Why isn't every manifold parallelizable, given that $TM = \bigsqcup_{p \in M}T_pM$?

Question

My text that I am self studying from says that a manifold $M$ is parallelizable if it has a trivial tangent bundle which means that there is an isomorphism $\varphi:M\times \mathbb{R}^n\rightarrow TM$. However, I am a bit confused as to why we don't have this isomorphism for every manifold. From what I understand if $M$ has dimension $m$ then at every point $p$, $T_pM$ is isomorphic to $\mathbb{R}^m$. If $TM = \bigsqcup_{p \in M}T_pM$ then why wouldn't $TM$ always be isomorphic to $M\times \mathbb{R}^m$. Thank you!

Answer 1

Probably the most intuitively convincing example of non-parallelizability is the Hairy Ball theorem. If $TS^2$ were parallelizable one could assign smoothly varying non-vanishing vectors to every point on the sphere, e.g. $p\mapsto\varphi(p,(1,0))$ in your notation, but that's impossible because "you can't comb a hairy ball".

"Smoothly varying" is key here, we can always assign such vectors if smoothness and continuity are not required, but that would not be germaine for manifolds. Parallelizability exactly means that not just smoothly varying non-vanishing vectors but entire $n$-frames of them can be assigned globally.