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):

enter image description here

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 $m$.

  • 2
    $\begingroup$ 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 Aug 12 '13 at 19:00
  • $\begingroup$ Have you thought about using shellability to get an answer in the general case? $\endgroup$ – John Shareshian Aug 12 '13 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 Aug 12 '13 at 19:41
  • 2
    $\begingroup$ 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 Aug 12 '13 at 20:57

Your Answer

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

Browse other questions tagged or ask your own question.