Suppose $G$ is a compact connected Lie group and $P$ is a $G$-bundle over $S^1$, $A$ is a connection. Then we can choose a frame such that $A = a d\theta$ where $a\in \mathfrak{g}$ is constant. My question is: can we choose the frame such that $a$ falls in a bounded set in $\mathfrak{g}$ for any connection $A$?
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Yes. Let $g\in G$ be the holonomy of the connection $A$; since $G$ is compact and connected, one can choose a frame (i.e., trivialization of $P$) such that $A = a\,\mathrm{d}\theta$ where $g = \exp(2\pi a)$. Since $G$ is assumed compact, there is a compact set $K\subset\frak{g}$ such that $\exp(2\pi K) = G$. Choose $a$ to lie in $K$ and satisfy $g = \exp(2\pi a)$; this does the job.
answered Mar 3, 2014 at 12:18
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$\begingroup$ Thanks! So what I was missing is your last statement. I know it is true but how can I prove the existence of the compact set $K$? $\endgroup$ Commented Mar 3, 2014 at 19:42
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$\begingroup$ Because $G$ is compact, it has a bi-invariant metric and the geodesics of such a metric are the 1-parameter subgroups. Since the exponential map from a point is onto when the manifold is compact, and since the diameter of $G$ is bounded, there is a upper bound $D$ on the length of a minimizing geodesic. Now let $K$ be the set of vectors in $\frak{g}$ of length at most $D/(2\pi)$. $\endgroup$ Commented Mar 3, 2014 at 22:08