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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

Pedro Perez
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  • What do you mean by a "continuous tangent vector bundle of dimension 1"? If you mean a rank 1 vector bundle, then every manifold has one: $M\times \mathbb{R}$. – Jason DeVito - on hiatus Sep 13 '13 at 00:31
  • @JasonDeVito I mean a rank 1 vector bundle in which the line assigned to the point s in the sphere lives in the tangent space of the point. – Pedro Perez Sep 13 '13 at 00:38

3 Answers3

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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.

Jason DeVito - on hiatus
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  • It will be terrific if you could give a reference. – Pedro Perez Sep 13 '13 at 01:09
  • If I remember correctly, this exact result is found in Milnor and Stasheff's characteristic classes book. I'll check tomorrow (if I remember) – Jason DeVito - on hiatus Sep 13 '13 at 01:33
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    This argument doesn't work. A 1-dimensional vector bundle is trivial if and only if it is orientable, so the only interesting case is when the 1-dimensional subbundle is non-orientable. In this case the Euler class is undefined so such an argument can never work. – PVAL-inactive May 03 '18 at 22:09
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    @PVAL-inactive: I see what you're saying, but as every vector bundle over $S^{2n}$ is orientable (since $S^{2n}$ is simply connected), I wouldn't say this argument doesn't work. But I would wholeheartedly agree that it is overkill and that your observation is the best way of proving it. – Jason DeVito - on hiatus May 04 '18 at 03:06
  • A great attack by @PVAL-inactive and a brilliant defense by Jason. This kind of duel contributes to what makes our site so wonderful! (+1 to both of them, of course) – Georges Elencwajg Apr 30 '21 at 13:31
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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 $M$.

Jason DeVito - on hiatus
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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).

Rob Arthan
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  • I am sorry, but I do not see why the theorem you mention gives an affirmative answer to my question. – Pedro Perez Sep 13 '13 at 00:28
  • The theorem says that the number of linearly independent vector fields on an $n$-sphere is $\rho(n+1)-1$, where $\rho(x)$ is the Radon-Hurwitz number of $x$ (see the paper by Adams and the papers it refers to or http://en.wikipedia.org/wiki/Vector_fields_on_spheres). $\rho(n+1) = 1$ when $n$ is even, giving us no non-zero vector fields at all on even-dimensional spheres. – Rob Arthan Sep 15 '13 at 00:39
  • @Rob: I don't think Pedro is interested in tangent vectors as much as the 1-d subspace they span. It seems a priori possible that one could potentially pick a field of $1$-d subspaces in $TS^{2n}$ but couldn't consistently choose a nonzero vector from each subspace. – Jason DeVito - on hiatus Sep 15 '13 at 02:22
  • For example, $S^{5}$ admits a field of $4$-planes, despite the fact that the maximum number of linearly independent vector fields it supports is $1$: Given a nonvanishing vector field $V$, let $E(p) = V(p)^\bot \subseteq T_p S^{5}$ (where $\bot$ is computed relative to, say, the standard Riemannian metric on $S^5$). – Jason DeVito - on hiatus Sep 15 '13 at 02:26