Timeline for Calabi-Yau theorem on arithmetic variety
Current License: CC BY-SA 4.0
18 events
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| Feb 15, 2020 at 16:49 | history | edited | YCor | CC BY-SA 4.0 |
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| Feb 26, 2016 at 17:05 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 26, 2016 at 16:52 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 20, 2016 at 1:31 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 15, 2016 at 16:08 | history | edited | user21574 |
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| Feb 14, 2016 at 11:15 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 14, 2016 at 1:46 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 13, 2016 at 15:01 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 12, 2016 at 19:48 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 10, 2016 at 6:28 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 9, 2016 at 19:21 | comment | added | user21574 | In Kahler current setting which is the singular version of Kahler form we have still Ricci flat Kahler current. In fact We can define first Chern class In the sense of current cohomology. We can say $Ric(\omega)=0$ In current sense | |
| Feb 9, 2016 at 18:51 | comment | added | Sebastian Goette | Please specify the kind of Calabi-Yau theorem you hope for. I thought you mean $\mathrm{Ric}=0$ on all of $\mathcal X(\mathbb C)$. This means, forget the Kähler current and find a Ricci-flat Kähler form in the specified Kähler class. If $\mathcal X(\mathbb C)$ is smooth, I don't see an obstacle. Or do you simply ask for the displayed formula, which does not look like a Calabi-Yau theorem to me? | |
| Feb 9, 2016 at 18:35 | comment | added | user21574 | I have used "Kahler current" and not Kahler form | |
| Feb 9, 2016 at 18:28 | comment | added | Sebastian Goette | I don't see the connection. You don't need specific properties properties of the divisor, just that it gives $0$ in cohomology. But you already assumed $c_1(\mathcal X)=0$ even in the arithmetic setting. | |
| Feb 9, 2016 at 18:22 | comment | added | user21574 | you mean mathoverflow.net/questions/195180/… ? | |
| Feb 9, 2016 at 17:04 | comment | added | Sebastian Goette | To the best of my knowledge, there is a forgetful functor from arithmetic cohomology to singular cohomology that respects Chern classes. E.g., if you represent $c_1(\mathcal X)$ by a divisor of the canonical bundle, you would obtain the Poincaré dual of that divisor. Assuming that $\mathcal X(\mathbb C)$ is smooth, the answer would then be "yes" by the classical Calabi-Yau theorem, because (as far as I know), arithmetic geometry does not impose any further constraint on the Kähler form. | |
| Feb 9, 2016 at 16:06 | history | edited | user21574 | CC BY-SA 3.0 |
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| Feb 9, 2016 at 16:00 | history | asked | user21574 | CC BY-SA 3.0 |