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I saw a statement in a question Non-compact Kähler manifolds which admit a positive line bundle, which says that any compact complex manifold that admits a positive line bundle must be a Kähler manifold automatically. Can someone explain to me why?

The second confusion appears in J.P. Demailly's book "Analytic Methods in Algebraic Geometry", Chapter 8, Corollary 8.3 here. How can he directly prove a compact complex manifold carrying a holomorphic big line bundle is a Moishezon manifold? There's not any hints and proof below Corollary 8.3 doesn't cover this statement.

(This is the first time I ask question on MathOverflow, I hope I can express my question accurately enough, and show my great gratitude to anyone who is willing to answer my question.)

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    $\begingroup$ Since this seems to be two questions, it might be better to ask them in two separate posts. (For example, you will eventually need to mark only one answer as accepted, and it is easier to do so when there is only one question to answer.) $\endgroup$
    – LSpice
    Commented Dec 10, 2023 at 19:08
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    $\begingroup$ Any complex manifold who admits a positive line bundle is automatically Kahler, compactness is unnecessary. $\endgroup$
    – Invariance
    Commented Apr 12 at 2:23

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The answer probably depends a lot on how exactly you define a Kähler manifold as there are multiple equivalent definitions. In any case, the definition of a positive line bundle is that it has a metric whose curvature 2-form is positive. The curvature 2-form is skew-Hermitian so this means that multiplying by $i$ to make it Hermitian gives a positive definite Hermitian form. This positive definite Hermitian form then gives the Kähler structure.

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  • $\begingroup$ Thank you for your answer! It indeed solves my first confusion. May I ask if you have any hint about the second confusion in the question? About existence of big line bundle imply Moishezon? $\endgroup$
    – Kenny S
    Commented Nov 8, 2023 at 2:45
  • $\begingroup$ @KennyS For a line bundle $L$ on $X$ a section of $H^0( X, L^n)$ divided by another section of $H^0(X,L^n)$ gives a meromorphic function. A big line bundle has a "lot" of sections and a Moishezon manifold has a "lot" of meromorphic functions, so roughly the point is to check that if you have enough sections then dividing them gives you enough meromorphic functions. But I can't think of exactly how to do this. $\endgroup$
    – Will Sawin
    Commented Nov 8, 2023 at 14:17
  • $\begingroup$ Very good intuition! I will think about it more..... And if you have any progress, please just reply me directly when you have time. $\endgroup$
    – Kenny S
    Commented Nov 9, 2023 at 3:00
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    $\begingroup$ The fact that a compact complex manifold is Moishezon if and only if it admits a big line bundle is proved for example in this textbook by Ma-Marinescu doi.org/10.1007/978-3-7643-8115-8 Theorem 2.2.15 $\endgroup$
    – YangMills
    Commented Dec 15, 2023 at 19:41
  • $\begingroup$ @YangMills Wow...Very detailed proof there, thank you! $\endgroup$
    – Kenny S
    Commented Apr 9 at 5:23
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About your second question: one equivalent definition of big line bundle is "the birational version of ample", meaning that some tensor power of it defines a rational map into projective space that is birational onto its image. In particular, consider how the fields of meromorphic functions and dimensions as complex manifolds of you initial manifold and the closure of its image in P^n compare, and conclude :)

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