Acknowledgement

I want to express my gratitude to the following persons and organizations. I want to thank ”The Fund for Scientific Research - Flanders (Belgium)” for their financial support during my research. I want to thank Prof. Dr. F. De Clerck, Prof. Dr. L. Storme, Prof. Dr. J. A. Thas and Prof. Dr. H. Van Maldeghem for the possibility to do research in Finite Geometry, and also for their helpful discussions and remarks concerning this research. In this context, I also want to thank Dr. C. O’Keefe, who spend one year at our department, and the finite geometers at the University of Colorado at Denver (Prof. Dr. S. E. Payne, Dr. M. Miller and Prof. Dr. W. Cherowitzo) who I was able to visit in the spring of 1999. And finally, I want to thank my supervisor Prof. Dr. F. De Clerck for the many hours he spend during the last three years and for the many remarks he made while I was writing this thesis.

Bart De Bruyn, December 1999

Preface

In 1980 Shult and Yanushka [82] introduced near polygons because of their connection with certain system of lines in the Euclidean space. These incidence structures can be seen as a generalization of a class of incidence structures, called generalized polygons, which themselves were introduced by Tits [92] in 1959 and are discussed in detail in [98]. In the beginning of the 80’s, near polygons were studied in detail, especially by Brouwer. Several regular near polygons were characterized by their parameters (e.g. [11], [12] and [24]). Very important was the characterization of classical near polygons by Cameron [22]. Since the very beginning ([82], see also [80]) it was clear that, under some mild conditions, these incidence structures had substructures, isomorphic to generalized quadrangles, called quads. Many results about generalized quadrangles are therefore very useful for the study of near polygons. In [17], it was proved that, under some mild conditions, every two points of a near polygon are contained in a sub near polygon.

Finding new regular near polygons is an extremely interesting problem, but also a very hard one. The search for new near polygons which have quads through every two points at distance 2, is still very interesting and often more successful. This thesis contains several new classes of such near polygons. They are all in Chapter 7 and constructed in a uniform way. The whole chapter 7 is devoted to these near polygons (we call them glued near polygons) and characterizations are given.

The importance of the search for new near polygons which have quads through every two points at distance 2 is also clear from the paper [14], containing a classification of all such near hexagons with 3 points on a line. In Chapter 8, we try to do the similar classification where all near hexagons have four points on a line. There remain however four open cases. The results of Chapter 7 are extremely important during this classification.

To develop the theory of glued near polygons, we had to derive some results concerning generalized quadrangles. All these results are contained in Chapter 2, together with a brief overview of the theory of these incidence structures. For more details, we often refer to the standard work [71], see also [88].

Chapter 3 contains an overview of the basic definitions and properties (some of them being new) of near polygons. In Chapter 4, we prove that association schemes can be associated to certain near polygons. In Chapter 5, we define i-neighbourhoods of a near polygon. We also present a new construction to obtain a class of near polygons and give a property that almost characterizes these near polygons. In Chapter 6, we discuss linear representations of near polygons. We try to classify all thick near hexagons which have a linear representation and discuss the open cases. This study led to the discovery of new near polygons which in turn led us to the bigger class of glued near polygons which are studied in Chapter 7. Finally, in an appendix we give the classification of all ovoids of the near hexagon related to the Steiner system S(5, 8, 24). This classification was done earlier [16], but a new proof is presented here.

Contents

1 Basic notions and properties 1 1.1 The notion of incidence structure . . . . . . . . . . . . . . . . 1 1.2 Near polygons . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Genesis . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.2.2 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Generalized polygons . . . . . . . . . . . . . . . . . . . . . . . 5 1.3.1 Definition . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3.2 Projections and projectivities . . . . . . . . . . . . . . 6 1.3.3 Classification results . . . . . . . . . . . . . . . . . . . 7

1.4 Projective spaces . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4.1 Caps . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4.2 The Steiner systems S(5, 6, 12) and S(5, 8, 24) . . . . . 9 1.4.3 Regular spreads in PG(3, q) . . . . . . . . . . . . . . . 11 1.4.4 The geometry H_{q}^{3} . . . . . . . . . . . . . . . . . . . . . 13 1.4.5 Affine representations . . . . . . . . . . . . . . . . . . . 14

2 Generalized quadrangles 15 2.1 Definitions and elementary properties . . . . . . . . . . . . . . 15 2.2 A description of some classes of GQ’s . . . . . . . . . . . . . . 18

2.2.1 The classical GQ’s . . . . . . . . . . . . . . . . . . . . 18 2.2.2 The GQ T_{2}^{∗}(H) . . . . . . . . . . . . . . . . . . . . . . 19 2.2.3 The GQ AS(q), q odd prime power . . . . . . . . . . . 19 2.2.4 The GQ T_{d}(O), d ∈ {2, 3} . . . . . . . . . . . . . . . . 19 2.2.5 The GQ S_{xy}^{−} . . . . . . . . . . . . . . . . . . . . . . . . 20

2.3 Classification and isomorphisms . . . . . . . . . . . . . . . . . 20 2.4 Some general constructions for generalized quadrangles …. 22

2.4.1 The Payne-construction . . . . . . . . . . . . . . . . . 22 2.4.2 The AT-construction . . . . . . . . . . . . . . . . . . . 23

2.5 The known GQ’s with an AT-model . . . . . . . . . . . . . . . 25 2.5.1 The trivial GQ’s . . . . . . . . . . . . . . . . . . . . . 26 2.5.2 The GQ P (W (q), x) . . . . . . . . . . . . . . . . . . . 26 2.5.3 The GQ T_{2}^{∗}(O) . . . . . . . . . . . . . . . . . . . . . . 27 2.5.4 The GQ Q(5, q) . . . . . . . . . . . . . . . . . . . . . . 28 2.5.5 The GQ (S_{xy}^{−})^{D} . . . . . . . . . . . . . . . . . . . . . . 28

2.6 Ovoids and spreads of GQ’s . . . . . . . . . . . . . . . . . . . 29 2.6.1 Ovoids . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.6.2 Spreads . . . . . . . . . . . . . . . . . . . . . . . . . . 30

2.7 Characterizations of GQ’s with an AT-model . . . . . . . . . . 34 2.7.1 Characterization in terms of a group of automorphisms 34 2.7.2 Characterizations in terms of a group of projectivities . 39

2.8 Automorphisms of a GQ fixing a spread . . . . . . . . . . . . 42 2.9 The nonexistence of spreads of symmetry in certain GQ’s of

order (s, s^{2}) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 2.9.1 Generalized quadrangles as group coset geometries… 46 2.9.2 Generalized quadrangles arising from q-clans . . . . . . 47 2.9.3 Ovoids of symmetry in GQ’s derived from q-clans … 49 2.9.4 The nonexistence of ovoids of symmetry in GQ’s in

nonclassical Kantor-Knuth quadrangles. . . . . . . . . 52

2.10 Generalized quadrangles of order (s, s + 2) with a lot of grids . 53 2.10.1 Amalgamations of projective planes . . . . . . . . . . . 53 2.10.2 GQ’s with a unique grid through every two intersecting

lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

3 Basic theory of near polygons 60 3.1 Definitions and line-line relation . . . . . . . . . . . . . . . . . 60 3.2 Examples of some regular near polygons . . . . . . . . . . . . 62

3.2.1 Near polygons of Hamming type . . . . . . . . . . . . . 62 3.2.2 Dual polar spaces . . . . . . . . . . . . . . . . . . . . . 63 3.2.3 The near hexagon related to S(5, 8, 24) . . . . . . . . . 65 3.2.4 The near hexagon related to the ternary Golay code . . 65 3.2.5 The near octagon related to the Hall-Janko group … 65

3.3 Characterizations of the Hamming near polygons in terms of

parallellism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 3.4 The nondegenerate support of a near polygon . . . . . . . . . 67 3.5 Quads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

3.5.1 Special point sets . . . . . . . . . . . . . . . . . . . . . 69 3.5.2 Local spaces . . . . . . . . . . . . . . . . . . . . . . . . 71

3.6 Dual polar spaces and classical near polygons . . . . . . . . . 71 3.7 Near polygons with quads . . . . . . . . . . . . . . . . . . . . 73 3.8 Near polygons with good quads . . . . . . . . . . . . . . . . . 75 3.9 Some characterizations for classical and regular near hexagons 76 3.10 A construction method for near polygons . . . . . . . . . . . . 78 4 Association schemes associated with near polygons 80

4.1 Association schemes and distance regular graphs . . . . . . . . 80 4.2 Some relations on the line set of a near hexagon . . . . . . . . 83 4.3 The numbers n_{i} and p_{jk}^{i} in the case of a regular near hexagon

with quads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 4.4 The case where R determines an association scheme . . . . . . 88 4.5 A characterization theorem . . . . . . . . . . . . . . . . . . . . 93 4.6 An association scheme derived from a GQ(s, s^{2}), s ≥ 1, having

a spread of symmetry . . . . . . . . . . . . . . . . . . . . . . . 94

5 The i-neighbourhood of a point 96 5.1 Definitions and properties . . . . . . . . . . . . . . . . . . . . 96 5.2 Construction of near polygons from partial linear spaces … 98

5.2.1 Main theorem . . . . . . . . . . . . . . . . . . . . . . . 98 5.2.2 Construction . . . . . . . . . . . . . . . . . . . . . . . 98

5.3 The 2-neighbourhood of a point . . . . . . . . . . . . . . . . . 102

6 Linear representations of near polygons 104 6.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 6.2 Linear representations of near polygons . . . . . . . . . . . . . 106 6.3 Linear representations of generalized quadrangles . . . . . . . 107 6.4 Some constructions of linear representations of near polygons . 107 6.5 Linear representations of near hexagons . . . . . . . . . . . . . 110

6.5.1 Necessary and sufficient conditions . . . . . . . . . . . 110 6.5.2 The known examples . . . . . . . . . . . . . . . . . . . 111 6.5.3 Conditions on the set K . . . . . . . . . . . . . . . . . 112 6.5.4 The near hexagons with a T_{2}^{∗}(O)-quad . . . . . . . . . 115 6.5.5 The near hexagons without a T_{2}^{∗}(O)-quad . . . . . . . 120

6.6 Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . 124 6.7 Some properties of the regular near hexagon with parameters

s = 2, t_{2} = 1, t = 11 . . . . . . . . . . . . . . . . . . . . . . . . 129 6.7.1 A general theorem on Cameron closed subspaces …. 129 6.7.2 The Cameron closed subspaces of the regular near hexagon

with parameters s = 2, t_{2} = 1, t = 11 . . . . . . . . . . 130 6.7.3 The ovoids of the regular near hexagon with parame-

ters s = 2, t_{2} = 1, t = 11 . . . . . . . . . . . . . . . . . 131

7 Glued near polygons 132 7.1 Constructions of glued near polygons . . . . . . . . . . . . . . 132 7.2 On the classification of glued near polygons . . . . . . . . . . 137 7.3 Sub near polygons . . . . . . . . . . . . . . . . . . . . . . . . 139 7.4 Isomorphisms between glued near hexagons . . . . . . . . . . . 142 7.5 The number of nonisomorphic glued near hexagons . . . . . . 145 7.6 Characterizations . . . . . . . . . . . . . . . . . . . . . . . . . 149

7.6.1 Local spaces with a thin point . . . . . . . . . . . . . . 149 7.6.2 Characterizations of glued near hexagons . . . . . . . . 150

8 Near polygons with four points on a line 154 8.1 The examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 8.2 The GQ’s with s = 3 and their ovoids . . . . . . . . . . . . . . 156 8.3 The classification for some special cases . . . . . . . . . . . . . 157

8.3.1 The case of classical near hexagons . . . . . . . . . . . 157 8.3.2 The case of glued near hexagons . . . . . . . . . . . . . 158 8.3.3 The case of regular near hexagons . . . . . . . . . . . . 159

8.4 The classification . . . . . . . . . . . . . . . . . . . . . . . . . 159 8.4.1 Case I: all quads of type Q(4, 3) are big . . . . . . . . . 159 8.4.2 Case II: there is a quad of type Q(4, 3) which is not big 161 8.4.3 The main theorem . . . . . . . . . . . . . . . . . . . . 170

A The classification of ovoids in the near hexagon associated with S(5, 8, 24) 171 A.1 The problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 A.2 The quads in S . . . . . . . . . . . . . . . . . . . . . . . . . . 172 A.3 The classification of the ovoids . . . . . . . . . . . . . . . . . . 173

Chapter 1

Basic notions and properties

1.1 The notion of incidence structure

An incidence structure S is a triple consisting of

(1) a nonempty set P, whose elements are called points;

(2) a (possible) empty set L, disjoint with P, whose elements are called lines;

(3) an incidence relation I ⊆ (P × L).

At some other places in the literature, I is regarded as a subset of (P × L) ∪ (L × P) and called the symmetrized incidence relation. One writes S = (P, L, I). Unless otherwise stated, all incidence structures considered in this thesis are finite, i.e. P and L are finite sets. If (p, L) ∈ I (we also write p I L), then we will say: the point p is incident with L, the point p lies on L, the line L goes through p, etc. Two different points of S are called collinear when there is a line through these points. An incidence structure is said to have order (s, t) if every line is incident with exactly s + 1 points and if every point is incident with exactly t + 1 lines. An incidence structure is called thick if every line is incident with at least three points and if every point is incident with at least three lines. With each incidence structure S, there is associated a graph Γ(S), called the collinearity graph or point graph of S. The vertices of Γ(S) are the points of S; two different vertices are adjacent whenever they are collinear. When no confusion is possible, we will write Γ instead of Γ(S). An incidence structure is called connected if its point graph is connected. With the graph Γ(S), there corresponds a distance function, denoted by d_{S}(·, ·) or shortly by d(·, ·) when no confusion is possible. If x is a point of S and if i ∈ ℕ, then Γ_{i}(x) denotes the set of all points at distance i from x (in the graph Γ) and we will write Γ(x) := Γ_{1}(x). If A and B are two sets of points, then we define d(A, B) := min d(x, y), where x and y range over A and B respectively. If A = {x}, we also write d(x, B) instead of d({x}, B). There is another graph related to an incidence structure S = (P, L, I), namely the incidence graph: the vertices of this graph are the elements of P ∪ L, two vertices being adjacent when they are incident. Two incidence structures S = (P, L, I) and S′ = (P′, L′, I′) are called isomorphic when there is a bijection α : P → P′ and a bijection β : L → L′ such that p I L ⟺ α(p) I′ β(L), ∀p ∈ P, ∀L ∈ L. An incidence structure is called a partial linear space when every line is incident with at least two points and when every two different points are incident with at most one line. If every two different points are incident with exactly one line, then the partial linear space is called a linear space. The partial linear space S′ = (P′, L′, I′) is called a subgeometry of the partial linear space S = (P, L, I) if P′ ⊆ P, L′ ⊆ L and if I′ is the restriction of I to P′ × L′. If S = (P, L, I) is an incidence structure without multiple lines, i.e. there are no two lines which are incident with the same set of points, then it is isomorphic to S′ = (P′, L′, I′), where P′ = P, L′ = ⋃_{L∈L}{{x|x I L}}, and where p′ I′ L′ ⟺ p′ ∈ L′ for all p′ ∈ P′ and all L′ ∈ L′. That is the reason why, at some places in this thesis, we will consider lines as sets of points and write p ∈ L instead of p I L. The reader is supposed to be familiar with some incidence structures like projective and affine spaces. Important examples of incidence structures which are in general not partial linear spaces are the designs. A t − (v, k, λ) design is an incidence structure S = (P, L, I) satisfying the following properties (for several reasons the elements of L will be called block instead of lines):

(1) there are exactly v points;

(2) each block is incident with k ≥ 2 points;

(3) every t different points are incident with exactly λ lines.

If λ = 1, then the design is called a Steiner system and denoted by S(t, k, v). We will meet some designs later in this thesis. In the following section, we will define the class of partial linear spaces which is the main object of study in this thesis.

1.2 Near polygons

1.2.1 Genesis

Consider the n-dimensional Euclidean space with origin O. A line system of type (a_{1}, a_{2}, …, a_{k}) is a set A of lines through O with the property that cos α ∈ {±a_{1}, ±a_{2}, …, ±a_{k}} for any two different lines of A at angle α. A system of vectors of type (a_{1}, a_{2}, …, a_{k}) is a set Σ of vectors satisfying:

(i) Σ = −Σ := {−x|x ∈ Σ},

(ii) the norm (x, x) of x is a nonzero constant c for all x ∈ Σ,

(iii) if x, y ∈ Σ and x ≠ ±y, then \frac{(x,y)}{c} ∈ \{±a_{1}, ±a_{2}, …, ±a_{k}\}.

Line systems of type (a_{1}, …, a_{k}) are clearly equivalent to systems of vectors of type (a_{1}, …, a_{k}). In [82] E. Shult and A. Yanushka discussed line systems of type (0, \frac{1}{3}) and showed that they were related to a class of geometries which they called near polygons. The angle α for which cos α = −\frac{1}{3} is the angle subtended by two chords drawn from the barycenter of a tetrahedron to two of its corner vertices. A line system of type (0, \frac{1}{3}) is called tetrahedrally closed if it has 0, 1, 2 or 4 lines in common with every set of four lines obtained by connecting the barycenter of an arbitrary tetrahedron centered at O with its four corner vertices. A system of vectors of type (0, \frac{1}{3}) is called a tetrahedrally closed (0, \frac{1}{3}) system if the corresponding line system is tetrahedrally closed.

Let Σ be a tetrahedrally closed (0, \frac{1}{3}) system of vectors of norm 3. Fix a vector u ∈ Σ and let Σ_{−1}(u) be the set of vectors of Σ having inner product -1 with u. If Σ_{−1}(u) ≠ ∅, then we can define the following incidence system S: the point set is equal to Σ_{−1}(u); a triplet of three vectors {y_{1}, y_{2}, y_{3}} ⊆ Σ_{−1}(u) is a line if and only if u, y_{1}, y_{2}, y_{3} define the four vertices of a tetrahedron centered at O.

Theorem 1.2.1 ([82], Proposition 3.10) If for every two points of Σ_{−1}(u) having inner product 1, there exists at least one point of Σ_{−1}(u) having inner product −1 with both points, then S is a connected partial linear space satisfying:

(a) the distance between two points is at most 3,

(b) for every point p and every line L, there exists a unique point on L nearest to p (with respect to the distance in the collinearity graph).

Two different points have distance 1, 2, respectively 3, if their inner product equals −1, +1, respectively 0.

Also, the converse of this theorem holds, as shown by the next theorem.

Theorem 1.2.2 ([82], Proposition 3.14) For every connected partial linear space S satisfying properties (a) and (b) of Theorem 1.2.1, there exists a tetrahedrally closed (0, \frac{1}{3}) system Σ of vectors of norm 3 and a vector u ∈ Σ such that

(i) Σ = {u, −u} ∪ (Σ_{−1}(u)) ∪ (−Σ_{−1}(u)) and

(ii) the partial linear space defined on Σ_{−1}(u) by the tetrahedra containing u is isomorphic to S.

1.2.2 Definition

A near polygon is a connected partial linear space satisfying the following property.

(NP) For each point p and every line L, there exists a unique point q on L nearest to p (with respect to the distance in the point graph).

This axiom is very important in this thesis, and we will often refer to it as (NP). If d is the diameter of the point graph, then the near polygon is called a near 2d-gon. A near 0-gon has one point, but no lines. A near 2-gon consists of exactly one line with a number (≥ 2) of points on it. The near quadrangles are known as the generalized quadrangles. These geometries have been widely studied. Because of their importance in the theory of near polygons, we devote Chapter 2 to these geometries.

At some places in this thesis, we will also meet the near (2d + 1)-gons, these are the connected partial linear spaces satisfying the following properties.

(1) The point graph has diameter d.

(2) For each point p and every line L with d(p, L) < d, there exists a unique point q on L nearest to p.

(3) There exists a point p and a line L such that d(p, L) = d.

Near 1-gons do not exist. The near 3-gons are exactly the linear spaces different from the point and the line. In this thesis we will meet a special type of near pentagon, namely the partial quadrangles. A partial quadrangle with parameters (s, t, μ) is a near pentagon of order (s, t) such that every two noncollinear points have exactly μ common neighbours. This thesis mainly treats the near 2d-gons. Unless otherwise stated, we always mean a near 2d-gon, when we talk about a near polygon. This convention is also used in the literature.

Remark. Properties (a) and (b) of Theorem 1.2.1 are shortly written as: S is a near 2d-gon with d ≤ 3.

1.3 Generalized polygons

These incidence structures were introduced by Tits in [92]. We will only give some definitions and properties related to these incidence structures; for a detailed survey, we refer to [98].

1.3.1 Definition

Let n ≥ 3, n ≠ 4, be an integer. A generalized n-gon S = (P, L, I) is a partial linear space satisfying the following axioms:

(GP1) S has no subgeometry which is an ordinary k-gon, 2 < k < n.

(GP2) Any two elements of P ∪ L are contained in some ordinary n-gon.

The above axioms also work in the case n = 4, but the geometries then obtained would not coincide with the generalized quadrangles which we defined above and which we will consider in the next chapter. However, only those generalized quadrangles which are degenerate (see next chapter for the definition) will not be covered by (GP1) and (GP2). Since we want that these degenerate examples are generalized polygons, we assumed n ≠ 4 in the above definition. The generalized 3-gons are exactly the (possible degenerate) projective planes. From (GP1) and (GP2), it follows that the dual of a generalized n-gon is again a generalized n-gon (if n ≠ 4). The reason why generalized n-gons are important in this thesis is because they are near n-gons (also for n odd).

Theorem 1.3.1 Every generalized n-gon S, n ≥ 3, is a near n-gon.

Proof. Suppose first that n = 2d, d ≥ 3. Every two points of S are contained in an ordinary 2d-gon which proves that the diameter of the point graph is at most d. Since the distance between two opposite points of such an ordinary 2d-gon is d (otherwise (GP1) would not be satisfied), the diameter is exactly d. Take now a point a and a line L, then there exists an ordinary 2d-gon through these elements which proves that every point of L at smallest distance from a has distance at most d − 1 to a. If L contains two points at smallest distance from a, then (GP1) would not be satisfied. Suppose next that n = 2d + 1, d ≥ 1. Every two points of S are contained in an ordinary (2d + 1)-gon which proves that the diameter of the point graph is at most d. Consider now an ordinary (2d + 1)-gon as subgeometry and take a point p and a line L which are opposite in this ordinary (2d+1)-gon, then d(p, L) = d otherwise (GP1) would not be satisfied. This also proves that the diameter of the point graph is equal to d. Now, take a point p and a line L, such that d(p, L) < d. The line L contains a unique point at distance d(p, L) from p, otherwise (GP1) would not be satisfied. □

Thick generalized polygons satisfy some nice properties.

Theorem 1.3.2 ([98], Theorem 1.3.2) A generalized polygon is thick if and only if it has an ordinary (n + 1)-gon as subgeometry.

Theorem 1.3.3 ([98], Corollary 1.5.3) Every thick generalized n-gon has order (s, t) with s, t ≥ 2; if n is odd, then s = t.

1.3.2 Projections and projectivities

Let S = (P, L, I) be a generalized n-gon and consider the incidence graph Δ with associated distance δ(·, ·). For a vertex x of Δ, let Δ(x) denote the set of elements of P ∪ L incident with x. Two elements a and b of P ∪ L are called opposite if δ(a, b) = n. If n is even, then two opposite elements have the same type (i.e. both lines or points); if n is odd, then two opposite elements have different types. If a and b are two different elements of P ∪ L which are not opposite, then there exists a unique element c of P ∪L incident with b and at distance δ(a, b) − 1 of a. This point c is called the projection of a onto b. Consider now two opposite elements a and b. Every element c ∈ Δ(a) has distance n − 1 from b and hence we can project it onto b. In this way, we find a bijection

[a, b] : Δ(a) → Δ(b),

called the perspectivity from a to b. Clearly [a, b] is a bijection, with inverse [b, a]. For elements a_{0}, …, a_{k} of P ∪ L with δ(a_{i}, a_{i+1}) = n for 0 ≤ i < k, the bijection

[a_{0}, a_{1}, …, a_{k}] := [a_{k−1}, a_{k}] ∘ … ∘ [a_{1}, a_{2}] ∘ [a_{0}, a_{1}]

is called a projectivity from Δ(a_{0}) to Δ(a_{k}). The set of all projectivities from Δ(a) to itself forms a permutation group Π(a), which is called the group of projectivities of a. If A is a set of mutual opposite elements of P ∪ L containing a_{0}, then, in the above definition of projectivity, one can require that all elements a_{i} belong to A. In this way we can define a permutation group Π_{A}(a_{0}), called the group of projectivities of a_{0} with respect to A.

1.3.3 Classification results

In this subsection, we give some theorems which should give the reader an idea what values of n, s and t are allowed for a generalized n-gon of order (s, t).

Theorem 1.3.4 ([40]) Let S be a generalized n-gon of order (s, t) with n ≥ 3, then at least one of the following holds.

(a) S is an ordinary n-gon.

(b) S is a nondegenerate projective plane.

(c) S is a generalized quadrangle.

(d) S is a generalized hexagon. Moreover, if S is thick, then st is a square.

(e) S is a generalized octagon. Moreover, if S is thick, then 2st is a square.

(f) S is a generalized 12-gon which is not thick.

Theorem 1.3.5 Let S be a thick generalized n-gon of order (s, t), n ≥ 4, then one of the following holds:

(a) n = 4, s ≤ t^{2} and t ≤ s^{2} ([47]);

(b) n = 6, s ≤ t^{3} and t ≤ s^{3} ([44]);

(c) n = 8, s ≤ t^{2} and t ≤ s^{2} ([47]).

For an overview of the possible orders of finite GQ we refer to Section 2.5 of Chapter 2. All known thick generalized hexagons have order (q, q), (q, q^{3}) or (q^{3}, q) with q a prime power. For (s, t) = (2, 2), there are exactly two dual examples; for (s, t) = (2, 8) (or (s, t) = (8, 2)), there is a unique example ([24]). Nothing on uniqueness is known about other orders. All known thick generalized octagons have order (2^{2e+1}, 2^{4e+2}) or (2^{4e+2}, 2^{2e+1}) with e ∈ ℕ (the so-called Ree-Tits octagons, [96]). Also here nothing is known about uniqueness. We also notice that every generalized polygon which is not thick can be obtained from some ordinary polygon or some thick generalized polygon ([95],[101]).

1.4 Projective spaces

As we already mentioned, the reader is supposed to be familiar with the (basic) theory of projective spaces. Nevertheless, we will summarize some definitions and results related to these projective spaces, especially those which will be used in this thesis and may not be known by a reader which is not too specialized in this subject.

1.4.1 Caps

A cap is a nonempty set of points no three of which are collinear. Caps in projective spaces of dimension d ≤ 1 are trivial objects.

Suppose d = 2 and let the projective plane have order q. If q is even, then a cap has size at most q + 2. If this upper bound is achieved, then the cap is called a hyperoval. If q is odd, then a cap has size at most q + 1. A cap of size q + 1 is called an oval (also if q is even). A famous theorem, due to Segre, says that every oval in PG(2, q), q odd, is a conic ([76] and [77]). If q is even, then every oval can be extended in a unique way to some hyperoval; the point that must be added is called the nucleus of the oval. In PG(2, q), q even, there is a special type of hyperoval, which consists of a conic together with its nucleus. Such a hyperoval is called regular, and any other hyperoval of PG(2, q) is called irregular. Each hyperoval in PG(2, q) with q ∈ {2, 4, 8} is regular ([78]). We refer to [89] for a survey of the known hyperovals. There is also an electronic update on the web available, see http://www-math.cudenver.edu/∼wcherowi.

Let H be a hyperoval in PG(2, 2^{h}), h ≥ 1, then we can take coordinates such that (1, 0, 0) ∈ H and (0, 1, 0) ∈ H. We then can write

H = H_{f} := {(1, 0, 0), (0, 1, 0)} ∪ {(f(λ), λ, 1)|λ ∈ GF(q)},

where f : GF(q) → GF(q) is some function satisfying the following two conditions:

(a) f is a bijection,

(b) \frac{f(λ_{1})−f(λ_{2})}{λ_{1}−λ_{2}} ≠ \frac{f(λ_{1})−f(λ_{3})}{λ_{1}−λ_{3}} for all λ_{1}, λ_{2}, λ_{3} ∈ GF(q) which are two by two distinct.

Using the remaining freedom in coordinatization, we even may suppose that f(0) = 0 and f(1) = 1. If we put f(λ) = λ^{2} for all λ ∈ GF(q), then H_{f} is a regular hyperoval. The function g_{i}(λ) = λ^{2^{i}}, where i ∈ {1, …, h − 1} and (i, h) = 1, also satisfies the above conditions (a) and (b). Every hyperoval projectively equivalent with such an H_{g_{i}} is called a translation hyperoval.

The maximal size of a cap in PG(3, q), q ≠ 2, is q^{2} + 1 ([8],[75]). A cap of size q^{2} + 1 in PG(3, q), q ≠ 2, is called an ovoid. An ovoid in PG(3, 2) is defined as an elliptic quadric. The only known examples of ovoids are the elliptic quadrics and the Tits ovoids. A Tits ovoid ([93]) is a set of points projectively equivalent with the following set of points in PG(3, 2^{2e+1}):

{(1, 0, 0, 0)} ∪ {(a^{2^{e+1}} + b^{2^{e+1}+2} + ab, 1, a, b)|a, b ∈ GF(2^{2e+1})}.

Every ovoid in PG(3, q), q odd, is an elliptic quadric ([4]) and every ovoid in PG(3, q), q ≤ 32 and even, is an elliptic quadric or a Tits ovoid ([59],[41],[55], [56],[57],[74]). If O is a nonsingular elliptic quadric or a Tits ovoid, then every other ovoid intersects it in an odd number of points (see Corollary 2 of [2]).

1.4.2 The Steiner systems S(5, 6, 12) and S(5, 8, 24)

E. Mathieu discovered the so-called Mathieu groups ([52],[53],[54]). Witt ([99],[100]) used these groups to construct the Steiner systems S(5, 6, 12) and S(5, 8, 24). Lüneburg ([51]) reversed this procedure and constructed first the designs (by ”extending” an affine or a projective plane) and defined then the Mathieu groups M_{12} and M_{24} as their automorphism groups. The uniqueness of both Steiner systems can be derived from this method. We will give Lüneburg’s construction for S(5, 8, 24) and give another construction for S(5, 6, 12).

Construction of S(5, 8, 24)

The projective plane Π = (X, B, I) of order 4 has the following properties.

(a) The set of all 168 hyperovals can be divided into three classes O_{1}, O_{2} and O_{3}, each of size 56. Two hyperovals are in the same class if and only if they intersect in an even number of points.

(b) The set of all 360 Baer-subplanes, i.e. subplanes of order 2, can be divided into three classes B_{1}, B_{2} and B_{3}, each of size 120. Two Baersubplanes are in the same class if and only if they intersect in an odd number of points.

(c) The indices i and j can be chosen in such a way that, for O ∈ O_{i} and for S ∈ B_{j}, we have |O ∩ S| is even if and only if i = j.

Construct now the following design. The point set is equal to X∪{∞_{1}, ∞_{2}, ∞_{3}}, where ∞_{1}, ∞_{2}, ∞_{3} ∉ X. There are four types of blocks:

(a) L ∪ {∞_{1}, ∞_{2}, ∞_{3}} where L ∈ B;

(b) (O ∪ {∞_{1}, ∞_{2}, ∞_{3}}) \ {∞_{i}} for each O ∈ O_{i};

(c) S ∪ {∞_{i}} for each S ∈ B_{i};

(d) L △ L′ (the symmetric difference of L and L′) for all L, L′ ∈ B, L ≠ L′.

Taking the natural incidence, we find then the unique Steiner system S(5, 8, 24). It has the following properties.

(a) Every two different blocks meet in 0, 2 or 4 points.

(b) If two blocks are disjoint, then the complement of their union is again a block.

(c) If two blocks intersect in 4 points, then their symmetric difference is again a block.

Construction of S(5, 6, 12)

Consider the following matrix over GF(3):

M = [ \begin{matrix}1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & −1 & −1 & 1 & −1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & −1 & −1 & −1 \\ 0 & 0 & 0 & 1 & 0 & 0 & −1 & 1 & 0 & 1 & −1 & −1 \\ 0 & 0 & 0 & 0 & 1 & 0 & −1 & −1 & 1 & 0 & 1 & −1 \\ 0 & 0 & 0 & 0 & 0 & 1 & 1 & −1 & −1 & 1 & 0 & −1 \end{matrix} ] .

The columns of M define a set K of 12 points in PG(5, 3). Any set of points projectively equivalent with K is called a Coxeter-cap as it was discovered by Coxeter in [25], see also [73]. It has the following properties.

(a) No five points of K are contained in a threedimensional space.

(b) Every hyperplane of PG(5, 3) which contains at least 5 points of K, contains exactly 6 points of K.

From (a) and (b) the following construction of the unique Steiner system S(5, 6, 12) follows. The points of the Steiner system are the points of K, the blocks are all the sets of 6 points which arise as intersection of K with a hyperplane.

Remark. The rows of the matrix M generate a 6-dimensional subspace of V (12, 3) (i.e. the 12-dimensional vector space over GF(3)). This subspace is called the extended ternary Golay code.

1.4.3 Regular spreads in PG(3, q)

In this section, we give properties about regular spreads in PG(3, q). We add proofs since they are short and easy. For more details, we refer to [48].

Definitions.

(1) Let K, L, M be three lines of PG(3, q), two by two disjoint. There are exactly q + 1 lines intersecting K, L and M. Every set of q + 1 lines obtained this way is called a regulus.

(2) A spread of PG(3, q) is a set of q^{2} + 1 lines partitioning the point set. A spread S of PG(3, q) is called regular if for every line M ∉ S the q + 1 lines of S intersecting M define a regulus. All regular spreads of PG(3, q) are projectively equivalent.

Lemma 1.4.1 Let S be a spread of PG(3, q) and let θ be an automorphism of PG(3, q) which fixes each line of S. If θ fixes one point of PG(3, q), then θ is the identity.

Proof. Suppose that x is a fixpoint of θ. Let y be another point of PG(3, q) such that xy ∉ S. Let L_{1} be the line of S through y and let L_{2} be a line of S meeting xy in a point different from x and y. Then y^{θ} ∈ L_{1}^{θ} ∩ ⟨x^{θ}, L_{2}^{θ}⟩ = L_{1} ∩ ⟨x, L_{2}⟩ = {y}. Hence y is fixed. Applying the same reasoning with y instead of x, we find that θ fixes each point of PG(3, q). □

The previous lemma says that there are at most q + 1 automorphisms of PG(3, q) fixing each line of a spread S. If there are exactly q + 1 such automorphisms, then S is regular as we will show in the following lemma.

Lemma 1.4.2 If there are at least three automorphisms of PG(3, q) fixing each line of a spread S, then S is regular.

Proof. Let θ_{1} and θ_{2} be two nontrivial automorphisms of PG(3, q) fixing each line of S. Let M be an arbitrary line of PG(3, q) not belonging to S and let L_{1}, L_{2}, …, L_{q+1} be the q + 1 lines of S meeting M. These q + 1 lines determine a regulus, since they meet the three lines M, M^{θ_{1}} and M^{θ_{2}}. □

Conversely, if S is a regular spread, then there are exactly q + 1 automorphisms fixing each line of S; these automorphisms will be given now.

The case of odd q

Let m be any nonsquare of GF(q). For (λ, μ) ∈ GF(q) × GF(q), we define

L_{λ,μ} = ⟨(λ, μ, 1, 0), (μ, mλ, 0, 1)⟩,

and we put

S = {⟨(1, 0, 0, 0), (0, 1, 0, 0)⟩} ∪ {L_{λ,μ}|λ, μ ∈ GF(q)}.

It is straightforward to check that S is a spread of PG(3, q). For a, b ∈ GF(q), let A(a, b) be the following matrix:

A(a, b) := ( \begin{matrix}a & b & 0 & 0 \\ bm & a & 0 & 0 \\ 0 & 0 & a & bm \\ 0 & 0 & b & a \end{matrix} ) .

If (a, b) ≠ (0, 0), then the automorphism x̄′ = A(a, b)x̄ of PG(3, q) fixes each line of S. We obtain in this way q + 1 (and hence all) automorphisms of PG(3, q) which fix each line of S. Hence S is regular. Clearly A(a, b) · A(c, d) = A(ac + bdm, ad + bc). Let \sqrt{m} be a square root of m in GF(q^{2}). Let A := {A(a, b)|(a, b) ∈ GF(q) × GF(q) \ {(0, 0)}}, then A is a multiplicative group. The map Ω : A → GF(q^{2}) \ \{0\}, A(a, b) → a + b\sqrt{m} is a homomorphism of multiplicative groups. Let α + β\sqrt{m} be a primitive element of GF(q^{2}), then Ω^{−1}[(α + β\sqrt{m})^{i}], i ∈ \{0, …, q\}, are the q + 1 matrices which define the q + 1 automorphisms of PG(3, q) fixing each line of S. The group of automorphisms of PG(3, q) which fix each line of S is hence isomorphic to the cyclic group C_{q+1}.

The case of even q

Let l be an element of GF(q) for which T r(l)=1. For (λ, μ) ∈ GF(q)×GF(q), we define

L_{λ,μ} = ⟨(λ, μ, 1, 0), (λ + lμ, λ, 0, 1)⟩,

and we put

S = {⟨(1, 0, 0, 0), (0, 1, 0, 0)⟩} ∪ {L_{λ,μ}|λ, μ ∈ GF(q)}.

It is straightforward to check that S is a spread of PG(3, q). For a, b ∈ GF(q), let A(a, b) be the following matrix:

A(a, b) := ( \begin{matrix}a + b & bl & 0 & 0 \\ b & a & 0 & 0 \\ 0 & 0 & a & bl \\ 0 & 0 & b & a + b \end{matrix} ) .

If (a, b) ≠ (0, 0), then the automorphism x̄′ = A(a, b)x̄ of PG(3, q) fixes each line of S. We obtain in this way q + 1 (and hence all) automorphisms of PG(3, q) which fix each line of S. Hence S is regular. Clearly A(a, b) · A(c, d) = A(ac+bdl, ad+bc+bd). Let i ∈ GF(q^{2}) be such that i^{2} +i = l. Let A := {A(a, b)|(a, b) ∈ GF(q) × GF(q) \ {(0, 0)}}, then A is a multiplicative group. The map Ω : A → GF(q^{2}) \ {0}, A(a, b) → a + bi is a homomorphism of multiplicative groups. Let α + βi be a primitive element of GF(q^{2}), then Ω^{−1}[(α + βi)^{j}], j ∈ {0, …, q}, are the q + 1 matrices which define the q + 1 automorphisms of PG(3, q) fixing each line of S. The group of automorphisms of PG(3, q) which fix each line of S is hence isomorphic to C_{q+1}.

1.4.4 The geometry H_{q}^{3}

Definition. Let L be a fixed line of PG(3, q), then the following geometry H_{q}^{3} can be defined. The points of H_{q}^{3} are the points of PG(3, q) not on L, the lines of H_{q}^{3} are the lines of PG(3, q) skew to L and incidence is the one derived from PG(3, q).

Every automorphism of PG(3, q) fixing L induces an automorphism of H_{q}^{3}. Also the converse is true.

Theorem 1.4.3 Every automorphism θ of H_{q}^{3} is induced by an automorphism ϕ of PG(3, q) fixing L.

Proof. Let α be a plane of PG(3, q).

(1) Suppose that α contains L and let a ∈ α \ L. We define α^{ϕ} = ⟨a^{θ}, L⟩. This is a good definition. If b is another point of α \ L, then a^{θ} and b^{θ} are not collinear in H_{q}^{3}, since a and b are not collinear in H_{q}^{3}. Hence ⟨b^{θ}, L⟩ = ⟨a^{θ}, L⟩.

(2) Suppose that α ∩ L = {r} and let a_{1}, a_{2}, a_{3} ∈ α \ {r} such that α = ⟨a_{1}, a_{2}, a_{3}⟩ and such that no three of the points r, a_{1}, a_{2}, a_{3} are collinear in PG(3, q). Since a_{1} is not incident with a_{2}a_{3} in H_{q}^{3}, a_{1}^{θ} is not incident with the line a_{2}^{θ}a_{3}^{θ} in H_{q}^{3}. We put α^{ϕ} = ⟨a_{1}^{θ}, a_{2}^{θ}, a_{3}^{θ}⟩. This is a good definition. For, we will prove now that a_{4}^{θ} ∈ ⟨a_{1}^{θ}, a_{2}^{θ}, a_{3}^{θ}⟩ for every point a_{4} ∈ α \ {r}. We may suppose that a_{4} ≠ a_{1} and a_{4} ≠ a_{2}. One of the lines a_{4}a_{1} or a_{4}a_{2}, say a_{4}a_{1}, does not meet L. The line a_{4}a_{1} meets a_{2}a_{3} in a point a′_{1}. Now, a_{4}^{θ} ∈ ⟨a_{1}^{θ}, a_{1}^{′θ}⟩ ⊆ ⟨a_{1}^{θ}, a_{2}^{θ}, a_{3}^{θ}⟩. It is also clear that α^{ϕ} does not contain L.

Let M be a line of PG(3, q).

(1) If M = L, then we put M^{ϕ} = L.

(2) If M does not meet L, then we put M^{ϕ} = M^{θ}.

(3) Suppose that M meets L in a point r. Let α_{1}, …, α_{q+1} be the q + 1 planes through M, then (M \ {r})^{θ} ⊆ α_{1}^{ϕ} ∩ α_{2}^{ϕ} ∩ ⋯ ∩ α_{q+1}^{ϕ}. Since (M \ {r})^{θ} is a set of q points, we have that α_{1}^{ϕ} ∩ α_{2}^{ϕ} ∩ ⋯ ∩ α_{q+1}^{ϕ} is a line meeting L (since one of the planes α_{i}^{ϕ} contains L). Put M^{ϕ} = α_{1}^{ϕ} ∩ α_{2}^{ϕ} ∩ ⋯ ∩ α_{q+1}^{ϕ}.

Let r be an arbitrary point of PG(3, q).

(1) If r ∉ L, then we put r^{ϕ} = r^{θ}.

(2) If r ∈ L, let α be a plane of PG(3, q) for which α ∩ L = {r}. We put {r^{ϕ}} = α^{ϕ} ∩ L. This is a good definition. For, let β be another plane of PG(3, q) for which β ∩ L = {r}. Then (α ∩ β)^{ϕ} ⊆ α^{ϕ} ∩ β^{ϕ}. Hence α^{ϕ} ∩ L = β^{ϕ} ∩ L = (α ∩ β)^{ϕ} ∩ L.

□

1.4.5 Affine representations

The incidence structure S = (P, L, I) is called affine or embedded in the finite affine space A if L is a set of lines of A, P is the union of all members of L and the incidence relation is the one induced by that of A. If A′ is the subspace of A generated by all points of P, then we say that A′ is the ambient space of S. A special type of affine embedding is the so-called linear representation. Let Π_{∞} be a projective space of dimension n ≥ 0 embedded as a hyperplane in the projective space Π and let K be a nonempty subset of the point set of Π_{∞}. The linear representation T_{n}^{∗}(K) is the geometry with points the affine points of Π (= the points not belonging to Π_{∞}). The lines of T_{n}^{∗}(K) are all the lines of Π, not contained in Π_{∞}, which intersect K. Incidence is the one derived from Π.

Generalized quadrangles

As we already mentioned, the theory of the generalized quadrangles is extremely important in the theory of the near polygons. That is the reason why we devote a whole chapter to these geometries. The generalized quadrangles are widely studied and we will give a survey of their theory, hereby restricting ourselves to those results which we will use later in this thesis and those results which were found by the author himself. For a more complete survey, we refer to [71] and [88].

2.1 Definitions and elementary properties

A generalized quadrangle (GQ for short) Q = (P, L, I) is a partial linear space satisfying the following properties:

(GQ1) there exists a point a and a line L such that (a, L) ∉ I,

(GQ2) for every point a and every line L not incident with a, there exists a unique point b on L collinear with a.

Generalized quadrangles were introduced in [92]. The point graph of a generalized quadrangle determines it completely, since the lines correspond to the maximal cliques in the graph. Sometimes one requires (like in [71] and [88]) that the geometry also must have an order to be a GQ. This is not very important since, as we shall see in Theorem 2.1.1, this additional property only excludes a few trivial cases, which we will define now. A generalized quadrangle is called trivial when it is isomorphic to one of the three examples given below; otherwise it is called nontrivial.

(a) Let t, n_{1}, …, n_{t+1} be nonzero positive integers. Put P = {(∞)} ∪ {x_{ij}|1 ≤ i ≤ t + 1, 1 ≤ j ≤ n_{i}}, L = {L_{1}, …, L_{t+1}}, let (∞) I L_{i}, ∀i ∈ {1, …, t+1}, and let x_{ij} I L_{k} if and only if i = k. The incidence structure Q = (P, L, I) is then a GQ. All GQ’s which can be constructed this way are called degenerate.

(b) Let s_{1} and s_{2} be nonzero positive integers. Put P = {x_{ij}|1 ≤ i ≤ s_{1} +1, 1 ≤ j ≤ s_{2} +1}, L = {L_{1}, …, L_{s_{1}+1}, M_{1}, …, M_{s_{2}+1}}, let x_{ij} I L_{k} if and only if i = k and let x_{ij} I M_{k} if and only if j = k. The incidence structure Q = (P, L, I) is then a GQ. The GQ’s arising this way are called grids, more precisely the GQ defined above is an (s_{1}+1)×(s_{2}+1)- grid. If s_{1} = s_{2}, then the grid is called symmetrical; otherwise it is called nonsymetrical.

(b) The dual grids are the point-line duals of the grids. A description follows by interchanging the roles of the points and lines in the above description of the grids.

We prove now the following known result about GQ’s.

Theorem 2.1.1 Every nontrivial GQ has an order (s, t) with s, t ≥ 2.

Proof. For a point x of a generalized quadrangle, let t_{x} + 1 denote the number of lines through it. The proof goes in several steps. Step 1. We prove that two noncollinear points x and y are incident with the same number of lines. To do this, we count in two different ways the pairs (L, M), where L and M are two intersecting lines incident with x, respectively y. By (GQ2), we have t_{x} + 1 = t_{y} + 1 or t_{x} = t_{y}. Step 2. We prove that a GQ is degenerate if there exists a point x which is incident with only one line L_{x}. For a point y not on L_{x}, let L_{y} be the unique line through it which intersects L_{x}, denote the intersection point by y′. Now x and y are not collinear, otherwise L_{x} and L_{y} are two different lines incident with x. By Step 1, L_{y} is the unique line through y. It suffices to prove now that y′_{1} = y′_{2} for every two points y_{1} and y_{2} not on L_{x}. If y_{2} I L_{y_{1}}, then L_{y_{1}} = L_{y_{2}} and y′_{1} = y′_{2}. Suppose therefore that y_{2} is not a point of L_{y_{1}}. The line L_{y_{2}} is the unique line through y_{2} which intersects L_{y_{1}} in a point z. The point z is incident with at least two lines, hence z I L_{x} and y′_{1} = y′_{2} = z. Step 3. We prove that all points are incident with the same number of lines if there exists a line which is incident with at least three points. Dually, all lines are incident with the same number of points if there exists a point which is incident with at least three lines. Let L be a line incident with at least three points and let x and y be two arbitrary points; we will prove that t_{x} = t_{y}. By Step 1, we may suppose that x and y are collinear. If x and y are incident with L, then there exists a point z not collinear with x and y; hence t_{x} = t_{z} = t_{y}. If exactly one of the two points, say x, is incident with L, let z ≠ x be another point of L. Then t_{x} = t_{z} = t_{y}. Finally, suppose that x and y are not incident with L, then there exists a point z on L not collinear with x and y. Once again t_{x} = t_{z} = t_{y}. Step 4. We prove that a GQ is a dual grid if all the lines are incident with exactly two points. Dually, a GQ is a grid if all the points are incident with exactly two lines. Let p be a fixed point of the GQ and partition the point set P in two sets X = {x_{1}, …, x_{s_{1}+1}} and Y = {y_{1}, …, y_{s_{2}+1}}. The elements of X have even distance to p and the elements of Y have odd distance to p. Let {z_{1}, z_{2}} be a line of the GQ; this line contains a unique point nearest to p. Hence z_{1} ∈ X, z_{2} ∈ Y or z_{1} ∈ Y, z_{2} ∈ X. This implies that the point graph Γ of the GQ is a bipartite graph. Now, let x_{i} ∈ X and y_{j} ∈ Y be arbitrary. If x_{i} and y_{j} are not collinear, then d(x_{i}, y_{j}) ≥ 3, a contradiction. Hence Γ is a complete bipartite graph. Putting L_{ij} the line through x_{i} and y_{j}, we find the dual of the description given above for the grids. □

Remark. The point-line dual of a nondegenerate GQ is again a GQ.

A GQ of order (s, t) is sometimes denoted by GQ(s, t). If s = t then we say that the GQ has order s; in this case the GQ is also denoted by GQ(s). A GQ of order (s, 1) is a symmetrical grid, dually a GQ of order (1, t) is a symmetrical dual grid. We have the following restrictions on the parameters (s, t).

Theorem 2.1.2 ([71]) Let Q = (P, L, I) be a GQ(s, t), then

(1) |P| = (s + 1)(st + 1) and |L| = (t + 1)(st + 1),

(2) s + t divides st(s + 1)(t + 1).

Let Q = (P, L, I) be a GQ(s, t). For x ∈ P, let x^{⊥} denote the set of all points collinear with x (hence x ∈ x^{⊥}). If A ⊆ P, then we define A^{⊥} = ⋂_{x∈A} x^{⊥}. If x and y are two different points, then the span of the pair {x, y} is defined as sp(x, y) := {x, y}^{⊥⊥} = ({x, y}^{⊥})^{⊥}. If x and y are not collinear, then sp(x, y) is also called the hyperbolic line through x and y. Each hyperbolic line contains at most t + 1 points. A pair {x, y} is called regular when

(i) x is collinear with y, or

(ii) x is not collinear with y and |sp(x, y)| = t + 1.

A point x is called regular if {x, y} is regular for each y ≠ x. Dually, the notion of regularity can be defined for lines of a GQ. A triad of points is a set of three pairwise noncollinear points. A center of a triad T is just a point of T^{⊥}.

Theorem 2.1.3 ([9],[21],[46],[47]) Let Q = (P, L, I) be a GQ(s, t). If s > 1 then t ≤ s^{2} and equality holds if and only if each triad of points has the same number of centers. This number of centers is then equal to s + 1. Dually, s ≤ t^{2} if t > 1.

2.2 A description of some classes of GQ’s

This section does not give a survey of all known GQ’s, just an overview of these GQ’s which we will meet at least once in this thesis.

2.2.1 The classical GQ’s

The examples given here were first recognized as GQ’s by Tits ([92]).

(a) Consider in PG(d, q), d ∈ {3, 4, 5}, a nonsingular quadric with projective index 1. The points of Q together with the lines of Q form then a GQ, denoted by Q(d, q). It has the following parameters:

Q(3, q): s = q, t = 1;

Q(4, q): s = t = q;

Q(5, q): s = q, t = q^{2}.

(b) Consider in PG(d, q^{2}), d ∈ {3, 4}, a nonsingular Hermitian variety H. The points of H together with the lines of H form then a GQ denoted by H(d, q^{2}). It has the following parameters:

H(3, q^{2}): s = q^{2}, t = q;

H(4, q^{2}): s = q^{2}, t = q^{3}.

(c) Consider in PG(3, q) a symplectic polarity ζ. The incidence structure with points the points of PG(3, q), with lines the totally isotropic lines of PG(3, q) with respect to ζ and with natural incidence, is then a GQ(q) denoted by W (q).

2.2.2 The GQ T_{2}^{∗}(H)

Let PG(2, 2^{h}), h ≥ 1, be embedded as a hyperplane Π_{∞} in PG(3, 2^{h}). For each hyperoval in Π_{∞}, the linear representation T_{2}^{∗}(H) is a GQ of order (2^{h} − 1, 2^{h} + 1), see [1] and [45].

2.2.3 The GQ AS(q), q odd prime power

The GQ AS(q), q odd prime power, is constructed as follows (see [1]). The points of AS(q) are the points of AG(3, q). The lines are the following curves of AG(3, q):

(i) x = σ, y = a, z = b;

(ii) x = a, y = σ, z = b;

(iii) x = cσ^{2} − bσ + a, y = −2cσ + b, z = σ.

Here the parameter σ ranges over GF(q) and a, b, c are arbitrary elements of GF(q). The incidence is the natural one. The GQ has order (q − 1, q + 1).

2.2.4 The GQ T_{d}(O), d ∈ {2, 3}

The examples of GQ’s which we will describe now were discovered by Tits and first appeared in Dembowski [37]. Let the projective space PG(d, q), d ∈ {2, 3}, be embedded as a hyperplane Π_{∞} in PG(d + 1, q) and consider an oval (in the case d = 2) or an ovoid (in the case d = 3) in Π_{∞}. We then can construct a GQ T_{d}(O) as follows. The points are of the following types:

(i) the points of PG(d + 1, q) \ Π_{∞},

(ii) the hyperplanes X of PG(d + 1, q) for which |X ∩ O| = 1;

(iii) one new symbol (∞).

The lines are of the following types:

(a) the lines of PG(d+1, q) which are not contained in Π_{∞} and which meet O;

(b) the points of O.

Incidence is defined as follows. A point of type (i) is incident only with lines of type (a); here incidence is the one from PG(d + 1, q). A point of type (ii) is incident with all lines of type (a) contained in it and with the unique element of O contained in it. The point (∞) is incident with no line of type (a) and all lines of type (b). The just defined incidence structure is a GQ with the following parameters:

T_{2}(O): s = t = q;
T_{3}(O): s = q, t = q^{2}.

2.2.5 The GQ S_{xy}^{−}

The generalized quadrangle S_{xy}^{−} first appeared in [60], but we give the description taken from [66]. Let H be a hyperoval in PG(2, 2^{h}), h ≥ 1, which is embedded as a hyperplane π in PG(3, 2^{h}). The following GQ of order (2^{h} +1, 2^{h} −1) can then be constructed. The points of S_{xy}^{−} are of three types:

(i) points of PG(3, 2^{h}) not contained in π;

(ii) planes through x not containing y;

(iii) planes through y not containing x.

The lines of S_{xy}^{−} are those lines of PG(3, 2^{h}) which are not contained in π and which intersect H \ {x, y}. A point of S_{xy}^{−} and a line of S_{xy}^{−} are incident if and only if they are incident as objects of PG(3, 2^{h}).

2.3 Classification and isomorphisms

The following theorem gives the (anti-)isomorphisms between some classes of GQ’s which we just defined. For proofs, we refer to [71].

{Theorem 2.3.1 \atop (1)} Q(3, q) is a grid.

(2) Q(4, q) is isomorphic to the dual of W (q). Moreover, Q(4, q) (or W (q)) is self-dual if and only if q is even.

(3) Q(5, q) is isomorphic to the dual of H(3, q^{2}).

(4) T_{2}(O) is isomorphic to Q(4, q) if and only if O is a conic, it is isomorphic to W (q) if and only if q is even and O is a conic.

(5) T_{3}(O) is isomorphic to Q(5, q) if and only if O is an elliptic quadric in PG(3, q).

(6) AS(3) is isomorphic to Q(5, 2).

For (anti-)isomorphisms between two GQ’s of type T_{2}(O), T_{2}^{∗}(O) or S_{xy}^{−}, we refer to [67]. Concerning isomorphisms between GQ′s of type T_{2}^{∗}(O), one can say the following.

Theorem 2.3.2 ([7]) Let O_{i}, i ∈ {1, 2}, be a hyperoval of π_{i}. Then T_{2}^{∗}(O_{1}) ≃ T_{2}^{∗}(O_{2}) if and only if there is an isomorphism θ of the plane π_{1} containing O_{1} onto the plane π_{2} containing O_{2} for which O_{1}^{θ} = O_{2}.

We give now classification results.

Let Q be a GQ of order (2, t). By Theorems 2.1.2 and 2.1.3, it follows that t = 1, t = 2 or t = 4. Clearly, there is a unique GQ of order (2, 1), namely the (3 × 3)-grid. It is also not so hard to prove that there is a unique GQ of order 2, see e.g. Theorem 5.2.3. of [71]. The uniqueness of the GQ of order (2, 4) was proved independently at least 5 times ([38], [42], [79], [81] and [85]). Hence summarizing we have the following theorem.

Theorem 2.3.3 There are three GQ’s with order (s, t) = (2, t), namely the following ones:

(1) the (3 × 3)-grid;

(2) W (2);

(3) Q(5, 2).

There is another interesting model for W (2) which goes back to Sylvester ([83]). A duad is an unordered pair ij = ji of distinct elements from the set {1, 2, 3, 4, 5, 6}. A syntheme is a set {ij, kl, mn} of three duads for which i, j, k, l, m, n are distinct. The incidence structure with points the duads, with lines the synthemes and with containment as incidence relation, is then the unique GQ of order 2.

Let Q be a GQ of order (3, t). By Theorems 2.1.2 and 2.1.3, we have that t ∈ {1, 3, 5, 6, 9}. In [38], it is proved that there is no GQ of order (3, 6). The determination of all GQ’s of order 3 was achieved in [38] and [62]. The uniqueness of the GQ of order (3, 5) was proved in [38]. The uniqueness of the GQ of order (3, 9) was independently proved by Dixmier and Zara [38] and Cameron (see [91]). All these proofs, some of them simplified or streamlined, are also contained in [71]. Hence summarizing we have the following theorem.

Theorem 2.3.4 There are five GQ’s with order (s, t) = (3, t), namely the following ones:

(1) the (4 × 4)-grid;

(2) W (3);

(3) Q(4, 3);

(4) T_{2}^{∗}(H) with H the, up to projective equivalence, unique hyperoval of PG(2, 4);

(5) Q(5, 3).

There is no complete classification for s ≥ 4. It is known that a GQ of order 4 must be isomorphic to W (4), see [63].

2.4 Some general constructions for generalized quadrangles

In Section 2.2, we defined several classes of generalized quadrangles. Sometimes, two or more of these classes can be obtained with the same construction. In this section, we will describe only two of these constructions. The AT-construction was found by the author himself and yields new models for several generalized quadrangles, being natural models for further applications (see Chapter 7).

2.4.1 The Payne-construction

If x is a regular point of a GQ Q of order s, then a GQ P (Q, x) of order (s − 1, s + 1) can be constructed as follows (see [60] and [71]). The points of P (Q, x) are the points of Q not collinear with x; the lines of P (Q, x) are the lines of Q not incident with x together with all the hyperbolic lines through x; incidence is the natural one. We give now some examples, see [71] for more details.

(1) Consider the GQ W (q) related to the symplectic polarity ζ of PG(3, q). Each point x of W (q) is regular and we find the following description for P (W (q), x). The points are the points of PG(3, q) not in the plane x^{ζ}; the lines are the lines of W (q) not in x^{ζ} together with those lines of PG(3, q) through x but not contained x^{ζ}; incidence is the natural one. If q is even then P (Q, x) is isomorphic to T_{2}^{∗}(O) with O a regular hyperoval; if q is odd then P (Q, x) is isomorphic to AS(q). If q is odd, then the hyperbolic lines of W (q) through x correspond with the lines of type (i) (see model given in Section 2.2.3).

(2) Consider the GQ T_{2}(O) with O an oval of PG(2, 2^{h}), h ≥ 1. The point (∞) is regular and P (T_{2}(O), (∞)) is isomorphic to T_{2}^{∗}(O ∪ {x}), where x is the nucleus of the oval O.

(3) Let H be a hyperoval of PG(2, 2^{h}) and let x, y ∈ H. The line y of T_{2}(H \ {x}) (i.e. a line of type (b) in the model given in Section 2.2.4) is regular and P (T_{2}(H \ {x}), y) ≃ S_{xy}^{−}, see [66].

2.4.2 The AT-construction

We describe now a procedure which will give generalized quadrangles. It was found by the author, see [27].

(1) Let D be a Steiner system S(2, k, v) with 1 < k ≤ v and let P be the set of its points. The number of blocks through a point is \frac{v−1}{k−1}. Hence, we can put k = s+1 and v = st+1 with s and t strict positive integers. The total number of blocks is equal to \frac{v(v−1)}{k(k−1)}, hence (s + 1)|t(t − 1).

(2) Find a group K of order s + 1 (with a multiplication operation, and identity e) and a map Δ : P × P ↦ K, (x, y) ↦ δ_{xy} such that the points x, y, z are collinear if and only if δ_{xy}δ_{yz} = δ_{xz}. If this holds, then we say that the triple (D, K, Δ) is admissible. Note that δ_{xx} = e and δ_{yx} = δ_{xy}^{−1} for all x, y ∈ P. Hence δ_{xy}δ_{yz}δ_{zx} = e for collinear points x, y, z ∈ P.

The following theorem says now that every admissible triple (AT for short) yields a generalized quadrangle.

Theorem 2.4.1 Suppose that (D, K, Δ) is an admissible triple. Let Γ be the graph with vertex set K × P; two different vertices (k_{1}, x) and (k_{2}, y) are adjacent whenever

(a) x = y and k_{1} ≠ k_{2}, or (b) x ≠ y and k_{2} = k_{1}δ_{xy}.

Then Γ is the collinearity graph of a GQ(s, t).

Proof. The graph Γ contains (1 + s)(1 + st) vertices and every vertex is adjacent to s(t+1) others. We will prove that every two adjacent vertices are contained in a unique maximal clique and that this clique contains exactly s + 1 elements. If we consider these cliques as the lines of a geometry Q with the vertices of Γ as points, then every point of Q is incident with t + 1 lines and the number of points at distance at most one to a fixed line is (s + 1) + (s + 1)ts = (s + 1)(1 + st). Hence Q is a GQ(s, t).

Now, suppose that p_{1} = (k_{1}, x) and p_{2} = (k_{2}, y) are two adjacent vertices of Γ; we determine how the common neighbours (k_{3}, z) look like. If x = y ≠ z, then k_{3} = k_{1}δ_{xz} = k_{2}δ_{xz}, implying that k_{1} = k_{2}, a contradiction. Hence if x = y, then p_{1} and p_{2} are in a unique maximal clique containing all the points (k, x) with k ∈ K. If x ≠ y, then also x ≠ z ≠ y and k_{3} = k_{1}δ_{xz} = k_{2}δ_{yz} = k_{1}δ_{xy}δ_{yz}. This implies that δ_{xz} = δ_{xy}δ_{yz} or that z ∈ xy. It follows now easily that p_{1} and p_{2} are contained in a unique maximal clique, namely {(k_{1}δ_{xz}, z)|z I xy}. □

If there is an admissible triple with D and K as components, then there are a lot of admissible triples with D and K as components, as we will show now; however all the corresponding GQ’s turn out to be isomorphic.

Theorem 2.4.2 Let (D, K, Δ) be an admissible triple. With every point x of D, we associate an element δ_{x} ∈ K. Put Δ′(x, y) = δ′_{xy} = δ_{x}^{−1}δ_{xy}δ_{y} for every two points x, y of D. Then

(1) (D, K, Δ′) is an admissible triple,

(2) the two corresponding GQ’s are isomorphic.

Proof. Part (1) is easily verified. Now, let Q_{1} and Q_{2} be the generalized quadrangles corresponding to (D, K, Δ) and (D, K, Δ′) respectively. The map (k, x) ↦ (kδ_{x}, x) defines then an isomorphism between Q_{1} and Q_{2}. □

Theorem 2.4.3 Let (D, K, Δ) be an admissible triple and let θ be an automorphism of the group K. Put Δ′′(x, y) = δ′′_{xy} = δ_{xy}^{θ} for every two points x, y of D. Then

(1) (D, K, Δ′′) is an admissible triple,

(2) the two corresponding GQ’s are isomorphic.

Proof. Part (1) is easily verified. Now, let Q_{1} and Q_{2} be the generalized quadrangles corresponding to (D, K, Δ) and (D, K, Δ′′) respectively. The map (k, x) ↦ (k^{θ}, x) defines then an isomorphism between Q_{1} and Q_{2}. □

Suppose that (D, K, Δ) is an admissible triple. If D is not a line (t > 1), then the fact that there are at least as many blocks as points (Fundamental Theorem of linear spaces, [26]) implies that t ≥ s + 1. If t = s + 1, then D is an S(2, s + 1, s^{2} + s + 1) and hence a projective plane, but we will prove that this is impossible. Hence s = 1, t = 1 or s + 2 ≤ t ≤ s^{2}.

Theorem 2.4.4 There are no admissible triples where the Steiner system is a projective plane.

Proof. If there was such an admissible triple, then there would exist a GQ of order (s, s + 1) with s ≥ 2, but this contradicts the condition (s + t) | st(s + 1)(t + 1) holding for a GQ(s, t), see Theorem 2.1.2. □

2.5 The known GQ’s with an AT-model

Every known GQ has one of the following parameters:

(I) (s, 1) or (1, t) with s, t ≥ 1;

(II) (q − 1, q + 1) or (q + 1, q − 1) with q a prime power;

(III) (q, q) with q a prime power;

(IV) (q^{2}, q^{3}) or (q^{3}, q^{2}) with q a prime power;

(V) (q, q^{2}) or (q^{2}, q) with q a prime power.

Only the following cases remain after checking the conditions (i) (s + 1) | t(t − 1) and (ii) s + 2 ≤ t ≤ s^{2} if s, t ≠ 1:

(a) (s, 1) or (1, t) with s, t ≥ 1;

(b) (q − 1, q + 1) with q a prime power;

(c) (q, q^{2}) with q a prime power.

There are examples of GQ’s with an AT-model for each of these parameters.

2.5.1 The trivial GQ’s

(1) Let D be a line of length s + 1 (t = 1) and let K be any group of order s + 1. For x, y ∈ P put δ_{xy} = e. The corresponding generalized quadrangle is an (s + 1) × (s + 1)-grid.

(2) Let D be an S(2, 2, t+1) (i.e. the complete graph K_{t+1}), let K = {e, a} be the group of order 2. Put δ_{xy} = e if x = y and δ_{xy} = a if x ≠ y. The corresponding generalized quadrangle is a dual grid.

2.5.2 The GQ P (W (q), x)

Let D be the Desarguesian affine plane AG(2, q) with point set P = {(x_{1}, x_{2})| x_{1}, x_{2} ∈ GF(q)}. Three points x = (x_{1}, x_{2}), y = (y_{1}, y_{2}) and z = (z_{1}, z_{2}) are collinear if and only if

| \begin{matrix}x_{1} & x_{2} & 1 \\ y_{1} & y_{2} & 1 \\ z_{1} & z_{2} & 1 \end{matrix} | = 0.

This condition can be rewritten as

| \begin{matrix}x_{1} & x_{2} \\ y_{1} & y_{2} \end{matrix} | + | \begin{matrix}y_{1} & y_{2} \\ z_{1} & z_{2} \end{matrix} | = | \begin{matrix}x_{1} & x_{2} \\ z_{1} & z_{2} \end{matrix} | .

If we take K equal to the additive group of GF(q) and δ_{xy} = | \begin{matrix}x_{1} & x_{2} \\ y_{1} & y_{2} \end{matrix} |, then

(D, K, Δ) is an admissible triple, which gives rise to a generalized quadrangle Q of order (q−1, q+1). The generalized quadrangle Q has points (X_{0}, X_{1}, X_{2}) with X_{0}, X_{1}, X_{2} ∈ GF(q); these are the points of AG(3, q). Two different points (X_{0}, X_{1}, X_{2}) and (Y_{0}, Y_{1}, Y_{2}) are collinear if and only if

(a) X_{0} ≠ Y_{0}, X_{1} = Y_{1}, X_{2} = Y_{2}, or

(b) Y_{0} = X_{0} + X_{1}Y_{2} − X_{2}Y_{1} and (X_{1}, X_{2}) ≠ (Y_{1}, Y_{2}).

This enables us to calculate the lines (see proof of Theorem 2.4.1).

(I) The above case (a) yields the lines {(σ, a, b)|σ ∈ GF(q)}, a, b ∈ GF(q).

(II) Consider the line {(a, σ)|σ ∈ GF(q)} of AG(2, q). Then the corre-

sponding lines of Q are the sets \{(b′ + | \begin{matrix}a & σ′ \\ a & σ \end{matrix} | , a, σ)|σ ∈ GF(q)\},

b′, σ′ ∈ GF(q), that is, the sets {(b + aσ, a, σ)|σ ∈ GF(q)}, b ∈ GF(q).

(III) Consider the line {(σ, aσ + b)|σ ∈ GF(q)} of AG(2, q). Then the cor-

responding lines of Q are the sets \{(b′ + | \begin{matrix}σ′ & aσ′ + b \\ σ & aσ + b \end{matrix} | , σ, aσ + b)|σ ∈

GF(q)}, b′, σ′ ∈ GF(q), that is, the sets {(c − bσ, σ, aσ + b)|σ ∈ GF(q)}, c ∈ GF(q).

Now, embed AG(3, q) in PG(3, q) (by adding a new coordinate X_{3} and identification of the points (X_{0}, X_{1}, X_{2}) of AG(3, q) and (X_{0}, X_{1}, X_{2}, 1) of PG(3, q)). Consider in PG(3, q) the generalized quadrangle W (q) arising from the symplectic polarity ζ determined by the equation X_{0}Y_{3} + X_{1}Y_{2} − X_{2}Y_{1} −X_{3}Y_{0} = 0. The point x = (1, 0, 0, 0) is a regular point of W (q) and we have Q ≃ P (W (q), x). Indeed, the lines of type (I) are the lines through x not contained in x^{ζ} and the lines of type (II) and (III) are the totally isotropic lines of ζ not contained in x^{ζ}. Hence we have the following conclusions.

(1) If q is odd, then Q ≃ AS(q).

(2) If q is even, then Q ≃ T_{2}^{∗}(O), with O a regular hyperoval.

2.5.3 The GQ T_{2}^{∗}(O)

There are other admissible triples related to the affine plane AG(2, q) and the additive group of GF(q). The affine plane AG(2, q) has q + 1 directions determined by the vectors ē_{1}, …, ē_{q+1}. For two different points x and y of AG(2, q), one has that \overline{xy} = δ_{xy}ē_{xy} with ē_{xy} ∈ \{ē_{1}, …, ē_{q+1}\} the vector corresponding to the direction of the line xy. This defines δ_{xy} when x ≠ y. If x = y, put δ_{xy} = 0. We determine now under which conditions we get an admissible triple. If x, y, z are three collinear points, then clearly δ_{xy} + δ_{yz} = δ_{xz}. Conversely, suppose that δ_{xy} + δ_{yz} = δ_{xz} with x, y, z three noncollinear points. Then ē_{xy}, ē_{yz}, ē_{xz} are three mutual different vectors which satisfy δ_{xy}ē_{xy} + δ_{yz}ē_{yz} = δ_{xz}ē_{xz} or ē_{xz} = λē_{xy} + (1 − λ)ē_{yz} for some λ ∈ GF(q). We hence obtain an admissible triple if and only if

ē_{k} − λē_{i} − (1 − λ)ē_{j} ≠ 0  (2.1)

holds for all λ ∈ GF(q) and any three mutual distinct vectors ē_{i}, ē_{j}, ē_{k}. Now, if a is a fixed point of AG(2, q), then H = {a, a + ē_{1}, …, a + ē_{q+1}} is a set of q + 2 points. If PG(2, q) is the projective completion of AG(2, q), then condition (2.1) holds if and only if H is a hyperoval of PG(2, q).

Theorem 2.5.1 If condition (2.1) holds, then the generalized quadrangle arising from the corresponding admissible triple is isomorphic to T_{2}^{∗}(H).

Proof. Let Q be the generalized quadrangle arising from the admissible triple. The points of Q are the elements of GF(q) × GF(q) × GF(q). A line of Q is of one of the following types.

(i) {(λ, x, y)|λ ∈ GF(q)} for fixed x, y ∈ GF(q).

(ii) {(a + λ, b + λe_{i1}, c + λe_{i2})|λ ∈ GF(q)} for fixed a, b, c ∈ GF(q) and fixed ē_{i} = (e_{i1}, e_{i2}).

Now, embed the points of Q in PG(3, q) as follows. Let the point (x, y, z) of Q correspond to the point (x, y, z, 1) of PG(3, q). The lines of Q intersect the plane at infinity of PG(3, q) (this is the plane with points (x, y, z, 0)) in one of the q + 2 points of the set {(1, 0, 0, 0)} ∪ {(1, e_{i1}, e_{i2}, 0)| i ∈ {1, …, q + 1}} and this latter set defines a hyperoval projectively equivalent to H. □

2.5.4 The GQ Q(5, q)

Let the vector space V (3, q^{2}) be equiped with a nonsingular Hermitian form (·, ·), i.e. (∑ μ_{i}v_{i}, ∑ λ_{j}w_{j}) = ∑ ∑ μ_{i}λ_{j}^{q}(v_{i}, w_{j}), and let U be the corresponding unital of PG(2, q^{2}). There is a Steiner system D = S(2, q + 1, q^{3} + 1) related to U (the blocks are the intersections of U with nontangent lines). Let K = {α ∈ GF(q^{2})|α^{q+1} = 1} with multiplication inherited from GF(q^{2}). Let z = ⟨ā⟩ be a fixed point of U. For two points x = ⟨b̄⟩ and y = ⟨c̄⟩ of U, we define

Δ(x, y) = −(ā, b̄)^{q−1}(b̄, c̄)^{q−1}(c̄, ā)^{q−1} ∈ K if x, y, z are mutually different;
= 1 otherwise.

This is a good definition. For, if we replace b̄ by μb̄ and c̄ by λc̄ with μ, λ ∈ GF(q^{2}) \ {0}, then the above value for Δ(x, y) is unaltered. We will prove in Section 2.7.1 that (D, K, Δ) is admissible and that the corresponding GQ is isomorphic to Q(5, q).

2.5.5 The GQ (S_{xy}^{−})^{D}

Let H be a hyperoval of PG(2, 2^{h}), h ≥ 1, and let x, y be two points of H. We can coordinatize PG(2, 2^{h}) in such a way that x ↔ (1, 0, 0), y ↔ (0, 1, 0) and such that the point with coordinates (0, 0, 1) belongs as well to H. We have then that H = {(1, 0, 0), (0, 1, 0)} ∪ {(f(λ), λ, 1)|λ ∈ GF(q)}, where f is a suitable function from GF(q) to GF(q). We have the following admissible triple. Let D be the affine plane AG(2, 2^{h}) with point set GF(2^{h}) × GF(2^{h}) and let K be the additive group of GF(2^{h}). For two points p_{1} = (α_{1}, β_{1}) and p_{2} = (α_{2}, β_{2}) of AG(2, 2^{h}), we put Δ(p_{1}, p_{2}) equal to \frac{[f(α_{1})−f(α_{2})][β_{1}−β_{2}]}{α_{1}−α_{2}} if α_{1} ≠ α_{2} and 0 otherwise. We will prove in Section 2.7.1 that (D, K, Δ) is admissible and that the corresponding GQ is isomorphic to (S_{xy}^{−})^{D}, the dual generalized quadrangle of S_{xy}^{−}.

Problem. Are there other (unknown) GQ’s derivable from an admissible triple?

2.6 Ovoids and spreads of GQ’s

Let Q = (P, L, I) be a GQ(s, t). An ovoid O of Q is a set of points such that every line of Q has exactly one point in common with O. The dual notion is that of spread. A spread S is a set of lines partitioning the point set. Every ovoid or spread of Q has 1 + st elements. Every set of mutually noncollinear points of Q has size at most 1 + st with equality if and only if this set is an ovoid. A rosette of ovoids is a set of s ovoids through a fixed point p of Q partitioning the set of points at distance 2 from p. A fan of ovoids is a set of s + 1 ovoids partitioning the point set of Q.

2.6.1 Ovoids

Some classes of GQ’s cannot have ovoids.

{Theorem 2.6.1 \atop (1)} W (q), q odd, has no ovoids ([87]).

(2) Every GQ of order (s, s^{2}), s ≠ 1, has no ovoids ([87], [80]).

For several GQ’s we will define those ovoids which we will meet later in this thesis.

The [(s + 1) × (s + 1)]-grid

There are exactly (s + 1)! ovoids.

Q(4, q)

Let Q be a nonsingular quadric in PG(4, q). An elliptic quadric Q^{−}(3, q) on Q is an ovoid of Q(4, q). Suppose now that q is even, then Q(4, q) ≃ W (q). Note that the points of W (q) are exactly the points of PG(3, q). If O is an ovoid of W (q), then it is also an ovoid of PG(3, q), see [84]. Conversely, if O is an ovoid of PG(3, q), then one can define a symplectic polarity ζ in PG(3, q) such that O is also an ovoid of the generalized quadrangle (≃ W (q)) related to ζ. A proof of this can be found in [48]. Consider now the case q = 2 and look at Sylvesters model of W (2) given in Section 2.3. For each i ∈ {1, …, 6}, the set {ij|j ≠ i} is an ovoid and all ovoids arise this way. Hence, there are exactly six ovoids and every two distinct ovoids meet in a point.

H(3, q^{2})

Let H be a nonsingular Hermitian variety in PG(3, q^{2}). Every nontangent plane intersects H in a Hermitian curve which is an ovoid of H(3, q^{2}).

P (Q, x)

Let Q be a GQ of order s having a regular point x. Then O is an ovoid of P (Q, x) if and only if O ∪ {x} is an ovoid of Q, see [71].

S_{xy}^{−}

Let H be a hyperoval of PG(2, 2^{h}), h ≥ 1, being embedded as a hyperplane α in PG(3, 2^{h}) and let x, y ∈ H.

(1) The set of planes through x not containing y forms an ovoid O_{xy} of S_{xy}^{−}. Similarly, the set of planes through y not containing x forms an ovoid O_{yx} of S_{xy}^{−}.

(2) Let π ≠ α be a plane through xy, then the set of affine points in π forms an ovoid O_{π} of S_{xy}^{−}.

2.6.2 Spreads

The above given examples of ovoids correspond to spreads in the dual GQ’s. We give now some additional examples of spreads.

(A) T_{2}^{∗}(O) Let O be a hyperoval of the plane PG(2, 2^{h}), h ≥ 1, which is embedded as a hyperplane α in PG(3, 2^{h}). Let o ∈ O be fixed. The lines of PG(3, 2^{h}) through o not contained in α defines then a spread S_{o} of T_{2}^{∗}(O).

(B) GQ’s from the Payne-construction Let Q be a GQ of order s having a regular point x, then the hyperbolic lines through x define a spread of P (Q, x).

(C) GQ’s from the AT-construction Let Q be the GQ arising from the AT (D, K, Δ). Let P denote the point set of D, then the set S = {L_{x}|x ∈ P} with L_{x} = {(k, x)|k ∈ K} is a spread of Q and called the associated spread of the admissible triple.

Definition. A partial spread of a Steiner system is a set of mutually disjoint blocks.

Theorem 2.6.2 With each partial spread S_{1} of D, there corresponds a spread S′_{1} of Q.

Proof. Let S_{1} = \{L_{1}, …, L_{r}\} be a partial spread of D (0 ≤ r ≤ \frac{1+st}{1+s}). For each i ∈ {1, …, r}, let x_{i} be a fixed point of L_{i}. If L_{(k,x_{i})} denotes the line {(kδ_{x_{i}y_{i}}, y_{i})|y_{i} I L_{i}} for each k ∈ K and each i ∈ {1, …, r}, then S′_{1} = {L_{(k,x_{i})}|k ∈ K; 1 ≤ i ≤ r} ∪ {L_{x}|x is not incident with L_{i}, ∀i ∈ {1, …, r}} is a spread of Q. □

A spread S of a GQ Q is called normal when each pair {L, M} ⊆ S is regular and when all the members of the hyperbolic line through {L, M} belong as well to S. Dually, one can define normal ovoids. Consider now the following problem. Given a nontrivial GQ Q, determine all normal spreads. In the next theorem, we will show a way how these spreads can be determined, but first we define the following linear space L(Q). The points of L(Q) are the lines of Q; the lines of L(Q) are the pencils (the sets of lines through a point of Q) together with the hyperbolic lines (on the set of lines of Q); incidence is the natural one.

Theorem 2.6.3 If S is a normal spread of a nontrivial GQ Q, then S = ⟨K, L, M⟩ for any three non-L(Q)-collinear points K, L, M ∈ S.

Proof. Let K, L, M ∈ S be three non-L(Q)-collinear points; then ⟨K, L, M⟩ ⊆ S contains at least s(s+1)+1 elements. If N ∈ S lies not in ⟨K, L, M⟩, then ⟨K, L, M, N⟩ ⊆ S contains at least s(s^{2} + s + 1) + 1 elements. Hence, 1 + st ≥ s^{3} + s^{2} + s + 1 or t ≥ s^{2} + s + 1, a contradiction, since t ≤ s^{2} holds for an arbitrary nontrivial GQ(s, t). □

A spread of symmetry is a spread S satisfying the following property.

For every K, L ∈ S and every two lines M and N meeting K and L, there exists an automorphism θ fixing each line of S such that M^{θ} = N.

If S is a spread of symmetry, then S is also a normal spread. Dually, one can define ovoids of symmetry.

We will give now all the normal spreads of the following GQ’s. It will turn out that these spreads are also spreads of symmetry.

Grids

The two spreads of a grid are spreads of symmetry.

T_{2}^{∗}(O)

Let O be a hyperoval of PG(2, 2^{h}), h ≥ 1. If h = 1, then T_{2}^{∗}(O) is a dual grid and all 24 spreads are spreads of symmetry. If h > 1, then we make use of Theorem 3.3.4 of [71]: the pair {L, M} of nonconcurrent lines of T_{2}^{∗}(O) is regular if and only if L and M define the same point of O. In this case {L, M}^{⊥⊥} consists of the 2^{h} lines of T_{2}^{∗}(O) which go through the point L∩M and are contained in the plane ⟨L, M⟩. Hence, the only normal spreads of T_{2}^{∗}(O), q ≠ 2, are the spreads S_{o} defined by a point o ∈ O. Take such a spread S_{o}. Consider two lines K, L ∈ S_{o} and two lines M, N ∈ {K, L}^{⊥}. There exists a translation of AG(3, 2^{h}) mapping M to N and having o as point at infinity (all lines of S_{o} are then fixed). Hence all normal spreads of T_{2}^{∗}(O) are also spreads of symmetry.

AS(q)

Consider the GQ AS(q), q odd. If q = 3, then AS(3) ≃ Q(5, 2) and we will treat this case later. Suppose therefore that q > 3. We make now use of Theorem 3.3.5 of [71]. The pair {L, M} of nonconcurrent lines of P (W (q), x) is regular if and only if one of the following holds: (i) L and M are hyperbolic lines of W (q) which contain x, or (ii) L and M are concurrent lines in W (q). Te hyperbolic lines of W (q) through x correspond to the lines of type (i) in the model of AS(q) given in Section 2.2.3. The span of two lines of type (i) consists of q lines of type (i). Hence, AS(q) with q > 3 has only one normal spread which consists of all lines of type (i). This spread is also a spread of symmetry. The required automorphisms are translations of the affine 3-space (see model given in Section 2.2.3).

Q(5, q)

Let S be a normal spread of Q(5, q). This dualizes to a normal ovoid of H(3, q^{2}). We will prove that O is the intersection of H(3, q^{2}) with a nontangent plane. Each pair {x, y} of noncollinear points of H(3, q^{2}) is regular and the hyperbolic line through them consists of the q+1 points in the intersection of the line xy with the Hermitian variety. Let x, y, z ∈ O, where z is not a member of the hyperbolic line through {x, y}. From Theorem 2.6.3, it follows that O is contained in the intersection of H(3, q^{2}) with the plane α through x, y, z. Hence α is nontangent and O = α ∩ H(3, q^{2}). Take coordinates such that H(3, q^{2}) has equation X_{0}^{q+1} + X_{1}^{q+1} + X_{2}^{q+1} + X_{3}^{q+1} = 0 and such that α is the plane X_{3} = 0. The ovoid O is now an ovoid of symmetry: the required automorphisms are of the form (X′_{0}, X′_{1}, X′_{2}, X′_{3}) = (X′_{0}, X′_{1}, X′_{2}, λX′_{3}), where λ is one of the q + 1 elements of GF(q^{2}) for which λ^{q+1} = 1.

(S_{xy}^{−})^{D}

Let H be a hyperoval of PG(2, 2^{h}), h ≥ 1, being embedded as a hyperplane α in PG(3, 2^{h}) and let x, y ∈ H. We may suppose that (S_{xy}^{−})^{D} is not isomorphic to T_{2}^{∗}(H), see [67] when this situation precisely occurs. In this case, O_{xy} and O_{yx} are the only normal spreads, see [70]. We will prove that these ovoids are also ovoids of symmetry. In order that O_{xy} is an ovoid of symmetry, it suffices that there are at least (and hence exactly) 2^{h} automorphisms of S_{xy}^{−} fixing each point of O_{xy} (see Theorem 2.7.1). These 2^{h} automorphisms are induced by the 2^{h} automorphisms of PG(3, 2^{h}) which fix each point of α and each line through x.

Let p be a point of a nontrivial GQ(s, t). A symmetry about p is an automorphism of GQ(s, t) fixing p and every point collinear with p. One can prove that there are at most t such symmetries; if there are exactly t symmetries, then p is called a center of symmetry. A center of symmetry is always a regular point. We mention now the following result proved in [36].

Theorem 2.6.4 ([36]) A GQ Q of order (s, s + 2) has a spread of symmetry if and only if Q ≃ P (Q′, (∞)) with Q′ a GQ(s + 1) and (∞) a center of symmetry of Q′. The spread of symmetry corresponds to the hyperbolic lines through (∞).

Corollary. With each AT (D, K, Δ), where D is an affine plane of order s + 1, there corresponds a GQ of order s + 1. This result was already in [18].

2.7 Characterizations of GQ’s with an ATmodel

2.7.1 Characterization in terms of a group of automorphisms

Let S = {L_{1}, …, L_{1+st}} be a spread of a generalized quadrangle Q of order (s, t) and let G_{S} be the group of automorphisms of Q fixing each line of S. If Q is an (s + 1) × (s + 1)-grid, then |G_{S}| = (s + 1)! for both spreads of Q.

Theorem 2.7.1 If Q is not a grid, then each nontrivial element of G_{S} has no fixpoints; hence |G_{S}| = \frac{s+1}{n} with n some nonzero integer.

Proof. Suppose that x is a fixpoint of θ ∈ G_{S}.

(a) Suppose y ∼ x with xy ∉ S. If L ∈ S is incident with y, then x I xy I y I L and x I xy^{θ} I y^{θ} I L implies that y^{θ} = y.

(b) Suppose y ≁ x, then x and y have a common neighbour z such that xz, yz ∉ S. From (a) it follows that z^{θ} = z and y^{θ} = y.

(c) Suppose that y ∼ x with xy ∈ S. Take a point z such that x ≁ z ∼ y. Then (b) implies that z^{θ} = z and (a) implies that y^{θ} = y.

Hence, the automorphism θ is trivial. The group G_{S} acts as a permutation group on the set of points of any line of S. If there are n orbits on such a line, then |G_{S}| = \frac{s+1}{n}. □

Corollary 2.7.2 If Q is not a grid, then S is a spread of symmetry if and only if |G_{S}| = s+1.

Theorem 2.7.3 S is a spread of symmetry of Q if and only if there exists an AT-model for Q with S as associated spread.

Proof. We may suppose that Q is not a grid. If (D, K, Δ) is an AT with Q as corresponding GQ and S as associated spread, then we may suppose that {(k, x)|k ∈ K; x is a point of D} is the set of points of Q and that S = {L_{x}|x is a point of D}, where L_{x} = {(k, x)|k ∈ K}. For each k′ ∈ K, the map (k, x) ↦ (k′k, x) is an automorphism of Q fixing each line of S. Hence |G_{S}| ≥ |K| = s + 1 and S is a spread of symmetry by Theorem 2.7.1 and Corollary 2.7.2. Conversely, suppose that S is a spread of symmetry (or |G_{S}| = s + 1).

(1) Construction of the Steiner system S(2,s+1,1+st). The points of the Steiner system are the lines of S. The blocks are obtained as follows. Take a line T ∉ S of Q. Let R_{1}, …, R_{s+1} be the lines of S intersecting T; then {R_{1}, …, R_{s+1}} is a line of the Steiner system. It remains to show that every two lines R_{1} and R_{2} of S are contained in a unique block. Every line intersecting R_{1} and R_{2} will yield such a block. If T is a line intersecting R_{1} and R_{2}, then A = {T^{h}|h ∈ G_{S}} is a set of s + 1 lines intersecting R_{1} and R_{2}, hence these are all the lines intersecting R_{1} and R_{2}. If L ∈ S intersects T in a point t, then L intersects T^{h} in t^{h}. Hence, every element of A will yield the same block through R_{1} and R_{2}.

(2) Construction of the group K and the map Δ. We take K = G_{S}. Let L_{1} ∈ S and let p be any point of L_{1}. Take two points u = L_{m} and v = L_{n} of the Steiner system. Let x_{1} be the projection of p on the line L_{m}, let x_{2} be the projection of x_{1} on the line L_{n} and finally let x_{3} be the projection of x_{2} on the line L_{1}. Now, there exists an element θ ∈ G_{S} such that x_{3} = p^{θ} and we define Δ(u, v) = δ_{uv} := θ^{−1}. It remains to show that the points x = L_{i}, y = L_{j}, z = L_{k} are collinear in the Steiner system if and only if δ_{xy}δ_{yz} = δ_{xz}. Put δ_{xy} = α^{−1}, δ_{yz} = β^{−1} and δ_{xz} = γ^{−1}. Denote by p_{l}(e) (l ∈ {1, i, j, k}) the projection of the point e on the line L_{l} in the generalized quadrangle Q. Put a = p_{i}(p), b = p_{j}(a), c = p_{k}(b), d = p_{k}(a), then p^{γ} = p_{1}(d). From p ∼ p_{j}(p) ∼ p_{k}p_{j}(p) ∼ p^{β} and p^{α} ∼ b ∼ c ∼ p_{1}(c), it follows that p^{βα} = p_{1}(c). Now,

δ_{xy}δ_{yz} = δ_{xz} ⇔ γ = βα
⇔ c = d
⇔ a, b, c are on a line T
⇔ x, y, z are collinear.

(3) Isomorphism between Q and the GQ from (D, K, Δ). Let z be an arbitrary point of Q. Let L_{z} denote the unique line of S incident with z and let z′ denote the projection of z on the fixed line L_{1}. There exists a unique g_{z} ∈ G_{S} such that z′ = p^{g_{z}}. We prove now that the map z ↦ (g_{z}^{−1}, L_{z}) defines an isomorphism between the GQ’s. It is clearly a bijection and since both geometries have the same parameters, it suffices to show that adjacency is preserved. So, let z_{1} and z_{2} be two adjacent points of Q and put x = L_{z_{1}} and y = L_{z_{2}}. From p^{g_{z_{1}}} ∼ z_{1} ∼ z_{2} ∼ p^{g_{z_{2}}}, it follows that g_{z_{2}} = δ_{xy}^{−1}g_{z_{1}} or g_{z_{2}}^{−1} = g_{z_{1}}^{−1}δ_{xy}. Hence (g_{z_{2}}^{−1}, y) ∼ (g_{z_{1}}^{−1}, x). □

Application. Using the previous theorem, we calculate an AT corresponding to Q(5, q) and to (S_{xy}^{−})^{D}.

(I) Q(5,q)

We use the dual representation. The points of H(3, q^{2}) are 1-dimensional subspaces of V (4, q^{2}). Consider a nonsingular Hermitian form (·, ·) in V (4, q^{2}), i.e. (∑_{i} λ_{i}v_{i}, ∑_{j} μ_{j}w_{j}) = ∑_{i} ∑_{j} λ_{i}μ_{j}^{q}(v_{i}, w_{j}), and let ζ be the corresponding polarity of PG(3, q^{2}). Take a nontangent plane π and let π^{ζ} = ⟨ū⟩. Take three arbitrary (but different) points ⟨ā⟩, ⟨b̄⟩, ⟨c̄⟩ of O = π ∩ H(3, q^{2}). The tangent plane at ⟨ā⟩ intersects π in a line ⟨ā, v̄⟩. Let L = ⟨ā, ū+λv̄⟩ be an arbitrary line of H(3, q^{2}) through ⟨ā⟩. Since (ū+λv̄, ū+λv̄) = 0, one finds that λ^{q+1} = −\frac{(ū,ū)}{(v̄,v̄)}. We determine the line L′ of H(3, q^{2}) through ⟨b̄⟩ intersecting

L. This line looks like ⟨b̄, ū+λv̄+βā⟩. An easy calculation yields β = −λ\frac{(v̄,b̄)}{(ā,b̄)}. Hence L′ = ⟨b̄, ū + λv̄′⟩ with ⟨v̄′⟩ ∈ π ∩ ⟨b̄⟩^{ζ} independent of λ. Similarly, if we project L′ to a line L′′ through ⟨c̄⟩ and finally L′′ to a line L′′′ through ⟨ā⟩, we will find that L′′′ = ⟨ā, ū + λv̄′′′⟩ with ⟨v̄′′′⟩ ∈ π ∩ ⟨ā⟩^{ζ} independent of λ. Now v̄′′′ = γ_{1}ā + γ_{2}v̄, hence L′′′ = ⟨ā, ū + λγ_{2}v̄⟩ where γ_{2} is independent of λ. As before, one has that (λγ_{2})^{q+1} = −\frac{(ū,ū)}{(v̄,v̄)} or γ_{2}^{q+1} = 1. We calculate now the corresponding admissible triple making use of the previous theorem. The ovoid O is just a unital in π and the Steiner system is an S(2, q + 1, q^{3} + 1) (the lines are the intersections of O with nontangent lines). If the above points ⟨ā⟩, ⟨b̄⟩, ⟨c̄⟩ are on a common line, then [⟨ā⟩, ⟨b̄⟩][⟨b̄⟩, ⟨c̄⟩][⟨c̄⟩, ⟨ā⟩] is the identical permutation and hence γ_{2} = 1. If the above points ⟨ā⟩, ⟨b̄⟩, ⟨c̄⟩ are not on a common line, then v̄ = λ_{1}ā + λ_{2}b̄ + λ_{3}c̄ with λ_{2}(b̄, ā) + λ_{3}(c̄, ā) = 0. A calculation yields

v̄′′′ = −λ_{3} \frac{(c̄, b̄)}{(ā, b̄)}ā + λ_{3} \frac{(c̄, b̄)(ā, c̄)}{(ā, b̄)(b̄, c̄)}b̄ − λ_{3} \frac{(c̄, b̄)(ā, c̄)(b̄, ā)}{(ā, b̄)(b̄, c̄)(c̄, ā)}c̄,

and hence γ_{2} = −(ā, b̄)^{q−1}(b̄, c̄)^{q−1}(c̄, ā)^{q−1}. When ⟨ā⟩, ⟨b̄⟩, ⟨c̄⟩ are on a common line, one calculates that −(ā, b̄)^{q−1}(b̄, c̄)^{q−1}(c̄, ā)^{q−1} = 1 = γ_{2}. By Theorem 2.4.3 (x ↦ x^{−1} is an automorphism of G_{O}), it follows that the triple defined in Section 2.5.4 is indeed admissible and that the corresponding GQ is isomorphic to Q(5, q).

(II) (S_{xy}^{−})^{D}

Let PG(2, q), q = 2^{h} with h ≥ 1, be embedded as a hyperplane π_{∞} in PG(3, q). Let O be a hyperoval in π_{∞}, let x, y be two points of O, and let Q be the generalized quadrangle S_{xy}^{−}. Consider now the following incidence structure P_{x}:

(1) the points of P_{x} are the planes through x,

(2) the lines of P_{x} are the lines through x,

(3) incidence is the natural one.

The incidence structure P_{x} is then a projective plane of order q. Consider now the line xy as line at infinity [∞] to construct the associated affine plane A_{xy}. The plane π_{∞} corresponds with a point (∞) on [∞]. The points of A_{xy} correspond with the planes through x not containing y, i.e. with the elements of O_{xy}. Let (X_{0}, X_{1}, X_{2}, X_{3}) denote coordinates in PG(3, q) and we may assume that x = (1, 0, 0, 0) and y = (0, 1, 0, 0). Every plane through x not containing y has equation X_{1} + a_{1}X_{2} + a_{2}X_{3}. Let this plane correspond with (a_{1}, a_{2}) ∈ GF(q) × GF(q). This correspondence yields an isomorphism between the affine planes A_{xy} and A, the latter plane being the one constructed in a well-known way on the set of points GF(q) × GF(q). We may assume that z = (0, 0, 1, 0) is a third point of O. Let o_{1} be a fixed point of O_{xy} containing z. We may assume that o_{1} has equation X_{1} = 0. The point o_{1} of S_{xy}^{−} is incident with q lines of S_{xy}^{−}, namely the lines L_{μ}, μ ∈ GF(q), where L_{μ} = {(μ, 0, λ, 1)|λ ∈ GF(q)}, we have also that O = {(1, 0, 0, 0), (0, 1, 0, 0)} ∪ {(f(λ), λ, 1, 0)|λ ∈ GF(q)}, where f : GF(q) → GF(q) is a bijection such that f(0) = 0. Now, let o_{2} : X_{1} + α_{1}X_{2} + β_{1}X_{3} = 0 and o_{3} : X_{1} + α_{2}X_{2} + β_{2}X_{3} = 0 be two arbitrary points of O_{xy}. The plane o_{2} intersects the hyperoval in two points; one of them is (1, 0, 0, 0), the other is (f(α_{1}), α_{1}, 1, 0). The plane o_{3} intersects the hyperoval in two points; one of them is (1,0,0,0), the other is (f(α_{2}), α_{2}, 1, 0). We suppose that o_{1} ≠ o_{2} ≠ o_{3} ≠ o_{1}.

Case I: 0 ≠ α_{1} ≠ α_{2} ≠ 0

Consider the line L_{u} through o_{1}. Its intersection with the plane o_{2} is equal to the point (u, 0, \frac{β_{1}}{α_{1}} , 1). The line L′_{u} through o_{2} intersecting L_{u} is then equal to

L′_{u} = \{(u + λf(α_{1}), λα_{1}, \frac{β_{1}}{α_{1}} + λ, 1)|λ ∈ GF(q)\}.

The line L′_{u} intersects the plane o_{3} in the point

(u + \frac{α_{1}β_{2} − β_{1}α_{2}}{α_{1}(α_{1} − α_{2})} f(α_{1}), \frac{α_{1}β_{2} − β_{1}α_{2}}{α_{1} − α_{2}} , \frac{β_{2} − β_{1}}{α_{1} − α_{2}} , 1).

The line L′′_{u} through o_{3} intersecting L′_{u} is then equal to

L′′_{u} = \{(u + \frac{α_{1}β_{2} − β_{1}α_{2}}{α_{1}(α_{1} − α_{2})} f(α_{1}) + λf(α_{2}), \frac{α_{1}β_{2} − β_{1}α_{2}}{α_{1} − α_{2}} + λα_{2}, \frac{β_{2} − β_{1}}{α_{1} − α_{2}} + λ, 1)|

λ ∈ GF(q)}.

The intersection of L′′_{u} with X_{1} = 0 yields λ = \frac{β_{1}α_{2}−α_{1}β_{2}}{α_{2}(α_{1}−α_{2})} . Hence L_{u} is mapped to L_{u′}, where

u′ = u + \frac{α_{1}β_{2} − β_{1}α_{2}}{α_{1}(α_{1} − α_{2})} f(α_{1}) + \frac{β_{1}α_{2} − α_{1}β_{2}}{α_{2}(α_{1} − α_{2})} f(α_{2})
= u + \frac{(α_{1}β_{2} − α_{2}β_{1})(α_{1}f(α_{2}) − α_{2}f(α_{1})}{α_{1}α_{2}(α_{1} − α_{2})}.

\underline{Conclusion:} Δ[(α_{1}, β_{1}), (α_{2}, β_{2})] = \frac{(α_{1}β_{2}−α_{2}β_{1})(α_{1}f(α_{2})−α_{2}f(α_{1})}{α_{1}α_{2}(α_{1}−α_{2})} Case II: α_{1} ≠ α_{2} = 0 Let L′′_{u} be as in the first case, i.e.

L′′_{u} = \{(u + \frac{β_{2}f(α_{1})}{α_{1}}, β_{2}, \frac{β_{2} − β_{1}}{α_{1}} + λ, 1)|λ ∈ GF(q)\}.

The plane through L′′_{u} and y intersects X_{1} = 0 in the line L_{u′}, where

u′ = u + \frac{β_{2}f(α_{1})}{α_{1}}.

\underline{Conclusion:} Δ[(α_{1}, β_{1}), (α_{2}, β_{2})] = \frac{β_{2}f(α_{1})}{α_{1}} Case III: 0 = α_{1} ≠ α_{2} Since Δ(p_{1}, p_{2}) = Δ(p_{2}, p_{1}), the solution follows from the second case. \underline{Conclusion:} Δ[(α_{1}, β_{1}), (α_{2}, β_{2})] = \frac{β_{1}f(α_{2})}{α_{2}} Case IV: α_{1} = α_{2} ≠ 0 Consider the line L′_{u} given in Case (I). The plane through L′_{u} and y has equation

\{(u + λf(α_{1}), λα_{1} + λ′, \frac{β_{1}}{α_{1}} + λ, 1)|λ, λ′ ∈ GF(q)\}.

The intersection of this plane with o_{3} yields the line

\{(u + λf(α_{1}), λα_{1} + β_{1} − β_{2}, \frac{β_{1}}{α_{1}} + λ, 1)|λ ∈ GF(q)\}.

Determining the intersection of this line with X_{1} = 0, we find λ = \frac{β_{1}−β_{2}}{α_{1}} . Hence the line L_{u} is mapped to the line L_{u′}, where u′ = u + \frac{β_{1}−β_{2}}{α_{1}} f(α_{1}). \underline{Conclusion:} Δ[(α_{1}, β_{1}), (α_{2}, β_{2})] = \frac{β_{1}−β_{2}}{α_{1}} f(α_{1})

Case V: α_{1} = α_{2} = 0

We have that

Δ[(α_{1}, β_{1}), (α_{2}, β_{2})] = 0.

We apply now Theorem 2.4.2. For every point (α, β) of the affine plane, we define δ_{(α,β)} := \frac{β}{α}f(α) if α ≠ 0 and δ_{(α,β)} := 0 otherwise. The new admissible triple is the one which was described in Section 2.5.5.

Remark. Let S be a spread of a GQ(s, t). If |G_{S}| > 1, then the condition (s + t) | st(s + 1)(t + 1) can be strengthened.

Lemma 2.7.4 ([5]) If f is the number of fixpoints of an automorphism θ and if g is the number of points x for which x^{θ} ≠ x ∼ x^{θ}, then (t + 1)f + g ≡ 1 + st (mod s + t).

Corollary 2.7.5 If |G_{S}| > 1, then (s + t) | s(s + 1)(t + 1).

Proof. Let θ be any nontrivial element of G_{S} and apply the previous lemma. □

2.7.2 Characterizations in terms of a group of projectivities

Let S = {L_{1}, …, L_{1+st}} be a spread of a generalized quadrangle Q of order (s, t). Every two lines L_{i} and L_{j} of S define a projectivity Δ_{ij} = [L_{j}, L_{1}][L_{i}, L_{j}][L_{1}, L_{i}] (= [L_{j}, L_{1}] ∘ [L_{i}, L_{j}] ∘ [L_{1}, L_{i}]) of L_{1}. The set A = {Δ_{ij}|1 ≤ i, j ≤ 1 + st} generates Π_{S}(L_{1}), the group of projectivities of L_{1} with respect to S.

Lemma 2.7.6 If Q ≃ W (q), then each Δ_{ij} is trivial or has order 2.

Proof. De generalized quadrangle Q is derived from a symplectic polarity ζ in PG(3, q). Let L_{1} = ⟨ā, b̄⟩, L_{i} = ⟨c̄, d̄⟩ and L_{j} = ⟨ē, f̄⟩ where 1 < i, j ≤ 1 + q^{2}, i ≠ j. Since L_{1} and L_{i} are skew, ē = αā + βb̄ + γc̄ + δd̄ and f̄ = α′ā + β′b̄ + γ′c̄ + δ′d̄ for certain α, …, δ′ ∈ GF(q). We may also suppose that ⟨c̄⟩ ∈ ⟨ā⟩^{ζ}. The plane ⟨c̄⟩^{ζ} = ⟨ā, c̄, d̄⟩ intersects L_{j} in the point ⟨ḡ⟩ = ⟨(β′α − βα′)ā + (β′γ − βγ′)c̄ + (β′δ − βδ′)d̄⟩ and the plane ⟨ḡ⟩^{ζ} = ⟨c̄, ē, f̄⟩ intersects L_{1} in ⟨(δ′α − δα′)ā + (δ′β − δβ′)b̄⟩, hence Δ_{ij}(ā) = ⟨(δ′α − δα′)ā + (δ′β − δβ′)b̄⟩. The plane ⟨ā⟩^{ζ} = ⟨ā, b̄, c̄⟩ intersects L_{j} in the point ⟨h̄⟩ = ⟨(δ′α−δα′)ā+(δ′β−δβ′)b̄+(δ′γ−δγ′)c̄⟩, the plane ⟨h⟩^{ζ} = ⟨ā, ē, f̄⟩ intersects L_{i} in k̄ = ⟨(β′γ−βγ′)c̄+(β′δ−βδ′)d̄⟩ and the plane ⟨k̄⟩^{ζ} = ⟨c̄, d̄, h̄⟩^{ζ} intersects L_{1} in ⟨(δ′α − δα′)ā + (δ′β − δβ′)b̄⟩. Hence Δ_{ij}(ā) = Δ_{ji}(ā) and Δ_{ij} = Δ_{ji} = Δ_{ij}^{−1}. □

If Q is a grid, then Π_{S}(L_{1}) is the trivial group; if Q is not a grid, then the following theorem says something about the number of elements of Π_{S}(L_{1}).

Theorem 2.7.7 Suppose that Q is not a grid. If x, y are points of L_{1}, then there exists an element a ∈ A, such that x^{a} = y. Hence |A|, |Π_{S}(L_{1})| ≥ s + 1.

Proof. Let L ≠ L_{1} be a fixed line of Q through y, let L_{i} be any line of S disjoint with L, let x′ be the projection of x on L_{i}, let x′′ be the projection of x′ on L and let L_{j} be the line of S through x′′. Now Δ_{ij} ∈ A maps x to y. □

Theorem 2.7.8 Let (D, K, Δ) be an AT with Q as corresponding GQ and S as associated spread. Let L_{1} be any line of S. If Q is not a grid, then Π_{S}(L_{1}) is isomorphic to K and |Π_{S}(L_{1})| = s + 1.

Proof. We have S = {L_{x}|x ∈ P} where L_{x} = {(k, x)|k ∈ K}. Let p ∈ P be fixed and put L_{1} := L_{p}. We can identify the point set of L_{1} with the elements of K (consider the bijection (k, p) ↦ k). With this identification the elements Δ_{ij} are right multiplications in the group K. For, let L_{i} = L_{x} and L_{j} = L_{y}. Then (k, p) ∼ (kΔ(p, x), x) ∼ (kΔ(p, x)Δ(x, y), y) ∼ (kΔ(p, x)Δ(x, y)Δ(y, p), p), hence Δ_{ij} corresponds to a (right) multiplication with Δ(p, x)Δ(x, y)Δ(y, p) ∈ K. Since |A| ≥ s + 1, we have Π_{S}(L_{1}) ≃ K. □

We will prove the converse theorem: if |Π_{S}(L_{1})| = s + 1, then Q has an AT-model with S as associated spread.

Lemma 2.7.9 Let X be a set of order n ≥ 1. If G is a regular group of permutations on X, then there are precisely n permutations of X which commute with every element of G. These n permutations form a regular group G̃ of permutations on X. Moreover G̃ ≃ G.

Proof. Let x ∈ X be fixed. If h commutes with every element of G, then hg(x) = gh(x) for all g ∈ G, hence the image of x under h determines h completely. Now, for every element y ∈ X, one can define the permutation h_{y} as follows:

h_{y}g(x) = g(y), ∀g ∈ G.

It is straightforward to check that h_{y} commutes with every element of G and that h_{y} is the trivial permutation if it has at least one fixpoint. Put G̃ = {h_{y}|y ∈ X}. For every element g of G, we associate an element g^{ϕ} of G̃ such that g^{−1}(x) = g^{ϕ}(x). Then

(gh)^{ϕ}(x) = (gh)^{−1}(x) = h^{−1}g^{−1}(x) = h^{−1}g^{ϕ}(x) = g^{ϕ}h^{−1}(x) = g^{ϕ}h^{ϕ}(x).

Now, let k ∈ G be arbitrary, then

(gh)^{ϕ}k(x) = k(gh)^{ϕ}(x) = kg^{ϕ}h^{ϕ}(x) = g^{ϕ}h^{ϕ}k(x);

hence (gh)^{ϕ} = g^{ϕ}h^{ϕ} for all g, h ∈ G and ϕ defines an isomorphism between G and G̃. □

Theorem 2.7.10 If θ ∈ G_{S} then θ induces a permutation θ̄ on the point set of L_{1} that commutes with each element of Π_{S}(L_{1}). Conversely, if a permutation ϕ on the point set of L_{1} commutes with each element of Π_{S}(L_{1}), then ϕ = θ̄ for some θ ∈ G_{S}.

Proof. First, we notice that ϕ = θ̄ determines θ completely. For, let x ∈ L_{i} be an arbitrary point of Q. From d(x, [L_{i}, L_{1}](x)) ≤ 1, it follows that d(θ(x), ϕ[L_{i}, L_{1}](x)) ≤ 1 and hence

θ(x) = [L_{1}, L_{i}]ϕ[L_{i}, L_{1}](x).  (2.2)

Next, let ϕ be a permutation on the point set of L_{1} and let us determine under which conditions the map θ defined by (2.2) is an automorphism. If x ∈ L_{i} and y ∈ L_{j} (with i ≠ j) are two collinear points, then y = [L_{i}, L_{j}](x) and

θ(x) ∼ θ(y) ⇔ [L_{1}, L_{i}]ϕ[L_{i}, L_{1}](x) ∼ [L_{1}, L_{j}]ϕ[L_{j}, L_{1}](y)
⇔ [L_{i}, L_{j}][L_{1}, L_{i}]ϕ[L_{i}, L_{1}](x) = [L_{1}, L_{j}]ϕ[L_{j}, L_{1}][L_{i}, L_{j}](x).

This holds for every x ∈ L_{i} if and only if

[L_{j}, L_{1}][L_{i}, L_{j}][L_{1}, L_{i}]ϕ = ϕ[L_{j}, L_{1}][L_{i}, L_{j}][L_{1}, L_{i}]
Δ_{ij}ϕ = ϕΔ_{ij}.

This holds for all i, j if and only if ϕ commutes with each element of Π_{S}(L_{1}). □

Theorem 2.7.11 If |Π_{S}(L_{1})| = s + 1 then also |G_{S}| = s + 1. Conversely, if |G_{S}| = s + 1 and if Q is not a grid, then |Π_{S}(L_{1})| = s + 1.

Proof. If |Π_{S}(L_{1})| = s + 1, then Q is not a grid and Theorem 2.7.7 implies that Π_{S}(L_{1}) is a regular group of permutations of L_{1}. Lemma 2.7.9 implies then that there are exactly s + 1 permutations of L_{1} which commute with each element of Π_{S}(L_{1}) and by Theorem 2.7.10 each of these permutations can be extended to an element of G_{S}. Hence |G_{S}| = s + 1. Conversely, suppose that Q is not a grid and that |G_{S}| = s + 1; then Theorem 2.7.1 implies that G_{S} induces a regular group of permutations on the point set of L_{1}. Lemma 2.7.9 and Theorem 2.7.10 imply then that |Π_{S}(L_{1})| ≤ s + 1 and the result follows by Theorem 2.7.7. □

We prove now the following theorem.

Theorem 2.7.12 If Π_{S}(L_{1}) is commutative then S is a spread of symmetry and Q is derivable from an admissible triple.

Proof. We may suppose that Q is not a grid. Theorems 2.7.7 and 2.7.10 imply that |G_{S}| ≥ |Π_{S}(L_{1})| ≥ s + 1, hence |G_{S}| = s + 1 by Theorem 2.7.1. So S is a spread of symmetry. □

2.8 Automorphisms of a GQ fixing a spread

With each normal spread S of a GQ(s, t), there is associated a linear space L(S): the points of L(S) are the st + 1 lines of S; the lines of L(S) are the sets {L, M}^{⊥⊥} with L and M two different lines of S; incidence is the natural one. If θ is an automorphism of the GQ fixing S, then θ induces an automorphism of L(S). If S is the associated spread of an AT (D, K, Δ), then L(S) ≃ D. Hence every automorphism of the GQ fixing S induces an automorphism of D.

Let Q = (P, L, I) be a GQ(s, t) and let S = {L_{1}, …, L_{1+st}} be a spread of Q (Notice that we number the lines of S). We call L_{1} the base line of the spread. The set of all automorphisms of Q fixing S is denoted by A_{S} and it is a subgroup of A = Aut(Q). The map Δ_{ij} := [L_{j}, L_{1}][L_{i}, L_{j}][L_{1}, L_{i}] (i, j ∈ {1, …, 1 + st}) defines a permutation on the point set of L_{1}; if β is a permutation of {1, …, 1 + st}, then we define Δ_{ij}^{β} := Δ_{β(i)β(j)}. If θ ∈ A_{S}, then the following permutations can be defined.

(a) A permutation β of {1, …, 1 + st}, such that

θ(L_{i}) = L_{β(i)},  (2.3)

for all i ∈ {1, …, 1 + st}.

(b) Permutations α_{i} of L_{1} such that

θ[L_{1}, L_{i}](x) = [L_{1}, L_{β(i)}]α_{i}(x),  (2.4)

for all i ∈ {1, …, 1 + st} and all x ∈ L_{1}.

Let [L_{1}, L_{i}](x) ∼ [L_{1}, L_{j}](x′) with x, x′ ∈ L_{1} and i ≠ j. Hence x′ = Δ_{ij}(x) and from [L_{1}, L_{β(i)}]α_{i}(x) ∼ [L_{1}, L_{β(j)}]α_{j}(x′) = [L_{1}, L_{β(j)}]α_{j}Δ_{ij}(x), it follows that α_{i} = Δ_{ji}^{β}α_{j}Δ_{ij}. Define

α := α_{1},  (2.5)

we get then that

α_{i} = Δ_{1i}^{β}α,  (2.6)

and the condition α_{i} = Δ_{ji}^{β}α_{j}Δ_{ij} becomes

(Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β})α = αΔ_{ij},  (2.7)

for all i, j ∈ {1, …, 1 + st}. So, we have proved the following theorem.

Theorem 2.8.1 If θ ∈ A_{S}, then we can define a permutation α of L_{1} and a permutation β of {1, …, 1 + st} by (2.3), (2.4), (2.5) and these permutations satisfy equation (2.7). Conversely, if (2.7) is satisfied by a permutation α of L_{1} and a permutation β of {1, …, 1 + st}, then (2.6) and (2.4) define a permutation θ ∈ A_{S}.

Let Ĝ be the set of all permutations α of L_{1} for which there exists a permutation β of {1, …, 1+st} such that (2.7) is satisfied for all i, j ∈ {1, …, 1+st}.

Theorem 2.8.2 If G is commutative, then Ĝ is a group.

Proof. Since Ĝ is finite, it suffices to prove that μ_{1}μ_{2} ∈ Ĝ for every two elements μ_{1}, μ_{2} ∈ Ĝ. Let β_{1} and β_{2} be the permutations such that

Δ_{j1}^{β_{1}}Δ_{ij}^{β_{1}}Δ_{1i}^{β_{1}} = μ_{1}Δ_{ij}μ_{1}^{−1},
Δ_{j1}^{β_{2}}Δ_{ij}^{β_{2}}Δ_{1i}^{β_{2}} = μ_{2}Δ_{ij}μ_{2}^{−1},

for all i, j ∈ {1, …, 1 + st}. One calculates then that

μ_{1}μ_{2}Δ_{ij}μ_{2}^{−1}μ_{1}^{−1} = Δ_{β_{2}(1)1}^{β_{1}}Δ_{j1}^{β_{1}β_{2}}Δ_{ij}^{β_{1}β_{2}}Δ_{1i}^{β_{1}β_{2}}Δ_{1β_{2}(1)}^{β_{1}},

and this is equal to Δ_{j1}^{β_{1}β_{2}}Δ_{ij}^{β_{1}β_{2}}Δ_{1i}^{β_{1}β_{2}} if G is commutative. □

Example 1: the GQ P (W (q), u)

We use the AT-model for P (W (q), u), q = p^{h}, given in Section 2.5.2. The associated spread consists of the lines L_{(x,y)} = {(z, x, y)|z ∈ GF(q)}, x, y ∈ GF(q). Consider L_{1} = L_{(0,0)} as base line. We can identify the point set of L_{1} with the elements of GF(q) and hence with the points of AG(1, q). The permutations Δ_{ij} of L_{1} are then translations of AG(1, q); indeed if L_{i} = L_{(x_{1},y_{1})} and L_{j} = L_{(x_{2},y_{2})}, then one easily checks that Δ_{ij} is a translation T (x_{1}y_{2}−x_{2}y_{1}) over x_{1}y_{2}−x_{2}y_{1}. Let θ ∈ A_{S}, then θ induces an automorphism of AG(2, q) (= D). Such an automorphism looks like

[ \begin{matrix}x′ \\ y′ \end{matrix} ] = [ \begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} ] [ \begin{matrix}x^{σ} \\ y^{σ} \end{matrix} ] + [ \begin{matrix}b \\ c \end{matrix} ] ,  (2.8)

where the map x ↦ x^{σ} is an automorphism of GF(q) and A = [ \begin{matrix}a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix} ] is

a nonsingular (2×2)-matrix over GF(q). One easily checks that Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β} = T (det(A)k^{σ}) if Δ_{ij} = T (k). From equation (2.7), it follows then that α(k) = det(A)k^{σ} + d, where d = α(0). Hence

α_{(x,y)}(k) = (ba_{21} − ca_{11})x^{σ} + (ba_{22} − ca_{12})y^{σ} + (a_{11}a_{22} − a_{12}a_{21})k^{σ} + d,

where α_{(x,y)} := α_{i} with i such that L_{i} = L_{(x,y)}. The automorphisms of A_{S} are hence the following maps:

[ \begin{matrix}z \\ x \\ y \end{matrix} ]^{θ} = [ \begin{matrix}a_{11}a_{22} − a_{12}a_{21} & ba_{21} − ca_{11} & ba_{22} − ca_{12} \\ 0 & a_{11} & a_{12} \\ 0 & a_{21} & a_{22} \end{matrix} ] [ \begin{matrix}z^{σ} \\ x^{σ} \\ y^{σ} \end{matrix} ] + [ \begin{matrix}d \\ b \\ c \end{matrix} ] ,

where σ is an arbitrary automorphism of GF(q) and a_{11}, a_{12}, a_{21}, a_{22}, b, c, d are arbitrary elements of GF(q) satisfying a_{11}a_{22} − a_{12}a_{21} ≠ 0. The knowledge of the orbit of S under A gives now a good idea of how close A is to A_{S}. The spread S is a normal spread and need to be mapped on a normal spread. If q ≠ 3 is odd, then S is the only normal spread; hence A = A_{S}. If q = 3, then AS(3) ≃ Q(5, 2) and A acts transitively on the set of all normal spreads. Suppose now that q is even and consider the model T_{2}^{∗}(O). If q ≠ 2, then the elements of A are induced by automorphisms of PG(3, q) which fix the hyperoval. If q = 2 or q = 4, then A acts transitively on the set of all normal spreads. If q ≥ 8, then S corresponds with the unique point a of O such that O \ {a} is a conic; hence A = A_{S}.

Example 2: the GQ Q(5, q)

We use the AT-model for Q(5, q), q = p^{h}, given in Section 2.5.4 and suppose that (ā, b̄) = ā^{T} b̄^{q} is the nonsingular Hermitian form in V (3, q^{2}). The associated spread consists of the lines L_{r} = {(k, r)|k ∈ K}, r ∈ U. Consider L_{1} = L_{⟨ā⟩} as base line. We can identify the points of L_{1} with the elements of K. The permutations Δ_{ij} are then multiplications: if L_{i} = L_{r} and L_{j} = L_{s}, then one easily verifies that Δ_{ij} is a multiplication M(Δ(r, s)) with Δ(r, s) ∈ K. Let θ ∈ A_{S}, then θ induces an automorphism of the design related to U. This automorphism is an automorphism of P G(2, q^{2}) fixing U, and hence is an element of P ΓU(3, q^{2}) (see [58]). Such an automorphism looks like ā ↦ Aā^{σ}, where σ is an automorphism of GF(q^{2}) and where A satisfies A^{T} A^{q} = αI with α^{q−1} = 1. One calculates that Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β} = M(k^{σ}) if Δ_{ij} = M(k). From equation (2.7), it follows then that α(k) = lk^{σ} where l = α(1). Hence A_{S} contains |K| × |P ΓU(3, q^{2})| = 2hq^{3}(q + 1)(q^{2} − 1)(q^{3} + 1) elements. Since A acts transitively on the set of q^{3}(q − 1)(q^{2} + 1) normal spreads, A contains 2hq^{6}(q^{2} − 1)(q^{3} + 1)(q^{4} − 1) elements. The group P ΓU(4, q^{2}) has indeed this order.

Example 3: the GQ T_{2}^{∗}(O)

Let O be a hyperoval of PG(2, q), q = 2^{h}, and let S be the spread of T_{2}^{∗}(O) determined by a ∈ O. Let θ ∈ A_{S}, then θ induces an automorphism of the affine plane D = AG(2, q) in the related AT-model (see Section 2.5.3). As before, let L_{1} denote the base line of S; the points of L_{1} can be identified with the elements of GF(q). For 1 ≤ i, j ≤ q^{2}, one easily checks that

Δ_{ij} = T (Δ(L_{1}, L_{i}) + Δ(L_{i}, L_{j}) + Δ(L_{j}, L_{1})),  (2.9)
Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β} = T (Δ(L_{β(1)}, L_{β(i)}) + Δ(L_{β(i)}, L_{β(j)}) + Δ(L_{β(j)}, L_{β(1)})).

(2.10)

By equation (2.7), Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β} is only dependent of Δ_{ij}. Now, let L_{i_{1}} and L_{j_{1}} be such that L_{1}, L_{i_{1}}, L_{j_{1}} are three noncollinear points of AG(2, q). Let (i_{k}, j_{k}), 1 ≤ k ≤ q − 1, be the q − 1 pairs such that

(A) L_{i_{k}} ≠ L_{1} ≠ L_{j_{k}},

(B) L_{1}L_{i_{k}} = L_{1}L_{i_{1}} and L_{1}L_{j_{k}} = L_{1}L_{j_{1}},

(C) L_{i_{k}}L_{j_{k}}∥L_{i_{1}}L_{j_{1}}.

If we put the (q −1) pairs (i_{k}, j_{k}) in equations (2.9) and (2.10), then one finds that Δ_{j1}^{β}Δ_{ij}^{β}Δ_{1i}^{β} = T (lk^{σ}) if Δ_{ij} = T (k). Here l is a fixed nonzero element of GF(q) and σ is the automorphism of GF(q) related (see equation (2.8)) to the above mentioned automorphism of AG(2, q). From equation (2.7), it follows then that α(k) = lk^{σ} +d with d = α(0). Hence |Ĝ| ≤ h(q −1)q. Now, if β corresponds with a homothetic transformation of AG(2, q) with factor l ≠ 0 and center L_{1}, then equation (2.7) has q solutions for α, namely the maps α(k) = lk + d with d ∈ GF (q). Hence |Ĝ| ≥ (q − 1)q. Since Ĝ is a group, |Ĝ| = h′(q − 1)q with h′ | h. The exact value for |Ĝ| depends on the kind of hyperoval we are working with.

2.9 The nonexistence of spreads of symmetry in certain GQ’s of order (s, s^{2})

2.9.1 Generalized quadrangles as group coset geometries.

Let G be a group of order s^{2}t, s > 1, t > 1, and let F be a family of t + 1 subgroups of G, each of order s. For each A ∈ F, let A^{∗} be a subgroup of G having st elements and containing A, and put F^{∗} := {A^{∗}|A ∈ F}. The following incidence structure S can then be defined. Points are of three types:

(i) elements of G;

(ii) right cosets of elements of F^{∗};

(iii) a symbol (∞).

Lines are of two types:

(a) right cosets of elements of F;

(b) symbols [A], where A ∈ F.

A point g of type (i) is incident with each line Ag of type (a). A point A^{∗}g of type (ii) is incident with [A] and with each line Ahg contained in it. The point (∞) is incident with all lines of type (b). There are no further incidences. The just defined geometry, denoted by S(G, F, F^{∗}), is introduced by Kantor ([49]), and called a group coset geometry. The following theorem was proved in [49].

Theorem 2.9.1 A group coset geometry S(G, F, F^{∗}) is a generalized quadrangle of order (s, t) if and only if the following two conditions are satisfied:

(K1) AB ∩ C = 1 whenever A, B, C are distinct elements of F;

(K2) A^{∗} ∩ B = 1 whenever A and B are distinct members of F.

If the above two conditions are satisfied, then for each A ∈ F, we have that

A^{∗} = ⋃{Ag|Ag = A or Ag ∩ B = ∅ for all B ∈ F},

and we can write S(G, F) := S(G, F, F^{∗}).

2.9.2 Generalized quadrangles arising from q-clans

The following definitions and results were taken from [50], [64] and [66]. Let q be a fixed prime power.

Definitions.

(1) A (2 × 2)-matrix A over GF(q) is called anisotropic matrix provided [α_{1} α_{2}]A[α_{1} α_{2}]^{T} = 0 if and only if α_{1} = α_{2} = 0.

(2) For each t ∈ GF(q), let A_{t} = ( \begin{matrix}x_{t} & y_{t} \\ 0 & z_{t} \end{matrix} ) be an upper triangular (2×2)-

matrix. The set C = {A_{t}|t ∈ GF(q)} is called a q-clan provided all pairwise differences A_{s} − A_{t} (s, t ∈ GF(q), s ≠ t) are anisotropic.

Suppose now that C = \{A_{t}|t ∈ GF(q)\} is a q-clan, where A_{t} = ( \begin{matrix}x_{t} & y_{t} \\ 0 & z_{t} \end{matrix} ).

Put K_{t} = A_{t} + A_{t}^{T} = ( \begin{matrix}2x_{t} & y_{t} \\ y_{t} & 2z_{t} \end{matrix} ). Let G denote the group consisting of

the set G = {(α_{1}, α_{2}, c, β_{1}, β_{2})|α_{1}, α_{2}, c, β_{1}, β_{2} ∈ GF(q)}, together with the binary operation

(α_{1}, α_{2}, c, β_{1}, β_{2}) · (α′_{1}, α′_{2}, c′, β′_{1}, β′_{2}) =

(α_{1} + α′_{1}, α_{2} + α′_{2}, c_{1} + c′_{1} + β_{1}α′_{1} + β_{2}α′_{2}, β_{1} + β′_{1}, β_{2} + β′_{2}).

We define now q + 1 subgroups of G having order q^{2}:

A(∞) = {(0, 0, 0, β_{1}, β_{2})|β_{1}, β_{2} ∈ GF(q)};
A(t) = {(α_{1}, α_{2}, α_{1}^{2}x_{t} + α_{1}α_{2}y_{t} + α_{2}^{2}z_{t}, 2α_{1}x_{t} + α_{2}y_{t}, α_{1}y_{t} + 2α_{2}z_{t})|
α_{1}, α_{2} ∈ GF(q)},

for all t ∈ GF(q), and we put F = {A(i)|i ∈ GF(q) ∪ {∞}}. We define now q + 1 subgroups of G having order q^{3}:

A^{∗}(∞) = {(0, 0, c, β_{1}, β_{2})|c, β_{1}, β_{2} ∈ GF(q)},
A^{∗}(t) = {(α_{1}, α_{2}, c, 2α_{1}x_{t} + α_{2}y_{t}, α_{1}y_{t} + 2α_{2}z_{t})|c, α_{1}, α_{2} ∈ GF(q)},

for all t ∈ GF(q), and we put F^{∗} = {A^{∗}(i)|i ∈ GF(q) ∪ {∞}}. With the above definitions, conditions (K1) and (K2) of Theorem 2.9.1 are satisfied. S(G, F) is hence a GQ of order (q^{2}, q). This GQ has the following properties:

(1) the point (∞) is a regular point;

(2) if S(G, F) is not isomorphic to H(3, q^{2}), then

(i) the point (∞) is fixed by each automorphism of S(G, F), see [72],

(ii) every ovoid of symmetry O contains the point (∞) (For, let θ be a nontrivial automorphism of S(G, F) which fixes each point of O, then the only fixpoints of θ are the elements of O.),

(iii) if S(G, F) has an ovoid of symmetry then the automorphism group of S(G, F) is transitive on the set of lines through (∞).

We refer to [89] for the known examples of q-clans. We only give four examples here. For each of these examples, Q denotes the corresponding GQ.

(1) q ≡ ±2 (mod 5), A_{t} = ( \begin{matrix}t & 5t^{3} \\ 0 & 5t^{5} \end{matrix} ), t ∈ GF(q). It was proved in [69]

that Aut(Q) has two orbits on the lines through (∞). Hence Q has no ovoid of symmetry.

(2) q = 5^{e} > 5, k is a nonsquare in GF(q) and A_{t} = ( \begin{matrix}t & t^{2} \\ 0 & k^{−1}t(1 + kt^{2})^{2} \end{matrix} ).

It was proved in [69] that Aut(Q) fixes the line [A(∞)]. Hence Q has no ovoid of symmetry.

(3) q = 3^{e}, e ≥ 3, m is a nonsquare in GF(q) and A_{t} = ( \begin{matrix}t & t^{3} \\ 0 & −\frac{t^{9}}{m} − mt \end{matrix} ).

It was proved in [69] that Aut(Q) fixes the line [A(∞)]. Hence Q has no ovoid of symmetry.

(4) Let q be odd, σ ∈ Aut[GF(q)], m be a nonsquare in GF(q) and A_{t} =

( \begin{matrix}t & 0 \\ 0 & −mt^{σ} \end{matrix} ). The quadrangles so-obtained are called Kantor-Knuth

quadrangles. The generalized quadrangle Q is classical (i.e. isomorphic to H(3, q^{2})) if and only if σ is the identity. We will prove in Section 2.9.4 that every nonclassical Kantor-Knuth quadrangle cannot have an ovoid of symmetry.

2.9.3 Ovoids of symmetry in GQ’s derived from q-clans

In this section, we suppose that Q is a GQ(q^{2}, q) derived from a q-clan

C = \{A_{t}|t ∈ GF(q)\}, where A_{t} = ( \begin{matrix}x_{t} & y_{t} \\ 0 & z_{t} \end{matrix} ). Since we already determined

all ovoids of symmetry of H(3, q^{2}), we suppose that Q is not isomorphic to H(3, q^{2}). Let O be an ovoid of symmetry of Q, then we know that (∞) ∈ O.

Definitions.

(1) A net of order k and degree r is a partial linear space satisfying

(i) each point is incident with r lines;

(ii) each line is incident with k points;

(iii) if x is a point and L is a line not incident with x, then there is a unique line M incident with x and not concurrent with L.

(2) Let x_{1} and x_{2} be two points of Q at distance two from (∞). If |{(∞), x_{1}}^{⊥}∩{(∞), x_{2}}^{⊥}| ≥ 2, then x_{2} ∈ {x_{1}, (∞)}^{⊥⊥} and {(∞), x_{1}}^{⊥} = {(∞), x_{2}}^{⊥} (since (∞) is regular). Hence we can define the following dual net N_{(∞)}^{∗}, see [71]:

• points are the elements of (∞)^{⊥} \ {(∞)};

• lines are the sets {x, (∞)}^{⊥} with d(x, (∞)) = 2;

• incidence is the natural one, i.e. containment.

N_{(∞)}^{∗} has q^{3} + q^{2} points, each line has q + 1 points and each point is incident with q^{2} lines. A spread of N_{(∞)}^{∗} is a set of lines partitioning the point set of N_{(∞)}^{∗}.

Theorem 2.9.2 Let S = {L_{1}, …, L_{q^{2}}} be a spread of N_{(∞)}^{∗}, then O_{S} = {(∞)}∪{x| d(x, (∞)) = 2 and {x, (∞)}^{⊥} ∈ S} is an ovoid of Q.

Proof. Let x_{1} and x_{2} be two collinear points of O_{S}, then {x_{1}, (∞)}^{⊥} and {x_{2}, (∞)}^{⊥} have at least one point in common. Hence {x_{1}, (∞)}^{⊥} = {x_{2}, (∞)}^{⊥} and x_{2} ∈ {x_{1}, (∞)}^{⊥⊥}. This implies that x_{1} and x_{2} are not collinear, a contradiction. O_{S} is a set of q^{3} + 1 points, no three of which are collinear. Hence O_{S} is an ovoid of Q. □

Definition. A spread S of N_{(∞)}^{∗} is called regular if for every line M_{1} of N_{(∞)}^{∗} not belonging to S (hence there exist q+1 lines L_{1}, …, L_{q+1} of S intersecting M_{1}), there are q other lines M_{2}, …, M_{q+1} such that:

• M_{i} is disjoint with M_{j} for all i, j ∈ {1, …, q + 1} with i ≠ j,

• M_{i} meets L_{j} in a point for all i, j ∈ {1, …, q + 1}.

Theorem 2.9.3 There exists a regular spread S of N_{(∞)}^{∗} such that O = O_{S}.

Proof. We prove that {o_{1}, (∞)}^{⊥} and {o_{2}, (∞)}^{⊥} are equal or disjoint for every two points o_{1}, o_{2} of O \ {(∞)}. Suppose x ∈ {o_{1}, (∞)}^{⊥} ∩ {o_{2}, (∞)}^{⊥} and let θ be a nontrivial automorphism of Q which fixes each point of O, then x, x^{θ} ∈ {o_{1}, (∞)}^{⊥} ∩ {o_{2}, (∞)}^{⊥} and hence {o_{1}, (∞)}^{⊥} = {o_{2}, (∞)}^{⊥} = {x, x^{θ}}^{⊥⊥}. Hence S = {{o, (∞)}^{⊥}|o ∈ O \ {(∞)}} is a spread of N_{(∞)}^{∗}. We prove now that S is regular. Let M_{1} be an arbitrary line of N_{(∞)}^{∗} not belonging to S and let M_{1} = {(∞), x_{1}}^{⊥}. Let {x_{1}, …, x_{q+1}} denote the orbit of x_{1} determined by the group of automorphisms fixing each point of O and put M_{i} = {(∞), x_{i}}^{⊥} for all i ∈ {1, …, q + 1}. The above conditions are then satisfied. □

Every automorphism θ of Q induces an automorphism θ̄ of N_{(∞)}^{∗}. If θ fixes every point of O, then θ̄ fixes every line of S. If L is a line of PG(3, q), then there is a geometry H_{q}^{3} with points those points of PG(3, q) not on L, see Section 1.4.4. If N_{(∞)}^{∗} ≃ H_{q}^{3}, then the spread S can be extended (by adding the line L) to a spread S̄ of PG(3, q).

Theorem 2.9.4 If N_{(∞)}^{∗} ≃ H_{q}^{3}, then S̄ is a regular spread of PG(3, q).

Proof. Let θ_{i}, i ∈ {1, …, q + 1}, denote the q + 1 automorphisms of Q fixing each point of O. The automorphism θ_{i} corresponds to an automorphism θ̄_{i} of N_{(∞)}^{∗} and an automorphism ¯ θ̄_{i} of PG(3, q), see Theorem 1.4.3. The automorphism ¯ θ̄_{i} fixes each line of S̄. Now, let M be any line of PG(3, q) not belonging to S̄, then M meets the lines L_{1}, …, L_{s+1} of S̄. Put {M_{1}, …, M_{q+1}} = {¯θ̄_{i}(M)|i ∈ {1, …, q + 1}}. Clearly L_{1}, …, L_{q+1} is a regulus of PG(3, q); hence S̄ is regular. □

Theorem 2.9.5 If N_{(∞)}^{∗} ≃ H_{q}^{3}, then the group of automorphisms of Q fixing each point of O is isomorphic to the cyclic group C_{q+1}.

Proof. Let θ_{i}, i ∈ {1, …, q +1}, denote the q +1 automorphisms of Q fixing each point of O. The automorphism θ_{i} corresponds to an automorphism θ̄_{i} of N_{(∞)}^{∗} and an automorphism ¯ θ̄_{i} of PG(3, q). The group {¯θ̄_{i}|i ∈ {1, …, q + 1}} is a subgroup of the group of automorphisms of PG(3, q) which fix each element of S̄. The result follows now by Section 1.4.3. □

Remark. By [90] we know that N_{(∞)}^{∗} ≃ H_{q}^{3} if and only if q is even or if Q is a Kantor-Knuth quadrangle. In the following section we will prove that only classical Kantor-Knuth quadrangles can have ovoids of symmetry. We give now a method for determining all ovoids of symmetry. As we already mentioned, every automorphism θ of Q induces an automorphism θ̄ of N_{(∞)}^{∗}.

Theorem 2.9.6 Let ϕ be an automorphism of N_{(∞)}^{∗} fixing each line of a spread S of N_{(∞)}^{∗}, then there exists at most one automorphism θ of Q such that the following two conditions are satisfied:

(i) θ fixes each point of O_{S},

(ii) θ̄ = ϕ.

Proof. If θ satisfies (i) and (ii), then it must be equal to the map which we will construct now. Put (∞)^{θ} = (∞). If x ∈ (∞)^{⊥} \ {(∞)}, then it corresponds with a point x′ of N_{(∞)}^{∗}. Let the point x^{′ϕ} correspond with the point y of Q, then x^{θ} = y. Finally, take z ∉ (∞)^{⊥} and let L_{1} and L_{2} be two lines through z. The line L_{i}, i ∈ {1, 2}, intersects (∞)^{⊥} in a point z_{i} and the ovoid O_{S} in a point o_{i}. Now, z^{θ} is the intersection of the lines z_{1}^{θ}o_{1} and z_{2}^{θ}o_{2}.

□

If we want to determine all ovoids of symmetry of Q, one can proceed as follows:

(1) Determine all regular spreads of N_{(∞)}^{∗}.

(2) For each regular spread S of N_{(∞)}^{∗}, determine the automorphisms of N_{(∞)}^{∗} fixing each line of S. There have to be at least q + 1 such automorphisms (including the identity), otherwise S cannot correspond to an ovoid of symmetry.

(3) Calculate the ovoid O_{S} of Q corresponding to S.

(4) Check for each of the automorphisms determined in (2) whether it corresponds with an automorphism of Q which fixes each point of O_{S}. If we find q + 1 such automorphisms, then O_{S} is an ovoid of symmetry.

2.9.4 The nonexistence of ovoids of symmetry in GQ’s in nonclassical Kantor-Knuth quadrangles.

Let q be an odd prime power, σ ∈ Aut[GF(q)], m be a nonsquare in GF(q). Let Q be the Kantor-Knuth quadrangle associated with these parameters. We suppose that Q is not isomorphic to H(3, q^{2}), or equivalently, that σ is not the identity. The choice of the nonsquare m in GF(q) is immaterial, see [65]. Also the GQ Q obtained by using a particular automorphism σ is isomorphic to the one obtained by using σ^{−1}. The Kantor-Knuth quadrangles have a lot of subquadrangles isomorphic to W (q). For each (γ_{1}, γ_{2}) ∈ GF(q) × GF(q) \ {(0, 0)}, we have the following subquadrangle S(γ_{1}, γ_{2}) isomorphic to W (q) ([65]). Points are of four kinds:

(a) (∞);

(b) A^{∗}(∞) · (kγ_{1}, k^{σ}γ_{2}, 0, 0, 0), k ∈ GF(q);

(c) A^{∗}(t) · (0, 0, 0, 2γ_{1}u, −2γ_{2}mu^{σ}), t, u ∈ GF(q);

(d) (kγ_{1}, k^{σ}γ_{2}, γ_{1}^{2}c − γ_{2}^{2}mc^{σ}, 2γ_{1}u, −2γ_{2}mu^{σ}), k, c, u ∈ GF(q).

Lines are of three kinds:

(a) [A(t)], t ∈ GF(q) ∪ {(∞)};

(b) A(∞) · (kγ_{1}, k^{σ}γ_{2}, γ_{1}^{2}c − γ_{2}^{2}mc^{σ}, 0, 0), k, c ∈ GF(q);

(c) A(r) · (0, 0, γ_{1}^{2}c − γ_{2}^{2}mc^{σ}, 2γ_{1}s, −2γ_{2}ms^{σ}), r, c, s ∈ GF(q).

Incidence is the one derived from Q.

Theorem 2.9.7 ([65]) Let σ ≠ 1 and let θ be any automorphism of Q. Then [S(1, 0)]^{θ} = S(1, 0) or [S(1, 0)]^{θ} = S(0, 1).

Theorem 2.9.8 If σ ≠ 1, then Q has no ovoid of symmetry.

Proof. Clearly q ≥ 5. Let x_{1}, …, x_{q−1} be the q − 1 points of S(1, 0) on the line [A(∞)] and different from (∞) and A^{∗}(∞)(0, 0, 0, 0, 0). Let θ_{1} and θ_{2} be two nontrivial automorphisms fixing each point of O. Let i ∈ {1, …, q − 1} be fixed. If θ_{1}(x_{i}) and θ_{2}(x_{i}) are no points of S(1, 0), then x_{i} ∈ {θ_{1}(x_{i}), θ_{2}(x_{i})}^{⊥⊥} ⊆ S(0, 1), a contradiction. Hence {θ_{1}(x_{i}), x_{i}}^{⊥⊥} ⊆ S(1, 0) for all i ∈ {1, …, q − 1}. Let o_{1}, o_{2} ∈ O \ {(∞)} such that o_{1} is collinear with x_{1} and o_{2} is collinear with x_{2}. Then o_{1} ∈ {x_{1}, θ_{1}(x_{1})}^{⊥} ⊆ S(1, 0). Similarly o_{2} ∈ S(1, 0). The points (∞), o_{1} and o_{2} belong to a subquadrangle isomorphic to W (q). These points have then at least one common neighbour, a contradiction. □

2.10 Generalized quadrangles of order (s, s+2) with a lot of grids

In this section we prove that there is a common construction for all GQ’s of order (s, s + 2) which have the property that every two intersecting lines are contained in a unique subquadrangle of order (s, 1).

2.10.1 Amalgamations of projective planes

All GQ’s of order n which have a regular point incident with some regular line can be constructed in the following way, see [61]. Let π_{1} and π_{2} be two projective planes of order n. Choose in π_{i}, i ∈ {1, 2}, an incident point-line pair ((∞)_{i}, [∞]_{i}). Let α be a bijection between the point sets of [∞]_{1} and [∞]_{2} such that (∞)_{1}^{α} = (∞)_{2}, and let β be a bijection between the set of lines through (∞)_{1} and the set of lines through (∞)_{2} such that [∞]_{1}^{β} = [∞]_{2}. Suppose also that the following configuration fails to exist:

(1) A_{i}, B_{i}, C_{i} (i ∈ {1, 2}) are points of π_{i} not on [∞]_{i};

(2) no three of the points (∞)_{i}, A_{i}, B_{i}, C_{i} (i ∈ {1, 2}) are collinear;

(3) A_{2} ∈ ((∞)_{1}A_{1})^{β}, B_{2} ∈ ((∞)_{1}B_{1})^{β}, C_{2} ∈ ((∞)_{1}C_{1})^{β},

(4) (A_{1}B_{1} ∩ [∞]_{1})^{α} = A_{2}B_{2} ∩ [∞]_{2}, (A_{1}C_{1} ∩ [∞]_{1})^{α} = A_{2}C_{2} ∩ [∞]_{2} and (B_{1}C_{1} ∩ [∞]_{1})^{α} = B_{2}C_{2} ∩ [∞]_{2}.

The following generalized quadrangle Q can then be constructed. The points are of two types:

(i) the points of π_{1},

(ii) pairs of lines (L_{1}, L_{2}), where L_{i} is a line of π_{i} different from [∞]_{i}, and (L_{1} ∩ [∞]_{1})^{α} = L_{2} ∩ [∞]_{2} ≠ (∞)_{2}.

The lines are of two types.

(a) lines of π_{2},

(b) pairs of points (x_{1}, x_{2}) with x_{1} ∉ [∞]_{1} a point of π_{1}, x_{2} ∉ [∞]_{2} a point of π_{2} and ((∞)_{1}x_{1})^{β} = (∞)_{2}x_{2}.

Incidence is as follows.

(i,a) A point Q of type (i) is incident with a line L of type (a) if and only if (Q ∈ [∞]_{1} and Q^{α} ∈ L) or ((∞)_{2} ∈ L and Q ∈ L^{β^{−1}}).

(i,b) A point Q of type (i) is incident with a line (R, S) of type (b) if and only if Q = R.

(ii,a) A point (L_{1}, L_{2}) of type (ii) is incident with a line L of type (a) if and only if L = L_{2}.

(ii,b) A point (L_{1}, L_{2}) of type (ii) is incident with a line (R_{1}, R_{2}) of type (b) if and only if R_{1} ∈ L_{1} and R_{2} ∈ L_{2}.

The generalized quadrangle Q has order n and is called an amalgamation of the projective planes π_{1} and π_{2}. The generalized quadrangle Q^{D} which is the dual of Q is also an amalgamation of two projective planes (namely of the dual planes). The point [∞]_{2} is a regular point of Q^{D} and we find the following description for P (Q^{D}, [∞]_{2}). The points of P (Q^{D}, [∞]_{2}) are the pairs of points (x_{1}, x_{2}) with x_{1} ∉ [∞]_{1} a point of π_{1}, x_{2} ∉ [∞]_{2} a point of π_{2} and ((∞)_{1}x_{1})^{β} = (∞)_{2}x_{2}. Lines are of three types:

(i) points of π_{1} not contained in [∞]_{1};

(ii) points of π_{2} not contained in [∞]_{2};

(iii) pairs of lines (L_{1}, L_{2}) such that L_{1} is a line of π_{1} not through (∞)_{1}, L_{2} is a line of π_{2} not through (∞)_{2} and (L_{1} ∩ [∞]_{1})^{α} = L_{2} ∩ [∞]_{2}.

Incidence is as follows. The point (x_{1}, x_{2}) is incident with the line x of type (i) if and only if x = x_{1}. The point (x_{1}, x_{2}) is incident with the line x of type (ii) if and only if x = x_{2} and the point (x_{1}, x_{2}) is incident with the line (L_{1}, L_{2}) of type (iii) if and only if x_{1} ∈ L_{1} and x_{2} ∈ L_{2}.

2.10.2 GQ’s with a unique grid through every two intersecting lines

The generalized quadrangle T_{2}^{∗}(O), with O a hyperoval in PG(2, 2^{h}), h ≥ 2, satisfies the following properties.

(P1) The order is (s, t) = (s, s + 2).

(P2) Every two intersecting lines are contained in a unique subquadrangle of order (s, 1), i.e. a grid.

Property (P2) holds because of the following reason: every subquadrangle of order (s, 1) is the set of affine points in a plane of PG(3, 2^{h}) which intersects O in exactly two points, see Theorem 3.3.4 of [71]. In this section, we try to prove a kind of converse theorem, namely the following one.

Theorem 2.10.1 If Q is a generalized quadrangle satisfying (P1) and (P2), then there exists a generalized quadrangle Q′ and a regular point x of Q′ such that Q is isomorphic to P (Q′, x), moreover Q′ is an amalgamated generalized quadrangle.

This theorem has the following corollary.

Corollary 2.10.2 If Q is a GQ(s, s + 2) with the property that every two lines are contained in a unique grid, then there exists a projective plane of order s + 1. Hence, there are no such GQ’s with s = 5, 9, . . ..

In order to prove the theorem, we start with the following lemma.

Lemma 2.10.3 Let Q be a GQ(s, t) satisfying (P2).

(A) There are \frac{t+1}{2} grids through every two points x and y at distance 2. Hence t is odd.

(B) If t ≠ 1, then t ≥ s + 2. If t = s + 2, then one of the following possibilities occurs for every two grids G_{1} and G_{2}:

(1) G_{1} = G_{2};

(2) G_{1} and G_{2} are disjoint;

(3) G_{1} and G_{2} intersect in a line;

(4) G_{1} and G_{2} intersect in an ovoid of G_{1}.

Proof.

(A) The grids through x and y partition the set of lines through x. For, every line L through x is contained in a unique grid through x and y, namely the grid containing L and the line through y meeting L.

(B) Let G_{1} be a fixed grid through a fixed point x. If G_{2} is another grid through x such that G_{1} and G_{2} have a common line L through x, then G_{1} and G_{2} have no other points in common than those on L. Now, let V be the set of the \frac{t(t+1)}{2} − 1 − 2(t − 1) = \frac{(t−2)(t−1)}{2} grids G_{2} ≠ G_{1} through x, such that G_{1} and G_{2} have no common line through x. For every G_{2} ∈ V, let A(G_{2}) be the set of the f(G_{2}) common points of G_{1} and G_{2} different from x. Suppose that α and β are two collinear points of A(G_{2}). Let γ denote that point on αβ collinear with x, then γx and αβ are contained in at least two grids, a contradiction. Hence A(G_{2}) is a set of mutual noncollinear points and f(G_{2}) ≤ s. Moreover, if f(G_{2}) = s, then G_{1} and G_{2} intersect in an ovoid of G_{1}. Counting the number of pairs (G_{2}, α) with G_{2} ∈ V and α ∈ A(G_{2}) yields

∑_{G_{2}∈V} f(G_{2}) = s^{2} \frac{t − 1}{2}.

Since |V | = \frac{(t−2)(t−1)}{2} and f(G_{2}) ≤ s for all G_{2} ∈ V, we have that s ≤ t − 2. If s = t − 2, then f(G_{2}) = s for all G_{2} ∈ V; hence G_{1} and G_{2} intersect in an ovoid of G_{1} for all G_{2} ∈ V.

□

From now on we suppose that Q is a GQ(s, t) satisfying (P1) and (P2).

Lemma 2.10.4 If the point p is not contained in the grid G, then p is contained in exactly two lines disjoint with G and exactly one grid disjoint with G.

Proof. The points of G collinear with p form an ovoid of G, hence there are exactly (s + 3) − (s + 1) = 2 lines, say L_{1} and L_{2}, through p disjoint with G. Every grid through p disjoint with G contains L_{1} and L_{2}. If G′ is the unique grid through L_{1} and L_{2}, then G′ cannot satisfy property (1), (3) or (4) of the previous lemma (with respect to G). Hence G and G′ are disjoint.

□

Lemma 2.10.5 Let R be the following relation on the line set:

L R M ⟺ \{ \begin{matrix}L = M, or \\ (L ∩ M = ∅) and (L and M are contained in a grid), \end{matrix}

then R is an equivalence relation.

Proof. It suffices to prove that R is transitive. So, let L, M and N be lines such that L R M and M R N. We may suppose that L ≠ M and M ≠ N. Let G_{1} be the grid through L and M, and G_{2} the grid through M and N. We may suppose that G_{1} ≠ G_{2}. Let x be an arbitrary point of N and let G_{3} be the unique grid through x and L. If G_{2} and G_{3} meet in a line, then this line is necessary N. In this case N R L. So, suppose that G_{2} and G_{3} are not meeting in a line, then G_{2} and G_{3} intersect in an ovoid of G_{2}. Hence, there is a point y of M which belongs to G_{3}, a contradiction. □

For every line L of Q, let E_{L} be the equivalence class (with respect to R) to which L belongs.

Lemma 2.10.6 E_{L} is a normal spread of Q for every line L.

Proof. If two lines M_{1} and M_{2} of E_{L} would meet, then M_{1} R M_{2} implies that M_{1} = M_{2}. If x is a point of Q not on L, consider then the grid through x and L. The line in this grid through x and not meeting L belongs to E_{L}. Hence E_{L} is a spread. Clearly E_{L} is a normal spread. □

Remark. The GQ has (s + 3)(s + 1)^{2} lines and each spread E_{L} has (s + 1)^{2} lines, hence there are at least s + 3 normal spreads.

Let L be a line of Q and consider the normal spread E_{L}. The incidence structure A_{L} with

(i) point set equal to E_{L},

(ii) lines of the form {K, M}^{⊥⊥} with K, M ∈ E_{L} and K ≠ M,

(iii) natural incidence,

is then an S(2, s + 1, (s + 1)^{2}) and hence an affine plane of order s + 1. Let L_{1}, L_{2}, …, L_{s+3} be lines of Q such that E_{L_{1}}, …, E_{L_{s+3}} are mutually different equivalence classes. Let x be a point of Q and let i, j ∈ {1, …, s + 3}, i ≠ j. Through x there is a unique line of E_{L_{i}} and a unique line of E_{L_{j}}; these two lines determine a grid, which we denote by G(x, L_{i}, L_{j}). The equivalence class E_{L_{i}}, i ∈ {1, …, s+3}, defines an affine plane A_{i} and a projective plane P_{i}.

We can define P_{1} as follows. The points of P_{1} are of two types.

(I) Lines L with L R L_{1}.

(II) Lines L_{j} with j ≠ 1.

The lines of P_{1} are of two types.

(A) Grids G(p, L_{1}, L_{j}) with j ≠ 1 and p a point of Q.

(B) [∞]_{1}.

A point of type (I) is incident with a grid G(p, L_{1}, L_{j}) if and only if it belongs to this grid. A point of type (I) is never incident with the line [∞]_{1}. A point L_{j} of type (II) is incident with a grid G(p, L_{1}, L_{k}) if and only if j = k and it is always incident with [∞]_{1}.

Similarly, we can define the points and lines of P_{2} as follows. The points of P_{2} are of two types.

(I) Lines L with L R L_{2}.

(II) Lines L_{j} with j ≠ 2.

The lines of P_{2} are of two types.

(A) Grids G(p, L_{2}, L_{j}) with j ≠ 2 and p a point of Q.

(B) [∞]_{2}.

Now, let α be the following bijection from the point set of [∞]_{1} to the point set of [∞]_{2}:

L_{2} → L_{1},
L_{i} → L_{i} if i ≠ 1, 2.

Let β be the following bijection from the set of lines through L_{2} to the set of lines through L_{1}:

G(p, L_{1}, L_{2}) → G(p, L_{1}, L_{2}),
[∞]_{1} → [∞]_{2}.

We prove now that Q admits a model like the one given at the end of Section 2.10.1. Let p be an arbitrary point of Q. Through p, there is a unique line A of E_{L_{1}} and a unique line B of E_{L_{2}}. We make the following identification: p ↔ (A, B). Clearly (L_{2}A)^{β} = L_{1}B, where L_{2}A is the line of P_{1} through L_{2} and A and where L_{1}B denotes the line of P_{2} through L_{1} and B. Let L be an arbitrary line of Q and let x ∈ L be fixed. If L ∈ E_{L_{1}}, then we regard L as point of P_{1}. If L ∈ E_{L_{2}}, then we regard L as point of P_{2}. If L ∈ E_{L_{i}} with i ≠ 1, 2, then we make the following identification: L ↔ (G(x, L_{1}, L_{i}), G(x, L_{2}, L_{i})). The intersection of the line G(x, L_{1}, L_{i}) with the line [∞]_{1} is equal to the point L_{i} (in P_{1}). Similarly, the intersection of the line G(x, L_{2}, L_{i}) with the line [∞]_{2} is equal to L_{i} (in P_{2}). Clearly (L_{i})^{α} = L_{i}.

Now, let p be an arbitrary point of Q and let L be an arbitrary line of Q. Let p ↔ (A, B). If L ∈ E_{L_{1}}, then p I L if and only if A = L. If L ∈ E_{L_{2}}, then p I L if and only if B = L. Finally, if L ∈ E_{L_{i}}, i ≠ 1, 2, then L ↔ (G(x, L_{1}, L_{i}), G(x, L_{2}, L_{i})) with x an arbitrary point of L. If p I L, then A is a point of G(x, L_{1}, L_{i}) (in P_{1}) and B is a point of G(x, L_{2}, L_{i}) (in P_{2}). Conversely suppose that A and B are points of G(x, L_{1}, L_{i}) and G(x, L_{2}, L_{i}) respectively, then L meets A and B; hence p I L. This proves the theorem. □

Remark. With the same notations as above we have that {E_{L_{1}}, …, E_{L_{s+3}}} is a partition of the line set in spreads. Moreover, each spread E_{L_{i}} is pivotal, i.e. E_{L_{i}} is normal and for every M, N ∈ E_{L_{i}}, M ≠ N, {M, N}^{⊥} ⊆ E_{L_{j}} for some j ∈ {1, …, s + 3}. Theorem 2.10.1 follows then from Section 3 of [70].

Basic theory of near polygons

This chapter contains the basic definitions and properties about near polygons which we will use through this thesis. It also contains some theorems and constructions found by the author himself. For proofs and a more complete survey, we refer to the literature about near polygons ([10],[11],[12],[13],[14], [15],[17],[80],[82]).

3.1 Definitions and line-line relation

A near polygon is a connected partial linear space satisfying the following property.

(NP) For every point p and every line L, there exists a unique point q on L nearest to p (with respect to the distance in the point graph).

The point q is called the projection of p on L. If d is the diameter of the point graph, then the near polygon is called a near 2d-gon. Hence we will talk about near 0-gons, near 2-gons, near quadrangles, near hexagons, near octagons, etc. The near quadrangles are just the generalized quadrangles discussed in the previous chapter. The point graph of a near polygon determines uniquely the near polygon. For, every two adjacent vertices of such graph are contained in a unique maximal clique and these maximal cliques necessary correspond with the lines of the near polygon. Property (NP) describes the point-line relation. Next theorem describes the line-line relation.

Theorem 3.1.1 ([17]) One of the following two cases occurs for any two lines L and M of a near polygon.

(1) There exists a unique point p on L and a unique point q on M such that d(l, m) = d(l, p) + d(p, q) + d(q, m) for all points l on L and m on M.

(2) There exists an i ∈ ℕ such that d(l, M) = d(m, L) = i for all points l on L and m on M.

Two lines are called parallel (∥) if they satisfy condition (2) of the previous theorem. The relation ∥ is most of the time not an equivalence relation, see Section 3.3 for characterizations related to this. A near 2d-gon S is called regular if the following two conditions are satisfied.

(a) S has order (s, t).

(b) There are constants t_{i} (0 ≤ i ≤ d) such that for every two points x and y at distance i, there are exactly t_{i} + 1 points in Γ(x) ∩ Γ_{i−1}(y).

Clearly t_{0} = −1, t_{1} = 0 and t_{d} = t. The parameters s, t, t_{i} (i ∈ {2, …, d−1}) are called the parameters of the regular near polygon. If x is a point of a near polygon, then easily counting yields that

|Γ_{0}(x)| = 1,
|Γ_{k}(x)| = s^{k} \frac{∏_{i=0}^{k−1}(t − t_{i})}{∏_{i=1}^{k}(1 + t_{i})} for all i ∈ \{1, …, d\}.

The question whether a certain regular near polygon is completely determined by its parameters is always an interesting problem (see e.g. next section). A thin near polygon is a near polygon with the property that every line is incident with exactly two points; we can consider these geometries as graphs. In this point of view the thin near polygons are exactly the connected bipartite graphs. A dual grid is a thin near polygon. If S_{1} = (P_{1}, L_{1}, I_{1}) and S_{2} = (P_{2}, L_{2}, I_{2}) are two partial linear spaces, then the direct product of S_{1} and S_{2} is the partial linear space S = (P, L, I) with P = P_{1} × P_{2} and L = (P_{1} × L_{2}) ∪ (L_{1} × P_{2}). The point (x, y) is incident with the line (a, L) ∈ P_{1} × L_{2} if and only if x = a and y I_{2} L and it is incident with the line (M, b) ∈ L_{1} × P_{2} if and only if y = b and x I_{1} M. We denote S also with S_{1} × S_{2}. Let d_{1}(·, ·), d_{2}(·, ·) and d(·, ·) denote the distances in S_{1}, S_{2} and S respectively, then d((x_{1}, y_{1}), (x_{2}, y_{2})) = d_{1}(x_{1}, x_{2}) + d_{2}(y_{1}, y_{2}). Hence if S_{i} (i ∈ {1, 2}) is a near 2d_{i}-gon then the direct product S = S_{1} × S_{2} is a near 2(d_{1} +d_{2})-gon. Since S_{1} ×S_{2} ≃ S_{2} ×S_{1} and (S_{1} ×S_{2})×S_{3} ≃ S_{1} ×(S_{2} ×S_{3}), also the direct product of k ≥ 1 partial linear spaces S_{1}, …, S_{k} is welldefined.

Theorem 3.1.2 ([17]) Let S be a near polygon with the property that every two points at distance 2 have at least two common neighbours. If lines of different length occur, then S is the direct product of a number of near polygons, each of which has a constant length for the lines.

We also prove the following result about near polygons.

Theorem 3.1.3 Let S = (P, L, I) be a partial linear space of order (s, t) ≠ (s, 0) satisfying

(1) for every point p and every line L not through p, there exists at most one point on L collinear with p,

(2) a = |Γ_{2}(x)| is independent of the point x ∈ P,

(3) d(x, L) ≤ 2 for all x ∈ P and L ∈ L,

then b = |Γ_{3}(x)| is also independent of x ∈ P and the following inequalities hold:

• a ≥ s^{2}t,

• b ≤ s(a − s^{2}t).

Moreover, S is a generalized quadrangle if and only if a = s^{2}t and S is a near hexagon if and only if a > s^{2}t and b = s(a − s^{2}t).

Proof. Clearly |Γ_{3}(x)| = |P| − 1 − s(t + 1) − |Γ_{2}(x)| is independent of x ∈ P. Take an arbitrary line L and let r be a point of L. There are a points in Γ_{2}(r), s^{2}t of these are contained in Γ_{1}(L). Hence a ≥ s^{2}t and Γ_{2}(L) ≤ (s + 1)(a − s^{2}t). If a = s^{2}t then Γ_{2}(L) = ∅ implies that S is a generalized quadrangle. So, suppose that a ≠ s^{2}t, then S is a near hexagon if and only if Γ_{2}(L) = (s+1)(a−s^{2}t). From |Γ_{2}(L)| = |P|−(s+1)−st(s+1) = a+b−s^{2}t, it follows that b ≤ s(a − s^{2}t) and equality appears if and only if S is a near hexagon. □

3.2 Examples of some regular near polygons

3.2.1 Near polygons of Hamming type

A near polygon is of Hamming type if it is the direct product of a number of lines. This yields the following description. With a collection of d ≥ 1 finite sets X_{1}, …, X_{d}, there is associated the following near 2d-gon: the points are the elements of the cartesian product X_{1} × … × X_{d} and the lines are of the form {x_{i}} × … × {x_{i−1}} × X_{i} × {x_{i+1}} × … × {x_{d}}, with i ∈ {1, …, d} and x_{j} ∈ X_{j}, j ∈ {1, …, d} \ {i}. Incidence is the natural one. If all the sets X_{i} have the same size n, then the near polygon is regular with parameters s = n − 1, t = d − 1 and t_{i} = i − 1. The regular near polygons of Hamming type are uniquely determined by their parameters, see Corollary 9.2.3 of [15].

3.2.2 Dual polar spaces

In this section, we mean with a projective space an incidence structure such that any three noncollinear points generate a (possible degenerate) projective plane. It is called irreducible when there are no lines of length 2; otherwise it is called reducible. Now, let (P_{i})_{i∈I} be a family of irreducible projective spaces whose point sets are pairwise disjoint. Then the union P of their point sets carries the structure of a projective space whose lines are the lines on each P_{i} on the one hand and all pairs {x_{i}, y_{j}} with x_{i} ∈ P_{i}, y_{j} ∈ P_{j}, i ≠ j, on the other hand. P is called the direct sum of all the P_{i}’s. It is know that a reducible projective space is the direct sum of irreducible projective spaces.

Definition. A polar space of rank n ≥ 1 is a set together with a collection of subsets, called subspaces, satisfying the following axioms.

(P1) Any proper subspace, together with the subspaces it contains, is a projective space of dimension at most n−1. This projective space may be reducible.

(P2) The intersection of two subspaces is a subspace.

(P3) If L is a subspace of dimension n − 1 and p is a point outside L, then there is a unique subspace M through p such that dim(L ∩ M) = n − 2; it contains all points q of L with the property that there is a onedimensional subspace through p and q.

(P4) There exist two disjoint subspaces of dimension n − 1.

Polar spaces of rank 2 are precisely the nondegenerate generalized quadrangles. With each polar space of rank n ≥ 1, there is associated a dual polar space as follows. The points of the dual polar space are the maximal subspaces of the polar space (i.e. the subspaces of dimension n − 1), the lines are the next to maximal subspaces (i.e. the subspaces of dimension n − 2) and incidence is reverse containment. The near polygons of Hamming type are examples of dual polar spaces. For each of the following polar spaces, the corresponding dual polar space is a regular near 2n-gon.

(a) The polar space W (2n − 1, q) associated with a nonsingular symplectic polarity in PG(2n − 1, q), n ≥ 2. Let W^{D}(2n − 1, q) denote the corresponding dual polar space.

(b) The polar space Q(2n, q) associated with a nonsingular quadric in PG(2n, q), n ≥ 2. Let Q^{D}(2n, q) denote the corresponding dual polar space.

(c) The polar space Q^{−}(2n + 1, q) associated with a nonsingular elliptic quadric in PG(2n + 1, q), n ≥ 2. Let [Q^{−}(2n + 1, q)]^{D} denote the corresponding dual polar space.

(d) The polar space Q^{+}(2n−1, q) associated with a nonsingular hyperbolic quadric in PG(2n − 1, q), n ≥ 2. Let [Q^{+}(2n − 1, q)]^{D} denote the corresponding dual polar space.

(e) The polar space H(2n, q) associated with a nonsingular Hermitian variety in PG(2n, q), q square and n ≥ 2. Let H^{D}(2n, q) denote the corresponding dual polar space.

(f) The polar space H(2n − 1, q) associated with a nonsingular Hermitian variety in PG(2n − 1, q), q square and n ≥ 2. Let H^{D}(2n − 1, q) denote the corresponding dual polar space.

The parameters are s = q^{e}, t_{i} = q^{i−1} + … + q^{2} + q (2 ≤ i ≤ n), t = t_{n}, where e is for each of the dual polar spaces as shown in the table.

\begin{matrix}S & e \\ W^{D}(2n − 1, q) & 1 \\ Q^{D}(2n, q) & 1 \\ [Q^{−}(2n + 1, q)]^{D} & 2 \\ [Q^{+}(2n − 1, q)]^{D} & 0 \\ H^{D}(2n, q) & \frac{3}{2} \\ H^{D}(2n − 1, q) & \frac{1}{2} \end{matrix}

If a regular near polygon has these parameters, then it is a dual polar space, see Corollary 3.7.5 and the classification of classical near polygons given in Section 3.6. Note that the dual polar spaces W^{D}(2n − 1, q) and Q^{D}(2n, q) have the same parameters, but they are isomorphic if and only if q is even.

3.2.3 The near hexagon related to S(5, 8, 24)

Consider the Steiner system S(5, 8, 24) which we defined in Section 1.4.2. The following near hexagon is associated with this Steiner system: the points are the blocks of the Steiner system, the lines are the sets of three mutually disjoint blocks and incidence is the natural one. The near hexagon is regular with parameters s = 2, t_{2} = 2, t = 14 and has 759 vertices. Two points of the near haxagon have distance 0, 1, 2, respectively 3 if the corresponding blocks intersect in 8, 0, 4, respectively 2 points. The near hexagon is completely determined by its parameters, see [12]. This regular near hexagon has the following interesting characterization.

Theorem 3.2.1 ([17]) Let S be a regular near hexagon with parameters s, t_{2} and t. Suppose that s > 1 and t_{2} > 0. Then S is the near hexagon related to S(5, 8, 24) if and only if 1 + t = (t_{2} + 1)(1 + st_{2}).

3.2.4 The near hexagon related to the ternary Golay code

Consider the ternary Golay code which we described in Section 1.4.2. This code is a subspace of 𝔽_{3}^{12}. The following near hexagon is associated with this code, see [15]:

(1) the points are the cosets of the code,

(2) the lines are the triples of cosets {C_{1}, C_{2}, C_{3}}, such that C_{i} and C_{j} have representatives which differ in only one position (C_{1}, C_{2} and C_{3} differ then in the same position).

(3) incidence is the natural one.

This near hexagon is regular with parameters s = 2, t_{2} = 1 and t = 11. The near hexagon is completely determined by its parameters, see [11]. This near hexagon has a linear representation T_{5}^{∗}(K) in AG(6, 3), where K is the Coxeter cap in PG(5, 3), see Chapter 6.

3.2.5 The near octagon related to the Hall-Janko group

The following near octagon was first constructed in [23]. Consider the sporadic simple Hall-Janko group of order 604800. Let V_{0} be the set of all 315 involutions, whose centralizer contains Sylow 2-subgroups, i.e. groups of order 128. Let Γ be the graph with vertex set V_{0}, two involutions being adjacent when they commute, or, equivalently, when their product is again an element of V_{0}. The graph Γ is then the point graph of a regular near octagon (the lines correspond with the maximal cliques). The parameters are s = 2, t_{2} = 1, t_{3} = 3 and t = 4 and the near octagon is uniquely determined by its parameters, see [24].

3.3 Characterizations of the Hamming near polygons in terms of parallellism

Recall that two lines L and M of a near polygon are called parallel if d(l, M) is independent of the point l I L.

Lemma 3.3.1 Let S_{i}, 1 ≤ i ≤ k, be a number of near polygons and let S be their direct product. Then parallelism (∥) is an equivalence relation in S if and only if it is an equivalence relation in each S_{i}.

Proof. We may suppose that k = 2, otherwise one can use induction. The result follows now by looking at the following remarks, which are readily obtained. (Here p, q denote points and L, M denote lines.)

• The lines (p, M) and (L, q) are never parallel.

• The lines (p, M) and (q, L) are parallel if and only if M and L are parallel.

• The lines (L, q) and (M, p) are parallel if and only if L and M are parallel.

□

Theorem 3.3.2 If S is a near polygon, then the following conditions are equivalent.

(1) S is a near polygon of Hamming type.

(2) Parallelism is an equivalence relation and every two points at mutual distance 2 have at least two common neighbours.

(3) For every point x and every line L, there is a unique line through x which is parallel to L.

Proof.

(1) ⇒ (2). Trivial.

(2) ⇒ (3). If there were two lines M and M′ through x parallel to L, then M∥M′, a contradiction. We only need to show that there is at least one line through x parallel to L. Since S is connected and ∥ is an equivalence relation, we may suppose that d(x, L) = 1. Let x′ be the unique point of L which is collinear with x and let y′ be any other point of L. If y is a common neighbour of x and y′, different from x, then xy∥L.

(3) ⇒ (1). We prove by induction that for every two points at mutual distance i, there are exactly i points in Γ_{i−1}(x) ∩ Γ(y). This is true for i = 0 and i = 1, so suppose i ≥ 2. Let z be a neighbour of x having distance i − 1 to y. Let y_{1}, …, y_{i−1} be the i − 1 neighbours of y having distance i − 2 to z. Let N be the line through y parallel with xz and y_{i} be the point on this line having distance i − 1 to x, then y_{j} (1 ≤ j ≤ i) are i different points collinear with y and having distance i − 1 to x. If y_{i+1} would be another point having these properties, then yy_{i+1}∥xz, so yy_{i+1} = yy_{i} and y_{i} = y_{i+1}, a contradiction. If all lines of S have a constant number of points, then the near polygon is regular and has the same parameters as a regular near polygon of Hamming type. Hence, S is a regular near polygon of Hamming type. If not all lines of S have a constant number of points, then the theorem follows from Lemma 3.3.1 and Theorem 3.1.2.

□

3.4 The nondegenerate support of a near polygon

A partial linear space is called degenerate if there is a point incident with exactly one line. Every such point is called superthin. Notice that the definition of degeneracy is not conform to other definitions given in the literature (e.g. degenerate polar spaces); however, this yields no confusion in the sequel. Although a near 0-gon has only one point and no lines, it is by our definition not degenerate. A near 2-gon consists of one line with a number of points on it, hence it is degenerate. The degenerate near quadrangles are just the degenerate generalized quadrangles.

Let S = (P, L, I) be a partial linear space. We give now three constructions slightly altering the structure of S. Constructions (1) and (2) increase the ”amount” of degeneracy, while the inverse construction (3) reduces this amount. Construction (3) can then be used to define the nondegenerate support of S.

(1) Assume p ∈ P, then we denote by S_{p} = (P_{p}, L_{p}, I_{p}) the following partial linear space:

P_{p} = P ∪ {(∞)}, where (∞) ∉ P;
L_{p} = L ∪ {[∞]}, where [∞] ∉ L;
I_{p} = I ∪ {(p, [∞]), ((∞), [∞])}.

(2) Assume L ∈ L, then we denote by S_{L} = (P_{L}, L_{L}, I_{L}) the following partial linear space:

P_{L} = P ∪ {(∞)}, where (∞) ∉ P;
L_{L} = L;
I_{L} = I ∪ {((∞), L)}.

(3) Assume that p ∈ P is a superthin point incident with the line L, then we denote by S^{p} = (P^{p}, L^{p}, I^{p}) the following partial linear space:

P^{p} = P \ {p};
L^{p} = L \ {L} if L has length 2, while L^{p} = L in the other case;
I^{p} = I ∩(P^{p} × L^{p}).

Suppose that p_{1} is a superthin point of a (finite) partial linear space S. If p_{2} is a superthin point of S^{p_{1}}, then (S^{p_{1}})^{p_{2}} is again a partial linear space. Repeating this argument, one finally finds a nondegenerate partial linear space (… (S^{p_{1}}) …)^{p_{l}}, which is called the nondegenerate support of S. We prove now that this is a good definition, i.e. that it is independent of the order in which the superthin points are considered. Proof. We may suppose that S is connected, otherwise one can reason on each connected component. Suppose that we obtain two nondegenerate partial linear spaces S′ = (… (S^{p_{1}}) …)^{p_{k}} and S′′ = (… (S^{q_{1}}) …)^{q_{l}}. Here p_{1} and q_{1} are superthin points of S and p_{i} (1 < i ≤ k) and q_{j} (1 < j ≤ l) are superthin points of (… (S^{p_{1}}) …)^{p_{i−1}} and (… (S^{q_{1}}) …)^{q_{j−1}} respectively. Suppose p_{i} ∉ {q_{1}, …, q_{l}} with i the smallest integer with that property, then p_{i} is not a superthin point of S′′. If p_{i} is contained in two lines L and M of S′′, then at least one of these lines, say L, is not a line of (… (S^{p_{1}}) …)^{p_{i−1}}. Let r ≠ p_{i} be a point of S′′ on L. Then r ∉ {q_{1}, …, q_{l}} and r = p_{i′} with i′ < i, a contradiction. Hence S′′ is the near 0-gon with unique point p_{i} (since there are no lines through p_{i}). Hence we may assume that S′ is not isomorphic to a near 0-gon and using the same reasoning as above, we may conclude that q_{j} ∈ {p_{1}, …, p_{k}} for all j ∈ {1, …, l}. This implies however that P − {p_{i}} = {q_{1}, …, q_{l}} ⊆ {p_{1}, …, p_{k}}, a contradiction. Hence {p_{1}, …, p_{k}} = {q_{1}, …, q_{l}} = A. Now S′ and S′′ have the same points (namely the points not in A) and the same lines (namely the lines of S having at least two points not in A), which implies that S′ and S′′ are isomorphic. □

If p is a superthin point of a near polygon S, then S^{p} is again a near polygon. Hence, the nondegenerate support of a near polygon is again a near polygon. If p is an arbitrary point of a near polygon S, then S_{p} is also a near polygon. Under certain circumstances, also construction (2) yields a near polygon.

Lemma 3.4.1 Let S = (P, L, I) be a near polygon. If L ∈ L is a line which is not parallel to any other line M ≠ L, then S_{L} is a near polygon.

Proof. Choose a point p and a line M of S_{L}. If p ≠ (∞), then the distance between p and any point of M is not changed by adding (∞) to P. If p = (∞), we may suppose that M ≠ L. Let α and β be the unique points of L and M such that d(m, l) = d(l, α) + d(α, β) + d(β, m) holds for all points l of L and m of M. The point β is now the unique point of M nearest (in S_{L}) to (∞). □

Corollary 3.4.2 If p ∈ P is a superthin point which is incident with the line L, then S_{L} is a near polygon.

Proof. If p is a superthin point of a near polygon which is incident with L, then every line M ≠ L is non-parallel to L. Indeed, if (p, q, …) is a shortest path from p to M, then p, q ∈ L and d(q, M) = d(p, M) − 1. □

3.5 Quads

3.5.1 Special point sets

In this section S = (P, L, I) denotes an arbitrary partial linear space. A subset X of P is called a subspace if all the points of a line L are in X as soon as two of its points are in X. Every such X induces a partial linear space S_{X} = (X, L′, I′). Here L′ denotes the set of lines having all their points in X and I′=I ∩(X ×L′). A nonempty subspace X is called Cameron closed if S_{X} is connected and if all the common neighbours of two different points of X are as well contained in X. A subset X of P is called geodetically closed if X is a subspace and if all the points of a shortest path between two points of X are as well contained in X. The set X is called a quad if X is geodetically closed and if X induces a nondegenerate generalized quadrangle. Since no confusion is possible in the sequel, this generalized quadrangle will also be called a quad. Finally, we call a subset X of the point set classical if for every point x there is a unique point π_{X}(x) ∈ X such that d(x, y) = d(x, π_{X}(x)) + d(π_{X}(x), y) for all y ∈ X. One easily checks that every classical set X is geodetically closed. We give now some properties of quads in the case that S is a near polygon.

Theorem 3.5.1 ([80] and [82]) If there is a quad through two points at distance 2, then this quad is unique. As a consequence, every two different quads are either disjoint or meet in a point or line of the near polygon.

Concerning the existence of quads, one can say the following.

Theorem 3.5.2 (Theorem of Yanushka ([80] and [82])) Let x and y be two points of a near polygon S at mutual distance 2. If x and y have two common neighbours c and d such that the line xc contains at least three points, then x and y are in a unique quad. This quad contains these points of the near polygon which have distance at most 2 from x, y, c and d.

We already have given the point-line relation (see property (NP)) and the line-line relation (see Theorem 3.1.1), we give now the point-quad relation.

Theorem 3.5.3 ([80] and [82]) Let p be a point of a near polygon S and let Q be a quad, then exactly one of the following cases occurs.

(1) There is a unique point q ∈ Q such that d(p, r) = d(p, q) + d(q, r) for all points r ∈ Q. In this case (p, Q) is called classical. The point q is called the projection of p on Q and is denoted by π_{Q}(p), or shortly by π(p).

(2) The points of Q which are nearest to p form an ovoid of Q. In this case (p, Q) is called ovoidal.

(3) The quad Q induces a dual grid. Let A be the set of points of Q at smallest distance k from p. Let B, respectively C, denote those points of Q, that have distance k + 1, respectively k + 2 to p. Then (a) |A| ≥ 2 and |C| ≥ 1,

(b) B and A ∪ C are the two maximal cocliques of the point graph of Q.

In this case (p, Q) is called thin ovoidal.

3.5.2 Local spaces

The notion of local space was already defined for near polygons which have quads through every two points at distance two. We slightly modify this definition such that local spaces exist for any near polygon. With this new definition, a local space is not necessary a linear space.

Let x be a point of a near polygon S. For a quad Q through x, let L_{Q} denote the set of lines through x which are also lines of Q. The following incidence structure S_{x}, called the local space at x, can then be defined:

• the point set of S_{x} is the set of lines of S through x;

• the line set of S_{x} is equal to {L_{Q}|Q is a quad through x};

• incidence is the natural one, i.e containment.

By Theorem 3.5.1, S_{x} is a partial linear space. If there are quads through every two points at distance 2, then S_{x} is a linear space.

3.6 Dual polar spaces and classical near polygons

Definition. A near (2d)-gon, d ≥ 1, is called classical if the following two conditions are satisfied:

(i) every two points at distance 2 are contained in a quad;

(ii) every point-quad relation is classical.

We have the following important theorem.

Theorem 3.6.1 ([22]) The dual polar spaces are precisely the classical near polygons.

Theorem 3.6.2 Let S be a near hexagon satisfying the following conditions:

(1) every two points at distance 2 are contained in a quad,

(2) all local spaces are (possible degenerate) projective planes,

then S is a classical near hexagon.

Proof. Let x be a point of S and let Q be a quad of S such that (x, Q) is not classical, then d(x, Q) = 2 and (x, Q) is ovoidal. Let y be a point of Q at distance 2 from x and let Q′ be the quad through x and y. The quads Q and Q′ determine now two mutually disjoint lines of the local space S_{y}, a contradiction. □

Given a number of classical near polygons then many others can be constructed. For, if S_{1} and S_{2} are classical near polygons, then S_{1} × S_{2} is also a classical near polygon. Given a number of polar spaces, then many others can be constructed as shown now. Let (Γ_{i})_{i∈I} be a family of polar spaces defined on the sets (P_{i})_{i∈I} and with (A_{i})_{i∈I} as collection of subspaces. We suppose that all the P_{i}’s are mutually disjoint. A new polar space, called the direct sum can then be constructed on the set P = ⋃_{i∈I} P_{i}. A general form of a subspace is as follows: ∪_{i∈I}a_{i}, where a_{i} ∈ A_{i}. The connection between the two constructions is given in the following theorem which is easy to verify.

Theorem 3.6.3 Let Γ_{1} and Γ_{2} be two polar spaces and let S_{1} and S_{2} be the dual polar spaces related to it. Let Γ be the direct sum of Γ_{1} and Γ_{2} and let S be the dual polar space related to it, then S ≃ S_{1} × S_{2}.

A polar space is called irreducible if it is not isomorphic to a direct sum of at least two polar spaces of rank ≥ 1. The classification of classical near polygons is equivalent to the classification of irreducible polar spaces of rank n ≥ 2. If n ≥ 3 and if all lines have length at least three, then the irreducible polar space is isomorphic to W^{D}(2n − 1, q), Q^{D}(2n, q), [Q^{−}(2n + 1, q)]^{D}, [Q^{+}(2n − 1, q)], H^{D}(2n, q) or H^{D}(2n − 1, q) ([94]). The case where there are some lines of length 2 is settled in the next theorem.

Theorem 3.6.4 ([19]) If Γ is an irreducible polar space of finite rank having some line of length 2, then Γ is a dualized projective space.

Remains to define what a dualized projective space ([19]) is.

Definition. Let P be an irreducible projective space of dimension d ≥ 1 and let P^{∗} be the dual of P whose points are the (d − 1)-dimensional subspaces of P and whose lines are the (d − 2)-dimensional subspaces of P. If d = 1, then we assume that P^{∗} is a copy of P disjoint from it. Consider now the following incidence structure S = (P, L, I). The point set P is the union of the point sets of P and P^{∗}. The lines have length 2 and look like {x, y}, where x is a point of P and where y is a hyperplane through x. The set P together with the subspaces of the geometry S form then a polar space of rank d + 1. Dual polar spaces obtained in the above described way are called dualized projective spaces.

3.7 Near polygons with quads

Let S = (P, L, I) be a near polygon such that

(1) every line has at least three points,

(2) every two points at distance 2 have at least two common neighbours.

By Yanushka’s Theorem (Theorem 3.5.2), every two points at distance 2 are contained in a unique quad. In the sequel, let Q denote a fixed quad. If x ∈ P, then the following possibilities can occur, see Theorem 3.5.3.

(a) (x, Q) is classical, in this case there is a unique point π(x) in Q nearest to x;

(b) (x, Q) is ovoidal.

We define now the following sets:

N_{i} := {x ∈ P| d(x, Q) = i},
N_{i,C} := {x ∈ N_{i}|x is classical with respect to Q},
N_{i,O} := {x ∈ N_{i}|x is ovoidal with respect to Q}.

Clearly N_{0} = Q, N_{1,O} = ∅, N_{d−1,C} = ∅ and N_{d} = ∅.

Theorem 3.7.1 ([17], p. 148) (1) If L is a line contained in N_{i} ∪ N_{i+1} for some i, then |L ∩ N_{i}| = 1.

(2) There are no edges between N_{i,O} and N_{i,C} for all i.

(3) If L is a line contained in N_{i,C} for some i, then π(L) is a line of Q parallel to L.

(4) If L is a line contained in N_{i,O} for some i, then the points of L determine a fan of ovoids in Q.

(5) If L is contained in N_{i,O} ∪ N_{i+1} for some i, then L is even contained in N_{i,O} ∪ N_{i+1,O}. In this case all points of L determine the same ovoid of Q.

(6) If L is contained in N_{i,C} ∪ N_{i+1,C} for some i, then all points of L determine the same point in Q.

(7) If L is contained in N_{i,C}∪N_{i+1,O} for some i, then the points of L∩N_{i+1,O} determine a rosette of ovoids in Q. The common point of all these ovoids is the point of Q determined by L ∩ N_{i,C}.

Let x be a fixed point of S. For every point y ∈ P, we define

S(x, y) = {L|x I L and L contains a point at distance d(x, y) − 1 from y}.

Theorem 3.7.2 ([17], Lemma 16 (i)) S(x, y) is a subspace of the local space S_{x} for all points y ∈ P.

Theorem 3.7.3 ([17], Theorem 4) Let x and y be two points at distance i, then there is a unique geodetically closed sub near 2i-gon H(x, y) containing x and y. Moreover

H(x, y) = {u|S(x, u) ⊆ S(x, y)}.

Suppose now that S is regular with parameters s, t, t_{i} (0 ≤ i ≤ d). The above conditions say that s ≥ 2 and t_{2} ≥ 1.

Theorem 3.7.4 ([17], p. 158)

|N_{i,C}| = (s + 1)(1 + st_{2})s^{i} \frac{∏_{j=2}^{i+1}(t − t_{j})}{∏_{j=2}^{i}(1 + t_{j})} for all i ∈ \{1, …, d − 1\},

|N_{i,O}| = s^{i}(s + 1)[t_{i+1} − t_{2}(t_{i} + 1)]\frac{∏_{j=2}^{i}(t − t_{j})}{∏_{j=2}^{i}(1 + t_{j})} for all i ∈ \{2, …, d − 1\}.

Corollary 3.7.5 (1) t_{i+1} ≥ t_{2}(t_{i} + 1) for all i ∈ {0, …, d − 1}.

(2) S is classical if and only if t_{i+1} = t_{2}(t_{i} + 1) for all i ∈ {0, …, d − 1}.

3.8 Near polygons with good quads

Definition. A quad Q of a near polygon is called good if every point of Q is incident with the same number of lines; we denote this number by t_{Q} + 1. Only quads isomorphic to nonsymmetrical dual grids are not good.

Theorem 3.8.1 ([17], Lemma 19) Let S be a near polygon with the property that every two points at distance 2 are contained in a good quad, then each point of S is incident with the same number of lines.

Proof. Let x and y be two collinear points. The point x (respectively y) is incident with t_{x} + 1 (respectively t_{y} + 1) lines. Now

t_{x} + 1 = 1 + ∑ t_{Q} = t_{y} + 1,

where the summation ranges over all quads Q through the line xy. Hence x and y are incident with the same number of lines and the result follows by connectedness of S. □

Remark. If S is a dualized projective space, then every two points at distance two are contained in a quad, but not every point is incident with the same number of lines; hence the condition that quads must be good is necessary.

If S = (P, L, I) is a near 2-gon or a good GQ, then |Γ_{i}(p)| (i ∈ {0, 1, 2}) is independent of p ∈ P. We derive a similar property for near hexagons.

Theorem 3.8.2 Let S = (P, L, I) be a near hexagon such that every two points at distance 2 are contained in a good quad, then |Γ_{i}(p)| (i ∈ {0, 1, 2, 3}) is independent of p ∈ P.

Proof. If not all lines of S are incident with the same number of points, then Theorem 3.1.2 implies that S is the direct product of a line with a good GQ. It is straightforward to check that the result is true in this case. Hence we may suppose that all lines are incident with s + 1 points. The previous theorem implies then that S has an order (s, t). Now, let p ∈ P be a fixed point and put n_{i} = |Γ_{i}(p)|. Then n_{0} = 1, n_{1} = s(t + 1). Let V be the set of quads through p. Counting points in Γ_{2}(p) we find

n_{2} = s^{2} ∑_{x∈V} t_{x}.  (3.1)

Counting edges between Γ_{2}(p) and Γ_{3}(p) we find that

n_{3}(t + 1) = s^{3} ∑_{x∈V} t_{x}(t − t_{x}).  (3.2)

Finally, counting triples (L_{1}, L_{2}, Q) where L_{1}, L_{2} are two different lines through p and Q is the quad through L_{1} and L_{2}, yields

t(t + 1) = ∑_{x∈V} t_{x}(t_{x} + 1).  (3.3)

Eliminating ∑ t_{x} and ∑ t_{x}^{2}, we find that n_{3} = s(n_{2} − s^{2}t). Together with v = n_{0} + n_{1} + n_{2} + n_{3} this gives

n_{2} = \frac{v}{s + 1} − 1 + st(s − 1),  (3.4)
n_{3} = s(\frac{v}{s + 1} − st − 1).  (3.5)
□

Corollary 3.8.3 If S is a near hexagon satisfying the property that every two points at distance 2 are contained in a quad of order (s, t_{1}) or (s, t_{2}), s ≥ 1 and 1 ≤ t_{1} < t_{2}, then for each i ∈ {1, 2}, the number of quads of order (s, t_{i}) through a point is independent of that point.

Proof. This follows from equations (3.1), (3.3) and (3.4). □

Remark. The previous corollary was proved in [14] in the case that s = 2, t_{1} = 1, t_{2} = 2 by using the same double countings as in the proof of Theorem 3.8.2.

3.9 Some characterizations for classical and regular near hexagons

In this section we suppose that S is a near hexagon of order (s, t) satisfying the property that every two points at distance 2 are contained in a good quad. Recall that every point of a good quad Q is incident with the same number of lines, this number being denoted by t_{Q} + 1. Let N_{x} be the total number of quads through a fixed point x, let v denote the total number of points and put M := \frac{n_{2}}{s^{2}} = \frac{1}{s^{2}} (\frac{v}{s+1} − 1 + st(s − 1)). For a point x of S, we define T_{x} + 1 := max |Γ(x) ∩ Γ(y)|, where y ranges over all n_{2} elements of Γ_{2}(x).

\begin{matrix}Theorem 3.9.1 \\ (i) \end{matrix} If α, β ∈ ℝ, then

∑(t_{Q}^{2} + αt_{Q} + β) = t(t + 1) + (α − 1)M + βN_{x},

where the summation ranges over all quads Q through x.

(ii) The inequality t(t + 1) ≥ 2M holds, with equality if and only if every quad is a grid.

(iii) One has the following lower bound for N_{x}:

N_{x} ≥ \frac{M^{2}}{t(t + 1) − M} ,

and equality holds if and only if every quad through x has the same order. Hence, S is a regular near hexagon if and only if the equality holds for every point of the near hexagon.

(iv) The inequality M ≥ t + 1 holds, with equality if and only if S is a Hamming near hexagon.

(v) The inequality [M −(t+1)]T_{x} ≥ t(t+1)−2M holds with equality if and only if the local space at x is a (possible degenerate) projective plane. If S is not a Hamming near hexagon, then the inequality is equivalent to

T_{x} ≥ \frac{t(t + 1) − 2M}{M − (t + 1)} .

Hence S is a classical near hexagon if and only if the equality holds for every point x of the near hexagon.

Proof.

(i) The result follows from equations (3.1), (3.3) and (3.4).

(ii) Counting edges between Γ(x) and Γ_{2}(x) yields

[s(t + 1)][st] ≥ 2n_{2} = 2s^{2}M,

and equality holds if and only if |Γ(x) ∩ Γ(y)| = 2 for every point y at distance 2 from x. The result follows now immediately.

(iii) From (i) it follows that ∑(t_{Q} − γ)^{2} = N_{x}γ^{2} − (2M)γ + t(t + 1) − M. If not all quads through x have the same order, then the above quadratic equation in γ cannot have a real root, hence the discriminant D = M^{2} + N_{x}M − t(t + 1)N_{x} must be negative. If all quads through x have order (s, t_{2}), then one calculates that N_{x} = \frac{t(t+1)}{t_{2}(t_{2}+1)}, M = \frac{t(t+1)}{1+t_{2}} and hence D = 0.

(iv) For every point y ∈ Γ_{2}(x), one has that |Γ(x) ∩ Γ(y)| ≤ t. Counting edges between Γ(x) and Γ_{2}(x) yields

[s(t + 1)][st] ≤ [s^{2}M]t,

from which the above inequality follows. If equality holds, then S is a regular near polygon with t_{2} = t − 1. The number of quads through a line is then equal to \frac{t}{t_{2}} = \frac{t}{t−1}, hence t = 2, t_{2} = 1 and S is a Hamming near hexagon.

(v) Applying the Fundamental Theorem for linear spaces, see [26], to the local space at x yields N_{x} ≥ t + 1. We have now the following inequalities:

0 ≥ ∑(t_{Q} − 1)(t_{Q} − T_{x})
= ∑(t_{Q}^{2} − (T_{x} + 1)t_{Q} + T_{x})
= t(t + 1) − (T_{x} + 2)M + T_{x}N_{x}
≥ t(t + 1) − (T_{x} + 2)M + T_{x}(t + 1),

from which the above equality readily follows. Suppose now that equality holds, then t_{Q} is equal to 1 or T_{x} for every quad through x and the local space at x is a possibly degenerate projective plane. Hence equality holds if and only if the local space at x is a (possible degenerate) projective plane and the theorem follows by Theorem 3.6.2.

□

3.10 A construction method for near polygons

The result of the following theorem is mentioned in [14] in the case where X_{1} and X_{2} are both classical.

Theorem 3.10.1 Let S_{1} = (P_{1}, L_{1}, I_{1}) and S_{2} = (P_{2}, L_{2}, I_{2}) be two near polygons. Let X_{1}, respectively X_{2}, be a classical subset of P_{1}, respectively a subspace of S_{2}. Denote by d_{i}(·, ·), i ∈ {1, 2} the distance in S_{i}. Suppose that there is a bijection α : X_{1} ↦ X_{2} such that d_{2}(α(x), α(y)) = d_{1}(x, y) for all points x, y ∈ X_{1}. Then the partial linear space obtained by identifying the corresponding points and lines in X_{1} and X_{2} is a near polygon, which we denote by S[S_{1}, S_{2}, X_{1}, X_{2}, α].

Proof. Choose in S[S_{1}, S_{2}, X_{1}, X_{2}, α] a point p and a line L. We want to prove that there is a unique point on L nearest to p. We define d(·, ·) as the distance in S[S_{1}, S_{2}, X_{1}, X_{2}, α]. We consider three cases.

(1) Assume that p and L both belong to the same near polygon S_{i} (i = 1, 2). As d_{2}(α(x), α(y)) = d_{1}(x, y) for all points x, y ∈ X_{1}, there exists for any two points a and b from P_{i} (i = 1, 2) a shortest path from a to b only containing points of S_{i} and hence d(a, b) = d_{i}(a, b) (i = 1, 2). This implies that L contains a unique point nearest to p.

(2) Assume that p ∈ P_{1} while L ∈ L_{2}. Let l_{1}, l_{2} be two points such that d(p, l_{1}) = d(p, l_{2}) = d(p, L). Since X_{1} is a classical subset, there exists a shortest path from p to l_{i}, that contains the point π_{X_{1}}(p). Hence l_{1} = l_{2} is the unique point of L nearest to π_{X_{1}}(p) (in S_{2}).

(3) Finally assume p ∈ P_{2} while L ∈ L_{1}. Let l_{1} and l_{2} be two different points of L such that d(p, l_{1}) = d(p, l_{2}) = d(p, L). Define s_{i} = π_{X_{1}}(l_{i}). Since X_{1} is a classical subset, there exists a shortest path from p to l_{i} containing the point s_{i}. If s_{1} = s_{2} then l_{1} and l_{2} are two different points on L, and both are nearest to s_{1}, a contradiction. If s_{1} ≠ s_{2} then, since X_{1} is classical, d(s_{1}, s_{2}) = 1 and d(l_{1}, s_{1}) = d(l_{2}, s_{2}). Denote by s_{3} the unique point of s_{1}s_{2} nearest to p, then s_{1} ≠ s_{3} ≠ s_{2}. Let l_{3} denote the unique point of L, for which d(l_{1}, s_{1}) = d(l_{2}, s_{2}) = d(l_{3}, s_{3}). Now, there exists a path of length d(p, l_{2}) − 1 from p to l_{3}, a contradiction.

□

Remark. In Section 5.2 we will describe a class of near polygons which can be obtained by successive application of the previous theorem.

Association schemes associated with near polygons

4.1 Association schemes and distance regular graphs

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

(1) R_{0} = {(x, x)|x ∈ X}.

(2) R_{i}^{T} ∈ {R_{0}, …, R_{n}} for all i ∈ {0, …, n} (R_{i}^{T} = {(y, x)|(x, y) ∈ R_{i}}).

(3) There exist integers p_{jk}^{i}, called the parameters of the association scheme, 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 (z, y) ∈ R_{k}.

If R_{i}^{T} = R_{i}, for all i ∈ {0, …, n}, then the association scheme is called symmetric and we have p_{jk}^{i} = p_{kj}^{i} for all i, j, k ∈ {0, …, d}. The number n_{i} = p_{ii}^{0} is called the valency of R_{i}. For more details on association schemes we refer to [3] or chapter 2 of [15].

Let Γ be a graph of diameter d and let d(·, ·) be the corresponding distance function. The graph Γ is called distance regular if and only if there exist parameters a_{i}, b_{i}, c_{i} (i ∈ {0, …, d}) such that the following equalities hold for every two vertices x and y of Γ (we put i = d(x, y) in each of these equalities):

(1) |Γ_{i}(x) ∩ Γ(y)| = a_{i}, (2) |Γ_{i+1}(x) ∩ Γ(y)| = b_{i},

(3) |Γ_{i−1}(x) ∩ Γ(y)| = c_{i}.

We call k := b_{0} the valency of Γ. Clearly a_{0} = 0, c_{0} = 0, c_{1} = 1, b_{d} = 0 and a_{i} + b_{i} + c_{i} = k for all i ∈ {0, …, d}. A distance regular graph of diameter 2 is called a strongly regular graph.

The relation between distance regular graphs and association schemes is given by the following theorem.

Theorem 4.1.1 (([15])) Suppose that Γ is a distance regular graph of diameter d and vertex set X. Let R_{i} = {(x, y) ∈ X × X| d(x, y) = i} for all i ∈ {0, …, d} and let R = {R_{0}, …, R_{d}}, then (X, R) is a symmetric association scheme.

The following known result gives the relation between distance regular graphs and regular near polygons.

Theorem 4.1.2 A near polygon is regular if and only if its point graph is a distance regular graph.

Proof. Let S be a regular near (2d)-gon with parameters s, t_{i} (i ∈ {0, …, d}) and t = t_{d}. Choose i ∈ {0, …, d} and consider two points x and y at distance i from each other. Put a_{i} = |Γ_{i}(x) ∩ Γ(y)|, b_{i} = |Γ_{i+1}(x) ∩ Γ(y)| and c_{i} = |Γ_{i−1}(x) ∩ Γ(y)|. We will prove that a_{i}, b_{i} and c_{i} are only dependent on i. Clearly c_{i} = t_{i} +1 and a_{i} +b_{i} +c_{i} = s(t+1). Now, if z ∈ Γ_{i}(x)∩Γ(y), then yz contains a unique point at distance i − 1 from x. There are now t_{i} + 1 lines through y containing a point at distance i − 1 from x and on each of these lines, there are s − 1 points in Γ_{i}(x) ∩ Γ(y). Hence a_{i} = (s − 1)(t_{i} + 1) and b_{i} = s(t + 1) − a_{i} − c_{i} = s(t − t_{i}). Conversely, suppose that Γ is the distance regular point graph of a near 2d-gon S. Every two adjacent vertices of Γ are then contained in a unique clique of size a_{1} + 2. Hence, every line of S has length s + 1 = a_{1} + 2. Clearly t_{i} + 1 = c_{i} for all i ∈ {0, …, d}. As a consequence S is regular. □

Corollary 4.1.3 With every regular near polygon, there is associated a symmetric association scheme.

If we have an association scheme (distance regular graph) with certain parameters, then there are conditions on the parameters of this association scheme (distance regular graph), see [3] and [15]. For instance, in the case of a regular near hexagon with parameters (s, t_{2}, t), s > 1, one of the Krein conditions yields the inequality

t ≤ s^{3} + t_{2}(s^{2} − s + 1).  (4.1)

This inequality is known as the Mathon bound (see also [43]).

Theorem 4.1.4 ([80]) Let S be a near hexagon with parameters s = 2, t_{2} and t. Then we have the following possibilities for (t_{2}, t):

\begin{matrix}(1) (t_{2}, t) = (0, 1), & (2) (t_{2}, t) = (0, 2), \\ (3) (t_{2}, t) = (0, 8), & (4) (t_{2}, t) = (1, 2), \\ (5) (t_{2}, t) = (1, 11), & (6) (t_{2}, t) = (2, 6), \\ (7) (t_{2}, t) = (2, 14), & (8) (t_{2}, t) = (4, 20). \end{matrix}

For every of these parameters, there is a unique near hexagon, except for case (2) where there is up to duality only one. The near hexagons with parameters as in (1), (2) or (3) are generalized hexagons. The near hexagons with parameters as in (4), (6) or (8) satisfy t = t_{2}(t_{2} + 1) and hence they are dual polar spaces. One can find a description of these near polygons in Chapter 3. The near hexagon with parameters (s, t_{2}, t) = (2, 1, 11) is the one related to the extended ternary Golay code. Finally, the near hexagon with parameters (s, t_{2}, t) = (2, 2, 14) is the one related to the Steiner system S(5, 8, 24).

Theorem 4.1.5 ([80]) Let S be a near hexagon with parameters s = 3, t_{2} and t. Then we have one of the following possibilities for (t_{2}, t):

\begin{matrix}(1) (t_{2}, t) = (0, 1), & (2) (t_{2}, t) = (0, 3), \\ (3) (t_{2}, t) = (0, 27), & (4) (t_{2}, t) = (1, 2), \\ (5) (t_{2}, t) = (1, 9), & (6) (t_{2}, t) = (1, 34), \\ (7) (t_{2}, t) = (3, 12), & (8) (t_{2}, t) = (3, 27), \\ (9) (t_{2}, t) = (3, 48), & (10) (t_{2}, t) = (9, 90). \end{matrix}

Near polygons with parameters as in (1), (2) or (3) are generalized hexagons. Except for case (1) (unique example), it is not known whether they are uniquely determined by their parameters. Near hexagons with parameters as in (4), (7) or (10) are dual polar spaces (since t = t_{2}(t_{2} + 1)). We refer to Chapter 3 for a description of these near hexagons. There is no near hexagon with parameters as in (5), see [10]. Whether there exists a near polygon with parameters as in (6) is still an open problem. It is also known that there exists no near hexagon with parameters as in (8) or (9), but this will also follow from the discussion given in Chapter 8, where we try to classify all near hexagons of order (3, t) which have a quad through every two points at distance 2.

4.2 Some relations on the line set of a near hexagon

If S = (P, L, I) is a near hexagon, then we will define the following relations on L:

R_{0} = {(L, L)|L ∈ L},
R_{1} = {(L, M)|L ≠ M and d(L, M) = 0},
R_{2} = {(L, M)|L∥M and d(L, M) = 1},
R_{3} = {(L, M)|L ∦M and d(L, M) = 1},
R_{4} = {(L, M)|L∥M and d(L, M) = 2}.

For a fixed line L define n_{i}(L) = |{M ∈ L|(L, M) ∈ R_{i}}| and for a fixed (L, M) ∈ R_{i} define p_{jk}^{i}(L, M) = |{N ∈ L|(L, N) ∈ R_{j} and (N, M) ∈ R_{k}}|. If n_{i}(L) is independant of L or if no confusion is possible we will write n_{i} instead of n_{i}(L). A similar remark holds for the coefficients p_{jk}^{i}(L, M). As we will see in the next section all the coefficients n_{i} and allmost all the coefficients p_{jk}^{i} are constant in the case of a regular near hexagon having quads through every two points at mutual distance 2.

4.3 The numbers n_{i} and p_{jk}^{i} in the case of a regular near hexagon with quads

Let S = (P, L, I) be a regular near hexagon with parameters (s, t_{2}, t). If t_{2} = 0, then S is a generalized hexagon. In this case R_{2} = ∅ and R = (R_{0}, R_{1}, R_{3}, R_{4}) defines an association scheme on the set of lines of S. This association scheme is in fact the association scheme associated with the dual of S (see Corollary 4.1.3). In this section we will suppose that there is a unique quad through every two points at mutual distance 2 (hence t_{2} ≥ 1). This assumption is always satisfied when s ≥ 2, t_{2} ≥ 1 (Yanushka’s Theorem) or s = t_{2} = 1.

If x is a fixed point of S, then one calculates that Γ_{0}(x) = 1, Γ_{1}(x) = s(t + 1), Γ_{2}(x) = \frac{s^{2}t(t+1)}{t_{2}+1}, Γ_{3}(x) = \frac{s^{3}t(t−t_{2})}{1+t_{2}}. Hence

v = |P| = (1 + s)(1 + st + \frac{s^{2}t(t − t_{2})}{1 + t_{2}})

and

b = |L| = (1 + t)(1 + st + \frac{s^{2}t(t − t_{2})}{1 + t_{2}}).

Theorem 4.3.1 The coefficients n_{i} are constant and equal to the numbers given below:

n_{0} = 1, n_{1} = (s + 1)t, n_{2} = st_{2}t,

n_{3} = st(s + 1)(t − t_{2}), n_{4} = \frac{s^{2}t(t − t_{2})^{2}}{1 + t_{2}} .

Proof. Let L be a fixed line of L. It is clear that n_{0} = 1. We have that n_{1} = (s + 1)t, since there are s + 1 choices for a point on L and t choices for a line M ≠ L through that point. Take a fixed point p on L. Counting pairs (q, M) with q ∈ Γ(p) ∩ M and (L, M) ∈ R_{2} yields n_{2} = st_{2}t. Counting triples (r, q, M) with r ∈ L, q ∈ Γ(p) ∩ M and (L, M) ∈ R_{3} yields n_{3} = st(s + 1)(t − t_{2}). Finally, we have that n_{4} = b − n_{0} − n_{1} − n_{2} − n_{3} = \frac{s^{2}t(t−t_{2})^{2}}{1+t_{2}} .

□

Corollary 4.3.2 The coefficients p_{ij}^{0} are constant and equal to δ_{ij}n_{i}.

Theorem 4.3.3 The coefficients p_{ij}^{1} are constant and equal to the numbers given below:

p_{00}^{1} = 0, p_{01}^{1} = p_{10}^{1} = 1, p_{02}^{1} = p_{20}^{1} = 0, p_{03}^{1} = p_{30}^{1} = 0, p_{04}^{1} = p_{40}^{1} = 0, p_{11}^{1} = t − 1,

p_{12}^{1} = p_{21}^{1} = st_{2}, p_{13}^{1} = p_{31}^{1} = s(t − t_{2}), p_{14}^{1} = p_{41}^{1} = 0, p_{22}^{1} = st_{2}(t_{2} − 1),

p_{23}^{1} = p_{32}^{1} = st_{2}(t − t_{2}), p_{24}^{1} = p_{42}^{1} = 0, p_{33}^{1} = s(t − t_{2})(st_{2} + t − t_{2} − 1),

p_{34}^{1} = p_{43}^{1} = s^{2}(t − t_{2})^{2}, p_{44}^{1} = \frac{s^{2}(t − t_{2})^{2}(t − t_{2} − 1)}{1 + t_{2}}.

Proof. We only need to prove the values for p_{11}^{1}, p_{12}^{1}, p_{14}^{1}, p_{22}^{1}, p_{24}^{1} and p_{33}^{1}. The remaining coefficients are then readily calculated. It is immediate that p_{11}^{1} = t − 1 and p_{14}^{1} = p_{24}^{1} = 0. Let K and L be two lines intersecting in a point p and let V_{ij} = {M ∈ L|(K, M) ∈ R_{i} and (L, M) ∈ R_{j}}. Counting pairs (q, M) where q ∈ K ∩ M and M ∈ V_{12} gives p_{12}^{1} = st_{2}. Counting tuples (q_{1}, q_{2}, q, M) with q_{1} ∈ K − {p}, q_{2} ∈ L − {p}, q ∈ Γ(q_{1}) ∩ Γ(q_{2}) ∩ M and M ∈ V_{22}, we find that p_{22}^{1} = st_{2}(t_{2} − 1). The value of p_{23}^{1} is then equal to n_{2} − p_{20}^{1} − p_{21}^{1} − p_{22}^{1} − p_{24}^{1} = st_{2}(t − t_{2}). If M ∈ V_{33}, let p_{1} (and p_{2}) be those points of M at distance 1 from K (and L). If p_{1} ≠ p_{2}, then (since d(p, p_{1}) = d(p, p_{2}) = 2) there exists a point p_{3} on M such that d(p, p_{3}) = 1, implying M ∈ V_{22}, a contradiction. Hence two cases can appear.

(1) The point p_{1}(= p_{2}) is collinear with p. There are s(t−1) choices for a point q ∈ Γ(p)\(K∪L) and through every such q, there are t lines not intersecting K or L. Hence, the contribution of this case to p_{33}^{1} equals A = st(t−1)−2p_{23}^{1} −p_{22}^{1} = s(t−t_{2})(t−t_{2} −1).

(2) The point p_{1}(= p_{2}) is not collinear with p. Counting tuples (q_{1}, q_{2}, r, N) with q_{1} ∈ K \ {p}, q_{2} ∈ L \ {p}, r ∈ N ∩ Γ(q_{1}) ∩ Γ(q_{2}) and N ∈ V_{33} yields that the contribution of this case to p_{33}^{1} equals B = s^{2}t_{2}(t − t_{2}).

We have now that p_{33}^{1} = A + B = s(t − t_{2})(st_{2} + t − t_{2} − 1). □

Theorem 4.3.4 The coefficients p_{ij}^{2} are constant and equal to the numbers given below:

p_{00}^{2} = 0, p_{01}^{2} = p_{10}^{2} = 0, p_{02}^{2} = p_{20}^{2} = 1, p_{03}^{2} = p_{30}^{2} = 0, p_{04}^{2} = p_{40}^{2} = 0, p_{11}^{2} = s + 1,

p_{12}^{2} = p_{21}^{2} = (s + 1)(t_{2} − 1), p_{13}^{2} = p_{31}^{2} = (s + 1)(t − t_{2}), p_{14}^{2} = p_{41}^{2} = 0,
p_{22}^{2} = st_{2}^{2} − st_{2} − t_{2} + s, p_{23}^{2} = p_{32}^{2} = 0, p_{24}^{2} = p_{42}^{2} = st_{2}(t − t_{2}),
p_{33}^{2} = (s + 1)(t − t_{2})(2st_{2} − 1), p_{34}^{2} = p_{43}^{2} = s(s + 1)(t − t_{2})(t − 2t_{2}),

p_{44}^{2} = s(t − t_{2})[\frac{st(t − t_{2})}{1 + t_{2}} − t_{2} − (s + 1)(t − 2t_{2})].

Proof. We only need to prove the values for p_{11}^{2}, p_{12}^{2}, p_{14}^{2}, p_{22}^{2}, p_{23}^{2} and p_{34}^{2}. The remaining coefficients are then readily calculated. It is immediate that p_{11}^{2} = s+1 and p_{14}^{2} = 0. Let K and L be two lines such that (K, L) ∈ R_{2} and let Q be the quad through K and L. Finally, let V_{ij} = {M ∈ L|(K, M) ∈ R_{i} and (L, M) ∈ R_{j}}. Counting pairs (p, M) with p ∈ K ∩ M and M ∈ V_{12} yields p_{12}^{2} = (s+1)(t_{2}−1). If M ∈ V_{22}, then every point of M is collinear with two points of Q and hence it is contained in Q. As a consequence p_{22}^{2} equals the number of lines in Q, not intersecting K or L. Hence p_{22}^{2} = st_{2}^{2}−st_{2}−t_{2}+s. Suppose M ∈ V_{23}. Let p be the unique point of M at distance 1 from L. As before, p must belong to Q. Now, every line through p parallel with K is contained in Q and hence it is impossible that (M, L) ∈ R_{3}. So V_{23} = ∅ and p_{23}^{2} = 0. Finally, counting tuples (p, q, M) with p ∈ K, q ∈ Γ(p)∩M, M ∈ V_{34} gives p_{34}^{2} = s(s + 1)(t − t_{2})(t − 2t_{2}). For, there are s + 1 choices for p, for fixed p, there are s(t − t_{2}) choices for q ∉ Q. Through every such q there are t + 1 lines, t_{2} of them are parallel with K and t_{2} + 1 others have relation R_{3} with L. □

Theorem 4.3.5 The coefficients p_{ij}^{3} are constant and equal to the numbers given below:

p_{00}^{3} = 0, p_{01}^{3} = p_{10}^{3} = 0, p_{02}^{3} = p_{20}^{3} = 0, p_{03}^{3} = p_{30}^{3} = 1, p_{04}^{3} = p_{40}^{3} = 0, p_{11}^{3} = 1,

p_{12}^{3} = p_{21}^{3} = t_{2}, p_{13}^{3} = p_{31}^{3} = st_{2} + t − t_{2} − 1, p_{14}^{3} = p_{41}^{3} = s(t − t_{2}), p_{22}^{3} = 0,

p_{23}^{3} = p_{32}^{3} = t_{2}(2st_{2} − 1), p_{24}^{3} = p_{42}^{3} = st_{2}(t − 2t_{2}),
p_{33}^{3} = (s − 1)(t − 2t_{2}) + 3st_{2}(t − 2t_{2}) + s^{2}(tt_{2} + t − t_{2} − 2t_{2}^{2}),
p_{34}^{3} = p_{43}^{3} = st(s + 1)(t − t_{2}) − p_{30}^{3} − p_{31}^{3} − p_{32}^{3} − p_{33}^{3},

p_{44}^{3} = \frac{s^{2}t(t − t_{2})^{2}}{1 + t_{2}} − p_{41}^{3} − p_{42}^{3} − p_{43}^{3}.

Proof. We only need to prove the values for p_{11}^{3}, p_{12}^{3}, p_{14}^{3}, p_{22}^{3}, p_{23}^{3} and p_{33}^{3}. The remaining coefficients are then readily calculated. Let K and L be two lines such that (K, L) ∈ R_{3} and let p and q denote these points of K and L such that d(p, q) = 1. Finally, let V_{ij} = {M ∈ L|(K, M) ∈ R_{i} and (L, M) ∈ R_{j}}. We immediately have that: (i) p_{11}^{3} = 1, (ii) p_{12}^{3} = t_{2}, (iii) p_{22}^{3} = 0 (since p_{23}^{2} = 0). Counting pairs (r, M) with r ∈ K ∩ M and M ∈ V_{14} yields p_{14}^{3} = s(t−t_{2}). Suppose M ∈ V_{23}. Let r ∈ M be the point at distance 1 from u ∈ L and let v ∈ K be the point collinear with r. There are four cases.

(1) p = v, u = q. The point r is then on the line pq. So, the contribution of this case to p_{23}^{3} is A = (s − 1)t_{2}.

(2) p = v, u ≠ q. There are s choices for u. For fixed u, there are t_{2} choices for r and through every such r, there are t_{2} lines parallel with K. Hence, the contribution of this case to p_{23}^{3} equals B = st_{2}^{2}.

(3) p ≠ v, u = q. Just like in the second case one calculates that the contribution is equal to C = st_{2}(t_{2} − 1).

(4) p ≠ v, u ≠ q. There exists a path of length 2 connecting u and v, a contradiction.

Hence p_{23}^{3} = A + B + C = t_{2}(2st_{2} − 1). Finally, suppose that M ∈ V_{33}. Let u and v be those points of M which are collinear with u′ ∈ K and v′ ∈ L. There are eight cases.

(1) u = v, u′ = p, v′ = q. As before, the contribution is easy to calculate and equal to A = (s − 1)(t − 2t_{2}).

(2) u = v, u′ = p, v′ ≠ q. The contribution is equal to B = st_{2}(t − 2t_{2}).

(3) u = v, u′ ≠ p, v′ = q. The contribution is again equal to C = st_{2}(t − 2t_{2}).

(4) u = v, u′ ≠ p, v′ ≠ q There exists a path of length 2 connecting u′ and v′, a contradiction.

(5) u ≠ v, u′ = p, v′ = q. Let Q_{1} (Q_{2}) be the quad through pq and K (pq and L). The line pu is not contained in Q_{1} or Q_{2}. The contribution of this case equals D = st_{2}(t − 2t_{2}). For, there are s(t − 2t_{2}) choices for u and for every such choice there are t_{2} choices for v (otherwise v and hence also M is contained in Q_{1} or Q_{2}).

(6) u ≠ v, u′ = p, v′ ≠ q. Since d(q, u) = d(q, v) = 2, there exists a point w on M collinear with q. Hence M is parallel to L, a contradiction.

(7) u ≠ v, u′ ≠ p, v′ = q. This is impossible again.

(8) u ≠ v, u′ ≠ p, v′ ≠ q. There are s choices for u′ and s choices for v′. For fixed u′ and v′, there are t − t_{2} points a collinear with u′, at distance 2 from v′ and not collinear with q. For fixed a, there are 1 + t_{2} points b collinear with a and v′. In this way we counted s^{2}(t − t_{2})(1 + t_{2}) lines ab, s^{2}t_{2}^{2} of them are parallel to K and the others are the lines we are looking for. Hence, the contribution of this case to p_{33}^{3} is E = s^{2}(tt_{2} + t − t_{2} − 2t_{2}^{2}).

Summing all the contributions, one finds the value of p_{33}^{3} mentioned in the theorem. □

Theorem 4.3.6 The coefficients p_{ij}^{4} with i ≤ 1 or j ≤ 1 are constant and equal to the numbers given below.

p_{00}^{4} = 0, p_{01}^{4} = p_{10}^{4} = 0, p_{02}^{4} = p_{20}^{4} = 0, p_{03}^{4} = p_{30}^{4} = 0, p_{04}^{4} = p_{40}^{4} = 1, p_{11}^{4} = 0,

p_{12}^{4} = p_{21}^{4} = 0, p_{13}^{4} = p_{31}^{4} = (s + 1)(t_{2} + 1), p_{14}^{4} = p_{41}^{4} = (s + 1)(t − t_{2} − 1).

Proof. We only need to prove the values of p_{11}^{4}, p_{12}^{4} and p_{13}^{4}. Clearly p_{11}^{4} = p_{12}^{4} = 0. Let K and L be two lines such that (K, L) ∈ R_{4}. Counting pairs (p, M) with p ∈ K ∩ M and (M, L) ∈ R_{3} gives p_{13}^{4} = (s + 1)(t_{2} + 1). □

Remarks.

(1) Some of the above values for the intersection numbers are only correct when S contains quads through every two points at mutual distance 2. If t_{2} ≥ 1 and S would contain two points x and y such that: (i) d(x, y) = 2, (ii) there is no quad through x and y, then s equals 2 (see [15] for examples). Then there exists three common neighbours a, b, c of x and y and a common neighbour z of a and b which is not collinear with c. Then (xb, yc), (xb, az) ∈ R_{2} and (az, yc) ∈ R_{3}, so p_{23}^{2}(xb, yc) ≠ 0 (we had p_{23}^{2} = 0 in Theorem 4.3.4).

(2) If the coefficients p_{ij}^{4} which are not mentioned in the above theorem are also constant, then R = (R_{0}, R_{1}, R_{2}, R_{3}, R_{4}) defines an association scheme. As we will see in the next paragraph it is sufficient for this that p_{22}^{4} is constant. This last condition is however not always satisfied.

4.4 The case where R determines an association scheme

Just like in the previous paragraph, let S be a regular near hexagon having quads through every two points at mutual distance 2. With the same notations as above, we have the following theorem.

Theorem 4.4.1 The following conditions are equivalent:

(1) R determines an association scheme,

(2) p_{22}^{4} is constant.

In this case, the remaining intersection numbers are equal to

p_{22}^{4} = \frac{(1 + t_{2})t_{2}^{2}}{t − t_{2}} , p_{23}^{4} = p_{32}^{4} = \frac{t_{2}(1 + t_{2})(s + 1)(t − 2t_{2})}{t − t_{2}},

p_{24}^{4} = p_{42}^{4} = \frac{n_{2}}{n_{4}} p_{44}^{2}, p_{33}^{4} = \frac{n_{3}}{n_{4}} p_{34}^{3}, p_{34}^{4} = p_{43}^{4} = \frac{n_{3}}{n_{4}} p_{44}^{3},

p_{44}^{4} = n_{4} − p_{04}^{4} − p_{14}^{4} − p_{24}^{4} − p_{34}^{4}.

Proof. Clearly (1) implies (2), so suppose that p_{22}^{4} is constant. Then p_{22}^{4} = \frac{n_{2}p_{24}^{2}}{n_{4}} = \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} . Let K and L be two lines such that (K, L) ∈ R_{4}. Counting triples (p, q, M) with p ∈ K, q ∈ Γ(p)∩Γ(L) and M∥K gives p_{23}^{4}+(s+1)p_{22}^{4} = (1 + s)(1 + t_{2})t_{2}. Hence p_{23}^{4} = p_{32}^{4} = \frac{(1+t_{2})t_{2}(s+1)(t−2t_{2})}{t−t_{2}}. If we are able to prove that p_{33}^{4} is constant, then the theorem is valid. Let M be a line such that (K, M), (L, M) ∈ R_{3}. Let u and v be those points that are collinear with u′ ∈ K and v′ ∈ L. We have the following cases.

(1) u ≠ v and d(u′, v′) = 2. Let Q be the quad through u′ and v′. Since u′ has distance 2 to v and v′, there exists a point on vv′ collinear with u. Hence v ∈ Q. Also u ∈ Q for similar reasons and hence M ⊆ Q. On the contrary, every line in Q not containing u′, v′ or any common neighbour of u′ and v′ has relation R_{3} with K and L. Hence, the contribution of this case to p_{33}^{4} is equal to A = (s + 1)[(1 + t_{2})(1 + st_{2}) − 2(1 + t_{2}) − (1 + t_{2})(t_{2} − 1)].

(2) u = v.

Let B denote the contribution of this case to p_{33}^{4}. Counting triples (p, q, N) with p ∈ K, q ∈ Γ(p) ∩ Γ(L) ∩ M and M ∩ (K ∪ L) = ∅ yields

(s + 1)(t_{2} + 1)(t − 1) = B + 2p_{23}^{4} + (s + 1)p_{22}^{4}.

Hence B is constant.

(3) u ≠ v and d(u′, v′) = 3. Let C denote the contribution of this case to p_{33}^{4}. Counting the tuples (p, q, r, N) with p ∈ K, q ∈ L∩Γ_{3}(p), r ∈ Γ(p)∩Γ_{2}(q)∩Γ_{1}(K)∩Γ_{2}(L)∩N and d(q, N) = 1 yields

(s + 1)s(t − t_{2} − 1)(t_{2} + 1) = C + sp_{23}^{4}.

Hence C is constant.

It is now clear that p_{33}^{4} = A + B + C is constant. □

Theorem 4.4.2 Let S be a regular near hexagon.

(1) If S is a dual polar space, then p_{22}^{4} is constant.

(2) If S is the near hexagon related to S(5, 8, 24), then p_{22}^{4} is constant.

(3) If S is a the near hexagon related to the extended ternary Golay code, then p_{22}^{4} is not constant.

Proof.

(1) If S has parameters (s, t_{2}, t), then we prove that p_{22}^{4} = t_{2} + 1. Let K and L be two lines such that d(K, L) = 2 and let k and l be two points on K and L such that d(k, l) = 2. If M is a line having relation R_{2} with K and M, then there is a point on M collinear with k and l. The quad through that point and the line K necessary contains M. Now, let m be one of the t_{2} + 1 common neighbours of k and l, and let Q be the quad through m and K. A point q ≠ l on L has distance 2 from two points of Q, hence there is a point q′ ∈ Q collinear with q. The line mq′ is then the unique line through m having relation R_{2} with K and L. This proves that p_{22}^{4} = t_{2} + 1.

(2) Let K = {K_{1}, K_{2}, K_{3}} and L = {L_{1}, L_{2}, L_{3}} be two lines of S such that d(K, L) = 2. Hence, we may assume that there is some partition {X_{1}, X_{2}, X_{3}, X_{4}, X_{5}, X_{6}, X_{7}, X_{8}, X_{9}} of the point set of S(5, 8, 24) such that

• |X_{1}| = |X_{5}| = |X_{9}| = 4, |X_{i}| = 2 for i ≠ 1, 5, 9;

• X_{1} ∪ X_{2} ∪ X_{3} = K_{1}, X_{4} ∪ X_{5} ∪ X_{6} = K_{2}, X_{7} ∪ X_{8} ∪ X_{9} = K_{3};

• X_{1} ∪ X_{4} ∪ X_{7} = L_{1}, X_{2} ∪ X_{5} ∪ X_{8} = L_{2}, X_{3} ∪ X_{6} ∪ X_{9} = L_{3}.

Suppose M = {M_{1}, M_{2}, M_{3}} is a line of S having relation R_{2} with K and L. We may assume that d(K_{i}, M_{i}) = d(M_{i}, L_{i}) = 1 for all i ∈ {1, 2, 3}. Now, let N be one of the three blocks of the Steiner system collinear with K_{1} and L_{1}. If we consider the possible intersections of N with K_{1}, K_{2}, K_{3}, L_{1}, L_{2} and L_{3}, we have the following cases.

(a) |N ∩ X_{5}| = |N ∩ X_{9}| = 4, |N ∩ X_{6}| = |N ∩ X_{8}| = 0. In that case N△L_{3} = X_{5}∪X_{6}∪X_{3} is also a block, a contradiction, since there is only one block through X_{5} ∪ X_{6}.

(b) |N ∩ X_{5}| = |N ∩ X_{9}| = 3, |N ∩ X_{6}| = |N ∩ X_{8}| = 1.

(c) |N ∩ X_{5}| = |N ∩ X_{6}| = |N ∩ X_{8}| = |N ∩ X_{9}| = 2.

So, the three common neighbours N_{1}, N_{2}, N_{3} of K_{1} and L_{1} must satisfy (b) or (c). Since |N_{i} ∩ N_{j}|=4 for i ≠ j, it is easy to check that one, say N_{1}, satisfies (c) and the two others satisfy (b). Now consider the possible intersections of M_{1} and M_{2} with X_{9}. Since M_{1} ∩ M_{2} = ∅, we must have that M_{1} = N_{1}. A similar reasoning holds for M_{2} and M_{3}. This proves that there is at most one line having relation R_{2} with K and L. If p_{22}^{4} is constant, we found earlier that p_{22}^{4} = \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} . If p_{22}^{4}

is not necessary constant, then \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} is the average value of all the

parameters p_{22}^{4}. Hence, since \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} = 1 in this case, one finds that p_{22}^{4} is constant and equal to 1.

(3) If p_{22}^{4} would be constant, then \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} = \frac{1}{5} would be integral, a contradiction.

□

The parameters of the association scheme related to S(5, 8, 24)

The association scheme is one on 3795 elements. One has that n_{0} = 1, n_{1} = 42, n_{2} = 56, n_{3} = 1008, n_{4} = 2688. The values of the coefficients p_{jk}^{i} are easily calculated and listed below.

p_{00}^{0} = 1, p_{01}^{0} = 0, p_{02}^{0} = 0, p_{03}^{0} = 0, p_{04}^{0} = 0, p_{10}^{0} = 0, p_{11}^{0} = 42, p_{12}^{0} = 0, p_{13}^{0} = 0, p_{14}^{0} = 0, p_{20}^{0} = 0, p_{21}^{0} = 0, p_{22}^{0} = 56, p_{23}^{0} = 0, p_{24}^{0} = 0, p_{30}^{0} = 0, p_{31}^{0} = 0, p_{32}^{0} = 0, p_{33}^{0} = 1008, p_{34}^{0} = 0, p_{40}^{0} = 0, p_{41}^{0} = 0, p_{42}^{0} = 0, p_{43}^{0} = 0, p_{44}^{0} = 2688, p_{00}^{1} = 0, p_{01}^{1} = 1, p_{02}^{1} = 0, p_{03}^{1} = 0, p_{04}^{1} = 0, p_{10}^{1} = 1, p_{11}^{1} = 13, p_{12}^{1} = 4, p_{13}^{1} = 24, p_{14}^{1} = 0, p_{20}^{1} = 0, p_{21}^{1} = 4, p_{22}^{1} = 4, p_{23}^{1} = 48, p_{24}^{1} = 0, p_{30}^{1} = 0, p_{31}^{1} = 24, p_{32}^{1} = 48, p_{33}^{1} = 360, p_{34}^{1} = 576, p_{40}^{1} = 0, p_{41}^{1} = 0, p_{42}^{1} = 0, p_{43}^{1} = 576, p_{44}^{1} = 2112, p_{00}^{2} = 0, p_{01}^{2} = 0, p_{02}^{2} = 1, p_{03}^{2} = 0, p_{04}^{2} = 0, p_{10}^{2} = 0, p_{11}^{2} = 3, p_{12}^{2} = 3, p_{13}^{2} = 36, p_{14}^{2} = 0, p_{20}^{2} = 1, p_{21}^{2} = 3, p_{22}^{2} = 4, p_{23}^{2} = 0, p_{24}^{2} = 48, p_{30}^{2} = 0, p_{31}^{2} = 36, p_{32}^{2} = 0, p_{33}^{2} = 252, p_{34}^{2} = 720, p_{40}^{2} = 0, p_{41}^{2} = 0, p_{42}^{2} = 48, p_{43}^{2} = 720, p_{44}^{2} = 1920, p_{00}^{3} = 0, p_{01}^{3} = 0, p_{02}^{3} = 0, p_{03}^{3} = 1, p_{04}^{3} = 0, p_{10}^{3} = 0, p_{11}^{3} = 1, p_{12}^{3} = 2, p_{13}^{3} = 15, p_{14}^{3} = 24, p_{20}^{3} = 0, p_{21}^{3} = 2, p_{22}^{3} = 0, p_{23}^{3} = 14, p_{24}^{3} = 40, p_{30}^{3} = 1, p_{31}^{3} = 15, p_{32}^{3} = 14, p_{33}^{3} = 258, p_{34}^{3} = 720, p_{40}^{3} = 0, p_{41}^{3} = 24, p_{42}^{3} = 40, p_{43}^{3} = 720, p_{44}^{3} = 1904, p_{00}^{4} = 0, p_{01}^{4} = 0, p_{02}^{4} = 0, p_{03}^{4} = 0, p_{04}^{4} = 1, p_{10}^{4} = 0, p_{11}^{4} = 0, p_{12}^{4} = 0, p_{13}^{4} = 9, p_{14}^{4} = 33, p_{20}^{4} = 0, p_{21}^{4} = 0, p_{22}^{4} = 1, p_{23}^{4} = 15, p_{24}^{4} = 40, p_{30}^{4} = 0, p_{31}^{4} = 9, p_{32}^{4} = 15, p_{33}^{4} = 270, p_{34}^{4} = 714, p_{40}^{4} = 1, p_{41}^{4} = 33, p_{42}^{4} = 40, p_{43}^{4} = 714, p_{44}^{4} = 1900.

The parameters of some association schemes related to classical near polygons

Consider the classical near hexagons W^{D}(5, q), Q^{D}(6, q), [Q^{−}(7, q)]^{D}, [Q^{+}(5, q)], H^{D}(6, q) or H^{D}(5, q) from Section 3.2.2. The parameters of these near hexagons look like s = q^{e}, t_{2} = q, t = q(q + 1), with q a prime power and e equal to 1, 1, 2, 0, \frac{3}{2} or \frac{1}{2}. It is now easy to calculate the parameters of the association scheme. The association scheme is one on (q^{2} + q + 1)(q^{2e+3} + q^{e+2} + q^{e+1} + 1) elements. One has that n_{0} = 1, n_{1} = q(q + 1)(q^{e} + 1), n_{2} = q^{e+2}(q + 1), n_{3} = q^{e+3}(q + 1)(q^{e} + 1), n_{4} = q^{2e+5}. The values of p_{jk}^{i} are listed below.

p_{00}^{0} = 1, p_{01}^{0} = 0, p_{02}^{0} = 0, p_{03}^{0} = 0, p_{04}^{0} = 0, p_{10}^{0} = 0, p_{11}^{0} = q(q + 1)(q^{e} + 1), p_{12}^{0} = 0, p_{13}^{0} = 0, p_{14}^{0} = 0, p_{20}^{0} = 0, p_{21}^{0} = 0, p_{22}^{0} = q^{e+2}(q + 1), p_{23}^{0} = 0, p_{24}^{0} = 0, p_{30}^{0} = 0, p_{31}^{0} = 0, p_{32}^{0} = 0, p_{33}^{0} = q^{e+3}(q + 1)(q^{e} + 1), p_{34}^{0} = 0, p_{40}^{0} = 0, p_{41}^{0} = 0, p_{42}^{0} = 0, p_{43}^{0} = 0, p_{44}^{0} = q^{2e+5}, p_{00}^{1} = 0, p_{01}^{1} = 1, p_{02}^{1} = 0, p_{03}^{1} = 0, p_{04}^{1} = 0, p_{10}^{1} = 1, p_{11}^{1} = q^{2} + q − 1, p_{12}^{1} = q^{e+1}, p_{13}^{1} = q^{e+2}, p_{14}^{1} = 0, p_{20}^{1} = 0, p_{21}^{1} = q^{e+1}, p_{22}^{1} = q^{e+1}(q − 1), p_{23}^{1} = q^{e+3}, p_{24}^{1} = 0, p_{30}^{1} = 0, p_{31}^{1} = q^{e+2}, p_{32}^{1} = q^{e+3}, p_{33}^{1} = q^{e+2}(q^{e+1} + q^{2} − 1), p_{34}^{1} = q^{2e+4}, p_{40}^{1} = 0, p_{41}^{1} = 0, p_{42}^{1} = 0, p_{43}^{1} = q^{2e+4}, p_{44}^{1} = q^{2e+4}(q − 1), p_{00}^{2} = 0, p_{01}^{2} = 0, p_{02}^{2} = 1, p_{03}^{2} = 0, p_{04}^{2} = 0, p_{10}^{2} = 0, p_{11}^{2} = q^{e} + 1, p_{12}^{2} = (q − 1)(q^{e} + 1), p_{13}^{2} = q^{2}(q^{e} + 1), p_{14}^{2} = 0, p_{20}^{2} = 1, p_{21}^{2} = (q − 1)(q^{e} + 1), p_{22}^{2} = q^{e+2} − q^{e+1} + q^{e} − q, p_{23}^{2} = 0, p_{24}^{2} = q^{e+3}, p_{30}^{2} = 0, p_{31}^{2} = q^{2}(q^{e} + 1), p_{32}^{2} = 0, p_{33}^{2} = q^{2}(q^{e} + 1)(2q^{e+1} − 1), p_{34}^{2} = q^{e+3}(q−1)(q^{e}+1), p_{40}^{2} = 0, p_{41}^{2} = 0, p_{42}^{2} = q^{e+3}, p_{43}^{2} = q^{e+3}(q−1)(q^{e}+1), p_{44}^{2} = q^{2e+5} − q^{2e+4} + q^{2e+3} − q^{e+4}, p_{00}^{3} = 0, p_{01}^{3} = 0, p_{02}^{3} = 0, p_{03}^{3} = 1, p_{04}^{3} = 0, p_{10}^{3} = 0, p_{11}^{3} = 1, p_{12}^{3} = q, p_{13}^{3} = q^{e+1} + q^{2} − 1, p_{14}^{3} = q^{e+2}, p_{20}^{3} = 0, p_{21}^{3} = q, p_{22}^{3} = 0, p_{23}^{3} = q(2q^{e+1} − 1), p_{24}^{3} = q^{e+2}(q − 1), p_{30}^{3} = 1, p_{31}^{3} = q^{e+1} + q^{2} − 1, p_{32}^{3} = q(2q^{e+1} − 1), p_{33}^{3} = q(q − 1)(q^{2e+1} + 3q^{e+1} + q^{e} − 1) + q^{2e+2}, p_{34}^{3} = q^{e+3}(q^{e+1}+q−2), p_{40}^{3} = 0, p_{41}^{3} = q^{e+2}, p_{42}^{3} = q^{e+2}(q−1), p_{43}^{3} = q^{e+3}(q^{e+1}+q−2), p_{44}^{3} = q^{e+3}(q^{e}+2), p_{00}^{4} = 0, p_{01}^{4} = 0, p_{02}^{4} = 0, p_{03}^{4} = 0, p_{04}^{4} = 1, p_{10}^{4} = 0, p_{11}^{4} = 0, p_{12}^{4} = 0, p_{13}^{4} = (q + 1)(q^{e} + 1), p_{14}^{4} = (q − 1)(q + 1)(q^{e} + 1), p_{20}^{4} = 0, p_{21}^{4} = 0, p_{22}^{4} = q+1, p_{23}^{4} = (q+1)(q−1)(q^{e}+1), p_{24}^{4} = (q+1)(q^{e+2}−q^{e+1}+q^{e}−q), p_{30}^{4} = 0, p_{31}^{4} = (q +1)(q^{e} +1), p_{32}^{4} = (q +1)(q −1)(q^{e} +1), p_{33}^{4} = q(q +1)(q^{e} +1)(q^{e+1} + q − 2), p_{34}^{4} = q(q + 1)(q^{e} + 1)(q^{e} + 2), p_{40}^{4} = 1, p_{41}^{4} = (q − 1)(q + 1)(q^{e} + 1), p_{42}^{4} = (q + 1)(q^{e+2} − q^{e+1} + q^{e} − q), p_{43}^{4} = q(q + 1)(q^{e} + 1)(q^{e} + 2), p_{44}^{4} = q^{2e+5} − q^{2e+2} − q^{2e+1} − q^{e+3} − 4q^{e+2} − 3q^{e+1} − 2q^{2} − q.

4.5 A characterization theorem

Theorem 4.5.1 If S is a regular near hexagon with parameters (s, t_{2}, t) and satisfying: (i) s ≥ t_{2}, (ii) p_{22}^{4} is constant, then S is one of the following near hexagons:

• a generalized hexagon of order (s, t),

• a classical near hexagon (a regular Hamming near hexagon, W^{D}(5, q), Q^{D}(6, q), [Q^{−}(7, q)]^{D} or H^{D}(6, q)),

• the near hexagon related to S(5, 8, 24),

• the near hexagon whose point graph is the incidence graph of the unique biplane of order 2.

Proof. Suppose that S is a near hexagon with parameters (s, t_{2}, t) and satisfying: (i) s ≥ t_{2}, (ii) p_{22}^{4} is constant. If t_{2} = 0, then S is a generalized hexagon. So, in the sequel we suppose that t_{2} ≥ 1. If s = 1 then t_{2} = 1, so there are always quads through every two points at mutual distance 2. If s = t_{2} = 1, then \frac{(1+t_{2})t_{2}^{2}}{t−t_{2}} ∈ ℕ implies that t ∈ \{2, 3\}. If t = 2, then S is a cube and hence a Hamming near hexagon. If t = 3, then S is the incidence graph of a square 2-design D (see [15], Theorem 1.3.1). Out of the parameters of S, we derive that D has 14 points, every block contains k = 4 points and every 2 points are contained in λ = 2 blocks. Hence D is a biplane of order n = k−λ = 2. This biplane is easily proved to be unique (see also [6] for more details about this). A description of the incidence graph Γ of D is as follows. Take an arbitrary vertix and label it ∞. The vertices in Γ_{1}(∞) are labeled with the singletons of X = {1, 2, 3, 4}. The vertices of Γ_{2}(∞) are labeled with the subsets of X having order 2 in such a way that {i} ∼ {j, k} ⟺ i = j or i = k. Finally one can label a point in Γ_{3}(∞) with the set of two vertices of Γ_{2}(∞), where that point is not adjacent with. With this description, it is now easy to prove that p_{22}^{4} is constant and equal to 1. For, let K and L be two lines such that d(K, L) = 2. We may suppose that K = {∞, {a}} and L = {{b, c}, {{a, b}, {c, d}}} with {a, b, c, d} = {1, 2, 3, 4}. It is immediate that M = {{c}, {a, c}} is the unique line having relation R_{2} with K and L. If s ≥ 2 and if S is no dual polar space, then 1 + t ≥ (1 + s)(1 + st_{2}) (see Theorem 3.7.4). Since s ≥ t_{2} we have that 1 + t ≥ (1 + t_{2})(1 + t_{2}^{2}), implying p_{22}^{4} ≤ 1. Hence 1 + t = (1 + st_{2})(1 + s) and s = t_{2}. Now, Theorem 3.2.1 implies that S is the regular near hexagon associated with S(5, 8, 24). □

4.6 An association scheme derived from a GQ(s, s^{2}), s ≥ 1, having a spread of symmetry

Suppose that Q is a nontrivial GQ(s, s^{2}), s ≥ 1, having a spread of symmetry S and let L be a fixed line of S. Put G = Π_{S}(L). Let X = S \ {L} and R = {R_{0}} ∪ {R_{g}|g ∈ G}; (M, N) ∈ R_{0} if and only if M = N, (M, N) ∈ R_{g} if and only if M ≠ N and [N, L] ∘ [M, N] ∘ [L, M] = g. In the following theorem, we will prove that (X, R) is an association scheme. We have that p_{0β}^{α} = p_{β0}^{α} = 1 or 0 depending on α = β or α ≠ β. For g, h ∈ G, we have that p_{gh}^{0} = n_{g} or 0 depending on gh = e or gh ≠ e. The other parameters together with the parameters n_{g} are given in the following theorem.

Theorem 4.6.1 (X, R) is an association scheme and the remaining parameters are (g, h, k denote elements of G, different from e):

n_{e} = s − 1;  n_{g} = s^{2} − 1;
p_{ee}^{e} = s − 2;  p_{ee}^{g} = 0;
p_{ge}^{e} = 0;  p_{eg}^{e} = 0;
p_{ge}^{g} = 0;  p_{eg}^{g} = 0;
p_{gh}^{e} = 0 if gh = e;  p_{gh}^{e} = s + 1 if gh ≠ e;
p_{eh}^{g} = 1 if h ≠ g;  p_{he}^{g} = 1 if h ≠ g;
p_{hk}^{g} = s + 1 if g ≠ hk; p_{hk}^{g} = 1 if g = hk.

Proof. Clearly R_{0}^{T} = R_{0} and R_{l}^{T} = R_{l^{−1}} for all l ∈ G. In this proof, we will several times make use of the fact that every three mutually noncollinear points of a GQ(s, s^{2}) have exactly s + 1 common neighbours, see Theorem 2.1.3. When we talk about a hyperbolic line, we always mean a hyperbolic line consisting of s + 1 lines of the normal spread S. Let x denote a fixed point of L.

We have that (M, N) ∈ R_{e} if and only if M ≠ N and L, M, N are contained in a hyperbolic line, hence n_{e} = s − 1, p_{ee}^{e} = s − 2, p_{ee}^{g} = p_{ge}^{e} = p_{eg}^{e} = 0.

Let g ∈ G \ {e} be fixed and let M ∈ S \ {L}. We prove that each hyperbolic line, different from the one through L and M, contains a unique line N such that (M, N) ∈ R_{g}. From this it follows that n_{g} = s^{2} − 1, p_{eg}^{g} = p_{ge}^{g} = 0 and p_{eh}^{g} = p_{he}^{g} = 1 if g ≠ h. So, choose such a hyperbolic line H and let N′ be the line of Q through x^{g} meeting each element of H. Let x′ be the unique point on M collinear with x, let x′′ denote the unique point of N′ collinear with x′ and let N be the unique line of H through x′′. Further, N is also the unique line of H with the property that (M, N) ∈ R_{g}.

Let g, h ∈ G \ {e}. Let (K, M) ∈ R_{e} and suppose that (K, N) ∈ R_{g} and (N, M) ∈ R_{h}. The points x^{gh[L,M]}, x^{[L,K]} and x^{g} have then a common neighbour on N. Indeed, let θ ∈ G_{S} such that x^{h^{−1}θ} = x^{g}, then x^{θh^{−1}} = x^{g} and hence x^{θ} = x^{gh}; x^{θ[L,M]} and x^{h^{−1}θ} have now a common neighbour on N. This is impossible if gh = e; if gh ≠ e, then there are s + 1 such common neighbours and each of them yields a good choice for N. Hence p_{gh}^{e} = s + 1 if gh = e.

Finally, let g, h, k ∈ G\{e}. Let (K, M) ∈ R_{g} and suppose that (K, N) ∈ R_{h} and (N, M) ∈ R_{k}; then N contains a common neighbour of x^{[L,K]}, x^{h} and x^{hk[L,M]}. If hk = g, then this common neighbour is equal to the projection of x^{h} on the line through x^{[L,K]} and x^{g[L,M]}. Hence p_{hk}^{g} = 1 if g = hk. If hk ≠ g, then there are s + 1 common neighbours and each of these yields a good element N. Hence p_{hk}^{g} = s + 1 if g ≠ hk. □

Remark. The above defined association scheme is symmetric if and only if g^{−1} = g for all g ∈ G. From ab = a^{−1}b^{−1} = (ba)^{−1} = ba, ∀a, b ∈ G, it follows that G is an elementary abelian group and hence isomorphic to the additive group of some finite field with characteristic 2.

The i-neighbourhood of a point

5.1 Definitions and properties

Let S = (P, L, I) be a near 2d-gon. For x ∈ P and i ∈ ℕ, we define P_{x}^{i} = {p ∈ P| d(x, p) ≤ i}. This set is always a subspace. Indeed, take a line L which has two points α and β in common with P_{x}^{i}. Since d(x, α) ≤ i and d(x, β) ≤ i, it follows that there must be a point γ on L such that d(x, γ) ≤ i − 1. Hence every point of L has distance at most i to x. The set P_{x}^{i} induces a partial linear space S_{x}^{i}, called the i-neighbourhood of x. The line set of S_{x}^{i} consists of those lines of S which have distance at most i − 1 to x. Clearly S_{x}^{0}, S_{x}^{1} and S_{x}^{d} are near polygons. We prove now that S_{x}^{2} is also a near polygon.

Theorem 5.1.1 If x is a point of a near polygon S, then S_{x}^{2} is a near polygon.

Proof. Let d_{2}(·, ·) denote the distance in S_{x}^{2}. Take any point p and any line L of S_{x}^{2}. Let p′ denote the unique point of L nearest to p (in S). If d_{2}(p, L) = 0, there is nothing to prove. If d_{2}(p, L) = 1, then d(p, L) = 1 and p′ is the unique point of L nearest to p (in S_{x}^{2}). If d_{2}(p, L) = 2, then d(p, L) = 2 and p′ is the unique point of L nearest to p (in S_{x}^{2}). So we may suppose that d_{2}(p, L) = 3. This implies that d(p, x) = 2 and d(x, L) = 1. Let α denote the point of L which is collinear with x. Any path of length ≤ 2 between p and α is contained in P_{x}^{2}, so d(p, α) = d_{2}(p, α) = 3. Let s denote another point of L for which d_{2}(p, L) = 3. Let (p, q, r, s) be a path in S_{x}^{2} connecting p and s. If d(x, q) = 1, then d(q, α) = 2 and d(q, s) = 2. So, there is a point β of L, collinear with q. The path (p, q, β) is contained in P_{x}^{2}, a contradiction. If d(x, r) = 1, then since d(p, x) = 2 = d(p, r) there is a point γ of xr which is collinear with p. Now d(γ, α) = d(γ, s) = 2, so there is a point δ of L collinear with γ. The path (p, γ, δ) is a path in P_{x}^{2}, a contradiction. So we may suppose that d(x, p) = d(x, q) = d(x, r) = 2. There exists a point a on pq and a point b on qr such that d(x, a) = d(x, b) = 1. Since d(b, s) = 2 and d(b, α) ≤ 2, there exists a point c on L collinear with b. Since d(p, b) = d(p, x) = 2, there exists a point d of xb collinear with p. If α = c then xb = xα, so d is collinear with α, contradicting d(p, α) = 3. Suppose α ≠ c. Since d(d, α) = d(d, c) = 2, there exists a point e of L which is collinear with d. The path (p, d, e) is contained in P_{x}^{2}, a contradiction. This proves that α is the unique point of L such that d_{2}(p, α) = 3. □

{Theorem 5.1.2 \atop (1)} If S is a thin near polygon, then S_{x}^{i} is a thin near polygon, for every integer i.

(2) If S is a generalized 2d-gon, then S_{x}^{i} is a near polygon for every i ∈ ℕ.

(3) If S is a dual polar space, then S_{x}^{i} is a near polygon for every i ∈ ℕ.

Proof.

(1) The point graph of S_{x}^{i} remains bipartite and connected.

(2) We may suppose that 0 < i < d. In that case, S_{x}^{i} is a degenerate near polygon, whose nondegenerate support is a point.

(3) Let d_{i}(·, ·) be the distance in S_{x}^{i}. It suffices to prove that d(p, q) = d_{i}(p, q) for all p, q ∈ P_{x}^{i}. Clearly d(p, q) ≤ d_{i}(p, q). Consider a path p = b_{0}, b_{1}, …, b_{k} = q of length k = d(p, q) such that d(x, b_{0}) + ⋯ + d(x, b_{k}) is minimal. If there exists an index j such that d(x, b_{j}) > i, then one of the following two cases certainly occurs.

• There exists an index l such that d(x, b_{l}) = m, d(x, b_{l+1}) = d(x, b_{l+2}) = m + 1. Define Q as the quad through b_{l}, b_{l+1} and b_{l+2} and define α as the unique point in Q nearest to x. It is impossible that α = b_{l}, otherwise d(x, b_{l+2}) = m + 2. The point α is collinear with b_{l} and has distance m − 1 to x. Call β the point on b_{l}α at distance 1 of b_{l+2}. The path (b_{l}, b_{l+1}, b_{l+2}) can be replaced by (b_{l}, β, b_{l+2}), a contradiction.

• There exists an l such that d(x, b_{l}) = d(x, b_{l+2}) = m, d(x, b_{l+1}) = m+1. Define Q as the quad through b_{l}, b_{l+1} and b_{l+2} and define α as the unique point in Q nearest to x. The point α is a common neighbour of b_{l} and b_{l+2}, so the path (b_{l}, b_{l+1}, b_{l+2}) can be replaced by (b_{l}, α, b_{l+2}), yielding a contradiction. □

One might wonder whether S_{x}^{i} is a near polygon for all choices of S, x and i. This is not the case, as shown in the following theorem. We refer to Section 3.2.5 for the construction of the regular near octagon involved.

Theorem 5.1.3 Let x be a point of the unique regular near octagon with parameters s = 2, t = 4, t_{2} = 0, t_{3} = 3, then S_{x}^{3} is not a near polygon.

Proof. Let d_{3}(·, ·) denote the distance in S_{x}^{3}. Let y ∈ Γ_{4}(x), {y_{1}, y_{2}} ⊂ Γ_{3}(x)∩Γ(y), y_{1} ≠ y_{2}, y_{3} ∈ Γ(y_{2})∩Γ_{2}(x) and let y_{4} be the third point of the line y_{3}y_{2}. Now y_{3} ≠ y_{1} ≠ y_{4}, d(y_{3}, y_{1}) ≠ 1 ≠ d(y_{4}, y_{1}) since t_{2} = 0, d(y_{1}, y_{3}) ≠ 2 since d(y_{1}, y_{4}) ≠ 1 and d(y_{1}, y_{4}) ≠ 2 since d(y_{1}, y_{3}) ≠ 1. So d(y_{1}, y_{3}) = d(y_{1}, y_{4}) = 3. There are four shortest paths (y_{1}, α, β, y_{i}) connecting y_{1} and y_{i} (i = 3, 4). Since t − t_{3} = 1, only two of these paths can have d(x, α) = 4 or d(x, β) = 4. This means that d_{3}(y_{1}, y_{3}) = d_{3}(y_{1}, y_{4}) = 3, but d_{3}(y_{1}, y_{2}) ≥ 3, so S_{x}^{3} is not a near polygon. □

Problem. Are there other classes of near polygons for which the i-neigh-bourhoods are again near polygons?

5.2 Construction of near polygons from partial linear spaces

5.2.1 Main theorem

In this section, we prove the following theorem.

Theorem 5.2.1 Let G be an arbitrary partial linear space, then there exists a near polygon S and a point (∞) of S such that the local space at (∞) is isomorphic to G.

In order to prove this theorem, we have to construct a near polygon from an arbitrary partial linear space. The near polygons from this construction have the property that there exists a point at distance at most 2 from all the other points and this property almost characterizes these near polygons as we shall see in Section 5.3.

5.2.2 Construction

Let G be an arbitrary partial linear space with point set A and line set B. We may and will suppose that the lines of B are sets of points. Let G^{∗} be the geometry obtained from G by deleting all the lines of length 2 and call C_{i}, 1 ≤ i ≤ n, the point sets of the connected components of G^{∗}.

Remark. Suppose that G is the local space at the point (∞) of a near polygon S. Every G-point p corresponds with an S-line through (∞), which is incident with s(p) + 1 S-points. Every G-line L corresponds with a quad through (∞). If L contains at least three G-points, then the corresponding quad is not a grid. Hence s(p) is independent of the point p of L; we define s(L) := s(p), where p is an arbitrary point of L. If we fix a certain connected component C_{i}, then s(p) is independent of the point p of C_{i}; we define s_{i} := s(p), where p is an arbitrary point of C_{i}. If L, |L| ≥ 3, is a G-line contained in C_{i}, then the quad defined by L is a dual grid or has order (s_{i}, |L| − 1). In both cases there exists a GQ of order (s_{i}, |L| − 1). This remark explains the following definitions.

For every i with 1 ≤ i ≤ n we can choose a natural number s_{i} = s(C_{i}) ≥ 1 as follows.

• If |C_{i}| = 1, choose s_{i} ≥ 1 at random.

• If |C_{i}| > 1, choose s_{i} ≥ 1 such that for every G-line L, |L| ≥ 3, contained in C_{i}, there exists a GQ(s_{i}, |L| − 1). Clearly s_{i} = 1 is always a feasible choice for s_{i}.

For every L ∈ B with |L| ≥ 3, we define s(L) = s(C_{i}), where C_{i} is the component containing L. For every p ∈ A, we define s(p) = s(C_{i}), where C_{i} is the component containing p.

Now we will define a partial linear space S = (P, L, I), where I is the inclusion. The elements of P are as follows:

(A) (∞);

(B) (p, i) with p ∈ A and 1 ≤ i ≤ s(p);

(C) (L, i) with L ∈ B and 1 ≤ i ≤ k(L), here k(L) ≥ 1 is at random if s(p) = 1 for all points p on L, k(L) = s(p_{1}) · s(p_{2}) if L = {p_{1}, p_{2}} and s(p_{1}) · s(p_{2}) > 1, k(L) = [s(L)]^{2}(|L| − 1) if |L| ≥ 3 and s(L) ≠ 1.

The lines of L are obtained as follows.

(1) The set {(∞)} ∪ {(p, i)|1 ≤ i ≤ s(p)} is a line for every p ∈ A.

(2) Consider a line L ∈ B with |L| = 2, L = {p_{1}, p_{2}}, s(p_{1}) · s(p_{2}) > 1. The set {(∞)}∪{(p_{1}, i)|1 ≤ i ≤ s(p_{1})}∪{(p_{2}, i)|1 ≤ i ≤ s(p_{2})}∪{(L, i)|1 ≤ i ≤ s(p_{1}) · s(p_{2})} contains (s(p_{1}) + 1) · (s(p_{2}) + 1) points, so we can give it the structure of an (s(p_{1}) + 1) × (s(p_{2}) + 1)-grid by introducing s(p_{1}) + s(p_{2}) + 2 lines; two of those lines were already defined in (1).

(3) Consider a line L ∈ B with |L| = t_{L} + 1 ≥ 3 and s(L) ≠ 1. The set {(∞)} ∪ {(p, i)|p ∈ L, 1 ≤ i ≤ s(L)} ∪ {(L, i)|1 ≤ i ≤ [s(L)]^{2}t_{L}} contains (1+s(L))·(1+s(L)t_{L}) elements, so we can give it the structure of a GQ(s(L), t_{L}) by introducing (1 + t_{L})(1 + s(L)t_{L}) lines; 1 + t_{L} of those lines were already defined in (1).

(4) Consider a line L ∈ B for which s(p) = 1 for every point p on L. The set {(∞)}∪{(p, 1)|p ∈ L}∪{(L, i)|1 ≤ i ≤ k(L)} can be given the structure of a dual grid with {(p, 1)|p ∈ L} and {(∞)} ∪ {(L, i)|1 ≤ i ≤ k(L)} as two maximal cocliques of the point graph.

If the below defined assumptions (AS1) and (AS2) are satisfied, then we can add additional points and lines; however we are not obliged to do this.

The first assumption is as follows.

(AS1) There exist a G-point p for which s(p) = 1.

The following graph Δ can then be defined. The vertices of Δ are G-points p for which s(p) = 1. Two G-points are adjacent when they are not collinear in G.

The second assumption is as follows.

(AS2) There is a set of (at least two) cliques in Δ, such that for every clique of this set, there exists another clique intersecting it in at least two points.

Let N_{1}, …, N_{l} denote such a set of l cliques. For each i ∈ {1, …, l}, let u_{i} ≥ 1 be arbitrary. We add now the following points:

(D) (N_{i}, j) with 1 ≤ i ≤ l and 1 ≤ j ≤ u_{i},

and the following lines:

(5) {(p, 1), (N_{i}, j)} with 1 ≤ i ≤ l, 1 ≤ j ≤ u_{i} and p ∈ N_{i}.

Theorem 5.2.2 S is a near polygon and the local space S_{(∞)} is isomorphic to G.

Proof. Let Q_{1}, …, Q_{r} be all the sets mentioned in (2) and (3) above, and which we gave the structure of a GQ. Put Q_{r+1} = {(∞)} ∪ {(p, 1)|s(p) = 1} ∪ {(L, i)|s(p) = 1, ∀p ∈ L and 1 ≤ i ≤ k(L)} ∪ {(N_{i}, j)|1 ≤ i ≤ l and 1 ≤ j ≤ u_{i}}. The set Q_{r+1} is a subspace inducing a point or a connected bipartite graph. We prove by induction that Q_{1} ∪ … ∪ Q_{i}, 1 ≤ i ≤ r + 1, is a subspace inducing a near polygon. Since P = Q_{1} ∪ … ∪ Q_{r+1}, the theorem is then proved. Suppose Q_{1} ∪ … ∪ Q_{i}, 1 ≤ i ≤ r, is a subspace inducing a near polygon S_{1} and let S_{2} be the near polygon induced by Q_{i+1}. We apply Theorem 3.10.1. Put X = X_{1} = X_{2} = (Q_{1} ∪ … ∪ Q_{i}) ∩ Q_{i+1} and let α be the identical map. The set X consists of a number of lines through (∞); hence d_{2}(α(x), α(y)) = d_{1}(x, y) for all points x, y ∈ X. Also, X is a subspace of S_{2}. It suffices to prove that X is a classical subset of S_{1}. Take x ∈ (Q_{1}∪…∪Q_{i})\X arbitrary. Suppose that there are two points x_{1}, x_{2} ∈ X at smallest distance from x (in S_{1}). By (NP), x_{1} and x_{2} are not collinear, hence x_{1} ≠ (∞) ≠ x_{2}. By (NP), we have d(x, (∞)) = 2. Let Q_{j} be the set for which x ∈ Q_{j} and let L_{i}, 1 ≤ i ≤ 2, be the line through (∞) and x_{i}. The lines L_{1} and L_{2} are contained in Q_{j} and Q_{i+1}. Let p_{i} be the G-point corresponding with the S-line L_{i}. If i < r, then p_{1} and p_{2} are contained in two G-lines, a contradiction. If i = r, then s(p_{1}) = s(p_{2}) = 1. Hence the G-line through p_{1} and p_{2} satisfies s(p) = 1 for all points of this line, but this is impossible since Q_{j} is not a dual grid.

By construction we have that S_{(∞)} ≃ G. (Remark that there are no quads through (∞) and a point of type (5) because of condition (AS2).) □

The above given proof is different from the one given in [31]. This latter proof, which will be given now, makes no use of Theorem 3.10.1.

Proof. Let x denote the point (∞). For every point y at distance 2 from x, we define the set S(x, y) = {L ∈ L|x ∈ L and L contains a point collinear with y}. Notice that if m is a point collinear with y but at distance 2 from x, then S(x, y) = S(x, m) is equal to the set of lines through x in the quad defined by x and y.

Let p be a point and let L be a line of S, then we have to prove that there exists a unique point on L nearest to p. This is certainly true if p ∈ L or p = x. So, suppose that p ≠ x and p ∉ L. The following statements are trivial or are readily obtained.

• If d(p, x) = 1 and x ∈ L, then x is the unique point of L nearest to p.

• If d(p, x) = 1 and x ∉ L, let a be the point of L at distance 1 from x and let b be any other point of L. If xp ∈ S(x, b), then there exists a unique point of L at distance 1 from p. If xp ∉ S(x, b), then a has distance 2 from p and every other point of L has distance 3 to p.

• If d(p, x) = 2, x ∈ L, L ∈ S(x, p) then there exists a unique point of L at distance 1 from p.

• If d(p, x) = 2, x ∈ L, L ∉ S(x, p), then x has distance 2 to p and every other point of L has distance 3 to p.

• If d(p, x) = 2, x ∉ L, let a be the point of L collinear with x and b any other point of L. If d(p, L) = 1, then there exists a unique point of L at distance 1 from p. Suppose therefore that d(p, L) ≥ 2. If S(x, b)∩S(x, p) ≠ ∅, then there exists a unique point of L at distance 2 from p, all the other points have distance 3 to p. If S(x, b)∩S(x, p) = ∅, then a has distance 3 to p and every other point of L has distance 4 to p.

□

Remarks.

(a) If G is a partial linear space without isolated points, then near polygons arising from the construction are nondegenerate; hence, the above construction can be used to construct many nondegenerate near polygons.

(b) The main property of the constructed near polygons is the existence of a point at distance at most 2 from all the other points. This property characterizes, up to degeneracy, all these near polygons as we shall show in the following section.

5.3 The 2-neighbourhood of a point

Let S = (P, L, I) be a near polygon and let x be a fixed point of P. Define the following sets:

V_{2} = {y ∈ P| d(x, y) = 2 and |Γ(x) ∩ Γ(y)| ≥ 2},
V_{1} = {z ∈ P|∃y ∈ V_{2} such that z ∈ Γ(x) ∩ Γ(y)},
P′ = {x} ∪ V_{1} ∪ V_{2}.

Theorem 5.3.1 The point set P′ is a subspace.

Proof. Consider a line L having two of its points, say α and β, in P′. We may suppose that L contains at least three points. If x is incident with L, then we may suppose that α ∈ V_{1}. So, there exists a point y ∈ V_{2} such that α is a common neighbour of x and y. By Yanushka’s theorem, the points x and y determine a quad Q. Hence, for every point γ ≠ x of L there exists a point y′ of V_{2} ∩ Q such that γ is a common neighbour of x and y′. If x is not incident with L, we may suppose that d(x, α) = 2. The points x and α determine a quad and once again it is clear that every point of L belongs to V_{1} or V_{2}. □

Let S′ be the partial linear space induced by P′.

Lemma 5.3.2 The nondegenerate support of S_{x}^{2} is S′, hence S′ is a near polygon.

Proof. If y ∈ P with d(x, y) = 2 and y ∉ V_{2}, then there is a unique line through y which belongs to L_{x}^{2}. Hence y does not belong to the nondegenerate support of S_{x}^{2}. If z ∈ P with d(x, z) = 1 and z ∉ V_{1}, then z does not belong to the nondegenerate support of S_{x}^{2} as otherwise this would imply the existence of a point u ∈ V_{2}, collinear with z. □

Theorem 5.3.3 If P′ ≠ {x}, then S′ is a near polygon which can be obtained by the construction given in the previous section.

Proof. Let G = (A, B, I) be the local space of S′ at the point x. Every point p of A corresponds to a line through x which is incident with s(p) + 1 points. Every line L of B corresponds to a quad through x; let k(L) be the number of points in the quad at distance 2 from x. Let y be a point of S′ at distance 2 from x with the property that there is no quad through x and y. Any common neighbour p of x and y satisfies s(p) = 1, hence we can define the graph Δ as above. The set of lines through x containing a point collinear with y defines then a clique N_{i} in Δ and we say that the clique N_{i} can be constructed u_{i} times in the just described way.

The point x of S′ is now the unique point of type (A), the points of S′ at distance 1 from x are the points of type (B), the points of S′ at distance 2 from x which are contained in a quad through x are the points of type (C), and the points of S′ at distance 2 from x which are not contained in a quad through x are the points of type (D). □

Corollary 5.3.4 If S is a near 2d-gon which satisfies d(x, y) ≤ 2 for all y ∈ P, then its nondegenerate support, if not a single point, can be obtained by the construction of Section 5.2.2.

Linear representations of near polygons

Throughout this chapter, Π_{∞} is a PG(n, q), n ≥ 0, which is embedded as a hyperplane in a projective space Π = PG(n + 1, q) and K is a nonempty set of points of Π_{∞}.

6.1 Generalities

With every point p ∈ Π_{∞}, we associate an element i_{K}(p) ∈ ℕ ∪ {+∞}, called the (generating) index of p with respect to K:

• if p ∉ ⟨K⟩, then i_{K}(p) = +∞,

• if p ∈ ⟨K⟩, then i_{K}(p) = m, where m is the smallest integer with the property that there are m points of K generating a subspace containing p.

Clearly i_{K}(p) = 1 if and only if p ∈ K.

Lemma 6.1.1 If x and y are 2 different points of T_{n}^{∗}(K) and if z is the intersection of xy with Π_{∞} then d(x, y) = i_{K}(z), where d(·, ·) denotes the distance in the point graph of T_{n}^{∗}(K).

Proof.

• i_{K}(z) ≥ d(x, y).

This is true for i_{K}(z) = 1 or i_{K}(z) = +∞. For 2 ≤ i_{K}(z) < +∞ we use induction. Assume i_{K}(z) = m and let a_{1}, …, a_{m} be m points of K such that z ∈ ⟨a_{1}, …, a_{m}⟩. The space ⟨y, a_{1}, …, a_{m}⟩ is then m-dimensional while ⟨y, a_{1}, …, a_{m−1}⟩ is (m − 1)-dimensional. Hence, the line xa_{m} intersects ⟨y, a_{1}, …, a_{m−1}⟩ in a point u. Now u ≠ y, so uy intersects Π_{∞} in a point of ⟨a_{1}, …, a_{m−1}⟩ and by induction d(u, y) ≤ m − 1. Together with d(x, u) = 1 this implies that d(x, y) ≤ m.

• d(x, y) ≥ i_{K}(z). We may suppose that d(x, y) < +∞. Let x = a_{0}, …, a_{l} = y be a path of length l = d(x, y) between x and y. The points a_{0}, …, a_{l} generate a subspace of Π which intersects Π_{∞} in a subspace β. Let b_{i}, i ∈ {1, …, l}, be the intersection of the line a_{i−1}a_{i} with Π_{∞}, then b_{i} ∈ K and ⟨a_{0}, b_{1}, b_{2}, …, b_{l}⟩ = ⟨a_{0}, a_{1}, …, a_{l}⟩. Hence β = ⟨b_{1}, b_{2}, …, b_{l}⟩ and since z ∈ β, we have i_{K}(z) ≤ l.

□

Corollary 6.1.2 T_{n}^{∗}(K) is connected if and only if K generates Π_{∞}.

Lemma 6.1.3 Consider in Π_{∞} two disjoint subspaces π_{1} and π_{2} with respective dimensions n_{1} and n_{2} such that ⟨π_{1}, π_{2}⟩ = Π_{∞} (n = n_{1} + n_{2} + 1). Let K_{i} (i = 1, 2) be a nonempty set of points of π_{i}. Then T_{n}^{∗}(K_{1} ∪ K_{2}) is isomorphic to the direct product T_{n_{1}}^{∗}(K_{1}) × T_{n_{2}}^{∗}(K_{2}).

Proof. Let Π_{i} (i = 1, 2) be an (n_{i}+1)-dimensional subspace of Π intersecting Π_{∞} in π_{i}. We can consider the points of T_{n_{i}}^{∗}(K_{i}) as the affine points of Π_{i}. Let a be an arbitrary point of T_{n}^{∗}(K_{1} ∪ K_{2}). The space ⟨a, π_{i}⟩ intersects Π_{3−i} in a point a_{3−i}. So, with every point a there corresponds a point (a_{1}, a_{2}) of the direct product T_{n_{1}}^{∗}(K_{1}) × T_{n_{2}}^{∗}(K_{2}). This correspondence is bijective. With every point (a_{1}, a_{2}) of the direct product there corresponds a unique point a: the point a is the point in the intersection of ⟨a_{1}, π_{2}⟩ with ⟨a_{2}, π_{1}⟩. One easily verifies that this bijection also determines a one to one mapping from the set of lines of T_{n}^{∗}(K_{1} ∪ K_{2}) to the set of lines of the direct product T_{n_{1}}^{∗}(K_{1}) × T_{n_{2}}^{∗}(K_{2}). □

Theorem 6.1.4 Two points of T_{n}^{∗}(K) at distance 2 of each other have at least two common neighbours.

Proof. Let x and y be two points of T_{n}^{∗}(K) at mutual distance 2 and let z be a common neighbour. Let u, respectively v, be the intersection point of zx, respectively zy, with Π_{∞}. The intersection point of the lines uy and vx is then a second common neighbour of x and y. □

6.2 Linear representations of near polygons

K satisfies the near polygon property if for every point p ∈ K and for every line L (of Π_{∞}) through p, there is a unique point q ∈ L \ {p} with smallest index.

Theorem 6.2.1 T_{n}^{∗}(K) is a near polygon if and only if K satisfies the near polygon property.

Proof. This is trivial for n = 0. So, we may suppose that n > 0. Suppose that K satisfies the near polygon property. Let a be a point and let L be a line of T_{n}^{∗}(K). We may suppose that a ∉ L, so a and L determine a plane α which intersects Π_{∞} in a line L′. Let p be the common point of L and L′ and let f : L → L′ be the projection such that f(x) = y if and only if x, y and a are collinear. Since d(a, x) = i_{K}(f(x)) for all x ∈ L \ {p}, there is a unique affine point b on L nearest to a. Since there are at least two affine points on L it follows that d(a, b) < +∞. Conversely, suppose that T_{n}^{∗}(K) is a near polygon. Take a point p of Π_{∞} and a line L of Π_{∞} through p. Let M be a line of T_{n}^{∗}(K) through p and let a be a point in the plane ⟨M, L⟩ but not on M or L. Using the projection from M to L, it is immediate that there is a unique point q ∈ L \ {p} with smallest index. □

Corollary 6.2.2 (1) T_{n}^{∗}(K) is a near 2d-gon if and only if

(a) for every point p ∈ K and every line L (of Π_{∞}) through p, there is a unique point r ≠ p on L with smallest index,

(b) the maximal index of a point of Π_{∞} is equal to d.

(2) T_{n}^{∗}(K) is a near (2d + 1)-gon if and only if

(a) the maximal index of a point of Π_{∞} is equal to d,

(b) there exists a point p ∈ K and a line L (of Π_{∞}) through p such that every point on L different from p has index d,

(c) for every point p ∈ K and every line L (of Π_{∞}) through p containing a point (≠ p) with index different from d, there is a unique point r ≠ p on L with smallest index.

Proof. Property (1) is trivial. The properties (a), (b) and (c) given in (2) are respectively equivalent to the properties (1), (3) and (2) from Section 1.2.2. □

Concerning the existence of quads, one can say the following.

Theorem 6.2.3 Suppose q ≥ 3. If T_{n}^{∗}(K) is a near polygon, then every two points at distance 2 are contained in a unique quad.

Proof. This follows immediate from Theorems 3.5.2 and 6.1.4. □

6.3 Linear representations of generalized quadrangles

Result 6.3.1 T_{n}^{∗}(K) is a generalized quadrangle if and only if every line of Π_{∞} intersects K in 0 or 2 points.

This yields the following examples.

• If n = 1, then Π is a PG(2, q), Π_{∞} is a line of Π and K is a set of 2 points of Π_{∞}. In this case T_{1}^{∗}(K) is a (q × q)-grid.

• If n = 2, then Π_{∞} is a PG(2, q) with q even and K is a hyperoval in Π_{∞}. In this case T_{2}^{∗}(K) is a generalized quadrangle of order (q − 1, q + 1).

• If Π_{∞} is a PG(n, 2) with n ≥ 3, then the complement of K is a subspace and hence a hyperplane, since |K| = 2^{n}.

• There are no examples if Π_{∞} is a PG(n, q) with n, q ≥ 3. Otherwise, there would exist a 3-dimensional subspace β of Π_{∞} such that β ∩ K is a cap of β with q^{2} + q + 2 elements, a contradiction (see Section 1.4.1).

Remark. It is possible to determine all affine embeddings of generalized quadrangles, see [86].

6.4 Some constructions of linear representations of near polygons

The following theorem illustrates how some linear representations of near polygons can be constructed.

Theorem 6.4.1 Consider in Π_{∞} two disjoint subspaces Π_{1} and Π_{2} of dimensions n_{1} ≥ 0 and n_{2} ≥ 0, such that Π_{∞} = ⟨π_{1}, π_{2}⟩. Let K_{i}, i ∈ {1, 2}, be a set of points in Π_{i} and put K = K_{1} ∪ K_{2}. If T_{n_{i}}^{∗}(K_{i}), i ∈ {1, 2}, is a near 2d_{i}-gon, then T_{n}^{∗}(K) is a near 2(d_{1} + d_{2})-gon.

Proof. This is immediate since T_{n}^{∗}(K) ≃ T_{n_{1}}^{∗}(K_{1}) × T_{n_{2}}^{∗}(K_{2}). □

Example. Consider in Π_{∞} a set K of n+1 points generating the space, then T_{n}^{∗}(K) is a regular near 2(n + 1)-gon of Hamming type.

Theorem 6.4.2 Assume n = 2k, k ≥ 1 and q = 2^{h} an even prime power. Take a fixed point p of Π_{∞} and consider k planes Π_{1}, …, Π_{k} through it which generate Π_{∞}. Let H_{i}, i ∈ {1, …, k}, be a hyperoval of Π_{i} containing p. Put K = H_{1} ∪ … ∪ H_{k}. Then T_{2k}^{∗}(K) is a near (2k + 2)-gon having the property that every two points at distance 2 are contained in a unique quad.

Proof. We use induction on k. The theorem is clearly true when k = 1, so suppose that k ≥ 1.

Let L be a line of Π_{∞} through p. We may suppose that L is not contained in ⟨Π_{1}, …, Π_{k−1}⟩ ∪ Π_{k}. The space ⟨L, Π_{k}⟩ intersects ⟨Π_{1}, …, Π_{k−1}⟩ in a line L_{1} through p and the space ⟨L, Π_{1}, …, Π_{k−1}⟩ intersects Π_{k} in a line L_{2} through p. Clearly ⟨L, L_{1}⟩∩Π_{k} = L_{2}. Let p_{i}, i ∈ {1, 2}, be the unique point of L_{i}\{p} with smallest index (hence p_{2} ∈ K). We now will prove that the point in the intersection of p_{1}p_{2} with L is the unique point of L \ {p} with smallest index. Let f be the smallest positive integer with the property that there exist f points a_{1}, …, a_{f} ∈ K such that ⟨a_{1}, …, a_{f}⟩ ∩ (L \ {p}) ≠ ∅. Clearly p ∉ {a_{1}, …, a_{f}}. Suppose that a_{1}, …, a_{l} ∈ H_{1}∪…∪H_{k−1} and a_{l+1}, …, a_{f} ∈ H_{k} for some l ∈ {1, …, f −1}. Every point r of ⟨a_{1}, …, a_{f}⟩∩(L\{p}) belongs to a line r_{1}r_{2} with r_{1} ∈ ⟨a_{1}, …, a_{l}⟩ and r_{2} ∈ ⟨a_{l+1}, …, a_{f}⟩. Clearly r_{1} ∈ L_{1}, r_{2} ∈ L_{2} and i_{K}(r) = f = l + (f − l) = i_{K}(r_{1}) + i_{K}(r_{2}). It follows immediately that L \ {p} contains a unique point with smallest index, namely the point L ∩ p_{1}p_{2}.

Let r ∈ H_{k} \ {p} and let L be a line of Π_{∞} through r. We may suppose that L is not contained in Π_{k}. The space ⟨L, Π_{k}⟩ intersects ⟨Π_{1}, …, Π_{k−1}⟩ in a line L′ through p. Let f be the smallest integer with the property that there exist f points a_{1}, …, a_{f} ∈ K such that ⟨a_{1}, …, a_{f}⟩ ∩ (L \ {r}) ≠ ∅. Clearly r ∉ {a_{1}, …, a_{f}}. Suppose that a_{1}, …, a_{l} ∈ (H_{1} ∪ … ∪ H_{k−1}) and a_{l+1}, …, a_{f} ∈ H_{k} \ {p} for some l ∈ {1, …, f}. If l = f, then L meets ⟨Π_{1}, …, Π_{k−1}⟩ in a point of L′. This point is a point of L \ {r} with smallest index. Suppose l ≠ f, then every point s of ⟨a_{1}, …, a_{f}⟩ ∩ (L \ {r}) belongs to a line s_{1}s_{2} with s_{1} ∈ ⟨a_{1}, …, a_{l}⟩ and s_{2} ∈ ⟨a_{l+1}, …, a_{f}⟩. Clearly s_{1} ∈ L′ and s_{1} ∉ L. The plane ⟨s_{1}, L⟩ meets Π_{k} in a line L′′ through r. Let s′_{1} be the second point of H_{k} on L′′. The line s_{1}s′_{1} meets L \ {r}. Hence f = l + 1 = i_{K}(s_{1}) + 1. The following conclusions are now straightforward. If L meets ⟨Π_{1}, …, Π_{k−1}⟩ in a point, then this point is the unique point of L\{r} with smallest index. If L does not meet ⟨Π_{1}, …, Π_{k−1}⟩, consider then the intersection L′ of ⟨L, Π_{k}⟩ with ⟨Π_{1}, …, Π_{k−1}⟩. Let s_{1} be the unique point of L′ \ {p} with smallest index. The plane ⟨s_{1}, L⟩ intersects Π_{k} in a line L′′ through r. Let s′_{1} be the second point of K on L′′. The intersection of the line s_{1}s′_{1} with L is then the unique point of L \ {r} with smallest index. So far, we have proved that T_{n}^{∗}(K) is a near polygon.

We prove now (again by induction on k) that k +1 is the maximum index of a point of Π_{∞}. This statement is true for k = 1; so suppose that k > 1. Take a point r ≠ p and consider the line pr. Above, we showed how we can construct the point of pr with smallest index. By induction we have that i_{K}(r) ≤ k + 1. Take now a point t_{1} of ⟨Π_{1}, …, Π_{k−1}⟩ with index k and let t_{2} ∈ H_{k} \ {p}. If t_{3} is a point of t_{1}t_{2} \ {t_{1}, t_{2}}, then i_{K}(t_{3}) = k + 1. This proves that T_{n}^{∗}(K) is a near (2k + 2)-gon.

If q ≥ 3, then we already know that T_{n}^{∗}(K) satisfies the property that every two points at distance 2 are contained in a unique quad. We construct now explicitely the quads and this will show that this property also holds if q = 2. Let x and y be two points of T_{n}^{∗}(K) at distance 2. The line xy intersects Π_{∞} in a point z. If z ∈ Π_{i}, i ∈ {1, …, k}, then the affine points in ⟨x, Π_{i}⟩ define a geodetically closed subspace of T_{n}^{∗}(K) inducing a GQ isomorphic to T_{2}^{∗}(H_{i}). If the above defined set would not be geodetically closed, then Π_{i} is contained in a 3-space β such that |β ∩ K| ≥ q + 4, a contradiction. If z ∈ z_{1}z_{2} with z_{1} ∈ H_{i} \ {p} and z_{2} ∈ H_{j} \ {p}, i ≠ j, then the affine points in the plane ⟨z_{1}, z_{2}, x⟩ define a geodetically closed subspace of T_{2k}^{∗}(K) inducing a grid. If the above defined set would not be geodetically closed, then z_{1}z_{2} is contained in a plane α such that |α ∩ K| ≥ 4, a contradiction. □

We construct some examples in the case q = 2.

Theorem 6.4.3 Let Π_{∞} be a PG(n, 2), n ≥ 0, embedded as a hyperplane in Π = PG(n+1, 2), let π be a hyperplane in Π_{∞} and let K be a set of points of Π_{∞} \ π generating Π_{∞}. Then T_{n}^{∗}(K) is a near polygon.

Proof. The geometry T_{n}^{∗}(K) is connected since K generates Π_{∞}. In the space Π there are two hyperplanes through π and different from Π_{∞}. Every line of T_{n}^{∗}(K) has one point in common with each of the two hyperplanes and as a consequence T_{n}^{∗}(K) is a thin near polygon since it is a connected bipartite graph. □

Now, let Π_{∞} be a PG(n, 2) with n ∈ ℕ \ {0}. Take a point p of Π_{∞} and consider k ≥ 1 subspaces π_{1}, …, π_{k} of dimension at least 1 through it which generate Π_{∞}. Let n_{i} (i ≥ 1) be the number of i-dimensional subspaces, hence ∑_{i≥1} n_{i} = k. We make the assumption that n = ∑_{i≥1} in_{i}. Now, choose in each π_{i} a hyperplane not through p and let V_{i} be the set of points of π_{i} not in that hyperplane. Finally, put K = V_{1} ∪ … ∪ V_{k}. Then T_{n}^{∗}(K) is a near (2k + 2)-gon. Indeed, in each of the k subspaces π_{k}, we had chosen a hyperplane. The subspace π of Π_{∞} generated by those hyperplanes is itself a hyperplane of Π_{∞} and the result follows from the previous theorem.

6.5 Linear representations of near hexagons

In this section, we suppose that n ≥ 2 and q ≥ 3. Our aim is to determine all sets K such that T_{n}^{∗}(K) is a near hexagon.

6.5.1 Necessary and sufficient conditions

From the results of Section 6.2, we may conclude the following result.

Result 6.5.1 T_{n}^{∗}(K) is a near hexagon if and only if the following conditions are satisfied.

(1) Every line of Π_{∞} intersects K in 0, 1 or 2 points, hence K is a cap of Π_{∞}.

(2) There is a tangent line and on every tangent line there is a unique point different from the intersection point, through which there is a secant line.

Proof. We prove that conditions (1) and (2) are equivalent to conditions (1a) and (1b) from Corollary 6.2.2 (the case d = 3). Suppose that (1) and (2) hold. Let p ∈ K and let L be any line through p. If L is a secant line, then by (1) there exists a unique point of K on L. This point is the unique point of L\{p} with smallest index (i.e. index 1). If L is a tangent line, then there exists a unique point of L \ {p} through which there is a secant line. This point is the unique point of L \ {p} with smallest index (i.e. index 2). Hence (a) is satisfied. Take now an arbitrary point r of Π_{∞}. We will prove that i_{K}(r) ≤ 3. We may suppose that r ∉ K. Let p ∈ K be arbitrary. If rp is a secant line, then i_{K}(r) = 2. If rp is a tangent line, then there exists a unique secant line p_{1}p_{2} (with p_{1}, p_{2} ∈ K) meeting rp in a point different from p. Since r ∈ ⟨p, p_{1}, p_{2}⟩, we have that i_{K}(r) ≤ 3. It suffices now to prove that there exists a point r with i_{K}(r) ≥ 3. Such a point always exists on every tangent line. Suppose now that properties (1a) and (1b) from Corollary 6.2.2 hold. If L is a line intersecting K in more than two points, then (a) would not be satisfied. Hence (1) is satisfied. Take now a point r with i_{K}(r) = 3, then rp is a tangent line for every p ∈ K. Let L be an arbitrary tangent line with tangent point p. There exists a unique point r in L \ {p} with smallest index. This index is necessary equal to 2 and r is the unique point through L \ {p} through which there is a secant line. Hence also (2) is satisfied. □

6.5.2 The known examples

If we consider only linear representations of near hexagons in AG(n + 1, q) with q > 2, then the following examples are known.

(1) Let Π_{∞} be a desarguesian projective plane PG(2, q) and let K be a set of three noncollinear points, then T_{2}^{∗}(K) is a near hexagon of Hamming type.

(2) Let Π_{∞} be a PG(3, q) with q even and let K consists of a hyperoval in a plane of Π_{∞} and one other point not in that plane, then T_{3}^{∗}(K) is a near hexagon which is a direct product of a line with a generalized quadrangle of order (q − 1, q + 1), see Lemma 6.1.3.

(3) Consider the near hexagon related to the extended ternary Golay code C, see Section 3.2.4. The points are the cosets from the 6-dimensional affine space 𝔽_{3}^{12}/C. Two cosets are adjacent whenever they contain vectors that differ in only 1 position, the line through them is the line of the affine space connecting them. The embedding of this affine space in a 6-dimensional projective space Π defines a hyperplane Π_{∞}. It is clear that there is a set K of 12 points in Π_{∞} such that the near hexagon is isomorphic to T_{5}^{∗}(K).

(4) Let Π_{∞} be a PG(4, q) with q even and let K be a union of two hyperovals whose carrying planes α and α′ meet each other in a point p of K. By Theorem 6.4.2, T_{4}^{∗}(K) is a near hexagon.

If there exists another linear representation of a near hexagon, then we will prove that the following holds for this linear representation.

(A) n ≥ 7

(B) q ≥ 16 and even

(C) There exists a quad isomorphic to a generalized quadrangle T_{2}^{∗}(O) and every such quad has a rosette of ovoids. Moreover, if there exists a quad isomorphic to T_{2}^{∗}(O) with O a regular hyperoval, then q ≥ 64 and a new type of ovoid in PG(3, q) exists.

6.5.3 Conditions on the set K

We suppose here that T_{n}^{∗}(K) is a near hexagon, although some of the results will also hold when we only require that T_{n}^{∗}(K) is a near polygon.

Lemma 6.5.2 Every plane of Π_{∞} intersects K in 0, 1, 2, 3 or q + 2 points. If there exists a plane α such that |α ∩ K| = q + 2, then α ∩ K is a hyperoval, hence q is even.

Proof. Assume p ∈ α ∩ K then every line of α through p has at most one point (≠ p) in common with K, hence |α∩K| ≤ q +2. Suppose 3 < |α∩K| < q + 2. Let a, b, c, d be four different points of α ∩ K. On a tangent line L in α through a, there are at least two points (≠ a) with smallest index, a contradiction. □

Definition. A plane of Π_{∞} is called thick whenever it intersects K in q + 2 points.

Lemma 6.5.3 If β is a 3-space through a thick plane, then |β ∩ K| ∈ {q + 2, q + 3}.

Proof. Let α denote the thick plane. Since β ∩ K is a cap of β, we have that |β ∩ K| ≤ q^{2} + 1. Suppose |β ∩ K| ≥ q + 4. Take two points of β ∩ K not contained in α. The line L through those points intersects α in a point a. Denote by M_{1}, …, M_{\frac{q+2}{2}} the secant lines in α through a. The plane through

L and M_{i} is thick, hence |β∩K| ≥ \frac{q+2}{2}q+2 = \frac{q^{2}+2q+4}{2}. Take a point b ∈ α∩K. Since q^{2}+q+1 > q^{2} ≥ |β∩K|−1, there exists a tangent line L′ of β through b. Every plane through L′ has at most three points in common with K. Hence |β ∩ K| ≤ 2(q + 1) + 1 = 2q + 3, contradicting |β ∩ K| ≥ \frac{q^{2}+2q+4}{2}. □

{Lemma 6.5.4 \atop (1)} Two thick planes cannot meet in a line.

(2) Two thick planes cannot meet in a point not belonging to K.

(3) If two thick planes α and β meet each other in a point of K, then the points of K in the 4-space through α and β are all contained in α ∪ β.

(4) If two thick planes α and β are disjoint, then the points of K in the 5-space through α and β are all contained in α or β.

Proof.

(1) If two thick planes meet in a line, then the 3-space through these planes has at least 2q + 2 points in common with K, a contradiction.

(2) Suppose that α and β are two thick planes meeting in a point p ∉ K. Take a line L through p, contained in α and having two points in common with K. Now, the 3-space through L and β has at least q + 4 points, a contradiction.

(3) Suppose that α and β are two thick planes meeting in a point p ∈ K. Suppose that the 4-space through α and β contains a point r ∈ K not contained in α∪β. Let Ω be the 3-space through α and r. This 3-space meets β in a line L through p. Hence, Ω contains at least q + 4 points, a contradiction.

(4) Let α and β be two disjoint thick planes. Suppose that the 5-space through α and β contains a point p ∈ K not contained in α ∪ β. Let r (respectively s) be the intersection of β (respectively α) with the 3space through p and α (respectively through p and β). It is impossible that r, s ∈ K. If one of the points, say r, is in K, then the 3-space through p and α contains at least q + 4 points of K, a contradiction. Hence r, s ∉ K, but this is also impossible, since rs \ {p} contains then two points with smallest index.

□

Theorem 6.5.5 Two points x and y at mutual distance 2 have 2 or q +2 common neighbours, x and y are in a unique quad isomorphic to a grid or a GQ of type T_{2}^{∗}(O). If there are no thick planes, then all quads are grids and the near hexagon is regular.

Proof. Since x and y are not collinear, the line xy intersects the hyperplane in a point z ∉ K. If there is no secant line through z, then d(x, y) = 3, a contradiction. From Lemma 6.5.2, it follows that there are two possibilities, either z is incident with one secant line or with \frac{q+2}{2} secant lines. Each secant line yields a contribution 2 to the number of common neighbours. For, if a secant line through z intersects K in u and v, then ux ∩ vy and uy ∩ vx are two common neighbours of x and y. If there is only one secant line L through z, then the affine points of the plane through L and x are the points of a grid Q. The set of the points of Q is geodetically closed. For, suppose that a, b, c is a path where a, b are noncollinear points of Q and c is no point of Q. The plane through a, b, c will intersect Π_{∞} in a secant line M. The plane through L and M is thick, contradicting the fact that there is only one secant line through z. If there are \frac{q+2}{2} secant lines through z, then these lines are contained in a thick plane α of Π_{∞}. The affine points of the space through x and α are the points of a generalized quadrangle Q. The set of points of Q is geodetically closed. For, suppose that a, b, c is a path where a, b are noncollinear points of Q and c is no point of Q. The plane through a, b, c intersects Π_{∞} in a secant line M. Now, the 3-space through α and M contradicts Lemma 6.5.3. □

Lemma 6.5.6 If k = |K| and if N is the total number of thick planes of Π_{∞}, then

\frac{1}{2}Nq(q^{2} − 1) + \frac{q^{n} − 1}{q − 1} − (k − 1) = \frac{1}{2}(q − 1)(k − 1)(k − 2).

Proof. Let p ∈ K be fixed. Let N_{1} be the total number of thick planes through p. Counting pairs (L, α) where L is a tangent line through p and α is a plane through L intersecting K in 3 points, yields

(N−N_{1})[(q^{2}+q+1)−(q+2)]\frac{q + 2}{2}+\frac{q^{n} − 1}{q − 1}−(k−1)−(N−N_{1})[(q^{2}+q+1)−(q+2)]

= (q − 1)[\frac{(k − 1)(k − 2)}{2} − N_{1} \frac{q(q + 1)}{2}],

from which the above equality follows. □

Corollary 6.5.7 The following congruences and inequalities are valid:

(1) k ≡ n + 1 (mod q − 1),

(2) k ≡ 2, 3 (mod q),

(3) k ≡ 1, 3 (mod q + 1) if n is even and q + 1 | (k − 2)^{2} if n is odd,

(4) \frac{q^{n}−1}{q−1} − (k − 1) ≤ \frac{1}{2}(q − 1)(k − 1)(k − 2),

(5) k ≥ \sqrt{2}q^{\frac{n−2}{2}} .

Proof. The equation of the previous lemma implies immediately (4). If we consider the equation as a congruence modulo q − 1 or q + 1, we find immediately (1) and (3). If we multiply both sides by 2 and then consider it as a congruence modulo q, we will find (2). From (4), it follows now that q^{n−1} ≤ \frac{1}{2}(q − 1)(k − 1)(k − 2) + (k − 1). Hence q^{n−1} ≤ \frac{1}{2}qk^{2}. □

6.5.4 The near hexagons with a T_{2}^{∗}(O)-quad

Assume that the near hexagon T_{n}^{∗}(K) has a thick plane, hence n ≥ 3 and q ≥ 4 is even. From Corollary 6.1.2 it follows that k = |K| ≥ q + n. We put k′ = k − q − 2.

Theorem 6.5.8 If n = 3, then K consists of a hyperoval in some plane and one other point not in that plane.

Proof. This is an immediate corollary of Lemma 6.5.3. □

Theorem 6.5.9 If n = 4, then K is the union of two hyperovals whose carrying planes meet each other in a point belonging to both hyperovals.

Proof. Let α denote a thick plane. Assume |Π_{∞} ∩ K| = q + 4 and let a and b be the two points of K not in α, then ab is skew to α (see Lemma 6.5.3). There are q^{3} − 1 lines through a skew to α and different from ab. On every such line there is a unique point (≠ a), through which there is a secant line. This secant line is one of the q + 2 lines through b and a point of α ∩ K. Hence q^{3} − 1 ≤ (q + 2)(q − 1), a contradiction. So, we may suppose that |Π_{∞} ∩ K| ≥ q + 5. Let a, b, c denote three points not contained in α. The plane β through these points meets α in a point d. If d does not belong to K, then da does not satisfy property (2) of Result 6.5.1. Hence d ∈ K and β is thick. By Lemma 6.5.4 all points of K are contained in α ∪ β. □

Theorem 6.5.10 There are no examples for n = 5.

Proof. Let n = 5 and assume K is a set of points in PG(5, q) yielding a near hexagon with a quad of type T_{2}^{∗}(O). We will find a contradiction. The proof uses several steps.

Step 1. There are at least two thick planes. Proof. Assume that there is only one thick plane. From Lemma 6.5.6, it follows that

2q^{4} + 2q^{3} + 2q^{2}(1 − k′) = k′(k′ − 1)(q − 1),

which implies q^{2} | k′ or q^{2} | k′ −1. Hence k′ ≥ q^{2} and 2q^{4} +2q^{3} +2q^{2}(1−q^{2}) ≥ q^{2}(q^{2} − 1)(q − 1), a contradiction. □

Step 2. Two thick planes α and β meet each other in a point of K. Proof. Suppose that α and β are disjoint, then by (4) of Lemma 6.5.4 all the points of K are contained in α ∪ β. Take a point b ∈ β \ K and a point a ∈ α ∩ K. Let L ≠ ab be a tangent line through a in the 3-space through b and α, then L does not satisfy property (2) of Result 6.5.1. □

Step 3. Through every point p of K, there are at most two thick planes. Proof. Suppose that there are three thick planes α, β and γ through p. The 4-space through α and β has no other points in common with K then those in α or β. Since every line of γ through p intersects K in a second point, γ must intersect the 4-space in the point p, a contradiction. □

In the sequel let α and β be two thick planes intersecting K in a point. Let Ω be the 4-space through α and β.

Step 4. If a and b are two points of K not in Ω then there exists a thick plane containing them. Proof. Let c be the intersection point of the line ab with Ω. The 3-space through α and c intersects β in a line L. Let x be the second point of K on L. The plane through L and c intersects α in a line M. Let y be the second point of K on M. If x, c and y are not collinear then the line cx does not satisfy property (2) of Result 6.5.1. Hence c, x, y are collinear and the plane containing x, c, y, a, b is thick. □

Step 5. There are at least two points of K not in Ω. Proof. It is impossible that ⟨K⟩ = Ω, since every near polygon is connected. Suppose r is the unique point of K not in Ω. Let s be a point of Ω through which there is no secant line. Then rs contradicts property (2) of Result 6.5.1. □

In the sequel, let r be a fixed point of K not in Ω. Then the points of K not in Ω ∪ {r} are partitioned by the thick planes through r.

Step 6: There is a unique thick plane through r. Proof. Suppose that γ and δ are thick planes through r. Let s (respectively t) be a point of γ ∩K (respectively δ∩K) not in Ω∪{r}. Now, the thick plane through s and t has at most four points (the intersections with α, β, γ, δ), a contradiction. □

Step 7. We derive the contradiction. From the previous steps it follows that all the points of K are contained in three thick planes α, β, γ. Through p there are q^{4} − q^{2} lines not contained in Ω and not intersecting γ. On every such line there is a unique point (≠ p) through which there is a secant. Hence q^{4} − q^{2} ≤ 2q^{2}, a contradiction. □

Lemma 6.5.11 Let Q be a quad of type T_{2}^{∗}(O). This quad determines a thick plane α in the hyperplane Π_{∞}. Then, the following statements are equivalent:

(1) Γ_{2}(Q) = ∅,

(2) Every 3-space of Π_{∞} through α contains q + 3 points of K,

(3) k = q + 2 + \frac{q^{n−2}−1}{q−1}.

Proof. Let x be an arbitrary point of S not in Q. The point x and the quad Q determine a 4-space in Π which intersects Π_{∞} in a 3-space through α. Now, there is a point of Q collinear with x if and only if this 3-space contains a point of K not in α. □

Theorem 6.5.12 Let Q be a quad of type T_{2}^{∗}(O). If n ≥ 6, then Γ_{2}(Q) ≠ ∅.

Proof. Suppose Γ_{2}(Q) = ∅, then by the previous lemma we have k = q+2+ \frac{q^{n−2}−1}{q−1}. Let x ∈ Γ(Q) be a point collinear with y ∈ Q. Let L_{1}, …, L_{q+2} be the q + 2 lines of Q through y. Since Γ_{2}(Q) = ∅, every line through x is contained in a quad through x and L_{i} for some i ∈ {1, …, q + 2}. Hence q + 2 + \frac{q^{n−2}−1}{q−1} ≤ 1 + (q + 1)(q + 2). This implies n ≤ 5, a contradiction. □

Corollary 6.5.13 If n ≥ 6 then every quad Q of type T_{2}^{∗}(O) has a rosette of ovoids.

Proof. Let x denote a point of Γ_{2}(Q), then x is ovoidal with respect to Q. Let y ∈ Γ(Q) be a point collinear with x and let y′ denote the point of Q collinear with y. By Theorem 3.7.1 all points (≠ y) of the line xy are ovoidal with respect to Q and the q − 1 ovoids defined by these points form a rosette of ovoids with y′ as common point. □

Corollary 6.5.14 If n ≥ 6 then q ≠ 4.

Proof. If q = 4, then the unique generalized quadrangle of order (3,5) would have a rosette of ovoids. In the model T_{2}^{∗}(O) an ovoid of GQ(3, 5) is the set of affine points in a plane that meets the carrying plane of O in a line exterior to O, see [68]. Hence two ovoids of GQ(3, 5) are disjoint or meet in 4 points. This proves the corollary. □

Lemma 6.5.15 The following inequality is valid:

(\frac{q^{n−2} − 1}{q − 1} − k′)q^{2}(q + 2) ≥ (q − 1)k′(k′ − q^{2} − 2q − 1).

Proof. Let Q be a quad of type T_{2}^{∗}(O). The points of Q are the affine points of a 3-space β which intersects Π_{∞} in a thick plane α. In Π_{∞}, there are (\frac{q^{n−2}−1}{q−1} − k′) 3-spaces β′ through α which contain only q + 2 points. All the q^{4} − q^{3} affine points of ⟨β, β′⟩ not in Q are points of Γ_{2}(Q) and all points of Γ_{2}(Q) are obtained this way. Hence, the number of points of Γ_{2}(Q) equals (\frac{q^{n−2}−1}{q−1} − k′)(q^{4} − q^{3}). Counting pairs (x, y) with x ∈ Γ(Q) and y ∈ Γ_{2}(Q) yields

(\frac{q^{n−2} − 1}{q − 1} − k′)(q^{4} − q^{3})q^{2}(q + 2) ≥ q^{3}(q − 1)^{2}k′(k′ − q^{2} − 2q − 1).

□

Corollary 6.5.16

k′ < \sqrt{q}(1 + \frac{3}{q})q^{\frac{n−2}{2}}

Proof. The inequality from the previous theorem is equivalent to

\frac{q^{n−2} − 1}{q − 1}q^{2}(q + 2) ≥ (q − 1)k′(k′ − q^{2} − 2q − 1 + \frac{q^{2}(q + 2)}{q − 1}).

From \frac{q^{2}(q+2)}{q−1} = q^{2} + 3q + 3 + \frac{3}{q−1} it follows that

q^{n} \frac{q + 2}{(q − 1)^{2}} > k′(k′ + q + 2).  (6.1)

If q ≠ 4, then 2q^{2} −11q +6 > 0, which is equivalent to \frac{1}{q} + \frac{6}{q^{2}} > \frac{q+2}{(q−1)^{2}} . Hence (6.1) implies q^{n−2}(q + 6) > k^{′2}, from which the stated inequality readily follows. □

Theorem 6.5.17 There are no examples for n = 6.

Proof. Assume there is an example, hence q ≥ 8. We shall derive a contradiction in several steps.

Step 1. One of the following three cases occurs:

\begin{matrix}(1) k = 2q^{2} + 3q + 2, & k′ = 2q^{2} + 2q, & N = 2q^{2} + 6q − 1, \\ (2) k = 3q^{2} + 2q + 2, & k′ = 3q^{2} + q, & N = 7q^{2} + q, \\ (3) k = 2q^{2} + 2q + 3, & k′ = 2q^{2} + q + 1, & N = 2q^{2} + 2q + 2. \end{matrix}

Proof. From the congruences in Corollary 6.5.7 it follows that k = mq(q^{2} − 1) + f(q) with m ∈ ℤ and f(q) ∈ {2q^{2} + 3q + 2, 3q^{2} + 2q + 2, q^{2} + 3q + 3, 2q^{2} + 2q + 3}. From f(q) < 7q^{2} and 7q^{2} < q(q^{2} − 1) it follows that m ≥ 0. Suppose now that m ≥ 1. Since f(q) ≥ q^{2} + 3q + 3, we have that k ≥ q^{3} + q^{2} + 2q + 3 or k′ ≥ q^{3} + q^{2} + q + 1, a contradiction since k′ ≤ q(q + 3)\sqrt{q} (see Corollary 6.5.16). We prove now that k = q^{2} + 3q + 3 cannot occur. As k ≥ \sqrt{2}q^{2} (see Corollary 6.5.7) q ≤ 8. For q = 8, k = q^{2} + 3q + 3 = 91 contradicts inequality (4) of the same corollary. □

Step 2. Let Γ denote the graph with vertices the thick planes. Two thick planes are adjacent whenever they are disjoint. Then Γ is the disjoint union of cliques. The size of each clique is at most q + 2. Proof. To prove that Γ is a disjoint union of cliques, it is enough to prove the following property: if α ∼ β and γ ≁ α then γ ≁ β. So, let α and β be two disjoint thick planes. The 5-space through α and β has no other point in common with K then those in α ∪ β. If γ intersects α in a point p, then p ∈ K. Now γ intersects the 5-space in a line L. Let p′ be the second point of K on this line, then p′ ∈ β. Hence γ meets β. Let C be a clique of Γ. Since k < N(q + 2), there exists a thick plane δ which is no member of C. Now, each member of C meets δ. Hence |C| ≤ q + 2. □

Step 3. Two thick planes meet each other. Proof. Suppose that there are two disjoint thick planes α and β and let C denote the component of α and β. If γ ∉ C is a thick plane then γ meets α and β. Hence N ≤ |C| + (q + 2)^{2} ≤ q^{2} + 5q + 6, a contradiction. □

Step 4. There are at most q + 2 thick planes through a point p. Proof. If all thick planes pass through p then k ≥ 1 + N(q + 1), a contradiction. So, let α be a thick plane not containing p. Now, every thick plane through p intersects α in a point of K. Hence there are at most q + 2 planes through p. □

Step 5. We derive the contradiction. Let p denote the intersection of two thick planes α and β. There are at most q + 2 thick planes through p. Every thick plane not through p has a point in common with α and β. Hence N ≤ (q + 2) + (q + 1)^{2}, a contradiction. □

Consider now the case n ≥ 7. Every quad of type T_{2}^{∗}(O) has a rosette of ovoids. By Section 2.6.1, there exists a set of q − 1 ovoids in T_{2}(O′), which two by two intersect in two points (O′ = O \ {x} where x is any point of O). Suppose now that O′ is a conic (hence O is a regular hyperoval), then T_{2}(O′) is isomorphic to W (q). An ovoid N in T_{2}(O′) is equivalent to an ovoid N′ in PG(3, q). If N′ is an elliptic quadric or a Tits ovoid, then every other ovoid intersects it in an odd number of points ([2]), a contradiction. Hence N′ is an ovoid different from an elliptic quadric or a Tits ovoid. This implies q ≥ 64. If q ≤ 8, then every hyperoval O is regular. Hence, there are no examples of linear representations of near hexagons for n ≥ 7 and q = 8.

6.5.5 The near hexagons without a T_{2}^{∗}(O)-quad

Theorem 6.5.18 If q ≠ 2 and if there is no thick plane, then one of the following two cases appears:

(1) n = 2 and K is a set of three noncollinear points,

(2) n = 5, q = 3 and K is a Coxeter cap in PG(5, 3).

Proof. From Lemma 6.5.6, it follows that k = |K| = ^{\frac{3q−5+\sqrt{8q^{n}+q^{2}−6q+1}}{2(q−1)}}. Hence

y^{2} = 8q^{n} + q^{2} − 6q + 1  (6.2)

for some y ∈ ℤ. From Theorem 6.5.5, it is clear that T_{n}^{∗}(K) is a regular near hexagon with parameters s = q − 1, t_{2} = 1, t = k − 1. The Mathon bound t + 1 ≤ (s^{2} − s + 1)(s + 1 + t_{2}), see equation (4.1), is then equivalent to

\sqrt{8q^{n} + q^{2} − 6q + 1} ≤ 2(q^{2} − 3q + 3)(q^{2} − 1) − (3q − 5).  (6.3)

If n ≥ 8, then one easily checks that (6.3) yields a contradiction; hence n ≤ 7. If n = 2 then 8q^{n} + q^{2} − 6q + 1 = (3q − 1)^{2} and |K| = 3. We allready met this example. It yields the regular near hexagon of Hamming type. So we may assume that 3 ≤ n ≤ 7. The equation (6.2) is equivalent to

8q^{n} − 8q^{2} = (y − 3q + 1)(y + 3q − 1),  (6.4)

so 8q^{2} | (y − 3q + 1)(y + 3q − 1). Let q = p^{h} with p prime. If p ≠ 2, then p is not a common divisor of y − 3q + 1 and y + 3q − 1. If p = 2, then 4 is not a common divisor of y − 3q + 1 and y + 3q − 1. Hence, in any case q^{2} is a divisor of one of the factors. Since y is determined up to its sign, we may suppose that y = xq^{2} − 3q + 1 with x ∈ ℤ. The equation (6.4) becomes

−8q^{n−2} + x^{2}q^{2} − 6xq + 2x + 8 = 0.  (6.5)

Suppose n = 3, then (6.5) becomes x^{2}q^{2} − (6x + 8)q + 2x + 8 = 0. If x = 0, then q = 1, a contradiction. Assume x ≠ 0, let z = |x|q and we have then that z^{2} − 6z − 8z − 2z + 8 ≤ 0. So z ≤ 15 and q ≤ 13. One finds no solutions for these values of q.

Suppose n = 4, then (6.5) becomes (x^{2} − 8)q^{2} − 6xq + 2x + 8 = 0. If |x| ≤ 3, then q ≤ 13 since q is a divisor of 2x + 8. If |x| ≥ 4 (so x^{2} − 8 ≥ \frac{x^{2}}{2} ) and if we put z = |x|q again, we find that \frac{z^{2}}{2} − 8z + 8 ≤ 0. So z ≤ 14 and again q ≤ 13. One finds no solutions for these values of q.

Now, since q is a divisor of 2x + 8, put x = \frac{uq}{2} − 4 with u ∈ ℤ. The equation (6.5) becomes

−8q^{n−3} + \frac{u^{2}q^{3}}{4} − 4uq^{2} + (16 − 3u)q + 24 + u = 0.  (6.6)

Suppose n = 5, then (6.6) becomes \frac{u^{2}q^{3}}{4} − (4u + 8)q^{2} + (16 − 3u)q + 24 + u = 0. If u = 0, then q ≤ 23, since q divides 24 + u. If u ≠ 0, put z = |u|q, we have then that \frac{z^{3}}{4} − 4z^{2} − 8z^{2} − 16z − 3z^{2} − 24z − z^{2} ≤ 0 or \frac{z^{2}}{4} − 16z − 40 ≤ 0. Hence z ≤ 66 and q ≤ 64. Only q = 3 gives a solution: q = 3, u = 0, x = −4, y = −44, k = 12. We will prove that K is a Coxeter cap in PG(5, 3), but we will first treat the cases n = 6 and n = 7.

Suppose n = 6, then (6.6) becomes (\frac{u^{2}}{4} −8)q^{3}−4uq^{2}+(16−3u)q+24+u = 0. If |u| ≤ 7, then q ≤ 31 since q is a divisor of 24 + u. If |u| ≥ 8 (so \frac{u^{2}}{4} − 8 ≥ \frac{u^{2}}{8} ) and if we put z = |u|q again, we find \frac{z^{3}}{8} − 4z^{2} − 16z − 3z^{2} − 24z − z^{2} ≤ 0 or \frac{z^{2}}{8} − 8z − 40 ≤ 0. Hence z ≤ 68 and q ≤ 8. Only q = 4 gives a solution: q = 4, u = 8, x = 12, y = 181, but in this case k is not an integer.

Suppose n = 7. Now, since q is a divisor of 24 + u, put u = vq − 24 with v ∈ ℤ. The equation (6.6) becomes

\frac{v^{2}q^{4}}{4} − (12v + 8)q^{3} + (144 − 4v)q^{2} + (96 − 3v)q + 88 + v = 0.  (6.7)

If v = 0, then q ≤ 83, since q divides 88 + v. If v ≠ 0, let z = |v|q, then \frac{z^{4}}{4} −12z^{3}−8z^{3}−144z^{2}−4z^{3}−96z^{2}−3z^{3}−88z^{2}−z^{3} ≤ 0 or \frac{z^{2}}{4} −28z−328 ≤ 0. Hence z ≤ 122 and q ≤ 121. This case gives no new solutions.

Hence, only the case n = 5, q = 3 and k = 12 remains to be considered. If the corresponding near hexagon exists, then it must be the regular near hexagon related to the extended ternary Golay code, since both near hexagons have the same parameters (there is only one near hexagon with these parameters, see [11]). The points of PG(5, 3) are the 1-spaces of V (6, 3). We call a point of PG(5, 3) of type (w, d) if w is the weight of the point and d is the difference between the number of coordinates equal to 1 and the number of coordinates equal to -1. So, points can be of the following 15 types: (1,1), (2,2), (2,0), (3,3), (3,1),(4,4), (4,2), (4,0), (5,5), (5,3), (5,1), (6,6), (6,4), (6,2), (6,0). Take now 6 linear independant points of K and choose coordinatization in such a way that these are the points of type (1,1). There is then a secant line through every point of weight 2. The determination of the set K will go on in two steps.

Step 1. K can have only points of weight 1 and 5. Proof. If there would be a point of weight 2 in K, then there exist 3 points which are collinear, a contradiction.

Let K contain a point a of weight 3. We may assume that a = ⟨e_{1}+e_{2}+e_{3}⟩, where e_{i} is the vector with all entries but the i-th one equal to 0, the i-th entry being equal to 1. Then the plane ⟨e_{1}, e_{2}, e_{3}⟩ is thick, as it has at least four points in K, namely a and ⟨e_{i}⟩ for i = 1, 2, 3, a contradiction. Thus, K contains no points of weight 3.

Let K contain a point a of weight 4. We may assume that a = ⟨v⟩ with v = e_{1} + e_{2} + e_{3} + e_{4}. The plane X = ⟨v, e_{1}, e_{2}⟩ meets the line ⟨e_{4}, e_{3}⟩ in b = ⟨e_{3} + e_{4}⟩. However, X is not thick. Hence the line ab is tangent to K at a. By (2) of Result 6.5.1, just one point of ab \ {a} belongs to a secant to K. However, ab \ {a} contains two such points, namely ⟨e_{1} + e_{2}⟩ and ⟨e_{3} + e_{4}⟩, a contradiction. Therefore, all points of K have weight either 1, 5 or 6.

Suppose now that there exists a point ⟨u⟩ of type 6 in K. We can choose coordinatization such that u = ∑_{i=1}^{6} e_{i}. We will prove that it is impossible that there are other points in K, hence we have a contradiction. Clearly no point of type (6,4) and (5,5) belongs to K, otherwise three points of K are collinear. Let a be a point of K of type (5,3). We may assume that a = ⟨v⟩ with v = e_{1} + e_{2} + e_{3} + e_{4} − e_{5}. Then a belongs to the plane ⟨u, e_{5}, e_{6}⟩ and the latter is thick, a contradiction. Thus, points of type (5,3) are ruled out from K.

Assume that a ∈ K is of type (5,1), for instance a = ⟨v⟩ with v = e_{1} + e_{2} + e_{3} − e_{4} − e_{5}. The plane X = ⟨v, u, e_{6}⟩ meets the line ⟨e_{4}, e_{5}⟩ in the point b = ⟨e_{4} + e_{5}⟩. As K contains no points of weight 2 or 3, the line L = ⟨e_{4} + e_{5}, e_{6}⟩ is tangent to K in ⟨e_{6}⟩. However, the line ⟨e_{4}, e_{5}⟩ is a secant for K and meets L in b. The line ⟨u, v⟩ is also secant and it meets L in ⟨e_{4} + e_{5} − e_{6}⟩. Thus, at least two points of L different from ⟨e_{6}⟩ belong to secant lines of K, contrary to (2) of Result 6.5.1. Therefore, points of type (5,1) are also ruled out from K.

Assume that a ∈ K is of type (6,2), for instance a = ⟨v⟩ with v = e_{1} + e_{2} + e_{3} + e_{4} − e_{5} − e_{6}. Then the plane X = ⟨u, e_{5}, e_{6}⟩ is thick, a contradiction.

Finally, let a ∈ K be of type (6,0), for instance a = ⟨v⟩ with v = e_{1} + e_{2} + e_{3} − e_{4} − e_{5} − e_{6}. The plane X = ⟨u, v, e_{4}⟩ meets the line ⟨e_{5}, e_{6}⟩ in b = ⟨e_{5} + e_{6}⟩. The line ab is tangent to K at a and it meets the line ⟨u, e_{4}⟩ in c = ⟨e_{1} + e_{2} + e_{3} − e_{4} + e_{5} + e_{6}⟩. Thus, b and c are distinct points of ab \ {a} belonging to secant lines of K, again contrary to (2) of Result 6.5.1.

As a consequence, we remark that if one takes 7 arbitrary points of K then there are 6 points between them which are in a hyperplane. Otherwise, one can choose the coordinatization in such a way that all points of type (1,1) and (6,6) are elements of K, but this is impossible by step 1.

Step 2. Construction of the set K. As above, we assume to have chosen coordinates in such a way that six of the 12 points of K correspond to the vectors e_{1}, e_{2}, ..., e_{6} of the canonical basis of V (6, 3). In every of the 6 remaining points there is a unique position where there is a “0”.

Assume there are two points a = ⟨u⟩ and b = ⟨v⟩ where this “0” takes the same position. We may assume that u = e_{1} + e_{2} + e_{3} + e_{4} + e_{5} and v = e_{1} + e_{2} + e_{3} + εe_{4} − e_{5}, with ε = 1 or −1. If ε = 1, then a, b and ⟨e_{5}⟩ are collinear points of K, a contradiction. Thus, ε = −1. Then a, b, ⟨e_{4}⟩ and ⟨e_{5}⟩ are points of K in the plane ⟨u, v, e_{4}⟩, which is forced to be thick, a contradiction. Therefore, no two points as above exist in K.

Hence, we may suppose that the coordinates of the 12 points look like (up to the sign):

(1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0),
(0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0), (0, 0, 0, 0, 0, 1),
(0, ∗, ∗, ∗, ∗, ∗), (∗, 0, ∗, ∗, ∗, ∗), (∗, ∗, 0, ∗, ∗, ∗),
(∗, ∗, ∗, 0, ∗, ∗), (∗, ∗, ∗, ∗, 0, ∗), (∗, ∗, ∗, ∗, ∗, 0).

Every ”*” in the above coordinates must be replaced by 1 or -1. Take now the 7 points (1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0), (0, ε_{1}, ε_{2}, ε_{3}, ε_{4}, ε_{5}), (ε′_{1}, 0, ε′_{2}, ε′_{3}, ε′_{4}, ε′_{5}), then there is a set A of 6 points between them which are in a hyperplane and the remaining point can not be contained in that hyperplane, since the 7 points above generate Π_{∞}. Clearly (1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, ε_{1}, ε_{2}, ε_{3}, ε_{4}, ε_{5}), (ε′_{1}, 0, ε′_{2}, ε′_{3}, ε′_{4}, ε′_{5}) ∈ A. If we suppose that (0, 0, 0, 0, 1, 0) ∉ A, then ε_{2} = −ε′_{2}, ε_{3} = −ε′_{3}, ε_{4} = ε′_{4}, ε_{5} = ε′_{5} or ε_{2} = ε′_{2}, ε_{3} = ε′_{3}, ε_{4} = −ε′_{4}, ε_{5} = −ε′_{5}, hence ε_{2}ε′_{2} + ε_{3}ε′_{3} + ε_{4}ε′_{4} + ε_{5}ε′_{5} = 0. A similar reasoning shows that any 2 of the points of weight 5 are orthogonal. We still have some freedom in coordinatization, for instance we may suppose that (0, −1, −1, −1, −1, −1) ∈ K. Since (0,-1,-1,-1,-1,-1) and (*,0,*,*,*,*) are orthogonal (and using the remaining freedom in coordinatization) we may suppose that (1, 0, 1, −1, −1, 1) ∈ K. Since (0,-1,-1,-1,-1,-1), (1,0,1,-1,-1,1) and (*,*,0,*,*,*) are orthogonal (and using the remaining freedom in coordinatization) we may suppose that (1, 1, 0, 1, −1, −1) ∈ K. From the orthogonality of the points of weight 5, the remaining points are readily obtained. These are the points (1,-1,1,0,1,-1), (1,-1,-1,1,0,1) and (1,1,-1,- 1,1,0). We did the coordinatization in such a way that the 12 points are just the 12 columns of the matrix that defined the extended ternary Golay code. Since we know that there is a linear representation of the near hexagon related to this code, we know that the set of 12 points given above satisfies the required conditions. From the previous reasoning, it also follows that all sets of points in PG(5, 3) which satisfy the required conditions are equivalent. □

Remark.

The proof of Theorem 6.5.18 can be shortened by using results of [20], moreover here q = 2 is allowed. Let p ∈ K and let α be a hyperplane of Π_{∞} not containing p, then the projection of K from p on α will yield a cap K′ with k − 1 points. It is easy to prove that T_{n−1}^{∗}(K′) is a partial quadrangle with parameters (s, t, μ) = (q − 1, k − 2, 2). In [20], an almost complete classification of partial quadrangles with a linear representation is given, see also [34] for a survey. The classification is complete for μ = 2. One only has to check whether these partial quadrangles are indeed coming from a projection of a near hexagon. Besides the projections of the examples mentioned in Theorem 6.5.18, there is one other example in the case q = 2 in which case K′ is a set of 5 points in PG(3, 2) no 4 of which are contained in a hyperplane, i.e. K′ is a frame of PG(3, 2). The set K, yielding the corresponding near hexagon T_{4}^{∗}(K), is a frame of PG(4, 2), i.e. a set of 6 points no 5 of which are contained in a hyperplane. Note that this proof relies on the classification of the linear representations of the partial quadrangles.

6.6 Characterizations

Definition.

(1) For m ∈ ℕ, let f_{m}(x, y) be the unique polynomial of degree m in x such that f_{m}(i, y) = \frac{y^{i−1}−1}{y−1} for all i ∈ \{1, …, m + 1\}. It is straightforward to check that f_{m}(x, y) = ∑_{j=1}^{m} \binom{x−1}{j}(y − 1)^{j−1} and that f_{m}(m + 2, y) = \frac{y^{m+1}−1}{y−1} − (y − 1)^{m}.

(2) For m ∈ ℕ \ \{0\}, let g_{m}(x, y) = \frac{1}{y} \binom{x−1}{m}(y − 1)^{m−1} + f_{m−1}(x, y).

Lemma 6.6.1 g_{m}(x, y) = \frac{1}{y}[f_{m}(x + 1, y) − 1].

Proof. A calculation shows that

f_{m}(x+1, y) = ∑_{j=1}^{m} \binom{x}{j}(y−1)^{j−1} = ∑_{j=1}^{m} \binom{x − 1}{j}(y−1)^{j−1}+∑_{j=1}^{m} \binom{x − 1}{j − 1}(y−1)^{j−1}

= 1 + ∑_{j=1}^{m} \binom{x − 1}{j}(y − 1)^{j−1} + ∑_{j=1}^{m−1} \binom{x − 1}{j}(y − 1)^{j} = yg_{m}(x, y) + 1.

□

For m ≥ 1, y > 1 and 1 ≤ x_{1} < x_{2}, one has that f_{m}(x_{1}, y) < f_{m}(x_{2}, y) and g_{m}(x_{1}, y) < g_{m}(x_{2}, y). Hence the inequalities of the following theorem yield upper bounds for the value k.

Theorem 6.6.2 Let V be a set of k > 0 points of PG(n, q) with the property that no l of them are contained in an (l − 2)-flat (n ≥ l − 2 ≥ 0).

(1) If l = 2m + 1, then f_{m}(k, q) ≤ \frac{q^{n}−1}{q−1} and equality holds if and only if T_{n}^{∗}(V) is a near 2(m + 1)-gon.

(2) If l = 2m and (n, m) ≠ (0, 1), then g_{m}(k, q) ≤ \frac{q^{n}−1}{q−1} and equality holds if and only if T_{n}^{∗}(V) is a near (2m + 1)-gon.

Proof. (1) Let l = 2m + 1, then m ≥ 1. If k ≤ 2m, then

f_{m}(k, q) ≤ f_{m}(2m, q) = ∑_{j=1}^{m} \binom{2m − 1}{j}(q − 1)^{j−1}

≤ ∑_{j=1}^{2m−1} \binom{2m − 1}{j}(q − 1)^{j−1} = \frac{q^{2m−1} − 1}{q − 1} ≤ \frac{q^{n} − 1}{q − 1}.

Suppose k ≥ 2m + 1. In what follows we will make use of the fact that no j points of V (with 1 ≤ j ≤ 2m + 1) are contained in a (j − 2)-flat. Now, fix a point p of V. A line L through p is said to be of type i ≥ 1 if i = min i_{V} (r), where r ranges over all points of L \ {p}. We prove by induction that for 1 ≤ i ≤ m, the number of lines through p with type at most i equals f_{i}(k, q). This is true for i = 1 since f_{1}(k, q) = k − 1. Suppose that 2 ≤ i ≤ m, then the number of lines through p with type at most i equals

f_{i−1}(k, q) + \binom{k − 1}{i} [\frac{q^{i} − 1}{q − 1} − f_{i−1}(i + 1, q)]

and hence

f_{i}(k, q) = f_{i−1}(k, q) + \binom{k − 1}{i}(q − 1)^{i−1}.

Now, since there are \frac{q^{n}−1}{q−1} lines through p, we must have \frac{q^{n}−1}{q−1} ≥ f_{m}(k, q). If T_{n}^{∗}(V) is a near 2(m + 1)-gon, then all the lines through p have type at most m, hence equality holds. Conversely, suppose that equality holds. If k ≤ 2m, then one calculates that m = n = 1 and k = 2. In this case T_{1}^{∗}(V) is a grid and hence a generalized quadrangle. Assume k ≥ 2m + 1, let p be a point of V and let L be an arbitrary line through p. All the points r ≠ p on L with smallest index satisfy i_{V} (r) ≤ m. Since no 2m + 1 points are contained in a (2m − 1)-flat, there exists a unique point on L with smallest index. Hence T_{n}^{∗}(V) is a near polygon. To prove that it is a near 2(m + 1)-gon, it suffices to prove the existence of a point of index m + 1. Such a point exists on a line of type m through p.

(2) Let l = 2m (then m ≥ 1) and (n, m) ≠ (0, 1). If k ≤ 2m − 1, then

g_{m}(k, q) ≤ g_{m}(2m−1, q) = \frac{1}{q}\binom{2m − 2}{m}(q −1)^{m−1} + ∑_{j=1}^{m−1} \binom{2m − 2}{j}(q −1)^{j−1}

≤ ∑_{j=1}^{m} \binom{2m − 2}{j}(q−1)^{j−1} ≤ ∑_{j=1}^{2m−2} \binom{2m − 2}{j}(q−1)^{j−1} ≤ \frac{q^{2m−2} − 1}{q − 1} ≤ \frac{q^{n} − 1}{q − 1}.

Suppose k ≥ 2m. Now, fix a point p of V. A line L through p is said to be of type i ≥ 1 if i = min i_{V} (r), where r ranges over all points of L \ {p}. As before, f_{m−1}(k, q) equals the number of lines through p with type at most m − 1 and the number of lines of type m through p equals at most

\frac{1}{q}\binom{k − 1}{m}[\frac{q^{m} − 1}{q − 1} − f_{m−1}(m + 1, q)] = \frac{1}{q}\binom{k − 1}{m}(q − 1)^{m−1}.

Hence \frac{q^{n}−1}{q−1} ≥ g_{m}(k, q). If T_{n}^{∗}(V) is a near (2m + 1)-gon, then every line through p is counted exactly once, hence equality holds. Conversely, suppose that equality holds. If k ≤ 2m − 1, then the above inequalities imply that n = 0, m = 1, k = 1, a contradiction. Assume k ≥ 2m + 1 and let p be an arbitrary point of V. Since equality holds, we have the following properties for lines L through p (i_{V} (L) denotes the type of L):

(a) i_{V} (L) ≤ m,

(b) if i_{V} (L) < m, then there exists a unique point r of L \ {p} such that i_{V} (r) = i_{V} (L),

(c) if i_{V} (L) = m, then i_{V} (r) = m for all points r of L \ {p},

(d) there exists an L with i_{V} (L) = m.

This implies that T_{n}^{∗}(V) is a near (2m + 1)-gon. □

Remark. Let V be a set of k > 0 points of PG(n, q) with the property that no 2m + 1 of them are contained in a (2m − 1)-flat (n ≥ 2m − 1, m ≥ 1). Fix v ∈ V and let PG(n − 1, q) be a hyperplane of PG(n, q) not through v. The projection (from v) of V onto PG(n − 1, q) will yield a set V′ with the property that no 2m points are contained in a (2m − 2)-flat. Hence g_{m}(k − 1, q) ≤ \frac{q^{n−1}−1}{q−1}. From Lemma 6.6.1, it follows then that f_{m}(k, q) ≤ \frac{q^{n}−1}{q−1}. Hence, one of the inequalities of the previous theorem follows from the other. It follows also that if T_{n}^{∗}(V) is a near 2(m + 1)-gon, then T_{n−1}^{∗}(V′) is a near (2m + 1)-gon, with an exception when n = 1, m = 1, |V | = k = 2.

Some special cases

The case l = 2

In this case there are no restrictions on the set V. Since g_{1}(k, q) = \frac{k−1}{q}, the inequality g_{1}(k, q) ≤ \frac{q^{n}−1}{q−1} is equivalent with k ≤ \frac{q^{n+1}−1}{q−1}.

The case l = 3

In this case we are looking for caps V in PG(n, q), n ≥ 1. For n = 1, we have k = |V | ≤ 2 and equality is always possible. For n = 2, we have k ≤ q + 2. Equality is possible when q is even and in this case V is a hyperoval. If q = 2 and n ≥ 1, then k ≤ 2^{n} and equality holds if and only if V is the complement of a hyperplane. For q ≠ 2 and n ≥ 3, k ≤ 1 + \frac{q^{n}−1}{q−1}. If equality holds then every line meets V in 0 or 2 points. If we take a 3-dimensional space β in PG(n, q) having at least one point in common with V, then this would imply the existence of a (q^{2} + q + 2)-cap in β, a contradiction.

The case l = 4

In this case we are looking for sets of points in PG(n, q), n ≥ 2, with the property that no four of them are contained in a plane. If the upper bound is achieved then T_{n}^{∗}(V) is a near pentagon. If x and y are two points of T_{n}^{∗}(V) at mutual distance 2, then the intersection point z of the line xy with PG(n, q) is contained in a unique 2-secant. Hence x and y have exactly two common neighbours. This implies that T_{n}^{∗}(V) is a partial quadrangle with parameters (s, t, μ) = (q − 1, k − 1, 2). From the classification given in [20] (see also [34]), it follows that there are only two possibilities.

(1) PG(n, q) = PG(3, 2) and V is projectively equivalent with {(1, 0, 0, 0), (0, 1, 0, 0), (0, 0, 1, 0), (0, 0, 0, 1), (1, 1, 1, 1)}.

(2) PG(n, q) = PG(4, 3) and V is an 11-cap which is the projection of a Coxeter cap in PG(5, 3).

With g_{2}(k, q) = \frac{(k−1)(k−2)(q−1)}{2q} + (k − 1), it is an easy calculation to show that the upper bound is indeed attained in both cases. Hence we have the following theorem.

Theorem 6.6.3 If V is a set of points in PG(n, q), n ≥ 2, with the property that no four

of them are contained in a plane, then k = |V | ≤ ^{\frac{(q−3)+\sqrt{8q^{n+1}+q^{2}−6q+1}}{2(q−1)}} and equality holds if and only if V is one of the two examples given above.

The case l = 5

In this case we are looking for sets of points in PG(n, q), n ≥ 3 with the property that no five of them are contained in a 3-dimensional space. If the upper bound is attained then T_{n}^{∗}(V) is a near hexagon. By the remark given after Theorem 6.6.2, it follows that the projection of V on a hyperplane is necessary one of the two examples from the case l = 4. We have then the following possibilities.

(1) PG(n, q) = PG(5, 3).

From Theorem 6.5.18, it follows that V is the Coxeter cap in PG(5, 3).

(2) PG(n, q) = PG(4, 2).

In this case V is projectively equivalent with {(1, 0, 0, 0, 0), (0, 1, 0, 0, 0), (0, 0, 1, 0, 0), (0, 0, 0, 1, 0), (0, 0, 0, 0, 1), (1, 1, 1, 1, 1)}. This case gives a thin near hexagon T_{4}^{∗}(V) with 32 points.

Hence, we have the following theorem.

Theorem 6.6.4 If V is a set of points in PG(n, q), n ≥ 3, with the property that no five of them

are contained in a 3-dimensional space, then k = |V | ≤ ^{\frac{(3q−5)+\sqrt{8q^{n+1}+q^{2}−6q+1}}{2(q−1)}} and equality holds if and only if V is one of the two examples given above.

The case l ≥ 6

We put l = 2m or l = 2m + 1 depending on l even or odd. Hence m ≥ 3. We will prove that equality can never hold when q ≠ 2, so let V be a counterexample. First of all we prove that every two points x and y of T_{n}^{∗}(V) at distance i ≤ m are always contained in a unique geodetically closed sub near 2i-gon. The intersection point of xy with PG(n, q) yields a point z and there is a unique set {v_{1}, …, v_{i}} of i points of V such that z ∈ ⟨v_{1}, …, v_{i}⟩. The linear representation T_{i−1}^{∗}({v_{1}, …, v_{i}}) (in the space ⟨x, v_{1}, …, v_{i}⟩) determines then a sub near 2i-gon of Hamming type through x and y. The condition that no l points of V are contained in an (l − 2)-flat implies that this sub near 2i-gon is geodetically closed. In this way, one can define quads and hexes and all the hexes are classical. By a remark in [13], page 195, T_{n}^{∗}(V) is a dual polar space and hence a classical near 2(m + 1)-gon which is regular with parameters (s, t, t_{2}, t_{3}, …, t_{m}) = (q − 1, k − 1, 1, 2, …, m − 1). From t = t_{m+1} = t_{2}(1 + t_{m}) = m, it follows that k = m + 1. Since f_{m}(k, q) = \frac{q^{n}−1}{q−1} and f_{m}(m + 1, q) = \frac{q^{m}−1}{q−1}, it follows that m = n, a contradiction since n ≥ 2m − 1. □

6.7 Some properties of the regular near hexagon with parameters s = 2, t_{2} = 1, t = 11

6.7.1 A general theorem on Cameron closed subspaces

Let Π_{∞} be a PG(n, q) which is embedded as a hyperplane in a PG(n + 1, q) Π. Let K be a nonempty set of points of Π_{∞} and let V be a Cameron closed subset of T_{n}^{∗}(K). Finally, denote by A the partial linear space induced by V.

Theorem 6.7.1 There exists an m-dimensional subspace α of Π_{∞} such that A ≃ T_{m}^{∗}(α ∩ K).

Proof. We prove by induction on k that all the affine points of ⟨x_{0}, …, x_{k}⟩ (k ≥ 0) are in V if x_{0}, …, x_{k} are k + 1 points of V such that d(x_{i}, x_{i+1}) = 1, i ∈ {0, …, k − 1}. This is true for k ≤ 1, hence suppose that k ≥ 2.

We define inductively the k + 1 points y_{0}, …, y_{k} as follows: y_{k} = x_{k}; for i ∈ {1, …, k − 1}, y_{i} = x_{i} whenever y_{i+1}, x_{i} and x_{i−1} are collinear, otherwise let y_{i} be the unique neighbour of y_{i+1} and x_{i−1} such that y_{i+1}x_{i}∩y_{i}x_{i−1} ⊆ Π_{∞} and y_{i+1}y_{i} ∩ x_{i}x_{i−1} ⊆ Π_{∞}; finally, put y_{0} = x_{0}. By induction, one can prove that y_{i} ∈ V for all i ∈ {0, …, k}. By induction, we also have that all affine points of ⟨x_{0}, …, x_{k−1}⟩ and ⟨y_{1}, …, y_{k}⟩ belong to V. Put {z_{i}} = Π_{∞} ∩x_{i}x_{i+1}, i ∈ {0, …, k −1}, then ⟨x_{0}, …, x_{k−1}⟩ = ⟨z_{0}, …, z_{k−2}, x_{k−1}⟩ and ⟨y_{1}, …, y_{k}⟩ = ⟨z_{0}, …, z_{k−2}, x_{k}⟩. We may also suppose that ⟨x_{0}, …, x_{k−1}⟩ ≠ ⟨x_{0}, …, x_{k}⟩. This implies that ⟨x_{0}, …, x_{k−1}⟩ ∩ ⟨y_{1}, …, y_{k}⟩ = ⟨z_{0}, …, z_{k−2}⟩. Now, let a be an arbitrary affine point of ⟨x_{0}, …, x_{k}⟩. The line az_{k−1} intersects ⟨x_{0}, …, x_{k−1}⟩ and ⟨y_{1}, …, y_{k}⟩ in points b and c respectively. Since b ≠ c and b, c ∈ V, all affine points of the line bc are in V, in particular a ∈ V.

Let β be the subspace of Π generated by the points of V and put α = β ∩ Π_{∞}. It is clear now that A is isomorphic to T_{m}^{∗}(α ∩ K), where m is the dimension of α. □

6.7.2 The Cameron closed subspaces of the regular near hexagon with parameters s = 2, t_{2} = 1, t = 11

We apply the previous theorem to determine all Cameron closed subspaces of the regular near hexagon which is the linear representation of the Coxeter cap. So, assume that K is a Coxeter cap in PG(5, 3). If V is a Cameron closed subspace of S = T_{5}^{∗}(K), then V is the set of affine points of an (m+1)- dimensional subspace of Π which intersects Π_{∞} in an m-dimensional subspace α (m ≥ −1). Since the partial linear space A induced by V is connected, one has |α ∩ K| ≥ m + 1 (see Corollary 6.1.2). Since distance in S_{V} is equal to the corresponding distance in S, A is a near 2k-gon with k ≤ 3. Moreover, A has a linear representation and all quads of A are grids. By Section 6.5, we have the following possibilities for V.

(1) V consists of one point of S (m = −1).

(2) V is the set of points on a line of S. This corresponds with the case m = 0 and |α ∩ K| = 1.

(3) V is a quad of S (i.e. a grid). This corresponds with the case m = 1 and |α ∩ K| = 2.

(4) V is the set of points of a near hexagon of Hamming type. This corresponds to the case m = 2 and |α ∩ K| = 3.

(5) V is the set of all points of S.

6.7.3 The ovoids of the regular near hexagon with parameters s = 2, t_{2} = 1, t = 11

Let K be a Coxeter cap in Π_{∞} = PG(5, 3) and let S = T_{5}^{∗}(K).

Theorem 6.7.2 If α is a hyperplane of Π such that α ∩ Π_{∞} is disjoint from K, then the set of affine points of α form an ovoid of S. Conversely, every ovoid is obtained in this way.

Proof. Let α be a hyperplane such that α ∩ Π_{∞} is disjoint from K. Since every line of S intersects α in 1 point, the set of affine points of α form an ovoid of S. Conversely, let O be an ovoid of S and let o_{1} and o_{2} be two points of O. We prove that all the affine points of the line o_{1}o_{2} are points of O. Assume d(o_{1}, o_{2}) = 2 and let x and y be those points of K such that xy and o_{1}o_{2} are concurrent. The points x and y are uniquely determined since no plane of Π_{∞} intersects K in more than four points. Consider now the grid T_{1}^{∗}({x, y}) (where the representation is in the plane ⟨x, y, o_{1}, o_{2}⟩). The ovoid O induces an ovoid in this grid. The grid has six ovoids, just like there are six lines in the plane ⟨x, y, o_{1}, o_{2}⟩ intersecting xy not in x or y. The affine points on every such line form an ovoid of the grid. This proves that all affine points of the line o_{1}o_{2} are points of O. Assume d(o_{1}, o_{2}) = 3, let x, y, z be three points of K such that ⟨x, y, z⟩ and the line o_{1}o_{2} meet each other (x, y, z are not necessary uniquely determined). Consider the near hexagon T_{2}^{∗}({x, y, z}) (the representation is in the 3-dimensional space ⟨x, y, z, o_{1}, o_{2}⟩). This near hexagon has 12 ovoids, just like there are 12 planes in ⟨x, y, z, o_{1}, o_{2}⟩, disjoint with {x, y, z}. As before, this implies that every affine point of o_{1}o_{2} is a point of O. The points of O generate now a space α and every affine point of α is a point of O. Since O is an ovoid (|O| = 243), α is a hyperplane of Π such that α ∩ Π_{∞} is disjoint with K. □

There are exactly 12 hyperplanes in PG(5, 3) carrying no point of K, see [97]. In the dual space of PG(5, 3), these hyperplanes define again a Coxeter-cap. Hence, there are 36 ovoids and 12 partitions in ovoids.

Glued near polygons

7.1 Constructions of glued near polygons

Let k be a nonzero positive integer. For every i ∈ {1, …, k}, let Q_{i} = (P_{i}, L_{i}, I_{i}) be a generalized quadrangle of order (s, t_{i}), let S_{i} = {L_{1}^{(i)}, …, L_{1+st_{i}}^{(i)}} be a spread of Q_{i} and let θ_{i} be a bijection from L_{1}^{(1)} to L_{1}^{(i)}. Notice that a special line L_{1}^{(i)} is taken in each spread S_{i}, i ∈ {1, …, k}; this line is called the base line of S_{i}. For i ∈ {1, …, k} and j ∈ {1, …, 1 + st_{i}}, let p_{j}^{(i)} be the projection on the line L_{j}^{(i)} in the generalized quadrangle Q_{i}.

Consider now the following graph Γ with vertex set L_{1}^{(1)} × S_{1} × … × S_{k}. Two different vertices (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) are adjacent if and only if the following two conditions are satisfied:

(i) there exists an l ∈ {1, …, k} such that i_{m} = j_{m} for all m ∈ {1, …, k}\ {l},

(ii) for every l as in (i), we have that p_{i_{l}}^{(l)} ∘ θ_{l}(x) and p_{j_{l}}^{(l)} ∘ θ_{l}(y) are collinear points in the generalized quadrangle Q_{l}.

If i_{m} = j_{m} for all m ∈ {1, …, k}, then the two vertices are adjacent (condition (ii) is satisfied for every l ∈ {1, …, k}).

Remarks.

(a) In the sequel, the composition f ∘ g of two functions f and g is shortly denoted by fg.

(b) In the above definition of Γ, we used a certain ordering in the set {Q_{1}, …, Q_{k}} of generalized quadrangles (the one induced by the indices). Γ can be retrieved using the same spreads and base lines, but a different ordering of the GQ’s; the maps θ_{i} must then be changed. Indeed, let σ be an arbitrary permutation of {1, …, k} and let Q_{σ(1)}, …, Q_{σ(k)} be the new ordering of the GQ’s. Let ϕ be an arbitrary bijection between L_{1}^{(σ(1))} and L_{1}^{(1)}. Then θ′_{σ(i)} = θ_{σ(i)}ϕ is a map from L_{1}^{(σ(1))} to L_{1}^{(σ(i))}. In the same way as we defined Γ, we can define now a graph Γ′ which is isomorphic to Γ; the isomorphism from Γ to Γ′ is given by (x, L_{1}, …, L_{k}) ↦ (ϕ^{−1}(x), L_{σ(1)}, …, L_{σ(k)}).

(c) If we take σ equal to the identity in (b), then we still can take ϕ arbitrarly. Hence, Γ can be retrieved if we choose θ_{1} equal to the identical permutation of L_{1}^{(1)}.

(d) Under certain conditions, see Theorem 7.1.3, the graph Γ can be retrieved by starting with k arbitrary base lines (using the same GQ’s and spreads).

Lemma 7.1.1 Through every two adjacent vertices of Γ, there is a unique maximal clique. This clique has size s + 1.

Proof. Let α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and β = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) be two adjacent vertices of Γ. Let l be an element of {1, …, k} such that i_{m} = j_{m} for all m ∈ {1, …, k} \ {l}. We determine now how the common neighbours γ = (z, L_{f_{1}}^{(1)}, …, L_{f_{k}}^{(k)}) look like. If i_{m} = j_{m} ≠ f_{m} for a certain element m ∈ {1, …, k} \ {l}, then i_{l} = f_{l} = j_{l} and p_{i_{m}}^{(m)}θ_{m}(x) ∼ p_{f_{m}}^{(m)}θ_{m}(z) ∼ p_{j_{m}}^{(m)}θ_{m}(y), implying that p_{i_{m}}^{(m)}θ_{m}(x) = p_{i_{m}}^{(m)}θ_{m}(y), x = y or α = β, a contradiction. Hence i_{m} = j_{m} = f_{m} for all elements m of {1, …, k} \ {l}. The condition p_{i_{l}}^{(l)}θ_{l}(x) ∼ p_{f_{l}}^{(l)}θ_{l}(z) ∼ p_{j_{l}}^{(l)}θ_{l}(y) holds if and only if p_{f_{l}}^{(l)}θ_{l}(z) is a point of the line connecting p_{i_{l}}^{(l)}θ_{l}(x) with p_{j_{l}}^{(l)}θ_{l}(y). Hence there are s − 1 solutions for z (or γ). These s − 1 solutions for γ form together with α and β the unique clique of size s + 1. □

Let S be the partial linear space with points the vertices of Γ, with lines the maximal cliques of Γ and with natural incidence. If S is a near polygon, then it is called a glued near polygon. We will examine now under what conditions this precisely happens. For i ∈ {1, …, k}, let G_{i} be the group of projectivities of L_{1}^{(i)} with respect to S_{i}. The group G_{i} is generated by the permutations Δ_{jk}^{(i)} : L_{1}^{(i)} → L_{1}^{(i)} : x ↦ p_{1}^{(i)}p_{k}^{(i)}p_{j}^{(i)}(x), 1 ≤ j, k ≤ 1 + st_{i}.

Remarks.

(1) If k = 1, then S ≃ Q_{1}. Indeed, the map (x, L_{j}^{(1)}) ↦ p_{j}^{(1)}θ_{1}(x) defines an isomorphism between S and Q_{1}.

(2) Let α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and β = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) be two points of S and suppose that l is an element of {1, …, k} such that i_{m} = j_{m} for all m ∈ {1, …, k} \ {l}, then α and β are collinear if and only if p_{i_{l}}^{(l)}θ_{l}(x) ∼ p_{j_{l}}^{(l)}θ_{l}(y) or if and only if

y = θ_{l}^{−1}Δ_{i_{l}j_{l}}^{(l)}θ_{l}(x).

In the next theorem 0 stands for the trivial group and [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] is the group generated by all commutators [θ_{i}^{−1}g_{i}θ_{i}, θ_{j}^{−1}g_{j}θ_{j}] with g_{1} ∈ G_{1} and g_{2} ∈ G_{2}.

Theorem 7.1.2 The partial linear space S is a glued near hexagon if and only if [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0 for all i, j ∈ {1, …, k} with i ≠ j.

Proof. We may suppose that k ≥ 2. First, suppose that S is a near polygon. Consider the point α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and the line L = {(y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)})|y I_{1} L_{1}^{(1)}}. Take two distinct indices m, n ∈ {1, …, k}. Suppose that i_{m} ≠ j_{m}, i_{n} ≠ j_{n} and i_{l} = j_{l} for all l ∈ {1, …, k} \ {m, n}. Clearly L contains no points collinear with α. Since L contains the points ((θ_{m}^{−1}Δ_{i_{m}j_{m}}^{(m)}θ_{m})(θ_{n}^{−1}Δ_{i_{n}j_{n}}^{(n)}θ_{n})(x), L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) and ((θ_{n}^{−1}Δ_{i_{n}j_{n}}^{(n)}θ_{n})(θ_{m}^{−1}Δ_{i_{m}j_{m}}^{(m)}θ_{m}) (x), L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) which have both distance 2 from α, it follows that

(θ_{m}^{−1}Δ_{i_{m}j_{m}}^{(m)}θ_{m})(θ_{n}^{−1}Δ_{i_{n}j_{n}}^{(n)}θ_{n})(x) = (θ_{n}^{−1}Δ_{i_{n}j_{n}}^{(n)}θ_{n})(θ_{m}^{−1}Δ_{i_{m}j_{m}}^{(m)}θ_{m})(x),

for all n, m ∈ {1, …, k} with m ≠ n and for all x I_{1} L_{1}^{(1)}. It follows now immediately that [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0 for all i, j ∈ {1, …, k} with i ≠ j.

Conversely, suppose that [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0 for all i, j ∈ {1, …, k} with i ≠ j.

Consider the point α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and the line L = {(y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)})|y I_{1} L_{1}^{(1)}}. If N is the number of positions in which (i_{1}, …, i_{k}) and (j_{1}, …, j_{k}) differ, then α has distance N from L. If the conditions of the theorem are satisfied, then L contains a unique point at distance N from α, namely the point (∏_{a∈A}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})(x), L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}). Here A denotes the set of all indices a ∈ {1, …, k} for which i_{a} ≠ j_{a} and ∏_{a∈A}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a}) denotes the composition of N = |A| permutations of L_{1}^{(1)}.

Now, suppose that M is a fixed line of Q_{1} not belonging to the spread S_{1}. Consider the point α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and the line L = {(y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)})| y I_{1} L_{1}^{(1)} and p_{j_{1}}^{(1)}θ_{1}(y) I_{1} M} (j_{2}, …, j_{k} are fixed). Let A denote the set of indices a ∈ {2, …, k} for which i_{a} ≠ j_{a}, then the set of points of L nearest to α is equal to the set of points of L nearest to (∏_{a∈A}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})(x), L_{i_{1}}^{(1)}, L_{j_{2}}^{(1)}, …, L_{j_{k}}^{(k)}). Let y′ ∈ L_{1}^{(1)} and j′_{1} ∈ {1, …, 1+st_{1}} be such that p_{j′_{1}}^{(1)}θ_{1}(y′) is the

unique point of M nearest to (p_{i_{1}}^{(1)}θ_{1})(∏_{a∈A}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})(x)), then (y′, L_{j′_{1}}^{(1)},

L_{j_{2}}^{(1)}, …, L_{j_{k}}^{(k)}) is the unique point of L nearest to α. □

Remark. The maximal distance between two points α and β of S is equal to k + 1. This distance is obtained if α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and β = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}), where i_{l} ≠ j_{l} for all l ∈ {1, …, k} and where y ≠ ∏_{a∈{1,...,k}} (θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})(x).

Conclusion. If S is a near polygon, then S is a near (2k + 2)-gon.

Theorem 7.1.3 If S is a near polygon, then S can be retrieved starting from any choice of the k base lines, with good chosen maps θ_{i}, i ∈ {1, …, k}.

Proof. Let L_{α_{1}}^{(1)}, …, L_{α_{k}}^{(k)} be the new base lines of the k spreads. With each point μ = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) of S, we associate the (k + 1)-tuple μ′ = (x′, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}), where x′ = p_{α_{1}}^{(1)} ∏_{a∈{1,...,k}}(θ_{a}^{−1}Δ_{i_{a}α_{a}}^{(a)}θ_{a})(x) ∈ L_{α_{1}}^{(1)}. If ν = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) is a second point of S, then μ and ν are adjacent if and only if y = θ_{l}^{−1}Δ_{i_{l}j_{l}}^{(l)}θ_{l}(x). Here l denotes an element of {1, …, k} with the property that i_{m} = j_{m} for all m ∈ {1, …, k} \ {l}. We rewrite this condition as follows:

y′ = p_{α_{1}}^{(1)} ( ∏_{a∈{1,...,k}}(θ_{a}^{−1}Δ_{j_{a}α_{a}}^{(a)}θ_{a})) (θ_{l}^{−1}Δ_{i_{l}j_{l}}^{(l)}θ_{l})x

= p_{α_{1}}^{(1)} (∏_{a<l}(θ_{a}^{−1}Δ_{i_{a}α_{a}}^{(a)}θ_{a})) (θ_{l}^{−1}Δ_{j_{l}α_{l}}^{(l)}Δ_{i_{l}j_{l}}^{(l)}θ_{l}) (∏_{a>l}(θ_{a}^{−1}Δ_{i_{a}α_{a}}^{(a)}θ_{a})) (x)
= p_{α_{1}}^{(1)}θ_{l}^{−1}Δ_{j_{l}α_{l}}^{(l)}Δ_{i_{l}j_{l}}^{(l)}Δ_{α_{l}i_{l}}^{(l)}θ_{l}p_{1}^{(1)}(x′)
= (p_{α_{1}}^{(1)}θ_{l}^{−1}p_{1}^{(l)})(p_{α_{l}}^{(l)}p_{j_{l}}^{(l)}p_{i_{l}}^{(l)})(p_{α_{l}}^{(l)}θ_{l}p_{1}^{(1)})(x′)
= (θ_{l}^{′−1}(p_{α_{l}}^{(l)}p_{j_{l}}^{(l)}p_{i_{l}}^{(l)})θ′_{l})(x′).

Hence, we can take the maps θ′_{l} : L_{α_{1}}^{(1)} → L_{α_{l}}^{(l)}, l ∈ {1, …, k}, as new maps between the new base lines. □

Remark. Suppose that S is a near (2k + 2)-gon with k ≥ 2. If we restrict ourselves to the generalized quagrangles Q_{1}, …, Q_{k−1} together with their corresponding spreads, base lines and maps, then by Theorem 7.1.2, we find a near polygon S′. If Q_{k} is a grid, then S is the direct product of S′ with a line of length s + 1.

In Chapter 6, we defined the following class of near polygons. Let Π_{∞} be a PG(2k, q), k ≥ 2, which is embedded as a hyperplane in Π. Consider in Π_{∞} k planes α_{1}, …, α_{k} intersecting in a point p and generating Π_{∞}. Let H_{i}, i ∈ {1, …, k}, be a hyperoval in α_{i} containing p. Then T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}) is a near polygon. We will prove now that this near polygon is glued.

Theorem 7.1.4 T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}) is a glued near polygon.

Proof. Let a be a fixed point of Π \ Π_{∞} and put A_{i} = ⟨a, α_{i}⟩. The set of affine points of A_{i} is a geodetically closed subspace and induces a generalized quadrangle Q_{i} isomorphic to T_{2}^{∗}(H_{i}). Let S_{i} be the spread of Q_{i} determined by the point p ∈ Π_{∞} (S_{i} is the set of lines of A_{i} through p), let L_{1}^{(i)} be the line pa and let θ_{i} be the identical map. The q translations of the affine space on A_{i}, which have p as corresponding point at infinity, induce q automorphisms of Q_{i} which fix each line of S_{i}; the group G_{i} is then equal to the group of the q translations of the affine line pa \ {p}. Hence, with the above choices for Q_{i}, S_{i}, L_{1}^{(i)} and θ_{i} (i ∈ {1, …, k}), we obtain a glued near polygon S.

We define now an isomorphism θ between T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}) and S. Let x be an arbitrary point of Π \ Π_{∞}.

(1) Let ϕ(x) denote the unique point of pa nearest to x (in the near polygon T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k})).

(2) For i ∈ {1, …, k}, the space ⟨x, α_{1}, …, α_{i−1}, α_{i+1}, …, α_{k}⟩ intersects ⟨a, α_{i}⟩ in a line ϕ_{i}(x) through p. Clearly ϕ_{i}(x) ∈ S_{i}.

Put θ(x) = (ϕ(x), ϕ_{1}(x), …, ϕ_{k}(x)). We prove now that θ is a bijection. Let (x, L_{1}, …, L_{k}) be an element of pa × S_{1} × … × S_{k}. The intersection of the k spaces ⟨α_{1}, …, α_{i−1}, L_{i}, α_{i+1}, …, α_{k}⟩, i ∈ {1, …, k}, is a line L through p. We prove now that pa and L are parallel lines of the near polygon T_{2k}^{∗}(H_{1}∪…∪H_{k}). We may suppose that pa ≠ L. The plane ⟨L, pa⟩ intersects Π_{∞} in a line L′ through p. Let z be the point of L′ \ {p} with smallest index. For every point u of pa, there exists a point u′ on L for which d(u, u′) = d(pa, L) (z, u and u′ are collinear), hence L and pa are parallel. Let y be the unique point of L nearest to x (in T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k})), then y is the unique point of Π \ Π_{∞} for which θ(y) = (x, L_{1}, …, L_{k}). Since S and T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}) have the same order, it suffices to show that θ preserves adjacency. Let x and y be two collinear points of T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}), the line xy meets then one of the hyperovals, say H_{l}. Clearly ϕ_{m}(x) = ϕ_{m}(y) for all m ∈ {1, …, k} \ {l}. We prove that x is classical with respect to Q_{l} and that the unique point x′ of Q_{l} nearest to x is contained in the line ϕ_{l}(x) of S_{l}. This is clear if x ∈ A_{l}; suppose therefore that x ∉ A_{l}. The 4-dimensional space ⟨x, A_{l}⟩ intersects Π_{∞} in a threedimensional space β through α_{l}, which itself intersects ⟨α_{1}, …, α_{l−1}, α_{l+1}, …, α_{k}⟩ in a line M through p. Let u be the unique point of M \ {p} with smallest index (say μ). The point u is then also the unique point of β \ α_{l} with smallest index. Hence, there is a unique point x′ in A_{l} nearest to x, namely the intersection of xu with A_{l}, and d(x, x′) = μ. The point x′ is clearly contained in ϕ_{l}(x). By the point-quad relation, see Theorem 3.5.3, we know that this point is equal to the projection of ϕ(x) on the line ϕ_{l}(x). Similarly, there is a unique point y′ in Q_{l} nearest to y. It remains to prove , see condition (ii) on page 132, that x′ and y′ are collinear points of Q_{l}. Since d(x, x′) = d(y, y′) = μ and d(x, x′) + d(x′, y′) = d(x, y′) ≤ d(x, y) + d(y, y′) = μ + 1, it follows that d(x′, y′) ≤ 1. If x′ = y′, then there is a point z on the line xy at distance μ − 1 from x′, a contradiction. □

7.2 On the classification of glued near polygons

Lemma 7.2.1 Let X be a set of order n ≥ 1 and let G_{1} and G_{2} be regular groups of permutations on X, then there exists a permutation θ ∈ Sym(X) such that G_{2} = θ^{−1}G_{1}θ if and only if G_{1} is isomorphic to G_{2}. Moreover, if ϕ is an isomorphism from G_{1} to G_{2}, then there exist n permutations θ such that ϕ(g_{1}) = θ^{−1}g_{1}θ for all g_{1} ∈ G_{1}.

Proof. If G_{2} = θ^{−1}G_{1}θ, then the map ϕ defined by ϕ(g_{1}) = θ^{−1}g_{1}θ, ∀g_{1} ∈ G_{1}, defines an isomorphism from G_{1} to G_{2}. Conversely, suppose that ϕ : G_{1} → G_{2} is an isomorphism. Let x ∈ X be fixed. The condition θ[ϕ(g_{1})](x) = g_{1}θ(x), ∀g_{1} ∈ G_{1} implies that θ is completely determined as soon as θ(x) is known. For all y, z ∈ X, we define

θ_{y}^{ϕ}(z) = [ϕ^{−1}(g_{2})](y),

where g_{2} is the unique element of G_{2} such that g_{2}(x) = z. It is straightforward to check that θ_{y}^{ϕ} satisfies the required conditions. □

We use the same notations as in the previous section and we assume that (1) S is a near polygon,

(2) k ≥ 2 (see Remark (1) on 134),

(3) none of the GQ’s Q_{1}, …, Q_{k} is a grid (see Remark on page 136).

{Theorem 7.2.2 \atop (1)} G_{i} ≃ G_{j} for all i, j ∈ \{1, …, k\}.

(2) S_{i} is a spread of symmetry of Q_{i}.

(3) G_{i} is commutative if k ≥ 3.

Proof. Let i, j be two distinct elements of {1, …, k}. Since [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0, see Theorem 7.1.2, every element θ_{i}θ_{j}^{−1}gθ_{j}θ_{i}^{−1} with g ∈ G_{j} will define an automorphism of Q_{i} fixing each line of S_{i} (see Theorem 2.7.10). By Theorems 2.7.1 and 2.7.7, |G_{j}| = s + 1 and S_{j} is a spread of symmetry of Q_{j}. This proves (2). Since G_{i} ≃ θ_{i}^{−1}G_{i}θ_{i}, G_{j} ≃ θ_{j}^{−1}G_{j}θ_{j}, (1) follows from Lemma 2.7.9. Now, suppose that k ≥ 3 and put G = θ_{1}^{−1}G_{1}θ_{1}. Since [G, θ_{i}^{−1}G_{i}θ_{i}] = 0 for all i ∈ {2, …, k}, θ_{i}^{−1}G_{i}θ_{i} = G̃. Since k ≥ 3, [G̃, G̃] = 0 implying that G̃ = G is commutative. □

Remark. Since S_{i} is a spread of symmetry of Q_{i}, there exists an AT-model (D_{i}, K_{i}, Δ_{i}) for Q_{i} with associated spread S_{i}. The group G_{i}, which is the group of projectivities of L_{1}^{(i)} with respect to S_{i}, is isomorphic to K_{i}, see Theorem 2.7.8.

We can construct glued near (2k + 2)-gons, k ≥ 2, in the following way.

(1) Take a group K of order s + 1 ≥ 2 for which there exists an admissible triple (D, K, Δ), where D is a linear space different from a line. If k ≥ 3, then K has to be commutative.

(2) Let (D_{i}, K, Δ_{i}), i ∈ {1, …, k}, be a sequence of k, not necessarly distinct, admissible triples. Let Q_{i} denote the corresponding generalized quadrangle and let S_{i} be the associated spread. Let L_{1}^{(i)} be an arbitrary line of S_{i} and let θ_{1} be an arbitrary permutation of L_{1}^{(1)}. As before, let G_{i} denote the group of projectivities of L_{1}^{(i)} with respect to S_{i}. Define G := θ_{1}^{−1}G_{1}θ_{1}. By Theorem 7.1.2, one has to find bijections θ_{i} : L_{1}^{(1)} → L_{1}^{(i)}, i ≥ 2, satisfying θ_{i}^{−1}G_{i}θ_{i} = G̃. We now prove that the number of possible θ_{i}’s is equal to (s + 1) × | Aut(K)|. Let ϕ_{i} denote an arbitrary bijection from L_{1}^{(1)} to L_{1}^{(i)}, then θ_{i}^{−1}G_{i}θ_{i} = G̃ is equivalent to (θ_{i}^{−1}ϕ_{i})(ϕ_{i}^{−1}G_{i}ϕ_{i})(ϕ_{i}^{−1}θ_{i}) = G̃. Since ϕ_{i}^{−1}G_{i}ϕ_{i} ≃ K and G̃ ≃ G ≃ G_{1} ≃ K, the equation has (s + 1) × | Aut(K)| solutions for ϕ_{i}^{−1}θ_{i} (see Lemma 7.2.1). As a consequence, the set of equations θ_{2}^{−1}G_{2}θ_{2} = G̃, …, θ_{k}^{−1}G_{k}θ_{k} = G̃ has (s + 1)^{k−1}| Aut(K)|^{k−1} solutions for (θ_{2}, …, θ_{k}).

Collecting the previous results, we have the following theorem.

Theorem 7.2.3 Let Q_{1}, …, Q_{k} be k generalized quadrangles having all the same line size s + 1. Suppose that none of these GQ’s is isomorphic to a grid, and let S_{i}, i ∈ {1, …, k}, be a spread of Q_{i}. Then there is a glued near polygon derived from the Q_{i}’s and the S_{i}’s if and only if there is a group K of order s + 1 such that

(1) The group of automorphisms of Q_{i}, i ∈ {1, …, k}, fixing each line of S_{i} is isomorphic to K.

(2) K is commutative if k ≥ 3.

Moreover, the number of nonisomorphic glued near polygons which can be obtained this way is at most (s + 1)^{k−1}| Aut(K)|^{k−1}.

Problem. What is the number of nonisomorphic glued near polygons, given the GQ’s and their spreads?

In order to tackle this problem, one must find conditions under which two glued near polygons are isomorphic. To determine these automorphisms, one can use the fact that substructures must be mapped to substrucures of the same kind. Good candidates for these substructures are the sub near polygons, which we will study in the next section. We will then emphasize on near hexagons in Sections 7.4 and 7.5. In fact, one can get better estimates than the one obtained in Theorem 7.2.3, see Theorem 7.5.1.

7.3 Sub near polygons

We use the same notations as in Section 7.1.

Theorem 7.3.1 Let S be a glued near polygon, then every two points at distance 2 have at least two common neighbours.

Proof. Let α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and β = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) be two points at distance 2. We may suppose that one of the following two cases occurs.

(1) If i_{1} ≠ j_{1}, i_{2} ≠ j_{2}, i_{m} = j_{m} for all m ∈ {3, …, k}, y = (∏_{a∈{1,2}}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)} θ_{a}))(x), then ((θ_{1}^{−1}Δ_{i_{1}j_{1}}^{(1)}θ_{1})(x), L_{j_{1}}^{(1)}, L_{i_{2}}^{(2)}, …) and ((θ_{2}^{−1}Δ_{i_{2}j_{2}}^{(1)}θ_{2})(x), L_{i_{1}}^{(1)}, L_{j_{2}}^{(2)}, …) are two common neighbours of α and β.

(2) Let i_{1} ≠ j_{1}, i_{m} = j_{m} for all m ∈ {2, …, k}, y ≠ (θ_{1}^{−1}Δ_{i_{1}j_{1}}^{(1)}θ_{1})(x). Since p_{j_{1}}^{(1)}θ_{1}(y) and p_{i_{1}}^{(1)}θ_{1}(x) are two noncollinear points of Q_{1}, they have t_{1} + 1 common neighbours. If p_{δ_{1}}^{(1)}θ_{1}(z_{1}) and p_{δ_{2}}^{(1)}θ_{1}(z_{2}) are two of these neighbours, then (z_{1}, L_{δ_{1}}^{(1)}, L_{i_{2}}^{(2)}, …) and (z_{2}, L_{δ_{2}}^{(1)}, L_{i_{2}}^{(2)}, …) are two common neighbours of α and β.

□

If s ≥ 2, then Theorem 3.7.3 implies the existence of geodetically closed sub near polygons. We will give now an explicit description of these sub near polygons, and this will prove that they also exist if s = 1. So, let α = (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) and β = (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) be two points at distance δ. Let H(α, β) be the unique smallest geodetically closed subspace containing α and β (i.e. the intersection of all geodetically closed subspaces containing α and β). We may suppose that one of the following cases occurs.

(1) Let i_{m} ≠ j_{m} for all m < δ, i_{m} = j_{m} for all m ≥ δ, y ≠ (∏_{a∈{1,...,δ−1}}(θ_{a}^{−1}

Δ_{i_{a}j_{a}}^{(a)}θ_{a})) (x). The point γ = (∏_{a∈{2,...,δ−1}}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})(x), L_{i_{1}}^{(1)}, L_{j_{2}}^{(2)}, …,

L_{j_{k}}^{(k)}) lies on a shortest path from α to β. Hence, H(α, β) contains H(β, γ). By remark (1) on page 134, we know that H(β, γ) is the quad consisting of all the points (z, L_{f_{1}}^{(1)}, L_{j_{2}}^{(2)}, …, L_{j_{k}}^{(k)}), where z I L_{1}^{(1)} and f_{1} ∈ {1, …, 1 + st_{1}}. Repeating the above argument, one finds {(z, L_{f_{1}}^{(1)}, …, L_{f_{δ−1}}^{(δ−1)}, L_{j_{δ}}^{(δ)}, …, L_{j_{k}}^{(k)})|z I L_{1}^{(1)} and f_{i} ∈ {1, …, 1 + st_{i}} for all i ∈ {1, …, δ−1}} ⊆ H(α, β), and we have equality since the first set is classical and hence geodetically closed. Moreover, H(α, β) induces a sub near (2δ)-gon which is also a glued one.

(2) Let i_{m} ≠ j_{m} for all m ∈ {1, …, δ}, i_{m} = j_{m} for all m ∈ {δ + 1, …, k}, y = (∏_{a∈{1,...,δ}}(θ_{a}^{−1}Δ_{i_{a}j_{a}}^{(a)}θ_{a})) (x). The point γ = ((θ_{1}^{−1}Δ_{i_{1}j_{1}}^{(1)}θ_{1})(x), L_{j_{1}}^{(1)},

L_{i_{2}}^{(2)}, …, L_{i_{k}}^{(k)}) lies on a shortest path from α to β. Since d(α, γ) = 1, every point of the line αγ is contained in H(α, β). Now, S_{1} is a normal spread since it is also a spread of symmetry. If L_{l_{1}}^{(1)} ∈ sp(L_{i_{1}}^{(1)}, L_{j_{1}}^{(1)}), i.e. if L_{l_{1}}^{(1)} belongs to the hyperbolic line through L_{i_{1}}^{(1)} and L_{j_{1}}^{(1)}, then ((θ_{1}^{−1}Δ_{i_{1}l_{1}}^{(1)}θ_{1})(x), L_{l_{1}}^{(1)}, L_{i_{2}}^{(2)}, …, L_{i_{k}}^{(k)}) ∈ H(α, β). Repeating this argument, one finds that H′(α, β) := { (∏_{a∈{1,...,δ}}(θ_{a}^{−1}Δ_{i_{a}l_{a}}^{(a)}θ_{a})(x), L_{l_{1}}^{(1)}, …, L_{l_{δ}}^{(δ)}, L_{i_{δ+1}}^{(δ+1)}, …, L_{i_{k}}^{(k)}) | L_{l_{μ}}^{(μ)} ∈ sp(L_{i_{μ}}^{(μ)}, L_{j_{μ}}^{(μ)}) for all μ ∈ {1, …, δ} } ⊆ H(α, β). We prove now that H′(α, β) is geodetically closed; hence H′(α, β) = H(α, β). Let γ_{3} = (x_{3}, L_{h_{1}}^{(1)}, L_{h_{2}}^{(2)}, …, L_{h_{k}}^{(k)}) be a point on a shortest path from γ_{1} = (x_{1}, L_{f_{1}}^{(1)}, …, L_{f_{k}}^{(k)}) ∈ H′(α, β) to γ_{2} =

(x_{2}, L_{g_{1}}^{(1)}, …, L_{g_{k}}^{(k)}) ∈ H′(α, β). Since x_{1} = (∏_{a∈{1,...,δ}}(θ_{a}^{−1}Δ_{i_{a}f_{a}}^{(a)}θ_{a})) (x)

and x_{2} = (∏_{a∈{1,...,δ}}(θ_{a}^{−1}Δ_{i_{a}g_{a}}^{(a)}θ_{a})) (x), it follows that x_{2} = _{(∏a∈{1,...,δ}}

(θ_{a}^{−1}Δ_{i_{a}g_{a}}^{(a)}Δ_{f_{a}i_{a}}^{(a)}θ_{a})) (x_{1}). Since S_{1} is a normal spread, x_{2} = _{(∏a∈{1,...,δ}}

(θ_{a}^{−1}Δ_{f_{a}g_{a}}^{(a)}θ_{a})) (x_{1}). For every point γ of S, let γ̄ denote the k-tuple

obtained by deleting the element (of L_{1}^{(1)}) in the first position of γ. Let A_{ij} ⊆ {1, …, k} be the set of positions in which γ̄_{i} and γ̄_{j} differ (1 ≤ i < j ≤ 3). Then |A_{12}| = d(γ_{1}, γ_{2}) = d(γ_{1}, γ_{3})+d(γ_{3}, γ_{2}) ≥ |A_{13}|+|A_{23}|. At the other hand, |A_{12}| ≤ |A_{13}| + |A_{23}|, implying A_{12} = A_{13} ∪ A_{23}, A_{13} ∩ A_{23} = ∅, d(γ_{1}, γ_{3}) = |A_{13}| and d(γ_{2}, γ_{3}) = |A_{23}|. Moreover h_{i} ∈ {f_{i}, g_{i}} for all i ∈ {1, …, k}, implying L_{h_{μ}}^{(μ)} ∈ sp(L_{i_{μ}}^{(μ)}, L_{j_{μ}}^{(μ)}) for all μ ∈

{1, …, δ}, and x_{3} = (∏_{1≤a≤δ}(θ_{a}^{−1}Δ_{f_{a}h_{a}}^{(a)}θ_{a})) (x_{1}) = (∏_{1≤a≤δ}(θ_{a}^{−1}Δ_{f_{a}h_{a}}^{(a)}

Δ_{i_{a}f_{a}}^{(a)}θ_{a})) (x) = (∏_{1≤a≤δ}(θ_{a}^{−1}Δ_{i_{a}h_{a}}^{(a)}θ_{a})) (x). Hence γ_{3} ∈ H′(α, β). It is also immediate that H(α, β) induces a sub near (2δ)-gon of Hamming type (by considering the map γ ↦ γ̄).

For every point γ_{1} not inside H(α, β), there exists a point γ_{2} in H(α, β) such that d(γ_{1}, γ_{2}) > δ. Collecting all the previous calculations, we have the following Theorem.

Theorem 7.3.2 Every two points of a glued near polygon at distance δ of each other are contained in a unique geodetically closed sub near (2δ)-gon. This sub near polygon is also a glued one.

Remark. Suppose that S is a glued near hexagon and suppose that none of the GQ’s Q_{1}, …, Q_{k}, k ≥ 2, is a grid. By Theorem 7.2.3 we know then that S_{i} is a spread of symmetry of Q_{i}, i ∈ {1, …, k}. This means that there are exactly s + 1 automorphisms of Q_{i} fixing each line of S_{i}. These automorphisms can be recovered as follows in the near polygon. We may assume that k = 2 (otherwise take a suitable sub near hexagon) and that i = 1. For fixed j ∈ {1, …, 1+st_{2}}, the set A_{j} = {(x, L_{l}^{(1)}, L_{j}^{(2)})|x I L_{1}^{(1)}, 1 ≤ l ≤ 1+st_{1}} is the point set of a quad A_{j} isomorphic to Q_{1} and since it is also a classical set, one can define projections [A_{j_{1}}, A_{j_{2}}] : A_{j_{1}} → A_{j_{2}} for all j_{1}, j_{2} ∈ {1, …, 1+st_{2}}. These projections preserve collinearity. It is straightforward to check that the map [A_{j_{2}}, A_{1}][A_{j_{1}}, A_{j_{2}}][A_{1}, A_{j_{1}}], 1 ≤ j_{1}, j_{2} ≤ 1 + st_{2}, defines an automorphism of A_{1} fixing each line of the spread {M_{1}, …, M_{1+st_{1}}}; here M_{l} denotes the line with point set {(x, L_{l}^{(1)}, L_{1}^{(2)})|x I L_{1}^{(1)}}.

7.4 Isomorphisms between glued near hexagons

Introduction. Let S be the glued near hexagon derived from Q_{1}, Q_{2}, S_{1} = {L_{1}^{(1)}, …, L_{1+st_{1}}^{(1)}}, S_{2} = {L_{1}^{(2)}, …, L_{1+st_{2}}^{(2)}}, θ_{1} and θ_{2}. Similarly, let S′ be the glued near hexagon derived from Q′_{1}, Q′_{2}, S′_{1} = {L_{1}^{′(1)}, …, L_{1+st_{1}}^{′(1)}}, S′_{2} = {L_{1}^{′(2)}, …, L_{1+st_{2}}^{′(2)}}, θ′_{1} and θ′_{2}. Suppose that S ≃ S′. By considering quads in S and S′, either (Q_{1} ≃ Q′_{1} and Q_{2} ≃ Q′_{2}) or (Q_{1} ≃ Q′_{2} and Q_{2} ≃ Q′_{1}). Since S can be obtained starting from any ordering of the GQ’s, we may suppose that Q_{1} = Q′_{1} and Q_{2} = Q′_{2}. Points of S are triples. If we take a quad A consisting of points with fixed third coordinate (A ≃ Q_{1}), and we intersect A with all quads consisting of points with fixed second coordinate, then one finds a spread of A equivalent to S_{1}. We may hence suppose that S_{1} = S′_{1} and S_{2} = S′_{2}. By Theorem 7.2.2, we may suppose that L_{1}^{(1)} = L_{1}^{′(1)} and L_{1}^{(2)} = L_{1}^{′(2)}. We may also suppose that θ_{1} and θ′_{1} are identical maps.

For i ∈ {1, 2}, let Q_{i} be a GQ different from a grid, let S_{i} be a spread of Q_{i} and let L_{1}^{(i)} be a base line of S_{i}. Finally, let θ_{1} be the identical map. For every map θ_{2} : L_{1}^{(1)} → L_{1}^{(2)}, we denote by S(θ_{2}) the partial linear space obtained by applying the construction given in Section 7.1. We consider now the following interesting problem.

Problem. Let θ and θ′ be two maps from L_{1}^{(1)} to L_{1}^{(2)}. Under which conditions are S(θ) and S(θ′) isomorphic near hexagons?

As before, let G_{i}, i ∈ {1, 2}, denote the group of projectivities of L_{1}^{(i)} with respect to S_{i}. Assume that S(θ) and S(θ′) are indeed near hexagons. Hence

[θ^{−1}G_{2}θ, G_{1}] = 0,  (7.1)

and

[θ^{′−1}G_{2}θ′, G_{1}] = 0.  (7.2)

We apply the following change of notation. The maps Δ_{ij}^{(1)}, i, j ∈ {1, …, 1 + st_{1}}, will shortly be denoted by δ_{ij}, while the maps Δ_{ij}^{(2)}, i, j ∈ {1, …, 1 + st_{2}}, will be denoted by Δ_{ij}. The notations δ_{ij}^{β} and Δ_{ij}^{β} (with β a suitable permutation) are then used to denote δ_{β(i)β(j)} and Δ_{β(i)β(j)} as we already did in Section 2.8. The points of S(θ) and S(θ′) are triples having three coordinates. A set of points with fixed third coordinate is a quad; quads arising this way are said to be of type I. Similarly, quads of type II are set of points with fixed second coordinate. The remaining quads (which are grids) are said to be of type III. If Φ is an isomorphism between S(θ) and S(θ′), then either Φ preserves the quad-type, or it interchanges quads of type (I) and (II).

(A) Let Φ be quad-type preserving, then there exist:

(a) a permutation β of {1, …, 1 + st_{1}},

(b) a permutation γ of {1, …, 1 + st_{2}},

(c) permutations α_{ij} of L_{1}^{(1)},

such that

Φ[(x, L_{i}^{(1)}, L_{j}^{(2)})] = (α_{ij}(x), L_{β(i)}^{(1)}, L_{γ(j)}^{(2)}),

for all x ∈ L_{1}^{(1)}, for all i ∈ {1, …, 1 + st_{1}} and for all j ∈ {1, …, 1 + st_{2}}. We define

α := α_{11}.

Consider in S(θ) the adjacent points (x, L_{i}^{(1)}, L_{j}^{(2)}) and (δ_{ik}(x), L_{k}^{(1)}, L_{j}^{(2)}) with i ≠ k. Then (α_{ij}(x), L_{β(i)}^{(1)}, L_{γ(j)}^{(2)}) ∼ (α_{kj}δ_{ik}(x), L_{β(k)}^{(1)}, L_{γ(j)}^{(2)}) in S(θ′) and hence

α_{ij} = δ_{ki}^{β}α_{kj}δ_{ik}.  (7.3)

Consider in S(θ) the adjacent points (x, L_{i}^{(1)}, L_{j}^{(2)}) and (θ^{−1}Δ_{jk}θ(x), L_{i}^{(1)}, L_{k}^{(2)}) with j ≠ k, then (α_{ij}(x), L_{β(i)}^{(1)}, L_{γ(j)}^{(2)}) ∼ (α_{ik}θ^{−1}Δ_{jk}θ(x), L_{β(i)}^{(1)}, L_{γ(k)}^{(2)}) in S(θ′) and hence

α_{ij} = θ^{′−1}Δ_{kj}^{γ}θ′α_{ik}θ^{−1}Δ_{jk}θ.  (7.4)

From (7.3) and (7.4), it follows that

α_{ij} = δ_{1i}^{β}α_{1j} = δ_{1i}^{β}θ^{′−1}Δ_{1j}^{γ}θ′α,  (7.5)

for all i ∈ {1, …, 1 + st_{1}} and for all j ∈ {1, …, 1 + st_{2}}. From (7.2), (7.3) and (7.5) it follows that

(δ_{k1}^{β}δ_{ik}^{β}δ_{1i}^{β})α = αδ_{ik},  (7.6)

for all i, k ∈ {1, …, 1 + st_{1}}. From (7.2), (7.4) and (7.5) it follows that

(Δ_{k1}^{γ}Δ_{jk}^{γ}Δ_{1j}^{γ})(θ′αθ^{−1}) = (θ′αθ^{−1})Δ_{jk},  (7.7)

for all j, k ∈ {1, …, 1 + st_{2}}.

(B) If Φ interchanges the quads of type (I) and (II), then we may suppose that Q_{2} = Q_{1}, S_{2} = S_{1} and L_{1}^{(2)} = L_{1}^{(1)}. Using another ordering of the GQ’s, we can consider Φ as a quad-type preserving isomorphism from S(θ^{−1}) to S(θ′). Equation (7.7) becomes then

(Δ_{k1}^{γ}Δ_{jk}^{γ}Δ_{1j}^{γ})(θ′αθ) = (θ′αθ)Δ_{jk},  (7.8)

for all j, k ∈ {1, …, 1 + st_{2}}.

We have now proved the following theorem.

{Theorem 7.4.1 \atop (1)} If the pairs (Q_{1}, S_{1}) and (Q_{2}, S_{2}) are not equivalent (i.e. if there exists no isomorphism from Q_{1} to Q_{2} mapping S_{1} to S_{2}), then S(θ) and S(θ′) are isomorphic glued near hexagons if and only if θ and θ′ satisfy (7.1), (7.2) and (7.7) for some permutation α of L_{1}^{(1)} satisfying (7.6), some permutation β of {1, …, 1 + st_{1}} and some permutation γ of {1, …, 1 + st_{2}}.

(2) If the pairs (Q_{1}, S_{1}) and (Q_{2}, S_{2}) are equivalent, then we may suppose that Q_{2} = Q_{1}, S_{2} = S_{1} and L_{1}^{(2)} = L_{1}^{(1)}. In this case S(θ) and S(θ′) are isomorphic glued near hexagons if and only if θ and θ′ satisfy (7.1), (7.2), (7.7) or (7.1), (7.2), (7.8) for some permutation α of L_{1}^{(1)} satisfying (7.6), some permutation β of {1, …, 1+st_{1}} and some permutation γ of {1, …, 1 + st_{2}}.

Theorem 7.4.2 Equations (7.1), (7.6) and (7.7) imply equation (7.2). Similarly, if Q_{2} = Q_{1}, S_{2} = S_{1} and L_{1}^{(2)} = L_{1}^{(1)}, then equations (7.1), (7.6) and (7.8) imply equation (7.2).

Proof. Suppose that (7.1), (7.6) and (7.7) hold, then

α^{−1}[θ^{′−1}G_{2}θ′, G_{1}]α =^{(7.6)}α^{−1}[θ^{′−1}G_{2}θ′, αG_{1}α^{−1}]α = [α^{−1}θ^{′−1}G_{2}θ′α, G_{1}]

=^{(7.7)}[α^{−1}θ^{′−1}(θ′αθ^{−1})G_{2}(θ′αθ^{−1})^{−1}θ′α, G_{1}] = [θ^{−1}G_{2}θ, G_{1}] =^{(7.1)}0.

The other statement is proved similarly. □

Remark. As in Section 2.8, one can define the sets Ĝ_{1} and Ĝ_{2}; equations (7.6), (7.7) and (7.8) become then α ∈ Ĝ_{1}, θ′αθ^{−1} ∈ Ĝ_{2} and θ′αθ ∈ Ĝ_{2}.

7.5 The number of nonisomorphic glued near hexagons

Let (D_{1}, K, Δ_{1}) and (D_{2}, K, Δ_{2}) be two admissible triples. Here D_{1} and D_{2} are two linear spaces with constant line size s + 1 and respective point sets P_{1} and P_{2}; D_{1} and D_{2} are supposed to be different from lines such that the corresponding GQ’s Q_{1} and Q_{2} are not grids. Recall that K is a group of order s + 1 and that Δ_{1} : P_{1} × P_{1} → K and Δ_{2} : P_{2} × P_{2} → K are two maps such that the admissibility condition is satisfied. As we have seen in Section 7.2, near hexagons can be derived from these two AT’s. Let S_{1} and S_{2} be the associated spreads of the AT’s. Let x_{1} ∈ P_{1} and x_{2} ∈ P_{2} be fixed and let L_{1}^{(1)} = {(k, x_{1})|k ∈ K} and L_{1}^{(2)} = {(k, x_{2})|k ∈ K}. With every bijection f from L_{1}^{(i)} to L_{1}^{(j)}, i, j ∈ {1, 2}, there corresponds a permutation f′ of K such that f((k, x_{i})) = (f′(k), x_{j}) for all k ∈ K. This can be applied to the maps θ, θ′, Δ_{ij}^{(1)}, Δ_{ij}^{(2)}, α, g_{1} ∈ G_{1}, g_{2} ∈ G_{2}, ĝ_{1} ∈ Ĝ_{1}, ĝ_{2} ∈ Ĝ_{2}, etc. In the sequel we will regard all these maps as permutations of K. Conditions (7.1), (7.2), (7.6), (7.7) and (7.8) remain then the same. To every permutation θ : K → K corresponds a map θ_{2} : L_{1}^{(1)} → L_{1}^{(2)} and hence a partial linear space S(θ) by Section 7.1 (suppose again that θ_{1} is the identical map). Let A be the set of all (s + 1)| Aut(K)| permutations θ of K for which S(θ) is a near hexagon. Let G be the group of permutations of K consisting of all right multiplications, then K ≃ G and G_{1} = G_{2} = G by Theorem 2.7.8. In Chapter 2, we saw an overview of all known GQ’s which have an AT-model. We have the following possibilities for K.

(1) K ≃ C_{q+1} with q a prime power.

(2) K ≃ C_{p} × … × C_{p} being the additive group of GF(p^{h}).

Sometimes classes (1) and (2) overlap; in this case h = 1 and q + 1 = p. This only happens if p is a prime of the form 2^{2^{l}} + 1 (a so-called Mersenne prime); the only known such primes are p = 3, p = 5, p = 17, p = 257 and p = 65537. In each of the above cases K and hence G is commutative. Equation (7.1) becomes then Gθ = θG and has as solution all the maps θ : K → K : x ↦ k · χ(x), with k ∈ K and χ an automorphism of K. The number of solutions for θ in both cases is then as follows.

(1) If K ≃ C_{q+1}, then there are (q + 1) ϕ(q + 1) solutions with ϕ the Euler indicator.

(2) If K is isomorphic to the additive group of GF(p^{h}), then there are p^{h}|GL(h, p)| solutions.

Let N denote the number of glued near hexagons arising from Q_{1}, Q_{2}, S_{1} and S_{2}. We will make estimates for N.

Theorem 7.5.1 If (Q_{1}, S_{1}) is equivalent to (Q_{2}, S_{2}), then

\frac{(s + 1)| Aut(K)|}{2|Ĝ_{1}||Ĝ_{2}|} ≤ N ≤ \frac{(s + 1)| Aut(K)|}{max(|Ĝ_{1}|, |Ĝ_{2}|)};

if not so, then we even have that

\frac{(s + 1)| Aut(K)|}{|Ĝ_{1}||Ĝ_{2}|} ≤ N ≤ \frac{(s + 1)| Aut(K)|}{max(|Ĝ_{1}|, |Ĝ_{2}|)}.

Proof. Let θ be one of the (s + 1)| Aut(K)| maps of A. There are at most |Ĝ_{1}||Ĝ_{2}| maps θ′ satisfying equations (7.6) and (7.7), and a same remark holds for equations (7.6) and (7.8). This proves the lower bounds for N. Now, consider again equations (7.6) and (7.7). Let α and β be identical permutations, and let θ′ be one of the |Ĝ_{2}| maps for which θ′θ^{−1} ∈ Ĝ_{2}, then θ′ ∈ A by Theorem 7.4.2. Hence N ≤ \frac{(s+1)| Aut(K)|}{|Ĝ_{2}|} and the upper bound follows by symmetry. □

Corollary 7.5.2 (1) If Q_{1} and Q_{2} are dual grids, then only one glued near hexagon can be constructed.

(2) Let p be a prime of the form 2^{2^{l}} + 1, then there is only one glued near hexagon arising from the generalized quadrangles AS(p) and Q(5, p−1).

Proof.

(1) Up to an automorphism there is only one spread of symmetry in Q_{1} and Q_{2}. For these spreads we have s + 1 = 2, |K| = 2, | Aut(K)| = 1 and |Ĝ_{1}| = |Ĝ_{2}| = 2; hence N = 1.

(2) Up to an automorphism there is only one spread of symmetry in each GQ. For these spreads the group K is cyclic of order p, e.g. K = GF(p) equiped with the addition of the field. By results of Section 2.8, Ĝ_{1} consists of all permutations K → K : x ↦ kx + k′ with k ∈ K \ {0} and k′ ∈ K. Again by Section 2.8, Ĝ_{2} consists of all permutations K → K : x ↦ kx^{2^{i}} with k ∈ K and i ∈ {0, …, 2^{l} − 1}. Hence |Ĝ_{1}| = p(p−1) and |Ĝ_{2}| = p2^{l}. By Theorem 7.5.1, 0 < N ≤ \frac{p(p−1)}{p(p−1)} = 1.

□

Remark. If p = 3, then AS(p) ≃ Q(5, p − 1) and the corresponding glued near hexagon was already known before (see [14]).

Sometimes the lower bound given in Theorem 7.5.1 can be sharpened.

Theorem 7.5.3 If Ĝ_{1} = Ĝ_{2} is a group, then 1 + μ\frac{(s+1)| Aut(K)|−|Ĝ_{1}|}{|Ĝ_{1}||Ĝ_{2}|} ≤ N. Here μ = \frac{1}{2} or μ = 1 depending on whether (Q_{1}, S_{1}) is equivalent to (Q_{2}, S_{2}) or not.

Proof. Let θ ∈ Ĝ_{1}, then also θ ∈ A (see equations (7.1) and (7.6)). Now, by equations (7.7) and (7.8), we have that S(θ′) ≃ S(θ) if and only if θ′ ∈ Ĝ_{1}. The two parts in the lower bound of N correspond to the two parts in the expression A = Ĝ_{1} ∪ (A \ Ĝ_{1}) (see also the proof of Theorem 7.5.1). □

Remark. Theorem 2.8.2 gives sufficient conditions for Ĝ_{1} and Ĝ_{2} to be a group.

Corollary 7.5.4 Let Q_{1} = Q_{2} = P (W (q), x), q = p^{h}, and let S_{1} = S_{2} be the spread consisting of all hyperbolic lines through x. Then N = 1 if q = 4 or if q is a prime; otherwise N > 1.

Proof. The group K is cyclic of order q, e.g. K = GF(q) equiped with the addition of the field. Applying Theorems 7.5.1 and 7.5.3 yields

1 + \frac{q|GL(h, p)| − h(q − 1)q}{2h^{2}(q − 1)^{2}q^{2}} ≤ N ≤ \frac{q|GL(h, p)|}{h(q − 1)q}.

Note that |GL(h, p)| = (p^{h} −1)(p^{h} −p) … (p^{h} −p^{h−1}). The group Ĝ_{1} consists of all maps K → K : x ↦ kx^{p^{i}} + k′ where k ∈ GF(q) \ {0}, k′ ∈ GF(q) and i ∈ {0, …, h−1}, see Section 2.8. If q = 4 or if q is prime, then N = 1 by the above inequalities. If h ≥ 2 and q ≠ 4, then |GL(h, p)| ≥ (p^{h} − 1)(p^{h} − p) > h(q − 1); hence N > 1. □

For fixed h, the lower bound in the previous theorem behaves like \frac{1}{2h^{2}} p^{h^{2}−3h} for great values of p; hence many glued near hexagons will arise. We treat now the case h = 2, p odd.

Glued near hexagons arising from P (W (p^{2}), x), p odd

Let Q_{1} = Q_{2} = P (W (p^{2}), x), p an odd prime, and let S_{1} = S_{2} be the unique spread of symmetry. The group K is cyclic of order p^{2}, e.g. K = GF(p^{2}) equiped with the addition of the field. The p^{2}(p^{2} − 1)(p^{2} − p) elements of A are the maps θ : z ↦ az + bz^{p} + c with a, b, c ∈ GF(p^{2}) and a^{p+1} ≠ b^{p+1}. If z ↦ a′z + b′z^{p} + c′ is the inverse map, then one easily checks that

a′ = \frac{a^{p}}{a^{p+1} − b^{p+1}} , b′ = −\frac{b}{a^{p+1} − b^{p+1}} , c′ = \frac{bc^{p} − a^{p}c}{a^{p+1} − b^{p+1}} .

The elements of Ĝ_{1} = Ĝ_{2} are the maps K → K : x ↦ kx^{p^{i}} + k′ where k ∈ GF(p^{2}) \ {0}, k′ ∈ GF(p^{2}) and i ∈ {0, 1}. Let θ : z ↦ a_{1}z + b_{1}z^{p} + c_{1} and θ′ : z ↦ a_{2}z + b_{2}z^{p} + c_{2} be elements of A. If a_{1} = 0 or b_{1} = 0, then S(θ) ≃ S(θ′) if and only if a_{2} = 0 or b_{2} = 0. If a_{1} ≠ 0 ≠ a_{2}, then one checks that S(θ) ≃ S(θ′) if and only if

(I) a_{2} ≠ 0 ≠ b_{2} and (\frac{a_{2}}{b_{2}} )^{p+1} = (\frac{a_{1}}{b_{1}} )^{p+1}; or

(II) a_{2} ≠ 0 ≠ b_{2} and (\frac{a_{2}}{b_{2}} )^{p+1} = (\frac{b_{1}}{a_{1}} )^{p+1}.

If θ ∈ A, then (i) a = 0 or b = 0, (ii) a ≠ 0 ≠ b and (\frac{a}{b})^{p+1} = −1, or (iii) a ≠ 0 ≠ b and (\frac{a}{b})^{p+1} ≠ ±1. Hence

N = 1 + 1 + \frac{p − 3}{2} = \frac{p + 1}{2}.

Glued near hexagons arising from Q(5, q)

Let Q_{1} and Q_{2} be generalized quadrangles isomorphic to Q(5, q), q = p^{h}. As we have seen in Section 2.6.2, there is up to an automorphism only one spread of symmetry in Q(5, q). For such a spread, we calculate the number of glued near hexagons arising. The group K is cyclic of order q + 1, e.g. K = {x ∈ GF(q^{2})|x^{q+1} = 1} equiped with the multiplication of the field GF(q^{2}). The (q + 1) ϕ(q + 1) elements of A are the maps x → lx^{n} with l ∈ K and (n, q+1) = 1. The elements of Ĝ_{1} = Ĝ_{2} are the maps K → K : k ↦ lk^{p^{σ}} with l ∈ K and σ ∈ ℕ. Let θ : k ↦ l_{1}k^{n_{1}} and θ′ : k ↦ l_{2}k^{n_{2}} be two elements of A satisfying (7.6),(7.7) or (7.6),(7.7). Let α(k) = mk^{p^{σ_{1}}} and μ(k) = nk^{p^{σ_{2}}}. If μ = θ′αθ^{−1}, then l_{2} = nl_{1}^{p^{σ_{2}}}(m^{n_{2}})^{−1} and n_{2}p^{σ_{1}} ≡ n_{1}p^{σ_{2}} (mod q + 1); similarly, if μ = θ′αθ, then n_{1}n_{2}p^{σ_{1}} ≡ p^{σ_{2}} (mod q + 1). Hence S(θ) ≃ S(θ′) if and only if at least one of the following conditions is satisfied:

(I) n_{1} ≡ n_{2}p^{η} (mod q + 1) for some η ∈ ℕ,

(II) n_{1}n_{2} ≡ p^{η′} (mod q + 1) for some η′ ∈ ℕ.

If (I) and (II) are satisfied, then n_{1}^{2} ≡ p^{η′′} (mod q + 1) for some η′′ ∈ ℕ. Suppose that there are R elements n ∈ {1, …, q}, (n, q + 1) = 1, such that n^{2} ≡ p^{η′′} (mod q + 1) for some η′′ ∈ ℕ, then the number of nonisomorphic glued near hexagons is equal to

N = \frac{(q + 1) ϕ(q + 1) − (q + 1)R}{4h(q + 1)} + \frac{(q + 1)R}{2h(q + 1)} = \frac{ϕ(q + 1) + R}{4h}.

The tabel below gives the value of N for all prime powers q ≤ 19.

\begin{matrix}q & N & q & N & q & N \\ 2 & 1 & 7 & 2 & 13 & 2 \\ 3 & 1 & 8 & 1 & 16 & 2 \\ 4 & 1 & 9 & 1 & 17 & 2 \\ 5 & 1 & 11 & 2 & 19 & 3 \end{matrix}

The other known glued near hexagons

The other known glued near hexagons arise if Q_{1} and Q_{2} are of one of the following two types (hence four possible combinations).

(1) T_{2}^{∗}(O) with O a hyperoval in PG(2, 2^{h}).

(2) (S_{xy}^{−})^{D} associated with a hyperoval in PG(2, 2^{h}).

The group K is isomorphic to the additive group of GF(2^{h}). By Theorems 7.5.1 and 7.5.3 estimates for the number of glued near hexagons can be made. We have s = 2^{h} − 1, | Aut(K)| = |GL(2, h)|; the values of |Ĝ_{1}| and |Ĝ_{2}| depend on the kind of hyperoval we are working with (|Ĝ| = h′2^{h}(2^{h} −1) with h′ | h in the case of T_{2}^{∗}(O), see Section 2.8). Again, lots of glued near hexagons can be constructed.

7.6 Characterizations

7.6.1 Local spaces with a thin point

Definition. For u, v ∈ ℕ \ {0}, let S_{u,v} = (P_{u,v}, L_{u,v}, I_{u,v}) be the following linear space:

• P_{u,v} = {α, β_{1}, …, β_{u}, γ_{1}, …, γ_{v}},

• L_{u,v} = {{α, β_{1}, …, β_{u}}, {α, γ_{1}, …, γ_{v}}} ∪ {{β_{i}, γ_{j}}|1 ≤ i ≤ u and 1 ≤ j ≤ v},

• I_{u,v} is the symmetrized containment.

S_{u,v} is a linear space with a thin point (i.e. a point incident with only two lines). Conversely, every linear space with a thin point is obtained in this way. If S is a glued near hexagon, then S_{x} ≃ S_{t_{1},t_{2}} for all points x of S. Recall that a quad is called good if all points are incident with the same number of lines.

Theorem 7.6.1 Let S be a near hexagon satisfying the following properties:

• every two points at distance 2 are contained in a quad,

• if all lines of S are thin, then all quads are good,

• there exists a point x of S such that S_{x} ≃ S_{1,r} for some r ∈ ℕ \ {0},

then S is the direct product of a line with a nondegenerate GQ.

Proof. If not all lines of S have the same number of points, then S is the direct product of a line with a GQ, see Theorem 3.1.2. Hence, by Theorem 3.8.1, we may assume that S has order (s, t) with t = r+1. Consider through x a quad R_{x} containing t lines through x and let L_{x} be the remaining line through x. Every point z of R_{x} is incident with exactly one line L_{z} which is not in R_{x}. Let y ∈ L_{x} \ {x} be fixed. Let M_{1} and M_{2} be two lines through y different from L_{x} and let R_{y} be the quad through M_{1} and M_{2}. The quad through M_{i} (i ∈ {1, 2}) and L_{x} intersects R_{x} in a line M′_{i}. Now, let u be one of the s^{2}(t − 1) points of R_{x} at distance 2 from x. Let u_{i} (i ∈ {1, 2}) be the unique point on M′_{i} collinear with u. The quad through uu_{i} and L_{u_{i}} is a grid. Let u′_{i} be the intersection of L_{u_{i}} with M_{i} and let v_{i} be the unique neighbour of u′_{i} and u different from u_{i}. The point v_{i} is then the unique point of L_{u} at distance 2 from y. This implies that v = v_{1} = v_{2}. Since v is collinear with the points u′_{1} and u′_{2} of R_{y}, v is itself contained in R_{y}. Hence |Γ_{2}(y) ∩ R_{y}| ≥ s^{2}(t − 1). This implies that R_{y} is a GQ of order (s, t − 1) containing all lines through y, except the line L_{x} and that R_{y} ≃ R_{x}. The result follows now immediately. □

7.6.2 Characterizations of glued near hexagons

Theorem 7.6.2 Let S = (P, L, I) be a near hexagon satisfying the following properties:

• every two points at distance 2 are contained in a quad,

• if all lines of S are thin, then all quads are good,

• there exists a point x such that S_{y} has a thin point for all y ∈ Γ(x), then S is the direct product of a line with a nondegenerate GQ or S is a glued near hexagon.

Proof. If not all lines of S have the same number of points, then S is the direct product of a line with a nondegenerate GQ. Hence, by Theorem 3.8.1 we may assume that S has an order (s, t). If S_{y} (with y ∈ P) is a linear space with a thin point, then we may suppose that S_{y} contains a unique thin point which we denote by L_{y}, otherwise the result would follow from Theorem 7.6.1. The line L_{y} is then contained in exactly two quads. The following properties hold.

(a) If y is a point for which S_{y} is a linear space with a thin point, then S_{y′} ≃ S_{y} and L_{y′} = L_{y} for all points y′ ∈ L_{y}. Indeed, suppose S_{y} ≃ S_{t_{1},t_{2}} with t_{1}, t_{2} > 1 and t = t_{1} + t_{2}. The point L_{y} of S_{y′} is contained in exactly two lines of S_{y′}, one line has t_{1} + 1 points, the other t_{2} + 1 points. Since there are exactly t_{1} + t_{2} + 1 points in S_{y′}, it follows that S_{y′} ≃ S_{t_{1},t_{2}}.

(b) If y_{1}, y_{2} are points such that S_{y_{1}}, S_{y_{2}} are linear spaces with a thin point, then L_{y_{1}} and L_{y_{2}} are equal or disjoint. (This follows immediately from (a).)

(c) There exists a point y ∈ Γ(x) such that x ∈ L_{y}. Indeed, suppose that this is not true. Let y ∈ Γ(x) be fixed. Let Q be the quad of order (s, t′) through xy and L_{y}. There are s(t′ + 1) points z_{i} ∈ Q collinear with x. These give rise to s(t′ + 1) lines L_{z_{i}} and all these lines are different and hence disjoint by (b). Suppose that L_{z} is not contained in Q for a certain z ∈ Γ(x)∩Q, then S_{z} contains at least three thick lines (namely the line defined by Q and the two lines of S_{z} through L_{z}); a contradiction, since S_{z} is a linear space with a unique thin point. Hence, all lines L_{z} are contained in Q and there are at least (s + 1)(st′ + s) points in Q, but this is again impossible.

Let y ∈ Γ(x) such that x ∈ L_{y}. Hence S_{x} is also a linear space with a unique thin point L_{x}. Let Q_{1} and Q_{2} be the two quads through L_{x} with respective orders (s, t_{1}) and (s, t_{2}). In Q_{i} (i = 1, 2), there are st_{i} points z collinear with x and not on L_{x}. These give rise to st_{i} disjoint lines L_{z} which together with L_{x} form a spread S_{i} of Q_{i}. Put S_{i} = {L_{1}^{(i)}, …, L_{1+st_{i}}^{(i)}} with L_{1}^{(i)} = L_{x}. Finally, let θ_{i}, i ∈ {1, 2}, be the identical permutation of L_{x}. By the construction given in Section 7.1, we can define now a partial linear space S′. We will prove that S ≃ S′, hence S is glued. First we prove that every point u of S has distance at most 1 to each Q_{i} (i ∈ {1, 2}). Let u′ be the unique point of L_{x} nearest to u; we may suppose that d(u, u′) = 2. Since S_{u′} ≃ S_{x}, it follows that the quad through u and u′ intersects each Q_{i} in a line. This proves that each Q_{i} contains a point collinear with u. For i ∈ {1, 2} and u ∈ P, let p_{i}(u) denote the unique point of Q_{i} nearest to u. Next we prove that all the local spaces S_{u} are isomorphic to S_{t_{1},t_{2}}. Since for all u ∈ Q_{i}, L_{u} is contained in exactly two quads (Q_{i} and another quad), we have that G_{u} ≃ S_{t_{1},t_{2}}. Let u be a point of S not contained in Q_{1} ∪ Q_{2}. Let u′ = p_{1}(u) and u′′ = p_{2}(u). The local space S_{u} contains t_{1} + t_{2} + 1 points, a line with t_{1} + 1 points (determined by the quad through uu′′ and L_{u′′}) and a line with t_{2} + 1 points (determined by the quad through uu′ and L_{u′}). From this it follows that S_{u} ≃ S_{t_{1},t_{2}}. Hence L_{u} is defined for all u ∈ P and all these lines determine a spread of S. Each L_{u} is contained in exactly two quads. One quad intersects Q_{2} in a line and is isomorphic to Q_{1}. The other quad intersects Q_{1} and is isomorphic to Q_{2}. Note that the isomorphisms are defined by the projections p_{i}, i ∈ {1, 2} . We consider now the following map Δ : P ↦ L_{x}×S_{1}×S_{2}, Δ(u) = (γ(u), δ_{1}(u), δ_{2}(u)), where γ(u) is the unique point of L_{x} nearest to u and δ_{i}(u) (i ∈ {1, 2}) is the unique line of S_{i} incident with p_{i}(u). Let (a, L_{1}, L_{2}) = (γ(u), δ_{1}(u), δ_{2}(u)) and put a_{i} (i ∈ {1, 2}) equal to the projection of a on the line L_{i} of Q_{i}. If L_{1} = L_{x}, then u = a_{2}; if L_{2} = L_{x}, then u = a_{1}; if L_{1} ≠ L_{x} ≠ L_{2}, then u is the common neighbour of a_{1} and a_{2} different from a. This proves that Δ is a bijection. Since S and S′ have the same order, it suffices to prove that Δ preserves adjacency in the point graph of the geometries. Let x and x′ be two adjacent points of S. If x and x′ are contained in a quad intersecting Q_{2}, then δ_{2}(x) = δ_{2}(x′). If δ_{1}(x) = L_{i}^{(1)} and δ_{1}(x′) = L_{j}^{(1)}, then p_{i}^{(1)}θ_{1}(γ(x)) = p_{1}(x) and p_{j}^{(1)}θ_{1}(γ(x′)) = p_{1}(x′). Since p_{1}(x) ∼ p_{1}(x′), p_{i}^{(1)}θ_{1}(γ(x)) ∼ p_{j}^{(1)}θ_{1}(γ(x′)). A similar reasoning holds if x and x′ are contained in a quad intersecting Q_{1}.

□

Theorem 7.6.3 Let S be a near hexagon satisfying the following properties:

• every two points at distance 2 are contained in a quad,

• if all lines of S are thin, then all quads are good,

• there exists a point x such that S_{x} has a thin point and such that S_{y} contains the same number of lines for all y ∈ Γ(x),

then S is the direct product of a line with a nondegenerate GQ or S is a glued near hexagon.

Proof. Just like before, we may suppose that S has an order (s, t). Theorem 3.8.2 implies that the number of points in Γ_{2}(y) is independent of the point y of S. For y ∈ Γ(x), let V_{y} denote the set of quads through y. Now,

∑_{Q∈V_{y}} 1, ∑_{Q∈V_{y}} s^{2}t_{Q}, ∑_{Q∈V_{y}} t_{Q}(t_{Q} + 1),

are respectively equal to the number of quads through y, the number of points in Γ_{2}(y) and t(t + 1), hence these expressions are independent of y ∈ Γ(x). Let L_{x} be a thin point of S_{x} and let Q_{1} and Q_{2} be the two quads through L_{x} with respective orders (s, t_{1}) and (s, t_{2}); t = t_{1} + t_{2}. Let z ≠ x be a second point of L_{x}. If y ∈ Q_{1} ∩ Γ(x), then

∑_{Q∈V_{y}}(t_{Q} − 1)(t_{2} − t_{Q}) = ∑_{Q∈V_{z}}(t_{Q} − 1)(t_{2} − t_{Q}) = (t_{1} − 1)(t_{2} − t_{1}).

Let V′_{y} = V_{y} \ {Q_{1}}, then

∑_{Q∈V′_{y}}(t_{Q} − 1)(t_{2} − t_{Q}) = 0.

Since there are only t + 1 lines through y and Q_{1} has t_{1} + 1 lines through y, it follows that 1 ≤ t_{Q} ≤ t_{2} for all Q ∈ V′_{y}. This implies that t_{Q} = 1 or t_{Q} = t_{2} for all Q ∈ V′_{y}. By Theorem 7.6.1, we may suppose that t_{1}, t_{2} ≠ 1. From

∑_{Q∈V_{y}} 1 = ∑_{Q∈V_{z}} 1,

and

∑_{Q∈V_{y}} t_{Q} = ∑_{Q∈V_{z}} t_{Q},

it follows now that the number of quads Q of V′_{y} with t_{Q} = t_{2} is equal to 1. This implies that S_{y} ≃ S_{t_{1},t_{2}} for all y ∈ Γ(x) ∩ Q_{1}. A similar reasoning shows that this is also true for y ∈ Γ(x) ∩ Q_{2}. The result follows now from the previous theorem. □

Near polygons with four points on a line

8.1 The examples

In this chapter we classify, up to four open cases, all near hexagons with four points on a line (s = 3) and having the property that every two points at distance two have at least two common neighbours. The corresponding classification with s = 2 was done in [14]. By Yanushka’s Theorem, these near hexagons satisfy the property that every two points at distance two are contained in a (necessary unique) quad. We describe now all known examples and divide them into the following three classes.

(A) The regular classical near hexagons. The direct product of three lines of length 4 is a classical near hexagon with parameters s = 3, t_{2} = 1 and t = 2. The other examples are the dual polar spaces W^{D}(5, 3), Q^{D}(6, 3), H^{D}(5, 9). The parameters of these regular near hexagons are:

\begin{matrix}Type & s & t_{2} & t \\ W^{D}(5, 3) & 3 & 3 & 12 \\ Q^{D}(6, 3) & 3 & 3 & 12 \\ H^{D}(5, 9) & 3 & 9 & 90 \end{matrix}

(B) The irregular classical near hexagons. These near hexagons are the direct product of a line of length 4 with one of the following GQ’s:

(1) W (3), (2) Q(4, 3),

(3) T_{2}^{∗}(H), with H the unique (up to projective equivalence) hyperoval in PG(2, 4),

(4) Q(5, 3).

Note that the direct product of a line of length 4 with a (4 × 4)-grid is a regular near hexagon of Hamming type.

(C) The nonclassical near hexagons. There are two examples known and both are glued near hexagons.

(I) The following example has been defined in Chapter 6, page 108. Consider in a PG(4, 4) two planes α_{1} and α_{2} intersecting in a point p. Let O_{i}, i ∈ {1, 2}, be a hyperoval in α_{i} containing p. The linear representation T_{4}^{∗}(O_{1} ∪ O_{2}) is then a near hexagon. Since there is a unique hyperoval in PG(2, 4) (up to projective equivalence) and since the stabilizer of such a hyperoval in the group of all automorphisms of PG(2, 4) acts transitively on the set of its points, there is up to isomorphism only one such near hexagon.

(II) Put K = {x ∈ GF(9)|x^{4} = 1}. Consider in V (3, 9) a nonsingular Hermitian form (·, ·) and let U be the corresponding unital in PG(2, 9). Let α = ⟨ā⟩ be a fixed point of U. For two points β = ⟨b̄⟩ and γ = ⟨c̄⟩ of U, we define

Δ(β, γ) = −(ā, b̄)^{2}(b̄, c̄)^{2}(c̄, ā)^{2} if α ≠ β ≠ γ ≠ α,
= 1 otherwise.

This is a good definition. Indeed, if we replace b̄ by μb̄ and c̄ by λc̄ with μ, λ ∈ GF(9) \ {0}, then the above value for Δ(β, γ) is not changed. We define now a graph Γ with vertex set K × U × U. Two different vertices (k_{1}, α_{1}, β_{1}) and (k_{2}, α_{2}, β_{2}) are adjacent if and only if one of the following conditions is satisfied:

(a) α_{1} = α_{2} and β_{1} = β_{2},

(b) α_{1} = α_{2}, β_{1} ≠ β_{2} and k_{2} = k_{1}Δ(β_{1}, β_{2}),

(c) α_{1} ≠ α_{2}, β_{1} = β_{2} and k_{2} = k_{1}Δ(α_{1}, α_{2}).

Every two adjacent vertices of Γ are contained in a unique maximal clique. The geometry, with points the vertices of Γ, with lines the maximal cliques of Γ, and with natural incidence, is then a near hexagon.

8.2 The GQ’s with s = 3 and their ovoids

We recall that there are five GQ’s with s = 3, namely the following ones.

(1) The (4 × 4)-grid

(2) W (3)

(3) Q(4, 3)

(4) T_{2}^{∗}(H) with H the unique (up to projective equivalence) hyperoval in PG(2, 4)

(5) Q(5, 3)

The following lemma settles the existence of ovoids and of rosettes and fans of ovoids in these GQ’s.

{Lemma 8.2.1 \atop (a)} W (3) and Q(5, 3) have no ovoids.

(b) T_{2}^{∗}(H) contains ovoids but no rosettes of ovoids.

(c) Q(4, 3) contains ovoids but no fans of ovoids.

Proof.

(a) This is a special case of Theorem 2.6.1.

(b) Let H be a hyperoval in π which is embedded as hyperplane in PG(3, 4). Consider a plane α in PG(3, 4) intersecting π in a line exterior to H. It follows that the points of α not in π form an ovoid of T_{2}^{∗}(H). Conversely, every ovoid of T_{2}^{∗}(H) arises this way (see [68]). Hence, two different ovoids are disjoint or intersect in four points. As a consequence no rosette of ovoids occurs.

(c) Let Q be a nonsingular quadric in PG(4, 3). If π is a hyperplane of PG(4, 3) intersecting Q in an elliptic quadric, then π ∩ Q is an ovoid of Q(4, 3). Conversely, every ovoid is obtained this way (see e.g. [17]). Let O_{1} and O_{2} be two ovoids and let α_{1} and α_{2} be the hyperplanes such that O_{1} = α_{1} ∩ Q and O_{2} = α_{2} ∩ Q. Now, O_{1} ∩ O_{2} = (α_{1} ∩ α_{2}) ∩ Q ≠ ∅; hence, Q(4, 3) has no fan of ovoids.

□

In the sequel, we will always assume that S is a near hexagon satisfying the following properties:

(1) all lines of S have 4 points;

(2) every two points at distance two have at least two common neighbours.

Definition. A quad of S is called big if every point of S has distance at most 1 from Q.

Lemma 8.2.1 has now the following corollaries.

Corollary 8.2.2 If Q is a quad of S which is not big, then Q is isomorphic to the (4 × 4)-grid or to Q(4, 3).

Proof. Let x be a point at distance 2 from Q, then (x, Q) is ovoidal. Since W (3) and Q(5, 3) have no ovoids, Q is not isomorphic to one of these GQ’s. Now, let L be a line through x having a point at distance 1 from Q, then L determines a rosette of ovoids, see Theorem 3.7.1. Hence Q is not isomorphic to T_{2}^{∗}(H). □

Corollary 8.2.3 If a quad Q of S is isomorphic to Q(4, 3), then no line of S is contained in Γ_{2}(Q).

Proof. If the line L would be contained in Γ_{2}(Q), then this line determines a fan of ovoids, see Theorem 3.7.1; a contradiction. □

8.3 The classification for some special cases

In this section we give the classification for some special cases. These results will be used during the classification in the following section and it will turn out that all known examples belong to one of the three special cases considered here.

8.3.1 The case of classical near hexagons

Supppose S is a classical near hexagon. By the classification of classical near polygons, see Section 3.6, it follows that S is one of the classical near hexagons mentioned in Section 8.1.

8.3.2 The case of glued near hexagons

Suppose now that S is a glued near hexagon arising from the GQ’s Q_{1} and Q_{2}, the spreads S_{1} and S_{2}, the base lines L_{1}^{(1)} and L_{1}^{(2)}, and the map θ_{2} : L_{1}^{(1)} → L_{1}^{(2)} (as explained in Chapter 7, we may suppose that θ_{1} is the identical permutation of L_{1}^{(1)}). From the classification results obtained in Chapter 7 we know that S is one of the following examples.

(1) Q_{1} is a (4 × 4)-grid and Q_{2} is a GQ of order (3, t) not isomorphic to Q(4, 3) (this GQ has no spread). In this case S is the direct product of Q_{2} with a line of size 4.

(2) Q_{1} and Q_{2} are isomorphic to T_{2}^{∗}(H). Up to an automorphism of T_{2}^{∗}(H), there is a unique choice for S_{1} and S_{2}. From Corollary 7.5.4 it follows that there is a unique example (up to isomorphism). This example is the nonclassical near hexagon (I) given on page 155.

(3) Q_{1} and Q_{2} are isomorphic to Q(5, 3). Up to an automorphism of Q(5, 3) there is a unique choice for S_{1} and S_{2}. From Section 7.5, it follows that there is a unique example. With the same notations as in Section 8.1, we can define a graph Γ′ with vertex set K × U. Two different vertices (k_{1}, α_{1}) and (k_{2}, α_{2}) are adjacent if and only if the following conditions are satisfied:

(a) α_{1} = α_{2};

(b) α_{1} ≠ α_{2} and k_{2} = k_{1}Δ(α_{1}, α_{2}).

Every two adjacent vertices of Γ′ are contained in a unique maximal clique. The geometry, with points the vertices of Γ′, with lines the maximal cliques of Γ′, and with natural incidence, is then a generalized quadrangle Q isomorphic to Q(5, 3), see Section 2.5.4. If we put L_{α} = {(k, α)|k ∈ K} for every α ∈ U, then S = {L_{α}|α ∈ U} is a spread of Q. The unique glued near hexagon is then obtained by putting

• Q_{1} = Q_{2} = Q;

• S_{1} = S_{2} = S;

• L_{1}^{(1)} = L_{1}^{(2)} = L_{⟨ā⟩};

• θ_{2} the identical map of L_{1}^{(1)}.

This example is the nonclassical near hexagon (II) given on page 155.

8.3.3 The case of regular near hexagons

In Chapter 4, we already gave the following classification result.

Theorem 8.3.1 ([80]) If S is a regular near hexagon with parameters s = 3, t_{2} > 0 and t, then one of the following possibilities occur:

(1) t_{2} = 1 and t = 2, (2) t_{2} = 3 and t = 12,

(3) t_{2} = 9 and t = 90, (4) t_{2} = 1 and t = 9,
(5) t_{2} = 1 and t = 34, (6) t_{2} = 3 and t = 27,
(7) t_{2} = 3 and t = 48.

The parameters given in (1), (2) and (3) satisfy t = t_{2}^{2} + t_{2} and hence correspond to classical near hexagons (see Corollary 3.7.5). It is known that no near hexagon exists with parameters as in (4) (see [10]). It is also known that there are no near hexagons with parameters as in (6) or (7), but this will also follow from the treatment given here. Whether there exists a regular near hexagon with parameters (s, t, t_{2}) = (3, 34, 1) is still an open problem.

8.4 The classification

By Theorem 3.8.1 every point of S is incident with the same number of lines, say t + 1 lines. We distinguish between two cases.

8.4.1 Case I: all quads of type Q(4, 3) are big

In this section we suppose that all quads of type Q(4, 3) are big. We first prove the following lemma. We refer to the definition of the linear spaces S_{u,v} given in Section 7.6.1, page 149.

Lemma 8.4.1 If G is a linear space satisfying the following properties:

(a) G has at least two lines,

(b) every line has size 2, 4, 6 or 10,

(c) if two lines are disjoint, then they have both size 2,

then G is one of the following examples:

(1) K_{n} with n ≥ 3,

(2) a projective plane of order 3, 5 or 9, (3) S_{t_{1},t_{2}} with t_{1}, t_{2} ∈ {1, 3, 5, 9}.

Proof.

• If all lines are thin, then we have example (1). If all lines have the same size α + 1 ≠ 2, then (a), (b) and (c) imply that G is a projective plane of order 3, 5 or 9.

• Suppose that the line sizes α+1 and β +1 occur with 2 < α+1 < β +1. Let L_{1} and L_{2} be two lines of sizes α + 1 and β + 1 respectively. If a is a point not on L_{1} (respectively L_{2}), then there are exactly α + 1 (respectively β + 1) lines through a. Since α + 1 ≠ β + 1, all points are contained in L_{1} ∪ L_{2} and hence G ≃ S_{α,β}.

• Suppose now that 2 and α + 1 ∈ {4, 6, 10} are the only line sizes that occur. Consider a line L of size α +1 and let N be the set of points not on L. If |N| = 1, then G ≃ S_{1,α}. If |N| > 1, then |N| ≥ α. If |N| = α, then G ≃ S_{α,α}. Suppose therefore that |N| > α. Let L′ ≠ L be any line of size α + 1 and let p ∉ L ∪ L′. Through p there are α or α + 1 lines of size α + 1. This implies that |N| = α^{2} − α + 1 or |N| = α^{2}. If |N| = α^{2}, then every line of G has size α + 1, a contradiction. If |N| = α^{2} − α + 1, then G induces a linear space A on the set N which is an S(2, α, α^{2} − α + 1) and hence a projective plane of order α − 1. The linear space A contains α^{2} − α + 1 lines and every such line determines a point on L. If L_{1} and L_{2} are two different lines of A, then they determine different points on L, hence α^{2} − α + 1 < α + 1, a contradiction.

□

Lemma 8.4.2 Every local space of S satisfies (a), (b) and (c) from the previous lemma.

Proof. Clearly (a) and (b) are satisfied. Let Q_{1} and Q_{2} be two quads intersecting only in a point x. Every point of Q_{2} at distance 2 from x has distance 2 from Q_{1}. Our assumption and Corollary 8.2.2 imply then that Q_{1} is the (4 × 4)-grid. A similar reasoning proves that Q_{2} is the (4 × 4)-grid. □

Corollary 8.4.3 If S contains a local space not isomorphic to a K_{n} with n ≥ 3, then t + 1 ∈ {5, 7, 9, 11, 13, 15, 19, 31, 91}.

Lemma 8.4.4 All local spaces of S are isomorphic.

Proof. If a local space is isomorphic to S_{1,α} with α ∈ {1, 3, 5, 9}, then by Theorem 7.6.1 S is the direct product of a GQ with a line and the lemma is true in this case. We therefore suppose that no local space is isomorphic to a S_{1,α} for some α ∈ {1, 3, 5, 9}. We may also suppose that there exists a point x for which S_{x} is not isomorphic to K_{t+1}. These assumptions imply that t + 1 ≠ 5. If t + 1 is equal to 7, 9, 11, 15, 19, respectively 31, then S_{x} ≃ S_{3,3}, S_{3,5}, S_{5,5}, S_{5,9}, S_{9,9}, respectively PG(2, 5). If y is a point collinear with x, then S_{y} is not isomorphic to K_{t+1}. Hence S_{y} ≃ S_{x} and the result follows by connectedness of S. If t + 1 = 91, then a similar reasoning yields that all local spaces are projective planes of order 9 and S is then classical by Theorem 3.6.2; hence all local spaces are isomorphic to PG(2, 9) (see Section 8.3.1). If t+1 = 13, then S_{x} ≃ S_{3,9} or S_{x} ≃ PG(3, 2). If y is a point collinear with x, then S_{y} ≃ S_{x} and the result follows once again by connectedness of S. □

The following theorem completes the classification (for Case I), up to the open case appearing in the regular case.

Theorem 8.4.5 S is a regular, a classical or a glued near hexagon.

Proof. If all local spaces are isomorphic to K_{t+1}, then S is regular with t_{2} = 1. If all local spaces are (possible degenerate) projective planes, then S is a classical near hexagon by Theorem 3.6.2. If all local spaces are isomorphic to S_{u,v} with u, v ≥ 2, then S is glued by Theorem 7.6.2. □

8.4.2 Case II: there is a quad of type Q(4, 3) which is not big

We suppose that there is a quad of type Q(4, 3) which is not big. If v is the number of vertices of S, then v > 40 + 120(t − 3), hence every quad of type W (3) or Q(4, 3) is not big. Since W (3) has no ovoids, quads of type W (3) do not occur. Some of the following lemmas are just adapted versions of results mentioned in [14].

Lemma 8.4.6 Let x be a point at distance 2 from a quad Q of type Q(4, 3), then x is contained in α = \frac{39−t}{2} grids meeting Q and β = \frac{t−19}{2} quads of type Q(4, 3) meeting Q. Hence 19 ≤ t ≤ 39 and t is odd.

Proof. The point x determines an ovoid O of Q. Let o and o′ be two different points of O. Let Q_{1} be the quad through x and o. The point o′ is ovoidal with respect to Q_{1}. Hence Q_{1} is either a (4 × 4)-grid or a GQ isomorphic to Q(4, 3), see Corollary 8.2.2. As a consequence, every quad through x meeting Q determines 2 or 4 lines through x. It is impossible that a line through x is contained in at least two such quads. If a line through x is not contained in one such quad, then this line determines a fan of ovoids in Q, contradicting Corollary 8.2.3. The lemma follows now from the following equalities:

α + β = 10,
2α + 4β = t + 1.
□

Lemma 8.4.7 There are no quads of type W (3), T_{2}^{∗}(H) and Q(5, 3).

Proof. As we already mentioned, quads of type W (3) cannot occur. Suppose that a quad of type Q(5, 3) occurs. Consider a local space S_{x} with a line L of size 10 and let p be a point of S_{x} not on L. Every line through p meets L and contains an even number of points. Since there are 10 such lines the number t must be even, contradicting Lemma 8.4.6. A totally similar argument yields that there are no quads of type T_{2}^{∗}(H). □

Lemma 8.4.8 There are constants a and b such that each point is contained in a grids and b quads of type Q(4, 3). One has that

a = \frac{t(t + 1)}{2} − 6b

and

v = 18t^{2} − 6t + 4 − 108b.

Proof. Let x be a any point of S and suppose that x is contained in a grids and b quads of type Q(4, 3). There are 9a + 27b points at distance 2 from x and v − 1 − 3(t + 1) − 9a − 27b points at distance 3 from x. Counting triples (L_{1}, L_{2}, Q), where L_{1} and L_{2} are two different lines through x and where Q is a quad through L_{1} and L_{2}, yields t(t + 1) = 2a + 12b or a = \frac{t(t+1)}{2} − 6b. Counting pairs (y, z) with d(y, z) = 1, d(x, y) = 2 and d(x, z) = 3 yields

((v − 1) − 3(t + 1) − 9a − 27b) (t + 1) = 27a(t − 1) + 81b(t − 3).

From a = \frac{t(t+1)}{2} − 6b, it follows that v = 18t^{2} − 6t + 4 − 108b. □

Lemma 8.4.9 t ≠ 39 Proof. Suppose t = 39 and let x be any point of S. The local space S_{x} has lines of cardinality two or four. We will call them 2-lines, respectively 4-lines. By Lemma 8.4.6, a 2-line and a 4-line of S_{x} always intersect. Now, take a 4-line L and a point u not on L. Through u, there is a 4-line M disjoint with L. Let v be a point not on L ∪ M and let N be a 4-line through v not meeting L ∪ M. Any 2-line of S_{x} will meet L, M and N; hence there are only 4-lines. S is then a regular near hexagon with parameters s = 3, t_{2} = 3 and t = 39, contradicting Theorem 8.3.1. □

Lemma 8.4.10 If t = 19, then b ≤ 13.

Proof. Let x be any point of S. From Lemma 8.4.6, it follows that every two different 4-lines of the local space S_{x} always meet. Consider now two different 4-lines L_{1} and L_{2} and let p be their common point. If all 4-lines go through p, then b ≤ 6. If there exists a line L not through p, then there are at most four 4-lines through p (since every such line meets L). Since every 4-line of S_{x} meets L_{1} and L_{2}, we have that b ≤ 4 + 3 × 3 = 13. □

Let N_{2} be the number of points which are ovoidal with respect to a quad of type Q(4, 3); one easily checks that

N_{2} = 18(t^{2} − 7t + 18 − 6b).

Lemma 8.4.11 Let L be a line of size 4 in the local space S_{x}. If t ≠ 21, then there are \frac{1}{2160}(t − 19)N_{2} lines of size 4 and \frac{1}{720}(39 − t)N_{2} lines of size 2 disjoint with L; hence these numbers are integers.

Proof. (a) Let Q be the quad corresponding with L. For a point z of Q, let A_{z} be the set of quads of type Q(4, 3) intersecting Q only in the point z. Let y be a point of Q at distance 2 from x. Counting pairs (R, S) with R ∈ A_{x}, S ∈ A_{y} and |R ∩ S| = 1 yields |A_{x}|\frac{t−21}{2} = |A_{y}|\frac{t−21}{2}. Hence |A_{x}| = |A_{y}|. By connectedness of the noncollinearity graph of Q(4, 3) it follows that |A_{z}| = |A_{x}| for all points z of Q. Since there are \frac{1}{27}(\frac{t−19}{2})N_{2} quads of type Q(4, 3) intersecting Q in only one point, it follows that |A_{x}| = \frac{1}{2160}(t−19)N_{2}.

(b) The numbers |Γ_{2}(z)|, |Γ_{2}(z) ∩ Q| and |Γ_{2}(z) ∩ Γ(Q)| are independent of z ∈ Q. Hence also |Γ_{2}(z) ∩ Γ_{2}(Q)| is independent of z ∈ Q. Hence |Γ_{2}(x) ∩ Γ_{2}(Q)| = \frac{1}{4}N_{2}. Counting |Γ_{2}(x) ∩ Γ_{2}(Q)| in another way yields 27\frac{1}{2160}(t − 19)N_{2} + 9δ = \frac{1}{4}N_{2}, where δ is the number of 2-lines disjoint from L. Hence δ = \frac{1}{720}(39 − t)N_{2}. □

Corollary 8.4.12 t ≠ 23, t ≠ 29 and t ≠ 35 Proof. If t ≠ 21, then \frac{1}{2160}(t − 19)N_{2} = \frac{1}{60}(t − 19)[\frac{t(t−7)}{2} + 9 − 3b] ∈ ℕ; hence 3 | t(t − 19)(t − 7), from which the corollary follows. □

Lemma 8.4.13

\frac{20t(t + 1) − (39 − t)(t^{2} − 7t + 18)}{6(t + 1)} ≥ b ≥ \frac{(t − 19)(t^{2} − 7t + 18) + 120}{6(t + 1)}

Proof. Let Q be a quad of type Q(4, 3). There are \frac{1}{54}(t − 19)N_{2} quads of type Q(4, 3) which intersect Q in exactly one point; this number is at most 40(b − 1), from which the lower bound for b follows. There are \frac{1}{18}(39 − t)N_{2} grids which intersect Q in exactly one point; hence this number is at most 40a, from which the upper bound for b follows. □

We also have the following lower bound for b.

Lemma 8.4.14

b ≥ \frac{(t + 1)(t^{3} − 26t^{2} + 151t − 702)}{6(t^{2} + 2t − 319)}

Proof. Let x be any point of S. For every point p of the local space S_{x}, let α_{p} denote the number of 4-lines through the point p. Elementary counting yields

∑ 1 = t + 1,
∑ α_{p} = 4b,
∑ α_{p}(α_{p} − 1) = b[(b − 1) − \frac{1}{2160}(t − 19)N_{2}],

where the summation ranges over all points p of S_{x}. The inequality ∑(α_{p} − \frac{4b}{t+1})^{2} ≥ 0 yields the bound for b. □

Remark. If t ≠ 19, 21, then the lower bound of b given in Lemma 8.4.14 is stronger then the one given in Lemma 8.4.13. Collecting the results of the above lemmas, we find the following bounds for b:

• t = 19: 1 ≤ b ≤ 13;

• t = 21: 6 ≤ b ≤ 27;

• t = 25 : 30 ≤ b ≤ 41; • t = 27 : 42 ≤ b ≤ 50;

• t = 31 : 67 ≤ b ≤ 71;

• t = 33 : 81 ≤ b ≤ 84;

• t = 37 : b = 113. Note that in each case the number of grids is equal to \frac{va}{16} while the number of quads of type Q(4, 3) is equal to \frac{vb}{40}. Hence these numbers are integers.

Lemma 8.4.15 The following congruences hold.

• If t ≠ 19, 21, then b ≡ t^{2} − 2t + 3 (mod 5).

• If t = 19 or t = 21, then b ≡ 0 (mod 5) or b ≡ t^{2} − 2t + 3 (mod 5).

• If t ≡ 1 (mod 8), then b ≡ 0 (mod 4).

• If t ≡ 3 (mod 8), then b is odd.

• If t ≡ 5 (mod 8), then b ≡ 2 (mod 4). Proof. From \frac{vb}{40} ∈ ℕ, it follows that b(3t^{2} − t − 1 − 3b) ≡ 0 (mod 5). Hence b ≡ 0 (mod 5) or 6t^{2} − 2t − 2 − 6b ≡ t^{2} − 2t + 3 − b ≡ 0 (mod 5). If t ≠ 19, 21, then (t − 19)N_{2} ≡ 0 (mod 5) and, since t ≠ 29, 39, we even can say N_{2} ≡ 0 (mod 5), from which it follows that b ≡ t^{2} − 2t + 3 (mod 5). From \frac{va}{16} ∈ ℕ, it follows that (\frac{9t^{2}−3t}{2} +1−27b)(\frac{t(t+1)}{2} −6b) ≡ 0 (mod 4) from which the remaining congruences readily follow. □ Corollary. From the above lemmas there remain the following possibilities:

(1) (t, b) = (19, 1);

(2) (t, b) = (19, 5);

(3) (t, b) = (19, 11);

(4) (t, b) = (21, 10);

(5) (t, b) = (21, 22);

(6) (t, b) = (27, 43);

(7) (t, b) = (31, 67).

The quad-quad relation For a quad Q of S, let Γ_{≤1}(Q) := Q∪Γ_{1}(Q). Now, let Q and Q′ be two quads. We will summarize the possibilities of Γ_{≤1}(Q)∩Q′ regarded as substructure of Q′. If a line of Q′ meets two points of Γ_{≤1}(Q), then it is completely contained in Γ_{≤1}(Q), see Theorem 3.7.1.

(1) Suppose that Q and Q′ are quads of type Q(4, 3). Since Q(4, 3) has no fan of ovoids, every line of S meets Γ_{≤1}(Q). Hence the following possibilities may occur (see also Theorem 2.3.1 of [71]): (A) Γ_{≤1}(Q) ∩ Q′ is an ovoid of Q′; (B) Γ_{≤1}(Q) ∩ Q′ consists of the four lines through a fixed point of Q′; (C) Γ_{≤1}(Q) ∩ Q′ is a grid of Q′; (D) Γ_{≤1}(Q) ∩ Q′ = Q′.

(2) Suppose that Q is a quad of type Q(4, 3) and that Q′ is a grid. With a same reasoning as above, we have the following possibilities for Γ_{≤1}(Q) ∩ Q′; (A) Γ_{≤1}(Q) ∩ Q′ is an ovoid of Q′; (B) Γ_{≤1}(Q) ∩ Q′ consists of the two lines through a fixed point of Q′; (C) Γ_{≤1}(Q) ∩ Q′ = Q′.

(3) Suppose that Q and Q′ are two grids. We find the following possibilities for Γ_{≤1}(Q) ∩ Q′: (A) Γ_{≤1}(Q) ∩ Q′ is a set of i noncollinear points and i ∈ {0, 1, 2, 4}; (B) Γ_{≤1}(Q) ∩ Q′ is a line of Q′; (C) Γ_{≤1}(Q) ∩ Q′ consists of two intersecting lines; (D) Γ_{≤1}(Q) ∩ Q′ = Q′. The possibility that Γ_{≤1}(Q) ∩ Q′ is a set of three noncollinear points is ruled out be a reasoning which one can find in [10].

Theorem 8.4.16 The following cases cannot occur:

(A) (t, b) = (19, 1),

(B) (t, b) = (19, 11),

(C) (t, b) = (21, 22),

(D) (t, b) = (31, 67).

Proof.

(A) The case (t, b) = (19, 1)

Let Q be any quad of type Q(4, 3). If G is a grid of S, then |G ∩ Γ_{2}(Q)| ∈ {0, 9, 12}. Let M_{i}, i ∈ {0, 9, 12}, be the number of grids G for which |G ∩ Γ_{2}(Q)| = i. Through every point of Γ_{1}(Q) there is one line meeting Q, four lines contained in Γ_{1}(Q) and 15 other lines. Hence M_{12} ≤ \frac{1}{4}|Γ_{1}(Q)|\frac{15·14}{2} = 50400. There are 10a − \frac{(39−t)N_{2}}{72} = 640 grids intersecting Q in a line and |Q|\frac{(t−3)(t−4)}{2} = 4800 grids intersecting Q in only one point. Counting triples (G, L_{1}, L_{2}), where L_{1}, L_{2} ⊆ Γ_{1}(Q) are two intersecting lines of the grid G, yields 16(M_{0} − 640 − 1) + (M_{9} − 4800) ≤ |Γ_{1}(Q)|\frac{4·3}{2} = 11520. Hence M_{9} ≤ 16320 and M_{0} ≤ 1361. Hence there are at most (50400+16320+1361)=68081 grids, a contradiction, since there are exactly \frac{va}{16} = 72220 grids.

(B) The case (t, b) = (19, 11)

Let x be a point of S and consider the local space S_{x}. By Lemma 8.4.11 every two 4-lines meet. We will suppose now that there is no point which is incident with exactly three 4-lines and derive a contradiction. There are at most four 4-lines through every point p of S_{x}. Indeed, there exists a 4line not containing p and every 4-line through p meets this line. Now, let M = {m_{1}, m_{2}, m_{3}, m_{4}} be a fixed 4-line. Since there are exactly eleven 4lines, we may suppose that m_{i}, i ∈ {1, 2, 3}, is incident with four 4-lines and that m_{4} is incident with two 4-lines, say M and T. Let L be a 4-line through m_{1}, different from M and let r be a point of L not on M or T. Every 4-line through r meets M. The line through r and m_{i}, i ∈ {1, 2, 3}, is a 4-line and the line through r and m_{4} is a 2-line, a contradiction. Hence there exists a line K through x, which is contained in exactly three quads of type Q(4, 3), say Q_{1}, Q_{2} and Q_{3}. Let Q_{4} be another quad of type Q(4, 3) through x and let y be a point of K different from x. There are 10 lines through y contained in Γ_{1}(Q_{4}); one of these lines, say S, is not contained in Q_{1} ∪ Q_{2} ∪ Q_{3}. Every quad of type Q(4, 3) through y, but not through K, meets Q_{i}, i ∈ {1, 2, 3}, in a line. Hence at least three of its lines through y are contained in Γ_{1}(Q_{4}).

From the quad-quad relation, it follows then that also the fourth line through y is contained in Γ_{1}(Q_{4}) and must coincide with S. Hence, there hence are at most 3+3 quads of type Q(4, 3) through y, a contradiction.

(C) The case (t, b) = (21, 22)

For a quad Q of type Q(4, 3) and a point x ∈ Q, let N_{x,Q} denote the number of quads of type Q(4, 3) intersecting Q only in the point x. We have

∑_{Q} ∑_{x} N_{x,Q} = ∑_{Q} \frac{1}{54}(t − 19)N_{2} = \frac{vb(t − 19)N_{2}}{2160} = 3vb.  (8.1)

On the other hand ∑_{Q} ∑_{x} N_{x,Q} = ∑_{x} ∑_{Q} N_{x,Q}. Now, let x be fixed and consider the local space S_{x}. For a 4-line L = {l_{1}, l_{2}, l_{3}, l_{4}} of S_{x}, we define α_{L} = α_{l_{1}} + α_{l_{2}} + α_{l_{3}} + α_{l_{4}}, with α_{l_{i}} the number of 4-lines through l_{i} in S_{x}.

Suppose that there exists a 4-line L = {l_{1}, l_{2}, l_{3}, l_{4}} of S_{x} for which α_{L} ≥ 23. Let M ≠ L be one of the (α_{L} − 4) 4-lines meeting L, say in the point l_{1}. There are at least

(α_{l_{2}} − 4) + (α_{l_{3}} − 4) + (α_{l_{4}} − 4) = α_{L} − α_{l_{1}} − 12 ≥ 23 − 7 − 12 = 4

4-lines disjoint with M. Hence

∑_{Q} N_{x,Q} ≥ (α_{L} − 4)4 ≥ 76 > 3b.  (8.2)

Suppose that α_{L} ≤ 22 for all 4-lines L of S_{x}. For every line L of S_{x}, there are (25 − α_{L}) 4-lines disjoint with L. Hence

∑_{Q} N_{x,Q} = ∑_{L}(25 − α_{L}) ≥ 3b.  (8.3)

From equations (8.1), (8.2) and (8.3) it follows that α_{L} = 22 for all 4-lines L of S_{x}. Once again let L = {l_{1}, l_{2}, l_{3}, l_{4}} be a 4-line of S_{x} and let M ≠ L be one of the 4-lines meeting L, say in l_{1}. There are at least

(α_{l_{2}} − 4) + (α_{l_{3}} − 4) + (α_{l_{4}} − 4) = 10 − α_{l_{1}}

4-lines disjoint with M. Hence α_{l_{1}} = 7. For similar reasons, we may suppose that α_{l_{2}} = α_{l_{3}} = 7 and α_{l_{4}} = 1. Hence, every 4-line contains a unique point p for which α_{p} = 1. Since there are 22 4-lines and 22 points, all points p of S_{x} satisfy α_{p} = 1, a contradiction.

(D) The case (t, b) = (31, 67)

Consider a local space S_{x}. With the same notations as above we have

∑ 1 = 32,
∑ α_{p} = 268,
∑ α_{p}(α_{p} − 1) = 2010,

where the summation ranges over all points p of S_{x}. For every point p of S_{x}, we have α_{p} ≤ 10. Suppose α_{p} = 10 for some point p of S_{x}; then there exists a unique point q such that p and q are not contained in a 4-line. For every line {p, p_{1}, p_{2}, p_{3}} through p, α_{p_{1}} + α_{p_{2}} + α_{p_{3}} = 24. As a consequence α_{q} = 268 − 10 × 24 − 10 = 18, contradicting α_{q} ≤ 10. Hence α_{p} ≤ 9 for every point p of S_{x}. Since the average value of the α’s is \frac{67}{8}, there exists a point p such that α_{p} = 9. Let q_{i}, i ∈ {1, 2, 3, 4}, be those points of S_{x} such that p and q_{i} are not contained in a 4-line. As before, one calculates that α_{q_{1}} + α_{q_{2}} + α_{q_{3}} + α_{q_{4}} = 34; hence α_{q_{i}} ∈ {7, 8, 9} for all i ∈ {1, 2, 3, 4}. For a line {p, p_{1}, p_{2}, p_{3}} through p, α_{p_{1}} +α_{p_{2}} +α_{p_{3}} = 25; hence α_{p_{i}} ∈ {7, 8, 9} for all i ∈ {1, 2, 3}. With N_{i} being the number of points r of S_{x} for which α_{r} = i, i ∈ {7, 8, 9}, we get

N_{7} + N_{8} + N_{9} = 32,
7N_{7} + 8N_{8} + 9N_{9} = 268,
42N_{7} + 56N_{8} + 72N_{9} = 2010.

Hence, N_{7} = 13, N_{8} = −6 and N_{9} = 25, a contradiction. □

Remark on the case (t, b) = (19, 5)

We want to make some remarks about the local spaces for this case. If x is a point of S, then any two 4-lines of the local space S_{x} always meet. These 4-lines induce one of the following configurations.

[figure: (I); (II); (III); (IV); (V)]

Suppose that Q is a quad of type Q(4, 3). It is impossible that all local spaces S_{y}, y ∈ Q, are of type (I). Otherwise, the lines of Q which are contained in five quads of type Q(4, 3) determine a spread of Q, but Q(4, 3) has no spread. We will prove now that no local space is of type (III). Let L be a line of S which is contained in exactly three quads of type Q(4, 3), say Q_{1}, Q_{2} and Q_{3}. We prove that S_{x} is of type (IV) for every point x of L. Let y be a second point of L and let Q_{4} be a quad of type Q(4, 3) through y, different from Q_{1}, Q_{2} and Q_{3}. There are 10 lines through x contained in Γ_{1}(Q_{4}); one of these lines, say K, is not contained in Q_{1}, Q_{2} or Q_{3}. Let Q_{5} be one of the two quads through x different from Q_{1}, Q_{2} and Q_{3}. At least three of the lines of Q_{5} through x are contained in Γ_{1}(Q_{4}). From the quad-quad relation, it follows that all four lines through x are contained in Γ_{1}(Q_{4}). Hence the line K is contained in two quads of type Q(4, 3), which proves that S_{x} is of type (IV).

8.4.3 The main theorem

By collecting the previous results we can now state the main theorem for this chapter.

Theorem 8.4.17 Let S be a near hexagon satisfying the following properties:

(1) all lines of S have 4 points;

(2) every two points at distance 2 have at least two common neighbours.

We distinguish between two cases.

(A) If S is classical or glued, then it is isomorphic to one of the ten examples described in Section 8.1.

(B) If S is not classical and not glued, then only quads isomorphic to the (4 × 4)-grid or to Q(4, 3) occur. Moreover, there are numbers a and b such that every point of S is contained in a grids and b quads isomorphic to Q(4, 3). Every point is contained in the same number of lines, say t + 1 lines. We have then the following possibilities for t, a, b and v (= the number of points):

• v = 5848, t = 19, a = 160, b = 5;

• v = 6736, t = 21, a = 171, b = 10;

• v = 8320, t = 27, a = 120, b = 43;

• v = 20608, t = 34, a = 595, b = 0.

It is still an open problem whether there exist near hexagons with parameters as in (B) of the previous theorem.

Appendix A

The classification of ovoids in the near hexagon associated with S(5, 8, 24)

A.1 The problem

In Chapter 1, we defined the Steiner system S(5, 8, 24), which is unique up to isomorphism. It has the following properties.

(1) Any two different blocks intersect in 0, 2 or 4 points.

(2) If two blocks are disjoint, then the complement of their union is again a block.

(3) If two blocks intersect in four points, then their symmetric difference is again a block.

As every t-design is also a (t − 1)-design, the number of blocks λ_{i} through i points (1 ≤ i ≤ 5) can be calculated, and we get the following table.

\begin{matrix}i & λ_{i} \\ 1 & 253 \\ 2 & 77 \\ 3 & 21 \\ 4 & 5 \\ 5 & 1 \end{matrix}

In Chapter 3, we associated a regular near polygon S with this Steiner system. The points are the blocks of this Steiner system while the lines are the triples of disjoint blocks. There are v = 759 vertices. Two blocks have distance 1, 2, respectively 3, when they meet in 0, 4, respectively 2 points. An ovoid O of S is a set of points such that every line of S has exactly one point in common with O. S has natural ovoids, as the following theorem shows. In this theorem and further on, the set X will denote the set of 24 points of the Steiner system S(5, 8, 24). If i ∈ ℕ, then \binom{X}{i} denotes the set of all subsets of X which have order i.

Theorem A.1.1 If x ∈ X, then the set O_{x} of all blocks through x defines an ovoid of S.

Proof. Let L be any line of S. Exactly one of the three blocks of L contains x, which proves the theorem. □

We will prove in this appendix that these natural ovoids are the only ovoids that can occur.

Theorem A.1.2 If O is an ovoid of the near hexagon S, then O = O_{x} for some point x ∈ X.

This theorem has already been proved before in [16]. The proof there is short, but relies on results from [39]. W give here a totally different proof. Although the proof is longer, it is more self-contained and many steps are very similar.

A.2 The quads in S

As we already mentioned in Chapter 3, S is regular with parameters (s, t, t_{2}) = (2, 14, 2). The quads are GQ’s of order 2 and so isomorphic to W (2). How does these quads appear in the near hexagon? Take therefore two blocks α_{1} and α_{2} at mutual distance 2 and let Q be the quad through these points. The set α_{1} ∩ α_{2} (which has order 4) is contained in 5 blocks, say α_{1}, …, α_{5}. Let A = {α_{1} ∩ α_{2}, α_{1} \ α_{2}, α_{2} \ α_{1}, α_{3} \ α_{1}, α_{4} \ α_{1}, α_{5} \ α_{1}}. By the above mentioned property (3) of the Steiner system S(5, 8, 24), the union of two elements of A is a block of S(5, 8, 24). Let B be the set of 15 blocks arising this way. Clearly B is a subspace and any two different blocks of B have distance 1 or 2. What is the partial linear space induced by B? By Sylvesters model of W (2), we know that B induces W (2). Since B contains α_{1} and α_{2}, it immediately follows that the 15 blocks in B are exactly the 15 points of the quad Q.

A.3 The classification of the ovoids

Let O be any ovoid of S.

Lemma A.3.1 The ovoid O has 253 blocks. If α is a block of O, then n_{2} = |Γ_{2}(α)∩O| = 140 and n_{3} = |Γ_{3}(α) ∩ O| = 112.

Proof. The number of lines of S is equal to

\frac{v(t + 1)}{s + 1} = (t + 1)|O|.

Hence |O| = 253. Counting the number of connections between points of Γ_{1}(α) and Γ_{2}(α) ∩ O yields

|Γ_{1}(α)|t = (t_{2} + 1)|Γ_{2}(α) ∩ O|,

from which it follows that n_{2} = 140 and n_{3} = |O| − 1 − n_{2} = 112. □

Lemma A.3.2 Four different elements of X are contained in 1 or 5 elements of O.

Proof. The five blocks through four elements of X form an ovoid O′ in a quad Q. Let O′′ be the ovoid of Q, induced by O, then O′ and O′′ have 1 or 5 elements in common (see the classification of the ovoids of W (2) given in Section 2.6). □

Lemma A.3.3 Three different elements of X are contained in at least five elements and at most 21 elements of O.

Proof. Take three elements a, b, c of X. Let d_{1} be a fourth element of X and let α be a block of O through {a, b, c, d_{1}}. Let d_{2} be an element of X not in α and let β be a block of O through {a, b, c, d_{2}}. There are now five blocks of O through α ∩ β. The upper bound is immediate since λ_{3} = 21. □

Lemma A.3.4 Three different elements of X are contained in 5 or 21 elements of O.

Proof. Let α ∈ \binom{X}{3} be contained in t_{α} blocks of O. Double counting yields

∑_{α∈\binom{X}{3}} 1 = \binom{24}{3},

∑_{α∈\binom{X}{3}} t_{α} = |O|\binom{8}{3},

∑_{α∈\binom{X}{3}} t_{α}(t_{α} − 1) = |O|n_{2}\binom{4}{3}.

From the previous lemma and ∑_{α}(t_{α} −5)(21−t_{α}) = 0, it follows that t_{α} = 5 or t_{α} = 21. □

Remarks.

(1) Let T_{1}, respectively T_{2}, be the number of α ∈ \binom{X}{3} for which t_{α} = 5, respectively t_{α} = 21. From

T_{1} + T_{2} = \binom{24}{3},
5T_{1} + 21T_{2} = |O|\binom{8}{3},

it follows that T_{1} = \binom{23}{3} and T_{2} = \binom{23}{2}.

(2) Let R_{1}, respectively R_{2}, denote the number of α ∈ \binom{X}{4}, which are contained in exactly 1, respectively exactly 5, elements of O. From

R_{1} + R_{2} = \binom{24}{4},
R_{1} + 5R_{2} = |O|\binom{8}{4},

it follows that R_{1} = \binom{23}{4} and R_{2} = \binom{23}{3}.

Lemma A.3.5 For every A = {p_{1}, p_{2}, p_{3}, p_{4}} ⊆ X, there exists an i ∈ {1, 2, 3, 4} such that A \ {p_{i}} is contained in exactly 5 blocks of O.

Proof. Suppose the lemma is not true. Let α and β be two different elements of O through A. Take two different points x and y in α\β and let z be a point of α∩β. Let γ ≠ α be a block of O through {x, y, z}, then γ intersects α in a fourth point u. Through {x, y, z, u}, there are five blocks of O. If u ∈ α \ β, then α \ {x, y, z, u} is contained in only 1 block of O, a contradiction. As a consequence u ∈ α ∩ β. Take a point v ∈ β \ α. Let D_{1} be the block through [(α ∩ β) − {z}] ∪ {v, y} and D_{2} be the block through [(α ∩ β) − {u}] ∪ {v, x}. D_{1} and D_{2} are two blocks of the ovoid meeting in four points. The block D_{1}△D_{2} is not an element of O, a contradiction, since the set {x, y, z, u} is contained in it. □

Lemma A.3.6 If {p_{1}, p_{2}, p_{3}, p_{4}} ⊆ X is contained in exactly 5 elements of O, then there are exactly three subsets of order 3 which are contained in 21 elements of O.

Proof. Counting pairs (A, B) with (i) A ∈ \binom{X}{3}, (ii) B ∈ \binom{X}{4}, (iii) A ⊆ B and (iv) A is contained in 21 elements of O, yields

21 T_{2} = ∑ k_{B},

where the sum ranges over all B ∈ \binom{X}{4}, through which there are 5 blocks of O and k_{B} denotes the number of subsets of order 3, through which there are 21 blocks of O. From this it follows that k̄_{B} = 3, and the result follows from the previous lemma. □

Lemma A.3.7 Every two elements x and y of X are contained in at least 21 elements and in at most 77 elements of O.

Proof. Let z ∈ X be a third element of X. From the proof of Lemma A.3.3, it follows that there exists a fourth element u ∈ X such that {x, y, z, u} is contained in five elements of O. From the previous lemma, it follows that {x, y, z} or {x, y, u} is contained in 21 blocks of O. The upper bound is immediate since λ_{2} = 77. □

Lemma A.3.8 Two different elements of X are contained in 21 or 77 blocks of O.

Proof. Let α ∈ \binom{X}{2} be contained in s_{α} blocks of O. Double counting yields

∑_{α∈\binom{X}{2}} 1 = \binom{24}{2},

∑_{α∈\binom{X}{2}} s_{α} = |O|\binom{8}{2},

∑_{α∈\binom{X}{2}} s_{α}(s_{α} − 1) = |O|n_{2}\binom{4}{2} + |O|n_{3}.

From the previous lemma and ∑_{α}(s_{α} − 21)(77 − s_{α}) = 0, it follows that s_{α} = 21 or s_{α} = 77. □

Remark. Let S_{1}, respectively S_{2}, be the number of α ∈ \binom{X}{2} for which s_{α} = 21, respectively s_{α} = 77. From

S_{1} + S_{2} = \binom{24}{2},
21S_{1} + 77S_{2} = |O|\binom{8}{2},

it follows that S_{1} = \binom{23}{2} and S_{2} = \binom{23}{1}.

Lemma A.3.9 For every A = {p_{1}, p_{2}, p_{3}} ⊆ X, there exists an i ∈ {1, 2, 3} such that A\{p_{i}} is contained in exactly 21 blocks of O.

Proof. Suppose that all blocks through {p_{1}, p_{2}} or through {p_{2}, p_{3}} belong to O. Choose q_{1}, q_{2} ∉ A and q_{3} not in the unique block through {p_{1}, p_{2}, p_{3}, q_{1}, q_{2}}. Let B_{1}, respectively B_{2}, be the block through {p_{1}, p_{2}, q_{1}, q_{2}, q_{3}}, respectively {p_{2}, p_{3}, q_{1}, q_{2}, q_{3}}. Then B_{1}△B_{2} ∉ O, since B_{1}, B_{2} ∈ O. Since {p_{1}, p_{3}} ⊆ B_{1}△B_{2}, the lemma is proved. □

Lemma A.3.10 If {p_{1}, p_{2}, p_{3}} ⊆ X is contained in 21 blocks of O, then there are exactly two subsets of order 2 which are contained in 77 blocks of O.

Proof. Counting pairs (A, B) with (i) A ∈ \binom{X}{2}, (ii) B ∈ \binom{X}{3}, (iii) A ⊆ B and (iv) A is contained in 77 elements of O, yields

22 S_{2} = ∑ l_{B},

where the sum ranges over all B ∈ \binom{X}{3}, through which there are 21 blocks of O and l_{B} denotes the number of subsets of order 2, through which there are 77 blocks of O. From this it follows that l̄_{B} = 2, and the result follows from the previous lemma. □

Lemma A.3.11 There are at least 77 blocks of O through every point.

Proof. Let x denote the point. The element x is contained in a set P ⊆ X of order 4, through which there are five elements of O. Let P′ ⊆ P be a set of order 3, containing x and contained in 21 blocks of O. Let P′′ ⊆ P′ be a set of order 2, containing x and contained in 77 blocks of O. This proves the lemma. □

Lemma A.3.12 Every point of X is contained in 77 or 253 blocks of O.

Proof. Let α ∈ X be contained in μ_{α} blocks of O. Double counting yields

∑_{α∈X} 1 = \binom{24}{1},

∑_{α∈X} μ_{α} = |O|\binom{8}{1},

∑_{α∈X} μ_{α}(μ_{α} − 1) = |O|n_{2}\binom{4}{1} + |O|n_{3}\binom{2}{1}.

From the previous lemma and ∑_{α}(μ_{α} − 77)(253 − μ_{α}) = 0, it follows that μ_{α} = 77 or μ_{α} = 253. □

Theorem A.3.13 There exists an x ∈ X, such that O = O_{x}.

Proof. It suffices to prove that there is a point x such that μ_{x} = 253 (the same notation as in the previous lemma). Let U_{1}, respectively U_{2}, be the number of x ∈ X for which μ_{x} = 77, respectively μ_{x} = 253. From

U_{1} + U_{2} = 24,
5U_{1} + 21U_{2} = |O|8,

it follows that U_{1} = 23 and U_{2} = 1. This proves the theorem. □

Samenvatting

1. Elementaire begrippen en eigenschappen

Dit hoofdstuk vangt aan met het begrip incidentiestructuur ([35]). De incidentiestructuren die in deze thesis behandeld worden, zijn steeds eindig verondersteld. Een schier veelhoek S is een incidentiestructuur die aan de volgende eigenschappen voldoet.

(1) Elke twee punten zijn bevat in ten hoogste een rechte, of equivalent, S is een partiële lineaire ruimte.

(2) Voor elk punt p en elke rechte L, bestaat er een uniek punt q op L dat het dichtst bij p gelegen is (met betrekking tot de afstand in de puntgraaf van de meetkunde).

Als d de diameter van de puntgraaf is, dan wordt de schier veelhoek een schier 2d-hoek genoemd. Een schier 0-hoek bestaat dus enkel uit één punt en een schier 2-hoek bestaat uit één rechte met een aantal punten (≥ 2) erop. Er bestaat ook zoiets als een schier (2d + 1)-hoek, dit is een incidentiestructuur met diameter d die voldoet aan (1), (2’) en (3).

(2’) Voor elk punt p en elke rechte L waarvoor d(p, L) < d, bestaat er een uniek punt q op L dat het dichtst bij p gelegen is.

(3) Er bestaat een punt p en een rechte L waarvoor d(p, L) = d.

De schier veelhoeken werden ingevoerd in [82] omwille van het verband met bepaalde rechtensystemen in de Euclidische ruimte. Voor een overzicht van wat gekend is over deze meetkunden refereren we naar het derde hoofdstuk van deze thesis of naar de referentie [35] die eveneens een uitgebreide literatuurlijst bevat. In dit inleidende hoofdstuk worden ook enkele definities en eigenschappen gegeven van andere klassen incidentiestructuren zoals de veralgemeende veelhoeken [98] en de projectieve ruimten.

2. Veralgemeende vierhoeken

Schier veelhoeken wiens puntgraaf diameter 2 hebben zijn beter gekend onder de naam van veralgemeende vierhoeken. Op enkele triviale uitzonderingen na hebben alle veralgemeende vierhoeken een orde (s, t), dit betekent dat elke rechte incident is met s + 1 punten en dat elk punt incident is met t + 1 rechten. Omdat deze incidentiestructuren zo belangrijk zijn in de theorie van de schier veelhoeken wordt een geheel hoofdstuk hieraan gewijd. Toch wordt niet teveel in detail gegaan, voor een uitgebreid overzicht van deze incidentiestructuren verwijzen we naar [71] en [89].

We beschrijven in dit hoofdstuk een nieuwe constructie voor veralgemeende vierhoeken welke gebruik maakt van de zogenaamde toelaatbare drietallen. Een drietal (D, K, Δ) wordt toelaatbaar genoemd als aan de volgende drie voorwaarden is voldaan.

(1) D is een lineaire ruimte met orde (s, t − 1) (we noteren de puntenverzameling met P).

(2) K is een groep van orde s + 1 (multiplicatieve notatie).

(3) De afbeelding Δ : P × P → K voldoet aan de voorwaarden dat de punten x, y en z collineair zijn als en slechts als Δ(x, y) Δ(y, z) = Δ(x, z).

De volgende graaf is dan de puntgraaf van een veralgemeende vierhoek Q van orde (s, t). De toppenverzameling is K × P en twee toppen (k_{1}, x) en (k_{2}, y) zijn adjacent als en slechts dan als (i) x = y en k_{1} ≠ k_{2}, of (ii) x ≠ y en k_{2} = k_{1} Δ(x, y). Van volgende veralgemeende vierhoeken is geweten dat zij op dergelijke wijze geconstrueerd kunnen worden.

(1) Het (s+1)×(s+1)-rooster. Hiervoor is D een rechte met s+1 punten en K een willekeurige groep van orde s + 1. Stel Δ(x, y) gelijk aan het eenheidselement voor alle punten x en y van D.

(2) Het duale van het (t + 1) × (t + 1)-rooster. Hiervoor is D de complete graaf op t + 1 toppen en K de groep van orde 2. Stel Δ(x, y) gelijk aan het eenheidselement als en slechts als x = y.

(3) De veralgemeende vierhoek P (W (q), x) ([71]). Hiervoor is D = AG(2, q) en K de additieve groep van het veld GF(q). Voor twee punten r_{1} = (x_{1}, y_{1}) en r_{2} = (x_{2}, y_{2}) van D stellen we Δ(r_{1}, r_{2}) = x_{1}y_{2} − x_{2}y_{1}.

(4) De veralgemeende vierhoek T_{2}^{∗}(H) ([71]). Deze veralgemeende vierhoek is de lineaire representatie van een hyperovaal H in PG(2, 2^{h}) (zie later). We nemen D = AG(2, 2^{h}) en K de additieve groep van het veld GF(2^{h}). Weze A een verzameling van q+2 punten in D, die opgevat als puntenverzameling van PG(2, 2^{h}), een hyperovaal oplevert in PG(2, 2^{h}) die equivalent is met H. Noem V de verzameling van de richtingsvectoren bepaald door twee punten van A. Voor elke twee punten x en y van D bestaat er een unieke Δ(x, y) ∈ GF(2^{h}) zodat \overline{xy} = Δ(x, y) v̄ voor een zekere v̄ ∈ V. Dit definieert de afbeelding Δ.

(5) De veralgemeende vierhoek Q(5, q) ([71]). Beschouw een niet-singuliere Hermitische vorm (·, ·) in V (3, q^{2}) en weze U de corresponderende unitaal in PG(2, q^{2}). Met deze unitaal is als volgt een lineaire ruimte D geassocieerd. De punten van D zijn de punten van U en de rechten van D zijn de verzamelingen van orde q+1 die onstaan als doorsnijding van U met rechten van het projectieve vlak. Stel K = {x ∈ GF(q^{2})|x^{4} = 1}. Weze α = ⟨ā⟩ een vast punt van U. Voor elke twee punten β = ⟨b̄⟩ en γ = ⟨c̄⟩ van U, definiëren we

Δ(β, γ) = −(ā, b̄)^{2}(b̄, c̄)^{2}(c̄, ā)^{2} als α ≠ β ≠ γ ≠ α,
= 1 in het andere geval.

Dit is een goede definitie (vervangen we b̄ door μb̄ en c̄ door λc̄ met μ, λ ∈ GF(q^{2}) \ {0}, dan blijft bovenstaande waarde voor Δ(β, γ) onveranderd).

(6) De veralgemeende vierhoek (S_{xy}^{−})^{D} ([66]). Deze is geassocieerd met een hyperovaal H in PG(2, 2^{h}) die de punten x en y bevatten. Neem coördinaten zodanig dat de punten (1, 0, 0), (0, 1, 0) en (0, 0, 1) bevat zijn in H. Er bestaat dan een bijectie f : GF(2^{h}) → GF(2^{h}) zodanig dat H = {(1, 0, 0), (0, 1, 0)} ∪ {(f(λ), λ, 1)|λ ∈ GF(2^{h})}. Weze D = AG(2, 2^{h}) en K de additieve groep van het veld GF(2^{h}). Voor twee punten p_{1} = (α_{1}, β_{1}) en p_{2} = (α_{2}, β_{2}) van AG(2, 2^{h}) stellen we Δ(p_{1}, p_{2}) gelijk aan \frac{[f(α_{1})−f(α_{2})][β_{1}−β_{2}]}{α_{1}−α_{2}} als α_{1} ≠ α_{2} en gelijk aan 0 in het andere geval.

Er worden eigenschappen gegeven van veralgemeende vierhoeken die geconstrueerd kunnen worden met behulp van toelaatbare drietallen. In het bijzonder wordt het verband uitgelegd met spreads van veralgemeende vierhoeken, dit zijn partities van de puntenverzameling in rechten. Beschikken we over een toelaatbaar drietal (D, K, Δ), dan is de verzameling S = {L_{x}|x ∈ P}, waarbij L_{x} = {(k, x)|k ∈ K}, een spread van Q, die de geassocieerde spread van het toelaatbaar drietal genoemd wordt. Een spread S van een veralgemeende vierhoek van orde (s, t) wordt normaal genoemd indien elke twee rechten van S bevat zijn in een veralgemeende vierhoek van orde (s, 1); S wordt een symmetriespread genoemd indien de automorfismegroep van de veralgemeende vierhoek minstens s + 1 elementen bevat die elke rechte van S fixeren. De spread S geassocieerd aan een toelaatbaar drietal is zowel een normale spread als een symmetriespread. De volgende karakteriseringen kunnen dan bekomen worden.

Stelling 1 Weze Q een veralgemeende vierhoek met orde (s, t).

(1) Q heeft een symmetriespread S als en slechts dan als Q afgeleid kan worden van een toelaatbaar drietal waarvan S de geassocieerde spread is.

(2) Q heeft een symmetriespread S als en slechts dan als de projectiviteitengroep van een rechte L ∈ S met betrekking tot S orde ten hoogste s + 1 heeft.

Van alle bovenvermelde veralgemeende vierhoeken (rooster, duaal rooster, P (W (q), x), T_{2}^{∗}(H), Q(5, q) en (S_{xy}^{−})^{D}) worden alle normale en alle symmetriespreads bepaald. Van bepaalde veralgemeende vierhoeken die afgeleid kunnen worden met behulp van de zogenaamde q-clans (bijvoorbeeld de niet-klassieke Kantor-Knuth veralgemeende vierhoeken) wordt bewezen dat ze geen symmetriespread hebben.

Er wordt een methode aangegeven hoe automorfismen van veralgemeende vierhoeken die een bepaalde spread fixeren kunnen bestudeerd worden. Dit wordt toegepast op sommige van de boven vermelde veralgemeende vierhoeken.

Alle veralgemeende vierhoeken van orde s, die een regulier punt hebben dat incident is met een reguliere rechte, kunnen geconstrueerd worden door amalgamatie van twee projectieve vlakken ([61]). Deze veralgemeende vierhoeken worden dan ook geamalgameerde veralgemeende vierhoeken genoemd. Is Q een veralgemeende vierhoek van orde s die een regulier punt x heeft, dan kan een nieuwe veralgemeende vierhoek P (Q, x) gevonden worden die orde (s − 1, s + 1) heeft ([60]). Deze constructie kan steeds toegepast worden op elke geamalgameerde veralgemeende vierhoek, daar deze steeds een regulier punt heeft. De volgende stelling wordt bewezen.

Stelling 2 Weze Q een veralgemeende vierhoek van orde (s − 1, s + 1) met de eigenschap dat elke twee snijdende rechten bevat zijn in een unieke veralgemeende vierhoek van orde (s, 1), dan bestaat er een geamalgameerde veralgemeende vierhoek Q′ en een regulier punt x van Q′, zodanig dat Q isomorf is met P (Q′, x).

3. Basistheorie van de schier veelhoeken

In dit hoofdstuk wordt de basistheorie van de schier veelhoeken besproken. Volgende schier veelhoeken komen veelvuldig in deze thesis voor.

(1) Klassieke schier veelhoeken of duale polaire ruimten Deze zijn geassocieerd met een polaire ruimte Γ. Weze d de maximale dimensie van een deelruimte van Γ. De punten, respectievelijk rechten, van de schier veelhoek S zijn de d-dimensionale, respectievelijk (d−1)-dimensionale deelruimten van Γ (met de natuurlijke incidentie). Voorbeelden van dergelijke klassieke schier veelhoeken worden bekomen door het direct product te nemen van een rechte van lengte s + 1 met een veralgemeende vierhoek Q. Hiertoe neemt men s + 1 copieën van Q. Men construeert echter additionele rechten die de corresponderende punten in de s + 1 copieën verbinden. De juist vermelde constructie werkt eveneens indien Q een willekeurige partiële lineaire ruimte is. Op inductieve wijze kan men dan het direct product definiëren van k ≥ 1 rechten. Zulke directe producten zijn eveneens klassieke schier veelhoeken, die Hamming schier veelhoeken genoemd worden.

(2) Schier zeshoek geassocieerd met S(5, 8, 24)

Neem het unieke Steiner systeem S(5, 8, 24). De incidentiestructuur S, met als punten de blokken van S(5, 8, 24) en als rechten de drietallen van onderling disjuncte blokken, is dan een schier zeshoek.

(3) Beschouw de Coxeter-cap K in PG(5, 3) ([25],[73]). De volgende incidentiestructuur T_{5}^{∗}(K) is dan een schier zeshoek. We bedden PG(5, 3) in als een hypervlak Π_{∞} in de ruimte Π = PG(6, 3). De punten van T_{5}^{∗}(K) zijn de punten van Π \ Π_{∞}. De rechten van T_{5}^{∗}(K) zijn al de rechten van Π, niet gelegen in Π_{∞}, die met K één punt gemeen hebben. Incidentie is de natuurlijke incidentie.

Een schier veelhoek wordt regulier genoemd indien zijn puntgraaf afstandsregulier ([15]) is. Een relatie ”parallellisme” kan gedefinieerd worden op de rechtenverzameling van elke schier veelhoek. Twee rechten L en M worden parallel genoemd als de afstand van een punt van L tot de rechte M onafhankelijk is van dat punt. We noteren dit ook nog als L∥M. De volgende karakterisering kan dan bewezen worden.

Stelling 3 Als S een schier veelhoek is, dan zijn volgende beweringen equivalent.

(1) S is een Hamming schier veelhoek.

(2) Parallellisme is een equivalentierelatie en elke twee punten op afstand twee hebben tenminste twee gemeenschappelijke buren.

(3) Voor elk punt x en elke rechte L bestaat er een unieke rechte door x parallel met L.

Een schier veelhoek wordt ontaard genoemd indien deze een superdun punt heeft, dit is een punt dat incident is met juist één rechte. Uit een ontaarde schier veelhoek S kunnen nieuwe schier veelhoeken geconstrueerd worden. Door opeenvolgende toepassing van deze constructies kan tenslotte een nietontaarde schier veelhoek bekomen worden, die uniek bepaald is, en die de niet-ontaarde ondersteuning van S genoemd wordt.

Eén van de belangrijkste begrippen in de theorie van de schier veelhoeken is het begrip quad. Een quad Q van een schier veelhoek S is een niet-ontaarde veralgemeende vierhoek die deelmeetkunde is van S en die bovendien voldoet aan de volgende eigenschap: elk punt van S dat collineair is met twee punten van Q is een punt van Q. De stelling van Yanushka geeft voldoende voorwaarden opdat twee punten bevat zouden zijn in zulk een quad. Deze quads kunnen dan aangewend worden om voor elk punt x van S een bepaalde partiële lineaire ruimte S_{x} te definiëren, de zogenaamde locale ruimte in het punt x.

In dit hoofdstuk worden er eveneens karakteriseringen gegeven van reguliere en klassieke schier zeshoeken in termen van het aantal quads door een punt en het aantal punten op afstand 2 van een vast punt.

4. Associatieschemas geassocieerd met schier veelhoeken

Is S = (P, L, I) een schier zeshoek, dan is L × L = R_{0} ∪ … ∪ R_{4}, waarbij

R_{0} = {(L, L)|L ∈ L},

R_{1} = {(L, M)|L ≠ M en d(L, M) = 0},
R_{2} = {(L, M)|L∥M en d(L, M) = 1},
R_{3} = {(L, M)|L ∦M en d(L, M) = 1},
R_{4} = {(L, M)|L∥M en d(L, M) = 2}.

Stel R := {R_{0}, R_{1}, R_{2}, R_{3}, R_{4}}. Voor elk koppel (L, M) ∈ R_{4} definiëren we p_{22}^{4}(L, M) als het aantal rechten K waarvoor (K, L), (K, M) ∈ R_{2}. De volgende stelling geldt dan.

Stelling 4 Veronderstel dat S een schier zeshoek is die aan de volgende eigenschappen voldoet.

(1) S is regulier met parameters (s, t_{2}, t), d.w.z. S heeft orde (s, t) en elke twee punten op afstand 2 hebben juist t_{2} + 1 gemeenschappelijke buren.

(2) Elke twee punten op afstand 2 zijn bevat in een unieke quad.

Dan bepaalt R een associatieschema ([3]) als en dan slechts als p_{22}^{4}(L, M) onafhankelijk is van (L, M) ∈ R_{4}.

Voorbeelden van dergelijke associatieschemas worden gegeven, evenals de volgende karakterisering.

Stelling 5 Weze S een reguliere schier zeshoek met parameters (s, t_{2}, t) zodanig dat

(i) s ≥ t_{2},

(ii) p_{22}^{4}(L, M) onafhankelijk is van (L, M) ∈ R_{4},

dan is S één van de volgende schier zeshoeken:

• een veralgemeende zeshoek,

• een reguliere klassieke schier zeshoek,

• de schier zeshoek geassocieerd met S(5, 8, 24),

• de schier zeshoek wiens puntgraaf de incidentiegraaf is van het unieke bivlak van orde 2 ([6]).

Tenslotte wordt ook bewezen dat er een (niet noodzakelijk symmetrisch) associatieschema verbonden is aan elke veralgemeende vierhoek van orde (s, s^{2}) die een symmetriespread heeft.

5. De i-buurmeetkunde van een punt

Veronderstel dat x een punt is van een schier veelhoek S = (P, L, I) en stel i ∈ ℕ. De i-buurmeetkunde van x is de deelmeetkunde S_{x}^{i} van S verkregen door alle punten te nemen met afstand ten hoogste i tot x, evenals alle rechten met afstand ten hoogste i−1 tot x. Het is duidelijk dat S_{x}^{i} een schier veelhoek is indien i ∈ {0, 1}. We hebben eveneens de volgende eigenschappen bewezen.

Stelling 6 Weze x een punt van een schier veelhoek S.

(1) S_{x}^{2} is een schier veelhoek.

(2) Is S een dunne schier veelhoek, dan is S_{x}^{i} eveneens een schier veelhoek voor elke i ∈ ℕ.

(3) Is S een veralgemeende veelhoek, dan is S_{x}^{i} eveneens een schier veelhoek voor elke i ∈ ℕ.

(4) Is S een klassieke schier veelhoek, dan is S_{x}^{i} eveneens een schier veelhoek voor elke i ∈ ℕ.

Een voorbeeld toont echter aan dat niet elke i-buurmeetkunde een schier veelhoek is. In dit hoofdstuk wordt eveneens de volgende stelling bewezen.

Stelling 7 Weze G een willekeurige partiële lineaire ruimte, dan bestaat er een schier veelhoek S en een punt (∞) van S zodanig dat de locale ruimte S_{(∞)} isomorf is met G.

Elke lineaire ruimte treedt dus op als locale ruimte. Om voorgaande stelling te bewijzen wordt een oneindige klasse Ω van meetkunden geconstrueerd. Elke meetkunde in Ω heeft de eigenschap dat er een punt bestaat dat afstand ten hoogste 2 heeft tot elk ander punt. De meetkunde S uit voorgaande stelling kan dan steeds in Ω gekozen worden. Ook de volgende stelling wordt in dit hoofdstuk bewezen.

Stelling 8 Weze (∞) een punt van een schier veelhoek S welke afstand ten hoogste 2 heeft tot elk ander punt. Is de niet-ontaarde ondersteuning van S verschillend van een punt, dan is deze bevat in Ω.

6. Lineaire representaties van schier veelhoeken

Veronderstel dat PG(n, q), n ≥ 0, ingebed is als een hypervlak Π_{∞} in Π = PG(n + 1, q) en weze K een niet ledige verzameling van punten in Π_{∞}. De lineaire representatie T_{n}^{∗}(K) is dan de volgende partiële lineaire ruimte. De punten van T_{n}^{∗}(K) zijn de punten van Π \ Π_{∞}; de rechten van T_{n}^{∗}(K) zijn deze rechten van Π, die niet bevat zijn in Π_{∞}, en die Π_{∞} snijden in een punt van K; incidentie is deze van Π. Dit hoofdstuk bestudeert het geval wanneer T_{n}^{∗}(K) een schier veelhoek is, in het bijzonder wordt ingegaan op het speciale geval van de schier zeshoeken. Een voorbeeld van een dergelijke schier zeshoek werd reeds gegeven op pagina 182; hier was K de Coxeter-cap in PG(5, 3).

Volgende stelling levert een belangrijke klasse van schier veelhoeken op.

Stelling 9 Weze n = 2k, k ≥ 1, en q = 2^{h} een even priemmacht. Beschouw door een vast punt p van Π_{∞} k vlakken Π_{1}, …, Π_{k} die Π_{∞} genereren. Weze H_{i}, i ∈ {1, …, k}, een hyperovaal van Π_{i} die p bevat. Als K = H_{1} ∪ … ∪ H_{k}, dan is T_{2k}^{∗}(K) een schier (2k + 2)-hoek die de eigenschap heeft dat elke twee punten op afstand 2 bevat zijn in een unieke quad.

Dit hoofdstuk heeft onder andere als doel alle schier zeshoeken te classificeren die een lineaire representatie hebben en dit in het geval q ≥ 3. Zulke schier zeshoeken hebben steeds quads door elke twee punten op afstand 2. De classificatie is partieel. Ze is wel compleet voor het geval n ≤ 6.

Stelling 10 Weze K een verzameling punten in PG(n, q), n ≤ 6 en q ≥ 3, dan is T_{n}^{∗}(K) een schier zeshoek als en slechts dan als één van de volgende gevallen zich voordoet.

(1) K is een verzameling van drie niet collineaire punten in een vlak (n = 2).

(2) K bestaat uit een hyperovaal plus een punt niet in het vlak van de hyperovaal (n = 3).

(3) K is de unie van twee hyperovalen wiens draagvlakken elkaar ontmoeten in een punt dat behoort tot beide hyperovalen (n = 4).

(4) K is de Coxeter-cap in PG(5, 3).

Geen voorbeelden zijn gekend voor n ≥ 7 en q ≥ 3. In dit geval zijn er nochtans wel enkele conclusies te trekken.

Stelling 11 Weze T_{n}^{∗}(K) een schier zeshoek en veronderstel dat n ≥ 7 en q ≥ 3, dan gelden volgende beweringen.

(1) q ≥ 2^{h} met h ≥ 4

(2) Elke quad is ofwel een rooster ofwel een lineaire representie van een hyperovaal. Quads van het laatste type treden zeker op, evenals een rosette van ovoiden in elk zulke quad. Als er een quad bestaat van type T_{2}^{∗}(H) met H een reguliere hyperovaal, dan geldt q ≥ 64 en een tot heden ongekende ovoide in PG(3, q) moet bestaan.

Stel f_{m}(x, y) = ∑_{j=1}^{m} \binom{x−1}{j}(y − 1)^{j−1} voor elke m ∈ ℕ en g_{m}(x, y) = \frac{1}{y} \binom{x−1}{m}(y − 1)^{m−1} + f_{m−1}(x, y) voor elke m ∈ ℕ \ \{0\}.

Stelling 12 Veronderstel dat K een verzameling is van k > 0 punten in PG(n, q) die aan de eigenschap voldoen dat geen l ervan bevat zijn in een (l − 2)-dimensionale ruimte (n ≥ l − 2 ≥ 0).

(1) Is l = 2m + 1, dan geldt f_{m}(k, q) ≤ \frac{q^{n}−1}{q−1} en de gelijkheid treedt op als en slechts dan als T_{n}^{∗}(K) een schier 2(m + 1)-hoek is.

(2) Is l = 2m en (n, m) ≠ (0, 1), dan geldt g_{m}(k, q) ≤ \frac{q^{n}−1}{q−1} en de gelijkheid treedt slechts op als en slechts dan als T_{n}^{∗}(K) een schier (2m + 1)-hoek is.

De gelijkheid in voorgaande stelling kan slechts optreden in de volgende gevallen.

l = 2: In dit geval bestaat K uit alle punten van de projectieve ruimte.

l = 3: Er zijn drie mogelijkheden.

(1) K is een verzameling van twee punten in PG(1, q).

(2) K is een hyperovaal in PG(2, q), q even.

(3) K is het complement van een hypervlak in PG(n, 2), n ≥ 3.

l = 4: Er zijn twee mogelijkheden.

(1) K is een verzameling van 11 punten in PG(4, 3) die als volgt bekomen wordt. Beschouw een punt x van de Coxeter-cap in PG(5, 3) en een hypervlak α in PG(5, 3) die x niet bevat. De projectie van de Coxeter-cap vanuit het punt x op α levert K op.

(2) K is een verzameling van 5 punten in PG(3, 2) waarvan geen 4 bevat zijn in een hypervlak (uniek bepaald op projectieve equivalentie na).

l = 5: Er zijn twee mogelijkheden.

(1) K is de Coxeter-cap in PG(5, 3).

(2) K is een verzameling van 6 punten in PG(4, 2) waarvan geen 5 bevat zijn in een hypervlak (uniek bepaald op projectieve equivalentie na).

Tenslotte wordt aan de hand van de lineaire representatie een classificatie bekomen van speciale puntenverzamelingen (ovoiden en Cameron gesloten verzamelingen) van de schier zeshoek die geassocieerd is met de Coxeter-cap.

7. Gelijmde schier veelhoeken

Weze k een positief natuurlijk getal. Voor elke i ∈ {1, …, k} beschouwen we de volgende objecten:

• een veralgemeende vierhoek Q_{i} = (P_{i}, L_{i}, I_{i}) van orde (s, t_{i}),

• een spread S_{i} = {L_{1}^{(i)}, …, L_{1+st_{i}}^{(i)}} van Q_{i},

• een bijectie θ_{i} : L_{1}^{(1)} → L_{1}^{(i)}.

In elke spread S_{i}, i ∈ {1, …, k}, wordt dus een speciale rechte L_{1}^{(i)} genomen, die de basisrechte van S_{i} genoemd wordt. De projectie op de rechte L_{j}^{(i)} in de veralgemeende vierhoek Q_{i} wordt genoteerd als p_{j}^{(i)}. De projectiviteitengroep van L_{1}^{(i)} met betrekking tot S_{i} wordt genoteerd als G_{i} en de samenstelling f ∘ g van twee functies f en g wordt genoteerd als fg.

Definieer nu de volgende graaf Γ met toppenverzameling L_{1}^{(1)} × S_{1} × … × S_{k}. Twee verschillende toppen (x, L_{i_{1}}^{(1)}, …, L_{i_{k}}^{(k)}) en (y, L_{j_{1}}^{(1)}, …, L_{j_{k}}^{(k)}) zijn adjacent als en slechts dan als aan de volgende twee voorwaarden voldaan is.

(i) Er bestaat een l ∈ {1, …, k} zodanig dat i_{m} = j_{m} voor elke m ∈ {1, …, k} \ {l}.

(ii) Voor elke l die aan (i) voldoet zijn p_{i_{l}}^{(l)}θ_{l}(x) en p_{j_{l}}^{(l)}θ_{l}(y) collineaire punten in de veralgemeende vierhoek Q_{l}. (Als i_{m} = j_{m} voor elke m ∈ {1, …, k}, dan kan l elke waarde van 1 tot en met k aannemen, maar aan voorwaarde (ii) is dan altijd voldaan.)

De volgende stelling kan dan bewezen worden. Hierin wordt de triviale groep met 0 genoteerd en [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] stelt de groep voor die gegenereerd wordt door alle commutatoren [θ_{i}^{−1}g_{i}θ_{i}, θ_{j}^{−1}g_{j}θ_{j}] waarbij g_{i} ∈ G_{i} en g_{j} ∈ G_{j}.

Stelling 13 Γ is de puntgraaf van een schier (2k + 2)-hoek S als en slechts dan als [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0 voor alle i, j ∈ {1, …, k} met i ≠ j.

Is S een schier veelhoek, dan zijn de rechten van S noodzakelijk de maximale cliques van Γ en S wordt een gelijmde schier veelhoek genoemd. Intuïtief kan men zeggen dat deze schier veelhoek bekomen kan worden door een heleboel veralgemeende vierhoeken aan elkaar te lijmen. Deze veralgemeende vierhoeken verschijnen dan als quads in de schier veelhoek. Is S een schier veelhoek, dan kan S bekomen worden

• voor elke keuze van de ordening van de veralgemeende vierhoeken,

• voor elke keuze van de basisrechten in de spreads,

• voor elke keuze van θ_{1}.

De schier veelhoeken T_{2k}^{∗}(H_{1} ∪ … ∪ H_{k}) die gedefinieerd werden in Stelling 9 zijn voorbeelden van gelijmde schier veelhoeken.

Er wordt ingegaan op de classificatie van gelijmde schier veelhoeken. De volgende stelling wordt bewezen.

Stelling 14 Veronderstel dat aan de volgende voorwaarden voldaan zijn:

(A) S is een schier veelhoek;

(B) k ≥ 2;

(C) de veralgemeende vierhoeken Q_{1}, …, Q_{k} zijn geen roosters.

Dan gelden de volgende beweringen:

(1) G_{i} ≃ G_{j} voor alle i, j ∈ {1, …, k},

(2) S_{i} is een symmetriespread van Q_{i}, (3) als k ≥ 3 dan is G_{i} commutatief.

Laten we nu de veralgemeende vierhoeken, de spreads en de basisrechten fixeren. Zoals eerder opgemerkt mogen we ook θ_{1} fixeren. In de thesis wordt aangetoond hoe de vergelijkingen [θ_{i}^{−1}G_{i}θ_{i}, θ_{j}^{−1}G_{j}θ_{j}] = 0, i, j ∈ {1, …, k} met i ≠ j, op te lossen naar (θ_{2}, …, θ_{k}). Het kan nu gebeuren dat twee verschillende oplossingen voor (θ_{2}, …, θ_{k}) toch nog isomorfe schier veelhoeken opleveren. In de thesis worden technieken besproken die het mogelijk maken om het aantal niet isomorfe gelijmde schier zeshoeken te bepalen of af te schatten. Dit wordt toegepast op enkele voorbeelden.

Tenslotte worden de volgende twee karakteriseringen bewezen.

Stelling 15 Veronderstel dat S een schier zeshoek is die aan de volgende voorwaarden voldoet.

(1) Elke twee punten op afstand 2 zijn bevat in een quad.

(2) Als alle rechten van S dun zijn, dan hebben alle quads een orde.

(3) Er bestaat een punt x zodanig dat alle locale ruimten S_{y}, y collineair met x, een dun punt bevatten.

Dan is S het direct product van een rechte met een niet-ontaarde veralgemeende vierhoek of S is een gelijmde schier zeshoek.

Stelling 16 Veronderstel dat S een schier zeshoek is die aan de volgende voorwaarden voldoet.

(1) Elke twee punten op afstand 2 zijn bevat in een quad.

(2) Als alle rechten van S dun zijn, dan hebben alle quads een orde.

(3) Er bestaat een punt x zodanig dat S_{x} een dun punt bevat en zodanig dat alle locale ruimten S_{y}, y collineair met x, eenzelfde aantal rechten bevatten.

Dan is S het direct product van een rechte met een niet-ontaarde veralgemeende vierhoek of S is een gelijmde schier zeshoek.

8. Schier zeshoeken met vier punten op een rechte

In [14] werden alle schier zeshoeken bepaald die aan de volgende voorwaarden voldoen:

(1) alle rechten hebben drie punten;

(2) elke twee punten op afstand 2 zijn bevat in een quad.

Het doel van dit hoofdstuk is de classificatie van alle schier zeshoeken die aan de volgende voorwaarden voldoen:

(1) alle rechten hebben vier punten;

(2) elke twee punten op afstand 2 zijn bevat in een quad.

De classificatie is echter niet compleet. Alle klassieke en gelijmde schier zeshoeken die aan bovenstaande voorwaarden voldoen zijn gekend. De klassieke voorbeelden zijn ofwel een direct product van een rechte met een veralgemeende vierhoek van orde (3, t) (vijf voorbeelden), ofwel zijn het de duale polaire ruimten geassocieerd met de volgende polaire ruimten:

(1) W (5, 3): de polaire ruimte geassocieerd met een symplectische polariteit van PG(5, 3);

(2) Q(6, 3): de polaire ruimte geassocieerd met een niet-singuliere kwadriek in PG(6, 3);

(3) H(5, 9): de polaire ruimte geassocieerd met een niet-singuliere Hermitische varieteit in PG(5, 9).

We beschrijven nu de gelijmde schier zeshoeken die niet klassiek zijn.

• De volgende gelijmde schier zeshoek is het unieke voorbeeld waarbij beide veralgemeende vierhoeken isomorf zijn met T_{2}^{∗}(H), waarbij H de unieke hyperovaal is in PG(2, 4). Beschouw twee vlakken α_{1} en α_{2} in PG(4, 4) die elkaar snijden in een punt p. Weze H_{i}, i ∈ {1, 2}, een hyperovaal in α_{i} die p bevat. De lineaire representatie T_{4}^{∗}(H_{1} ∪ H_{2}) is dan een schier zeshoek.

• De volgende gelijmde schier zeshoek is het unieke voorbeeld waarbij beide veralgemeende vierhoeken isomorf zijn met Q(5, 3). Stel K = {x ∈ GF(9)|x^{4} = 1}. Beschouw een niet singuliere Hermitische vorm (·, ·) in V (3, 9) en weze U de corresponderende unitaal in PG(2, 9).

Weze α = ⟨ā⟩ een vast punt van U. Voor elke twee punten β = ⟨b̄⟩ en γ = ⟨c̄⟩ van U, definiëren we

Δ(β, γ) = −(ā, b̄)^{2}(b̄, c̄)^{2}(c̄, ā)^{2} als α ≠ β ≠ γ ≠ α,
= 1 in het andere geval.

We definiëren nu een graaf Γ met toppenverzameling K × U × U. Twee verschillende toppen (k_{1}, α_{1}, β_{1}) en (k_{2}, α_{2}, β_{2}) zijn adjacent als en slechts dan als aan één van de volgende voorwaarden voldaan is:

(1) α_{1} = α_{2} en β_{1} = β_{2},

(2) α_{1} = α_{2}, β_{1} ≠ β_{2} en k_{2} = k_{1}Δ(β_{1}, β_{2}),

(3) α_{1} ≠ α_{2}, β_{1} = β_{2} en k_{2} = k_{1}Δ(α_{1}, α_{2}).

Elke twee adjacente toppen van Γ zijn bevat in een unieke maximale clique. De meetkunde waarvan de punten, respectievelijk de rechten, de toppen en de maximale cliques van Γ zijn, is dan een schier zeshoek.

Het is niet geweten of er voorbeelden zijn indien de schier zeshoek niet klassiek en niet gelijmd is. Toch zegt volgende stelling iets over hun structuur.

Stelling 17 Veronderstel dat S een schier zeshoek is die aan de volgende eigenschappen voldoet:

(1) alle rechten van S hebben vier punten,

(2) elke twee punten op onderlinge afstand 2 zijn bevat in een quad,

(3) S is niet klassiek en niet gelijmd,

dan zijn alle quads ofwel roosters ofwel isomorf met Q(4, 3). Er zijn constanten a en b zodanig dat elk punt van S bevat is in a roosters en b quads die isomorf zijn met Q(4, 3). Elk punt is incident met een constant (= t+1) aantal rechten. We hebben dan de volgende mogelijkheden voor t, a, b en v (= het aantal punten):

• v = 5848, t = 19, a = 160, b = 5;

• v = 6736, t = 21, a = 171, b = 10;

• v = 8320, t = 27, a = 120, b = 43;

• v = 20608, t = 34, a = 595, b = 0.

Appendix: De classificatie van de ovoiden in de schier zeshoek geassocieerd met S(5, 8, 24)

Een ovoide van een schier veelhoek is een verzameling punten met de eigenschap dat elke rechte een uniek punt gemeen heeft met deze verzameling. Weze S nu de schier zeshoek geassocieerd met het Steiner systeem S(5, 8, 24). De volgende classificatiestelling werd bewezen in [16].

Stelling 18 De verzameling van de blokken door een vast punt van S(5, 8, 24) is een ovoide van S. Omgekeerd wordt elke ovoide op deze wijze verkregen.

In de appendix van deze thesis wordt een bewijs hiervan geleverd dat verschillend is van het bewijs in [16].

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Index

admissible triple, 23 affine embedding, 14 amalgamation, 54 ambient space, 14 anisotropic matrix, 47 associated spread, 31 association scheme, 80 AT, 23

base line, 42, 132 block, 2

Cameron closed, 69 cap, 8 center of symmetry, 33 classical, 70 classical generalized quadrangle, 18 classical near polygon, 71 collinear, 1 collinearity graph, 1 connected, 1 Coxeter-cap, 10

degenerate, 67 degenerate generalized quadrangle,

16 design, 2 direct product, 61 direct sum of polar spaces, 72 direct sum of projective spaces, 63 distance regular graph, 80 duad, 21 dual grid, 16 dual polar space, 63 dualized projective space, 72

extended ternary Golay code, 11

fan of ovoids, 29

generalized polygon, 5 generalized quadrangle, 15 generating index, 104 geodetically closed, 69 glued near polygon, 133 GQ, 15 grid, 16 group coset geometry, 47 group of projectivities, 7

hyperbolic line, 17

hyperoval, 8

incidence graph, 2 incidence relation, 1 incidence structure, 1 irreducible polar space, 72 irreducible projective space, 63 irregular hyperoval, 8 isomorphism, 2

Kantor-Knuth quadrangles, 49

line system, 3

linear representation, 14

linear space, 2 local space, 71

Mathon bound, 82 multiple lines, 2 natural ovoid, 172 near polygon, 4, 60 near polygon of Hamming type, 62 near polygon property, 106 neighbourhood, 96 net, 49 nondegenerate support, 68 nontrivial generalized quadrangle,

15 normal ovoid, 31 normal spread, 31 nucleus, 8

opposite, 6 order, 1 oval, 8 ovoid, 9, 29, 172 ovoid of symmetry, 32 ovoidal, 70

parallel lines, 61 parameters of a near polygon, 61 parameters of an association scheme,

80 partial linear space, 2 partial quadrangle, 4 partial spread, 31 perspectivity, 6 pivotal, 59 point graph, 1 polar space, 63 projection, 6, 60 projectivity, 6

q-clan, 47 quad, 70

big, 157 good, 74 of type I, 143 of type II, 143 of type III, 143 reducible projective space, 63 Ree-Tits octagons, 7 regular hyperoval, 8 regular near polygon, 61 regular pair of points, 17 regular point, 18 regular spread, 11, 50 regulus, 11 rosette of ovoids, 29

span, 17 spread, 11, 29 spread of symmetry, 31 Steiner system, 2 strongly regular graph, 81 subgeometry, 2 subspace, 63, 69 superthin point, 67 symmetry, 33 syntheme, 21 system of vectors, 3

tetrahedrally closed, 3 thick incidence structure, 1 thick plane, 112 thin near polygon, 61 thin ovoidal, 71 thin point, 150 Tits ovoid, 9 translation hyperoval, 9 triad, 18 trivial generalized quadrangle, 15

valency, 80, 81