The flat connection on a trivial vector bundle

Prescribe $\omega =0$ in the constant frame. By [F1] the resulting derivative is precisely $du$ componentwise. For a scalar $f$, $X(f{u}^a)=X(f)u^a+fX(u^a)$ proves the section Leibniz identity, and $(fX)(u^a)=fX(u^a)$ proves direction-linearity. Since the components of $e_a$ are constant, every ${\mathrm{\nabla }}_X{e}_a$ is zero.

Applying the formula twice to a section gives the components of ${\mathrm{\nabla }}_X{\mathrm{\nabla }}_{Y}s-{\mathrm{\nabla }}_Y{\mathrm{\nabla }}_{X}s-{\mathrm{\nabla }}_{[X,Y]}s$ as $X(Y{u}^a)-Y(X{u}^a)-[X,Y]u^a=0$, by the defining commutator of vector fields. This is the stated direct meaning of flatness. For rank zero the formula is the unique zero operator; rank one gives $\mathrm{\nabla }u=du$. An empty base or a zero-dimensional base causes no exception. The product frame is given, so no choice of trivialization for an arbitrary bundle is involved.