# Kähler forms arising as the curvature form of a singular metric on a line bundle

The Fubini-Study metric on complex projective space $\mathbb{P}^n$ is a smooth metric $h = e^{-\phi}$ on the line bundle $\mathcal{O}(1)$ and it is a standard calculation to check that its curvature form $\frac{i}{\pi}\partial \overline{\partial} \phi$ is a Kähler form on $\mathbb{P}^n$.

I'm wondering whether this generalizes. To be precise: given a Kähler manifold $X$ with a Kähler form $\omega$, does there exist a holomorphic line bundle $L \to X$ and a singular metric $h = e^{-\phi}$ on $L$ such that $\omega = \frac{i}{\pi} \partial \overline{\partial} \phi$?

Aside from the example of $\mathbb{P}^n$, there is another class of examples that might lead to such a guess: if $X$ is smooth projective variety (over $\mathbb{C}$) and $L \to X$ is an ample line bundle, then the curvature form of any singular metric on $L$ is cohomologous to $c_1(L)$, which is a Kähler form.

If $\omega=\frac{i}{2\pi}\partial\bar\partial\phi$ for some holomorphic hermitian line bundle $(L,h)$ with $h=e^{-\phi}$, then necessarily $[\omega]\in H^{1,1}(X,\mathbb R)$ is an integral, i.e., represents class in $H^2(X,\mathbb Z)$, so unless $[\omega]$ is integral, such line bundle does not exist.
On the contrary, for any (1,1)-form $\omega$ with $[\omega]\in H^{1,1}(X,\mathbb R)\cap H^2(X,\mathbb Z)$ there exists a holomorphic hermitian line bundle $(L,h)$ such that its curvature form $\Theta(L,h)=\partial\bar\partial\phi$ equals $\frac{2\pi} {i}\omega$, see, e.g., Griffiths-Harris "Principles of algebraic geometry" $\S1.2$.