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I know that this question has been asked before but there are some doubts about it that I need to clarify, mainly what is the Frobenius theorem really saying.

The main goal is to prove the following :

If $\nabla$ is a flat connection on the vector bundle $(E,M,p)$ then around any point $p\in M$ we can find a basis of local flat sections.

So we start with an arbitrary basis of local sections $\{s_i\}_{i=1}^{r}$ with connection form $\omega$, and we want to find $s'=sA$ such that $\omega'=0$, where $\omega$ and $\omega'$ are the corresponding connection forms. Now basically we need to find $A\in C^{\infty}(U,GL(r))$. We know that $\omega'=A^{-1}\omega A+A^{-1}dA$ and so we would want that $\omega A+dA=0$.

Now this is where I am getting some trouble. Now we want to define an integrable distribution on $U\times GL(r)$ and then prove that it's involutive and use the Frobenius theorem . First I though we wanted to find $A$ and that is to solve the system of differential equations $\omega A+ dA=0$, and I don't quite see how the frobenius theorem helps. I know that it has something related to it , I saw this reading the Wikipedia page, but I don't see how it helps, what I mean is, this is the formulation of the frobenius theorem that I am aware of

Let $D$ be a smooth involute distribution in $M$ that is involutive then $D$ is integrable.

By integrable we mean that there exists a foliation $\mathcal{F}$ such that $T\mathcal{F}=D.$ And I don't see how we will get the $A$ we want .

Any help is aprecciated, thanks in advance.

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    $\begingroup$ You could use that a connection is uniquely determined by the so-called horizontal bundle, a subbundle of $TE$. An alternative proof, without the Frobenius theorem, is to prove that in a flat connection the parallel transport is independent by the curve $\endgroup$ Dec 22 '20 at 10:02
  • $\begingroup$ Hm thanks but I was trying to understand how the frobenius theorem can be used to solve systems of differential equations too. $\endgroup$
    – Something
    Dec 22 '20 at 10:31
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This is a method going back to Élie Cartan and appears in various textbooks. The goal is to find such a function $A$ on $U$ satisfying that differential equation by finding its graph as an integral manifold of the differential system on $U\times GL(r)$. Flatness gives the integrability condition, and a little bit of thought will show you that locally an integral manifold projects diffeomorphically to $U$ and hence is the graph of a smooth map. You will, of course, want to use the differential forms formulation of integrability: If you define the $\mathfrak{gl}(r)$-valued $1$-form $\phi = \omega A + dA$ on $U\times GL(r)$, then note that $d\phi = -\omega\wedge\phi$, which tells us that the differential system $\phi=0$ is integrable. (You can also get it quite prettily from the original $\eta = A^{-1}\omega A + A^{-1}\,dA$; then $d\eta = -\eta\wedge\eta$.)

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  • $\begingroup$ Thank you , I still have to think about some things that you mentioned, but could you give me a reference where I might find this ? @Ted Shifrin $\endgroup$
    – Something
    Dec 22 '20 at 22:04
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    $\begingroup$ Sure. Spivak's Diff Geo (vols. 1 and 2), Chern/Chen/Lam, Boothby (I think). Not in the context of general vector bundles, but in the context of proving that a Riemannian manifold with curvature $0$ is locally isometric to $\Bbb R^n$. The technique is also used to prove the fundamental theorem of hypersurfaces, etc. It is surely in classic books like Bishop/Crittenden, too. $\endgroup$ Dec 22 '20 at 22:07
  • $\begingroup$ I am confused about one thing. How can we define a priori the $\mathcal{gl}(r)$ value $1$-form $\phi$ if we don't know what $A$ is ? My understanding of things would be to define a differential ideal and then this will be the kernel of a distribution which will be integrable, but here it seems that we are defining the differential ideal using the solution. Since the $A$ is what we are trying to find. I guess I am confusing things I am not used to using the frobenius theorem to try and solve differential equations. $\endgroup$
    – Something
    Dec 22 '20 at 22:26
  • $\begingroup$ $A$ is the coordinate on $GL(r)$. :) Remember, you're looking for a graph of a function $U\to GL(r)$. $\endgroup$ Dec 23 '20 at 0:04

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