Let $T$ be a standard Young tableaux on $[n]$. Denote the RSK algorithm $\text{RSK}(w)=(P(T),Q(T))$ for $w\in\mathfrak{S}_n$, where $P(T)$ is the Schencted insertion tableaux.

For $1\leq i\leq j\leq n$. Let $T_{[i,j]}$ be the skew SYT by restricting $T$ to the segament $[i,j]$. For a skew shape $Y$, define the rectification of Y, $\text{Rect}(Y)$ to be applying *jeu de taquin* on $Y$ to obtain a standard shape. See section 2.1 of this paper, in which $\text{Rect}$ is denoted as $\text{std}$.

It is well known that

For $w\in \mathfrak{S}_n$, $T\in \text{SYT}_n$. If $P(w)=T$, then $$\text{Rect}(T_{[i,j]})=P(w_{[i,j]})$$ for all $[i,j]\subseteq [n]$, where $w_{[i,j]}$ means restricting the permutation to the subalphabet $[i,j]$, e.g. $126534_{[2,5]}=2534$.

My question is: is there a $K$-theoretic analog of this property, in terms of Hecke insertion, $K$-jeu-de-taquin of increasing tableaux?

Specifically, define $K$-rectification by replacing jdt with $K$-jdt, and denote $K$-$P(w)$ the Hecke-insertion tableau of the word $w$. We say a Tableau is *unique-rectification-target* if it's unique in its $K$-Knuth class. (Definition 3.15 of [BS13], see also section 4 of [PP14])

Let $T$ be an increasing Tableau (of alphabet $[n]$) and $Y=T_{[1,i]}$ such that $Y$ is an SYT and $1\cdots i\notin T\backslash Y$ (so that there is no ambiguity). If $K\text{-Rect}(T\backslash Y)$ is a URT, then is it always true that: $$K\text{-Rect}(T\backslash Y)=K\text{-}P(w_{[i+1,n]})$$? where $w$ is the row-reading word of $T$, or even all words such that $K\text{-}P(w)=T$.

In general, is it true that $$\{K\text{-Rect}(T\backslash Y)\}=\{K\text{-}P(w_{[i+1,n]}):K\text{-}P(w)=T\}$$

For example, Let $T=\begin{matrix}1&2&4\\3&5&6\\4&6&9 \end{matrix},Y=\begin{matrix}1&2\\3&\end{matrix}$. We have $$K\text{-Rect}\left(\begin{matrix}*&*&4\\*&5&6\\4&6&9 \end{matrix} \right)=\begin{matrix}4&5&6\\5&9&\\6&& \end{matrix}$$ The row reading word of $T$ is $w=\mathfrak{row}(T)=469356124$, and $$K\text{-}P(w_{[4,9]})=K\text{-}P(469564)= \begin{matrix}4&5&6\\5&9&\\6&& \end{matrix}$$

Thanks !

verywrong. $K$-Rect is not uniquely defined. See the edit of my question. $\endgroup$ – Sylvester W. Zhang Jun 18 '19 at 1:16