The Geometry of $SU(2,\KK'\otimes\KK)$

The groups in Table 1 are all orthogonal groups. What is the pattern to their signatures? Let's go ahead and label the rows with the split division algebras, as shown in Table 2. Let $\kappa=|\KK|=1,2,4,8$, and for $\KK'$ keep track separately of the number of positive-normed basis units, $\kappa'_+=1,1,2,4$, and negative-normed basis units, $\kappa'_-=0,1,2,4$, with $\kappa'_++\kappa'_-=|\KK'|$. Then the groups in the table are just $\SO(\kap)$! So we need a representation of $\SO(\kap)$. 1) The orthogonal groups are closely related to spinors and Clifford algebras, and we can translate this standard relationship into the language of division algebras.

We choose as a basis for the octonions $\OO$ the unit elements \begin{equation} \{e_\inda, \inda=1…8\} = \{1,i,j,k,k\ell,j\ell,i\ell,\ell\} \end{equation} and similarly for the split octonions $\OO'$ the unit elements \begin{equation} \{e_\inda, \inda=9…16\} = \{1,I,J,K,KL,JL,IL,L\} \end{equation} In what follows, the split quaternions $\HH'$ have basis $\{1,K,KL,L\}$, the split complex numbers $\CC'$ have basis $\{1,L\}$, and of course $\RR'\equiv\RR$.

We will work with the algebra $\OO'\otimes\OO$ and its subalgebras $\KK'\otimes\KK$, where $\KK$ is any of the division algebras $\RR$, $\CC$, $\HH$, $\OO$, and $\KK'$ any of their split versions, assumed to commute with each other. Elements of $\OO'\otimes\OO$ are simply (sums of) products of elements of $\OO'$ and $\OO$, such as “$jK$”. We assume that $\OO$ and $\OO'$ commute with each other, that is \begin{equation} aA = Aa \end{equation} for all $a\in\OO$, $A\in\OO'$, and we will therefore usually write elements of $\OO$ before elements of $\OO'$.

The Clifford Algebra $\Cl(\kap)$

We consider matrices of the form \begin{equation} \label{defx} \XX = \begin{pmatrix}A&\bar{a}\\a&-\star{A}\end{pmatrix} \end{equation} with $A\in\KK'$ and $a\in\KK$. Then $\XX$ can be written as \begin{equation} \XX = x^\inda\SIGMA_\inda \label{Xdef} \end{equation} where there is an implicit sum over the index $\inda$, which takes on values between $1$ and $16$ as appropriate for the case being considered. When not stated otherwise, we assume $\KK'=\OO'$ and $\KK=\OO$, as all other cases are special cases of this one. Equation (\ref{Xdef}) defines the generalized Pauli matrices $\SIGMA_a$, which are given this name because $\SIGMA_1$, $\SIGMA_2$, and $\SIGMA_9$ are just the usual Pauli spin matrices.

We define the determinant of $\XX$ in the obvious way, namely \begin{equation} \det\XX = -A\star{A}-\bar{a}a = -|A|^2-|a|^2 \label{det2} \end{equation} The collection of matrices of the form $\XX$ forms a normed vector space of signature $(\kap)$, using ($-1$ times) the determinant as the norm. We can therefore identify the resulting vector space with $\RR^{\kap}$. 2)

We now consider the $4\times4$ matrix \begin{equation} \PP = \begin{pmatrix}0&\XX\\\widetilde{\XX}&0\end{pmatrix} = x^\inda\GAMMA_\inda, \label{Pdef} \end{equation} where tilde represents trace reversal, \begin{equation} \widetilde{\XX} = \XX - \text{tr}(\XX)\,\II, \end{equation} and where the gamma matrices $\GAMMA_\inda$ are implicitly defined by (\ref{Pdef}). The only $\SIGMA_a$ which are affected by trace reversal are those containing an imaginary element of $\KK'$, which are imaginary multiples of the identity matrix. A straightforward computation using the commutativity of $\KK$ with $\KK'$ now shows that \begin{equation} \{\GAMMA_\inda,\GAMMA_\indb\} = 2g_{\inda\indb}\II \label{CliffordID} \end{equation} where $\II$ is the identity matrix and \begin{equation} g_{\inda\indb} = \begin{cases} 0&\inda\neq\indb\\ 1&1\le\inda=\indb<13\}\\ -1&13\le\inda=\indb\le16\} \end{cases} \end{equation} These are precisely the anticommutation relations necessary to generate a representation of the (real) Clifford algebra $\Cl(12,4)$ in the case of $\OO'\otimes\OO$, and $\Cl(\kap)$ in general.

However, Clifford algebras are associative, so our algebra must also be associative. Since the octonions are not associative, neither are matrix groups over the octonions, at least not as matrix groups. The resolution to this puzzle is to always consider octonionic “matrix groups” as linear transformations acting on a particular vector space, and to use composition, rather than matrix multiplication, as the group operation. This construction always yields an associative group, since composition proceeds in a fixed order, from the inside out.

Let's start again. The Clifford identity that we need to verify is not (\ref{CliffordID}) itself, but rather the action of (\ref{CliffordID}) on (linear combinations of) products of gamma matrices, such as $\GAMMA_\indc$, in which case we need to establish that \begin{equation} \GAMMA_\inda\left(\GAMMA_\indb\GAMMA_\indc\right) + \GAMMA_\indb\left(\GAMMA_\inda\GAMMA_\indc\right) = 2g_{\inda\indb}\GAMMA_c \label{CliffordID2} \end{equation} If the associator of the corresponding units vanishes, that is, if $[e_\inda,e_\indb,e_\indc]=0$, then (\ref{CliffordID2}) reduces to (\ref{CliffordID}), so we only need to consider cases where the associator is nonzero. But in this case an extra minus sign is required in each term on the left-hand side of (\ref{CliffordID2}) in order to factor out $\GAMMA_c$, which changes nothing since both sides are $0$. This argument establishes (\ref{CliffordID2}), and it is easy to see that the argument still holds if $\GAMMA_c$ is replaced by any product of gamma matrices. Thus, the linear transformations $\{\GAMMA_\inda\}$ do indeed generate a Clifford algebra, namely $\Cl(\kap)$.

The Orthogonal Group $\SO(\kap)$

Representations of Clifford algebras normally lead to representations of the corresponding orthogonal groups, using a well-known construction, so we expect $\Cl(\kap)$ to lead us to a representation of $\SO(\kap)$; we show here that it does. However, our use of nonassociative divisions algebras in our representations requires some modifications to the standard construction.

In the associative case, the homogeneous quadratic elements of $\Cl(\kap)$ act as generators of $\SO(\kap)$ via the map \begin{equation} \label{paction} \PP \longmapsto \MM_{\inda\indb}\PP\MM_{\inda\indb}^{-1} \end{equation} where \begin{equation} \MM_{\inda\indb} = \exp\left(-\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) \label{mdef} \end{equation} and $\PP = x^\inda\GAMMA_\inda$ as above. So we consider first the case where $[e_\inda,e_\indb,e_\indc]=0$, with $\inda,\indb,\indc$ assumed to be distinct. Then the original Clifford identity (\ref{CliffordID}) implies that \begin{align} \GAMMA_\inda\GAMMA_\inda &= \pm \II, \label{prop1}\\ (\GAMMA_\inda\GAMMA_\indb)\GAMMA_\indc &= \GAMMA_\indc(\GAMMA_\inda\GAMMA_\indb), \label{prop2}\\ (\GAMMA_\inda\GAMMA_\indb)\GAMMA_\indb &= (\GAMMA_\indb)^2\GAMMA_\inda = g_{\indb\indb}\GAMMA_\inda, \label{prop3}\\ (\GAMMA_\inda\GAMMA_\indb)\GAMMA_\inda &= -(\GAMMA_\inda)^2\GAMMA_\indb = -g_{\inda\inda}\GAMMA_\indb, \label{prop4}\\ (\GAMMA_\inda\GAMMA_\indb)^2 &= -\GAMMA_\inda^2\GAMMA_\indb^2 = \pm \II \label{prop5}, \end{align} With these observations, we compute \begin{equation} \label{action4} \MM_{\inda\indb}\PP\MM_{\inda\indb}^{-1} = \exp\left(-\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) \left(x^\indc\GAMMA_\indc\right) \exp\left(\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right). \end{equation} From (\ref{prop1}), if $\inda=\indb$, then $\MM_{\inda\indb}$ is a real multiple of the identity matrix, which therefore leaves $\PP$ unchanged under the action (\ref{paction}). On the other hand, if $\inda\neq\indb$, properties (\ref{prop2})–(\ref{prop4}) imply that $\MM_{\inda\indb}$ commutes with all but two of the matrices $\GAMMA_\indc$. We therefore have \begin{equation} \GAMMA_\indc\MM_{\inda\indb}^{-1} = \begin{cases} \MM_{\inda\indb}\GAMMA_\indc, &\indc=\inda\text{ or }\indc=\indb\\ \MM^{-1}_{\inda\indb}\GAMMA_\indc, &\inda\neq\indc\neq\indb \end{cases} \label{maction} \end{equation} so that the action of $\MM_{\inda\indb}$ on $\PP$ affects only the $\inda\indb$ subspace. To see what that action is, we first note that if $\AA^2 = \pm\II$ then \begin{equation} \label{euler} \exp\left(\AA\alpha\right) = \II\,c(\alpha) + \AA\,s(\alpha) = \begin{cases} \II\,\cosh(\alpha) + \AA\,\sinh(\alpha),&\AA^2 = \II\\ \II\,\cos(\alpha) + \AA\,\sin(\alpha),&\AA^2 = -\II \end{cases} \end{equation} where the second equality serves to define the functions $c$ and $s$. Inserting (\ref{maction}) and (\ref{euler}) into (\ref{action4}), we obtain \begin{align} \label{action} \MM_{\inda\indb} \left( x^\inda\GAMMA_\inda + x^\indb\GAMMA_\indb \right) \MM_{\inda\indb}^{-1} &= \left(\MM_{\inda\indb}\right)^2 \left(x^\inda\GAMMA_\inda + x^\indb\GAMMA_\indb\right) \nonumber\\ &= \exp\left(-\GAMMA_\inda\GAMMA_\indb\theta\right) \left(x^\inda\GAMMA_\inda + x^\indb\GAMMA_\indb\right) \nonumber\\ &= \bigl( \II\,c(\theta) - \GAMMA_\inda\GAMMA_\indb\,s(\theta) \bigr) \left(x^\inda\GAMMA_\inda + x^\indb\GAMMA_\indb\right) \nonumber\\ &= \left(x^\inda c(\theta) - x^\indb s(\theta)g_{\indb\indb}\right) \GAMMA_\inda + \left(x^\indb c(\theta) + x^\inda s(\theta)g_{\inda\inda}\right) \GAMMA_\indb. \qquad\quad \end{align} Thus, the action (\ref{paction}) is either a rotation or a boost in the $\inda\indb$-plane, depending on whether \begin{equation} (\GAMMA_\inda\GAMMA_\indb)^2 = \pm\II \end{equation} More precisely, if $\inda$ is spacelike ($g_{\inda\inda}=1$), then (\ref{paction}) corresponds to a rotation by $\theta$ from $\inda$ to $\indb$ if $\indb$ is also spacelike, or to a boost in the $\inda$ direction if $\indb$ is timelike ($g_{\indb\indb}=-1$), whereas if $\inda$ is timelike, the rotation (if $\indb$ is also timelike) goes from $\indb$ to $\inda$, and the boost (if $\indb$ is spacelike) is in the negative $\inda$ direction.

If $\KK'\otimes\KK=\HH'\otimes\HH$ (or any of its subalgebras), we're done: Since transformations of the form (\ref{paction}) preserve the determinant of $\PP$, it is clear from (\ref{det2}) that we have constructed $\SO(6,2)$ (or one of its subgroups).

What about the nonassociative case? We can no longer use (\ref{prop2}), which now contains an extra minus sign. A different strategy is needed.

If $e_\inda$, $e_\indb$ commute, then they also associate with every basis unit, that is \begin{equation} [e_\inda, e_\indb] = 0 \Longrightarrow [e_\inda, e_\indb,e_\indc]=0 \end{equation} and the argument above leads to (\ref{action}) as before. We therefore assume that $e_\inda$, $e_\indb$ anticommute, the only other possibility; in this case, $e_\inda$, $e_\indb$ are imaginary basis units that either both lie in in $\OO$, or in $\OO'$. As before, we seek a transformation that acts only on the $\inda\indb$ subspace. Consider the transformation \begin{equation} \label{flip} \PP \longmapsto e_\inda\PP e_\inda^{-1} \end{equation} which preserves directions corresponding to units $e_\indb$ that commute with $e_\inda$, and reverses the rest, which anticommute with $e_\inda$. We call this transformation a flip about $e_\inda$; any imaginary unit can be used, not just basis units. If we compose flips about any two units in the $\inda\indb$ plane, then all directions orthogonal to this plane are either completely unchanged, or flipped twice, and hence invariant under the combined transformation. Such double flips therefore affect only the $\inda\indb$ plane.

The rest is easy. We nest two flips, replacing (\ref{paction}) by \begin{equation} \PP \longmapsto \MM_2\left(\MM_1\PP\MM_1^{-1}\right)\MM_2^{-1} \label{nest} \end{equation} where \begin{align} \MM_1 &= -e_\inda\,\II \nonumber\\ \MM_2 &= \left(e_\inda\,c(\halfang)+e_\indb\,s(\halfang)\right)\,\II \nonumber\\ &= \begin{cases} \left(e_\inda\cosh(\halfang) + e_\indb\,\sinh(\halfang)\right)\II, &(e_\inda e_\indb)^2 = 1 \\ \left(e_\inda\cos(\halfang)+e_\indb\,\sin(\halfang)\right)\II, &(e_\inda e_\indb)^2 = -1 \end{cases} \label{M12} \end{align} Using the relationships \begin{align} \bigl(e_\inda c(\alpha)+e_\indb s(\alpha)\bigr)^2 = e_\inda^2 c^2(\alpha)+e_\indb^2 s^2(\alpha) &= e_\inda^2 = -g_{\inda\inda}\\ e_\inda^2 c^2(\alpha)-e_\indb^2 s^2(\alpha) &= -g_{\inda\inda} c(2\alpha) \\ 2s(\alpha)c(\alpha) &= s(2\alpha) \end{align} we now compute \begin{align} &\MM_2\left( \MM_1 \left( x^\inda\GAMMA_\inda + x^\indb\GAMMA_\indb \right) \MM_1^{-1} \right) \MM_2^{-1} = \MM_2\left( x^\inda\GAMMA_\inda - x^\indb\GAMMA_\indb \right) \MM_2^{-1} \nonumber\\ &\qquad\qquad = \left(e_\inda\,c(\halfang)+e_\indb\,s(\halfang)\right) \left( x^\inda\GAMMA_\inda - x^\indb\GAMMA_\indb \right) \left(e_\inda\,c(\halfang)+e_\indb\,s(\halfang)\right) (-g_{\inda\inda}) \nonumber\\ &\qquad\qquad = \left(x^\inda c(\theta) - x^\indb s(\theta)\, g_{\inda\inda}\,g_{\indb\indb}\right) \GAMMA_\inda + \left(x^\indb c(\theta) + x^\inda s(\theta)\right) \GAMMA_\indb. \end{align} and we have constructed the desired rotation in the $ab$ plane.

We also have \begin{equation} \GAMMA_\inda\GAMMA_\indb = -e_ae_b\,\II \qquad\qquad ([e_\inda,e_\indb]\ne0) \end{equation} so in the associative case (with $e_\inda$, $e_\indb$ anticommuting), we have \begin{equation} \MM_2\MM_1 = \left( g_{\inda\inda} c(\halfang) + e_\inda e_\indb s(\halfang) \right) \II = g_{\inda\inda} \exp\left( -g_{\inda\inda} \GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) \end{equation} which differs from $M_{\inda\indb}$ only in replacing $\theta$ by $-\theta$ (and an irrelevant overall sign) if $g_{\inda\inda}=-1$. In other words, the nested action (\ref{nest}) does indeed reduce to the standard action (\ref{paction}) in the associative case, up to the orientations of the transformations. In this sense, (\ref{nest}) is the nonassociative generalization of the process of exponentiating homogeneous elements of the Clifford algebra in order to obtain rotations in the orthogonal group.

We therefore use (\ref{paction}) if $e_\inda$ and $e_\indb$ commute, and (\ref{nest}) if they don't. Since both of these transformations preserve the determinant of $\PP$, it is clear from (\ref{det2}) that we have constructed $\SO(\kap)$ from $\Cl(\kap)$.

The Group $\SU(2,\KK'\otimes\KK)$

So far we have considered transformations of the form (\ref{paction}) and (\ref{nest}) acting on $\PP$. In light of the off-diagonal structure of the matrices $\{\GAMMA_\inda\}$, we can also consider the effect these transformations have on $\XX$. First, we observe that trace-reversal of $\XX$ corresponds to conjugation in $\KK'$, that is, \begin{equation} \widetilde{\SIGMA_\inda} = \star{\SIGMA}_\inda. \end{equation} The matrices $\GAMMA_\inda\GAMMA_\indb$ then take the form \begin{equation} \GAMMA_\inda\GAMMA_\indb = \begin{pmatrix} \SIGMA_\inda\star{\SIGMA}_\indb&0\\ 0&\star{\SIGMA}_\inda\SIGMA_\indb \end{pmatrix} \end{equation} and, in particular, \begin{equation} \exp\left(\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) = \begin{pmatrix} \exp\left( \SIGMA_\inda\star{\SIGMA}_\indb\>\frac{\theta}{2} \right)&0\\ \noalign{\smallskip} 0&\exp\left( \star{\SIGMA}_\inda\SIGMA_\indb\>\frac{\theta}{2} \right) \end{pmatrix}, \end{equation} so we can write \begin{align} &\exp\left(-\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) \>\PP\> \exp\left(\GAMMA_\inda\GAMMA_\indb\>\frac{\theta}{2}\right) \nonumber\\ &= \begin{pmatrix} 0&\exp\left( -\SIGMA_\inda\star{\SIGMA}_\indb\>\frac{\theta}{2} \right) \XX \exp\left( \star{\SIGMA}_\inda\SIGMA_\indb\>\frac{\theta}{2} \right)\\ \noalign{\smallskip} \exp\left( -\star{\SIGMA}_\inda\SIGMA_\indb\>\frac{\theta}{2} \right) \widetilde{\XX} \exp\left(\SIGMA_\inda\star{\SIGMA}_\indb\>\frac{\theta}{2} \right)&0 \end{pmatrix}. \qquad\quad \end{align} The $4\times4$ action (\ref{paction}) acting on $\PP$ is thus equivalent to the action \begin{equation} \XX \longmapsto \exp\left( -\SIGMA_\inda\star{\SIGMA}_\indb\>\frac{\theta}{2} \right) \XX \exp\left( \star{\SIGMA}_\inda\SIGMA_\indb\>\frac{\theta}{2} \right). \label{xaction} \end{equation} on $\XX$.

Transformations of the form (\ref{nest}) are even easier, since each of $\MM_1$ and $\MM_2$ are multiples of the identity matrix $\II$. These transformations therefore act on $\XX$ via \begin{equation} \XX \longmapsto \MM_2\left(\MM_1\XX\MM_1^{-1}\right)\MM_2^{-1} \end{equation} where $\MM_1$, $\MM_2$ are given by (\ref{M12}), but with $\II$ now denoting the $2\times2$ identity matrix.

Since $\XX$ is Hermitian with respect to $\KK$, and since that condition is preserved by (\ref{xaction}), we have realized $\SO(\kap)$ in terms of (possibly nested) determinant-preserving transformations involving $2\times2$ matrices over $\KK'\otimes\KK$. This $2\times2$ representation of $\SO(\kap)$ therefore deserves the name $\SU(2,\KK'\otimes\KK)$.

Magic Squares

This construction is reminiscent of the Freudenthal–Tits magic square of Lie algebras, which is discussed further in § 16.3. Indeed, Barton and Sudbery discuss the restriction of the Freudenthal–Tits magic square to precisely the case considered here, albeit at the Lie algebra level. Their restricted magic square contains the 16 Lie algebras \begin{equation} \alg_2(\KK_1,\KK_2) = sa(2,\KK_1\otimes\KK_2) \oplus \so(\Im\,\KK_1) \oplus \so(\Im\,\KK_2) \label{Vin2} \end{equation} where $\KK_1$, $\KK_2$ are division algebras (or possibly their split cousins), $sa(n,\KK_1)$ denotes the anti-Hermitian tracefree $n\times n$ matrices with elements in $\KK_1$ and $\so(\Im\,\KK_1)$ denotes the isometry Lie algebra of the imaginary elements of $\KK_1$. The algebras $\RR$ and $\CC$ have no such isometries, and \begin{align} \so(\Im\,\HH) &= \so(3) \\ \so(\Im\,\OO) &= \so(7) \end{align} We have argued here that the group version of (\ref{Vin2}) is \begin{equation} V_2(\KK_1,\KK_2) = \SU(2,\KK_1\otimes\KK_2) \end{equation} that is, that the groups in this $2\times2$ magic square deserve to be called $\SU(2,\KK_1\otimes\KK_2)$. We have presented the construction above only for the “half-split” case, with $\KK_1=\KK'$ and $\KK_2=\KK$, which, as already noted, has interesting applications to physics, but the other two cases (“non-split” and “double-split”) are similar.

$\RR$ $\CC$ $\HH$ $\OO$
$\RR'$ $\SO(2)$ $\SO(3)$ $\SO(5)$ $\SO(9)$
$\CC'$ $\SO(2,1)$ $\SO(3,1)$ $\SO(5,1)$ $\SO(9,1)$
$\HH'$ $\SO(3,2)$ $\SO(4,2)$ $\SO(6,2)$ $\SO(10,2)$
$\OO'$ $\SO(5,4)$ $\SO(6,4)$ $\SO(8,4)$ $\SO(12,4)$

Table 2: The “half-split” $2\times2$ magic square of Lie groups.

1) Throughout this section, we continue to use $\SO(\kap)$ to refer to its double cover, the spin group $\Spin(\kap)$.
2) There is a similar construction using matrices of the form \[ \XX' = \begin{pmatrix}A&\bar{a}\\a&\star{A}\end{pmatrix} \label{JordanK} \] which have some remarkable properties. As a vector space, of course, the collection of such matrices provides another representation of $\RR^{\kap}$. But matrices of the form (\ref{JordanK}), unlike those of the form (\ref{defx}), close under multiplication; not only do such matrices satisfy their characteristic equation, but the resulting algebra is a Jordan algebra.

Personal Tools