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.

  1. each element t of mH(2)m-1 can be expressed
    as t(x) = Ax for a non-singular matrix A,
  2. the related projective transformation t*([x])=[Ax] maps K to K,
    i.e. belongs to the projective symmetry group of K,
  3. every element of the projective symmetry group arises from
    an element of mH(2)m-1.

  1. 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].
  2. 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)).
  3. 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)).

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