Def/Sign of a permutation

{{Definition
 * Title=Sign of a permutation
 * Basic Information=Suppose that $X$ is a finite set, and $\sigma$ is a permutation of $X$. Then, the defines::sign of the permutation $\sigma$, written $sgn(\sigma)$ is either $1$ or $-1$, and can be defined by any of the following three methods:

The sign via derangements
One may define $sgn(\sigma)$ by the following process: $$T = \{ p \in P(X), \mbox{ such that } \vert p \vert = 2 \}.$$ $$Der_w(\sigma) = \{ t \in T, \mbox{ such that } (x,y \in t \wedge x < y) \Rightarrow \sigma(y) > \sigma(x) \}.$$ In other words, $Der_w(\sigma)$ consists of all pairs of distinct elements of $X$, whose order gets "switched" by $\sigma$.
 * Let $n = \vert X \vert$, and let $w \colon X \rightarrow \{ 0, \ldots, n-1 \}$ be a bijective function (since finite sets can be counted). Using $w$, we may define a total order on $X$, by defining: $x \leq y$ if and only if $w(x) \leq w(y)$.
 * Let $T$ be the set of two-element subsets of $X$:
 * Let $Der_w(\sigma)$ be the set of derangements (also called inversions) of $\sigma$. This is defined by:

The sign of $\sigma$ is now defined to be: $$sgn(\sigma) = (-1)^{\vert Der_w(\sigma) \vert }.$$

Observe that, since the order on $X$ depends upon the function $w$, and so $Der_w(\sigma)$ depends upon $w$, it is not obvious that $sgn(\sigma)$ does not depend upon $w$. This is the primary disadvantage of using derangements to define the sign, although combinatorial techniques can be used to demonstrate that the sign does not depend upon $w$.

The sign via transpositions
If $\sigma$ is a permutation of $X$, then it is possible to decompose $\sigma$ as a product of transpositions. Namely, there exists a natural number $k$, and a finite sequence $\tau_1, \ldots, \tau_k$ of transpositions of $X$, such that: $$\sigma = \prod_{i=1}^k \tau_i = \tau_1 \circ \tau_2 \circ \cdots \circ \tau_k.$$

The sign of $\sigma$ is now defined to be: $$sgn(\sigma) = (-1)^k.$$

As with the definition via derangements, the definition via transpositions does not obviously lead to a definition of $sgn(\sigma)$ that depends only upon $\sigma$. Namely, the process outlined above depends upon the choice of decomposition of $\sigma$ into transpositions, a choice for which there are many options.

The sign via cycles
Finally, if $\sigma$ is a permutation of $X$, then $\sigma$ determines a directed graph, the cycles of $\sigma$. This directed graph has the following properties:
 * The vertices are the elements of $X$.
 * For every $x \in X$, there is an edge from $x$ to $\sigma(x)$.
 * There is a unique partition $C$ of $X$, such that each element $c \in C$ is a cycle.

Let $n = \vert X \vert$. The sign of $\sigma$ can then be defined by: $$sgn(\sigma) = (-1)^{n - \vert C \vert}.$$

Observe that, with this definition, the sign depends only upon $\sigma$, and not upon further choices. Furthermore, this definition is quite suitable for practical compution. We follow cites::Bibliography:Nelson1987 in using this as our definition of the sign of a permutation, and we prove that it coincides with the two more common definitions given above.

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=Multiplicativity=

If $\sigma$ and $\tau$ are permutations of a finite set $X$, then: $$sgn(\sigma \circ \tau) = sgn(\sigma) \cdot sgn(\tau).$$

=Transpositions and Cycles=

=Derangements and Transpositions=