Let $a\_{n,k}$ be the coefficient of $$X\_1^{\frac{k(n-1)}{2}}X\_2^{\frac{k(n-1)}{2}}\cdots X\_n^{\frac{k(n-1)}{2}}$$ in the expansion of the real polynomial $$\left(\prod\limits\_{1\leq i<j\leq n}(X\_j-X\_i)\right)^k,$$where $n,k$ are positive integers such that $n>1$ and $k(n-1)\equiv0 \pmod 2$.


Since $$\prod\limits\_{1\leq i<j\leq n}(X\_j-X\_i)=\sum\limits\_{\sigma\in S\_n}sgn(\sigma)\prod\limits\_{i=1}^nX\_{\sigma\_i}^{i-1},$$
here the sum is computed over all permutations $\sigma$ of the set $\{1,2,\cdots,n\}$, we can get the following simple results:


$(1)a\_{n,1}=0$ for every odd positive integer $n>1$;


$(2)a\_{n,2}\neq 0$ for every positive integer $n>1$.


I want to ask whether $a\_{n,k}$ is equal to 0 or not when $k>2$.



The coefficient of a specific monomial in the expansion of the following polynomial
Let's call your polynomial $\Delta$. Then $\Delta$ is anti-symmetric in the $x\_i$ (if you switch $x\_i$ to $x\_j$ then $\Delta$ turns into $-\Delta$). If $k$ is odd, then $\Delta^k$ also has the property, so $a\_{n,k}=0$ whenever $k$ is odd.


I will now show that $(-1)^{\binom{n}{2} m} a\_{n,2m}>0$. Write $k=2m$. Let $T$ be the torus $\theta\_1 + \theta\_2 + \cdots + \theta\_n=0$ inside $(\mathbb{R}/2 \pi \mathbb{Z})^n$, equipped with the Haar measure $d V$ with volume one.


Since $\Delta^{k}$ is homogenous of degree $k \binom{n}{2}$, it cannot contain monomials of the form $(x\_1 x\_2 \cdots x\_n)^e$ for any $e$ other than $k \tfrac{n-1}{2}$. Therefore, we can extract $a\_{n,k}$ by a contour integral over $T$:


$$a\_{n,k} = \int\_T \Delta(e^{i \theta\_1}, \ldots, e^{i \theta\_n})^{2m} dV$$
$$=\int\_T \prod\_{a<b} \left( e^{i \theta\_a} - e^{i \theta\_b} \right)^{2m} dV$$
$$=\int\_T \prod\_{a<b} \left( e^{i2 \theta\_a} - 2 e^{i (\theta\_a+\theta\_b)} + e^{i 2 \theta\_b} \right)^{m} dV$$
$$=\int\_T e^{i (n-1) \sum \theta\_a} \prod\_{a<b} \left( e^{i (\theta\_a- \theta\_b)} - 2 + e^{i (\theta\_b-\theta\_a)} \right)^{m} dV$$
$$=(-2)^{\binom{n}{2} m} \int\_T \prod\_{a<b} \left( 1 - \cos (\theta\_a - \theta\_b) \right)^{m} dV.$$
Going from the second-to-last line to the last line, I used that $\sum \theta\_a=0$ on $T$.


The integrand is nonnegative (and positive except on a set of measure zero) so the final integral is positive.


