Suppose a collection $x_i$ defined on the line, and with ascending order, i.e. $x_1<x_2<x_3<\cdots$ and the definition $x_{ij} = x_i-x_j$. From the definition it is immediately clear that when $i>j$, then $x_i > x_j$ such that $x_{ij} >0$. I would now like to rewrite the expression \begin{align} \prod_1^N \prod_{j<i} x_{ji} \end{align} in terms of \begin{align} \prod_1^N \prod_{j\neq i} x_{ij}^{\frac12} \end{align} where the product over $j$ has indices $1\leq j\leq N$ barring the explicit constraint.
HEre is my attempt: \begin{align} \prod_1^N \prod_{j<i} x_{ji} &= \prod_1^N \left[\prod_{j<i} (-1)\right] \prod_{j<i} x_{ij} \\ &= \left[\prod_1^N (-1)^{i-1}\right] \prod_1^N\prod_{j<i} x_{ij} \\ &= (-1)^{N(N-1)/2} \prod_1^N\prod_{j<i} x_{ij}. \end{align}
Further, it is clear that all factors under the product are positive, and so we may perform the trivial operation $z_{ij} \mapsto (z_{ij}^{\frac12})^2$. Doing so for the expression we have just obtained, we get for example \begin{align} \prod_1^N\prod_{j<i} x_{ij} &= \prod_1^N \left[\prod_{j<i} x_{ij}^{\frac12} \right] \left[\prod_{j<i} x_{ij}^{\frac12} \right]\\ &= \prod_1^N \left[\prod_{j<i} x_{ij}^{\frac12} \right] \prod_1^N\left[\prod_{j > i} x_{ji}^{\frac12} \right]\\ &= \prod_1^N \left[\prod_{j<i} x_{ij}^{\frac12} \right] \prod_1^N\left[(-1)^{\frac12(i-1)} \prod_{j > i} x_{ij}^{\frac12} \right]\\ &= i^{N(N-1)/2} \prod_1^N \left[\prod_{j<i} x_{ij}^{\frac12} \right] \prod_1^N \prod_{j > i} x_{ij}^{\frac12} \end{align} after doing a relabeling $i\leftrightarrow j$.
When I calculate this for a few values of $N$ and a suitable set $x_i$, I find that in some cases I get an extra minus sign. How can this be? I can't find the mistake in the above.
I believe I have found the culprit:
In the third line of the second
align, I have used a plus sign in the power of (-1) under the product, i.e. the $(-1)^{1/2 * (i-1)}$. This is nonsense, we are effectively doing the reverse as in the first step (going from a root of a real number to a complex one). Using a minus sign reproduces a prefactor $(-1)^{N(N-1)/4}$, which numerically agrees for all values of $N$.