# Field strength vanishes iff $A_{\mu}$ is pure gauge

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.

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.

• 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
• In what sense could a topological obstruction make the claim $F=0$ $\rightarrow$$A= pure gauge invalid? If I had to guess, I would say you have instantons in mind. However, there one starts with a F=0 configuration at t= - \infty and ends with a F=0 configuration at t= -\infty. Both correspond to A= pure gauge. In between however, something happens (the instanton), such that F \neq 0 and thus, of course, A \neq pure gauge. – jak Dec 7 '17 at 8:53 • While the instanton corresponds to a non-pure-gauge configuration + F \neq 0 and connects inequivalent F=0 configurations, I don't see why it makes the claim F=0 \rightarrow$$A=$ pure gauge invalid... – jak Dec 7 '17 at 8:54