TITLE: Can I comb unoriented hair on a ball?
QUESTION [17 upvotes]: I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball.
(I am treating $S^2$ as a manifold without the ambient space $\mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.)
Is this still impossible if the hair is unoriented?
By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line.
That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)?
Geometrical intuition suggests that there is not, but that is far from proof.
I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.

REPLY [15 votes]: An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.