A thin near hexagon with 50 points

Rieuwert Blok, Bart De Bruyn, Ulrich Meierfankenfeld

Abstract

We show the existence and uniqueness of a thin near hexagon which has 50 points and an affine plane of order 3 as a local space.

1 Definitions and Motiviation

In this paper we solve the open problem mentioned at the end of [BDB]:

Does there exist a near hexagon which satisfies the conditions (C1), (C2) and (C3) below?

(C1) Every line is incident with exactly two points.

(C2) Every two points at distance 2 are contained in a good quad.

(C3) At least one local space is an affine plane of order 3.

We will prove that such a near hexagon exists and that it is unique. This completes the classification of all near hexagons with good quads and with at least one local space isomorphic to an affine plane, see [BDB]. Since the paper [BDB] is published in this issue of JCT-A, we will use it as a reference for some basic notions like ”near hexagon”, ”quad”, and so on. A quad Q is called good if every point of Q is incident with the same number of lines of Q; this number then is denoted by t_{Q} +1. From now on we assume that S is a near hexagon which satisfies (C1), (C2) and (C3). Since S has only lines of size 2, it can be regarded as a bipartite graph of diameter 3 (see Lemma 1 of [BDB]). The edges then play the role of the lines. For a set A of vertices, let A^{⊥} denote the set of all vertices adjacent to every element of A. If x and y are two points at distance 2 then Q(x, y) := {x, y}^{⊥} ∪({x, y}^{⊥})^{⊥} is the unique quad through x and y. The fact that this quad is good means that |{x, y}^{⊥}| = |{x, y}^{⊥⊥}|. For every point x, we can define the following linear space L(x): the points of L(x) are the elements of x^{⊥} and the lines are all the sets {a, b}^{⊥⊥} where a and b are two different vertices of x^{⊥}. The linear space L(x) is called the local space at x. In [BDB] another definition was given, but in this special case both definitions are easily seen to be equivalent. By Theorem 5 of [BDB], we know that S has precisely v = 50 points, that every point is incident with exactly t+1 = 9 lines and that every quad Q has order (1, t_{Q}) with t_{Q} ∈ {1, 2, 3}. Moreover, there exists a quad Q with t_{Q} = 3. The point set P of S can be partitioned into two subsets P_{+} and P_{−}, each of size 25, such that two points of the same subset are never collinear.

2 THE STRUCTURE OF S 2

We sometimes call a point of P_{ϵ} a point of type ϵ. The set P_{ϵ}, ϵ ∈ {+, −}, can be given the structure of a linear space S_{ϵ} by taking all sets {x, y}^{⊥⊥}, x, y ∈ P_{ϵ} and x ≠ y, as lines. Clearly L(a) is a subgeometry of S_{ϵ} for every a ∈ P_{−ϵ}.

2 The structure of S

Let n_{+} denote a point whose local space is an affine plane of order 3. We may suppose that n_{+} ∈ P_{+}. For every point x and every i ∈ {1, 2, 3}, let N_{i}(x) or shortly N_{i} denote the number of quads of order (1, i) through x, or equivalently, the number of lines of size i+1 in L(x). Since t + 1 = 9, N_{3} ∈ {0, 1, 2, 3}. Every two lines through x are contained in a unique quad; hence 72 = (t + 1)t = 12N_{3} + 6N_{2} + 2N_{1}. Since there are |P_{+}| − 1 = |P_{−}| − 1 = 24 points at distance 2 from x, 24 = 3N_{3} + 2N_{2} + N_{1}. These equations give

Lemma 2.1 For any point x one of the following holds:

(n) N_{3} = 0, N_{2} = 12, N_{1} = 0: x is called a nice point;

(e) N_{3} = 1, N_{2} = 9, N_{1} = 3; x is called an elusive point;

(g) N_{3} = 2, N_{2} = 6, N_{1} = 6: x is called a good point;

(b) N_{3} = 3, N_{2} = 3, N_{1} = 9: x is called a bad point. □

The local space of a nice point is a Steiner system S(2, 3, 9) and hence an affine plane of order 3.

Lemma 2.2 If n is a nice point, then d(n, R) ≤ 1 for every quad R of order (1, 2) or (1, 3).

Proof: Suppose that d(n, R) = 2. If R has order (1, 3), then the four quads through n meeting R contain 5 points of Γ_{2}(R), contradicting |Γ_{2}(R)| = |P| − |R| − |Γ_{1}(R)| = 50 − 8 − 40 = 2. Suppose therefore that R has order (1, 2). Let S_{i}, i ∈ {1, 2, 3}, denote the three quads through n meeting R, put {v_{i}} := S_{i} ∩ R, and let w_{i} denote the unique point of S_{i} at distance 2 from n and v_{i}. Let K_{1}, K_{2} and K_{3} denote the three lines through v_{1} not contained in S_{1} ∪ R. Let T_{i}, i ∈ {2, 3}, denote the unique quad through v_{1} and w_{i}. All the lines of T_{i} through v_{1} are contained in {K_{1}, K_{2}, K_{3}}. Hence, we may suppose that T_{i}, i ∈ {2, 3}, contains K_{1} and K_{i}. Every quad of order (1, 3) through v_{1} necessarily contains the lines K_{2} and K_{3}. Hence N_{3}(v_{1}) ≤ 1 and v_{1} is either a nice or an elusive point. Suppose that v_{1} is elusive and let T denote the unique quad of order (1, 3) through v_{1}. The quads T_{2} and T_{3} then have order (1, 1). Since each of the lines T ∩ S_{1} and T ∩ R is contained in a quad of order (1, 1), N_{1}(v_{1}) ≥ 4. This however contradicts our assumption that v_{1} is elusive. Hence v_{1} is nice and T_{2} = T_{3} has order (1, 2). The points v_{1}, w_{2} and w_{3} are all contained in the quad T_{2}. Similarly, v_{2} and v_{3} are nice and each of the sets {v_{2}, w_{1}, w_{3}} and {v_{3}, w_{1}, w_{2}} is contained in a quad of order (1, 2). The quad R′ through {v_{2}, w_{1}, w_{3}} has order (1, 2) and satisfies d(n, R′) = 2. By the same argument as above, w_{1} and w_{3} are nice. Clearly also w_{2} is nice. Notice that

A := {n, v_{1}, v_{2}, v_{3}, w_{1}, w_{2}, w_{3}}

together with the collection of quads of order (1, 2) intersecting A in at least two points has the structure of a Fano plane. Recall that there exists a quad Q of order (1, 3). Let z ∈ A. By the first paragraph of the proof, d(z, Q) = 1. Since Q is a quad, z is collinear with a unique point z̃ of Q. If z̃_{1} = z̃_{2} for two points z_{1}, z_{2} ∈ A, then also the third point of A in the quad through z_{1} and z_{2} is collinear with z̃_{1}. Let Ã = {z̃ | z ∈ A}. Since t_{Q} = 3, |Ã| ≤ 4 and hence every point of A is collinear with the same point q of Q. Since q is collinear with v_{1}, v_{2} and v_{3}, q belongs to R. Now d(n, q) = 1 contradicts d(n, R) = 2. □

Lemma 2.3 There are no elusive points. Exactly one point n_{−} in Γ_{1}(n_{+}) is nice while the other 8 are good. The 8 points of Γ_{2}(n_{+}) collinear with n_{−} are good, while the other 16 points are bad. Finally, all 16 points in Γ_{3}(n_{+}) are bad.

Proof: We first show

(I) Every point z ∈ Γ_{1}(n_{+}) is either nice or good.

The line n_{+}z is contained in 4 quads of order (1, 2), proving that z is not bad. Suppose z is elusive and let Q denote the unique quad of order (1, 3) through z. Since N_{1} = 3, there exists a line K of Q through z not contained in a quad of order (1, 1). The line K then is contained in 1 quad of order (1, 3) and \frac{5}{2} quads of order (1, 2), a contradiction.

(II) All 16 points in Γ_{3}(n_{+}) are bad, Γ_{1}(n_{+}) has one nice and eight good points.

Let z ∈ Γ_{3}(n_{+}). By Lemma 2.2 every quad of order (1, 2) or (1, 3) through z has a unique point in common with Γ_{1}(n_{+}). Thus N_{2} + N_{3} ≤ 9 and z is either of good or bad. Moreover, there are exactly (9 − N_{2} − N_{3}) quads of order (1, 1) through z which meet Γ_{1}(n_{+}). Let α_{i} and β_{i}, respectively, denote the number of points of type (i), i ∈ {n, e, g, b}, in Γ_{1}(n_{+}) and Γ_{3}(n_{+}). Counting the number of quads of order (1, 1) connecting a point of Γ_{1}(n_{+}) with a point of Γ_{3}(n_{+}), we find 6α_{g} = β_{g} + 3β_{b}. Counting the number of quads of order (1, 3) connecting a point of Γ_{1}(n_{+}) with a point of Γ_{3}(n_{+}), we find 2α_{g} = \frac{1}{3}(2β_{g} + 3β_{b}). Hence β_{g} = 0, β_{b} = 16, α_{g} = 8 and α_{n}=1.

Applying (I) and (II) to the unique nice point n_{−} in Γ_{1}(n_{+}), we find that Γ_{2}(n_{+}) contains 8 good points and 16 bad points. □

Lemma 2.4 The lines of size at least three in the local space of a good point determine a configuration as in (A) or (B). The lines of size at least three in the local space of a bad point determine a configuration as in (C). (In these pictures nice and good points are indicated with stars and squares respectively.)

[figure: ■; ⋆; (A); ■; ⋆; (B)]

[figure: ■; ■; ■; (C)]

Proof: Any two collinear good points are contained in a unique quad of order (1, 2) containing both nice points. Thus the local space of a good point g has one nice point n, two good points, and six bad points, and the nice and good points form a line.

Since quads are good, the four lines of length 3 on g in L(n) correspond to four lines of length 3 on n in L(g). Ignoring lines of size 2 the local space of g is a partial linear space having two lines of size 4 and six lines of size 3. This yields the configurations (A) and (B).

Let b ∈ P_{ϵ} be a bad point. The unique quad on b and n_{ϵ} has order (1, 2) and contains three good points. Conversely, if g, h are any two good points collinear to b, then the unique quad on g and h contains b and n_{ϵ}. Thus the local space of a bad point b contains three good points, all on one line, and six bad points.

Ignoring the lines of size 2 the local space of the bad point b is a partial linear space having three lines of size 4 and three lines of size 3. This yields configuration (C). Notice that the three good points are precisely the common neighbors of b and n_{ϵ}.

It remains to show that the nice and good points are as indicated by the stars and squares. For (C) this is obvious, as there are no nice points in L(b) and the good points form line of size three. Since all lines through the nice point in L(g) have size three, the nice point in (A) and (B) is uniquely determined. Let x be the unique point in L(g) such that {x, g} lies in two quads of order (1, 3). Then in L(x) there exist two lines of size four meeting in the good point g. So by (C), x cannot be bad. Thus x is good. □

We say that a line in some local space has type L_{ngg} if it consists of one nice point and two good points. Similarly we define L_{ggg}, L_{nbb}, L_{gbb}, and L_{gbbb}. By L_{ggg}^{⊥} = L_{nbb} we mean that if a line l has type L_{ggg}, then l^{⊥} has type L_{nbb}.

Lemma 2.5 (a) L_{ngg}^{⊥} = L_{ngg},

(b) L_{ggg}^{⊥} = L_{nbb}, L_{nbb}^{⊥} = L_{ggg},

(c) L_{gbb}^{⊥} = L_{gbb}, and

(d) L_{gbbb}^{⊥} = L_{gbbb}.

Proof: First note that if l is a line in some local space, then |l| = |l^{⊥}| since the quad l ∪l^{⊥} is good.

(a) Suppose that n_{ϵ} is the nice point on the line l of type L_{ngg}. Then n_{−ϵ} is a point in l^{⊥} and by looking at the affine plane L(n_{ϵ}) we find that l^{⊥} has type L_{ngg}.

(b) If the line l of type L_{ggg} belongs to P_{ϵ} then clearly n_{−ϵ} ∈ l^{⊥}. By (a) l^{⊥} cannot contain any good points. Thus L_{ggg}^{⊥} = L_{nbb}. Conversely, all common neighbours of a nice and a bad point are good. Hence L_{nbb}^{⊥} = L_{ggg}.

(c) Let l be a line of type L_{gbb}. Then l^{⊥} is a line of the local space of both the good and the bad points of l. By Lemma 2.4 l^{⊥} must have type L_{gbb}.

(d) By Lemma 2.4 all lines of length 4 are of type L_{gbbb}. □

A line of S is said to have valency 3^{k}4^{l} if it is contained in exactly k quads of order (1, 2) and l quads of order (1, 3). A point b of the local space L(a) is said to have valency v_{a}(b) = 3^{k}4^{l} if it is contained in exactly k lines of size 3 and l lines of size 4. Clearly v_{a}(b) is equal to the valency of the line ab. Hence v_{a}(b) = v_{b}(a). For every good point a of S, let \overline{a} denote the unique good point of valency 3^{1}4^{2} in L(a). This point exists; see (A) and (B) of Lemma 2.4. Similarly, for a bad point x, let \overline{x} denote the unique bad point of valency 3^{2}4^{1} in L(x). Also define \overline{n_{+}} = n_{−} and \overline{n_{−}} = n_{+}. Clearly \overline{\overline{a}} = a for every point a of S. For every good or bad point a of S there exists a unique nice point n at distance 2 from a. The unique point of {n, a}^{⊥⊥} different from n and a will be denoted by −a. Note that n, a, and −a all have the same type. Also define −n_{+} = n_{+} and −n_{−} = n_{−}. Clearly −(−a) = a for every point a of S. If a and b are distinct points in the affine plane L(n_{ϵ}), ϵ ∈ {+, −}, then the unique third point on the line ab will be denoted by a ∗ b. Also define a ∗ a = a. Clearly −a = n_{−ϵ} ∗ a for every point a of L(n_{ϵ}).

We record:

Lemma 2.6 Let a, b, x ∈ P_{ϵ} with a, b good and x bad. Then

(a) \overline{a} is the unique good point of valency 3^{1}4^{2} in L(a).

(b) a = \overline{\overline{a}}, that is a is the unique point of valency 3^{1}4^{2} in L(\overline{a}).

(c) {a, b}^{⊥⊥} = {a, b, a ∗ b} is the affine line in L(n_{−ϵ}) determined by a and b.

(d) {n_{ϵ}, a}^{⊥⊥} = {n_{ϵ}, a, −a} is the affine line in L(n_{−ϵ}) determined by n_{ϵ} and a.

(e) −(−a) = a and Q(a, n_{ϵ}) = \{n_{ϵ}, a, −a; n_{−ϵ}, \overline{a}, −\overline{a}\}.

(f) \overline{x} is the unique bad point of valency 3^{2}4^{1} in L(x). □

Since L(n_{ϵ}) is an affine plane with a distinguished point n_{−ϵ}, it carries the structure of a 2-dimensional vector space V_{−ϵ} over 𝔽_{3}. The vector addition and scalar multiplication are as follows.

(SM) 0a := n_{−ϵ}, 1a := a and (−1)a := −a for every point a of L(n_{ϵ}).

(VA) a + b := −(a ∗ b) for all points a and b of L(n_{ϵ}).

Lemma 2.7 We have \overline{−a} = −\overline{a} for every good point a of S.

Proof: Suppose that a ∈ P_{ϵ} with ϵ ∈ \{+, −\} and let Q = Q(a, n_{ϵ}). The point \overline{a} is a common neighbor of a and n_{ϵ} and hence belongs to Q. Since −a ∈ Q, also \overline{−a} ∈ Q and hence \overline{−a} ∈ \{\overline{a}, −\overline{a}\}. If \overline{−a} = \overline{a}, then a and −a are two points of valency 3^{1}4^{2} in L(\overline{a}), a contradiction to Lemma 2.4. Hence \overline{−a} = −\overline{a}. □

Lemma 2.8 Let x be a bad point and let a and c be the two good points of valency 3^{2}4^{1} in L(x); see (C) of Lemma 2.4. Then \overline{a ∗ c} = \overline{a} + \overline{c} = − \overline{a + c}.

Proof: Suppose that x ∈ P_{ϵ}. Then Q(a, \overline{x}) and Q(c, \overline{x}) are quads of order (1, 2) and Q(a ∗ c, \overline{x}) has order (1, 3). By Lemma 2.2 each of the quads Q(a, \overline{x}), Q(c, \overline{x}) and Q(a ∗ c, \overline{x}) contains a unique point collinear with n_{−ϵ}. The three good points so-obtained are precisely the three common neighbors of n_{−ϵ} and \overline{x}. The point Q(a, \overline{x}) ∩ n_{−ϵ}^{⊥} is a common neighbor of a, \overline{x}, and n_{−ϵ} and must therefore be equal to either \overline{a} or −\overline{a}. If \overline{a} ∈ Q(a, \overline{x}), then Lemma 2.4 applied to L(\overline{a}) implies that Q(a, \overline{x}) has order (1, 3), a contradiction. Hence Q(a, \overline{x}) ∩ n_{−ϵ}^{⊥} = \{−\overline{a}\}. Similarly Q(c, \overline{x}) ∩ n_{−ϵ}^{⊥} = \{−\overline{c}\}. Now, Q(a ∗ c, \overline{x}) is a quad of order (1, 3) through a ∗ c and so by definition of \overline{·} contains \overline{a ∗ c}. Since −\overline{a}, −\overline{c} and \overline{a ∗ c} are the three common neighbors of \overline{x} and n_{−ϵ}, we have \overline{a ∗ c} = (−\overline{a}) ∗ (−\overline{c}) = \overline{a} + \overline{c}. □

For a bad point x let a, c be as in the previous lemma. Define

F(x) = \{\overline{a}, \overline{c}\} and x̂ = \overline{a} + \overline{c}.

For a set of points A, let \overline{A} = \{ā | a ∈ A\} and −A = \{−a | a ∈ A\}.

¿From the proof of Lemma 2.8 we obtain:

Corollary 2.9 Let x be a bad point and F(x) = {a, c}. Then

(a) \overline{a} and \overline{c} are the two good points of valency 3^{2}4^{1} in L(x).

(b) \overline{x̂} is the unique good point of valency 3^{1}4^{1} in L(x).

(c) x̂ = a + c = \overline{\overline{a} ∗ \overline{c}}.

(d) ¯ x̂ = \overline{a} ∗ \overline{c} = −\overline{a} − \overline{c} = − \overline{a ∗ c}.

(e) \{n_{ϵ}, x, −x\}^{⊥} = \{\overline{a}, \overline{c}, \overline{a} ∗ \overline{c}\} = \overline{F(x) ∪ \{x̂\}}.

(f) \{n_{−ϵ}, \overline{x}, −\overline{x}\}^{⊥} = \{−a, −c, a + c\} = −F(x) ∪ \{x̂\}. □

Lemma 2.10 (a) If x is a bad point, then F(x) = F(−x) and x̂ = ˆ−x.

(b) Let x and y be bad points. Then x̂ = ŷ if and only if x = ±y.

(c) Every local space of a good point is isomorphic to the linear space (A) of Lemma 2.4.

Proof: Let x be a bad point in P_{ϵ}. By Corollary 2.9(e) applied to x and −x, F(x)∪{x̂} = F(−x)∪{ ˆ−x}. With F(x) = {u, v} and F(−x) = {u′, v′}, we have x̂ = −2 x̂ = −u−v−x̂ = −u′ − v′ − ( ˆ−x) = ˆ−x and hence also F(x) = F(−x). Now fix a good point a. Then by Corollary 2.9(b) a = x̂ if and only if \overline{a} is the unique good point of valency v_{x}(\overline{a}) = 3^{1}4^{1} and this happens if and only if v\overline{a}(x) = 3^{1}4^{1}. For each of the two possibilities for L(\overline{a}), see Lemma 2.4, there are exactly two solutions for x. This combined with (a) proves (b). Since by (b) both solutions must lie on the same line through the nice point we see that L(\overline{a}) cannot be isomorphic to the linear space (B) of Lemma 2.4. This proves (c). □

Lemma 2.11 There exists an 𝔽_{3}-linear isomorphism f_{+} : V_{+} → V_{−} with f_{+}(a) = ±\overline{a} for all a ∈ V_{+}.

Proof: Let N_{ϵ} be the set of lines of V_{ϵ} containing n_{ϵ}. By Lemma 2.7, \overline{·} induces a bijection from N_{+} to N_{−}. Let g be any 𝔽_{3}-linear isomorphism from V_{+} to V_{−}. Since GL(V_{−}) induces the full symmetric group on N_{−}, there exists an h ∈ GL(V_{−}) such that hg and \overline{·} induce the same map from N_{+} to N_{−}. Then f_{+} = hg works. □

The function f_{+} from the previous lemma is essentially unique. Only −f_{+} has the same property. Now, let f_{−} = f_{+}^{−1} and define f : V_{+} ∪ V_{−} → V_{+} ∪ V_{−}, a → f_{ϵ}(a), if a ∈ V_{ϵ}. Then f^{2} is the identity and for all good points a there exists a unique |a| ∈ 𝔽_{3} with f(a) = |a|\overline{a}. Since f(−a) = −f(a) and \overline{−a} = −\overline{a} we have | − a| = |a|. The element |a| is called the norm of a. Define |n| = 0 if n is a nice point. For all points a, b ∈ V_{ϵ}, we define ⟨a, b⟩ = \frac{1}{2}(|a + b| − |a| − |b|) = |a| + |b| − |a + b|.

Lemma 2.12 (a) If a is a good point, then |a| = |\overline{a}| = |f(a)|.

(b) For all good points a and b of the same type, ⟨a, b⟩ = ⟨f(a), f(b)⟩.

Proof: From a = f(f(a)) = f(|a|\overline{a}) = |a|f(\overline{a}) = |a| |\overline{a}| \overline{\overline{a}} = |a| |\overline{a}| a, it follows that |\overline{a}| = |a|. Since f(a) = ±\overline{a} we have |f(a)| = |\overline{a}| = |a|. If a and b are good points of the same type ϵ ∈ {+, −}, then ⟨a, b⟩ = |a| + |b| − |a + b| = |f(a)| + |f(b)| − |f(a + b)| = ⟨f(a), f(b)⟩. □

Lemma 2.13 Let V = V_{ϵ} for some ϵ.

(a) | · | is a non-degenerate quadratic form in V and ⟨·, ·⟩ is its associated symmetric bilinear form.

(b) Let a, c ∈ V^{#} := V \{0}. Then {a, c} = F(x) for a bad point x if and only if ⟨a, c⟩ = 0.

(c) Let b ∈ V^{#}. There exists a uniquely determined set F(b) := {a, c} with a, c ∈ V^{#}, a + c = b and ⟨a, c⟩ = 0.

(d) F(x) = F(x̂).

(e) F(\overline{x}) = −\overline{F(x)} and ˆ \overline{x} = \overline{x̂}.

Proof:

(a) Let l be a line of L(n_{−ϵ}) with n_{ϵ} ∉ l and let x denote a bad point in l^{⊥}. Putting F(x) = \{a, c\}, we have l = \{\overline{x̂}, \overline{a}, \overline{c}\} and x̂ = a + c. Thus by definition of | · |

|\overline{a}|f(\overline{a}) + |\overline{c}|f(\overline{c}) = a + c = x̂ = |\overline{x̂}| f(\overline{x̂}) = |\overline{x̂}| f(\overline{a} ∗ \overline{c}) = −|\overline{x̂}| f(\overline{a}) − |\overline{x̂}| f(\overline{c}).

Since \overline{a} ≠ ±\overline{c} also f(\overline{a}) ≠ ±f(\overline{c}) and hence |\overline{a}| = |\overline{c}| = −|\overline{x̂}|. As a consequence every line l of L(n_{−ϵ}), n_{ϵ} ∉ l, contains vectors of norm 1 and -1. Since L(n_{−ϵ}) ≃ AG(2, 3) and |a| = | − a| for every a ∈ V^{#} it is easy to verify that exactly four points in V^{#} have norm 1, while the other four have norm -1. This proves (a).

(b) If a = ±c then a and c are neither orthogonal nor is {a, c} of the form F(x). So we may assume that n_{ϵ} ∉ {a, c}^{⊥⊥}. Suppose {a, c} = F(x) for a bad point x, then by the proof of (a), |\overline{a}| = |\overline{c}|. Hence |a| = |c| = −|a + c| and ⟨a, c⟩ = |a| + |c| − |a + c| = 0. Conversely, suppose that ⟨a, c⟩ = 0 and let x denote a bad point of \{\overline{a}, \overline{c}\}^{⊥}. Then F(x) is a set of two orthogonal points. Since F(x) ⊆ \{a, c, \overline{\overline{a} ∗ \overline{c}}\}, we necessarily have F(x) = {a, c}.

(c) Readily verified.

(d) F(x) = {a, c} = F(a + c) = F(x̂).

(e) Applying Corollary 2.9(e) to x and Corollary 2.9(f) to \overline{x} we have \overline{F(x) ∪ \{x̂\}} = −F(\overline{x}) ∪ ˆ \overline{x}. Any affine line in V_{−ϵ} which is not incident with n_{−ϵ} contains a unique pair of orthogonal vectors. Thus \overline{F(x)} = −F(\overline{x}) and \overline{x̂} = ˆ \overline{x}. □

For a good or nice point a we define â = a. For any point a of S we call â the label of a. Note that â is a nice or good point of the same type as a.

Lemma 2.14 Let V = V_{ϵ} for some ϵ. Let b ∈ V^{#} be a good point and F(b) = {a, c}. Then the labels of the points in L(\overline{b}) are as follows:

[figure: −c; −a; ■; a + c; −a; −c; ■; −a − c; ⋆; n_{ϵ}; a + c; a + c]

Proof: Since a and c are orthogonal, there are a bad points x_{1} and −x_{1} satisfying F(x_{1}) = F(−x_{1}) = {a, c} (see Lemma 2.13(b) and Lemma 2.10(a)). Then x̂_{1} = ˆ−x_{1} = a + c = b and x_{1}, −x_{1} ∈ \overline{x̂_{1}}^{⊥} = \overline{b}^{⊥}. Since a + c and a − c are orthogonal, there are bad points x_{2} and −x_{2} satisfying F(x_{2}) = F(−x_{2}) = {a + c, a − c} and x̂_{2} = ˆ−x_{2} = (a + c) + (a − c) = −a. Clearly x_{2}, −x_{2} ∈ \overline{a + c}^{⊥} = \overline{b}^{⊥}. Similarly, since a + c and c − a are orthogonal, there are bad points x_{3} and −x_{3} in \overline{b}^{⊥} satisfying x̂_{3} = ˆ−x_{3} = (a + c) + (c − a) = −c. By Lemma 2.6(b), b is the unique good point of valency 3^{1}4^{2} in L(\overline{b}). If y is one of the two bad points of valency 3^{1}4^{1} in L(\overline{b}), then \overline{b} has valency 3^{1}4^{1} in L(y). So by Corollary 2.9(b), \overline{ŷ} = \overline{b} and ŷ = b. Hence the labels are as claimed. □

Corollary 2.15 (a) Let l be a line of size four in S_{ϵ}, ϵ ∈ {+, −}, and let a denote the unique good point on l. Then there exists a unique bad point b on l with b̂ = a. If c and d are the other two bad points on l, then ⟨ĉ, d̂⟩ = 0 and ĉ + d̂= −a.

(b) Let l be a line of size three in S_{ϵ}, ϵ ∈ {+, −}, with one good point a and two bad points b and c. Then {b̂, ĉ} = F(a).

Proof: Let f denote the unique good point in l^{⊥}. Applying the previous lemma to f^{⊥}, we immediately see that (a) and (b) hold. □

Lemma 2.16 Let x be a bad point and {a, c} = F(¯x̂). Then the labels of the points in L(x) are as follows:

[figure: ■; −a; ■; −c; a + c; a − c; −a + c; ■; a + c; −a; −a − c; −c]

Proof: By Corollary 2.9(e), \overline{F(x) ∪ \{x̂\}} is the set of good points in L(x). By parts (d) and (e) of Lemma 2.13 these points are −a, −c and \overline{x̂} = ˆ \overline{x} = a + c. By Corollary 2.9(b), \overline{x̂} has valency 3^{1}4^{1} in L(x). So all the good points are as claimed.

Also \overline{x} is labeled a + c and by Lemma 2.6(f) has valency 3^{2}4^{1} in L(x). Let f be the third point on \{−a, \overline{x}\}^{⊥⊥}. By Corollary 2.15(b) f̂ = a − c. Similarly the third point g on \{−c, \overline{x}\}^{⊥⊥} has label −a + c.

The two missing points on the line \{\overline{x}, a + c\}^{⊥⊥} need labels −a and −c. The missing points on the line {−a, g}^{⊥⊥} need labels −a and −a − c. So the intersection point needs to be labelled −a and all the labels are as claimed. □

Let x and y be bad points of opposite type. The preceding lemma combined with Lemma 2.10 tells us whether one of ±y is perpendicular to x. But if one of ±y is perpendicular we do not know yet which one it is. For this we introduce an equivalence relation on the set of bad points which helps us to distinguish y and −y.

For two bad points x and y of the same type we define x ∼ y if |{x, y}^{⊥⊥}| = 4 and ⟨x̂, ŷ⟩ = 0. Then ∼ is a symmetric relation. Let ≈ be the transitive extension of ∼. Let Δ_{ϵ} be the set of equivalence classes of ≈ on the bad points of type ϵ and put Δ = Δ_{+} ∪ Δ_{−}.

Lemma 2.17 Let x be a bad point of P_{ϵ}.

(a) Let y ∉ {x, −x} be a bad point of P_{ϵ}. Then x ∼ y if and only if |{x, −y}^{⊥⊥}| = 3.

(b) There exist exactly two bad points y, z in ∼-relation to x.

(c) There exists a bad point u ≠ x with y ∼ u ∼ z.

(d) [x] := {x, y, z, u} is the equivalence class of ≈ containing x.

(e) The labels of [x] consists of all the four good points of same norm and type as x̂. More precisely, x̂ = −û, ŷ = −ẑ and x̂ and ŷ are orthogonal.

(f) [−x] = −[x].

Proof:

(a) Assume that x ∼ y or that |{x, −y}^{⊥⊥}| = 3. Let a be the good point in {x, y}^{⊥} in the first case and in {x, −y}^{⊥} in the second case. Then x, y and −y are in a^{⊥}. A glance at Lemma 2.14 shows that |{x, −y}^{⊥⊥}| = 3 if and only if x̂ ⊥ ŷ and |{x, y}^{⊥⊥}| = 4.

(b) By (a) the number of bad points y with x ∼ y equals the number of quads of order (1, 2) through x different from Q(x, n_{ϵ}). By Lemma 2.1 and since x is bad, this number equals 2.

(c) Both of the lines {x, y}^{⊥} and {x, z}^{⊥} of L(x) have size 4 and hence intersect in a bad point d; see Lemma 2.4. Let F(¯x̂) = {a, c}. Now x, y and z are points of L(d) such that |{x, y}^{⊥}| = |{x, z}^{⊥}| = 4 and ⟨x̂, ŷ⟩ = ⟨x̂, ẑ⟩ = 0. By Lemma 2.16, it follows that x̂ = −a − c and {ŷ, ẑ} = {a − c, −a + c}. If v is the unique bad point of valency 3^{2}4^{1} in L(d), then yv and zv are lines of size 3. Hence by (a) u := −v satisfies the required properties.

(d) For each t ∈ {x, y, z, u}, {x, y, u, z} contains two and thus all points in ∼-relation with t.

(e) By the proof of (c), ŷ = −ẑ and û = v̂ = −x̂. Since x ∼ y, x̂ and ŷ are orthogonal.

(f) Both [−x] and −[x] consists of those four bad points outside [x] whose labels have the same norm and type as x. □

Lemma 2.18 Let x be a bad point.

(a) \{x, x̂\}^{⊥} = \{\overline{x}, ˆ \overline{x}\}^{⊥⊥} and the quad Q(x, x̂) has order (1, 3).

(b) Let t_{1}, t_{2} and s be bad points different from x such that t_{1}, t_{2} ∈ {x, x̂}^{⊥⊥}, t_{1} ≠ t_{2}, and x ∈ {s, ŝ}^{⊥⊥}. Then [t_{1}] = [t_{2}] = −[s].

(c) Define i · [x] := [t] where t is a bad point on {x, x̂}^{⊥⊥} different from x. Then i · [x] is well defined, that is does not depend on the choice of x in [x].

(d) |t̂| = −|x̂|.

(e) i · (i · [x]) = −[x].

(f) Let r ≠ \overline{x} be a bad point on \{x, x̂\}^{⊥}. Then [\overline{t}] = −[r].

(g) [\overline{x}] = \overline{[x]}(:= \{\overline{y} | y ∈ [x]\}).

(h) \overline{i · [x]} = −i · \overline{[x]}. Proof:

(a) By definition, \overline{x̂} (resp. \overline{x}) is the unique good (resp. bad) point of valency 3^{1}4^{1} (resp. valency 3^{2}4^{1}) in L(x). By Lemma 2.16, |\{\overline{x̂}, \overline{x}\}^{⊥⊥}| = 4. By definition of \overline{·} we have \overline{x} ∈ x^{⊥} and \overline{x̂} ∈ x̂^{⊥}. By Corollary 2.9(c), \overline{x̂} ∈ x^{⊥}. By Lemma 2.13(e) ˆ \overline{x} = \overline{x̂} and so \overline{x} ∈ x̂^{⊥}. This proves (a).

(b) By Corollary 2.15, t̂_{1} and t̂_{2} are orthogonal. Hence t_{1} ∼ t_{2} and [t_{1}] = [t_{2}]. By (a) and Lemma 2.4, the two lines {x, x̂}^{⊥} and {s, ŝ}^{⊥} of L(x) intersect in a bad point x′. The points x, x̂, t_{1}, t_{2}, s and ŝ are all contained in L(x′). By Lemma 2.16 and taking innerproducts into account, either |{t_{1}, s}^{⊥⊥}| = 3 or |{t_{2}, s}^{⊥⊥}| = 3. Thus either −s ∼ t_{1} or −s ∼ t_{2} and so −[s] = [−s] = [t_{1}] = [t_{2}].

(c) Note that by (b) the definition does not depend on the choice of t. Let y be a bad point with y ∼ x, so x̂ and ŷ are orthogonal. By Corollary 2.15 {x, y}^{⊥⊥} = {x, y, s, ŝ} for some bad point s. If t̃ is a bad point of {y, ŷ}^{⊥⊥} different from y, then by (b) applied twice, [t] = −[s] = [t̃]. Since ≈ is the transitive extension of ∼, (c) holds.

(d) Clear, for example by Corollary 2.15 and the fact that non-orthogonal vectors of V_{ϵ} have different norm.

(e) By (b) applied to t in place of x, i · [t] = −[x].

(f) Considering the line \{x, x̂\}^{⊥⊥} in L(\overline{x}) and using Lemma 2.16 we find t̂ ∈ \overline{x}^{⊥}. Hence \overline{x} ∈ \{t, t̂\}^{⊥} and so using (a), \overline{x} ∈ \{\overline{t},ˆ \overline{t}\}^{⊥⊥}. Now by (a) r is a bad point on \{\overline{x}, ˆ \overline{x}\}^{⊥⊥} different from \overline{x} and hence [\overline{t}] = −[r] by (b).

(g) Let x ∼ y and put {x, y}^{⊥⊥} = {x, y, s, ŝ}. Let u be a bad point on {x, y}^{⊥} other than \overline{s}. Applying (f) twice with s in place of x, [\overline{x}] = −[u] = [\overline{y}]. Since ≈ is the transitive extension of ∼, (g) holds.

(h) We have \overline{i · [x]} = \overline{[t]} = [\overline{t}] = −[r] = −i · [\overline{x}] = −i · \overline{[x]}. □

By part (g) we obtain a well-defined map \overline{·} : Δ → Δ, [x] → [\overline{x}]. For every δ ∈ Δ, we define |δ| := |x̂| for any x ∈ δ. Note that this is well-defined by (e) of Lemma 2.17. Put

B_{ϵ} = {(a, δ) | a ∈ V_{ϵ}^{#}, δ ∈ Δ_{ϵ}, |δ| = |a|}.

Lemma 2.19 The map Φ_{ϵ} : x → (x̂, [x]) from the set of bad points of type ϵ to B_{ϵ} is a bijection.

Proof: By (e) of Lemma 2.17, x̂ ≠ ŷ for any two distinct bad points with [x] = [y]. Thus Φ_{ϵ} is one to one. Since there are sixteen bad points of a given type and since also |B_{ϵ}| = 16, Φ_{ϵ} is a bijection. □

Let d ∈ P_{ϵ}. If d is bad, define Φ(d) = Φ_{ϵ}(d) = (d̂, [d]). If d is nice or good, let Φ(d) = d. We call Φ(d) the coordinates of d. For (a, δ) in B_{ϵ} we define \overline{(a, δ)} = (\overline{a}, \overline{δ}). By (a) and (g) of Lemma 2.18, Φ(\overline{d}) = \overline{Φ(d)}.

Lemma 2.20 Let a and c be orthogonal good ϵ-points and δ ∈ Δ_{ϵ} with |δ| = |a + c|. Let d and x be the points with coordinates \overline{a + c} and \overline{(a + c, δ)}, respectively. Then the coordinates of the points in L(d) and L(x) are as follows:

[figure: (−c, −i · δ); (−a, i · δ); ■; a + c; (−a, −i · δ); (−c, i · δ); ■; −a − c; ⋆; n_{ϵ}; (a + c, −δ); (a + c, δ); ■; −a; ■; −c; (a + c, δ); (a − c, −δ); (−a + c, −δ); ■; a + c; (−a, i · δ); (−a − c, −δ); (−c, i · δ)]

Proof: This easily follows from Lemma 2.14, Lemma 2.16 and Lemma 2.18. □

Remark. Let i ∈ 𝔽_{9} be satisfying i^{2} = −1 and let \overline{·} denote the unique nontrivial automorphism of 𝔽_{9}. Comparing the norms of ±[x], ±i [x], ±\overline{[x]}, and ±i \overline{[x]} using Lemma 2.18, one can define a map k : Δ → {1, −1, i, −i} such that the following holds for every δ ∈ Δ:

(i) |δ| = [k(δ)]^{2};

(ii) k(i · δ) = i k(δ);

(iii) k(\overline{δ}) = \overline{k(δ)} = [k(δ)]^{−1}.

3 Construction and uniqueness of the near hexagon

Based on the properties given in the previous section, we are now able to give a construction for S. Choosing a suitable base in V_{+}, we may suppose that V_{+} is the vector space 𝔽_{3}^{2} and that ⟨·, ·⟩ is the standard symmetric bilinear form on 𝔽_{3}^{2}, i.e. ⟨(a_{1}, b_{1}), (a_{2}, b_{2})⟩ = a_{1}a_{2} +b_{1}b_{2}. We choose for V_{−} the dual space of V_{+} and for f_{+} = f_{−}^{−1} the 𝔽_{3}-linear transformation V_{+} → V_{−} : a → ⟨a, ·⟩. Let 0_{ϵ} denote the 0-vector in V_{ϵ} and put V_{ϵ}^{#} = V_{ϵ} \ {0_{ϵ}}. For all a, b ∈ V_{−} define ⟨a, b⟩ = ⟨f_{−}(a), f_{−}(b)⟩. Notice that this makes f_{+} and f_{−} isometries. For every a ∈ V_{ϵ}, put |a| = ⟨a, a⟩ and define \overline{a} = |a|f_{ϵ}(a). Let B_{ϵ} = \{(a, δ) | a ∈ V_{ϵ}^{#}, δ ∈ 𝔽_{9}, δ^{4} = 1 and δ^{2} = |a|\}, and let P_{ϵ} = V_{ϵ} ∪ B_{ϵ}. For every (a, δ) ∈ B_{ϵ} define \overline{(a, δ)} = (\overline{a}, \overline{δ}) = (\overline{a}, δ^{−1}) ∈ B_{−ϵ}. Let Γ denote the bipartite graph whose vertices are the elements of P_{+} ∪ P_{−}, with two vertices d and e adjacent if and only if one of the following possibilities occurs for a certain ϵ ∈ {+, −}:

(1) d = 0_{−ϵ} and e ∈ V_{ϵ};

(2) d = \overline{a + b} for some a, b ∈ V_{ϵ}^{#} with ⟨a, b⟩ = 0 and e ∈ \{0_{ϵ}, a + b, −(a + b), (a + b, δ), (a+b, −δ), (−a, iδ), (−a, −iδ), (−b, iδ), (−b, −iδ)} for any δ ∈ {±1, ±i} satisfying δ^{2} = |a + b|;

(3) d = \overline{(a + b, δ)} and e ∈ \{−a, −b, a + b, (a + b, δ), (−a − b, −δ), (a − b, −δ), (−a + b, −δ), (−a, iδ), (−b, iδ)} for some a, b ∈ V_{ϵ}^{#} and δ ∈ {±1, ±i} satisfying ⟨a, b⟩ = 0 and δ^{2} = |a + b|.

By Lemma 2.20, any near hexagon satisfying (C1), (C2) and (C3) is isomorphic to the geometry G(Γ) whose points and lines are the vertices and edges, respectively, of the graph Γ. So, it remains to show that G(Γ) is a near hexagon in which conditions (C2) and (C3) hold.

Claim If (∗) |{x, y}^{⊥}| = |{x, y}^{⊥⊥}| holds for all vertices x and y of the same type, then G(Γ) is a near hexagon in which (C2) and (C3) hold. Proof: Let x and y be two vertices of the same type. Since {x, y} ⊆ {x, y}^{⊥⊥}, we have ({x, y}^{⊥⊥})^{⊥} ⊆ {x, y}^{⊥}. Clearly also {x, y}^{⊥} ⊆ ({x, y}^{⊥⊥})^{⊥} and hence ({x, y}^{⊥⊥})^{⊥} = {x, y}^{⊥}. Now suppose that condition (∗) holds. Since |{x, y}^{⊥}| = |{x, y}^{⊥⊥}| ≥ 2, Γ is a graph of diameter 3. As it is bipartite it is a near hexagon with lines of size 2. If {a, b} ⊆ {x, y}^{⊥}, then {x, y}^{⊥⊥} ⊆ {a, b}^{⊥} and {a, b}^{⊥⊥} ⊆ {x, y}^{⊥}. Applying condition (∗) to the pairs (x, y) and (a, b), we easily see that {x, y}^{⊥} ∪ {x, y}^{⊥⊥} = {a, b}^{⊥} ∪ {a, b}^{⊥⊥}. Similarly, if {a, b} ⊆ {x, y}^{⊥⊥}, then {x, y}^{⊥} ∪ {x, y}^{⊥⊥} = {a, b}^{⊥} ∪ {a, b}^{⊥⊥}. As a consequence, every two points of the same type are contained in a unique set of the form {x, y}^{⊥} ∪ {x, y}^{⊥⊥}. This proves (C2). By construction, L(0_{+}) and L(0_{−}) are affine spaces of order 3 and hence also (C3) holds. □

The verification of condition (∗) is straightforward but tedious and we leave it to the dedicated reader. The results of our verification are listed in the table below. In this table a, b and c are elements of V_{ϵ}^{#} satisfying b = a + c and ⟨a, c⟩ = 0. The elements δ, μ satisfy |δ| = |b| and |μ| = |a|. The notation L_{gbbb}(⋯) means that this line contains one good point and three bad points.

It might be helpful to observe that our object has an automorphism group of order 64. Indeed, let α ∈ GL(V_{+}) and σ ∈ F_{3}^{#} with ⟨α(a), α(b)⟩ = σ⟨a, b⟩ for all a, b ∈ V_{+}. Let λ ∈ 𝔽_{9} with λ^{2} = σ. Define a bijection of P as follows:

a → α(a), for all a ∈ V_{+}
(a, δ) → (α(a), λδ), for all (a, δ) ∈ B_{+}
a → a ∘ α^{−1}, for all a ∈ V_{−}
(a, δ) → (a ∘ α^{−1}, \overline{λ}δ), for all (a, δ) ∈ B_{−}

It is straightforward from the definitions that this map preserves the collinearity of points. This gives us 32 automorphisms. Since our description is symmetric in V_{+} and V_{−} there also exists an automorphism interchanging the types of points.

line l of S_{ϵ}  the points on l  line l^{⊥} of S_{−ϵ}  the points on l^{⊥}
L_{gbbb}(b, δ) b, (b, δ), (−a, iδ), (−c, iδ)  L_{gbbb}(\overline{b}, \overline{δ})  \overline{b}, (\overline{b}, \overline{δ}), (\overline{a}, i\overline{δ})(\overline{c}, i\overline{δ})
L_{ngg}(±a)  0_{ϵ}, a, −a  L_{ngg}(±\overline{a})  0_{−ϵ}, \overline{a}, −\overline{a}
L_{ggg}(b)  b, −a, −c  L_{nbb}(\overline{b})  0_{−ϵ}, (\overline{b}, \overline{δ}), (\overline{b}, −\overline{δ})
L_{nbb}(b)  0_{ϵ}, (b, δ), (b, −δ)  L_{ggg}(\overline{b})  \overline{b}, \overline{a}, \overline{c}
L_{gbb}(a, μ, c)  b, (a, μ), (c, −μ)  L_{gbb}(\overline{a}, \overline{μ}, \overline{c})  −\overline{b}, (\overline{a}, \overline{μ}), (\overline{c}, −\overline{μ})
L_{gb}(a, μ)  −a, (a, μ)  L_{gb}(−\overline{a}, −\overline{μ})  \overline{a}, (−\overline{a}, −\overline{μ})
L_{gb}(a, c, μ)  a, (c, μ)  L_{bb}(\overline{a} + \overline{c}, i\overline{μ}, −\overline{a} + \overline{c}) (\overline{a} + \overline{c}, i\overline{μ}), (\overline{a}, −\overline{μ})
L_{bb}(a, μ, c)  (a, μ), (−a + c, iμ)  L_{gb}(\overline{a} − \overline{c}, \overline{a} + \overline{c}, i\overline{μ})  \overline{a} − \overline{c}, (\overline{a} + \overline{c}, i\overline{μ})
L_{bb}(±a, μ)  (a, μ), (−a, μ)  L_{bb}(±\overline{c}, −\overline{μ})  (\overline{c}, −\overline{μ}), (−\overline{c}, −\overline{μ})
L_{bb}(±(a, μ))  (a, μ), (−a, −μ)  L_{bb}(±(\overline{a}, \overline{μ}))  (\overline{a}, \overline{μ}), (−\overline{a}, −\overline{μ})

The construction we gave for the near hexagon is one which is the result of a coordinatization process. There exists another construction using a so-called two-intersection set. This is a set of points in a projective space which intersects each hyperplane in either n_{1} or n_{2} points, n_{1} ≠ n_{2}.

Suppose that X is a set of 25 points in PG(3, 4) satisfying the following properties:

(I) every plane intersects X in either 5 or 9 points,

(II) there exists a point x ∈ X such that |l ∩ X| ∈ {1, 3} for every line l through x.

Call a plane which intersects X in 9 points a thick plane. Consider the following incidence structure S_{X}: the points of S_{X} are the elements of X together with the thick planes of PG(3, 4); the lines of S_{X} are the pairs {y, π} with y ∈ X and π a thick plane through y; incidence is containment. If λ_{l} denotes the number of thick planes through a line l of PG(3, 4), then 25 = |X| = |l ∩ X| + λ_{l}(9 − |l ∩ X|) + (5 − λ_{l})(5 − |l ∩ X|) or λ_{l} = |l ∩ X|. Hence every two points of the same type have at least two common neighbours and the maximal distance between two points of S_{X} is equal to 3. As a consequence S_{X} is a thin near hexagon. For every line l of PG(3, 4) with |l ∩ X| ≥ 2 we put Q_{l} := (l ∩ X) ∪ {π|π is a thick plane through l}. Clearly Q_{l} is a good quad of S_{X}. Moreover, every two points p_{1} and p_{2} of the same type are contained in a unique quad Q_{l}: if p_{1} and p_{2} are points of PG(3, 4), then l = p_{1}p_{2}; if p_{1} and p_{2} are thick planes, then l = p_{1} ∩ p_{2}. This proves (C2). By (II) the local space at x only contains lines of size three and hence is an affine plane of order 3. This proves (C3).

We will now give a set of 25 points satisfying (I) and (II). The set is related to a partial spread of PG(3, 4); that is, a set of pairwise disjoint lines. If we have a partial spread of size 12 in PG(3, 4), then the set X of points not contained in one of the lines of the partial spread satisfies (I): if a plane π contains a line of the partial spread, then |π ∩ X| = 5; otherwise |π ∩ X| = 9. By [LS] there are three nonequivalent partial spreads of size 12 which are not contained in a partial spread of size 13 (a so-called maximal partial spread). We verified that the sets X arising from these maximal partial spreads satisfy (II). We will give the set X corresponding to the first example of [LS]. If x = (1, 0, 1, 1) and

\begin{matrix}a_{1} = (1, 0, α, α) & a_{2} = (1, α, 0, 1) & a_{3} = (1, α^{2}, 0, α) & a_{4} = (1, α^{2}, 0, α^{2}) \\ b_{1} = (1, 0, α^{2}, α^{2}) & b_{2} = (0, 1, α^{2}, 0) & b_{3} = (1, 1, α^{2}, 0) & b_{4} = (0, 1, α, α^{2}) \\ a_{5} = (1, 1, 1, 0) & a_{6} = (1, α^{2}, 1, 0) & a_{7} = (1, 1, α, α^{2}) & a_{8} = (1, 1, α^{2}, α^{2}) \\ b_{5} = (1, α, 1, α^{2}) & b_{6} = (1, 1, 1, α^{2}) & b_{7} = (1, α, 0, α) & b_{8} = (0, 1, α, α) \\ a_{9} = (1, α, α, 1) & a_{10} = (1, α, α^{2}, α) & a_{11} = (1, α^{2}, α, α^{2}) & a_{12} = (1, 0, α^{2}, 0) \\ b_{9} = (1, α^{2}, 0, 1) & b_{10} = (0, 1, 1, α) & b_{11} = (0, 1, 1, α^{2}) & b_{12} = (1, 0, α, α^{2}) \end{matrix}

with α ∈ 𝔽_{4} \ {0, 1}, then X := {x, a_{1}, …, a_{12}, b_{1}, …, b_{12}} satisfies (I) and xa_{i} ∩ X = {x, a_{i}, b_{i}} for every i ∈ {1, …, 12}.

References

[BDB] Bart De Bruyn, Characterizations of near polygons by means of one local space, J. Comb. Theory Ser. A (This issue).

[LS] L. H. Soicher: http://www.maths.qmw.ac.uk/∼leonard/partialspreads/

Rieuwert Blok Department of Mathematics Michigan State University East Lansing, Michigan 48824 blokr@member.ams.org Bart De Bruyn Department of Pure Mathematics and Computer Algebra Ghent University B-9000 Gent bdb@cage.rug.ac.be

Ulrich Meierfrankenfeld Department of Mathematics Michigan State University East Lansing, Michigan 48824 meier@math.msu.edu