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If we consider a non-Abelian potential $A_{\mu}=A^{a}_{\mu}T^{a}$ of a general non-Abelian group $G$, satisfying the property $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}+[A_{\mu},A_{\nu}]=0$, is it possible to find a gauge transformation where the transformed gauge potential vanishes?

Please, I'm not asking if the condition $F_{\mu\nu}=0$ always implies the gauge potential $A_{\mu}$ to be a pure gauge, but if there exists at least one gauge transformation satisfying the conditions explained in the first paragraph.

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  • $\begingroup$ $F_{\mu\nu} = 0$ means that the the gauge potential is pure gauge, $A_\mu = - i C \partial_\mu C^{-1}$. So there definitely exists a gauge transformation you can do which sets $A_\mu = 0$. $\endgroup$
    – Prahar
    Oct 11, 2021 at 10:49
  • $\begingroup$ BTW, imposing the requirement that there exists a gauge transformation such that $A'_\mu = 0$ is the same thing as imposing the requirement that $A_\mu$ is pure gauge. $\endgroup$
    – Prahar
    Oct 11, 2021 at 10:50
  • $\begingroup$ It seems $F_{\mu\nu}=0$ doesn't imply $A_{\mu}$ is a pure gauge (physics.stackexchange.com/questions/98484/…). Then, is it always possible to vanish the gauge potential if it's not a pure gauge? $\endgroup$
    – Ruben29
    Oct 11, 2021 at 11:11
  • $\begingroup$ Well, there are definitely topological restrictions to this statement. Such issues are also there in the Abelian case, $dA = 0 \not \implies A = d \epsilon$. However, locally (due to the Poincar\'e lemma) this is always true. So at least in any given local patch, we can always write $A = - i C d C^{-1}$. To extend this statement globally, let $U_1$ and $U_2$ be two overlapping patches with $A_1 = A|_{U_1} = - i C_1 d C_1^{-1}$ and $A_2 = A|_{U_2} = - i C_2 d C_2^{-1}$. Then, in the overlapping patch there is a $g_{12}$ such that $C_1 = g_{12} C_2$. $\endgroup$
    – Prahar
    Oct 11, 2021 at 11:39
  • $\begingroup$ It follows that if the manifold is topologically non-trivial, then there does not generically exist a gauge transformation which takes $A = 0$ everywhere. However, it is definitely pure gauge everywhere. This is true as long as we are restricting ourselves to gauge transformations which are single-valued on $M$. If we allow topologically non-trivial gauge transformations, then it IS possible to set $A=0$ everywhere. $\endgroup$
    – Prahar
    Oct 11, 2021 at 11:39

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