In Cartan for Beginners, problem 2.4.3, the problem is that if $M$ is flat, show that there exist coordinates $x_1,x_2$ and an orthonormal adapted frame $(e_1,e_2,e_3)$ such that $\omega_1=dx_1, \omega_2=dx_2$.
My problem is with the hint given, which says that $\omega_1^2$ can be written as $du$ for some function $u:\mathcal{F} \rightarrow \mathbb{R}$ which I understand for a $M$ flat, and then it says, "Since $\omega_1^2$ is not semi-basic for the projection to $M$....
I'm trying to do this book self-study and I just can't figure out how they got that statement, or whether it's related to $M$ being flat, or is something I was supposed to have picked up previously about the connection forms.
Could somebody please explain it?
This is universally true of connection forms. As you change frame (move along the fiber), the transformation law has a $a^{-1}da$ term, which is vertical. I assume you can now use $u$ to find the appropriate frame (all locally)?