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Let $\varphi: X\longrightarrow \mathbb{R}^N$ be an submanifold immersed in $\mathbb{R}^N$, then everybody knows that $T\mathbb{R}^N\vert_X$= T(X)$\oplus N(X)$. It is clear that $T\mathbb{R}^N \vert_X$ is a trivial bundle over $X$. My question is: if we know that $N(X)$ is a trivial vector bundle over X does this implies that $TX$ is trivial?

I know that in the general case ($X$ a submanifold of a Riemannian manifold $M$) $N(X)$ trivial does not implies that $TX$ is trivial, my guess is that the answer to my question is NO, however the fact $T\mathbb{R}^N \vert_X$ is a trivial bundle and $N(X)$ is a sub bundle makes me feel that there might be a chance for $TX$ to be trivial.

Basic linear algebra shows that if {$v_1,..v_q$} is a global orthonormal frame of $N(X)$ then locally we can produce a local orthonormal frame for $TX$, however I think that there are problems if we try to construct a global frame for $TX$, however I am not sure if using that $T\mathbb{R}^N \vert_X$ is trivial we can construct such a global frame for $TX$.

Many thanks for you help math folks!!

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The normal bundle being trivial implies that the tangent bundle is stably trivial (which means, by definition, that its sum with some trivial bundle is trivial). This implies, for example, that all of its Stiefel-Whitney and Pontryagin classes vanish, but in general it does not imply that the tangent bundle is trivial.

The simplest example is $S^2 \subset \mathbb{R}^3$. The normal bundle is trivial because the outward-pointing normal is a trivialization of it, but the tangent bundle is still nontrivial; this is the hairy ball theorem. Slightly more generally, the normal bundle of the usual embedding of the closed orientable surface $\Sigma_g$ of genus $g$ in $\mathbb{R}^3$ is trivial (again the outward-pointing normal is a trivialization), but the tangent bundle of $\Sigma_g$ is trivial iff $g = 1$ (e.g. by computing its Euler class).

Qiaochu Yuan
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    It's worth noting that the existence of stably trivial bundles that aren't trivial implies that the operation of direct sum of vector bundles is not cancellative in the sense that $V_1 \oplus W \cong V_2 \oplus W$ does not imply $V_1 \cong V_2$. – Qiaochu Yuan Mar 05 '15 at 20:39