Line bundles over Kähler–Hodge manifolds 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.
 A: 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.
A: If we consider a usual sympleetic manifold with $[\omega]\in H^2(M,\mathbb Z)$ integral then the Weil's theorem guarantees that there exists a Hermitian line bundle $L$
over $X$ with a conneetion $V$ such that $V$ is compatible with the Hermitan strueture on $L$
and induces the symplectic form by $\omega = curv(V)$.
For the second part of your question ,
The proof is in section "12.2 Integral closed 2-forms and line bundles"
http://www.math.toronto.edu/~jeffrey/mat1312/lec14.gq.pdf
In third part of your question:
Manifolds which are not quantizable then there is no such line bundle $L$ which first chern class be proportional to kahler class
