# Major index generating polynomial for border-strip tableaux

The Question in its original form has been answered, but there is a follow-up, see the end of the post.

A border-strip is a skew Young diagram that does not contain a $$2 \times 2$$-box. A border-strip with $$n$$ boxes can be encoded as a binary string of length $$n-1$$ by starting from the lower left corner and writing $$0$$ for every right step and $$1$$ for every up step, see the example below. If $$\epsilon=\epsilon_1 \epsilon_2 \dotsb \epsilon_{n-1} \in \{ 0,1\}^{n-1}$$ is a binary string, then we denote the set of all standard Young tableau of shape $$\epsilon$$ as $$\textrm{SYT}(\epsilon)$$. One such standard Young tableau is seen below. On any Standard Young tableau $$T$$, one can define the major index $$\textrm{maj}(T)$$ as follows. A descent of $$T$$ is an entry $$i$$ such that $$i+1$$ appears strictly below $$i$$ in $$T$$. Define $$\textrm{maj}(T)$$ as the sum of all descents of $$T$$. For example, the major index of the standard Young tableau above is $$3+5+8=16$$. The major index generating polynomial of $$\epsilon$$ is then defined as

$$f^\epsilon(q)=\sum_{T \in \textrm{SYT}(\epsilon)}q^{\textrm{maj}(T)}.$$

Is anything known about major index generating polynomial of standard border-strip tableaux? If $$\epsilon$$ is a hook, that is $$\epsilon_1 = \dotsb = \epsilon_k =1$$ and $$\epsilon_{k+1} = \dotsb = \epsilon_{n-1}=0$$, then the answer is known by using the $$q$$-hook formula as

$$f^\epsilon(q)=\begin{bmatrix} n-1 \\ k \end{bmatrix}_q$$

However, I could not find any known theory about the general case. The particular problem I am interested in is the following: Let $$1 \leq i . If $$\epsilon_i=1$$ and $$\epsilon_{n-i-1}=0$$ then put $$\epsilon'=\epsilon_1\epsilon_2 \dotsb \epsilon_{i-1} 0 \epsilon_{i+1}\dotsb \epsilon_{n-i-2} 1 \epsilon_{n-i}\dotsb \epsilon_{n-1}$$. I conjecture that if $$\omega=e^{2\pi i/n}$$ is a primitive $$n$$:th root of unity, then $$f^\epsilon(\omega)f^{\epsilon'}(\omega)=1.$$

I have verified this conjecture for $$n \leq 4$$ and have verified several special cases for bigger $$n$$ as well.

UPDATE: My supervisor was able to prove an even stronger assertion. Suppose that we have a skew diagram $$\lambda/\mu$$ of size $$n$$, that is, $$|\lambda/\mu|=n$$. Then

$$f^{\lambda/\mu}(\omega)=\begin{cases}(-1)^{\textrm{ht}(\lambda/\mu)} & \text{if } \lambda/\mu \text{ is a border-strip}, \\ 0 & \text{otherwise}.\end{cases}$$

Here, the height $$\textrm{ht}(\lambda/\mu)$$ is the number of rows that $$\lambda/\mu$$ touches minus one. So the height of the border-strip in the Figure is $$3$$. He used the suggested Proposition 7.19.11 in Richard Stanley's Enumerative Combinatorics Volume 2 and the Murnaghan–Nakayama rule. It is not hard to see that this implies the original Conjecture (note that border-strips given by $$\epsilon$$ and $$\epsilon'$$ have the same height). However, I think it would be interesting to know if the equality above could be proved by a more straightforward, combinatorial argument.

• Have you tried Proposition 7.19.11 in Richard Stanley's Enumerative Combinatorics Volume 2? It reduces your question to computing $s_\epsilon\left(1,q,q^2,\ldots\right)$. This is an evaluation of a skew Schur function, but it has the shape of a ribbon so it may be easier to deal with than a usual Schur. – darij grinberg Mar 19 at 14:58
• I have not tried this but will look into it! When you say compute $s_\epsilon\left(1,q,q^2,\ldots\right)$, do you mean that one should try to find some closed form expression for $s_\epsilon\left(1,q,q^2,\ldots\right)$? Because this is not a polynomial but a formal power series, unless I am missunderstanding something. – Joakim Uhlin Mar 19 at 15:26
• Yeah, I'd try looking for a closed form as a formal power series. Maybe the Jacobi-Trudi determinant simplifies. – darij grinberg Mar 19 at 16:04
• $$s_\epsilon(1,q,q^2,\dots) = \frac{f^\epsilon(q)}{(1-q)(1-q^2)\cdots (1-q^n)}$$ – Ira Gessel Mar 21 at 14:51

What you're talking about are the "$$q$$-ribbon numbers", and there is a lot known about them. For a good explanation, see Section 2 of "0-Hecke algebra actions on coinvariants and flags" by Jia Huang: https://arxiv.org/abs/1211.3349v2.