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In $4$d Euclidean space, the Maldacena-Wilson line is defined as:

$$\mathcal{W}(C) = \frac{1}{N} \text{Tr} \left\lbrace \mathcal{P} \exp \int_C d\tau \left( i \dot{x}_\mu A_\mu^a(x) + \left| \dot{x} \right|\theta_i \phi^{i,a}(x) \right) \right\rbrace \tag{1}$$

where $\phi$ and $A_\mu$ are scalar and gluon fields respectively, $\tau$ is the $4$th component of $x$, $\dot{x}=(0,0,0,1)$, $\mathcal{P}$ refers to the path-ordering operator, $C$ is an infinite line on the $\tau$-direction, and the trace is meant to act on the color indices $a$, $b$ ($i$, $j$ are SO($6$) indices). When expanding the exponential to $2$nd order, we get for example the following term:

$$\frac{1}{2! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{2}$$

One can get rid of the path ordering by realizing the following relation:

$$\begin{align} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^{\tau_1} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} &= \text{Tr} \int_{-\infty}^\infty d\tau_2 \int_{\tau_2}^{\infty} d\tau_1 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \\ &= \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{\tau_1}^{\infty} d\tau_2 \left| \dot{x}_2 \right| \left| \dot{x}_1 \right| \theta_i \theta_j \phi_2^{i,a} \phi_1^{j,b} \\ &= \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{\tau_1}^{\infty} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{3} \end{align}$$

where in the last line I used the cyclicity of the trace. This implies:

$$\text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^{\tau_1} d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} = \frac{1}{2} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^\infty d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{4}$$

and thus

$$\frac{1}{2! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} = \frac{1}{2! N} \frac{1}{2} \text{Tr} \int_{-\infty}^\infty d\tau_1 \int_{-\infty}^\infty d\tau_2 \left| \dot{x}_1 \right| \left| \dot{x}_2 \right| \theta_i \theta_j \phi_1^{i,a} \phi_2^{j,b} \tag{5}$$

Now let us expand the Wilson line to $3$rd order. Is there a convenient formula such as $(5)$ to get rid of the path ordering at this order? Since the trace acts now on $3$ generators, I imagine it would be necessary to have something antisymmetric like $\epsilon (\tau_1 \tau_2 \tau_3)$ in the integral in order to change the sign depending on the path ordering. But how to set the limit of integration to $-\infty$ and $+\infty$ for all integrals?

Specifically I am searching for a formula for the following "mixed term":

$$\frac{1}{2! 1! N} \text{Tr}\ \mathcal{P} \int_C d\tau_1\ d\tau_2\ d\tau_3 \left| \dot{x}_1 \right| \dot{x}_{2\mu} \left| \dot{x}_3 \right| \theta_i \theta_j\ \phi_1^{i,a} A_{2\mu}^b \phi_3^{j,c} \tag{6}$$

As a response to the comment, a useful reference is probably this one, where an expression for $(6)$ involving a $\epsilon(\tau_1 \tau_2 \tau_3)$ is used in eq. $(13)$ without explaining how. I would like to know if I can always do that or if this is a special case.

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  • $\begingroup$ Could you give some references? $\endgroup$ – Nikita Mar 24 '20 at 14:30
  • $\begingroup$ @Nikita Sure, for what would you like a reference? $\endgroup$ – Jxx Mar 24 '20 at 16:20
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    $\begingroup$ @Nikita I have added a reference in which the authors make use of what I am looking for without explaining how they got there. I imagine this is pretty standard, but I was unsuccessful at deriving it so far. $\endgroup$ – Jxx Mar 24 '20 at 17:46
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Actually, it is very simple if I am not mistaken. The idea is that any odd permutation of the $\tau$'s on the Wilson line brings a minus sign because of interchanging two generators in the trace, while the limits of integration stay the same. Adding all the diagrams together is therefore the same as integrating over the whole domain, with the $\epsilon (\tau_1 \tau_2 \tau_3)$ accounting for the sign change mentioned above. And thus no prefactor is required.

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