By Irving Kaplansky

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A) Since k ≥ 1 and n − k ≥ 1, there exist both timelike and spacelike vectors in Rkn . Suppose that v is a unit timelike vector and w is a unit spacelike vector such that v, w = 0. Then v + w and v − w are both lightlike. By the hypothesis of the theorem, A(v + w) and A(v − w) are both lightlike. 1) 0 = A(v − w), A(v − w) = Av, Av − 2 Av, Aw + Aw, Aw . If we subtract the second equation from the first, we get Av, Aw = 0. 2) for some real number λ. Now suppose that {v1 , . . , vk , w1 , . . , wn−k } is an orthonormal basis for Rkn with v1 , .

It is clearly diffeomorphic to S 1 × S n , where S 1 is 2 the circle with equation x12 + xn+3 = 1 in the timelike plane spanned by e1 and en+3 , n and S is the n-sphere given by the equation 2 x22 + · · · + xn+2 = 1, in the Euclidean space R n+1 spanned by {e2 , . . , en+2 }. The manifold M n+1 is a double covering of Qn+1 , and the fiber containing the point x ∈ M n+1 is the orbit of x under the action of the group Z2 = {±I }. Suppose that x = (w, z) is an arbitrary point of S 1 × S n = M n+1 .

Xn+3 ) with x1 + x2 = 1, and let Ax = y = (y1 , . . , yn+3 ). 22) 42 3 Lie Sphere Transformations By our choice of A, we know that y1 + y2 = 1 also. Thus x and y are determined by X = (x3 , . . , xn+3 ) and Y = (y3 , . . 18). Since ai1 = ai2 and x1 + x2 = 1, we have n+3 yi = ai1 x1 + ai2 x2 + n+3 aij xj = ai1 + j =3 aij xj . 24) where B is the invertible linear transformation of R1n+1 represented by the matrix [aij ], 3 ≤ i, j ≤ n + 3, and C is the vector (a31 , . . , a(n+3)1 ). The fact that this transformation must preserve oriented contact of spheres implies further that T must preserve the relationship (X − Z, X − Z) = 0.