the hyperbolic-projective theorem

 The Hyperbolic-Projective Theorem If m is the bijection from the hyperbolic model to the minkowski model, and K the projective conic with equation x2+y2-z2 = 0, then the group mH(2)m-1 is isomorphic to the projective symmetry group of K. The proof is quite complicated. We proceed by stages. each element t of mH(2)m-1 can be expressed as t(x) = Ax for a non-singular matrix A, the related projective transformation t*([x])=[Ax] maps K to K, i.e. belongs to the projective symmetry group of K, every element of the projective symmetry group arises from an element of mH(2)m-1. Proof Suppose that P(α,β,γ) is on H. From the Poincare-Minkowski Map, we have formulae for the maps m and m-1. From the hyperbolic geometry page, we know the structure of any element h of H(2). Thus it is possible to compute the coordinates of mohom-1(P). By an explicit calculation, we find that the coordinates of the image are linear in α, β and γ. Hence, t = mohom-1 has the form t(x) = Ax for some matrix A. Since h, and hence t, are invertible, the matrix A is non-singular. It therefore corresponds to the projective transformation t*([x]) = [Ax]. Suppose the matrix A is such that, for x on H, Ax is also on H. As x, Ax are on H, xTKx = xTATKAx = -1, so x is on the locus xTMx = 0, where M = K-ATKA. Now, H contains many points on each plane which cuts H, and many planes cut H. Since M is a symmetric matrix, it defines a conic or degenerate projective conic. Since it contains collinear p-points, it is degenerate. As it contains p-points of more than two p-lines, the equation must be trivial, i.e. M is the zero matrix. Thus, ATKA = K. It follows that A defines a member of the symmetry group of K. A simple calculaution with determinants shows that det(A) = ±1. The matrices A and B define the same projective transformation if and only if B = kA. If the matrices have determinant ±1, then B = A or -A. Only one of these maps the sheet H to itself, the other will map it to the second sheet. Thus the maps found in (1) give distinct projective transformation, so mH(2)m-1 is certainly isomorphic to a subgroup of S(K,P(2)). We must now show that each element of s of S(K,P(2)) arises from an element of H(2) by conjugation by m. From our work on the projective symmetry group, we know that s can be written in the form s([x]) = [Ax], with ATMA = K, and det(A) = 1. Then, the map t(x) = Ax maps H to H or to -H, the second sheet of the hyperboloid. In the latter case, we may replace A by -A to get a map from H to H. Suppose that t(x) = Ax maps H to H, and that t(0,0,1) = PεH. Now, (0,0,1) corresponds to O(0,0) in D and P to Q on D. By the Origin Lemma, is an element h of H(2) mapping O to Q. Then the element b = mohom-1 maps Z(0,0,1) to P. By (1), b has the from b(x) = Bx, for a non-singular matrix B. Then c = b-1ot fixes Z. The map c has matrix C = B-1A. By a matrix calculation such a matrix defines a rotation about the z-axis or a reflection in a plane through the z-axis. It is geometrically clear that each type arises from a similar (euclidean) rotation or reflection in H(2), and so is in mH(2)m-1. Thus, t = boc is in mH(2)m-1. Each such t corresponds to a unique element of S(K,P(2)), so that, finally,                 mH(2)m-1 is isomorphic to S(K,P(2)).