Let $F=dA+A \wedge A$ be the field strength that solves vaccum Yang-Mills equation.

The question is: how to recover the gauge potential $A$? Is there any standard way? or any theorem stating the solvability? Suppose the metric is $g=g_{\mu \nu}dx^{\mu}dx^{\nu}=\eta_{ab}e^ae^b$, $e^a$ is tetrad basis.


3 Answers 3


Even if you only consider the gauge potentials up to gauge equivalence, you still can't recover a potential from a field in the non-abelian case. Not even locally! This is called the "field copy problem" in the literature, and goes back to Wu and Yang.

There are some special cases where the potential can be recovered from the field, e.g. if the holonomy group is as big as possible on every open subset.


You can't recover the gauge potential uniquely without specifying more information, because of gauge invariance. $A$ is not uniquely defined by $F$. For any gauge transformation $g: \Sigma \to G$, the transformed connection $A^g = g^{-1}A g + g^{-1}dg$ has curvature $F_{A^g} = g^{-1}F_A g$.

It's physically a bit weird to specify $F_A$ in non-Abelian gauge theories. Gauge transforms don't leave $F_A$ invariant, so $F_A$ isn't an observable. But if one insists that $F_{A^g} = F_A$, then all connections $A^g$ with $g$ in the centralizer of $F_A$ give the same curvature.


To supplement the answers that were posted earlier, here's an explicit example from ref 1. This example proves that $A$ is not always uniquely determined by $F$, not even up to gauge transformations, not even locally.

Work in four-dimensional spacetime with coordinates $(w,x,y,z)$, and take the structure group to be $SU(2)$. Let $T_1$, $T_2$, $T_3$ be the generators of $SU(2)$, and remember that the only element of $SU(2)$ that commutes with all of the $T_k$ is the identity element. Consider the gauge potential $$ A = T_1 w\,dx + T_2 y\,dz + \alpha T_1\,dz $$ where $\alpha$ is an arbitrary nonzero real number. The field strength is $F=dA+A\wedge A$ with \begin{align} dA &= T_1 dw\wedge dx + T_2 dy\wedge dz \\ A\wedge A &= [T_1,T_2]w y\,dx\wedge dz \\ &\propto T_3 w y\,dx\wedge dz. \end{align} Observe:

  • $F$ is independent of $\alpha$, so we have a family of gauge different potentials that all give the same $F$.

  • The three two-forms appearing in these expressions for $dA$ and $A\wedge A$ are linearly independent, and their coefficients are the three different generators of $SU(2)$.

Now, consider whether or not these potentials can be gauge-equivalent to each other (equal to each other up to gauge transformations). When we apply a gauge transform $g$ to $A$, its effect on $F$ is $F\to g^{-1} Fg$. But we already know that all of these $A$s give the same $F$, so we must have $g^{-1} F g=F$. Thanks to the observations highlighted above, this implies $g^{-1} T_k g=T_k$ for all $k$, which in turn implies $g=1$. Therefore, the potentials $A$ with different values of the coefficient $\alpha$ cannot be gauge-equivalent to each other, even though they all have the same field strength $F$.


  1. Mostow and Shnider (1983), "Counterexamples to Some Results on the Existence of Field Copies" (https://projecteuclid.org/euclid.cmp/1103940415)

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