Could someone explain to me why are line bundles over any real $n$-sphere trivial bundles ( $ n > 1 $ ) ?
Could someone tell me, in the case where $ n = 2k $ with $ k \in \mathbb{N}^* $ , why does the fact that a $n$-sphere admits a trivial line sub-bundle mean that it admits a section nowhere vanishing which doesn't exist ?
Have you any reference or any book treating that subject more deeply ?
Thanks a lot for your answers.
Here's one way to arrive at what you need to prove: (1) Any real line bundle is trivial if and only if it is orientable. (2) Any vector bundle on a simply connected manifold is orientable.
Now there are other ways to get this because you're working on $S^n$. You can trivialize a bundle on the sphere on each hemisphere (slightly thickened up), and so you need to see how to glue on the intersection of the two thickened hemispheres, which is homotopy equivalent to $S^{n-1}$. So you're looking at (up to homotopy) a map $S^{n-1}\to GL(1,\Bbb R) \cong \Bbb R^\times$, and such maps are trivial when $n>1$.
For your second question, it is badly phrased. You're referring to the tangent bundle of $S^n$. Generally, a nowhere vanishing section of a vector bundle gives a trivial line sub-bundle. (Take the line bundle to be spanned by the section.) As you noted, because $\chi(S^{2n}) = 2$, there can be no nowhere-vanishing vector field on $S^{2n}$.