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

• andresB
In summary, the non-Abelian Stokes theorem states that for a system where every element of the curvature 2-form F commutes with each other, the path-ordered exponential P e\oint\tilde{A} can be expressed as a surface ordered exponential e∫F. This can be proven by showing that the two expressions are equivalent, and it holds true as long as the connection and curvature satisfy the conditions of smoothness and closure.
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:
is true?The answer to your question is yes, your supposition is right. You can prove this fact by showing that the non-Abelian Stokes theorem holds for the system you are considering. This theorem states that given a smooth connection A and a corresponding curvature F, one can express the path-ordered exponential P e\oint\tilde{A} as a surface ordered exponential e∫F. In other words, the theorem states that the two expressions are equivalent. To prove this, one can begin by introducing a coordinate system and expressing the connection and curvature in terms of the coordinates. Then, one can use the definition of the exponential and the properties of the Lie derivative to calculate the path-ordered exponential. Finally, once the expression is obtained, it can be shown that it is equivalent to the surface ordered exponential. In order for the non-Abelian Stokes theorem to hold, the connection A and the curvature F must satisfy certain conditions. Firstly, the connection must be smooth, meaning that its components must be continuous and differentiable. Secondly, the curvature must be closed, meaning that the components of the curvature must satisfy the Bianchi identities. These conditions ensure that the path-ordered exponential can be expressed as a surface ordered exponential.

## 1. When is it not necessary to do surface ordering in the non-Abelian Stokes theorem?

In general, surface ordering is not necessary in the non-Abelian Stokes theorem when the surface in question is a simple closed curve, or if the path integral being calculated is independent of surface ordering. In both cases, the surface ordering does not affect the final result.

## 2. What is the purpose of surface ordering in the non-Abelian Stokes theorem?

Surface ordering is used to ensure that the path integral in the non-Abelian Stokes theorem is well-defined and independent of the choice of surface. It also allows for the calculation of the holonomy of a gauge field along a path.

## 3. How does surface ordering affect the calculation of the non-Abelian Stokes theorem?

Surface ordering can affect the calculation of the non-Abelian Stokes theorem by changing the value of the path integral or the holonomy of a gauge field along a path. This is why it is important to carefully consider and properly apply surface ordering in the theorem.

## 4. Are there any limitations to surface ordering in the non-Abelian Stokes theorem?

One limitation of surface ordering in the non-Abelian Stokes theorem is that it can only be applied to surfaces that are homotopically equivalent to a simple closed curve. It also does not account for possible topological obstructions in the path integral calculation.

## 5. Can surface ordering be extended to higher dimensions in the non-Abelian Stokes theorem?

Surface ordering can be extended to higher dimensions in the non-Abelian Stokes theorem by using higher-dimensional surfaces, such as surfaces with boundaries or higher-dimensional objects like manifolds. However, this can become more complicated and may require additional mathematical tools and techniques.

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