Timeline for Does the image of the exponential map generate the group?
Current License: CC BY-SA 3.0
10 events
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Apr 22, 2020 at 21:52 | comment | added | Andy Putman | @New_Topologist_On_The_Block: I don't think that Thurston ever wrote a paper with a complete proof of the simplicity of that group, but Banyaga's book "The structure of classical diffeomorphism groups" contains an account of it. | |
Apr 22, 2020 at 18:37 | comment | added | ABIM | @AndyPutman Do you have a reference to the paper saying that $Im(\exp)$ generates $Diff_0(M)$. | |
Sep 18, 2017 at 19:15 | vote | accept | Jarek Kędra | ||
Sep 18, 2017 at 19:15 | comment | added | Jarek Kędra | 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. | |
Sep 17, 2017 at 23:35 | comment | added | Andy Putman | @YCor: Absolutely. | |
Sep 17, 2017 at 23:14 | comment | added | YCor | Not the argument, but the description of the Lie algebra is certainly sensitive to the choice of topology. | |
Sep 17, 2017 at 22:14 | comment | added | Andy Putman | (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) | |
Sep 17, 2017 at 22:13 | comment | added | Andy Putman | @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. | |
Sep 17, 2017 at 21:41 | comment | added | YCor | $Diff_0$ means the unit component of the group of $C^\infty$ diffeomorphisms with the $C^\infty$ topology? | |
Sep 17, 2017 at 21:28 | history | answered | Andy Putman | CC BY-SA 3.0 |