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