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Let $X$ be a projective scheme over $\mathbb{C}$ and $\mathcal{F}$ a locally free sheaf on $X$ of rank $2$. Take a point $p$ in the support of $\mathcal{F}$. Suppose that there exists at least $3$ linearly independent global sections none of which vanish at this point $p$. Is it then true that the natural morphism from $H^0(\mathcal{F})$ to the stalk $H^0(\mathcal{F}_p)$ is surjective? If not, is there any bound on the number of linearly independent global sections as above which will ensure a positive answer?

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up vote 4 down vote accepted

I do not think this is true. Since the question is local, we may assume $X = \operatorname{Spec}(A)$ is affine and $\mathcal F$ is free on $X$. Thus, $\mathcal F = \mathcal O_X\oplus\mathcal O_X$, the global sections restrict to $H^0(X, \mathcal F) = A\oplus A$, and $\mathcal F_p = A_p\oplus A_p$. The natural map factors through $A\oplus A\to A_p\oplus A_p$, which is certainly not always surjective.

For example, if we start with $X = \mathbb P^1$ and $\mathcal F = \mathcal O(1)\oplus\mathcal O(1)$, then the support is all of $\mathbb P^1$ and there are linearly independent global sections $(x,0),(y,0),(0,x),(0,y)$, where $x,y$ are coordinates. But in an affine chart with coordinate $u$, setting $p = 0$, we end up with the map $k[u]\oplus k[u]\to k[u]_{(u)}\oplus k[u]_{(u)}$ which is not surjective.

Note that by playing this game with higher twists on $\mathbb P^1$, we can get any number of linearly independent global sections, but the local picture will remain unchanged.

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I guess by $k[u]$, you mean the $k$-vector space generated by $u$. But in that case I think $k[u]_{(u)}$ is isomorphic to $k$. Am I getting something wrong? – user46578 Feb 6 '14 at 16:37
Dear @user46578, no, I mean $A = k[u]$ is the affine coordinate ring of say $y\neq 0$ in $\mathbb P^1$. So $u = x/y$ is a coordinate on $\mathbb A^1$. Then the local ring at the origin $p = 0$ in this chart is given by the localization of $A$ at the prime ideal $(u)$. – Andrew Feb 6 '14 at 16:39
But space of global sections of a variety is finite dimensional $k$-vector space which $k[u]$ is not. Isnt this so? – user46578 Feb 6 '14 at 16:42
Yes, that is so. But I am localizing the global sections to an affine chart first, and then localizing again to the stalk. Maybe that gives an even easier argument though: global sections are finite dimensional, but $A_p$ is not. How can the localization then be surjective? – Andrew Feb 6 '14 at 16:46
Dear @user46578, what do you mean by a global section of $\mathcal F_p$? Is $\mathcal F_p$ not the stalk of the sheaf $\mathcal F$ at $p$ as usual? – Andrew Feb 6 '14 at 17:23

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