Dan, Indeed, that's very interesting. Observe that Spin(4) = Spin(3) x Spin(3) has an obvious centre of order 4 (namely {(1,1), (1,-1), (-1,1), (-1,-1)}). Every subgroup of this is normal (that's how centres work), so we get five nice normal subgroups. Considering their quotients, we have: Subgroup: {(1,1)} Quotient: Spin(4) = Spin(3) x Spin(3) Subgroup: {(1,1), (-1,1)} Quotient: Spin(3) x SO(3) Subgroup: {(1,1), (1,-1)} Quotient: SO(3) x Spin(3) Subgroup: {(1,1), (-1,-1)} Quotient: SO(4) Subgroup: {(1,1), (1,-1), (-1,1), (-1,-1)} Quotient: PSO(4) = SO(3) x SO(3) Consequently, SO(4) enjoys the status of being the double cover of SO(3) x SO(3), and having Spin(3) x Spin(3) as its own double cover! It seems really weird for a group like SO(4) to be neatly sandwiched between two direct squares in this way. Sincerely, Adam P. Goucher
Sent: Tuesday, July 29, 2014 at 12:49 AM From: "Dan Asimov" <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] sesqui-imponderables
Hello, Adam.
Noticed that CP^4 recently discussed SO(4) in terms of quaternions, and the fact that its double cover is S^3 x S^3.
Thought I'd mention an interesting fact about this (that had me confused for a while ca. 25 years ago):
SO(4) is actually homeomorphic is SO(3) x Spin(3), i.e., to P^3 x S^3. (For any g in SO(4), map it to g(1) (=g) in S^3. The stabilizer of this point is clearly a copy of SO(3). So topologically, SO(4) is an SO(3) bundle over S^3. This must be a trivial bundle over each D^3 hemisphere, so is determined by the identification of these two trivial bundles along the S^2 equator. But pi_2 of any Lie group is the trivial group, and so the bundle over S^3 is trivial to begin with.
But, as you may know, SO(4) is not isomorphic to SO(3) x Spin(3) as a Lie group.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun