向量丛平坦联络等价于可积分布？相关线性联络性质求证

Alright, let's tackle this problem connecting flat linear connections to integrable horizontal distributions—let's start by grounding ourselves in the definitions we're working with, then walk through both directions of the equivalence.

First, let's restate the setup clearly for reference:

• We have a vector bundle $E \xrightarrow{\pi} M$, where $\dim M = n$ and $\text{rank } E = r$.
• A linear connection $D$ on $E$ defines a rank-$n$ horizontal distribution $\mathcal{H}E$ on the total space $E$, with the key property: for any section $\sigma$ of $E$,
  $$(D\sigma)p = 0 \iff \text{im},\sigma{*,p} = \mathcal{H}_{\sigma(p)}$$
• The vertical subbundle $\mathcal{V}E$ is defined by $\mathcal{V}\xi E = \ker \pi{*,\xi}$ (tangent spaces to the fibers), and we have the direct sum decomposition $TE = \mathcal{V}E \oplus \mathcal{H}E$ (this is the Ehresmann联络 structure induced by $D$).

Our goal is to prove:

A linear connection $D$ is flat if and only if its horizontal distribution $\mathcal{H}E$ is Frobenius integrable.

First, recall the core definitions:

• A connection $D$ is flat if its curvature tensor $R^D$ vanishes. The curvature is defined for all vector fields $X,Y \in \Gamma(TM)$ and sections $\sigma \in \Gamma(E)$ by:
  $$R^D(X,Y)\sigma = D_X D_Y \sigma - D_Y D_X \sigma - D_{[X,Y]} \sigma$$
• A distribution $\mathcal{H}E$ is Frobenius integrable if the Lie bracket of any two horizontal vector fields is also horizontal: for all $H_1, H_2 \in \Gamma(\mathcal{H}E)$, $[H_1, H_2] \in \Gamma(\mathcal{H}E)$.

Since $TE = \mathcal{V}E \oplus \mathcal{H}E$, the differential $\pi_$ restricts to an isomorphism from $\mathcal{H}E$ to $TM$. This means every vector field $X \in \Gamma(TM)$ has a unique horizontal lift $^hX \in \Gamma(\mathcal{H}E)$ such that $\pi_(^hX) = X$.

Take two horizontal lifts $^hX$ and $^hY$. By the properties of $\pi_$ (it's a Lie algebra homomorphism), we have:
$$\pi_([^hX, ^hY]) = [\pi_(^hX), \pi_(^hY)] = [X,Y]$$
This tells us $[^hX, ^hY]$ can be written as $^h[X,Y] + V$, where $V \in \Gamma(\mathcal{V}E)$ is a vertical vector field. We need to show $V = 0$ when $R^D = 0$.

Now, consider a parallel section $\sigma$ (i.e., $D\sigma = 0$). By the defining property of $\mathcal{H}E$, $\sigma_{,p}(T_pM) = \mathcal{H}{\sigma(p)}$, so $\sigma(X) = ^hX|{\sigma(M)}$ for all $X$. For such a section, the curvature formula simplifies to:
$$R^D(X,Y)\sigma = -D{[X,Y]}\sigma$$
But $D_{[X,Y]}\sigma$ corresponds exactly to the vertical difference between $\sigma_([X,Y])$ and $^h[X,Y]|{\sigma(p)}$. If $R^D = 0$, then $D{[X,Y]}\sigma = 0$, so $\sigma_([X,Y]) = ^h[X,Y]|_{\sigma(p)}$.

Since $[\sigma_*X, \sigma_Y]|{\sigma(p)} = \sigma([X,Y])$, this means $[^hX, ^hY]|{\sigma(p)} = ^h[X,Y]|{\sigma(p)}$, so $V|_{\sigma(p)} = 0$. Because flat connections admit local parallel sections covering every point of $E$, $V$ must vanish everywhere. Thus $[^hX, ^hY] = ^h[X,Y]$, which is horizontal—so $\mathcal{H}E$ is integrable.

Now assume $\mathcal{H}E$ is integrable: for any horizontal vector fields $H_1, H_2$, $[H_1, H_2]$ is horizontal. For vector fields $X,Y \in \Gamma(TM)$, their horizontal lifts $^hX, ^hY$ have a Lie bracket that's horizontal, and since $\pi_*([^hX, ^hY]) = [X,Y]$, the uniqueness of horizontal lifts gives us:
$$[^hX, ^hY] = ^h[X,Y]$$

To show $R^D = 0$, take any section $\sigma \in \Gamma(E)$ and vector fields $X,Y$. Recall that $D_X\sigma$ is the vertical component of $\sigma_(X) - ^hX|{\sigma(p)}$ (since $\pi(\sigma_*(X) - ^hX) = X - X = 0$).

Using the Lie bracket identity for the tangent map $\sigma_*$:
$$[\sigma_*X, \sigma_Y] = \sigma_[X,Y] + \text{vertical component}$$
But $\sigma_*X = ^hX|{\sigma(M)} + (D_X\sigma)p$ (vertical term), so expanding the left-hand side:
$$[\sigma*X, \sigma*Y] = [^hX, ^hY] + \text{vertical terms}$$
Since $\mathcal{H}E$ is integrable, $[^hX, ^hY] = ^h[X,Y]$, so:
$$[\sigma_*X, \sigma_Y] = ^h[X,Y] + \text{vertical terms}$$
The right-hand side $\sigma_[X,Y] = ^h[X,Y] + (D_{[X,Y]}\sigma)p$. Equating the two expressions, the vertical terms must satisfy:
$$D_X D_Y \sigma - D_Y D_X \sigma = D{[X,Y]}\sigma$$
Substituting into the curvature definition gives $R^D(X,Y)\sigma = 0$ for all $X,Y,\sigma$—so $D$ is flat.

The curvature of a connection measures the "twist" between the Lie bracket of horizontal lifts and the horizontal lift of the Lie bracket. When this twist (a vertical vector field) vanishes, the horizontal distribution is integrable, and vice versa—this is exactly the equivalence between flatness and integrability here.

内容的提问来源于stack exchange，提问作者Louis