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Simple permutations proof

  1. Mar 16, 2007 #1
    [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.

    [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]
    Last edited: Mar 16, 2007
  2. jcsd
  3. Mar 17, 2007 #2


    User Avatar
    Gold Member

    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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