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I read many quiestion about $TS^{2}\ncong S^{2}\times\mathbb{R}^{2}$ where the hint is use Hairy ball theorem and directly is done.

My question is: how do I proof that $TS^{2}\ncong S^{2}\times\mathbb{R}^{2}$ only with theory of vector bundles?

If there is $F:S^{2}\times\mathbb{R}^{2}\longrightarrow TS^{2}$ diffeomorphism then exists a one-one correspondences between $\Gamma(S^{2}\times\mathbb{R}^{2})$ the set of sections on $S^{2}\times\mathbb{R}$ and $\mathfrak{X}(S^{2})$ the vector fields on $S^{2}$ where the Hairy ball theorem appears.

Donyarley
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    The Hairy Ball Theorem isn't part of the theory of vector bundles? – Robin Goodfellow Sep 29 '14 at 16:14
  • See here why you can't just use characteristic classes, so that the general theory of vector bundles seems difficult to use for solving that specific problem. The difficulty is that the tangent bundle to the sphere is stably trivial: adding a trivial line bundle to $TS^2$ gives a trivial bundle. – Georges Elencwajg Sep 29 '14 at 19:14
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    @Georges: Although you admit as much in the comments to your link, I'm puzzled that you exempt the Euler class from the list of characteristic classes. – Ted Shifrin Sep 30 '14 at 23:36

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There is a discussion in Hatcher's Vector Bundles, where the tangent bundle is constructed explicitly, it has Euler class=2, which means that the planes are twisted twice as you travel once around the equator.

7779052
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