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 !).