For a short <Wilson line>, a <gauge transformation> gives
$$
U'(x+an,x)=V(x+an)U(x+an,x)V^\dagger(x).
$$
Use $V(x+an)=1+i\alpha(x)+ia n^\mu\partial_\mu\alpha(x)+O(a^2,\alpha^2)$ and $U=1+iga n^\mu A_\mu+O(a^2)$. Keeping terms of order $a$, including $a\alpha$, gives
$$
U'=1+iga n^\mu
\left(A_\mu+\frac1g\partial_\mu\alpha+i[\alpha,A_\mu]\right)+O(a^2,\alpha^2).
$$
Since the <Lie bracket> is encoded by the <Lie algebra structure constants>, $i[\alpha,A_\mu]^a=f^{abc}A_\mu^b\alpha^c$, and therefore
$$
\boxed{(A^\alpha)_\mu^a=A_\mu^a
+\frac1g\partial_\mu\alpha^a
+f^{abc}A_\mu^b\alpha^c.}
$$
This is the infinitesimal form of the <Wilson-line gauge transformation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 304 / 4 / a / Solution

{c}

Expanding a short <Wilson line> determines the inhomogeneous gauge-potential transformation. For $V=1+i\alpha$ and Hermitian generators,
$$
A_\mu\mapsto A_\mu+\frac1g\partial_\mu\alpha+i[\alpha,A_\mu].
$$

Wilson-line gauge transformation

{c}
{wiki}

A Wilson line is the path-ordered parallel transporter
$$
U(y,x)=\mathcal P\exp\left(ig\int_x^y A_\mu(z)\,dz^\mu\right).
$$
Under a gauge transformation $V$, it transforms as $U(y,x)\mapsto V(y)U(y,x)V(x)^{-1}$.


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