Given $w\le w'$ (in Bruhat ordering), does $P_{x,w}(1)\le P_{x,w'}(1)$ (in usual ordering of $\mathbb{R}$), where $P_{x,w}(q)$ is the Kazhdan Lusztig polynomial?
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1$\begingroup$ No. Suppose that $W$ is finite and $w_0$ is the longest element. Then $P_{x,w_0} = 1$, but $w \le w_0$ for all $w$. Hence any nontrivial KL polynomial gives a counterexample. For example $P_{t,tsut} = 1 + q$ but $P_{t,stsuts} = 1$, when $s,t,u$ are the standard generators of $S_4$ (symmetric group on $4$ letters). $\endgroup$ – Geordie Williamson Nov 16 '18 at 12:33

$\begingroup$ Thank you. My follow up question would be: Given $I\subseteq \Delta$, consider the Weyl group generated by $I$, denote by $W_I$, let $w_I$ be the longest element of $W_I$ and ${}^IW$ denote the set of minimal length coset representatives of $W_I\backslash W$. Given $w_Iw\le w_Iw'$, does $P_{w_Ix, w_Iw}(1)\le P_{w_Ix, w_Iw'}(1)$ for $x,w,w'\in {}^IW$? $\endgroup$ – James Cheung Nov 16 '18 at 13:25

$\begingroup$ Doesn't $I = \emptyset$ yield your earlier question? So ... no again! $\endgroup$ – Geordie Williamson Nov 16 '18 at 20:51

$\begingroup$ by the way, the statement $P_{x',w}(1) \le P_{x,w}(1)$ is true for $x' \le x$. Perhaps this is what you meant? See the BradenMacPherson paper. It is known in general. This has been written down by Plaza. $\endgroup$ – Geordie Williamson Nov 17 '18 at 23:09