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Suppose $X$ is a complex manifold, we have the map $H^0(X,K_X^*/O_X^*)\to H^1(X,O_X^*)$, is the canonical line bundle $\wedge^n{\Omega}$ always in the image of the map?

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Did you mean to write $H^0(X,\mathcal{K}_X^*/\mathcal{O}_X^*)$? –  Jason Starr May 12 at 15:50
    
It's true if $X$ is algebraic. –  Ariyan Javanpeykar May 12 at 17:44

2 Answers 2

up vote 13 down vote accepted

The answer is no. The group $H^0(X, \mathcal{K}_X^*/\mathcal{O}_X^*)$ is isomorphic to the divisor group $\mathrm{Div}(X)$ (see e.g. Huybrechts' Complex Geometry, Prop. 2.3.9), so its image in $\mathrm{Pic}(X)$ is the group of line bundles associated to some divisor. Now there are complex compact surfaces which do not contain any curve but with a nontrivial canonical bundle, for instance the Inoue surfaces (see Barth et al., ch. V, §19).

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Perhaps I did not get you question. Anyway, $H^{1}(X,\mathcal{O}_X^{*})\cong Pic(X)$. If $X$ is an integral schemes the Picard group is isomorphic to the class group of Cartier divisors. Therefore one can interpret $\bigwedge^n\Omega_X$ both as a line bunlde on $X$ and as a Cartier divisor $K_X$ in $X$. Does this answer to your question?

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The question is about a complex manifold, not an integral scheme. –  abx May 12 at 18:34

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