Timeline for Proving Hodge decomposition without using the theory of elliptic operators?
Current License: CC BY-SA 2.5
12 events
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| Dec 31, 2019 at 8:45 | comment | added | user20948 | Your notes are no longer available. Is it copyrighted or there are some alternatives? | |
| Jun 24, 2010 at 13:23 | history | edited | Donu Arapura | CC BY-SA 2.5 |
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| Jun 22, 2010 at 23:32 | comment | added | T.. | Fantastic answer from Donu. Many thanks. In the case where the variety is smooth, projective and defined over a number field, is it presently unknown whether the (pure) Hodge structure can be defined algebraically? | |
| Jun 19, 2010 at 19:56 | history | edited | Donu Arapura | CC BY-SA 2.5 |
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| Jun 19, 2010 at 1:39 | comment | added | Deane Yang | I agree with the suggestion of starting with Riemann surfaces. The higher dimensional proof is more technical and involved, but involves no essential new ideas. | |
| Jun 18, 2010 at 19:11 | history | edited | Donu Arapura | CC BY-SA 2.5 |
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| Jun 18, 2010 at 19:06 | comment | added | Donu Arapura | I guess Kevin Lin answered the question, but the existence of a (mixed) Hodge structure on cohomology cannot be proved by pure algebra at present. | |
| Jun 18, 2010 at 19:00 | history | edited | Donu Arapura | CC BY-SA 2.5 |
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| Jun 18, 2010 at 7:17 | comment | added | Konrad Voelkel | You can, however, use Hironaka and Hodge theory for compact complex algebraic varieties to get a mixed hodge structure on the cohomology of noncompact singular complex varieties :-) | |
| Jun 17, 2010 at 20:43 | comment | added | Kevin H. Lin | @Boyarsky: I think you are correct. I think Hodge decomposition does not follow from degeneration. I think there is no purely algebraic proof of Hodge decomposition... | |
| Jun 17, 2010 at 18:42 | comment | added | Boyarsky | @Donu: so does your final paragraph mean that getting decomposition from degeneration isn't evident to you either (as it is not to me)? | |
| Jun 17, 2010 at 18:19 | history | answered | Donu Arapura | CC BY-SA 2.5 |