Timeline for Topological information via cohomology of sheaves
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 18, 2013 at 17:23 | comment | added | Entaou | Certainly we can see some topological structure of $X$ if it admits some $\mathcal F$ with special properties. For example, if $X$ admits a holomorphic line bundle with positive curvature,then it is a projective manifold, which will pose some topological restriction on $X$. But I think this is different thing. | |
| Oct 18, 2013 at 17:20 | comment | added | Entaou | The differential forms $\Omega^p$ are intrinsicly defined on $X$, and the Hodge decomposition reflects the relation of the complex structure and the topological structure of $X$. Yes a vector bundle if locally products of holomorphic functions, but the gluing process to a global bundle can be complicated, so $\mathcal F$ can be totally different form holomorphic functions and holomorphic forms. On the other hand, topological invariants are global object. | |
| Oct 18, 2013 at 16:29 | comment | added | N B | And any other coherent sheaf is locally modelled using few copies of the sheaf of holomorphic functions. So this doesn't look that unreasonable. | |
| Oct 18, 2013 at 16:26 | comment | added | N B | But you can do so considering the sheaf of holomorphic functions and "convolutions" (tensor products) with sheaves of holomorphic forms. So the question was mainly about analogs of sheaves of differential forms, allowing to "average" any other sheaf (or a vector bundle) to a topological invariant. | |
| Oct 18, 2013 at 10:47 | comment | added | Entaou | Sorry, in the last sentence, "eternal" should be external. | |
| Oct 18, 2013 at 10:39 | comment | added | Entaou | I don't think so. Since $\mathcal F$ is not an intrinsic object of $X$. So it is unreasonable to try to combine invariants of $\mathcal F$ to get a pure topological invariant of $X$. On the other hand, you can consider the category of all coherent sheaves on $X$, then you get some information of $X$, which is the content of K-theorey. In a word, you can not read out some topological invariants of $X$ just by considering a single eternal object $\mathcal F$. | |
| Oct 17, 2013 at 23:58 | comment | added | N B | This looks interesting, but it would be really enlightening to see if it is possible to combine all χ$(X, \Omega^p\otimes \mathcal F)$ on the left and respectively Chern classes in integrals on the right to something "without" $\mathcal F$. Thank you. | |
| Oct 17, 2013 at 19:52 | history | answered | Entaou | CC BY-SA 3.0 |