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 onthe sphere. Moreover, the tangent bundle of the sphere isnontrivial as a bundle, that is, it is not simply a product.

There are two proofs for this. The first proof is basedon the fact that the antipodal map on $S^{2n}$ is not homotopic to theidentity 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 $M$ withisolated zeroes, then $\\chi(M)=\\sum_{v\\in Z(X)}\\iota(v)$ where $Z(X)$ is the set of zeroes of $X$ , and $\\iota(v)$ isthe 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 vectorfields for all $n$ . A much harder result of Adams shows that thetangent 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 $-1$ andthe degree of the identity map is $+1$ . We therefore reject theassumption 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 characteristicof $S^{2n}$ 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.∎

单词	HairyBallTheorem
释义	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 onthe sphere. Moreover, the tangent bundle of the sphere isnontrivial as a bundle, that is, it is not simply a product. There are two proofs for this. The first proof is basedon the fact that the antipodal map on $S^{2n}$ is not homotopic to theidentity 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 $M$ withisolated zeroes, then $\\chi(M)=\\sum_{v\\in Z(X)}\\iota(v)$ where $Z(X)$ is the set of zeroes of $X$ , and $\\iota(v)$ isthe 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 vectorfields for all $n$ . A much harder result of Adams shows that thetangent 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 $-1$ andthe degree of the identity map is $+1$ . We therefore reject theassumption 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 characteristicof $S^{2n}$ 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.∎
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