Chapter 9: Groups over Other $\KK$

Some Lorentz Groups over Other Division Algebras

The construction in § 9.2 can be extended to the corresponding Lorentz groups.

We normally think of a vector in Minkowski spacetime in the form \begin{equation} \xx = \begin{pmatrix}t\cr x\cr y\cr z\cr\end{pmatrix} \end{equation} But there is another natural way to package these four degrees of freedom. Consider the complex matrix \begin{equation} \XX = \begin{pmatrix}t+z& x-iy\cr x+iy& t-z\cr\end{pmatrix} = t\,\II + x\,\SIGMA_x + y\,\SIGMA_y + z\,\SIGMA_z \label{spacetime} \end{equation} This matrix is Hermitian, that is \begin{equation} \XX^\dagger = \XX \end{equation} where the Hermitian conjugate of a matrix, denoted by a dagger, is the matrix obtained by taking both the transpose and the (complex) conjugate of the original matrix. Note that $2\times2$ complex Hermitian matrices have precisely 4 (real) independent components; the diagonal elements must be real, and the off-diagonal elements must be (complex) conjugates of each other.

As discussed in § 6.6, Lorentz transformations are transformations which preserve the “squared length” of $\xx$, \begin{equation} |\xx|^2 = x^2 + y^2 + z^2 - t^2 \end{equation} and these transformations form the Lorentz group, $SO(3,1)$.

How can we express this in terms of the matrix $\XX$? The beauty of this description lies in the fact that the norm of $\xx$ is just (minus) the determinant of $\XX$, that is \begin{equation} -\det\XX = |\xx|^2 = x^2 + y^2 + z^2 - t^2 \end{equation} We therefore seek linear transformations which preserve the determinant.

At first sight, this is easy. Since (complex) determinants satisfy \begin{equation} \det(\XX\YY) = (\det\XX) (\det\YY) \label{DetRule} \end{equation} it would seem that all we need to do is multiply $\XX$ (on either side) by a matrix with determinant $1$. There are two problems with this. First of all, there is no simple condition to ensure that the product $\XX\YY$ is Hermitian! Furthermore, this approach will fail over the other division algebras; (\ref{DetRule}) is not true for quaternionic matrices.

We already know how to solve these problems. We consider conjugation of $\XX$ by a matrix $\MM$, that is, we consider the transformation \begin{equation} \XX \longmapsto \MM\XX\MM^\dagger \label{Mtrans} \end{equation} Using the property \begin{equation} (\XX\YY)^\dagger = \YY^\dagger \XX^\dagger \end{equation} it is easy to check that $\MM\XX\MM^\dagger$ is Hermitian if $\XX$ is — with no such restriction on $\MM$. We therefore seek complex matrices $\MM$ such that \begin{equation} \det(\MM\XX\MM^\dagger) = \det(\MM) \det(\XX) \det(\MM^\dagger) = \det\XX \end{equation} or equivalently \begin{equation} \det(\MM\MM^\dagger) = \det(\MM) \det(\MM^\dagger) = 1 \nonumber\\ \end{equation} Since \begin{equation} \det\MM^\dagger=\bar{\det\MM} \end{equation} we must have \begin{equation} |\det\MM|=1 \end{equation} However, over $\CC$, we can assume without loss of generality that $\det\MM=1$, since we can always multiply $\MM$ by a complex phase without affecting anything else. We have in fact shown that the group of complex matrices with determinant $1$, written as $\SL(2,\CC)$, is (locally) the same as the Lorentz group. In fact, \begin{equation} \SL(2,\CC) \cong \Spin(3,1) \end{equation} the double cover of $\SO(3,1)$.

What do these transformations look like? What matrices $\MM$ correspond to spatial rotations? The spatial rotations can be built from the $\SU(2)$ transformations $R_x$, $R_y$, and $R_z$, as discussed in § 7.3. Because of the importance of the Lorentz group, we repeat that construction here, using slightly different notation.

First of all, a rotation by $2\theta$ in the $xy$-plane is given by \begin{equation} \RRR_z = \begin{pmatrix}e^{-i\theta}& 0\cr 0& e^{i\theta}\cr\end{pmatrix} \label{Rz} \end{equation} It is straightforward to check that if \begin{equation} \XX' = \RRR_z \XX \RRR_z^\dagger = \begin{pmatrix}t'+z'& x'-iy'\cr x'+iy'& t'-z'\cr\end{pmatrix} \end{equation} then \begin{align} t' &= t \nonumber\\ x' &= x\cos2\theta-y\sin2\theta \label{Lorentz4}\\ y' &= x\sin2\theta+y\cos2\theta \nonumber\\ z' &= z \nonumber \end{align} which corresponds to a counterclockwise rotation by $2\theta$ in the $xy$-plane as claimed. Of course, (\ref{Lorentz4}) can be put in matrix form, as \begin{equation} \xx' = {\LAMBDA} \xx \end{equation} from which the traditional representation of this Lorentz transformation as a $4\times4$ matrix $\LAMBDA$ can easily be determined. Similarly, rotations in the $yz$- and $zx$-planes are given, respectively, by \begin{equation} \RRR_x = \begin{pmatrix}   \cos\theta& -i\sin\theta\cr -i\sin\theta&   \cos\theta\cr \end{pmatrix} \qquad \RRR_y = \begin{pmatrix} \cos\theta& -\sin\theta\cr \sin\theta&   \cos\theta\cr \end{pmatrix} \end{equation} Any rotation can be built up out of these generators.

What about the other Lorentz transformations, namely the ones which “rotate” the time axis? These transformations, called boosts, are at the heart of special relativity; geometrically boosts are simply hyperbolic rotations. A boost in the $zt$-plane takes the form \begin{equation} \BB_z = \begin{pmatrix}e^\beta& 0\cr 0& e^{-\beta}\cr\end{pmatrix} \end{equation} It is straightforward to check that if \begin{equation} \XX' = \BB_z \XX \BB_z^\dagger \end{equation} then \begin{align} t' &= t\cosh2\beta + z\sinh2\beta \nonumber\\ x' &= x \\ y' &= y \nonumber\\ z' &= t\sinh2\beta + z\cosh2\beta \nonumber \end{align} corresponding to a boost in the $zt$-plane by $2\beta$, that is, to a relative speed of $c\tanh2\beta$. Similarly, boosts in the $xt$- and $yt$-planes are given, respectively, by \begin{equation} \BB_x = \begin{pmatrix} \cosh\beta& \sinh\beta\cr \sinh\beta& \cosh\beta\cr \end{pmatrix} \qquad \BB_y = \begin{pmatrix} \cosh\beta& -i\sinh\beta\cr i\sinh\beta&   \cosh\beta\cr \end{pmatrix} \end{equation}

What happens over the other division algebras? As in § 9.2, simply replace the complex number $x+iy$ in $\XX$ by a division algebra element $a$, so that \begin{equation} \XX = \begin{pmatrix}t+z& \bar{a}\cr a& t-z\cr\end{pmatrix} \end{equation} Since $a$ has 1, 2, 4, or 8 components depending on whether $a$ is in $\RR$, $\CC$, $\HH$, or $\OO$, respectively, $\XX$ corresponds to a vector in a spacetime with 3, 4, 6, or 10 dimensions. As before, the determinant gives the Lorentzian norm \begin{equation} - \det\XX = |a|^2 + z^2 - t^2 \end{equation} There is no problem defining the determinant here, since even in the octonionic case the components of $\XX$ lie in a complex subalgebra of $\OO$.

We are therefore led to seek transformations of the form (\ref{Mtrans}) which preserve the determinant. Even over the quaternions, however, we immediately have a problem: As already noted, (\ref{DetRule}) is not true for quaternionic matrices! Even worse, it is not at all obvious how to define the determinant in the first place for non-Hermitian matrices.

Fortunately, there is another identity which comes to the rescue here. \begin{equation} \det(\MM\XX\MM^\dagger) = \det(\MM^\dagger\MM) \det\XX \label{DetRule2} \end{equation} so we need to look for quaternionic matrices $\MM$ satisfying \begin{equation} \det(\MM^\dagger\MM) = 1 \label{detID} \end{equation} The order doesn't matter here, since \begin{equation} \det(\MM\MM^\dagger) \equiv \det(\MM^\dagger\MM) \end{equation} holds over $\HH$.

Over the octonions, the situation is even worse: because of the lack of associativity, (\ref{Mtrans}) is not well-defined — nor is it clear that the right-hand-side is Hermitian! We resolve this by restricting to those matrices $\MM$ for which $\MM\XX\MM^\dagger$ is well-defined for all Hermitian $\XX$. It turns out to be sufficient to assume that the components of $\MM$ lie in a complex subalgebra of $\OO$. 1) For such matrices, the transformation (\ref{Mtrans}) involves only two independent directions, and is therefore quaternionic. In particular, (\ref{DetRule2}) will hold.

It appears to be straightforward to generalize the rotations and boosts given above. $\RRR_y$ still rotates the real direction in $a$ with $z$; $\BB_z$ still yields a boost in the $z$-direction; $\BB_x$ still yields a boost in the $x$-direction. The remaining transformations are nearly as easy. $\BB_y$ yields a boost in the $i$-direction, with obvious generalizations to the other spatial directions obtained by replacing $i$ by $j$, $k$, etc. Similarly, $\RRR_x$ and $\RRR_z$ yield rotations in the plane defined by either the real part of $a$ or $z$, respectively, and the $i$-direction — again generalizing to $j$, $k$, etc.

Counting up what we've got, we see that we have boosts in all directions, as well as all rotations involving either $z$ or the real part of $a$. That's more than enough to generate all the Lorentz transformations except those involving two imaginary directions in $a$. But we know how to do these!

In §9.1.1, we obtained a rotation in the $jk$-plane for single quaternions by conjugating with $e^{i\theta}$. That works here as well! Conjugating $\XX$ by the matrix \begin{equation} \RRR_i = \begin{pmatrix}e^{i\theta}& 0\cr 0& e^{i\theta}\cr\end{pmatrix} = e^{i\theta}\II \end{equation} won't touch the diagonal of $\XX$, but will precisely rotate $a$ (and $\bar{a}$!) by $2\theta$ in the $jk$-plane. Similarly, replacing $i$ with $j$ and $k$ yields rotations in the other imaginary planes. The phase freedom in the complex case has become an essential ingredient in performing these “internal” rotations over $\HH$!

Over the octonions, we must be a bit more careful. As noted in §9.1.3 and §9.2, conjugation by $e^{i\theta}$, and hence by $\RRR_i$, rotates three planes, not just one. But again, we know how to solve this problem: Use flips! For instance, \begin{equation} \XX \longmapsto (i\cos\theta+j\sin\theta) i \XX i (i\cos\theta+j\sin\theta) \end{equation} is a rotation by an angle $2\theta$ in the $ij$-plane. Yet again, we see that the lack of associativity has come to the rescue; the ability to nest transformations is crucial to this construction.

We have therefore obtained an explicit form for the generators of Lorentz transformations in 3, 4, 6, and 10 dimensions. But we have actually shown more, namely that \begin{align} \SL(2,\RR) &\cong \Spin(2,1) \\ \SL(2,\CC) &\cong \Spin(3,1) \\ \SL(2,\HH) &\cong \Spin(5,1) \\ \SL(2,\OO) &\cong \Spin(9,1) \end{align} that is, $\SL(2,\KK)$ is the double cover of the Lorentz group $\SO(k+1,1)$, where $k=|\KK|=1,2,4,8$. However, the groups $\SL(2,\HH)$ and $\SL(2,\OO)$ require some explanation. The notation “$\SL$” normally means those matrices with determinant $1$, but the determinant of a quaternionic matrix is not well-defined. The generalization? To require precisely (\ref{detID})! Over the octonions, all we must do is to also add the restriction that (\ref{Mtrans}) be well-defined. 2)

In what sense do the Lorentz groups as defined here consist of “all matrices of determinant $1$”? In the complex case, multiplication of $M$ by an arbitrary phase $e^{i\theta}$ does not change the action (\ref{Mtrans}), so we can safely restrict to matrices with $\det M=1$. The quaternionic case is more subtle: Multiplication by $e^{i\theta}$ now corresponds to a rotation in the $jk$-plane, and therefore must be included as a Lorentz transformation even though its determinant is not real. However, we can rewrite such phase transformations as a product of two flips, each with determinant $-1$, and this construction carries over unchanged to the octonionic case. Furthermore, it is straightforward to rewrite the remaining Lorentz generators, which are already matrices with determinant $+1$, as the product of two matrices each with determinant $-1$.

Thus, the Lorentz groups could be defined for each of the division algebras as being generated by those transformations consisting of two complex matrices of determinant $-1$, which suitably generalizes the more traditional definition in terms of matrices of determinant $+1$. It is only in this nested sense that $\SL(2,\OO)$ consists of “all matrices of determinant $+1$.”

1) The only other possibility is for the columns of the imaginary part of $\MM$ to be real multiples of each other [arXiv:hep-th/9302044] .
2) Wait a minute, how can $SL(2,\OO)$ be a group when $\OO$ isn't associative? Simple; the multiplication in $SL(2,\OO)$ is not matrix multiplication, but composition, that is \begin{align} (\MM_1 \bullet \MM_2) [\XX] = \MM_1 \big[ \MM_2 [\XX] \big] = \MM_1 ( \MM_2\XX\MM_2^\dagger) \MM_1^\dagger \nonumber \end{align} where we have used square brackets to denote the action of an element of the group on a vector, and a bullet to indicate the group operation, which is associative. Similar comments apply to other groups constructed over $\OO$, such as $\SU(2,\OO)$.

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