The proof is trivial in the Abelian case by the Stokes' theorem.How to prove it in the non-Abelian case?
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1$\begingroup$ This follows immediately from the nonabelian Stokes theorem, see Theorem 4.4 and Lemma 4.5 in Schreiber and Waldorf's “Smooth functors vs. differential forms” (dx.doi.org/10.4310/HHA.2011.v13.n1.a7). $\endgroup$– Dmitri PavlovCommented Jul 8, 2015 at 16:06
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$\begingroup$ Dear Dimitri, I think this is not only too complicated, but also circular in the following sense: In order to prove Theorem 4.4. and Lemma 4.5 you have (at some point) to prove a (stronger) version of the assertion, see for example Proposition 3.8 in the mentioned paper. $\endgroup$– SebastianCommented Jul 10, 2015 at 8:24
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$\begingroup$ Dear QIAOJIAXIN, your question is not research level, and you should be able to extract the answer from any differential geometry book dealing with the Frobenius theorem, e.g. Lee's "Manifolds and Differential Geometry". $\endgroup$– SebastianCommented Jul 10, 2015 at 8:29
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A possible answer would be to invoke Ambrose-Singer Holonomy theorem:
Theorem (Ambrose-Singer) Let $M$ be a (smooth) connected manifold, $E\to M$ a vector bundle over $M$, and $\nabla$ a connection on $E$. Then, for each $x\in M$, $\mathfrak{hol}_x(\nabla)$ is a Lie subalgebra of $\text{End}(E_x)$ which, as a vector space, is spanned by all elements of $\text{End}(E_x)$ of the form $P_{\gamma}^{-1}[ (F_{\nabla})_y\cdot{(v\wedge w)}] P_{\gamma}$, where $\gamma:[0,1]\to M$ is a piece-wise smooth path with end points $\gamma(0) = x$ and $\gamma(1) = y$, $P_{\gamma}: E_x \to E_y$ is the associated parallel translation map, and $v,w\in T_yM$.
In particular, if the connection $\nabla$ is flat (i.e. if $F_{\nabla}=0$), then $\mathfrak{hol}_x(\nabla)=0$, and therefore the restricted holonomy group $\text{Hol}_x^0(\nabla)$ is trivial, for each $x\in M$. Recalling $$ \text{Hol}_x^0(\nabla)=\{P_{\gamma}: \gamma\text{ is a null-homotopic loop based at }x\}, $$ the conclusion you need follows. Indeed, given two homotopic paths $\gamma_1$ and $\gamma_2$ (I'm assuming this homotopy preserves the base-points), say with $\gamma_1(0)=\gamma_2(0)=x$, then you can consider the null-homotopic concatenation $\gamma_2^{-1}\cdot{\gamma_1}$, which is a (piece-wise smooth) loop based at $x$. By the above, we have that $1=P_{\gamma_2^{-1}\cdot{\gamma_1}} = P_{\gamma_2}^{-1}\circ P_{\gamma_1}$, i.e. $P_{\gamma_1}=P_{\gamma_2}$, as we wanted.
A principal bundle version of this theorem (together with a proof) can be found e.g. in Kobayashi&Nomizu's book Foundations of differential geometry, vol. I, Theorem 8.1, p. 89. The statement I give here was based on Theorem 2.4.3. (a) of Joyce's book Riemannian Holonomy Groups and Calibrated Geometry.
