Schutzenberger's evacuation and $\mu$-coefficient of Kazhdan–Lusztig polynomials

Question

$\def\SYT{\mathrm{SYT}}\def\RSK{\mathrm{RSK}}\DeclareMathOperator\evac{evac}$Let $\mathfrak{S}_n$ be the symmetric group, $\SYT_n$ be the set of standard young tableaux of size $n$.

For $u\in \mathfrak{S}_n$, let $\RSK:\mathfrak{S}_n\to \SYT_n^2$ denote the Robinson-Schensted-Knuth correspondance.

Let $P_{u,w}(q)$ be the Kazhdan-Lusztig polynomial and $\mu_{u,w}=[q^{{(l(w)-l(u)-1)}/{2} }]P_{u,w}(q)$.

Let $\evac:\SYT_n\to \SYT_n$ be Schutzenberger's involution.

Question: I am looking for a proof of the following proposition that Schutzenberger's involution preserves the $\mu$-coefficient:

Take $u,w\in \mathfrak{S}_n$, and let $(P_1,Q_1)=\RSK(u)$, $(P_2,Q_2)=\RSK(w)$. If we set \begin{gather*} u'=\RSK^{-1}(\evac(P_1),\evac(Q_1)) \\ w'=\RSK^{-1}(\evac(P_2),\evac(Q_2)), \end{gather*} then we have $$\mu(u,w)=\mu(u',w').$$

Answer 1

Let $P^* = evac(P)$. As noted above, if $RSK(u)=(P,Q)$, then $RSK(w_0uw_0) = (P^*,Q^*)$. Conjugation by $w_0$ induces an automorphism of the Hecke algebra sending $T_x \mapsto T_{w_0 x w_0}$ and $c_x \mapsto c_{w_0 x w_0}$, from which the result you want follows. However you might be interested that something stronger is true: if $u=RSK^{-1}(P,Q)$ and $\sigma(u) = RSK^{-1}(P^*,Q)$ then $\mu(u,v)=\mu(\sigma(u),\sigma(v))$. Similarly if $\rho(u) = RSK^{-1}(P,Q^*)$ then $\mu(u,v)=\mu(\rho(u),\rho(v))$.

This follows from a theorem of Mathas on the action of $T_{w_0}$ on cell representations. The relevant paper is Mathas, On the Left Cell Representations of Iwahori‐Hecke Algebras of Finite Coxeter Groups. Proposition 3.17 is the relevant result.