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Let $X$ be a compact complex manifold, and with the following sheaves

  • $\mathscr O$, the sheaf of holomorphic function,
  • $\mathscr O^*$, the sheaf of nonvanishing holomorphic function
  • $\mathscr K^*$ the sheaf of nonidentically zero meromorphic function.

A divisor is an element in $\Gamma(X, \mathscr K^*/\mathscr O^*)$ (Cartier divisor). The short exact sequence of sheaves $$ 0 \to \mathscr O^* \to \mathscr K^* \to \mathscr K^*/\mathscr O^* \to 0$$

induces a long exact sequence in cohomology

$$\cdots \to \Gamma(X, \mathscr K^*) \to \Gamma(X, \mathscr K^*/\mathscr O^*) \to H^1(X, \mathscr O^*) \to \cdots$$

and $\Gamma(X, \mathscr K^*/\mathscr O^*) \to H^1(X, \mathscr O^*)$ identify a divisor to a line bundle.

I suppose that the map is not in general surjective. Can anyone give an example?

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1 Answer 1

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The short paper "The sheaf of nonvanishing meromorphic functions in the projective algebraic case is not acyclic" (www.math.wustl.edu/~matkerr/MerR.pdf) explains that even for a projective algebraic variety $X/\mathbb{C}$, the sheaf $\mathcal{K}^*$ in the analytic topology is not acyclic in general, and in particular that as soon as $\dim(X)\geq 2$ and $H^1(X,\mathbb{Z})\neq 0$, one has $H^1(X,\mathcal{K}^*)\neq 0$ !

The proof is short and clear, so I will not reproduce it here. This provides plenty of examples, for instance any abelian variety of dimension $\geq 2$.

This does not contradict GAGA because $\mathcal{K}^*$ is not coherent.

There may be simpler, complex analytic counter-examples, and I hope others will provide some.

Edit: This does not answer the question, because in the projective case the map $H^1(X,\mathcal{O}^*_X)\rightarrow H^1(X,\mathcal{K}^*_X)$ is $0$. Hence one really need a non-projective example.

Edit 2: In fact this question has already been answered in math.SE : When is $CaCl(X) \to Pic(X)$ surjective? It turns out that a very general, non-algebraic complex torus of dimension 2 answers your question.

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    $\begingroup$ According to Huybrecht's textbook on Complex Geometry, Exercise 3.3.6, the complex torus $\mathbb{C}^2/\mathbb{Z}^4$ for the right choice of lattice provides a counterexample. I don't really know the methods of complex geometry, so if someone more knowledgeable wants to write up a proof of this, that would be great! $\endgroup$
    – Cass
    Commented Feb 19, 2015 at 13:51
  • $\begingroup$ Everything you wrote is correct, but I note that one (apparently) needs more information than just $H^1(K^*) \neq 0$ to answer the OP's question. What about $H^1(K^*/O^*)$? $\endgroup$
    – user64687
    Commented Feb 19, 2015 at 14:10
  • $\begingroup$ @AsalBeagDubh: I think $H^1(K^*) \neq 0$ will be good enough. $\endgroup$
    – user99914
    Commented Feb 19, 2015 at 14:38
  • $\begingroup$ Thanks a lot. This is exactly what I am looking for. $\endgroup$
    – user99914
    Commented Feb 19, 2015 at 14:40
  • $\begingroup$ Hi, after some thought it seems that the paper alone does not answer my question: $H^1(K^*)\neq 0$ does not imply, as @AsalBeagDubh mentioned, that $\Gamma(K^*/O^*) \to H^1(O^*)$ is onto. $\endgroup$
    – user99914
    Commented Feb 19, 2015 at 20:54

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