4 JOHN BAMBERG AND NICOLA DURANTE
(see [16, 3.2.3]). To construct
H(3,q2)
given a prime power q, we take a
non-degenerate Hermitian form such as
X, Y =
X0Y0q
+
X1Y1q
+
X2Y2q
+
X3Y3q
and the totally isotropic subspaces of the ambient projective space, with
respect to this form. Most of the material contained in this section can be
found in Barwick and Ebert’s book [2] and Hirschfeld’s book [11, Chapter
7].
Every line of
PG(3,q2)
is (i) a generator (i.e., totally isotropic line) of
H(3,q2),
(ii) meets
H(3,q2)
in one point (i.e., a tangent line), or (iii) meets
H(3,q2)
in a Baer subline (also called a hyperbolic line). A Baer subline of
the projective line
PG(1,q2)
is a subset of q + 1 points in
PG(1,q2)
which
form a GF(q)-linear subspace. We may also speak of Baer subplanes and
Baer subgeometries of
PG(3,q2)
as sets of points giving rise to projective
subgeometries isomorphic to PG(2,q) and PG(3,q) respectively. A Baer
subgenerator of
H(3,q2) is a Baer subline of a generator of H(3,q2). We
will often use the fact that three collinear points determine a unique Baer
subline ([2, Theorem 2.6]) and a planar quadrangle determines a unique
Baer subplane ([2, Theorem 2.8]). In particular, if b and b are two Baer
sublines of
PG(2,q2)
sharing a point, but not spanning the same line, then
there is a unique Baer subplane containing both b and b . We say that it is
the Baer subplane spanned by b and b .
One class of important objects for us in this paper will be the degenerate
Hermitian curves of rank 2. Suppose we have a fixed hyperplane, π : X3 = 0
say, meeting
H(3,q2)
in a Hermitian curve O. Let be a generator of
H(3,q2).
Then the polar planes of the points on meet π in the
q2
+ 1 lines through
L := ∩O. Now suppose we have a Baer subgenerator b contained in , and
containing the point L. Then the polar planes of the points of b meet π in
q + 1 lines through the point L giving a dual Baer subline of π with vertex
L. Moreover, the points lying on this dual Baer subline define a variety
with Gram matrix U; a Hermitian matrix of rank 2. So they correspond to
solutions of
XU(Xq)T
= 0 where U satisfies U
q
= U
T
. For example, if we
consider a point P in π, say (1,ω, 0, 0) where N(ω) = −1, and two points
A : (a0,a0ω, a2, 1), B : (b0,b0ω, b2, 1) spanning a line with P , then P, A, B
determine a Baer subline. In fact, if we suppose B = P + αA for some
α
GF(q2)∗,
then this Baer subline is {A}∪{p + t · αa | t GF(q)} where
A = a and P = p .
Let u be the polarity defining
H(3,q2).
Since P is precisely the nullspace
of U, and the tangent line P
u
π is contained in the dual Baer subline, it is
not difficult to calculate that U can be written explicitly as
U :





−δωq
δ −γω⎟
δq δqω
γ
−γqωq γq
0




,
δωq
=
δqω.
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