Vector field from Lamination. Let $S$ be a smooth closed (i.e. compact without boundary) surface. A geodesic lamination on $S$ is a nonempty closed subset of $S$ which is a disjoint union of geodesics. Suppose $\alpha$ is a geodesic lamination on $S$ such that for every $p\in S$ there is a geodesic of $\alpha$ containing $p$. 
Can we construct a smooth vector field in $S$ using $\alpha$? 
I was trying to use the tangents of the geodesics but I am not sure how to give the orientation on them.
P.S.: If necessary assume that $S$ is a hyperbolic surface.
Thanks in advance.    
 A: I just ran across this old problem and thought it would be good to give a definitive answer.
As the comments suggest the problem doesn't quite make sense, but there is one way for it to make sense: 

Does there exist a global vector field which is nowhere zero on the lamination and tangent to the lamination?

The answer is no. In fact, a positive answer to the previous question would imply a positive answer to the next question: 

Does there exist a vector field defined only on the lamination which is nowhere zero and is tangent to the lamination?

The answer to this question is also no. 
To see why, the question can be reformulated as an orientation problem. Given a geodesic lamination $\Lambda$ in a hyperbolic surface $S$, define its tangent bundle $T\Lambda$ to be the subset of $TS$ consisting of all $(p,v)$, $v \in T_p S$ such that $p \in \Lambda$ and $v$ is tangent to $\Lambda$. This is a 1-dimensional vector bundle over $\Lambda$. Mimicking the usual definition of orientability of a bundle, we define $\Lambda$ to be orientable if $T\Lambda$ is orientable, meaning that $T\Lambda$ has a continuous choice of orientation; and since this is a 1-dimensional vector bundle it is equivalent to the second question above. 
To get a counterexample, one uses a method pioneered by Nielsen and developed further, in a constructive manner, by Thurston. Thurston constructed examples $\Lambda \subset S$ where $S$ is a compact hyperbolic surface, such that if one lifts $\Lambda$ under the universal covering map $\mathbb H^2 \to S$, obtaining a geodesic lamination $\tilde\Lambda \subset \mathbb H^2$, there exists a triple of points $p,q,r \in \partial\mathbb H^2$ such that each of the three hyperbolic lines $\overline{pq}$, $\overline{qr}$, $\overline{rp}$ are leaves of $\tilde\Lambda$. The ideal triangle with vertices $q,p,r$ is called a Nielsen principal region of $\tilde\Lambda$.
If there existed an orientation on $\Lambda$ then it would lift to an orientation on $\tilde\Lambda$, and so each of the three lines $\overline{pq}$, $\overline{qr}$, $\overline{rp}$ would be oriented. We may assume, by reversing the orientation if necessary, that $\overline{pq}$ is oriented from $p$ to $q$, and now we work our way around the triangle by continuity. Using that the lines $\overline{pq}$ and $\overline{qr}$ approach each other exponentially close in the direction of $q$, and projecting to $S$ and using a continuity and compactness argument, one proves that $\overline{qr}$ is oriented from $r$ to $q$. Continuing around the triangle in the same fashion, next one concludes that $\overline{rp}$ is oriented from $r$ to $p$, and then that $\overline{pq}$ is oriented from $q$ to $p$. But that's a contradiction because we started with $\overline{pq}$ being oriented from $p$ to $q$.
