# Is analytic Quillen-Suslin simple?

This question is motivated by a sentence on the Wikipedia entry for Quillen-Suslin theorem. This theorem states that every algebraic vector bundle on affine space is trivial. The analogous result is also true for holomorphic vector bundles. Wikipedia states

A simple argument using the exponential exact sequence and the d-bar Poincaré lemma shows that it [affine space] also admits no non-trivial holomorphic vector bundles.

It sounds to me like they are thinking of showing that $H^1(\mathbb{C}^n, \mathcal{O}^*)$ vanishes. But that just shows that all line bundles are trivial. My understanding is that doing higher rank vector bundles requires proving Cartan's lemma, which is 9 pages of difficult analysis in Gunning and Rossi's book and I would not call simple.

I'm inclined to go edit Wikipedia, but I want to make sure that no one knows some simple argument which they could be referring to.

-
If the proof really takes "9 pages of difficult analysis", it would be rather ironic, because the eventual solution to Serre's problem on finitely generated projective modules over polynomial rings was much shorter than this. (So I am interested to hear answers to this question...) –  Pete L. Clark Apr 23 '11 at 5:48
Have a look into Forster's book on Riemann surfaces. In one of the last sections, he proves that any vector bundle on an opne Riemann surfaces is trivial, and this is derived from the triviality of line bundles. –  Johannes Ebert Apr 23 '11 at 10:49
Dear Johannes, you are quite right and I am very happy that you mention Forster. However, as I'm sure you know, these results definitely don't extend to higher dimensional manifolds. For those interested, I give an example at the end of my answer. –  Georges Elencwajg Apr 23 '11 at 12:40
I'm still tracking down the Griffiths reference, but I think I'll leave Wikipedia alone for now. If someone knows enough to go improve the article with confidence, I encourage you to do so. –  David Speyer Apr 27 '11 at 13:52

Dear David, I think there might be a slightly simpler proof of the analytic Quillen-Suslin theorem.

Given a holomorphic vector bundle $E$ on $\mathbb C^n$, it has a holomorphic connection since its Atiyah class $a(E)$ vanishes : indeed $a(E) \in H^1 (\mathbb C^n, \Omega^1_X\otimes \mathcal {End}E)$ , and the whole cohomology group is zero by Theorem B. And then I remember that a long tile ago, Otto Forster explained in a talk an argument by Griffiths that you could deduce from that connection the holomorphic triviality of $E$. Extend each vector $v\in E[0]$, the fiber of $E$ at zero , to to a global section $s_v \in \Gamma (\mathbb C^n, E)$ by first doing that on the restriction $E|L$ of $E$ to lines $L\subset \mathbb C^n$ and then show holomorphic dependency on $v$ by invoking a theorem of holomorphic dependency of a system of differential equations on its paramaters (here $v$).This gives a trivialization of $E$. Unfortunately I don't have a reference for this method of proof but I have no doubt you can fill in the details if you feel so inclined. However it is not clear how elementary this proof really is, since it invokes theorem B .

Finally, since you evoke the contrast for a manifold between having its holomorphic line bundles trivial and all its holomorphic vector bundles of any rank trivial, let me quote :
Theorem (Forster-Rammspott) On a Stein manifold of dimension $n$ every holomorphic vector bundle $E$ of rank $r\geq [n/2]$ can be decomposed as $E=F\oplus \mathcal O^{r-[n/2]}$ where $F$ is a holomorphic vector bundle of rank $[n/2]$ .

So we have the surprizing consequence that for a Stein surface the hypothesis that all holomorphic line bundles are trivial forces the conclusion that all holomorphic bundles, of any rank, are trivial.

An example (edit) In his comment Johannes remarks that on a non-compact Riemann surface, all holomorphic vector bundles are trivial. This is no longer true in higher dimensions in general : $\mathbb C^n$ is special in that respect.
For example take $X=\mathbb P^2(\mathbb C) \setminus C$ where $C$ is the conic $x^2+y^2+z^2=o$ . The surface $X$ is Stein (even affine) but its tangent bundle is not holomorphically trivial because it is not even topologically trivial. Since holomorphic line bundles are classified by $H^2(X,\mathbb Z)=\mathbb Z /2$ , as David pointed out, the holomorphic line bundles on $X$ aren't trivial either ( which you could predict from Forster-Ramspott 's theorem !).

-
Ah, I see that SGP has simultaneously posted a reference to Griffiths that might very well be what inspired Forster's talk. Unfortunately I don't have access to that book... –  Georges Elencwajg Apr 23 '11 at 11:46
Do you mean $rk(F) = [n/2]$? –  Henri Apr 23 '11 at 11:47
Dear Henri, yes of course. Thank you for catching this silly typo. –  Georges Elencwajg Apr 23 '11 at 19:57

The analytic case is simpler: Every holomorphic vector bundle on $C^n$ is trivial.

A short proof (3-4 pages) can be found in P. Griffiths, Topics in Algebraic and Transcendental Geometry, Princeton 1980 (Mathematical Notes 13).

-

I would be rather surprised to see a proof that is significantly simpler than a proof of Cartan's lemma.

Indeed, suppose you know that every vector bundle on $\mathbb{C}^n$ is trivial. Consider two open subsets U and V that cover $\mathbb{C}^n$ and pick a matrix A in $GL_n(\mathscr{O}(U\cap V))$. Now define a vector bundle $\mathscr{E}$ on $\mathbb{C}^n$ by glueing $\mathscr{O}^n$ on U with $\mathscr{O}^n$ on V using A. Take bases of global sections $(e_1,\ldots,e_n)$ and $(f_1,\ldots,f_n)$ of $\mathscr{O}^n$ over U and V. The vector bundle $\mathscr{E}$ is isomorphic to $\mathscr{O}^n$, by hypothesis. Now express a basis of global sections of $\mathscr{O}^n$ over $\mathbb{C}^n$ in terms of the $e_i$'s and $f_i$'s using the isomorphism. You end up with two matrices $A_U \in GL_n(\mathscr{O}(U))$ and $A_V \in GL_n(\mathscr{O}(V))$ such that $A_U A_V^{-1} = A$. This is quite close to Cartan's lemma.

As regards the answers by Georges Elencwajg and SGP, if I am not mistaken, both use the fact that $\mathbb{C}^n$ is Stein and I do not know how to prove this without using Cartan's lemma.

-
Actually, the proof in Griffiths' book is very special to $C^n$ because it uses the fact that $C^n$ is a homogeneous space for a Lie group. The proof does not suffice for a general Stein manifold. Anyway, it is best that you look at his book and decide for yourself! –  SGP Aug 17 '11 at 17:11