Dtyjlui

Good evening,

Could someone please demonstrate why this property is valid?

Given $sigmain S_n$

$$left|prod_{i<j} frac{sigma(j)-sigma(i)}{j-i}right|=1$$

This is because $sigma$ is a permutation, and therefore a one-to-one correspondence.

You can rewrite the product in terms of numerators and denominators by way of
$$
begin{align*}
leftlvertprod_{i<j}frac{sigma(j)-sigma(i)}{j-i}rightrvert&=frac{prodlimits_{i<j}leftlvertsigma(j)-sigma(i)rightrvert}{prodlimits_{i<j}(j-i)}.
end{align*}
$$
Re-index the product in the numerator by letting $h=sigma^{-1}(i)$ and $k=sigma^{-1}(j)$. Note that we can't assume $h<k$; however, we can still index the product in the numerator over all sets ${h,k}$ of two distinct integers in $[1,n]$.

This reindexing yields
$$
prod_{i<j}lvertsigma(j)-sigma(i)rvert=prod_{{h,k}}lvert k-hrvert=prod_{h<k}(k-h).
$$

Detailed proof: See Exercise 5.13 (a) in my Notes on the combinatorial fundamentals of algebra, 10th of January 2019. The claim I prove there is more general: I show that if $x_1, x_2, ldots, x_n$ are any $n$ complex numbers, and if $sigma$ is any permutation of $left{1,2,ldots,nright}$, then
begin{equation}
prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)
= left(-1right)^{sigma} cdot prod_{i < j} left(x_i - x_jright) ,
label{darij1.eq.1}
tag{1}
end{equation}
where $left(-1right)^{sigma}$ denotes the sign of the permutation $sigma$.
In order to obtain your equation from eqref{darij1.eq.1}, you have to set $x_i = i$ and take absolute values (so that the sign $left(-1right)^{sigma}$ disappears, since its absolute value is $1$).

Let me sketch how to quickly prove the weaker equality
begin{equation}
left|prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)right|
= left|prod_{i < j} left(x_i - x_jright)right|
label{darij1.eq.2}
tag{2}
end{equation}
(which is still sufficient for your purposes). This is what @NickPeterson has already suggested, but my more rigorous notations shall hopefully close the cracks which let confusion slip through.

First of all, the absolute value of a product equals the product of the absolute values of the factors; thus,
begin{align}
left|prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)right|
= prod_{i < j} left| x_{sigmaleft(iright)} - x_{sigmaleft(jright)} right| .
end{align}

Next, let $P$ be the set of all pairs $left(i, jright)$ of integers $i, j in left{1,2,ldots,nright}$ satisfying $i < j$; also, let $G$ be the set of all $2$-element subsets of $left{1,2,ldots,nright}$. Note that the product sign "$prodlimits_{i < j}$" is equivalent to "$prodlimits_{left(i, jright) in P}$".

The two sets $P$ and $G$ have the same size (namely, $dbinom{n}{2} = nleft(n-1right) / 2$), and this is no coincidence: There is a bijection from $P$ to $G$. This bijection simply maps each pair $left(i, jright)$ to the two-element set $left{i, jright}$. The inverse of this bijection maps each two-element set to the pair consisting of its smaller element and its larger element (in this order).

The permutation $sigma$ of $left{1,2,ldots,nright}$ gives rise to a permutation $sigma_*$ of the set $G$, which sends each two-element subset $I$ to $sigmaleft(Iright)$ (in other words, it sends each two-element subset $left{i,jright}$ to $left{sigmaleft(iright), sigmaleft(jright)right}$). Why is this a permutation of $G$? Well, again, its inverse is easy to find (it does the same thing, just with $sigma^{-1}$ instead of $sigma$). So $sigma_*$ is a permutation of $G$, i.e., a bijection from $G$ to $G$.

Now the crucial insight: If $left(i, jright) in P$, then the absolute value $left| x_i - x_j right|$ depends only on the set $left{i, jright} in G$ (not on the pair $left(i, jright) in P$). In other words, if $I in G$ is any two-element subset, then we can define a real number $f_I$ by setting
begin{align}
f_I = left| x_i - x_j right|,
qquad text{ where $I$ is written as $I = left{i, jright}$}.
end{align}
In order to formally prove this, you should recall that there are exactly two ways of writing $I$ as $I = left{i, jright}$, and check that these two ways lead to the same value of $left| x_i - x_j right|$ (easy: these two ways only differ in the order of elements, and we have $left| x_a - x_b right| = left| x_b - x_a right|$).

Note that every $left(i, jright) in P$ satisfies $sigma_* left( left{ i, j right} right) = left{ sigmaleft(iright), sigmaleft(jright) right}$ and thus
begin{align}
f_{sigma_* left( left{ i, j right} right)}
= f_{left{ sigmaleft(iright), sigmaleft(jright) right}}
= left| x_{sigmaleft(iright)} - x_{sigmaleft(jright)} right|
label{darij1.eq.3}
tag{3}
end{align}
(by the definition of $f_{left{ sigmaleft(iright), sigmaleft(jright) right}}$).

Now,
begin{align}
left|prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)right|
& = prod_{i < j} left| x_{sigmaleft(iright)} - x_{sigmaleft(jright)} right| \
& = prod_{left(i, jright) in P} underbrace{left| x_{sigmaleft(iright)} - x_{sigmaleft(jright)} right|}_{substack{ = f_{sigma_* left( left{ i, j right} right)} \ left(text{by eqref{darij1.eq.3}}right)}} \
& qquad left(text{since "$prodlimits_{i < j}$" is equivalent to "$prodlimits_{left(i, jright) in P}$"}right) \
& = prod_{left(i, jright) in P} f_{sigma_* left( left{ i, j right} right)} \
& = prod_{I in G} f_{sigma_* left(Iright)}
end{align}
(here, we have substituted $I$ for $left{ i, j right}$ in the product, since the map $G to P, left(i, jright) mapsto left{ i, j right}$ is a bijection).
Thus,
begin{align}
left|prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)right|
= prod_{I in G} f_{sigma_* left(Iright)} = prod_{I in G} f_I
label{darij1.eq.4}
tag{4}
end{align}
(here, we have substituted $I$ for $sigma_* left(Iright)$ in the product, since $sigma_*$ is a bijection).

Note that the right hand side of eqref{darij1.eq.4} does not depend on $sigma$. Applying the same reasoning to the permutation $operatorname{id}$ instead of $sigma$, we thus obtain
begin{align}
left|prod_{i < j} left(x_{operatorname{id}left(iright)} - x_{operatorname{id}left(jright)}right)right|
= prod_{I in G} f_I .
end{align}
In other words,
begin{align}
left|prod_{i < j} left(x_i - x_jright)right|
= prod_{I in G} f_I .
end{align}
Comparing this equality with eqref{darij1.eq.4}, we obtain
$left|prod_{i < j} left(x_{sigmaleft(iright)} - x_{sigmaleft(jright)}right)right|
= left|prod_{i < j} left(x_i - x_jright)right|$.
Thus, eqref{darij1.eq.2} is proven.