Consider symplectic vector fields $X,Y$ and a symplectic connection $\nabla$. Is $\nabla_{X}Y$ symplectic? Consider a symplectic manifold $(M,\omega)$, together with a symplectic connection $\nabla$, _i.e_. a torsion-free connection such that $\nabla{\omega} = 0$. Fix two symplectic vector fields $X$ and $Y$. Is it true that $\nabla_{X}Y$ is again a symplectic vector field? I tried with Cartan calculus but I'm stuck, not beeing able to collect the terms in a clever way. Any other suggestion is welcomed.

The answer is no. Consider as a counter-example $M = \mathbb{R}^2$, $\omega = dx \wedge dy$, $X = x \frac{\partial}{\partial y}$ (whose Hamiltonian function is $-x^2/2$), $Y = y \frac{\partial}{\partial x}$ (whose Hamiltonian function is $y^2 /2$) and the (symplectic) connection given by usual (i.e. component-wise) differentiation of vector fields. Then $\
abla_X Y = x \frac{\partial}{\partial x}$, which is not symplectic as it is not Hamiltonian (or alternatively because it doesn't have vanishing divergence).