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In the proof for the following theorem:

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where we defined a vector bundle to be flat if it is locally trivial to $M \times \mathbb{R}^n$, so there exists a parallel local frame.

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The rest of the proof is clear to me, but I don't understand why $\tilde{\Phi} = \Phi F^{-1}$ is a local frame. I think the author somehow implicitly uses the diffeomorphism here, but I'm not even sure what $F^{-1}$ is here, I think this just bad notation and he really means the inverse of the the map in $GL(k, \mathbb{R}^k)$ right?

And why does every local frame have this form?

eager2learn
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  • "Locally trivial to $M\times\Bbb R^n$"? No, all vector bundles are locally trivial. You need locally constant transition functions. ... To answer your question, I presume he means that you take the inverse of the matrix in $GL(k,\Bbb R)$ (you're thinking inverse function, which is not intended). Indeed, all through differential geometry you see a connection form appearing as $g^{-1}dg$ or $dg,g^{-1}$. This is always the inverse of the group element. – Ted Shifrin Jul 03 '19 at 20:29

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