# homomesy and asymptotic behaviour

For simplicity, consider an infinite locally-finite poset $\mathcal{P}$ with a unique bottom element $\perp$ whose finite order ideals obey a hook-length formula --- i.e. the number of linear extensions for each order ideal satisfy a (common) hook length formula. The examples I have in mind are

(1) the infinite square lattice: whose finite order ideals are exactly Young diagrams (the linear extensions of which are exactly Young Tableaux)

(2) the Young-Fibonacci lattice

(3) any infinite rooted tree: whose finite order ideals are among the finite rooted trees (the linear extensions of which are exactly the increasing trees)

Take a large but finite order ideal $I$ in $\mathcal{P}$ together with a large integer cut-off $0 << \, n \, << |I|$. Select with uniform probability a linear extension $l$ of $I$ and consider the order ideal $\text{res}_n(l)$ obtained by restricting it to the interval of values $[1 \dots n]$, namely

$$\text{res}_n(l) \, := \, \Big\{x \in I \, \Big| \, l(x) \in [1 \dots n] \Big\}$$

This restriction map induces a probability measure $\mu_n$ on the space $\mathcal{I}_n$ of all order ideals in $\mathcal{P}$ of size $n$.

Imagine now that $l$ is a fixed linear extension. Let $\sigma: \mathcal{L}_I \longrightarrow \mathcal{L}_I$ denote the Schützenberger promotion operator and form the promotion-orbit

$$\mathcal{O}_l \, := \, \Big\{\sigma^k \cdot l \, \Big| \, k \in \Bbb{Z} \Big\}$$

of $l$ and consider the restrictions $\text{res}_n \big( \sigma^k \cdot l \big)$ as $k$ varies; in this way we obtain another distribution $\rho_{n,l}$ on $\mathcal{I}_n$.

Question: What is the relationship between the distributions $\mu_n$ and $\rho_{n,l}$ as $|I| \rightarrow \infty$ in view of Propp's concept of homomesy (as manifest by promotion) ?

regards,

A. Leverkühn

dear A. Leverkühn,

I'm very sorry; the comments (which I'm striking out) are wrong and badly thought out.

There is no interesting asymptotic behaviour as long as the cut-off $n$ is fixed. Let $\Bbb{n}$ be the order ideal of consisting of all elements $x$ in $\mathcal{P}$ at distance $n$ from the bottom element $\perp$; note that $\Bbb{n}$ has finite size because $\mathcal{P}$ is locally finite. If $\mu_n^{\Bbb{n}}$ and $\mu^I_n$ denote the probability distributions on $\mathcal{I}_n$ induced from the restriction maps $\text{res}_n^{\Bbb{n}}: \mathcal{L}_{\Bbb{n}} \longrightarrow \mathcal{I}_n$ and $\text{res}_n^{\Bbb{n}}: \mathcal{L}_{I} \longrightarrow \mathcal{I}_n$, then $\mu^I_n$ and $\mu_n^{\Bbb{n}}$ will coincide provided $\Bbb{n} \subset I$ because the restriction of any linear extension of $I$ to $\Bbb{n}$ is independent of the extension's restriction to the complement $I - \Bbb{n}$, since the indices in the interval $[1 \dots n]$ are exhausted by $\Bbb{n}$. Consequently $\mu_n^{I_k} \rightarrow \mu_n^{\Bbb{n}}$ as $k \rightarrow \infty$ when $\Bbb{n} \subset I_k$ for $k >> 0$.

On way to create non-trivial asymptotics is to dialate an order ideal $I$ by a positive integer $s$ and examine the limit

$$\lim\limits_{s \to \infty} \mu^{s \cdot I}_{f(s,n)}$$

where $s \cdot I$ denote the dilation of $I$ by a factor of $s$ and $f(s,n) = \big| s \cdot J \big|$ for any order ideal $J$ of size $|J| = n$. Note that both the enveloping order ideal $I$ and the cut-off $n$ are being sent to infinity.

In the case of a Young diagram $\lambda$ the dilation $s\cdot \lambda$ is obtained by subdividing each box of $\lambda$ into $s^2$ squares; an so $f(s,n) = s^2n$.

In the case of a rooted tree $T$ the dilation $s \cdot T$ is the tree obtained by introducing a chain of $s-1$ intermediate vertices between each pair of vertices $x, y \in T$ for which there is an edge; here $f(s,n) = \big(|T| - 1\big)s \, + \, 1$. I'm not sure what would serve as an adequate notion of dilation for an order ideal of the Young-Fibonacci lattice.

yours,

Ines

p.s. Clearly dilation must satisfying the requirement that $\big| s \cdot I \big| \, = \, \big| s \cdot J \big|$ for any pair of order ideals $I$ and $J$ of the same size.

p.p.s. In case it helps, for a rooted finite tree $I$ together with a rooted subtree $J$ of sizes $m$ and $n$ respectively the probability $\mu_n^{I}\big(J \big)$ equals

$${\binom{m}{n}}^{-1} \prod_{j \in J} \, {h_I(j) \over {h_J(j)}}$$

where $h_I(j)$, respectively $h_J(j)$, is one plus the number of children of $j$ in $I$, respectively $J$. This is an easy consequence of the tree hook-length formula.

• Dear Ines, I edited your answer as {\it ... } doesn't work here: one needs to use _underscores before and after_ (which will result in: underscores before and after), and other Markdown formatting commands. – David Roberts Jan 21 '17 at 3:47