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Hello,

In the differentiable case it is quite easy to prove that vector bundles are equivalent to smooth maps to the Grassmannian $G_{k}(\mathbb{R}^N)$ for some integer $N>>0$. The proofs I have seen of this use a partition of unity on the base space to embed the bundle into a trivial bundle.

Now in the holomorphic case this is obviously impossible. Yet I would like to know if a similar result exists for holomorphic bundles, and if so, how to prove it. If not, what is the obstruction to this being possible ? Is there some kind of a Kodaira-type criterion ?

NB : I am assuming that the base space $X$ is NOT projective, but let's assume it is Kähler, if that helps.

Thanks !

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Youloush -- What is the precise statement you are trying to prove? –  Jason Starr Aug 6 '12 at 21:40
    
After a bit of thinking, i'm pretty sure hat I'm asking is the following : Let $E \rightarrow X$ be a holomorphic bundle over a Kähler manifold $X$, of rank $r$. Can we find an integer $N$ and a holomorphic map $f : X \rightarrow G_r(\mathbb{C}^N)$ Such that $E = f^*\mathcal{O}$, where $\mathcal{O}$ would be the tautological bundle. –  Youloush Aug 6 '12 at 21:53
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But when $X$ is projective, that would imply that $E$ is generated by its sections (using GAGA). I think that the statement you quote works in the $C^\infty$ category precisely because every $C^\infty$ vector bundle on a compact $C^\infy$ manifold is a direct summand of a trivial bundle (correct me if I misremember). –  Damian Rössler Aug 6 '12 at 21:58
    
I'm not extremely familiar with algebraic geometry, but I don't see what the problem is ? And yes that's how it's proven in the differential case (cf. my original post) but perhaps there is a holomorphic equivalent. –  Youloush Aug 7 '12 at 18:11
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2 Answers

up vote 4 down vote accepted

Here are some remarks; hope they will help.

Let us consider the algebraic case first: then we shall see what one can hope for in the analytic case. Let $X$ be a smooth compact complex algebraic variety and let $\mathcal{E}$ be a vector bundle on $X$. As Youloush points out, in general it is not true that $\mathcal{E}$ is obtained as a pullback of the universal quotient bundle over some Grassmannian: the universal quotient bundle has sections but it may happen that $\mathcal{E}$ doesn't.

Suppose however there is an ample line bundle $\mathcal{L}$ on $X$. Then, for some $n$, $\mathcal{E}\otimes\mathcal{L}^{\otimes n}$ is generated by global sections (e.g. Hartshorne, part II 7.6 and II 5.17), and as such, it is the quotient of a free sheaf of rank $k=\dim H^0(X,\mathcal{E}\otimes\mathcal{L}^{\otimes n})$ on $X$. In other words, $\mathcal{E}\otimes\mathcal{L}^{\otimes n}$ is the pullback of the universal quotient bundle on $G_{k-r}(\mathbb{C}^k),r=rank(\mathcal{E})$ under the map that takes an $x\in X$ to the subspace of $H^0(X,\mathcal{E}\otimes\mathcal{L}^{\otimes n})$ formed by the sections that vanish at $x$.

So every vector bundle on $X$, up to twisting by a power of $\mathcal{L}$, is in fact induced by a map from $X$ to a Grassmannian. However, in order for this to work, there must be at least one ample line bundle on $X$, which automatically makes $X$ projective (e.g. Hartshorne, ibid.). Something similar holds in the analytic case as well. Either there is a positive line bundle on $X$, in which case $X$ is projective by the Kodaira embedding theorem, or there isn't, in which case the above trick doesn't work, and I'm not sure there is one that does.

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I was about the post this as a comment, but...

Damian Rössler pointed out an essential obstruction which is in fact the only one: a holomorphic vector bundle $E$ is the pullback of the universal (quotient) bundle on a Grassmanian via a holomorphic map if and only if $E$ is generated by a finite number of holomorphic global sections. The point is that unlike the $C^\infty$ case, a holomorphic bundle need not have any nonzero global sections at all.

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