Let $G$ be a Lie group and $S=[0,1]^2$. Let $\omega$ be a connection $1$-form on the trivial principal $G-$bundle $P=S\times G$ over $S$. Let $(x_1,x_2)$ be coordinates on the base $S$.
We can choose a trivialization of $P$ which is parallel with respect to the $x_1-$direction over the $x_1-$axis (aka $[0,1]\times\{0\}$) and which is parallel in the $x_2-$direction at every point of $S$.
Then $\omega=A_1dx_1+A_2dx_2$ where $A_i\in\mathfrak{g}$ satisfy $A_2=0$ everywhere and $A_1|_{[0,1]\times\{0\}}=0$.
Denote the horizontal lift of a vector field $\chi$ on the base to $P$ as $\tilde{\chi}$. Let $\sigma$ be the vertical projection on $P$ given by $\omega$.
Let $\chi_i=\partial_{x_i}$ then $\tilde{\chi_2}=\partial_{x_2}$ and $\tilde{\chi_1}|_{[0,1]\times\{0\}}=\partial_{x_1}$.
How to show that $\sigma([\tilde{\chi_1},\tilde{\chi_2}])=-\partial A_1/\partial x_2$ at $(0,0)\in S$?
I mean, it doesn't make sense since $\omega|_{(0,0)}=0$, is it not? Where is my misunderstanding?
This is taken from page $73$ of the book "Gauge Theory and the Topology of Four-Manifolds" by Friedman and Morgan. Thank you.
The easiest way to calculate curvature is to note that the curvature satisfies the Cartan structure equation, $F_A = d\omega + [\omega, \omega]$. Note the first term! Just because $\omega$ is zero at some point doesn't mean its derivative is.
In particular, we see that at $(0,0)$ where $\omega = 0$, that $F_A = d\omega = (\partial A_2/\partial x_1 - \partial A_1/\partial x_2) dx_1 \wedge dx_2$; and applying this to the 2-vector $(\partial/\partial x_1, \partial/\partial x_2)$, because $A_2 = 0$, we get what you desire.