TITLE: Generalization of the hairy ball theorem.
QUESTION [5 upvotes]: The hairy ball theorem of states that there is no nonvanishing continuous tangent vector field on even dimensional n-spheres.
Can the hairy ball theorem be strengthened to say that there is no continuous tangent vector bundle of dimension 1 on even dimensional n-spheres?
Thanks

REPLY [2 votes]: Here's another approach which doesn't use characteristic classes.

Suppose $M$ has no connected double cover.  Then it has a non-vanishing vector field iff $TM$ admits a rank $1$ sub-bundle.

The hypothesis on $M$ is satisfied iff $\pi_1(M)$ has no subgroup of index $2$.  In particular, this applies to all simply connected manifolds.
Proof:  One direction doesn't depend on properties of $M$ at all: if $M$ admits a non-vanishing vector field $V$, then $\operatorname{span}\{V\}$ is a rank $1$ sub-bundle of $TM$.
Now we prove the more fun direction.  Suppose $L\subseteq TM$ is a rank $1$ sub-bundle.  Choose a background Riemannian metric on $M$ so that we may talk about lengths of vectors.  Let $\tilde{M} = \{(p,v)\in L: |v| = 1\}$.
Then $\pi:\tilde{M}\rightarrow M$ given by $\pi(p,v) = p$ is a $2$-fold cover.  By hypothesis, $M$ doesn't admit any connected $2$-fold covers, so $\tilde{M}$ must be disconnected.  Writing $\tilde{M} = M_1\coprod M_2$, it follows that $\pi:M_i\rightarrow M$ is $1$-sheeted, so is a diffeomorphism for each choice of $i$.
In fact, then the inverse $\pi^{-1}:M\rightarrow M_1\subseteq L\subseteq TM$ is a non-vanishing vector field on $M$.