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Jun 15, 2020 at 7:27 history edited CommunityBot
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Feb 1, 2011 at 18:28 comment added Qfwfq @Emerton: thanks for the explanation.
Feb 1, 2011 at 3:09 vote accept Daniel Briggs
Jan 31, 2011 at 17:10 comment added Daniel Briggs @Mark Sapir: No, I guess that does about wrap it up, without a need to change to the context of category theory. I will accept Emerton's answer later today.
Jan 31, 2011 at 14:23 comment added Emerton Dear unknowngoogle, it's not the case case that $SO(3) \times SO(3)$ is isomorphic to $SO(4)$; there is a central isogeny from the latter to the former, with kernel of order 2. (One has a succession of degree 2 central isogenies $SU(2) \times SU(2) \to SO(4) \to SO(3) \times SO(3)$. The latter isogeny induces the isomorphism $PSO(4) \to PSO(3) \times PSO(3) = SO(3) \times SO(3)$ which the OP is referring to.)
Jan 31, 2011 at 12:52 comment added Qfwfq Also, why are you using $PSO_3$, $PSO_4$ and not just $SO_3$, $SO_4$ ?
Jan 31, 2011 at 3:40 answer added Emerton timeline score: 20
Jan 30, 2011 at 23:39 comment added Qfwfq It's a standard (easy) fact that the quaternions that square to −1 form a 2-sphere in the purely imaginary quaternions. And automorphisms act transitively on that sphere...
Jan 30, 2011 at 23:39 comment added user6976 Is that true that the 2 lines of Mariano's comments are equivalent to the 40+ lines of the question? Or did I miss something important?
Jan 30, 2011 at 22:34 comment added Mariano Suárez-Álvarez Do you have an use for your "vision"? If what you want is simply to know whether "all units are created equal", then you can simply prove that the automorphism group of the algebra acts transitively on the set of elements which square to $-1$.
Jan 30, 2011 at 21:44 history edited Daniel Briggs CC BY-SA 2.5
added 365 characters in body
Jan 30, 2011 at 21:06 history edited Daniel Briggs CC BY-SA 2.5
changed category-theoretic context at end; added 13 characters in body
Jan 30, 2011 at 20:56 history asked Daniel Briggs CC BY-SA 2.5