Holonomy as integration of curvature for principal $G$-bundles?

Question

Holonomy and curvature may seem to be slightly advanced topics in geometry. However, their origins are easily imaginable. Namely, picture the surface of earth $S$, and pick an arbitrary contractible and piecewise smooth loop $\gamma$, parallel transport any vector from $x \in \gamma$ back to $x$, the difference in angle (holonomy) can be calculated by integrating the scalar curvature within the enclosed surface.

Several years after Gauss , things like the above were generalized into the setting of principal $G$-bundles. However, intuitive explanation became difficult to find. This post is looking for a formal yet intuitive statement like the above, and a proof for it.

Formal Setup

Let $G$ be a compact Lie group, $M$ a smooth manifold, and $P \xrightarrow{\Pi} M$ a smooth principal $G$-bundle (i.e. a bundle whose fibers are $G$-torsors; as usual the structure group is $G$ acting on the right). A connection of $\Pi$ is a $G$-invariant $\mathfrak{g}$-valued $1$-form on $P$ whose restriction to each fiber is the Maurer-Cartan $1$-form on $G$ torsor. By Cartan structure theorem, the curvature $2$-form $\Omega$ satisfies $$\Omega = d\omega + \frac{1}{2} [\omega, \omega].$$

A fact is that this setting vastly generalized the classical setting, where $M$ is a Riemannian manifold, $G$ is $O(dim(M))$, $\Pi$ is the underlying $G$-bundle for the tangent bundle, and the connection is the one associated to the Levi-Civita connection.

Wonder

I wonder if we can still make sense of the intuitive statement, that the holonomy along any contractible loop $\gamma$ is exactly the integration of the curvature over any surface which cobounds $\gamma$.

Naively, this does not make sense, as we need the curvature $2$-form to be on $M$, while the definition suggests that $\Omega$ is really a $2$-form on the bundle space $P$. There does not seem to be a natural $2$-form on $M$ too in this picture.

Q. How to make sense of it? And how to prove it?

Remarks

1. The Ambrose-Singer theorem should be weaker than what I'm hoping for. 2. That the curvature vanishes is equivalent to the connection is flat should also be weaker too.

After thoughts

• From Roberto Ladu's answer, a key to understand this is the Non-abelian Stoke's theorem. An illustrative and beginner case (dimension=$1+1$) is clearly given in [2]

Reference

• [1] Foundations of Differential Geometry I [Kobayashi and Nomizu], Chapter 2.
• [2] Non-Abelian Stokes theorem in action-Boguslaw Broda v3p12-16.

Answer 1

The curvature form descends to a genuine 2-form on the base space (unlike the connection 1-form). In fact, locally on the base space, we can pick a trivialization of the principal bundle and compute the curvature 2-form using the usual formula $dω+[ω,ω]/2$, where $ω$ denotes the connection 1-form on the base space. The result does not depend on the choice of a trivialization.

In addition to the references pointed in the other answer, a modern exposition can be found in Theorem 3.4 of

• Urs Schreiber, Konrad Waldorf. Smooth functors vs. differential forms, Homology, Homotopy and Applications 13:1 (2011), 143–203. arXiv:0802.0663.

The integration for is given in Corollary 3.6. The holonomy is the accumulated change along the curve $\gamma$, so it is an integration over $\gamma$. You need a path-ordered integration $Pexp \int^1_0 A_\Sigma$ because the changes (i.e. group elements) are not commutative. Then you pick any contraction of $\gamma$ ($s=1$) to the identity loop $id_{x}$ ($s=0$). Then obviously the final holonomy (at $s=1$) can be written as the accumulated change from $s=0$ to $s=1$. We only need to figure out what the change is from $s=s$ to $s = s + \Delta s$. Draw a thin moon shape and cut it into finitely little pieces; you can convince yourself that this change can be written in an integration in $t$ as well. However, the change at each $t$ isn't just the Lie-algebra element of the curvature 2-form at that point - you need to conjugate with the change coming from the path that starts from $s=s, t=0$. Therefore you get that $Ad^{-1}$ thing. When you $G$ is abelian, this is trivial, thus you get a simplified formula below the corollary.

Answer 2

It can be done, and I think that is known to many physicists. The main difficulty is to deal with a non-commutative structure group $G$.

Consider a loop $\gamma\subset M$ bounding a disk $D\subset M$, trivialize $P|_D$ so that we have get a preferred flat connection $A_0$ over $D$ now any connection on $P$ may be written over $D$ as $A_0 + A$ where $A \in \Omega^1(D, ad(P))$ (here $ad(P)$ is the vector bundle associated to the adjoint action of $G$ on its Lie algebra $\mathfrak{g}$).

If $G=U(1)$ then is not difficult to see that the holonomy around $\gamma$ may be written as $hol(A, \gamma) = e^{-\int_\gamma A}$, in this case $A\in \Omega^1(D, i\mathbb{R})$, and using Stokes theorem now gives that $hol(A, \gamma) = e^{-\int_D F_A}$ which is the kind of result we were looking for.

For general $G$ the holonomy around $\gamma$ may still be written with the help of the ordered exponential $hol(A,\gamma) = Pe^{-\int_\gamma A}$. An analogue of Stokes theorem (sometimes called non-abelian Stokes theorem) may be applied in this context yielding a similar result. For a precise statement and proof I refer you to Aref'eva, I.Y. Non-Abelian Stokes formula. Theor Math Phys 43, 353–356 (1980). The formula you are looking for is in Remark 2 of this paper. There are other references but this IMHO is the clearest one for mathematicians.

A good reference for the derivation of holonomy as ordered exponential is the book Baez, J. and Muniain, J. P. Gauge fields, knots and gravity.1995. (from pg. 231). There you can also find a proof that the curvature at a point approximates the holonomy around infinitesimal small square-loops up to second order in the length of the sides.