Let $G$ be a connected Fréchet-Lie group and let $\mathfrak g$ be its Lie algebra. Does the image $\exp(\mathfrak g) \subset G$ of the exponential map generate $G$?


I believe that this is open. In fact, even for the very special case where $G = \text{Diff}_0(M)$ for a smooth manifold $M$, the only proof I know that $G$ is generated by the image of the exponential map uses a very deep theorem of Thurston that says that in this case $G$ is a simple group (this implies that $G$ is generated by the image of the exponential map since the subgroup generated by the image of the exponential map is a nontrivial normal subgroup). Even giving a genuinely different proof in this special case would be very interesting.

  • $\begingroup$ $Diff_0$ means the unit component of the group of $C^\infty$ diffeomorphisms with the $C^\infty$ topology? $\endgroup$ – YCor Sep 17 '17 at 21:41
  • $\begingroup$ @YCor: Yes, that's correct (with the addendum that you need to take compactly supported diffeomorphisms if the manifold is not compact). The Lie algebra is the set of vector fields on the manifold (compactly supported if the manifold is not compact) with the usual bracket, and the exponential map is obtained by flowing along the vector field. $\endgroup$ – Andy Putman Sep 17 '17 at 22:13
  • $\begingroup$ (and just to emphasize, Thurston's theorem says that it is simple as an abstract group, so this argument is not really sensitive to the topology you take) $\endgroup$ – Andy Putman Sep 17 '17 at 22:14
  • $\begingroup$ Not the argument, but the description of the Lie algebra is certainly sensitive to the choice of topology. $\endgroup$ – YCor Sep 17 '17 at 23:14
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    $\begingroup$ Hi Andy, thanks for the answer. I also asked Karl-Hermann Neeb (by email) and he says that to the best of his knowledge the question is open even for a more general class of groups admitting the exponential map. I like your question about a "direct" proof for diffeomorphism groups. $\endgroup$ – Jarek Kędra Sep 18 '17 at 19:15

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