Dear  Joseph

Thank you very much for your answer.

I think there is no contradiction.

In fact TS^n  is a parallelizable manifold, for all n.

Because we need to prove that TTS^n is a trivial bundle over TS^n. But the later can deformation retracts to S^n. On the other hand the restriction of the second tangent bundle to the (zero section of TS^n)~S^n is isomorphic to TS^n + TS^n. The later is trivial bundle since the complexification of TS^n is a trivial bundle, the later statement is written iwithout proof n a very old paper by Swan. 

For a related discussion please see

https://mathoverflow.net/questions/151676/totally-non-parallelizable-manifold

Thank you for your attention to my question.

Yuri Ximenes Martins

I really appreciate your attention to my question. I am very happy you are interested in this question. Your idea is very interesting(using characteristic classes).

Since I asked this questions 2-3 years ago I do not remember the existence of a final answer to this question. I vaguely remember that (maybe) I saw an answer to this question somewhere(may be MO, MSE or RG. But I am not sure at all. May be I am mistaken, so the problem is worth of research.

Regarding embedding in Euclidean space, TS^n is orientable hence can be emmedded in dimension 4n-1. But your suggestion is very interesting:"What is the minimum possible dimension?? very interesting point1 thanks! I think more about it.

I have a remark (not an answer). $TS^n$ is diffeomorphic to the manifold of oriented lines (not necessarily passing through the origin) in $\mathbb{R}^{n+1}$. To see this, let $l$ be an oriented line in $\mathbb{R}^{n+1}$, and let $p \in l$ be the point on $l$ which is the closest to the origin (with respect to the Euclidean distance in $\mathbb{R}^{n+1}$). Let $v$ be a unit tangent direction to $l$, compatible with the orientation of $l$. Then $(v,p)$ define a point on $TS^n$, since $v$ and $p$ are orthogonal. This actually defines a diffeomorphism from the manifold of oriented lines in $\mathbb{R}^{n+1}$ and $TS^n$.

I think a generalization of the Borsuk-Ulam theorem may perhaps show that such a diffeomorphism is impossible. This is an idea.

Joseph Malkoun

Very interesting point! Thank you very much1