Let $\pi : E \to X$ be a smooth rank $k$ vector bundle on a manifold $X$ (I don't think my question depends upon the stipulations on the bundle, but I've just chosen smooth in case I'm incorrect). By definition, for every $x \in X$, there exists an open set $U \subseteq X$, with $x \in U$, such that $\pi^{-1}(U)$ is diffeomorphic to $U\times\mathbb{R}^k$; we say that $U$ trivialises the bundle.

The prototypical example of a smooth vector bundle on a manifold $X$ is the tangent bundle $TX$. For this bundle (in fact any tensor bundle), any coordinate neighbourhood on $X$ trivialises the bundle. My question is whether this happens for every vector bundle.

Let $\pi : E \to X$ be a smooth rank $k$ vector bundle on a manifold $X$. Does every coordinate neighbourhood on $X$ trivialise the bundle?

The only way I think this can fail is if there is a bundle such that, for a given coordinate neighbourhood $U$, you must pass to a smaller open set $U' \subset U$ in order to trivialise the bundle.

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    $\begingroup$ As student points out in his comment below, the answer is yes or no according as you consider that the target of a coordinate map is a ball or an arbitrary open subset of $\mathbb R^n$. $\endgroup$ – Georges Elencwajg Sep 8 '12 at 17:45
  • $\begingroup$ On the other hand, the comment "any coordinate neighborhood on $X$ trivializes the bundle" implies that the OP considers the target to be a ball. $\endgroup$ – Jason DeVito Sep 12 '12 at 19:58

Yes. Your question is the same as asking if there exist any non-trivial vector bundles on $\mathbb{R}^n$; there are not any (more generally there are not non-trivial vector bundles on contractible spaces).

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    $\begingroup$ I'm not sure the question immediately reduces to existence of non-trivial vector bundles on $\mathbb{R}^n$, because a coordinate neighborhood is only required to be homeomorphic to an open set, not necessarily a ball or $\mathbb{R}^n$. What if you choose a coordinate neighborhood homeomorphic to a non-contractible open set? $\endgroup$ – student Sep 8 '12 at 17:09
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    $\begingroup$ I guess I was working with the wrong notion of "coordinate neighborhood." If you allow non-contractible open sets, then the answer to OP's question is no (just take an open subset of $\mathbb{R}^2$ with a non-trivial line bundle on it, e.g. an open annulus), as the manifold, with the whole thing being a coordinate neighborhood. $\endgroup$ – user29743 Sep 8 '12 at 22:06
  • $\begingroup$ @countinghaus: You say that my question is the same as asking if there exists a non-trivial vector bundle on $\mathbb{R}^n$. Could you please explain why this is the case in your answer? $\endgroup$ – Michael Albanese Sep 11 '12 at 14:28
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    $\begingroup$ @MichaelAlbanese Again it depends if a coordinate neighborhood needs to look like $\mathbb{R}^n$ or just some open subset. If we mean the former, then $U$ and $\mathbb{R}^n$ admit the same isomorphism classes of smooth vector bundles (they are diffeomorphic). If we mean the latter, my answer is not correct (see student's comment above and my response to it). $\endgroup$ – user29743 Sep 12 '12 at 20:06

No, not every co-ordinate neighborhood trivialize a bundle. Trivial counterexample, take $X$ as the co-ordinate neighborhood of every point then the claim is that all vector bundles are trivial.

The point is given any point $x \in X$ you can trivialize the bundle in some open set around $x$. So the trivialization doesn't work globally it only works locally on (possibly very small open sets).

The Möbius strip (without boundary), seen as a vector bundle over the circle is perhaps the simplest non-trivial vector bundle. Try finding a open neighborhood of every point and you will see that those open neighborhoods do not extend to the whole space.

The idea of neighborhood is meant to convey something local (i.e. happening near a point). So any statement valid in neighborhoods are not meant to be global.


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