We define the octonionic conjugate $\bar{x}$ of an octonion $x$ as the (real) linear map which reverses the sign of each imaginary unit. Thus, \begin{equation} \bar{x} = x_1 - x_2 i - x_3 j - x_4 k - x_5 k\ell - x_6 j\ell - x_7 i\ell - x_8 \ell \end{equation} if $x=x_1+x_2i+x_3j+x_4k+x_5k\ell+x_6j\ell+x_7i\ell+x_8\ell$ as above. Direct computation shows that \begin{equation} \bar{xy} = \bar{y}\>\bar{x} \end{equation} The norm of an octonion $|x|$ is defined by \begin{equation} |x|^2 = x\bar{x} = x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2 + x_8^2 \label{onorm} \end{equation} Again, the only octonion with norm $0$ is $0$, and every nonzero octonion has a unique inverse, namely \begin{equation} x^{-1} = {\bar{x} \over  |x|^2} \end{equation} As with the other division algebras, the norm satisfies the identity \begin{equation} |xy| = |x| |y| \label{norm} \end{equation} Writing this out in terms of components yields the 8-squares rule, which is no longer at all obvious. The octonions therefore also form a division algebra.

A remarkable property of the octonions is that they are not associative! For example, compare \begin{align} (i j) (\ell) &= +(k)(\ell)  =  +k\ell \\ (i) (j \ell) &= (i)(j\ell)  =  -k\ell \end{align} However, the octonions are alternative, that is, products involving no more than 2 independent octonions do associate. Specifically, \begin{align} (xy)y &= xy^2 \\ (xy)x &= x(yx) \end{align} for any octonions $x$, $y$. Alternativity extends to products with conjugates, so that \begin{align} (xy)\bar{y} &= x|y|^2 \\ (xy)\bar{x} &= x(y\bar{x}) \end{align}

The commutator (4) of §3.3, defined for quaternions, extends naturally to the octonions as well. We define the associator of three octonions $x$, $y$, $z$ by \begin{equation} [x,y,z] = (xy)z - x(yz) \end{equation} which quantifies the lack of associativity. Alternativity can be phrased as \begin{equation} [x,y,x] = 0 = [x,y,y] \end{equation} More generally, both the commutator and associator are antisymmetric, that is, interchanging any two arguments changes the result by a minus sign; replacing any argument by its conjugate has the same effect, because the real parts don't contribute.

A consequence of alternativity is that the Moufang identities \begin{align} (xyx)z &= x\bigl(y(xz)\bigr) \\ z(xyx) &= \bigl((zx)y\bigr)x \\ (xy)(zx) &= x(yz)x \label{moufang} \end{align} are satisfied, all of which follow from the associator identity \begin{equation} [x,y,zx] = x[x,y,z] \end{equation}

Polarizing (\ref{onorm}), we obtain an inner product on $\RR^8$, namely \begin{equation} p\cdot q = \frac12 \left(|p+q|^2-|p|^2-|q|^2\right) = p\bar{q} + q\bar{p} = \bar{p}q + \bar{q}p \end{equation} where the last two equalities make use of the properties of these properties of the commutator, which imply that $[p,\bar{q}]=[\bar{p},q]$. If $v$ and $w$ are imaginary octonions, then \begin{equation} -v\cdot w = vw + wv = \{v,w\} \label{odot} \end{equation} where the last equality defines the anticommutator of $v$ and $w$. Thus, two imaginary octonions are orthogonal if and only if their anticommutator vanishes. Furthermore, if $v$ is an imaginary octonion, and $x$ is orthogonal to $v$, then $x\bar{v}+v\bar{x}=0$, that is \begin{equation} x\perp v \Longrightarrow xv = v\bar{x} \end{equation} for $x\in\OO$ and $v\in\Im\OO$, since $\bar{v}=-v$.

The analogy with vectors in three dimensions discussed in §3.3, also holds in seven dimensions. Identifying $v$, $w$ with $\vv$, $\ww$, not only is there a dot product, given either by (\ref{odot}) or equivalently by (3) of §3.3, but there is also a cross product, defined by by (2) of §3.3. The dot product exists in all dimensions, but the cross product does not. Which way should the cross product point in higher dimensions, where the “right-hand rule” isn't sufficient? The quaternionic and octonionic multiplication table provide an answer to this question, and it turns out that these are the only possibilities: The cross product exists only in three and seven dimensions!