The Poincare-Minkowski Map
- The map m is given by
m(x,y) = (2x/(1-x2-y2),2y/(1-x2-y2),
- The map m maps an h-line to the
intersection of H with a plane through O.
- The inverse is given by
(1) A point (x,y) on D corresponds to P(x,y,0) in three-dimensional space.
A point on PZ' has the form t(x,y,0)+(1-t)(0,0,-1) = (tx,ty,t-1) for some real t.
This meets H where (tx)2+(ty)2-(t-1)2 = -1,
i.e. 2t = t2(1-x2-y2).
The root t=0 corresponds to Z' (which lies on the other sheet of the hyperboloid).
The root t=2/(1-x2-y2) gives the required point m(x,y) of H.
(3) A point P(x,y,z) lies on H has z2 = 1+x2+y2.
Since z > 0, z = (1+x2+y2)½.
Each point on PZ' has the form t(x,y,z)+(1-t)(0,0,-1) = (tx,ty,tz-1+t).
This meets the xy-plane where t=1/(1+z), i.e. at (x/(1+z),y/(1+z),0),
which corresponds to the point (x/(1+z),y/(1+z)) of D. Note that we have
z = (1+x2+y2)½ > |x| and |y|, so that each point of H
a point of D.
(2) Suppose that the plane Π : ax+by+cz=0 actually cuts H.
By(3), these are the images of a non-empty set of points on D.
By (1), m(x,y) lies on Π if and only if
+c(1+x2+y2)/(1-x2-y2) = 0.
i.e. 2ax+2by+c(1+x2+y2) = 0.
If c = 0 this is the line ax+by = 0, so is an h-line.
If c ≠ 0, the locus has equation x2+y2+2(a/c)x+2(b/c)y+1 = 0.
The locus is non-empty, so (a/c)2+(b/c)2 > 1, so, by the
The circle C : x2+y2-2ax-2by+c = 0 is orthogonal to the circle
C : x2+y2 = 1
if and only if c = 1 and a2+b2 > 1.
the locus is the intersection of D with a circle orthogonal to C - an h-line.