Timeline for de Rham vs Dolbeault Cohomology
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2015 at 17:22 | comment | added | Qfwfq | I believe I've seen the term "holomorphic de Rham" used for the hypercohomology of the holomorphic de Rham complex. But in the Stein case it's the same due to vanishing of positive cohomology of analytic coherent sheaves, right? | |
| Sep 9, 2015 at 14:58 | comment | added | Steven Gubkin | Thanks! I was actually thinking along these lines, but didn't recognize $\frac{d\bar{z}}{\bar{z}}+\frac{dz}{z}$ as $d\log(|z|^2)$, although I knew it should be exact. Thanks! | |
| Sep 9, 2015 at 14:44 | comment | added | David E Speyer | @StevenGubkin No contradiction. The class represented by $\frac{d \bar{z}}{\bar{z}}$ is also represented holomorphically by $- \frac{dz}{z}$. (Note that the difference of these classes is $d(\log |z|^2)$, hence exact.) | |
| Sep 9, 2015 at 14:40 | comment | added | Steven Gubkin | Thanks! A quick question about this answer though: $\frac{1}{\bar{z}} d\bar{z}$ is closed, but not exact, so represents an element of $H^1(M)$, where $M$ is the punctured unit disk. The unit disc is an open Riemann surface, and hence Stein. Does this contradict "deRham = holomorphic deRham", or am I missing something obvious? I feel like I probably am... | |
| Sep 9, 2015 at 14:36 | comment | added | David E Speyer | @StevenGubkin Done! | |
| Sep 9, 2015 at 14:36 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Sep 9, 2015 at 5:13 | comment | added | Steven Gubkin | Can you update this link, if you are able? | |
| Apr 28, 2012 at 16:58 | vote | accept | Janos Erdmann | ||
| Apr 27, 2012 at 17:04 | history | answered | David E Speyer | CC BY-SA 3.0 |