nLab 11d SuGra from super C-field flux quantization -- section

11d-SuGra from Super C-Field Flux Quantization

We discuss (Thm. below, following GSS24, §3) how the equations of motion of D=11 supergravity — on an $11\vert\mathbf{32}$-dimensional super-torsion-free super spacetime $X$ with super vielbein $(e,\psi)$ (the graviton/gravitino-fields) — follow from just the requirement that the duality-symmetric super-C-field flux densities $(G_4^s, G_7^s) \,\in\, \Omega^4_{dR}(X) \times \Omega^7_{dR}(X)$:

1. satisfy their Bianchi identities

   (1)$$\begin{array}{l} \mathrm{d} \, G_4^s \;=\; 0 \\ \mathrm{d} \, G_7^s \;=\; \tfrac{1}{2} G_4^s \, G_4^s \end{array}$$

2. are on any super-chart $U \hookrightarrow X$ of the locally supersymmetric form

   (2)$$\begin{array}{l} G_4^s \;=\; \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} e^{a_1} \cdots e^{a_4} \,-\, \tfrac{1}{2} \big(\overline{\psi}\Gamma_{a_1 a_2} \psi\big) e^{a_1} \, e^{a_2} \\ G_7^s \;=\; \tfrac{1}{7!} (G_7)_{a_1 \cdots a_7} e^{a_1} \cdots e^{a_7} \,-\, \tfrac{1}{5!} \big(\overline{\psi}\Gamma_{a_1 \cdots a_5} \psi\big) e^{a_1} \cdots e^{a_5} \mathrlap{\,.} \end{array}$$

Up to some mild (but suggestive, see below) re-arrangement, the computation is essentially that indicated in CDF91, §III.8.5 (where some of the easy checks are indicated) which in turn is a mild reformulation of the original claim in Cremmer & Ferrara 1980 and Brink & Howe 1980 (where less details were given). A full proof is laid out in GSS24, §3, whose notation we follow here.

The following may be understood as an exposition of this result, which seems to stand out as the only account that is (i) fully first-order and (ii) duality-symmetric (in that $G_7$ enters the EoMs as an independent field, whose Hodge duality to $G_4$ is imposed by the Bianchi identity for $G_7^s$, remarkably).

Notice that the discussion in CDF91, §III.8 amplifies the superspace-rheonomy principle as a constraint that makes the Bianchi identities on (in our paraphrase) a supergravity Lie 6-algebra-valued higher vielbein be equivalent to the equations of motion of D=11 SuGra. But we may observe that the only rheonomic constraint necessary is that (2) on the C-field flux density — and this is the one not strictly given by rules in CDF91, p. 874, cf. around CDF91, (III.8.41) —; while the remaining rheonomy condition on the gravitino field strength $\rho$ is implied (Lem. below), and the all-important torsion constraint (3) (which is also outside the rules of rheonomy constraints, cf. CDF91, (III.8.33)) is naturally regarded as part of the definition of a super-spacetime in the first place (Def. below).

In thus recasting the formulation of the theorem somewhat, we also:

1. re-define the super-flux densities as above (2), highlighting that it is (only) in this combination that the algebraic form of the expected Bianchi identity (1) extends to superspace;

2. disregard the gauge potentials $C_3$ and $C_6$, whose role in CDF91, §III.8.2-4 is really just to motivate the form of the Bianchi identities equivalent to (1), but whose global nature is more subtle than acknowledged there, while being irrelevant for just the equations of motion.

Indeed, the point is that, in consequence of our second item above, the following formulation shows that one may apply flux quantization of the supergravity C-field on superspace in formally the same way as bosonically (for instance in Cohomotopy as per Hypothesis H, or in any other nonabelian cohomology theory whose classifying space has the $\mathbb{Q}$-Whitehead $L_\infty$-algebra of the 4-sphere), and in fact that the ability to do so implies the EoMs of 11d SuGra. Any such choice of flux quantization is then what defines, conversely, the gauge potentials, globally. Moreover, by the fact brought out here, that the super-flux Bianchi identity already implies the full equations of motion, this flux quantization is thereby seen to be compatible with the equations of motion on all of super spacetime.

For the present formulation, we find it suggestive to regard the all-important torsion constraint (3) as part of the definition of the super-gravity field itself (since it ties the auxiliary spin-connection to the super-vielbein field which embodies the actual super-metric structure):

Definition

(super-spacetime)
For

• $D \in \mathbb{N}_{\geq 1}$ a natural number

• $\mathbf{N} \in Rep_{\mathbb{R}}\big(Spin(1,D-1)\big)$ a real spin representation (“Majorana spinors”) of $\mathbb{R}$-dimension $N$

  whose $Spin(1,D)$-equivariant bilinear pairing we denote

  $$\overline{(\text{-})}(\text{-}) \;\colon\; \mathbf{N} \otimes \mathbf{N} \longrightarrow \mathbb{R}^{1,D-1} \,,$$

by a super-spacetime of super-dimension $D\vert \mathbf{N}$ we here mean:

1. a supermanifold

2. which admits an open cover by super-Minkowski supermanifolds $\mathbb{R}^{1,D-1\vert \mathbf{N}}$,

3. equipped with a super Cartan connection with respect to the canonical subgroup inclusion $Spin(1,D-1) \hookrightarrow Iso(\mathbb{R}^{1,D-1\vert\mathbf{N}})$ of the spin group into the super Poincaré group, namely:

   1. equipped with a super-vielbein $(e, \psi)$, hence on each super-chart $U \hookrightarrow X$

      $$\big( (e^a)_{a=0}^{D=1} ,\, (\psi^\alpha)_{\alpha=1}^N \big) \;\in\; \Omega^1_{dR}\big( U ;\, \mathbb{R}^{1,D-1\vert \mathbf{N}} \big)$$

      such that at every point $x \in \overset{\rightsquigarrow}{X}$ the induced map on tangent spaces is an isomorphism

      $$(e,\psi)_x \;\colon\; T_x X \overset{\sim}{\longrightarrow} \mathbb{R}^{1,10\vert \mathbf{N}} \,.$$

   2. and with a spin-connection $\omega$ (…),

4. such that the super-torsion vanishes, in that on each chart:

   (3)$$\mathrm{d} \, e^a - \omega^a{}_b \, e^b \;=\; \big( \overline{\psi} \,\Gamma^a\, \psi \big) \,,$$

   where $\Gamma^{(-)} \,\colon\, \mathbb{R}^{1,D-1} \longrightarrow End_{\mathbb{R}}(\mathbf{N})$ is a representation of $Pin^+(1,10)$, hence

   $$\Gamma_{a} \Gamma_b + \Gamma_{b} \Gamma_a \;=\; + 2\, diag(-, +, +, \cdots, +)_{a b} \,.$$

Definition

(the gravitational field strength)
Given a super-spacetime (Def. ), we say that (super chart-wise):

1. its super-torsion is:

   $$T^a \;\coloneqq\; \mathrm{d}\, e^a \,-\, \omega^a{}_b \, e^b \,-\, \big( \overline{\psi}\Gamma^a\psi \big)$$

2. its gravitino field strength is

   $$\rho \;\coloneqq\; \mathrm{d}\, \psi + \tfrac{1}{4} \omega_{a b}\Gamma^{a b}\psi \,,$$

3. its curvature is

   $$R^{a}{}_b \;\coloneqq\; \mathrm{d}\, \omega^{a}{}_b \,-\, \omega^a{}_c \, \omega^c{}_b \,.$$

Lemma

(super-gravitational Bianchi identities)
By exterior calculus the gravitational field strength tensors (Def. ) satisfy the following identities:

(4)$$\begin{array}{ccl} \mathrm{d} \, R^{a}{}_b &=& \omega^a{}_{a'} \, R^{a'}{}_b - R^{a}{}_{b'} \, \omega^{b'}{}_{b} \\ \mathrm{d} \, T^a &=& - R^{a}{}_b \ e^b + 2 \big( \overline{\psi} \,\Gamma^a\, \rho \big) \\ \mathrm{d} \, \rho &=& \tfrac{1}{4} R^{a b} \Gamma_{a b} \psi \end{array}$$

Remark

(role of the gravitational Bianchi identities)
Notice that the equations (4) are not conditions but identities satisfied by any super-spacetime (in the sense of Def. , hence even such that $T^a = 0$.) But conversely this means that when constructing a super-spacetime (say subject to further contraints, such as Bianchi identities for flux densities), the equations (4) are a necessary condition to be satisfied by any candidate super-vielbein-field, and as such they may play the role of equations of motion for the super-gravitational field, as we will see.

Write now $\mathbf{32} \in Rep_{\mathbb{R}}\big(Spin(1,10)\big)$ for the unique non-trivial irreducible real $Spin(1,10)$-representation.

Theorem

(11d SuGra EoM from super-flux Bianchi identity) Given

1. (super-gravity field:) an $11\vert\mathbf{32}$-dimensional super-spacetime $X$ (Def. ),

2. (super-C-field flux densities:) $(G^s_4,\, G^s_7)$ as in (2)

then the super-flux Bianchi identity (1) (the super-higher Maxwell equation for the C-field)

$$\begin{array}{l} \mathrm{d} \, G_4^s \;=\; 0 \\ \mathrm{d} \, G_7^s \;=\; \tfrac{1}{2} G_4^s \, G_4^s \end{array}$$

is equivalent to the joint solution by $\big(e, \psi, \omega, G_4^s,\, G_7^s\big)$ of the equations of motion of D=11 supergravity.

This is, in some paraphrase, the result of CDF91, §III.8.5, We indicate the proof broken up in the following Lemmas , , and .

In all of the following lemmas one expands the Bianchi identoties in their super-vielbein form components.

Remark

(Normalization conventions)
Our choice of prefactors and normalization follows CDF91 except for the following changes:

• our Clifford generators absorb a factor of $\mathrm{i}$: $\;\;\;\Gamma_a \;=\; \mathrm{i}\, \Gamma_a^{^{DF}}$

• our gravitinos absorb a factor of $\sqrt{2}$: $\;\;\;\psi \;=\; \sqrt{2}\psi^{^{DF}}$

• our 4-flux density absorbs a combinatorial factor of $1/2$: $\;\;\;G_4 = \tfrac{1}{2} R^{\Box}$

• our 7-flux density absorbs a combinatoiral factor of $1/5!$: $\;\;\;G_7 = \tfrac{1}{5!} R^{\otimes}$

Here:

• The first rescaling reflects that $\Gamma^{{}^{\mathrm{DF}}}$ is not actually a Majorana representation of $\mathrm{Pin}^+(1,10)$, but $\mathrm{i}\Gamma^{{}^{\mathrm{DF}}}$ is.

  This rescaling removes all occurrences of imaginary units in the Bianchi identities, as it should be for algebra over the real numbers with real fermion representations.

• The second rescaling has the effect that $\mathrm{d} e^a = \big(\overline{\psi} \Gamma^a \psi\big) + \cdots$ instead of $\mathrm{d}\, e^a = \tfrac{1}{2} \big(\overline{\psi} \Gamma^a \psi\big) + \cdots$.

Lemma

The Bianchi identity for $G^s_4$ (1) is equivalent to

1. the closure of the ordinary 4-flux density $G_4$

2. the following dependence of $\rho$ on $G_4$

shown in any super-chart:

(5)$$\begin{array}{l} \mathrm{d}\, G^s_4 \;=\; 0 \\ \;\Leftrightarrow\; \left\{ \begin{array}{l} \big( \nabla_{a} (G_4)_{a_1 \cdots a_4} \big) e^{a} \, e^{a_1} \cdots e^{a_4} \;=\; 0 \\ \rho \;=\; \rho_{a b} \, e^{a} \, e^b \,+\, \underset{ H_a }{ \underbrace{ \Big( \tfrac{1}{6} \, \tfrac{1}{3!} (G_4)_{a b_1 b_2 b_3} \,\Gamma^{a b_1 b_2 b_3}\, \, - \tfrac{1}{12} \, \tfrac{1}{4!} (G_4)_{b_1 \cdots b_4} \,\Gamma^{a b_1 \cdots b_4}\, \Big) } } \psi \, e^a \\ \Big( \tfrac{1}{4!} \psi^\alpha \nabla_\alpha (G_4)_{a_1 \cdots a_4} \;+\; \big( \overline{\psi} \Gamma_{a_1 a_2} \rho_{a_3 a_4} \big) \Big) e^{a_1} \cdots e^{a_4} \;=\; 0 \,. \end{array} \right. \end{array}$$

This is essentially the claim in CDF91 (III.8.44-49 & 60b); full proof is given in GSS24, Lem. 3.2.

Proof

The general expansion of $\rho$ in the super-vielbein basis is of the form

$$\rho \;:=\; \rho_{a b} \, e^a\, e^b + H_a \psi \, e^a + \underset{ = 0 }{ \underbrace{ \overline{\psi} \,\kappa\, \psi } } \,,$$

where the last term is taken to vanish.l (…).

Therefore, the Bianchi identity has the following components,

(6)$$\begin{array}{l} \mathrm{d} \Big( \, \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} \, e^{a_1} \cdots e^{a_4} - \tfrac{1}{2} \big( \overline{\psi} \Gamma_{a_1 a_2} \psi \big) \, e^{a_1}\, e^{a_2} \Big) \;=\; 0 \\ \;\Leftrightarrow\; \left\{ \begin{array}{l} \big( \nabla_{a} (G_4)_{a_1 \cdots a_4} \big) e^{a}\, e^{a_1} \cdots e^{a_4} \;=\; 0 \\ \Big( \tfrac{1}{4!} \psi^\alpha \big( \nabla_\alpha (G_4)_{a_1 \cdots a_4} \big) \;+\; \big( \overline{\psi} \Gamma_{a_1 a_2} \rho_{a_3 a_4} \big) \Big) e^{a_1} \cdots e^{a_4} \;=\; 0 \\ \tfrac{1}{3!} (G_4)_{a b_1 b_2 b_3} \big( \overline{\psi} \,\Gamma^a\, \psi \big) \, e^{b_1 b_2 b_3} + \big( \overline{\psi} \,\Gamma_{a_1 a_2}\, H_b \psi \big) e^{a_1} \, e^{a_2} \, e^b \;=\; 0 \,, \end{array} \right. \end{array}$$

where we used that the quartic spinorial component vanishes identically, due to a Fierz identity (here):

$$- \tfrac{1}{2} \big( \overline{\psi} \Gamma_{a_1 a_2} \psi \big) \big( \overline{\psi} \Gamma^{a_1} \psi \big) e^{a_2} \;=\; 0 \,.$$

To solve the second line in (6) for $H_a$ (this is CDF91 (III.8.43-49)) we expand $H_a$ in the Clifford algebra (according to this Prop.), observing that for $\Gamma_{a_1 a_2} H_{a_3}$ to be a linear combination of the $\Gamma_a$ the matrix $H_a$ needs to have a $\Gamma_{a_1}$-summand or a $\Gamma_{a_1 a_2 a_3}$-summand. The former does not admit a Spin-equivariant linear combination with coefficients $(G_4)_{a_1 \cdots a_4}$, hence it must be the latter. But then we may also need a component $\Gamma_{a_1 \cdots a_5}$ in order to absorb the skew-symmetric product in $\Gamma_{a_1 a_2} H_a$. Hence $H_a$ must be of this form:

(7)$$H_a \;=\; \mathrm{const}_1 \, \tfrac{1}{3!} (G_4)_{a b_1 b_2 b_3} \Gamma^{b_1 b_2 b_3} + \mathrm{const}_2 \, \tfrac{1}{4!} (G_4)^{b_1 \cdots b_4} \Gamma_{a b_1 \cdots b_4} \,.$$

With this, we compute:

(8)$$\begin{array}{ll} \big( \overline{\psi} \Gamma_{a_1 a_2} H_{a_3} \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} & =\; \mathrm{const}_1 \, \tfrac{1}{3!} (G_4)_{a_3 b_1 b_2 b_3} \, \big( \overline{\psi} \Gamma_{a_1 a_2} \Gamma^{b_1 b_2 b_3} \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} \\ & \;\;\;+\, \mathrm{const}_2 \, \tfrac{1}{4!} \, (G_4)^{b_1 \cdots b_4} \, \big( \overline{\psi} \Gamma_{a_1 a_2} \Gamma_{a_3 b_1 \cdots b_4} \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} \\ & \;=\; 1 \, \mathrm{const}_1 \, \tfrac{1}{3!} \, (G_4)_{a_3 b_1 b_2 b_3} \big( \overline{\psi} \,\Gamma_{a_1 a_2}{}^{b_1 b_ 2 b_3}\, \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} \\ & \;\;\;+\, 6 \, \mathrm{const}_1 \, \tfrac{1}{3!} \, (G_4)_{b_3 a_1 a_2 a_3} \big( \overline{\psi} \,\Gamma^{b_3}\, \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} \\ & \;\;\;+\, 8 \, \mathrm{const}_2 \, \tfrac{1}{4!} \, (G_4)^{b_1 \cdots b_3 a_3} \, \big( \overline{\psi} \Gamma^{a_1 a_2}{}_{b_1 \cdots b_3} \psi \big) e^{a_1} \, e^{a_2} \, e^{a_3} \,. \end{array}$$

Here the multiplicities of the nonvanishing Clifford-contractions arise via this Lemma:

$$\begin{array}{l} 1 \;=\; 0! \Big( {2 \atop 0} \Big) \Big( {3 \atop 0} \Big) \\ 6 \;=\; 2! \Big( {2 \atop 2} \Big) \Big( {3 \atop 2} \Big) \\ 8 \;=\; 1! \Big( {2 \atop 1} \Big) \Big( {4 \atop 1} \Big) \,, \end{array}$$

and all remaining contractions vanish inside the spinor pairing by this lemma.

Now using (8) in (6) yields:

$$\begin{array}{l} \mathrm{const}_1 = -1/6 \,, \\ \mathrm{const}_2 = - 4!/3! \, \mathrm{const}_1 / 8 = + 1/12 \,, \end{array}$$

as claimed.

Lemma

Given the Bianchi identity for $G^s_4$ (5), then the Bianchi identity for $G^s_7$ (1) is equivalent to

1. the Bianchi identity for the ordinary flux density $G_7$

2. its Hodge duality to $G_4$

3. another condition on the gravitino field strength

(9)$$\begin{array}{l} \mathrm{d} \, G^s_7 \;=\; \tfrac{1}{2} G^s_4 \, G^s_4 \\ \;\Leftrightarrow\; \left\{ \begin{array}{l} \big( \nabla_{a_1} \tfrac{1}{7!} (G_7)_{a_2 \cdots a_8} \big) e^{a_1} \cdots e^{a_8} \;=\; \tfrac{1}{2} \big( \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} \, \tfrac{1}{4!} (G_4)_{a_5 \cdots a_8} \big) e^{a_1} \cdots e^{a_8} \\ (G_7)_{a_1 \cdots a_7} \;=\; \tfrac{1}{4!} \epsilon_{a_1 \cdots a_b b_1 \cdots b_4} (G_4)^{b_1 \cdots b_4} \\ \Big( \tfrac{1}{7!} \psi^\alpha \nabla_\alpha (G_7)_{a_1 \cdots a_7} \psi^\alpha \;+\; \frac{2}{5!} \big( \overline{\psi} \Gamma_{a_1 \cdots a_5} \rho_{a_6 a_7} \big) \Big) e^{a_1} \cdots e^{a_7} \;=\; 0 \end{array} \right. \end{array}$$

This is essentially CDF91, (III.8.50-53).

Proof

The components of the Bianchi identity are

$$\begin{array}{l} \mathrm{d} \, G_4^s \;=\; 0 \\ \Rightarrow \left\{ \begin{array}{l} \mathrm{d} \Big( \tfrac{1}{7!} (G_7)_{a_1 \cdots a_7} \, e^{a_1} \cdots e^{a_7} - \tfrac{1}{5!} \big( \overline{\psi} \Gamma_{a_1 \cdots a_5} \psi \big) e^{a_1} \cdots e^{a_5} \Big) \\ \;=\; \tfrac{1}{2} \Big( \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} e^{a_1} \cdots e^{a_4} - \tfrac{1}{2} \big( \overline{\psi} \Gamma_{a_1 a_2} \psi \big) \Big) \Big( \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} e^{a_1} \cdots e^{a_4} - \tfrac{1}{2} \big( \overline{\psi} \Gamma_{a_1 a_2} \psi \big) \Big) \\ \;\Leftrightarrow\; \left\{ \begin{array}{l} \Big( \nabla_{a_1} \tfrac{1}{7!} (G_7)_{a_2 \cdots a_8} \;=\; \;\tfrac{1}{2}\; \tfrac{1}{4!} (G_4)_{a_1 \cdots a_4} \, \tfrac{1}{4!} (G_4)_{a_5 \cdots a_8} \Big) e^{a_1} \cdots e^{a_8} \\ \Big( \tfrac{1}{7!} \psi^\alpha \nabla_\alpha (G_7)_{a_1 \cdots a_7} + \frac{2}{5!} \big( \overline{\psi} \Gamma_{a_1 \cdots a_5} \rho_{a_6 a_7} \big) \Big) e^{a_1} \cdots e^{a_7} \;=\; 0 \\ \left. \begin{array}{l} \tfrac{1}{6!} (G_7)_{a_1 \cdots a_6 b} \big( \overline{\psi} \,\Gamma^b\, \psi \big) e^{a_1} \cdots e^{a_6} \\ \;\;\;+\, \tfrac{2}{12} \, \tfrac{1}{5!} \, \tfrac{1}{4!} \, (G_4)^{b_1 \cdots b_4} \big( \overline{\psi} \, \Gamma_{a_1 \cdots a_5} \, \Gamma_{a b_1 \cdots b_4}\, \psi \big) e^a \, e^{a_1} \cdots e^{a_5} \\ \;\;-\; \tfrac{2}{6} \tfrac{1}{5!} \tfrac{1}{3!} (G_4)_{a b_1 b_2 b_3} \big( \overline{\psi} \,\Gamma_{a_1 \cdots a_5}\, \Gamma^{b_1 b_2 b_3} \psi \big) e^{a} \, e^{a_1} \cdots e^{a_5} \\ \;\;\;-\, \Big( \tfrac{1}{2} \big( \overline{\psi} \Gamma_{a_1 a_2} \psi \big) e^{a_1} \, e^{a_2} \Big) \tfrac{1}{4!} (G_4)_{b_1 \cdots b_4} \, e^{b_1} \cdots e^{b_4} \;\;=\;\; 0 \,, \end{array} \right\} \Leftrightarrow (G_7)_{a_1 \cdots a_6 b} \;=\; \tfrac{1}{4!} \epsilon_{a_1 \cdots a_6 b b_1 \cdots b_4} (G_4)^{b_1 \cdots b_4} \end{array} \right. \end{array} \right. \end{array}$$

where:

(i) in the quadratic spinorial component we inserted the expression for $\rho$ from (5), then contracted $\Gamma$-factors using again this Lemma, and finally observed that of the three spinorial quadratic forms (see there) the coefficients of $\big(\overline{\psi}\Gamma_{a_1 a_2} \psi\big)$ and of $\big(\overline{\psi}\Gamma_{a_1 \cdots a_6} \psi\big)$ vanish identically, by a remarkable cancellation of combinatorial prefactors:

• $\underset{= 0 }{\underbrace{\bigg(- \frac{2}{12} \frac{1}{5!} \frac{1}{4!} 4! \Big( { 5 \atop 4 } \Big) \Big( { 4 \atop 4 } \Big) \;+\; \frac{2}{6} \frac{1}{5!} \frac{1}{3!} 3! \Big( { 5 \atop 3 } \Big) \Big( { 3 \atop 3 } \Big) \;-\; \frac{1}{2} \frac{1}{4!} \bigg) } } \; (G_4)_{a_2 \cdots a_5} \big( \overline{\psi} \,\Gamma_{a a_1}\, \psi \big) e^{a} \, e^{a_1} \cdots e^{a_6} \;\;\;$ (check)

• $\underset{ = 0 }{ \underbrace{ \bigg( \frac{2}{12} \frac{1}{5!} \frac{1}{4!} 2 \Big( { 5 \atop 2 } \Big) \Big( { 4 \atop 2 } \Big) \;-\; \frac{2}{6} \frac{1}{5!} \frac{1}{3!} 1 \Big( { 5 \atop 1 } \Big) \Big( { 3 \atop 1 } \Big) \bigg) } } \; (G_4)_{a_1 a_2 b_1 b_2} \big( \overline{\psi} \,\Gamma_{a_3 \cdots a_6}{}^{b_1 b_2}\, \psi \big) e^{a_1} \cdots e^{a_6} \;\;\;$ (check)

(ii) the quartic spinorial component holds identitically, due to the Fierz identity here:

$$-\tfrac{1}{4!} \big( \overline{\psi} \,\Gamma_{a_1 \cdots a_5}\, \psi \big) \big( \overline{\psi} \Gamma^{a_1} \big) e^{a_2} \cdots e^{a_5} \;=\; \tfrac{1}{8} \Big( \big( \overline{\psi} \,\Gamma_{a_1 a_2}\, \psi \big) e^{a_1} e^{a_2} \Big) \Big( \big( \overline{\psi} \,\Gamma_{a_1 a_2}\, \psi \big) e^{a_1} e^{a_2} \Big) \,.$$

Therefore the only spinorial component of the Bianchi identity which is not automatically satisfied is (with $\Gamma_{0 1 2 \cdots} = \epsilon_{0 1 2 \cdots}$, see there) the vanishing of

$$\tfrac{1}{6!} \Big( (G_7)_{a_1 \cdots a_6 b} - \tfrac{1}{4!} (G_4)^{b_1 \cdots b_4} \epsilon_{b_1 \cdots b_4 a_1 \cdots a_6 b} \Big) \big( \overline{\psi} \,\Gamma^b\, \psi \big) \,,$$

which is manifestly the claimed Hodge duality relation.

Lemma

Given the Bianchi identities for $G_4^s$ (5) and $G_7^s$ (9), the supergravity fields satisfy their Einstein equations with source the energy momentum tensor of the C-field:

(10)$$\begin{array}{l} \mathrm{d}\, G_4^s \;=\;0 \,, \;\;\; \mathrm{d}\, G_7^s \;=\; \tfrac{1}{2} G_4^s \, G_4^2 \\ \;\Rightarrow\; \left\{ \begin{array}{l} R^{a m}_{b m} - \tfrac{1}{2} \delta^a_b\, R^{m n}_{m n} \;=\; - \tfrac{1}{12} \Big( \, (G_4)^a{c_1 \cdots c_3} (G_4)_{b c_1 \cdots c_3} - \tfrac{1}{8} (G_4)^{c_1 \cdots c_4} (G_4)_{c_1 \cdots c_4} \delta^a_b \;\;\;\; ({\color{darkblue}\text{Einstein equation}}) \\ \Gamma^{b a_1 a_2} \rho_{a_1 a_2} \;=\; 0 \;\;\;\; ({\color{darkblue}\text{Rarita-Schwinger equation}}) \end{array} \right. \end{array}$$

Cf. e.g. CDF91, (III.8.54-60); full details are given in GSS24, Lem. 3.8.

In conclusion, the above lemmas give Thm. .