$\def\SYT{\mathrm{SYT}}\def\RSK{\mathrm{RSK}}\DeclareMathOperator\evac{evac}$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 $$\mu(u,w)=\mu(u',w').$$

  • 1
    $\begingroup$ In the title: "evacuation"? $\endgroup$ – paul garrett Dec 13 '18 at 22:42
  • 1
    $\begingroup$ Perhaps it follows from the fact that evacuation is basically conjugation by $w_0$, which is an automorphism of the Dynkin diagram? $\endgroup$ – Sam Hopkins Dec 13 '18 at 22:54
  • $\begingroup$ You say "For $u \in \mathfrak S_n$, let $RSK$ …", but your definition of RSK doesn't refer to $n$. $\endgroup$ – LSpice Dec 14 '18 at 1:23

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.