Trivilisations of Vector Bundles

Question

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.

Answer 1

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).

Answer 2

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.