Covariant derivative of a Wilson line

Question

Does the covariant derivative of a Wilson line given by $$W[A; z_0, z] = {\cal P}e^{-i\int^z_{z_0} dz ~A^af_{abc}}$$ vanish, i.e. $$D_zW[A; z_0, z] = 0~?$$

Answer 1

You would need $F=0$ so the Wilson line is path independent and only a function of the endpoint $z$. In that case $\nabla_X U=0$ where $U(z)= P\exp\{i \int^z A_\mu(x) dx^\mu\}\in G$

Answer 2

1. Let us write the Wilson-line of a simple open curve $\gamma: [s_i,s_f]\to \mathbb{R}^4$ as $$ U(s_f,s_i) ~=~ \mathcal{P}\exp \left[ i\int_{\gamma} A_{\mu}~ dx^{\mu} \right].\tag{1} $$

2. The path-ordering $\mathcal{P}$ becomes important if the gauge potential $$A_{\mu}~=~A^a_{\mu} T_a\tag{2}$$ is non-abelian. Here $T_a$ are the generators of the corresponding Lie algebra.

3. The Wilson-line has groupoid properties, e.g., $$U(s_3,s_2)U(s_2,s_1)~=~ U(s_3,s_1), \qquad U(s,s) ~=~ {\bf 1}.\tag{3}$$

4. If one differentiates wrt. the final point $s_f$, one gets $$\frac {dU(s_f,s_i)}{ds_f} ~=~ i\dot{\gamma}^{\mu}(s_f)~A_{\mu}(\gamma(s_f)) ~U(s_f,s_i). \tag{4}$$

5. Now let us return to OP's question. Eq. (4) can be written as $$ D_{\dot{\gamma}(s_f)}U(s_f,s_i)~=~0,\tag{5}$$ if we introduce the gauge covariant derivative as $$ D_{\mu}~=~\partial_{\mu}-iA_{\mu}. \tag{6}$$

6. For more information, see e.g. this & this related Phys.SE posts.