Sorry if I was unclear. Yes, S^3 here is the group of quaternions of norm one. And yes, the multiplication is in the quaternions. But what I left unsaid is that the y in the equation now labeled (*) below is a "pure" quaternion -- i.e., y has real part = 0. The pure quaternions H_0 form a copy of R^3. Then for any unit quaternion p, the map g(p): H_0 -> H_0 defined by g(p)(y) := p y p^(-1) acts as a rotation on the of pure quaternions, and so g(p) belongs to SO(3). --Dan Andy wrote: << I wrote: << On Fri, Nov 18, 2011 at 5:06 PM, Dan Asimov <dasimov@earthlink.net> wrote: Interesting subject. Yes, the map f: S^3 x S^3 -> SO(4) via f(p,q)(y) := p y q^(-1),

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So you're doing this multiplication in the quaternions, interpreting S^3 as the quaternions of norm 1, right? << Likewise, g: S^3 -> SO(3) via (*) g(p)(y) := p y p^(-1)

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But this has me baffled; how do you multiply a quaternion by a vector in R^3?

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