In § 7.2, the complex group $\SU(2)=\SU(2,\CC)$ was represented in terms of $2\times2$ matrices $R_x$, $R_y$, $R_z$, acting on complex matrices of the form 1) \begin{equation} \XX = \begin{pmatrix} 1+z & x-iy \\ x+iy & 1-z \end{pmatrix} \\ \end{equation} thus demonstrating the isomorphism $\SU(2)\cong\Spin(3)$, the double cover of $\SO(3)$. Recall that (setting $\beta=\frac\alpha2$ for convenience) \begin{equation} R_y = \begin{pmatrix} \cos\beta & -\sin\beta \\ \sin\beta & \cos\beta \end{pmatrix} \end{equation} is real, whereas $R_x$ and $R_z$ are complex. But $R_y$ has precisely the form of an element of $SO(2)$, as discussed in § 6.2. We therefore identify $\SO(2)$ as a “real unitary” matrix, that is, we write \begin{equation} \SU(2,\RR) = \SO(2) \end{equation}

Can we go in the other direction? Extending $\XX$ is easy; just replace the complex number by a division algebra element, so that now \begin{equation} \XX = \begin{pmatrix} 1+z & \bar{a} \\ a & 1-z \end{pmatrix} \\ \end{equation} with $a\in\KK$.

It is important to realize that $\XX$ is complex, in the sense that it involves only one octonionic direction. In particular, \begin{equation} \det\XX = 1-|a|^2-z^2 \end{equation} Since $\tr\XX=2$ and does not involve $z$ or $a$, any transformation that preserves both the determinant and trace of $\XX$ will also preserve the norm $|a|^2+z^2$. We therefore expect \begin{align} \SU(2,\HH) \cong \Spin(5) \label{spin5}\\ \SU(2,\OO) \cong \Spin(9) \label{spin9}\\ \end{align} where the $\Spin$ groups are of course the double-covers of the corresponding orthogonal groups. But which transformations are these?

Over $\HH$, this question is easy to answer. As in § 7.2, consider transformations of the form \begin{equation} \XX \longmapsto \MM\XX\MM^\dagger \end{equation} and build $\MM$ from $R_x$, $R_y$, and $R_z$, but allow $i$ to be replaced by $j$ or $k$ in $R_x$ and $R_y$. As is easily checked by direct computation, each of these 7 transformations preserves the determinant and trace of $\XX$. But there are ${5\choose2}=10$ independent generators of $\Spin(5)$. Which ones are we missing?

We've left out the transformations that mix up $i,j,k$. But we know from §9.1.1 how to implement these transformations using conjugation. Now, however, we must conjugate using multiples of the ($2\times2$) identity matrix $\II$, such as the phase transformation \begin{equation} \MM = e^{i\alpha} = e^{i\alpha}\II \label{Mphase} \end{equation} Since conjugation leaves real numbers alone, the diagonal of $\XX$ is not affected. These are our three missing generators, and we have indeed verified (\ref{spin5}).

The same procedure works over $\OO$ as well, but we must be careful about associativity. We now have seven versions of $R_x$ and $R_z$, which together with $R_y$ yield 15 generators. Which generators are we missing? The 21 $\SO(7)$ transformations that mix up (only) the imaginary units. But there are only seven phase transformations of the form (\ref{Mphase})!

The resolution to this apparent quandary is to recall the discussion in §9.1.3, where it was shown that two flips must be nested in order to generate rotations in just one plane; the same principle applies to phase transformations. Thus, there are in fact ${7\choose2}=21$ (nested!) phase transformations, just what we need to establish (\ref{spin9}).