Different derivations of the value of $prod_{0leq j<k<n}(eta^k-eta^j)$
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
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$
nt.number-theory gr.group-theory soft-question teaching elementary-proofs
add a comment |
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
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$
nt.number-theory gr.group-theory soft-question teaching elementary-proofs
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
5
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
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20
add a comment |
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
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$
nt.number-theory gr.group-theory soft-question teaching elementary-proofs
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
$$prod_{j<k}^{0,n-1}(eta^k-eta^j)=n^{frac{n}2}i^{T(n)}.$$
nt.number-theory gr.group-theory soft-question teaching elementary-proofs
nt.number-theory gr.group-theory soft-question teaching elementary-proofs
edited Dec 29 '18 at 6:35
asked Dec 28 '18 at 5:55
T. Amdeberhan
17.1k228126
17.1k228126
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
5
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
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20
add a comment |
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
5
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
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20
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
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
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
5
5
can you change to a more specific title?
– YCor
Dec 28 '18 at 7:52
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
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
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20
add a comment |
4 Answers
4
active
oldest
votes
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.
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
add a comment |
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
13 hours ago
add a comment |
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.
1
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
yesterday
@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
yesterday
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
add a comment |
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 $P$.
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
add a comment |
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4 Answers
4
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4 Answers
4
active
oldest
votes
active
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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.
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
add a comment |
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.
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
add a comment |
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.
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.
answered Dec 28 '18 at 7:13
community wiki
user44191
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
add a comment |
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
I appreciate for the technique.
– T. Amdeberhan
13 hours ago
add a comment |
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
13 hours ago
add a comment |
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
13 hours ago
add a comment |
Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.
Your are asking about determinant of the Schur Matrix. So you can use original Schur's article or another classical expositions mentioned at Mathworld.
answered Dec 28 '18 at 7:23
Alexey Ustinov
6,74245779
6,74245779
Thank you, indeed.
– T. Amdeberhan
13 hours ago
add a comment |
Thank you, indeed.
– T. Amdeberhan
13 hours ago
Thank you, indeed.
– T. Amdeberhan
13 hours ago
Thank you, indeed.
– T. Amdeberhan
13 hours ago
add a comment |
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.
1
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
yesterday
@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
yesterday
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
add a comment |
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.
1
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
yesterday
@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
yesterday
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
add a comment |
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.
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.
edited 18 hours ago
answered 2 days ago
Geoff Robinson
29.1k279108
29.1k279108
1
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
yesterday
@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
yesterday
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
add a comment |
1
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
yesterday
@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
yesterday
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
1
1
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
yesterday
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
yesterday
@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
yesterday
@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
yesterday
1
1
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
This is a very nice alternative. Thank you.
– T. Amdeberhan
13 hours ago
add a comment |
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 $P$.
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
add a comment |
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 $P$.
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
add a comment |
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 $P$.
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 $P$.
answered 2 days ago
François Brunault
12.7k23569
12.7k23569
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
add a comment |
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
Thank you for adding this variation in the proof.
– T. Amdeberhan
13 hours ago
add a comment |
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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
5
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
I think that "number theory" is more suitable here than "complex variables"
– Alexey Ustinov
Dec 29 '18 at 1:20