Acording to the non-Abelian stokes thoerem(adsbygoogle = window.adsbygoogle || []).push({});

http://arxiv.org/abs/math-ph/0012035

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

P e^{[itex]\oint[/itex][itex]\tilde{A}[/itex]}=Pe^{∫F}

whereFis some twisted curvature;F=U^{-1}FU, 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 (*)

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

Can you offer guidance or do you also need help?

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