Let $E \rightarrow F$ be a principal $G$-bundle. Let $\alpha$ be a connection 1-form with values in the Lie algebra of $G$. Let $\omega$ denote the curvature 2-form of connection $\alpha$.

We know that $\omega =0$ everywhere i.e., $\alpha$ is a flat connection if and only if the distribution $ker (\alpha)= \{v_x \in T_x E \ | \ \alpha (v_x)=0 \ ; x \in E \} $ on $E$ is completely integrable.

Now, suppose we have a norm on the space of 2-forms. We start with an "almost flat connection $\alpha$ " i.e., the curvature form satisfies inequality, $|\omega| < \epsilon$ everywhere, for sufficiently small $\epsilon$. Is it true that the distribution $\ker(\alpha)$ on $E$ is "close" to a completely integrable distribution? and "close" in what sense? We may assume that the base $F$ is compact.

I have a feeling that with appropriate notion of "closeness" of distributions the above question has an affirmative answer. While I am trying to show that it is so, I have a difficulty deducing any useful information about connection 1-form from bounds on curvature. Thanks in advance for bringing in any new insight.

I have posted this question on Math SE also. https://math.stackexchange.com/questions/47571/curvature-and-connections-in-principal-g-bundles

  • $\begingroup$ Dear Dheeraj: since there are lots of overlapping readership between Mathoverflow and Math.SE, it is generally discouraged to crosspost a question between the two within short period of time. Your question also appears as math.stackexchange.com/questions/47571/… on Math.SE. $\endgroup$ Commented Jun 25, 2011 at 18:22
  • $\begingroup$ ...and when you do, it is generally good etiquette to add a comment on both sites linking the two questions. $\endgroup$ Commented Jun 25, 2011 at 18:23
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    $\begingroup$ Dear Willie: Thanks for your comment. I have added a comment linking the two posts on both the sites. Actually, I was in two minds while posting the question, whether the question is of the level of MO or not. So I posted it first Math SE and just 3 hrs after that I changed my mind! I will try and avoid doing this in future. $\endgroup$ Commented Jun 25, 2011 at 20:15

1 Answer 1


The answer depends on the setting. If the group $G$ is compact, $M$ is compact and you are working modulo the gauge group, then the answer is positive and follows from Uhlenbeck's compactness theorem (first proven in her paper "Connections with $L^p$-bounds on curvature"). This theorem was extended by Katrin Wehrheim to the case when the base-manifold $M$ noncompact, see her book "Uhlenbeck compactness." In the case when $G$ is noncompact, I do not think there is a meaningful way to get a positive answer. For instance, $S^3$ admits "almost flat" symmetric affine connections, see Agaoka's example in http://www.jstor.org/stable/2161101 On the other hand, the frame bundle of $S^3$ admits unique flat connection, namely, the trivial one (up to gauge transformations), but it is highly nonsymmetric. Thus, there is no way you can approximate Agaoka's connections (lifted to the frame bundle) by flat ones. I am sure there are other examples.


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