----- Original Message ---- From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Saturday, January 12, 2008 2:21:36 PM Subject: Re: [math-fun] Moebius madness on Youtube On 1/12/08, Eugene Salamin <gene_salamin@yahoo.com> wrote:
----- Original Message ---- From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Friday, January 11, 2008 12:54:07 PM Subject: Re: [math-fun] Moebius madness on Youtube
Nicely done. And it grieves me to have to admit that, although I've been concerned with these groups for a good few years, I had no idea that the isomorphism of the Moebius group in n-space with the Euclidean group
in
(n+1)-space had such a straightforward demonstration! WFL
On 1/11/08, Henry Baker <hbaker1@pipeline.com> wrote:
Very cool animation of Moebius transformations:
Those two groups are not isomorphic. The Euclidean group has a normal subgroup (the translations), while the Mobius group is semisimple.
Gene
About ten years ago, I read somewhere (Coolidge?) that they were isomorphic (qua Lie groups, presumably); I remember being very surprised at the time, but never got around to investigating the matter until now. If this claim turns out to be wrong, that extricates me very conveniently from my elephant trap. I'd also have the consolation of company: Coolidge relates with sadistic relish how an unfortunate Ph. D. student by the name of Lohrl expended considerable time and effort in a futile attempt to prove that Moebius n-space and Lie-sphere (n-1)-space groups were isomorphic. [In that case at least, it seems perfectly obvious that they cannot possibly be; since only the latter group has a 1-parameter normal subgroup, to wit the offset transformations.] In the present case --- Moebius n-space and Euclidean (n+1)-space, continuous components say --- we have a transformation (stereographic projection) inducing a smooth map from isometries in (n+1)-space acting on the projecting sphere (though its north pole remains northwards), to conformalities in n-space. This map is a homomorphism, composition mapping to composition (I _think_ this is obvious!). It must surely be invertible, since its codomain contains generators for the whole Moebius group: translations, rotations, (proper) inversions. Therefore the groups are isomorphic ... no? Unless of course, there's a mistake somewhere in the reasoning above. And if there isn't, I'm right back there, still sitting in deep do-dos ... Fred Lunnon _______________________________________________ I should have been more specific. The Euclidean group in 3-space contains the 3-dimensional abelian normal subgroup of translations. The Mobius group on the 2-sphere (thanks, Dan), being semisimple cannot contain a solvable normal subgroup. The Mobius group has a 2-dimensional abelian subgroup (of translations preserving infinity), but it is not normal, since the fixed point, infinity, can be conjugated to any other point. In the YouTube video, translation of the sphere normal to the plane induced the magnification Mobius transformation. But magnification by factor m and translation by t do not commute: m(z+t) /= mz + t. So the YouTube map is not an isomorphism. (Is it of any interest to think about mappings between groups that are not homomorphisms?) Fred, what is a Lie sphere? While Googling for the answer, I discovered the following Indra's Pearls web site. http://klein.math.okstate.edu/IndrasPearls/ Gene ____________________________________________________________________________________ Never miss a thing. Make Yahoo your home page. http://www.yahoo.com/r/hs