When considering the geometrical meaning of the curvature on principal bundle $\pi: P \rightarrow M$, let us consider a coordinate system ${x_μ}$ on a chart $U$. Let $V = ∂/∂_{x_1}$ and $W = ∂/∂_{x_2}$. Take an infinitesimal parallelogram $γ$ whose corners are $O = \{0,0,...,0\}, P = \{ε,0,...,0\}, Q = \{ε,δ,0,...,0\}$ and $R = \{0,δ,0,...,0\}.$ Lift it to the horizontal space $\tilde{\gamma}(t)$. If $\tilde{\gamma}(0)= u$, let $X,Y \in H_u P$ such that $\pi_* X= \epsilon \cdot V$ and $\pi_*Y =\delta\cdot W $. We can show that $[X,Y]$ is vertical.
1) How to prove that if the $[X,Y]$ is nonzero then the horizontal lift $\gamma(t)$ fails to close? Equivalently, prove that a closed horizontal lift leads to horizontal $[X,Y]$. You may argue using integral surface. I am unfamiliar with it so far.
2)"This failure is proportional to the vertical vector $[X,Y]$ connecting the initial point and the final point on the same fibre." Is it literally what it means, "connecting the initial point and the final point"?
I apologise if they are too elementary to you. I appreciate it if you can help.
Source:GEOMETRY, TOPOLOGY and PHYSICS by MIKIO NAKAHARA.
The commutator of two vector fields can be expressed as differentiating the flows generated by the vector fields, see the second equation under http://en.wikipedia.org/wiki/Lie_bracket_of_vector_fields#Flows_and_limits
Thus in your case $[X, Y]$ coincides with the derivative of your parallelogram with respect to the lengths $\epsilon$ and $\delta$ (maybe up to a factor of $1/2$). Thus the endpoint of the lifted parallelogram equals the initial point for all $\epsilon$ and $\delta$ if and only if (the vertical part of) $[X, Y]$ vanishes.
Yes $[X, Y]$ is indeed the vector between the initial and the end point, at least if you consider your whole discussion infinitesimally (i.e. take the limit of $\epsilon$ and $\delta$ going to zero). It is the usual definition of a tangent vector as an equivalence class of curves, in your case you have the representative $\bar{\gamma}$.