Sign of a Particular Permutation

Question

Let $n,k\in\mathbb{N}$ be such that $k<n$. Define the permutation $\sigma\colon\{1,\ldots,n+1\}\to\{1,\ldots,n+1\}$ as \begin{align*} \sigma(j)=&~\begin{cases} j+k+1, & j\le n-k \\ j+k-n, & j>n-k \end{cases} \end{align*} What is the sign of $\sigma$?

This question arose since I am trying to find necessary and sufficient conditions on $n,k$ to guarantee $\det A=1$, where $A$ is the square matrix of size $n+1$ defined as \begin{align*} A=&~\begin{pmatrix}0 & I_{n-k} \\ I_{k+1} & 0\end{pmatrix} \end{align*} and $I_{n-k},I_{k+1}$ are identity matrices of size $n-k$ and $k+1$, respectively. By considering row swaps, this problem reduces to finding the sign of $\sigma$.

Answer 1

A matrix of the form

$$ {\displaystyle C={\begin{bmatrix}c_{0}&c_{n-1}&\dots &c_{2}&c_{1}\\c_{1}&c_{0}&c_{n-1}&&c_{2}\\\vdots &c_{1}&c_{0}&\ddots &\vdots \\c_{n-2}&&\ddots &\ddots &c_{n-1}\\c_{n-1}&c_{n-2}&\dots &c_{1}&c_{0}\\\end{bmatrix}}.} $$

is called circulant. Obviously, $A=\begin{pmatrix}0 & I_{n-k} \\ I_{k+1} & 0\end{pmatrix}$ is circulant with $c_j =\begin{cases}1:& j=n-k\\0:&\text{else}\end{cases}$

It satisfies the determinant formula $$ {\displaystyle \det(C)=\prod _{j=0}^{n-1}(c_{0}+c_{1}\omega _{j}+c_{2}\omega _{j}^{2}+\dots +c_{n-1}\omega _{j}^{n-1})=\prod _{j=0}^{n-1}f(\omega _{j}).}$$

where ${\displaystyle f(x)=c_{0}+c_{1}x+\dots +c_{n-1}x^{n-1}}$ and ${\displaystyle \omega _{j}=\exp \left(i{\tfrac {2\pi j}{n}}\right)}$ are the n-th roots of unity. Hence (mind the index shift $n\to n+1$):

$$\det(A) = \prod _{j=0}^{n} \exp \left(i{\tfrac {2\pi j}{n+1}}\right)^{n-k} = \exp\Big(2\pi i\frac{n-k}{n+1}\sum_{j=0}^{n}j \Big) =\exp\Big(2\pi i\frac{n-k}{n+1}\frac{n(n+1)}{2} \Big)$$

hence $\det(A) = \exp\Big(i\pi n(n-k)\Big) = \begin{cases}+1:&n(n-k) \text{ even} \\-1:&n(n-k)\text{ odd}\end{cases} = \begin{cases}+1:&\text{else} \\-1:&n\text{ odd and } k \text{ even}\end{cases} $

Answer 2

If $k+1$ and $n+1$ are co-prime, this is a cycle of length $n+1$, meaning it depends on $n+1\mod 2$

(this is a composition of $n$ cycles of length $1$, of the form: $$(1,k+2,2k+3,...,2-k)$$ (taking all the elements modulo n+1 and adjusting wherever needed..))

So you get an even permutation if $n+1$ is odd.

Otherwise, take $d:=\gcd (k+1,n+1)$, and then you get that this is a composition of $d$ disjoint cycles, each of length $\frac {n+1} d$, meaning it depends on $d\mod 2$ and $\frac {n+1} d\mod2$, where if $d$ is even, we have an even signature anyway, and if $\frac {n+1} d$ is odd we have we have that each cycle is even thus the composition is even.