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I want to understand the correspondence between locally constant sheaves and vector bundles with flat connection on a manifold $X$.

Given a local system $\mathcal L$, it is clear how to define a vector bundle $E$ with locally constant transition functions. This allows us to define a connection $\nabla$ on the sheaf of smooth sections of $L$, i.e. on $\mathcal L \otimes \mathcal C^\infty_X$ it by demanding that $\nabla$ vanishes on $\mathcal L$, i.e. locally $\nabla = d$.
This connection is clearly flat, as $d^2=0$, and $\mathcal L = \ker \nabla$.

But what about the other direction. Assume we have a vector bundle $E$ which admits a flat connection $\nabla$. I read, that then $\ker \nabla$ defines a local system $\mathcal L$, that is we consider the sheaf of solutions to the differential equation $\nabla s=0$

How do I show that $\mathcal L$ is locally constant?
For any $x_0\in X$, we can find a neighbourhood $U$, where $\nabla = d + A$ for some matrix $A$ of $1$-forms.
Thus there, a section $s$ we have$$s=(f_1,\dots, f_n) \in \ker \nabla ~~~~~~\Leftrightarrow ~~~~~ ~ df_i +\sum_j a_{ij} f_j=0, ~~\forall i.$$ Assuming that for any initial value $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$, there exists a unique solution to the above differential equation, we get an isomorphism $\mathcal L(U) \to \mathbb C^n$, which is consistent with restrictions.

But how can I show this? In the texts I read, it was always said that it follows from Frobenius theorem and then the proof was finished without expanding on it. So here is my question:

Why does the differential equation $$df_i +\sum_j a_{ij} f_j=0, ~~\forall i$$ have a unique solution for any initial value $s(x_0)=(f_1(x_0),\dots,f_n(x_0))\in \mathbb C^n$.
Why does it follow from Frobenius theorem and how do I use flatness of the bundle?

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1 Answer 1

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The short proof is:

$\ker\nabla$, the choice horizontal space, is a distribution on $E$. It is involutive by flatness of $\nabla$ (flat horizontal lifts preserves Lie brackets), so it can be integrated by Frobenius theorem.

It seems that you are not very comfortable with this, so as an alternative, let's try to do it without Frobenius.

Since this is local, assume your base is $(-1,1)^n$, and let's state the flatness condition as $[\nabla_i,\nabla_j]=0$ (where $\nabla_i=\nabla_{\partial/\partial x^i}$). We have a the starting value $f_{(0)}\in E_0$ at $0$.

  • Integrating $\nabla_1f=0$ along $x^1$-direction is no problem, gets a horizontal section $f_{(1)}$ on $(-1,1)\times\{0\}^{n-1}$.

  • Next, for each $x^1$, we integrate $\nabla_2f=0$ along the $x^2$-direction with initial value $f_{(1)}(x^1)$ at $(x^1,0,\dots,0)$. This gives $f_{(2)}$ that is horizontal in $x^2$-direction, but for $x^1$-direction? Away from $x^2=0$ we haven't guarantee $\nabla_1f_{(2)}=0$. For that, the flatness condition comes to rescue. Since $[\nabla_1,\nabla_2]=0$, we have $\nabla_2\nabla_1f_{(2)}=\nabla_1\nabla_2f_{(2)}=0$, i.e. $\nabla_1f_{(2)}$ is independent of $x^2$. So $\nabla_1f_{(2)}=0$ at all points $(x^1,x^2,0,\dots,0)$.

  • Similarly $x^3,\dots,x^n$. QED

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  • $\begingroup$ Could you make the short proof a little bit longer? Basicaly everything I know is the definitions: a distribution is a collection of subspaces $\mathcal D_p\subset T_p M$, one for each $p\in M$. It is integrable, if there is a submanifold $N\subset M$, such that $T_p N =\mathcal D_p$. If $\omega$ is a 1-form with $\dim \ker \omega_x$ constant, then $\ker \omega_x$ is a distribution. $\endgroup$
    – abdul
    May 26, 2019 at 12:25
  • $\begingroup$ So why is $\ker \nabla$ a distribution? As $\nabla: \mathcal A^0(E) \to \mathcal A^1(E)$, $\ker \nabla \subset \mathcal A^0(E)$, but we would need $\ker \nabla \subset TM$ in order for it to be a distribution. Do we consider something like $\bigcap_{s \in \mathcal A^0(E)} \nabla(s)$ instead?. Also, after applying frobenius and obtaining a submanifold $N\subset M$, why is $\ker \nabla$ (considered as $\subset \mathcal A^0(E)$) locally constant? $\endgroup$
    – abdul
    May 26, 2019 at 12:26
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    $\begingroup$ It is a distribution on $E$, not on $M$. Recall $\nabla$ chooses a horizontal subspace $H_p\subset T_pE$ as its kernel, complementary to the vertical subspace $V_p=\ker(T_p E\to T_{\pi(p)}M)$ $\endgroup$
    – user10354138
    May 26, 2019 at 12:27
  • $\begingroup$ So after applying Frobenius, you get in particular submanifolds $N_i\subset E$, passing through $e_i\in\mathbb{C}^n\cong E_{x_0}$. If you use the $N_i$ to define a local trivialization of $E\to M$ at $x_0$ you get $\ker\nabla$ is indeed constant on the patch. $\endgroup$
    – user10354138
    May 26, 2019 at 12:59
  • $\begingroup$ Hm, so far I only know Frobenius theorem for vector fields. Do you know where I can find its generalisation to sections of arbitrary vector bundles? How do you get the $N_i$? Do you take a section $s_i$ such that $(s_i)_{x_0} = e_i$ and observe $[s_i,s_i]=0 \in \left<s_i\right>$. But can't I always do this? But still, if I have $N_i$ and use it to define local trivializations. Why should the sheaf of sections of $E|_{N_i}$ be locally constant. Couldn't I then take $N=E$ and in this way deduce that the sheaf of sections of every manifold are locally constant? $\endgroup$
    – abdul
    May 26, 2019 at 15:07

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