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 $GL_{2}(\mathbb{C})$.

  • $\begingroup$ I have in mind the case $X=Tot_{\mathbb{P}^{1}}(\mathcal{O}(-2))$. $\endgroup$ – Oren Ben-Bassat Jun 6 '12 at 12:06
  • 7
    $\begingroup$ 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. $\endgroup$ – Jason Starr Jun 6 '12 at 12:09
  • $\begingroup$ 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. $\endgroup$ – Oren Ben-Bassat Jun 6 '12 at 12:22
  • $\begingroup$ Or not because $A^{2}(X)=\mathbb{Z}$ in my case. $\endgroup$ – Oren Ben-Bassat Jun 6 '12 at 16:52
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.