Supergravity on a three-torus: quantum linearization instabilities with a supergroup

In physics, the equations of interest are frequently non-linear and difficult to solve. Typically only a small number of solutions are known and these exploit special symmetries. To extract further physics from the known solutions, a common strategy is to perturbatively expand small deviations around the known background solutions. At lowest order in perturbation one obtains linear equations for the perturbations, which are typically easier to analyse. However, it is not guaranteed that all solutions to the linearized equations actually arise as first approximations to solutions of the non-linear equations of the system. Let us illustrate this point in a simple example [1]: for $\left(x,y\right)\in {\mathbb{R}}^{2}$, consider the algebraic system x(x² + y²) = 0, of which the exact solutions are (0, y), where y is any real number. On the other hand, if we consider the linearized equations for perturbations δx and δy around a background (x₀, y₀), these satisfy $\delta x\left({x}_{0}^{2}+{y}_{0}^{2}\right)+{x}_{0}\left(2{x}_{0}\delta x+2{y}_{0}\delta y\right)=0$. If we let the background be (x₀, y₀) = (0, 0), then any pairs (δx, δy) satisfy the linearized equation of motion, but those with δx ≠ 0 cannot have arisen as linearizations of solutions to the non-linear equation. One characterizes this phenomenon as the equation x(x² + y²) = 0 being linearization unstable at (0, 0).

A well-known field-theoretic system with linearization instabilities is electrodynamics in a 'closed Universe', i.e. in a spacetime with compact Cauchy surfaces [2, 3]. Consider, for example, the electromagnetic field coupled to a charged matter field in such a spacetime. Note that the conserved total charge of the matter field must vanish in this system. This fact is a simple consequence of Gauss's law $\to {\nabla }\cdot \to {E}\propto \rho$, where $\to {E}$ and ρ are the electric field and charge density, respectively. The integral over a Cauchy surface of the charge density ρ gives the total charge but the integral of $\to {\nabla }\cdot \to {E}$ vanishes because the Cauchy surface is compact. At the level of the linearized theory about vanishing background electromagnetic and matter field the theory is non-interacting. Since the matter field is non-interacting, there is no constraint on the total charge in the linearized theory, but a solution to the linearized (i.e. free) matter field equation does not extend to an exact solution to the full interacting theory unless its total charge Q_e vanishes. The linearization stability condition (LSC) in this case is Q_e = 0. If a solution to the linearized equations satisfies this condition, then it extends to an exact solution to the full theory. It is useful to note here that the charge Q_e generates the global gauge symmetry of the free charged matter field.

In gravitational systems, it is known that such linearization instabilities occur for any perturbations around a background which has both Killing symmetries and compact Cauchy surfaces [2, 4–12]. The LSCs which need to be imposed on such a closed Universe are that the generators Q of the background Killing symmetries must vanish when evaluated on any linearized solution. For example, if the spacetime possesses time and space translation symmetries, then the corresponding conserved charges are the energy and momentum, respectively.

In the late seventies Moncrief studied the rôle of these conditions in linearized quantum gravity in spacetimes with compact Cauchy surfaces. He proposed that they should be imposed as physical-state conditions as in the Dirac quantization [11, 12], i.e.

$$\begin{equation}Q\left\vert \mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}\right\rangle =0 .\end{equation} \tag{ 1.1 }$$

Since the conserved charges Q generate the spacetime symmetries (in the component of the identity), he concluded that quantum linearization stability conditions (QLSCs) imply that all physical states must be invariant under the spacetime Killing symmetries. He argued that these constraints can be viewed as a remnant of the diffeomorphism invariance of the non-linear theory.

Imposing the invariance of the physical Hilbert space under the full background symmetries (in the component of the identity) as required by the QLSCs would appear too restrictive. This problem is exemplified by de Sitter space, a spacetime which is physically relevant for inflationary cosmology. This spacetime has Cauchy surfaces with the topology of the three-dimensional sphere, which is compact. Therefore, the QLSCs imply that all physical states of linearized quantum gravity on a de Sitter background ought to be invariant under the full SO₀(4, 1) symmetry group, i.e. the component of the identity of SO(4, 1), of the spacetime. However, this would appear to exclude all states except the vacuum state, which would make the Hilbert space for the theory quite empty [3, 12].

However, there are non-trivial SO₀(4, 1) invariant states that have infinite norm and, hence, are not in the Hilbert space. Moncrief suggested that a Hilbert space consisting of these invariant states could be constructed by dividing the infinite inner product by the infinite volume of the group SO₀(4, 1). This suggestion was taken up in reference [13]. In that work a new inner product for SO₀(4, 1)-invariant states was defined by what would later be termed group-averaging, which is an important ingredient in the refined algebraic quantization [14] and has been well studied in the context of loop quantum gravity. (The group-averaging procedure was also proposed in [15].) In this approach, one defines the invariant states $\left\vert {\Psi}\right\rangle$ by starting with a non-invariant state $\left\vert \psi \right\rangle$ and averaging against the symmetry group G, assumed here to be an unimodular group such as SO₀(4, 1), to obtain

$$\begin{equation}\left\vert {\Psi}\right\rangle ={\int }_{G} \mathrm{d}g U\left(g\right)\left\vert \psi \right\rangle ,\end{equation} \tag{ 1.2 }$$

where U is the unitary operator implementing the symmetry on the states. The state |Ψ⟩ can readily be shown to be invariant, i.e. ⟨ϕ|U(g)|Ψ⟩ = ⟨ϕ|Ψ⟩ for any state |ϕ⟩ in the Hilbert space by the invariance of the measure dg. If the volume of the symmetry group is finite, the inner product of the invariant states |Ψ₁⟩ and |Ψ₂⟩ obtained from |ψ₁⟩ and |ψ₂⟩ as in (1.2) is

$$\begin{equation}\langle {{\Psi}}_{1}\vert {{\Psi}}_{2}\rangle ={V}_{G}{\int }_{G}\mathrm{d}g \langle {\psi }_{1}\vert U\left(g\right)\vert {\psi }_{2}\rangle ,\end{equation} \tag{ 1.3 }$$

where V_G is the volume of the group G. If G has infinite volume, e.g. if it is SO₀(4, 1), then one needs to redefine the inner product on the invariant states by removing a factor of the group volume to make them normalizable. Thus, one defines the inner product on the new Hilbert space by

$$\begin{equation}{\langle {{\Psi}}_{1} \vert {{\Psi}}_{2}\rangle }_{ga}={\int }_{G} \mathrm{d}g \left\langle {\psi }_{1}\right\vert U\left(g\right)\left\vert {\psi }_{2}\right\rangle .\end{equation} \tag{ 1.4 }$$

By this group-averaging procedure one obtains an infinite-dimensional Hilbert space of SO₀(4, 1) invariant states for linearized gravity in de Sitter space [3, 13]. The group-averaging procedure was carried out for this spacetime also for other free fields [16, 17] to obtain Hilbert spaces of invariant states. The QLSCs in de Sitter space were also studied in the context of cosmological perturbation [18, 19]. The group-averaging procedure has also been studied extensively in the context of constrained dynamical systems (see e.g. [20–23]). This method was extended to non-unimodular groups in [24].

The group-averaging procedure can also be explicitly carried out for perturbative quantum gravity in static space with topology of $\mathbb{R}{\times}{\mathrm{T}}^{3}$, where the spatial Cauchy surfaces are copies of T³, the three-dimensional torus, to find the states satisfying the QLSCs. The classical [2] and quantum theory [11, 25] of this model have been studied and the group-averaging procedure can be carried out to obtain a physical Hilbert space of states invariant under the $\mathbb{R}{\times}U{\left(1\right)}^{3}$ symmetry group of the background. Now, if one considers four-dimensional $\mathcal{N}=1$ simple supergravity [26, 27] on this background spacetime, it is not difficult to see that there are additional fermionic LSCs. The purpose of this paper is to find all states satisfying the bosonic and fermionic QLSCs and show that a Hilbert space of these states can be constructed using the group-averaging procedure over the supergroup of symmetries of the linearized theory.

Let us describe how the LSCs arise for supergravity in this spacetime. Recall that the energy and momentum of a system in general relativity can be expressed as an integral over a two-dimensional surface at infinity of a Cauchy surface (the ADM mass and momentum) in asymptotically-flat spacetime. In classical perturbation theory about Minkowski space, this fact implies that the total energy and momentum of the linearized fields can be expressed as a surface integral at infinity of perturbations of the next order. Then, one expects that in perturbation theory in the flat $\mathbb{R}{\times}{\mathrm{T}}^{3}$ background, the total energy and momentum of the linearized fields vanish because there is no spatial infinity. This is indeed the case, and the vanishing of the total energy and momentum of the linearized field is expressed as the LSCs. Note here that the expression for the total energy, or the Hamiltonian, in this spacetime is not positive definite and, therefore, can vanish for non-trivial field configurations. Now, in supergravity there is a spinor supercharge Q_α, α = 1, 2, 3, 4, associated with a global supersymmetry variation, and it is known that this supercharge can be written as an integral over two-dimensional surface at infinity of a Cauchy surface in asymptotically-flat spacetime [28]. This fact again implies that in static three-torus space there are quadratic constraints on the linearized theory corresponding to the vanishing of the supercharge, which are the fermionic LSCs. In this paper we study these fermionic LSCs together with the bosonic ones in linearized quantum supergravity.

The remainder of this paper is organized as follows. In section 2 we present a derivation of the (classical) bosonic and fermionic LSCs for $\mathcal{N}=1$ simple supergravity in the background of flat $\mathbb{R}{\times}{\mathrm{T}}^{3}$ spacetime. We show that these conditions are of the form that the conserved Noether charges of the linearized theory vanish. In section 3 we discuss linearized supergravity in this spacetime and express the LSCs in terms of the classical analogues of annihilation and creation operators. In section 4 we impose the bosonic QLSCs on the states and recall how this can be understood in the context of group-averaging over the bosonic symmetry group. In section 5 we describe how all physical states satisfying both the bosonic and fermionic QLSCs are found and how a Hilbert space of physical states is constructed. Then we show that the procedure of finding the physical Hilbert space can be interpreted as group-averaging over the supergroup of global supersymmetry. We summarize and discuss our results in section 6. In appendix A we present a gauge transformation of the vierbein field that is proportional to the Lie derivative of a vector field, though it is not necessary for this work. In appendix B we illustrate our derivation of the LSCs in the simple example of electrodynamics in flat $\mathbb{R}{\times}{\mathrm{T}}^{3}$ spacetime. In appendix C we present the proof of some identities used in this paper. In appendix D we discuss some aspects of the zero-momentum sector of the gravitino field. Appendix E presents an example of a two-particle state satisfying all QLSCs. We follow the conventions of reference [29] throughout this paper.

The action for the four-dimensional $\mathcal{N}=1$ simple supergravity [29, 30] with 8πG = 1 is

$$\begin{equation}S=\frac{1}{2}\int {\mathrm{d}}^{4}x e\left[R-{\overline{{\Psi}}}_{\mu }{\gamma }^{\mu \nu \rho }{D}_{\nu }{{\Psi}}_{\rho }+{X}_{\left(4{\Psi}\right)}\right] ,\end{equation} \tag{ 2.1 }$$

where X_(4Ψ) consists of terms quartic in Ψ_μ (see e.g. [30] for the explicit form of X_(4Ψ)). Here, ${e}_{\mu }^{a}$ are the vierbein fields, $e{:=}\mathrm{det}\left({e}_{\mu }^{a}\right)$, and Ψ_μα, α = 1, 2, 3, 4, is the gravitino field, which is a Majorana spinor. The 4 × 4 gamma matrices γ^a, a = 0, 1, 2, 3, satisfy the Clifford relation {γ^a, γ^b} = 2η^ab, where η^ab = diag(−1, 1, 1, 1). The indices a, b, c, ....are raised and lowered by the flat metric η_ab whereas the spacetime indices μ, ν, σ, ....are raised and lowered by the spacetime metric ${g}_{\mu \nu }={e}_{\mu }^{a}{e}_{a\nu }$. The matrices γ^a have the properties γ^0† = −γ⁰ and γ^i† = γⁱ, i = 1, 2, 3. One defines ${\gamma }^{\mu }{:=}{e}_{a}^{\mu }{\gamma }^{a}$ and ${\gamma }^{\mu \nu \rho }{:=}{\gamma }^{\left[\right. \mu }{\gamma }^{\nu }{\gamma }^{\rho \left.\right]}$, where [⋯] indicates total anti-symmetrization. The Riemann tensor is given by

$$\begin{equation}{{R}_{\mu \nu }}^{ab}{:=}{\partial }_{\mu }{\omega }_{\nu }^{ ab}-{\partial }_{\nu }{\omega }_{\mu }^{ ab}+{\omega }_{\mu }^{ ac}{\omega }_{\nu c}^{ b}-{\omega }_{\nu }^{ ac}{\omega }_{\mu c}^{ b} ,\end{equation} \tag{ 2.2 }$$

and the curvature scalar is $R{:=}{e}_{a}^{\mu }{e}_{b}^{\nu }{{R}_{\mu \nu }}^{ab}$. The spin connection ${\omega }_{\mu }^{ ab}={\omega }_{\mu }^{ \left[ab\right]}$ is expressed in terms of ${e}_{\mu }^{a}$ as

$$\begin{equation}{\omega }_{\mu }^{ ab}={e}^{a\nu }{\partial }_{\left[\right.\mu }{e}_{\nu \left.\right]}^{ b}-{e}^{b\nu }{\partial }_{\left[\right.\mu }{e}_{\nu \left.\right]}^{ a}-\frac{1}{2}\left({e}^{a\nu }{e}^{b\sigma }-{e}^{b\nu }{e}^{a\sigma }\right){e}_{\mu }^{c}{\partial }_{\nu }{e}_{c\sigma } .\end{equation} \tag{ 2.3 }$$

One also defines $\overline{{\Psi}}={{\Psi}}^{T}C$, where C is a unitary matrix satisfying C^T = −C and γ^μT = −Cγ^μC⁻¹. The Majorana condition satisfied by Ψ_μ reads ${\left({{\Psi}}_{\mu }^{{\dagger}}\right)}_{\alpha }=\mathrm{i}{\left(C{\gamma }^{0}{{\Psi}}_{\mu }\right)}_{\alpha }$ [30]. We later choose a Majorana representation for γ^a in which C = iγ⁰. In this representation we have ${\left({{\Psi}}_{\mu }^{{\dagger}}\right)}_{\alpha }={\left({{\Psi}}_{\mu }\right)}_{\alpha }$. The gravitino covariant derivative in (2.1) is given by

$$\begin{equation}{D}_{\mu }{{\Psi}}_{\nu }={\partial }_{\mu }{{\Psi}}_{\nu }+\frac{1}{4}{\omega }_{\mu ab}{\gamma }^{ab}{{\Psi}}_{\nu } .\end{equation} \tag{ 2.4 }$$

The action (2.1) is invariant under a local supersymmetry transformation of the form

$$\begin{equation}{{\Delta}}_{{\epsilon}}{e}_{\mu }^{a}=\frac{1}{2}\overline{{\epsilon}}{\gamma }^{a}{{\Psi}}_{\mu } ,\end{equation} \tag{ 2.5 }$$

$$\begin{equation}{{\Delta}}_{{\epsilon}}{{\Psi}}_{\mu }={D}_{\mu }{\epsilon}+{Y}_{\mu ab}{\gamma }^{ab}{\epsilon} ,\end{equation} \tag{ 2.6 }$$

where Y_μab is quadratic in Ψ_μ and where is a spacetime dependent Grassmann variable with four components. (See, e.g. [30] for the explicit form of Y_μab).

We consider this theory on a purely bosonic static background whose spatial sections are flat three-dimensional tori. For the background geometry we therefore take

$$\begin{equation}{\mathrm{d}\mathrm{s}}^{2}={\eta }_{\mu \nu }{\mathrm{d}\mathrm{x}}^{\mu }{\mathrm{d}\mathrm{x}}^{\nu }=-{\mathrm{d}\mathrm{t}}^{2}+ \mathrm{d}{x}^{2}+{\mathrm{d}\mathrm{y}}^{2}+{\mathrm{d}\mathrm{z}}^{2} ,\end{equation} \tag{ 2.7 }$$

where the spatial coordinates x, y and z are periodic with periods L₁, L₂ and L₃ respectively, and we let V = L₁L₂L₃ denote the spatial volume.

Let us first discuss the bosonic LSCs for this theory, which are well known as we described in section 1. The diffeomorphism transformation in the direction of a vector ζ^μ of the vierbein field ${e}_{\mu }^{a}$ and gravitino field Ψ_μ is²

$$\begin{equation}{\delta }_{\zeta }{e}_{\mu }^{a}={\zeta }^{\nu }{\partial }_{\nu }{e}_{\mu }^{a}+{e}_{\nu }^{a}{\partial }_{\mu }{\zeta }^{\nu } ,\end{equation} \tag{ 2.8 }$$

$$\begin{equation}{\delta }_{\zeta }{{\Psi}}_{\mu }={\zeta }^{\nu }{\partial }_{\nu }{{\Psi}}_{\mu }+{{\Psi}}_{\nu }{\partial }_{\mu }{\zeta }^{\nu } .\end{equation} \tag{ 2.9 }$$

Now, we write the vierbein field as

$$\begin{equation}{e}_{\mu }^{a}={\delta }_{\mu }^{a}+{\tilde {e}}_{\mu }^{a} .\end{equation} \tag{ 2.10 }$$

That is, we write ${e}_{\mu }^{a}$ as the sum of its background value ${\delta }_{\mu }^{a}$ and perturbation ${\tilde {e}}_{\mu }^{a}$. Then,

$$\begin{equation}{\delta }_{\zeta }{\tilde {e}}_{\mu }^{a}={\delta }_{\nu }^{a}{\partial }_{\mu }{\zeta }^{\nu }+{\zeta }^{\nu }{\partial }_{\nu }{\tilde {e}}_{\mu }^{a}+{\tilde {e}}_{\nu }^{a}{\partial }_{\mu }{\zeta }^{\nu } .\end{equation} \tag{ 2.11 }$$

It is important to note here that the part of ${\delta }_{\zeta }{\tilde {e}}_{\mu }^{a}$ that is independent of ${\tilde {e}}_{\mu }^{a}$ vanishes if ζ^μ is a constant vector, i.e. a Killing vector of the flat background³.

Since the action (2.1) is invariant under the diffeomorphism transformation given by (2.9) and (2.11), we have

$$\begin{equation}{\delta }_{\zeta }{\tilde {e}}_{\mu }^{a}\frac{\delta S}{\delta {\tilde {e}}_{\mu }^{a}}+{\delta }_{\zeta }{{\Psi}}_{\mu }\frac{\delta S}{\delta {{\Psi}}_{\mu }}={\partial }_{\mu }\left(\sqrt{-g}{J}_{\left(\zeta \right)}^{\mu }\right) ,\end{equation} \tag{ 2.12 }$$

for some vector field ${J}_{\left(\zeta \right)}^{\mu }$, which is linear in ζ^μ. Here the variation δS/δΨ_μ is a left-variation, i.e.

$$\begin{equation}\delta S=\int {\mathrm{d}}^{4}x \delta {{\Psi}}_{\mu }\frac{\delta S}{\delta {{\Psi}}_{\mu }} .\end{equation} \tag{ 2.13 }$$

We adopt left-variations for fermionic fields throughout this paper. We then expand ${\delta }_{\zeta }{\tilde {e}}_{\mu }^{a}$, δ_ζΨ_μ, $\delta S/\delta {\tilde {e}}_{\mu }^{a}$, δS/δΨ_μ and $\sqrt{-g}{J}_{\left(\zeta \right)}^{\mu }$ according to the order in the fields ${\tilde {e}}_{\mu }^{a}$ and Ψ_μ, i.e. the number of these fields in the product, as

$$\begin{equation}{\delta }_{\zeta }{\tilde {e}}_{\mu }^{a}={\delta }_{\zeta }^{\left(0\right)}{\tilde {e}}_{\mu }^{a}+{\delta }_{\zeta }^{\left(1\right)}{\tilde {e}}_{\mu }^{a} ,\end{equation} \tag{ 2.14 }$$

$$\begin{equation}{\delta }_{\zeta }{{\Psi}}_{\mu }={\delta }_{\zeta }^{\left(1\right)}{{\Psi}}_{\mu } ,\end{equation} \tag{ 2.15 }$$

$$\begin{equation}\frac{\delta S}{\delta {\tilde {e}}_{\mu }^{a}}={E}_{a}^{\left(1\right)\mu }+{E}_{a}^{\left(2\right)\mu }+\cdots ,\end{equation} \tag{ 2.16 }$$

$$\begin{equation}\frac{\delta S}{\delta {{\Psi}}_{\mu }}={\mathcal{E}}^{\left(1\right)\mu }+{\mathcal{E}}^{\left(2\right)\mu }+\cdots ,\end{equation} \tag{ 2.17 }$$

$$\begin{equation}\sqrt{-g} {J}_{\left(\zeta \right)}^{\mu }={J}_{\left(\zeta \right)}^{\left(0\right)\mu }+{J}_{\left(\zeta \right)}^{\left(1\right)\mu }+{J}_{\left(\zeta \right)}^{\left(2\right)\mu }+\cdots ,\end{equation} \tag{ 2.18 }$$

Recall that the background metric is flat. We note that ${E}_{a}^{\left(0\right)\mu }=0$ and ${\mathcal{E}}^{\left(0\right)\mu }=0$ because the flat spacetime with Ψ_μ = 0 satisfies the field equations. That is, $\delta S/\delta {\tilde {e}}_{\mu }^{a}=0$ and δS/δΨ_μ = 0 if ${\tilde {e}}_{\mu }^{a}=0$ and Ψ_μ = 0.

The identity (2.12) must be satisfied order by order. The first- and second-order equalities read

$$\begin{equation}\left({\delta }_{\zeta }^{\left(0\right)}{\tilde {e}}_{\mu }^{a}\right){E}_{a}^{\left(1\right)\mu }={\partial }_{\mu }{J}_{\left(\zeta \right)}^{\left(1\right)\mu } ,\end{equation} \tag{ 2.19 }$$

$$\begin{equation}\left({\delta }_{\zeta }^{\left(0\right)}{\tilde {e}}_{\mu }^{a}\right){E}_{a}^{\left(2\right)\mu }+\left({\delta }_{\zeta }^{\left(1\right)}{\tilde {e}}_{\mu }^{a}\right){E}_{a}^{\left(1\right)\mu }+\left({\delta }_{\zeta }^{\left(1\right)}{{\Psi}}_{\mu }\right){\mathcal{E}}^{\left(1\right)\mu }={\partial }_{\mu }{J}_{\left(\zeta \right)}^{\left(2\right)\mu } .\end{equation} \tag{ 2.20 }$$

We emphasize here that these identities hold for any vector ζ^μ and for any field configuration. Now, if the vector field ξ^μ is a constant vector, then ${\delta }_{\xi }^{\left(0\right)}{\tilde {e}}_{\mu }^{a}=0$ as can readily be seen from (2.11). Hence, by (2.19), the current ${J}_{\left(\xi \right)}^{\left(1\right)\mu }$ is conserved. Then the charge

$$\begin{equation}{P}_{\left(\xi \right)}^{\left(1\right)}{:=}\int {\mathrm{d}}^{3}\to {x} {J}_{\left(\xi \right)}^{\left(1\right)0} ,\end{equation} \tag{ 2.21 }$$

is conserved for any field configuration. In particular, it is conserved even if the fields are smoothly deformed to 0 in the past or future of the t = constant Cauchy surface where the integral is evaluated. Since ${P}_{\left(\xi \right)}^{\left(1\right)}=0$ if the fields vanish, the conservation of this charge implies that

$$\begin{equation}{P}_{\left(\xi \right)}^{\left(1\right)}=0 ,\end{equation} \tag{ 2.22 }$$

for any field configuration if ξ^μ is a constant vector. The compactness of the Cauchy surfaces is crucial for this conclusion because this charge is not necessarily conserved if the field configuration is time-dependent at infinity for the case where the Cauchy surfaces are non-compact.

Now, suppose one attempts to solve the field equations for ${\tilde {e}}_{\mu }^{a}$ and Ψ_μ order by order. Let $\left({\tilde {e}}_{\mu }^{\left(1\right)a},{{\Psi}}_{\mu }^{\left(1\right)}\right)$ be a solution to the linearized equations and $\left({\tilde {e}}_{\mu }^{\left(2\right)a},{{\Psi}}_{\mu }^{\left(2\right)}\right)$ be the second-order correction. Then,

$$\begin{equation}{E}_{a}^{\left(1\right)\mu }\left[{\tilde {e}}^{\left(1\right)}\right]=0 ,\end{equation} \tag{ 2.23 }$$

$$\begin{equation}{\mathcal{E}}^{\left(1\right)\mu }\left[{{\Psi}}^{\left(1\right)}\right]=0 ,\end{equation} \tag{ 2.24 }$$

$$\begin{equation}{E}_{a}^{\left(1\right)\mu }\left[{\tilde {e}}^{\left(2\right)}\right]+{E}_{a}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=0 ,\end{equation} \tag{ 2.25 }$$

$$\begin{equation}{\mathcal{E}}^{\left(1\right)\mu }\left[{{\Psi}}^{\left(2\right)}\right]+{\mathcal{E}}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=0 .\end{equation} \tag{ 2.26 }$$

Here ${\mathcal{E}}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ is the vector-spinor ${\mathcal{E}}^{\left(2\right)\mu }$ evaluated with $\left({\tilde {e}}_{\mu }^{a},{{\Psi}}_{\mu }\right)=\left({\tilde {e}}_{\mu }^{\left(1\right)a},{{\Psi}}_{\mu }^{\left(1\right)}\right)$, and similarly for the others. The identities (2.19) and (2.20) imply

$$\begin{equation}{\partial }_{\mu }\left({J}_{\left(\zeta \right)}^{\left(1\right)\mu }\left[{\tilde {e}}^{\left(2\right)}\right]+{J}_{\left(\zeta \right)}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]\right)=0 ,\end{equation} \tag{ 2.27 }$$

for any vector field ζ^μ as long as field equations (2.23)–(2.25) are satisfied. The charge corresponding to the conserved current ${J}_{\left(\zeta \right)}^{\left(1\right)\mu }\left[{\tilde {e}}^{\left(2\right)}\right]+{J}_{\left(\zeta \right)}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ must vanish for any ζ^μ because this charge is conserved even if the field ζ^μ is smoothly deformed to zero in the past or future and is evaluated there. Hence, if we define

$$\begin{equation}{P}_{\left(\zeta \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]{:=}\int {\mathrm{d}}^{3}\to {x} {J}_{\left(\zeta \right)}^{\left(2\right)0}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right] ,\end{equation} \tag{ 2.28 }$$

then

$$\begin{equation}{P}_{\left(\zeta \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=-{P}_{\left(\zeta \right)}^{\left(1\right)}\left[{\tilde {e}}^{\left(2\right)}\right] ,\end{equation} \tag{ 2.29 }$$

for any vector field ζ^μ as long as the fields $\left({\tilde {e}}_{\nu }^{\left(1\right)b},{{\Psi}}_{\nu }^{\left(1\right)}\right)$ and $\left({\tilde {e}}_{\nu }^{\left(2\right)b},{{\Psi}}_{\nu }^{\left(2\right)}\right)$ are perturbative solutions to the field equations. In particular, if ζ^μ = ξ^μ is a Killing vector, then

$$\begin{equation}{P}_{\left(\xi \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=0 ,\end{equation} \tag{ 2.30 }$$

because of (2.22). Equation (2.30) is a consequence of the requirement that the solution $\left({\tilde {e}}_{\nu }^{\left(1\right)b},{{\Psi}}_{\nu }^{\left(1\right)}\right)$ extend to an exact solution and does not follow from the linearized field equations. This equation is called a linearization stability condition (LSC) and has to be imposed on the solutions to the linearized equations. In appendix B we illustrate the argument leading to (2.30) for electrodynamics. The total charge of the linearized charged field must vanish in this model.

The LSCs from local supersymmetry can be derived in a similar manner. We expand the supersymmetry transformation given by (2.5) and (2.6) according to the number of fields in the product as

$$\begin{equation}{{\Delta}}_{{\epsilon}}{\tilde {e}}_{\mu }^{a}={{\Delta}}_{{\epsilon}}^{\left(1\right)}{\tilde {e}}_{\mu }^{a} ,\end{equation} \tag{ 2.31 }$$

$$\begin{equation}{{\Delta}}_{{\epsilon}}{{\Psi}}_{\mu }={{\Delta}}_{{\epsilon}}^{\left(0\right)}{{\Psi}}_{\mu }+{{\Delta}}_{{\epsilon}}^{\left(1\right)}{{\Psi}}_{\mu }+{{\Delta}}_{{\epsilon}}^{\left(2\right)}{{\Psi}}_{\mu }+\cdots .\end{equation} \tag{ 2.32 }$$

It is clear that ${{\Delta}}_{{\epsilon}}{\tilde {e}}_{\mu }^{a}={{\Delta}}_{{\epsilon}}{e}_{\mu }^{a}$ given by (2.5) has no field independent contribution. That is, ${{\Delta}}_{{\epsilon}}^{\left(0\right)}{\tilde {e}}_{\mu }^{a}=0$. The analogue of the identity (2.12) is

$$\begin{equation}{{\Delta}}_{{\epsilon}}{\tilde {e}}_{\mu }^{a}\frac{\delta S}{\delta {\tilde {e}}_{\mu }^{a}}+{{\Delta}}_{{\epsilon}}{{\Psi}}_{\mu }\frac{\delta S}{\delta {{\Psi}}_{\mu }}={\partial }_{\mu }\left(\sqrt{-g}{\mathcal{J}}_{\left({\epsilon}\right)}^{\mu }\right) .\end{equation} \tag{ 2.33 }$$

From this equation one finds

$$\begin{equation}\left({{\Delta}}_{{\epsilon}}^{\left(0\right)}{{\Psi}}_{\mu }\right){\mathcal{E}}^{\left(1\right)\mu }={\partial }_{\mu }{\mathcal{J}}_{\left({\epsilon}\right)}^{\left(1\right)\mu } ,\end{equation} \tag{ 2.34 }$$

$$\begin{equation}\left({{\Delta}}_{{\epsilon}}^{\left(0\right)}{{\Psi}}_{\mu }\right){\mathcal{E}}^{\left(2\right)\mu }+\left({{\Delta}}_{{\epsilon}}^{\left(1\right)}{\tilde {e}}_{\mu }^{a}\right){E}_{a}^{\left(1\right)\mu }+\left({{\Delta}}_{{\epsilon}}^{\left(1\right)}{{\Psi}}_{\mu }\right){\mathcal{E}}^{\left(1\right)\mu }={\partial }_{\mu }{\mathcal{J}}_{\left({\epsilon}\right)}^{\left(2\right)\mu } ,\end{equation} \tag{ 2.35 }$$

for any spinor field and any field configuration. Since ${{\Delta}}_{\varepsilon }^{\left(0\right)}{{\Psi}}_{\mu }={\partial }_{\mu }\varepsilon =0$ if ɛ is a constant spinor, the charge defined by

$$\begin{equation}{Q}_{\left(\varepsilon \right)}^{\left(1\right)}:=\int {\mathrm{d}}^{3}\to {x} {\mathcal{J}}_{\left(\varepsilon \right)}^{\left(1\right)0} ,\end{equation} \tag{ 2.36 }$$

vanishes for any field configuration if = ɛ is a constant spinor by the argument which led to (2.22). By the same argument as that led to (2.29), if we define

$$\begin{equation}{Q}_{\left({\epsilon}\right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]{:=}\int {\mathrm{d}}^{3}\to {x} {\mathcal{J}}_{\left({\epsilon}\right)}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right] ,\end{equation} \tag{ 2.37 }$$

then

$$\begin{equation}{Q}_{\left({\epsilon}\right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=-{Q}_{\left({\epsilon}\right)}^{\left(1\right)}\left[{{\Psi}}^{\left(2\right)}\right] ,\end{equation} \tag{ 2.38 }$$

for any spinor field if $\left({\tilde {e}}_{\nu }^{\left(1\right)b},{{\Psi}}_{\nu }^{\left(1\right)}\right)$ gives a solution to the linearized field equations, ${E}_{a}^{\left(1\right)\mu }\left[{\tilde {e}}^{\left(1\right)}\right]={\mathcal{E}}^{\left(1\right)\mu }\left[{{\Psi}}^{\left(1\right)}\right]=0$. In particular, if = ɛ is a constant spinor, since ${Q}_{\left(\varepsilon \right)}^{\left(1\right)}\left[{{\Psi}}^{\left(2\right)}\right]=0$ in this case, we must have

$$\begin{equation}{Q}_{\left(\varepsilon \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]=0 .\end{equation} \tag{ 2.39 }$$

These are the LSCs arising from local supersymmetry on static three-torus space.

Next, we shall derive the conserved currents ${J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ given by (2.20) and ${\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ given by (2.35) and the corresponding conserved charges ${P}_{\left(\xi \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ and ${Q}_{\left(\varepsilon \right)}^{\left(2\right)}\left[{\tilde {e}}^{\left(1\right)},{{\Psi}}^{\left(1\right)}\right]$ for a constant vector ξ^μ and a constant spinor ɛ. From now on we write ${\tilde {e}}_{\mu }^{\left(1\right)a}={\tilde {e}}_{\mu }^{a}$ and ${{\Psi}}_{\mu }^{\left(1\right)}={{\Psi}}_{\mu }$.

For either charge we only need the linearized field equations. To find ${E}_{a}^{\left(1\right)\mu }$ for ${\tilde {e}}_{\mu }^{a}$, it is useful to note that eR in the Lagrangian density in terms of the vierbein fields equals $\sqrt{-g} R$ given in terms of the metric tensor g_μν. Thus, we may vary $\sqrt{-g} R$ with respect to the metric tensor and then vary the metric tensor with respect to the vierbein fields. Writing g_μν = η_μν + h_μν, we find the perturbation h_μν in terms of ${\tilde {e}}_{\mu }^{a}$ at first order as

$$\begin{equation}{h}_{\mu \nu }={\delta }_{\mu }^{a}{\tilde {e}}_{a\nu }+{\tilde {e}}_{a\mu }{\delta }_{\nu }^{a} .\end{equation} \tag{ 2.40 }$$

Then we find

$$\begin{align}\hfill {E}_{a}^{\left(1\right)\mu }\left[\tilde {e}\right]& =\frac{1}{2}{\delta }_{a}^{\nu }\left[-{\partial }^{\mu }{\partial }^{\lambda }{h}_{\lambda \nu }-{\partial }_{\nu }{\partial }^{\lambda }{h}_{\lambda }^{ \mu }+{\partial }^{\mu }{\partial }_{\nu }h+\square {h}_{ \nu }^{\mu }+{\delta }_{\nu }^{\mu }{\partial }^{\lambda }{\partial }^{\sigma }{h}_{\lambda \sigma }-{\delta }_{\nu }^{\mu }\square h\right] ,\hfill \end{align} \tag{ 2.41 }$$

where h := η^μνh_μν and □ := ∂_λ∂^λ, with indices raised and lowered by η_μν. We readily find

$$\begin{equation}{\mathcal{E}}^{\left(1\right)\mu }\left[{\Psi}\right]=-C{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{{\Psi}}_{\rho } .\end{equation} \tag{ 2.42 }$$

To find the bosonic conserved current ${J}_{\left(\xi \right)}^{\left(2\right)\mu }$, we use (2.40)–(2.42) together with

$$\begin{equation}{\delta }_{\left(\xi \right)}^{\left(1\right)}{\tilde {e}}_{\mu }^{a}={\xi }^{\nu }{\partial }_{\nu }{\tilde {e}}_{\mu }^{a} ,\end{equation} \tag{ 2.43 }$$

$$\begin{equation}{\delta }_{\left(\xi \right)}^{\left(1\right)}{{\Psi}}_{\mu }={\xi }^{\nu }{\partial }_{\nu }{{\Psi}}_{\mu } ,\end{equation} \tag{ 2.44 }$$

in (2.20), recalling ${\delta }_{\left(\xi \right)}^{\left(0\right)}{\tilde {e}}_{\mu }^{a}=0$, to find

$$\begin{align}\hfill {\partial }_{\mu }{J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[\tilde {e},{\Psi}\right]& =\frac{1}{4}\left({\xi }^{\lambda }{\partial }_{\lambda }{h}_{\mu \nu }\right)\left[-{\partial }^{\mu }{\partial }^{\lambda }{h}_{\lambda }^{ \nu }-{\partial }^{\nu }{\partial }^{\lambda }{h}_{\lambda }^{ \mu }+{\partial }^{\mu }{\partial }^{\nu }h+\square {h}^{\mu \nu }\right.\hfill \\ \hfill & \left.\quad +{\eta }^{\mu \nu }{\partial }^{\lambda }{\partial }^{\sigma }{h}_{\lambda \sigma }-{\eta }^{\mu \nu }\square h\right]-\left({\xi }^{\lambda }{\partial }_{\lambda }{\overline{{\Psi}}}_{\mu }\right){\gamma }^{\mu \nu \rho }{\partial }_{\nu }{{\Psi}}_{\rho } .\hfill \end{align} \tag{ 2.45 }$$

The current ${J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$, which depends on ${\tilde {e}}_{\mu }^{a}$ only through h_μν, is identified with the Noether current for the spacetime translation in the direction of ξ^μ of the decoupled theory consisting of a Fierz–Pauli Lagrangian for the graviton and a Rarita–Schwinger Lagrangian for the Majorana gravitino [30]

$$\begin{align}\hfill \mathcal{L}& =-\frac{1}{4}{\partial }^{\mu }h{\partial }^{\nu }{h}_{\mu \nu }+\frac{1}{8}{\partial }_{\mu }h{\partial }^{\mu }h+\frac{1}{4}{\partial }_{\sigma }{h}^{\mu \sigma }{\partial }^{\rho }{h}_{\mu \rho }-\frac{1}{8}{\partial }_{\rho }{h}^{\mu \nu }{\partial }^{\rho }{h}_{\mu \nu }-\frac{1}{2}{\overline{{\Psi}}}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{{\Psi}}_{\rho } .\hfill \end{align} \tag{ 2.46 }$$

It can be given explicitly as

$$\begin{align}\hfill \quad {J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]& =-\left({\xi }^{\lambda }{\partial }_{\lambda }{h}_{\rho \sigma }\right)\frac{\partial \mathcal{L}}{\partial \left({\partial }_{\mu }{h}_{\rho \sigma }\right)}-\left({\xi }^{\lambda }{\partial }_{\lambda }{{\Psi}}_{\rho }\right)\frac{\partial \mathcal{L}}{\partial \left({\partial }_{\mu }{{\Psi}}_{\rho }\right)}+{\xi }^{\mu }\mathcal{L}\hfill \\ \hfill & =\frac{1}{4}\left[{\xi }^{\lambda }{\partial }_{\lambda }h{\partial }_{\nu }{h}^{\mu \nu }+{\xi }^{\lambda }{\partial }_{\lambda }{h}^{\mu \rho }{\partial }_{\rho }h-{\xi }^{\lambda }{\partial }_{\lambda }h{\partial }^{\mu }h\right.\hfill \\ \hfill & \quad \left.- 2{\xi }^{\lambda }{\partial }_{\lambda }{h}^{\mu \rho }{\partial }^{\sigma }{h}_{\rho \sigma }+{\xi }^{\lambda }{\partial }_{\lambda }{h}_{\rho \sigma }{\partial }^{\mu }{h}^{\rho \sigma }\right]\hfill \\ \hfill & \quad +\frac{1}{4}{\xi }^{\mu }\left[-{\partial }^{\rho }h{\partial }^{\sigma }{h}_{\rho \sigma }+\frac{1}{2}{\partial }_{\rho }h{\partial }^{\rho }h+{\partial }_{\sigma }{h}^{\rho \sigma }{\partial }^{\lambda }{h}_{\rho \lambda }-\frac{1}{2}{\partial }_{\lambda }{h}_{\rho \sigma }{\partial }^{\lambda }{h}^{\rho \sigma }\right]\hfill \\ \hfill & \quad +\frac{1}{2}\left({\xi }^{\sigma }{\overline{{\Psi}}}_{\nu }{\gamma }^{\nu \mu \rho }{\partial }_{\sigma }{{\Psi}}_{\rho }-{\xi }^{\mu }{\overline{{\Psi}}}_{\nu }{\gamma }^{\nu \sigma \rho }{\partial }_{\sigma }{{\Psi}}_{\rho }\right) .\hfill \end{align} \tag{ 2.47 }$$

Next we find the fermionic conserved current ${\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }$. We note that the first-order part of the global supersymmetry transformation is given by

$$\begin{equation}{{\Delta}}_{\left(\varepsilon \right)}^{\left(1\right)}{\tilde {e}}_{\mu }^{a}=\frac{1}{2}\overline{\varepsilon }{\gamma }^{a}{{\Psi}}_{\mu } ,\end{equation} \tag{ 2.48 }$$

$$\begin{equation}{{\Delta}}_{\left(\varepsilon \right)}^{\left(1\right)}{{\Psi}}_{\mu }=\frac{1}{4}\left({\partial }_{\mu }{\tilde {e}}_{a\nu }{\gamma }^{\nu a}+{\partial }_{\rho }{h}_{\mu \nu }{\gamma }^{\nu \rho }\right)\varepsilon .\end{equation} \tag{ 2.49 }$$

By substituting these formulas and equations (2.41) and (2.42) into (2.35), and using the identity [29]

$$\begin{equation}{\gamma }^{\rho \nu }{\gamma }^{\sigma \mu \kappa }=3\left({\eta }^{\nu \left[\right.\sigma }{\gamma }^{\mu \kappa \left.\right]\rho }-{\eta }^{\rho \left[\right.\sigma }{\gamma }^{\mu \kappa \left.\right]\nu }+{\eta }^{\nu \left[\right.\sigma }{\gamma }^{\mu }{\eta }^{\kappa \left.\right]\rho }-{\eta }^{\rho \left[\right.\sigma }{\gamma }^{\mu }{\eta }^{\kappa \left.\right]\nu }\right) ,\end{equation} \tag{ 2.50 }$$

we find

$$\begin{equation}{\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]=\frac{1}{4}\left({\partial }_{\rho }{h}_{\sigma \nu }\overline{\varepsilon }{\gamma }^{\rho \nu }{\gamma }^{\mu \sigma \kappa }{{\Psi}}_{\kappa }-{\delta }_{\rho }^{a}{\tilde {e}}_{a\nu }\overline{\varepsilon }{\gamma }^{\rho \nu }{\gamma }^{\mu \sigma \kappa }{\partial }_{\sigma }{{\Psi}}_{\kappa }\right) .\end{equation} \tag{ 2.51 }$$

Note that the second term vanishes if the (linear) local Lorentz invariance is fixed by requiring ${\tilde {e}}_{a\nu }={\delta }_{a}^{\mu }{\delta }_{\nu }^{b}{\tilde {e}}_{b\mu }$. (It vanishes by the linearized field equation for Ψ_μ as well.) With this condition imposed, the term which explicitly depends on _bν in the supersymmetry transformation ${{\Delta}}_{\left(\varepsilon \right)}^{\left(1\right)}{{\Psi}}_{\mu }$ given by (2.49) vanishes. With this choice the current ${\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$ and the charge ${Q}_{\left(\varepsilon \right)}^{\left(2\right)}\left[h,{\Psi}\right]$ are the Noether current and charge, respectively, for the supersymmetry transformation,

$$\begin{equation}{{\Delta}}_{\left(\varepsilon \right)}^{\left(1\right)}{h}_{\mu \nu }=\frac{1}{2}\left(\overline{\varepsilon }{\gamma }_{\mu }{{\Psi}}_{\nu }+\overline{\varepsilon }{\gamma }_{\nu }{{\Psi}}_{\mu }\right) ,\end{equation} \tag{ 2.52 }$$

$$\begin{equation}{{\Delta}}_{\left(\varepsilon \right)}^{\left(1\right)}{{\Psi}}_{\mu }=\frac{1}{4}{\partial }_{\rho }{h}_{\mu \nu }{\gamma }^{\nu \rho }\varepsilon ,\end{equation} \tag{ 2.53 }$$

of the linearized supergravity Lagrangian given by (2.46).

Thus, the classical LSCs are ${P}_{\left(\xi \right)}^{\left(2\right)}\left[h,{\Psi}\right]=0$ and ${Q}_{\left(\varepsilon \right)}^{\left(2\right)}\left[h,{\Psi}\right]=0$, where ${P}_{\left(\xi \right)}^{\left(2\right)}\left[h,{\Psi}\right]$ and ${Q}_{\left(\varepsilon \right)}^{\left(2\right)}\left[h,{\Psi}\right]$ are the conserved charges corresponding to the conserved currents ${J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$ in (2.47) and ${\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$ in (2.51) (without the second term) respectively (see (2.28) and (2.37)). In the next section we discuss linearized supergravity on static three-torus space at the classical level and express the LSCs in a form suitable for quantization.

The results of the previous section allow us to describe the LSCs entirely in terms of the linearized theory. By linearizing $\mathcal{N}=1$ simple supergravity in 4 dimensions about static three-torus background spacetime, we have the Lagrangian, which is given here again for convenience:

$$\begin{align}\hfill \mathcal{L}& =-\frac{1}{4}{\partial }^{\mu }h{\partial }^{\nu }{h}_{\mu \nu }+\frac{1}{8}{\partial }_{\mu }h{\partial }^{\mu }h+\frac{1}{4}{\partial }_{\sigma }{h}^{\mu \sigma }{\partial }^{\rho }{h}_{\mu \rho }-\frac{1}{8}{\partial }_{\rho }{h}^{\mu \nu }{\partial }^{\rho }{h}_{\mu \nu }-\frac{1}{2}{\overline{{\Psi}}}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{{\Psi}}_{\rho } .\hfill \end{align} \tag{ 3.1 }$$

A suitable real Majorana representation for the γ-matrices is given by

$$\begin{align}\hfill \begin{aligned}\hfill {\gamma }^{0}=\left(\begin{pmatrix}\hfill 0\hfill & \hfill 1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{pmatrix}\right),\quad & \hfill & \hfill {\gamma }^{1}=\left(\begin{pmatrix}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill \end{pmatrix}\right),\\ \hfill {\gamma }^{2}=\left(\begin{pmatrix}\hfill 0\hfill & \hfill {\sigma }^{1}\hfill \\ \hfill {\sigma }^{1}\hfill & \hfill 0\hfill \end{pmatrix}\right),\quad & \hfill & \hfill {\gamma }^{3}=\left(\begin{pmatrix}\hfill 0\hfill & \hfill {\sigma }^{3}\hfill \\ \hfill {\sigma }^{3}\hfill & \hfill 0\hfill \end{pmatrix}\right) ,\end{aligned}\end{align} \tag{ 3.2 }$$

where σ¹, σ² and σ³ are the standard Pauli matrices. In this representation the charge conjugation matrix takes the form C = iγ⁰. The action with the Lagrangian density (3.1) is invariant under the following gauge transformations:

$$\begin{equation}{h}_{\mu \nu }\to {h}_{\mu \nu }+{\partial }_{\mu }{\zeta }_{\nu }+{\partial }_{\nu }{\zeta }_{\mu } ,\end{equation} \tag{ 3.3 }$$

$$\begin{equation}{{\Psi}}_{\mu }\to {{\Psi}}_{\mu }+{\partial }_{\mu }{\epsilon} ,\end{equation} \tag{ 3.4 }$$

where ζ^μ is an arbitrary vector field and _α is an arbitrary Majorana spinor field satisfying ${{\epsilon}}_{\alpha }^{{\dagger}}={{\epsilon}}_{\alpha }$. This invariance is a remnant of the diffeomorphism invariance and local supersymmetry of the full theory. We shall fix this gauge freedom completely.

Working on the spatial torus and imposing periodic boundary conditions on the fields allows us to decompose each field as a Fourier series⁴. Due to the periodic boundary conditions, each field has a spatially constant zero-momentum component. The zero-momentum sector of the linearized theory will be seen to contain six bosonic and six fermionic degrees freedom and this sector forms an important ingredient in the QLSCs. Meanwhile, the non-zero momentum sector of the theory contains the usual gravitons and gravitinos, which each have two polarization states. Explicitly we expand the fields as

$$\begin{align*}\hfill {h}_{\mu \nu }\left(t,\to {x}\right)& =\frac{1}{\sqrt{V}}{h}_{\mu \nu }^{\left(0\right)}\left(t\right)+\frac{1}{\sqrt{V}}\sum _{\to {k}\ne 0}{\tilde {h}}_{\mu \nu }\left(t,\to {k}\right){\mathrm{e}}^{\mathrm{i}\to {k}\cdot \to {x}} ,\hfill \\ \hfill {{\Psi}}_{\mu }\left(t,\to {x}\right)& =\frac{1}{\sqrt{V}}{\psi }_{\mu }\left(t\right)+\frac{1}{\sqrt{V}}\sum _{\to {k}\ne 0}{\tilde {{\Psi}}}_{\mu }\left(t,\to {k}\right){\mathrm{e}}^{\mathrm{i}\to {k}\cdot \to {x}} ,\hfill \end{align*}$$

where the volume of the torus is V = L₁L₂L₃ and the L₁, L₂ and L₃ are the periods of the torus in the x-, y- and z-directions respectively. The periodic boundary condition implies $\to {k}=\left(\frac{2\pi }{{L}_{1}}{n}_{1},\frac{2\pi }{{L}_{2}}{n}_{2},\frac{2\pi }{{L}_{3}}{n}_{3}\right)$ with n₁, n₂ and n₃ integers, and reality restricts ${h}_{\mu \nu }^{\left(0\right)}$ to be real, ${\tilde {h}}_{\mu \nu }\left(t,-\to {k}\right)$ to be equal to ${\tilde {h}}_{\mu \nu }^{{\dagger}}\left(t,\to {k}\right)$, and similarly for ψ_μ and ${\tilde {{\Psi}}}_{\mu }\left(t,\to {k}\right)$.

The Lagrangian for the theory does not provide dynamical coupling between the modes with zero momentum and those with non-zero momentum $\to {k}$. Therefore, it is possible to analyse these separately. We begin with the zero-momentum sector of the theory. We write ${h}_{\mu \nu }^{\left(0\right)}\left(t\right)={h}_{\mu \nu }\left(t\right)$, dropping the superscript '(0)', in the rest of this section. The Lagrangian, i.e. the space integral of the Lagrangian density, for this sector reads

$$\begin{equation}{L}_{0}=-\frac{1}{8}{\partial }_{0}{{h}^{i}}_{i}{\partial }_{0}{{h}^{j}}_{j}+\frac{1}{8}{\partial }_{0}{h}^{ij}{\partial }_{0}{h}_{ij}+\frac{1}{2}{\overline{\psi }}_{i}{\gamma }^{0}{\gamma }^{ij}{\partial }_{0}{\psi }_{j} .\end{equation} \tag{ 3.5 }$$

The classical [31] and quantum [25] theory of the graviton contributions have been studied previously using the ADM formalism without fixing the gauge. Here we fix the gauge to extract the physical degrees of freedom in the linearized theory. (This gauge fixing has little to do with the imposition of LSCs discussed later).

The linearized Lagrangian does not provide equations of motion for h_μ0 and ψ_0α, and these are precisely the components which carry the gauge degrees of freedom. It is therefore possible to gauge-fix these components to vanish by solving h₀₀(t) = 2∂₀ξ₀(t), h_i0(t) = ∂₀ξ_i(t), i = 1, 2, 3 and ψ₀(t) = ∂₀(t). The elimination of h_0μ(t) and ψ₀(t) exhausts the gauge freedom in the zero-momentum sector, and all other components are physical. Thus, in the zero-momentum sector the physical degrees of freedom are contained in the symmetric tensor h_ij and vector-spinor ψ_iα, i = 1, 2, 3, each of which corresponds to six classical degrees of freedom.

The six bosonic degrees of freedom are contained in the symmetric tensor h_ij. To analyse this tensor it is convenient to break it up into the trace and trace-free sectors. Thus, we let

$$\begin{equation}{h}_{ij}\left(t\right)=\sqrt{\frac{2}{3}}{\delta }_{ij} c\left(t\right)+2\sum _{A=1}^{5}{T}_{ij}^{A}{c}_{A}\left(t\right) ,\end{equation} \tag{ 3.6 }$$

where the ${T}_{ij}^{A}$ are a set of five trace-free tensors which satisfy the orthonormality condition ${T}^{Aij}{T}_{ij}^{B}={\delta }^{AB}$. We choose them as

$$\begin{align}\hfill {T}^{1}& =\frac{1}{\sqrt{2}}\left(\begin{pmatrix}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{pmatrix}\right), {T}^{2}=\frac{1}{\sqrt{6}}\left(\begin{pmatrix}\hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill -2\hfill \end{pmatrix}\right),\hfill \\ \hfill {T}^{3}& =\frac{1}{\sqrt{2}}\left(\begin{pmatrix}\hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \end{pmatrix}\right), {T}^{4}=\frac{1}{\sqrt{2}}\left(\begin{pmatrix}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 0\hfill & \hfill 0\hfill \end{pmatrix}\right),\hfill \\ \hfill {T}^{5}& =\frac{1}{\sqrt{2}}\left(\begin{pmatrix}\hfill 0\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \end{pmatrix}\right).\hfill \end{align} \tag{ 3.7 }$$

In terms of the six variables c and c^A, the Lagrangian reads

$$\begin{equation}{L}_{0}=-\frac{1}{2}{\left({\partial }_{0}c\right)}^{2}+\frac{1}{2}\sum _{A=1}^{5}{\left({\partial }_{0}{c}_{A}\right)}^{2}+\frac{1}{2}{\overline{\psi }}_{i}{\gamma }^{0}{\gamma }^{ij}{\partial }_{0}{\psi }_{j} .\end{equation} \tag{ 3.8 }$$

The momentum conjugate to h_ij, i.e. p^ij = ∂L₀/∂(∂₀h_ij), is

$$\begin{align}\hfill {p}^{ij}& =-\frac{1}{4}\left({\partial }_{0}{{h}^{k}}_{k}\right){\delta }^{ij}+\frac{1}{4}{\partial }_{0}{h}^{ij}\hfill \\ \hfill & =\frac{1}{2}\sqrt{\frac{2}{3}}{\delta }^{ij} {c}_{P}+\frac{1}{2}\sum _{A}{T}^{Aij} {c}_{PA} ,\hfill \end{align} \tag{ 3.9 }$$

where c_P = −∂₀c and c_PA = ∂₀c_A are the momenta conjugate to c and c_A, respectively. The time derivative of the zero-momentum field h_ij is then

$$\begin{equation}{\partial }_{0}{h}_{ij}=2\sum _{A=1}^{5}{T}_{ij}^{A} {c}_{PA}-\sqrt{\frac{2}{3}}{\delta }_{ij} {c}_{P} .\end{equation} \tag{ 3.10 }$$

The canonical Poisson bracket relations for c, c_A, c_P and c_PA, and equivalently for h_ij and p^kl, are

$$\begin{equation}{\left\{c,{c}_{P}\right\}}_{P}=1,\quad {\left\{{c}_{A},{c}_{PB}\right\}}_{P}={\delta }_{AB} ,\end{equation} \tag{ 3.11 }$$

$$\begin{equation}{\left\{{h}_{ij},{p}^{kl}\right\}}_{P}=\frac{1}{2}\left({\delta }_{i}^{k}{\delta }_{j}^{l}+{\delta }_{i}^{l}{\delta }_{j}^{k}\right) .\end{equation} \tag{ 3.12 }$$

The fermionic part of L₀ is of first order in time derivatives and therefore already defines a constrained dynamical system [32, 33], and we need to use the Dirac bracket as the bracket to be 'promoted' to the anti-commutator upon quantization. The momentum variables ${\pi }_{\alpha }^{i}$ conjugate to ψ_iα are

$$\begin{align}\hfill {\pi }_{\alpha }^{i}& =\frac{\partial {L}_{0}}{\partial \left({\partial }_{0}{\psi }_{i\alpha }\right)}\hfill \\ \hfill & =-\frac{1}{2}{\left({\overline{\psi }}_{j}{\gamma }^{0}{\gamma }^{ji}\right)}_{\alpha } ,\hfill \end{align} \tag{ 3.13 }$$

where the derivative of L₀ with respect to ∂₀ψ_iα is the left derivative. These momenta are not invertible in terms of the coordinates and velocities, so there are twelve primary constraints,

$$\begin{equation}{\phi }_{\alpha }^{i}={\pi }_{\alpha }^{i}+\frac{1}{2}{\left({\overline{\psi }}_{j}{\gamma }^{0}{\gamma }^{ji}\right)}_{\alpha }\approx 0 .\end{equation} \tag{ 3.14 }$$

We note that the Poisson and Dirac brackets for two fermionic fields are symmetric under the exchange of arguments. That is, if the fields A and B are fermionic, then {A, B}_P = {B, A}_P, and similarly for the Dirac bracket.

For the associated primary Hamiltonian H_P we add arbitrary linear combinations of the primary constraints to the canonical Hamiltonian,

$$\begin{equation}{H}_{P}={\partial }_{0}{\psi }_{i}{\pi }^{i}-{L}_{0}+{\lambda }^{i}{\phi }_{i}={\lambda }^{i}{\phi }_{i} ,\end{equation} \tag{ 3.15 }$$

where the ${\lambda }_{\alpha }^{i}$ are arbitrary. Notice in particular that the primary Hamiltonian for these modes weakly vanishes, i.e. it vanishes if the constraints are satisfied. The constraints are to be preserved under evolution by the primary Hamiltonian. This requirement leads to consistency conditions

$$\begin{equation}{\partial }_{0}{\phi }^{i}\approx {\left\{{\phi }^{i},{H}_{P}\right\}}_{P}\approx 0 .\end{equation} \tag{ 3.16 }$$

To evaluate the Poisson bracket between the constraints, we use the canonical Poisson bracket,

$$\begin{equation}{\left\{{\psi }_{i\alpha },{\pi }_{\beta }^{j}\right\}}_{P}=-{\delta }_{j}^{i}{\delta }_{\alpha \beta } ,\end{equation} \tag{ 3.17 }$$

which allows us to compute the Poisson bracket between the constraints (and therefore the primary Hamiltonian) as

$$\begin{equation}{\left\{{\phi }_{\alpha }^{i},{\phi }_{\beta }^{j}\right\}}_{P}=-{\left(C{\gamma }^{0}{\gamma }^{ij}\right)}_{\alpha \beta } .\end{equation} \tag{ 3.18 }$$

Hence, we find that the consistency conditions (3.16) imply λⁱ = 0. Thus, the primary Hamiltonian H_P vanishes and, as a result, the fields ψ_i are time independent. This fact can also be deduced from the Euler–Lagrange equation resulting from (3.8). There are no secondary constraints and all the constraints are of second class. To obtain the bracket structure for the theory suitable for subsequent quantization, we compute the Dirac bracket

$$\begin{align}\hfill {\left\{{\psi }_{i\alpha },{\pi }_{\beta }^{j}\right\}}_{D}& ={\left\{{\psi }_{i\alpha },{\pi }_{\beta }^{j}\right\}}_{P}-{\left\{{\psi }_{i\alpha },{\phi }_{\gamma }^{k}\right\}}_{P}{\left({\left\{\phi ,\phi \right\}}_{P}\right)}_{kl\gamma \delta }^{-1}{\left\{{\phi }_{\delta }^{l},{\pi }_{\beta }^{j}\right\}}_{P}\hfill \\ \hfill & =\frac{1}{2}{\left\{{\psi }_{i\alpha },{\pi }_{\beta }^{j}\right\}}_{P}\hfill \\ \hfill & =-\frac{1}{2}{\delta }_{i}^{j}{\delta }_{\alpha \beta } .\hfill \end{align} \tag{ 3.19 }$$

On the Dirac bracket, we can impose the second-class constraints as strong conditions since the Dirac bracket between second-class constraints vanishes.

Working explicitly in the Majorana representation, where C = iγ⁰, the Dirac bracket for ψ_iα evaluates to

$$\begin{equation}{\left\{{\psi }_{i\alpha },{\psi }_{j\beta }\right\}}_{D}=-\frac{\mathrm{i}}{2}\left({\delta }_{ij}{\delta }_{\alpha \beta }-{\left({\gamma }_{ij}\right)}_{\alpha \beta }\right) .\end{equation} \tag{ 3.20 }$$

To provide some insight into these relations we write

$$\begin{equation}{\psi }_{i\alpha }=\frac{1}{\sqrt{6}}{\left({\gamma }_{i}\eta \right)}_{\alpha }+\sum _{A=1}^{2}{T}_{ij}^{A}{\left({\gamma }^{j}{\eta }^{A}\right)}_{\alpha } ,\end{equation} \tag{ 3.21 }$$

where η and η_A are Majorana spinors. The matrices T¹ and T² are given in (3.7). These symmetric and traceless matrices satisfy

$$\begin{equation}{T}^{Aij}{\gamma }_{j}{T}_{ik}^{B}{\gamma }^{k}={\delta }^{AB} .\end{equation} \tag{ 3.22 }$$

One can also show, by a component-by-component calculation,

$$\begin{equation}\sum _{A=1}^{2}{T}_{ik}^{A}{\gamma }^{k}{T}_{j\ell }^{A}{\gamma }^{\ell }=\frac{2}{3}{\delta }_{ij}-\frac{1}{3}{\gamma }_{ij} .\end{equation} \tag{ 3.23 }$$

These relations can be used to show that the Dirac bracket relations for ψ_iα are equivalent to

$$\begin{equation}{\left\{{\eta }_{\alpha },{\eta }_{\beta }\right\}}_{D}=\mathrm{i}{\delta }_{\alpha \beta } ,\quad {\left\{{\eta }_{\alpha }^{A},{\eta }_{\beta }^{B}\right\}}_{D}=-\mathrm{i}{\delta }^{AB}{\delta }_{\alpha \beta } ,\quad {\left\{{\eta }_{\alpha },{\eta }_{\beta }^{A}\right\}}_{D}=0 .\end{equation} \tag{ 3.24 }$$

Next, we briefly describe the mode expansion of the non-zero-momentum sector of this theory, which is well known. Letting the non-zero-momentum components of the fields be denoted by _μν and ${\hat{{\Psi}}}_{\mu }$, we can completely fix the gauge freedom in this sector by imposing the following conditions (see, for instance, [29, 30]):

$$\begin{equation}{\hat{h}}_{\mu 0}=\hat{h}={\partial }^{i}{\hat{h}}_{ij}=0 ,\end{equation} \tag{ 3.25 }$$

$$\begin{equation}{\gamma }^{i}{\hat{{\Psi}}}_{i}={\hat{{\Psi}}}_{0}={\partial }^{i}{\hat{{\Psi}}}_{i}=0 .\end{equation} \tag{ 3.26 }$$

Then we can write the expansion of the graviton as follows:

$$\begin{equation}{\hat{h}}_{ij}\left(\to {x},t\right)=\sqrt{\frac{2}{V}}\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\frac{1}{\sqrt{k}}\left[{H}_{ij}^{\lambda }\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right){\mathrm{e}}^{\mathrm{i}k\cdot x}+{H}_{ij}^{\lambda {\ast}}\left(\to {k}\right){a}_{\lambda }^{{\dagger}}\left(\to {k}\right){\mathrm{e}}^{-\mathrm{i}k\cdot x}\right] ,\end{equation} \tag{ 3.27 }$$

where $k{:=}\vert \to {k}\vert$ and ${k}^{\mu }=\left(k,\to {k}\right)$. We have let the complex conjugate of ${a}_{\lambda }\left(\to {k}\right)$ be denoted by ${a}_{\lambda }^{{\dagger}}\left(\to {k}\right)$, anticipating quantization. The symmetric and traceless polarization tensors are given by

$$\begin{equation}{H}_{ij}^{\lambda }\left(\to {k}\right)={{\epsilon}}_{i}^{\lambda }\left(\to {k}\right){{\epsilon}}_{j}^{\lambda }\left(\to {k}\right) ,\end{equation} \tag{ 3.28 }$$

where the polarization vectors ${{\epsilon}}_{i}^{\lambda }\left(\to {k}\right)$ are given by

$$\begin{equation}{{\epsilon}}_{i}^{{\pm}}\left(\to {k}\right)=\frac{1}{\sqrt{2}}\left({\hat{e}}_{i}^{\left(1\right)}\left(\to {k}\right){\pm}\mathrm{i}{\hat{e}}_{i}^{\left(2\right)}\left(\to {k}\right)\right) .\end{equation} \tag{ 3.29 }$$

The unit spatial vectors ${\hat{e}}_{i}^{\left(1\right)}\left(\to {k}\right)$, ${\hat{e}}_{i}^{\left(2\right)}\left(\to {k}\right)$ and ${\hat{e}}_{i}^{\left(3\right)}\left(\to {k}\right)=\to {k}/k$ form a right-handed orthonormal system in this order. The polarization vectors ${{\epsilon}}_{i}^{{\pm}}\left(\to {k}\right)$ satisfy ${{\epsilon}}^{\lambda {\ast}}\left(\to {k}\right)\cdot {{\epsilon}}^{{\lambda }^{\prime }}\left(\to {k}\right)={\delta }^{\lambda {\lambda }^{\prime }}$ and ${k}^{i}{{\epsilon}}_{i}^{\lambda }\left(\to {k}\right)=0$. As a result, ${H}_{ij}^{\lambda }\left(\to {k}\right)$ satisfy ${k}^{i}{H}_{ij}^{\lambda }\left(\to {k}\right)=0$ and ${H}_{ij}^{\lambda {\ast}}\left(\to {k}\right){H}^{{\lambda }^{\prime }ij}\left(\to {k}\right)={\delta }^{\lambda {\lambda }^{\prime }}$. The Lagrangian density for the non-zero-momentum sector of the graviton field in this gauge can be found from (3.1) as

$$\begin{equation}{\mathcal{L}}_{TT}=-\frac{1}{8}{\partial }_{\rho }{\hat{h}}^{ij}{\partial }^{\rho }{\hat{h}}_{ij} .\end{equation} \tag{ 3.30 }$$

The standard procedure to find the Poisson bracket relations for the coefficients ${a}_{\lambda }\left(\to {k}\right)$ and ${a}_{\lambda }^{{\dagger}}\left(\to {k}\right)$ leads to

$$\begin{equation}{\left\{{a}_{\lambda }\left(\to {k}\right),{a}_{\lambda }^{{\dagger}}\left({\to {k}}^{\prime }\right)\right\}}_{P}=-\mathrm{i}{\delta }_{\to {k},{\to {k}}^{\prime }}{\delta }_{\lambda {\lambda }^{\prime }} ,\end{equation} \tag{ 3.31 }$$

with all other brackets among ${a}_{\lambda }\left(\to {k}\right)$ and ${a}_{\lambda }^{{\dagger}}\left(\to {k}\right)$ vanishing.

The non-zero-momentum sector of the gravitino field can similarly be expanded into modes as

$$\begin{equation}{\hat{{\Psi}}}_{i\alpha }=\frac{1}{\sqrt{V}}\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\frac{1}{\sqrt{2k}}\left[{{\epsilon}}_{i}^{\lambda }\left(\to {k}\right){u}_{\alpha }^{\lambda }\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right){\mathrm{e}}^{\mathrm{i}k\cdot x}+{{\epsilon}}_{i}^{\lambda {\ast}}\left(\to {k}\right){u}_{\alpha }^{\lambda {\ast}}\left(\to {k}\right){b}_{\lambda }^{{\dagger}}\left(\to {k}\right){\mathrm{e}}^{-\mathrm{i}k\cdot x}\right] ,\end{equation} \tag{ 3.32 }$$

where ${u}^{{\pm}}\left(\to {k}\right)$ are eigenspinors of γ₅ = iγ⁰γ¹γ²γ³ and ${\gamma }^{0}\hat{k}\cdot \to {\gamma }$, where $\hat{k}{:=}\to {k}/k$, with eigenvalues ±1 and −1, respectively. We normalize them by requiring ${u}^{\lambda {\dagger}}\left(\to {k}\right){u}^{{\lambda }^{\prime }}\left({\to {k}}^{\prime }\right)=2k{\delta }^{\lambda {\lambda }^{\prime }}$. The fermionic coefficients ${b}_{\lambda }\left(\to {k}\right)$ and ${b}_{\lambda }^{{\dagger}}\left(\to {k}\right)$ satisfy the classical analogues of the anti-commutation relations for the annihilation and creation operators:

$$\begin{equation}{\left\{{b}_{\lambda }\left(\to {k}\right),{b}_{{\lambda }^{\prime }}^{{\dagger}}\left({\to {k}}^{\prime }\right)\right\}}_{D}=-\mathrm{i}{\delta }_{\to {k},{\to {k}}^{\prime }}{\delta }_{\lambda {\lambda }^{\prime }} ,\end{equation} \tag{ 3.33 }$$

with all other brackets among ${b}_{\lambda }\left(\to {k}\right)$ and ${b}_{\lambda }^{{\dagger}}\left(\to {k}\right)$ vanishing.

We quantize the linearized theory by promoting the physical degrees of freedom to operators acting on some Hilbert space, and we impose

$$\begin{equation}{\left[\left(\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{i}\right)-\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{m}\mathrm{u}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{r}\right]}_{{\pm}}=\mathrm{i}\hslash \left\{\mathrm{P}\mathrm{o}\mathrm{i}\mathrm{s}\mathrm{s}\mathrm{o}\mathrm{n}/\mathrm{D}\mathrm{i}\mathrm{r}\mathrm{a}\mathrm{c}\;\mathrm{B}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{k}\mathrm{e}\mathrm{t}\right\} ,\end{equation} \tag{ 4.1 }$$

as the algebraic relations between the operators, where [A, B]_± := AB ± BA. We work in units with ℏ = 1. Thus, equations (3.11) and (3.31) become

$$\begin{equation}{\left[c,{c}_{P}\right]}_{-}=\mathrm{i}, {\left[{c}_{A},{c}_{PB}\right]}_{-}=\mathrm{i}{\delta }_{AB} ,\end{equation} \tag{ 4.2 }$$

$$\begin{equation}{\left[{a}_{\lambda }\left(\to {k}\right),{a}_{{\lambda }^{\prime }}^{{\dagger}}\left({\to {k}}^{\prime }\right)\right]}_{-}={\delta }_{\lambda {\lambda }^{\prime }}{\delta }_{\to {k},{\to {k}}^{\prime }} ,\end{equation} \tag{ 4.3 }$$

respectively. On the other hand, the relations (3.24) and (3.33) become

$$\begin{equation}{\left[{\eta }_{\alpha },{\eta }_{\beta }\right]}_{+}=-{\delta }_{\alpha \beta }, {\left[{\eta }_{\alpha }^{A},{\eta }_{\beta }^{B}\right]}_{+}={\delta }^{AB}{\delta }_{\alpha \beta }, {\left[{\eta }_{\alpha },{\eta }_{\beta }^{A}\right]}_{+}=0 ,\end{equation} \tag{ 4.4 }$$

$$\begin{equation}{\left[{b}_{\lambda }\left(\to {k}\right),{b}_{{\lambda }^{\prime }}^{{\dagger}}\left({\to {k}}^{\prime }\right)\right]}_{+}={\delta }_{\lambda {\lambda }^{\prime }}{\delta }_{\to {k},{\to {k}}^{\prime }} ,\end{equation} \tag{ 4.5 }$$

respectively.

After imposing our gauge conditions and using the field equations, the time component of the conserved Noether current ${J}_{\left(\xi \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$ for the spacetime translation symmetry becomes

$$\begin{equation}{J}_{\left(\xi \right)}^{\left(2\right)0}=\frac{\sqrt{V}}{8}{\xi }_{0}\left[{\partial }_{0}{h}_{ij}{\partial }_{0}{h}^{ij}-{\partial }_{0}{{h}^{j}}_{j}{\partial }_{0}{{h}^{i}}_{i}\right]-\frac{1}{4}{\xi }_{\nu }{\partial }^{\nu }{\hat{h}}^{ij}{\partial }_{0}{\hat{h}}_{ij}+\frac{1}{8}{\xi }_{0}{\partial }_{\nu }{\hat{h}}_{ij}{\partial }^{\nu }{\hat{h}}^{ij}-\frac{\mathrm{i}}{2}{\xi }_{\nu }{\hat{{\Psi}}}_{i}^{T}{\partial }^{\nu }{\hat{{\Psi}}}^{i} ,\end{equation} \tag{ 4.6 }$$

where we have dropped the cross terms between the zero-momentum and non-zero-momentum sectors because they do not contribute to the space integral of ${J}_{\left(\xi \right)}^{\left(2\right)0}$, which gives the conserved charge. Let us write

$$\begin{equation}{\xi }_{0}H+{\xi }_{i}{P}^{i}=\int {\mathrm{d}}^{3}\to {x} {J}_{\left(\xi \right)}^{\left(2\right)0} .\end{equation} \tag{ 4.7 }$$

By substituting (3.9), (3.27) and (3.32) into (4.6) and integrating the result over space, one finds

$$\begin{equation}H=-\frac{1}{2}{c}_{P}^{2}+\sum _{A=1}^{5}\frac{1}{2}{c}_{PA}^{2}+\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\vert \to {k}\vert \left({a}_{\lambda }^{{\dagger}}\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right)+{b}_{\lambda }^{{\dagger}}\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right)\right) ,\end{equation} \tag{ 4.8 }$$

$$\begin{equation}\to {P}=\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\to {k}\left({a}_{\lambda }^{{\dagger}}\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right)+{b}_{\lambda }^{{\dagger}}\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right)\right) ,\end{equation} \tag{ 4.9 }$$

the bosonic sector of which agrees with the conserved charges given in [25]. Note that the zero-point energy which had to be renormalized away in the pure-gravity case is absent here because of supersymmetry. Note also that the Hamiltonian H is not positive definite.

The Hilbert space for the theory can be constructed as the tensor product of a non-zero momentum sector and a sector with zero momentum. In the sector with non-zero momentum, we have the usual Fock spaces for the gravitons and gravitinos with the number operators for the gravitons and gravitinos (with a given momentum and a polarization) being ${a}_{\lambda }^{{\dagger}}\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right)$ and ${b}_{\lambda }^{{\dagger}}\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right)$, respectively. A suitable basis of states is given by states with definite number of particles in each momentum and polarization. Thus, the normalized state with ${n}_{B\lambda }\left(\to {k}\right)$ gravitons and ${n}_{F\lambda }\left(\to {k}\right)$ gravitinos with momentum $\to {k}$ and polarization λ can be given as

$$\begin{equation}\left\vert \left\{{n}_{B}\right\}\right\rangle \otimes \left\vert \left\{{n}_{F}\right\}\right\rangle =\prod _{\to {k},\lambda ={\pm}}\left[\frac{1}{\sqrt{{n}_{B\lambda }\left(\to {k}\right)!}}{\left({a}_{\lambda }^{{\dagger}}\left(\to {k}\right)\right)}^{{n}_{B\lambda }\left(\to {k}\right)}{\left({b}_{\lambda }^{{\dagger}}\left(\to {k}\right)\right)}^{{n}_{F\lambda }\left(\to {k}\right)}\right]\vert 0\rangle ,\end{equation} \tag{ 4.10 }$$

where the vacuum state |0⟩ satisfies ${a}_{\lambda }\left(\to {k}\right)\vert 0\rangle ={b}_{\lambda }\left(\to {k}\right)\vert 0\rangle =0$ for all $\to {k}$ and λ. (For the gravitino ${n}_{F\lambda }\left(\to {k}\right)=0$ or 1, of course).

For the graviton zero-modes, we represent the commutation rules ${\left[c,{c}_{P}\right]}_{-}=\mathrm{i}$ and ${\left[{c}_{A},{c}_{PB}\right]}_{-}=\mathrm{i}{\delta }_{AB}$ on wave functions which are functions of the variables c and c_A by

$$\begin{equation}c{\mapsto}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{y}\;\mathrm{b}\mathrm{y}\;c,\quad {c}_{P}{\mapsto}-\mathrm{i}\frac{\partial }{\partial c},\end{equation} \tag{ 4.11 }$$

$$\begin{equation}{c}_{A}{\mapsto}\mathrm{m}\mathrm{u}\mathrm{l}\mathrm{t}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{y}\;\mathrm{b}\mathrm{y}\;{c}_{A},\quad {c}_{PA}{\mapsto}-\mathrm{i}\frac{\partial }{\partial {c}_{A}} .\end{equation} \tag{ 4.12 }$$

Therefore, if we take some state which is proportional to a single eigenstate of the number operators,

$$\begin{equation}\left\vert \mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\right\rangle ={\Psi}\otimes \left\vert \left\{{n}_{B}\right\}\right\rangle \otimes \left\vert \left\{{n}_{F}\right\}\right\rangle ,\end{equation} \tag{ 4.13 }$$

where Ψ is some wave function of c and c_A, then the Hamiltonian and momentum operators, (4.8) and (4.9), acting on such a state take the form

$$\begin{equation}H\left\vert \mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\right\rangle =\frac{1}{2}\left(+\frac{{\partial }^{2}}{\partial {c}^{2}}-\sum _{A=1}^{5}\frac{{\partial }^{2}}{\partial {c}_{A}^{2}}+{M}^{2}\right)\left\vert \mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\right\rangle ,\end{equation} \tag{ 4.14 }$$

$$\begin{equation}\to {P}\left\vert \mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\right\rangle =\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\to {k}\left({n}_{B\lambda }\left(\to {k}\right)+{n}_{F\lambda }\left(\to {k}\right)\right)\left\vert \mathrm{s}\mathrm{t}\mathrm{a}\mathrm{t}\mathrm{e}\right\rangle ,\end{equation} \tag{ 4.15 }$$

where we have defined a 'squared mass' for these states by

$$\begin{equation}{M}^{2}=2\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\vert \to {k}\vert \left({n}_{B\lambda }\left(\to {k}\right)+{n}_{F\lambda }\left(\to {k}\right)\right) .\end{equation} \tag{ 4.16 }$$

The operators H and $\to {P}$ do not contain the zero-momentum sector of the gravitino field. They appear in the global supercharges as we shall see in the next section.

As discussed in section 2, the conserved charges H and $\to {P}$ must vanish classically for the classical solutions to the linearized equations if they were to be extendible to exact solutions. Moncrief proposed that in quantum theory these linearization stability conditions (LSCs) should be imposed as constraints on the physical states. Thus, the constraints on the physical states |phys⟩ are

$$\begin{equation}H\left\vert \mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}\right\rangle =0,\quad \to {P}\left\vert \mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}\right\rangle =0 .\end{equation} \tag{ 4.17 }$$

In [25] the group-averaging procedure was used to find all states satisfying these constraints and define an inner product among these states for pure gravity on static three-torus space. We apply this procedure to $\mathcal{N}=1$ simple supergravity in this section.

We start with the Hilbert space ${\mathcal{H}}_{0}$ of superpositions of the states defined by (4.13). Take two states in this Hilbert space of the form |φ₁⟩ = Ψ₁(c, c_A) ⊗ |Φ₁⟩ and |φ₂⟩ = Ψ₂(c, c_A) ⊗ |Φ₂⟩, where |Φ₁⟩ and |Φ₂⟩ are states in the Fock space $\mathcal{F}$ of non-zero-momentum gravitons and gravitinos. The inner product between these states is

$$\begin{equation}{\langle {\varphi }_{1}\vert {\varphi }_{2}\rangle }_{{\mathcal{H}}_{0}}={\langle {{\Phi}}_{1}\vert {{\Phi}}_{2}\rangle }_{\mathcal{F}}\int \mathrm{d}c{\mathrm{d}}^{5}\to {c} {{\Psi}}_{1}^{{\ast}}{{\Psi}}_{2} ,\end{equation} \tag{ 4.18 }$$

where $\to {c}$ is the vector with components c_A, A = 1, 2, 3, 4, 5. Let us choose both |Φ₁⟩ and |Φ₂⟩ to have a definite value of M² defined by (4.16). (If these states have different values of M², then ${\langle {{\Phi}}_{1}\vert {{\Phi}}_{2}\rangle }_{\mathcal{F}}=0$.) Then, by expressing ${{\Psi}}_{I}\left(c,\to {c}\right)$, I = 1, 2, as a Fourier integral in the six-dimensional space with coordinates $\left(c,\to {c}\right)$, these states can be given as

$$\begin{equation}\vert {\varphi }_{I}\rangle =\int \frac{\mathrm{d}{p}^{0}}{2\pi }\frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}{F}_{I}\left({p}^{0},\to {p}\right){\mathrm{e}}^{-\mathrm{i}{p}^{0}c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert {{\Phi}}_{I}\rangle ,\end{equation} \tag{ 4.19 }$$

where $\to {p}$ is also a five-dimensional vector and $\to {p}\cdot \to {c}={p}^{A}{c}_{A}$. The inner product for these states is

$$\begin{equation}{\langle {\varphi }_{1}\vert {\varphi }_{2}\rangle }_{{\mathcal{H}}_{0}}={\langle {{\Phi}}_{1}\vert {{\Phi}}_{2}\rangle }_{\mathcal{F}}\int \frac{\mathrm{d}{p}^{0}}{2\pi }\frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}{F}_{1}^{{\ast}}\left({p}^{0},\to {p}\right){F}_{2}\left({p}^{0},\to {p}\right) .\end{equation} \tag{ 4.20 }$$

Since the operators H and $\to {P}$ are the generators of spacetime translations, we can construct states satisfying (4.17) by averaging the states |φ_I⟩ over this translation group as follows:

$$\begin{equation}\vert {\varphi }_{I}^{\left(B\right)}\rangle =\frac{1}{2 \mathrm{V}}{\int }_{-\infty }^{\infty } \mathrm{d}{\alpha }^{0}\left(\prod _{i=1}^{3}{\int }_{0}^{{L}_{i}}\mathrm{d}{\alpha }_{i}\right)\mathrm{exp}\left(\mathrm{i}{\alpha }^{0}H-\mathrm{i}\to {\alpha }\cdot \to {P}\right)\vert {\varphi }_{I}\rangle .\end{equation} \tag{ 4.21 }$$

The integral over space acts as the projector onto the sector of the Fock space with zero total momentum:

$$\begin{equation}\frac{1}{V}\left(\prod _{i=1}^{3}{\int }_{0}^{{L}_{i}}\mathrm{d}{\alpha }_{i}\right)\mathrm{exp}\left(-\mathrm{i}\to {\alpha }\cdot \to {P}\right)\vert {{\Phi}}_{I}\rangle ={\mathcal{P}}_{\to {P}=0}\vert {{\Phi}}_{I}\rangle .\end{equation} \tag{ 4.22 }$$

The integral over α⁰ can readily be evaluated using the representation (4.19) since

$$\begin{equation}H{\mathrm{e}}^{-\mathrm{i}{p}^{0}c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert {{\Phi}}_{I}\rangle =\frac{1}{2}\left[-{\left({p}^{0}\right)}^{2}+{\to {p}}^{2}+{M}^{2}\right]{\mathrm{e}}^{-\mathrm{i}{p}^{0}c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert {{\Phi}}_{I}\rangle .\end{equation} \tag{ 4.23 }$$

Thus we find, with the notation $\vert {{\Phi}}_{I}^{\left(\to {P}=0\right)}\rangle ={\mathcal{P}}_{\to {P}=0}\vert {{\Phi}}_{I}\rangle$,

$$\begin{align}\hfill \vert {\varphi }_{I}^{\left(B\right)}\rangle & =\int \frac{\mathrm{d}{p}^{0}}{2\pi }\frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}2\pi \delta \left({\left({p}^{0}\right)}^{2}-{\to {p}}^{2}-{M}^{2}\right){F}_{I}\left({p}^{0},\to {p}\right){\mathrm{e}}^{-\mathrm{i}{p}^{0}c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert {{\Phi}}_{I}^{\left(\to {P}=0\right)}\rangle \hfill \\ \hfill & =\int \frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}\left[{f}_{I}^{\left(+\right)}\left(\to {p}\right){\mathrm{e}}^{-\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}+{f}_{I}^{\left(-\right)}\left(\to {p}\right){\mathrm{e}}^{\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}\right]\otimes \vert {{\Phi}}_{I}^{\left(\to {P}=0\right)}\rangle ,\hfill \end{align} \tag{ 4.24 }$$

where $E\left(\to {p}\right)=\sqrt{{\to {p}}^{2}+{M}^{2}}$ and where

$$\begin{equation}{f}_{I}^{\left({\pm}\right)}\left(\to {p}\right)=\frac{{F}_{I}\left({\pm}E\left(\to {p}\right),\to {p}\right)}{2E\left(\to {p}\right)} .\end{equation} \tag{ 4.25 }$$

Although we obtained the invariant state $\vert {\varphi }_{I}^{\left(B\right)}\rangle$ by averaging |φ_I⟩ over the spacetime translation group, it is easy to show that any state with definite value of M² satisfying the constraints (4.17) is of this form. (The zero-momentum sector is a solution to the six-dimensional Klein–Gordon equation with mass M).

The states $\vert {\varphi }_{I}^{\left(B\right)}\rangle$ are indeed invariant, i.e. they satisfy the QLSCs given by (4.17). However, they have infinite norm and, hence, are not in the Hilbert space ${\mathcal{H}}_{0}$: the inner product ${\langle {\varphi }_{I}^{\left(B\right)}\vert {\varphi }_{I}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{0}}$ computed using (4.20) is infinite because of the δ-function in (4.24). The group-averaging inner product for the Hilbert space ${\mathcal{H}}_{B}$ of invariant states is defined by

$$\begin{align}\hfill \quad {\langle {\varphi }_{1}^{\left(B\right)}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}& =\frac{1}{2 \mathrm{V}}{\int }_{-\infty }^{\infty } \mathrm{d}{\alpha }^{0}\left(\prod _{i=1}^{3}{\int }_{0}^{{L}_{i}}\mathrm{d}{\alpha }_{i}\right){\langle {\varphi }_{1}\vert \mathrm{exp}\left(\mathrm{i}{\alpha }^{0}H-\mathrm{i}\to {\alpha }\cdot \to {P}\right)\vert {\varphi }_{2}\rangle }_{{\mathcal{H}}_{0}}\hfill \\ \hfill & ={\langle {\varphi }_{1}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{0}}\hfill \\ \hfill & =2\int \frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}E\left(\to {p}\right)\left[{f}_{1}^{\left(+\right){\ast}}\left(\to {p}\right){f}_{2}^{\left(+\right)}\left(\to {p}\right)+{f}_{1}^{\left(-\right){\ast}}\left(\to {p}\right){f}_{2}^{\left(-\right)}\left(\to {p}\right)\right]{\langle {{\Phi}}_{1}^{\left(\to {P}=0\right)}\vert {{\Phi}}_{2}^{\left(\to {P}=0\right)}\rangle }_{\mathcal{F}} .\hfill \end{align} \tag{ 4.26 }$$

Notice that, although this inner product is defined in terms of the 'seed states' |φ_I⟩, it depends only on the invariant states $\vert {\varphi }_{I}^{\left(B\right)}\rangle$. This inner product is equivalent to that proposed in [34] in the context of quantum cosmology.

Since the constraint H|phys⟩ = 0 is a Klein–Gordon equation, it is tempting to use the Klein–Gordon inner product for the Hilbert space ${\mathcal{H}}_{B}$:

$$\begin{equation}{\langle {\varphi }_{1}^{\left(B\right)}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{\mathrm{K}\mathrm{G}}=\mathrm{i}\int {\mathrm{d}}^{5}c\left({{\Psi}}_{1}^{{\ast}}\frac{\partial {{\Psi}}_{2}}{\partial c}-\frac{\partial {{\Psi}}_{1}^{{\ast}}}{\partial c}{{\Psi}}_{2}\right){\langle {{\Phi}}_{1}^{\left(\to {P}=0\right)}\vert {{\Phi}}_{2}^{\left(\to {P}=0\right)}\rangle }_{\mathcal{F}} .\end{equation} \tag{ 4.27 }$$

This inner product can readily be evaluated as

$$\begin{equation}{\langle {\varphi }_{1}^{\left(B\right)}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{\mathrm{K}\mathrm{G}}=2\int \frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}E\left(\to {p}\right)\left[{f}_{1}^{\left(+\right){\ast}}\left(\to {p}\right){f}_{2}^{\left(+\right)}\left(\to {p}\right)-{f}_{1}^{\left(-\right){\ast}}\left(\to {p}\right){f}_{2}^{\left(-\right)}\left(\to {p}\right)\right]{\langle {{\Phi}}_{1}^{\left(\to {P}=0\right)}\vert {{\Phi}}_{2}^{\left(\to {P}=0\right)}\rangle }_{\mathcal{F}} ,\end{equation} \tag{ 4.28 }$$

which is identical with the group-averaging inner product, ${\langle {\varphi }_{1}^{\left(B\right)}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}$, given by (4.26) except for the minus sign in the second term. Although the Klein–Gordon inner product is more widely used in quantum cosmology, it can be used only if the space of geometries considered has the structure of spacetime [35]. (In our example, it is the six-dimensional Minkowski space.) The group-averaging inner product, which we use in this paper, has wider applicability. For example, it can be used in recollapsing quantum cosmology as shown in [34].

In the next section we impose the fermionic QLSCs derived in the previous section. We shall find that the states satisfying these constraints and the inner product among them can be found by group-averaging over the relevant supergroup.

The time component of the conserved fermionic current ${\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)\mu }\left[h,{\Psi}\right]$ is

$$\begin{align}\hfill {\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)0}\left[h,{\Psi}\right]& =\frac{1}{4}\left[{\partial }_{0}h\overline{\varepsilon }{\gamma }^{i}{\psi }_{i}-{\partial }_{0}{h}_{ij}\overline{\varepsilon }{\gamma }^{i}{\psi }^{j}-{\partial }_{0}{\hat{h}}_{ij}\overline{\varepsilon }{\gamma }^{j}{\hat{{\Psi}}}^{i}+{\partial }_{\ell }{\hat{h}}_{ij}\overline{\varepsilon }{\gamma }^{0}{\gamma }^{\ell }{\gamma }^{j}{\hat{{\Psi}}}^{i}\right] ,\hfill \end{align} \tag{ 5.1 }$$

where we dropped the cross terms between the zero-momentum and non-zero-momentum sectors since they do not contribute to the space integral. By letting

$$\begin{equation}\overline{\varepsilon }Q=-\int {\mathrm{d}}^{3}\to {x}{\mathcal{J}}_{\left(\varepsilon \right)}^{\left(2\right)0} ,\end{equation} \tag{ 5.2 }$$

and substituting (3.10) and (3.21) into (5.1) and integrating over space, we find

$$\begin{equation}Q={Q}^{\left(0\right)}+\hat{Q} ,\end{equation} \tag{ 5.3 }$$

with the zero- and non-zero-momentum contributions respectively given by

$$\begin{equation}{Q}^{\left(0\right)}=\frac{1}{2}\left({c}_{P}\eta +\sum _{A=1}^{5}\sum _{B=1}^{2}{{T}^{Ai}}_{j}{T}_{ik}^{B}{\gamma }^{j}{\gamma }^{k}{c}_{PA}{\eta }^{B}\right) ,\end{equation} \tag{ 5.4 }$$

$$\begin{equation}\hat{Q}=\frac{1}{4}\int {\mathrm{d}}^{3}\to {x} \left({\partial }_{0}{\hat{h}}^{ij}{\gamma }_{j}{\hat{{\Psi}}}_{i}-{\hat{h}}^{ij}{\gamma }_{j}{\partial }_{0}{\hat{{\Psi}}}_{i}\right) ,\end{equation} \tag{ 5.5 }$$

where we have integrated by parts in the second term of $\hat{Q}$ and used ${\gamma }^{j}{\partial }_{j}{\hat{{\Psi}}}_{i}=-{\gamma }^{0}{\partial }_{0}{\hat{{\Psi}}}_{i}$.

One is readily able to find the mode expansion for $\hat{Q}$, which then yields

$$\begin{align}\hfill Q& =\frac{1}{2}\left({c}_{p}\eta +\sum _{A=1}^{5}\sum _{B=1}^{2}{{T}^{Ai}}_{j}{T}_{ik}^{B}{\gamma }^{j}{\gamma }^{k}{c}_{PA}{\eta }^{B}\right)\hfill \\ \hfill & \quad -\frac{\mathrm{i}}{2}\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\left[{{\epsilon}}^{\lambda }\left(\to {k}\right)\cdot \to {\gamma }{u}^{\lambda {\ast}}\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right){b}_{\lambda }^{{\dagger}}\left(\to {k}\right)-{{\epsilon}}^{\lambda {\ast}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{\lambda }\left(\to {k}\right){a}_{\lambda }^{{\dagger}}\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right)\right] .\hfill \end{align} \tag{ 5.6 }$$

This charge and the bosonic charges, ${P}^{\mu }=\left(H,\to {P}\right)$, satisfy the supersymmetry algebra. That is, ${\left[{P}^{\mu },Q\right]}_{-}=0$, ${\left[{P}^{\mu },{P}^{\nu }\right]}_{-}=0$ and

$$\begin{equation}{\left[{Q}_{\alpha },{Q}_{\beta }\right]}_{+}=\frac{\mathrm{1}}{2}{\left({\gamma }_{\mu }{\gamma }^{0}\right)}_{\alpha \beta }{P}^{\mu } .\end{equation} \tag{ 5.7 }$$

Since the contribution to Q with different $\to {k}$ anti-commute, equation (5.7) holds for each $\to {k}$ including $\to {k}=0$. In deriving (5.7) we have used the following identities for $\to {k}=0$ and $\to {k}\ne 0$ proved in appendix C:

$$\begin{equation}\sum _{B=1}^{2}\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{{T}^{Ai}}_{j}{T}_{ik}^{B}{T}_{{i}^{\prime }{j}^{\prime }}^{{A}^{\prime }}{{T}^{B{i}^{\prime }}}_{{k}^{\prime }}{\gamma }^{j}{\gamma }^{k}{\gamma }^{{k}^{\prime }}{\gamma }^{{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }}=\sum _{A=1}^{5}{\left({c}_{PA}\right)}^{2} ,\end{equation} \tag{ 5.8 }$$

$$\begin{equation}{\left[{{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}}\left(\to {k}\right)\right]}_{\alpha }{\left[{{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}{\ast}}\left(\to {k}\right)\right]}_{\beta }+\left(\alpha {\leftrightarrow}\beta \right)=2{\left(\gamma \cdot k{\gamma }^{0}\right)}_{\alpha \beta } .\end{equation} \tag{ 5.9 }$$

We also recall that γ^μ are real, γ₅ is purely imaginary and that ${{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)={{\epsilon}}^{\mp }\left(\to {k}\right)$ and ${u}^{{\pm}{\ast}}\left(\to {k}\right)={u}^{\mp }\left(\to {k}\right)$.

One can represent the anti-commutation relations (4.4) satisfied by the twelve fermionic operators η_α and ${\eta }_{\alpha }^{A}$ in the zero-momentum sector of the gravitino field by a 64-dimensional indefinite-metric Hilbert space (or Krein space) ${\mathcal{H}}_{0F}$ as shown in appendix D. The basis vectors of this space can be chosen such that 32 of them are normalized with positive norm and the other 32 are normalized with negative norm. [We say here that |v⟩ is normalized with positive (negative) norm if ⟨v|v⟩ = 1 (⟨v|v⟩ = −1).] An important property of ${\mathcal{H}}_{0F}$ used later is that the 16-dimensional subspace of ${\mathcal{H}}_{0F}$ of states annihilated by the operators d₁ and d₂ defined by

$$\begin{equation}{d}_{1}{:=}\left({\eta }_{1}+\mathrm{i}{\eta }_{2}\right)/\sqrt{2} ,\quad {d}_{2}=\left({\eta }_{3}+\mathrm{i}{\eta }_{4}\right)/\sqrt{2} ,\end{equation} \tag{ 5.10 }$$

is a positive-norm subspace because this subspace is spanned by the states of the form (D.5) with n₁ = n₂ = 0 (with M = 2) [see (D.6)]. The Hilbert space of states satisfying the bosonic QLSCs is in fact the tensor product ${\mathcal{H}}_{B}\otimes {\mathcal{H}}_{0F}$. We call this tensor product space also ${\mathcal{H}}_{B}$ from now on in order not to complicate the notation.

Now we are in a position to find the physical states |phys⟩ satisfying the fermionic QLSCs, Q_α|phys⟩ = 0, as well as the bosonic ones, P^μ|phys⟩ = 0. Since the operators P^μ commute with Q_α, we specialize to the Hilbert space ${\mathcal{H}}_{B}$ of states satisfying the bosonic QLSCs and identify P^μ with the null operator. Thus, we have ${\left[{Q}_{\alpha },{Q}_{\beta }\right]}_{+}=0$ in place of (5.7). Splitting the supercharge into zero- and non-zero-momentum contributions, these anti-commutation relations of the supercharge can be rewritten as

$$\begin{equation}{\left[{Q}_{\alpha }^{\left(0\right)},{Q}_{\beta }^{\left(0\right)}\right]}_{+}=-\frac{1}{4}{M}^{2}{\delta }_{\alpha \beta }=-{\left[{\hat{Q}}_{\alpha },{\hat{Q}}_{\beta }\right]}_{+} ,\end{equation} \tag{ 5.11 }$$

where M² is defined by (4.16). We specialize to an eigenspace of the operator M² without loss of generality. Thus, we treat M as if it were a non-negative number.

Assume M > 0. Then, in each common eigenspace of the operators c_P and c_PA with eigenvalues p₀ and p_A, respectively, satisfying ${p}_{0}^{2}-{\to {p}}^{2}={M}^{2}$, the operators ${Q}_{\alpha }^{\left(0\right)}$ given by (5.4) is of the form

$$\begin{equation}{Q}_{\alpha }^{\left(0\right)}=\frac{M}{2}\left({\pm}{\eta }_{\alpha } \mathrm{cosh} \theta +\sum _{B=1}^{2}\sum _{\beta =1}^{4}{{C}_{\alpha }}_{B}^{\beta }{\eta }_{\beta }^{B} \mathrm{sinh} \theta \right) ,\end{equation} \tag{ 5.12 }$$

where ${{C}_{\alpha }}_{B}^{\beta }$ can be regarded as a 4 × 8 matrix with the rows and columns labelled by α and (B, β), respectively. Here, cosh θ = |c_P|/M and $\mathrm{sinh} \theta =\vert {\to {c}}_{P}\vert /M$, where ${\to {c}}_{P}=\left({c}_{P1},{c}_{P2},{c}_{P3},{c}_{P4},{c}_{P5}\right)$. The four eight-dimensional column vectors of the matrix ${{C}_{\alpha }}_{B}^{\beta }$ are orthonormal, i.e.

$$\begin{equation}\sum _{B=1}^{2}\sum _{\beta =1}^{4}{{C}_{{\alpha }_{1}}}_{B}^{\beta }{{C}_{{\alpha }_{2}}}_{B}^{\beta }={\delta }_{{\alpha }_{1}{\alpha }_{2}} .\end{equation} \tag{ 5.13 }$$

Then by appendix D there is a unitary operator U, i.e. an operator preserving the inner product, on ${\mathcal{H}}_{0F}$ such that

$$\begin{equation}{Q}_{\alpha }^{\left(0\right)}=\left(M/2\right)U{\eta }_{\alpha }{U}^{{\dagger}} .\end{equation} \tag{ 5.14 }$$

Instead of directly working with ${Q}_{\alpha }^{\left(0\right)}$ and ${\hat{Q}}_{\alpha }$, it is more convenient to combine them into annihilation- and creation-type operators [33] as

$$\begin{equation}{a}_{1}=\frac{\sqrt{2}}{M}\left({Q}_{1}^{\left(0\right)}+\mathrm{i}{Q}_{2}^{\left(0\right)}\right),\quad {a}_{2}=\frac{\sqrt{2}}{M}\left({Q}_{3}^{\left(0\right)}+\mathrm{i}{Q}_{4}^{\left(0\right)}\right) ,\end{equation} \tag{ 5.15 }$$

$$\begin{equation}{b}_{1}=\frac{\sqrt{2}}{M}\left({\hat{Q}}_{1}+\mathrm{i}{\hat{Q}}_{2}\right),\quad {b}_{2}=\frac{\sqrt{2}}{M}\left({\hat{Q}}_{3}+\mathrm{i}{\hat{Q}}_{4}\right) ,\end{equation} \tag{ 5.16 }$$

provided that M > 0.⁵ These new operators satisfy the anti-commutation relations of Fermi oscillators (up to a sign), i.e. ${a}_{i}^{2}={b}_{i}^{2}={a}_{i}^{{\dagger}2}={b}_{i}^{{\dagger}2}=0$ and

$$\begin{equation}{\left[{a}_{i},{a}_{j}^{{\dagger}}\right]}_{+}=-{\delta }_{ij},\quad {\left[{b}_{i},{b}_{j}^{{\dagger}}\right]}_{+}={\delta }_{ij} .\end{equation} \tag{ 5.17 }$$

Now we construct all states $\vert {\varphi }^{\left(BF\right)}\rangle \in {\mathcal{H}}_{B}$ satisfying Q_α|φ^(BF)⟩ = 0, α = 1, 2, 3, 4. These constraints can be organized as

$$\begin{equation}\left({a}_{i}+{b}_{i}\right)\left\vert {\varphi }^{\left(BF\right)}\right\rangle =\left({a}_{i}^{{\dagger}}+{b}_{i}^{{\dagger}}\right)\left\vert {\varphi }^{\left(BF\right)}\right\rangle =0,\quad i=1,2 .\end{equation} \tag{ 5.18 }$$

Note that all states are linear combinations of the states each annihilated by either b₁ or ${b}_{1}^{{\dagger}}$ since $\vert {\varphi }^{\left(B\right)}\rangle ={b}_{1}^{{\dagger}}{b}_{1}\vert {\varphi }^{\left(B\right)}\rangle +{b}_{1}{b}_{1}^{{\dagger}}\vert {\varphi }^{\left(B\right)}\rangle$ for any state $\vert {\varphi }^{\left(B\right)}\rangle \in {\mathcal{H}}_{B}$. The same is true for each pair of operators, $\left({b}_{2},{b}_{2}^{{\dagger}}\right)$, $\left({a}_{1},{a}_{1}^{{\dagger}}\right)$ and $\left({a}_{2},{a}_{2}^{{\dagger}}\right)$. This means that a general state is a superposition of states, each belonging to a 16-dimensional Fock space built on a state |χ^(B)⟩ satisfying a_i|χ^(B)⟩ = b_i|χ^(B)⟩ = 0, i = 1, 2, by applying the creation-type operators ${a}_{i}^{{\dagger}}$ and ${b}_{i}^{{\dagger}}$. Notice that equations (5.14) and (5.15) imply a₁ = Ud₁U^† and a₂ = Ud₂U^†, where U is a unitary operator on ${\mathcal{H}}_{0F}$ and where d₁ and d₂ are defined by (5.10). Since the states annihilated by d₁ and d₂ have positive norm as we stated before, we have ${\langle {\chi }^{\left(B\right)}\vert {\chi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}{ >}0$.

We label the 16 possible states in such a Fock space as follows:

$$\begin{equation}{\left({a}_{1}^{{\dagger}}\right)}^{{m}_{1}}{\left({a}_{2}^{{\dagger}}\right)}^{{m}_{2}}{\left({b}_{1}^{{\dagger}}\right)}^{{n}_{1}}{\left({b}_{2}^{{\dagger}}\right)}^{{n}_{2}}\vert {\chi }^{\left(B\right)}\rangle =\vert {m}_{1}{m}_{2}{n}_{1}{n}_{2}\rangle ,\end{equation} \tag{ 5.19 }$$

with each m₁, m₂, n₁, n₂ being either 0 or 1. For example, we define

$$\begin{equation}\left\vert {\chi }^{\left(B\right)}\right\rangle =\left\vert 0000\right\rangle , {a}_{1}^{{\dagger}}\left\vert {\chi }^{\left(B\right)}\right\rangle =\left\vert 1000\right\rangle , {a}_{1}^{{\dagger}}{a}_{2}^{{\dagger}}{b}_{1}^{{\dagger}}{b}_{2}^{{\dagger}}\left\vert {\chi }^{\left(B\right)}\right\rangle =\left\vert 1111\right\rangle .\end{equation} \tag{ 5.20 }$$

Thus, we look for states satisfying the fermionic QLSCs in the form

$$\begin{equation}\vert {\varphi }^{\left(BF\right)}\rangle =\sum _{{m}_{1}=0}^{1}\sum _{{m}_{2}=0}^{1}\sum _{{n}_{1}=0}^{1}\sum _{{n}_{2}=0}^{1}{c}_{{m}_{1}{m}_{2}{n}_{1}{n}_{2}}\vert {m}_{1}{m}_{2}{n}_{1}{n}_{2}\rangle .\end{equation} \tag{ 5.21 }$$

We find

$$\begin{align}\hfill \left({a}_{1}+{b}_{1}\right)\vert {\varphi }^{\left(BF\right)}\rangle & =\sum _{{m}_{1}=0}^{1}\sum _{{m}_{2}=0}^{1}\sum _{{n}_{1}=0}^{1}\sum _{{n}_{2}=0}^{1}{c}_{{m}_{1}{m}_{2}{n}_{1}{n}_{2}}\left[-{m}_{1}\vert 0{m}_{2}{n}_{1}{n}_{2}\rangle +{\left(-1\right)}^{{m}_{1}+{m}_{2}}{n}_{1}\vert {m}_{1}{m}_{2}0{n}_{2}\rangle \right] .\hfill \end{align} \tag{ 5.22 }$$

The coefficient of the term |0m₂1n₂⟩ in this equation is $-{c}_{1{m}_{2}1{n}_{2}}$. Hence c_1m1n = 0 for all m and n. We can conclude similarly that c_0m0n = c_m0n0 = c_m1n1 = 0 by using the other constraints. Thus, ${c}_{{m}_{1}{m}_{2}{n}_{1}{n}_{2}}=0$ if m₁ = n₁ or m₂ = n₂. Hence the invariant state $\vert {\varphi }^{\left(BF\right)}\rangle \in {\mathcal{H}}_{B}$, i.e. the state satisfying the fermionic QLSCs, must be of the following form:

$$\begin{equation}\left\vert {\varphi }^{\left(BF\right)}\right\rangle =A\left\vert 1100\right\rangle +B\left\vert 0110\right\rangle +C\left\vert 1001\right\rangle +D\left\vert 0011\right\rangle ,\end{equation} \tag{ 5.23 }$$

where A, B, C and D are constants. Note that the four states |1100⟩, |0110⟩, |1001⟩ and |0011⟩ are mutually orthogonal and satisfy

$$\begin{align}\hfill \begin{aligned}\hfill \langle 1100\vert 1100\rangle & =\langle 0011\vert 0011\rangle ={\langle {\chi }^{\left(B\right)}\vert {\chi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}} ,\hfill \\ \hfill \langle 0110\vert 0110\rangle & =\langle 1001\vert 1001\rangle =-{\langle {\chi }^{\left(B\right)}\vert {\chi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}} .\hfill \end{aligned}\end{align} \tag{ 5.24 }$$

In particular, the two states |0110⟩ and |1001⟩ have negative norm.

By applying the constraints (5.18) we find the following unique solution up to an overall normalization:

$$\begin{equation}\left\vert {\varphi }^{\left(BF\right)}\right\rangle \propto \vert {\varphi }_{P}^{\left(BF\right)}\rangle :=\left\vert 1100\right\rangle -\left\vert 0110\right\rangle +\left\vert 1001\right\rangle +\left\vert 0011\right\rangle .\end{equation} \tag{ 5.25 }$$

Thus, there is precisely one possible combination which satisfies the fermionic QLSCs in each Fock space spanned by ${\left({a}_{1}^{{\dagger}}\right)}^{{n}_{1}}{\left({a}_{2}^{{\dagger}}\right)}^{{n}_{2}}{\left({b}_{1}^{{\dagger}}\right)}^{{n}_{1}}{\left({b}_{2}^{{\dagger}}\right)}^{{n}_{2}}\vert {\chi }^{\left(B\right)}\rangle$, where |χ^(B)⟩ is any state with fixed positive M², satisfying the bosonic QLSCs and the conditions

$$\begin{equation}\left({Q}_{1}^{\left(0\right)}+\mathrm{i}{Q}_{2}^{\left(0\right)}\right)\vert {\chi }^{\left(B\right)}\rangle =\left({Q}_{3}^{\left(0\right)}+\mathrm{i}{Q}_{4}^{\left(0\right)}\right)\vert {\chi }^{\left(B\right)}\rangle =0 ,\end{equation} \tag{ 5.26 }$$

$$\begin{equation}\left({\hat{Q}}_{1}+\mathrm{i}{\hat{Q}}_{2}\right)\vert {\chi }^{\left(B\right)}\rangle =\left({\hat{Q}}_{1}+\mathrm{i}{\hat{Q}}_{2}\right)\vert {\chi }^{\left(B\right)}\rangle =0 .\end{equation} \tag{ 5.27 }$$

Now, in section 4 it was shown that all states satisfying the bosonic QLSCs are obtained by group-averaging. Here we show that the same is true for the fermionic QLSCs. That is, the state |φ^(BF)⟩ in (5.23) is obtained as

$$\begin{equation}\vert {\varphi }^{\left(BF\right)}\rangle =-\int {\mathrm{d}}^{4}\theta {\mathrm{e}}^{-\overline{\theta }Q}\vert {\varphi }^{\left(B\right)}\rangle ,\end{equation} \tag{ 5.28 }$$

for some linear combination |φ^(B)⟩ of the states |m₁m₂n₁n₂⟩, where θ = (θ₁, θ₂, θ₃, θ₄) are Grassmann numbers and where $\overline{\theta }=\mathrm{i}{\theta }^{T}{\gamma }^{0}$ and d⁴θ = dθ₄dθ₃dθ₂dθ₁. (The minus sign in (5.28) has been introduced for convenience.) By the usual rules, ∫dθ_α = 0 and ∫dθ_αθ_α = 1, α = 1, 2, 3, 4, this equation becomes

$$\begin{align}\hfill \vert {\varphi }^{\left(BF\right)}\rangle & =-{Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\vert {\varphi }^{\left(B\right)}\rangle \hfill \\ \hfill & =\frac{1}{4}\left({Q}_{1}-\mathrm{i}{Q}_{2}\right)\left({Q}_{1}+\mathrm{i}{Q}_{2}\right)\left({Q}_{3}-\mathrm{i}{Q}_{4}\right)\left({Q}_{3}+\mathrm{i}{Q}_{4}\right)\vert {\varphi }^{\left(B\right)}\rangle \hfill \\ \hfill & =\frac{{M}^{4}}{16}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\vert {\varphi }^{\left(B\right)}\rangle .\hfill \end{align} \tag{ 5.29 }$$

One readily finds that Q₁Q₂Q₃Q₄|m₁m₂n₁n₂⟩ = 0 unless |m₁m₂n₁n₂⟩ = |1100⟩, |0110⟩, |1001⟩ or |0011⟩ and that

$$\begin{equation}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\left\vert 1100\right\rangle =\vert {\varphi }_{P}^{\left(BF\right)}\rangle ,\end{equation} \tag{ 5.30 }$$

$$\begin{equation}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\left\vert 0110\right\rangle =\vert {\varphi }_{P}^{\left(BF\right)}\rangle ,\end{equation} \tag{ 5.31 }$$

$$\begin{equation}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\left\vert 1001\right\rangle =-\vert {\varphi }_{P}^{\left(BF\right)}\rangle ,\end{equation} \tag{ 5.32 }$$

$$\begin{equation}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\left\vert 0011\right\rangle =\vert {\varphi }_{P}^{\left(BF\right)}\rangle ,\end{equation} \tag{ 5.33 }$$

where the state $\vert {\varphi }_{P}^{\left(BF\right)}\rangle$ is defined by (5.25). Thus, starting from any linear combination |φ^(B)⟩ of |m₁m₂n₁n₂⟩, we find

$$\begin{equation}-\int {\mathrm{d}}^{4}\theta {\mathrm{e}}^{-\overline{\theta }Q}\left\vert {\varphi }^{\left(B\right)}\right\rangle =-{Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\left\vert {\varphi }^{\left(B\right)}\right\rangle =\kappa \vert {\varphi }_{P}^{\left(BF\right)}\rangle ,\end{equation} \tag{ 5.34 }$$

with $\kappa \in \mathbb{C}$.

Now, an explicit computation shows that ${\langle {\varphi }_{P}^{\left(BF\right)}\vert {\varphi }_{P}^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{B}}=0$. In fact this can be deduced immediately from (5.34) because ${Q}_{\alpha }^{2}=0$ on the space ${\mathcal{H}}_{B}$ of states satisfying the bosonic QLSCs. However, the group-averaging formula (5.34) suggests that one can proceed in analogy with the bosonic case to define a new inner product. Thus, for two states $\vert {\varphi }_{1}^{\left(B\right)}\rangle ,\vert {\varphi }_{2}^{\left(B\right)}\rangle \in {\mathcal{H}}_{B}$ we obtain two states satisfying the fermionic QLSCs as

$$\begin{equation}\vert {\varphi }_{I}^{\left(BF\right)}\rangle =-{Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\vert {\varphi }_{I}^{\left(B\right)}\rangle ,\end{equation} \tag{ 5.35 }$$

and define the new inner product by

$$\begin{align}\hfill {\langle {\varphi }_{1}^{\left(BF\right)}\vert {\varphi }_{2}^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{BF}}& =-{\langle {\varphi }_{1}^{\left(B\right)}\vert {Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}\hfill \\ \hfill & ={\langle {\varphi }_{1}^{\left(B\right)}\vert {\varphi }_{2}^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{B}} .\hfill \end{align} \tag{ 5.36 }$$

Equations (5.30)–(5.33) and (5.25) imply that, if

$$\begin{equation}\vert {\varphi }_{I}^{\left(B\right)}\rangle ={\kappa }_{I}^{\left(1\right)}\vert 1100\rangle +{\kappa }_{I}^{\left(2\right)}\vert 0110\rangle +{\kappa }_{I}^{\left(3\right)}\vert 1001\rangle +{\kappa }_{I}^{\left(4\right)}\vert 0011\rangle ,\end{equation} \tag{ 5.37 }$$

then

$$\begin{equation}\vert {\varphi }_{I}^{\left(BF\right)}\rangle =\frac{{M}^{4}}{16}{\lambda }_{I}\vert {\varphi }_{P}^{\left(B\right)}\rangle ,\end{equation} \tag{ 5.38 }$$

where

$$\begin{equation}{\lambda }_{I}={\kappa }_{I}^{\left(1\right)}+{\kappa }_{I}^{\left(2\right)}-{\kappa }_{I}^{\left(3\right)}+{\kappa }_{I}^{\left(4\right)} ,\end{equation} \tag{ 5.39 }$$

and

$$\begin{equation}{\langle {\varphi }_{1}^{\left(BF\right)}\vert {\varphi }_{2}^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{BF}}=\frac{{M}^{4}}{16}{\lambda }_{1}^{{\ast}}{\lambda }_{2}\langle {\chi }^{\left(B\right)}\vert {\chi }^{\left(B\right)}\rangle .\end{equation} \tag{ 5.40 }$$

Thus, our new inner product ${\langle \cdot \vert \cdot \rangle }_{{\mathcal{H}}_{BF}}$ is positive definite⁶. Although it is defined in terms of the 'seed states' $\vert {\varphi }_{I}^{\left(B\right)}\rangle$, which do not satisfy the fermionic QLSCs, the new inner product depends only on the invariant states $\vert {\varphi }_{I}^{\left(BF\right)}\rangle$.

We constructed the states satisfying the bosonic and fermionic QLSCs step by step. We first found the states satisfying the bosonic QLSCs and an inner product among them in the previous section, following [25]. Then, we found the states satisfying the fermionic QLSCs as well among these states and defined an inner product for these invariant states. Our construction can in fact be understood as group-averaging over the supergroup generated by P^μ and Q_α. Starting from a state $\left\vert {\varphi }_{I}\right\rangle$, I = 1, 2, in the original Hilbert space ${\mathcal{H}}_{0}$, we define a state satisfying all constraints by integrating over the supergroup [37, 38]:

$$\begin{equation}\vert {\varphi }_{I}^{\left(BF\right)}\rangle =-\frac{1}{2 \mathrm{V}}\int {\mathrm{d}}^{4}\alpha ,{\mathrm{d}}^{4}\theta \mathrm{exp}\left(-\mathrm{i}\alpha \cdot P-\overline{\theta }Q\right)\vert {\varphi }_{I}\rangle ,\end{equation} \tag{ 5.41 }$$

where θ = (θ₁, θ₂, θ₃, θ₄) are Grassmann numbers and ${\alpha }^{\mu }=\left({\alpha }^{0},\to {\alpha }\right)$ are commuting numbers. As P^μ and Q_α commute, we can write this integral as

$$\begin{align}\hfill \vert {\varphi }_{I}^{\left(BF\right)}\rangle & =-\frac{1}{2 \mathrm{V}}\int {\mathrm{d}}^{4}\theta \mathrm{exp}\left(-\overline{\theta }Q\right)\int {\mathrm{d}}^{4}\alpha \mathrm{exp}\left(-\mathrm{i}\alpha \cdot P\right)\left\vert {\varphi }_{I}\right\rangle \hfill \\ \hfill & =-{Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\vert {\varphi }_{I}^{\left(B\right)}\rangle ,\hfill \end{align} \tag{ 5.42 }$$

where $\vert {\varphi }_{I}^{\left(B\right)}\rangle$ satisfies only the bosonic QLSCs:

$$\begin{equation}\vert {\varphi }_{I}^{\left(B\right)}\rangle =\frac{1}{2 \mathrm{V}}\int {\mathrm{d}}^{4}\alpha \mathrm{exp}\left(-\mathrm{i}\alpha \cdot P\right)\left\vert {\varphi }_{I}\right\rangle .\end{equation} \tag{ 5.43 }$$

We have shown that all states satisfying both the bosonic and fermionic QLSCs can be obtained in this manner. The inner product ${\langle \cdot \vert \cdot \rangle }_{{\mathcal{H}}_{BF}}$ we have defined is

$$\begin{align}\hfill {\langle {\varphi }_{1}^{\left(BF\right)}\vert {\varphi }_{2}^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{BF}}& =-\frac{1}{2 \mathrm{V}}\int {\mathrm{d}}^{4}\alpha {\mathrm{d}}^{4}\theta \left\langle {\varphi }_{1}\right\vert \mathrm{exp}\left(-\mathrm{i}\alpha \cdot P-\overline{\theta }Q\right){\left\vert {\varphi }_{2}\right\rangle }_{{\mathcal{H}}_{0}}\hfill \\ \hfill & =-\int {\mathrm{d}}^{4}\theta {\langle {\varphi }_{1}^{\left(B\right)}\vert \mathrm{exp}\left(-\overline{\theta }Q\right)\vert {\varphi }_{2}^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}} .\hfill \end{align} \tag{ 5.44 }$$

In appendix E we present an example of a state with two particles with non-zero momenta which satisfies all QLSCs.

In this paper we pointed out that there are fermionic linearization stability conditions as well as bosonic ones in four-dimensional $\mathcal{N}=1$ simple supergravity in the background of static three-torus space. Then we showed that states satisfying both fermionic and bosonic quantum linearization stability conditions (QLSCs) can be constructed by group-averaging over the supergroup of global supersymmetry and spacetime translation symmetry.

States satisfying the bosonic QLSCs have infinite norm in the original Hilbert space. This infinity results from the infinite volume of the symmetry group generated by the LSCs. Roughly speaking, this infinite volume is factored out in the group-averaging inner product. It is interesting that the inner product of states satisfying all QLSCs have zero norm in the Hilbert space of states satisfying only the bosonic ones. The finite group-averaging inner product is obtained by factoring out zero in this case.

As the bosonic QLSCs can be interpreted as a remnant of diffeomorphism invariance of the full generally covariant theory, one should be able to interpret the fermionic QLSCs as a remnant of full local supersymmetry in the context of canonical quantization [39, 40]. The bosonic QLSCs, H|phys⟩ = 0 and $\to {P}\vert \mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}\rangle =0$, have a natural physical picture. These conditions imply that all physical states are invariant under spacetime translations. This means that there is no meaning in the position and time coordinates of an event relative to the background spacetime of static three-torus. However, the physical states still encode relative positions and relative time differences between two or more events. Thus, the bosonic QLSCs can be seen as a manifestation of Mach's principle in quantum general relativity. (See e.g. [41] for a discussion of Mach's principle in general relativity.) On the other hand, it is not clear if there is a simple interpretation of the fermionic QLSCs. It would be interesting to find one.

It would also be interesting to investigate whether there are analogues of LSCs in string theory. A preliminary investigation in this direction [31] did not find such analogues in Bosonic string theory, but since string theory contains general relativity, we believe there should be analogues of LSCs in (super) string theory on any background spacetime with compact Cauchy surfaces.

The supergroup relevant to this work was a simple one with an abelian bosonic subgroup. It would be interesting to investigate the group-averaging procedure for general supergroups and establish general properties in analogy with the bosonic case studied in [24].

LS was supported by a PhD studentship from the Engineering and Physical Sciences Research Council (EPSRC) (EP/N509802/1).

In this appendix we show that one can modify the infinitesimal diffeomorphism transformation on ${e}_{\mu }^{a}$ by an infinitesimal local Lorentz transformation so that the transform is proportional to ∇_μζ_ν + ∇_νζ_μ. If we transform ${e}_{\mu }^{a}$ by diffeomorphism in the direction of ζ^μ and by a local Lorentz transformation, we have

$$\begin{equation}{\delta }^{\left(m\right)}{e}_{\mu }^{a}={\zeta }^{\rho }{\nabla }_{\rho }{e}_{\mu }^{a}+{e}_{\rho }^{a}{\nabla }_{\mu }{\zeta }^{\rho }-{s}^{ab}{e}_{b\mu } ,\end{equation} \tag{ A.1 }$$

where s^ab is anti-symmetric and where ${\nabla }_{\rho }{e}_{\mu }^{a}={\partial }_{\rho }{e}_{\mu }^{a}-{{\Gamma}}_{\rho \mu }^{\sigma }{e}_{\sigma }^{a}$ with the Levi-Civita connection ${{\Gamma}}_{\rho \mu }^{\sigma }$. Our task is to choose s^ab so that ${\delta }^{\left(m\right)}{e}_{\mu }^{a}$ is proportional to ∇_μζ_ν + ∇_νζ_μ. To this end, we require

$$\begin{equation}{e}_{a\left[\right.\mu }{\delta }^{\left(m\right)}{e}_{\nu \left.\right]}^{a}=\left({\zeta }^{\rho }{\nabla }_{\rho }{e}_{c\left[\right.\nu }\right){e}_{\mu \left.\right]}^{c}+{e}_{c\rho }{e}_{\left[\right.\mu }^{c}{\nabla }_{\nu \left.\right]}{\zeta }^{\rho }-{s}_{ab}{e}_{\left[\right.\mu }^{a}{e}_{\nu \left.\right]}^{b}=0 .\end{equation} \tag{ A.2 }$$

Since

$$\begin{equation}{D}_{\rho }{e}_{\nu }^{c}={\nabla }_{\rho }{e}_{\nu }^{c}+{{\omega }_{\mu }}^{cb}{e}_{b\nu }=0 ,\end{equation} \tag{ A.3 }$$

we find

$$\begin{equation}{s}_{ab}{e}_{\left[\right.\mu }^{a}{e}_{\nu \left.\right]}^{b}=-{\zeta }^{\rho }{\omega }_{\rho cb}{e}_{\left[\right.\mu }^{c}{e}_{\nu \left.\right]}^{b}+{\nabla }_{\left[\right.\nu }{\zeta }_{\mu \left.\right]} .\end{equation} \tag{ A.4 }$$

Thus,

$$\begin{equation}{s}^{ab}=-{\zeta }^{\rho }{{\omega }_{\rho }}^{ab}+{\nabla }_{\left[\right.\nu }{\zeta }_{\mu \left.\right]}{e}^{a\mu }{e}^{b\nu } .\end{equation} \tag{ A.5 }$$

Then

$$\begin{align}\hfill {\delta }^{\left(m\right)}{e}_{\mu }^{a}& ={\zeta }^{\rho }{\nabla }_{\rho }{e}_{\mu }^{a}+{e}_{\rho }^{a}{\nabla }_{\mu }{\zeta }^{\rho }+{\zeta }^{\rho }{{\omega }_{\rho }}^{ab}+\frac{1}{2}\left({\nabla }_{\lambda }{\zeta }_{\nu }-{\nabla }_{\nu }{\zeta }_{\lambda }\right){e}^{a\lambda }{e}^{b\nu }{e}_{b\mu }\hfill \\ \hfill & ={\zeta }^{\rho }{D}_{\rho }{e}_{\mu }^{a}+{e}_{\lambda }^{a}{\nabla }_{\mu }{\zeta }^{\lambda }+\frac{1}{2}\left({\nabla }_{\lambda }{\zeta }_{\mu }-{\nabla }_{\mu }{\zeta }_{\lambda }\right){e}^{a\lambda }\hfill \\ \hfill & =\frac{1}{2}\left({\nabla }_{\lambda }{\zeta }_{\mu }+{\nabla }_{\mu }{\zeta }_{\lambda }\right){e}^{a\lambda } .\hfill \end{align} \tag{ A.6 }$$

In this appendix we discuss the linearization stability condition for quantum electrodynamics in a static three-torus space. The Lagrangian density is

$$\begin{equation}\mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }+\overline{\psi }{\gamma }^{\mu }\left({\partial }_{\mu }-\mathrm{i}e{A}_{\mu }\right)\psi -m\overline{\psi }\psi ,\end{equation} \tag{ B.1 }$$

where F_μν = ∂_μA_ν − ∂_νA_μ and $\overline{\psi }=\mathrm{i}{\psi }^{{\dagger}}{\gamma }^{0}$. The field equations are

$$\begin{equation}\frac{\delta S}{\delta {A}_{\mu }}={\partial }_{\nu }{F}^{\nu \mu }-\mathrm{i}e\overline{\psi }{\gamma }^{\mu }\psi =0 ,\end{equation} \tag{ B.2 }$$

$$\begin{equation}\frac{\delta S}{\delta {\psi }^{{\dagger}}}=\mathrm{i}{\gamma }^{0}\left[{\gamma }^{\mu }\left({\partial }_{\mu }-\mathrm{i}e{A}_{\mu }\right)\psi -m\psi \right]=0 ,\end{equation} \tag{ B.3 }$$

$$\begin{equation}\frac{\delta S}{\delta \psi }=\left({\partial }_{\mu }+\mathrm{i}e{A}_{\mu }\right)\overline{\psi }{\gamma }^{\mu }+m\overline{\psi }=0 ,\end{equation} \tag{ B.4 }$$

where S is the action obtained by integrating $\mathcal{L}$ over spacetime. We have defined the left-hand side of (B.3) and (B.4) as the left functional derivative of the action S.

The field equations (B.3) and (B.4) imply that the current ${J}_{\left(E\right)}^{\mu }=\mathrm{i}e\overline{\psi }{\gamma }^{\mu }\psi$ is conserved. As a result, the total charge, which is the integral of ${J}_{\left(E\right)}^{0}$ over the space,

$$\begin{equation}{Q}_{\left(E\right)}=e\int {\mathrm{d}}^{3}\to {x} {\psi }^{{\dagger}}\psi ,\end{equation} \tag{ B.5 }$$

is time independent. This charge must vanish because the zeroth component of (B.2) implies ∂_iFⁱ⁰ = eψ^†ψ and hence

$$\begin{equation}{Q}_{\left(E\right)}=\int {\mathrm{d}}^{3}\to {x} {\partial }_{i}{F}^{i0} ,\end{equation} \tag{ B.6 }$$

which must vanish by Gauss's divergence theorem. (Recall our space is compact.) At linearized level, the fields A_μ and ψ are decoupled, and the condition Q_(E) = 0 does not automatically arise in the linearized theory. Hence this condition has to be imposed as a linearization stability condition on the linearized solution for it to extend to an exact solution. The aim of this appendix is to illustrate how the argument in section 3 works in this simple model.

The action is invariant under the gauge transformation δA_μ = ∂_μΛ, δψ = ieΛψ for any function Λ of the spacetime point. The argument in section 3 leads to the conclusion that

$$\begin{align}\hfill & {\partial }_{\mu }{\Lambda}\left({\partial }_{\nu }{F}^{\nu \mu }-\mathrm{i}e\overline{\psi }{\gamma }^{\mu }\psi \right)-\mathrm{i}e{\Lambda}\overline{\psi }\left[{\gamma }^{\mu }\left({\partial }_{\mu }-\mathrm{i}e{A}_{\mu }\right)\psi -m\psi \right]-\mathrm{i}e{\Lambda}\left\{\left[\left({\partial }_{\mu }+\mathrm{i}e{A}_{\mu }\right)\overline{\psi }\right]{\gamma }^{\mu }+m\overline{\psi }\right\}\psi \hfill \\ \hfill & \quad ={\partial }_{\mu }{J}_{\left({\Lambda}\right)}^{\mu }\left[A,\psi \right] ,\hfill \end{align} \tag{ B.7 }$$

for some current ${J}_{\left({\Lambda}\right)}^{\mu }\left[A,\psi \right]$ for any Λ. We indeed find that this equation is satisfied with

$$\begin{equation}{J}_{\left({\Lambda}\right)}^{\mu }\left[A,\psi \right]={\Lambda}\left({\partial }_{\nu }{F}^{\nu \mu }-\mathrm{i}e\overline{\psi }{\gamma }^{\mu }\psi \right) .\end{equation} \tag{ B.8 }$$

The identity (B.7) is satisfied order by order in the fields A_μ and ψ, i.e.

$$\begin{equation}{\partial }_{\mu }{\Lambda}{\partial }_{\nu }{F}^{\nu \mu }={\partial }_{\mu }{J}_{\left({\Lambda}\right)}^{\left(1\right)\mu }\left[A\right] ,\end{equation} \tag{ B.9 }$$

$$\begin{equation}-\mathrm{i}e{\partial }_{\mu }{\Lambda}\overline{\psi }{\gamma }^{\mu }\psi -\mathrm{i}e{\Lambda}\overline{\psi }{\gamma }^{\mu }{\partial }_{\mu }\psi -\mathrm{i}e{\Lambda}\left({\partial }_{\mu }\overline{\psi }\right){\gamma }^{\mu }\psi ={\partial }_{\mu }{J}_{\left({\Lambda}\right)}^{\left(2\right)\mu }\left[A,\psi \right] ,\end{equation} \tag{ B.10 }$$

where

$$\begin{equation}{J}_{\left({\Lambda}\right)}^{\left(1\right)\mu }\left[A\right]={\Lambda}{\partial }_{\nu }{F}^{\nu \mu } ,\end{equation} \tag{ B.11 }$$

$$\begin{equation}{J}_{\left({\Lambda}\right)}^{\left(2\right)\mu }\left[A,\psi \right]=-\mathrm{i}e{\Lambda}\overline{\psi }{\gamma }^{\mu }\psi .\end{equation} \tag{ B.12 }$$

Now, if Λ(x) = λ is constant, then the left-hand side of (B.9) vanishes. Hence the current ${J}_{\left(\lambda \right)}^{\left(1\right)\mu }\left[A\right]=\lambda {\partial }_{\nu }{F}^{\nu \mu }$ is conserved for any field configuration A_μ. In particular, we can smoothly deform A_μ to zero in the far future or past of any given Cauchy surface while keeping the value of the conserved charge

$$\begin{equation}{Q}_{\left(E\right)}^{\left(1\right)}\left[A\right]=\int {\mathrm{d}}^{3}\to {x} {\partial }_{i}{F}^{i0} ,\end{equation} \tag{ B.13 }$$

unchanged. It follows that ${Q}_{\left(E\right)}^{\left(1\right)}\left[A\right]=0$ for any field A_μ. Of course, one can readily verify this fact by Gauss's divergence theorem, as we observed before.

Now, consider solving the field equations order by order by letting

$$\begin{equation}{A}_{\mu }={A}_{\mu }^{\left(1\right)}+{A}_{\mu }^{\left(2\right)}+\cdots ,\end{equation} \tag{ B.14 }$$

$$\begin{equation}\psi ={\psi }^{\left(1\right)}+{\psi }^{\left(2\right)}+\cdots .\end{equation} \tag{ B.15 }$$

Then, with the definition ${F}_{\mu \nu }^{\left(I\right)}{:=}{\partial }_{\mu }{A}_{\nu }^{\left(I\right)}-{\partial }_{\nu }{A}_{\mu }^{\left(I\right)}$, I = 1, 2, the field equations (B.2)–(B.4) become

$$\begin{equation}{\partial }_{\nu }{F}^{\left(1\right)\nu \mu }=0 ,\end{equation} \tag{ B.16 }$$

$$\begin{equation}\mathrm{i}{\gamma }^{0}\left[{\gamma }^{\mu }{\partial }_{\mu }{\psi }^{\left(1\right)}-m{\psi }^{\left(1\right)}\right]=0 ,\end{equation} \tag{ B.17 }$$

$$\begin{equation}{\partial }_{\mu }\overline{{\psi }^{\left(1\right)}}{\gamma }^{\mu }+m\overline{{\psi }^{\left(1\right)}}=0 ,\end{equation} \tag{ B.18 }$$

$$\begin{equation}{\partial }_{\nu }{F}^{\left(2\right)\nu \mu }-\mathrm{i}e\overline{{\psi }^{\left(1\right)}}{\gamma }^{\mu }{\psi }^{\left(1\right)}=0 .\end{equation} \tag{ B.19 }$$

Then, equations (B.9) and (B.10) imply that the current

$$\begin{equation}{J}_{\left({\Lambda}\right)}^{\left(1\right)\mu }\left[{A}^{\left(2\right)}\right]+{J}_{\left({\Lambda}\right)}^{\left(2\right)\mu }\left[{A}^{\left(1\right)},{\psi }^{\left(1\right)}\right]={\Lambda}\left({\partial }_{\nu }{F}^{\left(2\right)\nu \mu }-\mathrm{i}e\overline{{\psi }^{\left(1\right)}}{\gamma }^{\mu }{\psi }^{\left(1\right)}\right) ,\end{equation} \tag{ B.20 }$$

is conserved for any Λ(x). This implies that the corresponding charge must vanish, which is rather obvious in this example because, in fact, ${J}_{\left({\Lambda}\right)}^{\left(1\right)\mu }\left[{A}^{\left(2\right)}\right]+{J}_{\left({\Lambda}\right)}^{\left(2\right)\mu }\left[{A}^{\left(1\right)},{\psi }^{\left(1\right)}\right]=0$ by (B.19). In particular, if we define

$$\begin{equation}{Q}_{\left(E\right)}^{\left(2\right)}\left[{\psi }^{\left(1\right)}\right]=-e\int {\mathrm{d}}^{3}\to {x} {\psi }^{\left(1\right){\dagger}}{\psi }^{\left(1\right)} ,\end{equation} \tag{ B.21 }$$

then, since $e\lambda {\psi }^{\left(1\right){\dagger}}{\psi }^{\left(1\right)}=\mathrm{i}e\lambda \overline{{\psi }^{\left(1\right)}}{\gamma }^{0}{\psi }^{\left(1\right)}=-{J}_{\left(\lambda \right)}^{\left(2\right)0}\left[{A}^{\left(1\right)},{\psi }^{\left(1\right)}\right]$, we find

$$\begin{equation}{Q}_{\left(E\right)}^{\left(2\right)}\left[{\psi }^{\left(1\right)}\right]=-{Q}_{\left(E\right)}^{\left(1\right)}\left[{A}^{\left(2\right)}\right]=0 ,\end{equation} \tag{ B.22 }$$

which is the linearization stability condition for electrodynamics in static torus space. In this example, it is the linearization of the condition (B.6) in the exact theory.

We first note that the matrices ${T}_{ij}^{B}$, B = 1, 2, are diagonal with ${{T}^{Bi}}_{k}={\delta }_{k}^{i}{q}_{k}^{B}$, where ${q}_{1}^{1}=-{q}_{2}^{1}=1/\sqrt{2}$, ${q}_{3}^{1}=0$. ${q}_{1}^{2}={q}_{2}^{2}=1/\sqrt{6}$ and ${q}_{3}^{2}=-2/\sqrt{6}$. Hence

$$\begin{align}\hfill X& {:=}\sum _{B=1}^{2}\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{T}_{ij}^{A}{{T}^{Bi}}_{k}{T}_{{i}^{\prime }{j}^{\prime }}^{{A}^{\prime }}{{T}^{B{i}^{\prime }}}_{{k}^{\prime }}{\gamma }^{j}{\gamma }^{k}{\gamma }^{{k}^{\prime }}{\gamma }^{{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }}\hfill \\ \hfill & =\sum _{k=1}^{3}\sum _{{k}^{\prime }=1}^{3}\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{T}_{kj}^{A}{T}_{{k}^{\prime }{j}^{\prime }}^{{A}^{\prime }}\sum _{B=1}^{2}{q}_{k}^{B}{q}_{{k}^{\prime }}^{B}{\gamma }^{j}{\gamma }^{k}{\gamma }^{{k}^{\prime }}{\gamma }^{{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }} .\hfill \end{align} \tag{ C.1 }$$

By a component-by-component calculation we find

$$\begin{equation}\sum _{B=1}^{2}{q}_{k}^{B}{q}_{{k}^{\prime }}^{B}={\delta }_{k{k}^{\prime }}-\frac{1}{3} .\end{equation} \tag{ C.2 }$$

By substituting this formula into (C.1) we obtain

$$\begin{equation}X=\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{{T}^{Ak}}_{j}{T}_{k{j}^{\prime }}^{{A}^{\prime }}{\gamma }^{j}{\gamma }^{{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }}-\frac{1}{3}\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{T}_{kj}^{A}{T}_{k{j}^{\prime }}^{{A}^{\prime }}{\gamma }^{j}{\gamma }^{k}{\gamma }^{{k}^{\prime }}{\gamma }^{{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime } .}\end{equation} \tag{ C.3 }$$

The second sum vanishes because the matrices ${T}_{ij}^{A}$ are symmetric and traceless so that ${T}_{kj}^{A}{\gamma }^{j}{\gamma }^{k}={T}_{jk}^{A}\left({\delta }^{jk}+{\gamma }^{jk}\right)=0$. Since the first sum is of the form S_jj'γ^jγ^j' where S_jj' is symmetric, we may replace γ^jγ^j' by δ^jj'. Hence

$$\begin{equation}X=\sum _{A=1}^{5}\sum _{{A}^{\prime }=1}^{5}{{T}^{Ak}}_{j}{T}_{k{j}^{\prime }}^{A}{\delta }^{j{j}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }}=\sum _{A=1}^{5}{\delta }^{A{A}^{\prime }}{c}_{PA}{c}_{P{A}^{\prime }}=\sum _{A=1}^{5}{\left({c}_{PA}\right)}^{2} ,\end{equation} \tag{ C.4 }$$

which is equation (5.8).

To show (5.9) we note that, since ${u}^{{\pm}}\left(\to {k}\right)$ is a simultaneous eigenspinor of ${\gamma }^{0}\hat{k}\cdot \to {\gamma }$ and γ₅ with eigenvalues −1 and ±1, respectively, with the normalization ${u}^{{\pm}{\dagger}}\left(\to {k}\right){u}^{{\pm}}\left(\to {k}\right)=2k$, the matrix ${M}_{\alpha \beta }={u}^{{\pm}}{\left(\to {k}\right)}_{\alpha }{u}^{{\pm}{\ast}}{\left(\to {k}\right)}_{\beta }$ is the product of the projection operator onto the eigenspace of ${\gamma }^{0}\hat{k}\cdot \to {\gamma }$ with eigenvalue −1 and that of γ₅ with eigenvalue ±1 (multiplied by 2k) since these projection operators commute. Hence, we have

$$\begin{align}\hfill {u}^{{\pm}}{\left(\to {k}\right)}_{\alpha }{u}^{{\pm}{\ast}}{\left(\to {k}\right)}_{\beta }& =\frac{k}{2}{\left[\left(1-{\gamma }^{0}\hat{k}\cdot \to {\gamma }\right)\left(1{\pm}{\gamma }_{5}\right)\right]}_{\alpha \beta }\hfill \\ \hfill & =\frac{1}{2}{\left[\left(1{\pm}{\gamma }_{5}\right)k\cdot \gamma {\gamma }^{0}\right]}_{\alpha \beta } ,\hfill \end{align} \tag{ C.5 }$$

where we have used $k\cdot \gamma =k\left(-{\gamma }^{0}+\hat{k}\cdot \to {\gamma }\right)$ so that $k\left(1-{\gamma }^{0}\hat{k}\cdot \to {\gamma }\right)=k\cdot \gamma {\gamma }^{0}$. Hence

$$\begin{equation}{\left[{{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}}\left(\to {k}\right)\right]}_{\alpha }{\left[{{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}{\ast}}\left(\to {k}\right)\right]}_{\beta }=\frac{1}{2}{\left[\left(1\mp {\gamma }_{5}\right)k\cdot \gamma \left({{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }\right)\left({{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }\right){\gamma }^{0}\right]}_{\alpha \beta } ,\end{equation} \tag{ C.6 }$$

where we have used the fact that k ⋅ γ, γ⁰ and γ₅ all anti-commute with ${{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }$ and ${{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }$. Now the identities $\left({{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }\right)\left({{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }\right)=1{\pm}{\gamma }^{0}\hat{k}\cdot \to {\gamma }{\gamma }_{5}$ and $k\cdot \gamma {\gamma }^{0}\hat{k}\cdot \to {\gamma }=k\cdot \gamma$ imply $k\cdot \gamma \left({{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \gamma \right)\left({{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \gamma \right)=k\cdot \gamma \left(1{\pm}{\gamma }_{5}\right)$. By substituting this identity into (C.6) we find

$$\begin{equation}{\left[{{\epsilon}}^{{\pm}{\ast}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}}\left(\to {k}\right)\right]}_{\alpha }{\left[{{\epsilon}}^{{\pm}}\left(\to {k}\right)\cdot \to {\gamma }{u}^{{\pm}{\ast}}\left(\to {k}\right)\right]}_{\beta }={\left[\left(1\mp {\gamma }_{5}\right)k\cdot \gamma {\gamma }^{0}\right]}_{\alpha \beta } .\end{equation} \tag{ C.7 }$$

Finally, by noting that ${\left[\left(1\mp {\gamma }_{5}\right)k\cdot \gamma {\gamma }^{0}\right]}_{\beta \alpha }={\left[\left(1{\pm}{\gamma }_{5}\right)k\cdot \gamma {\gamma }^{0}\right]}_{\alpha \beta }$ we find (5.9).

In this appendix we provide some details of the (indefinite-metric) Hilbert space describing the zero-momentum sector of the gravitino field. Some of the material presented here can be found in [33]. We make the discussion general and treat the fermionic algebra where the self-adjoint operators η_α, α = 1, 2, ..., 2M and ${\eta }_{\alpha }^{\left(+\right)}$, α = 1, 2, ..., 2N satisfying

$$\begin{equation}{\left[{\eta }_{\alpha },{\eta }_{\beta }\right]}_{+}=-{\delta }_{\alpha \beta } , {\left[{\eta }_{\alpha }^{\left(+\right)},{\eta }_{\beta }^{\left(+\right)}\right]}_{+}={\delta }_{\alpha \beta } , {\left[{\eta }_{\alpha },{\eta }_{\beta }^{\left(+\right)}\right]}_{+}=0 .\end{equation} \tag{ D.1 }$$

The zero-momentum sector of the gravitino field corresponds to M = 2 and N = 4.

We represent this algebra as follows. We define annihilation-type operators as

$$\begin{equation}{d}_{a}:=\frac{1}{\sqrt{2}}\left({\eta }_{2a-1}+\mathrm{i}{\eta }_{2a}\right) ,\quad a=1,2, \dots , M ,\end{equation} \tag{ D.2 }$$

$$\begin{equation}{d}_{a}^{\left(+\right)}:=\frac{1}{\sqrt{2}}\left({\eta }_{2a-1}^{\left(+\right)}+\mathrm{i}{\eta }_{2a}^{\left(+\right)}\right) ,\quad a=1,2, \dots , N .\end{equation} \tag{ D.3 }$$

These operators and their adjoint, the creation-type operators, have the following non-zero anti-commutators:

$$\begin{equation}{\left[{d}_{a},{d}_{b}^{{\dagger}}\right]}_{+}=-{\delta }_{ab} ,\quad {\left[{d}_{a}^{\left(+\right)},{d}_{b}^{\left(+\right){\dagger}}\right]}_{+}={\delta }_{ab} .\end{equation} \tag{ D.4 }$$

We define the state |0_0F⟩ by requiring ${d}_{a}\vert {0}_{0F}\rangle ={d}_{a}^{\left(+\right)}\vert {0}_{0F}\rangle =0$ for all a and ⟨0_0F|0_0F⟩ = 1. Then the 2^M+N-dimensional (indefinite-metric) Hilbert space representing the algebra (D.1) is spanned by the following orthogonal states:

$$\begin{equation}\vert \left\{n\right\},\left\{{n}^{\left(+\right)}\right\}\rangle =\prod _{a=1}^{M}{\left({d}_{a}^{{\dagger}}\right)}^{{n}_{a}}\prod _{b=1}^{N}{\left({d}_{b}^{\left(+\right){\dagger}}\right)}^{{n}_{b}}\vert {0}_{0F}\rangle ,\end{equation} \tag{ D.5 }$$

where n_a and ${n}_{b}^{\left(+\right)}$ are either 0 or 1 and where $\left\{n\right\}=\left\{{n}_{1},{n}_{2}, \dots , {n}_{M}\right\}$ and $\left\{{n}^{\left(+\right)}\right\}=\left\{{n}_{1}^{\left(+\right)},{n}_{2}^{\left(+\right)}, \dots , {n}_{N}^{\left(+\right)}\right\}$. One readily finds

$$\begin{equation}\langle \left\{n\right\},\left\{{n}^{\left(+\right)}\right\}\vert \left\{n\right\},\left\{{n}^{\left(+\right)}\right\}\rangle =\begin{cases}1\quad \hfill & \mathrm{i}\mathrm{f} {n}_{1}+{n}_{2}+\cdots +{n}_{M} \mathrm{i}\mathrm{s}\;\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{n},\hfill \\ -1\quad \hfill & \mathrm{i}\mathrm{f} {n}_{1}+{n}_{2}+\cdots +{n}_{N} \mathrm{i}\mathrm{s}\;\mathrm{o}\mathrm{d}\mathrm{d}.\hfill \end{cases}\end{equation} \tag{ D.6 }$$

If Ω is an anti-self-adjoint operator satisfying Ω^† = −Ω, then the operator exp Ω, which can be defined as a power series, is unitary in the sense that it preserves the inner product. Of particular interest are the following unitary operators:

$$\begin{equation}{U}_{\alpha \beta }\left(\theta \right){:=}\mathrm{exp}\left(\frac{\theta }{2}{\eta }_{\alpha }{\eta }_{\beta }\right)=\mathrm{cos}\frac{\theta }{2}+{\eta }_{\alpha }{\eta }_{\beta }\mathrm{sin}\frac{\theta }{2} , \alpha \ne \beta ,\end{equation} \tag{ D.7 }$$

$$\begin{equation}{V}_{\alpha \beta }\left(\theta \right){:=}\mathrm{exp}\left(\frac{\theta }{2}{\eta }_{\alpha }{\eta }_{\beta }^{\left(+\right)}\right)=\mathrm{cosh} \frac{\theta }{2}+{\eta }_{\alpha }{\eta }_{\beta }^{\left(+\right)}\mathrm{sinh} \frac{\theta }{2} ,\end{equation} \tag{ D.8 }$$

$$\begin{equation}{W}_{\alpha \beta }\left(\theta \right){:=}\mathrm{exp}\left(\frac{\theta }{2}{\eta }_{\alpha }^{\left(+\right)}{\eta }_{\beta }^{\left(+\right)}\right)=\mathrm{cos} \frac{\theta }{2}+{\eta }_{\alpha }^{\left(+\right)}{\eta }_{\beta }^{\left(+\right)}\mathrm{sin} \frac{\theta }{2} ,\quad \alpha \ne \beta .\end{equation} \tag{ D.9 }$$

These unitary operators act on η_α and ${\eta }_{\alpha }^{\left(+\right)}$ as follows:

$$\begin{equation}{U}_{\alpha \beta }\left(\theta \right){\eta }_{\alpha }{U}_{\alpha \beta }{\left(\theta \right)}^{{\dagger}}={\eta }_{\alpha } \mathrm{cos} \theta +{\eta }_{\beta } \mathrm{sin} \theta ,\end{equation} \tag{ D.10 }$$

$$\begin{equation}{W}_{\alpha \beta }\left(\theta \right){\eta }_{\alpha }^{\left(+\right)}{W}_{\alpha \beta }{\left(\theta \right)}^{{\dagger}}={\eta }_{\alpha }^{\left(+\right)}\mathrm{cos} \theta -{\eta }_{\beta }^{\left(+\right)} \mathrm{sin} \theta ,\end{equation} \tag{ D.11 }$$

$$\begin{equation}{V}_{\alpha \beta }\left(\theta \right){\eta }_{\alpha }{V}_{\alpha \beta }{\left(\theta \right)}^{{\dagger}}={\eta }_{\alpha } \mathrm{cosh} \theta +{\eta }_{\beta }^{\left(+\right)} \mathrm{sinh} \theta ,\end{equation} \tag{ D.12 }$$

$$\begin{equation}{V}_{\alpha \beta }\left(\theta \right){\eta }_{\beta }^{\left(+\right)}{V}_{\alpha \beta }{\left(\theta \right)}^{{\dagger}}={\eta }_{\beta }^{\left(+\right)}\mathrm{cosh} \theta +{\eta }_{\alpha } \mathrm{sinh} \theta .\end{equation} \tag{ D.13 }$$

Let us define $Y{:=}{\prod }_{\alpha =1}^{2M}{\eta }_{\alpha }{\prod }_{\beta =1}^{2N}{\eta }_{\beta }^{\left(+\right)}$. Then, Y is unitary, i.e. Y^† = Y⁻¹. The operator ${W}_{\alpha }{:=}Y{\eta }_{\alpha }^{\left(+\right)}$, α = 1, 2, ..., 2N, is also unitary. We find ${W}_{\alpha }{\eta }_{\alpha }^{\left(+\right)}{W}_{\alpha }^{{\dagger}}=-{\eta }_{\alpha }^{\left(+\right)}$ whereas ${W}_{\alpha }{\eta }_{\beta }^{\left(+\right)}{W}_{\alpha }^{{\dagger}}={\eta }_{\beta }^{\left(+\right)}$ if β ≠ α and ${W}_{\alpha }{\eta }_{\beta }{W}_{\alpha }^{{\dagger}}={\eta }_{\beta }$ for all β.

Since any product of U_αβ(θ), W_αβ(θ), V_α(θ) and W_α is unitary, we can conclude the following. Suppose that M ⩽ N and let

$$\begin{equation}{\tilde {X}}_{\alpha }={{A}_{\alpha }}^{\beta }{\eta }_{\beta } \mathrm{cosh} {\theta }_{\alpha }-{{C}_{\alpha }}^{\beta }{\eta }_{\beta }^{\left(+\right)} \mathrm{sinh} {\theta }_{\alpha } ,\end{equation} \tag{ D.14 }$$

where the index α = 1, 2, ..., 2M, is not summed over, and where $\left({{A}_{\alpha }}^{\beta }\right)$ and $\left({{C}_{\alpha }}^{\beta }\right)$ are a 2M × 2M matrix and a 2M × 2N matrix, respectively, satisfying ${{A}_{{\alpha }_{1}}}^{{\beta }_{1}}{{A}_{{\alpha }_{2}}}^{{\beta }_{2}}{\delta }_{{\beta }_{1}{\beta }_{2}}={{C}_{{\alpha }_{1}}}^{{\beta }_{1}}{{C}_{{\alpha }_{2}}}^{{\beta }_{2}}{\delta }_{{\beta }_{1}{\beta }_{2}}={\delta }_{{\alpha }_{1}{\alpha }_{2}}$ and $\mathrm{det}\left({{A}^{\alpha }}_{\beta }\right)=1$. ⁷Then there is a unitary operator U such that ${\tilde {X}}_{\alpha }=U{\eta }_{\alpha }{U}^{{\dagger}}$.

In this appendix we present a concrete example of a state satisfying all fermionic as well as bosonic QLSCs. We consider an example where there are two particles, one going in the positive x-direction and the other in the negative x-direction with momentum k > 0 and both with positive helicity. In this sector of the theory M² = 4k.

To find the non-zero-momentum contribution to the supercharge in this sector we need to find the spinors ${u}^{+}\left(\to {k}\right)$ with $\to {k}={\pm}k{\hat{e}}_{1}$, where ê₁ is the unit vector in the x-direction. It is an eigenspinor of γ⁰γ¹ and iγ²γ³ with eigenvalue −1 for both and, hence, an eigenspinor of γ₅ = iγ⁰γ¹γ²γ³ with eigenvalue 1. It is normalized as u^+†(±kê₁)u⁺(±kê₁) = 2k. We find from (3.2)

$$\begin{equation}{\gamma }^{0}{\gamma }^{1}=\left(\begin{pmatrix}\hfill 0\hfill & \hfill -1\hfill \\ \hfill -1\hfill & \hfill 0\hfill \end{pmatrix}\right) ,\quad \mathrm{i}{\gamma }^{2}{\gamma }^{3}=\left(\begin{pmatrix}\hfill {\sigma }_{2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill {\sigma }_{2}\hfill \end{pmatrix}\right) .\end{equation} \tag{ E.1 }$$

Hence we can take

$$\begin{equation}{u}^{+}\left(k{\hat{e}}_{1}\right)=\sqrt{\frac{k}{2}}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \end{pmatrix}\right) , {u}^{+{\ast}}\left(k{\hat{e}}_{1}\right)=\sqrt{\frac{k}{2}}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill \mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill \mathrm{i}\hfill \end{pmatrix}\right) .\end{equation} \tag{ E.2 }$$

Now, {ê₂, ê₃, ê₁} form a right handed basis so that we can choose

$$\begin{equation}{{\epsilon}}^{+}\left(k{\hat{e}}_{1}\right)\cdot \to {\gamma }=\frac{1}{\sqrt{2}}\left({\gamma }^{2}+\mathrm{i}{\gamma }^{3}\right)=\frac{1}{\sqrt{2}}{\gamma }^{2}\left(1+\mathrm{i}{\gamma }^{2}{\gamma }^{3}\right) .\end{equation} \tag{ E.3 }$$

Since u^+∗(kê₁) is an eigenspinor of iγ²γ³ with eigenvalue +1, we have

$$\begin{equation}{{\epsilon}}^{+}\left(k{\hat{e}}_{1}\right)\cdot \to {\gamma }{u}^{+{\ast}}\left(k{\hat{e}}_{1}\right)=\sqrt{k}{\gamma }^{2}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill \mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill \mathrm{i}\hfill \end{pmatrix}\right)=\sqrt{k}\left(\begin{pmatrix}\hfill \mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill \mathrm{i}\hfill \\ \hfill 1\hfill \end{pmatrix}\right) ,\end{equation} \tag{ E.4 }$$

and

$$\begin{equation}{{\epsilon}}^{+{\ast}}\left(k{\hat{e}}_{1}\right)\cdot \to {\gamma }{u}^{+}\left(k{\hat{e}}_{1}\right)=\sqrt{k}{\gamma }^{2}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \end{pmatrix}\right)=\sqrt{k}\left(\begin{pmatrix}\hfill -\mathrm{i}\hfill \\ \hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \\ \hfill 1\hfill \end{pmatrix}\right) .\end{equation} \tag{ E.5 }$$

The spinor u⁺(−kê₁) is an eigenspinor of γ⁰γ¹ and iγ²γ³ with eigenvalue +1 for both. We can choose

$$\begin{equation}{{\epsilon}}^{+}\left(-k{\hat{e}}_{1}\right)\cdot \to {\gamma }=\frac{1}{\sqrt{2}}{\gamma }^{2}\left(1-\mathrm{i}{\gamma }^{2}{\gamma }^{3}\right) .\end{equation} \tag{ E.6 }$$

Proceeding similarly as above, we find

$$\begin{equation}{{\epsilon}}^{+}\left(-k{\hat{e}}_{1}\right)\cdot \to {\gamma }{u}^{+{\ast}}\left(-k{\hat{e}}_{1}\right)=\sqrt{k}{\gamma }^{2}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill \mathrm{i}\hfill \\ \hfill -1\hfill \\ \hfill -\mathrm{i}\hfill \end{pmatrix}\right)=\sqrt{k}\left(\begin{pmatrix}\hfill -\mathrm{i}\hfill \\ \hfill -1\hfill \\ \hfill \mathrm{i}\hfill \\ \hfill 1\hfill \end{pmatrix}\right) ,\end{equation} \tag{ E.7 }$$

and

$$\begin{equation}{{\epsilon}}^{+{\ast}}\left(-k{\hat{e}}_{1}\right)\cdot \to {\gamma }{u}^{+}\left(-k{\hat{e}}_{1}\right)=\sqrt{k}{\gamma }^{2}\left(\begin{pmatrix}\hfill 1\hfill \\ \hfill -\mathrm{i}\hfill \\ \hfill -1\hfill \\ \hfill \mathrm{i}\hfill \end{pmatrix}\right)=\sqrt{k}\left(\begin{pmatrix}\hfill \mathrm{i}\hfill \\ \hfill -1\hfill \\ \hfill -\mathrm{i}\hfill \\ \hfill 1\hfill \end{pmatrix}\right) ,\end{equation} \tag{ E.8 }$$

Thus, the relevant part of the $\to {k}\ne 0$ contribution to the supercharge

$$\begin{equation}\quad \hat{Q}=-\frac{\mathrm{i}}{2}\sum _{\to {k}\ne 0}\sum _{\lambda ={\pm}}\left[{{\epsilon}}^{\lambda }\left(\to {k}\right)\cdot \gamma {u}^{\lambda {\ast}}\left(\to {k}\right){a}_{\lambda }\left(\to {k}\right){b}_{\lambda }^{{\dagger}}\left(\to {k}\right)-{{\epsilon}}^{\lambda {\ast}}\left(\to {k}\right)\cdot \gamma {u}^{\lambda }\left(\to {k}\right){a}_{\lambda }^{{\dagger}}\left(\to {k}\right){b}_{\lambda }\left(\to {k}\right)\right] ,\end{equation} \tag{ E.9 }$$

is, with the notation a_(±) = a₊(±kê₁) and b_(±) = b₊(±kê₁),

$$\begin{equation}\hat{Q}=\frac{\sqrt{k}}{2}\left(\begin{pmatrix}\hfill \left[{a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}+{a}_{\left(+\right)}^{{\dagger}}{b}_{\left(+\right)}\right]-\left[{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}+{a}_{\left(-\right)}^{{\dagger}}{b}_{\left(-\right)}\right]\hfill \\ \hfill -\mathrm{i}\left[{a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}-{a}_{\left(+\right)}^{{\dagger}}{b}_{\left(+\right)}\right]+\mathrm{i}\left[{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}-{a}_{\left(-\right)}^{{\dagger}}{b}_{\left(-\right)}\right]\hfill \\ \hfill \left[{a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}+{a}_{\left(+\right)}^{{\dagger}}{b}_{\left(+\right)}\right]+\left[{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}+{a}_{\left(-\right)}^{{\dagger}}{b}_{\left(-\right)}\right]\hfill \\ \hfill -\mathrm{i}\left[{a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}-{a}_{\left(+\right)}^{{\dagger}}{b}_{\left(+\right)}\right]-\mathrm{i}\left[{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}-{a}_{\left(-\right)}^{{\dagger}}{b}_{\left(-\right)}\right]\hfill \end{pmatrix}\right) .\end{equation} \tag{ E.10 }$$

Proceeding as in section 5 to construct a state satisfying the fermionic QLSCs we define

$$\begin{equation}{a}_{1}=\frac{\sqrt{2}}{M}\left({Q}_{1}^{\left(0\right)}+\mathrm{i}{Q}_{2}^{\left(0\right)}\right) ,\quad {a}_{2}=\frac{\sqrt{2}}{M}\left({Q}_{3}^{\left(0\right)}+\mathrm{i}{Q}_{4}^{\left(0\right)}\right) ,\end{equation} \tag{ E.11 }$$

as before and also

$$\begin{equation}{b}_{1}=\frac{1}{\sqrt{2k}}\left({\hat{Q}}_{1}+\mathrm{i}{\hat{Q}}_{2}\right)=\frac{1}{\sqrt{2}}\left({a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}-{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}\right) ,\end{equation} \tag{ E.12 }$$

$$\begin{equation}{b}_{2}=\frac{1}{\sqrt{2k}}\left({\hat{Q}}_{3}+\mathrm{i}{\hat{Q}}_{4}\right)=\frac{1}{\sqrt{2}}\left({a}_{\left(+\right)}{b}_{\left(+\right)}^{{\dagger}}+{a}_{\left(-\right)}{b}_{\left(-\right)}^{{\dagger}}\right) ,\end{equation} \tag{ E.13 }$$

in the relevant subspace. Then in the two-particle sector, M² = 4k, it is easy to see that

$$\begin{equation}{\left[{a}_{i},{a}_{j}^{{\dagger}}\right]}_{+}=-{\delta }_{ij} ,\quad {\left[{b}_{i},{b}_{j}^{{\dagger}}\right]}_{+}={\delta }_{ij} .\end{equation} \tag{ E.14 }$$

We define a state $\left\vert {\varphi }^{\left(B\right)}\right\rangle$ obeying the bosonic QLSCs as follows. We take the bosonic zero-momentum sector wave function to be given by the Gaussian

$$\begin{equation}{\Psi}\left(c,\to {c}\right)=N{e}^{-{c}^{2}}{e}^{-{\to {c}}^{2}} ,\end{equation} \tag{ E.15 }$$

where N is a constant, and the non-zero-momentum part of the state contains two positive helicity gravitons with momentum ±kê₁. Then the state obeying the bosonic QLSCs by group-averaging is given by (4.24), which reads

$$\begin{equation}\quad \left\vert {\varphi }^{\left(B\right)}\right\rangle =\int \frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}\left[{f}^{\left(+\right)}\left(\to {p}\right){\mathrm{e}}^{-\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}+{f}^{\left(-\right)}\left(\to {p}\right){\mathrm{e}}^{\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}\right]\otimes \left({a}_{\left(+\right)}^{{\dagger}}{a}_{\left(-\right)}^{{\dagger}}\left\vert 0\right\rangle \right) ,\end{equation} \tag{ E.16 }$$

where the state $\left\vert 0\right\rangle$ is the Fock vacuum in the non-zero-momentum sector, $E\left(\to {p}\right)=\sqrt{{\to {p}}^{2}+{M}^{2}}$ and

$$\begin{equation}{f}^{\left({\pm}\right)}\left(\to {p}\right)=\frac{{\pi }^{3}N{\mathrm{e}}^{-{M}^{2}/4}{\mathrm{e}}^{-{\to {p}}^{2}/2}}{2\sqrt{{\to {p}}^{2}+{M}^{2}}} .\end{equation} \tag{ E.17 }$$

The norm of this state in the Hilbert space ${\mathcal{H}}_{B}$ is then given by (4.26), which reads

$$\begin{equation}{\langle {\varphi }^{\left(B\right)}\vert {\varphi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}=\frac{{\pi }^{3}}{12}\vert N{\vert }^{2}{\mathrm{e}}^{-{M}^{2}/2}{\int }_{0}^{\infty }\mathrm{d}p \frac{{p}^{4}{\mathrm{e}}^{-{p}^{2}}}{\sqrt{{p}^{2}+{M}^{2}}}{< }\infty .\end{equation} \tag{ E.18 }$$

We choose the normalisation constant N so that ${\langle {\varphi }^{\left(B\right)}\vert {\varphi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}=1$.

Now we need to consider the fermionic zero-momentum sector. As we stated in section 5, we need to consider the tensor product of ${\mathcal{H}}_{B}$ and the 64-dimensional Hilbert space ${\mathcal{H}}_{0F}$ describing the fermionic zero-momentum sector. We call this tensor product ${\mathcal{H}}_{B}$ as in section 5. As in appendix D let |0_0F⟩ be the normalized state annihilated by ${d}_{1}=\left({\eta }_{1}+\mathrm{i}{\eta }_{2}\right)/\sqrt{2}$, ${d}_{2}=\left({\eta }_{3}+\mathrm{i}{\eta }_{4}\right)/\sqrt{2}$, ${d}_{1}^{B}=\left({\eta }_{1}^{B}+\mathrm{i}{\eta }_{2}^{B}\right)/\sqrt{2}$ and ${d}_{2}^{B}=\left({\eta }_{3}^{B}+\mathrm{i}{\eta }_{4}^{B}\right)/\sqrt{2}$, B = 1, 2. Let ā₁ and ā₂ be the operators a₁ and a₂, respectively, restricted to the eigenspace of c_P and c_PA with eigenvalues p₀ and p_A, respectively. Define $\vert {p}_{0},\to {p}\rangle \in {\mathcal{H}}_{0F}$ by $\vert {p}_{0},\to {p}\rangle {:=}{C}_{{p}_{0},\to {p}} {\bar{a}}_{1}{\bar{a}}_{2}{d}_{2}^{{\dagger}}{d}_{1}^{{\dagger}}\vert {0}_{0F}\rangle$, where ${C}_{{p}_{0},\to {p}}$ is a normalization constant such that ${\langle {p}_{0},\to {p}\vert {p}_{0},\to {p}\rangle }_{{\mathcal{H}}_{0F}}=1$. Then, ${\bar{a}}_{1}\vert {p}_{0},\to {p}\rangle ={\bar{a}}_{2}\vert {p}_{0},\to {p}\rangle =0$ because ${\bar{a}}_{1}^{2}={\bar{a}}_{2}^{2}=0$ and ā₁ā₂ = −ā₂ā₁. Now, we let

$$\begin{align}\hfill \left\vert {\varphi }^{\left(B\right)}\right\rangle & =\int \frac{{\mathrm{d}}^{5}\to {p}}{{\left(2\pi \right)}^{5}}\left[{f}^{\left(+\right)}\left(\to {p}\right){\mathrm{e}}^{-\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert E\left(\to {p}\right),\to {p}\rangle +{f}^{\left(-\right)}\left(\to {p}\right){\mathrm{e}}^{\mathrm{i}E\left(\to {p}\right)c+\mathrm{i}\to {p}\cdot \to {c}}\otimes \vert -E\left(\to {p}\right),\to {p}\rangle \right]\hfill \\ \hfill & \quad \otimes \left({a}_{\left(+\right)}^{{\dagger}}{a}_{\left(-\right)}^{{\dagger}}\left\vert 0\right\rangle \right) .\hfill \end{align} \tag{ E.19 }$$

Then, ${a}_{1}\left\vert {\varphi }^{\left(B\right)}\right\rangle ={a}_{2}\left\vert {\varphi }^{\left(B\right)}\right\rangle =0$ and ${\langle {\varphi }^{\left(B\right)}\vert {\varphi }^{\left(B\right)}\rangle }_{{\mathcal{H}}_{B}}=1$. Notice also that because $\left\vert {\varphi }^{\left(B\right)}\right\rangle$ contains only gravitons, it obeys ${b}_{1}^{{\dagger}}\left\vert {\varphi }^{\left(B\right)}\right\rangle ={b}_{2}^{{\dagger}}\left\vert {\varphi }^{\left(B\right)}\right\rangle =0$.

In the notation of section 5 we can thus take $\left\vert {\varphi }^{\left(B\right)}\right\rangle \propto \left\vert 0011\right\rangle$. Indeed, by the group-averaging procedure described in section 5, we obtain a state $\left\vert {\varphi }^{\left(BF\right)}\right\rangle \in {\mathcal{H}}_{BF}$ obeying the bosonic and fermionic QLSCs by applying −Q₁Q₂Q₃Q₄. That is,

$$\begin{equation}\left\vert {\varphi }^{\left(BF\right)}\right\rangle =-{Q}_{1}{Q}_{2}{Q}_{3}{Q}_{4}\left\vert {\varphi }^{\left(B\right)}\right\rangle \end{equation} \tag{ E.20 }$$

$$\begin{equation}=\frac{1}{4}\left({Q}_{1}-\mathrm{i}{Q}_{2}\right)\left({Q}_{1}+\mathrm{i}{Q}_{2}\right)\left({Q}_{3}-\mathrm{i}{Q}_{4}\right)\left({Q}_{3}+\mathrm{i}{Q}_{4}\right)\left\vert {\varphi }^{\left(B\right)}\right\rangle \end{equation} \tag{ E.21 }$$

$$\begin{equation}=\frac{{M}^{4}}{16}\left({a}_{1}^{{\dagger}}+{b}_{1}^{{\dagger}}\right)\left({a}_{1}+{b}_{1}\right)\left({a}_{2}^{{\dagger}}+{b}_{2}^{{\dagger}}\right)\left({a}_{2}+{b}_{2}\right)\left\vert {\varphi }^{\left(B\right)}\right\rangle .\end{equation} \tag{ E.22 }$$

By writing $\left\vert {\varphi }^{\left(B\right)}\right\rangle ={a}_{\left(+\right)}^{{\dagger}}{a}_{\left(-\right)}^{{\dagger}}\left\vert {{\Psi}}^{\left(B\right)}\right\rangle$ this can be evaluated as

$$\begin{align}\hfill \left\vert {\varphi }^{\left(BF\right)}\right\rangle & =\frac{{M}^{4}}{16}\left[{a}_{\left(+\right)}^{{\dagger}}{a}_{\left(-\right)}^{{\dagger}}\vert {{\Psi}}^{\left(B\right)}\rangle -\frac{1}{\sqrt{2}}\left({a}_{1}^{{\dagger}}-{a}_{2}^{{\dagger}}\right){a}_{\left(+\right)}^{{\dagger}}{b}_{\left(-\right)}^{{\dagger}}\vert {{\Psi}}^{\left(B\right)}\rangle \right.\hfill \\ \hfill & \quad \left.+\frac{1}{\sqrt{2}}\left({a}_{1}^{{\dagger}}+{a}_{2}^{{\dagger}}\right){b}_{\left(+\right)}^{{\dagger}}{a}_{\left(-\right)}^{{\dagger}}\vert {{\Psi}}^{\left(B\right)}\rangle -{a}_{1}^{{\dagger}}{a}_{2}^{{\dagger}}{b}_{\left(+\right)}^{{\dagger}}{b}_{\left(-\right)}^{{\dagger}}\vert {{\Psi}}^{\left(B\right)}\rangle \right] .\hfill \end{align} \tag{ E.23 }$$

In particular, by the group-averaging procedure, the norm of this state is given by

$$\begin{equation}{\langle {\varphi }^{\left(BF\right)}\vert {\varphi }^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{BF}}={\langle {\varphi }^{\left(B\right)}\vert {\varphi }^{\left(BF\right)}\rangle }_{{\mathcal{H}}_{B}}=\frac{{M}^{4}}{16},\end{equation} \tag{ E.24 }$$

as would be expected by the procedure presented in section 5.