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

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Conformal warped-product external Ricci trace

Statement

Let $D > 2$ , $1 \leq d < D$ , and $q = D - d$ . Let $U \subset R^{d}$ and $V \subset R^{q}$ be coordinate domains. Let $h_{μ ν} (x)$ be a smooth nondegenerate Lorentzian metric on $U$ with $Ric (h)_{μ ν} = λ h_{μ ν}$ for a constant $λ$ , and let ${\hat{g}}_{m n} (y)$ be a smooth positive-definite metric on $V$ . Use the product metric $\overset{ˉ}{g} = h \oplus \hat{g}$ on $U \times V$ . For $ω \in C^{\infty} (V, R)$ put $Ω = e^{ω}$ and $g = Ω^{2} \overset{ˉ}{g}$ , where all Greek and lower-case Latin indices range over $U$ and $V$ , respectively. Define $\hat{Δ} f = {\hat{g}}^{m n} (\partial_{m} \partial_{n} f - {\hat{Γ}}^{p}_{m n} \partial_{p} f)$ and $∣ d ω ∣_{\hat{g}}^{2} = {\hat{g}}^{m n} \partial_{m} ω \partial_{n} ω$ . In product coordinates use $Γ^{P}_{M N} = \frac{1}{2} g^{P Q} (\partial_{M} g_{N Q} + \partial_{N} g_{M Q} - \partial_{Q} g_{M N})$ and ${Ric}_{M N} = \partial_{P} Γ^{P}_{M N} - \partial_{N} Γ^{P}_{M P} + Γ^{P}_{P Q} Γ^{Q}_{M N} - Γ^{P}_{N Q} Γ^{Q}_{M P}$ . Then

$Ric (g)_{μ ν} = (λ - \hat{Δ} ω - (D - 2) ∣ d ω ∣_{\hat{g}}^{2}) h_{μ ν},$

and consequently

$\hat{Δ} ω + (D - 2) ∣ d ω ∣_{\hat{g}}^{2} = \frac{\hat{Δ} (Ω^{D - 2})}{(D - 2) Ω^{D - 2}} .$

These are local coordinate identities and use no field equation or physical premise.

Statement references

Maldacena and Nuñez, §6.1, equations (31)–(33)

Cecotti, Chapter 14, equations (14.4)–(14.6)

Facts & Assumptions

Given: The product-coordinate metric data, the explicit Christoffel and Ricci conventions in the Statement, and the smooth real function $ω$ depending only on $y$ .

[A1]

The external block $h_{μ ν}$ is independent of $y$ , the internal block ${\hat{g}}_{m n}$ is independent of $x$ , and $\partial_{μ} ω = 0$ .

Proof

1.1givenA1algebra

Put $u_{M} = \partial_{M} ω$ and $\overset{ˉ}{\nabla}$ for the product-metric connection. Direct substitution in the Christoffel formula gives $Γ^{P}_{M N} (g) - {\overset{ˉ}{Γ}}^{P}_{M N} = C^{P}_{M N} : = δ_{M}^{P} u_{N} + δ_{N}^{P} u_{M} - {\overset{ˉ}{g}}_{M N} {\overset{ˉ}{g}}^{P Q} u_{Q}$ . This tensor is symmetric in $M, N$ . Also, in $\nabla_{M}^{g} (e^{2 ω} {\overset{ˉ}{g}}_{N P})$ , the derivative of $e^{2 ω}$ contributes $2 e^{2 ω} u_{M} {\overset{ˉ}{g}}_{N P}$ and the two $C$ terms contribute its negative; the $\overset{ˉ}{\nabla} \overset{ˉ}{g}$ terms vanish. Thus these Christoffel symbols define the torsion-free metric-compatible connection of $g$ .

2.1givenA1step 1.1algebra

Inserting $Γ = \overset{ˉ}{Γ} + C$ into the Ricci formula in the Statement gives $Ric (g)_{M N} = Ric (\overset{ˉ}{g})_{M N} - (D - 2) {\overset{ˉ}{\nabla}}_{M} u_{N} - {\overset{ˉ}{g}}_{M N} {\overset{ˉ}{\nabla}}_{P} u^{P} + (D - 2) (u_{M} u_{N} - {\overset{ˉ}{g}}_{M N} ∣ u ∣_{\overset{ˉ}{g}}^{2})$ . The derivative contraction follows from ${\overset{ˉ}{\nabla}}_{P} C^{P}_{M N} = {\overset{ˉ}{\nabla}}_{M} u_{N} + {\overset{ˉ}{\nabla}}_{N} u_{M} - {\overset{ˉ}{g}}_{M N} {\overset{ˉ}{\nabla}}_{P} u^{P}$ and $C^{P}_{M P} = D u_{M}$ ; symmetry of the Hessian of $ω$ combines the first two terms. For the quadratic contraction, $C^{P}_{P Q} C^{Q}_{M N} = D (2 u_{M} u_{N} - {\overset{ˉ}{g}}_{M N} ∣ u ∣_{\overset{ˉ}{g}}^{2})$ and $C^{P}_{N Q} C^{Q}_{M P} = (D + 2) u_{M} u_{N} - 2 {\overset{ˉ}{g}}_{M N} ∣ u ∣_{\overset{ˉ}{g}}^{2}$ . The product Ricci block is $Ric (\overset{ˉ}{g})_{μ ν} = λ h_{μ ν}$ , while [A1] makes the external Hessian and external $u$ components zero. Substitution yields the asserted external Ricci formula.

3.1givenalgebra∎

For $f = e^{(D - 2) ω} = Ω^{D - 2}$ , the chain rule gives $\hat{Δ} f = (D - 2) f \hat{Δ} ω + (D - 2)^{2} f ∣ d ω ∣_{\hat{g}}^{2}$ . Divide by $(D - 2) f > 0$ to obtain the second identity.

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