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Say we are given a complex manifold $X$ and an $\mathcal{O}_X$-module $\mathcal{F}$. Assume that for any point $P\in X$ the stalk $\mathcal{F}_P$ is a free $(\mathcal{O}_X)_P$-module of finite rank. Does it imply that $\mathcal{F}$ is locally free? If not, what do you need to know additionally about $\mathcal{F}$ to make it true?

Note that if we were looking at the case of schemes then it would be wrong in general. Mathoverflow answer to a related question is here

Remark: As it was pointed out by Francesco Polizzi, this is true if $\mathcal{F}$ is coherent. What if we do not know it apriori?

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This is exercise II.5.7 in Hartshorne. – J.C. Ottem Sep 11 '11 at 19:12
Hartshorne's exercise is about coherent sheaves. As the remark above says, we do not assume this apriori. – maxim Sep 11 '11 at 21:01
up vote 6 down vote accepted

Just looking at stalks is not enough:

Suppose that $X$ is a nontrivial complex manifold. Let $i_x:x\to X$ denote the inclusion, and set $$\mathcal{F} =\bigoplus_{x\in X} i_{x*}\mathcal{O}_x$$ Notice that it is naturally an $\mathcal{O}_X$-module with $\mathcal{F}_x\cong \mathcal{O}_x$, and yet it is certainly not locally free.

Notes Rather than editing, I'll keep the original form of my answer in tact and add a few footnotes.

  1. Of course, this $\mathcal{F}$ is not coherent.

  2. (Re: UG's first comment.) I probably should have included the proof that $\mathcal{F}_x\cong \mathcal{O}_x$. Here it is. The left is the direct limit $$\varinjlim\bigoplus_{y\in U} \mathcal{O}_y$$ as $U$ shrinks to $x$. There is a projection $p$ to $\mathcal{O}_x$ which is surjective since it has a section. Suppose that $f=\sum f_y$ lies in the kernel of $p$. Shrink $U$ to avoid the support of $f$ (which excludes $x$). Then we see that the class of $f$ in the direct limit must be zero. (There is a reason I took the sum and not the product.)

  3. (Re: Laurent's comment.) By $i_{x*}\mathcal{O}_x$, I meant the skyscraper sheaf associated to $\mathcal{O}_x$.

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The stalk of your sheaf $\mathcal F$ at $x$ is much larger than $\mathcal O_x$. Rather, it's all germs around $x$ of "functions" associating to a point $y$ an element of $\mathcal O_y$. – John Pardon Sep 11 '11 at 22:07
Ah, yes, sorry. I was thinking direct product. – John Pardon Sep 11 '11 at 23:11
No problem. I had my doubts about it too. – Donu Arapura Sep 11 '11 at 23:30
I guess the sheaf you have in mind in the direct sum is not $i_*\mathcal{O}_x$, but the skyscraper sheaf at $x$ with stalk the free $\mathcal{O}_x$-module of rank one. If by $i$ you mean the inclusion of the point $x$, as a morphism of ringed spaces, then $i_* \mathcal{O}_x$ is the same skyscraper sheaf (of abelian groups), but as an $\mathcal{O}$-module it is killed by the maximal ideal of $x$. – Laurent Moret-Bailly Sep 12 '11 at 5:31
Yes, I meant the skyscraper the sheaf. I'll fix it in a bit. – Donu Arapura Sep 12 '11 at 12:31

This is a small modification of Donu's answer.

Let $\mathcal F=(i_x)_\ast\mathcal O_x$ (the skyscraper sheaf of $\mathcal O_x$ over $x$) for some $x\in X$. Then $\mathcal F$ is locally free of rank $1$ at $x$, and is locally free of rank $0$ everywhere else. Clearly $\mathcal F$ is not locally free near $x$, since it doesn't even have locally constant rank.

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The answer is yes, at least when $\mathcal{F}$ is a coherent sheaf.

This actually holds for any complex space. See [Grauert-Remmert, Coherent Analytic Sheaves, p. 90].

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Thank you for the answer! Right, this is because the support of a coherent sheaf is closed. What if the sheaf is not coherent? – maxim Sep 11 '11 at 17:57

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