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So, I want to show that a $n$-sphere $S$ is parallelizable iff it has trivial tangent bundle. For "$\Leftarrow$" I would like to take a trivialization $\varphi:S\times \mathbb{R}^n\rightarrow TS$ and consider the vector fields $v_i(x):=\varphi(x,e_i)$.

It seems to me like these vector fields are independent in every $x\in S$, but how can I deduce this exactly? The proof should somehow use that $\varphi$ is a homeomorphism, right? I don't see how to get from "homeomorphism" to linear independence.

Thanks for answers!

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1) The equivalence you mention has nothing to do with spheres: it is valid for any differential manifold. 2) The crucial point which solves your problem is that $\phi$ restricts at every point of the manifold to a vector space isomorphism between the fiber of the trivial bundle and that of the tangent bundle . – Georges Elencwajg Jan 15 '12 at 12:45
I thought that this is the definition of a parallelizable manifold. – HbCwiRoJDp Jan 15 '12 at 14:22
By the way, all spheres are not parallelizable. Actually, all but finitely many of them are not parallelizable. – HbCwiRoJDp Jan 15 '12 at 14:23
Thank you, Georges. That's what was missing! @ Karatug: Yes, I'm aware of Adam's theorem, that the only parall. spheres are in dimension 1,3,7. The definition I used: M is parall. if there exist n linear independent cross-sections. I just wanted to find out why this is equivalent to your definition. – user20388 Jan 15 '12 at 14:28

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