I was working on a problem when I found out I needed to know the signature of a permutation of the form :
\begin{equation} (\sigma_A,\sigma_B) \end{equation}
meaning that $(\sigma_A,\sigma_B)$ is defined from $A\times B$ where $A,B$ are finite ordered sets such that $(\sigma_A,\sigma_B)(a,b)=(\sigma_A(a),\sigma_B(b))$ where we consider the lexicographic order of $A\times B$. I intuitively find that the signature is :
\begin{equation} \varepsilon(\sigma_A)^{|B|} \varepsilon(\sigma_B)^{|A|} \end{equation}
Let me explain my reasoning. If I suppose $A=[|1,n|]$ and $B=[|1,m|]$, we get that the permutation couple is as follows :
\begin{equation}
\begin{pmatrix}
(1,1) & \cdots &(1,m) & (2,1) &\cdots & (2,m) & & \cdots& & (n,1) & \cdots & (n,m) \\
(\sigma_A(1),\sigma_B(1)) & \cdots &(\sigma_A(1),\sigma_B(m)) & (\sigma_A(2),\sigma_B(1)) &\cdots & (\sigma_A(2),\sigma_B(m)) & & \cdots& & (\sigma_A(n),\sigma_B(1)) & \cdots & (\sigma_A(n),\sigma_B(m))
\end{pmatrix}
\end{equation}
We can easily see then that :
\begin{equation} (\sigma_A,\sigma_B)=(\sigma_A,id_B)\circ (id_A,\sigma_B) \end{equation}
The first permutation $(\sigma_A,id_B)$ is :
\begin{equation}
\begin{pmatrix}
(1,1) & \cdots &(1,m) & (2,1) &\cdots & (2,m) & & \cdots& & (n,1) & \cdots & (n,m) \\
(\sigma_A(1),1) & \cdots &(\sigma_A(1),m) & (\sigma_A(2),1) &\cdots & (\sigma_A(2),m) & & \cdots& & (\sigma_A(n),1) & \cdots & (\sigma_A(n),m)
\end{pmatrix}
\end{equation}
This problem seems like the mini-problem :
\begin{equation}
\begin{pmatrix}
\text{Bloc 1} & \text{Bloc 2}\\
\text{Bloc 2} & \text{Bloc 1}
\end{pmatrix}
\end{equation}
where the two blocs are of the same size $b$ obviously, the signature of this permutation would be :
\begin{equation}
(-1)^b
\end{equation}
because we'd have to switch each element of the bloc with the other to return to the identity permutation, which is a total of $b$ inversions. Hence, I can easily extend it to a random permutation (justified by the factorisation theorem of permutations):
\begin{equation}
\begin{pmatrix}
\text{Bloc $1$} & \cdots & \text{Bloc $p$} \\
\text{Bloc $\sigma(1)$} & \cdots & \text{Bloc $\sigma(p)$}
\end{pmatrix}
\end{equation}
whose signature would be :
\begin{equation}
\varepsilon(\sigma)^b
\end{equation}
then $\varepsilon(\sigma_A,id_B)$ is : $\varepsilon(\sigma_A)^{|B|}$.
For the second permutation $(id_A,\sigma_B)$, it's a little bit easier :
\begin{equation}
\begin{pmatrix}
(1,1) & \cdots &(1,m) & (2,1) &\cdots & (2,m) & & \cdots& & (n,1) & \cdots & (n,m) \\
(1,\sigma_B(1)) & \cdots &(1,\sigma_B(m)) & (2,\sigma_B(1)) &\cdots & (2,\sigma_B(m)) & & \cdots& & (n,\sigma_B(1)) & \cdots & (n,\sigma_B(m))
\end{pmatrix}
\end{equation}
we can see that each block $(i,1) \cdots (i,m)$ is stable and has signature $\varepsilon(\sigma_B)$, then we'll have to multiply them all and get for the signature :
\begin{equation} \varepsilon(\sigma_B)^{|A|} \end{equation}
The first factorisation of $(\sigma_A,\sigma_B)$ shows the result.
The problem I find is that the formula isn't really nice for the usage I need. I'm working on figuring out the signature $\varepsilon_n(m)$ of the permutation :
\begin{equation}
\pi_n(m) :
\begin{cases}
\mathbb{Z}/n\mathbb{Z} \longrightarrow \mathbb{Z}/n\mathbb{Z} \\
x \longmapsto mx
\end{cases}
\end{equation}
where $n$ and $m$ are coprime. I'm thinking on focusing on $n$ being a prime number, and for that I'll need to decompose $\pi_{ab}$ using $\pi_a$ and $\pi_b$ for $a$ and $b$ coprime which can be done using the chinese factorisation theorem, where if $\varphi : \mathbb{Z}/a\mathbb{Z} \times \mathbb{Z}/b\mathbb{Z} \to \mathbb{Z}/ab\mathbb{Z}$ is the bijection $(x,y)\to z$ and $z$ is the unique solution mod $ab$ of the equation $(z,z)=(x \text{ mod $a$} , y \text{ mod $b$})$, we'll have for all $m$ coprime with $ab$ :
\begin{equation}
(\pi_a(m),\pi_b(m))=\varphi \circ \pi_{ab}(m) \circ \varphi^{-1}
\end{equation}
where the usage of the established result becomes clear. If $n=p_1^{\alpha_1} \cdots p_r^{\alpha_r}$, then I feel the expression of $\varepsilon_n(m)$ would be ugly using the formula, so I wonder if what I've done has been right?
2026-04-08 14:32:58.1775658778