LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-6-luna)verified 2026-11-04 (gpt-6.1-sol)

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Metric Variation of a Differential Form Kinetic Term

Statement

Let $U \subseteq R^{n}$ be an open oriented coordinate domain. For $s$ in an open interval about $0$ , let $g (s)$ be a jointly smooth family of smooth nondegenerate metrics on $U$ of the same signature as $g (0) = g$ , equal to $g$ outside one fixed compact set. Put $q^{M N} : = {\partial_{s} g (s)^{M N} ∣}_{s = 0}$ . Let $F$ be a fixed smooth covariant $p$ -form and let $A$ be a smooth scalar independent of $s$ . Define $∣ F ∣_{g}^{2} = F_{M_{1} \dots M_{p}} F^{M_{1} \dots M_{p}} / p!$ and $⟨ i_{M} F, i_{N} F ⟩_{g} = F_{M A_{2} \dots A_{p}} F_{N}^{A_{2} \dots A_{p}} / (p - 1)!$ for $p \geq 1$ . The pointwise derivative at $s = 0$ is

$δ (\sqrt{∣ \det g ∣} A ∣ F ∣_{g}^{2}) = \sqrt{∣ \det g ∣} A (⟨ i_{M} F, i_{N} F ⟩_{g} - \frac{1}{2} g_{M N} ∣ F ∣_{g}^{2}) q^{M N} .$

For $p = 0$ , the first term in parentheses is defined as zero and the same formula holds.

Statement references

Smooth manifolds and their smooth charts

A smooth tensor field

A smooth differential $k$ -form

The contraction of a mixed tensor

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$

Facts & Assumptions

Given: The metric, fixed covariant form $F$ , metric-independent scalar $A$ , and symmetric inverse-metric variation $q$ in the Statement.

[F1]

The derivative of the determinant along an invertible C¹ matrix curve: $(\det A)^{'} = \det A tr (A^{- 1} A^{'})$ : $δ \sqrt{∣ \det g ∣} = - \frac{1}{2} \sqrt{∣ \det g ∣} g_{M N} q^{M N}$ for $q^{M N} = δ g^{M N}$ .

[F2]

A smooth differential $k$ -form and The contraction of a mixed tensor: components of a $p$ -form are alternating covariant arrays; raising its indices uses one inverse metric in each covariant slot.

Proof

1.1F2givenalgebra

For $p \geq 1$ , write $∣ F ∣^{2} = (1 / p!) F_{M_{1} \dots M_{p}} F_{N_{1} \dots N_{p}} g^{M_{1} N_{1}} \dots g^{M_{p} N_{p}}$ . Differentiating the product of inverse metrics produces $p$ terms. Relabel the dummy indices in each term; antisymmetry of both copies of $F$ makes all $p$ terms equal, so $δ ∣ F ∣^{2} = F_{M A_{2} \dots A_{p}} F_{N}^{A_{2} \dots A_{p}} q^{M N} / (p - 1)! = ⟨ i_{M} F, i_{N} F ⟩ q^{M N}$ . When $p = 0$ , $F$ is a scalar and its norm has no metric dependence, giving zero.

2.1F1step 1.1algebra∎

The scalar $A$ and covariant components of $F$ are fixed, so the product rule and [F1] give $δ (\sqrt{∣ g ∣} A ∣ F ∣^{2}) = \sqrt{∣ g ∣} A [δ ∣ F ∣^{2} - \frac{1}{2} g_{M N} ∣ F ∣^{2} q^{M N}]$ . Substituting step 1.1 yields the asserted expression, including the $p = 0$ case.

Depends on

Dependency tree · 2 levels

Sources

Sean M. Carroll, Lecture Notes on General Relativity, arXiv:gr-qc/9712019v1