By Jean-Paul Pier
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Extra info for Amenable, locally compact groups
So we have nk − 1 = q(nk−1 − 1), from which it follows that nk = 1 + (n0 − 1)qk . So it is enough to prove the result for the case k = 0, that is, for a hyperbolic line. In the symplectic case, each of the q + 1 projective points on a line is isotropic. Consider the unitary case. We can take the form to be B((x1 , y1 ), (x2 , y2 )) = x1 y2 + y1 x2 , 42 where x = xσ = xr , r2 = q. So the isotropic points satisfy xy + yx = 0, that is, Tr(xy) = 0. How many pairs (x, y) satisfy this? If y = 0, then x is arbitrary.
Most are straightforward but n = 16 and n = 81 require some effort. 4 A technical result The result in this section will be needed at one point in our discussion of the unitary groups. It is a method of recognising the groups PSp(4, F) geometrically. Consider the polar space associated with PSp(4, F). Its points are all the points of the projective space PG(3, F), and its lines are the flat lines (those on which the symplectic form vanishes). We call them F-lines for brevity. Note that the F-lines through a point p of the porojective space form the plane pencil consisting of all the lines through p in the plane p⊥ , while dually the F-lines in a plane Π are all those lines of Π containing the point Π⊥ .
Moreover, any pair of distinct points spans either a flat subspace or a hyperbolic plane. Again, Witt’s Lemma shows that the group is transitive on the pairs of each type. ) Now a non-trivial equivalence relation preserved by G would have to consist of the diagonal and one other orbit. So to finish the proof, we must show: (a) if B(x, y) = 0, then there exists z such that B(x, z), B(y, z) = 0; (b) if B(x, y) = 0, then there exists z such that B(x, z) = B(y, z) = 0. This is a simple exercise. 2 Prove (a) and (b) above.