Generalization of the hairy ball theorem.
Sebastian Wright 
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
$\endgroup$ 23 Answers
$\begingroup$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.
$\endgroup$ 5 $\begingroup$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 $M$.
$\endgroup$ $\begingroup$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 (the statement of the result is visible on the first page, even if you don't have full access to the paper).
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