Physics theoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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Type IIB Source Free Bianchi Identities

Physical scope

Purely local source-free differential identities from smooth potentials; no magnetic sources, nontrivial gerbe patching, or global flux quantization.

Statement

On a smooth local patch with the potentials and field strengths of Type IIB Two Derivative Supergravity Pseudoaction,

$d H_{3} = 0, d G_{1} = 0, d {\tilde{G}}_{3} = H_{3} \land G_{1}, d {\tilde{G}}_{5} = H_{3} \land {\tilde{G}}_{3} .$

These are Bianchi identities from the local potential definitions, not field equations. A source or nontrivial global potential patch can change the hypotheses.

Statement references

Type IIB Two Derivative Supergravity Pseudoaction

Hamilton, §7.1, Theorem 7.2(f)

Facts & Assumptions

Given: Smooth local potentials $B_{2}, C_{0}, C_{2}, C_{4}$ and the field-strength definitions in [F1].

[F1]

Type IIB Two Derivative Supergravity Pseudoaction: $H_{3} = d B_{2}$ , $G_{1} = d C_{0}$ , $G_{3} = d C_{2}$ , $G_{5} = d C_{4}$ , ${\tilde{G}}_{3} = G_{3} - C_{0} H_{3}$ , ${\tilde{G}}_{5} = G_{5} - \frac{1}{2} H_{3} \land C_{2} + \frac{1}{2} G_{3} \land B_{2}$ .

[F2]

The exterior derivative squares to zero and The exterior derivative is a graded derivation: $d^{2} = 0$ and $d (α \land β) = d α \land β + (- 1)^{\deg α} α \land d β$ .

[F3]

The wedge product is associative and graded commutative: exchanging degree- $r$ and degree- $s$ forms contributes $(- 1)^{r s}$ .

Proof

1.1F1F2algebra

The first two identities follow directly from [F1]–[F2]: $d H_{3} = d^{2} B_{2} = 0$ and $d G_{1} = d^{2} C_{0} = 0$ . Also $d G_{3} = d^{2} C_{2} = 0$ and $d G_{5} = d^{2} C_{4} = 0$ .

2.1F1F2F3step 1.1algebra

By [F1]–[F2], $d {\tilde{G}}_{3} = d G_{3} - d C_{0} \land H_{3} - C_{0} d H_{3} = - G_{1} \land H_{3} = H_{3} \land G_{1}$ , where the last equality uses [F3] since the exchanged forms $G_{1}$ and $H_{3}$ have degrees one and three, so the exchange sign is $(- 1)^{1 \cdot 3} = - 1$ .

3.1F1F2F3step 1.1algebra∎

Applying [F2] to the five-form gives $d {\tilde{G}}_{5} = - \frac{1}{2} d (H_{3} \land C_{2}) + \frac{1}{2} d (G_{3} \land B_{2}) = \frac{1}{2} H_{3} \land G_{3} - \frac{1}{2} G_{3} \land H_{3} = H_{3} \land G_{3}$ by [F3]. Since $G_{3} = {\tilde{G}}_{3} + C_{0} H_{3}$ and $H_{3} \land H_{3} = 0$ for an odd-degree form by [F3], this equals $H_{3} \land {\tilde{G}}_{3}$ .

Depends on

Dependency tree · 2 levels

Sources

M. J. D. Hamilton, The field and Killing spinor equations of M-theory and type IIA/IIB supergravity in coordinate-free notation, arXiv:1607.00327v3