Let $\eta=e^{\frac{2\pi i}n}$, 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 $$\prod_{j<k}^{0,n-1}(\eta^k-\eta^j)=n^{\frac{n}2}i^{T(n)}.$$

  • $\begingroup$ How precise would you like the conclusion? Would getting the norm and the parity of $T(n)$ be enough, or do you want $T(n)$ exactly? $\endgroup$ – user44191 Dec 28 '18 at 6:04
  • $\begingroup$ It'd be nice to get it exactly, however even getting to what you described does add insight into the discussion. So, you're welcome to present it. $\endgroup$ – T. Amdeberhan Dec 28 '18 at 6:07
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    $\begingroup$ can you change to a more specific title? $\endgroup$ – YCor Dec 28 '18 at 7:52
  • $\begingroup$ I have changed the title to a more specific one, given that the question is now in HNQ. $\endgroup$ – Wojowu Dec 28 '18 at 15:22

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

  • $\begingroup$ Thank you, indeed. $\endgroup$ – T. Amdeberhan Jan 1 at 16:39

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 $A_n = \prod_{0\leq j<k<n} \eta'^{k + j} (\eta'^{k - j} - \eta'^{j - k})$. Note that the second term is a difference of (unequal) conjugates where the minuend has positive imaginary part (and the subtrahend therefore negative 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 $\sum_{0 \leq j < k < n} j + k$.

$\sum_{0 \leq j < k < n} j + k = \left(\sum_{0 \leq j < k < n} j\right) + \left(\sum_{0 \leq j < k < n} k\right)$

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

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

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

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.

  • $\begingroup$ I appreciate for the technique. $\endgroup$ – T. Amdeberhan Jan 1 at 16:39

Your determinant is essentially the Van der Monde determinant ${\rm det}(A),$ where $A$ is the $n \times n$ matrix $[\eta^{(j-1)(k-1)}].$

Note that $A$ is the character table of the cyclic group of order $n$ so that $A{\bar A}^{T}= nI_{n}$, using the orthogonality relations for group characters, and $|{\rm det A}| = n^{\frac{n}{2}}.$ One can continue in this vein, but I will sketch a more general calculation of the determinant of the character table of a general finite group, which simplifies considerably in the case of cyclic groups.

Let $G$ be a finite group with $t$ conjugacy classes, say with representatives $g_{1},g_{2}, \ldots g_{t}.$ Let us label these classes so that $1_{G} = g_{1}, g_{2},\ldots ,g_{s}$ are exactly those class representatives which are conjugate to their inverses in $G$ and so that $g_{s+2j} = g_{s+2j-1}^{-1}$ for $1 \leq j \leq \frac{t-s}{2}.$

Let $\chi_{1},\chi_{2}, \ldots \chi_{t}$ be the complex irreducible characters of $G.$

Let $B$ be the character table of $G$, which is the $t \times t$ matrix $[\chi_{j}(g_{k})].$

By the orthogonality relations for group characters, we see that ${\bar B}^{T}B$ is the diagonal matrix whose $j$-th diagonal entry is $|C_{G}(g_{j})|.$

Let $\pi \in {\rm S}_{t}$ be the permutation fixing $1,2,\ldots,s$ and interchanges $s+2j-1$ and $s+2j$ for $1 \leq j \leq \frac{t-s}{2}.$ Let $P$ be the associated permutation matrix.$

Note that $BP = {\bar B}$ since $BP$ has the same first $s$ columns as $B$ and has the columns corresponding to $g_{s+2j-1}$ and $g_{s+2j} = g_{s+2j-1}^{-1}$ interchanged for $1 \leq \frac{t-s}{2}$ (note that the first $s$ columns of $B$ are real as $g_{j}$ is conjugate to $g_{j}^{-1}$ for $1 \leq j \leq s.$

Hence ${\rm det}B^{2} = (-1)^{\frac{t-s}{2}} \prod_{j=1}^{t} |C_{G}(g_{j})|$ and ${\rm det}B = (i)^{\frac{t-s}{2}} \sqrt{\prod_{j=1}^{t} |C_{G}(g_{j})|}$ where $i = \sqrt{-1}.$

Edit following Mark Wildon's comment: In fact, although it looks as though there is a free choice of $\sqrt{-1}$ in the above, in the case that $G$ is cyclic of order $n$, if we define $\eta$ as in the question as $\eta = \exp{\frac{2 \pi i}{n}},$ then we have to use the other square root of $-1$ in the above expression for ${\rm det}(A),$ so ${\rm det}A = (-i)^{\frac{n-s}{2}} n^{\frac{n}{2}}$ where $s =1 $ if $n$ is odd and $s = 2$ if $n$ is even.

Even later edit: Here are some remarks about the general character table determinant which can be seen directly without making a choice for $\sqrt{-1}.$ Note that since ${\bar B} = BP$ and $\overline{{\rm det}B} = (-1)^{\frac{t-s}{2}}{\rm det}B$, we have that ${\rm det}(B) \in \mathbb{R}$ if and only if $t \equiv s$ (mod $4$). When $t \not \equiv s$ (mod $4$), we see that ${\rm det}B$ is pure imaginary. At present, I don't see a quick way (in the general case) to determine which choice of $\sqrt{-1}$ to use in the earlier formula once a choice of $i$ is fixed for the character table entries.

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    $\begingroup$ In the final step, isn't there a choice of square roots of $-1$? If so, $\det B$ is determined only up to a sign $(-1)^{(t-s)/2}$. Checking $n$ in $0,1,\ldots, 8$ using $t-s = n-1-[n$ is even$]$ shows that $(-i)^{(t-s)/2} = i^{T(n)} = i^{(3n-2)(n-1)/2}$ for all $n$, so it seems the other sign is correct. $\endgroup$ – Mark Wildon Dec 31 '18 at 13:20
  • $\begingroup$ @MarkWildon :Yes, there is a choice of sign to be made which is why I said I was "sketching a proof" and was a bit lazy- the question seems to allow the choice of a sign of $\sqrt{-1},$ but the definition of $\eta$ as $\exp(2 \pi i/n)$ in fact removes the freedom of choice, and your checking shows that you need to take the negative of the $i$ appearing in the exponential when taking the square root of a (sometimes) negative quantity. $\endgroup$ – Geoff Robinson Dec 31 '18 at 13:29
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    $\begingroup$ This is a very nice alternative. Thank you. $\endgroup$ – T. Amdeberhan Jan 1 at 16:40

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

Let $P= \displaystyle\prod_{\substack{j,k=0 \\ j<k}}^{n-1} (\eta^k-\eta^j)$. 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{2\pi i k}{n} - \sum_{a=1}^{n-1} (n-a) \mathrm{Li}_1(\eta^{-a}) \\ & = \frac{2\pi 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}) = 2\pi 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{3\pi 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 $P$.

  • $\begingroup$ Thank you for adding this variation in the proof. $\endgroup$ – T. Amdeberhan Jan 1 at 16:41

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