2.1 The Fano plane

The quaternions, $\H$, are a 4-dimensional algebra with basis $1,i,j,k$. To describe the product we could give a multiplication table, but it is easier to remember that:

• $1$ is the multiplicative identity,
• $i,j,$ and $k$ are square roots of -1,
• we have $ij = k$, $ji = -k$, and all identities obtained from these by cyclic permutations of $(i,j,k)$.

We can summarize the last rule in a picture:

When we multiply two elements going clockwise around the circle we get the next one: for example, $ij = k$. But when we multiply two going around counterclockwise, we get minus the next one: for example, $ji = -k$.

We can use the same sort of picture to remember how to multiply octonions:

This is the Fano plane, a little gadget with 7 points and 7 lines. The 'lines' are the sides of the triangle, its altitudes, and the circle containing all the midpoints of the sides. Each pair of distinct points lies on a unique line. Each line contains three points, and each of these triples has has a cyclic ordering shown by the arrows. If $e_i, e_j,$ and $e_k$ are cyclically ordered in this way then

$$e_i e_j = e_k, \qquad e_j e_i = -e_k .$$

Together with these rules:

• $1$ is the multiplicative identity,
• $e_1,\dots,e_7$ are square roots of -1,

the Fano plane completely describes the algebra structure of the octonions. Index-doubling corresponds to rotating the picture a third of a turn.

This is certainly a neat mnemonic, but is there anything deeper lurking behind it? Yes! The Fano plane is the projective plane over the 2-element field $\Z_2$. In other words, it consists of lines through the origin in the vector space $\Z_2^3$. Since every such line contains a single nonzero element, we can also think of the Fano plane as consisting of the seven nonzero elements of $\Z_2^3$. If we think of the origin in $\Z_2^3$ as corresponding to $1 \in \O$, we get the following picture of the octonions:

Note that planes through the origin of this 3-dimensional vector space give subalgebras of $\O$ isomorphic to the quaternions, lines through the origin give subalgebras isomorphic to the complex numbers, and the origin itself gives a subalgebra isomorphic to the real numbers.

What we really have here is a description of the octonions as a 'twisted group algebra'. Given any group $G$, the group algebra $\R[G]$ consists of all finite formal linear combinations of elements of $G$ with real coefficients. This is an associative algebra with the product coming from that of $G$. We can use any function

$$% latex2html id marker 1530 \alpha \maps G^2 \to \{ \pm 1 \}$$

to 'twist' this product, defining a new product

$$\star \maps \R[G] \times \R[G] \to \R[G]$$

by:

$$% latex2html id marker 1532 g \star h = \alpha(g,h) \; gh,$$

where $g,h \in G \subset \R[G]$. One can figure out an equation involving $\alpha$ that guarantees this new product will be associative. In this case we call $\alpha$ a '2-cocycle'. If $\alpha$ satisfies a certain extra equation, the product $\star$ will also be commutative, and we call $\alpha$ a 'stable 2-cocycle'. For example, the group algebra $\R[\Z_2]$ is isomorphic to a product of 2 copies of $\R$, but we can twist it by a stable 2-cocycle to obtain the complex numbers. The group algebra $\R[\Z_2^2]$ is isomorphic to a product of 4 copies of $\R$, but we can twist it by a 2-cocycle to obtain the quaternions. Similarly, the group algebra $\R[\Z_2^3]$ is a product of 8 copies of $\R$, and what we have really done in this section is describe a function $\alpha$ that allows us to twist this group algebra to obtain the octonions. Since the octonions are nonassociative, this function is not a 2-cocycle. However, its coboundary is a 'stable 3-cocycle', which allows one to define a new associator and braiding for the category of $\Z_2^3$-graded vector spaces, making it into a symmetric monoidal category [3]. In this symmetric monoidal category, the octonions are a commutative monoid object. In less technical terms: this category provides a context in which the octonions are commutative and associative! So far this idea has just begun to be exploited.

home