Vrftjkry

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?

For any Riemannian metric $g$ on a symplectic manifold $(M, omega)$ there exists (canonical) almost complex structure $J$ making these three tensors compatible (any one can be defined by other 2). It turns out that for a Levi-Civita connection $nabla$ the equation $nabla omega = 0$ is equivalent to $J$ being integrable thus making $(M,g,omega, J)$ to be Kähler manifold. And there is plenty of symplectic manifolds which are not Kähler.

I think I've got the answer, and it is "no", even in the local case.

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} & = yfrac{partial}{partial x} & nabla_{frac{partial}{partial x}}frac{partial}{partial y} & = 0 \
nabla_{frac{partial}{partial y}}frac{partial}{partial y} & = xfrac{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 zero 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 $T_{(0,0)}mathbb{R}^2$.