# Does the complex structure of a Kähler manifold preserves the Lie algebra of symplectic vector fields

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?

• 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.
– Danu
Mar 9 '20 at 15:51

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.