# Vector bundle as an extension of line bundles

This is a question about vector bundles on a smooth non-proper (non-projective) algebraic surface $X$ over $\mathbb{C}$. Are there any known examples of a non-proper surface $X$ and a rank two vector bundle $V$ on $X$ which is not an extension of line bundles, i.e. does not fit into a short exact sequence $0 \to L \to V \to M \to 0$ where $L$ and $M$ are line bundles on $X$?

Notice that the usual topological idea of using the second Chern class does not work here. I am asking about obstructions to restricting the structure group to the upper triangular subgroup of $GL_{2}(\mathbb{C})$.

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I have in mind the case $X=Tot_{\mathbb{P}^{1}}(\mathcal{O}(-2))$. – Oren Ben-Bassat Jun 6 '12 at 12:06
I do not see why the second Chern class idea cannot still work. For instance, if your non-proper surface is the complement of a closed point in an Abelian surface, then the Chow group of 0-cycles is enormous. It seems to me that it will contain elements which are not "pure" cup products. Using Serre's construction, presumably you can use that to construct a locally free sheaf of rank 2 whose second Chern class is not a "pure" cup product. – Jason Starr Jun 6 '12 at 12:09
Thanks, I was thinking of the Chern class as living in $H^{4}(X,\mathbb{C})=0$ but you are saying to consider its more natural target in $CH_{0}(X)$. Great idea, maybe it will work for my case. – Oren Ben-Bassat Jun 6 '12 at 12:22
Or not because $A^{2}(X)=\mathbb{Z}$ in my case. – Oren Ben-Bassat Jun 6 '12 at 16:52
First make an elementary transform of the vector bundle along a fiber F of projection to make the restriction to the zero section Z Cartier divisor trivial. If you can prove the transformed bundle is globally trivial, thus a pullback under projection, then also the inverse elementary transform is a pullback, i.e., the original bundle is a pullback. Now use the fact that the invertible ideal sheaf $I_Z$ of the zero section (and its powers) have vanishing cohomology (because the bundle is Tot of $\mathcal{O}(-2)$) to lift the trivializing global sections from $Z$ to the total space. – Jason Starr Jun 8 '12 at 11:04

Only a partial answer, starting with the disclaimer that it does not apply when the base is the total space of a vector bundle on a projective curve. ''If $X$ is a Stein manifold of dimension $2$, then each holomorphic vector bundle $V \to X$ of rank $r >1$ has a trivial one-dimensional subbundle''. In particular, it fits into a short exact sequence.

Proof: $X$ contains a $2$-dimensional CW-complex $K \subset X$ as a deformation retract. A generic (smooth) section of $V$ has a zero set of real dimension $4-2r < 2$, and moreover the zero set does not meet $K$ for dimensional reasons and by transversality. So $V|_K$ has a trivial complex line bundle as subbundle.

By homotopy invariance of vector bundles, this shows that $V$ has a trivial smooth subbundle. Now study the holomorphic fibre bundle $Mon(\mathbb{C};V)\to X$ (a point over $x$ is a complex linear monomorphism $\mathbb{C} \to V_x$). The fibre is the complex homogeneous space $Mon(\mathbb{C};\mathbb{C}^r)$, which is the quotient of $GL_r (\mathbb{C})$ by the stabilizer subgroup $G$ of the action of $GL_r (\mathbb{C})$ on $\mathbb{C}^{r} \setminus 0$.

The first paragraph says that there is a global smooth section $X \to Mon(\mathbb{C};V)$. Finding a holomorphic subbundle is the same as finding a holomorphic section.

Grauerts theorem (''Analytische Faserungen \"uber holomorph-vollst\"andigen R\"aumen'') says that for bundles of the above type over Stein manifolds, each smooth section is homotopic to a holomorphic section.

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I am not sure what surface you have in mind, but it will depend on the surface. For example, if $X=\mathrm{Spec}\,A$ is affine and Pic is trivial, but $A^2(X)\neq 0$, then you can always represent a non-zero class in $A^2(X)$ by a zero cycle defined by an ideal $I$ which is a local complete intersection. By Serre construction, one has an exact sequence, $0\to A\to P\to I\to 0$, where $P$ is a rank 2 vector bundle and its second Chern class is non-zero by construction. If $P$ was filtered by line bundles, since Pic is trivial, the vector bundle is trivial and thus second Chern class must be zero. Such examples are easy, by taking a general hypersurface of large degree in affine three space. (Technically, take a general one in projective three space of large degree and remove a hyperplane section).

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