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A manifold $M$ is said to be parallelizable if it admits $k$ linearly independent vector fields. I know that this is equivalent to the tangent space $TM$ being trivial. I am trying to show that $S^n\times \mathbb{R}$ is parallelizable, but have little idea how to start. One problem is that I'm not sure what the tangent space of this manifold looks like. I do know that I can write $T(S^n\times \mathbb{R}) = T(S^n)\times T(\mathbb{R})$, and I know that $S^n$ is not parallelizable in general.

I would appreciate any hints (rather than solutions).

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  • $\begingroup$ just for clarity (I'm a novice like you:) ) is $S^n \times \mathbb{R}$ the vector bundle to consider in the resolution or we have to consider the tangent bundle over it? $\endgroup$ Commented Mar 10, 2014 at 22:59

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Hint: Embed $S^n\times \mathbb{R}$ into a very very familiar space in a natural way. Then you can "see" a nonzero section.

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Hint: The sum of the tangent bundle of $S^n$ with the normal bundle of the embedding of $S^n$ in $\mathbb{R}^{n+1}$ is trivial.

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A tubular neighborhood of the sphere is parallelizable (it is an open set of an affine space). But a tubular neighborhood is diffeomorphic to the normal bundle, which is trivial. Thus the tubular nbhd is diffeomorphic to $\mathbb{S}^n\times\mathbb{R}$, hence the latter is parallelizable. Also (someone somewhere in this place) remarked that $(t,x)\mapsto e^tx$ is an open embedding of $\mathbb{S}^n\times\mathbb{R}$ into $\mathbb{R}^{n+1}$.

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