Timeline for Quotients of $K3$ surfaces by finite groups
Current License: CC BY-SA 4.0
9 events
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| Oct 30, 2019 at 21:20 | comment | added | Yosemite Stan | @EvgenyShinder Complex K3s do not admit rational dominant maps to bi-elliptic surfaces. One could show this as follows: Resolving the map would give a regular map from a simply connected surface (a blow-up of a K3 surface a finite # of times) to a bi-elliptic surface whose universal cover is $\mathbb{C}^2$. Lifting the map to the universal cover shows the map is constant. | |
| Oct 30, 2019 at 20:53 | answer | added | Zhiwei Zheng | timeline score: 7 | |
| Oct 23, 2019 at 16:48 | comment | added | Evgeny Shinder | Sorry, I forgot about other surfaces of Kodaira dimension zero: K3s definitely cover Enriques surfaces, and I do not know about whether K3 surfaces admit rational dominant maps to bi-elliptic surfaces. | |
| Oct 22, 2019 at 22:10 | comment | added | Evgeny Shinder | One starting point for a classification is to work birationally and to note that the resolutions of $S/G$ are all either K3 surfaces or rational surfaces (because K3 surfaces do not admit dominant rational maps into other types of surfaces). Then one would ask which K3s admit a generically finite (Galois) cover by another K3, and there is some work on this but probably no classification. | |
| Oct 15, 2019 at 13:44 | comment | added | abx | Up to deformation there are only a finite number of possible group actions, so a finite number of quotients. There is a huge literature about this, see for instance the work of Mukai. | |
| Oct 15, 2019 at 13:42 | history | edited | Basics | CC BY-SA 4.0 |
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| Oct 15, 2019 at 13:35 | history | edited | YCor |
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| Oct 15, 2019 at 13:30 | history | edited | Basics | CC BY-SA 4.0 |
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| Oct 15, 2019 at 13:23 | history | asked | Basics | CC BY-SA 4.0 |