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Has anyone studied continuous analogues of Schützenberger promotion, and in particular, a flow on (a suitable subset of) the order polytope of a poset?

Here’s what I have in mind: Given a poset $P$, with its order polytope ${\cal O}$ defined in be usual way as the set of labellings $L: P \rightarrow [0,1]$ satisfying $L(x) \leq L(y)$ whenever $x \preceq y$, define ${\cal O}_*$ as the set of all $L \in {\cal O}$ such that $L(x) < L(y)$ whenever $x \neq y$ [NOTE: Sam Hopkins points out that I meant $L(x) \neq L(y)$, not $L(x) < L(y)$], with “bottom face” ${\cal O}_0$ consisting of those $L$ in ${\cal O}_*$ such that $L(x) = 0$ for some $x \in P$ and with “top face” ${\cal O}_1$ consisting of those $L$ in ${\cal O}_*$ such that $L(x) = 1$ for some $x \in P$. We identify ${\cal O}_0$ with ${\cal O}_1$ using what could be called the “Schützenberger identification” (described in the next paragraph); then one can continuously flow points in ${\cal O}_*$ (at least those that aren’t in ${\cal O}_0 \cap {\cal O}_1$) in the (1,1,...,1) direction, reentering at the bottom face when one reaches the top face.

Given $L$ in ${\cal O}_1 \setminus {\cal O}_0$, we define $L’$ in ${\cal O}_0 \setminus O_1$ as follows. Take $x_0$ with $L(x_0) = 1$, and recursively take $x_i$ ($i = 1,2,3,\dots$) to be the child of $x_{i-1}$ with the largest $L$-label, until after $r$ steps (for some $r \geq 1$) one finds that $x_r$ is a minimal element of $P$; then one puts $L’(x_i) = L(x_{i+1})$ for $0 \leq i < r$, $L’(x_r) = 0$, and $L’(x) = L(x)$ for all other $x \in P$. Then $L \mapsto L’$ is the aforementioned identification of ${\cal O}_1$ with ${\cal O}_0$.

Example: Let $P = [2] \times [2]$, and take the point in ${\cal O}_*$ that, viewed as a labelling, sends (1,1) to 1/7, (1,2) to 2/7, (2,1) to 3/7, and (2,2) to 4/7. When we flow this point “upward” (i.e. in the (1,1,1,1) direction) for time 3/7, we reach the point $L$ in ${\cal O}_1$ that sends (1,1) to 4/7, (1,2) to 5/7, (2,1) to 6/7, and (2,2) to 7/7=1. After finding $x_0 = (2,2)$, $x_1 = (2,1)$, and $x_2 = (1,1)$, we see that $L$ in ${\cal O}_1$ is identified with $L’$ in ${\cal O}_0$ that sends (1,1) to 0, (1,2) to 5/7, (2,1) to 4/7, and (2,2) to 6/7. If we flow this point upward for time 1/7, we reach ${\cal O}_1$ and once again jump back to ${\cal O}_0$ under the Schützenberger identification. And so on.

The time-1 flow amounts to a permutation of the labels that coincides with "total promotion" on linear extensions of the poset. But for some dynamical concerns (especially homomesy) the discrete-time and continuous-time stories are different.

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  • $\begingroup$ $\mathcal{O}^*$ consists of $L\in \mathcal{O}$ such that $L(x)\neq L(y)$ for all $x,y\in P$ with $x \neq y$? $\endgroup$ – Sam Hopkins Apr 5 '19 at 15:34
  • $\begingroup$ If we define $\hat{\mathcal{O}}$ to be $\mathcal{O}^*\setminus (\mathcal{O}_0\cap\mathcal{O}_1)/R$ where $R$ is the equivalence between $\mathcal{O}_0$ and $\mathcal{O}_1$ you defined in terms of promotion, then I think $\hat{\mathcal{O}}$ just consists of M connected components, where M is the number of promotion orbits of linear extensions of $P$, and each connected component is a circle times an open ball. $\endgroup$ – Sam Hopkins Apr 5 '19 at 19:31
  • $\begingroup$ Sorry, what I'm saying is, I don't understand your definition of $\mathcal{O}_{*}$. If we only force $L(x)\neq L(y)$ for $x$ and $y$ which are comparable in $P$, then we can have $L(x) =1$ and $L(y)=1$ for two maximal elements in which case I don't think your identification of $\mathcal{O}_1$ with $\mathcal{O}_0$ is well-defined. $\endgroup$ – Sam Hopkins Apr 6 '19 at 3:51
  • $\begingroup$ Sam is right: I mis-defined what I called ${\cal O}_*$ and Sam calls ${\cal O}^*$. Sam’s amendsd definition is what I had in mind. I’ll add a note to that effect. $\endgroup$ – James Propp Apr 6 '19 at 3:55
  • $\begingroup$ I think the amended definition is now correct and yields a well-defined flow. $\endgroup$ – James Propp Apr 6 '19 at 4:02

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