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Let $(M,\omega, J, g)$ be a $4$ dimensional Kahler manifold. Put $\omega'=\star \omega$ where $\star$ is the Hodge operator associated the metric $g$.

Is $(M,\omega ')$ a symplectic manifold? Is it necessarilly symplectic equivalent to the original structure $(M,\omega)$?Namely, is thete a diffeomorphism $f$ which carries $\omega ' $ to $\omega$?

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    $\begingroup$ Yes, because $\omega '=\omega $. $\endgroup$
    – abx
    Commented Sep 17, 2019 at 15:55
  • $\begingroup$ @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. $\endgroup$
    – Ali Taghavi
    Commented Sep 17, 2019 at 16:42
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    $\begingroup$ Yes. You can see that (for instance) in Weil's "Variétés kählériennes". $\endgroup$
    – abx
    Commented Sep 17, 2019 at 17:54
  • $\begingroup$ @abx Thanks for this reference. So what about the same question without Kahlerian assumption? $\endgroup$
    – Ali Taghavi
    Commented Sep 17, 2019 at 18:24
  • $\begingroup$ I would guess it is the same — the equality is checked at each point, the fact that $\omega$ is closed plays no role. $\endgroup$
    – abx
    Commented Sep 18, 2019 at 7:22

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