By Flaass D.G.

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From these equations we deduce that b + b = c + c = 0, that is, b, c ∈ K, while a − a = d − d = 0, that is, a, d ∈ F0 . Choose a fixed element u ∈ K. Then λ ∈ K if and only if uλ ∈ F0 . Also, −1 u ∈ K. Hence the matrix P† = a ub u−1 c d belongs to SL(2, F0 ). Conversely, any matrix in SL(2, F0 ) gives rise to a matrix in SU(2, F0 ) by the inverse map. So we have a bijection between the two groups. It is now routine to check that the map is an isomorphism. Represent the points of the projective line over F by F ∪ {∞} as usual.

Then induction shows the result for higher rank spaces over GF(4). Again, the argument in 3 dimensions shows that transvections are commutators; the action on the points of the polar space is primitive; and so Iwasawa’s Lemma shows the simplicity. 64 6 Orthogonal groups We now turn to the orthogonal groups. These are more difficult, for two related reasons. First, it is not always true that the group of isometries with determinant 1 is equal to its derived group (and simple modulo scalars). Secondly, in characteristic different from 2, there are no transvections, and we have to use a different class of elements.

Over the real numbers, Sylvester’s theorem asserts that any quadratic form in n variables is equivalent to the form 2 2 , − . . − 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.

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