TITLE: global section vector bundle
QUESTION [8 upvotes]: 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

REPLY [8 votes]: 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.

REPLY [6 votes]: 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).

REPLY [6 votes]: 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.

REPLY [2 votes]: 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.