Intuition behind Non-Trivial Tangent Bundle

Question

I am fairly new to the notion of tangent bundle and vector bundles in general , and as a first glance intuitively I though that well for any manifold $M$ we would have that $TM\cong M\times \mathbb{R}^m$. As I looked more into the matter I found cases were this is not true however I can't have a good intuition behind the fact as to why this is not true . I would appreciate if someone could enlighten me on what can fail so that we dot not have that diffeomorphism without using very heavy machinery . I am just trying to understand the idea behind why this cannot be true in general .I guess the problem is that we only know that locally the tangent space at a point is the same as $\mathbb{R}^m$ and the problem would to do this globally but I am having trouble seeing where it fails. Thanks in advance.

Answer 1

There are plenty of fairly intuitive proofs of the Hairy Ball theorem, but I think it may be helpful in this case to paint an intuitive picture of why the tangent bundle of $S^2$ is non-trivial.

If you place any tangent vector at the North pole, and extend smoothly across the Northern hemisphere, to a tangent vector field, then do the same to the South pole and the Southern hemisphere, you will get a picture like this (looking down from above the North pole in both cases):

Notice that as you follow the red tangent field (defined on the equator) around the equator, on the Northern hemisphere the black arrow moves clockwise relative to the red arrow (looking from outside the sphere e.g. above), whereas on the Southern hemisphere the black arrow moves anticlockwise (looking from outside the sphere e.g. below) relative to the red arrow, as you follow the red arrow round the equator.

Thus they can never match up.