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Assume that $X$ is a Moishezon manifold, then there exists a modification $\pi:\tilde{X}\rightarrow X$, where $\tilde{X}$ is a projective algebraic manifold. Let $\tilde{w}$ be a Kahler metric on $\tilde{X}$, then we can construct a holomorphic line bundle $\tilde{L}$ whose first Chern class in terms of the curvature is $\tilde{w}$. Then we can find a divisor $\tilde{D}$ such that the line bundle associated to $\tilde{D}$ is $\tilde{L}$.

Now I consider the image of $\tilde{D}$ under the restriction of $\pi$ to $\tilde{X}\setminus \pi^{-1}(A)$, where $A$ is an analytic set of codimension$\ge 2$ such that $\pi:\tilde{X}\setminus \pi^{-1}(A)\rightarrow X\setminus A$ is a bihilomorphism. Then $\pi|_{\tilde{X}\setminus \pi^{-1}(A)}(\tilde{D})$ is a divisor in $X\setminus A$. By Remmert-Stein Theorem, $\pi|_{\tilde{X}\setminus \pi^{-1}(A)}(\tilde{D})$ can be extended to a divisor, say $D$, in $X$.

Now we denote $L$ the line bundle associated to $D$. Then I guess we can find a singular Hermitian metric $h$ such that the first Chern class of $L$ in terms of the curvature is $\pi_\ast\tilde{w}$?

I have tried to trace the correspondence among first Chern classes, line bundles and divisors, but I feel difficult solving this problem. Can you help me? Thanks in advance!

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    If you do not get any answers in a couple of days, I can move your question to MathOverflow. – Mariano Suárez-Álvarez Apr 17 '16 at 09:11
  • @MarianoSuárez-Alvarez Does this question deserve to post on MathOverflow?I guess by tracing the correspondence among first Chern classes, line bundles and divisors,we can figure it out.I'm just looking for better ideas. – user95633 Apr 17 '16 at 10:31

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