When does the gauge potential vanish?

Question

Question: Given a $U(1)$-principal bundle over the circle group $S^1$, when does the gauge potential vanish?

Attempt: Firstly, I believe that the principal bundle $P_M = U(1) \times S^1$ is trivial. This is because since $U(1)$ and $S^1$ are both represented by $e^{i \theta}$, where $\theta \in \mathbb{R}$, there is a homomorphism $e^{i (\theta_1 + \theta_2)} = e^{i \theta_1} e^{i \theta_2}$, and obviously there exists an inverse e^{i -\theta}$ such that it is a bijective mapping.

Because it's a trivial bundle, there is then a global section for which we can define the gauge potential $A = \sigma^\ast \omega$, where $\omega$ is the connection one-form. So, I believe that $A$ vanishes when the connection is flat, i.e. the curvature $\Omega^{\omega} = d\omega + \omega \wedge \omega = 0$. But here is where I'm stuck, how do we show that this is true? Also, am I even approaching the question in the right way? And is it necessary to show that there exists a global section?

Answer 1

First, we have to note that the gauge potential depends on a choice of connection and a choice of section, not just on the bundle. So the real question is, given a principal $U(1)$-bundle over $S^1$ endowed with a principal connection, how can we tell which local or global sections will yield a zero gauge potential?

The next thing to notice is that, as @TedShifrin pointed out in a comment, the curvature always vanishes because it's a $2$-form on $S^1$.

Third, it is always the case that there exists a local section in a neighborhood of each point that yields a zero gauge potential. To see this, choose any smooth local section $\sigma_0$ and let $A_0 = \sigma^*\omega$ be the associated gauge potential. Since $dA_0=0$, locally we can find an imaginary-valued smooth function $u$ such that $du = A_0$. (It can be chosen to be imaginary-valued because $A_0$ takes its values in the Lie algebra of $U(1)$, which is $i\mathbb R$.) Then the transformation law for the gauge potential guarantees that the gauge potential associated with the section $\sigma = e^{-u}\sigma_0$ satisfies $\sigma^*\omega = A_0 - du = 0$.

Finally, though, even for a flat connection on a trivial principal bundle, there might not be a global section that yields a zero gauge potential. It depends on the holonomy of the connection. For example, on the product bundle $S^1\times U(1)$ with the trivial global section $\sigma_0(e^{i\theta}) = (e^{i\theta},1)$, consider the connection given by the gauge potential $A_0 = \frac12 i d\theta$. The computation above shows that another section $\sigma = e^{-u}\sigma_0$ will yield a zero gauge potential if and only if $du = A_0$. But there is no global smooth function $u$ that satisfies this. What's going on here is that this flat connection has holonomy equal to the subgroup $\{\pm 1\}\subseteq U(1)$, whereas a connection that has a global section with zero gauge potential must have trivial holonomy.

Answer 2

You're close. The correct statement is that if the connection is flat, then it's possible to choose a local section $\sigma$ such that $A=\sigma^*\omega$ vanishes (and incidentally this holds for arbitrary principal $G$-bundles, not just trivial $U(1)$-bundles). To see this, suppose the connection is flat, and consider the horizontal distribution $H:=\ker \omega$. Taking $v, w$ to be vector fields in $H$, and applying $\Omega^\omega=0$ to $v,w$ gives $-\omega([v,w])=0$ (using $d\omega(v,w) = v(\omega(w))-w(\omega(v))-\omega([v,w]))$, implying $[v,w]$ is also a vector field in $H$. Hence $H$ is involutive, and so by Frobenius is integrable. Constructing a suitably small integral submanifold $S$ of $H$, and defining $\sigma:U\to S\subset P$ to be a local section mapping to $S$, it is true by construction that $A := \sigma^*\omega = 0$.

This is one way of interpreting geometrically what the curvature of a connection is: it measures the failure of the horizontal distribution $H=\ker\omega$ to be involutive, and hence integrable.

Edit: after reading Jack Lee's answer (posted while typing my own), I should have noticed that when the base manifold is 1-dimensional (as in your case), then the curvature automatically vanishes. The rest of my answer still applies.