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Is it true that the field strength $F_{\mu\nu}$ in a non-Abelian gauge theory with gauge group $G$ vanishes if, and only if, the gauge field $A_{\mu}$ is a pure gauge?

I can show one implication.

If $A_{\mu}=\frac{i}{g}U\partial_{\mu}U^{\dagger}$ where $U \in G$, then the field strength vanishes, but I am struggling with the other implication.

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up vote 6 down vote accepted

I) Vanishing field-strength $F=0$ does not imply that the gauge potential $A$ is pure gauge. It only holds locally. There could be global obstructions. In fact, topological obstructions could happen even if the gauge group $G$ is Abelian.

II) Let us sketched the proof of the local statement in a sufficiently small neighborhood $\Omega\subseteq M$ of a point $x_{0}\in M$.

  1. For a point $x\in \Omega$ choose a path/curve $C$ from $x_0$ to $x$.

  2. Define group element via a Wilson line $$\tag {1} U(x)~:=~P e^{\int_{C} \!A},$$ where $P$ denotes path ordering.

  3. Next use the non-Abelian Stokes' theorem to argue that this definition (1) does not depend on the curve $C$, because $F=0$.

  4. Finally, use the group-valued section (1) to gauge transform the gauge potential $A$ to be zero.

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Right, but it doesn't quite answer the question... Let's say I want to prove that locally it holds. Showing that $A_{\mu}$ pure gauge implies $F_{\mu\nu}=0$ is straightforward, but how does one work the viceversa? Also, I wonder if this fact is true. I can't find any reference actually showing the computation. – Federico Carta Feb 9 '14 at 17:43
I updated the answer with section II. – Qmechanic Feb 9 '14 at 18:27
Thanks, it is almost all clear to me. Just let me ask a small clarification, please. Is it correct to say that the the underlying idea of your proof is that if $F=0$ then it can be built a group element $U(x)$ appositely created to show that using that in the gauge transformation will lead to a vanishing $A$? If I understood right this makes sense, for it is a proof that in the $F=0$ case $A$ must be null, or gauge equivalent to $0$, i.e. in pure gauge. – Federico Carta Feb 9 '14 at 18:35
@Federico Carta: Right, it seems you got it. – Qmechanic Feb 9 '14 at 18:38

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