The Poincare-Klein Map
- The map k is given by k(x,y) = (2x/(1+x2+y2),2y/1+x2+y2)).
- The map k maps the h-line with boundary points AB to the
intersection of the euclidean line AB with D.
- The inverse is
Suppose that (x,y) is in D. This corresponds to P(x,y,0) in R3.
In vector notation, the line ZP has equation tp+(1-t)z = 0,
i.e. each point on ZP has the form (tx,ty,(1-t)), for some t.
This meets S where (tx)2+(ty)2+(1-t)2 = 1, which simplifies to
t2(x2+y2+1) = 2t. The solution t = 0 corresponds to Z. The
other solution, t = 2/(1+x2+y2), corresponds to P'. Thus we have
We obtain P" by replacing the z-coordinate by 0.
Thus, we have k(x,y) = (2x/(1+x2+y2),2y/(1+x2+y2)).
For part (2), we show that the points which map to a line segment
lie on an h-line. The point k(x,y) in D is on the line L : ax+by+c = 0
if and ony if 2ax/(1+x2+y2)+2by/(1+x2+y2)+c = 0,
i.e. 2ax+2by+c((1+x2+y2)) = 0.
If c = 0, this reduces to ax+by = 0, i.e.
If c ≠ 0, it is becomes x2+y2-2(-a/c)x-2(-b/c)y+1 = 0.
The distance of O from the line L is d = |c|/(a2+b2)½
Thus, L cuts D if and only if d < 1, i.e.
(-a/c)2+(-b/c)2 > 1.
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.
we see that the points on L lie on a circle C orthogonal to C.
Since k maps D to D, it maps the h-line
CnD to the segment LnD.
Finally, the map k is defined on the boundary C, and fixes each
point of C -
if P is on C, then P = P' = P" in the definition of k.
Thus, the boundary points ofthe h-line are the end-points of the
For part (3), we apply the projections in reverse order.
If P" is the point (x,y,0), the orthogonal projection maps it to the
point P'(x,y,z), where x2+y2+z2 = 1 and z < 0.
so z = (1-x2-y2)½.
For brevity, we will use z in our calculations. Then any point ZP'
has the form (tx,ty,tz+1-t). This cuts the xy-plane when t = 1/(1+z),
i.e. at the point (x/(1+z),y/(1+z)), as required.