### Sedenions

What happens if we continue this process? We define the sedenions by considering pairs of octonions, \begin{equation} s = (p,q) \end{equation} with $p,q\in\OO$. Sedenion multiplication is defined by the Cayley-Dickson process, with $\epsilon=1$, so we have \begin{align} \bar{(a,b)} &= (\bar{a},-b) \\ (a,b)(c,d) &= (ac-\bar{d}b,da+b\bar{c}) \\ (a,b)\bar{(a,b)} &= (|a|^2+|b|^2,0) \end{align} If we define the special element \begin{equation} e = (0,1) \end{equation} then we could also write \begin{equation} s = p + qe \end{equation} since \begin{align} (p,0)(1,0) &= (p,0) \\ (q,0)(0,1) &= (0,q) \end{align}

The sedenions possess some curious properties. They are of course neither commutative nor associative, since they contain a copy of the octonions. And they do possess a positive-definite inner product; the norm $s\bar{s}$ of any nonzero sedenion $s$ is strictly positive. However, the sedenions contain zero divisors, that is, nonzero elements whose product is nonetheless zero. For example, we have \begin{align} (i\ell+je)(j\ell+ie) &= (i\ell,j)(j\ell,i) \nonumber\\ &= \bigl( (i\ell)(j\ell) + ij, i(i\ell) - j(j\ell) \bigr) \nonumber\\ &= (-k+k,-\ell+\ell) = 0 \end{align} Thus, the sedenions are not a composition algebra, as they fail to satisfy the identity \begin{equation} |pq| = |p|\,|q| \end{equation} and they are not a division algebra, since, for example, zero divisors such as $i\ell+je$ can not have an inverse.

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