- #1

geodesic42

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I rarely hear a mathematician talk about a Lie group and its generating algebra as related via the exponential map, though I have seen it somewhere. When I do see it, it requires more background development than I have immediate time for.

I ALWAYS see physicists talk about this relation (indeed, almost all physics-oriented math texts define the algebra by starting with the group and writing down the exponential). So I've heard many, many times: the exponentiation of the algebra gives the connected part of the group.

I don't believe this (or at least am confused by this). At least at an abstract level. Given how universal it is, I'd like to be convinced otherwise.

My first and most nit-picky gripe is: multiplication isn't defined on the algebra; a bracket is, which becomes a commutator when we find a rep and thus specifies only half of the multiplication operator (i.e. exp(algebra) makes no sense until we find a rep).

Second, su(2)=so(3) (the algebras, not the groups). exp is a 1-1 map. So we should find exp(su(2))=exp(so(3))...but this isn't the case. Choose [tex]\frac{1}{2}\{\sigma^i\}[/tex] as our generators for the 2-D rep, and we indeed get SU(2)'s multiplication table. Choose the canonical generators for the 3-D rep and we get SO(3), which obviously does not have SU(2)'s multiplication table. SU(2) and SO(3) are both completely connected (though SU(2) is simply connected), so the whole "it generates the connected piece" bit doesn't solve anything. So how can we possibly say Group=exp(Algebra) if different groups have the same algebra? I'm pretty sure the answer must be that the group we get out of exponentiating is rep-dependent since the other half of the algebra's multiplication table isn't specified until we choose one anyway.

tl;dr: How can we possibly say Group=exp(Algebra) if different groups have the same algebra?

Thanks!