Subgroups of $SL(4, \mathbf{R})$

A subgroup of $GL(4, \mathbf{R})$ that leaves a nondegenerate antisymmetric covariant spinor $f$ invariant, namely

$$a^T f a = f, \qquad \det f \neq 0,$$ (3.73)

is contained in $SL(4, \mathbf{R})$. It is isomorphic to the symplectic group $Sp(4, \mathbf{R})$ and locally isomorphic to the anti-de Sitter group $SO(2, 3)$, namely to the subgroup of $SO(3,3)$ that leaves the 6-vector $f_u$ invariant.

All these symplectic subgroups are reciprocally conjugate and we shall use only the ones that contain the subgroup $SL(2, \mathbf{C})$ defined by eq. (3.27). They are defined by the condition (3.73) with

$$f = \cos \alpha C + \sin \alpha G, \qquad - \pi/2 < \alpha \leq \pi/2$$ (3.74)

and we indicate them by $Sp(4, \mathbf{R})_{\alpha}$. In particular, we call $Sp(4, \mathbf{R})_0 = Sp(4, \mathbf{R})_V$ the vector symplectic subgroup, because its Lie algebra contains, besides the six independent generators $\Sigma_{[ik]}$ of $SL(2, \mathbf{C})$, the four elements $\Sigma_ {[i5]}$ that transform as the components of a 4-vector. In a similar way we call $Sp(4, \mathbf{R})_{\pi/2} = Sp(4, \mathbf{R})_A$ the axial symplectic subgroup, because its Lie algebra contains, besides the generators of $SL(2, \mathbf{C})$, the four elements $\Sigma_ {[i4]}$ that transform as the components of an axial vector. We indicate by $SO(2, 3)_V$ and $SO(2, 3)_A$ the corresponding anti-de Sitter groups or, more exactly, their identity connected components. They leave invariant, respectively, the components $f_4$ and $f_5$ of a 6-dimensional vector $f_u$.

We indicate by $\mathcal{F}_0$ the subgroup of $GL(4, \mathbf{R})$ containing the transformations that do not mix the vertical and the horizontal subspaces and do not involve the parameter $\ell$. It is defined by the condition

$$a \gamma_5 = \pm \gamma_5 a$$ (3.75)

and it contains the subgroup $SL(2, \mathbf{C)}$, the space reflection represented by $a = \pm \gamma_0$, the dilatations of $\mathcal{T}$ and the one-parameter subgroup $U(1)_5$ generated by $\Sigma_{[45]}$ and described by the formulas

$$a = \exp(2^{-1}\alpha \gamma_5),$$ (3.76)

$$b^i \to b^i, \qquad b^{[ik]} \to \cos(\alpha) b^{[ik]} + 2^{-1} \sin(\alpha) \epsilon^{ik}{}_{jl} b^{[jl]}.$$ (3.77)

In particular, for $\alpha = \pm \pi$ we obtain the vertical reflection, represented by $a = \pm \gamma_5$, which changes the sign of the vertical vectors leaving the horizontal vectors unchanged. Note that

$$\gamma_0^T C \gamma_0 = C, \qquad \gamma_0^T G \gamma_0 = - G,$$

$$\gamma_5^T C \gamma_5 = - C, \qquad \gamma_5^T G \gamma_5 = - G,$$ (3.78)

and therefore

$$\gamma_0 \in Sp(4, \mathbf{R})_V, \qquad \gamma_0 \gamma_5 \in Sp(4, \mathbf{R})_A.$$ (3.79)

Since these groups are connected, a symmetry with respect to $\gamma_0$ or $\gamma_0 \gamma_5$ follows from the symmetry with respect to infinitesimal transformations of the corresponding group. Since space inversion is not a symmetry of nature, this remark gives an argument in favour of of the choice of $Sp(4, \mathbf{R})_A$ as a high symmetry group, as it was observed in refs. [6] and [12].

If, after the rescaling

$$\ell^{-1} \zeta^{[i4]} = \zeta_A^i, \qquad \ell^{-1} \zeta^{[i5]} = \zeta_V^i$$ (3.80)

of the parameters, we perform the limit $\ell \to 0$, we obtain a contraction [87,88] of the group $GL(4, \mathbf{R})$. The subgroup $\mathcal{F}_0$, that does not involve $\ell$ is not affected by the the contraction. The contracted group is a semi-direct product of $\mathcal{F}_0$ and a 8-dimensional commutative group parametrized by $\zeta_A^i$ and $\zeta_V^i$.

By taking the limit of eqs. (3.71) and (3.72), we find the transformation laws with respect to the commutative subgroup of the contracted group

$$p_i \to p_i + \zeta_V^k p_{[ik]} - 2^{-1} \zeta_A^j \epsilon_{ij}{}^{kl} p_{[kl]}, \qquad p_{[ik]} \to p_{[ik]},$$ (3.81)

$$b^i = b^i, \qquad b^{[ik]} \to b^{[ik]} + \zeta_V^i b^k - \zeta_V^k b^i - \zeta_A^j \epsilon^{ik}{}_{jl} b^l,$$ (3.82)

that also describe finite transformations. Note that the contracted group is a symmetry group of the normal wedge defined by eq. (3.5). The complete symmetry group of this wedge, however, is much larger.

In order to introduce the fundamental length $\ell$ by means of a symmetry group, in the same way as one introduces the fundamental velocity assuming the Lorentz symmetry, one has to use at least one of the symplectic groups $Sp(4, \mathbf{R})_{\alpha}$. We shall see in Chapter 7 that $Sp(4, \mathbf{R})_A$ is the best choice. This group is locally isomorphic to the anti-de Sitter group $SO(2, 3)_A$, and it is possible to develop a 5-dimensional tensor formalism based on this group, which was introduced in ref. [6].

We have already seen in Section 3.5 that an antisymmetric covariant spinor $f_{[AB]}$ is equivalent to a 6-dimensional $SO(3,3)$ vector $f_u$. If we consider only the subgroup $SO(2, 3)_A$, the quantities $f_u$, $u = 0,\ldots, 4$ are the components of a 5-dimensional vector and $f_5$ is a scalar. We have also seen that a contravariant symmetric spinor $b^{(AB)}$ is equivalent to a 6-dimensional self-dual antisymmetric tensor $b^{[uvw]}$, that is completely determined by its components of the form $b^{[uv5]}$, with $u, v = 0,\ldots, 4$. These components transform as a 5-dimensional antisymmetric tensor of rank 2 under $SO(2, 3)_A$, since the last index is not affected by this group. In a similar way, a contravariant spinor $h_{[AB]}$ is equivalent to a 5-dimensional antisymmetric tensor of rank 2 $h_{[uv5]}$.

In the transition from the 6-dimensional to the 5-dimensional tensor formalism, the following consequences of eq. (3.70) are useful:

$$p_{[uvw]} = 2^{-1} \epsilon_{uvw}{}^{xy} p_{[xy5]}, \qquad b^{[uvw]} = 2^{-1}\epsilon^{uvw}{}_{xy} b^{[xy5]},$$

$$u, v, w, x, y = 0,\ldots, 4.$$ (3.83)