Manifolds - 1D distribution on $S^2$ I'm trying to find a one dimensional smooth distribution on $\mathbb{S}^2$ - i.e. a map $p \mapsto D_p$, where $D_p$ is a one dimensional subspace of $T_p\mathbb{S}^2$, so that $D_p$ is locally generated by a smooth vector field. 
I know that such a distribution cannot be globally generated by a single vector field, since every smooth field on $\mathbb{S}^2$ vanishes. I can come up with various smooth vector fields on $\mathbb{S}^2$, e.g. $(1-z-x^2,-xy,x-xz)$, $(1-x^2, -xy, -xz)$, or $(0,-z,y)$. I thought I could patch these together somehow (partition of unity, maybe) to get a distribution on $\mathbb{S^2}$, but I haven't succeeded. Now I'm wondering if such a distribution exists. 
I'm trying this as part of a homework exercise. Any pointers would be appreciated. 
 A: Such a distribution cannot exist.
Let $S_g$ be a smooth closed connected surface of genus $g$, and $p\rightarrow D_p$ be a one dimensional distribution. Then $D_p$ is locally generated by a smooth vector field but, as you said, you might not be able to orient the lines $D_p$ in a coherent way to have a well defined smooth vector field on $S_g$ representing $D$.
So,


*

*Assume that the distribution comes from a vector field. Hence your surface admits a non-vanishing smooth vector field and its Euler Characteristic is $0$ due to the Poincaré-Hopf theorem. 

*if the distribution does not come from a vector field, then there exists a $2$-fold cover $\tilde S$ of $S$ and a lift $\tilde D$ of $p\mapsto D_p$ such that we can orient the lines of $\tilde D$ in the tangent space of $\tilde S$. This new distribution now comes from a non-vanishing vector field on $\tilde S$ so $0=\chi(\tilde S)=2\cdot \chi(S_g)$. 
In both cases, the Euler Characteristic of $S_g$ has to be $0$ i.e. $S_g$ is a torus.
In particular, there is no one-dimensional distribution on a $2$-sphere.
