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.

  • 1
    $\begingroup$ It is the Frobenius theorem. You can see a long discussion of different ways to write the Frobenius theorem in my lecture notes. github.com/Ben-McKay/… $\endgroup$
    – Ben McKay
    Commented Jul 4, 2020 at 13:16

2 Answers 2


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 $d\omega'(e_i,e_j)=L_{e_i}(\omega'(e_j))-L_{e_j}(\omega'(e_i))-\omega'([e_i,e_j])$.

  • $\begingroup$ I think there is one thing that you still need to prove. You have found that there exists some integral submanifold $Y\subset X$ with $T_x Y = V_x$ for any $x$, but not that this $Y$ is (at least locally) the image of a section of $p:X\rightarrow U$. In order to do this I guess you have to prove that at every point $p_*:V_x\rightarrow T_{p(x)} U$ is an isomorphism and then apply the inverse function theorem? $\endgroup$
    – G. Gallego
    Commented Jul 5, 2020 at 10:26
  • $\begingroup$ I guess my last statement is true since the form $da+\omega' a$ in $U\times GL_r$ is a pullback of a form in $U$, and $V_x$ is its kernel? $\endgroup$
    – G. Gallego
    Commented Jul 5, 2020 at 10:37
  • $\begingroup$ @G.Gallego: sorry I left so many gaps in the explanation; yes, the inverse function theorem, as you say. $\endgroup$
    – Ben McKay
    Commented Jul 5, 2020 at 10:39

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 $\Omega'=d\omega' + \omega' \wedge \omega'$.

  • $\begingroup$ Been trying to do this myself I get confused about something.To apply the frobenius theorem we need the functions $f$ to be defined a priori right ? I just don't see where that $f$ comes from. $\endgroup$
    – Someone
    Commented Jan 13, 2021 at 9:55
  • $\begingroup$ @Idon'tknow $f$ is the solution to the PDE, and the Frobenius theorem precisely gives you the existence of such $f$. In Spivak’s, my $f$ is denoted by $alpha$. $\endgroup$
    – G. Gallego
    Commented Jan 13, 2021 at 12:13
  • $\begingroup$ Hm okay, but then in your notation what is Spivak's $f$ ? I am just kinda confused since we want to determine $A$, but $A$ and $f$ are not defined a priori, so that $F$ is not defined a priori.@G. Gallego $\endgroup$
    – Someone
    Commented Jan 13, 2021 at 22:17
  • $\begingroup$ Spivak’s $f$’s are my $F$’s. You do not want to determine the $A$’s. The $A$’s are the components of the connection $1$-form in the initially chosen arbitrary frame $s’$. That is, they are already determined. $\endgroup$
    – G. Gallego
    Commented Jan 14, 2021 at 11:56
  • $\begingroup$ Maybe it can be confusing: what I now call $f$ is what in the original post I called $a$. Those $a$’s (now $f$’s) are the “variables” of the differential equation. These are “frame change functions” from the arbitrary initial frame $s’$ to a frame giving a flat structure (that is, a frame where the connection form vanishes). $\endgroup$
    – G. Gallego
    Commented Jan 14, 2021 at 12:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.