Timeline for Deforming to decompose vector bundles
Current License: CC BY-SA 3.0
19 events
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| May 29, 2013 at 16:24 | answer | added | David E Speyer | timeline score: 3 | |
| May 29, 2013 at 15:57 | comment | added | David E Speyer | Sam Payne proves that the moduli space of equivariant rank $3$ vector bundles on a toric variety, with fixed equivariant Chern class, can have arbitrarily many connected components. users.math.yale.edu/~sp547/pdf/Moduli-toric-vector-bundles.pdf That doesn't directly imply the analogous fact for the non-equivariant case, but it is strongly suggestive. | |
| May 29, 2013 at 15:39 | comment | added | Mohammad Farajzadeh-Tehrani | Another way one may look at this problem is whether all vector bundles with the same rank and chern character are deformation of each other? If true, then the proof would be supper easy. | |
| May 29, 2013 at 15:37 | comment | added | Francesco Polizzi | @M.Tehrani: Yes, I think you can. You can also do the proof in the general case by induction on the rank of $E$, since $\textrm{rank} M = \textrm{rank} E-1$. | |
| May 29, 2013 at 15:35 | comment | added | Piotr Achinger | I missed the discussion while typing an answer :(. How about deformations over a DVR or just irreducible base? In the approach of working step by step on a filtration with line bundle quotients, we basically get a deformation over a chain of lines... | |
| May 29, 2013 at 15:31 | answer | added | Piotr Achinger | timeline score: 6 | |
| May 29, 2013 at 15:31 | comment | added | Mohammad Farajzadeh-Tehrani | @ Polizzi: Can't you just continue this process (with M being of lower rank)? | |
| May 29, 2013 at 15:29 | comment | added | Francesco Polizzi | In my first comment of course I meant $E \otimes L$ instead of $E \times L$. | |
| May 29, 2013 at 15:28 | comment | added | Francesco Polizzi | Then one has a short exact sequence $0 \to \mathcal{O}_E \to E \otimes L \to M \to $, where $M$ is another line bundle (here one uses the fact that we are on a curve). Now deforming the extension class one shows that $E \otimes L$ can be deformed to $\mathcal{O}_E \oplus M$, hence $E$ can be deformed to $L^{-1} \oplus (M \otimes L^{-1})$. | |
| May 29, 2013 at 15:27 | comment | added | Francesco Polizzi | In general this is false. However, over curves this should be true, at least for rank $2$ vector bundles. In fact, assume that $E$ is a rank $2$ vector bundle on $X$ and take a sufficiently very ample divisor $L$ such that $E \times L$ is generated by global sections. | |
| May 29, 2013 at 15:26 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
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| May 29, 2013 at 15:22 | comment | added | Mohammad Farajzadeh-Tehrani | How is the proof over the curves? | |
| May 29, 2013 at 15:22 | comment | added | Mohammad Farajzadeh-Tehrani | You are right. What if the polynomial factors into integral linear terms? Anything known? | |
| May 29, 2013 at 15:19 | comment | added | Angelo | Yes, it is true over curves. | |
| May 29, 2013 at 15:18 | history | edited | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |
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| May 29, 2013 at 15:18 | comment | added | Angelo | No, this is false, a vector bundle can be indecomposable for simple numerical reason, i.e., because the Chern polynomial is not a product of linear factors (think of the tangent bundle to $\mathbb P^2$); deforming it does not change the Chern polynomial. | |
| May 29, 2013 at 15:17 | comment | added | Mohammad Farajzadeh-Tehrani | @Piotr: I basically need this over curves. That might simplify the situation a lot. | |
| May 29, 2013 at 15:14 | comment | added | Piotr Achinger | Your idea of deforming extension classes will work if the vector bundle has a filtration with line bundle quotients. Otherwise, I don't think it's possible in general, the most basic obstructions being the Chern classes, which stay constant in families of vector bundles. | |
| May 29, 2013 at 15:05 | history | asked | Mohammad Farajzadeh-Tehrani | CC BY-SA 3.0 |