Timeline for $\partial \bar{\partial}$ on a riemann surface
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2012 at 7:35 | vote | accept | william | ||
| Feb 9, 2012 at 7:35 | vote | accept | william | ||
| Feb 9, 2012 at 7:35 | |||||
| Feb 9, 2012 at 7:35 | vote | accept | william | ||
| Feb 9, 2012 at 7:35 | |||||
| Feb 8, 2012 at 14:27 | answer | added | marco | timeline score: 0 | |
| Feb 8, 2012 at 1:04 | answer | added | David E Speyer | timeline score: 7 | |
| Feb 8, 2012 at 0:59 | comment | added | David E Speyer | I am pretty sure what you want is that $\alpha$ is a closed $(1,1)$ form. In particular, this includes that $\bar{\partial} \alpha=0$, since $d \alpha = \partial \alpha + \bar{\partial} \alpha$ and the two summands, being a $(2,1)$ form and a $(1,2)$ form, cannot sum to zero unless they are both $0$. | |
| Feb 7, 2012 at 22:13 | comment | added | william | no i mean holomorphic ... i mean if its holomorphic then its also smooth. but does a solution $\varphi$ exist (on an open neighbourhood of $R$) ? | |
| Feb 7, 2012 at 22:12 | answer | added | Johannes Ebert | timeline score: 7 | |
| Feb 7, 2012 at 22:09 | comment | added | Ben McKay | You don't mean to say that $\alpha$ is holomorphic, just smooth. | |
| Feb 7, 2012 at 21:31 | history | asked | william | CC BY-SA 3.0 |