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 !



Holomorphic bundles and maps to the Grassmannian ?
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.


