How do connection 1-form and Ehresmann version of connections relate to each other?

Question

I am studying connections on abstract manifolds. So far, I have read several equivalent definitions but I can't establish the equivalence between them on my own.

The first definition is the Ehresmann connection that defines a connection on a manifold as a distribution of vector spaces completing the vertical space in the tangent space of the total space at each point.

The second definition defines a connection on a manifold as a covariant derivative, i.e. a map $$\nabla: \Gamma(E) \to \Gamma(T^*M\otimes E)$$ where $\pi: E \to M$ is a vector bundle and there is a version of the Leibnitz rule as follows: $$\nabla_X(fs)=df\otimes s + f \cdot\nabla_Xs$$ for any section $s$ and $f\in C^{\infty}(M)$.

I tried to write things in a chart to find out how the covariant derivation is induced from a given connection but I couldn't proceed forward. I think I haven't understood the definitions well. Would someone clarify how a distributional connection give us a connection one-form and how we can recover the horizontal space if we have a connection one-form?

Answer 1

Given a curve $\gamma$ and a section $X$ of $E$ defined at least along $\gamma$, we get a curve $\xi$ in $E$ defined by $\xi(t) = X(\gamma(t))$. We want to conflate the two notions of parallelism - that is, $\dot \xi \in H$ ($H$ our horizontal distribution) and $\nabla_{\dot \gamma} X = 0$.

Well, in one direction, we can just do it! Given a covariant derivative $\nabla$, we can define a corresponding $H$ as the set of all $\dot \xi$ satisfying $\nabla_{\dot \gamma}X=0$. To get a more concrete grasp on this, let's assume we have a chart $x^i$ and a local framing $e_\alpha$ of $E$ so that the covariant derivative takes the form $$\nabla_\dot\gamma X^\alpha = \dot\gamma^i X^\alpha_{,i} + \Gamma^\alpha_{i\beta}\dot\gamma^i X^\beta,$$ and we have coordinates $(x^i, e^\alpha)$ on the total space of $E$. ($e^\alpha$ is just the dual basis of $e_\alpha$.)

Since we can find $\xi$ with arbitrary initial position $\xi(0) = (p,X)$ and velocity $\dot \xi(0) = (v,W)$ in $E$, the horizontal subspace at $X \in E_p$ is (noting that the vertical component of the velocity is $W^\alpha = \dot\gamma^i X^\alpha_{,i}$)

$$ H_{(p,X)} = \left\{ v^i \frac{\partial}{\partial x^i} + W^\alpha \frac{\partial}{\partial e^\alpha} \Big|\ v \in T_p M,\ W^\alpha + \Gamma^\alpha_{i \beta} v^i X^\beta = 0 \right\}.$$

Solving this for $W^\alpha$ we see that for each element $(p,X) \in E$ and each direction $(p,v)$ in the base space we get a corresponding horizontal direction $v - \Gamma^\alpha_{i \beta}v^i X^\beta \partial/\partial e^\alpha$ in the total space. In particular, our coordinate system and framing gives us a basis

$$\frac{\partial}{\partial x^i} - \Gamma^\alpha_{i\beta}(p) X^\beta \frac{\partial}{\partial e^\alpha}$$

for $H_{(p,X)}$.

Note that we cannot reverse this process in general - locally we can choose arbitrary functions $\Gamma^\alpha_i$ on $E$ and get an Ehresmann connection with basis

$$\frac{\partial}{\partial x^i} - \Gamma^\alpha_i(p,X) \frac{\partial}{\partial e^\alpha}$$

which can only reduce to a covariant derivative/linear connection if $\Gamma^\alpha_i(p,X)$ is linear in $X$.