# Showing that a given vector bundle with connection is not trivial

Given the following exercise:

where d is the trivial connection.

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

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 $$(E, \tilde \nabla)$$?

• Do you know about curvature? – Ted Shifrin Dec 3 '18 at 21:59

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.