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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.

flag	asked Aug 21 2010 at 15:34 Vamsi 1,317●4●11

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 – Joel Fine Aug 21 2010 at 15:45

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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. – Per Vognsen Aug 21 2010 at 15:54

Daniel Pomerleano · Accepted Answer · 2010-08-21 19:02:22Z UTC

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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.

answered Aug 21 2010 at 19:02

Daniel Pomerleano
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I haven't seen Forster's paper, but are you sure the final sentence in this answer isn't circular reasoning? I would guess (since it is the natural argument coming to mind) that the way Forster proves the global sections of a vector bundle are a projective module is by proving it's a direct summand of a finite free module, and that in turn is proved by exhibiting the vector bundle as a direct summand of a finite free vector bundle (using Steinness). In the end you are getting a correct deduction, but it feels like the black box may rest on what you are deducing from it. Did you check? – BCnrd Aug 22 2010 at 5:58

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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. – Donu Arapura Aug 22 2010 at 13:48

Donu Arapura · Answer 2 · 2010-08-21 17:41:40Z UTC

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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.

edited Aug 21 2010 at 17:41

answered Aug 21 2010 at 17:29

Donu Arapura
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Thanks. Is it true for Stein manifolds? I mean, by Hormander's theorem, the global sections generate the vector bundle. So, if the global sections are finite dimensional it ought to be true, ought not it? – Vamsi Aug 21 2010 at 17:57

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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 $Ext^1(\mathcal{E},\ker f)=0$.) – Donu Arapura Aug 21 2010 at 18:08

I was thinking along the same lines(I thought it was Grauert:)) and my intuition is that it will work because in the algebraic category for an affine scheme, this is the same thing as the statement that a finite projective module is a direct summand of a free module. I'd have to go back and read those papers though. But I also wanted to ask as far as I understand Swan's theorem has to do with C^infinity manifolds. Then the point is basically the same as the affine scheme case. – Daniel Pomerleano Aug 21 2010 at 18:24

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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 ?) – Donu Arapura Aug 21 2010 at 19:03

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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 – BCnrd Aug 21 2010 at 19:23

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Holomorphic vector bundles and Swan’s theorem

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