What are the octonionic integers? Based on our discussion of the quaternionic case in § 12.5, we require the components of any such octonion to be either an integer or a half-integer, and the norm of any such octonion to be an integer. How many elements of this form are there?

First of all, we have the 16 elements \begin{equation} G = \{\pm1,\pm i,\pm j,\pm k,\pm k\ell,\pm j\ell,\pm i\ell,\pm\ell\} \end{equation} These 16 elements close under multiplication, and generate an eight-dimensional lattice which we will call the Gravesian integers, and which generalize the Lipschitz integers over $\HH$. 1) However, as is the case over $\HH$, the Gravesian integers are not maximal.

Next, we can consider the $2^8$ elements \begin{equation} \Omega = \left\{\frac12(\pm1\pm i\pm j\pm k\pm k\ell\pm j\ell\pm i\ell\pm\ell)\right\} \end{equation} where all combinations of signs are allowed. Remarkably, products of elements in $\Omega$ either have integer coefficients, such as \begin{align} &\frac12(1+i+j+k+k\ell+j\ell+i\ell+\ell) \nonumber\\ &\times \frac12(1+i+j+k+k\ell+j\ell+i\ell-\ell) \nonumber\\ &= -1+i+j+k \end{align} or have eight (nonzero) half-integer coefficients, such as \begin{align} &\frac12(1+i+j+k+k\ell+j\ell+i\ell+\ell) \nonumber\\ &\times \frac12(1-i-j+k+k\ell+j\ell+i\ell+\ell) \nonumber\\ &= \frac12(-1+i-j+k+k\ell+3j\ell+i\ell-\ell) \nonumber\\ &= j\ell + \frac12(-1+i-j+k+k\ell+j\ell+i\ell-\ell) \end{align} Thus, the eight-dimensional lattice generated by $G\cup\Omega$, that is, the set of all integer linear combinations of elements in $G\cup\Omega$, also closes under multiplication; these are the Kleinian integers. However, the Kleinian integers are also not maximal.

What happened to the Hurwitz integers? Surely we should include elements such as $q = \frac12(1+i+j+k)$ and, once we do so, also the “complement” of such elements, such as $\frac12(k\ell+j\ell+i\ell+\ell)$. There's only one problem: The set generated by the $2\times7\times2^4=224$ such elements, together with $G$ and $\Omega$, collectively known as the Kirmse integers, doesn't close under multiplication! For example, \begin{equation} \frac12(1+i+j+k) \times \frac12(1+k+k\ell+\ell) = \frac12(i+k+k\ell+j\ell) \end{equation} which has not yet been included (since the complement of $\{i,k,k\ell,j\ell\}$ is $\{1,j,i\ell,\ell\}$, which does not correspond to a quaternionic subalgebra of $\OO$). Even if we include such elements, expanding our 224 Kirmse units to a set of ${8\choose4}\times2^4=1120$ units, the product still doesn't close. For example, \begin{align} & \frac12(i+k+k\ell+j\ell) \times \frac12(1+i+j\ell+k\ell) \nonumber\\ &= \frac14(-3+i+j+k+k\ell+j\ell+k\ell-\ell) \end{align} which is clearly not in our pool of candidate integral octonions.

So how do we generalize the Hurwitz integers? Let's start again. Recall that the Hurwitz units over $\HH$ are \begin{equation} H = \{\pm1,\pm i,\pm j,\pm k,\frac12(\pm1\pm i\pm j\pm k)\} \end{equation} If we now include the Gravesian units, we can generate the complement of $H$ as $H\ell$, and the Kleinian elements $\Omega$ as $H\oplus H\ell$. The algebra generated by $H\cup G\cup \Omega$ closes, since all products that don't involve $\Omega$ lie entirely in $\HH$ or $\HH^\perp$. We will refer to this algebra as the double Hurwitzian integers generated by $\HH$. Any of the other six quaternionic triples can be used in place of $\{i,j,k\}$; there are seven different copies of the double Hurwitzian integers in $\OO$. However, the double Hurwitzian integers are still not maximal.

As discussed / in \cite{Conway}, / from [Conway and Smith] , the trick to obtaining a maximal set of integral octonions is to choose a preferred octonionic unit; we choose $\ell$. Looking at the octonionic multiplication table, there are three quaternionic triples containing $\ell$, so include the Hurwitz integers corresponding to those three quaternionic subalgebras, but not those corresponding to the other four quaternionic subalgebras. As before, include their complements—which can also be obtained as sums and differences of Hurwitz integers from different quaternionic subalgebras. We're almost there. Computing products, we discover that we have generated a different set of 224 “generalized Kirmse units”, namely \begin{align} \Omega_0 = &\Big\{ \frac12(\pm1\pm i\pm i\ell\pm \ell), \frac12(\pm1\pm j\pm j\ell\pm \ell), \frac12(\pm1\pm k\pm k\ell\pm \ell), \nonumber\\ &\frac12(\pm j\pm j\ell\pm k\pm k\ell), \frac12(\pm k\pm k\ell\pm i\pm i\ell), \frac12(\pm i\pm i\ell\pm j\pm j\ell), \nonumber\\ &\frac12(\pm1\pm i\ell\pm j\pm k), \frac12(\pm1\pm i\pm j\ell\pm k), \frac12(\pm1\pm i\pm j\pm k\ell), \nonumber\\ &\frac12(\pm i\pm j\ell\pm k\ell\pm \ell), \frac12(\pm i\ell\pm j\pm k\ell\pm \ell), \frac12(\pm i\ell\pm j\ell\pm k\pm \ell), \nonumber\\ &\frac12(\pm1\pm i\ell\pm j\ell\pm k\ell), \frac12(\pm i\pm j\pm k\pm \ell) \Big\} \end{align} Remarkably, the lattice generated by $G\cup\Omega\cup\Omega_0$ does close under multiplication; we call its elements the “$\ell$-integers”. There are $224+16=240$ units in the $\ell$-integers, since $\Omega$ is a sublattice of $\Omega_0$ (and its elements are not units anyway). Furthermore, the $\ell$-integers are maximal; no further elements can be added without giving up either closure under multiplication or our defining properties for candidate integers (namely integer norm and half-integer or integer coefficients).

This construction can of course be repeated replacing $\ell$ with any other imaginary octonionic unit; there are thus seven different maximal “orders” of integral octonions.

Further information about integral octonions can be found in [Conway and Smith] .