8 JOHN BAMBERG AND NICOLA DURANTE

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

curve

H(3,q2)

∩ π, where π is the hyperplane X3 = 0 of

PG(3,q2)

and let

GO be the stabiliser of O in PGU4(q). Fix a Baer subgenerator b0 of

H(3,q2)

with a point in O. Let b be a Baer subgenerator of

H(3,q2)

with a point in

O, and suppose MA is an element of GO such that b = b0

MA

. 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

GF(q2)∗

satisfying

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

GF(q2)

such that N(μ) = 1. Let O

be a Hermitian curve of

H(3,q2)

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-

plane.

(ii) For every point X ∈ O and for every aﬃne point Y in

Xu,

there

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

H(3,q2)\O,

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

∗

be

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

g

= b0

g

∈ Y

⊥

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

−ωq

1 0

.