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Feb 1, 2016 at 9:17 vote accept Alex M.
Jan 31, 2016 at 21:08 comment added Alex M. @DeaneYang: You mean, because $\omega$ is the pull-back of an exact form? In this case, yes, I see what you mean.
Jan 31, 2016 at 20:42 comment added Deane Yang Isn't $\omega$ being exact a necessary condition for the existence of an embedding?
Jan 31, 2016 at 20:42 answer added Danny Ruberman timeline score: 11
Jan 31, 2016 at 20:32 history edited Alex M. CC BY-SA 3.0
added 80 characters in body
Jan 31, 2016 at 20:22 history asked Alex M. CC BY-SA 3.0