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do non-zero global section always exist in a manifold $M$? If $M$ is compact I think they do because taking a partition of unity $\rho_{\alpha}$ subordinated to a finite covering, and defining local sections $s_{\alpha}$ in this finite covering I can take $s:=\sum s_{\alpha} \rho_{\alpha}$ Is this argument right? I guess it is not true for general $M$ thanks

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    Non-zero global section of what vector bundle exactly? It's certainly not true for every vector bundle - consider the Mobius bundle over the circle. (You can draw it in $\mathbb{R}^3$). The problem is that your partition of unity argument doesn't guarantee things won't cancel out at some points.2012-08-28

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You do not need compactness nor even paracompactness (i.e. no partition of unity is necessary).
Take any nowhere zero continuous section of the vector bundle on an open trivializing subset $U$ for the vector bundle , multiply it by a continuous plateau function with compact support in $U$ and extend by zero to the whole manifold: this yields a non-identically zero continuous section of the vector bundle.

NB I have interpreted your question as asking for non-identically zero continuous sections:they always exist.
In general it is however impossible to find a nowhere zero section of an arbitrary vector bundle on an arbitrary manifold, as shown in other answers.
But sometimes it is possible: on a contractible manifold like $\mathbb R^n$ all vector bundles are trivial and thus they certainly admit of nowhere zero continuous sections.

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Absolutely not! Take the tangent bundle over a manifold. A globally defined non-zero section is a non-singular vector field. The Poincaré-Hopf Theorem relates the topology of your surface with the existence of non-singular vector field. Consider, e.g. a sphere, there are no continuous, non-zero vector fields on the sphere. This is called the Hairy Ball Theorem. This is true of any compact, orientable surface with non-zero Euler Characteristic.

For further reading, take a look at Chern Classes (in the complex case) and Stiefel–Whitney classes (in the real case).

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No, every even dimensional sphere is a counter-example (cf. Hairy ball theorem).

Moreover, a closed orientable manifold admits a nowhere zero section of its tangent bundle iff its Euler class (and therefore also its Euler characteristic) vanishes.

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In general it's not true.

For instance take the tangent bundle of $\mathbb S^2$. Then you can't have a non-vanishing vector field defined globally.