10 JOHN BAMBERG AND NICOLA DURANTE
subline b containing Q and Q . Now the Baer subplane spanned by b and
b is fully contained if and only if b has a point T in O. This implies that
T and Z are points of O collinear on
H(3,q2);
a contradiction.
Note that P
u
O consists of the points of the form (1,δ, 0, 0) together
with (0, 1, 0, 0). Suppose (1,δ,η, 0) is an element of both dual Baer sublines.
That is,
(1,δ,η)U(1,δq,ηq)T
= 0 and (1,δ,η)U
(1,δq,ηq)T
= 0. Now
(1,δ,η)U(1,δq,ηq)T
= (1,δ,η)
0 0 −ω
0 0 1
−ωq 1 0
(1,δq,ηq)T
=
−ηωq
+
ηδq
+ (−ω +
δ)ηq
= T (η(δ
ω)q),
(1,δ,η)U
(1,δq,ηq)T
= (1,δ,η)
0 0 −γν
0 0 γ
−γqνq γq
c
(1,δq,ηq)T
=
−ηγqνq
+
ηγqδq
+ (−γν + δγ +
ηc)ηq
=
−(ηγqνq
+
ηqγν)
+
(ηγqδq
+
ηqγδ)
+
cηq+1
= T
(ηγq(δ

ν)q)
+ cN(η).
Since 1 + N(δ) + N(η) = 0, we see that our equations become
(*) T(η(δ
ω)q)
= 0 and
T(ηγq(δ

ν)q)
= c(1 + N(δ))
So since the dual Baer sublines defined by U and U share only the points
of P
u
O, then whenever condition (*) holds for a choice of δ, η, we will
have η = 0. Therefore, we must have a priori that c = 0 and γ / GF(q).
Let η = (γν
γqω)q
and
δ =
−ηq
+
ηq−1γq(ω

ν)q
γq γ
.
Then a straightforward calculation shows that 1+ N(δ)+ N(η) = 0, T(η(δ
ω)) = 0 and T(ηγ(δ ν)) = 0, so condition (*) holds, and hence η = 0.
Therefore, ν =
ωγq−1
and
U =
0 0
−γqω
0 0 γ
−γωq γq
0
.
We want to show that U is conjugate to U under some element of SU3(q).
Now the group SU2(q) of invertible 2 × 2 matrices with unit determinant,
and fixing the form
X0Y0q +X1Y1q
= 0 on
GF(q2)2,
has q+1 orbits on totally
isotropic vectors of
GF(q2)2.
Each orbit consists of vectors (x, y) where
y/xq
attains a common value. Therefore, there exists some element C0 of SU2(q)
such that C0(−γν, γ)T = (−ω, 1). Let
C :=


C0 0
0
0 0
1


.
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