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.
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 $d\eta = -\eta\wedge\eta$.)