Timeline for Surjectivity of $H^2(X,\mathbb C)\to H^2(X,\mathcal O)$
Current License: CC BY-SA 4.0
5 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Oct 6, 2022 at 12:42 | answer | added | Tom | timeline score: 0 | |
Oct 4, 2022 at 10:12 | comment | added | Tom | @Donu Arapura, According to your comment, I guess we only need the $(0,2)$ part of the condition of Frölicher spectral sequence degenerates at $E_1$, that is $H_{\bar\partial}^{0,2}(X)=\frac{H^2(X)}{F^1H^2(X)}$, where $F^1H^2(X)$ means $d$-closed $F^1A^2$ modulo $d$-exact ones in $F^1A^2$, so there is $\frac{Z_{\bar\partial}^{0,2}}{B_{\bar\partial}^{0,2}}=\frac{Z^2}{B^2}/\frac{F^1Z^2}{F^1B^2}$, by taking projection $\pi^{0,2}$, there is for any $\bar\partial$-closed $(0,2)$ form $\omega^{0,2}$, there is a closed $2$ form $\omega$ whose $(0,2)$ component is $\omega^{0,2}$? | |
Oct 3, 2022 at 13:14 | comment | added | Donu Arapura | You more or less answered your own question. If the $\partial\overline{\partial}$-lemma holds then the Frölicher (or Hodge to de Rham) spectral sequence degenerates. This implies that the so called edge map $H^p(X,\mathbb{C})\to H^p(X,\mathcal{O})$ is surjective. This can be made more explicit. | |
Oct 3, 2022 at 12:54 | answer | added | Francesco Polizzi | timeline score: 1 | |
Oct 3, 2022 at 10:25 | history | asked | Tom | CC BY-SA 4.0 |