I have a simple question about the generating function for reverse plane partitions:

$$\sum_{\pi \in RPP(\lambda)} z^{|\pi|}= \prod_{s \in \lambda} \frac{1}{1-z^{h_{\lambda}(s)}}$$

There's a natural refinement of the right hand side:

$$ \prod_{s \in \lambda} \frac{1}{1-t z_1^{a_{\lambda}(s)}z_2^{l_{\lambda}(s)}} $$

Or perhaps just with $t=z_1,z_2$.

I suspect there should be an equivalent left hand side to this identity - i.e. counting some "refined weight" of the reverse plane partition. Perhaps along diagonals? In a sense there has to be - I'm just not sure what the "statistic" is to count. I wondered if there is a known generating function?

If it helps the right hand side is something like $c_{\lambda}(q,t)$ from Macdonald polynomial theory.

Update: If I write this instead in terms of $w_1 = z_1/z_2$ and $w_2 = z_1z_2$ I actually expect the RHS to be a polynomial in $w_1$ as in something along the lines:

$$ \sum_{\pi \in RPP(\lambda)} w_2^{|\pi|}P_{|\pi|}(w_1) $$

Where $P(w_2)$ is a finite polynomial.


  • 2
    $\begingroup$ Don't we have $h_{\lambda}(s)=a_{\lambda}(s)+\ell_{\lambda}(s)+1$? So we need one more power on the bottom in the second product? $\endgroup$ Mar 10, 2020 at 13:35
  • $\begingroup$ Thanks yes, good spot! $\endgroup$ Mar 10, 2020 at 13:36
  • $\begingroup$ Not an answer to your question, but are you aware of Gansner's result which gives a refinement of this generating function by keeping track of the sum of each "diagonal"? See equation (5.6) of arxiv.org/abs/1503.05934 $\endgroup$ Mar 10, 2020 at 13:42
  • $\begingroup$ Thanks for your comment - I have seen this result, was part of my motivation for guessing it would count something along the diagonals. $\endgroup$ Mar 10, 2020 at 14:24
  • $\begingroup$ Gansner's result seem to give this as a special case, no? By specializing in an appropriate manner... $\endgroup$ Mar 10, 2020 at 21:56

1 Answer 1


Yes, there is a way to introduce certain statistics that lead to this refinement.

First, I'll assume partitions are given as collections of boxes with coordinates $(i,j)\in \mathbb N^2$. The content of the box $(i,j)$ is the quantity $i-j$. A border strip of a partition $\lambda$ is a subset of boxes of $\lambda$ which is a connected skew shape and contains no $2\times 2$ configuration of boxes. Let's call a border strip maximal if its box of largest content $(i_1,j_1)$ satisfies $(i_1+1,j_1)\notin \lambda$, and its box of smallest content $(i_2,j_2)$ satisfies $(i_2,j_2+1)\notin \lambda$. A skew shape $\lambda/\mu$ can be written as a disjoint union of maximal border strips in a unique way. Let $b(\lambda/\mu)$ be the number of border strips that appear in such a decomposition.

The height of a border strip is defined as one less than the number of rows it occupies (a statistic that should be familiar from the Murnaghan Nakayama rule, for example). The height of a skew shape, $\operatorname{ht}(\lambda/\mu)$ is defined as the sum of the heights of all the border strips that appear when writing $\lambda/\mu$ as a union of maximal border strips. Similarly we can define $\operatorname{ht}'(\lambda/\mu)$ by using columns instead of rows.

Now finally, when you have a reverse plane partition $\pi\in RPP(\lambda)$, you can picture it as a 3D stack of boxes. Each horizontal layer is a certain skew shape $\lambda/\mu_i$, for $i=1,2,\dots$. We define $\operatorname{ht}(\pi)=\sum_{i\geq 1} \operatorname{ht}(\lambda/\mu_i)$, $\operatorname{ht}'(\pi)=\sum_{i\geq 1} \operatorname{ht}'(\lambda/\mu_i)$ and $b(\pi)=\sum_{i\geq 1} b(\lambda/\mu_i)$. We can finally state the desired refined formula as $$\sum_{\pi\in RPP(\lambda)}z_1^{\operatorname{ht}(\pi)}z_2^{\operatorname{ht}'(\pi)}t^{b(\pi)}=\prod_{s \in \lambda} \frac{1}{1-t z_1^{a_{\lambda}(s)}z_2^{l_{\lambda}(s)}}.$$

  • $\begingroup$ This follows from Hillman-Grassl? $\endgroup$ Mar 12, 2020 at 3:47
  • $\begingroup$ @SamHopkins Yup :) $\endgroup$ Mar 12, 2020 at 3:49
  • $\begingroup$ This is great! Thanks. The expansion into skew shapes reminds me of expanding a Macdonald polynomial $P_{\lambda}(X,Y) = \sum_{\mu} P_{\lambda/\mu}(X)P_{\mu}(Y)$. I wonder if, since this is essentially a principally specialised Macdonald polynomial in one variable, we can also understand this generating function from that point of view - I guess the coefficient would be related to the Pieri coefficient - I'll have a think. I'm not an expert on enumerative combinatorics though! $\endgroup$ Mar 12, 2020 at 15:43

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