I have a mysterious symmetry that I have not managed to prove. First some definitions (see picture below)

Fix a partition that fit in a staircase shape with $n$ rows. There are $Catalan(n)$ such shapes. We can represent this with a diagram $D$, as below, where the gray squares is the partition. The yellow squares are enumerated $1,\dotsc,n$ from top to bottom, and thus any permutation in $S_n$ is seen as a labeling of the yellow squares.

Let $a+b=n$. Given a permutation in $S_n$ seen as a labeling of the yellow squares, the *first block* is the squares with labels $1,\dotsc,a$
and the *second block* is the remaining $b$ squares.
Thus, the blue labels is the first block, and the red labels is the second block in my example.

A permutation $\sigma \in S_n$ is called $(D,a,b)$-*good* if the following holds:

- The smallest label in each of the two blocks appear lowest in its block.
- If $i$ and $i+1$ are in the same block, and $i$ below $i+1$, then the square in the same row as $i+1$ and same column as $i$ must be white.

In the diagram, the permutation $342615$ is shown, and one can verify that it is $(D,4,2)$-good. The white squares that has to be white due to the second condition has been marked with bullets.

Let $Good(D,a,b)$ denote the set of $(D,a,b)$-good permutations.

**Warmup exercise**

Show that $|Good(D,a,b)|=|Good(D,b,a)|$.

Finally, we define the *ascent*, $asc_D$-statistic on permutations as follows:
For every white square $S$, we let $S_1$ be the index of the yellow square in the same row, and $S_2$ be the index of the yellow square in the same column.
Then $asc_D(\sigma)$ is the number of white squares $S$,
such that $\sigma(S_1)<\sigma(S_2)$.
In our diagram, $asc_D(342615) = 1+1+2+0+1 = 5$,
where the terms are contributions from each row.

**My problem**

Show (bijectively) that for every diagram $D$ and choice of $a+b=n$, $$ \sum_{\sigma \in Good(D,a,b)} q^{asc_D(\sigma)} =\sum_{\sigma \in Good(D,b,a)} q^{asc_D(\sigma)}. $$

For the diagram $D$ here, we have that $|Good(D,4,2)|=|Good(D,2,4)|=20$, and that both sums above become $$1 + 3 q + 4 q^2 + 4 q^3 + 4 q^4 + 3 q^5 + q^6.$$

*Comments:* For some diagrams $D$, it is straightforward
to produce a bijection, in particular the case when $D$ has no gray squares.
One would hope that a bijection would the number ascents 'within blocks',
that is, ascents where $S_1$ and $S_2$ belong to the same block.
However, this cannot be done for general $D$.

One can generalize the problem to permutations with more than two blocks, but the 2-block case implies the general case.

I am quite confident this result follows (non-bijectively) from a result by C. Athanasiadis, but it requires several messy steps.

**Motivation**

This is related to the $p_\lambda$-expansion of certain LLT polynomials.