Timeline for Action of orientation-preserving isometric involution on complex structure
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 9, 2020 at 0:17 | history | edited | Robert Bryant | CC BY-SA 4.0 |
correction of a few typos, careless phrasings and grammar mistakes
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| Sep 7, 2020 at 20:34 | vote | accept | CommunityBot | ||
| Sep 7, 2020 at 20:32 | comment | added | Robert Bryant | @JoeT: Well, no, because you can take the product of a hyperKähler counterexample with non-hyperKähler example for which there is such a decomposition in such a way that the product is neither hyperKähler nor does it have a decomposition of the kind you are asking for. If $M^n$ is holonomy-irreducible, then the only possibility for a counter-example to your decomposition is that it be hyperKähler, since, if it is holonomy-irreducible and Kähler but not hyperKähler, then the holonomy is either $\mathrm{U}(n)$ or $\mathrm{SU}(n)$ and, in that case, the only possibility is $\phi^*(J)=\pm J$. | |
| Sep 7, 2020 at 20:19 | comment | added | user164740 | is a counterexample necessarily hyperkähler? | |
| Sep 7, 2020 at 19:48 | history | answered | Robert Bryant | CC BY-SA 4.0 |