Section 7.2 From \(2\times2\) to \(3\times3\)
We showed in Section 7.1 that there are three families of orthogonal Lie algebras parameterized by a pair of (possibly split) division algebras, which deserve the name \(\su(2,\KK'\otimes\KK)\text{.}\) 1 We also showed that these Lie algebras act on vectors of the form given in (6.8.7). This action involves two-sided multiplication—as is standard for Clifford (sub)algebras acting on their elements of rank 1. What column vectors do these algebras act on? That is, what are their spinor representations?
The answer is straightforward. Each orthogonal Lie algebra in one of the \(2\times2\) magic squares has been written explicitly in terms of \(2\times2\) anti-Hermitian matrices over \(\KK'\otimes\KK\text{.}\) Such matrices can have any single element of \(\KK'\otimes\KK\) in one off-diagonal position, but the other off-diagonal component is then determined. And the diagonal elements must belong to \(\Im(\KK')\oplus\Im(\KK)\text{.}\) So it seems natural to act with such matrices on elements of \((\KK'\otimes\KK)^{2\times2}\text{.}\) 2 Since we have divided the original \(4\times4\) gamma matrices into \(2\times2\) blocks, effectively restricting the Clifford algebra to its even subalgebra, we conclude that the elements of \((\KK'\otimes\KK)^{2\times2}\) are Weyl spinors of \(\su(2,\KK'\otimes\KK)\text{.}\)
Not so fast. We haven't yet checked that our spinor representations are irreducible, and in most cases they are not. Recall that \(\frac12(1\pm L)\in\CC'\) act as projection operators. So if \(\CC'\subset\KK'\) we actually get two Weyl representations.
For example, the Lorentz algebra \(\so(3,1)\) corresponds to \(\su(2,\CC'\otimes\CC)\text{,}\) whose spinors are two-component columns whose elements are drawn from the four-dimensional algebra \(\CC'\otimes\CC\text{.}\) Thus, we have eight (real) degrees of freedom, which divide neatly into two four-component Weyl spinor representations, multiplied by \(\frac12(1\pm L)\text{,}\) respectively.
What happens if we consider \(3\times3\) matrices? We will see in Section 7.3 that there is indeed a \(3\times3\) magic square, and it includes the exceptional Lie algebras \(\ff_4\text{,}\) \(\ee_6\text{,}\) \(\ee_7\text{,}\) and \(\ee_8\text{.}\) Before doing so, we digress briefly to examine further the embedding of \(2\times2\) matrices into \(3\times3\) matrices.
One way to embed a \(2\times2\) matrix into a \(3\times3\) matrix is given by
Similarly, we can combine vectors and spinors by defining
where \(\theta\in(\KK'\otimes\KK)^{2\times2}\) denotes a spinor, and \(\phi\) denotes a new, scalar degree freedom. If \(M\) is an element of the group \(\SO(p,q)\text{,}\) what is the action of \(\MMM\) on \(\XXX\text{?}\) We have
which correctly reproduces the vector action of the group element \(M\) on \(X\text{,}\) while at the same time yielding the spinor action of \(M\) on \(\theta\)—and also leaving \(\phi\) alone (thus justifying having called it a scalar)!
These observations suggest that \(3\times3\) matrices of the form (7.2.2) might serve as the basic representation of some bigger Lie group containing to the group generated by \(3\times3\) matrices of the form (7.2.1). Unfortunately, this approach only works if one of the division algebras is \(\RR\) or \(\CC\text{.}\) In retrospect, this conclusion shouldn't be surprising, as only in such cases do all of the off-diagonal elements have the same number of (real) degrees of freedom. Thus, any attempt to expand the group beyond \(SO(p,q)\) will lead to mixing up “vector” and “spinor” components, thus forcing the inclusion of additional elements in the representation. Nonetheless, this approach is quite fruitful for studying the first two rows, as we will see again in Chapter 8.