Let's have a principal $G$ bundle $P \xleftarrow{\triangleleft G} P \xrightarrow{\pi} M$. At some point $p \in P$, let's consider the tangent space $T_p P$. We define the vertical tangent space as $V_p P \equiv ker(\pi^*)$ where $\pi^*: T_p P \rightarrow T_{\pi(p)} M$ is the pushforward of the projection $\pi: P \rightarrow M$. The data of a connection now chooses a subspace $H_p P$ such that $H_P \oplus V_p P = T_p P$ and $H_p \cap T_p = \{ 0 \}$. We now wish to show that this data is equivalent to having a Lie algebra valued one form one-form (the connection one-form) $\omega: T_p P \rightarrow \mathfrak g$, where $\mathfrak g$ is the Lie algebra of the Lie group $G$.
I believe the rough sketch of determining $\omega$ from $H_p P$ is as follows:
- Create an operator $\pi_{vert}: T_p P \rightarrow V_p P$ that projects onto the horizontal subspace. We can do this by eg. choosing bases for $H_p P, V_p P$ and then keeping only the components of the input vector that are in $V_p P$.
- We show that $V_p P$ is isomorphic to $\mathfrak g$, thereby letting us reinterpret $\pi_{vert}$ as $\pi_{vert}: T_p P \rightarrow V_p P \simeq \mathfrak g$.
I believe that one can embed $\mathfrak g$ into $V_p P$ by means of defining a map $i_p: \mathfrak g \rightarrow V_p P$ given by defining a curve $c^a_p: I \rightarrow P$
$$ i_p: \mathfrak g \rightarrow T_p P; i_p(a) = c^a_p \\ c^a_p: I \rightarrow P; c^a_p(t) \equiv \exp(p \triangleleft ta) $$
One can show $c^a_p$ the curve, interpreted as an element of the tangent space, does live in $T_p P$, and in fact lives in $V_p P$, and all that. (Intuitively, it lives in $V_p P$ because all the points on $c^a_p(t)$ lie in the fiber of $p$. Thus when being projected down onto $M$ by $\pi$, the curve becomes the constant curve $(\pi \circ c^a_p)(t) = \pi(p)$ and thus computes a zero directional derivative.)
What I don't know how to prove is that $i_p: \mathfrak g \rightarrow V_p P$ is surjective, and that thus $V_p P \simeq \mathfrak g$. I feel like I don't have any tools that let me relate $V_p P$ and $\mathfrak g$. I tried futzing around with dimensions, but was unable to come up with anything convincing. I'd appreciate all help in showing this.