In a groups course in my first year of university we defined the sign of a permutation by showing that every permutation can be written as a product of transpositions and defining the sign of the permutation as $(-1)^n$ with $n$ the number of transpositions in the product. Of course, one then has to show that $n$ always has the same parity, no matter how we write the permutation as a product of transpositions, so that this is well-defined. Once one has that it follows easily that sign is a homomorphism, but the proof that the parity is well-defined, though it wasn't too messy, seemed unmotivated/unilluminating/unnatural. It seemed to me there should be a "nicer" definition. I recently came across the definition related to inversions of a permutation, which was completely new to me, we hadn't been told anything about it in uni: place an arbitrary linear order on your finite set. An inversion is then a pair $\{i,j\}$ such that $i>j$ but $\sigma(i)<\sigma(j)$. The parity of a permutation is then the parity of the number of inversions of the permutation. Of course, one has again to show that this parity is well-defined, i.e. independent of the choice of linear order. And then one has to show that sign is a homomorphism, which isn't as obvious as it was with the first definition, though doable. But at least it's a different approach. What actually bothered me the most about this approach though was something else: why a linear order? Linear orders satisfy transitivity of $>$, and this seems to be unrelated/irrelevant to the parity of a permutation. So I spent some time thinking about this and generalised "linear order" to something I called a "choice". After some further thought I was eventually led to an alternative definition of the sign of a permutation along similar lines to that of the number of inversions. And because this definition seemed to be so related to the other, well-known one, I was convinced that it too had to be well-known. But multiple online searches revealed nothing similar to what I had thought of. So in the end I latexed a short, two and a half page detailed explanation. I personally think this definition puts it in a more elegant/natural way, but wanted to hold it up to criticism, so I decided to ask here . I also wanted to ask if what I write is already well-known, I find it hard to believe that it isn't, but again, couldn't find anything similar online. And if somehow it isn't, are there enough new/interesting ideas in it for it to be worth publishing on say arxiv? I'll copy the latex code for my explanation into here now so you can read it:
Let $X$ be a finite set with $|X|=n \geq 2$ and let Sym$(X) \cong S_n$ be the group of permutations of $X$ under composition. Let $X_2$ be the set of all subsets of $X$ of size 2. $|X_2|= \binom{n}{2}$. We define a \emph{choice} on $X$ to be a choice function on $X_2$, i.e. a selection, for each pair $\{i,j\} \in X_2$, of either $i$ or $j$. Note that a choice is a more general concept than a linear order - any linear order naturally induces a choice by selecting the larger of $i$ and $j$ within each pair $\{i,j\}$, but not all choices arise in this way, since a linear order must satisfy transitivity of $>$. Let $X_C$ be the set of all choices on $X$. $|X_C|= 2^{\binom {n}{2}} $. We define an equivalence relation $\sim$ on $X_C$ by $x \sim y$ iff $x$ and $y$ disagree on an even number of pairs in $X_2$. Reflexivity and symmetry are clear, we check transitivity. In general, let $x,y,z \in X_C$ and let $A\subseteq X_2$ and $B\subseteq X_2$ be the pairs on which $x$ and $y$ disagree and on which $y$ and $z$ disagree, respectively. Then $x$ and $z$ disagree on $(A \setminus B) \cup (B \setminus A) $. But \begin{equation} \label{} \begin{gathered} |A| + |B| = |A \setminus B| + |A \cap B| + |B \setminus A| + |A \cap B| \\ = |(A \setminus B) \cup (B \setminus A)| + 2|A \cap B| \equiv |(A \setminus B) \cup (B \setminus A)| \pmod{2} \end{gathered} \end{equation} \noindent (1) implies transitivity. We give $\sim$ a name: if $ x \sim y$ we say $x$ and $y$ have the same $\emph{orientation}$, otherwise we say $x$ and $y$ have opposite orientation. (1) also implies that if $x$ and $y$ have opposite orientation and $y$ and $z$ have opposite orientation, then $x$ and $z$ have the same orientation. We now show that $X_C$ has two equivalence classes. Pick a choice $x \in X_C$ and a pair $p \in X_2$. Let $y \in X_C$ be the choice that agrees with $x$ on all pairs except $p$. Then $x$ and $y$ have opposite orientation. Let $z \in X_C$. Either $x$ and $z$ have the same orientation or $x$ and $z$ have opposite orientation, in which case $y$ and $z$ have the same orientation. Hence $X_C$ has two equivalence classes, the one of $x$ and the one of $y$. Now let $\sigma \in$ Sym$(X)$. $\sigma$ naturally induces a permutation of $X_2$ by mapping $\{i,j\}$ to $\sigma\{i,j\} = \{\sigma(i),\sigma(j)\}$. $\sigma$ also naturally induces a permutation of $X_C$ as follows: map $x \in X_C$ to $\sigma x \in X_C$, where $\sigma x$ is the choice that selects $\sigma(i)$ from the pair $\{\sigma(i), \sigma(j)\}$, where $i$ is the element from $\{i,j\}$ selected by $x$. We abuse notation and use $\sigma$ to denote the permutation of $X$, $X_2$ and $X_C$. Now let $x,y \in X_C$ and let $A\subseteq X_2$ be the pairs on which $x$ and $y$ disagree. Then $\sigma x$ and $\sigma y$ disagree on $\sigma(A)$. Hence $x$ and $y$ have the same orientation iff $\sigma x$ and $\sigma y$ have the same orientation. In other words, $\sigma$ permutes the two equivalence classes of $X_C$. It follows that $\sigma$ is either \emph{orientation preserving}, meaning it maps every choice to a choice with the same orientation, or \emph{orientation reversing}, meaning it maps every choice to a choice with opposite orientation. It also follows that the composition of two orientation preserving permutations is orientation preserving, the composition of an orientation preserving and an orientation reversing permutation is orientation reversing, and the composition of two orientation reversing permutations is orientation preserving. In other words, the map $f: \ $Sym$(X) \to \{\pm1\}$, where $\{\pm1\}$ is the cyclic group of order two under multiplication, given by $f(\sigma)=1$ if $\sigma$ is orientation preserving and $f(\sigma)=-1$ if $\sigma$ is orientation reversing, is a group homomorphism. For $x \in X_C$ and $\sigma \in $ Sym$(X)$, we call a pair $\{i,j\}$ such that $x$ selects $i$ from $\{i,j\}$ but $\sigma(j)$ from $\{\sigma(i),\sigma(j)\}$ an \emph{inversion} of $\sigma$ with respect to $x$. Note that $x$ and $\sigma x$ disagree precisely on the pairs of the form $\sigma p$, with $p \in X_2$ an inversion of $\sigma$ with respect to $x$. The definition of orientation preserving and orientation reversing given before is independent of any particular $x \in X_C$, however, in light of the above, if we wish to determine whether $\sigma$ is orientation preserving or reversing, we can choose an arbitrary $x \in X_C$ and count the number of inversions of $\sigma$ with respect to $x$. $\sigma$ is orientation preserving if this number is even and orientation reversing if this number is odd. If the choice $x$ is induced by a linear order, this definition of inversion matches the usual definition, and we recover the usual criterion for determining the sign of a permutation based on the number of inversions as a special case of this criterion. For we now show that the orientation preserving permutations are precisely the even permutations and the orientation reversing permutations are precisely the odd permutations. We apply our criterion to a transposition $\sigma = (i \ j)$. We take a choice $x$ that selects an arbitrary element from pairs containing neither $i$ nor $j$, selects $i$ from a pair containing $i$ but not $j$, selects $j$ from a pair containing $j$ but not $i$, and selects (say) $i$ from the pair $\{i,j\}$. Then $\{i,j\}$ is the only inversion of $\sigma$ with respect to $x$, so $\sigma$ is orientation reversing. Together with the previous result that $f$ is a homomorphism, this shows that $f$ is the familiar signature homomorphism, so the orientation preserving permutations are the even permutations and the orientation reversing permutations are the odd permutations.
(sorry for this question being much longer than usual by the way, many thanks if you read all the way to the end)
