If we also suppose that the points of
π and
π are contained
in the dual Baer subline defined by U, then we can solve for α and γ (but
the expressions might be ugly!). Here we explore a simple example where
A : (0, 0, 1,ω). Now
∩π has points of the form (r, s, 0, 0), N(r)+N(s) = 0.
So if (r, s, 0, 0) also satisfies (r, s,
0)U(rq,sq, 0)T
= 0, then
(r, s,
0)U(rq,sq, 0)T
= 0 for every (r, s, 0, 0) satisfying N(r)+N(s) = 0.
In particular, δ is forced to be zero. Therefore, we can write
U :

0 0 −γω
0 0 γ
−γqωq γq

2.1. Proof of the first part of Theorem 1.2. Here we prove that
the incidence structure Γ of Theorem 1.2 is a generalised hexagon. Our
approach is to use a definition of a generalised hexagon which is equivalent
to the one stated in the introduction: (i) it contains no ordinary k-gon
for k {2, 3, 4, 5}, (ii) any pair of elements is contained in an ordinary
hexagon, and (iii) there exists an ordinary heptagon (see [22, §1.3.1]). A
thick generalised polygon has order (s, t) if every line has s + 1 points and
every point is incident with t + 1 lines. A counting argument shows that if
we know that the number of points and lines of a generalised hexagon are
(s + 1)(1 + st +
and (t + 1)(1 + st +
then the conditions (ii) and
(iii) automatically follow from the first condition.
Proof. First we show that Ω induces a point-partition of each generator
(minus its point in the Hermitian curve O). Let be a generator of
and let P be a point of \O. For a point X, we will let
be the q + 1
elements of Ω which lie on X. Consider the q + 1 elements P

of Ω on
P . Since P

covers the points of a Baer subplane, it follows that there is a
unique element of Ω contained in and containing P . Therefore Ω induces
a point-partition of each generator minus its point in the Hermitian curve
O. It follows immediately that Γ is a partial linear space (i.e., every two
points lie on at most one line).
is a generalised quadrangle, Γ has no triangles. So suppose
now that we have a quadrangle R, S, T , U of Γ. Then at least three of these
points are necessarily affine points. For example, if two of these points were
of type (a), two points of type (b), and with one line of type (i) and three
of type (ii) making up the quadrangle, the three lines of type (ii) would
yield a triangle of generators. So this case is clearly impossible. At least
three points, S, T , U say, are necessarily affine points and the lines of the
quadrangle are elements of Ω. Moreover, R is also an affine point, since if
R were a generator then S and U would lie on R and ST , TU, SU would
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