Generalization of the hairy ball theorem.

Question

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

Answer 1

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).

Answer 2

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.