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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 $\sigma$).

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    $\begingroup$ 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. $\endgroup$ Mar 17, 2012 at 4:19
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    $\begingroup$ @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. $\endgroup$ Mar 17, 2012 at 22:34

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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.

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  • $\begingroup$ 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. $\endgroup$ Jan 1, 2019 at 0:14

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