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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

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  • $\begingroup$ It is zero, for $d^2=0$ and the wedge product is antisymmetric. $\endgroup$ – Eduardo Longa Jul 27 '17 at 21:07
  • $\begingroup$ That's a very obscure way of thinking of $\tau$. As you can see from Lemma 3.3 in my differential geometry text, $\tau = d\theta$, where $\theta$ is the angle that you rotate one frame to get the other. (So, presumably, DoCarmo's $f$ and $g$ are $\cos\theta$ and $\sin\theta$. But, still, a bizarre way to think of it ...) So, of course $d\tau = d(d\theta) = 0$. $\endgroup$ – Ted Shifrin Jul 27 '17 at 22:45
  • $\begingroup$ @TedShifrin I thought about this as well. I came to the conclusion they were basically trig funcitions, but I had trouble justifying it. I wish I would've come across your text when I was trying to justify this days ago! $\endgroup$ – Blake Jul 27 '17 at 23:31
  • $\begingroup$ I should have said Lemma 3.3 of Chapter 3. If you're trying to learn moving frames style geometry from someone other than J.M. and Tom, you might try the new book of another Bryant student, Jeanne Clelland. It has some great stuff in it. That's a book I would love to write but never will :P $\endgroup$ – Ted Shifrin Jul 27 '17 at 23:33
  • $\begingroup$ @TedShifrin Thank you! I've always had trouble finding references for frames style geometry outside of Cartan for Beginners. $\endgroup$ – Blake Jul 27 '17 at 23:38
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From $f^2+g^2=1$, you have $fdf=-gdg$, and thus $$f^2df\wedge dg=0\\ g^2df\wedge dg=0$$ Thus $$0=(f^2+g^2)df\wedge dg=df\wedge dg$$

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  • $\begingroup$ Oh wow this is great. It was a deeper fact than I thought. Based on the way do Carmo justifies it I didn't expect it to use $f^2+g^2=1$. Thank you! $\endgroup$ – Blake Jul 27 '17 at 21:16

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