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.