# Two majs for standard Young tableaux?

Let $$\lambda$$ be a partition of $$n$$, and consider its set of standard Young tableaux (SYTs): bijective fillings of the Young diagram of $$\lambda$$, written in English notation, with the numbers $$1$$ through $$n$$ that strictly increase in rows and columns.

I am interested in two different but closely related notions of descents and major indices (or more precisely comajor indices) for these SYTs.

Let $$T$$ be such an SYT. A Schur descent of $$T$$ is an $$i$$ such that $$i+1$$ appears in a lower row than $$i$$. Let $$D_S(T)$$ denote the set of Schur descents of $$T$$. Define $$\mathrm{comaj}_S(T) := \sum_{i \in D_S(T)}(n-i)$$.

This is the usual notion of descents for SYT and we have $$\sum_{T} q^{\mathrm{comaj}_S(T)} = q^{b(\lambda)} \frac{(1-q^n)(1-q^{n-1})\cdots(1-q)}{\prod_{u \in \lambda} (1-q^{h(u)})}$$ where $$b(\lambda) := \sum_{i}(i-1)\lambda_i$$ and $$h(u)$$ is the hook length of box $$u$$.

Note. This is more commonly written $$\sum_{T} q^{\mathrm{maj}_S(T)} = q^{b(\lambda)} \frac{(1-q^n)(1-q^{n-1})\cdots(1-q)}{\prod_{u \in \lambda} (1-q^{h(u)})}$$ where $$\mathrm{maj}(T) := \sum_{i\in D_S(T)}i$$ (see Corollary 7.21.5 of Stanley's EC2). But we have $$\sum_{T} q^{\mathrm{comaj}_S(T)}=\sum_{T} q^{\mathrm{maj}_S(T)}$$ (see the proof of Proposition 7.19.11 in EC2) and for what I'm going to compare it to below the comajor index is more directly analogous. Ultimately the product formula for the $$\mathrm{comaj}_S$$ generating function follows from the formula for the principal specialization of the corresponding Schur function (see Corollary 7.21.3 of EC2) $$s_{\lambda}(1,q,q^2,\ldots) = q^{b(\lambda)} \frac{1}{\prod_{u \in \lambda} (1-q^{h(u)})}$$ and the fact that semistandard Young tableaux of shape $$\lambda$$ are $$(P,\omega)$$-partitions with respect to the Schur labeling.

Now let me define natural descent of $$T$$ to be an $$i$$ such that $$i+1$$ appears in a higher row than $$i$$. ("Schur" versus "natural" comes from "Schur labeling" versus "natural labeling," but note that the Schur descents are by far the more common notion of descent for SYT.) And define $$D_N(T)$$ to be the set of natural descents of $$T$$, and $$\mathrm{comaj}_N(T) := \sum_{i \in D_N(T)}(n-i)$$.

The generating function for reverse plane partitions of shape $$\lambda$$ is $$\sum_{\lambda} q^{|\lambda|} = \frac{1}{\prod_{u\in \lambda}(1-q^{h(u)})}$$ (see Theorem 7.22.1 of Stanley's EC2): the point is that we can turn an SSYT into a reverse plane partition by subtracting $$(i-1)$$ from the $$i$$th row. Then from the basic theory of $$(P,\omega)$$-partitions (as in Chapter 3.15 of Stanley's EC1) it follows that $$\sum_{T} q^{\mathrm{comaj}_N(T)} = \frac{(1-q^n)(1-q^{n-1})\cdots(1-q)}{\prod_{u\in \lambda}(1-q^{h(u)})}.$$

Question: Is there a simple bijection $$\phi$$ on the set of SYT of shape $$\lambda$$ for which $$\mathrm{comaj}_N(\phi(T)) = \mathrm{comaj}_S(T) - b(\lambda)$$?

Perhaps if you unwind all the theory of $$(P,\omega)$$-partitions you can extract some kind of bijection, but it feels like there should be something very simple...

Example. As a sanity check, consider $$\lambda=(2,1)$$. The two SYTs of shape lambda are $$T_1 = \begin{array}{c c} 1 & 2 \\ 3 \end{array} \qquad T_2 = \begin{array}{c c} 1 & 3 \\ 2 \end{array}$$ We have $$\mathrm{D}_S(T_1) = \{2\}$$, $$\mathrm{D}_S(T_2) = \{1\}$$, and so $$\sum_{T}q^{\mathrm{comaj}_S(T)} = q+q^2$$ while $$\mathrm{D}_N(T_1) = \varnothing$$, $$\mathrm{D}_N(T_2) = \{2\}$$, and so $$\sum_{T}q^{\mathrm{comaj}_N(T)} = 1+q.$$

Bonus question: for that matter, is there a simple bijection $$\psi$$ on SYTs of shape $$\lambda$$ for which $$\mathrm{maj}_S(\psi(T)) = \mathrm{comaj}_S(T)$$?

Note. Defining $$\mathrm{maj}_N(T) := \sum_{i\in D_N(T)}i$$, in general $$\sum_{T}q^{\mathrm{comaj}_N(T)} \neq \sum_{T}q^{\mathrm{maj}_N(T)}$$.

EDIT: Darij Grinberg suggested as an answer to the bonus question that evacuation of SYT should be the involution that sends $$\mathrm{comaj}_S$$ to $$\mathrm{maj}_S$$. For some reason I had previously convinced myself that evacuation did not work for this, but now thinking more carefully I believe Darij is right- in fact we should have $$i \in D_S(T)$$ if and only if $$(n-i) \in D_S(\mathrm{evac}(T))$$. I think this follows from the definition of evacuation where we rotate $$T$$ 180 degrees, swap $$i$$ for $$n+1-i$$, and then rectify, because jeu de taquin preserves the descent set. However, I can't find anywhere this property of evacuation is specifically written. Anyways, the main question about $$\mathrm{comaj}_S$$ vs. $$\mathrm{comaj}_N$$ remains.

• The bivariate polynomial that tracks both statistics is not symmetric in general; which hints that the bijection you seek is not an involution. Mar 2 at 19:26
• I have added these statistics to findstat. Apparently, the mystery map is not a combination of three maps known to findstat. Mar 4 at 10:48
• One observation from the database: The natural comajor index findstat.org/StatisticsDatabase/St001695 can be alternatively expressed as findstat.org/StatisticsDatabase/… Mar 4 at 12:03
• Two more observations: for two-rowed standard Young tableaux, the two statistics coincide. More interestingly, we can interpret also rectangular two-column tableaux as Dyck paths and feed findstat with the problem. It then comes up with the map findstat.org/MapsDatabase/Mp00127oMp00129oMp00120 Mar 5 at 8:07
• re the comment on evacuation: evacuation both reverses and complements the descent set of a tableau. This can be deduced from the fact that RSK recording tableaux have the property $D_S(w) = D_S(Q(w))$ and the fact that, for $w = w_1 ... w_n$ a permutation, the reverse complement $w' = n+1-w_n ... n+1-w_1$ has $Q(w')$ as the evacuation of $Q(w)$. May 4 at 0:59