Timeline for de-Rham moduli space over a compact Riemann surface
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 28, 2021 at 9:02 | comment | added | user131608 | @Tabes Bridges As i understand the comparison between de Rham and Dolbeault uses the fact the both these varieties are smooth varieties. I am considering here only the stable parts of the moduli. The semistable points can be singular. | |
| Jan 27, 2021 at 14:58 | answer | added | Nicolast | timeline score: 4 | |
| Jan 27, 2021 at 0:50 | comment | added | Tabes Bridges | @user131608 I assume you are familiar with Simpson's theorems relating Betti, de Rham, and Dolbeaut moduli spaces. Do none of his comparison results allow you to deduce smoothness of de Rham from smoothness of Dolbeaut? | |
| Jan 26, 2021 at 11:36 | comment | added | user131608 | @Piotr Achinger Although there is a concept of special bundles over a Riemann surface which comes from certain unitary representations of the punctured Riemann surface. These bundles will not have degree $0$. Similarly those representations in GL(n, C) give rise to Higgs bundles with non-zero degree. I am not sure about bundles with connections | |
| Jan 26, 2021 at 10:30 | comment | added | user131608 | @Piotr Achinger Thanks for the correction | |
| Jan 26, 2021 at 10:29 | history | edited | user131608 | CC BY-SA 4.0 |
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| Jan 26, 2021 at 10:16 | comment | added | Piotr Achinger | The degree of a bundle with connection is zero, so it cannot be coprime with the rank. Or am I missing something? | |
| Jan 26, 2021 at 9:51 | review | First posts | |||
| Jan 26, 2021 at 9:55 | |||||
| Jan 26, 2021 at 9:43 | history | asked | user131608 | CC BY-SA 4.0 |