Timeline for Explicit resolution of $\Omega^1_C$ for prestable curve $C$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2020 at 21:51 | comment | added | Tabes Bridges | @MohanSwaminathan yes, that works. Stability is equivalent to ampleness of the log dualizing sheaf. | |
| Jul 30, 2020 at 3:12 | comment | added | Mohan Swaminathan | Yes, this is pretty explicit. In the unstable case, I think we can just stabilize by adding marked points $p_1,\ldots,p_n$ and then take the ample line bundle $\omega_C(p_1 + \cdots + p_n)$ instead of $\omega_C$. | |
| Jul 30, 2020 at 2:50 | comment | added | Tabes Bridges | If $C$ is prestable then its dualizing sheaf is a line bundle; if $C$ is stable this bundle is ample so you can use it to embed in projective space. Now project from a well chosen linear space to obtain an immersion in $\mathbb P^2$, and blow up $\mathbb P^2$ at those singularities which arose from projection (as opposed to those which are part of the abstract pre-stable curve). Now you have an embedding of $C$ in a rational surface. How is that? | |
| Jul 30, 2020 at 1:30 | comment | added | Mohan Swaminathan | I suppose we could choose a 1 (complex) parameter deformation of C which resolves all the nodes simultaneously (xy = 0 --> xy = t for small t) and then the total space of this deformation could serve as $S$. | |
| Jul 30, 2020 at 1:27 | comment | added | Mohan Swaminathan | Maybe this is obvious, but is it clear that we can always embed into a smooth surface? Also, this is not canonical, but is there some explicit constructive way to get such an $S$ and an embedding? | |
| Jul 29, 2020 at 22:31 | comment | added | Jason Starr | ... "surface" --> "smooth surface". | |
| Jul 29, 2020 at 21:54 | comment | added | Jason Starr | Embed the prestable curve $C$ as a Cartier divisor in a surface $S$. Then the resolution is $\mathcal{O}_S(-\underline{C})|_C \to\Omega_S|_C$. | |
| Jul 29, 2020 at 20:23 | comment | added | Mohan | Here is a first thought. If it had a finite resolution as you write (canonical or otherwise), if $C$ is irreducible and sheaf of differential forms is torsion free, then it would be locally free by Auslanser-Buchsbaum. This will solve a long standing conjecture known as Berger's conjecture. So, the answer is most likely to be unknown in your case. | |
| Jul 29, 2020 at 20:03 | history | asked | Mohan Swaminathan | CC BY-SA 4.0 |