Let M be a 2-dimensional Riemannian manifold. Let $X_1,X_2$ be an orthonormal frame (w.r.t Riemannian metric) in an open set $U$. Let $(\omega_1,\omega_2)$ be the dual coframe to $(X_1,X_2)$.
a) Prove that there exists the unique 1-form $\omega_2^1$ such that $$d\omega_1=\omega_2^1\wedge \omega_2 \text{ and } d\omega_2=-\omega_2^1\wedge \omega_1$$
The form $\omega^1_2$ is called the canonical connection form of the frame $(X_1, X_2)$.
b) Let $\tilde{X_1},\tilde{X_2}$ be another orthonormal frame defining the same orientation on $U$. Let $\tilde{\omega_1}, \tilde{\omega_2}$ be the corresponding dual coframe. Let $\tilde{\omega}_2^1$ be the canonical connection form of the frame ($\tilde{X}_1,\tilde{X}_2$). Prove that $d\omega_2^1=d\tilde{\omega}_2^1$
c) Let $K$ and $\tilde{K}$ be two functions on U such that $$d\omega_2^1=K\omega_1\wedge \omega_2 \text{ and } d\tilde{\omega_2^1}=\tilde{K} \tilde{\omega}_1\wedge \tilde{\omega}_2$$
Show that $K=\tilde{K}$.
Solution:
a) Since $\omega_1, \omega_2$ form a coframe, $$\omega_2^1=f_1\omega_1+f_2\omega_2, \text{ where } f_1=\omega_2^1(X_1), f_2=\omega_2^1(X_2)$$
$d\omega_1=\omega_2^1\wedge\omega_2 \implies d\omega_1(X_1,X_2)=\omega_2^1(X_1)$
$d\omega_2=-\omega_2^1\wedge\omega_2 \implies d\omega_2(X_1,X_2)=\omega_2^1(X_2)$
Therefore $\omega_2^1=d\omega_1(X_1,X_2)\omega_1+d\omega_2(X_1,X_2)\omega_2$.
This shows the existence and uniqueness of the canonical connection form. Further, it is easily seen that the conditions are satisfied.
b) Since $X_1,X_2$ is a frame $$\tilde{X}_1=a_1X_1+a_2X_2 \text{ and } \tilde{X}_2=b_1X_1+b_2X_2$$
By orthonormality, $\langle\tilde{X_i},\tilde{X}_j\rangle=\delta_{i,j}$. We get $a_1^2+a_2^2=1=b_1^2+b_2^2$ and $a_1b_1+a_2b_2=0$. Assume $a_1=\cos\theta$ $a_2=\sin\theta$ and $b_1=\sin\phi$ $b_2=\cos\phi$. We get $\theta=-\phi$. So, $$\tilde{X}_1=\cos\theta X_1+\sin\theta X_2 \text{ and } \tilde{X}_2=-\sin \theta X_1+\cos\theta X_2; \text{ where } \theta \text { is a smooth function on U}$$
Similarly, $\tilde{\omega_i}(\tilde{X}_j)=\delta_{i,j}$ gives $$\tilde{\omega}_1=\cos\theta \omega_1+\sin\theta \omega_2 \text{ and } \tilde{\omega}_2=-\sin \theta \omega_1+\cos\theta \omega_2; $$
I don't know how to continue further. Any help is appreciated.
Thanks