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 = bMA. 0 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 b g 0 = 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 .
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