The quaternions, $\H,ドル are a 4-dimensional algebra with basis 1,ドルi,j,k$.
To describe the product we could give a multiplication
table, but it is easier to remember that:
1ドル$ is the multiplicative identity,
$i,j,$ and $k$ are square roots of -1,
we have $ij = k,ドル $ji = -k,ドル and all identities obtained
from these by cyclic permutations of $(i,j,k)$.
We can summarize the last rule in a picture:
\begin{figure} % latex2html id marker 258 \centerline{\epsfysize=1.5in\epsfbox{triangle.eps}}\end{figure}
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:
\begin{figure} % latex2html id marker 263 \centerline{\epsfysize=1.5in\epsfbox{fano.eps}}\medskip\end{figure}
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
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ドル \in \O,ドル we get the following picture of the
octonions:
\begin{figure} % latex2html id marker 270 \centerline{\epsfysize=1.5in\epsfbox{cube.eps}}\end{figure}
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,ドル 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
\begin{displaymath} % latex2html id marker 1532 g \star h = \alpha(g,h) \; gh, \end{displaymath}
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,ドル
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,ドル 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,ドル
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.