I'm learning the moving frame approach (with differential form) in surface theory from different books and papers (Cartan, O'Neill, Shifrin, Flanders and others).
Briefly: you define an adapted moving frame on the surface $(P, e_1, e_2, e_3)$, then you define the dual forms $\omega_i = dP\cdot e_i$ and the connection forms $\omega_{ij} = de_i \cdot e_j$. From Cartan's structure equations you find Gauss and Codazzi equations and other things. Next you prove the fundamental formula $d\omega_{12} = -Kd\sigma$ (where $K = detS$ is the gaussian curvature). Here in some texts follows two steps before arrive at theorem egregium:
1) You prove that if you consider another frame $(P, \bar e_1, \bar e_2, \bar e_3$) than $d\omega_{12} = d \bar \omega_{12}$
2) You prove that $\omega_{12}$ is the only 1-form which satisfy the two equations $d\omega_1 = -\omega_2 \wedge \omega_{12}$ and $d\omega_2 = \omega_1 \wedge \omega_{12}$ and so it is intrinsic
Are these steps necessary? I mean
1) It is not obvious that $d\omega_{12}$ does not depend on the frame since neither $K$ nor $d\sigma$ depend on the frame?
2) It is not obvious that $\omega_{12}$ is intrinsic since it's defined as $de_1\cdot e_2$?
Thanks in advance.
Best Answer
So, yes, we can use the extrinsic definition of $K$ (as the determinant of the shape operator), but then to infer the Theorema Egregium — i.e., that $K$ is intrinsic, i.e., dependent only on the induced metric and not on the particular embedding — we need to know that $d\omega_{12}$ is intrinsic. Note that the answer to your second question is no, as that dot product depends on the embedding of the surface in $\Bbb R^3$.