Argthtjtr

Let $eta=e^{frac{2pi i}n}$, an $n$-th root of unity. For pedagogical reasons and inspiration, I ask to see different proofs (be it elementary, sophisticated, theoretical, etc) for the following product evaluation.

If $T(n)=frac{(3n-2)(n-1)}2$ and $i=sqrt{-1}$ then
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$

We first find the norm; we then determine the argument.

Call the product you wrote $A_n$. Then $A_n^2 = prod_{j<k}^{0,n-1} (eta^k - eta^j)^2 = Disc(x^n - 1) = (-1)^{frac{n (n -1)}{2}}Res(x^n - 1, n x^{n - 1})$

$= (-1)^{frac{n(n-1)}{2}} n^n prod_{0 leq i < n, 0 leq j < n-1} (eta^i - 0)$

All terms in the expression except $n^n$ have norm $1$, so we have that $|A_n| = n^{frac{n}{2}}$. We therefore only need to figure out the argument of $A_n$.

Let $eta' = e^frac{2 pi i}{2n}$ be the square root of $eta$. We can rewrite $A_n = prod_{0leq j<k<n} eta'^{k + j} (eta'^{k - j} - eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates with positive imaginary part, and therefore will always have argument $frac{pi}{2}$. So let us concentrate on the argument of the first term, $prod_{0 leq j < k < n} eta'^{k +j}$. We can do this by finding $sum_{0 leq j < k < n} j + k$.

$sum_{0 leq j < k < n} j + k = left(sum_{0 leq j < k < n} jright) + left(sum_{0 leq j < k < n} kright)$

$= left(sum_{0 leq j <n} (n - j - 1)jright) + left(sum_{0leq k<n} k*kright)$

$= sum_{0 leq j < n} (n - j - 1)j + j*j = sum_{0 leq j < n} (n - 1)j$

$= (n - 1) frac{n (n - 1)}{2}$

We therefore end up with an argument of $frac{n(n - 1)}{2} frac{pi}{2} + frac{n (n - 1)^2}{2} frac{2 pi}{2n} = frac{(3n^2 - 5n + 2)pi}{4}$. We finally have that:

The norm of $A_n$ is $n^frac{n}{2}$, and the argument is $frac{(3n^2 - 5n + 2)pi}{4}$. Correspondingly, we have that $A_n = n^{frac{n}{2}} i^{T(n)}$, as desired.

Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.

Here is a proof using the logarithmic function $mathrm{Li}_1(z)=-log(1-z)$:

Let $P= displaystyleprod_{substack{j,k=0 \ j<k}}^{n-1} (eta^k-eta^j)$. Take the logarithm:
begin{align*}
log P & = sum_{j<k} log eta^k - sum_{j<k} mathrm{Li}_1(eta^{j-k}) \
& = sum_{k=0}^{n-1} k cdot frac{2pi i k}{n} - sum_{a=1}^{n-1} (n-a) mathrm{Li}_1(eta^{-a}) \
& = frac{2pi i}{n} sum_{k=0}^{n-1} k^2 - sum_{a=1}^{n-1} a mathrm{Li}_1(eta^a).
end{align*}
Call $S$ the second sum. We have
begin{align*}
S & = sum_{a=1}^{n-1} a mathrm{Li}_1(eta^a) = frac12 sum_{a=1}^{n-1} bigl(a mathrm{Li}_1(eta^a)+(n-a)mathrm{Li}_1(eta^{-a})bigr)\
& = frac{n}{2} sum_{a=1}^{n-1} mathrm{Li}_1(eta^a) + sum_{a=1}^{n-1} a cdot bigl(mathrm{Li}_1(eta^a)-mathrm{Li}_1(eta^{-a})bigr)
end{align*}
The first sum is easy to compute and is equal to $-log n$. Regarding the second sum, the classical Fourier expansion for the Bernoulli polynomial $B_1(x)=x-frac12$ on $(0,1)$ gives
begin{equation*}
mathrm{Li}_1(eta^a)-mathrm{Li}_1(eta^{-a}) = 2pi i bigl(frac12 - frac{a}{n}bigr).
end{equation*}
From there, it is not difficult to finish the computation
begin{align*}
log P = frac{n}{2} log n - frac{pi i}{4} n(n-1) + frac{3pi i}{n} sum_{k=1}^{n-1} k^2 = frac{n}{2} log n + frac{pi i}{4} (n-1)(3n-2)
end{align*}
which gives the desired value for $P$.