I'm reading about connections on principal bundles from Kobayashi-Nomizu, and I have a question about the "local picture." I explain the context (taken from chapter II.1 of K-N) and my question below.
Let $\pi: P \to M$ be a principal $G$-bundle. Then, by Proposition $1.1$ of K-N, a $\mathfrak{g}$-valued $1$-form $\omega$ is a connection form (i.e. its kernel is a $G$-connection on $P$) iff
- $\omega(A^*) = A$, for $A \in \mathfrak{g}$, where $A^*$ is the fundamental vector field associated to $A$;
- $(R_g)^* \omega = \textrm{Ad}(g^{-1}) \omega.$
Now, if $\{(U_\alpha, \psi_\alpha)\}$ a cover of $M$ by local trivializations, where $\psi_\alpha: \pi^{-1}(U_\alpha) \to U_\alpha \times G$ is of the form $p \mapsto (\pi(p),\varphi_\alpha (p))$, where $\varphi_\alpha(p \cdot g) = \varphi_\alpha(p) \cdot g$. Let $\psi_{\alpha \beta} : U_{\alpha \beta} \to G$, $x \mapsto \varphi_\alpha(\psi_\beta^{-1}(x,e))$, and let $\sigma_\alpha : U_\alpha \to P$, $x \mapsto \psi_\alpha^{-1}(x,e).$
Then, set $\theta_{\alpha \beta} := \psi_{\alpha \beta}^* \theta$, where $\theta$ is the fundamental form of $G$ and $\omega_\alpha := \sigma_\alpha^* \omega$. Proposition 1.4 in K-N says that \begin{equation} \omega_\beta = \textrm{Ad}(\psi_{\alpha \beta}^{-1})\omega_\alpha + \theta_{\alpha \beta} \end{equation} on $U_{\alpha \beta}$, and that if $\{\omega_\alpha\}$ is a family that satisfies this, there's a unique $G$-connection form $\omega$ such that $\omega_\alpha = \sigma_\alpha^* \omega$ for all indices $\alpha.$
I was able to prove that, given $\omega$, the family $\{\omega_\alpha\}$ really satisfies these equations, but not the other way around.
Here's what I tried: given $\{\omega_\alpha\}$, I set $$\omega|_{\pi^{-1}(U_\alpha)} := \pi^* \omega_\alpha + \varphi_\alpha^* \theta.$$ By the condition that this family satisfies, $\omega$ is well-defined, and since $d \pi(A^*) = 0$, it is easy to check that $$\omega(A^*) = A.$$However, I have no idea how to prove $(R_g)^* \omega = \textrm{Ad}(g^{-1}) \omega.$ The given condition only tells me the relationship between $\omega_\alpha$ and $\omega_\beta$, but nothing about them individually.
EDIT: here is a computation of the $\varphi_\alpha^* \theta$-part - I obtained that it is $0$, but it is quite possibly wrong.
\begin{align} ((R_g \circ \varphi_\alpha)^* \theta)_p (X) &= \theta_{\varphi_\alpha(p) \cdot g} ((dR_g)_{\varphi_\alpha(p)} \circ (d \varphi_\alpha)_p (X)) \\ &= (dL_{g^{-1} \varphi_\alpha(p)^{-1}})_{\varphi_\alpha(p) g} \circ (dR_g)_{\varphi_\alpha(p)} \circ (d \varphi_\alpha)_p (X) \\ &= (dL_{g^{-1}})_g \circ (dL_{\varphi_\alpha(p)^{-1}})_{\varphi_\alpha(p)g} \circ (dR_g)_{\varphi_\alpha(p)} \circ (d \varphi_\alpha)_p (X) \\ &= (dC_{g^{-1}})_e \circ (dL_{\varphi_\alpha(p)^{-1}})_{\varphi_\alpha(p)} \circ (d \varphi_\alpha)_p (X) \\ &= (dC_{g^{-1}})_e \circ (dc_e)_p (X) = 0, \end{align} where in the last line I used the fact that $L_{\varphi_\alpha(p)^{-1}} \circ \varphi_\alpha = c_e: \pi^{-1}(U_\alpha) \to G.$
For the $\pi^* \omega_\alpha$-part, I get $$(\pi \circ R_g)^* \omega_\alpha (X) = \omega_\alpha ((d \pi \circ dR_g)(X)) = \omega_\alpha ( d \pi(X)),$$since $\pi \circ R_g = \pi.$ Since I don't know anything about $\omega_\alpha$, I don't know what to do from here.