Is a Kähler manifold necessarily symplectic?

Question

Let $M$ be a Riemannian manifold. If we pick a basepoint $p \in M$, then for any smooth path $\gamma: [0, 1] \to M$, parallel transport along $\gamma$ induces an automorphism $g_\gamma \in \text{Aut}(T_pM) \cong \text{O}(2n, \mathbb{R})$. The holonomy group $G_p$ is the subgroup of $\text{Aut}(T_pM)$ generated by all such loops $\gamma$. One definition of a Kähler manifold is a $2n$-dimensional Riemannian manifold $M$ such that the holonomy group $G_p$ is conjugate into $\text{U}(n) \subset \text{O}(2n, \mathbb{R})$. My question is, is a Kähler manifold necessarily symplectic, i.e. is there a closed $2$-form $\omega$ such that $\omega^n$ is nowhere vanishing?

Answer 1

Yes. The holonomy description is equivalent to the existence of a complex structure $I$ and a riemannian metric $g$ such that $$\omega(X,Y):=g(IX,Y),$$ defines a closed 2-form. The non-degeneracy of $g$ implies that $\omega$ is non-degenerate.