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When is not necesary to do surface ordering in the non-Abelian Stokes theorem?

  1. Mar 18, 2012 #1
    Acording to the non-Abelian stokes thoerem


    I can transform a path ordered exponential to a surface ordered one.

    P e[itex]\oint[/itex][itex]\tilde{A}[/itex]= P eF

    where F is some twisted curvature;F=U-1FU, and U is a path dependet operator.

    So, I have a system where every element of the curvature 2-form F commute with each other, i just have the feeling that it is true that

    P e[itex]\oint[/itex][itex]\tilde{A}[/itex]=e∫F (*)

    where the RHS is just a ordinary surface exponential.

    So the questions are

    1)is my suposition right?
    2) if afirmative how cah i prove that fact?
    3) if negative, why? and what properties should have the conection A and/or the curvature F to ensure (*)
    Last edited: Mar 18, 2012
  2. jcsd
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