Timeline for Can a metric conformal to a Kahler metric be Kahler?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jul 8, 2011 at 14:10 | comment | added | Spiro Karigiannis | Looks like we were answering at the same time. At least I gave a slightly different explanation, so I might as well leave my answer there. | |
| Jul 8, 2011 at 14:06 | vote | accept | Gunnar Þór Magnússon | ||
| Jul 8, 2011 at 14:04 | comment | added | BS. | @Gunnar: definitely yes, $\omega$ is a 2-form, otherwise $d\omega$ wouldn't make much sense. The riemannian metric is $\omega(J.,.)$. | |
| Jul 8, 2011 at 13:57 | comment | added | Gunnar Þór Magnússon | Nice. Are we considering $\omega$ as a symplectic form here (instead of a metric)? | |
| Jul 8, 2011 at 13:57 | history | edited | BS. | CC BY-SA 3.0 |
added 241 characters in body
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| Jul 8, 2011 at 13:49 | history | answered | BS. | CC BY-SA 3.0 |