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Let $X$ be complex manifold, the image of $\text{Div}(X)\to \text{Pic}(X)$ is generated by those line bundles with $H^{0}(X,L) \ne 0$ ( i.e. it admits a global non trivial holomorphic section).

The proof for the image of a divisor $D$, gives a non trivial global section goes as follows:


Proof: Any divisor $D=\sum a_i\left[Y_i^i\right] \in \operatorname{Div}(X)$ can be written as $D=\sum a_i^{+}\left[Y_i^{+}\right]-\sum a_j^{-}\left[Y_j^{-}\right]$with $a_k^{\pm} \geq 0$. Hence, $\mathcal{O}(D) \cong \mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right) \otimes$ $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)^*$ and the line bundles $\mathcal{O}\left(\sum a_i^{+}\left[Y_i\right]\right)$ and $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ are both associated to effective divisors and, therefore, admit non-trivial global sections.


My question is we have proved that $\mathcal{O}\left(\sum a_j^{-}\left[Y_j\right]\right)$ admits non trivial global section however $\mathcal{O}(D)$ tensors a dual bundle of it, it's not clear that the dual bundle admits non trivial global section?

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  • $\begingroup$ And we know if $X$ is compact , if both line bundle and its dual admits non trivial global section then the bundle should be trivial? $\endgroup$
    – yi li
    Nov 20, 2022 at 12:33

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It is enough to consider effective divisors. Write a divisor $D$ as $D_1-D_2$, with $D_1,D_2$ both effective, and let $L_1:=\mathcal O(D_1)$, $L_2:=\mathcal O(D_2)$ be the associated bundles. Then $\mathcal O(D)=\mathcal O(D_1)\otimes \mathcal O(-D_2)= L_1\otimes L_2^* \in \langle L_1,L_2\rangle$, with $H^0(X,L_1)\neq 0$ and $H^0(X,L_2)\neq 0$, because $D_1,D_2$ are effective. Therefore, the image of the map $\mathrm{Div}(X)\to \mathrm{Pic}(X)$ is generated by line bundle having global sections, i.e. arising from effective diviors.

For the question in the comment, it is also correct: if both $L=\mathcal O(D)$ and $L^*= \mathcal O(-D)$ have non trivial global section, there exist divisor $A\in |D|$ and $B\in |-D|$. But $A+B$ is a principal divisor, and $A$ and $B$ are bot effective, whence $A=B=0$.

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  • $\begingroup$ thank you @Desperado , how to prove that $\mathrm{Pic}(X) $ is generated by line bundle having global section? $\endgroup$
    – yi li
    Nov 20, 2022 at 13:23
  • $\begingroup$ Sorry, I meant the image of the map $Div(X) \to Pic (X)$. I edit my answer accordingly $\endgroup$
    – Pomponazzo
    Nov 20, 2022 at 18:18
  • $\begingroup$ How to prove it then, that's what I want to prove? $\endgroup$
    – yi li
    Nov 21, 2022 at 0:27
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    $\begingroup$ A non effective divisor has no nontrivial global sections at all. But $L_1:=\mathcal O (\sum a_i^+[Y_i])$ and $L_2:=\mathcal O (\sum a_j^-[Y_j])$ are both effective and they do have nontrivial global sections. So $L=L_1\otimes L_2^*$ belongs to the subgroup of $\mathrm{Pic} (X)$ generated by $L_1$ and $L_2$, so in a subgroup generated by line bundles with nontrivial global sections. $\endgroup$
    – Pomponazzo
    Nov 21, 2022 at 19:40
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    $\begingroup$ thank you, Desperado, excellent solution! I didn't realize that as a subgroup $L_2$ will generate $L_2^*$. $\endgroup$
    – yi li
    Nov 22, 2022 at 0:23

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