Theorem   If is a vector field on , then 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 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 is a vector field on a compact manifold with isolated zeroes, then where is the set of zeroes of , and is the index of at , and is the Euler characteristic of .

It is not difficult to show that has non-vanishing vector fields for all . A much harder result of Adams shows that the tangent bundle of is trivial if and only if , corresponding to the unit spheres in the 4 real division algebras.

Proof. First, the low tech proof. Assume that has a unit vector field . Then the antipodal map is homotopic to the identity. But this cannot be, since the degree of the antipodal map is and the degree of the identity map is . We therefore reject the assumption that is a unit vector field.

This also implies that the tangent bundle of is non-trivial, since any trivial bundle has a non-zero section.

Proof. Now for the sledgehammer proof. Suppose is a nonvanishing vector field on . Then by the Poincaré-Hopf index theorem, the Euler characteristic of is . But the Euler characteristic of is . Hence must have a zero.