We can use the concept of degree to prove the (in)famous “hairy ball theorem”. That is, any smooth vector field defined on the whole sphere for even must vanish at some point. The name comes from thinking of a “hair” growing out of each point on the sphere, and trying to “comb” them all to lay flat — tangent — against the sphere. If this can be done, then the direction each hair points defines a unit vector at its point, and combing smoothly means we have a smooth vector field. The assertion is that such a “combing” is impossible.
First, I say that if is even then the antipodal map sending a point to is orientation-reversing. Indeed, we first extend to the larger space ; the antipodal map is clearly orientation-reversing, as we can see by taking its Jacobian. This matrix has eigenvalues, all equal to , so the determinant is .
If is the position vector of a point in , then we consider the canonical identification of with and calculate
So if is a positively-oriented basis of , then is a negatively-oriented basis of .
Now, the position vector field is -related to itself; if is used as the first vector in a positively-oriented basis of , then is the first vector in the antipodal basis, which we know is negatively-oriented. But this means that the positively-oriented basis of must be flipped to a negatively-oriented basis of .
So what does this mean? The identity map has degree , while we can calculate that the antipodal map has degree . Since these are different, the two maps must not act identically on homology, and therefore cannot be homotopic.
But now let’s assume that is an everywhere-nonzero vector field which, without loss of generality, we may assume to have constant length — use the inner product we get by taking and divide by the original length to normalize. For each point we can define the great circle
This is a curve that lies on the sphere for every . Indeed, we can check that
So if we define by , then we have a homotopy from the identity map on to the antipodal map, which is exactly what we just showed could not exist. Thus we conclude that no such non-vanishing vector field can exist on an even-dimensional sphere.