Timeline for When are parabolic Kazhdan-Lusztig polynomials nonzero?
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Dec 8, 2018 at 13:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Nov 8, 2018 at 13:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Oct 9, 2018 at 13:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Sep 9, 2018 at 12:58 | answer | added | James Cheung | timeline score: 1 | |
Mar 19, 2012 at 4:07 | history | edited | Jonah Blasiak |
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Mar 17, 2012 at 22:34 | comment | added | Jim Humphreys | @Jonah: As you observe, the question seems quite difficult. The answer might be of combinatorial interest, but I wonder whether it would have any impact on the subjects in which these polynomials arise most naturally: representation theory and algebraic geometry. (Deodhar's papers have had a lot of citations, in those directions especially.) Probably your question will have a reasonable answer mainly in very special cases. Anyway, it's a good idea to add broader tags such as co.combinatorics and coxeter-groups. | |
Mar 17, 2012 at 4:19 | comment | added | Alexander Woo | Known special cases: Pietro Mongelli has a paper on the ArXiv which calculates parabolic K-Ls for Boolean elements and includes an answer to this question. His paper extends an European J. Comb. paper by Marietti on parabolic K-Ls for Boolean elements for $S_n$. (For $S_n$, Boolean elements are those Bruhat smaller than the longest transposition, but the Coxeter generalization is not the obvious one.) It is probably possible to use the Lascoux-Schutzenberger formula for ordinary K-Ls to extend Marietti's answer (at least for the question of being 0 or not) to all covexillary permutations. | |
Mar 16, 2012 at 21:23 | history | asked | Jonah Blasiak | CC BY-SA 3.0 |