Timeline for Orientation-preserving isometric involution on compact Kähler manifold
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2020 at 0:24 | history | became hot network question | |||
| Sep 7, 2020 at 21:27 | comment | added | Paul Reynolds | @RobertBryant, I guess I wish I didn't have to keep writing 'assume full holonomy' or similar, as I think would make sense for the question above as well as other isometry questions. I won't argue with you though, because I would not win that one. | |
| Sep 7, 2020 at 18:46 | comment | added | Robert Bryant | @PaulReynolds: Really? You wouldn't want the Ricci-flat metric on K3 constructed by Yau to be called 'Kähler', or the product of two Kähler metrics to be called 'Kähler'? | |
| Sep 7, 2020 at 17:30 | comment | added | Paul Reynolds | I wish the standard definition of Kaehler was 'holonomy $U_n$'. So many questions would have neater answers. | |
| Sep 7, 2020 at 17:00 | vote | accept | CommunityBot | ||
| Sep 7, 2020 at 16:49 | comment | added | Robert Bryant | @BenMcKay: Actually, even if the holonomy is irreducible, the holonomy group may not determine the complex structure. For example, if the metric is hyperKähler, then there will be at least a 2-sphere of complex structures compatible with the metric. | |
| Sep 7, 2020 at 16:47 | comment | added | Robert Bryant | @JasonStarr: Your construction might not preserve the orientation of the product, since $i$ might not be orientation preserving. | |
| Sep 7, 2020 at 16:44 | answer | added | Robert Bryant | timeline score: 4 | |
| Sep 7, 2020 at 16:44 | comment | added | Ben McKay | The holonomy of the metric preserves the complex structure, since it is Kaehler. By the de Rham splitting theorem, and the Berger classification of holonomy groups, either the universal covering space splits isometrically into a product, or else the holonomy group determines the complex structure, i.e. does not sit inside two conjugates of the unitary group inside the rotation group. | |
| Sep 7, 2020 at 16:43 | comment | added | Jason Starr | Let $X$ be a Kaehler manifold with an antiholomorphic involution $i$ that is isometric. On the product $X\times X$, consider the product of the identity map and $i$. | |
| Sep 7, 2020 at 16:20 | history | asked | user164740 | CC BY-SA 4.0 |