Relation symplectic form and Riemannian metric

Consider a symplectic manifold. If we have also any Riemannian metric $g$ , it turns out that the symplectic gradient and the usual gradient are orthogonal: remember that given a function $H$ the gradient is the only vector field $\nabla H$ such that $d H = g (\nabla H, -)$ . Therefore

g (\nabla H, X_{H}) = d H (X_{H}) = ω (X_{H}, X_{H}) = 0

Also, the Poisson bracket can be computed as

{A, B} = ω (X_{A}, X_{B}) = d A (X_{B}) = g (\nabla A, X_{B}) = \nabla A \cdot X_{B}

So the Poisson bracket can be seen as the dot product of both gradients, for any Riemannian metric!!

This also should also has to do with the note Hermitian form, inner product and symplectic form relationship.