By Peter J. Cameron

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**Extra info for Notes on Classical Groups [Lecture notes]**

**Sample text**

We say that a quadratic form q is nondegenerate if q(x) = 0 & (∀y ∈ V )B(x, y) = 0 ⇒ x = 0, where B is the associated bilinear form; that is, if the form q is non-zero on every non-zero vector of the radical. If the characteristic is not 2, then non-degeneracy of the quadratic form and of the bilinear form are equivalent conditions. Now suppose that the characteristic is 2, and let W be the radical of B. Then B is identically zero on W ; so the restriction of q to W satisfies q(x + y) = q(x) + q(y), q(λx) = λ2 q(x).

Consider a parallel class of hyperplanes in the affine space whose hyperplane at infinity is p⊥ . , p. 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.

The corresponding geometry is called a unital. For q = 2, the group has order 72, and so is soluble. In fact, it is sharply 2-transitive: a unique group element carries any pair of points to any other. 1 (a) Show that the unital associated with PSU(3, 2) is isomorphic to the affine plane over GF(3), defined as follows: the points are the vectors in a vector space V of rank 2 over GF(3), and the lines are the cosets of rank 1 subspaces of V (which, over the field GF(3), means the triples of vectors with sum 0).