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Let $X$ be a Lie group, $Aut(X)$ be the Lie automorphism group of $X$ (group automorphisms which are also diffeomorphisms), and $Homeo(X)$ be the homeomorphism group of the underlying manifold. For any $f\in Homeo(X)$, does there exist some $g\in Aut(X)$ such that $f$ and $g$ are isotopic?

I hesitate to post this on this board as I realize that most questions are of a very high standard; however, after consulting other sources I can't seem to find any similar results.

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Obviously not. If $G=X$ is not connected, then the connected components are the cosets of $G^0$ and therefore all homeomorphic. In particular any permutation of these components can be realized by an homeomorphism of $G$ (and isotypic homeomorphism induce the same permutation), while every Lie group automorphism induces a group automorphism of the quotient $G/G^0$. Hence every disconnected Lie group (e.g. $O(n)$ or every finite group) is a counterexample.

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  • $\begingroup$ And if I restricted the question to connected Lie groups? $\endgroup$ – Joseph Zambrano Apr 9 '14 at 21:15
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Every automorphism of the simply conected cover $G$ of $SO(3)$ preserves orientation, so no orientation reversing diffeo of $G$, which is a sphere, is isotopic to an automorphism.

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  • $\begingroup$ Thank you (as well as Johannes) for the examples. If its not too much trouble to ask, is there anything we can say about equivalence classes of Lie automorphisms up to isotopy? The only thing I can think of is that inner automorphisms from elements of the connected component of the identity are isotopic. $\endgroup$ – Joseph Zambrano Apr 9 '14 at 21:37

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