Section 6.5 Clifford Rotations
The rotations \(\SO(p,q)\) act on \(\RR^{p,q}\text{.}\) But a copy of this vector space is contained in \(\Cl(p,q)\text{,}\) namely as its elements of rank 1. Can we use the Clifford algebra to describe \(\SO(p,q)\) and/or its Lie algebra \(\so(p,q)\text{?}\)
A vector in \(\RR^{p,q}\subset\Cl(p,q)\) has the form \(v=v^m\gamma_m\text{.}\) Assuming that both of \(\gamma_i\) and \(\gamma_j\) are spacelike (\(\gamma_m^2=+\one\)), a rotation in the \((i,j)\)-plane has the form
where \(i,j,k\) are assumed to be distinct. The infinitesimal form of this rotation is therefore
which is obtained by differentiating with respect to \(\alpha\) and setting \(\alpha=0\text{.}\) If \(\gamma_j\) is instead timelike, then \(\sin\) and \(\cos\) need to be replaced by \(\sinh\) and \(\cosh\text{,}\) respectively, in (6.5.1), and the minus sign removed from (6.5.1)–(6.5.2). Direct computation shows that
which is enough to establish that the infinitesimal rotation or boost in the \((i,j)\)-plane is generated by commuting with the element \(\frac12\gamma_j\gamma_i\text{.}\) 1 Since commutation is an operation in \(\Cl(V)\text{,}\) we have indeed obtained an action of the Lie algebra \(\so(p,q)\) on vectors, all within \(\Cl(p,q)\text{.}\) Direct computation now verifies that the \(n\choose 2\) elements of rank 2 indeed have the same commutators as the corresponding infinitesimal rotations. Thus, \(\so(p,q)\) itself is contained within \(\Cl(p,q)\text{!}\)
Furthermore, we can exponentiate this description of \(\so(p,q)\subset\Cl(p,q)\) to obtain the group \(\SO(p,q)\subset\Cl(p,q)\text{.}\) Explicitly, reversing the construction above yields the rotation matrices
which correctly contain circular or hyperbolic trigonometric functions depending on the character of \(\gamma_i\) and \(\gamma_j\text{.}\)