First, some background: I will assume the anglophone conventions for Young tableaux in what follows. Given a standard Young tableau $T$ of shape $\lambda$, we can define the cocharge tableau $C(T)$ as follows. Let $n(i)$ be a labeling of the boxes of a $\lambda$-diagram such that $n(i)$ is the box of $T$ containing entry $i$. Let the $n(1)$ box (ie the top-left box) of $C(T)$ be $0$. If entry $n(i)$ of $C(T)$ is $c$, we obtain the next entry as follows: if the $n(i+1)$-th entry of $T$ is below the $n(i)$-th entry, fill box $n(i+1)$ of $C(T)$ with $c+1$. If not, fill it with entry $c$. In other words, every time we have an inversion in $T$, we increase the number we are filling $C(T)$ with. Let $\sum C(T)$ denote the sum of the entries of $C(T)$.

It is clear that the minimum value for $\sum C(T)$ is obtained from the Young tableaux $T$ with first row $1,2,\dots, \lambda_1$, second row $\lambda_1+1, \dots$, and so on. (Filling first left to right, then down.) The maximum value is obtained from the Young tableaux with first column $1,2,\dots, \lambda_1$, and so on. (Filling first top to bottom, then leftwards.) Let the minimum value be $m$ and maximum value be $M$.

My question: I believe that for any $0\leq i\leq M-m$, the number of standard Young tableaux $T$ with $\sum C(T)=m+i$ is exactly equal to the number of standard Young tableaux with $\sum C(T) = M-i$. I've checked this for all partitions of $n$ up to 7 or 8. Does anyone know if there is a reference that has a proof of this, or does anyone have a good explanation for why this should be true?

EDIT: I'll give an example to make clear what I'm asking. Consider, for example, the partition $(3,1,1)$ of 5. There are 6 standard Young tableaux of this shape. Consider the two below $$ T_1 = \begin{pmatrix} 1 & 2 & 3 \\ 4 \\ 5 \end{pmatrix} \hspace{20pt} T_2 = \begin{pmatrix} 1 & 4 & 5 \\ 2 \\ 3 \end{pmatrix} $$ These lead to cocharge tableaux $$ C(T_1) = \begin{pmatrix} 0 & 0 & 0 \\ 1 \\ 2 \end{pmatrix} \hspace{20pt} C(T_2) = \begin{pmatrix} 0 & 2 & 2 \\ 1 \\ 2 \end{pmatrix} $$ with $\sum C(T_1)=3$ and $\sum C(T_2) = 7$. You can show that each of these is the unique tableaux leading to the sum. What I claim, and what is in fact true in this case, is that there is in fact the same number of tableaux with cocharge summing to 4 as there is with cocharge summing to 6. (In this case, this is trivial to verify.)


1 Answer 1


Just a sketch of a proof; sorry, this is all I can do with the time I have right now.

If $m\in\mathbb N$ is arbitrary, then an $m$-reciprocal polynomial means a polynomial $p \in \mathbb Z\left[q\right]$ whose coefficient before $q^i$ equals its coefficient before $q^{m-i}$ for every $i\in\mathbb Z$ (this implies that the degree of $p$ is $\leq m$).

Let $\operatorname*{SYT}\left(\lambda\right)$ denote the set of all standard tableaux of shape $\lambda$. You want to prove that the generating function $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\sum C(T)} \in \mathbb Z\left[q\right]$ is an $m$-reciprocal polynomial, where $m = \dbinom{n}{2} + \sum\limits_i\dbinom{\lambda_i}{2} - \sum\limits_i\dbinom{\lambda^{\prime}_i}{2}$ (where we use standard notations: $\lambda^{\prime}$ is the conjugate partition of $\lambda$, and $\mu_i$ denotes the $i$-th entry of $\mu$). Sorry that my $m$ is not your $m$ !

If $T$ is a standard Young tableau of size $n$, we denote by $D\left(T\right)$ the set of all descents of $T$, that is, the set of all $i\in\left\{1,2,...,n-1\right\}$ such that the entry $i+1$ appears in $T$ in a lower row than the entry $i$ appears in. The major index of the standard Young tableau $T$ is defined as $\sum\limits_{i\in D\left(T\right)} i$, and denoted by $\operatorname*{maj}T$. The comajor index of the standard Young tableau $T$ of size $n$ is defined as $\sum\limits_{i\in D\left(T\right)} \left(n-i\right)$, and denoted by $\operatorname*{comaj}T$. In the proof of Proposition 7.19.11 in Richard Stanley's Enumerative Combinatorics, volume 2, it is shown that $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{comaj} T} = \sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ (actually, it is shown that this holds for skew shapes as well). But it is easily seen that every standard Young tableau $T$ satisfies $\sum C(T) = \operatorname*{comaj} T$, and so this becomes $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\sum C(T)} = \sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$. Hence, it remains to prove that the polynomial $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ is $m$-reciprocal for $m = \dbinom{n}{2} + \sum\limits_i\dbinom{\lambda_i}{2} - \sum\limits_i\dbinom{\lambda^{\prime}_i}{2}$.

But this should follow easily from plugging $q^{-1}$ instead of $q$ into the formula for $\sum\limits_{q\in\operatorname*{SYT}\left(\lambda\right)} q^{\operatorname*{maj} T}$ obtained by using Proposition 7.19.11 in Stanley's above-mentioned book and then simplifying the Schur function using (7.104) (again, in Stanley's book).


Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.