I am wondering if there are necessary and sufficient conditions under which an one-dimensional subbundle of $TM$ has a nowhere vanishing vector field.


More precisely let $M$ be a compact smooth manifold. 


a. When dose there exist a one-dimensional (smooth or continuous) subbundle $L\subset TM$?


b. If $L\subset TM$ is a continuous/smooth line subbundle of $TM$, does there exist a nowhere vanishing continuous/smooth section $X:M\to L$? If so, the euler characteristic of $L$ should be zero.


This is related to the partially hyperbolic system $f:M\to M$ and $TM=E^s\oplus E^c\oplus E^u$. I am curious if there is a 'center flow' if $\dim E^c=1$. 




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To Ryan: Am I right to say the following about your answers: 


1. If there exists an 1-dimensional subbundle $L$ of $TM$, then $\chi(M)=0$. This is independent of the case whether $M$ is orientiable or not.
2. If $M$ is orientable, then there always exists an orientable 1-dimensional subbundle $L$ of $TM$.


Another question is, when is an 1-dimensional subbundle $L\subset TM$ orientable? Is it sufficient to assume that $M$ is orientable?




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Thank you all. I did not formulate some questions properly. What I really mean is: 


I. For a given line bundle $L\subset TM$, what is the obstruction for $L$ beging orientable? (or equivalently trivial according to Georges)


For example let $L\_{\mathbb{C}}$ be a complex line bundle over a complex manifold $M$, if the top Chern class $c\_1(L\_{\mathbb{C}})$ does not vanish, then $L\_{\mathbb{C}}$ can not be trivial. Is there some similar results in the real case?


II. Is there an example such that $M$ is orientable and has a non-orientable line bundle $L\subset TM$?



nowhere vanishing vector field on a manifold
A bundle is orientable if and only if its first Stiefel-Whitney class is 0 (one can see the first Stiefel-Whitney class as the function $w\_1: H\_1(M)\rightarrow \mathbb{Z}\_2$ which associate to a loop the sign of the determinant of the monodromy).


As mentionned by Ryan, if a line bundle is non-orientable then there is a two sheeted cover of $M$ which orients $L$ (the covering correspond exactly to the index two subgroup $ker(w\_1)$) this implies that if an oriented bundle admits a 1-dimensionnal sub-bundle the its Euler class has to be $0$ (regardless if the bundle is the tangent bundle or not).


Finally one can easily see that $T(T^2)$ is the sum of two non-trivial line bundle:


The canonical line bundle $\gamma$ on $\mathbb{R}P^1$ is non trivial but $\gamma\oplus\gamma^\*$ is (it is oriented, 2 dimensionnal and admit a section given by the trace map). Pulling back this bundle by the projection $T^2\rightarrow S^1$ you get a trivial bundle (hence the tangent bundle) on $T^2$ written as a sum of two non trivial line bundle ($w\_1$ is non zero on each summand).


