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Suppose $M$ is a smooth and metrizable manifold. Then $\operatorname{Isom}{(M)}$ can be given the structure of a Lie group, so that the action of $\operatorname{Isom}{(M)}$ on $M$ is still smooth.

I found that as a sidenote somewhere. I really would like to see a prove of the above. So if somebody happens to know an online (free accessible) source, please tell me. I know $\operatorname{Isom}{(M)}$ is locally compact w.r.t. the compact-open topology. Is this also the topology of the Lie-Group?

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Certainly this doesn't even come close to answering the question, but you should think about the fact that the Lie algebra will be $\Gamma(TM)$, the vector fields on $M$ (if you haven't already). –  Aaron Mazel-Gee May 1 '12 at 14:47

1 Answer 1

How about this, this or this?

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thank you very much –  fk44 May 3 '12 at 16:12
    
@fk44: I should mention that those links were obtained by a very simple Google search (I certainly have no expertise in this field). It is always a good idea to do a thorough search before asking a question on this and similar sites. –  Martin Wanvik May 3 '12 at 16:30

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