{L} the line class containing the non-isolated line L.

We start with some observations similar to the ones in Step 0 of the proof of Theorem 3.3.1. Let p be a point at distance 5 from the line L, and put |p| = k. Then the following cases can occur (see Figure 3.4).

• δ(K_{p}, L) = 4.

(a) If there exists a line at distance 3 from p and equivalent with L, then |L| = |p| = k.

(b) If no such line exists (for example when the line L is isolated), then |L| = |p| + 1 = k + 1.

• δ(K_{p}, L) = 6 or the point p is isolated. (c) If there exists a line at distance 3 from p and equivalent with L, then |L| = |p| − 1 = k − 1.

(d) If no such line exists (for example when the line L is isolated), then |L| = |p| = k.

These observations will be used throughout, without further reference to them.

Lemma 3.4.1 If a point class K and a line class α lie at distance 5 from each other, then every line of α lies at distance 4 from K, and dually every point of K lies at distance 4 from α.

Proof. Let K be a point class, K = {r_{1}, r_{2}, …, r_{t+1}}, and α a line class at distance 5 from K and 4 from r_{1}. Suppose by way of contradiction that α contains a line L at distance 6 from K. Put k = |r_{1}|. Then |L| = k − 1 and |z| ∈ {k − 1, k}, for z ∈ K \ {r_{1}}.

Claim 1 All points of K have degree k. Suppose first that K contains at least two points r_{2} and r_{3} of degree k−1. Let x be a point on L. Without loss of generality, we can assume N := proj_{x}r_{2} is a line. If r_{2} ⋈x would be a line class, then |r_{1}| = |N| = |r_{3}|, a contradiction. If N would be isolated, then |r_{1}| + 1 = |N| = |r_{3}| + 1, again a contradiction. So the line N is non-isolated, and the line class α_{N} lies at distance 5 from K. Since |r_{1}| = k, |N| ∈ {k, k+1}, and since |r_{3}| = k−1, |N| ∈ {k, k−1}. Hence |N| = k, there exists a line N′ ∈ α_{N} at distance 3 from r_{1}, and all lines of α_{N} lie at distance 5 from r_{3}. Since |r_{2}| = k − 1 and δ(r_{2}, N) = 3, |N′| = k − 1. But since the point r_{3} has degree k − 1 and does not lie at distance 3 from a line equivalent with N′, we obtain |N′| = k, a contradiction.

Suppose now that K contains a unique point r_{3} of degree k − 1. Then all the points of K different from r_{3} lie at distance 4 from α. Put N_{i} = proj_{α}r_{i}, i = 1, …, t + 1, i ≠ 3. Note that |N_{i}| = k. The lines incident with a point r_{i}, i ≠ 3, are isolated. Indeed, suppose that a line M through r_{1} is not isolated, and let M′ ∈ α_{M} \ {M}. Then |r_{2}| = |M′| = |r_{3}|, a contradiction. With a dual argument, one sees that the points incident with the lines N_{i} are isolated. Let now R be a line at distance 4 from K and at distance 3 from r_{3} for which proj_{R}r_{3} is a point. We claim that R is isolated (∗). Suppose by way of contradiction that R′ is a line equivalent with R, R′ ≠ R. If δ(R′, K) = 6, then |r_{3}| − 1 = |R′| ∈ {|r_{2}|, |r_{2}| − 1}, a contradiction. So δ(R′, K) = 4. Without loss of generality, we can assume that δ(r_{2}, R′) = 3. But now we obtain |r_{3}| = |R′| ∈ {|r_{1}|, |r_{1}| + 1}, again a contradiction, which shows the claim. Note also that |R| = |r_{1}| + 1 = k + 1. Now by (∗) and the fact that every point is incident with at least 3 lines, it is clear that we can always find an isolated line of degree k + 1 at distance 5 from the isolated point x_{i} := r_{i} ⋈α, i = 1, 2, hence |x_{i}| = k + 1. From this follows that also the lines through r_{1} or r_{2}, and the points on the lines N_{1} and N_{2} have degree k + 1.

We now show that the class K contains exactly k + 1 points. Let y be a point on the line N_{2} different from x_{2}, and A a line through y, A ≠ N_{2}. Then |A| ∈ {|x_{1}|, |x_{1}| − 1} = {k + 1, k}. If δ(A, K) = 6, then |A| ≤ |r_{3}| = k − 1, a contradiction. Hence each line through y lies at distance 4 from K, implying that |K| ≥ |y| = k + 1. Since also every point of K lies at distance 3 from a unique line through y, |K| = k + 1.

Consider an arbitrary point p on the line L. Since p lies at distance 4 from every point of K, we need at least k+1 elements of L∪C_{P} incident with p and at distance 4 from K. Since the line L lies at distance 6 from K, |p| ≥ k + 1 if p is non-isolated, and |p| ≥ k + 2 if p is isolated. If p is isolated, then |p| = |r_{1}x_{1}| = k + 1, a contradiction, so we can assume that p is non-isolated. Without loss of generality, we can assume that proj_{p}r_{2} ≠ K_{p}. Note also that proj_{r_{2}}p ≠ K, since L and K lie at distance 6. If δ(K_{p}, x_{2}r_{2}) = 4, then |p| = |x_{2}r_{2}| − 1 = k, which is too small. So δ(K_{p}, x_{2}r_{2}) = 6, implying that |p| = |x_{2}r_{2}| = k + 1. Since |K| = k + 1 and δ(L, K) = 6, this implies that the class K_{p} lies at distance 4 from K. Then necessarily, the projection of K_{p} onto K is the point r_{3} (indeed, if proj_{K}K_{p} = r_{i}, i ≠ 3, then any line through r_{i} different from proj_{r_{i}}p would contain |p|+1 = k+2 points, a contradiction). Now let B be the projection of p onto r_{3}, and B′ a line intersecting B in a point not belonging to K or K_{p}. Note that by (∗), the line B′ is isolated and has degree k + 1. Then |p| = |B′| − 1 = k, the final contradiction. Hence we have shown that all points of K have the same degree k. ◇

Now it follows that every point of K lies at distance 4 from the class α, and hence at distance 3 from a line of degree k belonging to α. Put x_{i} = r_{i} ⋈ x_{i} and N_{i} = proj_{α}r_{i}, i = 1, …, t + 1. As before, the points on the lines N_{i} are isolated.

Claim 2 The points x_{i} have degree k + 1. We claim that |x_{i}| ∈ {k, k + 1}. Indeed, let A be a line at distance 3 from r_{1} such that K ≠ proj_{r_{1}}A ≠ proj_{r_{1}}x_{1} and such that the projection of r_{1} onto A is a point. Then |A| ∈ {|r_{2}|, |r_{2}| + 1} = {k, k + 1}, from which follows that |x_{1}| ∈ {k, k + 1, k + 2}. If |x_{1}| = k + 2, then |A| = k + 1, there exists a line A′ equivalent with A at distance 3 from x_{1} but every line of α_{A} different from A lies at distance 5 from all points of K. So δ(K, A′) = 6. But now |r_{1}| − 1 = |A′| = |r_{2}|, a contradiction. Hence the claim.

Suppose |x_{1}| = k. We look for a contradiction. Let M be an arbitrary line intersecting x_{1}r_{1} in a point z different from x_{1} and r_{1}. Suppose M is not isolated, and let M′ be a line equivalent with M, different from M. Then |M′| = |x_{1}| − 1 = k − 1. This implies that the line M′ lies at distance 6 from K, and (since δ(r_{2}, M′) = 5 and |M′| = |r_{2}|−1) that there is a line M′′ of α_{M} at distance 3 from r_{2}. Then |x_{1}| − 1 = |M′′| = |r_{1}|, a contradiction. Hence the line M is isolated and has degree k + 1 (∗∗). From this easily follows that every point on one of the lines N_{i}, different from x_{1} has degree k + 1.

We next show that K contains exactly k + 1 points. Let R be a line intersecting N_{1} in a point z′ different from x_{1}. Let y be a point on N_{1} different from x_{1} and not on R. Note that |y| = k + 1. If R is non-isolated, then |y| − 1 = |R′| = |x_{1}| − 1, with R′ a line belonging to α_{R} \ {R}, a contradiction. Hence R is isolated and has degree k + 1, so R lies at distance 4 from K. It now follows that |K| = |z′| = k + 1.

Let p be an arbitrary point on the line L. As before, |K| = k + 1 implies that |p| ≥ k + 2 if p is isolated, and |p| ≥ k + 1 if p is non-isolated. If p would be isolated, then |p| = k + 1 by (∗∗), so p is non-isolated. Again by (∗∗), |p| ∈ {k, k + 1}. Hence |p| = k + 1. This implies that δ(K_{p}, K) = 4. Since the line N_{1} has degree k, every point of K_{p} different from p has degree k. Suppose first δ(K_{p}, r_{i}) = 3, for some i, i ≠ 1. Consider the line A = r_{i}x_{i}. Let y be an arbitrary point on N_{1} different from x_{1}. If A would be isolated, then |x_{1}| = |A| = |y|, a contradiction, so A is non-isolated. Then |A| ∈ {|x_{1}|, |x_{1}| − 1}, and |A| ∈ {|y|, |y| − 1}, implying |A| = k. But also |A| ∈ {|p|, |p| + 1}, a contradiction. So necessarily δ(K_{p}, r_{1}) = 3. Let p′ be a point of K_{p} different from p and proj_{K_{p}}r_{1}. The line x_{1}r_{1} is not isolated (indeed, otherwise |p| = |x_{1}r_{1}| = |p′|). Now clearly (by comparing with the degrees of p and p′), |x_{1}r_{1}| = k + 1. This implies that there is a line B equivalent with x_{1}r_{1} at distance 3 from p, and no line of α_{x_{1}r_{1}} lies at distance 3 from p′. Now |B| = |r_{2}| = k, but also |B| = |p′|+1 = k+1, a contradiction. This shows that |x_{1}| = k + 1. ◇

Claim 3 Every point on a line N_{i} has degree k + 1. Let y be a point on the line N_{1}, y different from x_{1}. Let R be a line through the point r_{1} different from x_{1}r_{1}. If R is isolated, then |y| = |R| = |x_{2}| = k+1. Suppose R is not isolated. Since |α_{R}| ≥ 3, there exists a line R′ ∈ α_{R} at distance 5 from both y and r_{1}. Hence |R′| = |r_{2}| and |y| ∈ {|R′|, |R′| + 1} = {k, k + 1}. Suppose by way of contradiction that |y| = k. Let z be a point on N_{1} different from x_{1} and y, and A a line through z different from N_{1}. Since the degrees of x_{1} and y are different, it is easy to see that the line A is necessarily isolated, and has degree |x_{2}| = k + 1. From this follows that every point on one of the lines N_{i}, i ≠ 1 has degree k + 1, and that every line through z lies at distance 4 from K. We now show that |z| = k + 1, which will imply that the class K contains exactly k + 1 points. If |z| ≠ k + 1, then as before, |z| = k. Consider a line B at distance 3 from y and at distance 4 from N_{1} for which the projection of y onto B is a point. Because the degrees of x_{1} and z are different, the line B is non-isolated. Choose a line B′ equivalent with B at distance 5 from x_{1} and different from B. Then |y| − 1 = |B′| ∈ {|x_{1}|, |x_{1}| − 1}, a contradiction. So |z| = |K| = k + 1. Completely similar as in the previous claim, one shows that this contradicts the degree of a point p on the line L. ◇

Claim 4 The class K contains k + 1 points Let y be a point on the line N_{1} different from x_{1}. Then |K| ≤ |y| = k + 1. Suppose by way of contradiction that |K| < k + 1. Then there exists a line Y through y at distance 6 from K. Since |Y | ∈ {|x_{2}|, |x_{2}| − 1} and also |Y | ∈ {|r_{1}|, |r_{1}| − 1}, we obtain |Y | = k. This implies that Y is non-isolated, all the lines of α_{Y} lie at distance 5 from any point of K, and for every point z on the line N_{2}, there exists a line Y_{z} equivalent with Y and at distance 3 from z (since z is an isolated point with degree k + 1). Note that Y_{z} ≠ x_{2}r_{2} (indeed, δ(r_{2}, Y_{z}) = 5), so all these lines Y_{z} are different. Hence we need at least k (=|N_{2}|) lines in α_{Y} different from Y. Now every line of α_{Y} \ {Y } lies at distance 4 from a unique point on L. Hence |L| ≥ k, a contradiction. So |K| = k + 1. ◇

We can now finish the proof of the lemma. Similarly as in the proof of Claim 1, the fact |K| = k + 1 implies that the degree of a point p on the line L is at least k + 1 if p is non-isolated, and at least k + 2 if p is isolated. Let R be a line concurrent with N_{1}, not through the point x_{1}. Note that if R is isolated, then |R| = |x_{2}| = k + 1, and if R is non-isolated, then |R′| = |x_{1}| − 1 = k, for any line R′ of α_{R} \ {R}. Suppose p is isolated. If R is isolated, then |p| = |R| = k + 1, a contradiction. If R is non-isolated, then let R′ be a line of α_{R} different from R at distance 5 from p. This implies |p| ≤ |R′|+1, a contradiction. So the point p is non-isolated. If R is isolated, then |p| ≤ |R| = k + 1. If R is non-isolated, then |p| ≤ |R′| + 1 = k + 1, with R′ a line of α_{R} \ {R} at distance 5 from p. So in both cases, |p| = k + 1, implying that δ(K, K_{p}) = 4. Without loss of generality, we can assume δ(r_{1}, K_{p}) = 3. Let M be the line intersecting both K and K_{p}, and p′ an arbitrary point of K_{p} different from p and not on M. Put p′′ the point of K_{p} on M. Let N be a line intersecting M in a point and at distance 4 from K and K_{p}. Note that |p′| = |N_{1}| = k. The line N is non-isolated since otherwise |p| + 1 = |N| = |p′| + 1, a contradiction. By comparing the degrees of p and p′, we see that |N| = k + 1, implying that there exists a line N′ equivalent with N at distance 3 from p, and that every line of α_{N} lies at distance 5 from p′. Let N′′ ∈ α_{N} \ {N, N′}. Then δ(p, N′′) = δ(p′′, N′′) = 5, but since the point p′ above was chosen arbitrarily in K_{p} \ {p, p′′}, N′′ also lies at distance 5 from every point of K_{p} \ {p, p′′}. Hence |p| − 1 = |N′′| = |p′|, the final contradiction. Now the lemma is proved. □

Corollary 3.4.2 Let K be a point class containing a point p with degree k.

(i) If there exists a line class α at distance 5 from K, then |K| = |α| and every point of K and every line of α have the same degree k.

(ii) All points in the same point class have the same degree. Dually, all lines in the same line class have the same degree.

Proof. Observation (i) follows immediately from Lemma 3.4.1. Now let K be a point class, and {r_{1}, r_{2}, r_{3}} ⊆ K. Let L be a line at distance 4 from K and 3 from r_{3}. If L is isolated, then |r_{1}|+1 = |L| = |r_{2}|+1, hence |r_{1}| = |r_{2}|. If L is non-isolated, then |r_{1}| = |L| = |r_{2}| by (i). So any two points of K have the same degree, showing (ii). □

Proposition 3.4.3 (i) All non-isolated points and lines have the same degree k.

(ii) An isolated element has degree k or k + 1.

(iii) Any point class or line class has size at most k + 1.

Proof. Let K be an arbitrary point class, and k the degree of the points of K. Let L be an arbitrary non-isolated line, and L′ ∈ α_{L} \ {L}. If L intersects K, then |L′| = k, hence by Corollary 3.4.2 (ii) also |L| = k. If L lies at distance 4 from K, then by Lemma 3.4.1, |L| = k. If finally L lies at distance 6 from K, then also α_{L} lies at distance 6 from K, and |L| = |p| = k, with p an arbitrary point of K. So all non-isolated lines have the same degree k, which is equal to the degree of the points of K. Dually all non-isolated points have the same degree k. This shows (i). Now let p be an isolated point and suppose |p| ∉ {k, k + 1}. Let L be a line at distance 5 from p. If L is non-isolated (and thus has degree k) then p ∈ {|L|, |L| + 1}, a contradiction. Hence every line at distance 5 from p is isolated and has degree |p|. Dually, every point at distance 5 from a line at distance 5 from p is isolated. It is now easy to deduce that all elements are isolated, so Γ is not forgetful at all. Hence (ii). Now let K be an arbitrary point class, and r a point at distance 5 from K. Since each point of K lies at distance 4 from r, |K| is at most the number of elements of L ∪ C_{P} incident with r, showing (iii). □

Theorem 3.4.4 If Γ contains at least one point class and at least one line class of size k+1, then Γ arises from a finite generalized pentagon, and hence cannot exist.

Proof. Let K be a point class of size k + 1, and A a line class of size k + 1. Let p be a point at distance 5 from K. Since p lies at distance 4 from every point of K, and since p is incident with at most k + 1 elements of L ∪ C_{P}, p

Figure 3.5: Proof of Theorem 3.4.4.

is incident with exactly k + 1 elements of L ∪ C_{P} and every element incident with p lies at distance 4 from K. This shows that each line and each point class lies at distance at most 4 from K. Also, every isolated point at distance 5 from K necessarily has degree k + 1. Dual results hold for the line class A. Now we show that any point class K′ of size ≤ k lies at distance 3 from A. By Corollary 3.4.2 (i), δ(K′, A) ≠ 5. Suppose δ(K′, A) = 7. Let L be a line belonging to A, and p a point of K′. Then δ(p, L) = 5. Let R be a line through p at distance 4 from L. Since every line of A lies at distance 4 from R, and since R is incident with at most k + 1 elements of P ∪ C_{L}, the point p lies at distance 3 from a line of A, a contradiction. So δ(K′, A) = 3, showing that every class of size ≤ k is intersected by a line of A.

Now suppose by way of contradiction that there exists a point class K with |K| ≤ k. Let x be the point of K at distance 3 from K, and y the point of K collinear with x. Put N_{1} = xy, and let N_{2} be an arbitrary line through y different from N_{1}. Let N′ be a line intersecting N_{1} in a point different from x or y. Then N′ is isolated. Indeed, if not, then |K| = |α_{N′}| = |K| by Corollary 3.4.2 (i), a contradiction. Note that |N′| = k + 1 (∗).

Let r be an arbitrary point on N_{2} different from y. We claim that every line L through r different from N_{2} lies at distance 4 from K. Indeed, if L is isolated, then |L| = k + 1 (because δ(L, K) = 4) and hence, noting that |x| = k, δ(K, L) = 4. Suppose L is non-isolated. Then |α_{L}| = |K| = k + 1 (by Corollary 3.4.2 (i)). It then follows as in the first paragraph of the proof that K is intersected by a line of α_{L}. Hence the claim. If r were isolated, then |r| = k + 1 by (∗), hence |K| = k + 1, a contradiction. So the point r is non-isolated and has degree k. Since |K| ≥ |r|, we conclude that K contains exactly k points, and that the classes K and K_{r} mutually lie at distance 6. But now the class K_{r} also contains at most k points (since otherwise δ(K, K_{r}) = 4 by the first paragraph of the proof). By symmetry, we now obtained the following situation. Let N_{1}, …, N_{k} be the lines through the point y. Then each point r on one of the lines N_{i}, r ≠ y, belongs to a class K_{r}^{i} of size k, and K_{r}^{i} lies at distance 6 from every class K_{r′}^{j} intersecting a line N_{j}, i ≠ j, with K_{r′}^{j} ≠ K (see Figure 3.5). Now consider the classes K_{r}^{i} of size k intersecting the line N_{i}. Note that N_{i} ∉ A, since otherwise any class intersecting N_{j}, j ≠ i, not through y would have size |A| = k +1. So we need at least k − 1 lines in A intersecting one of the classes K_{r}^{i} (keeping in mind that every point class of size ≤ k is intersected by a line of A). Note that these k − 1 lines do not intersect the classes K_{r′}^{j}, for j ≠ i, since the classes K_{r}^{i} and K_{r′}^{j} lie at distance 6 from each other. Because |y| ≥ 3, we need at least 3(k − 1) lines in A, the final contradiction.

Hence each point class (and dually also each line class) has size k+1, implying that the distance between two point classes and between a line and a point class is at most 4 (and dually for the line classes). Now it immediately follows that the geometry (P ∪ C_{L}, L ∪ C_{P}, I) is a finite generalized pentagon, and hence does not exist. □

Remarks

• All non-classified semi-planes have the property that the size of any equivalence class is at most the largest occuring degree. Hence the analogon of the extra assumption in Theorem 3.4.4 for square forgetful 3-gons would kill all non-classified semi-planes.

• The major problem for generalizing the results above to square forgetful n-gons with n ≥ 7 is that the proof of Claim 4 in Lemma 3.4.1 does not go through.

3.5 Classification results for n even

Let for the rest of this section, Γ be a forgetful n-gon, n even, admitting non-isolated points.

Lemma 3.5.1 Every line contains the same number l of points, and l ≥ g.

3.5 Classification results for n even 79

Proof. Let L and L′ be two arbitrary lines. Note that δ(L, L′) ≤ n by axiom (FP2). If δ(L, L′) = n, then the projection map defines a bijection between P_{L} and P_{L′} (by axiom (FP1)), hence |L| = |L′|. If L and L′ meet in a point p, then consider a line M at distance n−1 from p for which L ≠ proj_{p}M ≠ L′ (such a line exists by axiom (FP3)). Then δ(L, M) = δ(L′, M) = n, hence |L| = |M| = |L′|. If finally 2 < δ(L, L′) < n, then let N be a line at distance n − δ(L, L′) from L such that proj_{L}N ≠ proj_{L}L′, for which the path between L and N only contains points and lines. Then |L| = |N|; and because δ(N, L′) = n we also have |N| = |L′|. So all lines contain the same number of points. Now let G be a class of size g, and L a line at distance n from G. Since for every point x of G, there is a unique point x′ on L for which δ(x, x′) = n − 2, and since all these points are different, we obtain l ≥ g. □

Lemma 3.5.2 Let K be a class of size at least 3. Then all the points in K have the same degree, or K is a forgetful quadrangle of type (III).

Proof. Let K be a class of size at least 3, and suppose that there are two points p_{1}, p_{2} ∈ K for which |p_{1}| ≠ |p_{2}|. Suppose K′ is a class different from K and x a point of K′. Since every point of K lies at distance < n from any line through x, δ(K, K′) ≤ n + 2. Suppose δ(K, K′) = n + 2. Since the map

\begin{aligned}σ : & L_{x} & → & L_{p_{i}} \\ & L & → & L′, & with δ(L, L′) = n − 2, \end{aligned}

is a bijection, we have |x| = |p_{i}|, i = 1, 2, a contradiction. Suppose δ(K, K′) = n, and δ(p_{1}, K′) = δ(p_{2}, K′) = n + 1. Let y be a point of K′ at distance n − 1 from K. Then the map

σ′ : L_{y} \ {proj_{y}K} → L_{p_{i}}
L  → L′, with δ(L, L′) = n − 2,

is a bijection, hence |y| = |p_{i}| + 1, i = 1, 2, a contradiction. Suppose δ(K, K′) = n, and δ(p_{1}, K′) = δ(p_{2}, K′) = n − 1. Let r_{i} = proj_{K′}p_{i}, i = 1, 2, and let p_{3} ∈ K \ {p_{1}, p_{2}}. If δ(p_{3}, K′) = n + 1, then |r_{1}| − 1 = |p_{3}| = |r_{2}| − 1. If δ(p_{3}, K′) = n − 1, then |r_{1}| = |p_{3}| = |r_{2}|. Since |p_{1}| = |r_{2}| and |p_{2}| = |r_{1}|, we obtain a contradiction in both cases. We conclude that δ(K, K′) < n, or δ(K, K′) = n but then exactly one of p_{1} and p_{2} lies at distance n − 1 from K′. Completely similar, one shows that any isolated point w lies at distance at most n − 1 from K, and that if δ(w, K) = n − 1, then either δ(w, p_{1}) = n − 2 or δ(w, p_{2}) = n − 2.

Suppose first that no point at distance n − 1 from K is isolated. Let p_{3} ∈ K \ {p_{1}, p_{2}}. Without loss of generality, we can assume |p_{2}| ≠ |p_{3}|. Let K′ be a class at distance n from K for which δ(p_{1}, K′) = n − 1 and put r_{1} = proj_{K′}p_{1}. Then δ(p_{2}, K′) = n + 1. If δ(p_{3}, K′) = n + 1, then |p_{3}| = |r_{1}| − 1 = |p_{2}|, a contradiction, hence δ(p_{3}, K′) = n − 1. Put r_{3} = proj_{K′}p_{3}. Now |p_{1}| = |r_{3}| = |p_{2}| + 1. Let K′′ be a non-trivial class at distance n from K for which δ(p_{2}, K′′) = n − 1, and put r_{2} = proj_{K′′}p_{2}. Let finally r be a point of K′′ different from r_{2}. If δ(r, K) = n − 1, then |p_{2}| = |r| = |p_{1}| + 1. If δ(r, K) = n + 1, then |p_{1}| = |r| = |p_{2}| − 1. So in both cases, we obtain a contradiction with |p_{2}| + 1 = |p_{1}|.

So we may assume that there exists an isolated point w at distance n − 1 from K for which δ(p_{1}, w) = n − 2. Then |w| = |z| + 1, for all points z of K different from p_{1}. Let v be a point at distance n−2 from p_{2} and n−1 from K for which proj_{v}p_{2} is a line. Let p_{3} be an arbitrary point of K different from p_{1} and p_{2}. If v would be isolated, then we obtain |p_{1}| + 1 = |v| = |p_{3}| + 1, hence (since |p_{2}| = |p_{3}|) also |p_{1}| = |p_{2}|, a contradiction. So v is contained in a class K′. By the first paragraph of the proof, δ(p_{1}, K′) = n + 1 and hence |v| = |p_{1}| + 1. Since the degree of an arbitrary point z of K different from p_{1} is |p_{2}|, such a point z lies at distance n − 1 from K′ (if not, the degree of z would be |v| − 1 = |p_{1}|). Also, it is now clear that |v| = |p_{3}|, hence |p_{3}| = |p_{1}| + 1. Note that any isolated point at distance n − 1 from K necessarily lies at distance n − 2 from p_{1}. Now let w′ be an arbitrary point at distance n − 1 from K and at distance n − 2 from p_{1} such that proj_{w′}p_{1} is a line. We show that w′ is isolated. Suppose by way of contradiction that w′ is equivalent with a point w′′, w′′ ≠ w′. Since w′′ does not lie at distance n−2 from any point of K, we see that |p_{2}| = |w′′| = |p_{1}|−1, a contradiction. Finally, we claim that every point u of K′ (with K′ as above) lies at distance n−2 from a point of K\{p_{1}}, and |u| = |p_{2}|. Suppose by way of contradiction that the class K′ contains a point v′ at distance n + 1 from K. Then |v′| = |p_{2}|−1 = |p_{1}|. Since K′ contains at least two points of degree |p_{2}| (namely the projections of p_{2} and p_{3} onto K′), every isolated point at distance n − 1 from K′ has to lie at distance n − 2 from v′. Let γ be a fixed n-path between p_{1} and v′, and x the element of γ at distance n/2 + 1 from v′. If x is not a point of degree two or a class containing only two points, then consider a point y at distance n/2−1 from x for which proj_{y}p_{1} is a line and proj_{x}p_{1} ≠ proj_{x}y ≠ proj_{x}v′. The point y is isolated (since δ(y, K) = n − 1 and δ(y, p_{1}) = n − 2) and lies at distance n − 1 from K′ (if δ(y, K′) = n + 1, all the points of K′ would have equal degree) but at distance n from v′, a contradiction. If x is a class or a point of degree two, then let R be a line at distance n/2 − 1 from x′ = proj_{x}p_{1} for which proj_{x′}p_{1} ≠ proj_{x′}R ≠ proj_{x′}v′. Then the line R contains at least two isolated points, hence at least one isolated point at distance n from v′, a contradiction. This shows the claim.

So we obtained the following situation (◇): if K′ is an arbitrary non-trivial class at distance n from K, then |K′| = |K| − 1 and each point of K′ lies at distance n − 2 from a unique point of K \ {p_{1}}. The degree of a point in K′ is equal to |z| = |p_{1}| + 1, with z an arbitrary point of K \ {p_{1}}. A point w′ at distance n − 1 from K for which proj_{w′}K is a line is isolated if and only if δ(w′, p_{1}) = n − 2. Moreover, |K| = g = l. Indeed, consider a line L at distance n from K. Suppose l > |K|. Then L contains a point y which lies at distance n from all the points of K. This contradicts the observations at the beginning of the proof. Consequently, l = |K| = g.

Suppose n = 4. We show that Γ is of type (III). Note that every non-trivial class different from K lies at distance 4 from K and hence has size g −1. It is also easy to see that, if two classes K′ and K′′ lie at distance 4, K′ ≠ K ≠ K′′, then every point of K′ lies at distance 3 from a point of K′′ and conversely. Indeed, let z ∈ K′, δ(z, K′′) = 3, and suppose y is a point of K′′ at distance 5 from K′. Then |y| = |z| − 1, a contradiction with the fact that all points in K′ and K′′ have degree |p_{1}| + 1. We define the following equivalence relation ∼_{C} on the classes of size g − 1:

K_{1} ∼_{C} K_{2} ⇔ δ(K_{1}, K_{2}) = 6.

The transitivity of ∼_{C} is shown as follows: suppose K_{1} ∼_{C} K_{2}, K_{1} ∼_{C} K_{3}, but δ(K_{2}, K_{3}) = 4. Let L be a line intersecting both K_{2} and K_{3}. Every point of K_{1} has to lie at distance 2 from a unique point of L, not belonging to K_{2} or K_{3}, hence |L| ≥ |K_{1}| + 2 = g + 1, a contradiction. We associate a symbol ∞_{i}, i = 1, …, s to each equivalence class C_{i} of ∼_{C}. Now define the following geometry Δ = (P′, L′, I′). A point of Δ is either a point of Γ or a symbol ∞_{i}, i = 1, …, s. A line of Δ is either a line of Γ; the set K (with K the unique class of size g); the set of points of a class of size g − 1 together with the symbol of its equivalence class, or the set of points {∞_{1}, …, ∞_{s}, p_{1}}. Incidence is the incidence of Γ if defined, or symmetrized containment otherwise. Then it is easy to see that Δ is a finite generalized quadrangle of order (s, k), by checking the main axiom. We illustrate this for two cases. First, let p be a point of Γ, p ∉ K, and L a line of Δ containing the points of a class K′ and the symbol ∞_{j}, p ∉ K′. Suppose p is not collinear in Γ with any point of K′. If p is not isolated, then it is contained in a class K′′ of size g − 1 which necessarily lies at distance 6 from K′, hence K′ ∼_{C} K′′. So p is collinear in Δ with the point ∞_{j}. If p is isolated, then p is collinear in Γ with p_{1}, see (◇). Since no point of K′ is collinear with p_{1}, and |K′| = |pp_{1}| − 1, p has to be collinear in Γ with a point of K′. Secondly, let p be a symbol ∞_{j}, and L a line of Γ. Let K′ be a class such that K′ ∪ {∞_{j}} is a line of Δ. Suppose K′ does not intersect L. By projecting the points of K′ onto L, we see that there is a unique point y on L not collinear with any of the points of K′. If y would be isolated, then y ⊥ p_{1} and we obtain a contradiction by projecting the points of K′ onto the line yp_{1}, hence y is contained in a non-trivial class K′′. If |K′′| = g, then y = p_{1} and the result follows. If |K′′| = g − 1, then K′ ∼_{C} K′′, hence the result. Now clearly, Γ is a forgetful quadrangle of type (III).

Suppose n ≥ 6. We look for a contradiction. Let K′ be a class at distance n from K (such a class exists by (◇)), and z ∈ K′. We construct a line M for which δ(M, K) = n − 2, δ(M, K′) = n, δ(M, p_{1}) = n − 3 and such that there exists a point r ∈ K′ for which δ(proj_{M}p_{1}, r) = n − 2. Fix a line L through p_{1} and put z′ = proj_{L}z. If n ≡ 2 mod 4, let m be the point of [z, z′] at distance n/2 − 3 from z′ and let M be a line at distance n/2 − 2 from m for which proj_{m}p_{1} ≠ proj_{m}M ≠ proj_{m}z. (Note that, for n = 6, z′ = m, but since |z′| ≥ |z′′| = |p_{1}| + 1 ≥ 3, with z′′ ∈ K′ \ {z}, the line M exists.) If n ≡ 0 mod 4, consider the element N of [z, z′] at distance n/2 − 3 from z′. If N is a line or a class containing at least three points, then let M be a line at distance n/2 − 2 from N such that proj_{N}p_{1} ≠ proj_{N}M ≠ proj_{N}z. If N is a class of size 2, then let N′ be a line at distance n/2 − 3 from x = proj_{N}z′ such that proj_{x}p_{1} ≠ proj_{x}N′ ≠ proj_{x}z. Fix a point z′′ ∈ K′ \ {z} and put y = proj_{N′}z′′. Note that if proj_{y}z′′ is a line, then |y| ≥ |z| = |p_{1}| + 1 ≥ 3. Hence it is possible to choose a line M through y different from N′ or proj_{y}z′′. So in each case, we constructed a line M as claimed. Now all the points on M different from proj_{M}p_{1} are isolated (see ◇). Since |K′| = |M|−1, there exists a point a on M which lies at distance n from all the points of K′. Hence |a| = |r| = |p_{2}|, the final contradiction since a is isolated (which implies |a| = |p_{2}| + 1). □

From now on, we assume that two points belonging to a class of size at least 3, have the same degree.

Lemma 3.5.3 The points of a class K of size 2 have the same degree.

Proof. Let K = {p_{1}, p_{2}} and suppose by way of contradiction that |p_{1}| ≠ |p_{2}|. For an arbitrary class K′, one shows similarly as in the proof of Lemma 3.5.2 that δ(K, K′) ≤ n, that δ(K, K′) = n implies that |K′| = 2 and that the two points of K′ have different degrees. Also, any isolated point lies at distance at most n − 1 from K. We first claim that the degrees of p_{1} and p_{2} differ by one. Let L be a line at distance n from K. Since l ≥ 3, there is at least one point x on L at distance n from p_{1} and p_{2}. Since x is not isolated, this point is contained in a class K′ of size 2 at distance n from K. Without loss of generality, we can assume that the point y of K′ different from x lies at distance n − 2 from p_{2}. From this follows that |p_{2}| − 1 = |x| = |p_{1}| = |y| − 1, hence the claim. From now on, we assume that |p_{2}| = |p_{1}| + 1. Now the following observation (♢) can easily been shown. If K_{1} = {q_{1}, q_{2}} and |q_{1}| ≠ |q_{2}|, then (up to interchanging q_{1} and q_{2}) |q_{1}| = |q_{2}| − 1, and for any class K_{2} = {r_{1}, r_{2}} at distance n from K_{1}, we have (up to interchanging r_{1} and r_{2}) either δ(q_{1}, r_{1}) = δ(q_{2}, r_{2}) = n − 2 with |q_{1}| = |r_{2}| and |q_{2}| = |r_{1}| or δ(q_{2}, r_{2}) = n − 2, δ(q_{1}, K_{2}) = δ(r_{1}, K_{1}) = n + 1 and |q_{i}| = |r_{i}|, i = 1, 2.

First consider the case n = 4. We start by showing that there are no isolated points. Let L be an arbitrary line not intersecting K, and put r_{i} = proj_{L}p_{i}, i = 1, 2. Let z be a point on L, r_{1} ≠ z ≠ r_{2}. Then z is not isolated (because of the observations at the beginning of the proof), hence z is contained in a class K_{z} = {z, z′} at distance 4 from K and by (♢), |z| = |p_{1}| and |z′| = |p_{2}|. If r_{2} would be isolated, then |p_{1}| + 1 = |r_{2}| = |z′| + 1, a contradiction since |z′| = |p_{1}| + 1, hence r_{2} is contained in a class K_{r_{2}} = {r_{2}, r′_{2}}, and |r_{2}| = |p_{2}| (indeed, if |r_{2}| = |p_{1}|, then the two points r_{2} and z of degree |p_{1}| would be collinear, contradicting (♢)). If r_{1} would be isolated, then |r′_{2}| + 1 = |r_{1}| = |z′| + 1, again a contradiction. Now it is clear that no point of Γ is isolated. Hence every point is contained in a class of size 2, and has degree |p_{1}| or |p_{2}|. Put h = \frac{|P|}{2}. We count the number of pairs (p, L), L a line of Γ through the point p, with p a point of degree |p_{i}|. Since by (♢) every line contains at most 1 point of degree |p_{1}|, we obtain

h|p_{1}| ≤ |L| (i = 1),
h(|p_{1}| + 1) ≥ 2|L| (i = 2).

Hence |p_{1}| ≤ 1, the final contradiction.

Now consider the case n ≥ 6 (in fact, the argument below also works for n = 4 except when l = 3). Choose an element x at distance n/2 + 1 from p_{1} and at distance n/2 + 2 from K such that x is a line if n ≡ 0 mod 4. Suppose first that, if n = 6, the point x can be chosen such that either x is isolated (implying |x| ≥ 3) or proj_{x}p_{1} is a class. Let M_{1} and M_{2} be two lines at distance n from K and at distance n/2 − 2 from x such that proj_{x}M_{1} ≠ proj_{x}M_{2} (note that such lines exist because of the assumptions just made). On each of the lines M_{i}, i = 1, 2, there is at least one point m_{i} that lies at distance n + 1 from the class K. Because of the first paragraph of the proof, the point m_{i} is contained in a class K_{i} of size 2, and by (♢), |m_{1}| = |p_{1}| = |m_{2}| (recall that |p_{1}| = |p_{2}| − 1). But since δ(m_{1}, m_{2}) = n − 2, this is a contradiction with (♢). Suppose now that n = 6 and that we cannot choose a point x as above. This implies in particular that every point collinear with p_{1} and at distance 3 from K is isolated. Then let again x be a point at distance 5 from K for which δ(x, p_{1}) = 4, M_{1} a line through x, M_{1} ≠ proj_{x}K and m_{1} a point on M_{1} at distance 6 from both p_{1} and p_{2}. Again, the point m_{1} is not isolated, and the class K_{1} containing m_{1} lies at distance 6 from K. Hence K_{1} = {m_{1}, m′_{1}} and by (♢), also |m_{1}| = |m′_{1}| − 1. Now the point x⋈p_{1} lies at distance 5 from K_{1} and 4 from m_{1}, and has degree at least 3. Hence we can apply the argument of the general case above (with K_{1} in the role of K and x⋈p_{1} in the role of x) to obtain a contradiction with |m_{1}| ≠ |m′_{1}| □

Lemma 3.5.4 (i) If two classes K and K′ lie at distance n, then every point of K lies at distance n − 1 from K′ and vice versa, hence |K| = |K′|.

(ii) All non-isolated points have the same degree k.

Proof. Let K and K′ be two classes for which the points have degree k and k′ respectively. Suppose first that δ(K, K′) = n + 2. Let x ∈ K and x′ ∈ K′. Then k = |x| = |x′| = k′. Suppose now that δ(K, K′) = n. Then there are points x ∈ K and x′ ∈ K′ such that δ(x, x′) = n − 2. If there exist points y ∈ K, y ≠ x and y′ ∈ K′ such that δ(y, y′) = n − 2, then |x′| = |y|, hence k = k′. But if no such points exist, then, for an arbitrary point y ∈ K, y ≠ x, and a point y′ ∈ K′, y′ ≠ x′, we have |y′| = |y| = |x′|−1, contradicting |x′| = |y′|. Hence k = k′. Note that if K would contain a point z at distance n + 1 from K′, then k = |z| = |x′| − 1 = k − 1, a contradiction. This shows (i). Now choose a class K′ at minimal distance from K (if such a class does not exist, (ii) is proved). We show that the points in K and the points in K′ have the same degree. We can assume δ(K, K′) ≤ n − 2. Let X be the element at distance \frac{δ(K,K′)}{2} from both K and K′ (note that X cannot be a class because of the minimality of δ(K, K′)). Consider a point x at distance n − 1 − \frac{δ(K,K′)}{2} from X such that proj_{X}K′ ≠ proj_{X}x ≠ proj_{X}K (such a point exists since again by the minimality of δ(K, K′), X is not a class of size 2 or a point of degree 2) and such that proj_{x}X is not a class. If x is isolated, then it is easy to see that k = k′ (indeed, then |x| = |z| + 1, for z an arbitrary point of K or K′ different from the projection of X onto K or K′). If x is contained in a non-trivial class K′′, then K′′ lies at distance n from both K and K′, hence the result. Now (ii) easily follows. □

Lemma 3.5.5 One of the following situations occurs:

(i) There is a unique isolated point of degree k. In this case, Γ is a generalized quadrangle of type (II), with |X_{1}| = s.

(ii) Any isolated point has degree k + 1, and lies at distance at most n − 1 from any class.

Proof. Let w be an isolated point. We first prove that |w| ∈ {k, k + 1}. For an arbitrary class K, we have δ(w, K) ≤ n + 1. If there is a class K at distance n+1 from w, then |w| = k, and if there is a class K at distance n−1 from w, then |w| = k + 1. Now choose a class K at minimal distance from w. We can assume δ(w, K) ≤ n − 3. Put v = proj_{K}w. Let X be the element on the shortest path between v and w at distance \frac{δ(v,w)}{2} from w. Then X is not a class, hence it is possible to choose a point x at distance n − \frac{δ(v,w)}{2} from X such that proj_{X}w ≠ proj_{X}x ≠ proj_{X}v. If x is isolated, then (since δ(x, K) = n ± 1), |x| = k or |x| = k + 1. Since opposite isolated points have the same degree, also |w| ∈ {k, k + 1}. If x is not isolated, then, with K′ the class containing x, δ(w, K′) = n ± 1, hence also |w| ∈ {k, k + 1}. So we conclude that every isolated point has degree k or k + 1; if it has degree k + 1, then it cannot lie at distance n + 1 from any class; if it has degree k, then it cannot lie at distance n − 1 from any class.

\underline{n = 4} Suppose there exists an isolated point w of degree k. We show that all other isolated points have degree k + 1. Note that all points collinear with w are isolated. Let K be an arbitrary class, and x, y ∈ K. Let L_{1} and L_{2} be two different lines through w and put x_{i} = proj_{L_{i}}x, y_{i} = proj_{L_{i}}y, i = 1, 2. Then x_{i} and y_{i} are isolated points of degree k + 1 (since they lie at distance 3 from K). If w′ is a second isolated point of degree k, then w′ is collinear with the points x_{i} and y_{i}, i = 1, 2 (since isolated points at distance n = 4 have the same degree), hence w = w′. So w is the unique isolated point of degree k. From this it immediately follows that a point is isolated if and only if it is collinear with w. Let G be a class of size g. Since every point of a line L through w, different from w, is collinear with a unique point of G, it follows that l = g + 1. Note that this implies that all classes have size g. Indeed, for an arbitrary non-trivial class K, every point of L different from w has to be collinear with a unique point of K and vice versa, hence |K| = g. Now we define the following equivalence relation ∼_{C} on the classes of size g:

K_{1} ∼_{C} K_{2} ⇔ δ(K_{1}, K_{2}) = 6.

The transitivity of ∼_{C} is shown as follows: suppose K_{1} ∼_{C} K_{2}, K_{1} ∼_{C} K_{3}, but δ(K_{2}, K_{3}) = 4. Let L be a line intersecting both K_{2} and K_{3}. Then each point of K_{1} has to be collinear with a unique point of the line L, not belonging to K_{2} or K_{3}, hence l ≥ g + 2, a contradiction. We associate a symbol ∞_{i}, i = 1, …, s to each equivalence class C_{i} of ∼_{C}. Now define the following geometry Δ = (P′, L′, I′). A point of Δ is either a point of Γ or a symbol ∞_{i}, i = 1, …, s. A line of Δ is either a line of Γ, the set of points of a class of C together with the symbol ∞_{j} of its equivalence class, or the set of points {∞_{1}, …, ∞_{s}, w}. Incidence is the incidence of Γ if defined, or symmetrized containment otherwise. Then it is easy to see that Δ is a finite generalized quadrangle of order (s, k). Hence Γ is a forgetful quadrangle of type (II), with |X_{1}| = s.

\underline{n ≥ 6} We show that an isolated point of degree k cannot exist. So let by way of contradiction, w be an isolated point of degree k. Let S be the set of points x at distance n − 2 from w for which proj_{x}w is a line. Clearly, S is nonempty and consists of isolated points. We first show that all the points of S have degree k + 1. Suppose by way of contradiction that S contains a point x of degree k. Since opposite isolated points have the same degree, it is easy to see that all the points of S then have the same degree k. We can now always find a point y of S at distance n − 1 from a certain class, which is a contradiction. Indeed, let K be an arbitrary non-trivial class at minimal distance from w. If δ(w, K) = n + 1, let v be a point of K and γ a fixed n-path between v and w. If δ(w, K) < n + 1, let v = proj_{K}w and γ = [v, w]. Let X be the element of γ at distance \frac{δ(v,w)}{2} from both v and w. Since X cannot be a class or a non-isolated point, it is possible to choose a point y at distance n − 2 − \frac{δ(v,w)}{2} from X such that proj_{X}w ≠ proj_{X}y ≠ proj_{X}v and proj_{y}X is not a class. Now the point y belongs to S and lies at distance n−1 from K, the contradiction. We conclude that all the points of S have degree k + 1.

Now let x be a point at distance n from w. Then x cannot be isolated. Indeed, if x is isolated, then |x| = |w| = k, but it is easy to see that there exists a point y of S opposite x, hence |x| = |y| = k + 1, a contradiction. Also, the class K_{x} containing x cannot lie at distance n − 1 from w. Let L be a line through w. Since δ(w, K_{x}) = n + 1, projecting the points of K_{x} onto L shows that l ≥ |K_{x}| + 1. Let γ be a fixed n-path between x and w, and X the element of γ at distance n/2 + 1 from x. If n > 6 and X is not a class of size two, then consider a line M at distance n/2 − 2 from X for which proj_{X}x ≠ proj_{X}M ≠ proj_{X}w. Since the points of M, different from proj_{M}w, are contained in S, they all have to lie at distance n − 1 from K_{x}, hence |K_{x}| = |M| = l, a contradiction. If n = 6, then the same argument can be applied except if X is a point of degree 2 for which both proj_{X}x and proj_{X}w are lines. In this case, consider a line M at distance 3 from w and at distance 6 from the class K′ containing X. Then similarly as above, we obtain |K′| = |M| = l, but this is a contradiction since by Lemma 3.5.4(i), |K′| = |K_{x}| ≤ l − 1. Finally, if X is a class containing exactly two points, we proceed as follows. Let Y be the element of γ at distance \frac{δ(w,X)−3}{2} from w and, if Y is not a class of size 2, M a line at distance n − \frac{δ(w,X)+3}{2} from Y such that proj_{Y} w ≠ proj_{Y} M ≠ proj_{Y} X. Note that δ(w, M) = n − 3 and δ(M, X) = n. The points of M different from proj_{M}w (and there are at least 2 of them) are contained in S, hence lie at distance n − 2 from a point of X (different from proj_{X}w). This is a contradiction, since |X| = 2. If Y is a class of size 2, then we repeat the argument above with Y in the role of X. In this way, we obtain that there are no isolated points of degree k, and the lemma is proved. □

From now on, we assume that any isolated point has degree k + 1.

Lemma 3.5.6 If there exists a class X, with 1 < |X| < g, then Γ is a forgetful quadrangle of type (II), with X = X_{2} and 1 < |X_{1}| < s.

Proof.

\underline{n = 4} Let X be a class of size < g, and G a class of size g. Note that δ(X, G) = 6, because of Lemma 3.5.4(i). By projecting the points of G onto a line L intersecting X, we see that |L| ≥ g + 1. Suppose there is a second class X′ with |X′| < g. If δ(X, X′) = 4, then let (X, x, M, x′, X′) be a 4-path between X and X′. Every point of G is collinear with a point of M different from x or x′, and all these points are isolated (indeed, if proj_{M}p, with p ∈ G, is not isolated, then the class K′ containing proj_{M}p would satisfy |G| = |K′| = |X| because of Lemma 3.5.4(i)). Hence there are at least g isolated points on M. Now let L be a line through x different from M, and assume there is a class K intersecting L, but not containing x. Because of Lemma 3.5.5(ii), every isolated point of M is collinear with a point of K, different from proj_{K}x. Hence |K| ≥ g + 1, a contradiction. So all the points on L different from x are isolated (this makes at least g isolated points on L). But since, again by Lemma 3.5.5(ii), every isolated point of L is collinear with a point of X′, |X′| ≥ g + 1, a contradiction. So all points at distance 3 from X are isolated, and δ(X, X′) = 6. Let N be a line intersecting X. Since N contains at least g isolated points, |X′| ≥ g, a contradiction. Hence X is the unique class of size < g, and a point is isolated if and only if it lies at distance 3 from X. By projecting the points of a class G of size g onto a line L intersecting X, we see that l = g + 1. As in the proof of Lemma 3.5.5, it is now possible to define the following equivalence relation on the classes of size g:

K_{1} ∼_{C} K_{2} ⇔ δ(K_{1}, K_{2}) = 6.

Note that it is possible to find two classes of size g at distance 4 (indeed, consider the points on a line at distance 4 from X), hence ∼_{C} defines at least two equivalence classes. We associate a symbol ∞_{i}, i = 1, …, r to each equivalence class C_{i} of ∼_{C}. Now define the following geometry Δ = (P′, L′, I′). A point of Δ is either a point of Γ or a symbol ∞_{i}, i = 1, …, r. A line of Δ is either a line of Γ, the set of points of a class of size g together with the symbol ∞_{j} of its equivalence class, or the set of points {∞_{1}, …, ∞_{r}} ∪ X. Incidence is the incidence of Γ if defined, or symmetrized containment otherwise. Then it is easy to see that Δ is a finite generalized quadrangle of order (s, k) with s = r + |X| − 1, hence Γ is a forgetful quadrangle of type (II), with 1 < |X_{1}| < s. We check for example the main axiom for a ‘point’ ∞_{i} and a line L of Γ, L not intersecting X. Suppose ∞_{i} is not collinear in Δ with any point of L. Let K be a class belonging to the equivalence class of ∼_{C} with symbol ∞_{i}. Then K does not intersect L. Every point x of L is collinear with a point of K. (Indeed, this is clear if x is isolated. If x belongs to a non-trivial class K′, then δ(K, K′) = 4, hence x is collinear with a point of K′.) Hence |K| ≥ |L| = g + 1, a contradiction.

\underline{n = 6} We treat this case separately, because here the reasoning is slightly different from the general case. Let by way of contradiction X be a class for which |X| < g. Let G be a class of size g. Then δ(X, G) ∈ {4, 8}. If δ(X, G) = 8, then l ≥ g + 1. Indeed, choose x ∈ X and L a line through x. By projecting the points of G onto the line L, we see that |L| ≥ g + 1.

We next show that, if δ(X, G) = 4, then l = g. Let (X, x, M, y, G) be the 4path between X and G. Consider a line N concurrent with M not through x or y. Then the points on N different from proj_{M}N are isolated (indeed, a class K intersecting M but not containing proj_{M}N would satisfy |X| = |K| = |G| by Lemma 3.5.4(i)). So N contains at least g − 1 isolated points. Now let L be a line through y different from M, and suppose there is a class K intersecting L, different from G. Since δ(X, K) = 6, |K| = |X| < g. By Lemma 3.5.5(ii), there is a bijection between the points of K different from proj_{L}y and the points on N different from proj_{M}N. This implies that |K| = l, hence |K| = g = l, a contradiction. So all the points on L are isolated. Now let y′ be a point of G different from y, and L′ a line through y′. Not all points on L′ different from y′ can be isolated (since otherwise Lemma 3.5.5(ii) would imply |X| = |L′| ≥ g), so there is a class K′ intersecting L′, K′ ≠ G. Then |K′| = |L| ≥ g, and hence |K′| = |L| = g.

First, suppose δ(X, G) = 8 (so l ≥ g + 1 in this case), and let γ = (x, …, y) be a fixed 6-path between points x ∈ X and y ∈ G. Let M be the element of γ at distance 3 from both x and y. Suppose first M is a class. If |M| = g, then δ(X, M) = 4 implies l = g. If |M| < g, then δ(M, G) = 4 implies l = g. Hence we obtain a contradiction in both cases, so M is necessarily a line. Clearly, the points on M different from the projection of x or y onto M are isolated. Also the point z = proj_{M}x is isolated. Indeed, this point cannot be contained in a class of size < g (since such a class would lie at distance 6 from G). But z cannot lie in a class of size g either, since such a class would lie at distance 4 from X, implying l = g. Hence we have at least g isolated points on M. Now let N be a line through y different from proj_{y}M, and suppose there is a class K intersecting N in a point different from y. Since there is a bijection between the points of M and the points of K, we obtain |K| = g + 1, a contradiction. Hence N contains at least g isolated points, but this implies (using Lemma 3.5.5(ii)) that |X| ≥ g, a contradiction. This shows that δ(X, G) ≠ 8.

Next, suppose δ(X, G) = 4 (so l = g in this case). Let again (X, x, M, y, G) be the 4-path between X and G. Let L′′ be a line at distance 3 from y such that proj_{y}L′′ is a line different from M. Suppose there is a class K intersecting L′′, but not containing the point proj_{L′′}y. Since δ(G, K) = 6, |K| = g, but this contradicts δ(K, X) = 8 and the previous paragraph. Hence all g − 1 points of L′′ different from proj_{L′′}y are isolated. Consequently |X| = g (again by Lemma 3.5.5(ii)), a contradiction. We conclude that all classes have size g. Hence X cannot exist.

\underline{n > 6} Let by way of contradiction X be a class for which |X| < g. Let G be a class of size g. Then δ(X, G) ≤ n + 2 and δ(X, G) ≠ n.

We first claim that if δ(X, G) = n + 2, γ = (x, …, y) is an arbitrary n-path between points x ∈ X and y ∈ G, and M is the element of γ at distance n/2 from both x and y, then either n ≡ 2 mod 4 and M is a class of size two, or n = 8 and there does not exist a line through M different from proj_{M}x and proj_{M}y. (∗) So assume δ(X, G) = n + 2, and suppose by way of contradiction that there is a path γ = (x, …, y) between points x ∈ X and y ∈ G such that the element M (with M as above) is not a class of size two if n ≡ 2 mod 4, or such that there does exist a line through M different from proj_{M}x and proj_{M}y if n = 8. By projecting the points of G onto the line proj_{x}M, we obtain l ≥ g + 1. Let L be a line at distance n/2 − 3 from M such that proj_{M}x ≠ proj_{M}L ≠ proj_{M}y (L exists because of the assumptions on M). Then the points on L different from proj_{L}M are isolated (indeed, a class K intersecting L but not containing proj_{L}M would satisfy δ(X, K) = δ(G, K) = n, so |X| = |K| = |G|, a contradiction). Now let N be a line through y different from proj_{y}M. Since a class K′ intersecting N, K′ ≠ G, would satisfy |K′| = |L| ≥ g + 1 (using Lemma 3.5.5(ii)), every point on N is isolated. But (again using Lemma 3.5.5(ii)) this implies |X| ≥ |N| − 1 ≥ g, a contradiction. This shows the claim.

We next show that δ(X, G) ≠ j, with j ≤ n/2 + 1.

Suppose that δ(X, G) = 4, and let (X, x, xy, y, G) be a 4-path between X and G. Let L be a line at distance n − 3 from y such that proj_{y}L ≠ xy, and such that the element M of the path [x, L] at distance n/2 from x is not a class of size 2 if n ≡ 2 mod 4, or such that there can be chosen a line through M not belonging to [x, L] if n = 8. Let r be a point on L, r ≠ proj_{L}y. Then r is isolated. Indeed, suppose by way of contradiction that r is contained in a class K. Since δ(G, K) = n, |K| = |G| = g. This implies that δ(X, K) = n + 2. But now, by considering the n-path between x ∈ X and r ∈ K containing L, we see that K cannot exist because of (∗). So the line L contains at least g − 1 isolated points. By Lemma 3.5.5(ii), there is a bijection between the points of X \ {x} and the points on L different from proj_{L}y, a contradiction with |X| < g.

We proceed by induction on δ(X, G). Let 4 ≤ k < n/2 + 1 and suppose that δ(X, G′) > k, for any class G′ of size g. Then we first claim that there does not exist a class G′′ of size g at distance n+2−k from X such that the element of the path [X, G′′] at distance n/2 − k + 1 from G′′ is not a class of size 2 if n ≡ 2 mod 4 (∗∗). Suppose by way of contradiction a class G′′ as above does exist. Let z be a point of G′′ not belonging to [X, G′′], and L a line at distance k−3 from z such that proj_{z}L ≠ G′′. All the points on L different from proj_{L}z are isolated. Indeed, if K′ would be a class intersecting L, proj_{L}z ∉ K′, then |K′| < g contradicts δ(K′, G′′) ≤ k and the previous paragraph, but |K′| = g contradicts δ(K′, X) = n + 2 and (∗). So there are at least g − 1 isolated points on L. By Lemma 3.5.5(ii), there is a bijection between the points of L \ {proj_{L}z} and the points of X \ proj_{X}G′′. Hence |X| = g, a contradiction. This shows (∗∗). Now we show that δ(X, G′′) ≠ k + 2, k < n/2, for any class G′′ of size g. So suppose by way of contradiction that δ(X, G′′) = k + 2, and let y be the element of G′′ at distance k+1 from X. Let L be a line at distance n − 1 − k from y, and n − k from G′′ such that proj_{y}L ≠ proj_{y}X and such that the element of the path [G′′, L] at distance n/2 − k + 1 from G′′ is not a class of size 2 if n ≡ 2 mod 4. Then every point r on L different from proj_{L}y is isolated. Indeed, suppose by way of contradiction that r is contained in a class K. Since δ(K, G′′) = n + 2 − k, (∗∗) implies that |K| = g. But if |K| = g, then δ(X, K) = n + 2, which contradicts (∗). Hence r is isolated. Now again by Lemma 3.5.5(ii), there is a bijection between the points of L and X, implying |X| ≥ g, a contradiction.

So we obtained the following: if δ(X, G) = j, then j > n/2 + 1 and the element of a j-path between X and G at distance j − (n/2 + 1) from G is a class K of size two. But now, applying this result on the classes G and K (note that |K| < |G| and δ(K, G) < n/2 − 1 if j ≠ n + 2) leads to the final contradiction. This shows that X cannot exist.

□

We summarize the situation reached so far. If Γ is a finite forgetful n-gon, n even, then Γ is either a generalized n-gon, a forgetful quadrangle of type (II) (with 1 < |X_{1}| < s + 1) or type (III), or there exist parameters g, k, d such that the following axioms are satisfied:

(S1) Every isolated point is incident with exactly k + 1 lines, every nonisolated point is incident with exactly k lines (k ≥ 2).

(S2) Every class has the same size g, g > 1.

(S3) Every line contains g + d points, d ≥ 0.

Let now Γ be a finite forgetful n-gon, n even, satisfying axioms (S1), (S2) and (S3). The parameter d is called the deficiency of Γ.

Lemma 3.5.7 If d = 0, then Γ is a forgetful n-gon of type (I).

Proof. Suppose d = 0. Define the geometry Δ = (P, L ∪ C, I). Then Δ is a generalized n-gon. Indeed, we only have to check that a point p and a class K lie at distance at most n − 1 from each other. Suppose by way of contradiction that δ(p, K) = n+1. But then projecting the points of K onto an arbitrary line L through p shows that |L| ≥ g + 1, a contradiction. □

Lemma 3.5.8 If n = 4 and d = 1, then Γ is a forgetful quadrangle of type (II), with |X_{1}| ∈ {1, s + 1}.

Proof. Suppose d = 1, so l = g + 1. As in Lemma 3.5.6, we define an equivalence relation ∼_{C} on the classes (which are all of size g), and associate a symbol ∞_{i}, i = 1, …, r to each equivalence class C_{i} of ∼_{C}.

Suppose first there are no isolated points. Then define the following geometry Δ = (P′, L′, I′). A point of Δ is either a point of Γ or a symbol ∞_{i}, i = 1, …, r. A line of Δ is a line of Γ, the set of points of a class K together with the symbol of its equivalence class, or the set of points {∞_{1}, …, ∞_{r}}.

Incidence is the incidence of Γ if defined, or symmetrized containment otherwise. Similarly as in Lemma 3.5.6, one shows that Δ is a finite generalized quadrangle of order (g, k), hence Γ is a forgetful quadrangle of type (II), with |X_{1}| = s + 1.

Suppose now there is an isolated point w. We show that any two classes of size g lie at distance 6 (hence ∼_{C} has only one equivalence class). Note first that no line through w only contains isolated points. Indeed, if L_{0} would be a line through w full of isolated points, then for a class G of size g (which necessarily lies at distance 3 from w), Lemma 3.5.5(ii) implies that |G| = |L_{0}| = g + 1, a contradiction. Now let L_{0}, …, L_{k} be the lines through w and X_{0} a class of size g intersecting L_{0}. Since l = g + 1, there is a unique point x_{i} on L_{i}, i = 1, …, k that is not collinear with any point of X_{0}. Hence by lemmas 3.5.5(ii) and 3.5.4(i), x_{i} is contained in a class X_{i} of size g for which δ(X_{0}, X_{i}) = 6. Since ∼_{C} is an equivalence relation, also δ(X_{j}, X_{j′}) = 6, for j, j′ ∈ {1, …, k}, j ≠ j′. Suppose now there exists a class K of size g, K ≠ X_{i}, i = 0, …, k. Then without loss of generality, we can assume that K intersects L_{0} in the point y. Because of the construction of the classes X_{i}, the point y lies at distance 3 from every class X_{i}, i = 0, …, k. Hence there exists a line N_{i} through y, i = 0, …, k such that δ(N_{i}, X_{i}) = 2 (note that all these lines N_{i} are different because the classes X_{i} mutually lie at distance 6). But since y is not isolated, it has degree k, a contradiction. So ∼_{C} has a unique equivalence class with associated symbol ∞. Now define the following geometry Δ = (P′, L′, I′). A point of Δ is either a point of Γ or the symbol ∞. A line of Δ is either a line of Γ or the set of points of a class K together with the symbol ∞. Incidence is the incidence of Γ if defined, or symmetrized containment otherwise. Then it is easy to see that Δ is a finite generalized quadrangle of order (g, k), hence Γ is a forgetful quadrangle of type (II), with |X_{1}| = 1. □

A forgetful n-gon, n even, satisfying (S1), (S2) and (S3) with d ≥ 2 if n = 4 and d ≥ 1 if n ≥ 6, is called a short forgetful n-gon. This name refers to both the short classes and the short memory of these objects. (Indeed, their memory seems to be too short to prove that they arise from generalized polygons.)

We now obtained the following theorem:

Theorem 3.5.9 A finite forgetful n-gon, n even, is either a generalized ngon, a forgetful n-gon of type (I), (II) or (III), or a short forgetful n-gon.

Proposition 3.5.10 Let Γ be short forgetful polygon.

(i) Γ contains at least two classes.

(ii) Γ contains either 0 or at least two isolated points.

Proof. Let Γ be a short forgetful polygon. Suppose there is a unique nontrivial class K (K ≠ P). Then by Lemma 3.5.5(ii), any line at distance n from K contains g points, so d = 0, a contradiction. This shows (i). Suppose now Γ contains a unique isolated point w. Let K be a class at distance n − 1 from w, L_{1} the line intersecting K at distance n − 3 from w, and L_{2} an arbitrary line through proj_{K}w, different from L_{1}. Let S be the set of points at distance n+1 from K. Since d ≠ 0, S ≠ ∅. Each line at distance n−2 from L_{1} (L_{2}) and at distance n from K contains d points of S. Conversely, every point of S is on a unique line lying at distance n from K and at distance n−2 from L_{1} (L_{2}). Hence, denoting by R_{i} the set of lines at distance n from K and n − 2 from L_{i}, i = 1, 2, |S| = d|R_{i}|. But |R_{1}| − |R_{2}| = 1, since the point w is incident with k + 1 lines, but every point at distance n − 3 from L_{2} is incident with k lines. This is a contradiction, showing (ii) . □

3.6 Short forgetful quadrangles

3.6.1 General properties

Let Γ be a finite short forgetful quadrangle, with parameters (g, k, d). Recall that an isolated point of Γ lies at distance 3 from any class, and that, if two classes K_{1} and K_{2} lie at distance 4, then any point of K_{1} (K_{2}) lies at distance 3 from K_{2} (K_{1}).

Lemma 3.6.1 Every line of Γ contains a constant number ρ of isolated points.

Proof. Suppose there is a line L of Γ only containing isolated points. Let K be any class. Then the map

\begin{aligned}σ : & P_{L} & → & K \\ & x & → & y, & with δ(x, y) = 2, \end{aligned}

is a bijection between the points on L and the points of K, so l = g, a contradiction with l ≥ g + 2. Hence every line intersects at least one class.

Let K be a fixed class of size g, and L_{1} (L_{2}) a line intersecting K and containing ρ_{1} (ρ_{2}) isolated points. Each point of Γ at distance 5 from K lies at distance 3 from L_{1} and L_{2}, and each line at distance 4 from K contains exactly d points at distance 5 from K. So counting the number of points of Γ at distance 5 from K, we obtain

ρ_{1}kd + (g + d − ρ_{1} − 1)(k − 1)d = ρ_{2}kd + (g + d − ρ_{2} − 1)(k − 1)d,

hence ρ_{1} = ρ_{2}. Now let K_{1} (K_{2}) be a class of size g such that every line intersecting K_{1} (K_{2}) contains exactly ρ_{1} (ρ_{2}) isolated points. We count the number of points of Γ:

\begin{aligned}|P| & = & g + gk(g + d − 1) + ρ_{1}kd + (g + d − ρ_{1} − 1)(k − 1)d \\ & = & g + gk(g + d − 1) + ρ_{2}kd + (g + d − ρ_{2} − 1)(k − 1)d, \end{aligned}

implying ρ_{1} = ρ_{2}. This shows the lemma. □

Lemma 3.6.2 (i) Either ρ = 0 or ρ = g − (d − 1)(k − 1).

(ii) If ρ ≠ 0, then |L| = gk(k+1). If ρ = 0, then |L| = k((d−1)(k−1)+gk).

Proof. Let I be the set of isolated points. If K is a class of size g, then every isolated point lies at distance 3 from K, hence |I| = gkρ. Also, every line of Γ intersects K, or lies at distance 3 from a fixed point of K, hence

|L| = gk + k(ρk + (g + d − ρ − 1)(k − 1)).

Counting the number of pairs (i, L), i ∈ I, L ∈ L, iIL, we obtain:

gkρ(k + 1) = (gk + k(ρk + (g + d − ρ − 1)(k − 1))ρ.

If ρ ≠ 0, this simplifies to ρ = g − (d − 1)(k − 1), showing (i). Now by using this in the expression for |L| above, we obtain (ii). □

Define the following graph G_{Γ}. The vertices of G_{Γ} are the classes of Γ. Two vertices are adjacent if and only if the corresponding classes lie at distance 6.

Lemma 3.6.3 (i) If ρ ≠ 0, then G_{Γ} is a

srg((k + 1)(kd + 1 − k), kd, k − 1, d))
and G_{Γ}^{C} is a

srg((k + 1)(kd + 1 − k), k^{2}(d − 1), (d − 1)(k^{2} + 1) − dk, k(k − 1)(d − 1)).

(ii) If ρ = 0, then, with f = \frac{d(d−1)(k−1)}{g}, G_{Γ} is a srg(1 + k(g + d − 1) + (k − 1)d + f, (k − 1)d + f, k − d − 1 + f, f) and G_{Γ}^{C} is a

srg(1 + k(g + d − 1) + (k − 1)d + f, k(g + d − 1), d − 1 + k(g − 1), kg).

Proof. We determine the parameters of G_{Γ}^{C}. The number of classes follows from |P| = g+gk(g+d−1)+(g+d−1−ρ)(k−1)d+ρkd and |I| = gkρ, with I the set of isolated points. Now let K be a fixed class, and r ∈ K. A class K′ lies at distance 4 from K if and only if K′ contains a point collinear with r, hence there are k(g + d − 1 − ρ) classes lying at distance 4 from K. Let K′ be a fixed class, with δ(K, K′) = 4, and (K, r, R, r′, K′) a 4-path between K and K′. A class K′′ lies at distance 4 from both K and K′ if and only if K′′ intersects R (not in r or r′ of course), or K′′ intersects a line L through r, L ≠ R, in a point v (v ≠ r) that lies at distance 3 from K′. Note that every isolated point on such a line L necessarily lies at distance 3 from K′. Hence there are g + d − 2 − ρ + (k − 1)(g − 1 − ρ) classes at distance 4 from K and K′. Let finally \overline{K} be a class at distance 6 from K. A class K′′ lies at distance 4 from both K and \overline{K} if and only if there exists a line N through r such that K′′ intersects N in a point at distance 3 from \overline{K}. Since any line N through r contains exactly g − ρ non-isolated points at distance 3 from \overline{K}, we have in total k(g − ρ) classes at distance 4 from K and \overline{K}. □

If ρ ≠ 0, then g|d(d − 1)k is a necessary condition for the existence of Γ. Also, the fact that the multiplicities of the eigenvalues of G_{Γ} are integers, gives necessary conditions on (g, k, d) for the existence of G_{Γ}, and hence for Γ itself.

? △ Is there is similar structure (a distance-regular graph for example) on the classes of a short forgetful hexagon without isolated points ?

Let Γ be a short forgetful quadrangle without isolated points. For a point x of Γ, we denote by K_{x} the class containing x. We define the following relations R = (R_{0}, R_{1}, R_{2}, R_{3}, R_{4}) on P.

R_{0} = {(x, x)|x ∈ P}
R_{1} = {(x, y) ∈ P^{2}|x ⊥ y}
R_{2} = {(x, y) ∈ P^{2}|δ(x, y) = 4 and δ(K_{x}, K_{y}) = 4}
R_{3} = {(x, y) ∈ P^{2}|δ(x, y) = 4 and δ(K_{x}, K_{y}) = 6}

Lemma 3.6.4 The pair (P, R) is an association scheme.

Proof. We have to prove that the intersection numbers p_{jk}^{i} are defined. These numbers are easily determined using Lemma 3.6.3. (We only mention the non-zero intersection numbers for which j ≤ k).

p_{00}^{0} = 1, p_{11}^{0} = k(g + d − 1), p_{22}^{0} = g − 1, p_{33}^{0} = k(g − 1)(g + d − 1), p_{44}^{0} = (k − 1)d(g + d − 1), p_{01}^{1} = 1, p_{11}^{1} = g + d − 2, p_{13}^{1} = (k − 1)(g − 1), p_{14}^{1} = d(k −1), p_{23}^{1} = g −1, p_{33}^{1} = (g −1)(gk −2k +d), p_{34}^{1} = d(g −1)(k −1), p_{44}^{1} = d(d − 1)(k − 1), p_{02}^{2} = 1, p_{13}^{2} = k(g + d − 1), p_{22}^{2} = g − 2, p_{33}^{2} = k(g − 2)(g + d − 1), p_{44}^{2} = (k − 1)(d^{2} + dg − d), p_{03}^{3} = 1, p_{11}^{3} = k − 1, p_{12}^{3} = 1, p_{13}^{3} = (k − 1)(g − 2) + g + d − 2, p_{14}^{3} = d(k − 1), p_{23}^{3} = g − 2, p_{33}^{3} = (g − 2)(gk − 2k + d) + (g − 1)(k − 1), p_{34}^{3} = d(g − 1)(k − 1), p_{44}^{3} = d(d − 1)(k − 1), p_{04}^{4} = 1, p_{11}^{4} = k, p_{13}^{4} = k(g −1), p_{14}^{4} = k(d−1), p_{24}^{4} = g −1, p_{33}^{4} = k(g −1)^{2}, p_{34}^{4} = k(d − 1)(g − 1), p_{44}^{4} = g(k − d − 1) + d(d − 1)(k − 1). □

The intersection matrix L_{3} of the association scheme (P, R) has five distinct eigenvalues. The corresponding normalized eigenvectors u_{0}, u_{1}, …, u_{4} are:

u_{0} = (1, 1, 1, 1, 1)

u_{1} = (1, \frac{1}{k}, \frac{−1}{g − 1}, \frac{−1}{k(g − 1)}, 0)

u_{2} = (1, \frac{−1}{g + d − 1}, 1, \frac{−1}{g + d − 1}, \frac{g}{d(g + d − 1)})

u_{3} = (1, \frac{−1}{g + d − 1}, \frac{−1}{g − 1}, \frac{1}{(g − 1)(g + d − 1)}, 0)

u_{4} = (1, \frac{d − 1}{k(g + d − 1)}, 1, \frac{d − 1}{k(g + d − 1)}, \frac{−g}{(k − 1)(g + d − 1)}).

Theorem 3.6.5 If Γ is a short forgetful quadrangle without isolated points, then k ≤ (l − 1)^{2}.

Proof. This follows from the Krein condition q_{332} ≥ 0, with q_{332} = ∑_{l} p_{ll}^{0}(u_{2})_{l}(u_{3})_{l}^{2}. □

The multiplicities of the association scheme (X, R) are the following: f_{0} = 1

f_{1} = \frac{(g + d)(kg + (d − 1)(k − 1))k(g − 1)}{g(k + g + d − 1)}

f_{2} = \frac{d(g + d − 1)(kg + (d − 1)(k − 1))}{g(d + k − 1)}

f_{3} = \frac{(g − 1)(g + d)(g + d − 1)(kg + (d − 1)(k − 1))}{g(g + d + k − 1)}

f_{4} = \frac{(g + d)k(k − 1)(g + d − 1)}{g(k + d − 1)}.

The fact that these multiplicities are integers, gives additional necessary conditions on (g, k, d) for the existence of Γ.

3.6.2 Examples of short forgetful quadrangles

Subquadrangle type Let Δ be a finite generalized quadrangle of order (s, t), having an ideal (possibly thin) subquadrangle Δ′ of order (s′, t), s′ ≥ 1. Then we define the following geometry Γ = (P, L, I, ∼). The points of Γ are the points of Δ not contained in the subquadrangle Δ′. The lines of Γ are the lines of Δ that do not intersect Δ′. Incidence is the incidence of Γ. Two points of Γ are equivalent if and only if they are on a line of Δ′. So the equivalence classes correspond to the sets Δ_{1}(L) \ Δ′_{1}(L), with L a line of Δ that is also a line of the subquadrangle Δ′. It is now easy to see that Γ is a short forgetful quadrangle, with parameters g = s−s′, k = t, d = s′ +1 and ρ = s−s′t (the value of ρ follows from the proof of Theorem 1.2.5(ii)). Note that in this example, G_{Γ} corresponds with the line graph of Δ′. If a short forgetful quadrangle Γ arises from this construction, we say that Γ is of subquadrangle type.

One has the following examples of this construction in the classical case.

Δ Δ′ Γ(g, k, d)
W(q)  dual grid (q − 1, q, 2), ρ = 0
H(3, q^{2}) dual grid (q^{2} − 1, q, 2), ρ = q^{2} − q
H(3, q^{2}) W(q)  (q^{2} − q, q, q + 1), ρ = 0
H(4, q^{2})^{D} H(3, q^{2})^{D} (q^{3} − q, q^{2}, q + 1), ρ = 0

Ovoid type Let Δ be a finite generalized quadrangle of order (s, t), admitting a regular ovoid O. Then we define the following geometry Γ = (P, L, I, ∼). The points of Γ are the points of Δ not belonging to O. The lines of Γ are the lines of Δ. Two points of Γ are equivalent if and only if they are contained in a set {o_{1}, o_{2}}^{⊥}, for o_{1}, o_{2} ∈ O. Incidence is the incidence of Δ. Then Γ is a forgetful quadrangle with parameters g = k = t + 1, d = s − t − 1 and ρ = 0. Indeed, we check that for a non-incident point-line pair (p, L) of Γ, there exists a unique point rIL, r collinear or equivalent with p. Note that there exists a path [p, L] = (p, L′, p′, L) in the generalized quadrangle Δ. If p′ ∈ O, then p′ is not collinear in Δ with any point of L. Let in this case o be a point of O collinear with p, o ≠ p′. The point r = proj_{L}o is then the unique point on L equivalent with p, since {p, r} ⊆ {o, p′}^{⊥}. If p′ ∉ O, then p′ is the unique point on L collinear in Δ with p. If p would be equivalent with a point p′′IL, the point p′ would be a point of the ovoid O, a contradiction. This shows that Γ is a forgetful quadrangle. If a short forgetful quadrangle Γ arises from this construction, we say that Γ is of ovoid type.

Note that a forgetful quadrangle of ovoid type has the following property:

(O) If there is a line intersecting three different (possibly trivial) classes K, K′ and K′′, then any line intersecting two of these classes, also intersects the third one.

The known regular ovoids giving rise to a short forgetful quadrangle (thus with d ≥ 2) are the following.

• Let Δ be the (q + 1) × (q + 1)-grid (so a thin generalized quadrangle of order (q, 1)), q ≥ 4, and O the points on one of the diagonals. Then the associated short forgetful quadrangle has parameters g = k = 2 and l = q. The graph G_{Γ}^{C} is the triangular graph T(q + 1).

• Let Δ be the generalized quadrangle H(3, q^{2}), and O the points of a Hermitian curve H lying on H(3, q^{2}). Then the associated short forgetful quadrangle has parameters g = k = q + 1 and l = q^{2}. The vertices of the graph G_{Γ} correspond to the lines intersecting H in q + 1 points, and two vertices of G_{Γ} are adjacent if the corresponding lines intersect in a point not belonging to H.

Application. The other known examples of regular ovoids all occur in generalized quadrangles of order (q + 1, q − 1) (see for instance De Bruyn [16] section 2.6.2). By applying the above construction on a generalized quadrangle Δ of order (q + 1, q − 1) admitting a regular ovoid O, one obtains a forgetful quadrangle Γ with g = k = q and d = 1. By Lemma 3.5.8, Γ then arises from a generalized quadrangle Δ′ by applying construction (II), with |X_{1}| = s + 1. Clearly, Δ′ has order (q, q). Since property (O) holds, the line L of Δ′ corresponding to the set X_{1} has to be regular. One can now easily see that the quadrangle Δ arises by applying the construction of Payne on the generalized quadrangle Δ′ with L as regular line. The regular ovoid O corresponds with the points of type (B) (using the notation of section 1.4.4). So any generalized quadrangle of order (q+1, q−1) having a regular ovoid O, arises from the construction of Payne. This result can be found in Payne [33], section 3.

3.6.3 Characterization results

Lemma 3.6.6 Let Γ be a short forgetful quadrangle with k ≤ g and satisfying property (O). Then Γ is of ovoid type.

Proof. Let Γ be a short forgetful quadrangle satisfying the conditions of the lemma. We first claim that Γ does not contain isolated points. Indeed, suppose first 0 < ρ < l − 1 and let w be an isolated point. Let L be a line through w. Then there are at least two non-trivial classes K and K′ intersecting L. Since K and K′ lie at distance 4, there is a line N different from L intersecting K and K′, hence, by property (O), N is incident with w. Now a ‘triangle’ arises, the contradiction. Suppose ρ ≥ l − 1. Let L be a line at distance 4 from a non-trivial class K. Note that l − 1 ≥ g + 1. Now by Lemma 3.5.5, every isolated point of L lies at distance 3 from K, implying |K| ≥ g + 1, again a contradiction. Hence ρ = 0.

We now prove that g = k. Let L_{1} be a line of Γ and K_{1}, …, K_{l} the classes intersecting L_{1}. Then by Lemma 3.5.4(i) there exists a set of lines S = {L_{1}, …, L_{g}} such that every line of S intersects K_{1}, …, K_{l}. Since every line of Γ intersects at most one line of S (otherwise a ‘triangle’ would arise), there are (g + d)g(k − 1) lines intersecting a line of S. Hence g + (g + d)g(k − 1) ≤ |L| = k((d − 1)(k − 1) + gk). Using k ≠ 1, this simplifies to

(g − k)(g + d − 1) ≤ 0,

hence g ≤ k. Since also k ≤ g, we obtain g = k. Note that in particular, this implies that every line of Γ intersects exactly one line of S.

Define the following equivalence relation on the set of lines of Γ. Two lines L_{1} and L_{2} of Γ are equivalent if and only if there exist at least two classes intersecting both L_{1} and L_{2} (the fact that this is an equivalence relation immediately follows from property (O)). To each equivalence class C_{i} of lines of Γ, we associate a symbol ∞_{i}, i = 1, …, r. Then define the following geometry Δ = (P, L, I). A point of Δ is either a point of Γ or a symbol ∞_{i}. The lines of Δ are the lines of Γ. Incidence is the incidence of Γ if defined, and symmetrized containment otherwise. Then Δ is a generalized quadrangle of order (l, k−1). Indeed, we check the main axiom for a non-incident point-line pair (p, L) of Δ.

(i) p a point of Γ. Either p is collinear in Γ (and hence in Δ) with a point p′ of L, or p is equivalent with a point p′ of L. In the former case, no line through p can be equivalent with L (otherwise p would be equivalent with a point of L), so p′ is the unique point of Δ on L that is collinear in Δ with p. In the latter case, let p′′ be a point on L, different from p′. The point p lies at distance 3 from the class K containing p′′, hence the line R through p intersecting K is equivalent with the line L. Now the symbol ∞_{i} corresponding to the equivalence class containing R and L is the unique point of Δ collinear with p and incident in Δ with L.

(ii) p = ∞_{i}. Let C_{i} be the equivalence class of lines corresponding to ∞_{i}. By the second paragraph of the proof, every line not belonging to C_{i} intersects a unique line of C_{i}. This implies that the line L is concurrent in Δ with a unique line of Δ through the point ∞_{i}.

The points ∞_{i} form an ovoid O of Δ. Moreover, O is a regular ovoid. Indeed, let ∞_{i} and ∞_{j} be two different equivalence classes of lines. Let L and L′ be two lines of the class ∞_{i}. Then by the second paragraph of the proof, L (L′) meets a unique line M (M′) of the class ∞_{j} in a point r (r′). Note that M ≠ M′, since otherwise a ‘triangle’ would arise. Suppose that r and r′ are not equivalent. Let K be the class containing r. Since L ∼ L′, K intersects the line L′ in a point a, a ≠ r, and since M ∼ M′, K intersects M′ in a point b, r ≠ b ≠ r′. But now a ‘triangle’ arises through the points a, b and r′, a contradiction. Hence the lines of ∞_{i} and ∞_{j} meet in the set of points of a class K. From {∞_{i}, ∞_{j}}^{⊥} = K follows easily that the ovoid O is regular. Now clearly, Γ is of ovoid type. □

Theorem 3.6.7 Let Γ be a short forgetful quadrangle without isolated points, such that g = k and l ≥ (g − 1)^{2}. Then Γ is of ovoid type.

Proof. Let Γ be a short forgetful quadrangle without isolated points, such that g = k =: q + 1, and l = q^{2} + r, r ≥ 0. We show that Γ has property (O).

Suppose first g = 2. Let L be a line of Γ, and K_{i} = {r_{i}, r′_{i}}, i = 1, 2, 3, three different classes intersecting the line L in the point r_{i}. Since g = 2, r′_{1}r′_{2}, r′_{1}r′_{3} and r′_{2}r′_{3} are lines. This gives rise to a triangle, unless r′_{1}r′_{2}r′_{3} is a line. So in this case, property (O) is satisfied. From now on, we assume g ≥ 3. Let K be a fixed class, and put K = {a_{0}, …, a_{q}}. Let L be a fixed line through a_{0} and C = {K_{1}, …, K_{q^{2}+r−1}} the set of classes different from K intersecting L. Note that, since δ(K, K_{i}) = 4, i = 1, …, q^{2} + r − 1, every point of K is collinear with exactly one point of each class K_{i}, i = 1, …, q^{2} +r −1. Put b_{i}, i = 0, …, q the point of K_{1} collinear with the point a_{i} of K. We claim that if at least one class of C intersects a line concurrent with K, then at least q +r −1 classes of C intersect this line. Indeed, consider the class K_{1} and the line a_{1}b_{1}. Let V be the set of the q^{2} +r−2 points of K_{2}, …, K_{q^{2}+r−1} that are collinear with b_{1} (these points exist, since δ(K_{1}, K_{i}) = 4, i = 2, …, q^{2}+r−1). No point of V lies on a line a_{i}b_{i}, i = 0, 2, …, q, since these lines already contain a point that is equivalent with b_{1} (otherwise a ‘triangle’ would arise). Hence every point of V either lies on a_{1}b_{1}, or on a line through a_{i}, different from a_{i}b_{i}, with 2 ≤ i ≤ q. Since a line through a_{i}, 2 ≤ i ≤ q, different from a_{i}b_{i} can contain at most one point of V, the line a_{1}b_{1} contains at least q^{2} + r − 2 − q(q − 1) = q + r − 2 points of V. So at least q + r − 1 classes of C intersect the line a_{1}b_{1}. This shows the claim. Note also that, if the line a_{1}b_{1} contains exactly q + r − 2 points of V, then each line through a_{i}, 2 ≤ i ≤ q, different from a_{i}b_{i}, contains exactly one point of V (∗).

Suppose first that there is a line M concurrent with K that is intersected by exactly q + r − 1 classes of C. Without loss of generality, we can assume that M = a_{1}b_{1}. By (∗), this means that every line through a_{i}, i = 2, …, q, intersects at least one, and hence at least q + r − 1 classes of C. So in total, a point a_{i}, 2 ≤ i ≤ q, lies at distance 3 from at least (q + 1)(q + r − 1) classes of C. Hence q^{2} + r − 1 ≥ (q + r − 1)(q + 1), implying that r = 0 and that every line through a_{i}, 2 ≤ i ≤ q, intersects exactly q − 1 classes of C. By symmetry, also every line through a_{1} intersects exactly q − 1 classes of C. Let, without loss of generality, C′ = {K_{1}, …, K_{q−1}} be the set of q − 1 classes of C intersecting the line a_{1}b_{1}. The point b_{1} has to be collinear with a point p_{i} ∈ K_{i}, for q ≤ i ≤ q^{2} − 1 (note that these K_{i} are the classes of C \ C′). By (∗) each line through a_{i}, 2 ≤ i ≤ q, different from a_{i}b_{i} contains a point collinear with b_{1} and belonging to a class of C \ C′. Let C_{1} be the set of q − 1 classes of C intersecting the line a_{2}b_{2}, and C_{2} the set of q^{2} − q classes intersecting a_{2}b_{2}, different from K and not belonging to C_{1}. Since exactly q − 1 classes intersecting a_{2}b_{2} different from K also intersect the line a_{0}b_{0} (namely the classes of C_{1}), every line concurrent with K but not through a_{2} is intersected by exactly q − 1 classes of C_{1} ∪ C_{2} (this follows from the argument above, applied on C_{1} ∪ C_{2} instead of C). Now since K_{1} lies at distance 4 from every class of C_{2}, there is a set V′ = {v_{1}, …, v_{q^{2}−q}} of q^{2} − q points collinear with b_{1} and belonging to a class of C_{2}. These points cannot be collinear with a_{i}, i ≥ 2 (since the points collinear with b_{1} on lines through a_{i}, i ≥ 2, belong to classes of C, and C ∩ C_{2} = ∅). So all the points of V′ are collinear with a_{0} (and there are at most q such points), or lie on the line a_{1}b_{1}. So at least q^{2} − 2q + 1 classes of C_{1} ∪ C_{2} intersect the line a_{1}b_{1}. This implies q^{2} − 2q + 1 ≤ q − 1, hence q = 1 (then g = 2, a contradiction) or q = 2. But if q = 2, then l = 4, hence d = l − g = 1, a contradiction.

We may now assume that if at least one class of C intersects a line concurrent with K, then at least q + r classes of C intersect this line. For each point a_{i}, 2 ≤ i ≤ q, let A_{i} be the number of lines through a_{i} that do not intersect any class of C. If A_{i} = 0 for some i, then every line through a_{i} contains at least q + r points belonging to one of the q^{2} + r − 1 classes of C. This would imply that (q +1)(q +r) ≤ q^{2} +r −1, a contradiction. Hence M := min_{2≤i≤q}A_{i} ≠ 0. We next claim that every line through a_{i}, 2 ≤ i ≤ q, that intersects at least one class of C, intersects at least (q − 1)(M + 1) + r classes of C. Let V again be the set of the q^{2} + r − 2 points of K_{2}, …, K_{q^{2}+r−1} that are collinear with b_{1}. Now the number of points of V that do not lie on the line a_{1}b_{1} is at most (q − 1)(q − M), hence there are at least q^{2} + r − 1 − (q − 1)(q − M) = (q − 1)(M + 1) + r classes of C that intersect the line a_{1}b_{1}. This shows the claim. Now consider a point a_{j} of K, 2 ≤ j ≤ q, for which A_{j} = M. Then there are q + 1 − M lines through a_{j} such that each of these lines intersects at least (q − 1)(M + 1) + r classes of C, hence

(q + 1 − M)(qM + q + r − M − 1) ≤ q^{2} + r − 1,

implying

M ≤ \frac{−r}{q − 1} or M ≥ q.

Since 0 < M ≤ q, we have M = q, meaning that if a line concurrent with K intersects a class of C, it intersects every class of C. This is exactly property (O), so the conditions of Lemma 3.6.6 are satisfied, and Γ is of ovoid type.

□

Corollary 3.6.8 If Γ is a short forgetful quadrangle without isolated points, satisfying k = g ≠ 2, then l ≤ (g − 1)^{2}.

Proof. Suppose by way of contradiction that Γ is a short forgetful quadrangle without isolated points, with k = g =: q + 1 and l = (g − 1)^{2} + r = q^{2} + r, r ≥ 1. Then because of Theorem 3.6.7, Γ is of ovoid type, and there exists a generalized quadrangle of order (l, q), with l > q^{2}, a contradiction since q ≠ 1. □

Lemma 3.6.9 Let Γ be a short forgetful quadrangle for which G_{Γ} is the line graph of a generalized quadrangle. Then Γ is of subquadrangle type.

Proof. Let Γ be a short forgetful quadrangle for which G_{Γ} is the line graph of a generalized quadrangle Δ′ of order (s, t). Then one calculates (using Lemma 3.6.3) that s = d − 1, t = k and that, if ρ = 0, g = (k − 1)(d − 1). Each point of Δ′ corresponds to a (maximal) clique of size k + 1 in the graph G_{Γ}, so to k + 1 classes lying at distance 6 from each other. Also, every two classes at distance 6 from each other are contained in a unique clique of size k + 1, and every class belongs to exactly d (k + 1)-cliques. We denote the points of Δ′ by ∞_{i}, i = 1, …, v. Now define the following geometry Δ = (P, L, I). The points of Δ are the points of Γ and the symbols ∞_{i}. There are two types of lines of Δ. The lines of type (A) are the lines of Γ. A line of type (B) consists of the points of a class K of Γ, together with the symbols ∞_{1}, …, ∞_{d} of the (k + 1)-cliques containing K. Incidence is the incidence of Γ if defined, and symmetrized containment otherwise. Then Δ is a generalized quadrangle of order (l − 1, k). Indeed, we check the main axiom for a non-incident point-line pair (p, L) of Δ.

(i) p a point of Γ, L type (A). Immediate.

(ii) p a point of Γ, L type (B). Let K be the class of Γ such that all its points are incident with L. If p is isolated, then p is collinear with a unique point of K, and p is not incident with any line of type (B). Hence the path [p, L] exists and is unique in Δ. If p is non-isolated and is collinear in Γ with a (necessarily unique) point of K, then the class K′ containing p lies at distance 4 from K, hence K′ is not contained in any (k + 1)-clique containing K, so [p, L] is unique in Δ. If finally p is non-isolated and lies at distance 5 from K, then the class K′ containing p belongs to a unique (k + 1)-clique through K, so p is collinear in Δ with exactly one of the symbols ∞_{i} on L.

(iii) p = ∞_{j}, L type (A). There are exactly gk(k + 1) lines intersecting one of the k + 1 classes of the clique corresponding to ∞_{j}. Since this is equal to the number of lines of Γ, the path [p, L] exists and is unique in Δ.

Figure 3.6: The complement of the Shrikhande graph.

(iv) p = ∞_{j}, L type (B). This follows immediately from the fact that Δ′ is a generalized quadrangle.

Since Δ′ is an ideal subquadrangle of Δ, Γ is of subquadrangle type. □

Remark. Let Γ be a short forgetful quadrangle with parameters (g, k, d) such that, if ρ = 0, g = (k − 1)(d − 1). If every two adjacent vertices of G_{Γ} are contained in a clique of size k + 1, then G_{Γ} is the line graph of a generalized quadrangle (see Brouwer, Cohen & Neumaier [8] Lemma 1.15.1), hence Γ is of subquadrangle type by Lemma 3.6.9.

Theorem 3.6.10 Let Γ be a short forgetful generalized quadrangle with parameters (g, k, 2).

(i) If Γ contains isolated points, then Γ is of subquadrangle type.

(ii) If g = k − 1, then Γ is of subquadrangle type.

Proof. Let Γ be a short forgetful quadrangle with d = 2, and suppose that either Γ contains isolated points, or the parameters of Γ satisfy g = k − 1 (which implies that there are no isolated points because of Lemma 3.6.2(i)). Put l = s + 1, g = s − 1 and k = t. Then in both cases, G_{Γ} is a srg ((t + 1)^{2}, 2t, t − 1, 2).

Suppose first t ≠ 3. Then G_{Γ} is the (t + 1) × (t + 1)-grid (or equivalently, the line graph of a thin generalized quadrangle of order (1, t)) (see Bose [5], or De Clerck & Van Maldeghem [17], Theorem 4). By Lemma 3.6.9, Γ is of subquadrangle type, which proves the theorem in this case. Suppose now t = 3. Then G_{Γ} is a srg (16,6,2,2). Any strongly regular graph with these parameters is either the 4 × 4-grid or the Shrikhande graph (see for example Brouwer, Cohen & Neumaier [8], Theorem 3.12.4). In the former case, the theorem again follows as before, so assume that G_{Γ} is the Shrikhande graph. We show that this leads to a contradiction. For convenience, we work with the complementary graph G_{Γ}^{C}. Label the classes of G_{Γ}^{C} with K_{1}, …, K_{16} as in Figure 3.6. Remember that Γ is a short forgetful quadrangle with g = s − 1, l = s + 1, k = 3 and ρ = s − 3, so every line contains exactly 4 non-isolated points. Also, two vertices are adjacent in G_{Γ}^{C} if and only if the corresponding classes lie at distance 4 from each other. We now make the following observations.

(a) Every vertex v of G_{Γ}^{C} is contained in exactly 3 maximal cliques of size 4 (which only intersect in the vertex v). This implies the following property for Γ: if there exists a line of Γ intersecting the four classes K_{i}, K_{j}, K_{m} and K_{n}, then there exist exactly g = s−1 lines intersecting the classes K_{i}, K_{j}, K_{m} and K_{n}.

(b) If p is an isolated point of Γ, then p lies at distance 3 from every nontrivial class, so the point p will determine four cliques in G_{Γ}^{C}, each of size 4 (corresponding to the four lines through p). Hence p will determine a partition of the vertices of G_{Γ}^{C} into four disjoint maximal cliques.

We will call a line intersecting the four classes K_{i}, K_{j}, K_{m} and K_{n}, an (i, j, m, n)-line.

Case g = 2^{1}. Put K_{i} = {a_{i}, b_{i}}. Without loss of generality, we can assume a_{1} ⊥ a_{2} ⊥ a_{6} ⊥ a_{5} Because δ(K_{2}, K_{5}) = 4, the point a_{2} is collinear with b_{5}. This implies that b_{5} ⊥ b_{1} and a_{5} ⊥ a_{1}. We can argue similarly for the ‘squares’ A_{13}, A_{22}, A_{31} and A_{33} (labelling the ‘squares’ in figure 3.6 as the elements of a 3 × 3matrix). Hence we can choose the notation such that we obtain the lines a_{5}a_{6}a_{7}a_{8}, a_{9}a_{10}a_{11}a_{12}, a_{2}a_{6}a_{10}a_{14} and a_{3}a_{7}a_{11}a_{15}. Also, a_{3} ⊥ a_{4}, a_{13} ⊥ a_{14}, a_{15} ⊥ a_{16}, a_{4} ⊥ a_{8}, a_{9} ⊥ a_{13} and a_{12} ⊥ a_{16}. Since δ(K_{3}, K_{10}) = 4, the point a_{3} is collinear with b_{10}. This implies that a_{2} ⊥ a_{3}. But now there is no room

1We advise the patient reader to make a large picture similar to Figure 3.6, but with the vertices replaced by classes containing two points.

Figure 3.7: Case g = 4 in the proof of Theorem 3.6.10

any more for the (1,6,11,16)-line through a_{1} (indeed, a_{1} ⊥ b_{6}, a_{1} ⊥ b_{11}, hence b_{6} ⊥ b_{11}, a contradiction since δ(a_{6}, a_{11}) = 4).

Case g = 4. Note that through every point of K_{1}, there is a line of type (1, 2, 3, 4), (1, 5, 9, 13) and (1, 6, 11, 16). Put K_{1} = {x, y, z, v}. For w ∈ {x, y, z, v}, let L_{H}^{w} (L_{V}^{w}, L_{D}^{w}) be the (1,2,3,4)-line ((1,5,9,13)-line, (1,6,11,16)-line respectively) through w, and let b_{i}^{w} be the point of the class K_{i} that is collinear with w, for i ∈ {2, 3, 4, 5, 9, 13, 6, 11, 16}. The point b_{2}^{x} has to be collinear with a point b_{5}^{w} of K_{5} and a point b_{6}^{w′} of K_{6}, w, w′ ∈ {y, z, v}. Without loss of generality, we can assume w = y. We first show that also w′ = y. Suppose by way of contradiction that w′ ≠ y. The line b_{2}^{x}b_{6}^{w′} has to contain a point r collinear with y. But r I/L_{H}^{y} since b_{2}^{x} ∼ b_{2}^{y}, r I/L_{D}^{y} since b_{6}^{w′} ∼ b_{6}^{y} and r I/L_{V}^{y} since b_{2}^{x} ⊥ b_{5}^{y}. Since k = 3, there is no room for the point r, a contradiction. Hence we can assume b_{2}^{x} ⊥ b_{6}^{y}. Note that the projections of z and v on the line b_{2}^{x}b_{5}^{y} are both isolated, and are incident with the lines L_{D}^{z} and L_{D}^{v} respectively. Similarly the projections of z and v on the line b_{2}^{x}b_{6}^{y} are both isolated, and are incident with the lines L_{V}^{z} and L_{V}^{v} respectively. Now the point b_{3}^{x} is collinear with a point b_{11}^{w} of K_{11} and a point b_{13}^{w} of K_{13}, w ∈ {y, z, v}. We show that y ≠ w. Suppose by way of contradiction that w = y. Then as before, the projections of z (v) onto the lines b_{3}^{x}b_{11}^{y} and b_{3}^{x}b_{13}^{y} are isolated and are incident with L_{V}^{z} and L_{D}^{z} (L_{V}^{v} and L_{D}^{v}) respectively. The point b_{4}^{x} is collinear with a point b_{9}^{u} of K_{9}, and a point b_{16}^{u} of K_{16}, u ∈ {y, z, v}. The line b_{4}^{x}b_{9}^{u} has to contain an isolated point incident with one of the lines L_{D}^{z} or L_{D}^{v}, a contradiction since the four isolated points on these lines are already collinear with b_{2}^{x} or b_{3}^{x}. Hence we can assume without loss of generality that w = z (so b_{3}^{x} ⊥ b_{11}^{z} and b_{3}^{x} ⊥ b_{13}^{z}) and that b_{4}^{x} ⊥ b_{9}^{v} and b_{4}^{x} ⊥ b_{16}^{v}. (This is the situation drawn in Figure 3.7. When a point has the label i in this figure, it belongs to the class K_{i}.) Let p and r (p′ and r′) be the two isolated points on the line L_{H}^{x} (L_{H}^{y}). Since through both p and r, there is a line intersecting K_{5} and K_{6} (by observation (b)), and since p and r lie at distance 4 from the points b_{5}^{y} and b_{6}^{y}, we can assume that rr′b_{5}^{z}b_{6}^{v} and pp′b_{5}^{v}b_{6}^{z} are lines. But now one easily shows that there is no room any more for the line through r intersecting the classes K_{9} and K_{11}.

Case g ≠ 2, 4. Note that through every point of K_{1}, there is a line of type (1, 2, 3, 4), (1, 5, 9, 13) and (1, 6, 11, 16). Since g ≠ 2, the geometry Γ contains isolated points. Let a_{1} ∈ K_{1} and let r be an isolated point on the (1,2,3,4)-line L through a_{1}. Each line through r contains s − 3 isolated points, which necessarily lie at distance 3 from K_{1}, and hence each line through r different from L contains exactly 2 non-isolated points at distance 3 from K_{1}. So in total there are 6 non-isolated points collinear with r, at distance 3 from K_{1} and not incident with L. These 6 points belong to the classes K_{5}, K_{6}, K_{9}, K_{11}, K_{13} and K_{16} (since according to observation (b), the point r determines the partition (1, 2, 3, 4), (5, 6, 7, 8), (9, 10, 11, 12), (13, 14, 15, 16)). Hence we need at least 6 lines concurrent with K_{1}, not through a_{1} and not of type (1,2,3,4), implying that g ≥ 4. Now we claim that every line through r different from L intersects g − 3 lines of type (1, 2, 3, 4) different from L. Indeed, consider for example the (5,6,7,8)-line R through r. The points b_{5} and b_{6} of K_{5} and K_{6} on R lie at distance 3 from K_{1}, hence R has to intersect a (1,5,9,13)-line and a (1,6,11,16)-line concurrent with K_{1}. But R cannot intersect two (1,5,9,13)-lines concurrent with K_{1}, since this would give rise to a ‘triangle’. Similarly R cannot intersect two (1,6,11,16)-lines concurrent with R. Hence every line concurrent with both R and K_{1} and not through b_{5} or b_{6} is necessarily a (1,2,3,4)-line. This shows the claim. So each of the three lines through r different from L intersects g − 3 lines of type (1, 2, 3, 4) different from L. Since all these lines of type (1,2,3,4) at distance 3 from r are different, 3(g − 3) ≤ g − 1, implying that g = 4. This case was excluded before. □

Application 3.6.11 Let Γ be a short forgetful quadrangle with k = 2. If Γ contains isolated points, then Γ is of subquadrangle type. If Γ does not contain isolated points, then Γ is of ovoid type. In both cases, the corresponding generalized quadrangles are uniquely determined.

Proof. Let Γ be a short forgetful quadrangle with k = 2. Suppose first that Γ contains isolated points. Since every two adjacent vertices of G_{Γ} are contained in a clique of size k + 1 = 3 (see Lemma 3.6.3(i)), Γ is of subquadrangle type by Lemma 3.6.9. Using the notation of section 3.6.2, the associated generalized quadrangle Δ has order t = 2. Hence Δ is isomorphic with H(3, 4). Suppose now Γ does not contain isolated points. If g = k = 2, then Γ is of ovoid type, by Lemma 3.6.6. Again using the notation of section 3.6.2, we see that the associated generalized quadrangle Δ is an (l + 1) × (l + 1)-grid. We now show that the case g > 2 leads to a contradiction.

g = 3

Let G be a fixed class of size 3, put G = {r_{1}, r_{2}, r_{3}}, and let R_{i} and R′_{i} be the two lines through the point r_{i}, i = 1, 2, 3. Let a_{1} be a point on R_{1} different from r_{1}, and L the line through a_{1} different from R_{1}. Let A be the class containing a_{1}. Without loss of generality, we can assume that L intersects the lines R_{2} and R_{3} in a point of the classes B and C respectively. Put A = {a_{1}, a_{2}, a_{3}}, B = {b_{1}, b_{2}, b_{3}} and C = {c_{1}, c_{2}, c_{3}}, with x_{i} ⊥ r_{i}, i = 1, 2, 3 and x ∈ {a, b, c}. Note that b_{1}, c_{1} ∈ R′_{1}, a_{2}, c_{2} ∈ R′_{2} and b_{3}, a_{3} ∈ R′_{3}. Also, since δ(A, B) = δ(A, C) = δ(B, C) = 4, we obtain b_{1} ⊥ a_{2}, c_{1} ⊥ a_{3} and c_{2} ⊥ b_{3}. Let d_{1} be a point on R_{1} different from a_{1} and r_{1}. Then the class D = {d_{1}, d_{2}, d_{3}} cannot intersect R′_{2}. Indeed, if d_{2} ∈ R′_{2}, then d_{3} ∈ R′_{3} (using δ(A, D) = 4), but then δ(B, D) = 4, so b_{2} ⊥ d_{1} or b_{2} ⊥ d_{2}, a contradiction. Hence d_{2} ∈ R_{2} and d_{3} ∈ R_{3}. Since δ(B, D) = δ(C, D) = 4, we have d_{1}Ic_{2}b_{3}, hence d_{1} is determined on R_{1}, implying |R_{1}| = 3, a contradiction with l = g + d ≥ 5.

g = 4

Similarly to the previous case.

g ≥ 5

Let G = {r_{1}, r_{2}, …, r_{g}} be a class and L a line at distance 4 from G. For i = 1, …, g, put a_{i} = proj_{L}r_{i}, K_{i} the class containing a_{i}, R_{i} = a_{i}r_{i} and R′_{i} the line through r_{i} different from R_{i}. Clearly, K_{2} intersects the lines R′_{i}, i ≠ 2, K_{3} intersects the lines R′_{i}, i ≠ 3, and K_{4} intersects the lines R′_{i}, i ≠ 4. Since K_{3} and K_{4} both have a point on R′_{1}, δ(K_{3}, K_{4}) = 4. Let a′_{3} be the point of K_{3} on the line R′_{4}, and a′_{4} the point of K_{4} on the line R′_{3}. The line L and the lines R′_{i}, i ≠ 3, 4, intersect both K_{3} and K_{4}. So we already have g −1 lines intersecting K_{3} and K_{4}. Since none of these lines contains the points a′_{3} or a′_{4}, a′_{3} and a′_{4} have to be collinear. Put L′ = a′_{3}a′_{4}. Clearly, the line L′ intersects R_{1}, say in a point of a class B. Note that B ≠ K_{i}, i = 1, …, g, and that B does not intersect the line L. Since L′ intersects the lines R_{2} and R_{i}, i = 5, …, g, the class B intersects the lines R′_{2}, R_{3}, R_{4} and R′_{i}, i = 5, …, g. Now δ(B, K_{2}) = 4, since these classes have collinear points on the line R′_{5}. This is a contradiction, since the point a_{2} is not collinear with any point of B (indeed, the line L through a_{2} does not intersect R′_{i}, i = 1, …, g). □

3.6.4 Short forgetful quadrangles ‘arising’ from generalized quadrangles.

We say that a forgetful quadrangle Γ arises from a generalized quadrangle Δ if the points of Γ are points of Δ, the lines of Γ are (parts of) lines of Δ, and each class of Γ is a subset of the point set of a line of Δ. Note that all examples of forgetful quadrangles, except possibly from the short forgetful quadrangles of ovoid type, arise in this way.

Let Δ be a finite generalized quadrangle of order (s, t), and W a set of points in Δ satisfying the following conditions.

(W1) There exist constants g and l, g ≥ 2, g + 2 ≤ l ≤ s + 1, such that every line of Δ intersects W in either s + 1, s, s + 1 − g or s + 1 − l points (these lines are called respectively W -, T -, C-, and F -lines).

(W2) If a point w ∈ W lies on an F -line L, then every line through w different from L is a W -line.

(W3) There exists a constant k, 2 ≤ k ≤ t such that any point p not belonging to W either lies on exactly k F -lines and one C-line, or on exactly (k+1) F -lines and no C-line (so each point not contained in W lies on exactly (t − k) T -lines).

Then we construct a short forgetful quadrangle Γ = (P, L, I, ∼) in the following way. The points of Γ are the points of Δ not belonging to W. The lines of Γ are the F -lines. Incidence is the incidence of Δ. Two points of Γ are equivalent if they lie on a C-line. Clearly, Γ is a short forgetful quadrangle with parameters (g, k, l − g). Also, every short forgetful quadrangle arising from a generalized quadrangle can be constructed in this way. We give two examples of this construction.

1. Let Δ be the unique finite generalized quadrangle of order (4,2). Let L_{1}, L_{2} be two opposite lines of Δ, and S = {L′_{0}, …, L′_{4}} the five lines concurrent with both L_{1} and L_{2}. Let W be the set of points lying on L_{1} ∪ L_{2} ∪ L′_{0} ∪ … ∪ L′_{4}. Then every line of Δ not contained in W intersects W in 1 or 3 points. Indeed, a line concurrent with L_{1} or L_{2} different from L′_{i}, i = 0, …, 4 intersects W in 1 point. A line opposite both L_{1} and L_{2} and intersecting at least 2 lines of S, intersects exactly three lines of S, since every triad of lines in Δ has exactly 3 centers (see Theorem 1.2.5(i)). From this, it easily follows that every line opposite both L_{1} and L_{2} intersects W in exactly 3 points. Now it is readily seen that the set W satisfies properties (W1), (W2) and (W3). The corresponding short forgetful quadrangle Γ has k = 2 and does not contain isolated points, hence is of ovoid type and already classified in application 3.6.11 (Γ arises from the 5 × 5-grid).

2. Let Δ be a generalized quadrangle of order (q, q), and Δ′ a subquadrangle of Δ of order (1, q). Then Δ\Δ′ defines a short forgetful quadrangle Γ of subquadrangle type. Let Δ_{1} be a generalized quadrangle of order (q^{2}, q) containing Δ, and Δ_{2} a generalized quadrangle of order (q^{2}, q^{3}) containing Δ_{1}. Then the set W = (Δ_{2} \ Δ) ∪ Δ′ in Δ_{2} satisfies conditions (W1), (W2) and (W3): the C-lines are the lines of Δ intersecting Δ′, the F -lines are the lines of Δ not intersecting Δ′, the T -lines are the lines of Δ_{2} intersecting Δ \ Δ′ in a unique point. Note that also the set W′ = (Δ_{1} \ Δ) ∪ Δ′ satisfies properties (W1), (W2) and (W3) (and gives rise to the same forgetful quadrangle as the set W).

Classifying all short forgetful quadrangles arising from a generalized quadrangle boils down to classifying all sets W satisfying the conditions (W1), (W2) and (W3). The following results can easily be obtained:

• If l = s + 1 and no T -lines exist, then Γ is of subquadrangle type.

• If l = s, then k = 2 hence Γ is classified in 3.6.11.

Classifying all sets W seems to be quite difficult. Note that example 2 above shows that the set W can be ‘large’ compared with the short forgetful quadrangle obtained from it. Also, different sets W can give rise to the same short forgetful quadrangle. So restricting the classification of all short forgetful quadrangles to the ones arising from a generalized quadrangle does not seem to make the question easier.

? △ Do the short forgetful quadrangles of ovoid type (or at least the ‘classical examples’) arise from generalized quadrangles ?

Distance-preserving maps

4.1 Introduction

Any isomorphism between two generalized n-gons preserves all distances and conversely, every bijective map between two generalized n-gons preserving all distances, defines an isomorphism. The aim of the present chapter is to weaken that condition. The inspiration for this problem came from the the theorem of Beckman and Quarles (see for instance [3]) stating that a permutation of the point set of a Euclidean real space preserves distance i between points (for some positive real number i) if and only if it preserves all distances.

Let us be a bit more precise. In fact, we consider two versions of the problem. In the first version, we look at surjective maps between the point sets of two generalized n-gons if i is even, and between the point sets and the line sets of two generalized n-gons if i is odd, preserving distance i. With ‘preserving distance i’, we mean that two elements are at distance i if and only if their respective images are at distance i. The question now reads: does this map extend to an isomorphism? We show that the answer to the question is ‘yes’ if i is not equal to the maximal distance. In case i is the maximal distance, a counterexample arises for the split Cayley hexagon. Two natural problems turn up: do only hexagons give rise to counterexamples - and is the split Cayley hexagon the only hexagon giving rise to a counterexample? If one restricts to the finite case, we prove that counterexamples can only occur for hexagons of order (s, s). In the infinite case, if there is enough transitivity around, the only counterexample-hexagon is the one described for the classical hexagon. We use the arguments of these proofs in two applications. First, we determine the intersection of the line sets of two classical generalized hexagons living on the same quadric in 6-space. Secondly, we prove the (well-known) maximality of the group G_{2}(q) in O_{7}(q) in an entirely geometric way. In the second version of the problem, we consider maps between the flags of two generalized n-gons, and ask the same question. For this problem, the answer is ‘yes’ up to one counterexample arising in the smallest generalized quadrangle. A variation on this problem can be obtained by asking that a certain Coxeter distance between the flags is preserved. Here, the counterexample has to give up and we obtain a ‘yes’ in all cases. We conclude with some words about the proofs of the main theorems of this chapter. We are given a map preserving a certain distance i. In fact, this means that we look at our polygon with a pair of glasses that only allows us to see whether two elements lie at distance i or not. The aim is to find a property that distinguishes the collinear points among all the other pairs of points, and that one can see with such a pair of glasses. However, the quote ‘Mathematics is as love, the idea is easy but it can become difficult’ really applies to this chapter. Indeed, the proofs are quite technical, since they are for general n. We therefore advise the reader to keep an appropriate n in mind when going through them (and to make a lot of pictures...). To make some arguments a bit more explicit, a few specific examples are included before the proofs. The results of this chapter are contained in Govaert & Van Maldeghem [28] and [29].

4.2 Main theorems and some words about the proof

Theorem 4.2.1 (Point-Line Theorem)

• Let Γ and Γ′ be two generalized n-gons, n ≥ 4, let i be an even integer satisfying 1 ≤ i ≤ n − 1, and let α be a surjective map from the point set of Γ onto the point set of Γ′. If for every two points a, b of Γ, we have δ(a, b) = i if and only if δ(a^{α}, b^{α}) = i, then α extends (in a unique way) to an isomorphism from Γ to Γ′.

• Let Γ and Γ′ be two generalized n-gons, n ≥ 2, let i be an odd integer satisfying 1 ≤ i ≤ n − 1, and let α be a surjective map from the point set of Γ onto the point set of Γ′, and from the line set of Γ onto the line set of Γ′. If for every point-line pair {a, b} of Γ, we have δ(a, b) = i if and only if δ(a^{α}, b^{α}) = i, then α defines an isomorphism from Γ to Γ′.

Theorem 4.2.2 (Flag Theorem) Let Δ and Δ′ be two generalized m-gons, m ≥ 2, let r be an integer satisfying 1 ≤ r ≤ m, and let α be a surjective map from the set of flags of Δ onto the set of flags of Δ′. If for every two flags f, g of Δ, we have δ(f, g) = r if and only if δ(f^{α}, g^{α}) = r, then α extends to an (anti)isomorphism from Δ to Δ′, except possibly when Δ and Δ′ are both isomorphic to the unique generalized quadrangle of order (2, 2) and r = 3.

Theorem 4.2.3 (Special Flag Theorem) Let (W, S) be the Coxeter system associated with the dihedral group W = D_{2m} of order 2m. Let Δ and Δ′ be two generalized m-gons, m ≥ 2, let w be a non-trivial element of W \ S, and let α be a surjective map from the set of flags of Δ onto the set of flags of Δ′. Denote by δ^{∗} the Coxeter distance between flags in both Δ and Δ′. If for every two flags f, g of Δ, we have δ^{∗}(f, g) = w if and only if δ^{∗}(f^{α}, g^{α}) = w, then α extends to an (anti)isomorphism from Δ to Δ′. If moreover, the length of w is not maximal in W, then α extends to an isomorphism from Δ to Δ′.

If i = 1 in Theorem 4.2.1, then the result is obvious from the definition of isomorphism. The case i = 2 is exactly Lemma 1.3.14 in Van Maldeghem [57].

The case r = 1 in Theorem 4.2.2 can be found in Tits [50], Theorem 3.21, and the case r = m was proved in Abramenko & Van Maldeghem [1], Corollary 5.2 (in the latter paper, the case r = m is proved for spherical and twin buildings). Also, Abramenko & Van Maldeghem recently proved the analogue of Theorem 4.2.3 for buildings.

Point-Line Theorem

We first show that the map α from Γ to Γ′ is necessarily a bijection. Let a and b be two points of Γ for which a^{α} = b^{α}, and suppose α preserves distance i (note that we can assume i ≥ 3). Since any element of Γ′ at distance i from a^{α} is also at distance i from b^{α}, the set of elements of Γ at distance i from a coincides with the set of elements at distance i from b. This is easily seen to be a contradiction. Indeed, let γ be a fixed minimal path joining a and b. If δ(a, b) ≠ 2i, let c be an element at distance i from a such that c belongs to γ if i ≤ δ(a, b), or such that the path [a, c] contains γ if i > δ(a, b).

Then δ(b, c) ≠ i. If k = 2i, let m be the element of γ at distance i from a, m′ = proj_{m}a, and c an element incident with m′, m ≠ c ≠ proj_{m′}a. Then also δ(a, c) = i but δ(b, c) ≠ i.

Now clearly, it is enough to prove that α preserves collinearity. The result will then follow from Van Maldeghem [57] Lemma 1.3.14.

This will be done as follows. If α preserves a certain distance i, then we look for a property P that characterizes distance 2 in terms of distance i and ‘not distance i’. So two points a and b of Γ lie at distance 2 if and only if P (a, b) is satisfied. If the same characterization of collinearity holds in the polygon Γ′, then we are sure that also the images a^{α} and b^{α} lie at distance 2. Hence α preserves collinearity, and we are done.

Flag Theorem

Let Γ and Γ′ be the doubles of Δ and Δ′, respectively. Put n = 2m. Then Γ and Γ′ are generalized n-gons, n ≥ 6, with thin points (corresponding to the flags of Γ and Γ′) and with thick lines (corresponding to the points an lines of Γ and Γ′). Put 2r = i. The map α induces a surjective map (which we may also denote by α) from the point set of Γ to the point set of Γ′ preserving distance i. As in the previous case, one shows that α is a bijection. So in section 4.4.6, we in fact prove the following:

If α is a bijection from the point set of Γ to the point set of Γ′ such that for every two points a, b of Γ, we have δ(a, b) = i if and only if δ(a^{α}, b^{α}) = i, then α preserves collinearity, except possibly when Γ and Γ′ are isomorphic to the unique generalized octagon of order (2, 1) and i = 6.

The result will then follow from Theorem 3.21 in Tits [50].

In this way, we reduced Theorem 4.2.2 to a particular case of Theorem 4.2.1 for weak polygons with thin points and thick lines. We will not gain so much by doing that, because a separate proof remains necessary. But the intuition is easier.

Special Flag Theorem

As above, we again consider the doubles Γ, Γ′ and the associated map α from Γ to Γ′. Let i′ be the length of the element w of W, and i = 2i′ (note that i′ ≠ 1 since w ∉ S). If i′ is even, then two flags of Δ lie at Coxeter distance w if and only if they lie at distance i′, hence in this case, we are back to the situation of Theorem 4.2.2. Also if i′ = m, it is clear that Theorem 4.2.3 adds nothing new. Suppose i′ is odd, i′ ≠ m. Then, for two points a and b lying at distance i in Γ, either both proj_{a}b and proj_{b}a correspond to points of Δ, or both proj_{a}b and proj_{b}a correspond to lines of Δ. In the former case,

Figure 4.1: Example 1

we say that δ(a, b) = i_{p}, in the latter that δ(a, b) = i_{L}. So we know that α preserves either distance i_{p} or distance i_{L}. Without loss of generality, we assume that α preserves distance i_{p}. So in section 4.7.4, we will in fact prove the following result:

If α is a bijection from the point set of Γ to the point set of Γ′ such that for every two points a, b of Γ, we have δ(a, b) = i_{p} if and only if δ(a^{α}, b^{α}) = i_{p}, 2 < i < n, i ≡ 2 mod 4 , then α preserves collinearity.

From Theorem 3.21 in Tits [50] then follows that Δ and Δ′ are (anti)isomorphic. But for 2 < i < n, i ≡ 2 mod 4, the fact that distance i_{p} is preserved, contradicts Δ and Δ′ being anti-isomorphic. Hence in this case, Δ and Δ′ are necessarily isomorphic.

4.3 Some examples by pictures

Example 1: Point-Line Theorem, n = 9, i = 5 For two points a and b of Γ, we first consider the set of lines T_{a,b} = Γ_{5}(a) ∩ Γ_{5}(b). Put k = δ(a, b) and m = a⋈b. From Figure 4.1, one sees that for any line L of T_{a,b}, the paths [a, L] and [b, L] both contain m, except if k = 8.

Now let S be the set of pairs of points (a, b) for which there exists a point c different from a and b satisfying T_{a,b} ⊆ Γ_{5}(c). If δ(a, b) = 2, then clearly such a point c cannot exist. If δ(a, b) ∈ {4, 6}, the possibilities for the point c are indicated in Figure 4.1. Note that these are the only possible positions for the point c. For example, if δ(a, b) = 4, it is easy to see that the point c has to lie at distance 3 from any line through m different from am or bm. Hence c has to be collinear with m and lies on am or bm. Similarly if δ(a, b) = 6. Finally if δ(a, b) = 8, the point c cannot exist. Indeed, c would have to lie at distance 5 every line through m different from the projections of a and b onto m. Since there are at least two such lines (noting that both s and t are infinite in this case), c lies at distance 4 from m and either proj_{m}a = proj_{m}c or proj_{m}b = proj_{m}c. But now it is easy to see that c cannot lie at distance 5 from the line X (see picture). So the set S contains exactly the pairs of points at mutual distance 4 or 6.

Next, let S′ be the set of pairs of points (a, c) for which there exists a point b satisfying (a, b) ∈ S and T_{a,b} ⊆ Γ_{5}(c) (so in fact, we collect all the pairs (a, c) and (b, c) indicated in Figure 4.1). Clearly, the set S′ contains exactly the pairs of points at mutual distance 2, 4 or 6. Now S′ \ S is exactly the set of pairs of collinear points. This example corresponds with Case 4.4.3 in the proof of Theorem 4.2.1.

Example 2: Point-Line Theorem, n = 9, i = 8 Let a and b be two points at distance k ≠ 8. We again consider the set T_{a,b} = Γ_{8}(a) ∩ Γ_{8}(b). For the cases k = 2, 4, all possible positions of a point belonging to T_{a,b} are indicated in Figure 4.2. For k = 6, a subset of T_{a,b} is indicated. Now we look for the pairs of points (c, c′), c and c′ distinct from a and b such that

T_{v,v′} ⊆ Γ_{8}(w) ∪ Γ_{8}(w′), whenever {v, v′, w, w′} = {a, b, c, c′}.

It is easy to see that if k = 2, 4, the possibilities (c_{1}, c′_{1}) and (c_{2}, c′_{2}) indicated on the picture indeed satisfy the condition above. In the proof of Theorem 4.2.1, Case 4.4.5, we will show that these are the only possibilities for the pair (c, c′) if k = 2, 4, and that if k = 6 and a pair (c, c′) with the above properties exists, the points c and c′ are necessarily as in the picture.

Now for two points a and b at distance 2,4 or 6, we collect all the points c and c′ as above in a set C_{a,b}. If k = 2 or 4, it is impossible to find a point x at distance 8 from all the points of C_{a,b} ∪ {a, b}. If k = 6, one can find such a point (see for instance the point x on the picture). So we can characterize ‘being at distance 2 or 4’. Now if δ(a, b) = 4, all points of C_{a,b} lie at mutual distance 2 or 4, which is not true for δ(a, b) = 2. Hence we can distinguish distance 2. This example corresponds with Case 4.4.5 in the proof of Theorem 4.2.1.

Figure 4.2: Example 2: the set T_{a,b}

Figure 4.3: Example 2: the set C_{a,b}

Example 3: Flag Theorem, n = 24, i = 16 We first look for the pairs (a, b), a, b points of Γ, for which the set T_{a,b} := Γ_{i}(a) ∩ Γ_{i}(b) is empty. By Figure 4.4, it is intuitively clear that this will be the case if and only if δ(a, b) ∈ {4, 8, 12}. Indeed, if δ(a, b) = k ≡ 2 mod 4, one can find a point c ∈ T_{a,b} at distance 16 − k/2 from a line at distance k/2 from both a and b. If δ(a, b) = k ≡ 0 mod 4 and c ∈ T_{a,b}, a circuit of length at most k + 2i (≥ 48) arises (noting that the paths [a, c] and [b, c] cannot meet in an element ‘in the middle’ of a and b, since points are thin), hence k ≥ 16. So we can characterize the set of distances κ = {4, 8, 12}.

Next, for two points a and b lying at a distance contained in κ, we consider the set R_{a,b} = Γ_{κ}(a) ∩ Γ_{κ}(b). For two points at distance 12, which is the biggest distance in κ, a point of R_{a,b} will necessarily lie on the path [a, b], while this is not the case for smaller κ-distances (see Figure 4.5). We will then characterize distance 12 as the κ-distance for which |R_{a,b}| is minimal.

Finally, we show that two points a and b are opposite if and only if the set Γ_{12}(a) ∩ Γ_{12}(b) contains exactly two points c, d, and moreover, Γ_{12}(c) ∩ Γ_{12}(d) = {a, b}. Hence we recovered opposition, and the theorem follows from Abramenko & Van Maldeghem [1]. This example corresponds with Figure 4.4: Example 3: the set T_{a,b}

Figure 4.5: Example 3: the set R_{a,b}

cases 4.7.2 and 4.7.3 in the proof of Theorem 4.2.2.

4.4 Proof of the Point-Line Theorem

Throughout, we put T_{a,b} := Γ_{i}(a) ∩ Γ_{i}(b), for points a, b of Γ.

4.4.1 Case i < \frac{n}{2}

Let a and b be two points at distance k, and γ a fixed k-path joining a and b. Denote by m the element of γ at distance \frac{k}{2} from both a and b.

Claim 1. The set T_{a,b} is empty if and only if k > 2i. Proof. Suppose first δ(a, b) > 2i and let by way of contradiction x be an element of T_{a,b}. Note that x cannot lie on γ, nor do the paths [a, x] and [b, x] meet on γ. Hence there arises a circuit of length at most k + 2i < 2n (determined by γ and the paths [a, x] and [x, b]), a contradiction. Suppose now δ(a, b) ≤ 2i. Then any element x at distance i − \frac{k}{2} from m for which proj_{m}a ≠ proj_{m}x ≠ proj_{m}b belongs to T_{a,b} (note that such an element exists because Γ is thick). ◇

Claim 2. Suppose k = 2. Then there does not exist a point c, c ≠ a, b at distance i from every element of T_{a,b}. Proof. Suppose c is a point for which T_{a,b} ⊆ Γ_{i}(c). Let x ∈ T_{a,b}, and put [a, x] = (a = x_{0}, x_{1}, x_{2}, …, x_{i} = x). Note that necessarily x_{1} = ab and x_{2} ≠ b. Let y be an element incident with x_{i−1} different from x_{i−2} and x. Then y ∈ T_{a,b}. Since δ(c, x) = i = δ(c, y), necessarily δ(c, x_{i−1}) = i − 1. Proceeding like this we obtain δ(c, x_{j}) = j, for all j, hence c ∈ {a, b}. ◇

Claim 3. Suppose 2 < k ≤ 2i. Then there exists a point c, c ≠ a, b at distance i from every element of T_{a,b}. Proof. Let x ∈ T_{a,b} and define v as [a, v] = [a, b] ∩ [a, x]. Let j = δ(a, v). Combining [b, v] and [v, x], we obtain a path of length ℓ = k + i − 2j between b and x. If ℓ > n, then by combining this path with the i-path joining b and x, a circuit of length at most k + 2i − 2j < 2n arises, a contradiction. Hence ℓ ≤ n, so ℓ = δ(b, x) = i. This implies j = k/2, hence every element x of T_{a,b} lies at distance i − \frac{k}{2} from m, with proj_{m}a ≠ proj_{m}x ≠ proj_{m}b. Now clearly, a point c on the line N := proj_{a}b different from a and from the projection of b onto N satisfies T_{a,b} ⊆ Γ_{i}(c). ◇

Let S be the set of pairs of points (a, b) for which T_{a,b} ≠ ∅, and such that there does not exist a point c different from a and b satisfying T_{a,b} ⊆ Γ_{i}(c). Then by the claims above, S is the set of pairs of collinear points. This ends the proof of the case i < \frac{n}{2}.

4.4.2 Case i = \frac{n}{2} + 1, i even

Since the case n = 6 and i = 4 is considered in 4.4.4, we can assume n > 6. Let S be the set of pairs of distinct points (a, b) such that δ(a, b) ≠ i and the set T_{a,b} contains at least two points at distance i from each other. We claim that a pair (a, b) belongs to S if and only if δ(a, b) < i. Suppose first that 0 ≠ δ(a, b) = k, k < i and put m = a ⋈ b. Consider a point c at distance i − k/2 from m such that proj_{m}a ≠ w := proj_{m}c ≠ proj_{m}b (note that δ(c, w) = i − k/2 − 1 > 0). Let v be the element of [c, w] at distance i/2 from c (such an element exists since i/2 ≤ i − k/2 − 1). Consider a point c′ at distance i/2 from v such that proj_{v}m ≠ proj_{v}c′ ≠ proj_{v}c. The points c and c′ are both points of T_{a,b} and lie at distance i from each other.

Now let δ(a, b) = k > i and suppose by way of contradiction that c, c′ ∈ T_{a,b} with δ(c, c′) = i. If proj_{a}c ≠ proj_{a}c′, then we have a path of length 2i between c and c′ containing a. This implies that δ(c, c′) ≥ 2n−2i > i, a contradiction. Suppose now that proj_{a}c = proj_{a}c′. Define v as [a, c] ∩ [a, c′] = [a, v]. If we put δ(a, v) = j, then there is a path of length ℓ = 2i − 2j ≤ n between c and c′, hence ℓ = δ(c, c′). Now ℓ = i implies j = i/2, hence v = a ⋈ c = a ⋈ c′. Similarly, b ⋈ c = b ⋈ c′ =: v′. Now there arises a circuit of length at most 2i < 2n (determined by the paths [v, c], [c, v′], [v′, c′] and [c′, v]) unless v = v′. But this implies there arises a path between a and b of length at most i, the final contradiction. Our claim is proved.

Put κ = {2, …, i − 2}. Then (a, b) ∈ S if and only if δ(a, b) ∈ κ.

We claim that two distinct points a and b are collinear if and only if δ(a, b) ∈ κ and Γ_{κ}(a) ⊆ Γ_{κ}(b) ∪ Γ_{i}(b). Indeed, if δ(a, b) = 2 and x is a point at distance j ∈ κ from a, then δ(b, x) ∈ {j − 2, j, j + 2} ⊆ κ ∪ {i}. Now suppose δ(a, b) = k, 2 < k < i. Then k = i − j, 0 < j < i − 2. Consider a point c at distance j + 2 from a such that proj_{a}c ≠ proj_{a}b. Then δ(a, c) ∈ κ, but δ(b, c) = i + 2 ∉ κ ∪ {i}. This shows the claim. Hence we can distinguish distance 2 and so α preserves collinearity of points.

4.4.3 Case \frac{n}{2} < i < n − 2

Let a and b be points of Γ at distance k, and m an element at distance k/2 from both a and b. Note that T_{a,b} is never empty. For the case i = \frac{n}{2}+1, i even and k = n, we assume that |Γ_{1}(m)| ≥ 4 (so s ≥ 3). This can be done, since this case was in fact already handled without assumptions on the order in the previous paragraph^{1}. We first prove the following claims. For later purposes, we remark that these claims also hold for the case i ∈ {n − 2, n − 1, n}, with a similar proof^{2}.

Claim 1. Every element y of T_{a,b} lies at distance i−k/2 from m with proj_{m}a ≠ proj_{m}y ≠ proj_{m}b if and only if k < 2(n − i).

Proof. Suppose first that 2(n − i) ≤ k. Consider an apartment Σ through a and b containing m, and let m′ be the element of Σ opposite m. Then any element y at distance i − (n − \frac{k}{2}) from m′ with proj_{m′}a ≠ proj_{m′}y ≠ proj_{m′}b belongs to T_{a,b} and proj_{a}y ≠ proj_{a}m (hence y is not as in the claim above). Now suppose k < 2(n − i) and let y ∈ T_{a,b}. Let j be the length of the path [a, b] ∩ [a, y]. Then there is a path of length ℓ = k − 2j + i between b and y. If ℓ ≤ n, then ℓ = i and so necessarily j = k/2. Hence y is an element as claimed. If ℓ > n, then δ(b, y) ≥ 2n − ℓ > i, a contradiction. ◇

1The reason we want to include this case in the proof for \frac{n}{2} < i < n − 2 is that in this way, we can characterize distance 2 by the same property for all i, \frac{n}{2} < i < n − 2 if s, t ≥ 3. This will be useful in the proof of Application 4.4.1

2The only difference is that in the proof of Claim 2 for the case i = n − 1, one considers the set T′ ∪ {x ∈ T_{a,b} | δ(x, m) = i − k/2 + 2 and proj_{m}a ≠ proj_{m}x ≠ proj_{m}b}.

Claim 2. Suppose there exists a point c at distance i from every element of T_{a,b}. Then c lies at distance k/2 from m, and proj_{m}a = proj_{m}c or proj_{m}b = proj_{m}c.

Proof. Suppose c is a point at distance i from every element of T_{a,b}. Consider the set

T′ = {x ∈ T_{a,b} | δ(x, m) = i − k/2 and proj_{m}a ≠ proj_{m}x ≠ proj_{m}b}.

Then we may assume that T′ contains at least two elements y and y′ at distance 2 from each other. Indeed, this is clear if i− \frac{k}{2} > 1 or if |Γ_{1}(m)| ≥ 4. If i = \frac{k}{2} + 1, then either i = \frac{n+1}{2} (then necessarily n odd, hence s, t ≥ 3), or i = \frac{n}{2} + 1 (thus k = n). In the latter case, if |Γ_{1}(m)| = 3, i is odd by assumption (thus t = 2), and we can apply the dual reasoning. (This is, we consider exactly the same arguments as given here, but with a and b lines instead of points. Note that this is allowed since by the fact that i is odd, α is also defined on the line set - see also the last paragraph of 4.4.3.)

Put w = y ⋈y′. Then δ(c, w) = i − 1 (noting that i ≠ n − 1). Put γ = [c, w]. We show that γ contains m. Suppose by way of contradiction that this is not true. Define the element z as [w, c] ∩ [w, m] = [w, z]. Put γ′ = [z, c]. An element y′′ of T′ either lying on γ′ (if δ(c, m) ≥ i − k/2) or such that the path [y′′, m] contains γ′ (otherwise), clearly does not lie at distance i from c, a contradiction. So the point c has to lie at distance k/2 from m. But if proj_{m}a ≠ proj_{m}c ≠ proj_{m}b, then similarly we can find an element of T′ not at distance i from c. The assertion follows. ◇

Claim 3. Suppose k = 2 or k ≥ 2(n − i). Then there does not exist a point c, c ≠ a, b at distance i from every element of T_{a,b}.

Proof. If k = 2, then this follows immediately from Claim 2. So suppose k ≥ 2(n − i). We first show the following property for an element v of the path [a, proj_{m}a].

(*) There exists y ∈ T_{a,b} such that [a, y] ∩ [a, m] = [a, v] if and only if δ(a, v) ≤ j := i − n + k/2.

Indeed, let v be an element of the path [a, proj_{m}a], put j′ = δ(a, v). If there is an element y ∈ T_{a,b} such that [a, y] ∩ [a, m] = [a, v], then there arises a circuit of length at most k + 2i − 2j′, so necessarily, j′ ≤ j. Suppose now j′ ≤ j. Let Σ be an apartment through v and b containing m, but not the element proj_{v}a, and let m′ be the element of Σ at distance n + j′ − k/2 from b for which proj_{b}m′ ≠ proj_{b}m. Let finally y be an element at distance i − (n + j′ − k/2) from m′ such that proj_{m′}b ≠ proj_{m′}y ≠ proj_{m′}v. Then y is as in (*).

Now suppose c is a point at distance i from every element of T_{a,b}. We may assume that, if δ(m, c) ≠ n, then proj_{m}a = proj_{m}c. If δ(a, c) ≠ n, then we define the element z as [m, c] ∩ [m, a] = [m, z]; otherwise we define z as [proj_{m}a, c] ∩ [proj_{m}a, a] = [proj_{m}a, z]. Note that by Claim 2, δ(c, z) = δ(a, z) =: ℓ.

Suppose first ℓ > j, and consider an element y of T_{a,b} such that [a, y]∩[a, m] = [a, v], with δ(a, v) = j. Then we obtain a path of length d = i+2l−2j between y and c. Clearly, d ≠ i. But if d > n, δ(y, c) = i implies there arises a circuit of length at most 2(i + l − j) < 2n, a contradiction. Similarly, if ℓ ≤ j, then an element y ∈ T_{a,b} such that proj_{z}c ∈ [a, y], does not lie at distance i from c, the final contradiction. ◇

Now let S be the set of pairs of points (a, b) for which there exists a point c, a ≠ c ≠ b, such that T_{a,b} ⊆ Γ_{i}(c). Then a pair (a, b) with δ(a, b) = k belongs to S if and only if 2 < k < 2(n − i) (note that there are always even numbers k satisfying these inequalities because i < n−2). Indeed, if k = 2 or k ≥ 2(n−i), this is Claim 3. If 2 < k < 2(n−i) then by Claim 1 it suffices to consider a point c at distance k/2 from m with proj_{m}c ∈ {proj_{m}a, proj_{m}b}.

Define S′ = {(a, c) | ∃b ∈ P such that T_{a,b} ⊆ Γ_{i}(c)}. Suppose (a, c) ∈ S′, and b is such that T_{a,b} ⊆ Γ_{i}(c). Then (a, b) ∈ S and by Claim 2, δ(a, c) ≤ δ(a, b). Hence S′ is exactly the set of pairs of points at distance < 2(n − i) from each other. Now clearly S′ \ S is precisely the set of all pairs of collinear points.

So we found a property characterizing distance 2 between points, which is independent of the order of Γ except for the case i = \frac{n}{2} + 1 and i odd. Consider this case. There, if t ≥ 3, there was no problem, but if t = 2, the property we found was exactly the dual of the original one, and hence is valid for all orders of Γ except for s = 2. Now we still have to distinguish the cases s = 2 and t = 2 (indeed, it might be possible that the order of Γ is (2, t), while the order of Γ′ is (s′, 2)). This is done as follows. Define the set S′ as above. If s = 2, t > 2, hence S′ is exactly the set of pairs of collinear points. If t = 2, it is easy to see that all the pairs of collinear points are included in S′. Now consider the set S′′ of pairs (a, b) of S′ for which there exist at least two points x, x′ such that (a, x), (b, x), (a, x′) and (b, x′) belong to S′. If s = 2, then S′′ is empty. If t = 2, then s ≥ 3, hence S′′ is nonempty (considering points a, b, x, x′ on the same line). This distinguishes the cases s = 2 and t = 2. The theorem is now proved for this case.

4.4.4 Case i = n − 2

Case n = 6

Let C be the set of pairs of points (a, b), δ(a, b) ≠ 4, such that for every point y in T_{a,b}, there exists a point y′ in T_{a,b}, y′ ≠ y and δ(y, y′) ≠ 4, with the property that Γ_{4}(y) ∩ T_{a,b} = Γ_{4}(y′) ∩ T_{a,b} (**). We show that C is the set of pairs of collinear points.

If a and b are collinear points, a point y of T_{a,b} lies at distance 3 from ab and b ≠ proj_{ab}y := x ≠ a. Then any point y′ on yx, y′ ≠ x, y satisfies Γ_{4}(y) ∩ T_{a,b} = Γ_{4}(y′) ∩ T_{a,b}, hence (a, b) ∈ C.

Suppose now that (a, b) ∈ C with δ(a, b) = 6. We look for a contradiction. Put M = Γ_{3}(a) ∩ Γ_{3}(b). If x is a point of T_{a,b}, then either x lies on a line of M, or x is a point at distance 3 from a line A through a and from a line B through b, with A opposite B. Let y be a point of T_{a,b} on a line M of M, and y′ a point such that (**) is satisfied. Clearly, y′ cannot be incident with an element of M. Let A = proj_{a}y′ and B = proj_{b}y′. If A ≠ proj_{a}y and B ≠ proj_{b}y, then the point proj_{M}y′ lies at distance 4 from y′ but not from y, a contradiction. Suppose A = proj_{a}y. Let M′ be the line of M concurrent with B. Then the point proj_{M′}y lies at distance 4 from y but is opposite y′, a contradiction. Hence (a, b) ∉ C.

So we distinguished distance 2 and the theorem follows.

Case n ≠ 6

Step 1: the set S_{a,b} For two points a and b, we define

S_{a,b} = {x ∈ P | Γ_{n−2}(x) ∩ T_{a,b} = ∅}.

Note that by symmetry, x ∈ S_{a,b} implies a ∈ S_{b,x} and b ∈ S_{a,x}. We claim the following:

(i) If δ(a, b) = 2 and s ≥ 3, then S_{a,b} = (Γ_{2}(a) ∪ Γ_{2}(b)) \ Γ_{1}(ab). If δ(a, b) = 2 and s = 2, then S_{a,b} = (Γ_{2}(a) ∪ Γ_{2}(b)) \ {a, b}.

(ii) If δ(a, b) = 4, then {a⋈b} ⊆ S_{a,b} ⊆ {a⋈b} ∪ [Γ_{2}(a⋈b) ∩ Γ_{4}(a) ∩ Γ_{4}(b)]. If t ≥ 3, then S_{a,b} = {a⋈b}.

(iii) If k := δ(a, b) ∉ {2, 4, n}, then every x ∈ S_{a,b} lies at distance k/2 from a ⋈ b =: m with proj_{m}a ≠ proj_{m}x ≠ proj_{m}b. If moreover s > 2 and k ≡ 2 mod 4, or t > 2 and k ≡ 0 mod 4, then S_{a,b} = ∅.

(iv) If δ(a, b) = n, then let γ be an arbitrary path of length n joining a and b, let m be the middle element of γ and put v_{a} = proj_{m}a, v_{b} = proj_{m}b. Then

S_{a,b} ⊆ (Γ_{n/2}(m) ∩ Γ_{n/2+1}(v_{a}) ∩ Γ_{n/2+1}(v_{b}))
⋃(Γ_{n/2+1}(v_{a}) ∩ Γ_{n/2+2}(m) ∩ Γ_{n}(a))
⋃(Γ_{n/2+1}(v_{b}) ∩ Γ_{n/2+2}(m) ∩ Γ_{n}(b)).

If moreover s > 2 and n ≡ 2 mod 4, or t > 2 and n ≡ 0 mod 4, then

S_{a,b} ⊆ (Γ_{n/2+1}(v_{a}) ∩ Γ_{n/2+2}(m) ∩ Γ_{n}(a))
⋃(Γ_{n/2+1}(v_{b}) ∩ Γ_{n/2+2}(m) ∩ Γ_{n}(b)).

We prove these claims.

(i) Suppose δ(a, b) = 2. Since any point y of T_{a,b} lies at distance n−3 from the line ab, with proj_{ab}y ∉ {a, b}, it follows that every point collinear with a or b, not on the line ab, belongs to S_{a,b}. Also, if s = 2, then the unique point of ab different from a and b is an element of S_{a,b}. Let x be an arbitrary point in S_{a,b}. Put j = δ(x, a). If j = s = 2, then there is nothing to prove, so we may assume (j, s) ≠ (2, 2). Suppose first there exists a j-path γ between a and x containing ab, but not the point b. Let v be the element on γ at distance j/2 from a, and consider an element y at distance n − 2 − j/2 from v such that proj_{v}y ∉ {proj_{v}a, proj_{v}b, proj_{v}x}. Note that such an element v exists because (j, s) ≠ (2, 2). Then y lies at distance n − 2 from a, b and x, a contradiction. So we can assume that proj_{ab}x = a. If j = 2, then again, there is nothing to prove. So we may assume 2 < j < n (the case j = n is contained in the previous case, or can be obtained from the present case by interchanging the roles of a and b). Let v be an element at distance n−j −1 from the line ab such that a ≠ proj_{ab}v ≠ b. Note that v and x are opposite and δ(a, v) = n−j. Consider an element v′ incident with v, different from proj_{v}a, and let v′′ be the projection of x onto v′. Let w be the element of [x, v′′] at distance j/2 − 2 from v′′. An element y at distance j/2 − 2 from w such that proj_{w}x ≠ proj_{w}y ≠ proj_{w}v′′ lies at distance n − 2 from a, b and x, a contradiction. Claim (i) is proved.

(ii − iv) We proceed by induction on the distance k between a and b, the case k = 2 being Claim (i) above. Suppose δ(a, b) = k > 2 and let m be an element at distance k/2 from both a and b. Note that, if δ(a, b) = 4, the point a ⋈ b indeed belongs to S_{a,b} (indeed, in this case, every element of T_{a,b} either lies at distance n − 4 from a⋈b or lies opposite a⋈b in an apartment containing a, b). Suppose x is an arbitrary element of S_{a,b} and put ℓ = δ(x, m).

Suppose first that, if ℓ ≠ n, proj_{m}a ≠ proj_{m}x ≠ proj_{m}b. Then we have the following possibilities:

1. Suppose ℓ < k/2. Then δ(a, x) < k and we apply the induction hypothesis on S_{a,x}. Since b ∈ S_{a,x} ≠ ∅, δ(a, x) < n and m ≠ a⋈x, we have δ(a, x) ∈ {2, 4}. Hence either δ(a, b) = 4 and x = a ⋈ b (which is a possibility mentioned in (ii)), or δ(a, b) = 6 and x lies on m, or δ(a, b) = 8 and x = m. But in these last two cases, the “position” of b contradicts the induction hypothesis applied on S_{a,x}.

2. Suppose ℓ ≥ k/2. Let γ′′ be an ℓ-path between m and x containing neither proj_{m}a nor proj_{m}b. Put γ′ = [a, m] ∪ γ′′. Let w be the element on γ′ at distance (ℓ + k/2)/2 from both x and a. Note that w belongs to the path γ′′. If ℓ = k/2 (and hence w = m) and either k ≡ 2 mod 4 and s = 2, or k ≡ 0 mod 4 and t = 2, then there is nothing to prove. Otherwise, there exists an element y of Γ at distance n − 2 − (k/2 + ℓ)/2 from w such that proj_{w}a ≠ proj_{w}y ≠ proj_{w}x and proj_{w}b ≠ proj_{w}y. Now y lies at distance n − 2 from a, b and x, a contradiction.

Suppose now x is a point of S_{a,b} at distance ℓ from m, 0 < ℓ < n, for which proj_{m}x = proj_{m}a. Let [a, m] ∩ [x, m] = [v, m], and put i′ = δ(v, a). (Note that ℓ and k/2 have the same parity.) We have the following possibilities:

1. Suppose ℓ = k/2+2 and i′ < k/2−1 or ℓ ≤ k/2. Again δ(a, x) < k and applying the induction hypothesis on S_{a,x}, we obtain a contradiction as in Case 1 above.

2. Suppose n > ℓ > k/2 + 2. Let z be an element at distance n − ℓ from m with proj_{m}a ≠ proj_{m}x ≠ proj_{m}b, and Z an element incident with z and different from proj_{z}m. Put h = proj_{Z}x. Then δ(h, x) = n−2 and δ(m, h) = n−ℓ+2. Let j = n−2−δ(h, m)−k/2. Let h′ be the element on the (n − 2)-path between x and h at distance j/2 from h. An element y at distance j/2 from h′ such that proj_{h′}x ≠ proj_{h′}y ≠ proj_{h′}h lies at distance n − 2 from a, b and x, a contradiction.

3. Suppose ℓ = k/2 + 2, i′ = k/2 − 1 and k < n − 1. Then δ(b, x) = k + 2 and v lies at distance k/2 + 1 from both b and x. Let Σ be an apartment containing x, b and v, and let v′ be the element in Σ opposite v. Let w = proj_{v}a, w′ = proj_{v′}w (note that δ(w, v′) = n−1 and w′ ≠ proj_{v′}x, proj_{v′}b) and d the length of the path [w, a]∩ [w, w′]. Note that d ≤ k/2 − 2. For an element y not opposite w′, let w′′_{y} be the element such that [w, w′] ∩ [y, w′] = [w′′_{y}, w′]. Consider now an element y such that δ(w′′_{y}, w′) = k/2 − d − 2 and δ(w′′_{y}, y) = d. Then y lies at distance n − 2 from a, b and x, a contradiction.

4. If ℓ = k/2 + 2, i′ = n/2 − 1 and k = n, there is nothing to prove.

5. Suppose finally ℓ = k/2 + 2, i′ = k/2 − 1 and k = n − 1. Then δ(b, x) = n − 1. Let b′ and x′ be the elements of the path [b, x] at distance (n − 1)/2 − 1 from b and x, respectively. Since a ∈ S_{b,x}, either δ(a, b′) = (n + 1)/2 or δ(a, x′) = (n + 1)/2 (this is what we proved up to now for the “position” of a point of S_{b,x}). But since we obtain a path between a and b′ (x′) of length d = (3n − 5)/2 (passing through proj_{m}a), the triangle inequality implies δ(a, b′), δ(a, x′) ≥ 2n − d > (n + 1)/2, a contradiction.

This completes the proof of our claims.

Step 2: the sets O and \overline{O}. For a point c ∈ S_{a,b}, we define the set

C_{a,b;c} = {c′ ∈ S_{a,b}|S_{c,c′} ∩ {a, b} ≠ ∅}.

Now let O be the set of pairs of points (a, b), δ(a, b) ≠ n − 2 for which |S_{a,b}| > 1 and |C_{a,b;c}| > 1, ∀c ∈ S_{a,b}. We claim that O contains only pairs of collinear points and pairs of opposite points, and all pairs of collinear points are included in O.

Suppose first s ≥ 3 and t ≥ 3. Then by Step 1, only pairs of collinear points and pairs of opposite points can satisfy |S_{a,b}| > 1. If a and b are collinear points then, for a point c ∈ S_{a,b} ∩ Γ_{2}(a), every point y collinear with a not on the lines ab or ac belongs to C_{a,b;c} (since a ∈ S_{c,y}), hence the claim. Note that if n is odd, there are no pairs of opposite points. So in this case, O is the set of pairs of collinear points, which proves the theorem for the case n odd.

Let now s = 2 or t = 2 (so n is even).

Suppose first δ(a, b) = 2 and let c be a point of S_{a,b}. If c is collinear with a, then any point y′ collinear with a not on the lines ac or ab belongs to C_{a,b;c} (since a ∈ S_{c,y′}), hence (a, b) ∈ O.

Suppose δ(a, b) = 4. If (a, b) ∈ O, then (because of the condition |S_{a,b}| > 1 and (ii) in Step 1) necessarily t = 2. But for a point c ∈ S_{a,b} different from a⋈b, we obtain C_{a,b;c} = {a⋈b}, a contradiction.

Suppose finally δ(a, b) = k, 4 < k < n−2 and suppose by way of contradiction that (a, b) belongs to O. Put m = a⋈b and let x be a fixed point of S_{a,b}. Note that by (iii) in Step 1, |Γ_{1}(m)| = 3. Let x′ be an element of S_{a,b} different from x. Because |Γ_{1}(m)| = 3, we have proj_{m}x = proj_{m}x′, so δ(x, x′) ≤ k − 2 < n. But now δ(a, x ⋈ x′) = δ(b, x ⋈ x′) ≥ k/2 + 1 > δ(x, x′)/2, so neither a nor b belongs to S_{x,x′}, the final contradiction. Hence the claim.

Let \overline{O} be the set of pairs of points (a, b) satisfying δ(a, b) ≠ n − 2, (a, b) ∉ O and such that there exists a point c ∈ S_{a,b} for which (a, c) and (b, c) both belong to O. Then by considering the results of Step 1, one easily obtains that \overline{O} contains only pairs (a, b) of points at mutual distance 4 or n, and that all distance-4-pairs are contained in \overline{O} (indeed, consider the point c = a⋈b).

Step 3: the set of pairs of collinear points

Case n = 8 and s = 2 Note that t ≥ 4. For a point c ∈ S_{a,b}, we define the set

R_{c} := {x ∈ S_{a,b}|x ≠ c and a ∈ S_{x,c}}.

Let O′ be the subset of O of pairs of points (a, b) for which there exists a point c ∈ S_{a,b} with the following property:

(∗) every point r for which R_{c} ⊆ Γ_{6}(r) satisfies (r, a) ∈ \overline{O}.

We claim that O′ is exactly the set of pairs of collinear points. Let first (a, b) ∈ O with δ(a, b) = 2. Let c be a point collinear with a not on the line ab. Now one easily checks that R_{c} contains the set of points collinear with a not on the lines ab or ac. Hence clearly, a point r lying at distance 6 from all the points of R_{c} lies at distance 4 from a, so (a, r) ∈ \overline{O}. Let now (a, b) ∈ O with δ(a, b) = 8, m a point at distance 4 from both a and b, a′ = proj_{m}a and b′ = proj_{m}b. Then a point c of S_{a,b} lies at distance 6 from m, 5 from a′ and 8 from a, or at distance 6 from m, 5 from b′ and 8 from b. Suppose first δ(a′, c) = 5. Suppose c′ is a point of R_{c}. If δ(c′, a′) = 5 then, since s = 2, the projections of c and c′ onto a′ coincide. From this, it is easily seen (using the results of Step 1) that a ∉ S_{c,c′}. Hence all points of R_{c} lie at distance 5 from b′. But now the point y := proj_{b′}b lies at distance 6 from all points of R_{c} and from a, hence (a, y) ∉ \overline{O} and property (∗) is not satisfied for the point c. Suppose now δ(b′, c) = 5. Then as above, one easily sees that a point of R_{c} lies at distance 5 from a′. The point y′ := proj_{a′}a lies at distance 6 from all points of R_{c}, but is collinaer with a, hence (y′, a) ∉ \overline{O} and property (∗) is not satisfied for the point c. This shows that (a, b) ∉ O′, hence the claim. Next we distinguish this case from the case s > 2. For s > 2, one can also consider the set O′ defined as above. It is easy to see that all pairs of collinear points are included in O′. Now let O′′ be the subset of O′ consisting of these pairs (a, b) for which there exist at least two points x, x′ different from a and b such that (a, x), (a, x′), (b, x) and (b, x′) belong to O′. If s = 2, O′′ is empty (since x and x′ should be two different points collinear with the collinear points a and b), but if s ≠ 2, every pair of collinear points is included in O′′. This distinguishes the cases s = 2 and s > 2. Hence from now on, we can assume that, if n = 8, then s > 2.

The general case Let (a, b) be a pair of points of O, and c, c′ points of Γ. Consider the following conditions:

(T1) T_{a,b} ⊆ Γ_{n−2}(c) ∪ Γ_{n−2}(c′)

(T2) (c, y), (c′, y) ∉ O, ∀y ∈ T_{a,b}, if n ≥ 12, (c, y), (c′, y) ∉ \overline{O}, ∀y ∈ T_{a,b}

(T3) (x, y) ∈ O, for any two distinct points x, y ∈ {a, b, c, c′}

(T3′) (x, y) ∈ \overline{O}, for any two distinct points x, y ∈ \{a, c, c′\} or x, y ∈ \{b, c, c′\}

(T4) T_{c,c′} ⊆ Γ_{n−2}(a) ∪ Γ_{n−2}(b)

(T5) ∃!x_{1} ∈ S_{a,b} : ∀x′ ∈ S_{a,b}, |S_{x_{1},x′} ∩ {a, b}| = 1, (x_{1}, v) ∈ O for any point v ∈ {a, b, c, c′}, (x_{1}, y) ∉ O, ∀y ∈ T_{a,b}, x_{1} = S_{a,c} ∩ S_{a,c′} ∩ S_{b,c} ∩ S_{b,c′} ∩ S_{c,c′}.

Let T be the set of conditions T1, T2, T3 and T4. Let T′ be the set of conditions T1, T2, T3, T3′ and T5.

Claim 1. Let δ(a, b) = 2. If s ≥ 3, there exist points c, c′ such that conditions T are satisfied, but there do not exist points c, c′ for which conditions T′ hold. If s = 2 and n > 8, there exist points c, c′ such that conditions T′ hold.

Proof. Suppose first s ≥ 3. Choose c and c′ on the line ab, different from a and b. Then it is easy to see that the conditions T hold (note that (T2) is satisfied because n ≠ 6). Clearly, there does not exist a point y in S_{a,b} for which both (a, y) and (b, y) belong to O, hence (T5) cannot be satisfied. Suppose now s = 2. Let x_{1} be the unique point on ab, different from a and b, and c and c′ points on two different lines (different from ab) through the point x_{1}. Then it is easy to see that the conditions T′ hold (the condition (c, y) ∉ O, ∀y ∈ T_{a,b} does not hold if n = 8, which is the reason we treated the octagons separately). ◇

Claim 2. Let δ(a, b) = n. If s ≥ 3, there do not exist points c, c′ such that conditions T are satisfied, and if moreover n > 8, there do not exist points c, c′ such that conditions T′ hold. If s = 2 and n > 8, there do not exist points c, c′ such that conditions T′ are satisfied. Proof. Suppose by way of contradiction a, b, c and c′ are such that s ≥ 3 and conditions T hold or n > 8 and conditions T′ hold. Let m ∈ Γ_{n/2}(a)∩Γ_{n/2}(b), a′ = proj_{m}a and b′ = proj_{m}b. For an element x at distance j from m, 0 ≤ j ≤ n/2 − 3, such that a′ ≠ proj_{m}x ≠ b′, define the following set:

T_{x} = {y ∈ T_{a,b}|δ(x, y) = n/2 − 2 − j, proj_{x}a ≠ proj_{x}y ≠ proj_{x}b}.

Note that T_{x} is the subset of T_{a,b} of elements y for which the path [a, y] contains x. We first prove that for any set T_{x},

(◇) there does not exist a point v ∈ {c, c′} such that T_{x} ⊆ Γ_{n−2}(v).

Put M = Γ_{n/2}(a)∩Γ_{n/2}(b). Suppose T_{m} ⊆ Γ_{n−2}(v), for a point v ∈ {c, c′}. It is easy to see that δ(v, m) = n/2 and proj_{m}a = proj_{m}v or proj_{m}b = proj_{m}v. Without loss of generality, we assume proj_{m}a = proj_{m}v, hence δ(a, v) ≤ n − 2. Suppose first s ≥ 3 and conditions T hold. But then by condition (T3), δ(a, v) ≤ n − 2 implies δ(a, v) = 2 and v is a point at distance n/2 from m lying on the line L = proj_{a}m. This implies that for an arbitrary element m′ of M, m′ ≠ m, T_{m′} ∩ Γ_{n−2}(v) = ∅ (note that T_{m} ∩ T_{m′} = ∅), so T_{m′} ⊆ Γ_{n−2}(v′), with {v, v′} = {c, c′}. We obtain a contradiction by considering a third element of M. Suppose now conditions T′ hold. Note that n > 8 by assumption. In this case, δ(a, v) ≤ n − 2 implies δ(a, v) = 4 (see Condition (T3′)). Then since x_{1} ∈ S_{a,v}, x_{1} is either the point a ⋈ v or lies on the projection of b onto a⋈v (and the latter can only occur if t = 2). In any of these cases, δ(x_{1}, b) ∉ {2, n}, contradicting condition (T5). This shows (◇) for the case j = 0.

Let x be an element at distance j = 1 from m such that proj_{m}a ≠ x ≠ proj_{m}b. Suppose T_{x} ⊆ Γ_{n−2}(v), with v ∈ {c, c′}. Then again it is easy to show that δ(v, x) = δ(x, a) = n/2 + 1 and proj_{x}v = proj_{x}a = m. If proj_{m}a = proj_{m}v or proj_{m}b = proj_{m}v, then we are back in the previous case, which led to a contradiction, so suppose proj_{m}a ≠ proj_{m}v ≠ proj_{m}b. Consider the n-path between a and v that contains m. Then we can find a point y of T_{a,b} on this path that is collinear with v, in contradiction with condition (T2). This shows (◇) for the case j = 1. Note that thus no element of {c, c′} lies at distance n/2 from m.

We now proceed the proof of (◇) by induction on the distance j between x and m. Let j > 1. Consider an element x at distance j from m such that proj_{m}a ≠ proj_{m}x ≠ proj_{m}b. Suppose by way of contradiction that T_{x} ⊆ Γ_{n−2}(v), with v ∈ {c, c′}. Let x′ = proj_{x}m. Then it is again easy to show that δ(v, x) = δ(a, x) = n/2 + j and proj_{x}v = x′. Remark that proj_{x′}a ≠ proj_{x′}v, since otherwise T_{x′} ⊆ Γ_{n−2}(v) (since δ(v, x′) = n/2 + j − 1 and δ(x′, w) = n/2−j −1, with w ∈ T_{x′}), in contradiction with the induction hypothesis. Suppose first that if j = 2 we do not have the case t = 2 and n ≡ 0 mod 4 or s = 2 and n ≡ 2 mod 4. Consider now an element z incident with the element w = proj_{x′}a, but different from proj_{w}a, from proj_{w}b and from x′ (such an element exists, because of the restrictions above). But then we have δ(v, w′) = n, for every element w′ of T_{z}, so T_{z} ⊆ Γ_{n−2}(v′), with {c, c′} = {v, v′}, a contradiction with the induction hypothesis.

Let j = 2, t = 2 and n ≡ 0 mod 4 (note that n ≥ 12 since j ≤ \frac{n}{2} − 3). Let L = mx and let γ′ be the path of length n + 2 between a and v consisting of the paths [a, L] and [L, v]. Then the element of γ′ at distance 4 from v is contained in T_{a,b}, contradicting (T2).

Let j = 2, s = 2 and n ≡ 2 mod 4. Note that m and x are lines, and we assumed conditions T′. The arguments given above for t = 2 and n ≡ 0 mod 4 also work for this case, except when n = 10. We give a general argument. Put a′ = proj_{m}a and b′ = proj_{m}b. By Claim (iv) of Step 1, there are essentially two possibilities for x_{1}. First, suppose the point x_{1} lies at distance n/2 + 2 from m and at distance n/2 + 1 from a′. Then there arises an n-path γ′ between a and x_{1} sharing the path [a, a′] with γ. Let a′′ be the projection of x_{1} onto a′. Since a′′ is a line at distance n/2 from both a and x_{1}, and v ∈ S_{x_{1},a} (by Condition (T5)), either the distance between v and a′′ is n/2 (which is not true), or the distance between v and a′′ is n/2 + 2, which is again impossible. Secondly, x_{1} cannot lie at distance n/2 from m, since this would contradict condition (T5) (x_{1} would be collinear with a point of T_{a,b}).

This completes the proof of (◇).

Suppose s ≥ 3. Consider now a line L at distance j = n/2 − 3 from m, such that proj_{m}a ≠ proj_{m}L ≠ proj_{m}b. The points on L different from the projection of m onto L are points of T_{L}. By (◇), we know that T_{L} ⊈ Γ_{n−2}(v), for v ∈ {c, c′}. Since s ≥ 3, T_{L} contains at least 3 points, so we may suppose that at least two points of them are contained in Γ_{n−2}(v), with v ∈ {c, c′}. This implies that v is at distance n − 3 from L, so at distance n − 4 from a unique point x of L. If x = proj_{L}a, then T_{L} ⊆ Γ_{n−2}(v), a contradiction, so we can assume that x ≠ proj_{L}a. Let first n ≠ 8 or t ≥ 3. Then consider a line L′ incident with proj_{L}a, L′ ≠ L, at distance n−3 from both a and b (such a line always exists because of our assumptions). Now T_{L′} ∩ Γ_{n−2}(v) = ∅ (because all points of T_{L′} lie opposite v), so T_{L′} is contained in Γ_{n−2}(v′), with {v, v′} = {c, c′}, contradicting (◇).

Let now n = 8 and t = 2 (hence we assume conditions T hold). Then δ(v′, x) = 6, with {v, v′} = {c, c′}, T_{L} ⊈ Γ_{6}(v′) and δ(v, v′) ∈ {2, 8}. Now for each potential v′, it is possible to construct a point of T_{v,v′} not at distance n − 2 from a nor from b, a contradiction with condition (T4). For example, let us do the case δ(v, v′) = 2 in detail. Since v does not lie at distance 6 from a or b, we obtain δ(v, a) = δ(v, b) = 8, hence δ(a, vv′) = δ(b, vv′) = 7, v′ ≠ proj_{vv′}a ≠ v and v′ ≠ proj_{vv′}b ≠ v. Also proj_{vv′}a ≠ proj_{vv′}b, since otherwise we would obtain a point of T_{a,b} not at distance 6 from v nor from v′. Now let N be the line at distance 3 from b and at distance 4 from vv′. Then the points of N different from proj_{N}v are points of T_{v,v′} \ Γ_{6}(b), but not all these points lie at distance 6 from a, a contradiction.

This ends the proof of Claim 2 for the case s ≥ 3.

Suppose s = 2. Note that we assume n > 8 and conditions T′ hold. We keep the same notation as in the previous paragraph. Now the only possibility (to rule out) that we have not considered yet (because it does not occur in the previous case) is the case that c and c′ both lie at distance n − 2 from different points u and u′ on L, δ(c, L) = δ(c′, L) = n − 1 and u and u′ different from the projection w of a onto L.

Suppose n > 10 (otherwise some of the notations introduced below don’t make sense). Put L′ = proj_{w}a and l′ = proj_{L′}a. Suppose first the unique point z on L′ at distance n − 2 from c is not l′. Then consider a line K through z, different from L′ and from proj_{z}c. Because c is at distance n from all the points of K different from z (which are elements of T_{a,b}) we conclude that T_{K} ⊆ Γ_{n−2}(c′), a contradiction to (◇). So [c, L′] contains l′. Define the element p as [l′, m] ∩ [l′, c] = [l′, p]. Suppose p ≠ m and let j = δ(l′, p). Consider the element z′ on [c, p] at distance j + 3 from p. Note that z′ is a line at distance n−3 from both a and b. Since c is not at distance n−2 from any of the points of T_{z′}, we conclude that T_{z′} ⊆ Γ_{n−2}(c′), a contradiction to (◇). If p = m, but a′ ≠ proj_{m}c ≠ b′, we obtain a similar contradiction considering the line z′ at distance n/2 − 3 from m on the path [c, m] (note that this path does not contain a′ or b′). So the path [c, l′] contains a′ or b′ (hence δ(c, m) = n/2 + 4). Suppose without loss of generality [c, l′] contains a′. Consider now the element q defined by [m, c] ∩ [m, a] = [m, q]. Then we first show that q coincides either with a′ (Case 2 below), or with the element a′′ = proj_{a′}a (Case 1 below). Indeed, if not, then δ(a, c) < n, which implies that δ(a, c) = 4 (by Condition (T3′)) and x_{1} = a ⋈ c. Since (b, x_{1}) ∈ O, δ(b, x_{1}) is then equal to n. Now it is easy to see that there exists an element of T_{a,b} for which the projection onto ax_{1} is different from a and from x_{1}, a contradiction (such an element would be at distance n − 2 from x_{1}, which would imply that x_{1} ∉ S_{a,b}). One checks that in the case n = 10, we end up with the same possibilities.

Case 1 Consider the element m′ ∈ [a′′, c] that is at distance 2 from a′′. By Step 1, a point of S_{a,c} lies at distance n/2 or n/2 + 2 from m′. Because of the conditions, x_{1} ∈ S_{a,c}. We now check the different positions of x_{1} (using x_{1} ∈ S_{a,b} and Step 1). If x_{1} lies at distance n/2 + 1 from a′, then δ(x_{1}, m′) = n/2 + 4, a contradiction. If x_{1} lies at distance n/2 + 1 from b′, there arises a path of length n/2 + 6 between x_{1} and m′, which is again a contradiction, since n > 8. Note that x_{1} cannot lie at distance n/2 from m because (x_{1}, y) ∉ O for y ∈ T_{a,b}.

Case 2 Suppose first x_{1} lies at distance n/2 + 1 from b′. Note that as in Case 1, δ(x_{1}, m) ≠ n/2. Let b_{0} be the projection of x_{1} onto b′. Then a point of S_{x_{1},b} lies at distance n/2 or n/2 + 2 from b_{0}. Because of the conditions, c ∈ S_{x_{1},b}. But we have a path of length n/2 + 6 between c and b_{0} (containing [c, a′]), a contradiction since n ≠ 8. So we know that x_{1} lies at distance n/2 + 1 from a′. Let a_{0} = proj_{a′}x_{1}. Suppose the projections of c and x_{1} onto a′ are not equal (which only occurs if n ≡ 2 mod 4, since s = 2). Since c ∈ S_{x_{1},a}, the distance between c and a_{0} is either n/2 or n/2 + 2, a contradiction (δ(c, a_{0}) = n/2 + 4). So the projection of c onto a′ is the element a_{0}. Suppose proj_{a_{0}}c ≠ proj_{a_{0}}x_{1}. Since the distance between c and a_{0} is n/2 + 2, and c ∈ S_{a,x_{1}}, the point c has to lie at distance n/2 + 1 from either a′ or proj_{a_{0}}x_{1}, which is not true. So proj_{a_{0}}c = proj_{a_{0}}x_{1} := h. Note that the projections of c and x_{1} onto h are certainly different, since we know that the distance between c and x_{1} is either n or 2, and the last choice would contradict the fact that a ∈ S_{x_{1},c}. Now consider the projection m′ of c onto h. This is an element at distance n/2 from both c and x_{1}. Now δ(b, m′) = n/2 + 4, which contradicts the fact that b ∈ S_{c,x_{1}}.

This ends the case s = 2 and the proof of Claim 2. ◇

By Claims 1 and 2 above, no pairs (a, b) ∈ O satisfy conditions T′ if n > 8 and s > 2, while if n > 8 and s = 2, all pairs of collinear points belong to O and satisfy conditions T′. This distinguishes the cases s = 2 and s > 2. If s = 2, let O′ be the subset of O of pairs (a, b) for which there exist points c and c′ such that conditions T′ hold. If s ≥ 3, let O′ be the subset of O of pairs (a, b) for which there exist points c and c′ such that conditions T hold. Now it is clear that O′ is the set of pairs of collinear points. Hence α preserves collinearity.

4.4.5 Case i = n − 1

We can obviously assume n ≥ 6. If n = 6 and s = t = 2, then an easy counting argument yields the result, so we exclude this case in the following.

Step 1: the set O_{a,b} For two points a and b with δ(a, b) ≠ n − 1, let O_{a,b} be the set of pairs of points {c, c′}, c and c′ different from a and from b, for which

T_{v,v′} ⊆ Γ_{n−1}(w) ∪ Γ_{n−1}(w′),

whenever {a, b, c, c′} = {v, v′, w, w′}. For a pair {c, c′} ∈ O_{a,b}, we claim the following:

(i) If δ(a, b) = 2, then either c and c′ are different points on the line ab (distinct from a and b), or, without loss of generality, c is a point on ab and c′ ∈ Γ_{3}(ab) with proj_{ab}c′ ∉ {a, b, c}. Moreover, all the pairs {c, c′} obtained in this way are elements of O_{a,b}.

(ii) If δ(a, b) = 4, then either c and c′ are collinear points on the lines am or bm (where m = a⋈b) different from m, or c and c′ are points collinear with m, at distance 4 from both a and b, and at distance 4 from each other. Again, all the pairs {c, c′} obtained in this way, are elements of O_{a,b}.

(iii) Let 4 < δ(a, b) = k ≠ n − 1 and m an element at distance k/2 from both a and b. Then c and c′ are points at distance k/2 from m, at distance k from both a and b, and at distance k from each other (but such pairs {c, c′} do not necessarily belong to O_{a,b}).

If δ(a, b) = 2, then an element x of T_{a,b} is either opposite the line ab, or lies at distance n − 3 from a unique point on ab, different from a and from b. If δ(a, b) = 4, then an element x of T_{a,b} either lies at distance n − 1 or n − 3 from m = a ⋈ b with proj_{m}a ≠ proj_{m}x ≠ proj_{m}b or lies at distance n − 3 from a point x′ on am or bm, x′ ∉ {a, b, m} with am ≠ proj_{x′}x ≠ bm. It is now easy to see that the given possibilities for c and c′ in (i) and (ii) indeed satisfy the claim for δ(a, b) = 2 and δ(a, b) = 4, respectively. Note that if s = 2, only the second possibility of (ii) remains.

Let δ(a, b) = k ≠ n − 1 and let m be a fixed element at distance k/2 from both a and b. Suppose {c, c′} ∈ O_{a,b}. For an element y with δ(m, y) = j ≤ n − k/2 − 2 and proj_{m}a ≠ proj_{m}y ≠ proj_{m}b, we define the following set:

T_{y} = {x ∈ T_{a,b} | δ(x, y) = (n±1)−j−k/2 and proj_{y}x ≠ proj_{y}m if δ(x, y) ≠ n}.

For an element y with δ(m, y) = n − k/2 − 1 and proj_{m}a ≠ proj_{m}y ≠ proj_{m}b, we define T_{y} as the set of elements at distance 2 from y, not incident with proj_{y}m. For an element y with δ(m, y) = n − k/2 and proj_{m}a ≠ proj_{m}y ≠ proj_{m}b, we define T_{y} as the set of elements incident with y, different from proj_{y}m. Note that T_{y} ⊆ T_{a,b}.

First we make the following observation. Let y be an element for which the set T_{y} is defined, and for which δ(m, y) ≤ n − k/2 − 2. Then there exists an element v ∈ {c, c′} such that T_{y} ⊆ Γ_{n−1}(v) if and only if δ(v, y) = δ(a, y) and proj_{y}v = proj_{y}a or proj_{y}v = proj_{y}b.

Now we prove claims (i), (ii) and (iii) above by induction on the distance k between a and b. Let k ≥ 2. In the sequel, we include the proof for the case k = 2 in the general case.

Suppose first there exists an element v ∈ {c, c′} such that T_{m} ⊆ Γ_{n−1}(v). Then, by the previous observation, δ(v, m) = δ(m, a) = k/2 and we may assume that proj_{m}v = proj_{m}a. This implies that δ(a, v) ≤ k − 2, so we can apply the induction hypothesis on T_{a,v}. Put {c, c′} = {v, v′}. If k = 2, we obtain a = v, a contradiction. If k = 4, then v is a point on the line am, v ≠ m, and the only remaining possibility, considering the induction hypothesis and the condition T_{a,v} ⊆ Γ_{n−1}(b) ∪ Γ_{n−1}(v′) is that v′ is also a point on am, different from m. This is indeed a possibility mentioned in (ii). If k > 4, the position of b contradicts again the fact that T_{a,v} ⊆ Γ_{n−1}(b)∪Γ_{n−1}(v′) and the induction hypothesis. Indeed, the element at distance δ(a, v)/2 from both a and v belongs to the path [a, proj_{m}a] and hence does not lie at distance δ(a, v)/2 from b. In this way, we described all the possibilities for the points c and c′ in case there is a point v ∈ {c, c′} for which T_{m} ⊆ Γ_{n−1}(c). So from now on, we assume that there does not exist an element v ∈ {c, c′} such that T_{m} ⊆ Γ_{n−1}(v).

Let l be any element incident with m, different from the projection of a or b onto m. Suppose there exists a point v ∈ {c, c′} such that T_{l} ⊆ Γ_{n−1}(v). Then δ(v, l) = δ(l, a) = k/2+1 and we can assume that proj_{l}v = proj_{l}a = m. Since T_{m} ⊈ Γ_{n−1}(v), we also know that proj_{m}a ≠ proj_{m}v =: w ≠ proj_{m}b. Put {v, v′} = {c, c′}.

Suppose first k = 2. Then v is a point on the line ab. We now show that the point v′ lies at distance 2 or 4 from v such that proj_{v}v′ = m. Indeed, suppose proj_{v}v′ ≠ m or δ(v, v′) = n. If δ(v, v′) ≠ n, put γ′ = [v, v′]. If δ(v, v′) = n, let γ′ be an arbitrary n-path between v and v′ not containing m. Let x be an element of T_{a,b} at distance n − 3 from v such that either x lies on γ′, or [v, x] contains γ′. Then x is an element of T_{a,b} not at distance n − 1 from v or v′, a contradiction. So we can assume that δ(v, v′) < n and proj_{v}v′ = m. Suppose first 4 < δ(v, v′). Let Σ be an arbitrary apartment through v and v′. Then the unique element of Σ at distance n − 3 from v and belonging to T_{a,b}, does not lie at distance n − 1 from v′, a contradiction, so the distance between v and v′ is 2 or 4. Suppose now δ(v, v′) = 4 and proj_{ab}v′ = b. Then we obtain a contradiction interchanging the roles of b and v (noting that T_{a,v} ⊆ Γ_{n−2}(b) ∪ Γ_{n−2}(v′)). So v′ is a point on ab, or v′ is a point at distance 3 from ab for which the projection onto ab is different from a, b or v, as claimed in (i).

Suppose now k ≠ 2. Let w′ = proj_{w}v. Since the distance between v and any element of T_{w′} is less than or equal to n−3, we have that T_{w′} ⊆ Γ_{n−1}(v′), from which follows that δ(v′, w′) = δ(w′, a) = k/2 + 2 and proj_{w′}v′ = proj_{w′}a = w. Since T_{m} ⊈ Γ_{n−1}(v′), we either have that v′ is a point at distance k/2 from m for which the projection onto m is different from w and proj_{m}a ≠ proj_{m}v′ ≠ proj_{m}b (as required in (ii) and (iii)), or v′ is a point at distance k/2 + 2 from m for which the projection onto m is w. In the latter case, let z be the projection of v′ onto w (then δ(v, z) = δ(v′, z) = k/2) and consider an element x at distance n−1−k/2−2 from z such that proj_{z}v ≠ proj_{z}x ≠ proj_{z}v′. Then x is an element of T_{a,b} at distance n − 3 from both v and v′, a contradiction. In this way, we described all the possibilities for the points c and c′ in case there is a point v ∈ {c, c′} and an element l as above for which T_{l} ⊆ Γ_{n−1}(c). So from now on, we assume that there does not exist an element v ∈ {c, c′} such that T_{l} ⊆ Γ_{n−1}(v), for any l as above.

We now prove that (under the assumption just made)

(◇) if y is an element for which the set T_{y} is defined, with δ(m, y) > 1, then there does not exist a point v ∈ {c, c′} such that T_{y} ⊆ Γ_{n−1}(v).

This is done by induction on the distance j between y and m.

So let by way of contradiction l be an element at distance j from m, j > 1, for which the set T_{l} is defined and such that there exists an element v ∈ {c, c′} with T_{l} ⊆ Γ_{n−1}(v). Put {v, v′} = {c, c′}. Let first j < n − k/2 − 1. Then δ(v, l) = δ(l, a) = k/2 + j and w := proj_{l}v = proj_{l}a but by the induction hypothesis, u := proj_{w}v ≠ proj_{w}a. Let w′ = proj_{u}v. Note that the distance between w′ and an element of T_{w′} is (n ± 1) − k/2 − (j + 1), so an element of T_{w′} lies at distance at most n − 3 from v. We conclude that T_{w′} ⊆ Γ_{n−1}(v′), from which follows that δ(v′, w′) = δ(a, w′) = k/2 + j + 1 or δ(v′, w′) = n − 3 (the latter is possible only if j = n − k/2 − 2), and proj_{w′}v′ = proj_{w′}a = u. Let proj_{w}a = u′. First suppose δ(v′, w′) ≠ n − 3. From the assumptions, it follows that proj_{w}v′ ≠ u′. Depending on whether the projection of v′ onto w is u or not, the distance between v′ and u′ is k/2+j +2 or k/2+j. Note that δ(v, u′) = k/2+j. Now consider an element x at distance (n − 1) − (k/2 + j) from u′ such that proj_{u′}x ≠ w, and such that x either lies on [u′, b], or [u′, x] contains [u′, b]. Then x is an element of T_{v,v′} not contained in Γ_{n−1}(a) ∪ Γ_{n−1}(b), a contradiction. If δ(v′, w′) = n − 3, then we similarly obtain a contradiction. So T_{y} ⊈ Γ_{n−1}(v), for any element y at distance j from m.

Now let j = n−k/2−1. Note that T_{l} consists of all elements at distance 2 from l, not incident with l′ = proj_{l}m. Then δ(v, l) = n−1 or δ(v, l) = n−3, and in both cases, proj_{l}v = proj_{l}a. If δ(v, l) = n − 1(= δ(a, l)), we proceed as in the previous paragraph and end up with a contradiction. So let δ(v, l) = n − 3. Suppose n = k = 6 and s = 2. Then l′ is the unique point on m different from proj_{m}a and proj_{m}b. Since no line through l′ lies at distance 5 from v, every line through l′ distinct from m has to lie at distance 5 from v′, hence δ(v′, l′) ∈ {4, 6}. Now for each of the positions of v and v′, it is easy to construct a line at distance 5 from exactly two points of {a, b, v, v′}, contradicting the initial conditions. So we can assume (n, s) ≠ (6, 2). First suppose that proj_{l′}v ≠ proj_{l′}a = w. Now consider an element w′ incident with w, l′ ≠ w′ ≠ proj_{w}a and w′ ≠ proj_{w}b (such an element always exists by the restrictions made above). Then T_{w′} ⊆ Γ_{n−1}(v), a contradiction since δ(m, w′) = j − 1. So proj_{l′}v = w. Let [u, m] = [v, m] ∩ [w, m] and put u′ = proj_{u}v. Suppose first that proj_{m}a ≠ u′ ≠ proj_{m}b and v ≠ m (v = m can occur only if k = 4). Then T_{u′} ⊆ Γ_{n−1}(v′). Indeed, if we put i = δ(u, l), then δ(v, u′) = n − 4 − i and δ(m, u′) = n − k/2 − i. So the distance between u′ and an element of T_{u′} is i ± 1, and the distance between v and an element of T_{u′} is at most n − 3. So T_{u′} is contained in Γ_{n−1}(v′), which is a contradiction since δ(m, u′) < j (indeed, i ≥ 2). Suppose finally u′ = proj_{m}a or v = m. If k = 2, we end up with a point v lying on ab (namely v = proj_{m}l). But then, if s ≠ 2, for an arbitrary point x on m, different from a, b and v, we have that T_{x} ⊆ Γ_{n−1}(v), in contradiction with our assumptions. If s = 2, we end up with a point v′ at distance 3 from m and collinear with a or b, which contradicts T_{a,v} ⊆ Γ_{n−1}(b) ∪ Γ_{n−1}(v′). If k = 4, we end up with v = m, but then the position of b contradicts the fact that T_{a,v} ⊆ Γ_{n−1}(b) ∪ Γ_{n−1}(v′) and the (general) induction hypothesis. Finally, if k > 4, then δ(v, a) ≤ δ(v, proj_{m}a) + δ(a, proj_{m}a) = k − 4. Now the position of b contradicts again the fact that T_{a,v} ⊆ Γ_{n−1}(b)∪Γ_{n−1}(v′) and the (general) induction hypothesis.

Let finally j = n − k/2. Note that T_{l} consists of all elements incident with l, different from the projection l′ of m onto l. Then δ(v, l′) = n − 1 or δ(v, l′) = n − 3. Note that, in both cases, proj_{l′}v ≠ proj_{l′}a. Indeed, proj_{l′}v = proj_{l′}a would imply that T_{l′} ⊆ Γ_{n−1}(v), a contradiction with our assumptions. Suppose first δ(v, l′) = n − 3. Let l′′ = proj_{l′}v. Since no element incident with l′′ is at distance n−1 from v, we have T_{l′′} ⊆ Γ_{n−1}(v′), which implies that δ(v′, l′) is either n−3 or n−1 and proj_{l′}v′ ≠ proj_{l′}a. Consider now the element on [a, l′] at distance 2 from l′. This is an element of T_{v,v′} which is at distance n − 3 from both a and b, a contradiction. Suppose now δ(v, l′) = n − 1. Let x be the element on [v, l′] at distance 2 from l′. Since x is the only element of T_{l′} not at distance n − 1 from v, this element x lies at distance n − 1 from v′. But then δ(v′, l′) is either n − 1 or n − 3. If proj_{l′}v′ = proj_{l′}a, then T_{l′} ⊆ Γ_{n−1}(v′), a contradiction with our assumptions. If proj_{l′}v′ ≠ proj_{l′}a, then again the element on [a, l′] at distance 2 from l′ is an element of T_{v,v′} at distance n − 3 from both a and b, the final contradiction. This proves (◇).

So we can now assume T_{y} ⊈ Γ_{n−1}(v) for all v ∈ {c, c′} and for any appropriate element y. Consider an element l at distance n − k/2 from m such that the projection of l onto m is different from the projections of a and b onto m. Let u be the projection of m onto l. Since T_{l} ⊈ Γ_{n−1}(c) and T_{l} ⊈ Γ_{n−1}(c′), there is an element x incident with l, different from u, at distance n − 1 from c but not from c′, and an element y incident with l, different from u, at distance n − 1 from c′ but not from c. So δ(x, c′) = n − 3 = δ(y, c) and proj_{x}c′ ≠ l ≠ proj_{y}c. But from this follows that, for an arbitrary element l′ incident with u, l ≠ l′ ≠ proj_{u}a, we have T_{l′} ⊆ Γ_{n−1}(c), a contradiction. This proves the claims (i), (ii) and (iii).

Step 2: the set C_{a,b} For two points a, b, let C_{a,b} be the set containing a, b, and all points c for which there exists a point c′ such that {c, c′} ∈ O_{a,b}. Now let S be the set of pairs of points (a, b), δ(a, b) ≠ n − 1, for which there does not exist an element at distance n − 1 from all the points of C_{a,b}. We claim that, if s > 2, S contains exactly the pairs of points (a, b) for which δ(a, b) = 2 or δ(a, b) = 4 and if s = 2, S = ∅.

First assume δ(a, b) = 2. If s = 2, then clearly (a, b) ∉ S, so suppose s > 2. Let by way of contradiction w be an element at distance n − 1 from all points of C_{a,b}. Since all the points of the line ab are contained in C_{a,b}, w lies opposite ab. If v is an arbitrary point on ab, different from a and from b, then the element on [v, w] that is collinear with v, is contained in C_{a,b}, but lies at distance n − 3 from w, a contradiction. Suppose now δ(a, b) = 4. If s = 2, consider an element w at distance n − 2 from the line am, with a ≠ proj_{am}w ≠ m. This element lies at distance n − 1 from all points of C_{a,b;c} hence (a, b) ∉ S. Suppose s > 2. Let by way of contradiction w be an element at distance n−1 from all points of C_{a,b}. Then w lies at distance n−1 from all the points collinear with m = a ⋈ b, which is not possible. Finally suppose 4 < δ(a, b) = k ≠ n − 1. Let a′ be the element on a fixed k-path joining a and b at distance k/2 − 1 from a, and x an element at distance (n − 1) − (k/2 − 1) from w with proj_{a′}a ≠ proj_{a′}w ≠ proj_{a′}b. Then w lies at distance n − 1 from all points of C_{a,b}. Our claim is proved.

Step 3: the set S′ of pairs of collinear points Suppose s > 2. Let S′ be the subset of S containing all the pairs (a, b) with the property that there exist points x and x′ belonging to C_{a,b} such that (x, x′) ∉ S. Then S′ contains exactly the pairs of collinear points. Indeed, if δ(a, b) = 2, we can find points x and x′ in C_{a,b} at distance 6 from each other, while if δ(a, b) = 4, then C_{a,b} ⊆ Γ_{2}(m). If s = 2, then, since both s and t are infinite for n odd, n is even and hence t > 2. In this case, we dualize the arguments given above, i.e. we consider a and b lines instead of points (this is allowed since i is odd). Now we only have to distinguish between the case s = 2 and s > 2. By Step 2, if s = 2, the set S is empty while if s ≠ 2, S contains all pairs of collinear points.

This completes the proof of the case i = n − 1.

4.4.6 Case i = n/2

Let a and b be two elements at distance k, k even, and m an element at distance k/2 from both a and b. Then it is easy to see that, if k ≠ n, an arbitrary element x ∈ T_{a,b} lies at distance n/2 − k/2 from m such that proj_{m}a ≠ proj_{m}c ≠ proj_{m}b. Now we define the set S_{a,b} as the set of elements c, a ≠ c ≠ b, for which T_{a,b} ⊆ Γ_{n/2}(c).

Suppose first i is odd, and s > 2. Let S be the following set:

S = {(a, b) ∈ P^{2} ∪ L^{2} : |S_{a,b}| ≥ 2 and ∃c, d ∈ S_{a,b} : T_{a,b} ≠ T_{c,d}}.

Then a pair (a, b) of points or lines belongs to S if and only if 2 < δ(a, b) < n. Indeed, suppose to fix the ideas that a and b are points. If 2 ≠ δ(a, b) ≠ n, then consider two points c and d on the line L = proj_{a}b, different from a or proj_{L}b (this is possible since s > 2). If δ(a, b) = 2, then S_{a,b} = ∅. Let finally δ(a, b) = n. Then a point c for which T_{a,b} ⊆ Γ_{n/2}(c) is necessarily contained in any set Γ_{n/2}(x) ∩ Γ_{n/2}(y), with x, y ∈ T_{a,b}. We now show that for points c, d satisfying T_{a,b} ⊆ T_{c,d}, necessarily T_{a,b} = T_{c,d} (and hence (a, b) ∉ S). Suppose by way of contradiction c and d are such that T_{a,b} ⊆ T_{c,d}, but R is an element belonging to T_{c,d} for which δ(a, R) ≠ n/2. Define v as [proj_{c}R, a] ∩ [proj_{c}R, d] = [proj_{c}R, v] and put δ(c, v) = j. Note that j < n/2. Define w as [proj_{a}v, b] ∩ [proj_{a}v, c] = [proj_{a}v, w] and put δ(a, w) = j′. Let R′ be the element of T_{a,b} on the n-path between a and b containing proj_{a}v. Suppose first R′ lies on the path [a, w]. Then, joining the paths [d, v] and [v, R′], one obtains a path of length ℓ = \frac{3n}{2} − 2j between d and R′. Note that δ(d, R′) = n/2. If ℓ ≤ n, this implies that ℓ = n/2, hence j = n/2, a contradiction. If ℓ > n, there arises a circuit of length at most 2n − 2j < 2n, again a contradiction. So R′ does not belong to the path [a, w], implying j′ < n/2. Now similarly, one obtains a contradiction with δ(c, R′) = n/2. Hence T_{a,b} = T_{c,d} for this case. So we obtained that (a, b) ∈ S if and only if δ(a, b) ∈ κ = {4, …, n − 2}^{3}. If n = 6, then S is the set of all pairs of elements of Γ at distance 4 from each other, which ends the proof in this case (because of Paragraph 4.4.4). So assume n ≥ 10.

Define the following sets S′ and S′′:

S′ = {(p, L) ∈ P × L|Γ_{n/2}(p) ⊆ Γ_{κ}(L)},
S′′ = {(a, b) ∈ P^{2}|∃L ∈ L : (a, L), (b, L) ∈ S′}.

We claim that (p, L) ∈ S′ if and only if δ(p, L) ≤ n/2 − 4. Suppose δ(p, L) = k ≤ n/2 − 4. A line X at distance n/2 from p lies at distance at most k + n/2 ≤ n − 4 from L. A line L′ concurrent with L lies at distance k + 2, k or k − 2 from p, hence δ(L′, p) ≠ n/2. This shows that Γ_{n/2}(p) ⊆ Γ_{κ}(L). If δ(p, L) = n/2 − 2, a line L′ concurrent with L for which proj_{L}p ≠ proj_{L}L′ lies at distance n/2 from p, but δ(L, L′) ∉ κ, hence (p, L) ∉ S′. If finally k ≥ n/2, then one can easily find a line at distance n/2 from p and opposite L. This shows the claim. Now it immediately follows that a pair (a, b) ∈ S′′ if and only if δ(a, b) ≤ n−8. Then S′′ \(S ∩S′′) is the set of pairs of collinear points, which concludes the proof.

Let now s = 2. The case n = 6 and t = 2 follows from an easy counting, so we can assume t > 2. Then for two points a and b, δ(a, b) = 4 if and only if |S_{a,b}| = 2. Indeed, if δ(a, b) = 4, then the points on am and bm, different from a, b or m, are exactly the elements of S_{a,b}. If δ(a, b) = 2, then S_{a,b} = ∅. If 4 < δ(a, b) = k < n, then any point x at distance k/2 from m for which proj_{m}a = proj_{m}x or proj_{m}b = proj_{m}x belongs to S_{a,b}. If finally δ(a, b) = n, then the set Γ_{n/2}(x) ∩ Γ_{n/2}(y), with x, y ∈ T_{a,b} contains only one element different from a and b (since i is odd and s = 2), hence |S_{a,b}| ≤ 1. So we recovered distance 4 and by Subsections 4.4.3 and 4.4.4, the result follows in this case.

3Alternatively, one can now argue as in Case 4.4.3.

If i is even and s ≠ 2, the proof is similar to the case i odd (the only difference is that for the sets S and S′, we consider pairs of points). Let finally i be even and s = 2. Then one easily shows that for two points a and b, δ(a, b) = n−2 if and only if |T_{a,b}| = 1, which again ends the proof (see case Subsection 4.4.4).

Now we still have to distinguish the cases s = 2 and s > 2. Let R be the set of pairs of points (a, b) for which |S_{a,b}| = 2 and, putting S_{a,b} = {x, y}, either |S_{a,x}| ≠ 2 or |S_{a,y}| ≠ 2. Suppose for two points a and b, |S_{a,b}| = 2. Then either δ(a, b) = 4, s = 2 and x, y are the unique points on the lines am and bm different from a, m and b, or (possibly) δ(a, b) = n. In the former case, a is collinear with x or y, hence S_{a,x} = ∅ or S_{a,y} = ∅, implying (a, b) ∈ R. In the latter case, putting S_{a,b} = {x, y}, it is easy to see that S_{a,b} = S_{a,v} for v ∈ {x, y}, (since we already showed that T_{a,b} = T_{a,v} = T_{x,y} for v ∈ {x, y}), hence (a, b) ∉ R. So if s = 2, then R ≠ ∅, while if s > 2, R = ∅. This distinguishes these cases and ends the proof. □

Application 4.4.1 • Let Γ, Γ′ be a generalized n-, respectively a generalized m-gon, n ≠ m and n, m ≥ 4. Let i be an even integer satisfying 1 ≤ i ≤ n, 1 ≤ i ≤ m. Furthermore, suppose that the orders of Γ and Γ′ do not contain 2. Then there does not exist a surjective map α from the point set of Γ onto the point set of Γ′ preserving ^{4} distance i.

• Let Γ, Γ′ be a generalized n-, respectively a generalized m-gon, n ≠ m and n, m ≥ 4. Let i be an odd integer satisfying 1 ≤ i ≤ n, 1 ≤ i ≤ m. Furthermore, suppose that the orders of Γ and Γ′ do not contain 2. Then there does not exist a surjective map from the point set of Γ onto the point set of Γ′, and from the line set of Γ onto the line set of Γ′ preserving distance i.

Proof. This application follows by combining parts of the proof of Theorem 4.2.1. As before, one shows that α is bijective. If in the proof of Theorem 4.2.1 collinearity of points is characterized by the same property for Γ and Γ′, then it follows that α would preserve distance 2. This is a contradiction with n ≠ m. Indeed, without loss of generality, we can assume n < m. Let Σ be an ordinary n-gon in Γ. Then, if α preserves collinearity, the points of Σ are mapped onto the points of a (stammering) closed path of length at most 2n in Γ′, hence all these points must be sent to Γ′_{1}(R), for some line R of Γ′. Since any two points of Γ are contained in an apartment, this implies that all the points of Γ are mapped onto points of R, contradicting the bijectivity of α.

4‘preserving’ means as before that for any two points a, b of Γ, we have δ(a, b) = i if and only if δ(a^{α}, b^{α}) = i

So we only have to consider the cases where the characterization of distance 2 is not the same for Γ and Γ′ (and this roughly corresponds with the subsections in the proof of Theorem 4.2.1). Therefore we start with some observations. Let Γ be a generalized n-gon, n ≥ 4, for which the order (s, t) satisfies s, t ≥ 3 and i an integer, 2 ≤ i ≤ n.

(1) There exist points a, b of Γ for which T_{a,b} = ∅ if and only if i < \frac{n−1}{2}.

(2) There exist points a, b, c for which T_{a,b} ⊆ Γ_{i}(c) if and only if i < n − 2. Proof. Suppose first i < n − 2 and let a, b be points at distance 4. Then each element x of T_{a,b} lies at distance i − 2 from w := a ⋈ b, with proj_{w}a ≠ proj_{w}x ≠ proj_{w}b. A point c on the line aw, a ≠ c ≠ w satisfies T_{a,b} ⊆ Γ_{i}(c). Suppose now i ≥ n − 2 and let a, b be two points at distance k. Let w be a fixed element at distance k/2 from both a and b. Since |Γ_{1}(w)| ≥ 4, we can argue as in subsection 4.4.3, Claims 2 and 3 to obtain that a point c as above cannot exist. ◇

Define S_{a,b} = {x ∈ P | Γ_{i}(x) ∩ T_{a,b} = ∅}.

(3) If i = n − 2, there exist points a, b for which S_{a,b} ≠ ∅, if i ∈ {n − 1, n}, S_{a,b} = ∅ for any two points a and b. Proof. The claim for i = n − 2 was shown in Step 1 of Case 4.4.4. Suppose now i ∈ {n − 1, n}. Let a, b be points at distance k, and w an element at distance k/2 from both a and b. Let, by way of contradiction, c be a point of S_{a,b}. Suppose first there exists a shortest path between c and w not containing proj_{w}a or proj_{w}b. Put j = δ(c, w). If j = k/2, then any point x of T_{a,b} at distance i − \frac{k}{2} from w for which proj_{w}x ∉ \{proj_{w}a, proj_{w}b, proj_{w}c\} lies at distance i from c, a contradiction (note that x exists since both s, t ≥ 3). If j < k/2, let X be the element of the path [a, c] at distance j/2 + k/4 from both a and c. Note that X belongs to the path [a, w]. Now for an arbitrary element x at distance i−(j/2+k/4) from X with proj_{X}a ≠ proj_{X}x ≠ proj_{X}c (note that such a point x belongs to T_{a,c}) there arises a path of length ℓ = k + i − 2(j/2 + k/4) between b and x (consisting of the paths [b, X] and [X, x]). Since ℓ ≥ n − 1, we can choose the element x to lie at distance i from b, a contradiction with c ∈ S_{a,b}. If j > k/2, let X be the element of the path [c, w] at distance j/2 + k/4 from c. Since j/2 + k/4 ≤ n − 1, it is possible to find a point x at distance i − (j/2 + k/4) from X for which proj_{X}a ≠ proj_{X}x ≠ proj_{X}c. Such a point x lies at distance i from a, b and c, a contradiction. So we can assume that a shortest path between w and c contains proj_{w}a. Define the element v as [w, a] ∩ [w, c] = [w, v]. Let j = δ(v, c) and ℓ = δ(v, a). Let γ′ be the path between a and c obtained by joining [a, v] and [v, c], and X the element of γ′ at distance \frac{ℓ+j}{2} from both a and c. Let x be an element at distance i − \frac{ℓ+j}{2} from X such that proj_{X}x ∉ {proj_{X}a, proj_{X}b, proj_{X}c} (note that such a point x belongs to T_{a,c}). There arises a path of length ≥ n−1 between b and x (consisting of the paths [b, X] and [X, x]). Hence we can choose the point x to lie at distance i from b, the final contradiction. ◇

(4) If i = n = m − 1, then α cannot exist. Proof. Suppose i = n = m − 1. Let for two points a, b of Γ or Γ′, O_{a,b} be the set of pairs of points {c, c′}, c and c′ different from a and b, for which T_{v,v′} ⊆ Γ_{i}(w) ∪ Γ_{i}(w′) whenever {a, b, c, c′} = {v, v′, w, w′}, and C_{a,b} the set containing a, b and all points c for which there exists a point c′ such that {c, c′} ∈ O_{a,b}. Let S^{Γ} be the set of pairs of points (a, b) of Γ with δ(a, b) ≠ i, for which there does not exist an element at distance i from all the points of C_{a,b}. Similarly, we define the set S^{Γ′} of pairs of points of Γ′. In the proof of Theorem 4.2.1, Case 4.4.5, it was shown that S^{Γ′} contains exactly the pairs of points at mutual distance 2 or 4. We claim that S^{Γ} contains the set of pairs of collinear points, and that for two collinear points a and b of Γ, one has C_{a,b} = Γ_{1}(ab) (and to prove this, similar arguments as in the proof of Case 4.4.5 are used).

So let a and b be two collinear points of Γ, i = n and suppose {c, c′} ∈ O_{a,b}. For an element y at distance 0 ≤ j ≤ n − 2 from m := ab, with proj_{m}y ∉ {a, b}, define the set T_{y} = {x ∈ T_{a,b}|δ(x, y) = n−j −1}. It is easy to see that there does not exist a point x distinct from a and b such that T_{m} ⊆ Γ_{n}(x). Let y be a point incident with m, y ≠ a, b and suppose there is an element v ∈ {c, c′} for which T_{y} ⊆ Γ_{n}(v). Then it is easy to see that v is necessarily a point on m different from a, b and y. Since such a point v is not opposite any point of T_{v}, T_{v} ⊆ Γ_{n}(v′), with {v, v′} = {c, c′}. This implies that also the point v′ is incident with m, and v′ is different from a, b and v. Noting that every point of T_{a,b} lies at distance n − 2 from a unique point of m, it is now easy to see that such a pair {c, c′} indeed belongs to O_{a,b}. From now on, we assume that no point of {v, v′} satisfies T_{y} ⊆ Γ_{n}(v), for a point y incident with m. We proof by induction on δ(m, y) that no point of {v, v′} satisfies T_{y} ⊆ Γ_{n}(v), for any element y for which the set T_{y} is defined.

Consider an element y for which the set T_{y} is defined, with δ(y, m) = j > 1. Suppose there is an element v ∈ {c, c′} for which T_{y} ⊆ Γ_{n}(v). Then δ(y, v) = δ(y, a) and proj_{y}a = proj_{y}v = y′ but, using the induction hypothesis, proj_{y′}a ≠ proj_{y′}v =: y′′. Note that T_{y′′} ∩ Γ_{n}(v) = ∅, hence T_{y′′} ⊆ Γ_{n}(v′), with {v, v′} = {c, c′}. This implies that δ(y′′, v′) = δ(y′′, a) and proj_{y′′}a = proj_{y′′}v′ = y′ but proj_{y′}a ≠ proj_{y′}v′. Now let z be an element at distance n−δ(v, y′) (= n−δ(v′, y′)) from y′ such that either the path [y′, z] contains the path [y′, a], or z belongs to the path [y′, a]. Then z belongs to T_{v,v′}, but does not lie at distance n from a or b, a contradiction. So we have now shown that T_{y} ⊈ T_{v}, for any element y and v ∈ {c, c′}. Let L be a line at distance n − 2 from ab such that proj_{ab}L ∉ {a, b}. Since neither c nor c′ lies opposite all the points of T_{L}, we can assume that δ(c, L) = δ(c′, L) = n − 3 and the projections of c and c′ onto L are different from proj_{L}ab, and different from each other. Now the element on the path [L, ab] at distance 3 from L belongs to T_{c,c′} but is not opposite a or b, the final contradiction. This shows the claim concerning S^{Γ}.

Let (a, b) be a pair of points of Γ belonging to S^{Γ} for which δ(a, b) = 2. Then by the above, for any two distinct points c, c′ belonging to C_{a,b}\{a, b}, one has T_{a,b} ⊆ Γ_{n}(c) ∪ Γ_{n}(c′), hence (c, c′) ∈ O_{a,b}. However, no two points in S^{Γ′} have this propery. Indeed, if (a, b) ∈ S^{Γ} with δ(a, b) = 2, then consider a point c on ab, and a point c′ collinear with c not incident with ab; if δ(a, b) = 4, consider a point c on the line am (with m = a ⋈ b) distinct from m, and a point c′ collinear with m not on the lines am or bm. This shows that α cannot exist. ◇

Now let Γ and Γ′ be as above. Without loss of generality, we can assume n < m. If i ≤ \frac{n−1}{2} (hence also i ≤ \frac{m−1}{2}), we can characterize distance 2 with the same property for Γ and Γ′ (because of subsection 4.4.1). The case i ≥ \frac{n−1}{2} but i < \frac{m−1}{2} would contradict (1). Indeed, for a pair (a, b) of points of Γ′ for which T_{a,b} = ∅, one would obtain, with (a′, b′) = (a^{α^{−1}}, b^{α^{−1}}), T_{a′,b′} ≠ ∅, contradicting the fact that α preserves the cardinality of T_{a,b}. Hence we can assume from now on that i ≥ \frac{n−1}{2} and i ≥ \frac{m−1}{2}.

From (2), we deduce that either \frac{n−1}{2} ≤ i < n − 2 and \frac{m−1}{2} ≤ i < m − 2 or i ≥ n − 2 and i ≥ m − 2. The case i = m − 2 and i ∈ {n − 1, n} contradicts (3), so we can assume i = n = m−1 or \frac{n−1}{2} ≤ i < n−2 and \frac{m−1}{2} ≤ i < m−2. The first case is exluded by (4), so consider the latter case. If i ≥ \frac{n+1}{2} and i ≥ \frac{m+1}{2}, we can apply the result of Case 4.4.3. So the remaining cases are i ∈ \{\frac{m−1}{2}, \frac{m}{2}\} and either i ≥ \frac{n+1}{2} or i ∈ \{\frac{n−1}{2}, \frac{n}{2}\}.

• i = \frac{m−1}{2} In this case |T_{a,b}| = 1 for two points a and b of Γ′ at distance m − 1 from each other. If i ≥ \frac{n+1}{2} no points a′, b′ of Γ for which |T_{a′,b′}| = 1 can be found. If i = \frac{n}{2}, then points a′, b′ of Γ for which |T_{a,b}| = 1 only exist if s = 2 or t = 2, contradicting the assumption on the order of Γ.

• i = \frac{m}{2} Note that this implies i ≥ \frac{n+1}{2}. Let S_{Γ} (S_{Γ′}) be the set of pairs of points (a, b) of Γ (Γ′) for which there exist two elements c, d satisfying T_{a,b} ⊆ Γ_{i}(c) ∩ Γ_{i}(d) and T_{a,b} ≠ T_{c,d}. We claim that (a, b) ∈ S_{Γ} (S_{Γ′}) if and only if 2 < δ(a, b) < 2(n − i) (2 < δ(a, b) < m). If a and b are points of Γ′, this was shown in Case 4.4.6. If a and b belong to Γ, this follows from Case 4.4.3. Now one can proceed similarly as in the proof of Case 4.4.3 (defining the set S′) to characterize distance 2 in the same way for Γ and Γ′, from which the result.

This proves the application. □

? △ Is the same result true if n > m and we do not require that the map α is surjective ?

4.5 Some exceptions and applications to the Point-Line Theorem

Counterexample for the case i = n

Let H(𝕂) be the split Cayley hexagon defined on the quadric Q(6, 𝕂). Now we choose an automorphism α of Q(6, 𝕂) which does not preserve the line set of H(𝕂). Such an automorphism α induces a permutation of the points of H(𝕂). Because opposition and non-opposition in the hexagon corresponds with non-collinearity and collinearity respectively on the quadric, α has the property that δ(x, y) = 6 if and only if δ(x^{α}, y^{α}) = 6, for any two points x, y of H(𝕂). But clearly, α does not preserve collinearity. Hence we have produced a counterexample to Theorem 4.2.1 for i = n = 6.

Note that the previous class of counterexamples contains finite hexagons (putting 𝕂 equal to any finite field). We now show that, for the finite case, the only counterexamples must be hexagons of order (s, s). If there is enough transitivity around, then these are the only counterexamples (see below for a precise statement).

Theorem 4.5.1 Let Γ and Γ′ be two finite generalized n-gons of order (s, t) and (s′, t′), respectively, let α be a bijection between the points of Γ and Γ′, and fix an even number i, 2 < i ≤ n. If for every two points x and y of Γ, δ(x, y) = i ⟺ δ(x^{α}, y^{α}) = i, then either α extends to an isomorphism between Γ and Γ′, or else we have n = i = 6 and s = t = s′ = t′.

Proof. By Theorem 4.2.1 we may assume that i = n. First consider the case n = i = 6. We may assume s ≠ t. Then clearly, also s′ ≠ t′. Let a, b be two points of Γ. If δ(a, b) = 2, then |Γ_{6}(a) ∩ Γ_{6}(b)| = s^{2}t^{2}(s − 1). If δ(a, b) = 4, then |Γ_{6}(a) ∩ Γ_{6}(b)| = st(t − s + st(s − 1)). These two numbers are different since s ≠ t. Hence either two points at distance 4 are always mapped onto collinear points, or two points at distance 4 are always mapped onto points at distance 4. In the latter case, the theorem is proved. In the former case, we obtain by counting the number of points collinear with a fixed point in Γ - and this should be equal to the number of points at distance 4 from a fixed point in Γ′ - that s(t + 1) = (t′ + 1)s′^{2}t′ and similarly s′(t′ + 1) = (t + 1)s^{2}t. Combining these, we obtain the contradiction sts′t′ = 1.

Next consider the case n = i = 8. We first prove that (s, t) = (s′, t′). Indeed, we already have (1 + s)(1 + st)(1 + s^{2}t^{2}) = (1 + s′)(1 + s′t′)(1 + s′^{2}t′^{2}), and also, looking at the number of points opposite a given point, s^{4}t^{3} = s′^{4}t′^{3}. Suppose st ≠ s′t′. Putting X = st and X′ = s′t′ in the first equation (thus eliminating t and t′) and then substituting s′ = sX^{3}/X′^{3} in the equation obtained, we get, after dividing by X − X′ the following quadratic equation in X:

(X′^{3}−sX′^{2}−sX′−s)X^{2}+(X′^{4}+X′^{3}−sX′^{2}−sX′)X+(X′^{5}+X′^{4}+X′^{3}−sX′^{2}) = 0.

Note that the expression X′^{3} − sX′^{2} − sX′ − s is always positive for X′ = s′t′ ≥ 8 (this can been shown using s ≤ t^{2} and hence s^{4} = \frac{s′^{4}t′^{3}}{t^{3}} ≤ \frac{s′^{4}t′^{3}}{s^{3/2}} , implying s ≤ s′^{8/11}t′^{6/11}). From this follows that the quadratic equation above has no positive solutions. This proves st = s′t′ and hence (combined with s^{4}t^{3} = s′^{4}t′^{3}) s = s′ and t = t′. Now let a, b be two points of Γ. If δ(a, b) = 2, then ℓ_{2} := |Γ_{8}(a) ∩ Γ_{8}(b)| = (s − 1)s^{3}t^{3}. If δ(a, b) = 4, then ℓ_{4} := |Γ_{8}(a) ∩ Γ_{8}(b)| = s^{2}t^{2}(st(s − 1) + t − s). These two numbers are different because s ≠ t. Let ℓ_{6} = |Γ_{8}(a) ∩ Γ_{8}(b)| with δ(a, b) = 6. Notice that ℓ_{6} is a constant, independent of a, b. If ℓ_{6} ≠ ℓ_{2}, then clearly α must preserve collinearity. Likewise, if ℓ_{6} ≠ ℓ_{4}, then α must preserve distance 4. The result now follows from Theorem 4.2.1. □

Theorem 4.5.2 Let Γ and Γ′ be two generalized n-gons, n ∈ {6, 8}, and suppose that Γ′ has an automorphism group acting transitively on the set of pairs of points at mutual distance n − 2 (this is in particular satisfied if Γ′ is a Moufang n-gon, or if Γ′ arises from a BN-pair). Suppose there exists a bijection α from the point set of Γ to the point set of Γ′ such that, for any pair of points a, b of Γ, we have that a is opposite b if and only if a^{α} is opposite b^{α}. If α is not an isomorphism, then Γ ≅ Γ′ ≅ H(𝕂) and for any isomorphism β : Γ → Γ′, the permutation of the points of Γ defined by αβ^{−1} arises as in the counterexample above.

Proof. Let first n = 6. Let x and y be two collinear points for which x′ := x^{α} and y′ := y^{α} lie at distance 4 (these exist since otherwise α is an isomorphism by Lemma 1.3.14 of [57]). We look for the image of the line L := xy. Note that a point z, x ≠ z ≠ y, lies on L if and only if there is no point of Γ opposite exactly one point of the set {x, y, z} (see for instance [1]). Since this property is preserved by α, it is easy to check that a point z of the line L has to be mapped onto a point of the distance-2 hyperbolic line H := H(x′, y′). Now we claim that H is a long distance-2 hyperbolic line. Indeed, let K be a line of Γ′ at distance 5 from all the points of H, and suppose that the projection of H onto K is not surjective. This would imply that there is a point opposite all the points of H, so in particularly opposite all the points of L^{α}. Applying α^{−1}, we see that there would be a point opposite all the points of L, a contradiction. Our claim follows. (In fact, the very same argument shows that L^{α} = H.) So Γ′ contains a long hyperbolic line. The transitivity condition on the group of automorphisms of Γ′ now easily implies that all distance-2 hyperbolic lines are long. From Theorem 1.8.5 (i) then follows that Γ′ ≅ H(𝕂), and we may actually put Γ′ = H(𝕂).

Moreover, since the map α preserves distance 6, we obtain a representation of Γ on Q(6, 𝕂) with the property that opposition in Γ coincides with opposition in Q(6, 𝕂) (the latter viewed as a polar space: opposite points are just noncollinear points). Now, it is easy to see that, if x is any point of Γ (whose point set is identified with the point set of Q(6, 𝕂)), then the set Γ_{2}(x) is contained in a plane π_{x} of Q(6, 𝕂) (indeed, the space generated by Γ_{2}(x) in PG(6, 𝕂) is a singular subspace of Q(6, q)). If a point y of π_{x} would be at distance 4 from x in Γ, then y would be at distance ≤ 4 from all points in Γ_{1}(x), a contradiction. Hence we can apply Theorem 1.2 of Cuypers & Steinbach [14] to obtain Γ ≅ H(𝕂). It is clear that, for a given isomorphism β : Γ → H(𝕂) = Γ′, the map αβ^{−1} can be seen as a permutation of the point set of Q(6, 𝕂) preserving opposition and collinearity, hence it is an isomorphism of Q(6, 𝕂). The result follows.

Let now n = 8. Let x and y be two collinear points in Γ for which x′ = x^{α} and y′ = y^{α} lie at distance 4 or 6. Completely similar as above, one shows that the image of L = xy is the long distance-2 hyperbolic line or the long distance-3 hyperbolic line defined by x′ and y′. The transitivity condition now implies that either all distance-2 hyperbolic lines or all distance-3 hyperbolic lines are long. This contradicts Theorem 1.3 resp. Theorem 2.6 of van Bon, Cuypers & Van Maldeghem [55].

The theorem is proved. □

Remark. The previous theorem means in fact that, for hexagons and octagons with a fairly big automorphism group, Theorem 4.2.1 remains true if we rephrase the conclusion as: “ … then Γ and Γ′ are isomorphic”, and if we do not insist on the fact that α defines that isomorphism. Also, we have only considered the important values n = 6, 8. Using the results of van Bon, Cuypers & Van Maldeghem [55], we can allow for more (though all odd) values, such as n = 5, 7, 9. Indeed, for example for n = 9 (under the same transitivity conditions as in the theorem above) the existence of α would imply that either all distance-2, distance-3 or distance-4 hyperbolic lines are long. But each of these values contradicts Theorem 2.6 in [55].

? △ Given two finite generalized hexagons Γ and Γ′ of order (q, q) and a bijection from the points of Γ to the points of Γ′ preserving opposition but not collinearity. Is Γ ≅ Γ′ ?

Lemma 4.5.3 Let Γ be a generalized hexagon, and let α be a permutation of the point set of Γ preserving the opposition relation. Then the set S of lines L of Γ such that Γ_{1}(L)^{α} = Γ_{1}(M), for some line M of Γ, is a dual ovoidal subspace in Γ.

Proof. We have to show that every line of Γ not in S is concurrent with a unique line of S. We first claim that

(a) if L and L′ are two lines of S at distance 4, then also the line L ⋈ L′ belongs to S,

(b) if L and L′ are two concurrent lines of S, then all the lines concurrent with both L and L′ belong to S.

Indeed, let first δ(L, L′) = 4, with L, L′ ∈ S. Let M and M′ be the lines of Γ incident with all the images (under α) of L and L′, respectively. All points of L except for proj_{L}L′ are opposite all points of L′ except for proj_{L′}L; hence all elements of Γ_{1}(M)\(proj_{L}L′)^{α} are opposite all elements of Γ_{1}(M′)\ (proj_{L′}L)^{α}. Hence M and M′ must be at distance 4 from each other, and x := (proj_{L}L′)^{α} must be collinear with x′ := (proj_{L′}L)^{α}. Consequently the points of the line L ⋈ L′ are mapped onto the points of the line xx′. This proves (a). A similar argument shows (b).

Now suppose L is a line of Γ not belonging to S. We know that, by the proof of Theorem 4.5.2, the image under α of Γ_{1}(L) is a certain distance-2 hyperbolic line H(x, y). Put a := x⋈y. The point a′ := a^{α^{−1}} is not opposite any element of Γ_{1}(L), hence it is collinear with a unique point b ∈ Γ_{1}(L). It now easily follows that the line a′b belongs to S. This shows that L is concurrent with at least one line L′ belonging to S. By (a) and (b), this line is unique. □

Application 1. The intersection of the line sets of two generalized hexagons Γ ≅ H(𝕂) and Γ′ ≅ H(𝕂) on the same quadric Q(6, 𝕂) is a dual ovoidal subspace in both these hexagons.

Proof. Denote by S the intersection of the line set of the two hexagons Γ and Γ′ living on the same quadric Q(6, q). By a simple change of coordinates, one easily verifies that for both Γ and Γ′, coordinates can be chosen as in section 1.9.2. Hence there exists an automorphism θ of the quadric Q(6, 𝕂) mapping Γ to Γ′. (This also follows directly from Tits’ classification of trialities in [49].) Now θ preserves the opposition relation in the hexagons. Applying Lemma 4.5.3, we obtain that θ^{−1}(S) is a dual ovoidal subspace in Γ, so S is a dual ovoidal subspace in Γ′. Applying θ^{−1}, we conclude that S is also a dual ovoidal subspace in Γ. □

Remark. A similar result is true for the symplectic quadrangle W(𝕂) over some field 𝕂. But there, the proof is rather easy, because the intersection of the line sets of two symplectic quadrangles naturally represented in PG(3, 𝕂) boils down (dually using the Klein correspondence) to the intersection of a non-singular quadric Q(4, 𝕂) in PG(4, 𝕂) with a hyperplane. Hence this intersection is always a dual geometric hyperplane (of classical type).

? △ Determine the intersection of the set of reguli of two generalized hexagons Γ ≅ H(𝕂) and Γ′ ≅ H(𝕂) lying on the same quadric Q(6, 𝕂).

The aim of the second application is to prove the maximality of the group G_{2}(q) in O_{7}(q) in an entirely geometric way. We first prove some general results about dual ovoidal subspaces in the hexagon H(q). Note that, if Γ and Γ′ are two hexagons isomorphic to H(q) such that the line sets of Γ and Γ′ intersect in a dual ovoidal subspace which is a spread S, then S is necessarily Hermitian. Indeed, if two lines belong to S, then clearly so do all lines of the regulus defined by those two lines on Q(6, q). By Theorem 1.7.1, the spread S is Hermitian. In the following, we say that a dual ovoidal subspace of H(q) is of type S′ if it is a Hermitian spread.

Lemma 4.5.4 Let N_{X} be the number of dual ovoidal subspaces of type X in H(q), then we have

\{ \begin{matrix}N_{P} = q^{5} + q^{4} + q^{3} + q^{2} + q + 1, \\ N_{H} = \frac{q^{3}(q^{3}+1)}{2}, \\ N_{S′} = \frac{q^{3}(q^{3}−1)}{2}. \end{matrix}

The automorphism group G_{2}(q) of H(q) acts transitively on the dual ovoidal subspaces of type P, H and S′ respectively.

Proof. A dual ovoidal subspace of type P (H) is determined by one point (a pair of opposite points), hence the result follows for these types of subspaces. In Theorem 2.5.4, it was shown that there are exactly \frac{q^{2}−q}{2} Hermitian spreads containing a fixed line regulus. Noting that there are q^{4}(q^{4}+q^{2}+1) line reguli, and that each Hermitian spread contains \frac{q^{2}(q^{3}+1)}{q+1} line reguli, N_{S′} follows. The stabilizer in G_{2}(q) of a Hermitian spread S is the group U_{3}(q), hence the length of the orbit of S is \frac{|G_{2}(q)|}{|U_{3}(q)|}, which is equal to N_{S′}. We conclude that G_{2}(q) acts transitively on the set of Hermitian spreads. □

Lemma 4.5.5 Let Γ = H(q) be defined on Q(6, q). Then there are exactly q + 1 copies of Γ on Q(6, q) containing a given dual ovoidal subspace of Γ of type S′, exactly q containing one of type P, and exactly q − 1 containing one of type H.

Proof. Let O = Γ_{1}(p) ∪ Γ_{3}(p) be a dual ovoidal subspace of type P in Γ, and suppose Γ′ is a copy of H(q) also containing O. Let M be a line at distance 3 from p. Suppose there exists a point x on M at distance 4 from p for which Γ_{1}(x) = Γ′_{1}(x). We show that this implies Γ = Γ′. Therefore, it suffices to prove that Γ and Γ′ share at least one apartment, and all lines concurrent with one of three consecutive concurrent lines L_{1}, L_{2}, L_{3} of that apartment (see Van Maldeghem [57], proof of Theorem 6.3.1). Let Σ be an apartment of Γ containing x and p and denote by z the point of Σ opposite x. Let p′ be the point of Σ opposite p, let L be the line of Σ through z different from pz and put y = proj_{L}x (everything in the hexagon Γ). Since y is the intersection of the line L (which also belongs to the hexagon Γ′) with the tangent hyperplane of Q in x, the point y is also the unique point on L at distance 4 from x in the hexagon Γ′. Let N be the projection of y onto x in Γ. Since N is the unique line through x containing no points opposite y (opposition seen on the quadric), N is also the projection of y onto x in Γ′. Now the point p′ is the unique point on the line N not opposite z. Hence p′y is a line of Γ′, so Γ and Γ′ share the apartment Σ. Completely similar, one shows that if Γ and Γ′ share two opposite lines X, X′, and all lines through a certain point v of X, then they also share all lines through the unique point on X′ not opposite v. Let a be a point on the line proj_{p}M, different from p and p ⋈ x, and a′ = proj_{p′y}a in Γ. Note that Γ and Γ′ share the path [a, a′] and all lines concurrent with the line through a′ and a ⋈ a′ (because this line is opposite pz). Hence Γ and Γ′ share the lines concurrent with the line M. We conclude that Γ = Γ′. This shows that a hexagon containing O is completely determined by the choice of a plane x^{⊥} on Q through the line M, different from the plane containing p. Hence there are at most q such hexagons.

Let O be a dual ovoidal subspace of type S′ in Γ and suppose Γ′ is a copy of H(q) also containing O. Let L_{0} be a fixed line of the spread O and suppose Γ and Γ′ share the lines through a certain point x_{0} on L_{0}. We show that Γ = Γ′. Let L_{0}, L_{1}, …, L_{q} be the lines of a regulus contained in O through L_{0}. Then as in the first paragraph of the proof, it follows that on each line L_{i}, i > 0, there is a point x_{i} not opposite x_{0} such that Γ_{1}(x_{i}) = Γ′_{1}(x_{i}) (and these points x_{i} mutually lie at distance 4 in both Γ and Γ′). Now let N be a line of O opposite every line L_{i}. Because the spread has property (◇) (see section 2.5), the projections of the points x_{i} on the line N are all different. Hence each line concurrent with N belongs to both Γ and Γ′. Since every line of O \ {N} is opposite N, it follows that every line concurrent with a spread line belongs to both Γ and Γ′. We conclude that Γ = Γ′. Since there are q + 1 choices for the plane x_{0}^{⊥} through L_{0}, this shows that there are at most q + 1 hexagons containing O.

Let O be a dual ovoidal subspace of type H in Γ, and suppose Γ′ is a copy of H(q) also containing O. Let Σ be a fixed apartment contained in O, and L a line of Σ. Then there are at least two points x, y on L for which the planes x^{⊥} and y^{⊥} are the same for Γ and Γ′. If Γ and Γ′ also share a plane z^{⊥}, zIL, z ≠ x, y (and there are q − 1 choices for such a plane), then Γ and Γ′ coincide. Indeed, if we denote by Π the set of planes of Q containing the line L, then the map

\begin{aligned}σ : & Γ_{1}(L) & → & Π \\ & p & → & π_{p}, \end{aligned}

with π_{p} the plane corresponding to p^{⊥} in Γ, defines a projectivity of the line L. So σ is completely determined by the choice of the planes in x, y and z. Hence the two hexagons share all lines concurrent with any line of Σ, implying Γ = Γ′. So there are at most q − 1 copies of Γ containing O in this case.

In total, we obtain at most N := (q − 1)N_{P} + (q − 2)N_{H} + qN_{S′} hexagons on Q, different from Γ and intersecting Γ in a dual ovoidal subspace. By application 1, the number N + 1 has to be equal to \frac{|PGO_{7}(q)|}{|G_{2}(q)|}, from which the result. □

Application 2. The group G_{2}(q) is maximal in O_{7}(q).

Proof. Recall that O_{7}(q) is the derived group of PSO_{7}(q) and is simple. It coincides with PSO_{7}(q) if q is even, and has order |PSO_{7}(q)|/2 if q is odd. Let g be any element of O_{7}(q) not belonging to the automorphism group G_{2}(q) of H(q). Let G be the group generated by G_{2}(q) and g. We show that G = O_{7}(q). Clearly it suffices to show that |G| = |O_{7}(q)|. To that end, we look at the orbit O of H(q) under G. This orbit contains images of H(q) the line set of which intersect H(q) in dual ovoidal subspaces. By the transitivity of G_{2}(q) on the three types of dual ovoidal subspaces of H(q), there are a constant number of elements of O meeting H(q) in each of the three types of dual ovoidal subspaces. Hence we may assume that there are exactly k elements of O whose line set contains a given dual ovoidal subspace of type P of H(q). Similarly we define the numbers ℓ and m for type H and type S′, respectively. Hence in total, we have

N := 1 + k(q^{5} + q^{4} + q^{3} + q^{2} + q + 1) + ℓ\frac{q^{3}(q^{3} + 1)}{2} + m\frac{q^{3}(q^{3} − 1)}{2}

elements in O, with k ≤ q −1, with ℓ ≤ q −2 and with m ≤ q. We know that N|G_{2}(q)| (= |G|) divides the order of O_{7}(q), in particular, it divides the order of PSO_{7}(q), which is q^{3}(q^{4} − 1)|G_{2}(q)|. Hence N divides q^{3}(q^{4} − 1). Since (k, ℓ, m) ≠ (0, 0, 0), we see that N > q^{5}. Hence q must divide N, implying q divides 1 + k. Since 0 ≤ k ≤ q − 1, this means that k = q − 1. Hence

N = q^{6} + ℓ\frac{q^{3}(q^{3} + 1)}{2} + m\frac{q^{3}(q^{3} − 1)}{2}

divides q^{3}(q^{4} − 1). We may write N = abcd, where a divides q^{3}, where b divides q^{2} + 1, where c divides q + 1 and where d divides q − 1. If q is even, then a, b, c, d are unique, since every two of the numbers q^{3}, q^{2} + 1, q + 1 and q −1 are relatively prime. For q odd, there may be different possibilities, and we will make advantage of that below.

First suppose that q is even. Then both c and d are odd, and hence one can divide by 2 modulo c or d. We have 0 ≡ N mod c ≡ 1 + m mod c and 0 ≡ N mod d ≡ 1 + ℓ mod d. Hence m ≥ c − 1 and ℓ ≥ d − 1. Since ab ≤ q^{3}(q^{2} + 1), we also have

(q^{2} + 1)cd − c\frac{q^{3} − 1}{2} − d\frac{q^{3} + 1}{2} ≥ 0.

This implies

d(c(q^{2} + 1) − \frac{q^{3} + 1}{2}) ≥ c\frac{q^{3} − 1}{2},

which on its turn implies that c(q^{2} + 1) − \frac{q^{3}+1}{2} ≥ 0. Hence c > \frac{q−1}{2}. Similarly d > \frac{q−1}{2}. Since d divides q − 1, we necessarily have d = q − 1 = ℓ + 1. Also, c ∈ \{\frac{q+1}{2}, q + 1\} (or c = \frac{q}{2} = 1, hence m = 2 and N = q^{3}(q^{4} − 1), so we are done). If c = \frac{q+1}{2}, then m ∈ \{\frac{q−1}{2}, q\}. But clearly, m = \frac{q−1}{2} leads to a contradiction (the N derived from that value does not divide q^{3}(q^{4} − 1), because it is bigger than half of that number, and not equal to it). Hence m = q and therefore c = q+1. We obtain N = q^{3}(q^{4}−1) and so |G| = |O_{7}(q)|. This completes the case q even.

Now suppose that q is odd. We essentially try to give a similar proof as for q even, but the arguments need a little more elementary computations. Note that for q odd, |O_{7}(q)| = \frac{q^{3}(q^{4}−1)}{2}|G_{2}(q)|. Hence, we may choose c in such a way that it divides \frac{q+1}{2}. We easily compute N ≡ 1 + m mod c. Similarly, we obtain N ≡ 1 + ℓ mod d/i, where i ∈ {1, 2}, depending on the fact whether d divides \frac{q−1}{2} (i = 1) or not (i = 2). In any case, estimating cd as for q even, we obtain d > \frac{q−1}{2} and c > \frac{q−1}{2i}. For i = 1, this is a contradiction (because d cannot exist!). Hence i = 2 and d = q − 1. Consequently ℓ ∈ \{q − 2, \frac{q−3}{2}\}. Also, c ∈ \{\frac{q+1}{4}, \frac{q+1}{2}\} and hence m ∈ \{\frac{q−3}{4}, \frac{q−1}{2}\}. Clearly ℓ = q − 2 leads to an order of G which is bigger than |O_{7}(q)|. And m = \frac{q−3}{4} leads to an order of G that is bigger than half the order of O_{7}(q). Hence (ℓ, m) = (\frac{q−3}{2}, \frac{q−1}{2}) and this implies that |G| = |O_{7}(q)|. The application is proved. □

4.6 The exception in the Flag Theorem

Let W(2) be the symplectic quadrangle of order (2, 2). Its automorphism group is isomorphic to the symmetric group S_{6}, which is isomorphic to the

W(2)  {1, …, 6}  PG(1, 9)
point  (ij)  Baer subline of one orbit under PSL_{2}(9)
line  (ij)(kl)(mn) Baer subline of the other orbit under PSL_{2}(9)
incidence containment disjoint sublines

Table 4.1: Representations of W(2)

linear group PΣL_{2}(9). It is well known that the duads of a 6-set correspond to one orbit under PSL_{2}(9) of the set of Baer sublines of PG(1, 9), and that the synthemes of a 6-set correspond to the other orbit (see [12], page 4). Since the duads and the synthemes of a 6-set are the points and lines of W(2), one obtains a representation of W(2) on the projective line PG(1, 9) (see Table 4.1). We now investigate how one can recognize the flags of W(2) in this representation. In the following, we use the notation [a, b, c, d] for the Baer subline through the points of PG(1, 9) with affine coordinates (a), (b),

(c) and (d). Also, we denote the Baer subline corresponding to the point p of W(2) with B_{p} (and refer to this subline as ‘the point B_{p}’). Similarly for the lines of W(2).

Now fix two points of PG(1, 9) for which we choose coordinates (∞) and (0). Assume the Baer subline [∞, 0, 1, −1] corresponds with a point of W(2). Then also the Baer subline [∞, 0, i, −i] is a point. The other Baer sublines containing (∞) and (0) (namely [∞, 0, 1 + i, −1 − i] and [∞, 0, 1 − i, i − 1]) are lines. So fixing the subline B_{p} = [∞, 0, 1, −1], there are 12 Baer sublines corresponding to lines and intersecting B_{p} in exactly two points of PG(1, 9). The other three Baer sublines corresponding to lines are disjoint from B_{p} (these are the sublines B_{L_{1}} = [i, −i, 1+i, 1−i], B_{L_{2}} = [1+i, 1−i, i−1, −i−1] and B_{L_{3}} = [i − 1, −i − 1, i, −i]). So there are 180 pairs of Baer sublines (B_{p}, B_{L}) for which B_{p} is a point, B_{L} is a line and |B_{p} ∩ B_{L}| = 2, and 45 pairs of sublines (B_{p}, B_{L}) for which B_{p} is a point, B_{L} is a line and B_{p} and B_{L} (as subsets of the point set of PG(1, 9)) are disjoint. Since the automorphism group of W(2) acts transitively on both the set of flags and the set of antiflags (and there are respectively 45 and 180 of them), we deduce that a point p and a line L of W(2) are incident precisely when the Baer sublines B_{p} and B_{L} are disjoint. Hence we may identify a flag of W(2) with the pair of points of PG(1, 9) not contained in either of the two disjoint Baer sublines. This identification is bijective since there are 45 flags and 45 pairs of points, and every pair of points occurs by the 2-transitivity of PSL_{2}(9).

Let p, L_{1}, L_{2}, L_{3} be as above. It is now clear that the flags (p, L_{i}) and (p, L_{j}) correspond with disjoint pairs whose union forms the Baer subline L_{k}, (i, j, k) = (1, 2, 3). Now let (p, L) and (p′′, L′′) be two flags at distance 4 from each other, and (p, L), (L, p′), (p′, L′), (L′, p′′), (p′′, L′′) a 4-path between these two flags. Choose B_{p} = [∞, 0, 1, −1], B_{L} = [i, −i, 1 + i, 1 − i], B_{p′} = [∞, −1, i−1, −i−1] and B_{L′} = [0, 1, i, 1+i]. The cross-ratio of the pairs corresponding to the flags (p, L) and (p′, L′) is a square in GF(9) \ GF(3). Now B_{p′′} has to be disjoint from B_{L′} and intersects B_{L} in two points. Since the pair of the flag (p′′, L′) is different from the pair corresponding to the flag (p, L) (and using the fact that through two points of PG(1, 9), there are exactly two Baer sublines of point-type), these two pairs necessarily meet in exactly one point. From the observations above, we deduce that

(1) flags at distance 1 correspond to disjoint point pairs whose union forms a Baer subline (the latter corresponds to the unique element of W(2) which, together with the intersection of the two flags, forms again a flag distinct from both original flags);

(2) flags at distance 2 correspond to disjoint point pairs {a, b} and {c, d} such that the cross-ratio (a, b; c, d) is a square in GF(9) \ GF(3);

(3) flags at distance 3 correspond to non-disjoint pairs of points;

(4) flags at distance 4 correspond to disjoint point pairs {a, b} and {c, d} such that the cross-ratio (a, b; c, d) is a non-square in GF(9).

It is now clear that an arbitrary permutation of the points of PG(1, 9), which does not belong to PΓL_{2}(9), preserves the set of flags of W(2), even preserves the distance 3, but does not extend to an (anti)automorphism of W(2). Note that W(2) does not provide a counterexample to Theorem 4.2.3. Remark. Our description makes it obvious that the graph on the flags of W(2) where adjacency is being at distance 3, is the strongly regular graph with parameters (v, k, λ, μ) = (45, 16, 8, 4) obtained from a 10-set by taking as vertices the pairs of points and adjacency being non-disjoint. We give an explicit example of a bijection of the flags of W(2) preserving distance 3, but not preserving distance 1. Let F be a fixed flag of W(2) and G a flag at distance 3 from F. Then we define the flags G_{F} and G′_{F} as follows. Suppose to fix the ideas that F = (p, L) and G = (p′, L′) with L and L′ concurrent lines (see Figure 4.6). Let m = proj_{L}p′. Let L′′ be the unique line through m, different from L and L′, and x the point on L′′ of the trace containing p and p′. Let x′ be the unique point on L′′ different from x and m. Then we define G_{F} = (x, L′′) and G′_{F} = (x′, L′′). Dually if p and p′ are collinear points. Now we define the following map θ_{F} between the flags of W(2).

θ_{F} (G) = \{ \begin{matrix}G_{F} & if δ(F, G) = 3 \\ G & if δ(F, G) ≠ 3. \end{matrix}

Figure 4.6: A counterexample to Theorem 4.2.2.

Then it is easy to check that θ_{F} preserves distance 3 between the flags of W(2), but not distance 1. (In fact, for a flag G not at distance 3 from F, it suffices to show that the set T_{F,G} is preserved.) By choosing G′_{F} instead of G_{F}, one obtains another map with this property.

Exceptions to the Special Flag Theorem

We explain the restriction r ∉ S in Theorem 4.2.3. Suppose Γ is a generalized n-gon, n ≥ 3 and S = {s_{p}, s_{L}}. Suppose, to fix the ideas, that n is even and that two adjacent flags lie at Coxeter distance s_{p} if and only they have a point in common. Now the following relation is easily seen to be an equivalence relation on the set F of flags of Γ:

F ∼ F′ ⟺ δ^{∗}(F, F′) = s_{p}, with F, F′ ∈ F.

Let p and p′ be two opposite points of Γ, and denote by F_{p} (F_{p′}) the set of flags containing p (p′). For a flag F in F_{p}, define F^{α} to be the unique flag of F_{p′} at distance n − 1 from F. Similarly, one defines the map α on F_{p′}. For a flag F not contained in F_{p} ∪ F_{p′}, we define F^{α} = F. Now α is a bijection on the set of flags of Γ preserving Coxeter distance s_{p}, but clearly not s_{L}, hence α does not extend to an (anti)automorphism of Γ.

4.7 Proof of the Flag Theorem

4.7.1 Case i < n/2

Let S be the set of pairs of points (a, b) of Γ satisfying δ(a, b) ≠ i and T_{a,b} = ∅. We claim that a pair (a, b) belongs to S if and only if δ(a, b) > 2i or δ(a, b) = k < 2i, with k ≡ 0 mod 4 and 0 ≠ k ≠ i. Indeed, let (a, b) be an arbitrary pair of points of Γ. We distinguish the following possibilities.

(i) δ(a, b) = k > 2i. Suppose by way of contradiction that c ∈ T_{a,b}. If, for k < n, proj_{a}c ≠ proj_{a}b, then there arises a circuit of length at most k + 2i < 2n, a contradiction. So we can assume that, for k < n, proj_{a}c = proj_{a}b and symmetrically, proj_{b}c = proj_{b}a. But then again, a circuit of length < 2n arises, unless the paths [a, c] and [b, c] meet on [a, b] (with [a, b] any n-path between a and b if k = n). Clearly, this contradicts i < k/2, so T_{a,b} = ∅.

(ii) δ(a, b) = k < 2i, with k ≡ 0 mod 4 and 0 ≠ k ≠ i. Note that a ⋈ b is a point. Suppose by way of contradiction that c ∈ T_{a,b}. If proj_{a}c ≠ proj_{a}b, then we obtain a path of length k + i between b and c (consisting of the paths [b, a] and [a, c]). Since this cannot be the i-path between b and c, there arises a circuit of length < 2n, a contradiction. Hence we may assume proj_{a}c = proj_{a}b and proj_{b}c = proj_{b}a. In this case, since there are no circuits of length < 2n, the paths [a, c] and [b, c] must meet on [a, b], necessarily in a ⋈ b (and c ≠ a⋈b since i > k/2). This is impossible since a⋈b is a thin point.

(iii) δ(a, b) = k < 2i, with k ≡ 2 mod 4 and k ≠ i. Any point c at distance i− \frac{k}{2} from M := a⋈b with proj_{M}a ≠ proj_{M}c ≠ proj_{M}b belongs to T_{a,b} (since M is thick, such a point c can be found). So (a, b) ∉ S.

(iv) The cases δ(a, b) = 0, i, 2i are trivial.

This shows the claim. We put κ = {δ(a, b) | (a, b) ∈ S} (hence κ = {k ∈ ℕ | 2i < k ≤ n or k < 2i, k ≡ 0 mod 4 and 0 ≠ k ≠ i}).

Case i ≡ 0 mod 4

Let S′ be the set of pairs (a, b) of distinct points of Γ such that i ≠ δ(a, b) ∉ κ and Γ_{i}(a)∩Γ_{κ}(b) = ∅. We claim that S′ is exactly the set of pairs of collinear points of Γ. Indeed, let (a, b) be an arbitrary pair of distinct points of Γ, i ≠ δ(a, b) ∉ κ. There are three possibilities.

(i) δ(a, b) = 2. Every point at distance i from a but not at distance i from b lies at distance i ± 2 ≡ 2 mod 4 from b, which is not a distance belonging to κ. Hence (a, b) ∈ S′.

(ii) δ(a, b) = k ≡ 2 mod 4, 2 < k < 2i. Let L be the line of [a, b] at distance k/2 − 2 from a, and let c be any point at distance i − (k/2 − 2) from L such that proj_{L}a ≠ proj_{L}c ≠ proj_{L}b. Then δ(a, c) = i and δ(b, c) = i + 4. The latter is a multiple of 4. So, if i ≠ 4, then 4 + i < 2i and δ(b, c) ∈ κ. If, on the other hand, i = 4, then necessarily k = 6. In this case, re-choose the point c at distance 4 from a with proj_{a}c ≠ proj_{a}b. Then δ(b, c) = 10 ∈ κ. Hence in both cases (a, b) ∉ S′.

(iii) δ(a, b) = 2i. Let L be the unique line of [a, b] at distance i/2 − 1 from a and let c be any point at distance i/2 + 1 from L such that proj_{L}a ≠ proj_{L}c ≠ proj_{L}b. Then δ(a, c) = i and δ(b, c) = 2i + 2, hence δ(b, c) ∈ κ. Consequently c ∈ Γ_{i}(a) ∩ Γ_{κ}(b), implying (a, b) ∉ S′.

Our claim is proved.

Case i ≡ 2 mod 4

We proceed similarly as above. Now S′ is the set of pairs (a, b) of distinct points of Γ such that i ≠ δ(a, b) ∉ κ and Γ_{i}(a) ∩ Γ_{≠i}(b) ⊆ Γ_{κ}(b) and we again claim that S′ is exactly the set of pairs of collinear points of Γ. So let (a, b) be an arbitrary pair of distinct points of Γ, i ≠ δ(a, b) ∉ κ. There are three possibilities.

(i) δ(a, b) = 2. Every point at distance i from a but not at distance i from b lies at distance i ± 2 from b, which is a distance belonging to κ. Hence (a, b) ∈ S′.

(ii) δ(a, b) = k ≡ 2 mod 4, 2 < k < 2i. We consider a point c as in 4.7.1(ii). Then δ(b, c) = i + 4 implies δ(b, c) ∉ κ.

(iii) δ(a, b) = 2i. Let L be the unique line of [a, b] at distance i/2 from a and let c be any point at distance i/2 from L such that proj_{L}a ≠ proj_{L}c ≠ proj_{L}b. Then δ(a, c) = i and δ(b, c) = 2i, hence δ(b, c) ∉ κ. Consequently c is in Γ_{i}(a) ∩ Γ_{≠i}(b), but not in Γ_{κ}(b), implying (a, b) ∉ S′.

This shows the claim and completes the proof of Case i < n/2.

4.7.2 Case i = n/2

n = 8 and Γ has order (2, 1)

Note that also Γ′ has order (2, 1) by the bijectivity of α. We first distinguish distance 6. Let a and b be two different points of Γ. If δ(a, b) = 2, then a point x belongs to T_{a,b} if and only if x lies at distance 3 from the line ab, hence |T_{a,b}| = 2. If δ(a, b) = 6, then T_{a,b} = {c}, with c the unique point on the line a ⋈ b not collinear with a or b. If finally δ(a, b) = 8, the points of T_{a,b} necessarily lie on one of the 8-paths between a and b, hence |T_{a,b}| = 2. So δ(a, b) = 6 if and only if δ(a, b) ≠ i and |T_{a,b}| = 1. Unfortunately, all straightforward counting arguments do not lead to a distinction between points at distance 2 or 8. Hence we give a more sophisticated reasoning. Let a, b be points of Γ at distance 2 or 8 from each other. Put T_{a,b} = {c, d} and S = {a, b, c, d}. We claim that there is a unique point x such that

(*) Γ_{4}(x) ∩ S = ∅ and Γ_{6}(x) ∩ S = ∅.

Indeed, if a and b are collinear, then c and d are collinear points such that the line cd meets the line ab in a point x ∉ S. One can easily check that x is the only point of Γ that satisfies (*). If δ(a, b) = 8, then S is contained in the unique apartment through a and b. Note that Γ is the double of the unique generalized quadrangle W(2) of order 2. In W(2) the points a, b, c, d correspond to flags whose union is an apartment Σ in W(2) (see Figure 4.7). There is a unique point u (respectively a unique line U) in W(2) opposite every point (respectively line) of Σ and u is incident with U. The flag {u, U} corresponds in Γ with the unique point x satisfying (*). This proves our claim. Now if δ(a, b) = 8, then there exists a point y of T_{a,x} at distance 6 from b (see Figure 4.7, in fact, every point of T_{a,x} has this property) while if δ(a, b) = 2, every point of T_{a,x} is collinear with b. Hence we can distinguish distance 2 and the theorem follows.

Figure 4.7: Case i = n/2 and Γ the double of W(2).

The general case

Here we assume that, if n = 8, then Γ contains lines with more than 3 points. Note also that, since i is even, necessarily n ≡ 0 mod 4.

In this case, we show that we can recover opposition. Let a, b be points of Γ. We claim that δ(a, b) = n if and only if

(**) |T_{a,b}| = 2 and, putting T_{a,b} = {c, d}, T_{c,d} = {a, b}.

Obviously, if a and b are opposite, then they satisfy (**). So we may assume that δ(a, b) =: k < n. We distinguish three cases.

(i) k ≡ 0 mod 4, k ≠ n. We show that T_{a,b} = ∅. Suppose by way of contradiction that c ∈ T_{a,b}. Assume first that proj_{a}b = proj_{a}c. Note that the path [a, c] does not contain [a, b], and since k ≠ 2i, the point c does not belong to [a, b]. Hence we can define the line L as [a, b] ∩ [a, c] = [a, L]. Let j = δ(a, L). There is a path of length k + i − 2j between b and c, consisting of the paths [b, L] and [L, c]. If k + i − 2j ≤ n, then δ(b, c) = i implies j = k/2, hence L = a⋈b, contradicting the fact that L is a line. Hence k + i − 2j > n. But now δ(b, c) = i implies there is a circuit of length at most (k + i − 2j) + i = n + k − 2j < 2n, a contradiction.

The case proj_{a}b ≠ proj_{a}c corresponds with j = 0 in the previous argument.

(ii) k = n − 2.

Let c be an arbitrary element of T_{a,b} (T_{a,b} is easily seen to be nonempty; this will also follow from our next argument). Similarly as in (i) above, one shows that [a, b] ∩ [a, c] ∩ [b, c] = a ⋈ b =: L. But then cIL and proj_{L}a ≠ c ≠ proj_{L}b. So if (a, b) satisfies (**), then L contains 4 points c, d, proj_{L}a, proj_{L}b. But every point on M := proj_{a}b distinct from proj_{M}b belongs to T_{c,d}. Similarly for M′ := proj_{b}a. Note that M ≠ L and M′ ≠ L since n − 2 ≠ 2. Hence, since also M′ ≠ M, we conclude by thickness of those lines that |T_{c,d}| ≥ 4. So (a, b) does not satisfy (**).

(iii) k ≡ 2 mod 4 and k ≠ n − 2. Every point c at distance \frac{n−k}{2} from the line L := a ⋈ b with proj_{L}a ≠ proj_{L}c ≠ proj_{L}b belongs to T_{a,b}. So if |T_{a,b}| = 2, then necessarily \frac{n−k}{2} = 3 and both L and proj_{c}L are incident with exactly 3 points (note that \frac{n−k}{2} = 1 corresponds with case (ii) above). We put T_{a,b} = \{c, d\}. As in (ii) above, |T_{c,d}| ≥ 4 whenever proj_{a}b ≠ proj_{b}a. Hence we may assume that a and b are incident with L and that k = 2 and i = 4. But this is Case 4.7.2.

So we obtained that α preserves opposition. By Abramenko & Van Maldeghem [1], Corollary 5.2, this completes the proof of Case i = n/2.

4.7.3 Case n/2 < i < n − 2

Let S be the set of pairs of points (a, b) of Γ such that T_{a,b} = ∅. Put κ = {k ∈ ℕ | 0 < k ≤ 2n − 2i − 4 and k ≡ 0 mod 4}. We claim that (a, b) ∈ S if and only if δ(a, b) ∈ κ. Indeed, let a, b be points of Γ. Put k = δ(a, b).

(i) 0 < k ≤ 2n − 2i − 4 and k ≡ 0 mod 4. Similarly as in 4.7.1(ii), one shows that T_{a,b} = ∅ in this case.

(ii) k ≤ 2n − 2i − 2 and k ≡ 2 mod 4. Here, a point c ∈ T_{a,b} can be found similarly as in 4.7.1(iii).

(iii) k ≥ 2n − 2i.

Let c′ be a point opposite b and at distance n−k from a (c′ lies in some apartment containing a, b). Let X be a line incident with c′, distinct from proj_{c′}a if k ≠ n. Clearly, there is a point xIX, x ≠ c′, with x opposite b. Then δ(c′, x) = 2, and an inductive argument shows that there is a point c′′ opposite b with δ(c′, c′′) = k − 2n + 2i and with proj_{c′}a ≠ proj_{c′}c′′ if k ≠ n. Note that δ(a, c′′) = 2i − n ≠ 0. Let c ∈ Γ_{i}(b) ∩ Γ_{n−i}(c′′) be such that proj_{c′′}c ≠ proj_{c′′}a (c is the point at distance i from b lying on the n-path between b and c′′ not containing proj_{c′′}a). Clearly, c belongs to T_{a,b}.

This shows our claim.

Case i ≡ 0 mod 4 and i ≤ 2n − 2i − 4

In this case i precisely belongs to κ. We claim that two distinct points a and b are collinear in Γ if and only if δ(a, b) ∉ κ and R := Γ_{i}(a) ∩ Γ_{≠i}(b) ∩ Γ_{κ}(b) is empty. Indeed, let a, b be two arbitrary distinct points of Γ, δ(a, b) ∉ κ.

(i) δ(a, b) = 2. This is similar to 4.7.1(i).

(ii) δ(a, b) ≡ 0 mod 4. Note that i < k := δ(a, b) < 2i. Let c ∈ Γ_{i}(a) ∩ Γ_{k−i}(b) (choose c on a k-path between a and b). Then c ∈ R because δ(b, c) = k − i is distinct from i, it is a multiple of 4 and it is at most 2n − 2i − 4 (for i ≤ 2n − 2i − 4 < 2n − k − 4).

(iii) 2 ≠ δ(a, b) ≡ 2 mod 4. First let i < k := δ(a, b) < 2i − 2. Let L ∈ Γ_{i−1}(a) ∩ Γ_{k−i+1}(b) and let cIL with proj_{L}a ≠ c ≠ proj_{L}b. Then we show that c ∈ R. Indeed, δ(b, c) = k − i + 2, so δ(b, c) = i implies k/2 + 1 = i, a contradiction. Also, δ(b, c) ≡ 0 mod 4 and the inequalities i ≤ 2n − 2i − 4 and k ≤ 2i − 6 imply δ(b, c) ≤ 2n − 2i − 4. Consequently δ(b, c) ∈ κ.

Now suppose k = 2i − 2. This implies, since 2i ≥ n + 2, that k ≥ n, hence k = n and n = 2i − 2. Let L ∈ Γ_{i−3}(a) ∩ Γ_{n−i+3}(b) and let c ∈ Γ_{3}(L) with proj_{L}a ≠ proj_{L}c ≠ proj_{L}b. Then c ∈ Γ_{i}(a)∩Γ_{≠i}(b) (because δ(b, c) = n − i + 6 = (2i − 2) − i + 6 ≠ i). Also, δ(b, c) is a multiple of 4. If n ≥ 22, then one verifies that δ(b, c) = n − i + 6 ≤ 2n − 2i − 4, hence c ∈ R. If n < 22, then, since i is a multiple of 4, the only possibility is (n, k, i) = (14, 14, 8). But then κ = {4, 8} and we can distinguish distance 4 in Γ; hence also distance 2 by Subsection 4.7.1.

Finally let k := δ(a, b) < i. Put L = proj_{b}a. Let c ∈ Γ_{i−k+1}(L) with proj_{L}a ≠ proj_{L}c ≠ proj_{L}b. As above, one checks that c ∈ R.

This shows our claim.

Case i ≡ 2 mod 4 and i ≤ 2n − 2i − 4

Here, we claim that two distinct points a, b of Γ are collinear if and only if i ≠ δ(a, b) ∉ κ and Γ_{i}(a) ∩ Γ_{≠i}(b) ⊆ Γ_{κ}(b). Indeed, let a, b be two arbitrary distinct points of Γ, i ≠ δ(a, b) ∉ κ. We have the following cases.

(i) δ(a, b) = 2.

This is similar to 4.7.1(i).

(ii) δ(a, b) = k ≡ 0 mod 4, k > 2n − 2i − 4. Note that i < k < 2i. Let c be the point of a fixed k-path between a and b at distance i from a. Then i ≠ δ(b, c) = k − i ≡ 2 mod 4, hence δ(b, c) ∉ κ.

(iii) 2 ≠ δ(a, b) = k ≡ 2 mod 4. (a) Suppose first k > i. Let L be the line of a fixed k-path between a and b at distance i − 1 from a. If i ≠ k/2 + 1, let c be a point incident with L, proj_{L}a ≠ c ≠ proj_{L}b. Then i ≠ δ(b, c) = k − i + 2 ∉ κ. If i = k/2 + 1, let L′ be the line concurrent with L and closest to a, and c a point at distance 3 from L′ for which proj_{L′}a ≠ proj_{L′}c ≠ proj_{L′}b. Then δ(a, c) = i and δ(b, c) = k − i + 6 ≡ 2 mod 4 (note that i ≥ 6), hence i ≠ δ(b, c) ∉ κ.

(b) Suppose now k < i. In this case, a point c at distance i − k + 1 from the line R := proj_{b}a for which proj_{R}a ≠ proj_{R}c ≠ b will do the job.

This shows the claim.

Case i ≥ 2n − 2i − 2

We claim that two points a, b of Γ are at distance 2n − 2i − 4 from each other if and only if δ(a, b) ∈ κ and T_{a,b}^{κ} := Γ_{κ}(a) ∩ Γ_{κ}(b) contains exactly (2n − 2i − 8)/4 =: ℓ elements. Indeed, let a, b be two distinct points of Γ such that δ(a, b) ∈ κ. We distinguish the following cases.

(i) δ(a, b) = 2n − 2i − 4. Note that there are exactly ℓ elements of T_{a,b}^{κ} contained in [a, b] (indeed, every point x of [a, b] different from a, b for which δ(a, x) ≡ 0 mod 4 belongs to T_{a,b}^{κ}.) Conversely, we show that every element of T_{a,b}^{κ} is contained in [a, b]. Suppose c ∈ T_{a,b}^{κ} and c not on the path [a, b]. If proj_{a}c ≠ proj_{a}b, we obtain a circuit of length ≤ 3(2n − 2i − 4) < 2n (indeed, 3i ≥ 2n − 2), a contradiction. So we can assume proj_{a}c = proj_{a}b and proj_{b}c = proj_{b}a. Let in this case, [a, c] ∩ [a, b] = [a, L] and r = δ(a, L). Since c ∈ T_{a,b}^{κ}, δ(a, c) ≡ 0 mod 4. We obtain a path of length d = δ(a, b)+δ(a, c)−2r between b and c (joining the paths [b, L] and [L, c]). If d > n, δ(b, c) ∉ κ since otherwise we obtain a circuit of length < 2n. If d ≤ n, then δ(b, c) = d ≡ 2 mod 4, hence δ(b, c) ∉ κ.

(ii) δ(a, b) := k ∈ κ \ {2n − 2i − 4}. On the path [a, b], we already find k/4 − 1 members of T_{a,b}^{κ}. Now let h ∈ κ with h > k. Then every point x ∈ Γ_{h}(a) ∩ Γ_{h−k}(b) belongs to T_{a,b}^{κ}. Now for each such h, we find at least two such points. Indeed, consider points x at distance h − k from b for which proj_{b}a ≠ proj_{b}x. The number of choices for h is \frac{2n−2i−4}{4} − \frac{k}{4} = ℓ + 1 − \frac{k}{4}, hence together with the points of T_{a,b}^{κ} on [a, b] we obtain at least 2ℓ + 1 − k/4 elements of T_{a,b}^{κ}. This number is bigger than ℓ, since ℓ > k/4 − 1.

So we can recover distance 2n − 2i − 4. By the previous cases, this is enough to recover collinearity. This completes the proof of Case n/2 < i < n − 2.

4.7.4 Case i = n − 2

It is convenient to treat the cases n = 6, 8 separately.

Case n = 6

Here, i = 4, so we only have to distinguish distance 2 from 6. But for opposite points a, b, the set T_{a,b} contains points at mutual distance i = 4, while this is not the case for collinear points a, b. Hence in this case α preserves collinearity.

Case n = 8

First suppose that Γ is the double of a quadrangle Δ of order (2, t) (with t automatically finite). Then t = 2, 4 and Δ is unique. Notice that by the bijectivity of α, in this case Δ and Δ′ have the same order. If t = 2, then there is nothing to prove (this was the exception). Suppose t = 4. For two points a, b at distance k, put l_{k} = |T_{a,b}|. Then it is easily verified that l_{2} = 8 if ab contains 3 points, l_{2} = 24 if ab contains 5 points, l_{4} = 8 and l_{8} = 16. Hence α preserves opposition and we are done.

So from now on we may assume that all lines of Γ have at least 4 points. We claim that two distinct points a, b of Γ are collinear if and only if δ(a, b) ≠ 6 and there are no distinct points c, c′ ∈ Γ_{6}(b)∩Γ_{≠6}(a) satisfying T_{a,b} ⊆ Γ_{6}(c)∪ Γ_{6}(c′). Indeed, if δ(a, b) = 4, then we take two different points c, c′ (unequal a) on the unique line through a at distance 5 from b; if δ(a, b) = 8, then we take {c, c′} = Γ_{2}(a) ∩ Γ_{6}(b). In these cases one easily checks that T_{a,b} ⊆ Γ_{6}(c)∪Γ_{6}(c′). Now let δ(a, b) = 2. Suppose by way of contradiction that there do exist two points c, c′ as above. Let L be an arbitrary but fixed line meeting the line ab but not through a or b. Then the set of points R = Γ_{3}(L) \ Γ_{1}(ab) is contained in T_{a,b} and hence is a subset of Γ_{6}(c) ∪ Γ_{6}(c′). Either δ(a, c) = 4 or δ(a, c) = 8 (and similarly for c′). First suppose δ(a, c) = 4. Clearly, for any line M ≠ ab meeting L, there is exactly one point xIM at distance 6 from c. Hence there are at least 2 points of T_{a,b} on M and opposite c, implying that the line M must be at distance 5 from c′. Since there are at least 3 such lines M, we similarly have that δ(c′, L) = 3, and δ(c′, ab) = 1 (because proj_{L}c′ cannot be on a line M, so must be incident with ab), contradicting δ(b, c′) = 6.

So we showed that δ(a, c) = 8 and symmetrically, also δ(a, c′) = 8. So δ(c, L) = δ(c′, L) = 7, and hence, since δ(c, ab) = 7, there must be a unique line M_{c} ≠ ab meeting L having distance 5 to c. Similarly, there is such a line M_{c′} at distance 5 from c′. Now let M ∈ Γ_{2}(L) \ {M_{c}, M_{c′}, ab}. Since δ(c, M) = δ(c′, M) = 7, at most two points on M are covered by Γ_{6}(c)∪Γ_{6}(c′), a contradiction with the fact that the line M contains at least 3 points of T_{a,b}. This proves our claim. So α preserves collinearity and the theorem follows.

Case n > 8

Suppose first that, up to duality, Δ (or Δ′) has order (2, t) with t finite (hence n ∈ {12, 16}), or has order (3, 3) (and then n = 12). Then the same holds for Δ′ (or Δ). We now give a similar counting argument as in 4.7.4. Let first n = 12 and Δ a hexagon of order (s, t). Then, with as before l_{k} := |T_{a,b}| for δ(a, b) = k, a rather easy counting in Γ shows the following:

l_{2} = (s − 1)s^{2}t^{2} if |ab| = s + 1 l_{2} = (t − 1)s^{2}t^{2} if |ab| = t + 1 l_{4} = s^{2}t^{2} l_{6} = st(2s(t − 1) + (s − 1)(s + t)) if |ab| = s + 1 l_{6} = st(2t(s − 1) + (t − 1)(s + t)) if |ab| = t + 1 l_{8} = st(s + t − 2)^{2} l_{12} = (s + t − 2)((s − 1)(t − 1)^{2} + (t − 1)(s − 1)^{2} + 2s(t − 1) + 2t(s − 1)).

Now (s, t) ∈ {(2, 2), (2, 8), (3, 3)}. In any of these cases, one checks that l_{12} ∉ {l_{2}, l_{4}, l_{6}, l_{8}}, hence we can distinguish opposition. Similarly if n = 16 and Δ has order (2, 4).

So from now on we may assume that Δ has order (s, t) ≠ (3, 3) with s, t ≥ 3, or {s, t} = {2, ∞}. We divide the proof in several steps.

Step 1: the set S_{a,b} For any three points a, b, c of Γ, define T_{a,b,c} := Γ_{n−2}(a) ∩ Γ_{n−2}(b) ∩ Γ_{n−2}(c). Let a, b be two arbitrary points of Γ not at distance n − 2, then we define

S_{a,b} = {c ∈ Γ_{≠(n−2)}(a) ∩ Γ_{≠(n−2)}(b) | T_{a,b,c} = ∅}.

Note that, by symmetry, c ∈ S_{a,b} implies b ∈ S_{a,c} and a ∈ S_{b,c}.

We will prove the following claims (where w = a⋈b whenever defined).

Claim 1. δ(a, b) = 2.

If the line ab contains at least 4 points, then S_{a,b} = Γ_{3}(ab). Otherwise, S_{a,b} = Γ_{{1,3,7}}(ab) \ ({a, b} ∪ Γ_{6}(a) ∪ Γ_{6}(b)).

Claim 2. δ(a, b) = 4.

Here, S_{a,b} = Γ_{1}(aw) ∪ Γ_{1}(bw) ∪ Γ_{4}(a) ∪ Γ_{4}(b) \ ({a, b} ∪ Γ_{4}(w)).

Claim 3. 2 ≠ δ(a, b) = k ≡ 2 mod 4, k ≤ n − 4. Here S_{a,b} ⊆ {x ∈ Γ_{≤k/2+2}(w) | proj_{w}a ≠ proj_{w}x ≠ proj_{w}b}. If k = 6, then no point incident with w belongs to S_{a,b}. Also, if w contains at least 4 points, then no point of Γ_{k/2}(w) belongs to S_{a,b}.

Claim 4. 4 ≠ δ(a, b) = k ≡ 0 mod 4, k ≤ n − 4, n ≠ 12. Put A = proj_{w}a and B = proj_{w}b. Also, define

S′_{a,b} = {x ∈ Γ_{{k/2−1,k/2+1}}(A) | proj_{A}a ≠ proj_{A}x ≠ w}
∪{x ∈ Γ_{{k/2−1,k/2+1}}(B) | proj_{B}b ≠ proj_{B}x ≠ w}.

If k ≠ 8, then S_{a,b} ⊆ S′_{a,b}. If k = 8 and if both proj_{a}b and proj_{b}a contain at least 4 points, then {w} ⊆ S_{a,b} ⊆ S′_{a,b} ∪ {w}. If k = 8 and either proj_{a}b or proj_{b}a contains exactly three points (and suppose without loss of generality that proj_{b}a has size 3), then {w, e} ⊆ S_{a,b} ⊆ S′_{a,b} ∪ {w, e}, where e is incident with proj_{b}a and distinct from both b and b⋈w.

Claim 5. δ(a, b) = k = 8 and n = 12.

Here, with the notation of Claim 4, we have, if s, t ≥ 3, then w ∈ S_{a,b} ⊂ S′_{a,b} ∪ (Γ_{8}(a) ∩ Γ_{8}(b)) ∪ {w}. If {s, t} = {2, ∞} (and we may assume without loss of generality that A′ := proj_{a}b is incident with infinitely many points), then {w, e} ⊆ S_{a,b} ⊂ S′_{a,b} ∪ (Γ_{8}(a) ∩ Γ_{8}(b)) ∪ S′′_{a,b} ∪ {w, e}, where S′′_{a,b} = {x ∈ Γ_{11}(B) | proj_{B}b ≠ proj_{B}x ≠ w} ∪ {x ∈ Γ_{7}(A′) | proj_{A′}w ≠ proj_{A′}x ≠ a}.

We will prove these claims by induction on δ(a, b).

Claim 1. Let c be an arbitrary point of Γ, a ≠ c ≠ b. Note that the points of T_{a,b} all lie at distance n − 3 from the line ab. First assume that proj_{ab}c = a. Put j = δ(c, a). If j = 2, then δ(c, x) = n, for all x ∈ T_{a,b}, hence c ∈ S_{a,b}. Suppose j > 2. Let y be a point at distance n − j − 1 from ab for which b ≠ proj_{ab}y ≠ a. Then δ(c, y) = n. Let Y be the line incident with y and different from proj_{y}a. On the line Y, there is at least one point opposite c and at distance n − j + 2 from a, b. Proceeding like this, one constructs a point c′ opposite c with δ(a, c′) = δ(b, c′) = n−4. Now let L be a line incident with c′ and different from proj_{c′}a. Then proj_{L}c ∈ T_{a,b,c}, hence c ∉ S_{a,b}.

So we may assume proj_{ab}c ∉ {a, b}. Put j + 1 = δ(c, ab). If j = 0, then clearly c ∈ S_{a,b} if and only if ab is incident exactly three points. If j = 2, then clearly c always belongs to S_{a,b}. Suppose j > 2. Let L ∈ Γ_{j−1}(c)∩Γ_{2}(ab). If j = 4, then clearly there are points at distance n − 5 from L which belong to T_{a,b,c}. If j = 6 and |Γ_{1}(ab)| = 3, then one verifies c ∈ S_{a,b}. If |Γ_{1}(ab)| > 3, then similarly as in the previous paragraph, we find a point x ∈ T_{a,b,c} with proj_{ab}x ∉ {proj_{ab}c, a, b}. Finally if j > 6, then, as before, we find a point x in T_{a,b,c} with δ(x, L) = n − 5, and with proj_{L}x ∉ {proj_{L}c, proj_{L}ab}.

Claim 2. Note that all points of T_{a,b} lie opposite w in an apartment through a and b. Let c be an arbitrary point of Γ distinct from a, b and put j = δ(w, c). Without loss of generality, we may assume that a minimal path from c to w contains aw, except if c = w. But in the latter case, clearly c ∈ S_{a,b}. So from now on c ≠ w. If j = 2, then clearly c is opposite every point of T_{a,b}, hence c ∈ S_{a,b}. Now suppose j > 2. Let Σ be an apartment containing b, c. Suppose first j ≡ 0 mod 4 and let M be the line of Σ at distance n − 1 − j/2 from both b, c and at distance n + 1 − j/2 from w. If j = 4 and δ(a, c) = 4, then proj_{M}a ∈ T_{a,b,c}. If j = 4 and δ(a, c) = 2, then c ∈ S_{a,b} would imply b ∈ S_{a,c}, contradicting Claim 1. So we can assume j ≥ 8. Note that δ(a, M) = n + 3 − j/2. We construct a point of T_{a,b,c} as follows. Let x′ be a point at distance j/2 − 3 from M for which proj_{M}x′ does not belong to Σ. Then a and x′ are opposite. Let X′ be a line through x′, X′ ≠ proj_{x′}M and x = proj_{X′}a. Then x ∈ T_{a,b,c}, showing that c ∉ S_{a,b}. Suppose now j ≡ 2 mod 4. If j ≠ n, then we consider an apartment Σ′ containing [b, proj_{c}w], but not containing c. If j = n, then we consider an apartment Σ′ containing [b, L], with L the line of [aw, c] at distance 1 from c, and containing the projection of b onto L (note that we can assume proj_{L}b ≠ c since δ(b, c) ≠ n − 2 for a point c ∈ S_{a,b}). In this way we obtain a path of length h = 2n − j ≡ 2 mod 4 between b and c (combining the path [c, L] with the path between b and L contained in Σ′, but not containing w). We now argue similarly as before. Let M be the line of Σ′ at distance n − j/2 from both b and c, and at distance n − j/2 + 2 from w. Suppose j ≥ 10. Note that δ(a, M) = n − j/2 + 4. Let x′ be a point at distance j/2 − 4 from M for which proj_{M}x′ does not belong to Σ′. Then x′ and a are opposite. As before, the projection of a onto a line X through x′ different from proj_{x′}M belongs to T_{a,b,c}, showing c ∉ S_{a,b}. Now let j = 6 = δ(a, c). In this case, the projection of c onto M belongs to T_{a,b,c}, hence also c ∉ S_{a,b}. If finally j = 6 and δ(a, c) = 4 then c ∈ S_{a,b}. Indeed, let (a, L, p, L′, c) be the 4-path between a and c, and x an arbitrary point of T_{a,b}. Then either proj_{L}x ≠ p, implying δ(x, c) = n, or proj_{L}x = p, implying δ(x, c) ≤ n − 4. This shows Claim 2. We now proceed by induction on δ(a, b). Claim 3. Let c be any point of Γ. Suppose proj_{w}c = proj_{w}a. As before, put δ(w, c) = j (note that j is odd since w is a line). If j > k/2 + 2, then we can find a point in T_{a,b,c} at distance n − 2 − k/2 from w. Indeed, let x be a point at distance n−j from w for which proj_{w}a ≠ proj_{w}x ≠ proj_{w}b. Then c and x are opposite, but δ(a, x) = δ(b, x) < n − 2. Put h = n − 2 − δ(a, x). It is easy to see that one can find a point x′ at distance h − 2 from x, proj_{x}x′ ≠ proj_{x}w, and x′ opposite c. Now the projection of c onto any line X incident with x′ and different from proj_{x′}w is a point of T_{a,b,c}. Hence c ∉ S_{a,b}. If j ≤ k/2 + 2, then one calculates δ(a, c) ≤ k/2 − 2 + j − 2 < k. Now if c would be in S_{a,b}, then b ∈ S_{a,c}. We check that this contradicts the induction hypothesis. So suppose b ∈ S_{a,c}. It is easy to verify that δ(a, c) ∉ {2, 4}. Hence, for any element x of S_{a,c} (so also for b), we have δ(x, a ⋈ c) ≤ \frac{δ(a,c)}{2} + 2 < \frac{k}{2} + 2 by the induction hypothesis. Clearly, c cannot lie on the path [w, a]. If the path [w, c] contains the path [w, a], then we obtain a path of length \frac{3k}{4} + \frac{j}{2} between b and a ⋈ c, a contradiction. So suppose this is not the case. Put [w, a] ∩ [w, c] = [w, L] and r = δ(w, L). Then we obtain a path of length k/2 + j − 2r < n between a and c. Since \frac{k}{4} + \frac{j}{2} − r ≤ \frac{k}{2} − r + 1 = δ(a, L) + 1, the element a ⋈ c belongs to [a, b], or is incident with an element of [a, b]. Hence δ(b, a⋈c) ≥ k/2 + 2, a contradiction. Now suppose proj_{w}a ≠ proj_{w}c ≠ proj_{w}b and j ≥ k/2+4. If j ≠ k/2+6, then similarly as before, we can find a point c′ ∈ T_{a,b,c} with |[w, c′] ∩ [w, c]| = 3. Suppose j = k/2 + 6. Let w′ be the element of [a, b] at distance k/2 − 2 from a. Suppose first n ≥ 12. Let Z be a line at distance n − k/2 − 7 from w′ for which proj_{w′}a ≠ proj_{w′}Z ≠ proj_{w′}w. Note that δ(a, Z) = n − 9 and δ(c, proj_{Z}a) = n. Now consider the path γ′ consisting of the union of the paths [a, Z] and [Z, c] (which has length 2n − 10). Let M be the line of γ′ at distance n − 5 from both a and c, and M′ a line intersecting M, δ(M′, a) = δ(M′, c) = n − 3. Now the point proj_{M′}b belongs to T_{a,b,c}; showing c ∉ S_{a,b}. If n = 10, then necessarily k = 6 and δ(a, c) = 8 (but in this case, c ∉ S_{a,b}) or δ(a, c) = 10. In the latter case, we argue similarly as above, choosing for γ′ the 10-path between a and c containing proj_{a}b. The assertions for k = 6 and |Γ_{1}(m)| ≥ 4 are easy and left for the reader.

Claim 4-5. Let c again be an arbitrary point of Γ. If c = w, then c ∈ S_{a,b} implies b ∈ S_{a,c}, and by the induction hypothesis this only happens if k = 8 (and in this case one easily verifies that indeed w ∈ S_{a,b}). So we may assume that c ≠ w and, without loss of generality, that there is a minimal path from c to w containing A. Put j = δ(c, w) and let ℓ be the distance from w to the unique element X of [a, w] closest to c.

\underline{j < k/2 + 2ℓ} In this case, we can apply the induction hypothesis. Indeed, the condition above implies that the path between a and c consisting of [a, X] and [X, c] (which has length k/2 + j − 2ℓ) is a path of length less than k, so we can apply the induction hypothesis on S_{a,c}. Suppose first the path [w, c] contains the path [w, a]. Then similarly as in the proof of Claim 3, one shows that b ∈ S_{a,c} would contradict the induction hypothesis on S_{a,c}. So we can assume ℓ < k/2. Suppose now j ≤ k/2. This condition implies that the element a ⋈ c belongs to [a, b] (since δ(a, a ⋈ c) = k/4 + j/2 − ℓ ≤ k/2 − ℓ), so δ(b, a ⋈ c) ≥ k/2 + ℓ. Suppose δ(a, c) ≠ 2, 4. Then the fact that b ∈ S_{a,c} implies δ(b, a ⋈ c) ≤ δ(a, c)/2 + 2 = k/4 + j/2 − ℓ + 2. This can only be satisfied if j ≥ k/2 + 4ℓ − 4, which is again only satisfied if ℓ < 2. Hence ℓ = 1 and necessarily j = k/2, which is one of the cases mentioned in Claim 4. Clearly, δ(a, c) ≠ 4 unless k = 8 and c = w. If δ(a, c) = 2, then δ(b, ac) = k − 1, which is only possible if (s, t) = (2, ∞), k = 8, proj_{a}b is a line containing exactly three points and c is the unique point on the line proj_{a}b at distance 8 from b (this gives the exception mentioned in Claim 4). Suppose now k/2 < j < k/2 + 2ℓ (which implies that a ⋈ c is either a line of [X, c] or a point on a line of this path). Note that δ(a, c) ≠ 2, 4. We obtain a path of length d = k/4 + j/2 + l between b and a ⋈ c. If d ≤ n and l > 1, then d > δ(a, c)/2 + 2, a contradiction. But d ≤ n and ℓ = 1 implies k/2 < j < k/2 + 2, a contradiction. If finally d > n, then δ(b, a⋈c) ≥ 2n − d > δ(a, c)/2 + 2, again a contradiction.

So we may assume that j ≥ k/2 + 2l. Let L be the line of [w, b] at distance 3 from w, and x a point on L different from the projections of w and b onto L.

\underline{k/2 + 2l ≤ j ≤ n − 4} We use the same method as in the proof of Claim 2. Let y be a point at distance n − j − 3 from L such that the projection onto L is x, and Y the line through y different from the projection of x onto y. Note that δ(c, Y) = n − 1. We obtain a path γ′ between b and c (consisting of the paths [b, Y] and [Y, c]) of length d = 2n + k/2 − j − 6. Suppose d ≡ 2 mod 4. Consider the line M of γ′ at distance d/2 from both b and c. Let ℓ′ be the length of the path between a and proj_{M}b consisting of [a, L] and [L, proj_{M}b]. Then ℓ′ = n + k/4 − j/2 + 2. If j ≥ k/2 + 4 (which is certainly satisfied if ℓ > 1), then d/2 ≤ n − 5 and ℓ′ ≤ n (and d/2 = n − 5 if and only if ℓ′ = n). Suppose first ℓ′ < n. Then it is possible to find a point z at distance n − 4 − d/2 from M with proj_{M}a ≠ proj_{M}z ≠ proj_{M}b that is opposite a. If Z is the line through z different from proj_{z}b, then there is a unique point on Z (and thus contained in T_{b,c}) at distance n − 2 from a, hence c ∉ S_{a,b}. Suppose now ℓ′ = n (then d/2 = n − 5 and j = k/2 + 4). If we find a point z′ on M, proj_{M}a ≠ z′ ≠ proj_{M}b, that lies opposite a, then the projection of a onto the line through z′ different from M is a point of T_{b,c} at distance n − 2 from a, implying c ∉ S_{a,b}. If we cannot find such a point z′, then M is a line containing three points and proj_{M}a ≠ proj_{M}c ≠ proj_{M}b. A point v on the line W through proj_{M}a, different from M, proj_{W} c ≠ v ≠ proj_{W} a, then is an element of T_{b,c} at distance n − 2 from a, hence c ∉ S_{a,b}. The case ℓ = 1 and j = k/2 + 2 is the second remaining case mentioned in Claim 4. Suppose d ≡ 0 mod 4. Consider a point on Y different from y or proj_{Y} c, and the line Y′ through this point different from Y. Then joining the paths [b, Y′] and [Y′, c] gives a path γ′ between b and c of length ≡ 2 mod 4. Then we proceed similarly as in the previous case. We again obtain the possibility ℓ = 1 and j = k/2 + 2 mentioned in Claim 4.

\underline{j = n − 2} If it is possible to choose the point x on L such that δ(c, x) = n, we proceed as in the previous paragraph. If it is not possible to choose x as above, L contains exactly 3 points, and proj_{L}b ≠ proj_{L}c = x ≠ proj_{L}a. In this case, let γ′ be the union of the paths [b, x] and [x, c]. If this gives a path of length ≡ 0 mod 4, then let y′ be a point at distance 3 from the line L′ = proj_{x}c, x ≠ proj_{L′}y′ ≠ proj_{L′}c, and Y′′ the line through y′ different from proj_{y′}b. Put γ′′ the path (of length ≡ 2 mod 4) joining [b, Y′′] and [Y′′, c]. Now proceeding similarly as before with the path γ′ or γ′′, this gives one of the possibilities mentioned in Claim 5 for the case (s, t) = (2, ∞).

\underline{j = n} A similar reasoning as before gives the other exceptions mentioned in Claim 5.

The claims are proved.

In order to make future arguments uniform, we redefine the set S_{a,b} for two points a, b of Γ in the case n = 12 as follows. Put

S̃_{a,b} = S_{a,b} \ {x ∈ S_{a,b} | Γ_{10}(x) ∩ S_{a,b} ≠ ∅}.

If δ(a, b) = 2 with |ab| = ∞ or δ(a, b) = 4, then S̃_{a,b} = S_{a,b}. If δ(a, b) = 2 with |ab| = 3, then S̃_{a,b} = Γ_{1}(L), with L the unique line concurrent with ab not through a, b. For the cases n = 6 or n = 8 and {s, t} ≠ {2, ∞}, we content ourselves with the observation S̃_{a,b} ⊆ S_{a,b}. Suppose finally δ(a, b) = 8 and {s, t} = {2, ∞}. Now with the notation of Claim 5, if x ∈ S_{a,b} ∩Γ_{8}(a)∩Γ_{8}(b), then δ(x, e) = 10, hence x ∉ S̃_{a,b}. If x ∈ S_{a,b} ∩ Γ_{7}(A′), δ(x, w) = 10, hence x ∉ S̃_{a,b}. If finally x ∈ S_{a,b} ∩ Γ_{11}(B) then since δ(x, b) ≠ n − 2, δ(x, e) = 10, so x ∉ S̃_{a,b}. We conclude that S̃_{a,b} ⊆ S′_{a,b} ∪ {w, e}. We write S_{a,b} for S̃_{a,b} from now on.

Step 2: the set C_{a,b;c} Let c ∈ S_{a,b}. We keep the same notation as in Step 1. Then we define C_{a,b;c} = {c′ ∈ S_{a,b} | S_{c,c′} ∩ {a, b} ≠ ∅}.

For δ(a, b) = k ≡ 2 mod 4 and k ∉ {2, n − 2, n}, we will prove that C_{a,b;c} is always empty, except possibly in the following cases (with w := a⋈b):

(1) δ(c, w) = k/2 − 2. Here, a point c′ ∈ C_{a,b;c} lies at distance k/2 − 2 from w, with proj_{w}c ≠ proj_{w}c′.

(2) δ(c, w) = k/2 + 2. Here, a point c′ ∈ C_{a,b;c} lies at distance k/2 + 2 from w and either proj_{w}c ≠ proj_{w}c′ or proj_{w}c = proj_{w}c′ =: z (and let {w, Z} = Γ_{1}(z)) but proj_{Z}c ≠ proj_{Z}c′; if {s, t} = {2, ∞} and k = 6, then there is an extra possibility (*) for c′ described below.

Indeed, let δ(c, w) = j and suppose c′ ∈ C_{a,b;c}, δ(c′, w) = j′.

Suppose first proj_{w}c = proj_{w}c′. Then δ(c, c′) ≤ j + j′ − 4 ≤ k (because j, j′ ≤ k/2 + 2 by Claim 3 above). Without loss of generality we may assume a ∈ S_{c,c′}. Then, if δ(c, c′) ∉ {2, 4},

δ(a, c⋈c′) ≤ \frac{δ(c, c′)}{2} + 2 ≤ \frac{k}{2} + 2.

Since clearly δ(a, c⋈c′) ≥ k/2 + 2 (c⋈c′ lies on [c, c′]!), this implies j = j′ = k/2 + 2. Using Claim 2 and 3 above, one checks that δ(c, c′) ≠ 4 (indeed, the only possibility would be k = 6 and c or c′ a point on a ⋈ b, but this is excluded by Claim 3). If δ(c, c′) = 2, then it is easy to see that we necessarily have {s, t} = {2, ∞}, k = 6 and

(*) c, c′ are collinear points on a line incident with exactly 3 points and both c, c′ are at distance 5 from w.

These are some of the possibilities mentioned in (2).

Suppose now proj_{w}c ≠ proj_{w}c′. Here, δ(c, c′) = j + j′ ≤ k + 4. We may again assume a ∈ S_{c,c′}. Then, if δ(c, c′) = 2, we must have k = 6 by Claim 1 above (noting that the line w contains at least 4 points in this case). But this contradicts c ∈ S_{a,b} and Claim 3. Also, it is easily verified that δ(c, c′) ≠ 4. Now for δ(c, c′) ∉ {2, 4}, we obtain the following possibilities.

(a) j + j′ ≡ 2 mod 4.

By Claim 3 above, w = c⋈c′ and j = j′. Since a ∈ S_{c,c′} and c, c′ ∈ S_{a,b}, we have k/2−2 ≤ j ≤ k/2+2. The case j = j′ = k/2+2 corresponds to the remaining part of possibility (2). The case j = j′ = k/2 contradicts Claim 3 above (noting w contains at least 4 points here). Finally, the case j = j′ = k/2 − 2 corresponds to possibility (1).

(b) j + j′ ≡ 0 mod 4. Without loss of generality we may assume j > j′. By a ∈ S_{c,c′} and Claim 4, c⋈c′ = proj_{w}c and hence j = j′ + 2. Furthermore, k/2 = (j + j′)/2 ± 1. This implies that either j or j′ is equal to k/2, contradicting c, c′ ∈ S_{a,b} and Claim 3.

This proves (1) and (2).

Step 3: the sets D_{2} and D_{4} if s, t ≥ 3 for both Δ and Δ′ The aim of Step 3 is to construct sets D_{2} and D_{4} consisting of all pairs of points of Γ at mutual distance 2 and 4, respectively, possibly containing some pairs of opposite points as well. Therefore, we first define the sets D′_{2} and D′_{4}, as follows.

A pair (a, b) of points of Γ belongs to D′_{2} if

(1) |S_{a,b}| > 1 and δ(a, b) ≠ n − 2;

(2) |C_{a,b;c}| > 1, for all c ∈ S_{a,b};

Figure 4.8: Proof of Step 3.

(3) there exists a point c ∈ S_{a,b} such that c itself and all points c′ ∈ C_{a,b;c} satisfy Property P(c) and P(c′) respectively, with

P(z) If y ∈ C_{a,b;z} and x ∈ T_{y,z}, then x is at distance n − 2 from all points of C_{a,b;z} ∪ {a} but exactly one;

(4) for all c ∈ S_{a,b} and all c′, c′′ ∈ C_{a,b;c} we have S_{c,c′} ∩ {a, b} = S_{c,c′′} ∩ {a, b} and C_{a,b;c} \ {c′} = C_{a,b;c′} \ {c}.

A pair (a, b) of points of Γ belongs to D′_{4} if

(1′) |S_{a,b}| > 1 and δ(a, b) ≠ n − 2;

(2′) there exists a point c ∈ S_{a,b} such that C_{a,b;c} ≠ ∅ and such that no point of Γ is at distance n − 2 from all the points of C_{a,b;c}.

We show the following assertions.

If δ(a, b) = 2, then (a, b) ∈ D′_{2} \ D′_{4}.

Proof. Clearly, (1) holds. For c ∈ S_{a,b}, one easily sees C_{a,b;c} = Γ_{1}(proj_{c}ab) \ {c, proj_{ab}c}. Now (2) and (4) are clear, while (2′) cannot be satisfied. Every point c ∈ S_{a,b} collinear with a (such c exists) satisfies P(c), whence (3). ◇

If δ(a, b) = 4, then (a, b) ∈ D′_{4} \ D′_{2}.

Proof. Clearly, (1′) holds. Now we put c = a ⋈ b. Then C_{a,b;c} = Γ_{1}(ac) ∪ Γ_{1}(bc) \ {a, b, c}. So it is clear that (2′) is satisfied, but (4) does not hold. Indeed, let c′ ∈ Γ_{1}(ac) and c′′ ∈ Γ_{1}(bc), c′, c′′ ∉ {a, b, c}. Then S_{c,c′} ∩ {a, b} = {b} and S_{c,c′′} ∩ {a, b} = {a}. ◇

If δ(a, b) ≡ 2 mod 4 with 2 ≠ δ(a, b) < n − 2, then (a, b) ∉ D′_{2} ∪ D′_{4}.

Proof. Put w = a ⋈ b. Suppose by way of contradiction that (a, b) ∈ D′_{2}. Let c be a point for which (3) holds and c′, c′′ two distinct arbitrary elements of C_{a,b;c}. By Step 2 the paths [w, c′] and [w, c′′] have at most 3 elements in common with [w, c] and δ(w, c′) = δ(w, c) = δ(w, c′′). Since, by the last part of (4), c′′ ∈ C_{a,b;c} \ {c′} implies c′′ ∈ C_{a,b;c′} \ {c}, it follows from Step 2 that also the paths [w, c′] and [w, c′′] have at most 3 elements in common. Put {c, c′, c′′} = {c_{1}, c_{2}, c_{3}}. In the following, we construct a point x for which δ(a, x) ≠ n−2, and such that x lies at distance n−2 from exactly 2 points of {c_{1}, c_{2}, c_{3}}. Put j = δ(w, c_{1}). Suppose first proj_{w}c_{3} ∉ {proj_{w}c_{1}, proj_{w}c_{2}}. Let x be a point at distance n − 2 − j from w such that the path [w, c_{3}] contains the path [w, x] or vice versa. Then δ(a, x) ≠ n−2 and Γ_{n−2}(x)∩{c_{1}, c_{2}, c_{3}} = {c_{1}, c_{2}}. Assume now proj_{w}c_{1} = proj_{w}c_{2} = proj_{w}c_{3}. Note that j = k/2 + 2. Let x be a point at distance n−k/2 from w such that the path [w, x] contains [w, c_{3}]. Then again δ(a, x) ≠ n − 2 and Γ_{n−2}(x) ∩ {c_{1}, c_{2}, c_{3}} = {c_{1}, c_{2}}. So the point x is as claimed. Now the existence of x contradicts (3). Indeed, suppose Γ_{n−2}(x) ∩ {c_{1}, c_{2}, c_{3}} = {c_{1}, c_{2}}. Then property P(c_{1}) (with y = c_{2}) is not satisfied, since x does not lie at distance n − 2 from the two points a and c_{3} of C_{a,b,c_{1}}. So (a, b) ∉ D′_{2}.

Now suppose by way of contradiction that (a, b) ∈ D′_{4}. Let c ∈ S_{a,b} be as in (2′) and Σ any apartment through a, b. By Step 2, the points of C_{a,b;c} all lie at the same distance from w. Hence a point x of Σ at distance n−2−δ(w, c) from w lies at distance n−2 from all elements of C_{a,b;c}, contradicting (2′) ◇.

For a pair (a, b) of points of Γ, we define

\overline{S}_{a,b} = \{x ∈ S_{a,b} | (a, x), (b, x) ∈ D′_{2} ∪ D′_{4}\}.

Now a pair (a, b) of points of Γ belongs to D_{2} (respectively D_{4}) if

(1′′) (a, b) ∈ D′_{2} ((a, b) ∈ D′_{4} respectively);

(2′′) |\overline{S}_{a,b}| > 1;

(3′′) for any point x of Γ, there are at least 2 points of \overline{S}_{a,b} not lying at distance n − 2 from x.

We show the following assertions.

If δ(a, b) = 2, then (a, b) ∈ D_{2}; if δ(a, b) = 4, then (a, b) ∈ D_{4}. Proof. If δ(a, b) = 2, then clearly S_{a,b} = \overline{S}_{a,b}; if δ(a, b) = 4, then (putting w = a⋈b) Γ_{1}(aw) ∪ Γ_{1}(bw) ⊆ \overline{S}_{a,b} ∪ \{a, b\}. Hence (2′′) and (3′′) are satisfied.

◇

If δ(a, b) ≡ 0 mod 4 with 4 ≠ δ(a, b) < n − 2, then (a, b) ∉ D_{2} ∪ D_{4}. Proof. If δ(a, b) > 8, then Claim 4 of Step 1 implies that for any c ∈ S_{a,b} either δ(a, c) ≡ 2 mod 4 or δ(b, c) ≡ 2 mod 4 (and also, these distances are not equal to 2,4 or n); hence \overline{S}_{a,b} = ∅ (and (2′′) is not satisfied). If δ(a, b) = 8 and n ≠ 12, then similarly \overline{S}_{a,b} = \{a⋈b\} (and again (2′′) is not satisfied). If δ(a, b) = 8 and n = 12, then \overline{S}_{a,b} ⊆ (Γ_{8}(a) ∩ Γ_{8}(b)) ∪ \{a ⋈ b\}. But then, if (2′′) holds, then (3′′) cannot be satisfied by considering the point a⋈(a⋈b).

◇

Hence we have shown that D_{2} consists of all pairs of collinear points of Γ and some (or possibly no) pairs of opposite points; likewise D_{4} consists of all pairs of points of Γ at mutual distance 4 and some (or possibly no) pairs of opposite points.

Step 4: the set Ω of pairs of collinear points if s, t ≥ 3 for both Δ and Δ′ We define the set Ω of pairs of points of Γ as follows. A pair (a, b) belongs to Ω if it belongs to D_{2} and if there exists a pair of points (c, c′) ∈ D_{2}, with {a, b} ∩ {c, c′} = ∅, satisfying

(1) whenever {a, b, c, c′} = {v, v′, w, w′}, then T_{v,v′} ⊆ Γ_{n−2}(w) ∪ Γ_{n−2}(w′);

(2) for any two distinct points x, y ∈ {a, b, c, c′}, we have (x, y) ∈ D_{2};

(3) whenever {a, b, c, c′} = {v, v′, w, w′}, then for all z ∈ T_{v,v′}, we have (w, z), (w′, z) ∉ D_{2} ∪ D_{4}.

We claim that Ω is precisely the set of pairs of collinear points of Γ. Indeed, let (a, b) ∈ D_{2} be arbitrary.

First suppose δ(a, b) = 2. Then we can choose two distinct points c, c′ on the line ab (with {a, b} ∩ {c, c′} = ∅). It is easy to check that (c, c′), which obviously belongs to D_{2}, satisfies (1), (2) and (3) above. We now show for later purposes that, if (c, c′) ∈ D_{2} satisfies (1), (2) and (3), then both c and c′ are incident with the line ab. First assume c ∈ Γ_{2}(a). If c is not incident with the line ab, then δ(b, c) = 4 and so (b, c) ∉ D_{2}. Hence cIab. If c′ is not incident with ab, then it must be opposite a, b and c, and hence proj_{ab}c′ ∉ {a, b, c}. But then the point y collinear with c′ on the path [c′, ab] belongs to T_{a,b} and contradicts (3) since (c′, y) ∈ D_{2}. So we may assume that both c, c′ are opposite a, b. But then again the point y collinear with c′ on the path [c′, ab] contradicts (3) since (c′, y) ∈ D_{2}.

Hence we have shown that

(∗) if (a, b) ∈ Ω and δ(a, b) ≠ 2, then for any pair of distinct points c, c′ ∈ D_{2} satisfying (1), (2) and (3), we must have δ(x, y) = n, for any two distinct points x, y in {a, b, c, c′}.

Indeed, if two elements of {a, b, c, c′} would be collinear, then we can let them play the roles of a and b in the previous paragraph and obtain a contradiction (by remarking that all conditions (1) up to (3) are symmetric in a, b, c, c′).

Now suppose δ(a, b) = n. We must show (a, b) ∉ Ω. Suppose by way of contradiction that there exists a pair of points (c, c′) ∈ D_{2}, with {a, b} ∩ {c, c′} = ∅, and satisfying conditions (1), (2) and (3). If n ≡ 2 mod 4, we choose a fixed line M of Γ at distance n/2 from both a and b. If n ≡ 0 mod 4, we choose a fixed line M at distance n/2 + 1 from both a and b (such a line can be obtained as follows: fix a line A through a and let B be the line through b opposite A; let a′ be a point on A, a ≠ a′ ≠ proj_{A}b, and put b′ = proj_{B}a′; let then M be the line of [a′, b′] at distance n/2 − 1 from both a′ and b′). In both cases (by possibly interchanging the roles of the two lines through a, and hence also of those through b), we may assume that M contains more than four points (this follows from our assumption that at most one of the parameters s, t is equal to 3). Let Y be a line at distance j from M, 0 ≤ j ≤ n − 3 − δ(a, M), with proj_{M}b ≠ proj_{M}Y ≠ proj_{M}a (note that δ(a, M) < n − 3 since n > 8). Define the following sets T_{Y}:

T_{Y} := {x ∈ P | δ(x, Y) = (n−2)−δ(a, M)−j and proj_{Y} a ≠ proj_{Y} x ≠ proj_{Y} b}.

Note that T_{Y} ⊆ T_{a,b}, hence by (1), T_{Y} ⊆ Γ_{n−2}(c) ∪ Γ_{n−2}(c′). We first prove, by induction on j = δ(Y, M), that T_{Y} ⊈ Γ_{n−2}(v), v ∈ {c, c′}, for all lines Y for which the set T_{Y} is defined.

First let j = 0. Then Y = M. Suppose T_{M} ⊆ Γ_{n−2}(c). Then it is easy to see that δ(a, M) = δ(c, M) and proj_{M}a = proj_{M}c or proj_{M}b = proj_{M}c. Assume proj_{M}a = proj_{M}c. This implies that δ(a, c) ≤ n − 2, so (since (a, c) ∈ D_{2} by (2)), δ(a, c) = 2, contradicting (∗). Hence T_{M} ⊈ Γ_{n−2}(v) for any v ∈ {c, c′}. Now let j = 2. So let N be a line concurrent with M, not through the projection of a or b onto M. Suppose T_{N} ⊆ Γ_{n−2}(c). Then δ(c, N) = δ(a, N) = δ(a, M) + 2 and proj_{N}c = proj_{N}a but proj_{M}a ≠ proj_{M}c ≠ proj_{M}b (because otherwise T_{M} ⊆ Γ_{n−2}(c)). Now δ(a, M) = δ(c, M). If δ(a, M) = n/2, then the point y on [M, c] collinear with c belongs to T_{a,b}. If δ(a, M) = n/2 + 1, then the point y on [M, c] at distance 4 from c belongs to T_{a,b}. In both cases, (c, y) ∈ D_{2} ∪ D_{4}, contradicting (3). Hence T_{N} ⊈ Γ_{n−2}(v), v ∈ {c, c′} for all lines N concurrent with M, not through the projection of a or b onto M.

Now let j ≥ 4 be arbitrary, j ≤ n−3−δ(a, M) and let Y be a line at distance j from M with proj_{M}b ≠ proj_{M}Y ≠ proj_{M}a. Suppose T_{Y} ⊆ Γ_{n−2}(c). Let [Y, M] =: (Y, p, Y′, p′, Z, …, M) (with possibly Z = M). Then δ(c, Y) = δ(a, Y) = δ(a, M) + j and proj_{p}c = proj_{p}a = Y′ but proj_{p′}a ≠ proj_{p′}c (otherwise T_{Y′} ⊆ Γ_{n−2}(c), contradicting the induction hypothesis). Let Y′′ be the line through proj_{Y′}c = p′′, different from Y′. Now it is readily checked that T_{Y′′} ∩Γ_{n−2}(c) = ∅, so (1) implies T_{Y′′} ⊆ Γ_{n−2}(c′). Since also δ(Y′′, M) = j, we have that δ(c′, Y′′) = δ(a, Y′′) = δ(a, M) + j, proj_{p′′}c′ = proj_{p′′}a = Y′ but proj_{p′}a ≠ proj_{p′}c′. Let X be a line concurrent with Z, not through p′ or the projection of a or b onto Z. Consider a line L at distance n−1−δ(a, M)−j from X with proj_{X}M ≠ proj_{X}L (then all the points of L except from proj_{L}a are points of T_{a,b}). Since δ(c, L) = δ(c′, L) = n − 1, there is exactly one point of L at distance n − 2 from c, and the same for c′. This is a contradiction with (1), since L contains at least 3 points of T_{a,b}. Hence T_{Y} ⊈ Γ_{n−2}(v), v ∈ {c, c′}, for all lines Y for which the set T_{Y} is defined.

Now consider a line K at distance n − 5 − δ(a, M) from M for which the set T_{K} is defined. Let R, R′ and R′′ be three different lines concurrent with K at distance n − 3 − δ(a, M) from a (such lines exist because s, t ≥ 3 and since, if K = M, which occurs if n = 10 or n = 12, then M contains at least three points different from proj_{M}a and proj_{M}b by assumption). We already know that T_{R} ⊈ Γ_{n−2}(v), v ∈ {c, c′} , so the only remaining possibility for the points c and c′ is that (since T_{R} contains at least 3 points and necessarily T_{R} ⊆ Γ_{n−2}(c) ∪ Γ_{n−2}(c′)) up to interchanging c and c′, the point c lies at distance n − 4 from a point r on R, r not on K, with proj_{r}c ≠ R. Because then c is opposite all but one point of T_{R′}, we must have that the point c′ lies at distance n − 4 from a point r′ on R′, r′ not on K, with proj_{r′}c′ ≠ R′. But now at most two points of T_{R′′} will be contained in Γ_{n−2}(c) ∪ Γ_{n−2}(c′), a contradiction with (1) and the fact that T_{R′′} contains at least 3 points. This shows that the points c, c′ cannot exist, so (a, b) ∉ Ω.

This shows that α preserves collinearity if both the orders of Δ and Δ′ do not contain a 2, and completes the proof in this case.

Step 5: the sets D_{2}, D′_{2} and D_{4} if {s, t} = {2, ∞} for both Δ and Δ′ The aim of Step 5 is to construct sets D_{2}, D′_{2} and D_{4} (for the case {s, t} = {2, ∞}) consisting of all pairs of points of Γ at mutual distance 2 (and the joining lines have infinitely many points or exactly three points, for D_{2} and D′_{2} respectively) and 4, respectively, possibly containing some pairs of opposite points as well. Therefore, we first define the sets E_{2} and E_{4}, as follows.

A pair (a, b) of points of Γ belongs to E_{4} if

(1) |S_{a,b}| > 1 and δ(a, b) ≠ n − 2;

(2) there is a point c ∈ S_{a,b} such that |C_{a,b;c}| = ∞ and such that no point x of Γ satisfies {a, c} ∪ C_{a,b;c} ⊆ Γ_{n−2}(x).

A pair (a, b) of points of Γ belongs to E_{2} if

(1′) |S_{a,b}| > 1 and δ(a, b) ≠ n − 2;

(2′) no point lies at distance n − 2 from all elements of S_{a,b};

(3′) for every point c ∈ S_{a,b} we have |C_{a,b;c}| = 1, and, putting C_{a,b;c} = {c′}, we must have (c, c′) ∈ E_{4}.

Note that the sets E_{2} and E_{4} are disjoint, because of properties (2) and (3′). We show the following assertions.

If two points a, b are collinear in Γ and the line ab contains exactly three points, then (a, b) ∈ E_{4} and (a, b) ∉ E_{2}.

Proof. Note that Γ_{3}(ab) ⊆ S_{a,b}, hence (1) holds. Let e be the unique point on ab different from a and b, and c a point of Γ_{3}(ab) collinear with e. Then Γ_{1}(ec) ⊆ C_{a,b;c} ∪ {c}; showing that (2) holds for this point c. So (a, b) ∈ E_{4} \ E_{2}. ◇

If two points a, b are collinear in Γ and the line ab contains infinitely many points, then (a, b) ∈ E_{2} and (a, b) ∉ E_{4}.

Proof. In this case, S_{a,b} = Γ_{3}(ab), hence (1′) and (2′) hold. For any point c ∈ S_{a,b}, the set C_{a,b;c} contains exactly one point, namely the unique point on the line L = proj_{c}ab different from c and proj_{L}a. This shows that (3′) holds, hence (a, b) ∈ E_{2} \ E_{4}. ◇

If two points a, b are at mutual distance 4 in Γ, then (a, b) ∈ E_{4} and (a, b) ∉ E_{2}.

Proof. Put c = a ⋈ b. Then C_{a,b;c} contains the set Γ_{1}(ac) ∪ Γ_{1}(bc) \ {a, b, c}. This shows that (2) holds for the point c, hence (a, b) ∈ E_{4} \ E_{2}. ◇

If δ(a, b) ≡ 2 mod 4 with 2 ≠ δ(a, b) < n − 2, then (a, b) ∉ E_{2} ∪ E_{4}.

Proof. Suppose by way of contradiction that (a, b) ∈ E_{4}, and let c be a point in S_{a,b} satisfying (2). But since any element of C_{a,b;c} lies at distance δ(a ⋈ b, c) from a ⋈ b, one can easily find a point at distance n − 2 from all points of C_{a,b;c} ∪ {a, c}; a contradiction, hence (a, b) ∉ E_{4}. Suppose now (a, b) ∈ E_{2}, and let c ∈ S_{a,b} with C_{a,b;c} = {c′}. Put w := a ⋈ b. If δ(w, c) = k/2 − 2 = δ(w, c′), then δ(c, c′) = k − 4 ≡ 2 mod 4. Hence (c, c′) ∈ E_{4} implies δ(c, c′) = 2, so k = 6 and c, c′ are incident with m. This contradicts Claim 3 in Step 1. If δ(w, c) = k/2 + 2 = δ(w, c′), then either δ(c, c′) = k + 4 ≡ 2 mod 4 or δ(c, c′) = k ≡ 2 mod 4 (and in both cases, we obtain a contradiction with (c, c′) ∈ E_{4}) or k = 6 and c, c′ are collinear points on a line with exactly 3 points (this is the exception mentioned in (*)). But in the latter case, all points of S_{a,b} lie at distance k/2 + 2 from a⋈b, and we easily find a point at distance n − 2 from every point of S_{a,b}, contradicting (2′). Hence (a, b) ∉ E_{2}. ◇

Now we define

\overline{S}_{a,b} = \{x ∈ S_{a,b} | (a, x), (b, x) ∈ E_{2} ∪ E_{4}\}.

Completely similar as in Step 3, one shows that if two points a, b satisfy 4 < δ(a, b) = k ≡ 0 mod 4, k < n − 2, then |\overline{S}_{a,b}| ≠ ∞. Also, for two points a, b with δ(a, b) ∈ \{2, 4\}, |\overline{S}_{a,b}| = ∞.

By definition, a pair (a, b) of points of Γ belongs to D_{2} if (a, b) ∈ E_{2} and |\overline{S}_{a,b}| = ∞. Also, a pair (a, b) of points of Γ belongs to D′_{2} if (a, b) ∈ E_{4}, |\overline{S}_{a,b}| = ∞ and there are some c, c′ ∈ S_{a,b} such that (a, c), (b, c′) ∈ D_{2}. Finally, D_{4} consists precisely of those pairs (a, b) of points of E_{4} \ D′_{2} that satisfy |\overline{S}_{a,b}| = ∞. We conclude that D_{2} consists of all pairs (a, b) of collinear points with |Γ_{1}(ab)| = ∞, possibly together with some pairs of opposite points; D′_{2} consists of all pairs (a, b) of collinear points with |Γ_{1}(ab)| = 3, possibly together with some pairs of opposite points; D_{4} consists of all pairs (a, b) of points at mutual distance 4, possibly together with some pairs of opposite points.

Step 6: the set Ω of pairs of collinear points if {s, t} = {2, ∞} for both Δ and Δ′ Note that n ≡ 0 mod 4 (indeed, remember that (s, t) was the order of the corresponding n/2-gon Δ). We first pin down the set \overline{Ω} of pairs (a, b) of collinear points with |Γ_{1}(ab)| = ∞. Therefore we define V_{a,b}, for two arbitrary points a, b of Γ, as V_{a,b} = Γ_{n−2}(a) \ Γ_{n−2}(b). Now let \overline{Ω} be the set of pairs (a, b) of D_{2} such that there exist points c, c′, c′′ in Γ, all distinct from a and from b, with the following properties.

(1) V_{a,b} is the disjoint union of the sets V_{a,b} ∩ Γ_{n−2}(c), V_{a,b} ∩ Γ_{n−2}(c′) and V_{a,b} ∩ Γ_{n−2}(c′′);

(2) (a, c′), (a, c′′) ∈ D′_{2}; (b, c′), (b, c′′) ∈ D_{4} and (a, c), (b, c) ∈ D_{2};

(3) no point x in Γ_{n−2}(a) ∩ Γ_{n−2}(c) satisfies (b, x) ∈ D_{2} ∪ D′_{2}; likewise no point x in Γ_{n−2}(b) ∩ Γ_{n−2}(c) satisfies (a, x) ∈ D_{2} ∪ D′_{2};

(4) a ∈ S_{b,c′} ∩ S_{b,c′′};

(5) if {u, v, w} = {c, c′, c′′}, then v, w ∈ S_{a,u} ∩ S_{b,u}.

Figure 4.9: (I) : (a, b) ∈ \overline{Ω} (II) : The sets T_{L_{j}} for n = 12.

We now show that \overline{Ω} is the set of pairs (a, b) of collinear points with |Γ_{1}(ab)| = ∞. Clearly, if δ(a, b) = 2 and |Γ_{1}(ab)| = ∞, then choosing c ∈ Γ_{1}(ab) \ {a, b} arbitrarily, and putting \{c′, c′′\} = Γ_{2}(a) \ Γ_{1}(ab), we see that (a, b) ∈ \overline{Ω} (see Figure 4.9 (I)).

So there remains to show that no pair of opposite points belongs to \overline{Ω}. By way of contradiction, let (a, b) be a pair of opposite points of Γ belonging to \overline{Ω}. Let c, c′, c′′ be as in (1) up to (5) above.

We claim that δ(a, x) = n = δ(b, x), for x ∈ {c, c′, c′′}. Indeed, by (2), we already know that δ(a, c′) and δ(a, c′′) are either 2 or n. Suppose δ(a, c′) = 2. Then δ(b, ac′) = n − 1. But b ∈ S_{a,c′} (which is Condition (4)) implies δ(b, ac′) ≤ 7 (by Claim 1 of Step 1), a contradiction. Similarly for (a, c′′). Also, δ(b, c′) and (b, c′′) are either 4 or n. Suppose δ(b, c′) = 4 and put w = b⋈c′. Then δ(a, bw) = n − 1. Condition (4) states that a ∈ S_{b,c′}, hence by Claim 2 of Step 1, δ(a, bw) ≤ 7, a contradiction. Similarly for (b, c′′). By (2), we also know that δ(a, c) and δ(b, c) are either 2 or n. If δ(a, c) = 2 (and so δ(b, c) = n), then the point collinear with b at distance n − 3 from the line ac lies at distance n − 2 from both a and c, contradicting Condition (3). Similarly for δ(b, c). This show the claim.

Form now on until the end of the proof, we assume that n is “large enough” (the generic case) in certain arguments. When n is too small, then either the given argument can be skipped or a separate but easier argument can be given (and we do not do that explicitly).

Let γ be the path of length n between a and b for which the line of γ through b contains exactly three points. Denote by L_{j} the line of γ at distance j (j is odd!) from a and define for n/2 − 1 ≤ j ≤ n − 3,

T_{L_{j}} = {x ∈ P|δ(x, L_{j}) = n − 2 − j, proj_{L_{j}}a ≠ proj_{L_{j}}x}.

Note that the sets T_{L_{j}} are subsets of V_{a,b}, and that these sets consist of unions of certain sets Γ_{1}(L) \ {proj_{L}a}, with |Γ_{1}(L)| = ∞ (see Figure 4.9 (II)).

For an element z at distance ≤ n − 2 − j from L_{j} for which proj_{L_{j}}a ≠ proj_{L_{j}}z ≠ proj_{L_{j}}b, we define the set

T_{z} = {x ∈ P|δ(x, z) = n − 2 − j − δ(z, L_{j}), proj_{z}a ≠ proj_{z}x}.

Note that T_{z} is the subset of T_{L_{j}} containing the points x for which [x, L_{j}] contains z.

Let Z be a line for which the set T_{Z} is defined. We first show by induction on i_{Z} := n − δ(a, Z) that

(◇) for such a line Z there exist points v, v′ ∈ {c, c′, c′′} such that T_{Z} ⊂ Γ_{n−2}(v) ∪ Γ_{n−2}(v′). Moreover, for any two points z′, z′′ ∈ Γ_{1}(Z) \ {proj_{Z}a, proj_{Z}b}, we have that T_{z′} ⊆ Γ_{n−2}(v) ∪ Γ_{n−2}(v′), with T_{z′} ∩ Γ_{n−2}(v) ≠ ∅ ≠ T_{z′} ∩ Γ_{n−2}(v′), implies T_{z′′} ⊆ Γ_{n−2}(v) ∪ Γ_{n−2}(v′), with

T_{z′′} ∩ Γ_{n−2}(v) ≠ ∅ ≠ T_{z′′} ∩ Γ_{n−2}(v′).

Suppose first i_{Z} = 3. Then Z is a line at distance n − 3 from a containing an infinite number of points, and T_{Z} = Γ_{1}(Z) \ {proj_{Z}a}. Condition (1) implies that there exists a point x ∈ {c, c′, c′′} at distance n − 2 from at least two points of T_{Z}, hence at distance n − 2 from all but at most one point of T_{Z}. This shows (◇) for the line Z.

Now we assume i_{Z} = 5. Then necessarily |Γ_{1}(Z)| = 3. If Z = L_{n−5}, then T_{Z} = T_{N}, with N the unique line concurrent with Z and different from proj_{Z}a and proj_{Z}b, and (◇) follows. So suppose Z ≠ L_{n−5}. Let r and r′ be the two points on Z different from proj_{Z}a, and let R, R′ be the lines through r respectively r′ different from Z. Put Z′ the line through proj_{Z}a, different from Z. We claim that

(*) no point v ∈ {c, c′, c′′} is at distance n − 2 from exactly one point of Γ_{1}(R) \ {proj_{R}a} and from exactly one point of Γ_{1}(R′) \ {proj_{R′}a}.

Indeed, suppose some v ∈ {c, c′, c′′} is at distance n−2 from exactly one point of T_{R} and from exactly one point of T_{R′}. Since by Condition (1) every point of V_{a,b} lies at distance n−2 from exactly one point of {c, c′, c′′}, there exists a point v′ ∈ {c, c′, c′′}\{v} such that δ(v′, R) = n−3 and proj_{R}v = proj_{R}v′. But then proj_{R′}v ≠ proj_{R′}v′ (because δ(v′, proj_{R′}v′) = n − 2 and proj_{R′}v ∈ V_{a,b}) and so the unique point of {c, c′, c′′} \ {v, v′} lies at distance n − 2 from all but two or three points of R′, which is impossible. Our claim is proved.

If R is not contained in Γ_{n−2}(v) for a point v ∈ {c, c′, c′′}, then some point w ∈ {c, c′, c′′} is at distance n − 4 from exactly one point of T_{R}, and at distance n − 2 from all the other points of T_{R}. But then there is exactly one point of T_{R′} at distance n − 2 from w. So the only possibility to satisfy Condition (1) is that a point w′ ∈ {c, c′, c′′} \ {w} is at distance n − 4 from exactly one point of T_{R′} (namely proj_{R′}w) and at distance n − 2 from all the other points of this set and at distance n − 2 from proj_{R}w. Whence (◇).

If T_{R} ⊂ Γ_{n−2}(v) for some v ∈ {c, c′, c′′}, then δ(v, r) = n − 4 and proj_{r}c = Z. If proj_{Z}v = r′, then we consider a line Z′′ concurrent with Z′, different from Z and not through proj_{Z′}a or proj_{Z′}b (this is possible since |Z′| = ∞). But now δ(v, Z′′) = n − 1 and δ(v, proj_{Z′′}Z) = n − 2, so v is at distance n − 2 from exactly two non-collinear points of T_{Z′′}, contradicting (*). So proj_{Z}v = proj_{Z}a and T_{Z} ⊆ Γ_{n−2}(v).

This shows (◇) for the line Z.

Now suppose i_{Z} > 5. Put j = n − i_{Z} = δ(a, Z). Suppose first that |Γ_{1}(Z)| = 3. If Z = L_{j} (i.e., if Z belongs to γ), then T_{L_{j}} = T_{L}, with L the unique line concurrent with L_{j} and not contained in γ, and with i_{L} = i_{Z} −2. So the result follows from the induction hypothesis. Hence we may assume that Z does not belong to γ. Put Γ_{1}(Z) = {x, x_{1}, x_{2}} with x = proj_{Z}a, put L = proj_{x}a and let X_{i} be the line through x_{i} distinct from Z, i = 1, 2. By the induction hypothesis, there are two cases to consider.

(i) There exists v ∈ {c, c′, c′′} such that T_{X_{1}} ⊆ Γ_{n−2}(v). We show that T_{X_{2}} ⊆ Γ_{n−2}(v). Indeed, δ(v, x_{1}) = j + 1 and proj_{x_{1}}v = Z. If proj_{Z}v ≠ x_{2}, then clearly T_{X_{2}} ⊂ Γ_{n−2}(v). If proj_{Z}v = x_{2}, then consider an arbitrary point p at distance n − 4 − j from L for which proj_{L}p ∉ {x, proj_{L}a, proj_{L}b}. The point p lies at distance n − 2 from v and a contradiction to (*) arises in the set T_{p} (indeed, we find exactly two non-collinear points of T_{p} lying at distance n − 2 from v).

(ii) Suppose now that we are not in case (i) and there exist v, v′ ∈ {c, c′, c′′}, v ≠ v′, such that T_{X_{1}} ⊆ Γ_{n−2}(v) ∪ Γ_{n−2}(v′), with T_{X_{1}} ∩ Γ_{n−2}(v) ≠ ∅ ≠ T_{X_{1}} ∩ Γ_{n−2}(v′). From the proof of the case i_{Z} = 5 now follows that δ(v, x_{1}) = δ(v′, x_{1}) = δ(a, x_{1}) + 6 = j + 7. (Indeed, let Y be a line at distance n − 5 from a for which the path [a, Y] contains the path [a, Z] and let Y_{1}, Y_{2} be the two lines concurrent with Y and at distance n − 3 from a. Then the proof of the case i_{Z} = 5 shows that we can assume δ(v, Y_{1}) = n − 3, δ(v′, Y_{2}) = n − 3, δ(v, Y) = δ(v′, Y) = n − 1 and, with Y′ the line concurrent with Y at distance n − 7 from a, proj_{Y′}v = proj_{Y′}a = proj_{Y′}v′.) If δ(v, x_{2}) = δ(v′, x_{2}) = j + 7, then T_{X_{2}} ⊆ Γ_{n−2}(v) ∪ Γ_{n−2}(v′) with T_{X_{2}} ∩ Γ_{n−2}(v) ≠ ∅ ≠ T_{X_{2}} ∩ Γ_{n−2}(v′). Suppose now by way of contradiction that δ(v, x_{2}) = j + 5. Then we consider a point p at distance n − (j + 8) from L such that proj_{L}a ≠ proj_{L}p ≠ x and proj_{L}p ≠ proj_{L}b. Put Γ_{1}(p) = {proj_{p}a, R}. Note that v and p are opposite points of Γ. Put [R, v] = (R, p′, R′, p′′, …, v), and let r be any point incident with R′, p′ ≠ r ≠ p′′. Then considering T_{r} and v, we obtain a contradiction to (*).

This shows (◇) for the case |Γ_{1}(Z)| = 3.

Suppose now |Γ_{1}(Z)| = ∞. Suppose first Z ∉ γ. Let x be any point on Z different from proj_{Z}a. By the induction hypothesis, there are two cases.

(i) There exists v ∈ {c, c′, c′′} such that T_{x} ⊆ Γ_{n−2}(v). Similarly as above, one shows that in this case T_{y} ⊆ Γ_{n−2}(v), for all y ∈ Γ_{1}(Z) \ {proj_{Z}a}, except possibly for one point x^{∗} ∈ Γ_{1}(Z) \ {proj_{Z}a}, in which case T_{x^{∗}} ∩ Γ_{n−2}(v) = ∅.

(ii) There exists v ∈ {c, c′, c′′} such that T_{x} ≠ T_{x} ∩ Γ_{n−2}(v) ≠ ∅. Again similarly as above, one shows that in this case T_{y} ≠ T_{y} ∩ Γ_{n−2}(v) ≠ ∅ for all y ∈ Γ_{1}(Z) \ {proj_{Z}a}, except possibly for one point x^{∗} ∈ Γ_{1}(Z) \ {proj_{Z}a}, in which case T_{x^{∗}} ∩ Γ_{n−2}(v) = ∅.

Combining (i), (ii) and |Γ_{1}(Z)| = ∞ (and using the fact that the three sets T_{Z} ∩ Γ_{n−2}(v), v ∈ {c, c′, c′′} are disjoint), we readily deduce (◇). If Z ∈ γ, then a similar reasoning shows the result.

So we have shown (◇) for all appropriate lines Z.

Suppose now that there exists v ∈ {c, c′, c′′} with T_{L_{n−3}} ⊆ Γ_{n−2}(v). We look for a contradiction. Note that δ(v, L_{n−3}) = n − 3 and proj_{L_{n−3}}a = proj_{L_{n−3}}v. Define j ∈ ℕ as [a, L_{n−3}] ∩ [v, L_{n−3}] = [L_{j}, L_{n−3}]. Suppose first n/2 < j ≤ n − 5. Then v lies at distance j from L_{j}. Consider a point p at distance n − j − 4 from L_{j−2} satisfying proj_{L_{j−2}}a ≠ proj_{L_{j−2}}p ≠ proj_{L_{j−2}}b. Then p lies at distance n − 2 from v and at distance n − 6 from a, and we obtain a contradiction to (*) by considering T_{p} and v. Suppose now j ≤ n/2 − 1. Then δ(a, v) ≤ δ(a, L_{j}) + δ(L_{j}, v) ≤ n − 2, the final contradiction (since δ(a, v) = n, for v ∈ {c, c′, c′′}).

Hence, since at least one element of {c, c′, c′′} must be at distance n − 2 from infinitely many points of L_{n−3}, there exists v ∈ {c, c′, c′′} satisfying δ(v, L_{n−3}) = n − 3 and proj_{L_{n−3}}v ∉ γ (remembering v is opposite b). Now v lies at distance n − 2 from exactly one point of T_{L_{n−5}}, so there is a v′ ∈ {c, c′, c′′}\{v} such that T_{L_{n−5}} ⊆ Γ_{n−2}(v)∪Γ_{n−2}(v′). Hence v′ lies at distance n − 1 from L_{n−5} and proj_{L_{n−5}}v′ ∉ γ. Note that both v and v′ are opposite the point w := proj_{L_{n−7}}b, hence δ(v, L_{n−7}) = n − 1 = δ(v′, L_{n−7}). Let j be defined as [L_{n−7}, a] ∩ [L_{n−7}, v] = [L_{n−7}, L_{j}] and let j′ be defined as [L_{n−7}, a] ∩ [L_{n−7}, v′] = [L_{n−7}, L_{j′}] (these are well-defined since a ∉ [L_{n−7}, v] ∪ [L_{n−7}, v′]). Then δ(v, L_{j}) = j + 6 and δ(v′, L_{j′}) = j′ + 6, with proj_{Lj}v, proj_{Lj′}v′ ∉ γ.

Suppose first n/2 − 2 < j and, if n ≡ 0 mod 8, j ≠ n/2 − 1.

(i) If |Γ_{1}(L_{j})| = 3, then, because of the conditions on j, the line L_{j−2} has infinitely many points and the set T_{L_{j−2}} is defined. We proceed similarly as in (ii) of the proof of (◇), case i_{Z} > 5 and |Γ_{1}(Z)| = 3 (see above) to obtain a contradiction with (*).

(ii) If |Γ_{1}(L_{j})| = ∞, then let x = proj_{L_{j}}v. Calculating distances, it is easy to check that T_{x} ∩ Γ_{n−2}(v) = ∅ and T_{x′} ∩ Γ_{n−2}(v) ≠ ∅, for all points x′ ∈ Γ_{1}(L_{j}) \ {x, proj_{L_{j}}a, proj_{L_{j}}b}. This contradicts (◇).

We now treat the remaining cases. Note that in the foregoing, we may interchange the roles of j and j′.

(iii) If n ≡ 0 mod 8 and j ≤ n/2 − 1, then j′ ≤ n/2 − 1 and |Γ_{1}(L_{n/2−1})| = 3. Note that {j, j′} ⊆ {n/2 − 1, n/2 − 3} (since both v and v′ are opposite a). If j = j′ = n/2 − 1, then T_{L_{n/2−1}} ∩ Γ_{n−2}(v) = ∅ = T_{L_{n/2−1}} ∩ Γ_{n−2}(v′), so T_{L_{n/2−1}} ⊂ Γ_{n−2}(v′′), with {v, v′, v′′} = {c, c′, c′′}. But this implies that δ(v′′, L_{n/2−1}) = n/2 − 1 and proj_{L_{n/2−1}}v′′ ∈ γ. So, calculating distances, we see that either δ(a, v′′) < n or δ(b, v′′) < n, a contradiction.

Suppose j = n/2 − 1 and j′ = n/2 − 3, or j = j′ = n/2 − 3. Let a′ and a′′ be the two points on the line L_{3} at distance n − 2 from b. Then at least one of these two points lies at distance n − 2 from the points v, v′ and b, contradicting Condition (5) (namely v′ ∈ S_{b,v}). This concludes the case n ≡ 0 mod 8.

(iv) If n ≡ 4 mod 8 and j ≤ n/2−3, then again the case j < n/2−3 can not occur. So j = n/2−3, and by symmetry, also j′ = n/2−3. We proceed similarly as in the last part of (iii) above to obtain a contradiction with Condition (5).

This shows that a pair of opposite points never belongs to \overline{Ω}. Hence \overline{Ω} consists precisely of all pairs (a, b) of collinear points with Γ_{1}(ab) = ∞.

Finally, we define

\overline{\overline{Ω}} = \{(a, b) ∈ D′_{2}|(∀z ∈ Γ_{n−2}(b))((a, z) ∉ \overline{Ω})\}.

Clearly, if a and b are collinear points with |ab| = 3, then (a, b) ∈ \overline{\overline{Ω}}. But if δ(a, b) = n then, with L the line through a containing infinitely many points, the point proj_{L}b lies at distance n − 2 from b, and (a, proj_{L}b) ∈ \overline{Ω}. Hence (a, b) ∉ \overline{\overline{Ω}}, and \overline{\overline{Ω}} consists precisely of all pairs (a, b) of collinear points with Γ_{1}(ab) = 3.

Now Ω := \overline{Ω} ∪ \overline{\overline{Ω}} is the set of all pairs of collinear points of Γ. This shows that α preserves collinearity in case both the orders of Δ and Δ′ contain a 2.

Step 7: Distinction between the orders

By the results of the previous steps, we know that the given bijection extends to an isomorphism between Δ and Δ′ if the orders of Δ and Δ′ both contain a 2, or if they both do not contain a 2. We now want to exclude the remaining case. So suppose by way of contradiction that every line of Γ contains at least 4 points, and that Γ′ has lines containing exactly 3 points. Note that we can assume that both Γ and Γ′ are infinite, and n ≡ 0 mod 4. In the case of Γ′, the set \overline{Ω} defined in Step 6 is non-empty (since it contains all pairs of collinear points a, b for which |Γ′_{1}(ab)| = ∞). We now show that the set \overline{Ω} (with \overline{Ω} as in Step 6) defined for the polygon Γ is empty. Since the size of the set \overline{Ω} is preserved by α, this is a contradiction. First, one has to determine the sets D_{2}, D′_{2} and D_{4} as defined in Step 5 for the polygon Γ. It is easy to check that D_{2} and D′_{2} can only contain pairs of points (a, b) for which δ(a, b) = n. The set D_{4} consists of all pairs of points (a, b) for which δ(a, b) = 4, or for which δ(a, b) = 2 such that the line ab is concurrent with a line containing infinitely many points, possibly together with some pairs of opposite points. Now suppose (a, b) ∈ \overline{Ω}. Then we have δ(a, b) = n, and there exist points c, c′, c′′ in Γ such that the conditions (1) up to (5) listed in Step 6 are satisfied. Similarly as in Step 6, one proves that δ(a, v) = δ(b, v) = n, for v ∈ {c, c′, c′′}.

Let γ be a fixed n-path between a and b such that the line of γ through a contains infinitely many points. Let L_{0} be the line of γ at distance \frac{n}{2} − 1 from a. Let L_{j} be a line at distance j from L_{0}, 0 ≤ j ≤ \frac{n}{2} − 2, for which proj_{L_{0}}a ≠ proj_{L_{0}}L_{j} ≠ proj_{L_{0}}b. For such a line L_{j}, we define the set T_{L_{j}} as follows:

T_{L_{j}} = {x ∈ P|δ(x, L_{j}) = n − 2 − j − δ(a, L_{j}), proj_{L_{j}}a ≠ proj_{L_{j}}x ≠ proj_{L_{j}}b}.

The sets T_{L_{j}} are subsets of the set V_{a,b} and consist of unions of sets Γ_{1}(L) \ {proj_{L}a}. Note that by the choice of γ, these sets Γ_{1}(L) contain infinitely many points.

We prove by induction on j that there does not exist a point v ∈ {c, c′, c′′} and a line L_{j} for which the set T_{L_{j}} is defined such that T_{L_{j}} ⊆ Γ_{n−2}(v). This is done in three parts. In (A), we handle the case j = 0 or j = 2, in (B) the case 2 < j ≤ \frac{n}{2} − 4 and in (C) the case j = \frac{n}{2} − 2.

(A) Let L_{2} be a line concurrent with L_{0}, proj_{L_{0}}a ≠ proj_{L_{0}}L_{2} ≠ proj_{L_{0}}b. Suppose there exists a point v ∈ {c, c′, c′′} such that T_{L_{2}} ⊆ Γ_{n−2}(v). Then it is easy to see that δ(v, L_{0}) = δ(a, L_{0}), implying δ(a, v) < n, a contradiction.

(B) Suppose there exists a line L_{j}, 2 < j ≤ \frac{n}{2} − 4 for which the set T_{L_{j}} is defined, and a point v ∈ {c, c′, c′′} such that T_{L_{j}} ⊆ Γ_{n−2}(v). Denote by (L_{0}, a_{1}, L_{2}, …, a_{j−1}, L_{j}) the j-path between L_{0} and L_{j}. Then δ(a, L_{j}) = δ(v, L_{j}), proj_{L_{j}}v = a_{j−1} but a_{j−3} ≠ proj_{L_{j−2}}v =: z (since otherwise, we obtain a contradiction with the induction hypothesis). Let Z be the projection of v onto z (note that δ(Z, L_{0}) = j). Then it is easily verified that T_{Z} ∩Γ_{n−2}(v) = ∅ hence, with {v, v′, v′′} = {c, c′, c′′}, T_{Z} ⊆ Γ_{n−2}(v′) ∪ Γ_{n−2}(v′′). Since for any line Z′ concurrent with L_{j−2} and different from Z and L_{j−4}, T_{Z′} ⊆ Γ_{n−2}(v), one also has T_{Z′} ∩ (Γ_{n−2}(v′) ∪ Γ_{n−2}(v′′)) = ∅. Let N be a line at distance ℓ, δ(a, Z) ≤ ℓ ≤ n − 3, from a for which the path [a, N] contains the path [a, Z] (implying that the set T_{N} is defined and contained in T_{Z}). We claim that there does not exist a point v′ ∈ {c, c′, c′′} \ {v} for which T_{N} ⊆ Γ_{n−2}(v′). This is shown by induction on ℓ. In (B1), we consider the case ℓ = δ(a, Z) and in (B2) the case δ(a, Z) < ℓ ≤ n − 3.

(B1) Suppose N = Z and T_{Z} ⊆ Γ_{n−2}(v′). Then as before, δ(a, Z) = δ(v′, Z), proj_{Z}v′ = proj_{Z}a = z but proj_{L_{j−2}}a ≠ proj_{L_{j−2}}v′ =: z′. Now for a line Z′′ concurrent with L_{j−2} and not through the points z, z′ or a_{j−3}, the set T_{Z′′} is contained in both Γ_{n−2}(v) and Γ_{n−2}(v′), contradicting condition (1) (note that such a line Z′′ exists since we are in the case s, t ≥ 3).

(B2) Suppose δ(a, Z) < ℓ ≤ n − 3 and T_{N} ⊆ Γ_{n−2}(v′). Let N′ be the line of the path [N, a] concurrent with N, and d the intersection point of N and N′. Then δ(a, N) = δ(v′, N), proj_{N}v′ = proj_{N}a = d but proj_{N′}a ≠ proj_{N′}v′ =: d′. Put B = proj_{d′}v′. It is easy to check that T_{B} ∩ Γ_{n−2}(v′) = ∅, implying T_{B} ⊆ Γ_{n−2}(v′′) (with {v, v′, v′′} = {c, c′, c′′}). But now we obtain as before a contradiction with condition (1), by considering the set T_{B′} for a line B′ concurrent with N′ not through the points d, d′ or proj_{N′}a. This shows that there does not exist a point v′ ∈ {c, c′, c′′} \ {v} for which T_{N} ⊆ Γ_{n−2}(v′).

Now let M be a line at distance n−5 from a for which the path [a, M] contains the path [a, Z]. Let M_{1}, M_{2} and M_{3} be three distinct lines concurrent with M and at distance n−3 from a. Since T_{M_{1}} is not contained in one of the sets Γ_{n−2}(v′) or Γ_{n−2}(v′′), we can assume without loss of generality that v′ lies at distance n − 3 from the line M_{1}, and that y = proj_{M_{1}}v′ is a point of T_{M_{1}}. Then v′ lies at distance n − 2 from exactly one point of T_{M_{2}}, hence the point v′′ lies at distance n − 3 from the line M_{2}, and y′ = proj_{M_{2}}v′′ is contained in T_{M_{2}}. But now the set T_{M_{3}} cannot be covered by Γ_{n−2}(v′) ∪ Γ_{n−2}(v′′), the final contradiction. So we have shown that there does not exist a line L_{j} with j ≤ \frac{n}{2} − 4 and a point v ∈ \{c, c′, c′′\} such that T_{L_{j}} ⊆ Γ_{n−2}(v).

(C) Suppose j = \frac{n}{2} −2 and T_{N} ⊆ Γ_{n−2}(v). Hence N lies at distance n−3 from a. Let M be the line of [a, N] concurrent with N and put a_{M} the projection of a onto M. For an arbitrary point x_{i} on M different from a_{M}, we denote by M_{i} the line through x_{i} different from M. Without loss of generality, we choose N = M_{1}. Since proj_{M_{1}}v = x_{1} and proj_{M}v = x_{2} (with x_{2} ≠ a_{M} by the induction hypothesis), Γ_{n−2}(v) ∩ T_{M_{2}} = ∅. Hence there exists a point v′ ∈ {c, c′, c′′}, v′ ≠ v lying at distance n−3 from the line M_{2}. Since v covers every set T_{M_{i}}, i ≠ 2, it is clear that proj_{M_{2}}v′ ≠ x_{2}. Hence v′ lies at distance n − 2 from a unique point z of T_{M_{1}}, a contradiction.

Let M and the lines M_{i} be defined as in the previous paragraph. We now have the following situation. Since the set T_{M_{1}} contains infinitely many points, there is a point v ∈ {c, c′, c′′} at distance n − 2 from all but one point of T_{M_{1}}. Note that this point v lies at distance n − 2 from exactly one point of the sets T_{M_{i}}, i ≠ 1. We deduce that the line M contains exactly 4 points, and that there are points v′, v′′ of {c, c′, c′′}, with {v, v′, v′′} = {c, c′, c′′} lying at distance n − 2 from all but one point of T_{M_{2}}, T_{M_{3}} respectively. But now the projections of v and v′ onto the line M_{3} have to coincide, so we obtain a point at distance n − 2 from a, b, v and v′, contradicting Condition (5). This ends the proof of Step 7 and hence the case i = n − 2.

The theorem is now proved. □

4.8 Proof of the Special Flag Theorem

We first introduce some notation. A line in the thin polygon Γ (Γ′) corresponding with a point in Δ (Δ′) will be called a p-line, a line in Γ (Γ′) corresponding with a line in Δ (Δ′) will be called an L-line. Two points of the thin polygon Γ (Γ′) at distance k, k ≠ n and k ≡ 2 mod 4 are said to be at distance k_{p} (k_{L}) if both proj_{a}b and proj_{b}a are p-lines (L-lines). For the rest of this section, δ will always refer to this ‘extended’ distance function. Recall that we only have to consider the case i ≡ 2 mod 4, 2 < i < n. Furthermore, put T_{a,b} := Γ_{i_{p}}(a) ∩ Γ_{i_{p}}(b)

4.8.1 Case i_{p} ≤ n/2

Let S be the set of pairs of points (a, b) of Γ satisfying δ(a, b) ≠ i_{p} and T_{a,b} = ∅. We claim that

(a, b) ∈ S ⟺ \{ \begin{matrix}δ(a, b) ≥ 2i \\ δ(a, b) = k < 2i, & k ≡ 0 mod 4 \\ δ(a, b) = k_{L} < 2i, & k ≡ 2 mod 4. \end{matrix}

Indeed, let (a, b) be an arbitrary pair of points of Γ. We distinguish the following possibilities.

(i) δ(a, b) > 2i or δ(a, b) = k < 2i with k ≡ 0 mod 4. If i ≠ n/2 then in Case 4.7.1 (i) and (ii) in the proof of Theorem 4.2.2 it was shown that Γ_{i}(a) ∩ Γ_{i}(b) = ∅, hence also T_{a,b} = ∅. If i = n/2, then it is easy to see that Γ_{i}(a) ∩ Γ_{i}(b) ≠ ∅ if and only if δ(a, b) = n, which is case (ii) below.

(ii) δ(a, b) = 2i. Suppose first i = n/2. In this case, Γ_{i}(a) ∩ Γ_{i}(b) contains exactly two points x and y, namely lying on one of the two distance-n-paths between a and b. But since δ(a, b) ≡ 0 mod 4, the lines proj_{a}x and proj_{b}x cannot be of the same type. Similarly for y. Hence neither x nor y is contained in T_{a,b}. Suppose now i < n/2. Without loss of generality, we can assume proj_{a}b is a p-line and proj_{b}a is an L-line. Suppose by way of contradiction T_{a,b} contains a point x. Clearly, x ≠ a ⋈ b and proj_{a}b = proj_{a}x. Put [a, R] = [a, b] ∩ [a, x] and j = δ(a, R). Since proj_{b}x ≠ proj_{b}a (indeed, the line proj_{b}x is necessarily a p-line), there arises a circuit of length at most 2(2i − j) < 2n, a contradiction.

(iii) δ(a, b) = k_{L} < 2i with k ≡ 2 mod 4. Suppose x ∈ T_{a,b}. Clearly, proj_{a}b ≠ proj_{a}x and proj_{b}a ≠ proj_{b}x. So there arises a circuit of length at most k + 2i < 4i ≤ 2n, the contradiction.

(iv) δ(a, b) = k_{p} < 2i with k ≡ 2 mod 4. In this case, any point x at distance i − k/2 from M := a⋈b for which proj_{M}a ≠ proj_{M}x ≠ proj_{M}b belongs to T_{a,b}.

This shows the claim. Put κ = {δ(a, b) | (a, b) ∈ S}.

Now let S′ be the set of pairs (a, b) of distinct points of Γ such that i_{p} ≠ δ(a, b) ∉ κ and Γ_{i_{p}}(a)∩Γ_{≠i_{p}}(b) ⊆ Γ_{κ}(b). We claim that (a, b) ∈ S′ if and only if δ(a, b) = 2_{p}. Let (a, b) be an arbitrary pair of distinct points of Γ satisfying i_{p} ≠ δ(a, b) ∉ κ. Hence i_{p} ≠ δ(a, b) = k_{p}, k < 2i. There are two possibilities.

(a) δ(a, b) = 2_{p}. Any point x of Γ_{i_{p}}(a) ∩ Γ_{≠i_{p}}(b) lies at distance i − 2 ≡ 0 mod 4 from b, hence δ(b, x) ∈ κ.

(b) i_{p} ≠ δ(a, b) = k_{p}, 2 < k < 2i. Put M := a⋈b and M′ the line concurrent with M at distance k/2 − 2 from a. Let x be a point at distance i − (k/2 − 2) from M′ for which proj_{M′}b ≠ proj_{M′}x ≠ proj_{M′}b. Then δ(a, x) = i_{p} and the length of the path consisting of [b, M′] and [M′, x] is i + 4 ≤ n, hence δ(b, x) = i + 4. Since i + 4 ≡ 2 mod 4 and i + 4 < 2i, δ(b, x) ∉ κ.

This shows the claim.

Let finally S′′ be the set of pairs (a, b) of distinct points of Γ with δ(a, b) ≠ i_{p} and for which there exists a point c such that either δ(a, c) = 2_{p} and δ(b, c) = i_{p} or δ(a, c) = i_{p} and δ(b, c) = 2_{p}. Then S′′ is the set of pairs of points at distance i − 2 from each other. Indeed, let (a, b) ∈ S′′ and put k = δ(a, b). Clearly, k ∈ {i − 1, i_{L}, i + 2}. Without loss of generality we can assume there is a point c such that δ(a, c) = 2_{p} and δ(b, c) = i_{p}. This is easily seen to be a contradiction unless k = i − 2.

By Theorem 4.2.2, α preserves collinearity. This ends the case i_{p} < n/2.

4.8.2 Case n/2 < i_{p} < n − 2

Let S be the set of pairs of points (a, b) of Γ satisfying δ(a, b) ≠ i_{p} and T_{a,b} = ∅. We claim that

(a, b) ∈ S ⟺ \{ \begin{matrix}δ(a, b) = k ≤ 2n − 2i, & k ≡ 0 mod 4 \\ δ(a, b) = k_{L} ≤ 2n − 2i − 2, & k ≡ 2 mod 4. \end{matrix}

Indeed, let (a, b) be an arbitrary pair of points of Γ. We distinguish the following possibilities.

(i) δ(a, b) = k_{p}, k ≡ 2 mod 4. Any point x at distance i − k/2 from M := a ⋈ b for which proj_{M}a ≠ proj_{M}x ≠ proj_{M}b is contained in T_{a,b}.

(ii) δ(a, b) = k_{L}, k ≡ 2 mod 4, k ≤ 2n − 2i − 2. Suppose x ∈ T_{a,b}. Then proj_{a}b ≠ proj_{a}x and proj_{b}a ≠ proj_{b}x. Hence there arises a circuit of length at most k + 2i < 2n, a contradiction.

(iii) δ(a, b) = k_{L}, k ≡ 2 mod 4, k > 2n − 2i − 2. Let Σ be an apartment containing a and b, and X the element of Σ at distance n − k/2 from both a and b for which proj_{a}X and proj_{b}X are p-lines. Any point x at distance i − (n − k/2) from X for which proj_{X}a ≠ proj_{X}x ≠ proj_{X}b belongs to T_{a,b}.

(iv) δ(a, b) = k, k ≡ 0 mod 4, k ≤ 2n − 2i. It is easy to see that Γ_{i}(a) ∩ Γ_{i}(b) ≠ ∅ if and only if k = 2n − 2i. But if k = 2n − 2i, the only points of Γ_{i}(a) ∩ Γ_{i}(b) lie in an apartment containing a and b and opposite the element a⋈b. Hence either proj_{a}x or proj_{b}x is an L-line, showing that T_{a,b} = ∅.

(v) δ(a, b) = k, k ≡ 0 mod 4, k > 2n − 2i. Fix a k-path γ between a and b. Without loss of generality, we can assume that the element X of γ incident with a is a p-line. Let x be a point on X different from a and from the projection of b onto X. Let Σ be an apartment containing x and b, and Y the line of Σ at distance \frac{2n−k}{2} + 1 from b for which proj_{b}Y is a p-line. Note that also δ(a, Y) = \frac{2n−k}{2} + 1 and proj_{a}Y is a p-line. Now any point y at distance i − (\frac{2n−k}{2} + 1) from Y for which proj_{Y} a ≠ proj_{Y} y ≠ proj_{Y} b belongs to T_{a,b}.

This shows the claim. Put κ = {δ(a, b) | (a, b) ∈ S}.

Case i ≤ \frac{2n+2}{3}

Let S′ be the set of distinct points of Γ such that δ(a, b) ∉ κ and Γ_{i_{p}}(a) ∩ Γ_{≠i_{p}}(b) ⊆ Γ_{κ}(b) and symmetrically Γ_{i_{p}}(b) ∩ Γ_{≠i_{p}}(a) ⊆ Γ_{κ}(a).

We claim that (a, b) ∈ S′ if and only if δ(a, b) = 2_{p}. Let (a, b) be an arbitrary pair of distinct points of Γ for which δ(a, b) ∉ κ. Then the following cases can occur.

(i) δ(a, b) = k_{p}, k ≡ 2 mod 4. Similarly as in Case 4.8.1, (a)-(b), one shows that (a, b) ∈ S′ if and only if δ(a, b) = 2_{p}.

(ii) δ(a, b) = k_{L}, k ≡ 2 mod 4, k ≥ 2n − 2i + 2. Let Σ be an apartment through a and b, and X the line of Σ at distance \frac{2n−k}{2} − 2 from a for which proj_{a}X is a p-line. Let x be a point at distance i − δ(a, X) from X for which proj_{X}a ≠ proj_{X}x ≠ proj_{X}b. Then δ(b, x) = j_{p} with j = i + 4. Hence δ(b, x) ∉ κ.

(iii) δ(a, b) = k, k ≡ 0 mod 4, k ≥ 2n − 2i + 4. Fix a k-path γ between a and b. Without loss of generality, we can assume that the element of γ incident with a is a p-line. Consider a point x at distance i_{p} from b and opposite a. (One can construct such a point x as follows. Let x′ be a point at distance n − k from b for which proj_{b}x is a p-line. Let X be a line incident with x′ and different from proj_{x′}b if k ≠ n, and X the p-line through b if k = n. Since δ(a, X) = n − 1, it is possible to choose a point x′′IX at distance n−k +2 from b and opposite a. Proceeding like this, we obtain a point x as claimed). Now δ(a, x) ∉ κ.

This shows the claim. Similarly as in Case 4.8.1 it now follows that α preserves collinearity, which ends the proof.

Case \frac{2n+2}{3} < i < \frac{3n}{4}

Noting that i ≡ 2 mod 4, the condition above implies n ≥ 30. Let S′ be the set of distinct points of Γ such that δ(a, b) ∈ κ and Γ_{i_{p}}(a) ∩ Γ_{κ}(b) ≠ ∅ or Γ_{i_{p}}(b) ∩ Γ_{κ}(a) ≠ ∅.

We claim that (a, b) ∈ S′ if and only if δ(a, b) = k ≡ 0 mod 4, 3i − 2n + 2 ≤ k ≤ 2n − 2i (note that this interval is nonempty, since n ≥ 8). Let (a, b) be an arbitrary pair of distinct points of Γ for which δ(a, b) ∈ κ. Then the following cases can occur.

(i) δ(a, b) = k_{L}, k ≡ 2 mod 4, k ≤ 2n − 2i − 2. Let x be a point at distance i_{p} from a. If i + k ≤ n, then δ(x, b) = i + k ∉ κ (indeed, k + i ≤ 2n − 2i implies 3i ≤ 2n − k, but we assumed 3i > 2n + 2). If i + k > n and δ(b, x) ∈ κ, we obtain a circuit of length at most i + k + 2n − 2i < 2n (noting that k < i), a contradiction. Hence Γ_{i_{p}}(a) ∩ Γ_{κ}(b) = ∅. Symmetrically, also Γ_{i_{p}}(b) ∩ Γ_{κ}(a) = ∅. So (a, b) ∉ S′.

(ii) δ(a, b) = k, k ≡ 0 mod 4, k ≤ 2n − 2i. Without loss of generality, we can assume proj_{a}b is a p-line and R := proj_{b}a is an L-line. As in (i) above, it follows that Γ_{i_{p}}(b) ∩ Γ_{κ}(a) = ∅. Note that i > k. Suppose first k ≥ 3i − 2n + 2. Let x be a point at distance i − k + 1 from R for which proj_{R}a ≠ proj_{R}x ≠ b. Then δ(a, x) = i_{p} and δ(b, x) = i−k+2 ≡ 0 mod 4. Since i−k+2 ≤ 2n−2i, δ(b, x) ∈ κ, hence (a, b) ∈ S′. Suppose now k < 3i − 2n + 2 and let x be a point at distance i_{p} from a. Suppose by way of contradiction that δ(b, x) ∈ κ. If [a, x] contains [a, b], then δ(b, x) = j_{p} with j = i − k, hence δ(b, x) ∉ κ. Put [a, x] ∩ [a, b] = [a, R′] and r = δ(a, R′). Note that we obtain a path of length k + i − 2r between b and x (consisting of the paths [b, R′] and [R′, x]). If k + i − 2r ≤ n, then δ(b, x) ∈ κ implies k + i − 2r ≤ 2n − 2i. But k + i − 2r > i − k > 2n − 2i, a contradiction. If k + i − 2r > n, then we obtain a circuit of length at most k + i − 2r + 2n − 2i < 2n, a contradiction. So (a, b) ∉ S′.

This shows the claim. Put λ = {δ(a, b) | (a, b) ∈ S′}. Hence λ = {k ∈ ℕ | k ≡ 0 mod 4 and 3i − 2n + 2 ≤ k ≤ 2n − 2i}.

Let S′′ be the set of distinct points of Γ such that δ(a, b) ∈ λ and |Γ_{λ}(a) ∩ Γ_{λ}(b)| is finite. Define T_{a,b}^{λ} := Γ_{λ}(a) ∩ Γ_{λ}(b).

We claim that (a, b) ∈ S′′ if and only if δ(a, b) = k ≡ 0 mod 4, 4n − 5i + 2 ≤ k ≤ 2n − 2i (note that this interval is nonempty, since i ≥ \frac{2n+2}{3}). Let (a, b) be an arbitrary pair of distinct points of Γ for which δ(a, b) ∈ λ. Then we distinguish the following cases.

(i) δ(a, b) = k ≤ 4n − 5i − 2 Let x be a point a distance 2n − 2i from a for which the path [a, x] contains the path [a, b]. Then δ(b, x) = 2n − 2i − k ≥ 3i − 2n + 2, so clearly, δ(b, x) ∈ λ. Since 2n − 2i − k ≥ 4 and not both the p-lines and the L-lines are finite (indeed, n > 16), we obtain that T_{a,b}^{λ} is infinite.

(ii) δ(a, b) = k ≥ 4n − 5i + 2 We show that the points of T_{a,b}^{λ} all belong to the path [a, b], which implies that this set is finite. Let x be a point for which δ(a, x) ∈ λ. Suppose first proj_{a}b ≠ proj_{a}x. Then we obtain a path of length k + δ(a, x) between b and x. If k + δ(a, x) ≤ n, then δ(b, x) = k + δ(a, x) ≥ (4n − 5i + 2) + (3i − 2n + 2) = 2n − 2i + 4, showing that δ(b, x) ∉ λ. If k + δ(a, x) > n, then δ(b, x) ∈ λ implies there is a circuit of length at most k + δ(a, x) + 2n − 2i ≤ 6n − 6i < 2n, a contradiction. Suppose now that proj_{a}b = proj_{a}x. If the path [a, x] contains the path [a, b], then δ(b, x) = δ(a, x) − k ≤ (2n − 2i) − (4n − 5i + 2) = 3i − 2n − 2, showing δ(b, x) ∉ λ. Now suppose x does not lie on the path [a, b] and put [a, x] ∩ [a, b] = [a, R′] and r = δ(a, R′). If k + δ(a, x) − 2r ≤ n, then δ(b, x) = k + δ(a, x) − 2r ≡ 2 mod 4, hence δ(b, x) ∉ λ. If k + δ(a, x) − 2r > n, then δ(b, x) ∈ λ implies there is a circuit of length at most k + δ(a, x) − 2r + 2n − 2i ≤ 6n − 6i − 2r < 2n, the final contradiction.

This shows the claim.

Now put A := max{|T_{a,b}^{λ}| | (a, b) ∈ S′′}. Since for a pair (a, b) ∈ S′′, the points of T_{a,b}^{λ} all lie on [a, b], the number of points in T_{a,b}^{λ} only depends from the distance between a and b. Note that A ≠ 0. We claim that for a pair (a, b) ∈ S′′, |T_{a,b}^{λ}| = A if and only if δ(a, b) = 2n − 2i. Indeed, let (a, b) be a pair of points contained in S′′ with δ(a, b) = k ≠ 4n − 5i + 2 (this is possible since 4n − 5i + 2 ≠ 2n − 2i) and for which |T_{a,b}^{λ}| ≠ 0. We show that |T_{a,b}^{λ}| > |T_{a,b′}^{λ}|, with b′ the point of [a, b] at distance 4 from b. Clearly, we may assume |T_{a,b′}^{λ}| ≠ 0. Let x be a point of T_{a,b′}^{λ}. Then x also belongs to T_{a,b}^{λ} because δ(x, b) = δ(x, b′) + 4, δ(x, b′) ∈ λ and δ(x, b′) < 2n − 2i (indeed, δ(x, b′) = 2n − 2i would imply δ(a, x) = 0). This shows that |T_{a,b}^{λ}| ≥ |T_{a,b′}^{λ}|. Let y be the point of T_{a,b′}^{λ} at minimal distance from a. Now let y′ be the point of [a, b] at distance δ(a, y) from b. Then δ(b, y′) ∈ λ and δ(a, y′) = δ(b′, y) + 4 ∈ λ. So y′ ∈ T_{a,b}^{λ} but y′ ∉ T_{a,b′}^{λ} since δ(b′, y′) = δ(b, y′) − 4 = δ(a, y) − 4 ∉ λ since δ(a, y) was chosen to be minimal in λ. We thus showed that |T_{a,b}^{λ}| > |T_{a,b′}^{λ}|. Now the claim immediately follows.

We thus recovered distance 2n − 2i. By Theorem 4.2.2, this ends the proof in this case.

Case \frac{3n}{4} ≤ i < n − 2

Note that 4(n − i) ≤ n. If 4(n − i) < n, let S′ be the set of pairs (a, b) of distinct points of Γ such that δ(a, b) ∉ κ and |Γ_{κ}(a) ∩ Γ_{κ}(b)| = 1. Let S′′ be the set of pairs of points (a, c) for which there exists a point b such that (a, b) ∈ S′ and Γ_{κ}(a) ∩ Γ_{κ}(b) = {c}. If 4(n − i) = n, let S′ be the set of pairs (a, b) of distinct points of Γ such that δ(a, b) ∉ κ and Γ_{κ}(a) ∩ Γ_{κ}(b) = {c, d}, with Γ_{κ}(c) ∩ Γ_{κ}(d) = {a, b}. Let S′′ be the set of pairs of points (a, c) for which there exists a point b such that (a, b) ∈ S′ and c ∈ Γ_{κ}(a) ∩ Γ_{κ}(b). We claim that in both cases S′′ is exactly the set of pairs of points at distance 2n − 2i from each other. We have the following possibilities to consider.

(i) δ(a, b) = k_{p}, k < 2n − 2i. Every point x at distance 2n−2i−k/2 from the line M := a⋈b for which proj_{M}a ≠ proj_{M}x ≠ proj_{M}b belongs to Γ_{κ}(a) ∩ Γ_{κ}(b). Moreover, since 2n − 2i − k/2 ≥ 3, there are at least two collinear points c, d contained in Γ_{κ}(a) ∩ Γ_{κ}(b). So for these two points c, d, Γ_{κ}(c) ∩ Γ_{κ}(d) ≠ {a, b}. We conclude that (a, b) ∉ S′.

(ii) δ(a, b) = k, k > 4(n − i). In this case, it is easy to see that Γ_{κ}(a)∩Γ_{κ}(b) = ∅ (indeed, if not, then necessarily 4(n − i) = n, contradicting k > 4(n − i)). So (a, b) ∉ S′.

(iii) δ(a, b) = k ≡ 0 mod 4, 2(n − i) < k ≤ 4(n − i). Suppose first k < 4(n − i). Then the point c (d) of [a, b] at distance 2n − 2i from a (b) belongs to Γ_{κ}(a) ∩ Γ_{κ}(b). If a ∈ Γ_{κ}(c) ∩ Γ_{κ}(d), then clearly, every point on the line R′ := proj_{a}b different from proj_{R′}b also belongs to Γ_{κ}(a)∩Γ_{κ}(b), hence Γ_{κ}(c)∩Γ_{κ}(d) ≠ {a, b}. So if k < 4(n−i), (a, b) ∉ S′. Now consider the case k = 4(n − i). If 4(n − i) < n, then it is easy to see that Γ_{κ}(a)∩Γ_{κ}(b) = {c}, with c = a⋈b, so (a, b) ∈ S′ and c lies at distance 2n − 2i from both a and b. If 4(n − i) = n, then it is easy to see that Γ_{κ}(a) ∩ Γ_{κ}(b) = {c, d}, with c and d the unique points on the two n-paths joining a and b. Hence also Γ_{κ}(c) ∩ Γ_{κ}(d) = {a, b} (so (a, b) ∈ S′), and both c, d lie at distance 2n − 2i from a, b.

(iv) δ(a, b) = k ≡ 2 mod 4, 2(n − i) < k < 4(n − i). Let R_{a} (R_{b}) be the line of [a, b] at distance 2n − 2i − 1 from a (b), and c (d) a point on R_{a} (R_{b}) different from both proj_{R_{a}}a and proj_{R_{a}}b (proj_{R_{b}}a and proj_{R_{b}}b). In this way, we obtain two distinct points c, d of Γ_{κ}(a)∩Γ_{κ}(b) for which Γ_{κ}(c)∩Γ_{κ}(d) ≠ {a, b}, except if k = 4(n−i)−2 (then R_{a} = R_{b} = a ⋈ b) and a ⋈ b contains exactly 3 points. In the latter case, it is easily seen that the unique point c on M := a ⋈ b distinct from proj_{M}a and proj_{M}b is the unique point of Γ_{κ}(a) ∩ Γ_{κ}(b) (hence (a, b) ∈ S′), and c lies at distance 2n − 2i from both a and b.

This shows the claim. We thus recovered distance 2n − 2i, and the result follows by Theorem 4.2.2.

4.8.3 Case i_{p} = n − 2

Note that n ≡ 0 mod 4. Let S be the set of pairs of points (a, b) of Γ satisfying δ(a, b) ≠ i_{p} and T_{a,b} = ∅. We claim that (a, b) ∈ S if and only if δ(a, b) ∈ {2_{L}, 4}. Indeed, we distinguish the following cases.

(i) δ(a, b) = k_{p}, k ≡ 2 mod 4 Any point x at distance i − k/2 from the line M := a ⋈ b for which proj_{M}a ≠ proj_{M}x ≠ proj_{M}b belongs to T_{a,b}.

(ii) δ(a, b) = k_{L}, k ≡ 2 mod 4 If k = 2_{L}, then any point x of Γ_{i}(a) ∩ Γ_{i}(b) lies at distance n − 3 from the line ab, hence x ∉ T_{a,b}. If k ≥ 6, let Σ be an apartment through a and b, and M′ the line of Σ opposite a⋈b. Then any point x at distance n−2−(\frac{2n−k}{2}) from M′ for which proj_{M′}a ≠ proj_{M′}x ≠ proj_{M′}b belongs to T_{a,b} (and such points exist because k ≥ 6).

(iii) δ(a, b) = k, k ≡ 0 mod 4 If k = 4, then it is easy to see that any point x of Γ_{i}(a) ∩ Γ_{i}(b) lies opposite a ⋈ b in an apartment containing a and b. But then either proj_{a}x or proj_{b}x is an L-line, hence x ∉ T_{a,b}. If k ≥ 8, then let R be the p-line through a. Without loss of generality, we can assume that R = proj_{a}b if k ≠ n. Let a′ be a point on R, a ≠ a′ ≠ proj_{R}b. Let Σ be an apartment containing b and a′, and R′ the line of Σ at distance \frac{2n−k+2}{2} from b for which proj_{b}R′ is a p-line (note that R′ is indeed a line since 2n − k + 2 ≡ 2 mod 4). Then any point x at distance n−2−(\frac{2n−k+2}{2}) from R′ for which proj_{R′}a ≠ proj_{R′}x ≠ proj_{R′}b belongs to T_{a,b} (and such points exist because k ≥ 8).

This shows the claim. Put κ = {2_{L}, 4}. Now it is easy to verify that two points a and b lie at distance 2_{L} if and only if δ(a, b) ∈ κ, and there exist points c, c′ ∈ Γ_{κ}(a) ∩ Γ_{κ}(b), c ≠ c′, for which δ(c, c′) ∉ κ. (Indeed, if δ(a, b) = 2_{L}, one can consider points c, c′ lying on a line M intersecting ab in a point x different from a or b, c ≠ x ≠ c′. If δ(a, b) = 4, then assume that proj_{b}a is an L-line. Since n > 6, the only points in Γ_{κ}(a) ∩ Γ_{κ}(b) lie on proj_{b}a, hence lie at distance 2_{L} from each other.) So we can distinguish distance 4. By Theorem 4.2.2 this ends the proof in this case. □

Appendix A

Minimal generating sets in H(q)^{D}

A.1 Introduction

In this chapter, the following question keeps us puzzling: how many points do you need to generate the whole point set of one of the classical hexagons (in the sense that two distinct collinear points generate the points on the joining line)? We tackle this question for the finite dual split Cayley hexagon, since in this case, some strong results are available in Thas & Van Maldeghem [48]. Indeed, they show that our hexagon admits an embedding in 13-dimensional projective space, implying that we will need at least 14 points to generate the whole point set. They also prove that the point set of this hexagon is generated by the point sets of three thin full subhexagons lying ‘close’ to each other. In the first section, we show how one can select 15 points to generate these three subhexagons. Hence 15 points are always enough. If the underlying field is of prime order, then it follows that one can do with 14 points. Now the question becomes: is it possible in the general case to delete one point of this generating set of 15 points, and still to generate the whole point set? We let the computer work for us, and obtain that this is indeed possible for q = 4, 8, 9, 16. Now it is up to us to be smarter and faster than the computer, and beat it by proving the result for all values of q. But

196 MINIMAL GENERATING SETS IN H(q)^{D}

Figure A.1: A generating set of 9 points for Γ(R_{1}, R_{2})

for the moment, it seems that we are losing this competition. So the partial result reads : 14 ⇒ (1 + q)(1 + q^{2} + q^{4}) if q is prime or q ∈ {4, 8, 9, 16}...

Throughout this chapter, we put Γ ≅ H(q)^{D}. Denote by m the size of a minimal set of points M of Γ such that the points of M generate Γ (in the sense that two distinct collinear points generate all points on the joining line).

A.2 14 ≤ m ≤ 15

The generalized hexagon H(q)^{D} admits an embedding in PG(13, q) (for an explicit description, see Thas & Van Maldeghem [48]). This implies that the point set of Γ generates PG(13, q), hence m ≥ 14. For the hexagon H(2)^{D}, there exists a generating set of 14 points (see Thas & Van Maldeghem [48] or Cooperstein [13]). So from now on, we assume that q > 2.

In Thas & Van Maldeghem [48], the following useful property is proved.

Theorem A.2.1 Let L_{1} and L′_{1} be two arbitrary opposite lines of Γ. Let L_{2} and L′_{2} be two distinct lines of the regulus R(L_{1}, L′_{1}) both distinct from L_{1} and L′_{1}. Then the union of the point sets of the three thin full subhexagons Γ(L_{1}, L′_{1}), Γ(L_{2}, L′_{2}) and Γ(L_{1}, L_{2}) generates the point set of Γ.

Consider a thin full subhexagon Γ(R_{1}, R_{2}) of Γ. Note that Γ(R_{1}, R_{2}) is the double of a desarguesian projective plane Δ ≅ PG(2, q) (where the lines of Γ(R_{1}, R_{2}) correspond with the points and lines of Δ, and the points of Γ(R_{1}, R_{2}) correspond with the flags of Δ). Let x, y, z be three different points of the regulus ⟨R_{1}, R_{2}⟩, and define v_{i} = proj_{R_{i}}v, with v ∈ {x, y, z}. Let finally

A.2 14 ≤ m ≤ 15 197

w′ be a point on the line xx_{1} different from x or x_{1}, and w = w′ ⋈proj_{zz_{2}}w′. Now we consider the subset Γ′ of the point set of Γ(R_{1}, R_{2}) generated by the set N = {x, x_{1}, x_{2}, y, y_{1}, y_{2}, z, w} (see Figure A.1). Denote by Δ′ the subset of the plane Δ corresponding with the set Γ′. Then Δ′ is a subplane of Δ. Indeed, if p_{1} and p_{2} are two points of Δ′, then these two points correspond with two lines M_{1} and M_{2} of Γ(R_{1}, R_{2}) that are generated by N. Since M_{1} and M_{2} necessarily lie at distance 4, the line M_{1} ⋈ M_{2} of Γ(R_{1}, R_{2}) is also generated, and corresponds with the line p_{1}p_{2}. So the line p_{1}p_{2} is contained in Δ′. Similarly, every two lines of Δ′ meet. The ordinary octagon in Γ′ through the points w′, w, proj_{zz_{2}}w, z_{2}, y_{2}, y, y_{1}, x_{1} corresponds with a quadrangle in Δ′, so Δ′ is a projective plane isomorphic to PG(2, q′). Now we distinguish two cases.

• q is prime.

In this case, Δ = Δ′, so Γ(R_{1}, R_{2}) is generated by the 8 points of N. Note that N is a minimal generating set for Γ(R_{1}, R_{2}), since there actually exists an embedding of Γ(R_{1}, R_{2}) in PG(7, q) (for a description, see for instance Thas & Van Maldeghem [47]).

• q is not prime. Choose a point u of the regulus ⟨R_{1}, R_{2}⟩, different from x, y, z and such that, with u_{1} = proj_{R_{1}}u, the cross-ratio (x_{1}, y_{1}; z_{1}, u_{1}) generates the field GF(q). Again as before, the set generated by N ∪{u} corresponds with a subplane Δ′′ of Δ, but because of the choice of u, Δ = Δ′′ (indeed, Δ′′ ≅ PG(2, q′′), but GF(q′′) contains a generating element of the field GF(q)). So Γ(R_{1}, R_{2}) is generated by the 9 points of N ∪ {u}. As in the previous case, this is a generating set of minimal size, because there exists an embedding of Γ(R_{1}, R_{2}) in PG(8, q) (for a description, see again Thas & Van Maldeghem [47]).

Now choose a set M′ of 15 points in Γ as indicated in Figure A.2, with (p_{0}, p_{1}; p_{2}, p_{3}) a generating element of the field GF(q). Then by the previous observations, these points generate the thin full subhexagons Γ(p_{0}p_{3}, r_{0}r_{3}), Γ(r_{0}r_{3}, s_{0}s_{3}) and Γ(s_{0}s_{3}, t_{0}t_{3}). Theorem A.2.1 implies that M′ generates Γ. Note that if q is prime, the set M′ \ {u} still generates Γ. So we obtained the following theorem.

Theorem A.2.2 If q is prime, then m = 14. If q is not prime, then 14 ≤ m ≤ 15.

198 MINIMAL GENERATING SETS IN H(q)^{D}

Figure A.2: A generating set of 15 points for H(q)^{D}

A.3 m = 14 if q ∈ {4, 8, 9, 16} 199

A.3 m = 14 if q ∈ {4, 8, 9, 16}

Put q = p^{h}, p prime. Let M′ be a generating set as in figure A.2. We want to delete one point of M′ to obtain a minimal generating set. More precisely, we delete the point u. Put M = M′ \ {u}. Note that M certainly generates three subhexagons Γ′(p_{0}p_{3}, r_{0}r_{3}), Γ′(r_{0}r_{3}, s_{0}s_{3}) and Γ′(s_{0}s_{3}, t_{0}t_{3}) of order (p, 1). We now investigate whether it is possible to select the points a, b, c and t_{0} such that the set M still generates Γ. Our tools are coordinates and the computer.

choice of coordinates Let the apartment through x, y, p_{0} and r_{0} be the hat-rack of the coordinatization, with x = (∞), xp_{0} = [∞], y = (00000), xr_{0} = [0]. Without loss of generality, we can choose xs_{0} = [1]. Further we choose xt_{0} = [k], z = (00100), a′ = (a), b′ = (0b) and c′ = (1c).

idea of the computer program The first idea is of course to give the computer the coordinates of the 14 points we selected, and let it do the generating work. This is, whenever two collinear points are already in the set of generated points, then all the points on the joining line are added to this set. If at the end, the program tells us that (q + 1)(q^{4} + q^{2} + 1) distinct points were generated, we are done. Alternatively, we give the computer the coordinates of the 9 lines in the set \overline{M} = \{p_{0}x, p_{3}y, p_{1}z, r_{0}r_{3}, s_{0}s_{3}, t_{0}t_{3}, aa′, bb′, cc′\} (note that these lines are certainly generated by M). The job then becomes: whenever two lines L and L′ at distance 4 from each other are contained in \overline{M}, add the line L⋈L′ to \overline{M}. If at the end, the set \overline{M} contains (q + 1)(q^{4} + q^{2} + 1) distinct lines, then the set M of 14 points we started from, is indeed a generating set for Γ. We give a method to make the program a bit faster. Denote by Γ′_{1}, Γ′_{2} and Γ′_{3} the three thin subhexagons Γ′(p_{0}p_{3}, r_{0}r_{3}), Γ′(r_{0}r_{3}, s_{0}s_{3}) and Γ′(s_{0}s_{3}, t_{0}t_{3}). Then each Γ′_{i} is contained in a full subhexagon Γ_{i} of order (q, 1). Dually, in H(q), these Γ_{i} correspond to ideal subhexagons lying in a hyperplane γ_{i} of PG(6, q). Suppose that at a certain moment, a line L is generated for which the corresponding point of Γ^{D} lies in a hyperplanes γ_{j} of PG(6, q), j ∈ {1, 2, 3}, and such that L lies at distance 4 from a line L′ contained in the thin subhexagon Γ′_{j}. Then in the following step, the line L ⋈ L′ of Γ_{j} will be generated. Now let d_{0}, d_{1}, d_{2} be the three points on L belonging to Γ′_{j}. If proj_{L′}L ∉ {d_{0}, d_{1}, d_{2}}, and the cross-ratio (d_{0}, d_{1}; d_{2}, proj_{L′}L) generates GF(q), then we know by the previous section that the thin full subhexagon Γ_{j} can be generated. This implies that it is possible to generate the whole point regulus R(x, y), hence the generated set contains the set M′ we started from, and we are done.

Figure A.3: numbering of the lines M_{i}

In the following, we give a generating set obtained by this method for the cases q = 4, 8, 16, 9 respectively. Put M_{1} = xp_{0}, M_{2} = p_{0}p_{3}, M_{3} = yp_{3}, M_{5} = r_{0}r_{3}, M_{8} = s_{0}s_{3} and M_{11} = t_{0}t_{3}. We further label the lines of the three subhexagons Γ′(M_{2}, M_{5}) (= Γ′_{1}), Γ′(M_{5}, M_{8}) (= Γ′_{2}) and Γ′(M_{8}, M_{11}) (= Γ′_{3}) with M_{1}, …, M_{k}, k = 34 for p = 2 and k = 68 for p = 3. The lines M_{i}, i = 1, …, 34 are numbered as in Figure A.3. The subhexagons Γ_{i} of order (q, 1) containing Γ′_{i} correspond with the following hyperplanes γ_{i} of PG(6, q):

\begin{aligned}γ_{1} & : & X_{3} = 0 \\ γ_{2} & : & X_{3} − X_{5} = 0 \\ γ_{3} & : & kX_{1} + (1 + k)X_{3} + X_{5} = 0. \end{aligned}

The computer results are given by the diagrams, where the numbers refer to the lines M_{i}, and where, if a line N splits into two lines N_{1} and N_{2}, this has

Figure A.4: a minimal generating set for H(4)^{D}

to be read as N = N_{1} ⋈N_{2}.

Case H(4)^{D} Let α be a generating element of GF(4) (satisfying α^{2} = 1 + α). We choose

\{ \begin{matrix}a = α^{2} \\ b = 1 \\ c = α^{2} \\ k = α. \end{matrix}

\begin{matrix}M_{11} = [α, 0, 0] & L_{1} = [1, α^{2}, 1, 0, 1] \\ M_{17} = [α^{2}, 0] & L_{2} = [α^{2}, 0, α^{2}, α, 1] \\ M_{18} = [0, α, 0, 1, 0] & L_{3} = [α^{2}, α, α^{2}, α] \\ M_{24} = [1, 1, 1, 0, 1] & L_{4} = [1, 1, 1, 0, 0] \\ M_{25} = [1, 1, 1, 1, 1] & L_{5} = [α, 0, 0, α^{2}, α^{2}] \\ M_{29} = [1, α^{2}, 1] & L_{6} = [0, 0, 1, 0, 0] \\ M_{31} = [α, α, α^{2}, α^{2}, 1] & L_{7} = [α^{2}, 0, 0, 1] \\ M_{32} = [α, α, α^{2}] & L_{8} = [α, α, α^{2}, 0, 1] \end{matrix}

The line L_{8} is contained in Γ_{2} and lies at distance 4 from the line M_{8} of Γ′_{2}. Also, L_{8} lies opposite the lines M_{7} = [1], M_{9} = [1, 0, 0, 0, 0] and M_{15} = [1, 0, 0, 1, 0], which are the lines of Γ′_{2} intersecting M_{8}. This means that a line of Γ_{2} not contained in Γ′_{2} will be generated. Since GF(4) contains no subfield different from GF(2), we are done.

Figure A.5: a minimal generating set for H(8)^{D}

Case H(8)^{D} Let α be the generating element of GF(8) (satisfying α^{3} = 1+α). We choose

\{ \begin{matrix}a = α^{2} \\ b = 1 \\ c = α \\ k = α. \end{matrix}

M_{17} = [α^{2}, 0]  L_{1} = [α, α, α^{5}, α^{2}, α^{4}]
M_{18} = [0, α^{5}, 0, 1, 0] L_{2} = [α^{4}, 0, α, α^{5}, α^{4}]
M_{19} = [0, α^{5}, 0, 0, 0] L_{3} = [α^{4}, 0, α, α^{4}, α^{4}]
M_{24} = [1, 1, 1, 0, 1]  L_{4} = [α^{3}, α, α^{4}, 1, α^{5}]
M_{32} = [α, α, α^{5}]  L_{5} = [α^{4}, 0, α]
M_{33} = [1, α, α^{5}, 0, α^{2}] L_{6} = [α^{2}, 0, α^{2}, α^{3}]
M_{34} = [1, α, α^{5}, 1, α^{2}] L_{7} = [α^{4}, 0, α, α^{3}, α^{2}]
L_{8} = [α^{2}, 1, α^{4}, 1, α^{4}]

The line L_{8} is contained in Γ_{2}, lies at distance 4 from the line M_{9} = [1, 0, 0, 0, 0] and lies opposite the lines M_{6} = [0, 0, 0, 0, 0], M_{8} = [1, 0, 0] and M_{28} = [0, 1, 1, 1, 1]. Similarly as in the previous case, we are done.

Case H(16)^{D} Let α be the generating element of GF(16) (satisfying α^{4} = 1 + α). We choose

\{ \begin{matrix}a = α^{14} \\ b = α^{8} \\ c = α^{8} \\ k = α. \end{matrix}

Figure A.6: a minimal generating set for H(16)^{D}

M_{18} = [0, α, 0, 1, 0]  L_{1} = [α^{5}, α^{10}, α^{9}, α^{9}, α^{4}]
M_{20} = [0, α, 0]  L_{2} = [α^{5}, α^{10}, α^{9}, α^{7}, α^{4}]
M_{21} = [α^{14}, 0, 1, 0]  L_{3} = [α^{5}, α^{10}, α^{9}]
M_{22} = [α^{14}, 0, 0, 0]  L_{4} = [α^{5}, α^{10}, α^{9}, α^{12}, α^{13}]
M_{23} = [0, α^{8}, α^{8}]  L_{5} = [1, α^{8}, α^{14}, α^{7}, α^{14}]
M_{24} = [1, α^{7}, α^{7}, α^{3}, α^{7}] L_{6} = [0, α, 0, α^{10}, α^{9}]
M_{29} = [1, α^{8}, α^{14}]  L_{7} = [α^{2}, α^{8}, α^{7}, α^{6}, α^{6}]
M_{30} = [α^{7}, α^{4}, α^{10}, α^{12}, α] L_{8} = [0, α^{8}, α^{8}, α^{10}, α^{5}]
M_{31} = [α^{7}, α^{4}, α^{10}, α^{11}, α] L_{9} = [α^{2}, α^{5}, α^{13}, α^{8}]
M_{33} = [1, α^{8}, α^{14}, α, α^{5}] L_{10} = [α^{4}, α^{9}, α, α^{11}, α^{9}]
L_{11} = [1, α^{8}, α^{14}, α^{9}, α^{10}]
L_{12} = [α^{9}, α^{3}, 1, α^{13}, α^{6}]
L_{13} = [α^{5}, α, α^{10}, α^{3}, α^{8}]

The line L_{13} is contained in Γ_{2} and lies at distance 4 from the line M_{26} = [1, α^{7}, α^{7}] of Γ′_{2} and opposite the lines M_{7} = [1], M_{24} = [1, α^{7}, α^{7}, α^{3}, α^{7}] and M_{25} = [1, α^{7}, α^{7}, α^{14}, α^{7}] (which are the lines of Γ′_{2} intersecting M_{26}).

Denote by p_{0}, p_{1}, p_{2}, p_{3} the projections of respectively M_{7}, M_{24}, M_{25} and L_{13} on the line M_{26}. Then these points have coordinates:

Figure A.7: a minimal generating set for H(9)^{D}

\{ \begin{matrix}p_{0} = (1, α^{7}) \\ p_{1} = (1, α^{7}, α^{7}, α^{3}) \\ p_{2} = (1, α^{7}, α^{7}, α^{14}) \\ p_{3} = (1, α^{7}, α^{7}, α^{7}). \end{matrix}

In PG(6, q), the point p_{i}, i = 0, …, 3, corresponds with the line pr_{i}, with p = (0, 1, 0, 1, 0, 1, α^{7}) and

\{ \begin{matrix}r_{0} = (1, 0, 0, 0, 0, 0, 1) \\ r_{1} = (1, 0, 1, α^{7}, 1, α^{7}, α^{3}) \\ r_{2} = (0, 0, 1, α^{7}, 1, α^{7}, α^{14}) \\ r_{3} = (α, 0, 1, α^{7}, 1, α^{7}, α^{7}). \end{matrix}

The points r_{0}, r_{1}, r_{2} and r_{3} are collinear. It is now easy to calculate that (r_{0}, r_{1}; r_{2}, r_{3}) equals α, hence we are done.

Case H(9)^{D} Let α be the generating element of GF(9) (satisfying α^{2} = 1−α). We choose

\{ \begin{matrix}a = α \\ b = α \\ c = α^{5} \\ k = α \end{matrix}

In this case the subhexagons Γ′_{i}, i = 1, 2, 3, are of order (3, 1), the union of their line sets contains 68 lines M_{i}, i = 1, …, 68. We again choose the lines M_{1}, …, M_{34} as in Figure A.3.

M_{6} = [0, 0, 0, 0, 0]  M_{30} = [α, α^{5}, α^{3}, α^{4}, α^{5}] L_{1} = [α, 0, 1, α^{6}]
M_{7} = [1]  M_{31} = [α, α, α^{7}, 1, α]  L_{2} = [α^{5}, α^{5}, α^{7}, α^{7}, α^{2}]
M_{17} = [α, 0]  M_{35} = [0, α^{7}, 0, α^{4}, 0]  L_{3} = [α^{7}, α^{5}, α^{5}, α^{4}, 1]
M_{20} = [0, α^{7}, 0]  M_{36} = [1, α, α^{3}, 1, α]  L_{4} = [1, α^{3}, α^{7}, 0, α^{6}]
M_{21} = [α, 0, 1, 0]  M_{37} = [α^{5}, 0, 0, 0]  L_{5} = [α^{3}, 1, α^{7}, α^{6}, α^{2}]
M_{25} = [1, α^{3}, α^{7}, α^{6}, α^{7}] M_{38} = [1, α^{3}, α^{7}]

M_{28} = [0, α, α, α^{6}, α^{5}]

The line L_{5} belongs to the subhexagon Γ_{2}, and lies at distance 4 from the line M_{28} of this Γ′_{2}. The lines of Γ′_{2} intersecting M_{28} are the lines M_{23} = [0, α, α], M_{9} = [1, 0, 0, 0, 0], M_{24} = [1, α^{7}, α^{3}, α, α^{3}] and M_{39} = M_{28} ⋈ M_{38} = M_{28} ⋈ (M_{7} ⋈ M_{25}) = [1, α^{3}, α^{7}, α^{7}, α^{7}]. Since L_{5} lies opposite all the lines of Γ′_{2} intersecting M_{28}, we are done.

A.4 Comments

Our initial aim was to use the computer results to see how one can construct a generating set for H(q)^{D}, for all values of q. One thing we did was checking whether for example diagram A.5 also goes through in other fields of characteristic 2. So we imitated the calculations of the computer, but now with general coordinates a, b, c and k. The conclusion was that this ‘path’ only works for the field GF(8). Another attempt was to concentrate on the values of a, b, c and k instead of on the diagrams, and see similarities between the results for GF(4) and GF(8). For these fields, the computer program does not need much time, so we could really try all possible values of a, b, c and k. We then tried to see ‘symmetries’ in the good choices, and made a guess for what a good choice in the field GF(16) could be. We seem not to be very good clairvoyants, since our predictions turned out to be wrong... So looking at the computer data, no answers, but all the more questions turned up. For example, sometimes the generated set of lines was not the whole line set of the hexagon, and not the line set of the three subhexagons over the prime field, but something in between. Moreover, the size of such a set was often the same for distinct choices of a, b, c and k. We would of course like to explain these mysteries in a geometric way. But perhaps this puzzle was not designed for geometers at all, and came into our hands only by accident?

206

Appendix B

Veelhoeken in Veelvoud

B.1 Van veelhoek naar veralgemeende veelhoek

Veralgemeende veelhoeken zijn meetkundige structuren die werden ingevoerd door de (van oorsprong Belgische) wiskundige Jacques Tits in 1959, in een appendix van het artikel ‘Sur la trialité et certains groupes qui s’en déduisent’. Oorspronkelijk stonden de veralgemeende veelhoeken ten dienste van de groepentheorie, maar al gauw begon men zich te interesseren voor deze structuren op zich. Zoals de naam al laat vermoeden, zijn veralgemeende veelhoeken opgebouwd uit veel gewone veelhoeken. We geven nu een precieze definitie.

Een veralgemeende n-hoek, n ≥ 2, is een meetkunde Γ bestaande uit een verzameling punten P, een verzameling rechten L, en een relatie I, de zogenaamde ‘incidentierelatie’ die beschrijft wanneer een punt op een rechte ligt (of een rechte door een punt gaat), zodat aan de volgende axioma’s voldaan is:

(i) in de meetkunde Γ zijn geen gewone k-hoeken, met k < n te vinden,

(ii) door elke twee punten, twee rechten, of een punt en een rechte is steeds een gewone n-hoek te vinden,

(iii) er is ergens in de meetkunde een gewone (n + 1)-hoek te vinden.

Axioma (iii) kan men ook vervangen door het volgende axioma:

(iii)′ Elke rechte bevat minstens 3 punten, en door elk punt gaan minstens 3 rechten.

Gebruik makende van deze axioma’s kan men aantonen dat elke rechte eenzelfde aantal (= s + 1) punten bevat, en dat door een punt eenzelfde aantal (= t + 1) rechten gaan. We zeggen dan dat Γ orde (s, t) heeft. Zijn er slechts een eindig aantal punten en rechten, dan wordt Γ een eindige veralgemeende veelhoek genoemd. De veralgemeende 3-hoeken zijn juist de projectieve vlakken, en werden al uitgebreid bestudeerd vóór de andere leden van de veralgemeende veelhoeken-familie het levenslicht zagen.

Er bestaan voorbeelden van veralgemeende n-hoeken voor elk natuurlijk getal n. Merkwaardig genoeg bestaan eindige veralgemeende n-hoeken enkel voor n ∈ {3, 4, 6, 8}. Voor deze waarden van n zijn er wat we noemen klassieke voorbeelden, d.w.z. voorbeelden die nauw verwant zijn met favoriete objecten van meetkundigen, zoals kwadrieken in projectieve ruimtes.

Een belangrijk begrip in een veralgemeende veelhoek is de afstand. Een pad in een veralgemeende veelhoek bestaat uit een opeenvolging van punten en rechten die incident zijn. De lengte van een pad wordt gedefinieerd als het aantal stappen dat je moet zetten om van het begin naar het einde te wandelen. Een pad (p, L, p′, L′, p′′) tussen de punten p en p′ heeft dus lengte 4. Omdat elke twee elementen van een veralgemeende n-hoek bevat zijn in een gewone n-hoek, is het duidelijk dat, om van een element in een ander element te geraken, hoogstens n stappen nodig zijn. De afstand tussen twee elementen is dus hooguit n. Elementen op afstand n worden tegenvoeters genoemd.

Veralgemeende veelhoeken zijn niet enkel het speelgoed van meetkundigen, maar duiken ook op in eerder algebraïsch of groep-theoretisch onderzoek. In deze thesis gaan we echter de meetkundige toer op. De bedoeling is enkele (los van elkaar staande) problemen te bekijken, en zo ‘onze veelhoekjes’ bij te dragen aan de grote veelhoek-puzzel.

Hoofdstuk 1 bundelt de gebruikte definities en stellingen. Voor een uitgebreidere kennismaking met de theorie van de veralgemeende veelhoeken verwijzen we naar het standaardwerk Generalized Polygons, Van Maldeghem [57].

B.2 Ken uw klassiekers

Voor de klassieke voorbeelden van veralgemeende veelhoeken vertaalt het verband met meetkundige objecten als kwadrieken zich in ‘mooie’ eigenschappen van deze veelhoeken. Nog mooier is het als die eigenschappen het klassieke voorbeeld in kwestie karakteriseren, d.w.z. van zodra een willekeurige veralgemeende veelhoek die eigenschap heeft, is hij klassiek. In hoofdstuk 2 worden een aantal karakteriseringen van klassieke veralgemeende zeshoeken gegeven. Hierbij zijn de volgende begrippen in een veralgemeende zeshoek essentieel (zie ook de figuur):

• Neem twee tegenvoetse punten a en b. Op elke rechte door a ligt een uniek punt op afstand 4 van b. Deze verzameling punten wordt het spoor a^{b} met basispunt a genoemd. Als alle sporen van een veralgemeende zeshoek Γ (dus met willekeurig basispunt) zich gedragen zoals rechten (d.w.z. twee verschillende sporen snijden in ten hoogste 1 punt), dan wordt Γ punt-2-regulier genoemd.

• Neem twee tegenvoetse rechten L en M. De verzameling punten op afstand 3 van L en M wordt de regulus bepaald door L en M genoemd. Als de reguli van een veralgemeende zeshoek Γ zich gedragen zoals rechten (d.w.z. twee verschillende reguli snijden in ten hoogste 1 rechte), dan wordt Γ 3-regulier genoemd.

Het begrip regulariteit ligt aan de grondslag van heel wat meetkundige karakteriseringen van veralgemeende zeshoeken. Een belangrijke karakterisering is deze van Ronan [35], die zegt dat, van zodra een veralgemeende zeshoek punt-2-regulier is, de zeshoek noodzakelijk klassiek is.

In een 2-reguliere zeshoek wordt dus een voorwaarde opgelegd op de doorsnede van elke twee sporen met zelfde basispunt. Een aantal karakteriseringen pogen die voorwaarde te verzwakken. Een intersectieblok is een doorsnede van twee sporen a^{b} en a^{c}, maar waarbij de punten b en c zodanig gekozen zijn dat er een punt bestaat collineair met b en c, en op afstand 4 van het punt a (zie figuur). Dit begrip werd ingevoerd door Ronan (zie [37]). In een klassieke zeshoek van de orde (q, t) die niet punt-2-regulier is, bevat zo’n intersectieblok juist t/q + 1 punten. De vierde tekening in de figuur geeft aan hoe het begrip intersectieblok iets kan veralgemeend worden: we eisen nu dat er een rechte bestaat op afstand 3 van a, b en c. In dit geval zeggen we dat b en c in (3, 4)-positie liggen ten opzicht van a. Dit geeft de volgende karakterisering.

Stelling Een eindige veralgemeende zeshoek van de orde (q, t) is duaal klassiek als en slechts als |a^{b} ∩ a^{c}| ≤ t/q + 1, voor elk drietal punten a, b en c zodat b en c in (3, 4)-positie liggen ten opzichte van a.

We geven nu nog enkele voorbeelden van karakteriseringen uit hoofdstuk 2.

Stelling Zij Γ een eindige veralgemeende zeshoek. Dan is Γ isomorf met de klassieke zeshoek H(q) of T(q^{3}, q), beide met q even, als en slechts als voor elk punt x, en elke twee tegenvoetse rechten L en M, er steeds een punt in de regulus bepaald door L en M bestaat dat niet tegenvoets x ligt.

(C) Veronderstel dat een punt a op afstand 4 ligt van juist één punt r van een regulus R, en tegenvoets alle andere punten van die regulus, dan liggen alle punten van R \ {r} op afstand 4 van het unieke punt dat collineair is met a en r.

Stelling Zij Γ een eindige veralgemeende zeshoek van de orde (s, s^{3}) of (s′^{3}, s′) die aan voorwaarde (C) voldoet. Dan is Γ isomorf met één van de klassieke zeshoeken T(s, s^{3}) of T(s′^{3}, s′), met s′ oneven.

Stelling Zij Γ een eindige veralgemeende zeshoek van de orde (q, t), q even. Dan voldoet Γ aan eigenschap (C) als en slechts als Γ isomorf is met één van de duaal klassieke zeshoeken H(q)^{D}, q niet deelbaar door 3 of T(q, q^{3}).

B.3 Vergeethoeken

De bedoeling van hoofdstuk 3 is een soort veralgemening van het begrip veralgemeende veelhoek te definiëren. Daarvoor inspireerden we ons op de definitie van een (duaal) semi-affien vlak, een structuur ingevoerd door Dembowski. Bij een affien vlak is er voor elk niet-incident punt-rechte paar {p, L} juist één rechte door p ‘parallel’ met L. Door ‘juist één’ te vervangen door ‘hoogstens één’, bekomt men de definitie van een semi-affien vlak. Aangezien parallellisme een equivalentierelatie definieert, kunnen we ook zeggen dat elke twee rechten snijdend of parallel zijn. We bekijken nu de duale structuur, d.w.z. met een equivalentierelatie op de punten in plaats van op de rechten. In de bekomen structuur zullen dus elke twee rechten snijden, en elke twee punten collineair of equivalent zijn. Sommige rechten zijn dus ‘vergeten’, en vervangen door equivalentieklassen van punten (de ‘gaten’ in het geheugen van de veelhoek).

Om dit te veralgemenen naar n-hoeken starten we met een meetkunde Γ = (P, L, I), en een equivalentierelatie op de puntenverzameling. Een gewone n-vergeethoek is dan een gewone n-hoek in deze meetkunde, maar waarbij sommige zijden vervangen kunnen zijn door equivalentieklassen van punten. De meetkunde Γ wordt een n-vergeethoek genoemd als de volgende axioma’s voldaan zijn:

(i) in de meetkunde Γ zijn geen gewone k-vergeethoeken, met k < n te vinden,

(ii) door elke twee punten, twee rechten, of een punt en een rechte is steeds een gewone n-vergeethoek te vinden,

(iii) elke rechte bevat minstens drie punten, door elk punt gaan minstens twee rechten, door een punt dat enkel equivalent is met zichzelf gaan minstens drie rechten.

Duidelijkerwijs voldoet een veralgemeende n-hoek aan bovenstaande axioma’s, door elke equivalentieklasse van grootte 1 te nemen. De eindige duale semiaffiene vlakken (dus de 3-vergeethoeken) zijn geclassificeerd; ze ontstaan uit een projectief vlak (dus een veralgemeende 3-hoek) door (stukken van) rechten te vervangen door equivalentieklassen. Dit brengt ons op ideeën om n-vergeethoeken te constueren. Zo kunnen we bijvoorbeeld starten met een veralgemeende n-hoek Δ, en een verzameling niet-snijdende rechten in Δ. Door elk van deze rechten te vervangen door een equivalentieklasse, bekomen we een n-vergeethoek. Een ander voorbeeld verkrijg je door te starten van een veralgemeende vierhoek Δ, en hieruit één punt p weg te laten. Elke rechte door p wordt een equivalentieklasse, het resultaat is een 4-vergeethoek waarbij elke equivalentieklasse juist 1 punt minder bevat dan een willekeurige rechte.

De vraag is nu of elke eindige vergeethoek eigenlijk niets anders is dan een vergeetachtig geworden veralgemeende veelhoek. In hoofdstuk 3 wordt de volgende stelling aangetoond:

Stelling Een eindige n-vergeethoek, met n oneven, wordt steeds bekomen uit een eindige veralgemeende n-hoek, en bestaat dus enkel voor n = 3.

Een vergeethoek wordt kort van geheugen genoemd als er parameters (g, k, d) bestaan zodat:

• door elk punt juist k rechten en één niet-triviale klasse, of k +1 rechten en geen niet-triviale klasse gaan, • elke niet-triviale klasse juist g punten bevat,

• elke rechte g + d punten bevat, d ≥ 1.

Bij een vergeethoek die kort van geheugen is zijn de rechten dus ‘langer’ dan de klassen.

Stelling Een eindige n-vergeethoek, met n even, wordt bekomen uit een eindige veralgemeende n-hoek, of is kort van geheugen.

In hoofdstuk 3 concentreren we ons dan verder op de 4-vergeethoeken die kort van geheugen zijn. We tonen aan dat, als d = 1, deze ook bekomen kunnen worden uit een veralgemeende 4-hoek, en geven voorbeelden en karakteriseringen van 4-vergeethoeken die werkelijk zeer kort van geheugen zijn. Hierbij blijft het wel een vraagteken of deze voorbeelden ook bekomen kunnen worden uit een veralgemeende 4-hoek op de gebruikelijke manier, d.w.z. door het vergeten van rechten.

B.4 Veelhoeken door een speciale bril

Zoals reeds gezegd is in een veralgemeende veelhoek een afstand gedefinieerd. Als tussen twee veralgemeende veelhoeken een afbeelding bestaat die alle afstanden bewaart, dan is die afbeelding een isomorfisme. De vraag is nu of we deze voorwaarde kunnen verzwakken. Veronderstel dus dat tussen twee veralgemeende n-hoeken een afbeelding bestaat die een bepaalde afstand i bewaart: is deze afbeelding een isomorfisme? In hoofdstuk 4 tonen we het volgende aan:

Stelling

• Zij Γ en Γ′ twee veralgemeende n-hoeken, n ≥ 4, i een even getal, 2 ≤ i ≤ n − 1 en α een surjectieve afbeelding van de puntenverzameling van Γ naar de puntenverzameling van Γ′. Als voor elke twee punten a en b van Γ, δ(a, b) = i als en slechts als δ(a^{α}, b^{α}) = i, dan is α uit te breiden tot een isomorfisme tussen Γ en Γ′.

• Zij Γ en Γ′ twee veralgemeende n-hoeken, n ≥ 2, i een oneven getal, 1 ≤ i ≤ n − 1 en α een surjectieve afbeelding van de puntenverzameling van Γ naar de puntenverzameling van Γ′ en van de rechtenverzameling van Γ naar de rechtenverzameling van Γ′. Als voor elk punt-rechte paar {a, b} van Γ, δ(a, b) = i als en slechts als δ(a^{α}, b^{α}) = i, dan definieert α een isomorfisme tussen Γ en Γ′.

De idee van het bewijs van deze stelling is de volgende. De bedoeling is aan te tonen dat α collineariteit bewaart (dan volgt het gestelde uit een reeds bestaande karakterisering van isomorfismen). We kijken nu naar de gegeven veelhoek met een speciale bril, die ons enkel toelaat te zien of twee elementen al dan niet op afstand i gelegen zijn. Door deze bril zien Γ en Γ′ er hetzelfde uit, wegens de afbeelding α. Als we nu een manier kunnen vinden om met deze bril toch collineariteit te ontdekken, dan weten we dus dat α ook collineariteit bewaart. Het bewijs reduceert zich dus tot het zoeken naar een eigenschap waarin twee collineaire punten zich onderscheiden van twee niet-collineaire punten, en die geformuleerd kan worden door enkel gebruik te maken van ‘afstand i’ en ‘niet afstand i’.

Voor i = n zal het niet mogelijk zijn de stelling te bewijzen, aangezien er een tegenvoorbeeld bestaat voor één van de klassieke zeshoeken. We kunnen natuurlijk hopen dat er niet al te veel tegenvoorbeelden zijn. Zo vragen we ons in hoofdstuk 4 af of misschien alleen zeshoeken aanleiding geven tot tegenvoorbeelden. Voor eindige veralgemeende veelhoeken kunnen we inderdaad bewijzen dat alleen zeshoeken van de orde (q, q) problemen geven. Een andere vraag is of onder de zeshoeken alleen die éne klassieke zeshoek een tegenvoorbeeld is. Hier krijgen we een positief antwoord op voorwaarde dat de automorfismengroep van de zeshoek in kwestie voldoende groot is.

Een variant op bovenstaande stelling verkrijgen we door te gaan kijken naar afbeeldingen gedefinieerd op de vlaggenverzameling van een veralgemeende n-hoek. In dit geval kunnen we bewijzen dat er maar één tegenvoorbeeld is, namelijk voor de kleinste veralgemeende vierhoek.

B.5 Strategische verzamelingen in een klassieke zeshoek

In appendix A concentreren we ons op de klassieke zeshoek H(q)^{D}. Veronderstel dat een bericht moet doorgegeven worden naar alle punten van deze zeshoek, en dat, van zodra twee collineaire punten van het bericht op de hoogte zijn, ook alle punten op de rechte door deze twee punten kunnen geïnformeerd worden. Hoeveel startpunten zijn minimaal nodig om op die manier alle (q + 1)(q^{4} + q^{2} + 1) punten op de hoogte te brengen? Zo’n verzameling startpunten die de hele zeshoek voortbrengen, noemen we een strategische verzameling. De uitdaging is nu een strategische verzameling met een minimaal aantal punten te vinden. Een stelling bewezen door Thas & Van Maldeghem (zie [48]) leert dat zeker een strategische verzameling van 15 punten bestaat, en dat je altijd minstens 14 punten zal nodig hebben. Als q een priemgetal is, dan kan het steeds met 14 punten. In appendix A proberen we nu uit zo’n strategische verzameling van 15 punten, er één te selecteren, zodat de verzameling zonder dit punt nog steeds strategisch blijft. Om een idee te hebben van het te bewijzene (14 is mogelijk, of 14 is zeker niet mogelijk) pakten we de kleinste gevallen aan met de computer. In deze gevallen bleek er inderdaad een ‘overbodig’ punt te zitten in de verzameling van 15 punten. Of en hoe dit te veralgemenen is voor een willekeurige waarde van q, blijft voorlopig een goed bewaard ‘militair’ geheim van deze zeshoek, dus het gedeeltelijk resultaat dat we bekomen is: 14 ⇒ (q + 1)(q^{4} + q^{2} + 1) als q priem is, of q ∈ {4, 8, 9, 16}...

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220

Index of Notations

Γ_{i}(x) the set of elements of Γ at distance i from x
x^{⊥}  the set of points collinear with the point x
x^{⊥ ⊥}  the set of points not opposite the point x
x_{[i]}^{y}  Γ_{i}(x) ∩ Γ_{n−i}(y), with δ(x, y) = n
x^{y}  x_{[2]}^{y}, with δ(x, y) = n
⟨x, y⟩ x_{[ \frac{n}{2} ]}^{y}, with δ(x, y) = n
R(x, y) (in 3-regular hexagon) the regulus containing x and y
a⋈b  the unique element at distance \frac{δ(a,b)}{2} from a and b (if defined)
[x, y]  the shortest path between x and y (if defined)
[x; y]  projectivity from Γ_{1}(x) to Γ_{1}(y)
H(q)  finite split Cayley hexagon of order q
H(q)^{D} finite dual split Cayley hexagon of order q
T(q^{3}, q) finite twisted triality hexagon of order (q^{3}, q)
T(q, q^{3}) finite dual twisted triality hexagon of order (q, q^{3})

Index

(3,4)-position, 42

adjacent flags, 2 anti-automorphism, 5 anti-isomorphism, 5 anti-regular

2n-gon, 46 hexagon, 44

antiflag, 2 apartment, 3 association scheme, 23, 95 automorphism, 5

BN-pair, 14, 146

circuit, 2 collineation, 5 coordinatization, 20 Coxeter distance, 14, 113, 155

deficiency, 91 distance, 2 distance-i-regular, 6 distance-i-trace, 6 distance-preserving map, 111 double of a geometry, 2 dual geometric hyperplane, 148 dual of a geometry, 2 dual ovoidal subspace, 15, 51, 147

elation, 13 embedding, 5 extremal hexagon, 6, 16, 41

finite classical hexagons, 11 finite dual classical hexagons, 11 finite geometry, 2 flag, 2 forgetful polygon, 61 forgetful quadrangle

arising from generalized quad-

rangle, 109 full subpolygon, 4

G_{2}(𝕂), 14, 151 generalized n-gon, 3 generating set, 196 geometric hyperplane, 15, 148 geometry, 2 girth, 2 grid, 8 group of projectivities, 13

with respect to spread, 54

H(𝕂), 10

hat-rack of the coordinatization, 18 Hermitian spread, 15, 51 homology, 13 hyperbolic line, 16, 146

long, 17

ideal subpolygon, 4, 97 imaginary line, 16, 36 incidence geometry, 2 incidence graph, 2 incidence relation, 2 intersection set

(i, j)-intersection set, 30 half (i, i)-intersection set, 32

in generalized hexagon, 16 intersection-regular, 50 isolated point, 61 isomorphism, 5

k-path, 2 Krein conditions, 23

length of a path, 2 line blocking set, 50 line pencil, 4 line regulus, 7 little projective group, 13

Moufang hexagons, 13 Moufang path, 13 Moufang polygon, 13

opposite elements, 4 order, 2 ordinary n-gon, 2 ovoid

of generalized hexagon, 15 of generalized polygon, 14 of generalized quadrangle, 14 regular ovoid, 15, 97

ovoidal subspace, 15, 47

parameters of a polygon, 3 path, 2

forgetful, 61 perp, 4 point regulus, 7 point row, 4 projection, 4 projectivity, 13

R(u, v), 7 Ree-Tits octagons, 13 regular element, 6 regulus, 7

semi-affine plane, 59 semi-plane, 70 short forgetful polygon, 92 of ovoid type, 98 of subquadrangle type, 97

span-regular, 17

split Cayley hexagon, 10

spread Hermitian, 15 of polygon, 14

square forgetful polygon, 70

strongly regular graph, 22, 94, 154 subgeometry, 2 subpolygon, 4

T(𝕂, 𝕂^{(σ)}, σ), 11 thick geometry, 2 thin element, 2 Tits polygon, 14 Tits system, 14 trace, 6 triad of points, 6 triality, 10 twisted triality hexagon, 11

weak anti-regular, 44 weak generalized n-gon, 3 Weyl group, 14