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Let $(M, \omega, g, J)$ be a Kähler manifold with symplectic form $\omega$, Riemannian metric $g$ and complex structure $J$.

Question: If $X$ is a symplectic vector field, is $JX$ also symplectic?

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    $\begingroup$ I see no reason why this would be the case and I would be rather surprised if it were, but I don't have time to think of a counterexample right now. $\endgroup$
    – Danu
    Mar 9 '20 at 15:51
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This is almost never the case. Given a smooth function $H : M \to \mathbb{R}$, the Riemannian gradient $\nabla H$ and the symplectic gradient $X_H$ are related by $X_H = J \nabla H$ (up to a multiplicative sign depending on convention). Since $H$ is a Lyapunov function for $\nabla H$, the flow of $\nabla H$ is not volume-preserving in general, hence does not consist in symplectomorphisms.

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