3
$\begingroup$

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.

$\endgroup$
4
  • 5
    $\begingroup$ 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? $\endgroup$ Commented Nov 13, 2019 at 16:29
  • 1
    $\begingroup$ 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. $\endgroup$ 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

4
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ 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-- $\endgroup$ Commented Nov 13, 2019 at 23:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.