# Existence of flat connections and the application of the Frobenius theorem

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.

• 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 Dec 22 '20 at 10:02
• Hm thanks but I was trying to understand how the frobenius theorem can be used to solve systems of differential equations too. Dec 22 '20 at 10:31

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 $$d\eta = -\eta\wedge\eta$$.)
• 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. Dec 22 '20 at 22:07
• 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. Dec 22 '20 at 22:26
• $A$ is the coordinate on $GL(r)$. :) Remember, you're looking for a graph of a function $U\to GL(r)$. Dec 23 '20 at 0:04