8
$\begingroup$

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.

$\endgroup$
11
$\begingroup$

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.

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.