Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6-luna)

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Type Iia RR Bianchi From Potentials

Physical scope

Local, source-free differential-form identities in the massless Hamilton potential convention. No global twisted cohomology, magnetic NS5 source, or sourced RR equation is asserted.

Statement

On a smooth contractible source-free ten-dimensional patch $U$ , let $B_{2} \in Ω^{2} (U)$ , $C_{1} \in Ω^{1} (U)$ , $C_{3} \in Ω^{3} (U)$ be smooth real differential-form potentials, $H_{3} = d B_{2}$ , $G_{2} = d C_{1}$ , $G_{4} = d C_{3}$ , and ${\tilde{G}}_{4} = G_{4} - C_{1} \land H_{3}$ . If $d H_{3} = 0$ , then $d G_{2} = 0$ and $d {\tilde{G}}_{4} = - H_{3} \land G_{2}$ . A smooth polyform is an element of $Ω^{∙} (U) : = ⨁_{p = 0}^{10} Ω^{p} (U)$ . Give this space the parity grading $Ω^{\overline{0}} = ⨁_{p even} Ω^{p}$ and $Ω^{\overline{1}} = ⨁_{p odd} Ω^{p}$ . An odd differential means a real-linear parity-reversing map whose square is zero. For $D_{H} : = d + H_{3} \land$ , $D_{H}^{2} = d H_{3} \land$ ; hence $D_{H}$ is an odd differential when $d H_{3} = 0$ . No degree-one operator for the integer grading is asserted. The minus sign follows from the declared potential convention and does not agree with the plus sign printed in Hamilton’s Theorem 6.2(e).

Theorem references

M. J. D. Hamilton, §6.1, definitions and Theorem 6.2(e)

Joseph Polchinski, §12.1.5 and §12.1.8

Type Iia Two Derivative Supergravity Action

Facts & Assumptions

Given: The smooth local potentials and $d H_{3} = 0$ above, and the differential-form conventions in Supergravity Conventions And Normalizations.

[A1]

Supergravity Conventions And Normalizations (current batch-56 carrier and current ready Step 1 record inspected; its stated hypotheses match this use): signature $(- + \dots +)$ , standard wedge product, and ten-dimensional form norm $∣ F_{p} ∣^{2} = F_{M_{1} \dots M_{p}} F^{M_{1} \dots M_{p}} / p!$ . The argument below uses only the wedge convention.

[A2]

Type Iia Two Derivative Supergravity Action adopts $H_{3} = d B_{2}$ , $G_{2} = d C_{1}$ , $G_{4} = d C_{3}$ , and ${\tilde{G}}_{4} = G_{4} - C_{1} \land H_{3}$ , on a local patch with no magnetic source.

[F1]

The exterior derivative squares to zero: $d^{2} ω = 0$ for every smooth differential form $ω$ .

[F2]

The exterior derivative is a graded derivation: $d (α \land β) = d α \land β + (- 1)^{\deg α} α \land d β$ for homogeneous forms.

Proof

1.1givenA2F1def. field-strength definition

By [A2] and [F1], $d G_{2} = d (d C_{1}) = 0$ .

2.1givenA1A2F1F2step 1.1algebra

Apply [F1] and [F2] to ${\tilde{G}}_{4}$ : $d {\tilde{G}}_{4} = d^{2} C_{3} - d (C_{1} \land H_{3}) = - G_{2} \land H_{3} + C_{1} \land d H_{3} .$ The second term vanishes by the hypothesis $d H_{3} = 0$ . Since $\deg G_{2} = 2$ and $\deg H_{3} = 3$ , the wedge convention in [A1] and graded commutativity give $G_{2} \land H_{3} = H_{3} \land G_{2}$ , so $d {\tilde{G}}_{4} = - H_{3} \land G_{2}$ .

3.1givenF2algebrastep 2.1∎

For any smooth polyform $ω$ , expand $D_{H}^{2} ω = d (H_{3} \land ω) + H_{3} \land d ω + H_{3} \land H_{3} \land ω .$ By [F2], the first two terms sum to $d H_{3} \land ω + ((- 1)^{3} + 1) H_{3} \land d ω = d H_{3} \land ω$ . Graded commutativity also gives $H_{3} \land H_{3} = - H_{3} \land H_{3}$ , hence $H_{3} \land H_{3} = 0$ over $R$ . Therefore $D_{H}^{2} = d H_{3} \land$ , and $D_{H}^{2} = 0$ under the stated closedness hypothesis. Both summands of $D_{H}$ reverse parity (their integer-degree shifts are 1 and 3), so this is an odd differential in the defined parity grading.

Type Iia RR Bianchi From Potentials

Statement

Theorem references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources