A torus may be brushed smooth but a sphere can not. Shaggy dog theorem

By Hubert Shutrick June 19, 2008

The sphere does not admit a continuous unit vector field.

In math folklore the theorem reads:

However you comb a shaggy dog, there is always a tuft or a whorl.

Consider the unit sphere |v| = 1 in three dimensional space. Let N be the point (0, 0, 1), S be (0, 0, -1) and E be the equatorial circle where z = 0. N and S are the North and South poles, respectively. The unit vector from the center of the sphere to the North pole is denoted n = (0, 0, 1). In what follows, a map means a continuous mapping. (See the discussion on the various names given to functions in different circumstances and an article that enumerates the orthogonal vector fields on the spheres in many dimensions.)

A continuous vector field would be given by a continuous unit vector function u(v) such that v and u(v) are orthogonal. If one exists, consider the map

ft(v) = v cos(t π) + u(v) sin(t π).

For each 0 ≤ t ≤ 1, it defines a map of the sphere to itself and t = 0 gives the identity while t = 1 gives the antipodal map v to -v. It is called a homotopy between the identity and the antipodal map since it deforms from one to the other continuously. It is free in this case because, in homotopy theory, it is usual to have a preferred point in each space, called the base point, and insist that all maps take base point to base point for each t value. We can make ft into a proper homotopy by continuously rotating the sphere about the axis orthogonal to n and u(n) as t varies so that N stays put. The new f1 is the composite of the antipodal map with a rotation of π taking the equatorial circle E to itself but reversing its sense.

Consider the sphere as consisting of the semicircular longitudes joining N and S each defined by a vector e on the equator E: it is said to be the suspension of the circle E. Any map g of E to itself can be extended to a map of the sphere to itself by suspending it, that is, each latitude goes to itself in such a way that the point on the longitude e goes to the point on the g(e) longitude. If g winds E n times round itself, then its suspension will wrap the sphere round n times leaving N and S fixed. In the case of the circle E, it is well known that deforming g continuously leaves the integer n fixed so the set of equivalence classes of maps under homotopy from the circle to itself is in 1-1 correspondence with the set of integers. Also, deforming g continuously deforms its suspension continuously. Suspension gives then a mapping m of the integers to the set of equivalence classes of maps of the sphere into itself under homotopy.

To prove that it is injective requires some basic homotopy theory and is usually proved using the Hopf fibration which is a map from the three dimensional sphere onto the sphere such that the inverse image of each point is a circle. In fact the homotopy classes of maps of the sphere to itself form an abelian group and it can be proved that m is a group isomorphism.

This method of proof applies to all even-dimensional n-spheres none of which admit a unit vector field because the antipodal map changes the orientation of the equatorial (n-1)-sphere. Odd dimensional spheres can admit several independent (at each point) unit vector fields. The formula for how many is very elegantly deduced in Husemoler's Fibre Bundles, McGraw-Hill, 1966, using Clifford algebras. They give the vector fields but it was Frank Adams: Ann. Math. 75, 603-632 (1962) who proved that there weren't any others. The 3-sphere admits three independent fields: it is parallelizable. I had the very good fortune to be in Manchester in the sixties when Adams was there and attended the seminars he arranged and got to know him quite well.

Note: In case it would be of interest in itself, here is the Hopf fibration:

The Hopf fibration is defined by considering the 3-sphere as the subspace of C² defined by

|z1|² + |z2|² = 1.

The map takes (z1, z2) to z1/z2 in the complex projective line, which is the 2-sphere. It takes (z'1, z'2) to (z1, z2) if and only if there is an angle θ such that

z'1 = z1exp(iθ) and
z'2 = z2exp(-iθ)

Hence the circle. Elements of Topology   