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I know there is no non-vanishing vector field on $S^2$, so I cannot comb the hair on a ball. (I am treating $S^2$ as a manifold without the ambient space $\mathbb R^3$, which amounts to demanding that the vector field is tangential to $S^2$ at every point if you prefer the Euclidean point of view.) Is this still impossible if the hair is unoriented? By unoriented hair I mean that instead of looking at a tangent vector at every point, I am looking at a tangent line. That is, is there a rank one subbundle of $TS^2$ (which is a line bundle on $S^2$)? Geometrical intuition suggests that there is not, but that is far from proof.

I am mostly interested in $S^2$, but I expect the answer is the same for all even-dimensional spheres just like in the oriented case.

Joonas Ilmavirta
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  • I just found this older question on the topic. Should I close this as duplicate? – Joonas Ilmavirta May 02 '18 at 17:55
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    OT: holy semantic overloading batman – Kuba hasn't forgotten Monica May 02 '18 at 18:54
  • Presumably, you want a continuously varying subbundle. In which case, wouldn't picking a "forward" direction on one tangent line allow one to extend that orientation throughout the sphere? – Acccumulation May 02 '18 at 21:06
  • @Acccumulation Not necessarily. You can always locally orient the line bundle and thus get a non-vanishing vector field as you describe. This does not always happen globally, as the orientations might not match for topological reasons. The simplest example I could think of is considering a line bundle over the Möbius strip with all lines being "across" the strip. No local orientation of this line bundle can be extended to the whole manifold. – Joonas Ilmavirta May 02 '18 at 21:11
  • @JoonasIlmavirta But it does seem like, for orientable surfaces, unoriented hair can be combed only if oriented hair can be. – Acccumulation May 02 '18 at 21:29
  • @Acccumulation That would be an interesting result if it's true. I really don't know in general. Do you want to write that up as an answer? Or ask as a follow-up question? – Joonas Ilmavirta May 02 '18 at 21:33

2 Answers2

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An alternative way to see why this isn't true is to fix a Riemannian metric. If such a subbundle existed, then restricting to those vectors of norm $1$ would give a two-sheeted covering of $S^2$. If it were disconnected, there would be a nowhere zero vector field on $S^2$, a contradiction. If it were connected, $S^2$ would admit a connected two-sheeted covering, a contradiction with it being simply connected.

Aloizio Macedo
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No. Line bundles on $S^2$ are classified by $H^1(S^2;\mathbb{Z}/2)=0$ which means that there is only the trivial line bundle $L$ on $S^2$. But then $L\oplus L$ is the trivial bundle hence not the tangent bundle of $S^2$.

An interesting question is if $TS^n$ splits as a direct sum $E\oplus F$ of low dimensional bundles. If $n$ is odd the Euler characteristic is zero, so there is a non-zero vector field. So let us assume $n$ is even.

If $E$ and $F$ are required to be orientable bundles this is not possible: The Euler class is multiplicative and $e(TS^n)$ is twice the generator of $H^n(S^n;\mathbb{Z})$ if $n$ is even and in particular it is non-zero. But if $E$ and $F$ are proper subbundles the euler classes $e(E)$ and $e(F)$ lie in some $H^i(S^n;\mathbb{Z})$ for $0<i<n$ hence $e(E)\cup e(F)=0$ which contradicts the fact that $e(E)\cup e(F)=e(TS^n)\not=0$.

A bundle is orientable iff its determinant bundle has a section, i.e. is a trivial line bundle. But we have seen that the sphere only admits the trivial line bundle. This implies that every bundle on $S^n$ is orientable and the previous argument shows that the tangent bundle cannot split at all.

This discussion btw does not imply that there are no non-trivial bundles on $S^n$ of rank $r<n$, and indeed there are some.

Thomas Rot
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