TITLE: The determinant of a $4\times4$ matrix associated to some specific polynomial as follow
QUESTION [7 upvotes]: Let $f\in \mathbb{R}[x_1,x_2,x_3,x_4]$ defined by
$$f_a(x_1,x_2,x_3,x_4)=\prod_{1\leqslant i<j\leqslant4}(x_i-x_j)^{2a_{ij}}$$
where $a=(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34})\in \mathbb{N}^6$.
Define a $4\times4$ matrix $A_f$ as follow:
$$A_{f_a}=\begin{bmatrix}
 L(1) & L(\frac{x_2}{x_1}) & L(\frac{x_4}{x_3}) & L(\frac{x_2x_4}{x_1x_3})\\
 L(\frac{x_1}{x_2}) & L(1) & L(\frac{x_1x_4}{x_2x_3}) & L(\frac{x_4}{x_3})\\
 L(\frac{x_3}{x_4}) & L(\frac{x_2x_3}{x_1x_4}) & L(1) & L(\frac{x_2}{x_1})\\
 L(\frac{x_1x_3}{x_2x_4}) & L(\frac{x_3}{x_4}) & L(\frac{x_1}{x_2}) & L(1)
\end{bmatrix}$$
where $L(\frac{x_{i_1}\cdots x_{i_k}}{x_{j_1}\cdots x_{j_k}})$ denotes the coefficient of 
$$\left(\prod_{1\leqslant i<j\leqslant4}(x_ix_j)^{a_{ij}}\right)\frac{x_{i_1}\cdots x_{i_k}}{x_{j_1}\cdots x_{j_k}}$$
in the expansion of $f$.
For example, when $a_1=(a,0,0,0,0,b)$,
\begin{align}
f_{a_1}(x_1,x_2,x_3,x_4)&=(x_1-x_2)^{2a}(x_3-x_4)^{2b},\\
\\
(-1)^{a+b}A_{f_{a_1}}&=
\begin{bmatrix}
 \binom{2a}{a}\binom{2b}{b} & -\binom{2a}{a-1}\binom{2b}{b} & -\binom{2a}{a}\binom{2b}{b-1} & \binom{2a}{a-1}\binom{2b}{b-1}\\
 -\binom{2a}{a-1}\binom{2b}{b} & \binom{2a}{a}\binom{2b}{b} & \binom{2a}{a-1}\binom{2b}{b-1} & -\binom{2a}{a}\binom{2b}{b-1}\\
 -\binom{2a}{a}\binom{2b}{b-1} & \binom{2a}{a-1}\binom{2b}{b-1} & \binom{2a}{a}\binom{2b}{b} & -\binom{2a}{a-1}\binom{2b}{b}\\
 \binom{2a}{a-1}\binom{2b}{b-1} & -\binom{2a}{a}\binom{2b}{b-1} & -\binom{2a}{a-1}\binom{2b}{b} & \binom{2a}{a}\binom{2b}{b}
\end{bmatrix}.
\end{align}
It is not difficult to verify that the determinant of $A_{f_{a_1}}$ is nonzero. In fact,$$\det(A_{f_{a_1}})=\binom{2a}{a}^4\binom{2b}{b}^4\frac{(2a+1)^2(2b+1)^2}{(a+1)^4(b+1)^4}.$$
Moreover, I verified that $\det(A_{f_a})\neq0$ for many simple $a\in \mathbb{N}^6$. 
My question
Does $\det(A_{f_a})\neq0$ hold for all $a\in \mathbb{N}^6$? Is there any significance for the matrix $A_{f_a}$? Any idea is welcome!

REPLY [10 votes]: For a given monomial $Y=\frac{x_{i_1}\cdots x_{i_k}}{x_{j_1}\cdots x_{j_k}}$ the coefficient $L(Y)$ multiplied by the constant $(-1)^{\sum_{i<j} a_{ij}}$ equals $$[Y]\prod_{i,j}(1-x_i/x_j)^{a_{ij}}=\int Y^{-1}d\mu,$$
where $d\mu$ is the measure on the $4$-dimensional torus $\mathbb{T}^4=\{(x_1,x_2,x_3,x_4)\in \mathbb{C}^4:|x_1|=|x_2|=|x_3|=|x_4|=1\}$ which density w.r.t. normalized Lebesgue measure equals to $\prod_{i,j}(1-x_i/x_j)^{a_{ij}}$ (the key observation is that this is always real and almost always positive). Your matrix is then (up to aforementioned sign)  Gram matrix of the functions $1,x_2/x_1,x_4/x_3,x_2x_4/x_1x_3$ in $L^2(\mu)$. They are linearly independent, hence the determinant is strictly positive.