In the smooth setting, we define a connection in a principal $G$-bundle $(P, \pi, M)$ as a smooth assignment to each point $p \in P$ of a subspace $H_p P$ of $T_p P$ such that the following two conditions hold:
- $T_p P \cong V_p P \oplus H_pP$,
- $\delta_{g^*}(H_pP) = H_{p \cdot g}P$,
where $V_pP = \{ \tau \in T_pP | \pi_* \tau = 0\}$ is called the vertical subspace of $T_pP$ and $\delta_g(p) = p \cdot g$ denotes the right action of $G$ on $P$.
There is an equivalent formulation based on differential forms. Given an element $A \in \mathrm{Lie}(G) \cong T_e G$, we can define a vector field $X^A$ defined by
$$ X_p^A(f) = \frac{d}{dt}_{\Big|t=0} f(p \cdot \mathrm{exp}(tA)). $$
A connection form $\omega$ is a $\mathrm{Lie}(G)$-valued one-form on $P$ such that the following hold:
- $\omega_p(X^A) = A, \quad \forall p\in P, \forall A \in \mathrm{Lie}(G)$,
- $\delta_g^*\omega = \mathrm{Ad}_{g^{-1}}\omega, \quad \forall g \in G$.
I want to prove the following theorem : There is a one-to-one correspondence between connections and connection forms.
Given a connection form $\omega$, the hint given is to set $H_pP = \mathrm{ker}(\omega)$. I do not see why this gives a connection.
Best Answer
Hint: The main point that seems to be missing in your descirption is that the map $A\mapsto X^A_p$ defines a linear isomorphism from the Lie algebra of $G$ onto the subspace $V_pP$ (which usually is called the vertical subspace and not the horizontal subspace). This is just an infinitesimal version of the fact that the orbites of the principal right action $\delta$ are the fibers of $G$. Using this, you conclude that $\omega_p$ restricts to a linear isomoprhism on $V_pP$ and then the hint you already got should help.