Symplectic connections are (locally) Levi-Civita connections I was wondering... Is every symplectic connection $\nabla$
on some symplectic manifold $(M,ω)$ the Levi-Civita connection of some metric $g$ on $M$? What about the local statement? 
 A: I think I've got the answer, and it is "no", at least for the global question.
When I started to try to understand the problem, I realized that to get some structure that is invariant by the connection we can (have to) fix the structure at some point and use parallel transport through picewise smooth paths to extend it. The resulting structure will be well-defined iff the initially fixed structure is invariant by the holonomy group at the point. The idea is then to construct some connection whose holonomy group at some point preserves some symplectic structure on the tangent space of the point but does not preserve any metric on such tanget space. To get the second feature, we may try to construct the connection in such a way that its holonomy group is unbounded. After some tries, I got the following:
On $\mathbb{R}^2$, we consider the connection $\nabla$ given by \begin{align*} \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial x} & = y\frac{\partial}{\partial x}  &  \nabla_{\frac{\partial}{\partial x}}\frac{\partial}{\partial y} & = 0 \\
\nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial y} & = x\frac{\partial}{\partial y}  & \nabla_{\frac{\partial}{\partial y}}\frac{\partial}{\partial x} & = 0.
\end{align*}
The curvature $R$ of $\nabla$ at $(0,0)$ is given by  $R(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}) = \begin{bmatrix} 
-1 & 0 \\
0 & 1
\end{bmatrix} \in GL(\mathbb{R}^2) = GL(T_{(0,0)}\mathbb{R}^2)$, and all its covariant derivatives are of the form $\begin{bmatrix} 
-c & 0 \\
0 & c
\end{bmatrix}$ at such point. Because of that and of the connectedness of $Hol(\nabla, (0,0))$ (simply-connectedness of $\mathbb{R}^2$), Ambrose-Singer's theorem implies that the canonical symplectic form of $T_{(0,0)}\mathbb{R}^2 = \mathbb{R}^2$ is invariant by $Hol(\nabla, (0,0))$. On the other hand, $Hol(\nabla, (0,0))$ contains the exponential of $tR(\frac{\partial}{\partial x}, \frac{\partial}{\partial y})$ for every $t \in \mathbb{R}$, and is therefore an unbounded subset of $GL(T_{(0,0)}\mathbb{R}^2)$. As such, it doesn't fix any metric on $T_{(0,0)}\mathbb{R}^2$.

$ \textit{edit:}$ using Levi-Civita's formula, it is easy to conclude that the connection $\nabla$ defined above is not the Levi-Civita connection of any metric, even locally.
With that in mind, Darboux's theorem shows us that for every symplectic manifold there exist some open subset $U$ of the manifold and some symplectic connection $\nabla$ defined on $U$ such that $\nabla$ is not the Levi-Civita connection of any metric on $U$.
By using partitions of unity, we conclude, then, that $\textbf{every symplectic manifold admits}$ $\textbf{a symplectic connection that is  not a}$ $\textbf{Levi-Civita connection.}$
A: For any Riemannian metric $g$ on a symplectic manifold $(M, \omega)$ there exists (canonical) almost complex structure $J$ making these threetensors compatible (any one can be defined by other 2). compatible with $\omega$, which means that $\omega(\_, J\_)$ defines a Riemannian metric $\widetilde{g}$. For such a compatible triple and the Levi-Civita connection $\widetilde{\nabla}$ the equation $\widetilde{\nabla} \omega = 0$ is equivalent to $J$ being integrable thus making $(M,\widetilde{g},\omega, J)$ into Kähler manifold. There is plenty of symplectic manifolds which are not Kähler. So if there is metric $g$ for a non-Kähler Fedosov manifold $(M, \omega, \nabla)$ whose Levi-Civita connection is $\nabla$ then it cannot be compatible with $\omega.$
