Why is not every vectorbundle trivial and we only need local trivialization?

Question

Why is should the trivialization only need to be local for vectorbundles? For example the tangent bundle. If the tangent space at a point p of a manifold M is identified with let's say $\mathbb{R}^n$ , why can't we just identify the total space of the vectorbundle with M x $\mathbb{R}^n$ ? My questions could be formulated as, why is not every vectorbundle a trivial one? Since I have the feeling that all the fibers can be identified with $\mathbb{R}^n$ and therefore you don't have only local trivialization, but everywhere.

Answer 1

The same reason not all manifolds are copies of $\mathbb{R}^n$ --- just because we're gluing things together to locally preserve structure, that doesn't mean the global result has that same structure. (If it did, then differential topology wouldn't be such an interesting area of study!)

Here are two examples of nontrivializable vector bundles.

1. Mobius band. This is a 1-dimensional real vector bundle over $\mathbb{S}^1$ and it is not orientable, hence not diffeomorphic to a cylinder (and therefore not trivializable).

2. Tangent bundle to $\mathbb{S}^2$. By the hairy ball theorem, this has no globally nonzero sections, hence it cannot be trivializable.