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Given a vector bundle $\pi\colon E \rightarrow B$ equipped with a connection $\nabla$, it is well known that a basis of flat sections $s_i$ ($i=1,\dots,\text{rank}(E)$) (i.e. $\nabla_X s_i = 0$ for all vector fields $X \in \Gamma(TB)$) locally exists if and only if the corresponding curvature 2-form $\Omega_\nabla$ vanishes.

I would now like to relax the condition $\nabla s_i = 0$ to hold not necessarily for all tangent vectors but only on a subbundle of $TB$. So given a distribution $\xi \subset TB$, I want to know if there locally exists a basis of sections $s_i$ of $E$ such that $$ \nabla_X s_i = 0 \quad\text{for all}\quad X \in \Gamma(\xi) \,. $$ My guess is that $$ \Omega_\nabla(X,Y) = 0 \quad\text{for all}\quad X,Y \in \Gamma(\xi) $$ is a necessary and sufficient condition. Is this true? And if yes, is there an easy way to see it (e.g. from the above statement)?

Furthermore, I am interested in "non-trivial" solutions. By this I mean the existence of $n = \dim(B) - \dim(\xi)$ local frames $s^{(1)}_i, \dots, s^{(n)}_i$ such that for each $i$ the $\nabla s^{(k)}_i$ are linearly independent (I am not 100% sure if this is the correct way to phrase it...).

Let me illustrate this for the case of a trivial line bundle with trivial connection. Here, the connection is flat and solutions always exists and are given by constant functions. However, if I demand the existence of $n$ independent solutions, the problem is answered by Frobenius' theorem, which requires $\xi$ to be involutive (i.e. $\xi$ must be closed with respect to the Lie bracket). Is it possible to do a similar statement in the general case, i.e. for a non-trivial, non-flat connection?

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    $\begingroup$ In your first sentence, I believe you mean a vector bundle and a covariant constant section. Also, note that you don't need your $X$ to be vector fields; vectors will do. Finally the result you quote is not quite right: flatness of the connection is equivalent to the local existence of ${\rm rank}(E)$ linearly independent covariant constant sections. $\endgroup$ – Oliver Nash Jun 20 '18 at 10:31
  • $\begingroup$ Yes, I am mainly interested in vector bundles, but I assumed that I could easily generalize to fibre bundles. Regarding your second remark, it is indeed plausible that the existence of one section is not enough to deduce the vanishing of the curvature tensor. However, this seems to be in contradiction to Def. 5.1 and Theorem 5.15 of these lecture notes. $\endgroup$ – Severin Jun 20 '18 at 10:47
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    $\begingroup$ There is no contradiction. Note that Def. 5.1 requires the existence of a covariant-constant section taking every possible value in the fibre over $x$. This is equivalent to requiring the existence of covariant-constant sections forming a basis. This is a much stronger condition than existence of a single covariant-constant section. Regarding generalizing to any fibre bundle, this doesn't quite make sense: a connection is something that exists for a principal bundle, it corresponds to a covariant derivative for associated vector bundles but a general fibre bundle has no such concept. $\endgroup$ – Oliver Nash Jun 20 '18 at 10:59
  • $\begingroup$ Ok, thanks for the clarification. I will think how to rephrase my question, I hope it makes still some sense... I am still a little bit confused, as before mentioned lecture notes seem to be talking about connections on fibre bundles. $\endgroup$ – Severin Jun 20 '18 at 11:42
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    $\begingroup$ @Oliver Nash Well, for a general fiber bundle one still has the notion of a connection which yields a notion of a parallel transport. Then a "covariantly constant" section would simply mean a parallel section (along certain curves, certain submanifolds). So the question seems to make still sense when interpreted in this way. A "basis" of sections requires of course a linear structure on the fibers i.e. a vector bundle... $\endgroup$ – Stefan Waldmann Jun 21 '18 at 8:57

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