On 2/13/08, Dan Asimov <dasimov@earthlink.net> wrote:
I don't think I ever considered this before, but it seems I just figured out all possible ways such that, given a real number c > 0, one can construct a bijection of the hyperbolic plane H^2 to itself that multiplies all distances by c:
f: H^2 -> H^2
such that dist(f(x),f(y)) = c dist(x,y) for all x, y in H^2.
(If the word isn't "homothety", then I withdraw it.)
I wonder if anyone else would care to tackle the same question.
For basic stuff see URL http://en.wikipedia.org/wiki/Poincar%C3%A9_disc_model Consider the Poincar\'e model of hyperbolic space, qua disc of unit radius centred at origin in the complex plane, using polar coordinates z = r exp(i t). Suppose there is a hyperbolic homothety z -> z' fixing the origin. We may assume it is isotropic at 0 (why?), that is fixing t: then t' = t. The infinitesimal distance element is ds = (2 r)/(1 - r^2) dt; so to give homothety scale factor c, we require (2 r')/(1 - r'^2) = c (2 r)/(1 - r^2), and solving r' = ( - (1 - r^2) + sqrt((1 - r^2)^2 + (2 c r)^2) ) / (2 c r). However, this gives only a necessary (tangential) condition on z -> z'. To incorporate radial distance, it is further required that arccosh(1 + q') = c arccosh(1 + q), where q = q(u, v) = 2 |u - v|^2 /(1 - |u|^2)(1 - |v|^2). Checking the specific case c = 1/2, u = 0, v = 1/2, we find r = 1/2, r' = (sqrt(13) - 3)/2, q = 0.6666666667, q' = 0.2018504248, arccosh(1 + q') / arccosh(1 + q) = 0.5690316077 which sadly fails to equal to c. The conclusion is that there are no nontrivial hyperbolic homotheties --- which is not too surprising, since (a) we would otherwise surely have heard of them already; and (b) there are none in spherical space. Fred Lunnon