Timeline for Is it always possible to extend a closed (1, 1)-form on a divisor to a closed (1, 1)-form on a tubular neighbourhood?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2021 at 23:13 | comment | added | Misha Verbitsky | yes, precisely this | |
| Oct 20, 2021 at 7:51 | comment | added | red_trumpet | Do you mean $\omega'|Z = \omega_0$? | |
| Mar 25, 2011 at 16:33 | vote | accept | user3566 | ||
| Mar 25, 2011 at 16:33 | history | bounty ended | user3566 | ||
| Mar 19, 2011 at 22:13 | comment | added | Misha Verbitsky | No, it's not obvious, in fact, it can be false - the existence of a Kaehler class which restricts to $[\omega_0]$ is not guaranteed. However, the proof of the above theorem works to extend the form to a tubular neighbourhood. | |
| Mar 16, 2011 at 13:24 | comment | added | user3566 | Let \alpha be a closed (1, 1)-form on Z. Add a big multiple of a Kahler form \omega' on Z to \alpha, so that k\omega'+\alpha is a positive (1, 1)-form on Z for some large positive real number k. How do I know that there exists a Kahler class [\omega] on M that restricts to the Kahler class [k\omega'+\alpha] on Z? This may be obvious... | |
| Mar 15, 2011 at 20:55 | history | edited | Misha Verbitsky | CC BY-SA 2.5 |
deleted 9 characters in body
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| Mar 15, 2011 at 5:30 | history | answered | Misha Verbitsky | CC BY-SA 2.5 |