Timeline for How can one show that orbit closures in representations of a linear quiver don't have small resolutions?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 24, 2014 at 3:56 | comment | added | Hugh Thomas | There are people from the finite-dimensional algebra world who have studied these singularities, though so far as I found, they did not discuss smallness of resolutions; and in any case, it seems like you understand the geometry yourself and want input of a different kind. In case it might be useful, though, some references: Bobinski and Zwara, Schubert varieties and representations of Dynkin quivers, Colloquium Mathematicum 94 (2002), no. 2, 285-309; and Kavita Sutar, arXiv:1111.1179, Resolutions of defining ideals of orbit closures for quivers of type $A_3$. | |
| May 11, 2014 at 12:25 | answer | added | Nicolas Perrin | timeline score: 2 | |
| May 11, 2014 at 8:18 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
added 545 characters in body
|
| May 10, 2014 at 19:52 | comment | added | Ben Webster♦ | @AllenKnutson I hadn't been thinking about that perspective, but I don't know why it should change my mind about how hard this problem is. I suppose the resolutions I'm discussing are (some?) Bott-Samelson resolutions in this case, but I don't know if there's some reasonable obstruction to them being small. I'll note, I'm not necessarily looking for a perfect criterion, but something like torsion which will help me to find counterexamples. | |
| May 10, 2014 at 14:44 | comment | added | Allen Knutson | Equioriented type A quiver cycles are Kazhdan-Lusztig varieties (intersections of Schubert varieties with opposite Bruhat cells), by the Zelevinskii isomorphism. Do you expect these varieties to be easier than general K-L varieties? (I don't, really.) | |
| May 10, 2014 at 13:32 | history | asked | Ben Webster♦ | CC BY-SA 3.0 |