Let $(M^{2n}, \omega)$ be a (simply connected if it helps) symplectic manifold which is not Kähler. What can we say about its holonomy with respect to a compatible metric $g_\omega$?
I am thinking of the theorem here, which says that the holonomy of a Riemannian manifold $(M^n, g)$ is contained in (a conjugate of) a group $G \subseteq O(n)$ iff $M$ admits a torsion-free $G$-structure.
I believe that a torsion-free $\mathrm{Sp}(2n, \mathbb{R})$-structure is the same thing as a symplectic form (i.e. an almost-symplectic structure with a closed symplectic form).
However, $\mathrm{Sp}(2n, \mathbb{R})$ is not one of the Lie groups appearing in the classification of special holonomy groups - does this mean the holonomy group is a proper subgroup of $\mathrm{Sp}(2n, \mathbb{R})$? If so, what is it?