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Young's lattice $Y$ is a graded poset and a distributive lattice whose elements are all the partitions of $n$ for $n \in \mathbb{N}$ with the poset relation coming from inclusion of Young diagrams. Here is Wikipedia's picture with the smallest few elements that shows only the adjacent order relations (i.e., one needs transitive closure to get the full picture):

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Fix integers $0 < m \leq n$ and let $Y(m,n)$ denote the sub-poset of $Y$ "between" the partitions of $m$ and $n$, both inclusive. So, $Y(3,6)$ is the sub-poset generated by partitions of $3, 4, 5$ and $6$, but nothing else. Since each poset corresponds to a simplicial order complex whose $d$-simplices are precisely chains of length $d+1$, each $Y(m,n)$ may be viewed as a simplicial complex. Here's my question:

For each dimension $d$, what is known about the $d$-th Betti number of $Y(m,n)$ as a function of $m$ and $n$?

If there is no simple formula, I'd be happy with asymptotics as $n$ is sent to $\infty$ for each fixed $m$. Note that for $m = 1$ the whole thing is contractible to the unique minimal element, but things (appear to) get more complicated for larger $m$.

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    $\begingroup$ I think that if you fix $m$ and let $n$ increase, the order complex will eventually become contractible. The poset $Y(m,n)$ becomes a lattice upon the addition of a minimum element and a maximum element. Given $m$, define the partition $\lambda=(\lambda_1,\ldots,\lambda_m)$ by letting $\lambda_i$ be the ceiling of $m/i$. Then every partition of $m$ has Young diagram fitting in that of $\lambda$. Thus if $n$ is larger than something like $m \log m$, then the join of all the atoms in the given lattice is not the maximum element. It follows that the order complex of $Y(m,n)$ is contractible. $\endgroup$
    – John Shareshian
    Commented Aug 12, 2013 at 19:00
  • $\begingroup$ Have you thought about using shellability to get an answer in the general case? $\endgroup$
    – John Shareshian
    Commented Aug 12, 2013 at 19:03
  • $\begingroup$ @JohnShareshian fantastic, I wonder if $n = m\log(m)$ is a threshold for contractibility. Thank you also for the shellability suggestion for the general formula, is it clear in general that $Y(m,n)$ is shellable for arbitrary $m$ and $n$? $\endgroup$
    – Vidit Nanda
    Commented Aug 12, 2013 at 19:41
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    $\begingroup$ Take a partition $\mu$ whose Young diagram contains those of all partitions of $n$. The interval $\left[\emptyset,\mu\right]$ in Young's lattice is a sublattice of a distributive lattice and therefore a (finite) distributive lattice. Every finite distributive lattice has shellable order complex. Now $Y(m,n)$ is a rank-selected subposet of a poset with shellable order complex and therefore has shellable order complex. $\endgroup$
    – John Shareshian
    Commented Aug 12, 2013 at 20:57
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    $\begingroup$ I believe it can be shown using the theory of P-partitions that if we fix $k$, then the least $m$ for which $Y(m-k,m)$ is nonacyclic is $m=\lfloor \frac{(k+3)^2}{4}\rfloor$. $\endgroup$
    – Richard Stanley
    Commented Mar 2 at 20:05

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The question was answered in the comments by John Shareshian. I'm just posting this to have this old question marked as answered. The comments by John explain in detail the answers to the specific questions, but the upshot is: because Young's lattice $Y$ is a (locally finite) distributive lattice, its homological properties are easy to work out explicitly using the standard tools of poset topology.

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