Cyclic action on Kreweras walks

Martin Rubey and I solved my conjecture.

The basic idea of the proof is as follows. First to a Kreweras word $w$ we associate what we call its bump diagram, which is just the union of the two noncrossing partial matchings of $\{1,2,...,3n\}$ associated to $w$ (the one for the A's and B's and the one for the A's and C's), drawn as a graph in the obvious way. For example, with $w=AABBCACCB$ its bump diagram is

We also think of this diagram as a set of ordered pairs ('arcs'); in this example that set is $$\{ (1,4),(1,8),(2,3),(2,5),(6,7),(6,9)\} $$

We extract a permutation $\sigma_w$ of $\{1,2,...,3n\}$ from the bump diagram as follows.

For $i=1,2,...,3n$, we take a trip from position $i$. We start traveling from position $i$ along the unique arc ending at $i$ (if $w_i=B$ or $C$), or the "shorter arc" beginning at $i$ (if $w_i=A$), and we continue until we reach a "crossing" of arcs. When we hit a crossing of arcs $(i,k)$ and $(j,\ell)$ with $i \leq j < k < \ell$ (note that we allow the case $i=j$), we follow the following "rules of the road":

• if we were heading towards $i$, then we turn right and head towards $\ell$;

• if we were heading towards $\ell$, then we turn left and head towards $i$;

• otherwise we continue straight to where we were heading.

When we finish our trip at position $j$ then we define $\sigma_w(i) := j$.

For example, to compute $\sigma_w(3)$: we start traveling along the arc $(2,3)$ heading towards $2$; then we come to the crossing of $(2,3)$ and $(2,5)$ and we turn right, heading towards $5$; then we come to crossing of $(1,4)$ and $(2,5)$ and we turn left, heading towards $1$; then we come to the crossing of $(1,4)$ and $(1,8)$ and we turn right, heading towards $8$; then we come to the crossing of $(1,8)$ and $(6,9)$, but we just continue straight to $8$; and so we finish our trip at $8$. So $\sigma_w(3)=8$.

Or to compute $\sigma_w(7)$: we start traveling along the arc $(6,7)$ heading towards $6$; then we come to the crossing of $(6,7)$ and $(6,9)$ and we turn right, heading towards $9$; then we come to the crossing of $(1,8)$ and $(6,9)$ and we turn left, heading towards $1$; and then we come to the crossing of $(1,4)$ and $(1,8)$, but we just continue straight to $1$; and so we finish our trip at $1$. So $\sigma_w(7)=1$.

We could compute the whole permutation is $\sigma_w = [4,3,8,5,2,7,1,9,6]$.

You might notice that this example $w$ is the same as the original post and that this permutation $\sigma_w$ is the same as the "permutation" $p$ defined in terms of the cylindrical rotation array.

Indeed, this always happens (that the trip permutation is the same as the permutation from the cylindrical rotation array). It follows from the main lemma behind the overall proof, which is

Lemma. If $w'$ is the rotation of $w$, then $\sigma_{w'} = c^{-1} \sigma_w c$ where $c= (1,2,3,...,3n)$ is the "long cycle."

As a remark, these trip permutations come from the theory of plabic graph (cf. Section 13 of Postnikov's paper https://arxiv.org/abs/math/0609764).

Since $\sigma_w$ does not completely determine $w$, to finish the proof we need to keep track of a little more data. For that purpose, we define $\varepsilon_w=(\varepsilon_w(1),...,\varepsilon_w(3n))$, a sequence of $3n$ letters which are B's or C's, defined by $$ \varepsilon_w(i) := \begin{cases} w_{\sigma(i)} &\textrm{if $w_{\sigma(i)}\neq A$} \\ w_{\sigma(\sigma(i))} &\textrm{if $w_{\sigma(i)}= A$}. \end{cases} $$ Similarly to the previous lemma, we can show

Lemma. If $w'$ is the rotation of $w$, then $\varepsilon_{w'} = (\varepsilon_w(2),\varepsilon_w(3),...,\varepsilon_w(3n),-\varepsilon_w(1))$ with the convention that $-B=C$ and vice-versa.

The above lemmas easily imply that the $3n$th rotation of $w$ is obtained from $w$ by swapping B's and C's.

Martin and I will post a preprint to the arXiv with all the details soon.

EDIT: The paper is now on the arXiv at https://arxiv.org/abs/2005.14031.

This is not an answer, but too long for a comment.

This promotion operator is (apparently) governed by local rules, similar to https://arxiv.org/abs/1804.06736, as follows:

• regard each path as a sequence of coordinates, that is, $A$ adds $(1,1)$, $B$ adds $(-1,0)$ and $C$ adds $(0,-1)$ to the current coordinate

• create a cylindrical array from each promotion orbit, for example, for the path $AABBCACCB$ ${\scriptstyle\begin{array}{llllllllllllllllllll} 0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\ &0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 0,0 \\ &&0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 3,1 & 3,0 & 2,0 & 1,0 & 0,0 \\ &&&0,0 & 1,1 & 1,0 & 2,1 & 2,0 & 3,1 & 2,1 & 1,1 & 0,1 & 0,0 \\ &&&&0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 2,2 & 1,2 & 0,2 & 0,1 & 0,0 \\ &&&&&0,0 & 1,1 & 1,0 & 2,1 & 1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\ &&&&&&0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 0,1 & 0,0 \\ &&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 2,3 & 2,2 & 2,1 & 1,1 & 1,0 & 0,0 \\ &&&&&&&&0,0 & 1,1 & 2,2 & 3,3 & 3,2 & 3,1 & 2,1 & 2,0 & 1,0 & 0,0 \\ &&&&&&&&&0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0 \\ &&&&&&&&&&0,0 & 1,1 & 1,0 & 2,1 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 0,0 \\ &&&&&&&&&&&0,0 & 1,1 & 2,2 & 1,2 & 2,3 & 1,3 & 0,3 & 0,2 & 0,1 & 0,0 \\ &&&&&&&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 1,0 & 0,0 \\ &&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 1,2 & 2,3 & 2,2 & 2,1 & 2,0 & 1,0 & 0,0 \\ &&&&&&&&&&&&&&0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0 \\ &&&&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 3,1 & 2,1 & 1,1 & 1,0 & 0,0 \\ &&&&&&&&&&&&&&&&0,0 & 1,1 & 1,0 & 2,1 & 3,2 & 2,2 & 1,2 & 1,1 & 0,1 & 0,0 \\ &&&&&&&&&&&&&&&&&0,0 & 1,1 & 2,2 & 3,3 & 2,3 & 1,3 & 1,2 & 0,2 & 0,1 & 0,0 \end{array}}$

• consider any square of four coordinate in this array \begin{array}{ll} \lambda & \nu\\ \kappa & \mu \end{array} and let $\tilde\mu = \kappa + \nu - \lambda$. Then, apparently, we have $ \mu = \begin{cases} \tilde\mu &\text{if $\tilde\mu$ has positive coordinates}\\ \tilde\mu + (2,1) &\text{if the first coordinate of $\tilde\mu$ is negative}\\ \tilde\mu + (1,2) &\text{if the second coordinate of $\tilde\mu$ is negative} \end{cases} $

Possibly we can get a proof that the occurrences of $\tilde\mu$ with a negative coordinate yield a permutation, assuming that the local rules are correct.

First we paste the triangular region to the right of the first $3n$ rows into the empty region below the diagonal (and removing the final $3n-1$ rows). The pasting must be done in a way such that there is precisely one $(0,0)$ coordinate in each row and column: ${\scriptstyle\begin{array}{llllllllllllllll} 0,0 & 1,1 & 2,2 & 1,2 & 0,2 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 \\ 1,0 & 0,0 & 1,1 & 0,1 & 1,2 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 \\ 2,0 & 1,0 & 0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 3,1 & 3,0 & 2,0 \\ 2,1 & 1,1 & 0,1 & 0,0 & 1,1 & 1,0 & 2,1 & 2,0 & 3,1 & 2,1 \\ 2,2 & 1,2 & 0,2 & 0,1 & 0,0 & 1,1 & 2,2 & 2,1 & 3,2 & 2,2 \\ 1,1 & 0,1 & 1,2 & 1,1 & 1,0 & 0,0 & 1,1 & 1,0 & 2,1 & 1,1 \\ 1,2 & 0,2 & 1,3 & 1,2 & 1,1 & 0,1 & 0,0 & 1,1 & 2,2 & 1,2 \\ 0,1 & 1,2 & 2,3 & 2,2 & 2,1 & 1,1 & 1,0 & 0,0 & 1,1 & 0,1 \\ 1,1 & 2,2 & 3,3 & 3,2 & 3,1 & 2,1 & 2,0 & 1,0 & 0,0 & 1,1 \\ 0,0 & 1,1 & 2,2 & 2,1 & 2,0 & 1,0 & 2,1 & 1,1 & 0,1 & 0,0 \end{array}}$

(This array does not satisfy the local rules along the diagonal.)

Now we consider one square of four, but instead of its corners label the four edges with $\lambda-\kappa$, $\nu-\lambda$, $\mu-\kappa$ and $\nu-\mu$. There are 11 different squares occurring, two of which correspond to a $b$ or $c$ respectively. For these two, the labels on parallel edges are different, for the others, they are same. Put a bullet in the squares whose parallel edges have distinct labels.

In the case at hand, we obtain ${\def\x{\huge\bullet} \scriptstyle\begin{array}{llllllllllllllllll} & A & & A & & B & & B & & C & & A & & C & & C & & B &\\ B & & A & & A & & A & \x & B & & B & & B & & B & & B & & B\\ & B & & A & & B & & A & & C & & A & & C & & C & & B &\\ B & & B & & A & \x & B & & B & & B & & B & & B & & B & & B\\ & B & & B & & A & & A & & C & & A & & C & & C & & B &\\ C & & C & & C & & A & & A & & A & & A & & C & \x & C & & C\\ & B & & B & & C & & A & & C & & A & & C & & A & & B &\\ C & & C & & C & & C & & A & \x & C & & C & & C & & C & & C\\ & B & & B & & C & & C & & A & & A & & C & & A & & B &\\ A & & A & \x & B & & B & & B & & A & & A & & A & & A & & A\\ & B & & A & & C & & C & & B & & A & & C & & A & & B &\\ C & & C & & C & & C & & C & & C & & A & \x & C & & C & & C\\ & B & & A & & C & & C & & B & & C & & A & & A & & B &\\ A & \x & B & & B & & B & & B & & B & & A & & A & & A & & A\\ & A & & A & & C & & C & & B & & C & & B & & A & & B &\\ B & & B & & B & & B & & B & & B & & B & & B & & A & \x & B\\ & A & & A & & C & & C & & B & & C & & B & & B & & A &\\ A & & A & & A & & A & & A & & A & \x & C & & C & & C & & A\\ & A & & A & & C & & C & & B & & A & & B & & B & & C & \end{array}}$

It remains to show that in each row of "vertical" labels only $A$ and one other letter occurs, and in each column of "horizontal" labels, only $A$ and one other letter occurs, except that for the "horizontal" labels below the diagonal, we have to swap $B$ and $C$.

I believe this follows from the local rules.