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:

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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.

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    $\begingroup$ Formatting tip: use markdown code for non-math formatting, instead of using MathJax. $\endgroup$ – Zev Chonoles Mar 9 '13 at 4:42
  • $\begingroup$ Thanks for improving the post. $\endgroup$ – Cass Mar 9 '13 at 4:49
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    $\begingroup$ I think your last two paragraphs are identifying complementary ways of thinking about the same issue. What forces nearby points to be identified with $E_0$ in similar ways is that their respective parallel transport maps proceed along similar paths. If you try to do this construction with a non-contractible manifold, you will no longer be able to guarantee this: you'll be forced to, at some point, choose paths that traverse very different paths but come to nearby points. At that point, there's no longer any reason that the identification with $E_0$ should be continuous, let alone smooth. $\endgroup$ – Micah Mar 9 '13 at 4:51
  • $\begingroup$ @Micah That's good. You should make it into an answer. $\endgroup$ – Neal Mar 9 '13 at 6:51
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    $\begingroup$ Possible duplicate of A fiber bundle over Euclidean space is trivial. $\endgroup$ – Alex M. Oct 5 '18 at 6:09

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 $E_{x}$.

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    $\begingroup$ At the point in Nicolaescu's book at which the question is asked, the only vector bundles which we know for sure have flat connections are trivial bundles, so I would not be able to simply assume that E has a flat connection. Is there an easy way of telling when the bundle has a flat connection? Does it only depend on the manifold? $\endgroup$ – Cass Mar 9 '13 at 6:21
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    $\begingroup$ Also, I don't think this solves my problem. You're saying that parallel transport depends on the homotopy type of the path. Ok, but in my situation with the torus, I've already fixed my paths. So fixing a different set of paths may give me a different way of identifying fibers with the fiber over $0$, but I need to know why none of these ways would be smooth. $\endgroup$ – Cass Mar 9 '13 at 6:27
  • $\begingroup$ The problem is not that parallel transport depends on the path -- the problem is that in a non-contractible manifold there's no way to continuously choose paths for each point that all end at some fixed chosen point -- because that's exactly what it means for it to be contractible. $\endgroup$ – Pedro Mar 15 '16 at 21:52
  • $\begingroup$ Flat connection does not always exist. See link. You can identify fibers with the fiber over 0 for individual fiber. But you can't do it in a consistent way for the whole manifold. In the torus case for which flat connection exists, if you Travel along a non trivial loop and identify parallel transport with the fiber at 0, at last you will find isomorphism over the base point is not well-definied (not continuous)for the reason santharo explained. $\endgroup$ – Dai Apr 23 '17 at 21:41

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