Question: I am trying to compute the pullback of the Maurer-Cartan form for a Darboux Frame of a surface in $\mathbb{E^3}$, and I seem to have gotten a bit stuck. Here is the question:
Let $F: U^2 \rightarrow E(3)$ be the Darboux lift of a map $f:U^2 \rightarrow \mathbb{E}^3$ where $E(3)$ is the Euclidean group. Compute $F^*(\omega^2_1)$.
My attempt: I am going to use $\bar{\omega}$ to stand for $F^*(\omega)$ here to avoid cumbersome notation. Since we are using a Darboux frame, we know that $$\bar{\omega}^3_1 = \kappa_1 \bar{\omega}^1\\ \bar{\omega}^3_2 = \kappa_2\bar{\omega}^2.$$
Taking the derivatives of these functions and using the Cartan structure equations we can get that $$(\kappa_2-\kappa_1)\bar{\omega}^2_1\wedge\bar{\omega}^1 = d\kappa_2\wedge\bar{\omega}^2\\ (\kappa_2-\kappa_1)\bar{\omega}^2_1\wedge\bar{\omega}^2 = d\kappa_1\wedge\bar{\omega}^1$$
We can then rearrange the above equations to get the following: $$((\kappa_1-\kappa_2)\bar{\omega}^2_1 + d\kappa_1)\wedge \bar{\omega}^1 + ((\kappa_1-\kappa_2)\bar{\omega}^2_1 + d\kappa_2)\wedge \bar{\omega}^2 = 0$$
So, using Cartan's lemma, we know that there exist functions $h_{ij}$ such that $$\begin{pmatrix}(\kappa_1-\kappa_2)\bar{\omega}^2_1 + d\kappa_1\\ (\kappa_1-\kappa_2)\bar{\omega}^2_1 + d\kappa_2\end{pmatrix} = \begin{pmatrix}h_{11} & h_{12}\\ h_{12} & h_{22}\end{pmatrix}\begin{pmatrix}\bar{\omega}^1\\ \bar{\omega}^2\end{pmatrix}.$$
However, solving this system to determine my functions $h_{ij}$ is giving me a bit of trouble. I feel like I should be able to explicitly determine these functions, but the method to do it is escaping me currently. Any help on this would be appreciated. Thank you!
I figured it out! You just need to decompose these into components. Let's note that $$\bar{\omega}^2_1 = \lambda_1\bar{\omega}^1 + \lambda_2\bar{\omega}^2$$ and similarly, decompose $$d\kappa_i = \kappa_{i,1}\bar{\omega}^1 + \kappa_{i,2}\bar{\omega}^2$$ Then we can reduce the equations $$(\kappa_2-\kappa_1)\bar{\omega}^2_1\wedge\bar{\omega}^1 = d\kappa_2\wedge\bar{\omega}^2\\ (\kappa_2-\kappa_1)\bar{\omega}^2_1\wedge\bar{\omega}^2 = d\kappa_1\wedge\bar{\omega}^1$$ to $$(\kappa_2-\kappa_1)\lambda_2\bar{\omega}^2\wedge\bar{\omega}^1 = \kappa_{2,1}\bar{\omega}^1\wedge\bar{\omega}^2\\ (\kappa_2-\kappa_1)\lambda_1\bar{\omega}^1\wedge\bar{\omega}^2 = \kappa_{1,2}\bar{\omega}^2\wedge\bar{\omega}^1.$$ These equations are easy to reduce to get that $$\bar{\omega}^2_1 = \frac{\kappa_{1,2}\bar{\omega}^1 + \kappa_{2,1}\bar{\omega}^2}{\kappa_1-\kappa_2}.$$