How Does the Sign of a Permutation Product Relate to Its Components?

[bomba923]

Consider
$$S_N = \left\{ {\left. {\sigma :\left\{ {1, \ldots ,N} \right\} \to \left\{ {1, \ldots ,N} \right\}} \right|\sigma {\text{ is a bijection}}} \right\}$$
i.e., the set of all permutations on 'N' values.

Define
$$\Delta \left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_i - x_j } \right)}$$
and, for $$\sigma \in S_N$$,
$$\sigma \left( \Delta \right)\left( {x_1 , \ldots ,x_N } \right) = \prod\limits_{i < j} {\left( {x_{\sigma \left( i \right)} - x_{\sigma \left( j \right)} } \right)}$$

Also, define $${\mathop{\rm sgn}} : S_N \to \left\{ {\pm 1} \right\}$$ as
$${\mathop{\rm sgn}} \left( \sigma \right) = \left\{ \begin{array}{l} 1,\;\sigma \left( \Delta \right) = \Delta \\ - 1,\;\sigma \left( \Delta \right) = - \Delta \\ \end{array} \right.$$

How do I prove that, for $$\sigma ,\pi \in S_N$$,
$${\mathop{\rm sgn}} \left( {\sigma \circ \pi } \right) = {\mathop{\rm sgn}} \left( \sigma \right){\mathop{\rm sgn}} \left( \pi \right) \; ?$$

 

[MathematicalPhysicist]

sgn(sigma o pi)=1
then sigma(pi(delta))=delta
if pi(delta)=delta then sigma(delta)=delta so both sgn are 1.
if pi(delta)=-delta sigma(-delta)=delta, which is the same as sigma(delta)=-delta, you can see it by obserivng that sigma(delta)+sigma(-delta)=0.
the same goes when sgn (sigma(pi))=-1.

 

[tacman]

To prove this, we will use the fact that the sign of a permutation is determined by the number of inversions it has. An inversion in a permutation is defined as a pair of elements that are in the opposite order in the permutation as they are in the original sequence.

First, let's consider the permutation \sigma \circ \pi. This permutation is obtained by first applying \pi and then applying \sigma. So, if we have a pair of elements x_i and x_j such that x_i < x_j in the original sequence, then after applying \pi, we will have x_{\pi(i)} < x_{\pi(j)}. Similarly, after applying \sigma, we will have x_{\sigma(\pi(i))} < x_{\sigma(\pi(j))}. This shows that the number of inversions in \sigma \circ \pi is the same as the number of inversions in \sigma and in \pi separately.

Next, we will show that the sign of a permutation is determined by the number of inversions it has. Recall that if a permutation has an even number of inversions, its sign is 1, and if it has an odd number of inversions, its sign is -1.

Now, let's consider \sigma. If \sigma has an even number of inversions, then applying \sigma does not change the number of inversions, and therefore, \sigma \left( \Delta \right) = \Delta. This means that {\mathop{\rm sgn}} \left( \sigma \right) = 1. Similarly, if \sigma has an odd number of inversions, then applying \sigma will change the number of inversions by 1, and therefore, \sigma \left( \Delta \right) = - \Delta. In this case, {\mathop{\rm sgn}} \left( \sigma \right) = -1.

Now, let's consider \pi. We can apply the same logic as above to show that {\mathop{\rm sgn}} \left( \pi \right) = 1 if \pi has an even number of inversions and {\mathop{\rm sgn}} \left( \pi \right) = -1 if \pi has an odd number of inversions.

Combining these results, we can see that if \sigma and \pi both have an even number