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I posed myself a question a while ago: are there any non-commuting one parameter Lie groups?

I'm thinking that there are no non-commuting connected Lie groups (not sure how to proceed to the disconnected case yet), but I want to be sure. My argument was based on classification of connected one 1-manifolds. My thought was that any connected one-parameter Lie group has to be homeomorphic to the line or the circle. I figured that since there are unique differential structures on each and since we know that the Lie group has to have a differential structure, and since the line and the circle are not diffeomorphic to each other, that the Lie group has to diffeomorphic to either the line or the circle. That would suggest that every connected one-parameter Lie group is equivalent (diffeomorphic and homomorphic at the same time, whatever fancy morphism that would be) to either good ol' translation in one dimension or SO2. In either case, the group commutes. Have I made any fallacies? I want to double check because I'm new to Lie groups and differential topology.

I'm fairly sure you can prove my conjecture (no noncommuting connected one-parameter Lie groups) easily from Lie algebra axioms and the correspondence theorems, but my intuitions on Lie algebras aren't that strong yet and I haven't yet proved the correspondence theorems, so I stuck to what I know about Lie groups. That said, if you have said proof, I'd appreciate seeing it.

Finally, if you know how I can extend this to the disconnected case, I would love some hints (no spoilers please!). I would think that there would be constraints on how disconnected a Lie group can be, but my topology and algebra aren't strong enough to say exactly what.

Thanks!

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  • $\begingroup$ I'm not sure how you get this sentence "That would suggest that every connected one-parameter Lie group is equivalent (diffeomorphic and homomorphic at the same time, whatever fancy morphism that would be) to either good ol' translation in one dimension or SO2." But that may be my own fault $\endgroup$
    – Tim kinsella
    Apr 19, 2015 at 16:53
  • $\begingroup$ Yeah, I realized after I posted this that just because the Lie group is diffeomorphic to another Lie group doesn't necessarily mean its equivalent. It could have the same differential structure and different group structures, hypothetically. This brought me back to a question that was asked here three years ago math.stackexchange.com/questions/44464/… $\endgroup$
    – Twigg
    Apr 19, 2015 at 16:59

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One elementary way to see this is that the exponential map gives a smooth homomorphism (a "Lie group homomorphism" to answer your terminological question) from the Lie algebra to the group, G. The derivative of this map at 0 is nonzero so, by the inverse function theorem, exp is a diffeomorphism in a neighborhood of 0. So the image of exp contains a neighborhood of the identity in G. Its not difficult to see that any neighborhood of the identity generates a connected Lie group. Therefore G is generated by a subset all of whose elements commute with eachother. So G is abelian.

The analogous statement about disconnected groups is false, as you can see by multiplying G by any nonabelian discrete group.

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  • $\begingroup$ The first part is a very clear explanation. Thanks! I just want to make sure I understand the second part. Suppose the noncommutative discrete group has order n. Then the product group would be a one-paramter Lie group with n components? $\endgroup$
    – Twigg
    Apr 19, 2015 at 17:04
  • $\begingroup$ You're welcome. and yes, that's right $\endgroup$
    – Tim kinsella
    Apr 19, 2015 at 17:05

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