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Is every real vector bundle over the circle necessarily trivial? If yes - could you please point to a reference. If no - what are sufficient conditions?

I am particularly concerned with the case of a smooth map $\gamma:S^1\rightarrow Q$ and the vector bundle $\gamma^* TQ$.

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The Moebius band is a non-trivial vector bundle: it does not have a non-zero section. – Mariano Suárez-Alvarez Sep 7 '10 at 14:49
... and if you embed the circle in the M\"obius band as the zero section, you get a nontrivial bundle as the pullback of a tangent bundle. In general, the pullback of a tangent bundle is non-trivial iff the loop is orientation-reversing. – algori Sep 7 '10 at 15:07
See also:… – Qfwfq Sep 7 '10 at 15:21
Although @Mariano's and @algori's comments do make (the main part of) this question "no longer relevant", I disagree with the votes to close this question. Rather, Mariano should leave his comment as an answer. Moreover, the question may have more content: "what are sufficient conditions" could have an interesting answer, especially for a young researcher starting to get a feel for the area. – Theo Johnson-Freyd Sep 7 '10 at 19:12
Theo, in deference to your request, I am not voting to close, but "young researcher starting to get a feel for the area" is a grossly inaccurate description. Perhaps, "a student learning about vector bundles for the first time" (and who can't be bothered to read a textbook on the subject) is closer to the mark. – Victor Protsak Sep 8 '10 at 2:37
up vote 6 down vote accepted

In the spirit of Theo's comment, I'll say something about sufficient conditions.

A real vector bundle over the circle is trivial if and only if it is orientable. I discussed this a bit here.

The main point is that up to isomorphism, every real vector bundle over the circle is either trivial, or the Whitney sum of a trivial bundle with the Mobius bundle. The latter is not orientable.

The other answers at the question I linked to above may also be helpful to you.

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For example, for (real) line bundles this is very easy: without loss of generality we may assume that local trivilialization charts are $(0,1)\times \mathbb{R}^1$ and that the transition functions are $\pm 1$. Then by going over the circle counterclockwise we may "fix" all transition functions to be 1 (by changing the local trivialization frame $e\to-e$ if need be), except possibly for the last one -- thus there are only two line bundles. – Paul Yuryev Sep 7 '10 at 21:14
More generally, $G$-bundles on $S^n$ are classified by the elements of $\pi_{n-1}(G):$ after trivializing over two enlarged hemispheres, the transition function becomes a map from the equatorial $S^{n-1}$ into $G.$ – Victor Protsak Sep 8 '10 at 2:40

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