# Holomorphic vector bundles and Swan's theorem

Is every holomorphic vector bundle a direct summand of a trivial vector bundle on submanifolds of C^n? What about projective varities? I believe Swan's theorem says something about the first question. But I wanted to make sure.

• A link to the Wikipedia page for the Serre-Swan theorem for those who want to know more: en.wikipedia.org/wiki/Serre–Swan_theorem Aug 21, 2010 at 15:45
• You need a compactness assumption. By looking at Chern classes, you can show that the canonical line bundle on $\mathbb{C}P^\infty$ isn't a direct summand of a trivial vector bundle. Aug 21, 2010 at 15:54

The statement for Stein manifolds follows indeed from the analogue of the Serre-Swan theorem for Stein manifolds, which was proven first in 1967 in "Zur Theorie der Steinschen Algebren un Moduln" by O. Forster. The situation is a bit more complicated than the affine scheme or manifold case, but the final result relevant for the purposes of the question is the same The category of locally free sheaves of finite rank is the same as the category of finitely generated projective modules over the global sections $\Gamma(O_X)$. Then one notes that a f.g. projective module is always a direct summand of a finite free module.
• I was also curious, so I looked up the paper. My German is very poor, so I may be misunderstanding, but it seems in Satz 6.2, Forster proves directly that locally free implies projective, by first getting a surjection $\mathcal{O}_X^n\to \mathcal{E}$ etc. as you say. However, in any case, the original question has a positive answer either by Forster, or by the arguments outlined in the comments below. Aug 22, 2010 at 13:48
Yes, for $\mathbb{C}^n$ itself, since vector bundles are (holomorphically) trivial. See Griffiths and Adams " Topics in Algebraic and Analytic Geometry" p 209. I would need to think about the case of submanifolds, before giving an answer. But definitely NO for nontrivial projective varieties: an ample line bundle won't be a summand of a trivial vector bundle. Proof: If it were, then its dual would be generated by global sections, and this is impossible.
• I think it's Cartan, but yeah I was thinking along those lines. The space of sections would be infinite dimensional, so you need to be careful, but it seems conceivable that a finite set of sections generates. If you can do that, then you be done. (You would get a surjection $f:\mathcal{O}_X^N\to \mathcal{E}$ onto your locally free sheaf, which would split because $Ext^1(\mathcal{E},\ker f)=0$.) Aug 21, 2010 at 18:08
• Vamsi, yes I should have been more clear. Let $K=ker(f)$. Then $Ext^1(\mathcal{E}, K)= H^1(X,\mathcal{E}^*\otimes K)=0$. The first equality is an algebraic formality, the second is by Cartan B (or is it Gauss ?) Aug 21, 2010 at 19:03
• Vamsi, for any ringed space $X$, loc. free sheaf $V$ of finite rank, $O_X$-mod $F$, and $i \ge 0$, ${\rm{Ext}}^i(V,F) = {\rm{H}}^i(X, F \otimes V^{\ast})$. This vanishes if $F$ coherent, $X$ Stein, and $i > 0$. To prove finite generation, assume $X$ finite-dim'l (e.g., irreducible). The irreducible components $X_i$ are loc. finite in $X$, so can find $x_i \in X_i$ not in any other $X_j$. By Steinness, if $V$ has rank $n$ can find global sections $s_1,\dots,s_n$ generating $V$ near each $x_i$. Restrict $V$ to support of cokernel of $O_X^n \rightarrow V$, induct on dimension, use Nakayama. QED Aug 21, 2010 at 19:23