# When are parabolic Kazhdan-Lusztig polynomials nonzero?

Let $W$ be a Coxeter group with simple reflections $S$ and let $J \subseteq S$. Let $P^J_{\tau, \sigma}$ be the parabolic Kazhdan-Lusztig polynomials in the case $u = q$ in the sense of On Some Geometric Aspects of Bruhat Orderings II. The Parabolic Analogue of Kazhdan-Lusztig Polynomials by Deodhar. These give the transition matrix between a canonical basis and standard basis $\{T_w\}$ of $M^J$, where $M^J \cong \text{Ind}_{W_J}^W \text{ triv}$. This canonical basis is like the $C_w$ basis, not the $C'_w$ basis --the trivial module is a cellular quotient, not a cellular submodule.

When is $P^J_{\tau, \sigma}$ nonzero? This question seems quite difficult, and I am wondering if there has been any work done on it or if it is equivalent to some well-known problem in Kazhdan-Lusztig theory that is known to be difficult.

Unlike ordinary Kazhdan-Lusztig polynomials which are nonzero if and only if $\tau \leq \sigma$, these are nonzero only if $\tau \leq \sigma$, but can often be $0$ when $\tau < \sigma$. For example, in the type A case and the case that $J$ is maximal parabolic, which $P^J_{\tau, \sigma}$ are nonzero is easily described in terms of the $sl_2$ graphical calculus (the number of nonzero $P^J_{\tau, \sigma}$ for fixed $\sigma$ is $2^k$ where $k$ is the number of arcs in the diagram corresponding to $\sigma$).

• 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 17, 2012 at 4:19
• @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 22:34

There are two types of parabolic Kazhdan-Lusztig polynomial, namely $P_{u,v}^{I,q}$ and $P_{u,v}^{I,-1}$.

The question: "When is $P_{u,v}^{I,q}=0$ for the Hermitian symmetric pairs?" is fully discussed in the papers Parabolic Kazhdan–Lusztig R-polynomials for Hermitian symmetric pairs https://msp.org/pjm/2002/207-2/pjm-v207-n2-p01-s.pdf

and Parabolic Kazhdan–Lusztig polynomials for Hermitian symmetric pairs

https://www.ams.org/journals/tran/2009-361-04/S0002-9947-08-04458-9/S0002-9947-08-04458-9.pdf

For $P_{u,v}^{I,-1}$, see corollary 3.11 in the paper On some geometric aspects of Bruhat orderings II. The parabolic analogue of Kazhdan-Lusztig polynomials https://www.sciencedirect.com/science/article/pii/0021869387902328

It is always nonzero since it has a constant term 1.

• If I recall correctly, there are very few cases of Hermitian symmetric spaces, so your "answer" is pretty far from answering the question. But the references may be useful. Jan 1, 2019 at 0:14