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For a holomorphic vector bundle $E$ over a complex manifold $M$, we denote its space of smooth sections by $\Gamma^{\infty}(E)$, and its space of holomorphic sections by $\Gamma^{hol}(E)$. Now I've been looking at the line bundles $L_k$ over the complex projective spaces ${\bf C} P^N$, and I have managed to show that $\Gamma^{\infty}(L_k)$ is generated as a $C^{\infty}({\bf C} P^N)$-module by $\Gamma^{hol}(E)$, which is to say that every element $\Gamma^{\infty}(E)$ is a sum of elements of the form $ef$, where $e \in \Gamma^{hol}(E)$, and $f \in C^{\infty}({\bf C} P^N)$.

I am guessing that this result is extremely well known, and an example of a well understood general phenomenon. So I would like to ask if there is a characterization of the manifolds for which this result holds, for both the case of line bundles alone, and holomorphic vector bundles of general dimension?

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up vote 5 down vote accepted

Swan has proved that taking global section gives an anti-equivalence between finitely generate projective $\Gamma^{\infty}(M)$-modules and $C^{\infty}$ vector bundles on $M$; this correspondence is functorial in $M$. Hence a set of section of a $C^{\infty}$ vector bundle $E$ on $M$ generated $\Gamma^{\infty}(E)$ if and only if it generates each fiber of $E$. So if $E$ is holomorphic, the holomorphic sections generate if and only if $E$ is globally generated. In particular, this is always true if $M$ is Stein. In the case of $L_k$ on $\mathbb{CP}^N$, this is true if and only if $k ≥ 0$.

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Excuse my ignorance, but what is a globally generated vector bundle? – Jean Delinez Feb 10 '13 at 15:56
It is a vector bundles whose global sections generate each fiber. – Angelo Feb 10 '13 at 19:47
Sorry, but I would like to get this totally clear for my mind. What you are saying is that there exists for $v \in E_p$, for any $p \in M$, an element $s \in \Gamma^{\infty}(E)$ of the smooth sections (or global sections as you call them), such that $v = s(p)$. Is this correct? – Jean Delinez Feb 11 '13 at 16:06
.... but I can't see why every vector is not globally generated. Surely, for $v \in E_p$ as above, we have a local section for which $s(p) = v$, why can one not just use a partition of unity argument to extend $s$ to a global section, and hence conclude that your vector bundle is globally generated? I can't see the problem in my logic here. – Jean Delinez Feb 11 '13 at 16:13
A holomorphic vector bundle is globally generated when for every point $p$ of $M$ and every $v \in E_p$ there exists a holomorphic global section $s$ with $s(p) = v$. There are lots of holomorphic vector bundles which have no non-zero sections, for example, the line bundles on $\mathbb{CP}^N$ with negative degree, so they can't be globally generated. – Angelo Feb 11 '13 at 17:08

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