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# hairy ball theorem

###### Theorem.

If $X$ is a vector field on $S^{{2n}}$, 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 hairy ball theorem 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 $\chi(M)=\sum_{{v\in Z(X)}}\iota(v)$ where $Z(X)$ is the set of zeroes of $X$, and $\iota(v)$ is the index of $x$ at $v$, and $\chi(M)$ 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. 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. ∎

## Mathematics Subject Classification

57R22*no label found*

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