Timeline for The kernel of $H^{\bullet,\bullet}_{\bar\partial}(X)\to H_A^{\bullet,\bullet}(X)$
Current License: CC BY-SA 4.0
7 events
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| Sep 15, 2022 at 13:26 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Sep 15, 2022 at 11:57 | comment | added | Tom | Thanks, I think the answer is totally clear now! | |
| Sep 15, 2022 at 11:56 | vote | accept | Tom | ||
| Sep 15, 2022 at 11:47 | comment | added | Michael Albanese | Your question is about exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$. Replacing $f : B^{\bullet,\bullet} \to H^{\bullet,\bullet}_{\bar{\partial}}$ with any map which has the same image will preserve exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$. That is, exactness at $H^{\bullet,\bullet}_{\bar{\partial}}$ doesn't force the previous term to be $B^{\bullet,\bullet}$. I assume that the specific form of $B^{\bullet,\bullet}$ is needed to get exactness of the full sequence. | |
| Sep 15, 2022 at 3:07 | comment | added | Tom | The explanation is precise! But why the denominator of $B^{\bullet,\bullet}$ should be $\text{im }\partial\bar\partial$? | |
| Sep 14, 2022 at 16:20 | history | edited | Michael Albanese | CC BY-SA 4.0 |
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| Sep 14, 2022 at 16:11 | history | answered | Michael Albanese | CC BY-SA 4.0 |