Simple permutations proof

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

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

Also, define [tex]{\mathop{\rm sgn}} : S_N \to \left\{ {\pm 1} \right\} [/tex] as
[tex]{\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.[/tex]

How do I prove that, for [tex]\sigma ,\pi \in S_N [/tex],
[tex]{\mathop{\rm sgn}} \left( {\sigma \circ \pi } \right) = {\mathop{\rm sgn}} \left( \sigma \right){\mathop{\rm sgn}} \left( \pi \right) \; ?[/tex]
 
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Answers and Replies

  • #2
MathematicalPhysicist
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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.
 

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