Continuous analogues of Schützenberger promotion

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.

• $\mathcal{O}^*$ consists of $L\in \mathcal{O}$ such that $L(x)\neq L(y)$ for all $x,y\in P$ with $x \neq y$? Apr 5 '19 at 15:34
• 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. Apr 5 '19 at 19:31
• 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. Apr 6 '19 at 3:51
• 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. Apr 6 '19 at 3:55
• I think the amended definition is now correct and yields a well-defined flow. Apr 6 '19 at 4:02