A surface connection form depends on its frame

Take $M={\mathbb{R}}^2$ with its standard flat Riemannian metric, its standard orientation, $U={\mathbb{R}}^2$, and the standard frame $E_1={\mathrm{\partial }}_x$, $E_2={\mathrm{\partial }}_y$, which is positively oriented and orthonormal. By [F3] each ${\mathrm{\nabla }}_{{\mathrm{\partial }}_i}{\mathrm{\partial }}_j$ vanishes on $U$, so $\omega (X)=g({\mathrm{\nabla }}_X{E}_1,E_2)=0$ for every $X$; that is, $\omega =0$ identically.

Put $\phi (x,y)=x\in C^{\mathrm{\infty }}(U)$ and define the smooth positive frame $E_1^{\mathrm{\prime }}=\cos \phi E_1+\sin \phi E_2$, $E_2^{\mathrm{\prime }}=-\sin \phi E_1+\cos \phi E_2$. It is orthonormal with the same orientation as $(E_1,E_2)$ because a rotation by $\phi$ preserves the metric and the orientation of the plane. The function $\phi$ is a global smooth angle lift of this rotation, so [F1] applies on all of $U$ and gives ${\omega }^{\mathrm{\prime }}=\omega +d\phi =dx$. Since $dx$ is not the zero form and $\omega =0$, the two frames have different connection forms; the asserted identity ${\omega }^{\mathrm{\prime }}=\omega$ fails.

The exterior derivative is nevertheless frame-independent here: by [F2] the structure form $-KdA$ equals $d\omega =0$ on the flat plane. Since $d\phi =dx$, direct differentiation gives $d{\omega }^{\mathrm{\prime }}=d(dx)=0$. Thus the failure of frame-independence is exactly the exact-form ambiguity $d\phi$ in [F1], and the invariant object is $d\omega$, not $\omega$ itself.