5
$\begingroup$

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

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

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

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

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

$\endgroup$
1
$\begingroup$

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

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

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

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

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – David Roberts Jan 21 '17 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.