We have argued in § 14.5–14.7 that the ordinary momentum-space (massless and massive) Dirac equation in $3+1$ dimensions can be obtained via dimensional reduction from the Weyl (massless Dirac) equation in $9+1$ dimensions. This latter equation can be written as the eigenvalue problem \begin{equation} \label{Dirac} \tilde{\PPP}\psi = 0 \end{equation} where $\PPP$ is a $2\times2$ octonionic Hermitian matrix corresponding to the 10-dimensional momentum and tilde again denotes trace reversal. The general solution of this equation is \begin{eqnarray} \PPP &=& \pm\theta\theta^\dagger \\ \psi &=& \theta\xi \end{eqnarray} where $\theta$ is a 2-component octonionic vector whose components lie in the same complex subalgebra of $\OO$ as do those of $\PPP$, and where $\xi\in\OO$ is arbitrary. (Such a $\theta$ must exist since $\det(\PPP)=0$.)

It is then natural to introduce a 3-component formalism; this approach was used by Schray [arXiv:hep-th/9407045] for the superparticle. Defining \begin{equation} \Psi = \begin{pmatrix}\theta\\ \bar\xi\\\end{pmatrix} \end{equation} we have first of all that \begin{equation} \Pcal = \Psi \Psi^\dagger = \begin{pmatrix}\PPP& \psi\\ \psi^\dagger&|\xi|^2\\\end{pmatrix} \end{equation} so that $\Psi$ combines the bosonic and fermionic degrees of freedom. Lorentz transformations can be constructed by iterating (“nesting”) transformations of the form [arXiv:hep-th/9302044] \begin{eqnarray} \PPP &\mapsto& \MM\PPP\MM^\dagger \label{MTrans}\\ \psi &\mapsto& \MM\psi \label{VTrans} \end{eqnarray} which can be elegantly combined into the transformation \begin{equation} \label{Trans} \Pcal \mapsto \MMM\Pcal\MMM^\dagger \end{equation} with \begin{equation} \MMM = \begin{pmatrix}\MM& 0\\ 0& 1\\\end{pmatrix} \end{equation} This in fact shows how to view $SO(9,1)$ as a subgroup of $E_6$; the rotation subgroup $SO(9)$ lies in $F_4$. It turns out that the Dirac equation (\ref{Dirac}) is equivalent to the equation \begin{equation} \Pcal*\Pcal=0 \label{DiracF} \end{equation} which shows both that solutions of the Dirac equation correspond to the Cayley plane and that the Dirac equation admits $E_6$ as a symmetry group. Using the particle interpretation from § 14.7 then leads to the interpretation of (part of) the Cayley plane as representing 3 generations of leptons.

We therefore refer to the 3-component octonionic column $\Psi$ as a Cayley spinor. This name is a bit misleading, as $\Psi$ is not a “spinor” in the classical sense of belonging to an appropriate representation of an orthogonal group (more precisely, of a spin group). However, $\Psi$ does bear a similar relationship to the “vector” $\Pcal$ as the true spinor $\psi$ does to the vector $\PPP$.

We emphasize that not all 3-component octonionic columns are Cayley spinors. For $\Psi$ to be a Cayley spinor, $\Psi\Psi^\dagger$ must satisfy the Dirac equation in the form (\ref{DiracF}), which, as we saw in § 12.5, forces the components of $\Psi\Psi^\dagger$ to lie in a quaternionic subalgebra of $\OO$. As in the above construction, we assume that the components of $\Psi$ also lie in this quaternionic subalgebra.

We conclude with a bit of speculation. We refer to $\Pcal$ as a “1-square”, since it is a primitive idempotent (squares to itself and has trace 1). In general, we refer to decompositions of the form (\ref{Decomp}) as $p$-square decompositions, where $p$ is the number of nonzero eigenvalues, and hence the number of nonzero primitive idempotents in the decomposition. If $\det(\AAA)\ne0$, then $\AAA$ is a 3-square. If $\det(\AAA)=0\ne\sigma(\AAA)$, then $\AAA$ is a 2-square. Finally, if $\det(\AAA)=0=\sigma(\AAA)$, then $\AAA$ is a 1-square (unless also $\tr(\AAA)=0$, in which case $\AAA\equiv0$). It is intriguing that, since $E_6$ preserves both the determinant and the condition $\sigma(\AAA)=0$, $E_6$ therefore preserves the class of $p$-squares for each $p$. If, as argued above, 1-squares correspond to leptons, is it possible that 2-squares are mesons and 3-squares are baryons?