# hairy ball theorem

###### Theorem.

If $X$ is a vector field on ${S}^{\mathrm{2}\mathit{}n}$, then $X$ has a zero. Alternatively, there are no continuous unit vector field on the sphere. Moreover, the tangent bundle of the sphere is nontrivial as a bundle, that is, it is not simply a product.

There are two proofs for this. The first proof is based
on the fact that the antipodal map on ${S}^{2n}$ is not homotopic^{} to the
identity map. The second proof gives the
as a corollary of the Poincaré-Hopf index theorem.

Near a zero of a vector field, we can consider a small sphere around the zero, and restrict the vector field to that. By normalizing, we get a map from the sphere to itself. We define the index of the vector field at a zero to be the degree of that map.

###### Theorem (Poincaré-Hopf index theorem).

If $X$ is a vector field on a compact manifold $M$ with
isolated zeroes, then $\chi \mathit{}\mathrm{(}M\mathrm{)}\mathrm{=}{\mathrm{\sum}}_{v\mathrm{\in}Z\mathit{}\mathrm{(}X\mathrm{)}}\iota \mathit{}\mathrm{(}v\mathrm{)}$
where $Z\mathit{}\mathrm{(}X\mathrm{)}$ is the set of zeroes of $X$, and $\iota \mathit{}\mathrm{(}v\mathrm{)}$ is
the index of $x$ at $v$, and $\chi \mathit{}\mathrm{(}M\mathrm{)}$ is the Euler characteristic^{} of $M$.

It is not difficult to show that ${S}^{2n+1}$ has non-vanishing vector
fields for all $n$. A much harder result of Adams shows that the
tangent bundle of ${S}^{m}$ is trivial if and only if $n=0,1,3,7$,
corresponding to the unit spheres in the 4 real division algebras^{}.

###### Proof.

First, the low tech proof. Assume that ${S}^{2n}$ has a unit vector field $X$. Then the antipodal map is homotopic to the identity (http://planetmath.org/AntipodalMapOnSnIsHomotopicToTheIdentityIfAndOnlyIfNIsOdd). But this cannot be, since the degree of the antipodal map is $-1$ and the degree of the identity map is $+1$. We therefore reject the assumption that $X$ is a unit vector field.

This also implies that the tangent bundle of ${S}^{2n}$ is non-trivial,
since any trivial bundle^{} has a non-zero section^{}.
∎

###### Proof.

Now for the sledgehammer proof. Suppose $X$ is a nonvanishing vector field on ${S}^{2n}$. Then by the Poincaré-Hopf index theorem, the Euler characteristic of ${S}^{2n}$ is $\chi (X)={\sum}_{v\in {X}^{-1}(0)}\iota (v)=0$. But the Euler characteristic of ${S}^{2k}$ is $2$. Hence $X$ must have a zero. ∎

Title | hairy ball theorem |
---|---|

Canonical name | HairyBallTheorem |

Date of creation | 2013-03-22 13:11:33 |

Last modified on | 2013-03-22 13:11:33 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 12 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 57R22 |

Synonym | porcupine theorem |

Synonym | PoincarÃÂ©-Hopf theorem |

Defines | Poincaré-Hopf index theorem |