I've trying to figure this out for a while and I am finally desperate enough to post to stack exchange
In do Carmo's Differential Forms and Applications Proposition 2 on pg 92 (this is the proof that the Gaussian curvature is well-defined, independent of choice frame and coframe). I post the statement and proof for context:
Proposition 2 Let $M^2$ be a Riemannian manifold of dimension two. For each $p\in M$, we define a number $K(p)$ by choosing a moving [orthonormal] frame $\{e_1,e_2\}$ around $p$ and setting $$ d\omega_{12}(p):=-K(p)(\omega_1\wedge\omega_2)(p). $$ [Here $\{\omega_1,\omega_2\}$ is the coframe associated to $\{e_1,e_2\}$.] Then $K(p)$ does not depend on the choice of frames, and is called the Gaussian curvature of $M$ at $p$.
Proof. Let $\{\bar{e}_1,\bar{e}_2\}$ be another moving [orthonormal] frame around $p$. Assume first that the orientations of the two moving frames are the same. Then $$ \omega_{12}=\bar{\omega}_{12}-\tau. $$ [Here $\tau=fdg-gdf$, where $f$ and $g$ are differentiable functions such that $f^2+g^2=1$; this was shown in a earlier lemma -- Lemma 4 on pg 90.] Since $\tau =fdg-gdf$, $d\tau =0$, hence $d\omega_{12}=d\bar{\omega}_{12}$ [this is the part I don't understand; I will elaborate afterwards]. It follows that $$ -K\omega_1\wedge\omega_2=d\omega_{12}=d\bar{\omega}_{12}=-\bar{K}\bar{\omega}_{1}\wedge\bar{\omega}=-\bar{K}\omega_1\wedge\omega_2 $$ hence $K=\bar{K}$, as we wished.
If the orientations are opposite, we obtain $$ d\omega_{12}=-d\bar{\omega}_{12},\hspace{.2 in}\omega_1\wedge\omega_{2}=-\bar{\omega}_1\wedge\bar{\omega}_2 $$ and the same conclusion holds.
This is slightly embarrassing since I have been working with differential forms for a few years now, but I don't understand why $d\tau=0$. In my line of thinking (interpreting $fdg$ and $gdf$ as wedge products between 0-forms and 1-forms and writing out in full detail) \begin{align*} d\tau &=d(fdg-gdf)\\ &=df\wedge dg+f d^2g-dg\wedge df-gd^2f=2df\wedge dg\\ &=df\wedge dg+f\cdot 0-dg\wedge df-g\cdot 0\\ &=df\wedge dg-dg\wedge df\\ &=2df\wedge dg \end{align*} which is not zero, unless there is some extra information about $f$ and $g$ I don't know about. It does seem like Differential Forms and Applications has quite a few typos, so I was thinking it was supposed to be $fdg+gdf$, but I went through the work where this "$\tau$" first popped up, and it seems like this is the correct form to be working with.
Does anyone have any insight on what could be going wrong here? Thanks
From $f^2+g^2=1$, you have $fdf=-gdg$, and thus $$f^2df\wedge dg=0\\ g^2df\wedge dg=0$$ Thus $$0=(f^2+g^2)df\wedge dg=df\wedge dg$$