By P. M. Cohn
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Extra info for Lie Groups
D) Any three non-collinear points lie in a unique set p⊥ . 5 Prove conditions (b)–(d). Conditions (a)–(d) guarantee that the geometry of F-lines and H-lines is a projective space, hence is isomorphic to PG(3, F) for some (possibly non-commutative) field F. Then the correspondence p ↔ p⊥ is a polarity of the projective space, such that each point is incident with the corresponding plane. By the Fundamental Theorem of Projective Geometry, this polarity is induced by a symplectic form B on a vector space V of rank 4 over F (which is necessarily commutative).
Xr+s x12 + . . + xr2 − xr+1 for some r, s with r + s ≤ n. If the form is non-singular, then r + s = n. If both r and s are non-zero, there is a non-zero singular vector (with 1 in positions 1 and r + 1, 0 elsewhere). 8 If V is a real vector space of rank n, then an anisotropic form on V is either positive definite or negative definite; there is a unique form of each type up to invertible linear transformation, one the negative of the other. The reals have no non-identity automorphisms, so Hermitian forms do not arise.
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. If y = 0, then a fixed multiple of x is in the kernel of the trace map, a set of size q1/2 (since Tr is GF(q1/2 )-linear). , q1/2 + 1 projective points. Finally, consider the orthogonal case. The quadratic form is equivalent to xy, and has two singular points, (1, 0) and (1, 0) .