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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

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    $\begingroup$ You can't make a proof which works for every odd $n$, since it's only true for $n=1$, $n=3$ and $n=7$. en.wikipedia.org/wiki/Parallelizable_manifold $\endgroup$ Commented Feb 1, 2011 at 17:38
  • $\begingroup$ This 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 $S^7$. $\endgroup$ Commented Feb 3, 2011 at 0:44

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You want to show that the tangent bundle T(S^3) is a trivial bundle. It is a (not so hard) theorem that a vector bundle being trivial is equivalent to the existence of a (global) basis of sections. A section of the tangent bundle is another word for a vector field.

For S^1, considered as the unit circle in the complex plane, you probably have considered the vector field $z\mapsto iz$, which forms a basis. For S^3, you can do something similar, just like Mariano has described: now you consider S^3 as the unit sphere in the quaternions, and look at the vector fields $V_i:z\mapsto iz$ $V_j:z\mapsto jz$ $V_k:z\mapsto kz$

[Lack of reputation prevents me from being able to place this as a comment. I wanted to point out a similarity between the S^1 and S^3 case.]

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Identify the sphere with the set $S\subset\mathbb H$ of quaternions of norm $1$. Pick an ordered basis $(u_1,u_2,u_3)$ of the tangent space $T_1S$ to $S$ at the point $1$, and now consider the vector fields $$X_i:p\in S \mapsto pu_i\in T_pS,$$ where on the right hand side $pu_i$ is the quaternion product of the element $p$ with the quaternion $u_i$ (recall that one can identify the tangent space at $T_1S$ with a subspace of $\mathbb H$) You need to check that this makes sense: in particular, that $pu_i$ is, in fact, in the tangent space $T_pS$ to $S$ at $p$.

This construction gives you three non-zero vector fields $X_1$, $X_2$ and $X_3$. Check that at each point they are a basis. Voilà.

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It does not work for any odd dimension, if I recall correctly it will only work for 1,3 and 7 which correspond to real-division algebras...

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Actually, doesn't every S^(2n-1) admit the nowhere-zero tangent vector field:

(-x2,x1,-x4,x3,..,-x2n )? (using that S^(2n-1)={ x in R^2n , ||x||=1} )

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    $\begingroup$ Yes. But it doesn't admit a basis of nowhere zero vector fields. I.e. you can't find 2n-1 examples like this, which are everywhere independent. $\endgroup$ Commented Feb 3, 2011 at 12:03
  • $\begingroup$ are there any S^(2n-1) 's that have trivial bundles other than for S^3,S^5 and S^7 as Lie groups? Is there some general result in this area? $\endgroup$
    – user6600
    Commented Feb 4, 2011 at 8:44
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    $\begingroup$ Did you miss Hans Lundmark's comment? The only parallelizable S^n's are those with n=1,3,7. $\endgroup$ Commented Feb 4, 2011 at 14:19

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