I know using natural coordinate to compute the relation ship between $M,g $ and $ (M,\phi^2g)$; using the orthornormal frame can also do the computation, both using the kozul formula.
However , i want to find the third different proof of the relation ship between the connecton form with respect ot conformal metric and the original metric.
what i have done is as follows:
$ M,e_1,\cdot\cdot\cdot e_n $ be the orthornormal frame of $ g $, $ M,,\omega^1 \cdot\cdot\cdot \omega^n $ be the orthornormal dual frame of $ g $ $ M,\frac{1}{\phi}e_1,\cdot\cdot\cdot \frac{1}{\phi}e_n $ be the orthornormal frame of $ \phi ^2g $, $ M,,\tilde{\omega}^1 \cdot\cdot\cdot \tilde{\omega}^n $ be the orthornormal dual frame of $ g $, we know that $$ \tilde{\omega}^i=\phi \omega^i $$ then we use the cartan's structure equartion,
$$d\omega^i=\omega_j^i\wedge \omega^j $$ $$d\tilde{\omega}^i=\tilde{\omega}_j^i\wedge \tilde{\omega}^j $$
so, we get
\begin{align} d\tilde{\omega}^i&=d(\phi \omega^i)\\ & =d\phi\wedge \omega^i+\phi d\omega^i\\ &=d\phi \wedge\omega^i+\phi \omega_j^i\wedge \omega^j\\ &=\tilde{\omega}_j^i\wedge \tilde{\omega}^j\\ &=\phi \tilde{\omega}_j^i\wedge \omega^j \end{align} thus ,we get \begin{align} \phi (\tilde{\omega}_j^i- \omega_j^i)\wedge \omega^j=d\phi \wedge \omega^i \end{align}
Now , I have used the cartan lemma, find the result is far away the correct answer. I have also using the formula action on the bivector $e_i \wedge e_j$ ,find alway lost the symmetric part
using the moving frame, what I what to find is :
$$\tilde{\omega}_j^i- \omega_j^i=? $$
recently , i can just find the two term of the right answer, lost one term. I have doubt that the moving frame cannot be used to solve this problem or I have made some mistakes.
Any answer is quite hepful and will be appriciated.I have suffered a lot.strong text