Some questions about Wilson loops

• Let me assume the bundle is trivial. Then $A$ is a $\mathfrak{g}$-valued 1-form on the base $M$. Let $\gamma(t):[0,1]\rightarrow M$ be a parametrization of $C$. A horizontal lift $\tilde{\gamma}:[0,1]\rightarrow P\cong G\times M$ is just a pair $\tilde{\gamma}(t)=(\gamma(t),g(t))$, where $g(t)\in G$. The parallel transport equation then says $g^*\theta = \gamma^* A$, where $\theta$ is the Maurer-Cartan form on $G$. Explicitly, for matrix groups the equation reads $$g(t)^{-1}dg(t) = A(\gamma(t)).$$ The solution to this equation is by definition the path-ordered exponential $$g(t)=\mathcal{P}\exp\left(\int_0^1 \gamma^*A\right)g(0)=\mathcal{P}\exp\left(\oint_C A\right)g(0).$$ This works for general groups as well if you use the exponential map.

For an abelian group $G$ one has $F=dA$, and so you can use Stokes theorem to write the exponent as $\int_D F$. Note, that any connected compact abelian group is isomorphic to $U(1)^n$. The only difference for non-abelian groups $G$ is that you cannot reduce the integral of the connection to an integral of the curvature.

• To evaluate a Wilson loop, you need $R$ to be a representation of $\mathfrak{g}$, it does not have to exponentiate to $G$. It works as follows: you first use your representation to make $A$ a matrix-valued 1-form and then use the ordinary matrix path-ordered exponential. If the representation exponentiates, it coincides with the usual definition, where you first use the exponential map and then take the trace.

• I understand that Witten's remark as follows. Since your action is $\frac{1}{e^2}\int F\wedge *F$, to use perturbation series you rescale $A\rightarrow e A$. That means you have to expand the exponential for the Wilson loop in Taylor series. If your connection is not coupled to anything, all your diagrams are photons emitted and absorbed by the Wilson loop.

For example, at order $e^2$ you would have a process like $\langle Tr(AA)\rangle$, which you can think of as a Wilson loop together with a photon propagator attached, where the photon carries the representation indices away. So, half of the Wilson loop "carries" a representation (it corresponds to the trace), while the other half has the zero representation (since you are multiplying $A$'s). I hope it's clear without a picture.

Possible answer to your first bullet: Stokes theorem. If $G=U(1)$, then $F=dA$ (not $dA+[A,A]$) and so $\int_{\partial D}A=\int_{D}dA=\int_D F.$ Further, if memory serves, $U(1)$ is the only Abelian Lie algebra (thinking of the Cartan classification) so $G$ is Abelian means $G=U(1)$ anyway.

EDIT: Coordinate expression for the Holonomy. $\mathrm{Hol}(A,C)$ is defined to be $\mathcal{P}\exp(\int_{C}A)$. The connection is usually written as a 1-form $A_{\mu}=A_{\mu}^a\tau^a$ that takes values in $\mathrm{Lie}(G)$ with generators $\tau^a$. Then the holonomy is

\begin{equation} \mathrm{Hol}(A,C)=\mathcal{P}\exp(-ig \int_C A_\mu dx^\mu). \end{equation}

Here $g$ is the coupling constant and the $-i$ is to make it look like QFT rather then LQG :-) To get the gauge-invariant Wilson loop, expand the exponential and trace over the group indices. Check out Peskin and Schroeder (or any other QFT book) for more details.