S^2 isn't a special case though: Brouwer's showed the theorem can be easily extended to high dimensions, hence today we usually consider the more general statement that there is a nonzero tangent vector field on the n-sphere S^n iff n is odd.

Not only does it generalize to higher n, it also shows a bit more: not only that the lack of such vector field for an even n, but the also the existence of such for odds.

▲

xyzzyz 3 days ago | parent [-]

It’s really easy to see that such a vector field exists on odd dimensional spheres, though, by extending the construction on S^1: f(x, y) = (-y, x). In higher dimensions, you do the same thing, swap elements pair wise and multiply one of the elements of the pair by -1. This works in odd dimensional sphere because you can pair up coordinates.

	▲	xanderlewis 2 days ago \| parent [-]
		Another 'reason' it works for odd-dimensional spheres is that the (2n - 1)-sphere can be identified with a certain subset of C^n (n-dimensional complex coordinate space) where your 'swap elements and multiply one by -1' idea is just multiplication by i, which, when you think of your vectors as being back in R^2n again, always produces something orthogonal to the original vector. Even better, the (4n - 1)-sphere (so think of S^3, S^7, S^11, ...) can be thought of as a certain subset of H^n (same thing as before but with quaternions instead of complex numbers), where multiplication by i, j and k are available! And now in this case you have not only one nowhere-vanishing vector field on the sphere, but three everywhere pairwise orthogonal vector fields. This in particular shows that S^3 is 'parallelisable' — a property it shares with S^1 and means that there exists a continuous global choice of basis for each tangent space.