# Different derivations of the value of $\prod_{0\leq j<k<n}(\eta^k-\eta^j)$

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

• 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? – user44191 Dec 28 '18 at 6:04
• 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. – T. Amdeberhan Dec 28 '18 at 6:07
• can you change to a more specific title? – YCor Dec 28 '18 at 7:52
• I have changed the title to a more specific one, given that the question is now in HNQ. – 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.

• Thank you, indeed. – 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

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

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 $$A_n = \prod_{0\leq j. Note that the second term is a difference of (unequal) conjugates where the minuend has positive imaginary part (and the subtrahend therefore negative imaginary part), and therefore will always have argument $$\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

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

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

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.

• I appreciate for the technique. – 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. • 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. – Mark Wildon Dec 31 '18 at 13:20 • @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. – Geoff Robinson Dec 31 '18 at 13:29
• This is a very nice alternative. Thank you. – T. Amdeberhan Jan 1 at 16:40

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

Let $$P= \displaystyle\prod_{\substack{j,k=0 \\ j. Take the logarithm: \begin{align*} \log P & = \sum_{j Call $$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$$.

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