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For a positive integer $n$, a derangement of $\{1,\ldots,n\}$ is a permutation $\tau$ of $\{1,\ldots,n\}$ with $\tau(j)\not=j$ for all $j=1,\ldots,n$. For convenience, we let $D(n)$ denote the set of all derangements of $\{1,\ldots,n\}$.

I have the following conjecture on identities involving both derangements and roots of unity.

Conjecture. Let $n>1$ be an integer and let $\zeta$ be a primitive $n$-th root of unity.

(i) If $n$ is even, then $$\sum_{\tau\in D(n)}\prod_{j=1}^n\frac1{1-\zeta^{j-\tau(j)}}=\frac{((n-1)!!)^2}{2^n}.\tag{1}$$ If $n$ is odd, then $$\sum_{\tau\in D(n-1)}\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac1n\left(\frac{n-1}2!\right)^2.\tag{2}$$

(ii) If $n$ is even, then $$\sum_{\tau\in D(n)}\mathrm{sign}(\tau)\prod_{j=1}^n\frac1{1-\zeta^{j-\tau(j)}}=(-1)^{n/2}\frac{((n-1)!!)^2}{2^n}.\tag{3}$$ If $n$ is odd, then $$\sum_{\tau\in D(n-1)}\mathrm{sign}(\tau)\prod_{j=1}^{n-1}\frac1{1-\zeta^{j-\tau(j)}}=\frac{(-1)^{(n-1)/2}}n\left(\frac{n-1}2!\right)^2.\tag{4}$$

My numerical computation suggests that the conjecture should be true. The first assertion in part (i) of the conjecture appeared in my recent preprint available from arXiv:2108.07723. Part (ii) of the conjecture involves determinants, hence it might be not so difficult.

QUESTION. How to prove the above conjecture? Any ideas?

Your comments are welcome!

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  • $\begingroup$ I know that there is a closed formula for $\det[\frac1{x_j-y_k}]_{1\le j,k\le n}$ due to Cauchy. $\endgroup$ Sep 2, 2021 at 10:11
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    $\begingroup$ I have just proved $(3)$ by making use of a known result of Calogero and Perelomov on eigenvalues. $\endgroup$ Sep 4, 2021 at 7:52
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    $\begingroup$ $(1)$, $(2)$ and $(4)$ have just been proved by Prof. Xuejun Guo and his three PhD students, see arxiv.org/abs/2206.02592 . $\endgroup$ Jun 7 at 5:18
  • $\begingroup$ Apologies for my silly question, I'm still at the beginning of my journey, but shouldn't the odd and even cases evaluate to the same value if you replace "n" by "n-1" in the right hand side of (2) ? Since the calculation we do for the odd case is exactly the calculation for the even case of the number one less than n? So if you name the sum/product S(n) then S(5) should equal S(4) for example? Or what am I missing? $\endgroup$
    – Carlyle
    Jul 12 at 15:20

1 Answer 1

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My idea is to define a $(n-1)\times(n-1)$ Hermitian matrix with diagonal elements equal to zero and off-diagonal elements $a_{ij}= 1/(1-x_{i-j})$, where $x_k=\zeta^k$. So the left-hand side of (4) is equal to $\det(A)$. Multiply $A$ from right by a $(n-1)\times(n-1)$ diagonal matrix $b_{ii}=(1-x_{is})$ for any fixed $s\in\{-(n-1)/2,\ldots,-1,1,(n-1)/2\}$. We can show the resulting matrix $C_s$ has an eigenvalue $s$ with eigenvector whose $j$-th element is equal to $x_{j((n-1)/2-s)}$. The difficulty is to show such an eigenvalue of $C_s$ is also an eigenvalue of $C_1$ for all s, even though numerical evaluation has suggested it. Prof. Fedor Petrov has a very elegant proof of this. See here.

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    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 1 at 4:32
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    $\begingroup$ "the Hermitian matrix" - what Hermitian matrix? And why is there a $j$ in your formula as a lower index? What is it doing there? Please edit your post and clarify it, otherwise it is at risk of being deleted. $\endgroup$
    – Alex M.
    Jun 1 at 10:39
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    $\begingroup$ $(1)$, $(2)$ and $(4)$ have just proved by Prof. Xuejun Guo and his three PhDstudents, $\endgroup$ Jun 7 at 5:17
  • $\begingroup$ Thanks for the beautiful proofs of such problems by Prof. Guo and Prof. Sun. My strategy implies a conjecture about the integer eigenvalues of a related class of matrices. For more details, please see: mathoverflow.net/questions/424138/… $\endgroup$
    – KLiu
    Jun 7 at 6:22

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