Normed division algebras Why not a three-dimensional algebra?

Hamilton spent years seeking a number system with three real components before finding the quaternions. To understand why three components are not enough, we can use a result from topology.

The hairy ball theorem (L. E. J. Brouwer, 1912) states that the surface of a sphere admits no continuous, nowhere-zero tangent vector field. We can picture the vectors as hairs lying flat against the surface: combing the sphere means arranging them so that their direction changes continuously from point to point. The theorem says that we cannot do this without at least one of these vectors vanishing.

For example, if we arrange the hairs along the meridians, from the north pole towards the south pole, we find no well-defined tangent direction at the poles. The theorem ensures that other arrangements also leave at least one point where the field vanishes; here we accept this result without proving it.

Suppose a three-dimensional normed division algebra exists, with the Euclidean modulus defined in the chapter. Each of its numbers has three real components: we can represent it as a point in space or as the vector from the origin to that point. The numbers of modulus 1 form the surface of a sphere centred at the origin. A vector tangent to the sphere at a point is perpendicular to the vector locating that point.

The point $1$ lies on the sphere. There we plant the first hair: we choose any tangent direction, that is, a non-zero vector $v$ perpendicular to $1$. It is the only choice we make by hand; all the other hairs will come from this one.

Now take any point of the sphere, that is, a number $u$ of modulus 1, and consider the operation that to every $x$ assigns $xu$. It is linear, because multiplication distributes over addition and real factors come out. And it does not change the moduli, because $|xu|=|x|\,|u|=|x|$: having taken the modulus equal to the distance from the origin, it preserves all distances. It is therefore a rigid transformation of space, and as such it preserves angles too. It carries the sphere into itself, and carries the point $1$ to the point $u$.

At the point $u$ let us then plant the hair $vu$, the product of $v$ by $u$, which is the image of $v$ under that transformation. It lies flat: $v$ is perpendicular to $1$ and angles are preserved, hence $vu$ is perpendicular to $u$. It is never missing: if $vu$ were zero it would have modulus zero, but $|vu|=|v|\neq 0$. And it changes continuously, because $vu$ depends on $u$ linearly. As $u$ runs over the whole sphere the hairs are all there, lying flat and continuous: the sphere is combed, with no tonsure.

This contradicts the hairy ball theorem. A three-dimensional normed division algebra with these properties therefore cannot exist.

On a circle, by contrast, we can arrange tangent vectors all pointing the same way round, with none vanishing. In dimension 2 this obstruction is absent, but its absence alone does not prove that the algebra exists.

We have proved the connection between the existence of the algebra and that of a continuous, nowhere-zero tangent field on the sphere. The hairy ball theorem and Hurwitz’s theorem, which gives the table of the four algebras, remain results accepted without proof.