Section 6.9 (Lack of) Associativity
The careful reader will have noticed that we have glossed over the issue of associativity in Section 6.8. How can the octonions, a non-associative algebra, be used to construct a Clifford algebra, which is associative?
The answer is rather subtle: The action of octonionic matrices on a representation is associative, even though octonionic matrix multiplication itself is not!
In other words, when discussing Lie or Clifford algebras over the octonions, we must always bear in mind that these algebras act on representations. Recall from Section 3.1 that a representation of a Lie algebra \(\gg\) on a vector space \(V\) is a map
that takes an element \(P\in\gg\) to a linear map on \(V\text{,}\) that is, to a (real!) matrix \(\rho(P)\) acting on vectors \(v\in V\text{.}\) For matrix Lie algebras, it is common to omit explicit reference to the map \(\rho\text{,}\) but doing so hides some important properties, especially when using non-real matrices to represent real algebras. In particular, if we compose two such actions, we multiply the corresponding matrices, that is,
where the operation on the right is matrix multiplication. But composition of actions is always associative, because the order of operations is fixed (namely from right to left).
Put differently, a collection of octonionic matrices must always be regarded as acting on something. So “multiplication” of such matrices really refers to composition, that is
which may or may not be the same operation as \((PQ)v\text{.}\) There is no way to disentangle \(v\) from this expression. This realization resolves the associativity problem, since, as already noted, composition is associative. Mind you, certain properties, notably the Jacobi identity, must be reinterpreted slightly so as to refer to composition, rather than multiplication.
Returning to the construction of \(\Cl(12,4)\) in Section 6.8, the elements of rank 2 are block diagonal, with blocks of the form \(\sigma_m\tilde{\sigma}_n\text{,}\) where each of \(p\text{,}\) \(q\) are chosen from the units of \(\OO\oplus\OO'\text{.}\) We have claimed that each block acts as \(\so(12,4)\) on the corresponding Weyl spinor representation. Rather than check this claim directly, we consider the action of such a block on vectors such as \(X\) in (6.8.7).
Recall that orthogonal groups \(G\) act on such vectors via the two-sided action
for \(M\in G\text{.}\) Ths operation preserves the determinant, and hence the inner product. At the Lie algebra level, we therefore have
where as usual \(A\in\gg\) is the infinitesimal version of \(M\text{.}\) But “\(M\)” now refers to the composition of two operations, so we must perform them in order. Furthermore, the vectors are themselves elements of the Clifford algebra, so we need to imagine that they, too, are acting on something.
To make a long story short, 1 we need to examine the nested action of \(\so(p,q)\) on vectors of the form \(X\text{,}\) which in turn act on spinors \(v\text{.}\) Remarkably, the alternativity of the octonions comes to the rescue, allowing us to express the Clifford identity in the reduced form
which can be polarized to yield
In principle, if we want to consider the action of \(\sigma_m\circ\tilde{\sigma}_n\in\so(p,q)\) on \(X\text{,}\) we should really examine the action
But (6.9.6) allows us to consider the simpler action
Furthermore, so long as we work with orthogonal generators such as \(\sigma_m\text{,}\) we can (mostly) ignore the tilde operation, which merely contributes an overall sign in some computations.
With that lengthy preamble, we are finally in a position to verify that the action of \(\so(p,q)\) on vectors \(X\) is given (up to sign) by
where in both cases the order of operations is from right to left. But the rest of the argument is straightforward, using the anticommutativity of the \(\sigma_m\text{.}\) So long as \(m\ne n\text{,}\) the action (6.9.10) mixes up \(\sigma_m\) and \(\sigma_n\) in \(X\text{,}\) but leaves the \(\sigma_r\) term alone if \(r\ne m,n\text{.}\) Furthermore, \(\sigma_m\) and \(\sigma_n\) transform with the appropriate signs for an infinitesimal rotation or boost, depending on the sign of the squares of \(\sigma_m\) and \(\sigma_n\text{.}\)
Obtaining the corresponding group elements requires preserving this notion of composition. The answer, as given in [5] based on earlier work by Manogue and Schray [6], is to introduce the notion of flips.
Associativity is only an issue for operations that involve more than two elements from the same (possibly split) division algebra. So generalized rotations between octonionic and split-octonionic directions, or that involve the “\(x\)” or “\(z\)” directions can be handled as usual, leaving only the “transverse” rotations in \(\Im(\OO)\) or \(\Im(\OO')\text{.}\) Consider the operation
where \(u\) is an imaginary unit element of either \(\OO\) or \(\OO'\text{.}\) This operation preserves the \(u\) direction, but multiplies all orthogonal (imaginary) directions (in the same algebra) by \(-1\text{,}\) thus flipping them. It is well known that a second flip about some other imaginary unit element \(w\) (of the same algebra) yields a rotation in the \(uw\)-plane—and of course flips the remaining (imaginary) directions back to where they started.
For example, to generate a rotation through an angle \(\alpha\) in the \(ij\)-plane, first flip around \(i\text{,}\) and then around \(i\,\cos\left(\frac\alpha2\right)+j\,\sin\left(\frac\alpha2\right)\text{.}\) In the associative case, we could collapse these two operations as
so that such nested flips would yield the expected answer (possibly up to sign). However, in the non-associative case, \(e^{k\theta}\) would rotate all three planes orthogonal to \(k\text{,}\) whereas the nested flip implicit on the right-hand side rotates only the \(ij\)-plane. Furthermore, the infinitesimal versions (derivative at zero) of such nested flips agree with the nested generators \(\sigma_m\circ\sigma_n\) considered earlier, which we have therefore succeeded in exponentiating.