Timeline for closed dual of vector fields
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2012 at 20:11 | comment | added | Igor Khavkine | Now I don't feel so bad for posting the same "easy answer" and then deleting it. It looks like at least three other people fell into the same trap! :-) | |
| Sep 20, 2012 at 19:46 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 20, 2012 at 19:11 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 20, 2012 at 19:07 | comment | added | Liviu Nicolaescu | @ all commenters: Apologies for the dumb mistake. Please read the Mea Culpa included in my updated answer. | |
| Sep 20, 2012 at 19:03 | history | edited | Liviu Nicolaescu | CC BY-SA 3.0 |
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| Sep 20, 2012 at 18:57 | comment | added | Matthias Ludewig | Greg: Riemannian metrics always exist. | |
| Sep 20, 2012 at 18:41 | comment | added | Qfwfq | As soon as I read the question I was about to post exactly the same answer, but then I asked myself if the resulting form would be closed or not. It may depend on the metric, so probably a (nontrivial?) differential equation would be involved. | |
| Sep 20, 2012 at 18:09 | comment | added | Andreas Blass | Another answerer and I had written this and then deleted it, because I (and presumably the other answerer) see no reason for $\alpha$ to be closed. My answer actually included a second wrong approach, namely to find such $\alpha$'s locally and patch them together with a partition of unity. Again, the result need not be closed. | |
| Sep 20, 2012 at 17:44 | comment | added | Greg Friedman | Assuming you can construct a Riemmanian metric! (though I grant that one can in most cases of interest) | |
| Sep 20, 2012 at 17:38 | history | answered | Liviu Nicolaescu | CC BY-SA 3.0 |