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Hi, as the title says I'm looking for a way to see the projection $\pi:H^1(X,\mathcal{O}_X)\rightarrow \operatorname{Pic}^{0}(X)$ in terms of holomorphic structures on $X\times\mathbb{C}$. ($X$ is a compact complex manifold and $\operatorname{Pic}^0(X)$ is the kernel of $c_1:H^1(X,\mathcal{O}_X^{ * })\rightarrow H^2(X,\mathbb{Z})$ in the long exact sequence induced by the exponential sequence). Since $$H^1(X,\mathcal{O}_X) \simeq H_{\overline{\partial}}^{0,1}$$ by the Dolbeault isomorphism, I take $[c]\in H^1(X,\mathcal{O}_X)$, so I take the corresponding $[\gamma]\in H_{\overline{\partial}}^{0,1}$ and a representative $\gamma$ of the class $[\gamma]$. My wrong thought was that $\pi([c])=(X\times\mathbb{C},\overline{\partial}+\gamma)$, i.e. to associate to $[c]$ the trivial bundle with the "holomorphic structure" $\overline{\partial}+\gamma$, but it is not a holomorphic structure unless $\gamma\wedge\gamma=0$! So how can I see explicitly (if it is possible) the map $\pi$ in terms of holomorphic structures on the trivial line bundle?

Thank you in advance.

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I have edited some TeX, as well as correcting capitalization and a few typos. If I have changed any meaning (or made any other errors) I apologize, and feel free to roll back the edits. – Theo Johnson-Freyd Aug 25 '10 at 5:20
I couldn't read the question before. Now that I can, I will point out that $\gamma\wedge \gamma$ is zero (it's a $1$-form). So your initial idea should work. – Donu Arapura Aug 25 '10 at 12:13
@Donu: you're definitely right and i'm definitely stupid:-) thank you! – Italo Aug 25 '10 at 12:44
This happens to everyone. Perhaps I might also to mention that a similar (conjugate) description was employed by Green and Lazarsfeld in their work (Invent '87, JAMS '90). – Donu Arapura Aug 25 '10 at 13:04

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