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Given the following exercise:

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where d is the trivial connection.

We defined an isomorphism between two vector bundles with connection in the following way:

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I'm not sure what I have to show. Can I use that any isomorphism between E and $M \times \mathbb{R}^2$ must be the identity and then show that the identity is not parallel? So that there exist $X_0 \in \Gamma (TM), \psi_0 \in \Gamma(E)$ such that: $d_{X_0} \psi_0 = \nabla_{X_0}\psi_0$?

If that's the case could I use the matrix of 1-forms and the standard frame $(e_1, e_2)$ on E to write out the connection $d - \nabla$ and then choose as $\psi_0$ the section $e_1$, which would yield a term that would not be equal to 0 for some $p \in M$.

For the second part of the exercise, we have shown that a vector bundle with connection is trivial if and only if there exists a parallel frame field. In the proof for this, it was shown how to construct an isomorphism given the parallel frame field. Can someone give a hint on how to construct a parallel frame field for $(E, \tilde \nabla)$?

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  • $\begingroup$ Do you know about curvature? $\endgroup$ – Ted Shifrin Dec 3 '18 at 21:59
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Here's a hint: If you have an orthonormal frame field $e_1,e_2$ for the bundle with connection form $\omega_{12}$ and rotate the frame field by considering $e_1' = \cos\theta e_1+\sin\theta e_2$, $e_2'=-\sin\theta e_1+\cos\theta e_2$ for some function $\theta$, then $\omega_{12}' = \omega_{12}+d\theta$. If $\omega_{12}=df$, then of course you can make $\omega_{12}' = 0$ by choosing $\theta=-f$. Then $e_1', e_2'$ will be parallel frame fields.

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