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I'm trying to understand the proof of this version of the fundamental lemma of Riemannian geometry.

Let $\pi : \mathcal{F}_{on}(M) \rightarrow M $ the orthonormal frame bundle of an n-dimensional Riemannian manifold $M$ and let $\eta$ ∈ $\Omega^1(\mathcal{F}_{on}(M), \mathbb{R}^n )$ be the tautological form. Then there exists a unique $\mathfrak{so}(V)$-valued 1-form $\theta \in \Omega^1(\mathcal{F}_{on}(M), \mathbb{R}^n )$ such that $$ d\eta= \theta \wedge \eta $$ or equivalently, if we call $\theta= (e_i \otimes \eta^j) \otimes \eta^i_J$ where $e_i$ is the dual of $\eta^i$ we have $$ d \eta^i=\eta^i_j \wedge \eta^j $$ $$ \eta^i_j+\eta^j_i=0 $$ The proof is the following:

On $M$, let $\bar{\eta}$ be an orthonormal coframe, giving a local trivialization of $\mathcal{F}_{on}(M)$ such that $\eta = g^{−1}\pi^∗\bar{\eta}$ for some $O(\mathbb{R^n})$-valued function $g$ on the frame bundle. Then $$ d\eta = −g^{−1}dg ∧ η + g^{−1}\pi^∗d\bar{\eta}. $$ So, there exist 1-forms $\alpha^i_{j}$ on the frame bundle such that $d\eta^i=\alpha^i_j \wedge \eta^j$ [...]

(source : Cartan for beginners, second edition lemma 3.1.4 )

i don't understand why this statement is true. How is the term $g^{−1}\pi^∗d\bar{\eta}$ handled?

Edit

As suggested by Ted Shifrin in the comments i looked up to a comprehensive introduction to differential geometry and i found It a more "beginners friendly" book

Marco
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  • What is your source for this? – Deane Aug 29 '23 at 00:22
  • @Deane Cartan for beginners lemma 3.1.4 – Marco Aug 29 '23 at 00:56
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    It must be the 2016 edition. In the 2003 edition that I bought recently it is Lemma 2.6.5 and the proof is a bit different. – Kurt G. Aug 29 '23 at 17:52
  • @KurtG. you are right, i quoted the wrong edition in the comment. Any idea about why this is true? because i think it is widely assumed also in the exercises. – Marco Aug 29 '23 at 17:55
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    I am not quite sure if you ask about Ivey & Landsberg 2003 Lemma 2.6.5 or 2016 Lemma 3.1.4 which seems to have given a slightly different proof. In case you are asking about moving frames in general: $d\eta^i=\eta^i_j\wedge \eta^j$ is Cartan's first structure equation. You can see it in action here. This page is also highly recommended. – Kurt G. Aug 29 '23 at 18:14
  • @KurtG. thank you for the links, my doubt was about why it looks like the term $g^{-1}\pi^* d\bar{\eta}=\gamma^i_j \wedge g^{-1}\pi^*(\bar{\eta})$, for some one form $\gamma^i_j$. I think it has something to do with the trivialization of $\mathcal{F}{on}(M)$, but i can't figure out why does it holds. The book seems to define the 1-connection form without assuming that there is an affine connection, but only using the tautological form of $\mathcal{F}{on}(M)$ – Marco Aug 29 '23 at 18:42
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    That's definitely not the notation they use in my older edition of the book so I cannot comment. Much of the bundle stuff I still need to practice. – Kurt G. Aug 29 '23 at 18:47
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    So, working downstairs on $M$, what is $d\tilde\eta$? We’re then going to pull back by $\pi$. – Ted Shifrin Aug 30 '23 at 02:52
  • @TedShifrin $d \bar{\eta}=f_{ij} \bar{\eta}^i \wedge \bar{\eta}^j$ for some function. When i pull back i should obtain $\pi^*(f_{ij}\bar{\eta}^i) \wedge \eta^j$. Is it correct? – Marco Aug 30 '23 at 12:41
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    Correct except for missing $g$. – Ted Shifrin Sep 01 '23 at 04:43
  • @TedShifrin thank you very much. I was overcomplicating things – Marco Sep 01 '23 at 15:07
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    You might find the exposition in volume 2 of Spivak a bit more explicit on this stuff. Although I know the authors of the book you’re working on, I do not know the book. You might also look at Jeanne Clelland’s new book on moving frames. – Ted Shifrin Sep 01 '23 at 17:50

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