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I am an optical engineer, so please forgive any ignorance my questions betoken. I am interested in whether one can tear down the manifold of a finite dimensional Lie group, leaving an abstract group, and then give the group another manifold structure that makes it again into a Lie group and get anything essentially different in the process. I know the manifold replacement can be done in some special cases. Here are the examples I have been thinking about and could come up with answers to. In all examples I can make progress on, the Lie groups arising from the different manifolds built on the same set are isomorphic (as Lie groups that is: i.e. they have the same Lie algebra, fundamental group and connected components) - they are of course isomorphic as abstract groups!!!. I have a hunch that this is generally true, otherwise one might get weird examples where an abstract group might have several different Lie algebras, and I'm sure I'd have read about that somewhere by now! If someone knows anything about the general case, I'd be most interested to hear about it.

Example : Take the additive group $\left(\mathbb{R},\cdot +\right)$, it is its own Lie algebra. As the lie algebra I denote it $\mathbf{g}$ and the Lie group $G$ - the exponential map is the identity map. Now take one of the everywhere discontinuous, bijective solutions $\phi$ of the Cauchy functional equation $f\left(x\right) + f\left(y\right) = f\left(x + y\right)$ as detailed in the end of Chapter 2 of Hewitt and Stromberg "Real and Abstract Analysis"; $\phi$ is now an exponential map from the $g$ onto a new Lie group $G^\prime$, to wit the same group $\left(\mathbb{R},\cdot +\right)$ but now given a new topology generated by images $\phi\left(U\right)$ of open intervals in $\mathbb{R}$. This topology is of course totally disconnected in the group topology for $G$; however $\phi$ is continuous, indeed $C^\infty$ when thought of as a map from $g$ to $G^\prime$, the latter with the new topology. Indeed, the Lie groups arising from the two topologies are isomorphic, their topologies are homoemorphic.

I have a University of Pittsburgh 2007 PhD. Thesis "On the Uniqueness of Polish Group Topologies" by Bojana Pejic that I believe may be relevant, but so far have not made great headway in understanding it. If someone could point me to other material on this question, I'd be grateful.

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Linus Kramer has a recent paper that answers the question in your title (at least for semi-simple Lie groups) arxiv.org/abs/1009.5457 –  Theo Buehler Apr 20 '11 at 7:57
    
Have we had a similar question to this before on MO? algori's answer (below) is ringing a faint bell... –  Yemon Choi Apr 20 '11 at 9:08
    
A closely related is the question when a topological group is actually a Lie group. See en.wikipedia.org/wiki/Hilbert%27s_fifth_problem –  Vít Tuček Apr 20 '11 at 9:39
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Can't any Lie group be given the discrete topology, making it a 0-dimensional Lie group? –  Qiaochu Yuan Apr 20 '11 at 18:24
    
Nice simple example Qiaochu Yuan: Shows that I should have stated that I was looking for examples that leave the connectedness of the group unchanged. So, "How can a Connected Lie group as an abstract group ... " would be a better question. –  WetSavannaAnimal aka Rod Vance Apr 22 '11 at 8:26
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1 Answer

up vote 17 down vote accepted

To answer the question of the title: given a group, there may be infinitely many different topologies that render it a Lie group: $\mathbb{R}^n$ and $\mathbb{R}^m$ (where $m$ and $n$ are positive integers) are isomorphic as groups since they are $\mathbb{Q}$-vector spaces of the same dimension, hence they are isomorphic to the direct sum of continuously many copies of $\mathbb{Q}$. But they are not isomorphic as topological spaces, let alone Lie groups.

However, given two semi-simple Lie groups, I think they are isomorphic as Lie groups as soon as they are isomorphic just as groups.

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Many thanks, Algori. So, if I'm reading this correctly, some abstract groups can have many different Lie algebras: $\mathbb{R}^n$ can be a Lie algebra for $\mathbb{R}^m$ as a groupm for any integers n and m? –  WetSavannaAnimal aka Rod Vance Apr 20 '11 at 11:01
    
You are most welcome! Abstract groups do not in general have a Lie algebra attached to them, or they may have several, like a sum of continuously many copies of $\mathbb{Q}$. –  algori Apr 20 '11 at 11:27
    
Well, there are uniform ways to attach Lie algebras to abstract groups, but it's not quite the way a Lie algebra comes from a Lie group. For example, the lower central series of a discrete group is a filtered pro-nilpotent group, and its associated graded abelian group naturally admits a Lie-ring structure. Of course, this is not what OP is after. –  Theo Johnson-Freyd Apr 20 '11 at 14:10
    
@algori: Are you claiming that every group isomorphism of semisimple Lie groups is automatically smooth, or just using the classification result to assert that if the groups are isomorphic qua groups, then there exists a smooth isomorphism? –  Theo Johnson-Freyd Apr 20 '11 at 14:12
    
Theo -- I am under the impression that the first of these two statements should be true (although I do not remember how to prove it). –  algori Apr 20 '11 at 14:28
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