# Holomorphic bundles and maps to the Grassmannian ?

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 !

• Youloush -- What is the precise statement you are trying to prove? Aug 6, 2012 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. Aug 6, 2012 at 21:53
• 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). Aug 6, 2012 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. Aug 7, 2012 at 18:11

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.
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.