coefficients in the expansion of multivariable expression

Question

Consider the expansion of the following $N$ variable expression $$ D_N(z_1,\ldots,z_N)=\prod_{1\leq j<k\leq N}\left(1-\frac{z_j}{z_k}\right)\left( 1-\frac{z_k}{z_j} \right) $$ For example, in the case of $N=3$ we have $$ \begin{align*} D_3(z_1,z_2,z_3) & = \left(1-\frac{z_1}{z_2}\right)\left( 1-\frac{z_2}{z_1} \right) \left(1-\frac{z_2}{z_3}\right)\left( 1-\frac{z_3}{z_2} \right) \left(1-\frac{z_3}{z_1}\right)\left( 1-\frac{z_1}{z_3} \right) = \\ & = 6 -2\frac{z_1}{z_2}-2\frac{z_2}{z_1} -2\frac{z_2}{z_3}-2\frac{z_3}{z_2} -2\frac{z_3}{z_1}-2\frac{z_1}{z_3} + \\ & +2\frac{z_1z_2}{z_3^2} +2\frac{z_3^2}{z_1z_2} +2\frac{z_2z_3}{z_1^2} +2\frac{z_1^2}{z_2z_3} +2\frac{z_3z_1}{z_2^2} +2\frac{z_2^2}{z_3z_1} - \\ & -\frac{z_1^2}{z_2^2} -\frac{z_2^2}{z_1^2} -\frac{z_2^2}{z_3^2} -\frac{z_3^2}{z_2^2} -\frac{z_3^2}{z_1^2} -\frac{z_1^2}{z_3^2} \end{align*} $$

It is known that the constant term of $D_N$ is $N!$ (this can be proved using the formula for the Dyson integral). I am interested in finding out if there are any patterns that describe the coefficients of the other (non-constant) terms in the expansion of $D_N$ ?

As an analogy for what I'm hoping to obtain, in the case of the well-kown multinomial expansion $$ (z_1+z_2+\ldots+z_M)^N $$ the coefficient of any term $z_1^{k_1}z_2^{k_2}\ldots z_M^{k_M}$ can be simply expressed as $$ \frac{N!}{k_1! k_2!\ldots k_M!} $$

Answer 1

Sills and Zeilberger have a paper, Disturbing the Dyson Conjecture (in a GOOD Way), where they talk about precisely what happens when you want to describe other coefficients in your product, other than the constant term. The bad news is that the formulas seem to get very complicated very fast, which makes it unlikely to hope for an elegant uniform formula. The good news is that these coefficients will always satisfy a recurrence relation (exploited by Good in one of the original proofs of Dyson's conjecture) and this can be used to guess and prove a formula for specific coefficient in the product.

As an illustration of this, in the paper Two Coefficients of the Dyson Product, Lv, Xin, and Zhou give formulas for the coefficient of $x_r^2/x_s^2$ and $x_r^2/x_sx_t$.

Answer 2

You may play with this method and find formulae for many coefficients. To be more concrete, below I prove using it that the coefficient of $(z_p/z_q)^c$ equals $-(N-c) (N-2)!$ and the coefficient of $z_1^c(z_2\dots z_{c+1})^{-1} $ equals $(-1)^c c! (N-c)!$ for all $c=1,2,\dots,N-1$.

Assume without loss of generality that $c_1\geqslant c_2\geqslant \ldots \geqslant c_N$, $\sum c_i=0$, and we look for $M:=[\prod z_i^{c_i}]\prod_{i<j} (1-z_i/z_j)(1-z_j/z_i)$. We have $$M=\left[z_1^{N-1+c_1}z_2^{N-1+c_2}\ldots z_N^{N-1+c_N}\right]F,\quad F=\prod_{i<j} (z_j-z_i)(z_i-z_j-\lambda_{ij}),$$ for any constants $\lambda_{ij}$.

Denote $A_i=[0,1,\dots,N-1+c_i]$. Look how can it happen that $z_i\in A_i$ and $F\ne 0$. At first, all $z_i's$ must be distinct. Choose all $\lambda_{ij}=1$. Consider the set $Z=\{z_1,\dots,z_N\}\subset A_1$. It consists of several segments of consecutive integer points, and in each segment the indices of corresponding $z_i$'s must increase (else some bracket $z_i-z_j-1$, with $i<j$, does vanish).

If all $c_i$'s are equal to 0, this already implies that $z_i=i-1$, so the RHS of CN formula has unique summand and we get $N!$ formula for the constant term. If $c_1=c>0$, $c_{N}=c_{N-1}=\dots =c_{N-c+1}=-1$, the same choice of $\lambda$'s gives you unique non-zero value, that yields the aforementioned answer $(-1)^c c! (N-c)! $.

Now consider, say, $c_1=c>0$, $c_2=\dots=c_{N-1}=0$, $c_N=-c$, as I said in the beginning. (This includes, in particular, the case $c_1=2,c_{N}=-2$ mentioned in the answer of Gjergji Zaimi.)

It makes sense to change $\lambda$'s: take $\lambda_{iN}=1$ for $i=1,\dots,N-1$; $\lambda_{1j}=c$ for $j=2,3,\dots,N-1$, other $\lambda$'s are equal to 0. We have $z_{N}\leqslant N-1-c$. It implies that $z_{N}+1\notin Z$. Therefore $Z=\{0,1,\dots,N-1\}\cup z_1\setminus (z_N+1)$ and $z_1\in \{N,N+1,\dots,N+c-1\}$. But $z_1-c\notin \{z_2,z_3,\dots,z_{N-1}\}$. It may happen only if $z_1-c=z_N$ or $z_1-c=z_N+1$. But $z_1-c\geqslant N-c\geqslant z_N+1$, thus we must have $z_N=N-1-c$, $z_1=N$, other $z_i$'s form a permutation of $\{0,1,\dots,N-1\}\setminus \{N-c\}$. All permutations give the same contribution to the CN formula, and after careful but straightforward cancellation we get the above answer.