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Sep 18, 2019 at 7:22 comment added abx I would guess it is the same — the equality is checked at each point, the fact that $\omega$ is closed plays no role.
Sep 17, 2019 at 18:24 comment added Ali Taghavi @abx Thanks for this reference. So what about the same question without Kahlerian assumption?
Sep 17, 2019 at 17:54 comment added abx Yes. You can see that (for instance) in Weil's "Variétés kählériennes".
Sep 17, 2019 at 16:42 comment added Ali Taghavi @abx Thank you! I am aware of Darbeux charts for symplectic manifolds. But is there a Kahler chart for Kahlerian manifolds? That is a chart which preserves all 3 structures in their standard Euclidean forms.
Sep 17, 2019 at 16:20 review Close votes
Oct 18, 2019 at 3:05
Sep 17, 2019 at 15:55 comment added abx Yes, because $\omega '=\omega $.
Sep 17, 2019 at 15:13 history edited Ali Taghavi CC BY-SA 4.0
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Sep 17, 2019 at 14:51 history asked Ali Taghavi CC BY-SA 4.0