Timeline for Non-Kahler manifolds and the dd^c-lemma
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 22, 2011 at 12:59 | comment | added | Donu Arapura | It's possible that Craig intended the statement only for (1,1) forms as in the follow up discussion, it's hard to say. However, I am sure that the stronger statement is false. The reason I think so is that by Deligne-Griffiths-Morgan-Sullivan, a compact complex manifold satisfying the $dd^c$-lemma should be formal. Now, I think one can use a twistor construction (Taubes, JDG 1992) to build a compact complex manifold which is not formal but with $b_1=0$. | |
| Apr 22, 2011 at 2:28 | comment | added | YangMills | so it seems that $S^3\times S^3$ does not satisfy the $dd^c$-lemma after all, since $b_1\neq 2h^{0,1}$. Also, it has $h^{1,1}=1$, see the same paper of Hofer, so there are nontrivial $\overline{\partial}$-closed $(1,1)$-forms | |
| Apr 22, 2011 at 2:26 | comment | added | Joel Fine | @Yang-Mills, from what you say it follows that $h^{0,1} = 0$ DOES imply the $dd^c$-lemma for $(1,1)$-forms. For if $h^{0,1}=0$ then it follows that $b_1=0$ hence certainly $b_1 = 2h^{0,1}$. But there is a much more prosaic way to see it. Let $a = db$ be an exact real (1,1) form. Then $\bar{\partial}b^{0,1} = 0$ so $b^{0,1} = \bar{\partial} f$. From here it follows that $a = 2 i \partial\bar{\partial} g$ where $g$ is the imaginary part of $f$. | |
| Apr 22, 2011 at 2:12 | comment | added | Joel Fine | I think that perhaps $H^{0,1}=0$ implies the d-dbar lemma for (1,1)-forms, but I can't see how it would help for higher degrees. | |
| Apr 21, 2011 at 16:03 | comment | added | Donu Arapura | The exercise is not easy for me. The $dd^c$ lemma has very strong consequences, such as degeneration of various spectral sequences, formallity etc. So I find the last statement surprising. | |
| Apr 21, 2011 at 7:56 | history | answered | Craig | CC BY-SA 3.0 |