I've been following Friedrich Schuller's Lectures on the Geometrical Anatomy of Theoretical Physics, and specifically on the lecture on parallel transports, I am trying to prove a theorem whose proof is on a problem sheet that unfortunately isn't available online.
The setup for it is that we have a principal G-bundle, with a connection one form $\omega$, and we want to find an explicit expression for the horizontal lift $\tilde\gamma$ in an Principal bundle of a curve $\gamma$ in the manifold. The strategy being to first start with an arbitrary curve $\delta$ in the principal bundle that projects down to $\gamma$ and via right action of a curve g in the Lie group, transform $\delta$ into $\gamma$. i.e $\tilde\gamma(t) = \delta(t) \triangleleft g(t)$
Then the statement of the theorem is that we locally solve for g by a first order ordinary differential equation:
$$ Ad_{g^{-1}(t)*}(\omega_{\delta(t)}(\delta'(t))) + \Xi_{g(t)} (g'(t)) = 0\tag{1}$$
Where $\Xi$ is the Maurer-Cartan form. The hint given was that proving it amounts to showing that the $\delta\triangleleft g$ satisfies the conditions for a horizontal lift.
My first thought was that this kind of looks similar to the equation for a local trivialization of a principal bundle $h^*\omega$
\begin{equation} \left(h^* \omega\right)_{(m, g)}(v, \gamma)=\left(\operatorname{Ad}_{g^{-1}}\right)_*\left(\omega^U(v)\right)+\Xi_g(\gamma), \end{equation}
but that equation involves a Yang Mills $\omega^U = \sigma^*\omega$ in place of the $\omega$, which also means a section is given. So instead I tried showing that $\delta\triangleleft g$ has the property of a horizontal lift where its tangent vector has no vertical component. Equivalent to this statement is that the vector is entirely horizontal, and therefore must be in the kernel of connection one-form $\omega$. So right now I'm trying to prove that
$$ \omega_{\delta\triangleleft g(t)}((\delta\triangleleft g)'(t)) = 0\tag{2} $$
and showing that it somehow is equivalent to equation 1. But I have no clue how to prove this or if it is the correct direction to go.