Can a Lie group as an abstract group be given more than one topology making it a Lie group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:14:35Z http://mathoverflow.net/feeds/question/62385 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/62385/can-a-lie-group-as-an-abstract-group-be-given-more-than-one-topology-making-it-a Can a Lie group as an abstract group be given more than one topology making it a Lie group? WetSavannaAnimal aka Rod Vance 2011-04-20T07:52:50Z 2011-04-20T09:36:22Z <p>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.</p> <p>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. </p> <p>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.</p> http://mathoverflow.net/questions/62385/can-a-lie-group-as-an-abstract-group-be-given-more-than-one-topology-making-it-a/62388#62388 Answer by algori for Can a Lie group as an abstract group be given more than one topology making it a Lie group? algori 2011-04-20T08:10:33Z 2011-04-20T08:43:42Z <p>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.</p> <p>However, given two semi-simple Lie groups, I think they are isomorphic as Lie groups as soon as they are isomorphic just as groups.</p>