# 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.

TL;DR: 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.

More details:

1. The starting point is a connected (but not necessarily simply connected) gauge Lie group $$G$$ and a globally defined gauge potential $$A$$ on a connected (but not necessarily simply connected) spacetime manifold $$M$$. In this answer the covariant derivative is by convention $$\mathrm{D}=\mathrm{d}-A$$, i.e. $$A$$ is typically an anti-Hermitian matrix-valued 1-form. A gauge transformation takes the form $$A^{\prime}~=~-U(\mathrm{D} U^{-1}), \qquad U~\in~G.\tag{1}$$

2. Let us warm up by reviewing the easy way. If $$A^{\prime}$$ is pure gauge $$A^{\prime}=-U(\mathrm{d} U^{-1})$$, then there exists a gauge transformation (1) such that the new gauge potential $$A=0$$ vanishes identically, and hence the (new and old) field strengths $$F^{\prime}=UFU^{-1}=0$$ vanish identically.

3. Next let us return to OP's question and sketched the proof of the opposite implication in a simply connected region $$\Omega\subseteq M$$ containing a fiducial point $$x_{0}\in M$$:

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

• Define group element via a Wilson line$$^1$$ $$U(x,x_0)~:=~P e^{\int_{C} \!A},\tag{2}$$ where $$P$$ denotes path ordering.

• Next use the non-Abelian Stokes' theorem to argue that this definition (2) does not depend on the curve $$C$$, because $$F=0$$.

• Finally, use the group-valued section (2) to gauge transform the gauge potential $$A$$ to be zero.

4. Example: Consider the punctured plane $$M=\mathbb{R}^2\backslash\{(0,0)\}$$ with coordinates \begin{align}x~=~&r\cos\theta, \qquad y~=~r\sin\theta, \cr \theta~\sim~&\theta+2\pi,\qquad r~>~0;\end{align}\tag{3} and with Abelian gauge group $$G=U(1)$$.
Let the (imaginary valued) gauge potential 1-form be $$-iA~=~\frac{x\mathrm{d}y-y\mathrm{d}x}{x^2+y^2}~=~\mathrm{d}\theta~=~-iU^{-1}\mathrm{d}U,\tag{4}$$ where $$U(x,y)~=~e^{i\theta(x,y)}~\in~G\tag{5}$$ is a globally well-defined group-valued section. The field strength $$F$$ vanishes, so the gauge potential (4) is pure gauge. However, if we scale $$A\to \lambda A$$ in eq. (4) with a non-integer constant $$\lambda\in\mathbb{R}\backslash\mathbb{Z}$$, then $$A$$ will no longer be pure gauge (because the corresponding $$U=e^{i\lambda\theta}$$ becomes multivalued), but $$F$$ will still be zero.

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$$^1$$ If $$G$$ is not simply connected then work in the universal covering group $$\tilde{G}$$. We can always later project down to $$G$$.

• 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. Feb 9, 2014 at 18:35
• @Federico Carta: Right, it seems you got it. Feb 9, 2014 at 18:38
• In what sense could a topological obstruction make the claim $F=0$ $\rightarrow$$A= pure gauge invalid? If I had to guess, I would say you have instantons in mind. However, there one starts with a F=0 configuration at t= - \infty and ends with a F=0 configuration at t= -\infty. Both correspond to A= pure gauge. In between however, something happens (the instanton), such that F \neq 0 and thus, of course, A \neq pure gauge. – jak Dec 7, 2017 at 8:53 • While the instanton corresponds to a non-pure-gauge configuration + F \neq 0 and connects inequivalent F=0 configurations, I don't see why it makes the claim F=0 \rightarrow$$A=$ pure gauge invalid...
– jak
Dec 7, 2017 at 8:54
• @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? Jul 28, 2020 at 1:44