Tangent Bundle on S^3 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
 A: 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. 
A: 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...
A: 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.]
A: 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} )
