Timeline for Singular cohomology to cohomology of quasi-coherent sheaf
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2023 at 4:38 | comment | added | dhy | To be more explicit: For a complex variety, two sensible cohomology theories you can consider are a) the sheaf cohomology (in the Zariski topology) of quasi-coherent sheaves and b) singular cohomology, which can be identified with the sheaf cohomology (in the complex topology) of the constant sheaf. You seem to be trying to mix the two, considering the (Zariski) sheaf cohomology of the constant sheaf - as Zhen Lin mentioned, this only gives you trivial cohomology groups. However, there is a more indirect link given by the Hodge-de Rham spectral decomposition. | |
| Jan 1, 2023 at 22:19 | comment | added | Zhen Lin | If you want to think about varieties as compact orientable manifolds you should not be talking about arbitrary algebraically closed fields $k$, and you should specify what topology you are asking about. | |
| Jan 1, 2023 at 17:55 | comment | added | locally trivial | So the singular cohomology of any variety (for me, varieties are irreducible) vanishes in cohomological degree greater than or equal to 1. But if we take singular homology with coefficients in a field $k$, by (usual) Poincaré duality for a compact, orientable manifold, won't we have $H_0(X, k)$ and [after applying duality, if there is no torsion] $H_n(X,k)$ are non-zero [or, if we like, both $H^0(X,\underline{k})$ and $H^n(X,\underline{k})$ are non-zero]. How does this not contradict cohomology of the constant sheaf $\underline{k}$ vanishing in cohomological degrees $\geq 1$? | |
| Jan 1, 2023 at 5:28 | review | Close votes | |||
| Jan 6, 2023 at 3:04 | |||||
| Jan 1, 2023 at 2:31 | comment | added | Zhen Lin | The sheaf cohomology for constant sheaves on irreducible spaces is basically trivial: see here. | |
| Dec 31, 2022 at 23:39 | history | asked | locally trivial | CC BY-SA 4.0 |