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.
On
As Jordan Payette correctly pointed out, your statement is in general not true. However, $$ \nabla_X Y - \nabla_Y X $$ is indeed symplectic: since the torsion vanishes, it is equal to $ [X,Y] $ and $$ \mathcal L_{[X,Y]} \omega = \mathcal L_X \mathcal L_Y \omega-\mathcal L_Y \mathcal L_X \omega=0\,. $$
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 $\nabla_X Y = x \frac{\partial}{\partial x}$, which is not symplectic as it is not Hamiltonian (or alternatively because it doesn't have vanishing divergence).