Generalization of the hairy ball theorem. 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
 A: I'm interpreting "continuous tangent vector bundle of dimension 1" to mean a rank $1$ subbundle of $TS^{2n}$, i.e., a $1-$dimensional distribution.
In this case, we have the following result (I do not know who first proved it).

If $E\subseteq TS^{2n}$ is a rank $k$ subbundle, then either $k=0$ or $k=2n$.  In other words, $TS^{2n}$ has no non-trivial subbundles.

Proof:  Suppose $E\subseteq TS^{2n}$ is a subbundle.  Since $TS^{2n}$ admits a fiberwise metric, $E$ has a complement $E^\bot$ in the sense that $E\oplus E^\bot \cong TS^{2n}$.
By the Whitney sum formula for the Euler Class, we know $e(TS^{2n}) = e(E)\cup e(E^\bot)$.  Identifying the Euler class with the Euler characteristic, we have $e(TS^{2n}) = 2\in H^{2n}(S^{2n})$, so $e(E)\cup e(E^\bot) = 2$.  Since $H^k(S^{2n}) = 0$ unless $k = 0$ or $k=2n$, the result follows.
A: 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 $M$.
A: A classic result of Frank Adams on vector fields on spheres gives the full story on generalisations of the hairy ball theorem and answers your question in the affirmative. See http://www.jstor.org/stable/1970213 (the statement of the result is visible on the first page, even if you don't have full access to the paper).
