Lorentz group $SO(9,1)$ via the identification of their double-covers, the spin groups $\hbox{Spin}(d,1)$, namely [^7] $$\hbox{Spin}(3,1) = SL(2,{{\mathbb C}}) \subset SL(2,{{\mathbb O}}) = \hbox{Spin}(9,1)$$

Therefore, all it takes to break 10 spacetime dimensions to 4 is to choose a preferred octonionic unit to play the role of the complex unit. We choose $\ell$ rather than $i$ to fill this role, preferring to save $i$, $j$, $k$ for a (distinguished) quaternionic triple. The projection $\pi$ from ${{\mathbb O}}$ to ${{\mathbb C}}$ is given by $$\pi(q) = {1\over2} (q + \ell q {\overline}\ell)$$ and we thus obtain a preferred $SL(2,{{\mathbb C}})$ subgroup of $SL(2,{{\mathbb O}})$, corresponding to the “physical” Lorentz group.

Spin {#SPIN}
----

Since we now have a preferred 4-d Lorentz group, we can use its rotation subgroup $SU(2)\subset SL(2,{{\mathbb C}})$ to define spin. However, care must be taken when constructing the Lie algebra $su(2)$, due to the lack of commutativity.

Under the usual action of $M\in SU(2)$ on a Hermitian matrix $Q$ (thought of as a spacetime vector via (\[Qdef\])), namely $$Q \mapsto MQM^\dagger$$ we can identify the basis rotations as usual as $$R_z = \pmatrix{e^{\ell{\phi\over2}}& 0\cr
        \noalign{\smallskip} 0&
        e^{-\ell{\phi\over2}}\cr}
\qquad
R_y = \pmatrix{~~\cos{\phi\over2}& \sin{\phi\over2}\cr
        \noalign{\smallskip}
        -\sin{\phi\over2}& \cos{\phi\over2}\cr}
\qquad
R_x 