Curvature as infinitesimal holonomy 2

Question

This question may be seen as a follow up of this original question. I'm learning Cheeger-Simons differential characters (reading Differential Characters of Bär and Becker). If I understand correctly, the idea is based on the fact that something like this is true:

Let $P \to M$ be a principal $G$-bundle where $G = U(1)$ and let $\omega$ be a connection in $P$. Denote by $F_\omega \in \Omega^2(M, \mathfrak{g}_P)$ the curvature of $\omega$.

Let $S \subseteq M$ be an oriented $2$-dimensional submanifold with boundary $\gamma = \partial S$. Then $$\mathit{hol}_\omega (\gamma) = \exp\left(\int_S F_\omega\right) \in G\,.$$

(In the book, there is $2i\pi$ in front of the integral, but I guess it's because they choose an identification $\mathfrak{u}(1) \approx \mathbb{R}$).

Why is this identity true? Does some version of this hold for more general $G$? If that were the case, it would seem like a better version of Ambrose-Singer / a pretty good answer to the original question.

Answer 1

As it is stated, the formula is not true (even up to some normalisation factor). Consider a connection $d+\omega$ on a complex line bundle, where $\omega\in\Omega^1(M,\mathbb C).$ The curvature is just $d\omega$, and the parallel transport along a curve $\gamma$ is just $exp(-\int_\gamma\omega)$. Thus, if you can apply Stokes theorem you can derive the holonomy formula. But this is not always the case, as you see from the example of the punctured disc (with boundary $S^1$) and the flat connection $d+a\frac{dz}{z}.$

There is no direct generalisation to non-abelian Lie groups $G$. For example, there exists flat $SU(2)$-connections on the 1-holed torus with non-trivial monodromy along the boundary.