Sheaf with free stalks - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T12:08:05Z http://mathoverflow.net/feeds/question/75150 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75150/sheaf-with-free-stalks Sheaf with free stalks maxim 2011-09-11T16:52:26Z 2011-09-12T15:19:34Z <p>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?</p> <p>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 <a href="http://mathoverflow.net/questions/44839/wikipedias-definition-of-locally-free-sheaf" rel="nofollow">here</a></p> <p><strong>Remark:</strong> As it was pointed out by Francesco Polizzi, this is true if $\mathcal{F}$ is coherent. What if we do not know it apriori?</p> http://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75152#75152 Answer by Francesco Polizzi for Sheaf with free stalks Francesco Polizzi 2011-09-11T17:24:14Z 2011-09-11T17:24:14Z <p>The answer is <strong>yes,</strong> at least when $\mathcal{F}$ is a <em>coherent</em> sheaf.</p> <p>This actually holds for any complex space. See [Grauert-Remmert, Coherent Analytic Sheaves, p. 90].</p> http://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75161#75161 Answer by Donu Arapura for Sheaf with free stalks Donu Arapura 2011-09-11T21:38:12Z 2011-09-12T15:19:34Z <p>Just looking at stalks is not enough:</p> <p>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.</p> <p><hr> <strong>Notes</strong> Rather than editing, I'll keep the original form of my answer in tact and add a few footnotes.</p> <ol> <li><p>Of course, this $\mathcal{F}$ is not coherent.</p></li> <li><p>(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 <code>$$\varinjlim\bigoplus_{y\in U} \mathcal{O}_y$$</code> 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.)</p></li> <li><p>(Re: Laurent's comment.) By $i_{x*}\mathcal{O}_x$, I meant the skyscraper sheaf associated to $\mathcal{O}_x$.</p></li> </ol> http://mathoverflow.net/questions/75150/sheaf-with-free-stalks/75167#75167 Answer by unknown (google) for Sheaf with free stalks unknown (google) 2011-09-11T22:10:54Z 2011-09-11T22:10:54Z <p>This is a small modification of Donu's answer.</p> <p>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.</p>