Hyperbolic Geometry by Birger Iversen

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10 A vector v E E gives rise to a transformation Ov E O(F) given by Ov(z,O = (z,( - ) ; z E E, E 02 Direct verification shows that Ov(e) = e + v, e E E. 2° A real scalar A E Jr gives rise to a transformation sA E O(F) given by sA(z,() = (z,aS) ; z E E, r; E R Direct verification gives us s'A(e) = A e, e E E. 3° An orthogonal transformation o E O(E) defines v E 0(F) given by v(z,C) = (o(z),C) ; z E E, S E R Another direct verification gives us o- = 0. We leave it to the reader to show that any orthogonal transformation of F can be decomposed into a product of transformations of the three types we have just considered.

10 that the last three conditions are equivalent. To proceed, introduce normal vectors H and K for two bounding spheres. 7 identifies the Mobius inversion r with the Lorentz reflection rK along the vector K. From the formula 7K(H) = H + 2K we conclude that rKK(H) = ± H if and only if = 0. This shows that condition 1° is equivalent to condition 3°. 5 The geometric meaning of this concept will be given in next section. 15 Let A and B be points of t conjugated with respect to the sphere I (by this we mean that A and B are points outside Y interchanged by inversion in Y).

N = a,,, By assumption the form is positive but singular. We shall eventually prove that F is parabolic. ,n. From the inequality Q(E xses) = E r,sarsxrxs >_ E r,s arslxr1Ixsl = Q(E Ix5Ies) it follows that we can assume that all coefficients of n are positive. ,n. 2 that the isotropic vector n is orthogonal to any vector: ; s = 0.... ,es> In particular we conclude that a1. = = 0 for s 0 J, contradicting the hypothesis that the matrix A is indecomposable. Let us prove that F l has dimension 1.