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
\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
 A: 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.
