TITLE: Line bundles over Kähler–Hodge manifolds
QUESTION [6 upvotes]: A Kähler–Hodge manifold $M$ can be defined as a Kähler manifold whose Kähler form $\omega$ is integral, namely $\omega\in H^{2}(M,\mathbb{Z})$. It is known then that there always exists a Hermitian line bundle $L\to M$ whose Chern class satisfies $c_{1}(L) = [\omega]$. I have the following questions:

Does anyone know a reference where the statements above are proven?
How can this line bundle $L$, given a particular integral Kähler form $\omega$, be explicitly constructed?
What happens if $\omega$ is not integral? Can one still construct some kind of $\mathbb{R}$-bundle whose curvature is $\omega$?

Thanks.

REPLY [6 votes]: At least when $M$ is compact, which I assume here,
the standard reference for this subject is probably the book

Algebraic Geometry, Griffiths & Harris 

but the following books are excellent references as well :

Differential analysis on complex manifolds, Wells
Complex Analytic and Differential Geometry, Demailly
Hodge Theory and Complex Algebraic Geometry, Voisin 

A proof could run as follows:

Hodge's theorem implies that there exists a complex line bundle $L$
on $M$ with first Chern class $c_1(L)=\omega$;
Endow $L$ with an hermitian metric $(L,\|\cdot\|_0)$; its curvature form 
$c_1(L,\|\cdot\|_0)$ is a $(1,1)$-form cohomologous to $\omega$. By the $dd^c$-lemma, there exists a function $\phi$ such that $\omega= c_1(L,\|\cdot\|_0)+dd^c\phi$; then $\|\cdot\|=e^{-\phi}\|\cdot\|_0$ is a hermitian metric on $L$ with curvature form equal to $\omega$.

For an explicit construction of $L$: the standard proof of Hodge's theorem is cohomological, and uses a long exact sequence in cohomology. Tracing down the arguments gives an actual construction if one is given an open covering of $M$ by contractible open sets.
The arguments really require that $\omega$ be integral. A generic complex torus of dimension $>1$ carries no complex line bundle, but has many Kähler forms.