Timeline for indecomposable vector bundles having proper sub-bundles.
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 17, 2011 at 14:57 | comment | added | Arend Bayer | 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 $\mathrm{Hom}(B, A \oplus B)$. | |
| Nov 8, 2010 at 21:52 | comment | added | Jim Bryan | 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. | |
| Nov 8, 2010 at 21:41 | comment | added | Mariano Suárez-Álvarez | 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. | |
| Nov 8, 2010 at 20:59 | history | answered | Jim Bryan | CC BY-SA 2.5 |