An operator derived from the divided difference operator $\partial_{w_0}$

Some main definitions and basic facts of divided differences:

In the symmetric group $S_n$, let $s_i$ denote the adjacent transposition $(i,i+1),i\in \{1,2,\cdots,n-1\}.$ Since $S_n$ is generated by adjacent transpositions, any permutation $w\in S_n$ can be written as $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ and an expression $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ of minimal possible length $l$ is called a reduced decomposition, $l=\ell(w)$ is the length of $w$. In fact, $\ell(w)=\#\{(i,j):1\leq i<j\leq n\ \text{and}\ w(i)>w(j)\}$. So $w_0$, given by $w(i)=n+1-i$, is the longest element in $S_n$ whose length is $\frac{n(n-1)}{2}$.

Let $f=f(x_1,x_2,\cdots,x_n)$ be a polynomial in $\mathbb{Z}[x_1,x_2,\cdots,x_n]$. For $w\in S_n$, denote $w(f)=f(x_{w^{-1}(1)},x_{w^{-1}(2)},\cdots,x_{w^{-1}(n)})$. The divided difference operator $\partial_i$ is then defined by
For any permutation $w\in S_n$, define the operator $\partial_w$ by
$$\partial_w=\partial_{i_1}\partial_{i_2}\cdots \partial_{i_l},$$
where $w=s_{i_1}s_{i_2}\cdots s_{i_l}$ is a reduced decomposition and $\partial_w$ does not depend on the choice of such reduced decomposition.


1. For any polynomial $f$,
$$\partial_{w_0}(f)=\left(\sum\limits_{w\in S_n}(-1)^{\ell(w)}w(f)\right)\cdot\prod\limits_{1\leq i<j\leq n}(x_i-x_j)^{-1}.$$
2. Let $f=\sum\limits_{i=1}^na_ix_i$ be a polynomial of degree one, then for any permutation $w$ and polynomial $g$, we have
$$\partial_w(fg)=w(f)\partial_w(g)+\sum\limits_{1\leq i<j\leq n}(a_i-a_j)\partial_{wt_{ij}}(g),$$
where the sum is over all $1\leq i<j\leq n$ such that $\ell(wt_{ij})=\ell(w)-1$ and $t_{ij}$ is the permutation that interchanges $i$ and $j$.

Note: The content above are all come from this article.

For any $\frac{n(n-1)}{2}\times n$ matrix $A=(a_{ij})$ with entries from $\mathbb{Z}$, let $f_A$ be the following polynomial
then we can get the following corollary which follows from the above proposition.

For simplicity, $\frac{n(n-1)}{2}$ hereinafter is denoted by $m$.

Corollary For any matrix $A=(a_{ij})\in M_{m\times n}(\mathbb{Z})$,
where the sum is over all $1\leq i_1<j_1\leq n,1\leq i_2<j_2\leq n,\cdots,1\leq i_m<j_m\leq n$ such that $t_{i_1j_1},t_{i_2j_2},\cdots,t_{i_mj_m}$ satisfy $\ell(t_{i_1j_1}\cdots t_{i_kj_k})=k,\ k=1,2,\cdots,m$.

For simplicity, the subscript of the sum above hereinafter is denoted by $T_n$.

Define a operator
\alpha:M_{m\times n}(\mathbb{Z})&\longrightarrow \mathbb{Z},\\
A=(a_{ij})&\mapsto \partial_{w_0}(f_A)=\sum\limits_{T_n}(a_{1i_1}-a_{1j_1})(a_{2i_2}-a_{2j_2})\cdots(a_{mi_m}-a_{mj_m}),
then $\partial_{w_0}(f_A)=\alpha(A)$ and I can get some properties of the operator $\alpha$ as follows.

1. $\alpha(cA)=c^m\alpha(A)$, $\forall c\in \mathbb{Z}$ and $A\in M_{m\times n}(\mathbb{Z})$.

2. Viewing an matrix in $M_{m\times n}(\mathbb{Z})$ as being composed of $m$ rows, $\alpha$ is an $m$-linear function.

3. Interchanging any pair of rows of $A$ does not change $\alpha(A)$, $\forall A\in M_{m\times n}(\mathbb{Z})$.

4 Interchanging any pair of columns of $A$ multiplies $\alpha(A)$ by −1, $\forall A\in M_{m\times n}(\mathbb{Z})$. Therefore, if two columns of $A$ are identical, $\alpha(A)=0$.

I have two questions as follows.


1. The properties $1,2$ above can easily follow from the definition of the operator $\alpha$ and I got the properties $3,4$ by using the equality $\partial_{w_0}(f_A)=\alpha(A)$ and the property of $\partial_{w_0}$. Take the property $4$ for example, for any $1\leq i<j\leq n$ and $A\in M_{m\times n}(\mathbb{Z})$, let $A’$ be the matrix obtained from $A$ by interchanging the $i$th and $j$th columns of $A$, then we have
My first question is how to get the properties $3,4$ by using the definition of $\alpha$.

2. When $\alpha(A)\neq 0$? What I am asking for is a sufficient condition or necessary and sufficient condition of $A\in M_{m\times n}(\mathbb{Z)}$ that $\alpha(A)\neq 0$.


Source : Link , Question Author : user173856 , Answer Author : Community

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