2
$\begingroup$

In Demailly's book p.272, Theorem 13.9, there is:

Let $X$ be an arbitrary complex manifold. (b) Let $\omega$ be a $\mathcal C^∞$ closed real (1, 1)-form such that ${ω}\in H^2_{dR}(X,\mathbb R)$ is the image of an integral class. Then there exists a hermitian line bundle $E\to X$ such that $\frac{i}{2π}Θ(E) = ω$.

It should be noted that the author dose not assume $X$ to be a compact Kähler manifold, while in Voisin's book Hodge theory and complex algebraic geometry. I, p.163-164, $X$ is assumed to be a compact Kähler manifold, which is used to get the $\partial\bar\partial$-lemma, and use it to deduce that $\omega-\omega_{L,h}=\frac{1}{2\pi i}\partial\bar\partial\phi$, where $\omega_{L,h}$ is the curvature form of a Hermitian holomorphic line bundle $L$.

So, my question is that: is the compact and Kähler (or $\partial\bar\partial$-) assumptions necessary to make that there is a holomorphic line bundle satisfying $\frac{i}{2π}Θ(E) = ω$?

By the way, I find Demailly'proof a bit hard to understand since there are some typos, for example, I can't find Th. I-3.35 in his book, so can anybody help me figure out whether his statement is right or wrong?

$\endgroup$
3
  • 4
    $\begingroup$ The inter-chapter references are a little borked, but Prop. III-1.20 should be a local $\partial\bar\partial$ lemma, valid on any complex manifold, and Th. I-3.35 should be Th. I-5.16. The statement (originally a lemma due to Weil) is correct as written; the proof is basically "get a Cech cocycle that represents the class, find local holomorphic representatives, lift those via the exponential function, and use those to cook up the transition functions for the line bundle". $\endgroup$ Sep 21, 2022 at 13:56
  • 2
    $\begingroup$ @GunnarÞórMagnússon Please write that up as an answer! $\endgroup$ Sep 21, 2022 at 14:26
  • $\begingroup$ @GunnarÞórMagnússon, thanks for pointing out that "Th. I-3.35 should be Th. I-5.16" and ensure the statement in the question is true, they are so helpful, it gives me more courage to try to understand Demailly's proof. $\endgroup$
    – Tom
    Sep 21, 2022 at 15:13

1 Answer 1

1
$\begingroup$

Thanks to the aid of @Gunnar Þór Magnússon, I will write down my understanding of Demailly's proof, if there is anything unclear, please comment below.

Let $\cup_{i\in I}U_i$ be a covering of $X$ such that $U_i\cap U_j$ is simply connected. Since $\omega$ is a closed real (1,1) form on $X$, by local $\partial\bar\partial$-lemma (see Demailly's book mentioned in the question, p.135. Prop 1.19), in $U_i$, there is a real-valued $\mathcal C^{\infty}$ function $\phi_i$ such that $\frac{i}{2\pi}\partial\bar\partial\phi_i=\omega$.

Note that $\phi_i-\phi_j\in \mathcal C^{\infty}(U_i\cap U_j)$ is $\partial\bar\partial$-closed, by Th. I-5.16 ( in Demailly's book p.42 ), there exists a holomorphic function $f_{ij}\in \mathcal O(U_i\cap U_j)$ satisfying $\phi_i-\phi_j=2\text{Re}f_{ij}$.

Let $g_{ij}=e^{2\pi if_{ij}}\in\mathcal O^*(U_i\cap U_j)$, it is easy to check $g_{ij}\cdot g_{jk}\cdot g_{ki}=1$ and $g_{ij}\cdot g_{ji}=1$. So the transition functions $\{g_{ij}\}$ determin a holomorphic line bundle $L$ over $X$. And $e^{\phi_i}$ is the Hermitian metric of $L$ which satsfies $\frac{i}{2\pi}\partial\bar\partial\phi_i=\omega$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.