The problem comes from Liviu Nicolaescu's book Lectures on the Geometry of Manifolds. He asks the reader to prove that any vector bundle $E$ over $\mathbb{R}^n$ is trivializable. The idea he gives is to fix a connection $\nabla$ on $E$ and use the parallel transport to identify fibers over nonzero points with the fiber over $0$.
Searching around, I've found proofs online that proceed in this way, using specifically the parallel transport over lines through the origin. I've also found a proof sketch on page 15 of John D. Moore's Lectures on Seiberg-Witten Invariants, showing that any vector bundle over a contractible manifold is trivializable:
This sketch also uses the parallel transport along lines.
The trouble I'm having has to do with the smoothness of this procedure. Perhaps it follows from the theory of differential equations that fibers over points which are close together will be identified with $E_0$ in a similar (smooth) way. I can accept that, even if I don't fully understand (my background in differential equations is basically nil).
But then I still struggle with the question of why this doesn't work for any path connected manifold. For example, given a bundle over the torus, say, why can't I fix a connection, fix a point $0$, fix a collection of curves starting at $0$ and indexed by their ending point $m$, and then use the parallel transport to identify all fibers with the fibers over $0$? What breaks down here that doesn't break down in the special case of $\mathbb{R}^n$ with lines? Thanks in advance for any help.
When your manifold is not contractible, the parallel transport depends on the path you choose. If you take a flat connection, it only depends on the homotopy class of the path you choose. The parallel transport is always smooth because you solve localy a second degree differential equation...
If you want to see it in the case of the torus: take a vector bundle $E$ and a flat connection (with non trivial holonomy) you see that if you take a small loop $ \gamma$ around a given point $x$, the parallel transport will give the identity of $E_{x}$ but if you follow the meridian or the longitude you will have a non trivial Automorphism of $E_{x}$.