On Aug 29, 2014, at 10:47 AM, Adam P. Goucher <apgoucher@gmx.com> wrote:
Yes, but they differ in how we associate (some compactification of) R^n with (some quotient of) S^n.
Not a problem. The projective group and the conformal group both *naturally* reside on (the round) S^n. That's where they begin life. So both groups are naturally subgroups of the diffeomorphism group Diffeo^w(S^n) (here w indicates that the diffeomorphism are real analytic). Hence elements of both groups can be naturally composed with each other So it makes perfect sense to ask what subgroup of Diffeo^w(S^n) is generated by the two groups together -- as Fred asked. Neither group naturally resides on R^n. That is, some -- in fact almost all -- group elements of either group on S^n fail to take R^n to itself. To get the effect of any element g of either group on R^n, we just restrict it to S^n - {oo}. But then we also have to be careful to also remove any point taken to oo by g. (When we do this, we identify S^n - {oo} with R^n by inverse stereographic projection, which is conformal.) Which means neither group acts honestly on R^n in the first place. The best we can do is let them be pseudogroups on R^n. (We can always restrict either group to those transformations on S^n that carry oo to itself, and those elements do form a group on R^n.)
There is, however, a rather nice fact:
Let G be the group of projective transformations of R^(n+1) which preserve S^n. Let H be the group of conformal transformations of R^(n+1) which preserve S^n.
Then G and H are isomorphic, and we can choose the isomorphism such that they act on S^n in the same way.
I've been suspecting that in fact the conformal and projective groups on S^n, as I've described them, are identical. Have not proved this yet. But if it's true, it must be well-known. --Dan