klein view page

A disk model for euclidean geometry - mappings and equations

We can actually produce a mapping from the usual euclidean plane E to the disk D
in which the images of the euclidean lines appear as the e-lines. The diagram below
shows how this can be done.
We embed E as the plane z=-1 in R³. We embed D as the disk x²+y² < 1, z=0.
O is the origin (0,0,0), Z the point (0,0,1) and S the lower half of the unit sphere,
i.e. S is x²+y²+z²=1, z<0.
Take P on E, and project centrally onto P' on S i.e. P' is where OP meets S.
Now project P' to P" on D using Z, i.e. P" is where ZP' cuts the plane z=0.
We then define a map f from E to D taking P to P". It is clearly bijective.
In the picture, each point for E has the form (x,y,-1), and each point for D
the form (x,y,0) with x²+y²<1. We simply omit the z-coordinates to get a
map f from the plane E to the disk D. We can use real or complex coordinates.

geometrical description
Let L be a line on E, and let S* be the unit sphere centred at O.
The central projection maps L to the circle C on S* where the plane through
L and O cuts S*. This great circle cuts the equator C: x²+y²=1, z=0 in
diametrically opposite points A,B. Now the polar (stereographic) projection
from S to D maps the arc of C on S to an arc of an i-line through A and B,
i.e. to an e-line. See the stereographic projection page.

algebraic description
theorem
For (x,y)εE, f(x,y) = (x/(1+A),y/(1+A)), where A = (1+x²+y²)^½,
For (x,y)εD, f^-1(x,y) = (2x/B,2y/B), where B = 1-x²-y².
proof
Take P=(x,y,-1). The central projection maps P to P'=(tx,ty,-t) on S.
Hence t²(x²+y²+1) = 1. Also as P' has z< 0, we need the positive root.
Thus, P' = (x/A,y/A,-1/A).
The second, polar, projection maps (x,y,z) (z<0) to (sx,sy,sz+1-s) on the
plane z=0. This requires s = 1/(1-z), so we get (x/(1-z),y/(1-z),0).
Composing there two mappings and ignoring the z-coordinate gives f.
Verifying that the inverse is as stated is simply a matter of composing
f with the putative f^-1 and checking that the result is the identity.
corollary
If K is the euclidean line ax+by+c=0, (a,b)≠(0,0), then f(K) is the e-line
L_nD, where L is the i-line c(x²+y²)-2ax-2by-c=0.
proof
Observe that (x,y)εF(K) if and only if f^-1(x,y)εK. From the theorem,
this is equivalent to 2ax+2by+c(1-x²-y²) = 0, as required.
If c=0, then L is 2ax+2by=0, which clearly cuts C diametrically.
If c≠0, then L meets C where 2ax+2by=0, again cutting C diametrically.
Thus, we get an e-line in every case.

main model page

We can actually produce a mapping from the usual euclidean plane E to the disk D in which the images of the euclidean lines appear as the e-lines. The diagram below shows how this can be done. We embed E as the plane z=-1 in R³. We embed D as the disk x²+y² < 1, z=0. O is the origin (0,0,0), Z the point (0,0,1) and S the lower half of the unit sphere, i.e. S is x²+y²+z²=1, z<0. Take P on E, and project centrally onto P' on S i.e. P' is where OP meets S. Now project P' to P" on D using Z, i.e. P" is where ZP' cuts the plane z=0. We then define a map f from E to D taking P to P". It is clearly bijective. In the picture, each point for E has the form (x,y,-1), and each point for D the form (x,y,0) with x²+y²<1. We simply omit the z-coordinates to get a map f from the plane E to the disk D. We can use real or complex coordinates. geometrical description Let L be a line on E, and let S* be the unit sphere centred at O. The central projection maps L to the circle C on S* where the plane through L and O cuts S. This great* circle cuts the equator C: x²+y²=1, z=0 in diametrically opposite points A,B. Now the polar (stereographic) projection from S to D maps the arc of C on S to an arc of an i-line through A and B, i.e. to an e-line. See the stereographic projection page.
algebraic description theorem For (x,y)εE, f(x,y) = (x/(1+A),y/(1+A)), where A = (1+x²+y²)^½, For (x,y)εD, f^-1(x,y) = (2x/B,2y/B), where B = 1-x²-y². proof Take P=(x,y,-1). The central projection maps P to P'=(tx,ty,-t) on S. Hence t²(x²+y²+1) = 1. Also as P' has z< 0, we need the positive root. Thus, P' = (x/A,y/A,-1/A). The second, polar, projection maps (x,y,z) (z<0) to (sx,sy,sz+1-s) on the plane z=0. This requires s = 1/(1-z), so we get (x/(1-z),y/(1-z),0). Composing there two mappings and ignoring the z-coordinate gives f. Verifying that the inverse is as stated is simply a matter of composing f with the putative f^-1 and checking that the result is the identity. corollary If K is the euclidean line ax+by+c=0, (a,b)≠(0,0), then f(K) is the e-line L_nD, where L is the i-line c(x²+y²)-2ax-2by-c=0. proof Observe that (x,y)εF(K) if and only if f^-1(x,y)εK. From the theorem, this is equivalent to 2ax+2by+c(1-x²-y²) = 0, as required. If c=0, then L is 2ax+2by=0, which clearly cuts C diametrically. If c≠0, then L meets C where 2ax+2by=0, again cutting C diametrically. Thus, we get an e-line in every case.