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[ELIA]:

What Dan wrote about jeu de taquin and LPP reminded me of this: applying RSK and then the Schützenberger involution (to both the lower and upper trapezoidal parts of the matrix) is the same thing as reversing rows and columns of the input matrix and then applying RSK. Namely, we have that $SK = KRC$, defining the following operators on (rectangular) matrices: $S$ is the Schützenberger involution, $K$ is the RSK, $R$ is the row reversion, and $C$ is the column reversion.

The above is related to what Dan wrote in two (related, I think) ways:

1. The Schützenberger involution is intimately related to the jeu de taquin. More precisely, it can be obtained by applying several jeu de taquin operations.
2. Reversing rows and columns of a matrix of exponential random variables with a zero in the bottom right corner gives exactly what Dan has called $(X'_{i,j})$.

Let's consider, as an example, a matrix with i.i.d. Exp(1) entries except a zero in the bottom-right corner, like \[ X= \begin{bmatrix} X_{1,1} &X_{1,2} &X_{1,3} \\ X_{2,1} &X_{2,2} &X_{2,3} \\ X_{3,1} &X_{3,2} &0 \end{bmatrix} \, . \] If we apply RSK to this matrix, we will obtain something of the form \[ Z= \begin{bmatrix} Z_{1,1} &Z_{1,2} &Z_{1,3} \\ Z_{2,1} &Z_{2,2} &Z_{2,3} \\ Z_{3,1} &Z_{3,2} &Z_{3,2}\vee Z_{2,3} \end{bmatrix} \, , \] where $Z_{2,3}$ and $Z_{3,2}$ are the LPP times for paths starting from $(1,1)$ and ending at $(2,3)$ and $(3,2)$ respectively. Notice, however, that not all $Z_{i,j}$'s are the LPP times from $(1,1)$ to $(i,j)$! The bottom right corner is the LPP time on the whole array, i.e. the maximum of $Z_{2,3}$ and $Z_{3,2}$.

Let us now consider the input matrix with rows and columns reversed: \[ X' = (X'_{i,j})_{1\leq i,j\leq 3} =\begin{bmatrix} 0 &X_{3,2} &X_{3,1} \\ X_{2,3} &X_{2,2} &X_{2,1} \\ X_{1,3} &X_{1,2} &X_{1,1} \end{bmatrix} \, . \] If we apply RSK to this matrix, we will obtain a certain matrix $Z'=(Z'_{i,j})_{1\leq i,j\leq 3}$.

For the property of the Schützenberger involution mentioned above, we have that $S(Z') = Z$. In words:

1. Take a matrix, like $X'$, of i.i.d. Exp(1) variables, except for the top-left corner which is a zero.
2. Apply the RSK correspondence.
3. Apply the Schützenberger involution.
4. What you get in positions $(2,3)$ and $(3,2)$ are the two LPP times we're looking for, i.e. $T_{2,3}$ and $T_{3,2}$.

There is a clear analogy – which is probably not a coincidence – with what Dan does, i.e.:

1. Take a matrix, like $X'$, of i.i.d. Exp(1) variables, except for the top-left corner which is a zero.
2. Compute the matrix of LPP times $T'=(T'_{i,j})_{1\leq i,j\leq 3}$. Notice that $T_{1,1}=0$, as $X'_{1,1}=0$.
3. Perform a jeu de taquin sliding operation, more precisely an inward slide into cell $(1,1)$, obtaining $T''=(T''_{i,j})_{1\leq i,j\leq 3, (i,j)\neq (3,3)}$.
4. What you get in positions $(2,3)$ and $(3,2)$, i.e. $T''_{2,3}$ and $T''_{3,2}$, are equal in distribution to the two block finishing times $V^5_2$ and $V^5_3$.

Step 1 is the same in both procedures. Step 2 is not exactly the same, but it's similar, in the sense that $Z'_{i,j} = T'_{i,j}$ for $i=3$ or $j=3$ (however, the other $Z'_{i,j}$'s do not coincide with the $T_{i,j}$'s: the latter are just LPP times on $X'$, while the former can be interpreted as “LPP on non-intersecting paths” on $X'$). Step 3, again, is not the same, but the analogy is evident: the Schützenberger involution can be obtained by applying several jeu de taquin operations. If one can make sense of steps 2 and 3 by concluding that what you obtain in positions $(2,3)$ and $(3,2)$ in both procedures is the same (or at least distributed in the same way), then this would prove the weakened version of the conjecture.

Page/line numbers refer to the version submitted to FPSAC on Oct 25 2019. Add any typo/correction ideas below.

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