how to show T(S^3) isomorphic to S^3 cross R^3? so can I say it for every odd dimension?I have shown it for n=1
Tangent Bundle on S^3
4
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differential-geometry
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3You can't make a proof which works for every odd $n$, since it's only true for $n=1$, $n=3$ and $n=7$. http://en.wikipedia.org/wiki/Parallelizable_manifold – 2011-02-01
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0This is a fact about Lie groups. You can use the group structure to translate a local trivialization around the manifold. However, it also works for $H$-spaces such as $S^7$. – 2011-02-03