Let $M$ be a complex manifold, and $\omega$ a closed $2$-form. When is $\omega$ a Kähler form? I mean, when does there exist a Kähler metric for which $\omega$ is the corresponding form.


I would (wildly) guess that necessary and sufficient conditions might be got from the Kähler identities.



When is a Form a Kähler Form?
I decided to make my comment into a more detailed answer. When $M$ has an almost complex structure $J$, then one can talk about smooth complex-valued differential forms of type $(p,q)$ in the usual way. A complex valued $2$-form $\omega$ is type $(1,1)$ if and only if it satisfies $\omega(JX,JY) = \omega(X,Y)$ for all smooth vector fields $X$ and $Y$ on $M$. If $\omega$ is a *real* $2$-form of type $(1,1)$, which means that $\overline \omega = \omega$, and if we define $g(X,Y) = \omega(X, JY)$, then it is easy to show that $g$ is a smooth, symmetric bilinear form on $M$. So it is a Riemannian metric if and only if it is positive definite. This is the definition of a *positive* $(1,1)$-form (that the associated $g$ is positive definite.)


The triple of data $(J, \omega, g)$, where $J$ is an almost complex structure, $\omega$ is a real positive $(1,1)$-form, and $g$ is the associated Riemannian metric as defined above together define an *almost Hermitian manifold*. Now the condition for $M$ to be Kaehler is that $M$ be complex ($J$ is integrable) and that $d\omega = 0$. (These two conditions can be packaged together as $\nabla \omega = 0$ or $\nabla J = 0$, where $\nabla$ is the Levi-Civita connection of $g$.) Hence, if one is starting out with a complex manifold $M$, together with a *closed* real $2$-form, the only additional condition required to ensure that it defines a Kaehler metric is that it be a positive $(1,1)$-form.


