Journal of Algebraic Combinatorics 11 (2000), 211–226
c 2000 Kluwer Academic Publishers. Manufactured in The Netherlands.
°
On Near Hexagons and Spreads
of Generalized Quadrangles
BART DE BRUYN∗
bdb@cage.rug.ac.be
Department of Pure Mathematics & Computer Algebra, Galglaan 2, Ghent, Belgium, B-9000
Received September 15, 1998; Revised March 23, 1999
Abstract. The glueing-construction described in this paper makes use of two generalized quadrangles with a
spread in each of them and yields a partial linear space with special properties. We study the conditions under
which glueing will give a near hexagon. These near hexagons satisfy the nice property that every two points at
distance 2 are contained in a quad. We characterize the class of the “glued near hexagons” and give examples,
some of which are new near hexagons.
Keywords:
1.
spread, generalized quadrangle, near polygon
Definitions
An incidence structure is a triple S = (P, L, I) with P (the point set) a nonempty set and L
(the set of lines) a (possibly empty) set and I a symmetric incidence relation between those
sets. Although the incidence relation is symmetric, we will write, in order not to overload
the notation, I ⊆ P × L or even use “∈” as incidence relation. The incidence structures
which we will consider here are all finite. If x is a point, then 0i (x) denotes the set of all
points at distance i from x (in the point graph). We will denote 0(x) = 01 (x).
1. An incidence structure is called a partial linear space if the following conditions are
satisfied.
(a) Every line L ∈ L is incident with at least two points.
(b) Two different points are incident with at most one line.
A linear space is a partial linear space with the property that every two points are
collinear.
2. An incidence structure of points and lines is connected if its point graph is connected.
3. A connected partial linear space is called degenerate if there is a point incident with
exactly one line.
4. A near polygon S is a connected partial linear space satisfying the following conditions.
(a) The diameter of the point graph 0 of S is finite.
(b) For every point p and every line L, there is a unique point q on L, nearest to p
(nearest with respect to the distance d(. , .) in 0).
∗ Research
Assistant of the Fund for Scientific Research - Flanders (Belgium).
212
5.
6.
7.
8.
9.
DE BRUYN
If d is the diameter of 0 then S is called a near 2d-gon. A near 0-gon has only one
point and no lines and a near 2-gon consists of one line with a number (≥2) of points
on it. The near quadrangles are just the generalized quadrangles. A generalized quadrangle (GQ for short) is called degenerate if there is a point incident with exactly
one line. The point-line dual of a nondegenerate GQ is again a nondegenerate GQ.
If a nondegenerate GQ is neither a grid nor a dual grid, then it must have an order
(s, t).
A GQ is called bad when it is degenerate or when it is a nonsymmetrical dual grid;
otherwise it is called a good GQ. If Q is a good GQ, then every point of it is incident
with the same number of lines, this number being denoted by tQ + 1.
An ovoid of a generalized quadrangle Q is a set O of points such that every line of Q is
incident with exactly one element of O. If Q has order (s, t), then |O| = 1 + st. A set
of 1 + st mutually noncollinear points of Q is always an ovoid of Q. The dual notion
is that of a spread. A spread is a set of lines of Q such that every point is incident
with exactly one line of the set. For more details on generalized quadrangles, we refer
to [6].
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 A0 is the subspace of A generated by all
points of P, then we say that A0 is the ambient space of S.
Q
A special type of affine embedding is the so-called linear representation. Let ∞ be
aQprojective space of dimension n ≥ 0 embedded as a hyperplane
in the projective space
Q
and let K be a nonempty subset of the point set ofQ ∞ . The linear representation
of (= the Q
points not belonging to
Tn∗ (K) is the geometry with points the affine points
Q
Q
∗
).
The
lines
of
T
(K)
are
all
the
lines
of
which
intersect
n
∞
∞ in a (unique) point
Q
of K. Incidence is the one derived from .
If S1 = (P1 , L1 , I1 ) and S2 = (P2 , L2 , I2 ) are two partial linear spaces, then the direct
product of S1 and S2 is the partial linear space S = (P, L, I) with P = P1 × P2 and
L = (P1 × L2 ) ∪ (L1 × P2 ). The point (x, y) is incident with the line (a, L) ∈ P1 × L2
if and only if x = a and y I2 L and it is incident with the line (M, b) ∈ L1 × P2 if and
only if y = b and x I1 M. We denote S also with S1 × S2 . Since S1 × S2 ' S2 × S1
and (S1 × S2 ) × S3 ' S1 × (S2 × S3 ), also the direct product of k ≥ 1 partial linear
spaces S1 , . . . , Sk is well-defined. If Si (i ∈ {1, 2}) is a near 2di -gon, then one can
easily prove that S1 × S2 is a near 2(d1 + d2 )-gon.
Let S = (P, L, I) be a partial linear space. A set X ⊆ P is called a subspace whenever
all the points of a line are in X as soon as two of them are in X . Every such subspace
induces a partial linear space S X = (X, L X , I0 ) where L X is the set of all lines of L
which have all their points in X and I0 is the restriction of I to X × L X . A subspace X
is called geodetically closed when all points of a shortest path between two points of X
are also contained in X . A quad is a geodetically closed subset of P which induces a
nondegenerate GQ. Since no confusion will be possible in the sequel, the GQ induced
by a quad will also be called a quad. If a quad Q contains a unique point nearest a fixed
point x, then this point is called the projection of x on Q.
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
2.
213
Some theorems
Theorem 2.1 ([7, 8]) Let x and y be two points of a near polygon 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 (necessarily good ) quad.
Theorem 2.2 Let S be a near polygon and let x be a point at distance at most 1 from a quad
Q, then there exists a unique point x 0 of Q nearest to x and d(x, y) = d(x, x 0 ) + d(x 0 , y)
for all points y of Q. Hence, if L is a line of Q, then the unique point of L nearest to x is
also the unique point of L nearest to x 0 .
Proof:
This follows from the fact that Q is geodetically closed.
2
Corollary 2.3 Let Q be a quad of a near polygon S and let x and y be two collinear
points of S such that the line x y is disjoint with Q. If x, respectively y, is collinear with
x 0 ∈ Q, respectively y 0 ∈ Q, then d(x 0 , y 0 ) = 1.
Proof:
By Theorem 2.2, we have that 2 = d(x 0 , y) = d(x 0 , y 0 ) + d(y 0 , y) = 1 + d(x 0 , y 0 ).
2
Theorem 2.4 ([3]) 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
tx + 1 (respectively t y + 1) lines. Now
tx + 1 = 1 +
X
tQ = t y + 1,
where the summation ranges over all quads Q through the line x y. Hence x and y are
incident with the same number of lines and the result follows by connectedness of S. 2
Theorem 2.5 ([3]) Let S be a near polygon satisfying the following properties:
(a) every two points at distance 2 have at least two common neighbours,
(b) there are lines incident with a different number of points,
then S is the direct product of a number of near polygons, each of which has a constant
length for the lines.
If S = (P, L, I) is a near 2-gon or a good GQ, then |0i ( p)| (i ∈ {0, 1, 2}) is independent
of p ∈ P. We derive a similar property for near hexagons.
Theorem 2.6 Let S = (P, L, I) be a near hexagon such that every two points at distance
2 are contained in a good quad, then |0i ( p)| (i ∈ {0, 1, 2, 3}) is independent of p ∈ P.
214
DE BRUYN
Proof: If not all lines of S are incident with the same number of points, then Theorem 2.5
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. Theorem 2.4 implies then that S has an order (s, t). Now, let p ∈ P be a
fixed point and put n i = |0i ( p)|. Then n 0 = 1, n 1 = s(t + 1). Let V be the set of quads
through p. Counting points in 02 ( p) we find
X
n2 = s 2
tx .
(1)
x∈V
Counting edges between 02 ( p) and 03 ( p) we find that
X
n 3 (t + 1) = s 3
tx (t − tx ).
(2)
x∈V
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
X
t (t + 1) =
tx (tx + 1).
(3)
x∈V
P
P
Eliminating tx and tx2 , we find that n 3 = s(n 2 − s 2 t). Together with v = n 0 + n 1 + n 2
+ n 3 this gives
n2 =
v
− 1 + st (s − 1),
s +1
(4)
¶
v
− st − 1 .
s +1
(5)
µ
n3 = s
2
Corollary 2.7 If S is a near hexagon satisfying the property that every two points at
distance 2 are contained in a quad of order (s, t1 ) or (s, t2 ), s ≥ 1 and 1 ≤ t1 < t2 , then
for each i ∈ {1, 2}, the number of quads of order (s, ti ) through a point is independent of
that point.
Proof: This follows from Eqs. (1), (3) and (4).
2
Remark The previous corollary was proved in [2] in the case that s = 2, t1 = 1, t2 = 2
by using the same double countings as in the proof of Theorem 2.6.
Theorem 2.8 Let S = (P, L, I) be a partial linear space of order (s, t) 6= (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 = |02 (x)| is independent of the point x ∈ P,
3. d(x, L) ≤ 2 for all x ∈ P and L ∈ L,
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
215
then b = |03 (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 |03 (x)| = |P| − 1 − s(t + 1) − |02 (x)| is independent of x ∈ P. Take an
arbitrary line L and let r be a point of L. There are a points in 02 (r ), s 2 t of these are contained
in 01 (L). Hence a ≥ s 2 t and 02 (L) ≤ (s + 1)(a − s 2 t). If a = s 2 t then 02 (L) = ∅ implies
that S is a generalized quadrangle. So, suppose that a 6= s 2 t, then S is a near hexagon if and
only if 02 (L) = (s + 1)(a − s 2 t). From |02 (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. 2
3.
A possible construction for near hexagons
(i)
Let Qi = (Pi , Li , Ii ) (for each i ∈ {1, 2}) be a GQ of order (s, ti ), let Si = {L (i)
1 , . . . , L 1 + sti }
(1)
(2)
⊂ Li be a spread of Qi and let θ be a bijection from L 1 to L 1 (here we suppose that every
line is a subset of the point set).
(i)
For every i ∈ {1, 2} and every j ∈ {1, . . . , 1 + sti }, 8(i)
j : Pi 7→ L j is defined such that
(i)
x ∈ Pi is mapped to the unique point of L j nearest to x (in the generalized quadrangle Qi ).
(2)
Let 0(Q1 , Q2 , S1 , S2 , L (1)
1 , L 1 , θ ) (0 for short if no confusion is possible) be the graph
(1)
(1)
(2)
with vertex set L 1 × S1 × S2 . Two different points (x, L i(1) , L (2)
j ) and (y, L k , L l ) are
adjacent whenever at least one of the following two conditions are satisfied:
(1) j = l and 8i(1) (x), 8(1)
k (y) are collinear points in Q1 ,
◦
θ(x),
8l(2) ◦ θ(y) are collinear points in Q2 .
(2) i = k and 8(2)
j
If i = k and j = l, then both (1) and (2) are satisfied. It is clear that 0(Q1 , Q2 , S1 , S2 , L (1)
1 ,
(2)
(1)
(1)
(2)
(2)
(1)
−1
L (2)
,
θ)
'
0(Q
,
Q
,
S
,
S
,
L
,
L
,
θ
).
For,
1
:
(x,
L
,
L
)
→
7
(θ
(x),
L
,
L
2
1
2
1
i
j
j
i )
1
1
1
defines an isomorphism. The definition of 0 is hence symmetric in Q1 and Q2 .
Remark In the sequel, we will not write the symbol “◦” between functions, i.e. with f g
we mean the function f ◦ g.
Lemma 3.1 Through every two adjacent vertices of 0, there is a unique maximal clique.
This clique has size s + 1.
(1)
(2)
Proof: Let a = (x, L i(1) , L (2)
j ) and b = (y, L k , L l ) be two fixed adjacent vertices; we
(1)
(2)
determine what the common neighbours (z, L m , L n ) look like. If i = k 6= m, then j =
(1)
n = l and 8i(1) (x) ∼ 8(1)
m (z) ∼ 8i (y) implies that x = y and hence a = b, a contradiction.
(2)
(2)
Similarly, j = l 6= n is impossible. If i = k = m, then 8(2)
j θ (x) ∼ 8n θ (z) ∼ 8l θ (y)
(2)
(2)
(2)
implies that 8n θ(z) is an element of the line of Q2 through 8 j θ (x) and 8l θ (y). This
216
DE BRUYN
yields s − 1 common neighbours of a and b and they are all mutually adjacent. Together
with the vertices a and b, they yield a clique of size s + 1. A similar reasoning holds in the
case j = l = n.
2
(2)
Let S(Q1 , Q2 , S1 , S2 , L (1)
1 , L 1 , θ ) be the partial linear space with points the vertices of
0 and with lines the maximal cliques of 0. The incidence is the natural one. Again, we
will write S when no confusion is possible.
Definition 3.2 A line L is said to be of type I, if there exists a fixed j, such that every
point of L is of the form (x, L i(1) , L (2)
j ). A line M is said to be of type II, if there exists
a fixed i, such that every point of M is of the form (x, L i(1) , L (2)
j ). Remark that there are
)
| x ∈ L (1)
lines which are of both types, namely the lines {(x, L i(1) , L (2)
j
1 }, where i and j
are fixed. These lines partition the point set of S (hence they form a spread of S).
Lemma 3.3
(1)
(a) For a fixed j ∈ {1, . . . , 1 + st2 }, the set {(x, L i(1) , L (2)
j ) | x ∈ L 1 , 1 ≤ i ≤ 1 + st1 }
is a quad isomorphic to Q1 .
(1)
(b) For a fixed i ∈ {1, . . . , 1 + st1 }, the set {(x, L i(1) , L (2)
j ) | x ∈ L 1 , 1 ≤ j ≤ 1 + st2 }
is a quad isomorphic to Q2 .
(1)
Proof: The isomorphisms are given by 11 : (x, L i(1) , L (2)
j ) 7→ 8i (x) for (a) and 12 :
(1)
(2)
(2)
2
(x, L i , L j ) 7→ 8 j θ(x) for (b).
Definition 3.4
(1) The previous lemma shows that several GQ’s (isomorphic to Q1 or Q2 ) are glued
together to form the geometry S. For this reason the above construction is called
glueing and S will be called a glued geometry.
(2) A quad of type I, respectively II is a quad that arises like in (a), respectively (b) of the
previous lemma. The following properties hold then.
•
•
•
•
•
Every line contained in a quad of type A (∈ {I, II}) is also of type A.
Two quads of the same type are equal or disjoint.
Two quads of different type meet each other in a line which is of both types.
Through every point of S, there is a unique quad of each type.
Every line of type A (∈ {I, II}) is contained in a unique quad of type A.
Lemma 3.5
S has order (s, t1 + t2 ) and satisfies properties 1 and 3 of Theorem 2.8.
Proof: Let p be an arbitrary point of S. The quad of type I (respectively type II) through
p contains t1 + 1 (respectively t2 + 1) lines through p and both quads have exactly one line
in common. Hence S has order (s, t1 + t2 ).
Property 1 clearly holds by Lemma 3.1, so let x and M be a point and a line of S, both
arbitrarily chosen. Through M, there is a quad R1 of type A ∈ {I, II}. Take the unique quad
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
217
R2 through p of type B such that {A, B} = {I, II}. On the intersection line of R1 and R2
there is a unique point nearest to x. This point has distance at most 1 to x and M. This
proves the lemma.
2
Definition 3.6
• For all i, j ∈ {1, . . . , 1 + st1 }, φi,(1)j is the permutation of L (1)
1 equal to the restriction of
(1) (1)
(1)
(1)
(1)
8(1)
1 8 j 8i to L 1 . The group of permutations of L 1 generated by the elements φi, j
is denoted by G 1 .
• For all i, j ∈ {1, . . . , 1 + st2 }, φi,(2)j is the permutation of L (2)
1 equal to the restriction of
(2) (2)
(2)
(2)
8
8
to
L
.
The
group
of
permutations
of
L
generated
by the elements φi,(2)j
8(2)
j
i
1
1
1
is denoted by G 2 .
Remark
(1)
(2)
, φi,i
are identity permutations,
• φi,i
(k)
• φi, j and φ (k)
j,i (k ∈ {1, 2}) are inverse permutations.
Theorem 3.7 S is a near hexagon if and only if [G 1 , θ −1 G 2 θ ] = 0. (Here 0 stands for
the trivial group and [G 1 , θ −1 G 2 θ] is the group generated by all commutators [g1 , θ −1 g2 θ ]
with g1 ∈ G 1 and g2 ∈ G 2 .)
Proof: Suppose that S is a near hexagon. It suffices to prove that φi,(1)j commutes with
(2)
θ for all possible i, j, k, l with i 6= j and k 6= l. If x ∈ L (1)
θ −1 φk,l
1 , then we have the
following adjacencies:
¡ (1) (1) (1) −1 (2) (2) (2)
(2) ¢
81 8 j 8i θ 81 8l 8k θ (x), L (1)
j , Ll
¡
(2) (2)
(1)
(2) ¢
∼ θ −1 8(2)
1 8l 8k θ(x), L i , L l
¢
¡
∼ x, L i(1) , L (2)
k
¡
(1) (1)
(1)
(2) ¢
∼ 8(1)
1 8 j 8i (x), L j , L k
¡
(2) (2)
(1) (1) (1)
(1)
(2) ¢
.
∼ θ −1 8(2)
1 8l 8k θ81 8 j 8i (x), L j , L l
(1)
(2)
(1)
Let p be the point (x, L i(1) , L (2)
k ) and L be the line {(x, L j , L l ) | x ∈ L 1 } (this is a line
of type I and of type II). Since there is only one point of L at distance 2 from p, it follows
that
(2)
(2)
θφi,(1)j = φi,(1)j θ −1 φk,l
θ.
θ −1 φk,l
Conversely, suppose that [G 1 , θ −1 G 2 θ ] is the trivial group. Let x be an arbitrary point
of S. Through x, there is a unique quad R1 of type I and a unique quad R2 of type II. In
R1 ∪ R2 , there are s 2 (t1 + t2 ) points of 02 (x). The points of S not in R1 ∪ R2 are partitioned
218
DE BRUYN
by s 2 t1 t2 lines which have both types. The previous reasoning shows that each of these lines
contains a unique point at distance 2 from x. Hence a = |02 (x)| = s 2 (t1 t2 + t1 + t2 ) is independent of the point x. From this it follows that b = |03 (x)| = (s + 1)(st1 + 1)(st2 + 1)
− 1 − |01 (x)| − |02 (x)| = s 3 t1 t2 . Since a > s 2 (t1 + t2 ) and b = s(a − s 2 (t1 + t2 )), it
follows from Theorem 2.8 that S is a near hexagon.
2
(2)
(1)
Above, we defined S = S(Q1 , Q2 , S1 , S2 , L (1)
1 , L 1 , θ ). Take now an arbitrary line L i in
(2)
(2)
(1)
S1 and an arbitrary line L j in S2 . If we define θi, j as the restriction of 8 j θ 81 to L i(1) ,
then we can define
¡
¢
Si, j = S Q1 , Q2 , S1 , S2 , L i(1) , L (2)
j , θi, j .
Theorem 3.8 If S is a near hexagon, then Si, j is isomorphic to S for all i ∈ {1, . . . , 1 + st1 }
and all j ∈ {1, . . . , 1 + st2 }.
(1)
(1)
(2)
(1) (1)
Proof: We prove that 1 : L (1)
1 × S1 × S2 7→ L i × S1 × S2 , (x, L k , L l ) 7→ (8i φk,i
(2)
(1)
(2)
θ −1 φl, j θ(x), L k , L l ) is an isomorphism between S and Si, j . This map is clearly a bijection and it suffices to prove that adjacency is preserved in the point graph of the geometries.
(2)
(1)
(2)
Consider the two adjacent vertices a = (x, L (1)
k , L l ) and b = (y, L k , L m ) in S, then
−1 (2)
y = θ φl,m θ(x) and
¡
(1) −1 (2)
(2) ¢
θ φl, j θ(x), L (1)
,
1(a) = 8i(1) φk,i
k , Ll
¢
¡ (1) (1) −1 (2)
(1)
(2)
1(b) = 8i φk,i θ φm, j θ(y), L k , L m .
Now, 1(a) ∼ 1(b) (in Si, j ) if and only if
(1) (1) (1) −1 (2)
(1) (1) (1) −1 (2)
(2) (2)
8l(2) 8(2)
j θ81 8i φk,i θ φl, j θ (x) ∼ 8m 8 j θ 81 8i φk,i θ φm, j θ (y)
(1) −1 (2)
(1) −1 (2)
(2) (2)
8l(2) 8(2)
j θφk,i θ φl, j θ (x) ∼ 8m 8 j θ φk,i θ φm, j θ (y)
(2)
(1)
(1)
(2) (2) (2)
8l(2) 8(2)
j φl, j θφk,i (x) ∼ 8m 8 j φm, j θ φk,i (y)
(1)
(1)
(x) ∼ 8(2)
8l(2) θφk,i
m θ φk,i (y)
(2)
(1)
(1)
θφk,i
(x) = φk,i
(y)
θ −1 φl,m
(2)
(1)
(1) −1 (2)
θφk,i
(x) = φk,i
θ φl,m θ (x).
θ −1 φl,m
(2)
(2)
(1)
Consider the two adjacent vertices a = (x, L (1)
k , L l ) and b = (y, L m , L l ) in S, then
(1)
y = φk,m (x) and
¡
(1) −1 (2)
(2) ¢
θ φl, j θ(x), L (1)
,
1(a) = 8i(1) φk,i
k , Ll
¢
¡ (1) (1) −1 (2)
(2)
.
1(b) = 8i φm,i θ φl, j θ(y), L (1)
m , Ll
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
219
Now, 1(a) ∼ 1(b) (in Si, j ) if and only if
(1) (1) −1 (2)
(1) (1) (1) −1 (2)
8(1)
k 8i φk,i θ φl, j θ(x) ∼ 8m 8i φm,i θ φl, j θ (y)
−1 (2)
(1) −1 (2)
8(1)
k θ φl, j θ(x) ∼ 8m θ φl, j θ (y)
(1) −1 (2)
θ φl, j θ(x) = θ −1 φl,(2)j θ (y)
φk,m
(1) −1 (2)
(1)
θ φl, j θ(x) = θ −1 φl,(2)j θ φk,m
(x).
φk,m
2
Theorem 3.9
a quad.
If S is a near hexagon, then any two points at distance 2 are contained in
(1)
(2)
Proof: Let p1 = (x, L i(1) , L (2)
j ) and p2 = (y, L k , L l ) denote the two points at distance 2. If i = k (respectively j = l), then p1 and p2 are contained in a quad of type II
(respectively I). If i 6= k and j 6= l, then the adjacencies of Theorem 3.7 show that p1 and
p2 have two common neighbours p3 and p4 . If s ≥ 2, then Theorem 2.1 implies that p1
and p2 are contained in a quad (which is a (s + 1) × (s + 1)-grid in this case). If s = 1,
then p1 and p2 are contained in a quad, since { p1 , p2 , p3 , p4 } is geodetically closed and
induces a (2 × 2)-grid.
2
Definition 3.10 Suppose S is a near hexagon. The quads in S, different from the above defined quads of type I and II are called the quads of type III. These quads are (s + 1) × (s + 1)grids.
Remarks
(a) For i ∈ {1, 2} fixed, let Qi be an (s + 1) × (s + 1)-grid and Si be one of the two spreads
of Qi . Since φ (i)
j,k is the identity permutation for all j, k ∈ {1, . . . , 1 + s}, one has that
−1
[G 1 , θ G 2 θ] = 0, hence S is a near hexagon. It is straightforward to check that S is
the direct product of Q3−i with a line of size s + 1.
(b) For every t ∈ N\{0}, there is a unique GQ of order (1, t). This GQ contains several spreads which are all equivalent. Since G 2 is a commutative group, the above
construction with s = 1, t1 , t2 ≥ 1 will yield a thin near hexagon.
(c) In the next sections we will construct near hexagons using two generalized quadrangles
(Q1 and Q2 ) and certain spreads in them (S1 and S2 respectively). In the definition of
(2)
S, we took in each spread two special lines (namely L (1)
1 and L 1 ). Theorem 3.8 says
(in the case that S is a near hexagon) that those special lines are in fact not so special.
One can obtain the same near hexagon starting with two arbitrary lines (one in each
spread) by taking a suitable θ.
(d) We will not study the problem of determining suitable spreads and suitable maps θ .
Also, the above construction can be generalized to obtain other near polygons (e.g. near
octagons). These two problems will be considered in forthcoming papers.
220
4.
4.1.
DE BRUYN
A new construction for (T2∗ (O1 ), T2∗ (O2 ))
The generalized quadrangle T2∗ (O)
Consider a hyperoval O in PG(2, q) with q even. Embed PG(2, q) as a hyperplane in
PG(3, q), then T2∗ (O) is a generalized quadrangle of order (q − 1, q + 1), see [1, 5, 6]. Let
p be a fixed point of O, then the set of lines of PG(3, q) intersecting O in p defines a spread
S of T2∗ (O). Consider now the model of PG(3, q) where the points are the 1-dimensional
subspaces of V (4, q) and let L 1 , L 2 , L 3 denote three arbitrary (but different) lines of S.
The plane hL i , L j i (i 6= j and i, j ∈ {1, 2, 3}) intersects PG(2, q) in a line through p. Let
hc̄i j i denote the second point of O on that line. Take ā, b̄ ∈ V (4, q) such that p = hāi and
L 1 = hā, b̄i and let x = hα ā + b̄i with α ∈ Fq be an arbitrary point of L 1 . The projection
(in T2∗ (O)) of x on L 2 is equal to 82 (x) = hα ā + b̄ + β c̄12 i where β ∈ Fq is independent
of α. In the same way, we will find that 81 83 82 (x) = hα ā + b̄ + β c̄12 + γ c̄23 + δ c̄31 i
where γ , δ are independent of α. Now β c̄12 + γ c̄23 + δ c̄31 = µā where µ is independent
of α. Hence the map φ2,3 (which is equal to the restriction of 81 83 82 to L 1 ) maps the
point hα ā + b̄i to h(α + µ)ā + b̄i where µ is independent of α ∈ Fq .
4.2.
The near hexagon (T2∗ (O1 ),T2∗ (O2 ))
Q
In [4] the following near hexagon was described. Let Q∞ be a PG(4, q),
Q with q even,
embedded as a hyperplane in the 5-dimensional space . Consider in ∞ two planes
α1 and α2 meeting each other in a point p and consider in αi (i = 1, 2) a hyperoval Oi
containing p. It was proved in [4] that T4∗ (O1 ∪ O2 ) is a near hexagon and it was denoted
there by (T2∗ (O1 ), T2∗ (O2 )).
Theorem 4.1 The near hexagon (T2∗ (O1 ), T2∗ (O2 )) is glued.
Q
Proof: Let
Qa be a fixed affine point of and put Ai = ha, αi i (i ∈ {1, 2}). For every affine
point x ∈ , we define Qi (x) (i ∈ {1, 2}) as the GQ with the affine points of hx, αi i as
points, two points are collinear in the GQ whenever they are collinear in T4∗ (O1 ∪ O2 ). These
GQ’s are quads of T4∗ (O1 ∪ O2 ) and each point of T4∗ (O1 ∪ O2 ) has distance at most one to
each such quad. For i = {1, 2}, let Qi = Qi (a), let Si be the set of lines of Ai intersecting
Q
(1)
(2)
∞ in p, let L 1 = L 1 = pa and finally let θ be the identity map. In the previous
paragraph we determined what φi,(1)j and φi,(2)j look like. We can conclude that [G 1 , G 2 ] = 0,
hence we can define a near hexagon S = S(Q1 , Q2 , S1 , S2 , pa, pa, θ ). We will construct
∗
now
Q an isomorphism 1 between T4 (O1 ∪ O2 ) and S. Let x be an arbitrary affine point of
. The quad Q1 (x) (respectively Q2 (x)) intersects Q2 (respectively Q1 ) in a line δ2 (x)
(respectively δ1 (x)) of S2 (respectively S1 ). We put γ (x) equal to the unique point of pa
nearest to x (in T4∗ (O1 ∪ O2 )). The point of Qi nearest to x is then equal to the projection (in
Qi ) of γ (x) on the line δi (x) ∈ Si , see Theorem 2.2. If we put 1(x) = (γ (x), δ1 (x), δ2 (x)),
then we will prove that 1 is an isomorphism. Let (a, L 1 , L 2 ) = (γ (x), δ1 (x), δ2 (x)) and
put ai (i ∈ {1, 2}) equal to the projection of a on the line L i of Qi . If L 1 = pa, then
x = a2 ; if L 2 = pa, then x = a1 ; if L 1 6= pa 6= L 2 , then x is the common neighbour
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
221
of a1 and a2 (in T4∗ (O1 ∪ O2 )) different from a. This proves that 1 is a bijection. Since
both geometries have the same order, it suffices to prove that 1 preserves adjacency in the
point graph of theQ
geometries. Let r and r 0 be two adjacent points of T4∗ (O1 ∪ O2 ). If the
line rr 0 intersects ∞ in a point of Oi , then δ3−i (r ) = δ3−i (r 0 ) and the result follows from
2
Corollary 2.3 by considering the projection on the quad Qi .
5.
New example related to Q(5, q)
The generalized quadrangle Q(5, q) is the GQ of the points and the lines of a nonsingular
elliptic quadric in PG(5, q). Its order is (q, q 2 ). The corresponding dual generalized quadrangle is the GQ of the points and the lines of a nonsingular Hermitian variety H (3, q 2 ) in
PG(3, q 2 ), see [6]. If we intersect this variety with a nontangent plane, then we get a set O
of q 3 + 1 mutually noncollinear points in H (3, q 2 ), hence O is an ovoid of H (3, q 2 ). This
ovoid O dualizes to a spread S of Q(5, q).
Take now Q = Q(5, q) and let L be an arbitrary line of S. The following theorem holds
then (1 L denotes the identity permutation of the set of points of L).
Theorem 5.1 S = S(Q, Q, S, S, L , L , 1 L ) is a near hexagon.
Proof: We determine the permutations φi,(1)j = φi,(2)j while reasoning in the dual GQ.
2
a nonsingular
The points of H (3, q 2 ) are 1-dimensional
Psubspaces
P of V (4, q ).
PConsider
P
q
2
Hermitian form (· , ·) in V (4, q ), i.e. ( i λi vi , j µ j w j ) = i j λi µ j (vi , w j ), and
let ζ be the corresponding polarity of PG(3, q 2 ). Take now a nontangent plane π and let
π ζ = hūi. Take three arbitrary (but different) points hāi, hb̄i, hc̄i of O = π ∩ H (3, q 2 ).
The tangent plane at hāi intersects π in a line hā, v̄i. Let L = hā, ū + λv̄i be an arbitrary
ū)
.
line of H (3, q 2 ) through hāi. Since (ū + λv̄, ū + λv̄) = 0, one finds that λq+1 = − ((ū,
v̄,v̄)
0
2
We determine the line L of H (3, q ) through hb̄i intersecting L. This line looks like
v̄,b̄)
hb̄, ū + λv̄ + β āi. An easy calculation yields β = −λ ((ā,
. Hence L 0 = hb̄, ū + λv̄ 0 i
b̄)
0
ζ
with hv̄ i ∈ π ∩ hb̄i independent of λ. Similarly, if we project L 0 to a line L 00 through
hc̄i and finally L 00 to a line L 000 through hāi, we will find that L 000 = hā, ū + λv̄ 000 i with
hv̄ 000 i ∈ π ∩ hāiζ independent of λ. Now v̄ 000 = γ1 ā + γ2 v̄, hence L 000 = hā, ū + λγ2 v̄i where
q+1
ū)
or γ2 = 1. It is
γ2 is independent of λ. Just like before, one has that (λγ2 )q+1 = − ((ū,
v̄,v̄)
2
now clear that [G 1 , G 2 ] = 0, hence S is a near hexagon.
6.
New example related to AS(q)
For every odd prime power q, there exists a generalized quadrangle of order (q − 1, q + 1)
denoted by AS(q), see [1, 6]. The points of AS(q) are the points of the affine space
AG(3, q). The lines of AS(q) are the following curves of AG(3, q):
(1) x = σ, y = a, z = b;
(2) x = a, y = σ, z = b;
(3) x = cσ 2 − bσ + a, y = −2cσ + b,
z =σ.
222
DE BRUYN
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 set S which consists of all lines of type (1) is a spread of
AS(q). If L is an arbitrary line of S, then we have the following theorem.
Theorem 6.1 S = S(AS(q), AS(q), S, S, L , L , 1 L ) is a near hexagon.
Proof: Let a, b, c, d ∈ GF(q) be fixed. Consider then the lines M = {(σ, a, b) | σ ∈
GF(q)} and N = {(σ, c, d) | σ ∈ GF(q)}. Let p = (α, a, b) be an arbitrary point of M and
let p0 = (β, c, d) be its projection on N . If b = d, then β = α and there is a line of type
(2) through p and p 0 . If b 6= d, then the line through p and p 0 must necessarily be of type
(3). Let x = mσ 2 − lσ + k, y = −2mσ + l, z = σ be that line. Then we get the following
equations:
α
β
a
c
= mb2 − lb + k,
= md 2 − ld + k,
= −2mb + l,
= −2md + l.
Since b 6= d, m and l are completely determined by a, b, c and d. We have that β =
α + m(d 2 − b2 ) + l(b − d).
It is now clear that the maps φi,(1)j = φi,(2)j are translations of the line L. This proves that
2
[G 1 , G 2 ] = 0, hence S is a near hexagon.
Remark All the near hexagons with lines of size 3 and quads through every two points at
distance 2 were classified in [2]. The near hexagons derived here from AS(3) and Q(5, 2)
are both isomorphic to example (vi) of [2]. (Notice that AS(3) ' Q(5, 2).)
7.
7.1.
Characterizations
The local space
Let S = (P, L, I) be a near hexagon satisfying the property that every two points at
distance 2 are contained in a unique quad. For x ∈ P, we define the following incidence
structure Sx .
• The points of Sx are the lines of S through x.
• A line of Sx is the set of lines of S through x in a quad on x.
• Incidence is the symmetrized containment.
The space Sx is linear and is called the local space at x.
For u, v ∈ N\{0}, let Su,v = (Pu,v , Lu,v , Iu,v ) be the following linear space:
• Pu,v = {α, β1 , . . . , βu , γ1 , . . . , γv },
• Lu,v = {{α, β1 , . . . , βu }, {α, γ1 , . . . , γv }} ∪ {{βi , γ j } | 1 ≤ i ≤ u and 1 ≤ j ≤ v},
• Iu,v is the symmetrized containment.
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
223
Su,v is a linear space with a thin point (namely α). Conversely, every linear space with a
thin point is obtained in this way. If S is a glued near hexagon, then Sx ' St1 ,t2 for all
points x of S.
Theorem 7.1 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 Sx ' S1,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 2.5. Hence, by Theorem 2.4, we may assume that S has
order (s, t) with t = r + 1. Consider through x a quad Rx containing t lines through x
and let L x be the remaining line through x. Every point z of Rx is incident with exactly
one line L z which is not in Rx . Let y ∈ L x \{x} be fixed. Let M1 and M2 be two lines
through y different from L x and let R y be the quad through M1 and M2 . The quad through
Mi (i ∈ {1, 2}) and L x intersects Rx in a line Mi0 . Now, let u be one of the s 2 (t − 1) points
of Rx at distance 2 from x. Let u i (i ∈ {1, 2}) be the unique point on Mi0 collinear with u.
The quad through uu i and L u i is a grid. Let u i0 be the intersection of L u i with Mi and let
vi be the unique neighbour of u i0 and u different from u i . The point vi is then the unique
point of L u at distance 2 from y. This implies that v = v1 = v2 . Since v is collinear with
the points u 01 and u 02 of R y , v is itself contained in R y . Hence |02 (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
2
line L x and that R y ' Rx . The result follows now immediately.
7.2.
Characterizations of the new class of near polygons
Theorem 7.2 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 ∈ 0(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 2.4 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 7.1. The line L y is then contained in exactly two quads. The following
properties hold now.
(a) If y is a point for which S y is a linear space with a thin point, then S y 0 ' S y and
L y 0 = L y for all points y 0 ∈ L y .
Proof: Suppose S y ' St1 ,t2 with t1 , t2 > 1 and t = t1 + t2 . The point L y of S y 0 is
contained in exactly two lines of S y 0 , one line has t1 + 1 points, the other t2 + 1 points.
2
Since there are exactly t1 + t2 + 1 points in S y 0 , it follows that S y 0 ' St1 ,t2 .
224
DE BRUYN
(b) If y1 , y2 are points such that S y1 , S y2 are linear spaces with a thin point, then L y1 and
L y2 are equal or disjoint.
Proof:
This follows immediately from (a).
2
(c) There exists a point y ∈ 0(x) such that x ∈ L y .
Proof: Suppose that this is not true. Let y ∈ 0(x) be fixed. Let Q be the quad of
order (s, t 0 ) through x y and L y . There are s(t 0 + 1) points z i ∈ Q collinear with x.
These give rise to s(t 0 + 1) lines L zi and all these lines are different and hence disjoint
by (b). Suppose that L z is not contained in Q for a certain z ∈ 0(x) ∩ Q, then Sz
contains at least three thick lines (namely the line defined by Q and the two lines of Sz
through L z ), a contradiction since Sz 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 0 + s) points in Q, but
this is again impossible.
2
Let y ∈ 0(x) such that x ∈ L y . Hence Sx is also a linear space with a unique thin point L x .
Let Q1 and Q2 be the two quads through L x with respective orders (s, t1 ) and (s, t2 ). In Qi ,
there are sti points z collinear with x and not on L x . These give rise to sti disjoint lines L z
(i)
(i)
which together with L x form a spread Si of Qi . Put Si = {L (i)
1 , . . . , L 1+sti } with L 1 = L x .
Finally, let θ be the identity permutation of L x . We prove now that S ' S(Q1 , Q2 ,
(2)
S1 , S2 , L (1)
1 , L 1 , θ).
First we prove that every point u of S has distance at most 1 to each Qi (i ∈ {1, 2}).
Let u 0 be the unique point of L x nearest to u; we may suppose that d(u, u 0 ) = 2. Since
Su 0 ' Sx , it follows that the quad through u and u 0 intersects each Qi in a line. This proves
that each Qi contains a point collinear with u. For i ∈ {1, 2} and u ∈ P, let pi (u) denote
the unique point of Qi nearest to u.
Next we prove that all the local spaces Su are isomorphic to St1 ,t2 . Since for all u ∈ Qi ,
L u is contained in exactly two quads (Qi and another quad), we have that G u ' St1 ,t2 . Let u
be a point of S not contained in Q1 ∪ Q2 . Let u 0 = p1 (u) and u 00 = p2 (u). The local space
Su contains t1 + t2 + 1 points, a line with t1 + 1 points (determined by the quad through
uu 00 and L u 00 ) and a line with t2 + 1 points (determined by the quad through uu 0 and L u 0 ).
From this it follows that Su ' St1 ,t2 . 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
Q2 in a line and is isomorphic to Q1 . The other quad intersects Q1 and is isomorphic to
Q2 . Note that the isomorphisms are defined by the projections pi , i ∈ {1, 2}.
We consider now the following map 1 : P 7→ L x × S1 × S2 , 1(u) = (γ (u), δ1 (u),
δ3 (u)), where γ (u) is the unique point of L x nearest to u and δi (u) (i ∈ {1, 2}) is the unique
line of Si incident with pi (u). By Theorem 2.2, it follows that pi (u) is the projection (in
Qi ) of γ (u) on the line δi (u). Let (a, L 1 , L 2 ) = (γ (u), δ1 (u), δ2 (u)) and put ai (i ∈ {1, 2})
equal to the projection of a on the line L i of Qi . If L 1 = L x , then u = a2 ; if L 2 = L x ,
then u = a1 ; if L 1 6= L x 6= L 2 , then u is the common neighbour of a1 and a2 different
from a. This proves that 1 is a bijection. Since both geometries have the same order, it
suffices to prove that 1 preserves adjacency in the point graph of the geometries. Let x
and x 0 be two adjacent points. If x and x 0 are contained in a quad intersecting Q2 , then
ON NEAR HEXAGONS AND SPREADS OF GENERALIZED QUADRANGLES
225
δ2 (x) = δ2 (x 0 ) and the result follows from Corollary 2.3 by projection on the quad Q1 . If
x and x 0 are contained in a quad intersecting Q1 , then δ1 (x) = δ1 (x 0 ) and the result follows
2
from Corollary 2.3 by projection on the quad Q2 .
Theorem 7.3 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 Sx has a thin point and such that S y contains the same
number of lines for all y ∈ 0(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 2.6 implies
that the number of points in 02 (y) is independent of the point y of S. For y ∈ 0(x), let Vy
denote the set of quads through y. Now,
X
X
X
1,
s 2 tQ ,
tQ (tQ + 1),
Q∈Vy
Q∈Vy
Q∈Vy
are respectively equal to the number of quads through y, the number of points in 02 (y) and
t (t + 1), hence these quantities are independent of y ∈ 0(x). Let L x be a thin point of Sx
and let Q1 and Q2 be the two quads through L x with respective orders (s, t1 ) and (s, t2 ).
One has that t = t1 + t2 . Let z 6= x be a second point of L x . If y ∈ Q1 ∩ 0(x), then
X
X
(tQ − 1)(t2 − tQ ) =
(tQ − 1)(t2 − tQ ) = (t1 − 1)(t2 − t1 ).
Q∈Vy
Q∈Vz
Let Vy0 = Vy \{Q1 }, then
X
(tQ − 1)(t2 − tQ ) = 0.
Q∈Vy0
Since there are only t + 1 lines through y and Q1 has t1 + 1 lines through y, one has that
1 ≤ tQ ≤ t2 for all Q ∈ Vy0 . This implies that tQ = 1 or tQ = t2 for all Q ∈ Vy0 . By
Theorem 7.1, we may suppose that t1 , t2 6= 1. From
X
X
1=
1,
Q∈Vy
Q∈Vz
and
X
Q∈Vy
tQ =
X
tQ ,
Q∈Vz
it follows now that the number of quads Q of Vy0 with tQ = t2 is equal to 1. This implies
that S y ' St1 ,t2 for all y ∈ 0(x) ∩ Q1 . A similar reasoning shows that this is also true for
2
y ∈ 0(x) ∩ Q2 . The result follows now from the previous theorem.
226
DE BRUYN
Acknowledgment
I want to thank J.A. Thas for some discussions about generalized quadrangles and my
supervisor F. De Clerck for some advice while I was writing this manuscript.
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