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

1. Mar 18, 2012

### andresB

Acording to the non-Abelian stokes thoerem

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

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

P e$\oint$$\tilde{A}$= 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$\oint$$\tilde{A}$=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