Timeline for Kahler version of Darboux's Theorem
Current License: CC BY-SA 3.0
11 events
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| Nov 15, 2016 at 23:45 | history | edited | user44191 | CC BY-SA 3.0 |
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| Nov 14, 2016 at 22:28 | vote | accept | user44191 | ||
| Nov 14, 2016 at 20:55 | answer | added | Malkoun | timeline score: 10 | |
| Nov 14, 2016 at 20:19 | comment | added | user44191 | @LiviuNicolaescu My apologies; I see that I forgot a key phrase. I've now edited to include it. Both of your answers seem to suggest that the Riemannian curvature is the tensor I'm looking for; would you mind expanding on why the vanishing of the Riemannian curvature allows for coordinates that also work for the symplectic form? (copied because I could only notify one user per comment) | |
| Nov 14, 2016 at 20:19 | comment | added | user44191 | @Malkoun My apologies; I see that I forgot a key phrase. I've now edited to include it. Both of your answers seem to suggest that the Riemannian curvature is the tensor I'm looking for; would you mind expanding on why the vanishing of the Riemannian curvature allows for coordinates that also work for the symplectic form? | |
| Nov 14, 2016 at 20:17 | history | edited | user44191 | CC BY-SA 3.0 |
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| Nov 13, 2016 at 11:06 | comment | added | Liviu Nicolaescu | No. It has to do with the fact that a Kahler form is more than a symplectic form. It also defines a metric. For a Kahler form to be locally canonic we need the associated metric to be locally Euclidean. And we know from the early work of Riemann that there is obstruction to that happening and it is called curvature. | |
| Nov 13, 2016 at 8:38 | comment | added | Malkoun | A Kahler metric can be thought of as consisting of 2 pieces, a metric $g$ and a complex structure $J$, with the two being compatible, in the sense that $J$ is $g$-orthogonal, and such that the Kahler form is closed. I will assume that your local coordinates $x_i$ and $y_i$ are the real and imaginary parts of some holomorphic local coordinates $z_i$. The vanishing of the Riemann curvature is a necessary condition for the Kahler form to be of that form (because $g$ is then locally Euclidean). It is also sufficient, because it forces the (local) holonomy to be trivial. | |
| Nov 13, 2016 at 7:58 | history | edited | user44191 | CC BY-SA 3.0 |
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| Nov 13, 2016 at 7:57 | comment | added | Malkoun | You mean in your first sentence symplectic manifold, and not symmetric manifold. | |
| Nov 13, 2016 at 7:16 | history | asked | user44191 | CC BY-SA 3.0 |