# Holomorphic vector bundles over $\mathbb{C}^{n}\setminus 0$

Is it true that every holomorphic vector bundle over $\mathbb{C}^{n}\setminus 0$ is trivial? If not true, how can one construct a counterexample?

And just a small note here (wrong):

• For $n\leq 2$, we can push the bundle $\mathcal{F}$ over $\mathbb{C}^{n}\setminus 0$ to be defined on $\mathbb{C}^{n}$, denoted by $\bar{\mathcal{F}}$. By taking a double dual $\bar{\mathcal{F}}^{**}$ from which we can get reflexive sheaf which is isomorphic to $\mathcal{F}$ away from $0$. For a coherent analytic reflexive sheaf, we know that the set where it fails to be locally free has codimension $\geq 3$. So for $n\leq 2$, $\bar{\mathcal{F}}^{**}$ is indeed a bundle. But for $n\geq3$, I have no clue.

This note turns to be wrong because I mistook $\bar{\mathcal{F}}$ to be a coherent sheaf, while the behavior of $\mathcal{F}$ near 0 is unknown.

A further comment here: For the line bundle case, based on the reference provided in the comments by pgraf, for $q\geq 1$, $H^{q}(\mathbb{C}^{n}\setminus 0, \mathcal{O})\neq 0$ iff $q=n-1$ which implies there exists nontrivial line bundle on $\mathbb{C}^{2}\setminus 0$ which cannot be extended to be on $\mathbb{C}^{2}$. In other cases, all holomorphic line bundles over $\mathbb{C}^{n}$ for $n\geq 3$ and $n=1$ would be trivial. Note when $n=1$, $\mathbb{C}^{*}$ is a Stein manifold, or we can directly solve $\bar{\partial}$ equation by integration.

• Something seems to be wrong here. For $X = \mathbb C^2 \setminus \{ 0 \}$, the exponential sequence tells us that $\mathrm{Pic}(X)$ is isomorphic to $H^1(X, \mathcal O_X)$, which is nonzero by a Čech computation (see e.g. Grauert-Remmert, Theorie der Steinschen Räume, p. 135). So not every holomorphic line bundle on $X$ extends to $\mathbb C^2$. – pgraf May 24 '16 at 11:07
• $\def\CC\mathbb{C}$To make @pgraf's comment concrete, take trivial bundles on $\CC \times \CC_{\neq 0}$ and $\CC_{\neq 0} \times \CC$ and glue them by $\exp(x^{-1} y^{-1})$ on the overlap. The resulting bundle is not holomorphically trivial. – David E Speyer May 24 '16 at 13:19
• @pgraf Yes. You are right. I made a mistake in my argument. $\bar{\mathcal{F}}$ might not be coherent. The subtle point here is about the behavior of $\mathcal{F}$ near $0$. Thanks for correcting this. – user40184 May 24 '16 at 17:32

## 1 Answer

For $n\geq 3$, there exists such non-trivial bundles. For example, take the bundle on $\mathbb{C}^n-\{0\}$ given by the quotient of $\mathcal{O}^n$ by the subbundle generated by the vector $(x_1,\ldots,x_n)$, where the $x_i$s are the coordinate functions. Then, for $n\geq 3$, this bundle is not trivial. The non-triviality will essentially follow from an argument like Hartog's theorem.