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Let $A$ be a flat connection on a principal $G$-bundle $G\hookrightarrow P\to M$. Consider an homotopically trivial loop $\gamma \subset M$. For simplicity, suppose $\gamma = \partial D$ is the boundary of an smoothly embedded disk. I was trying to prove that the holonomy of $A$ along $\gamma$ is trivial (must be by flatness). The holonomy of $A$ along $\gamma$ is given by $$\exp^{-\int_\gamma A} \quad ( \ = 1 \text{ by flatness}) $$ Therefore applying Stokes, we get ($F_A=0$) $${\int_\gamma A}=\int_D dA = \int_D (F_A - \frac 1 2 [A\wedge A])= \int_D - \frac 1 2 [A\wedge A]$$

Supposing that $G$ is not abelian, then I would like to understand why $\int_D [A\wedge A]$ lies in the kernel of $\exp$.

I know other proofs of this fact, e.g flatness implies the horizontal distribution is trivial hence we can use charts where $A$ is identically zero. But I hope there is a simpler explanation for this, i.e. we can prove that trivial loops have trivial holonomy just using that $F_A=0$ instead of the deeper/equivalent integrability of the horizontal distribution.

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    $\begingroup$ If $G$ is non-abelian, the holonomy is given by the path-ordered exponential of $A$, and cannot simply be written as $\exp(-\int_\gamma A)$. So you can't simply apply Stoke's theorem as you did. $\endgroup$
    – user17945
    Commented Nov 12, 2019 at 17:47
  • $\begingroup$ @user17945 I'm sorry, I'm not following you. Since $D$ is contractible we can work in a single chart. Once we fix a basis for $\mathfrak{g}= Lie(G)$ we can compute he horizontal lift of the path $\gamma$, call it $\tilde{\gamma}$. The vertical coordinates of $\tilde{\gamma}(t)$ are given by$-\int_{\gamma([0,t])} A\in \mathfrak{g}$. Now, in our trivialization the element of $G$ corresponding to this coordinates is just the exponential of the Lie group. Where am I using that $G$ is abelian? $\endgroup$
    – Overflowian
    Commented Nov 13, 2019 at 16:29
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    $\begingroup$ I was wondering also, why do we need to use the ordered path integral? The exponential map is defined over all $\mathfrak{g}$, $\exp:\mathfrak{g}\to G$. If $G$ is connected, compact is also surjective. Once we compute $\int_\gamma A$ we can just compute its exponential. The only possible issue I see, is that maybe that the the exponential map doesn't provide an atlas as I am implicitely using it. $\endgroup$
    – Overflowian
    Commented Nov 13, 2019 at 16:51
  • $\begingroup$ The holonomy is given by solving the differential equation $(g\circ\gamma)'(s) = -(g\circ \gamma)(s)\cdot A_{\gamma(s)}$, and evaluating at $s=t$ (the endpoint of the loop $\gamma$). If the elements $A\in\mathfrak{g}$ commute everywhere along $\gamma$ (in particular, if $G$ is abelian), the solution to this equation is $g(s) = \exp(-\int_{\gamma\vert_{[0,s]}} A)$, but in general it is given by the path-ordered exponential - one needs to take into account the non-commutativity of $A$ along $\gamma$. $\endgroup$
    – user17945
    Commented Nov 13, 2019 at 17:06

1 Answer 1

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The Stokes theorem must be modified first to deal with the nonabelian case.

See http://arxiv.org/abs/0802.0663, Section 3.2, Theorem 3.4 and the displayed formula on top of page 48 for an appropriate formulation.

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  • $\begingroup$ Sorry, Dmitri, $\int_\gamma A$ is just a vector of integrals (lying in $\mathfrak{g}$), each component of $A$ is a $1$ form. I don't see any possible issue with applying the usual Stokes theorem to this integral. Can you help me in figuring out where the problem is, please? $\endgroup$
    – Overflowian
    Commented Nov 13, 2019 at 16:44
  • $\begingroup$ @WarlockofFiretopMountain: As it was pointed out to you already, the formula for holonomy that you used (with ∫_γ A) only works for abelian Lie groups. For nonabelian groups the formula is not correct and must be replaced by a different (path-ordered) formula. $\endgroup$
    – Dmitri Pavlov
    Commented Nov 13, 2019 at 17:35
  • $\begingroup$ At this point I'll review my computations. I guess my error is to assume that the local frame for the vertical bundle induced by the $\exp$ map, i.e. $\{\exp_* e_i\}$ for $e_i$ basis of $\mathfrak{g}$ is the same of the frame of infinitesimal generators $\hat{e_i}$ for the right action. Thank you. $\endgroup$
    – Overflowian
    Commented Nov 13, 2019 at 22:02
  • $\begingroup$ I explained above what your error was, and it's not what you describe here. Think about why the solution to $\dot{g}(t) = -A(t)g(t)$ for $A(t)$ a commuting (or scalar) quantity is $g(t) = \exp(-\int_0^t A(s) ds)$ (hint: consider the power series for $\exp$). Now think about whether your derivation still works if $A(t)$ becomes non-commuting for different $t$ (hint: it doesn't). $\endgroup$
    – user17945
    Commented Nov 13, 2019 at 22:20
  • $\begingroup$ Dear user17945, it seems to me you have just stated two times that the formula $\exp -\int_\gamma A$ works only for the abelian case, your second comment doesn't add much to the first (for which I thank you of course). But now, you are assuming that I computed the holonomy from $\dot{g}= -Ag$ which I did not. Instead, I took a trivializing chart $P|_U \simeq U\times G$ where $G$ was locally image of $\exp$ and computed the equations for the horizontal lift of $\gamma$. I mistakenly assumed that my local frame was also the one induced by the right action. --continue-- $\endgroup$
    – Overflowian
    Commented Nov 13, 2019 at 23:02

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