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I know that for a given $\sigma \in \mathfrak{S}_n$, we have $$ \displaystyle \varepsilon (\sigma )=\prod _{1\leq i<j\leq n}{\frac {\sigma (j)-\sigma (i)}{j-i}}=\prod _{\{i,j\}\in {\mathcal {P}}}{\frac {\sigma (j)-\sigma (i)}{j-i}} $$ where $$\displaystyle {\mathcal {P}}=\{\{i,\,j\}\mid 1\leq i\leq n{\mbox{ et }}1\leq j\leq n{\mbox{ et }}i\neq {j}\}$$ but I don't understand two linked things :

-> How do we move from $1 \leq i <j \leq n$ to $\mathcal {P}$ ? If I understood correctly $\left(2,1\right) \in \mathcal{P}$ but $\left(2,1\right) \notin \left\{1 \leq i < j \leq n\right\}$

-> Then we also have $$ \displaystyle \prod _{\{i,j\}\in {\mathcal {P}}}\left|\sigma (j)-\sigma (i)\right|=\prod _{\{k,l\}\in {\mathcal {P}}}\left|k-l\right| $$ " because $\sigma$ is a bijection", this is not obvious to me that this equality is true. Ok $\sigma$ is a bijection from $\left\{1, 2, \dots, n\right\}$ to $\left\{1, 2, \dots, n\right\}$ but how is it that it is the same value ?

Even with $n=3$ I have difficulties writing things correctly. Can someone give me intermediate steps or explanations please ?

Atmos
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  • The definition of $\mathcal{P}$ is simply wrong. Maybe there is a typo in the textbook where you took the definition of $\mathcal{P}$ from? With this definition of $\mathcal{P}$, $\varepsilon(\sigma)$ would always turn out to be $1.$ – Reinhard Meier Oct 23 '23 at 17:24
  • It stems from the associated French wiki signature d'une permutation : https://fr.m.wikipedia.org/wiki/Signature_d%27une_permutation I even took the latex code from the php page and forgot to change the "et" instead of and... – Atmos Oct 23 '23 at 21:13
  • Apparently, there is a mistake on the French wiki. Such things happen. – Reinhard Meier Oct 24 '23 at 09:03

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In the second one, the terms in the product are the same, just in a different "order" (the products aren't ordered, but you get the idea). Maybe start by showing (or at least understanding) that $\{(\sigma(j),\sigma(i)) : i,j \in \mathcal{P}\} = \mathcal{P}$.

Mark S Davis
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