Let $\pi:P\rightarrow M$ be a principal bundle and let $\omega\in \Omega(P;\mathfrak{g})$ be a one form satisfying
$\omega(\sigma(X))=X$ and $R_g^*\omega=\text{Ad}_{g^{-1}}\circ\omega$
Then $H_p:=\ker\omega_p$ is a connection on P. That is,
1) $H_p$ depends smoothly on $p$
2) $T_p P=H_p \oplus V_p = H_p\oplus \ker \pi_{*,p}$
3) $(R_g)_*H_p=H_{pg}$
I was wondering how to show properties 1) and 2) above? I am able to prove property 3). For property 2)... is the below argument missing a crucial step? Write $X\in T_p P$ as
$X=X-\omega^i(X)\sigma(V_i)+\omega^i(X)\sigma(V_i)$
Where $(V_i)$ is a basis for $\mathfrak{g}$ and writing $\omega$ as $\omega=\omega^iV_i$ and $\sigma:\mathfrak{g}\rightarrow V_p$ is the map taking $X\in \mathfrak{g}$ to its fundamental vector field.
It can be shown that $\omega(X-\omega^i(X)\sigma(V_i))=0\,$ and thus, $X-\omega^i(X)\sigma(V_i)\in\ker \omega_p=H_p$.
Furthermore, it can be shown that $\sigma(Y)\in V_p$ for any $Y\in\mathfrak{g}$. So doesn't the above decomposition of $X$ prove that any vector $X\in T_p P$ can be written as a sum of vectors in $H_p$ and $V_p$, thus $T_p P=H_p \oplus V_p$.
I've just started studying this things so there are some things which aren't quite clear to me too but I'll try to give you some ideas:
$(i)$ $H_p\cap V_p=\{0\}$.
Recall we have an isomorphism $V_p\longrightarrow \mathfrak{g}$. If $v\in V_p$ is non-zero then $\omega_p(v)\neq 0$ for the image of $v$ by $\omega_p$ coincides with the image of $v$ by the isomorphism $V_p\longrightarrow \mathfrak{g}$. This shows $$V_p\setminus\{0\}\cap H_p=V_p\setminus\{0\}\cap \textrm{Ker}(\omega_p)=\phi\Rightarrow V_p\cap H_p=\{0\}.$$
$(ii)$ Hint for showing $T_pP=H_p+V_p$:
Take $X\in T_p P$ and associate a vertical vector field $\tilde{X}$ on $P$. Then $\tilde{X}_p\in V_p$. Since $$X=(X-\tilde{X}_p)+\underbrace{\tilde{X}_p}_{\in V_p},$$ it suffices showing $X-\tilde{X}_p\in \textrm{Ker}(\omega_p)$.
I'll try to improve the proof later.