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Let $G$ be the Lie group of a given theory with the Lie algebra $\mathfrak{g}$.

According to the Wikipedia article, a Wilson line is of the form \begin{equation} W[x_i,x_f]= P e^{i \int_{x_i}^{x_f} A} \end{equation} where $A$ is the $\mathfrak{g}$-valued connection.

Then, I guess that this Wilson line $W$ must be $G$-valued, but the path-ordering operator $P$ in front of the exponential seems like an issue...

Could anyone please confirm or correct my understanding?

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    $\begingroup$ Path ordering is just a prescription for the order in which the group elements are multiplied. Since a group is closed under multiplication, the path ordering does not pose any issue. $\endgroup$
    – Prahar
    Commented Jun 12 at 18:58
  • $\begingroup$ @Prahar The exponent consists of Lie "algebra" elements, whose "ordinary" exponential is supposed to yield an Lie "group" element. That is why I am confused. $\endgroup$
    – Keith
    Commented Jun 12 at 19:17
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    $\begingroup$ Usually one takes trace so as to get a gauge invariant expression. The result will be a complex number. $\endgroup$
    – mike stone
    Commented Jun 12 at 19:24
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    $\begingroup$ Then the line is a non-gauge-invariant element of the gauge group in whatever representation the $A$'s live in. $\endgroup$
    – mike stone
    Commented Jun 12 at 19:28
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    $\begingroup$ @Keith - You break up the path $\gamma$ into infinitesimal segments of length $dx$. You define a Wilson line for each segment. The Wilson line on each segment is given by $e^{i d x \,t^\mu(x) A_\mu(x)}$ and is clearly group-valued ($t^\mu$ is the unit normalized tangent vector to $\gamma$ at the point $x$). The full Wilson line is then given by a path-ordered product over all the segments, followed by a limit $dx \to 0$. It should be clear from this that $W$ is group valued. $\endgroup$
    – Prahar
    Commented Jun 12 at 19:54

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