How to tell the tangent bundle of $S^2$ from the bundle $S^2\times$ $y-z$ plane?

Asked Mar 20 '23 at 10:05

Active Mar 20 '23 at 17:29

Viewed 81 times

Intuitively, I would like to say that the tangent bundle of $S^2$ and the bundle ($S^2\otimes \rm{y-z\ plane}$) is different. By the latter I mean a product bundle, embedding in $R^3$ it is equivalent to attaching a yz plane at each point of $S^2$.

Is there a type of characteristic class that tells these two apart? The Pontryagin class for both is $0$ as it is $2d$ by $2d$.

edited Mar 20 '23 at 17:29

Deeponjit Bose

asked Mar 20 '23 at 10:05

Alex

3

The latter has a nonzero vanishing vector field, the former has not. – Didier Mar 20 '23 at 10:17
@Didier what is a nonzero vanishing vector field? You mean nowhere-vanishing section? Is there a type of characteristic class that can tell these apart? – Alex Mar 20 '23 at 11:14
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The euler class measures that. – Arctic Char Mar 20 '23 at 11:53
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@Alex Sorry, I meant a nowhere-vanishing vector field – Didier Mar 20 '23 at 13:00
@ArcticChar could you please elaborate? – Alex Mar 20 '23 at 13:33
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See here. If the bundle has a nowhere-vanishing section, it's euler class is zero. – Arctic Char Mar 20 '23 at 16:06
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As Didier hints, nonisomorphism follows from the Hairy Ball Theorem. In fact more is true: The total spaces of the bundles are not even diffeomorphic: https://math.stackexchange.com/questions/1323740/are-t-mathbbs2-and-mathbbs2-times-mathbbr2-different – Travis Willse Mar 20 '23 at 23:29

How to tell the tangent bundle of $S^2$ from the bundle $S^2\times$ $y-z$ plane?

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