# Sign of a Particular Permutation

Let $$n,k\in\mathbb{N}$$ be such that $$k. Define the permutation $$\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 $$\sigma$$.

• The sign of $\sigma$ is $\left(-1\right)^{\ell\left(\sigma\right)}$, where $\ell\left(\sigma\right)$ is the length (= number of inversions) of $\sigma$. But the inversions of $\sigma$ are easy to describe: They are all the pairs $\left(i, j\right)$ with $i \leq n-k$ and $j > n-k$. Thus a very simple answer. – darij grinberg Aug 14 at 21:58

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

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.