# Refined reverse plane partition generating function

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.

Thanks

• 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? Mar 10, 2020 at 13:35
• Thanks yes, good spot! Mar 10, 2020 at 13:36
• 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 Mar 10, 2020 at 13:42
• Thanks for your comment - I have seen this result, was part of my motivation for guessing it would count something along the diagonals. Mar 10, 2020 at 14:24
• Gansner's result seem to give this as a special case, no? By specializing in an appropriate manner... Mar 10, 2020 at 21:56

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 $$\sum_{\pi\in RPP(\lambda)}z_1^{\operatorname{ht}(\pi)}z_2^{\operatorname{ht}'(\pi)}t^{b(\pi)}=\prod_{s \in \lambda} \frac{1}{1-t z_1^{a_{\lambda}(s)}z_2^{l_{\lambda}(s)}}.$$
• 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! Mar 12, 2020 at 15:43