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Source-Free Eleven-Dimensional CJS Bosonic Field Equations

Physical scope

Conditional classical Euler-Lagrange equations of the adopted source-free two-derivative bosonic CJS action. No source currents, boundaries, higher-derivative terms, quantum corrections, or empirical claims are included.

Statement

Let $M$ be an oriented, boundaryless Lorentzian eleven-manifold, let $g$ be its smooth metric, let $C_{3} \in Ω^{3} (M)$ be the global potential of Eleven-Dimensional CJS Supergravity Bosonic Action, and set $F_{4} = d C_{3}$ . Interpret variations locally, with compact support, so the action difference is finite even when the unvaried total action is not. Metric variations are smooth one-parameter variations that remain Lorentzian for sufficiently small parameter. A stationary configuration of the source-free two-derivative action under all such compactly supported variations of $C_{3}$ and the inverse metric $g^{M N}$ satisfies

$d *_{g} F_{4} + \frac{1}{2} F_{4} \land F_{4} = 0, d F_{4} = 0,$

and

$R_{M N} = \frac{1}{12} (F_{M A B C} F_{N}^{A B C} - \frac{1}{12} g_{M N} F_{A B C D} F^{A B C D}) .$

Here $R_{M N}$ and $R = g^{M N} R_{M N}$ use the curvature convention $R^{P}_{Q M N} = \partial_{M} Γ^{P}_{N Q} - \partial_{N} Γ^{P}_{M Q} + Γ^{P}_{M R} Γ^{R}_{N Q} - Γ^{P}_{N R} Γ^{R}_{M Q}$ , $R_{Q N} = R^{P}_{Q P N}$ , from the adopted CJS convention. The signed form norm is $∣ F_{4} ∣_{g}^{2} = (1 / 4!) F_{A B C D} F^{A B C D}$ . These are conditional classical deductions from the adopted action, not empirical laws or claims about source-bearing or corrected theories.

Facts & Assumptions

Given: The adopted action, a smooth field configuration, and stationarity under every compactly supported variation of $C_{3}$ and $g^{M N}$ .

[A1]

Eleven-Dimensional CJS Supergravity Bosonic Action: $S = (2 κ_{11}^{2})^{- 1} \int (R {vol}_{g} - \frac{1}{2} F_{4} \land *_{g} F_{4} - \frac{1}{6} C_{3} \land F_{4} \land F_{4})$ , $F_{4} = d C_{3}$ ; it is the source-free two-derivative bosonic truncation, with compactly supported variations and no boundary terms.

[F1]

Supergravity Conventions And Normalizations: the current carrier fixes signature $(- + \dots +)$ , $∣ F_{p} ∣^{2} = (1 / p!) F_{M_{1} \dots M_{p}} F^{M_{1} \dots M_{p}}$ , and the named CJS convention. Its current Step 1 record is ready at hash $41 d 207 e 61 d b e f 5573 f 48 b 5 c f 2 e 70 c 2 c 19 a 136 e c 755 a 3 c 8047 c 832905 c 0 b 550$ ; the proof uses only these inspected conventions.

[F2]

Eleven-Dimensional Oriented Lorentzian Geometry defines the oriented Lorentzian Hodge star and signed form pairing.

[F3]

Levi-Civita Connection on an Eleven-Dimensional Lorentzian Manifold gives the unique connection used for the scalar curvature and metric variation.

[F4]

Compactly Supported Metric Variation of a Weighted Scalar-Curvature Action: for $n \geq 2$ , a Lorentzian metric in a chart, and compactly supported inverse-metric variation $q^{M N}$ , the compactly supported variation of $\int \sqrt{∣ \det g ∣} R d^{n} x$ has coefficient $R_{M N} - \frac{1}{2} g_{M N} R$ against $q^{M N}$ . Its complete current proof derives the formula in the general $n$ -dimensional chart case; here $n = 11$ .

[F5]

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ : for a differentiable invertible matrix curve $A (s)$ , $(\det A)^{'} = \det A t r (A^{- 1} A^{'})$ .

[F6]

Compactly Supported Tests Determine Continuous Local Coefficients: if a continuous coefficient $A_{i}$ on an open coordinate domain pairs to zero against every smooth compactly supported component test with positive continuous density $w$ , then $A_{i}$ vanishes pointwise. Its complete current proof uses a bump supported where a nonzero coefficient has fixed sign.

[F7]

A compactly supported primitive has zero total derivative integral: on an oriented boundaryless manifold, the integral of the exterior derivative of a compactly supported primitive vanishes.

[F8]

The exterior derivative by the invariant vector-field formula, The local coordinate formula for the exterior derivative, The exterior derivative is a graded derivation, Clairaut--Schwarz theorem for continuous second partial derivatives: for a smooth form, $d$ has the alternating coordinate-derivative formula and graded product rule; smooth mixed partials commute.

Proof

technique · vary the potential and metric independently, integrate compactly supported exact terms, and solve the resulting tensor equation

1.1A1F8givenalgebra

With $g$ fixed and $C_{3}$ varied by a compactly supported $δ C_{3}$ , $δ F_{4} = d δ C_{3}$ . Also $d F_{4} = d^{2} C_{3} = 0$ : in coordinates each component of $d^{2} C_{3}$ is an alternating sum of second partial derivatives, and terms cancel in pairs after exchanging the two derivative indices by [F8].

1.2A1F1F2F3F4F5algebra

Now vary $g^{M N}$ with $C_{3}$ , hence $F_{4}$ , fixed. The Einstein–Hilbert contribution is $(R_{M N} - \frac{1}{2} g_{M N} R) δ g^{M N} {vol}_{g}$ by [F4], using the Levi-Civita curvature convention of [F3]. Differentiating the four inverse metrics in $∣ F_{4} ∣_{g}^{2} = (1 / 4!) F_{A B C D} F^{A B C D}$ gives $δ ∣ F_{4} ∣_{g}^{2} = \frac{1}{3!} F_{M A B C} F_{N}^{A B C} δ g^{M N} .$ Indeed, each of the four differentiated inverse-metric factors gives the same contraction after relabeling the antisymmetric indices, so the factor is $4 / 4! = 1 / 3!$ . Differentiating $g^{M A} g_{A N} = δ^{M}_{N}$ gives $δ g_{M N} = - g_{M A} g_{N B} δ g^{A B}$ . The determinant derivative in [F5] then yields $δ {vol}_{g} = \frac{1}{2} g^{M N} δ g_{M N} {vol}_{g} = - \frac{1}{2} g_{A B} δ g^{A B} {vol}_{g}$ . Therefore $δ (- \frac{1}{2} ∣ F_{4} ∣_{g}^{2} {vol}_{g}) = (- \frac{1}{12} F_{M A B C} F_{N}^{A B C} + \frac{1}{4} g_{M N} ∣ F_{4} ∣_{g}^{2}) δ g^{M N} {vol}_{g} .$

2.1A1F2F7F8step 1.1algebra

Since $g$ is fixed, the Hodge star is fixed. By symmetry of the metric-induced pairing in [F2], $δ (F_{4} \land *_{g} F_{4}) = 2 d δ C_{3} \land *_{g} F_{4}$ , so the kinetic variation is $- d δ C_{3} \land *_{g} F_{4}$ . The product rule in [F8] gives $d (δ C_{3} \land *_{g} F_{4}) = d δ C_{3} \land *_{g} F_{4} - δ C_{3} \land d *_{g} F_{4}$ . Its primitive is compactly supported, so [F7] yields $\int_{M} d δ C_{3} \land *_{g} F_{4} = \int_{M} δ C_{3} \land d *_{g} F_{4}$ .

2.2A1F7F8step 1.1algebra

The Chern–Simons variation is $- \frac{1}{6} \int_{M} (δ C_{3} \land F_{4} \land F_{4} + C_{3} \land d δ C_{3} \land F_{4} + C_{3} \land F_{4} \land d δ C_{3}) .$ The two last terms agree because both $F_{4}$ and $d δ C_{3}$ have degree $4$ . By $d F_{4} = 0$ and the product rule, $d (C_{3} \land δ C_{3} \land F_{4}) = F_{4} \land δ C_{3} \land F_{4} - C_{3} \land d δ C_{3} \land F_{4} .$ Since $F_{4}$ commutes past $δ C_{3}$ with sign $(- 1)^{12} = 1$ , the product rule in [F8] gives the displayed derivative identity, and the primitive $C_{3} \land δ C_{3} \land F_{4}$ has compact support. Thus [F7] gives $\int_{M} C_{3} \land d δ C_{3} \land F_{4} = \int_{M} δ C_{3} \land F_{4} \land F_{4}$ . The Chern–Simons contribution is $- \frac{1}{2} \int_{M} δ C_{3} \land F_{4} \land F_{4}$ .

2.3A1F2F5F6F8step 1.2algebra

The Chern–Simons 11-form $C_{3} \land F_{4} \land F_{4}$ contains no metric, inverse metric, Hodge star, or metric volume form, so its metric variation is zero. Adding the two metric-dependent contributions from step 1.2, stationarity gives $\int_{M} E_{M N} q^{M N} {vol}_{g} = 0$ for every admissible compactly supported symmetric inverse-metric variation $q$ , where $E_{M N} : = R_{M N} - \frac{1}{2} g_{M N} R - \frac{1}{12} F_{M A B C} F_{N}^{A B C} + \frac{1}{4} g_{M N} ∣ F_{4} ∣_{g}^{2}$ . First, $E_{M N}$ is symmetric: the Christoffel formula and [F5] give $Γ^{P}_{P M} = \frac{1}{2} g^{P Q} \partial_{M} g_{P Q} = \partial_{M} \log \sqrt{∣ \det g ∣}$ , and the curvature convention yields $R_{M N} - R_{N M} = - \partial_{N} \partial_{M} \log \sqrt{∣ \det g ∣} + \partial_{M} \partial_{N} \log \sqrt{∣ \det g ∣} - Γ^{P}_{N R} Γ^{R}_{P M} + Γ^{P}_{M R} Γ^{R}_{P N} = 0.$ The partial derivatives commute by [F8], and relabeling $P, R$ and using symmetry of the lower Christoffel indices makes the two quadratic sums equal; the flux contraction is symmetric as well. We now verify that every smooth compactly supported symmetric $q$ is admissible. On its compact support $K$ , at each point the inverse metric has a positive ten-dimensional subspace and a negative one-dimensional subspace. In a local trivialization, continuity preserves strict positivity on that fixed ten-plane and strict negativity on that line in a neighborhood of the point, and sufficiently small perturbations preserve both inequalities. These dimensions sum to $11$ , so the perturbed form remains nondegenerate with inertia $(10, 1, 0)$ . Compactness of $K$ gives a finite such cover and a uniform $ε > 0$ ; hence $g_{t}^{- 1} = g^{- 1} + t q$ is Lorentzian for $∣ t ∣ < ε$ and equals $g^{- 1}$ off $K$ . Its inverse is a smooth metric, so this is an admissible variation. For a coordinate test function $ψ$ with compact support, take $q^{i i} = ψ$ and all other components zero to test $E_{i i}$ ; for $i < j$ take $q^{i j} = q^{j i} = ψ$ and all other components zero, which tests $2 E_{i j}$ . Since $E$ is continuous, [F6] applied with the positive coordinate density makes every component of $E$ vanish pointwise.

3.1A1F2F6step 2.1step 2.2algebra

Combining steps 2.1 and 2.2 gives $δ_{C_{3}} S = - \frac{1}{2 κ_{11}^{2}} \int_{M} δ C_{3} \land (d *_{g} F_{4} + \frac{1}{2} F_{4} \land F_{4}) .$ In any chart, choose $δ C_{3}$ with one arbitrary compactly supported component at a time. The positive coordinate density $\sqrt{∣ \det g ∣}$ and [F6] imply every component of the 8-form coefficient vanishes. This is the first field equation.

3.2F1step 2.3algebra

Contract $E_{M N} = 0$ with $g^{M N}$ . Since $g^{M N} R_{M N} = R$ , $g^{M N} g_{M N} = 11$ , and $g^{M N} F_{M A B C} F_{N}^{A B C} = F_{A B C D} F^{A B C D} = 24 ∣ F_{4} ∣_{g}^{2}$ , the trace is $- \frac{9}{2} R + (- 2 + \frac{11}{4}) ∣ F_{4} ∣_{g}^{2} = 0,$ so $R = \frac{1}{6} ∣ F_{4} ∣_{g}^{2}$ . Substituting this value back into $E_{M N} = 0$ yields $R_{M N} = \frac{1}{12} F_{M A B C} F_{N}^{A B C} - \frac{1}{6} g_{M N} ∣ F_{4} ∣_{g}^{2} = \frac{1}{12} (F_{M A B C} F_{N}^{A B C} - \frac{1}{12} g_{M N} F_{A B C D} F^{A B C D}) .$

4.1A1step 1.1step 3.1step 3.2∎

Step 1.1 gives $d F_{4} = 0$ , step 3.1 gives the four-form equation, and step 3.2 gives the Einstein equation. Each is conditional on stationarity of the specified source-free bosonic action and the stated compact-support hypotheses; no source or higher-derivative equation follows.

Remarks

The normalization $∣ F_{4} ∣^{2} = F_{A B C D} F^{A B C D} / 4!$ is essential to the displayed Einstein coefficient. The independent Gauntlett–Pakis component equation in §2, equation (2.2), matches the result after this normalization. Their supersymmetry formula is treated separately in Supersymmetric Background in Eleven-Dimensional CJS Supergravity.

Source-Free Eleven-Dimensional CJS Bosonic Field Equations

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources