# 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?

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$$^3$$.

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$$^1$$ 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.

$$^3$$ While we're at it let us calculate the Wilson line along an infinitesimal plaquette using the same notation: \begin{align} Te^{-\oint_{\Gamma}\!A}~=~&e^{-\int_4\!A}e^{-\int_3\!A}e^{-\int_2\!A}e^{-\int_1\!A}\cr ~=~&e^{bA_y(4)}e^{aA_x(3)}e^{-bA_y(2)}e^{-aA_x(1)}\cr ~=~&e^{b(A_y-\frac{a}{2}\partial_xA_y)}e^{a(A_x+\frac{b}{2}\partial_yA_x)}e^{-b(A_y+\frac{a}{2}\partial_xA_y)}e^{-a(A_x-\frac{b}{2}\partial_yA_x)}\cr ~=~&\ldots~=~1+ab(\partial_{[y}A_{x]}+[A_y,A_x])\cr ~=~&1+ab F_{yx}~=~e^{-ab F_{xy}}~=~e^{-\iint_{\Sigma}\!F},\end{align} where the non-abelian field strength is $$F_{\mu\nu}~=~\partial_{[\mu}A_{\nu]}+[A_{\mu},A_{\nu}].$$