Integrability condition for flat connections

Question

I am reading Kobayashi's book "Differential geometry of complex vector bundles". More precisely, I am on section 2 of chapter 1, page 5.

Kobayashi is trying to prove that if $E$ is a vector bundle on some manifold $M$, with a flat connection $D$, then it admits a "flat structure" $\{U,s_U\}$ which consists on an open cover of $M$ and a local frame of $E$ such that the transition functions are locally constant.

In order to do this, he starts with some arbitrary local frame $s'$ and looks for functions $a:U \rightarrow GL(r,\mathbb{C})$ such that in the frame $s_U= s' a$ the connection $1$-form is $0$.

Therefore, if $\omega'$ is the connection $1$-form in the frame $s'$, what he is trying to do is solve the following equation for $a$

$$ \omega' a + da = 0. $$

He claims that solutions exists since the "integrability condition" for this equation is obtained by differentiating it

$$ 0=(d\omega') a -\omega' \wedge da = (d\omega')a + (\omega' \wedge \omega')a = \Omega' a, $$

which is true since we assumed that the connection is flat.

My question is what does he mean by the "integrability condition". Moreover, why is that the integrability condition for that equation? And, also, why can he use the fact $da=-\omega' a$ when computing it?

I think he might be using some form of the Frobenius theorem, since I know that it is what you use from a "global" point of view.

In any way, I want to know precisely in this context what he means by that "integrability condition", maybe it is just something basic or standard that I am missing.

Answer 1

On the manifold $X=U\times \operatorname{GL}_r$, with points written $x=(m,a)$, each tangent space $T_x X$ contains a linear subspace $V_x$ consisting of tangent vectors on which $a^{-1}da=-\omega'$. The problem is to prove that these $V_x$ spaces form a smooth subbundle $V \subset TX$, and that this subbundle is closed under bracket. For this we can take any framing $e_1,\dots,e_n$ of tangent vector fields on $U$, and associate to each the vector field $e'_i$ which projects to $e_i$ and satisfies $da=-\omega'a$. These are a basis for $V_x$, for each $x$, so $V$ is a smooth subbundle. Note that the $e_i'$ project to the $e_i$, so brackets project to the brackets. Bracket closure (i.e. we can apply the Frobenius theorem to $V$) is precisely flatness of the connection, using the equations $L_v \xi=d(i_v \xi)+i_vd\xi$ and $d\omega'(e_i,e_j)=L_{e_i}(\omega'(e_j))-L_{e_j}(\omega'(e_i))-\omega'([e_i,e_j])$.

Answer 2

I think I can give a more explicit proof of this fact.

Let us take $x^\nu$ coordinates on $U$ and write $\omega'=\sum_\nu A_\nu dx^\nu$ and $df = \sum_\nu \partial_\nu f dx^\nu$. Our differential equation then becomes the PDE

$$ \partial_\nu f(x) + f(x) A_\nu (x) = 0. $$

Now, if we write $F_\nu (x,y) = -yA_\nu(x)$, we can regard our equation as

$$ \frac{\partial f}{\partial x^\nu} = F_\nu (x,f(x)). $$

This is the kind of equation that appears in the classical form of the Frobenius theorem (see Spivak, theorem 1 in chapter 6). The integrability condition for this equation is

$$ \partial_\nu F_\mu - \partial_\mu F_\nu + \sum_k \partial_{y^k} F_\mu F_\nu^k - \sum_k \partial_{y^k} F_\nu F_\mu^k=0. $$

But this precisely means

$$ \partial_\nu A_\mu - \partial_\mu A_\nu - \sum_k A_{k,\mu} A^k_{\nu} + \sum_k A_{k,\nu} A^k_{\mu}=0, $$

which is the $dx^\nu \wedge dx^\mu$ component of the form $\Omega'=d\omega' + \omega' \wedge \omega'$.