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.
- 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
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
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
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)).