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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Federico Carta Feb 9 '14 at 18:35
  • $\begingroup$ @Federico Carta: Right, it seems you got it. $\endgroup$ – Qmechanic Feb 9 '14 at 18:38
  • $\begingroup$ In what sense could a topological obstruction make the claim $F=0$ $\rightarrow$$A=$ pure gauge invalid? If I had to guess, I would say you have instantons in mind. However, there one starts with a $F=0$ configuration at $t= - \infty$ and ends with a $F=0$ configuration at $t= -\infty$. Both correspond to $A=$ pure gauge. In between however, something happens (the instanton), such that $F \neq 0$ and thus, of course, $A \neq$ pure gauge. $\endgroup$ – jak Dec 7 '17 at 8:53
  • $\begingroup$ While the instanton corresponds to a non-pure-gauge configuration + $F \neq 0$ and connects inequivalent $F=0$ configurations, I don't see why it makes the claim $F=0$ $\rightarrow$$A=$ pure gauge invalid... $\endgroup$ – jak Dec 7 '17 at 8:54
  • $\begingroup$ @Qmechanic Could you please explain what the global obstructions are for both Abelian and non-Abelian theories? Do you mean that if the manifold on which the theory is defined has nontrivial first cohomology group or you have something else in mind? $\endgroup$ – QGravity Jul 28 '20 at 1:44

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