The above lemma allows us to attach a value to a Baer subgenerator
that is an invariant for the action of SU3.
Definition 2.4 (Norm of a Baer subgenerator). Let O be the Hermitian
π, where π is the hyperplane X3 = 0 of
and let
GO be the stabiliser of O in PGU4(q). Fix a Baer subgenerator b0 of
with a point in O. Let b be a Baer subgenerator of
with a point in
O, and suppose MA is an element of GO such that b = b0
. Then the norm
of b is
b := det(A).
Moreover (by Lemma 2.3), the map b b induces a group homomorphism
φ from GO to the multiplicative subgroup of elements of
N(x) = 1.
Note that the kernel of φ is SU3. The homomorphism φ is surjective and
hence there is a natural partition of Baer subgenerators with a point in O
into q + 1 classes. Each orbit of SU3 consists of Baer subgenerators with a
common value for their norm.
Lemma 2.5. Let μ be an element of
such that N(μ) = 1. Let O
be a Hermitian curve of
defined by X3 = 0, and let Ω be a set of
Baer subgenerators with a point in O which have norm equal to μ. Then:
(i) Every affine point is on q + 1 elements of Ω covering a Baer sub-
(ii) For every point X O and for every affine point Y in
is a unique element of Ω through X and Y .
Proof. Recall that Ω is an orbit of SU3 on Baer subgenerators and SU3
acts transitively on the affine points
and so clearly every affine
point is on q+1 elements of Ω. Moreover, such a set of q+1 elements of Ω will
cover a Baer subplane, as we show now. Let Y be an affine point, let Y

the set of q +1 elements of Ω through Y and let b0 be one particular element
of Y
Then every other element of Y

is in the orbit of b0 under the stabiliser
of Y in SU3. Now for every g (SU3)Y , we know that b0
= b0

and so every element of Y

lies in the plane Y
At infinity, Y

meets O in
a Baer subline and so we have a triangle of Baer sublines spanning a Baer
subplane of Y
and it is covered completely by the elements of Y
To complete the proof, we need only prove (ii). Since the stabiliser of
a point in O is transitive on the set of affine points in the perp of that
point, we can assume that X = (1,ω, 0, 0) and Y = (0, 0, 1,ω) for some ω
satisfying N(ω) = −10. We have already seen, in the proof of Lemma 2.3,
that X and Y lie on a Baer subgenerator, which we can assume without loss
of generality, is in Ω. This Baer subgenerator is uniquely defined by a 3 × 3
Hermitian matrix U of rank 2 and the generator spanning X and Y , and
we assume (as before) that U has the form
U :=
0 0 −ω
0 0 1
1 0
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