This chap seems to have encountered some glitch in the system --- Rich? --- so I'm forwarding his contribution, without prejudice. WFL ---------- Forwarded message ---------- From: jean gallier <jean@cis.upenn.edu> Date: Jan 13, 2008 4:47 AM Subject: PSL(2, C) versus SO_0(1,3) To: Fred lunnon <fred.lunnon@gmail.com> Cc: Eugene Salamin <gene_salamin@yahoo.com>, Dan Asimov <dasimov@earthlink.net> Dear Colleagues Even though I receive e-mail from math-fun@mailman.xmission.com regularly, it seems that I do not have permission to post messages so I'm emailing you directly. Actually, Mobius(2) = PSL(2, C) = SL(2, C)/{-I, I} is isomorphic to the Lorentz group SO_0(1, 3), the connected component of the identity in SO(1, 3). It is wrong to quotient SO(1, 3) by the subgroup {-I, I}. This can be seen by defining a homomorphism from SL(2, C) to SO_0(1, 3) and then proving that this map is onto with kernel {-I, I}. Such a definition is not entirely trivial. We can proceed as follows: 1. Define a bijection h between R^4 and the 2 x 2 Hermitian matrices, Herm(2), as follows: t + x y - iz h: (t, x, y, z) |-> y + iz t - x (Note a linear combination of the Pauli matrices!) 2. Then, for every matrix A in SL(2, C), define the map l_A: Herm(2) -> Herm(2) by l_A(S) = ASA^* with S in Herm(2). We get a map, lor_A: R^4 -> R^4, via lor_A = h^{-1} o l_A o h. Since det(A) = 1 and det h(t, x, t, z) = t^2 - x^2 - y^2 - z^2, we deduce that lor_A in O((1, 3). We get a homomorphism, A |-> lor_A, from SL(2, C) to O(1, 3). However, it is continuous and SL(2, C) is connected so its range is in fact SO_0(1, 3). 3. Proving that this homomorphism is onto is not entirely obvious. I have a proof in some course notes (Proposition 4.21, see Chapter 4 in http://www.cis.upenn.edu/~jean/gbooks/manif.html.) I hope this helps clarify matters. -- Jean Gallier