# indecomposable vector bundles having proper sub-bundles.

Over rational curve we know that any vector bundle is decomposable to direct sum of line bundles.

In higher dimensions there are examples of indecomposable bundles.

some indecomposable vector bundles have might have proper sub-bundles (all bundles and sub bundles here are in holomorphic category and not topological)

First Question: Over curves, can we have a indecomposable bundle having a proper sub bundle? For rational curves the answer is negative obviously, what about elliptic curves, and higher genus curves?

Second: same question for Calabi-Yau 3-folds and K3-surfaces? for example for Quintic.

Please provide examples (or give reference) if you know any.

• One way to phrase your question is: do vector bundles on a curve give a semisimple category? – Mariano Suárez-Álvarez Nov 8 '10 at 17:46

## 5 Answers

Over a curve any rank $2$ bundle has a rank $1$ subbundle: Choose a subbundle defined over a Zariski dense open set, and then extend it over the missing points by observing that locally the problem of making such an extension is the problem of extending a map into $\mathbb P^1$.

• Similarly, by the schematic closure method (which preserves flatness over a Dedekind base) this provides a full flag for any vector bundle over any smooth algebraic curve. – BCnrd Nov 8 '10 at 17:45
• Or more geometrically you get a map into a flag space over some open subset which extends because the base is Dedekind and the flag space proper. – Torsten Ekedahl Jan 16 '11 at 5:21

If $E$ is an indecomposable vector bundle and $S\subset E$ is a subbundle, then the sequence $$0\to S\to E\to Q\to 0$$ gives a non-trivial element in $Ext^1(Q,S) = H^1(Q^* \otimes S)$. So a concrete way to construct the bundles you want is to find bundles $Q$ and $S$ such that $H^1(Q^*\otimes S)$ is non-trivial. On a curve $C$ of genus 1 or bigger, we can take $S=Q=\mathcal{O}_C$ (in genus one, the resulting non-trivial bundle is called the Atiyah bundle).

To use this to get an example on a $K3$ surface, suppose that we have a $K3$ surface $S$ with an effective smooth genus 0 curve $C\subset S$. By the adjunction formula, such a curve has a degree -2 normal bundle and so using Riemann-Roch (or just the exact sequence of a divisor) one can see that the line bundle $L=\mathcal{O}_S(2C)$ has non-trivial $H^1$. Thus we get a non-trivial extension $$L \to E \to \mathcal{O}_S.$$ One can see that $E$ is indecomposable by numerical considerations: by looking at Chern characters, the only possible decomposition of $E$ would be $L\oplus \mathcal{O}$, but this can't be since by construction, the extension class is non-trivial.

• Can't it happen that E is isomorphic to $L\oplus\mathcal O$ yet the extension does not split? We had an example of that in modules a little while ago. – Mariano Suárez-Álvarez Nov 8 '10 at 21:41
• In the case at hand, that can't happen since $Hom(L,L\oplus \mathcal{O})=H^(\mathcal{O}\oplus L^{-1})=\mathbb{C}$ and so the only map is the one that splits. More generally, I thought it was the case that if $E$ is a non-trivial extension of two bundles, then $E$ is not isomorphic to the direct sum of the bundles, but your comment gives me pause. Let me think about it. – Jim Bryan Nov 8 '10 at 21:52
• I think Jim is right, and in fact this holds in any category with finite-dimensional Hom-spaces: the short exact sequence $0 \to A \to E \to B \to 0$ induces a long exact Hom-sequence [ 0 \to \mathrm{Hom}(B, A) \to \mathrm{Hom}(B, E) \to \mathrm{Hom}(B, B) \to^\delta \mathrm{Ext}^1(B, A) ] The identity in $\mathrm{Hom}(B, B)$ gets mapped to the class in $Ext^1$ defining the extension $E$, and so the dimension of $\mathrm{Hom}(B, E)$ is strictly smaller than the dimension of $\mathrm{Hom}(B, A \oplus B)$. – Arend Bayer Jan 17 '11 at 14:57

Similarly to Tom's answer any vector bundle of rank $r$ on a variety of dimension $n<r$ contains a non-trivial subbundle. In fact it always contains a non-trivial sub line bundle (a sub-bundle not sub-sheaf). The proof of this fact is a variant of the proof of Bertini's theorem. Twist your vector bundle so that it is generated by global sections and then find a section that is not zero at any point.

Suppose we are over the complex numbers and that $f:S\to C$ is a smooth morphism from a smooth surface $S$ to a smooth curve $C$ of genus at least $2$, with non-constant classifying map $C\to M_g$, where $g$ is the genus of the generic fiber. (These exist, and are sometimes called Kodaira surfaces.) There is a tautological exact sequence $0\to f^*\Omega^1_C\to\Omega^1_S\to\Omega^1_{S/C}\to 0$; this is not split, since the classifying map is non-constant, and so has non-zero derivative, while $f^*\Omega^1_C$ is a sub-bundle of $\Omega^1_S$.

Now suppose that $\Omega^1_S = O(A)\oplus O(B)$. The de Franchis-Bogomolov lemma gives that each subsheaf $O(A), O(B)$ of $\Omega^1_S$ has Kodaira dimension at most $1$. On the other hand, $f^*\Omega^1_C$ embeds into each of $O(A)$ and $O(B)$ (else the tautological sequence splits). So we have $A\sim f^*K_C+D$ and $B\sim f^*K_C+E$ with $D,E\ge 0$. Since $K_C$ is ample on $C$, each effective class $D,E$ must be vertical (supported on fibers of $f$), by dF-B. But then $K_S\sim 2f^*K_C+D+E$ and so is also vertical, which contradicts the easy observation that $S$ is of general type.

[If these Kodaira surfaces seem too special, then use any semi-stable family of curves with base and fiber of genus at least $2$, and replace the bundles $\Omega^1_C$ and $\Omega^1_S$ by bundles of $1$-forms wioth appropriate log poles.]

Stable rank 2 bundles over Riemann surfaces of genus greater than 1 do not split. See the discussion in Bott's Stable Bundles Revisited'', p.3 inSurveys in Differential Geometry', a supplement from 1981 to the Journal of Differential Geometry. [If this is not readily available there are discussions of stable bundles in Donaldson-Kronheimer, or in Atiyah-Bott Yang-Mills on Riemann Surfaces'' which probably get to that point.