Expand an infinitesimal Wilson loop I have a question about expanding an infinitesimal Wilson loop operator to get the field tensor $F_{\mu \nu}$ in chapter 3 of Fradkin's notes Classical Symmetries and Conservation Laws. For a infinitesimal Wilson loop
$$\widehat{W}_{\Gamma} = \widehat{P} \exp[ig \oint_{\Gamma} dz^\mu A_\mu (z)],\tag{3.108}$$
we have
$$
\widehat{W}_{\Gamma} \approx I+i g \widehat{P} \oint_{\Gamma} d z^{\mu} A_{\mu}(z)+\frac{(i g)^{2}}{2 !} \widehat{P}\left(\oint_{\Gamma} d z^{\mu} A_{\mu}(z)\right)^{2}+\cdots\tag{3.112}
$$
The term $\oint_{\Gamma} dz^{\mu} A_{\mu}$ can be written as
$$
\oint_{\Gamma} d z_{\mu} A^{\mu}(z)=\iint_{\Sigma} d x^{\mu} \wedge d x^{\nu} \frac{1}{2}\left(\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}\right)\tag{3.113}
$$
by the Stokes’ theorem.
However, I don't understand the following result:
$$
\frac{1}{2 !} \widehat{P}\left(\oint_{\Gamma} d z^{\mu} A_{\mu}(z)\right)^{2} \equiv \frac{1}{2} \iint_{\Sigma} d x^{\mu} \wedge d x^{\nu}\left(-\left[A_{\mu}, A_{\nu}\right]\right)+\cdots \tag{3.114}$$
How can we convert the path-ordered product to an integral over 2-form?
 A: For what it's worth, here is a physicist's heuristic derivation of eq. (3.114). We may w.l.o.g. pick a coordinate system so that the infinitesimal loop is parallel to the $xy$-plane.
  y
  ^
  |
  |    
  |----<----
  |    3    |
  |         |
b v 4       ^ 2
  |         |
  |    1    |
  ----->----------------> x
       a

$\uparrow$ Fig. 1. A counterclockwise infinitesimal rectangular loop in the $xy$-plane.
The circulation of the 1-form $A$ on the closed path $\Gamma$ is approximated by$^1$
$$ \oint_{\Gamma}A~=~a A_x(1) + bA_y(2) - a A_x(3) - bA_y(4), $$
cf. Fig. 1. The time-ordered$^2$ second power is approximated by
$$\begin{align} 
T\left(\oint_{\Gamma}A\right)^2~=~&16 \text{ terms}\cr
~=~& \underbrace{\oint_{\Gamma}A}_{=0}a A_x(1)~-~b A_y(4)\underbrace{\oint_{\Gamma}A}_{=0} \cr \cr
&~+~  bA_y(2)\{aA_x(1)~+~  bA_y(2)\}\cr 
&~-~ \{a A_x(3)~+~bA_y(4)\} bA_y(2) \cr
&~-~  a A_x(3)\{aA_x(1)~+~  bA_y(2)\}\cr 
&~+~ \{a A_x(3)~+~bA_y(4)\} a A_x(3) \cr
~=~&0 ~+~  bA_y(2)~aA_x(1) ~-~ a A_x(3)~bA_y(2) \cr
& ~-~ aA_x(3)~bA_y(2) ~+~  bA_y(4)~a A_x(3) ~+~ 0 \cr
~=~& 2ab [A_y,A_x]\cr
~=~&2\iint_{\Sigma} \!\mathrm{d}x \wedge \mathrm{d}y~[A_y,A_x]\cr
~=~&\iint_{\Sigma} \!\mathrm{d}x^{\mu} \wedge \mathrm{d}x^{\nu}~[A_{\nu},A_{\mu}].
\end{align} \tag{3.114}$$
For more information, see e.g. this related Phys.SE post.
--
$^1$ In this answer we ignore at various stages subleading contributions not relevant to the final answer (3.114).
$^2$ We prefer to use time-ordering rather than path-ordering, since the latter is ambiguous.
