Holonomy of a curve in case of principal $U(1)$ bundle Suppose $\pi : P\rightarrow M$ is principal $U(1)$ bundle. Let $\gamma$ be a loop in $M$ based at $x_0$ and write $iA$ as connection 1-form on $P$ where $A\in \Omega(P)$. Now define $hol_{\gamma}(A)\in U(1)$ as follows: let $\tilde{\gamma}$ be a horizontal lift of $\gamma$ with $\tilde{\gamma}(0)=p_0$ and $\pi(p_0)=x_0$. Then there is a unique element $hol_{\gamma}(A)$ in $U(1)$ such that $\tilde{\gamma}(1)=p_0\cdot hol_{\gamma}(A)$ because the action is free and transitive. 
I now want to calculate explicitly what is this number $hol_{\gamma}(A)$. I know already it has to be $hol_{\gamma}(A)=e^{i\int_{\gamma}A}$, but I have no idea how this is calculated as I am new to holonomy. Any hint is useful for me. Thanks
 A: Given a loop $\gamma: S^1 \to M$, we can form the pullback bundle $\gamma^*P$ with pullback connection $\gamma^*A$. The holonomy of the pullback along the identity loop $S^1 \to S^1$ is the same as the holonomy of the original connection along the original loop $\gamma$. The bundle is now trivial (there's only one $U(1)$-bundle over $S^1$) and the connection form we'll still call $iA$. Consider the torus for now as $\Bbb R^2/2\pi\Bbb Z^2$. 
If $iA$ is explicitly $f(t)dt + d\theta$, $d\theta$ the canonical 1-form on the circle fibers, the kernel at $T_{t,\theta}$ is those points of the form $(s,-f(t)s)$. We can then see explicitly that a horizontal lift is given by a curve with derivative $\eta'(t) = (1,-f(t))$; and we can explicitly write this down as $\eta(t) = (t, -\int_{0}^t f(s)ds)$. Thus $\eta(1) = (1, - \int_0^1 f(t)dt)$. Now this says that the holonomy is $$- \int_0^1 f(t)dt \pmod 1.$$ But that's just because here I'm considering the circle fiber to be $\Bbb R/2\pi \Bbb Z$; the canonical isomorphism $\Bbb R/2\pi \Bbb Z \to S^1$ is precisely the exponential map, giving you the desired holonomy $$e^{-\int_\gamma iA}.$$
