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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Maldacena–Nuñez constraints on Minkowski warping and de Sitter compactifications

Physical scope

Conditional two-derivative Einstein-matter model with positive scalar/form kinetic terms, nonpositive potential, smooth compact internal space, invariant Minkowski or de Sitter external metric, and only the explicitly stated smooth or warp-boundary singular branch. No higher-curvature terms, extra source/boundary terms, positive-potential sector, or time-dependent external geometry is covered.

Statement

Use natural units $ℏ = c = 1$ , length-valued coordinates, and dimensionless metric components. Let $D > 2$ and $2 \leq d < D$ . Let $X$ be a nonempty smooth compact manifold without boundary of dimension $D - d$ , with a smooth positive-definite metric $\hat{g}$ . Let $(Y_{d}, g_{d})$ be either Minkowski spacetime $(R^{d}, η)$ , where $η = diag (- 1, 1, \dots, 1)$ , or de Sitter space $d S_{d} (H_{d}) = {z \in R^{1, d} : - z_{0}^{2} + \sum_{i = 1}^{d} z_{i}^{2} = H_{d}^{- 2}}$ with the induced Lorentzian metric, where $H_{d} > 0$ has units of inverse length. Thus $R_{μ ν} (g_{d}) = (R_{d} / d) (g_{d})_{μ ν}$ , with scalar curvature $R_{d} = 0$ in the Minkowski case and $R_{d} = d (d - 1) H_{d}^{2} > 0$ in the de Sitter case. On the smooth branch, let $Ω \in C^{\infty} (X, (0, \infty))$ ; compactness makes it bounded above and below by positive constants. On the singular branch, let $Z \subset X$ be a nonempty closed proper set, write $X_{r e g} = X ∖ Z$ , and let $Ω \in C^{0} (X, [0, \infty))$ be smooth and positive on $X_{r e g}$ , with zero set $Z$ . Assume the Axiom of Choice on this branch: every family of nonempty sets admits a choice function. On either branch use the $D$ -dimensional metric on its regular locus $d s_{D}^{2} = Ω (y)^{2} (d s_{d}^{2} + {\hat{g}}_{m n} (y) d y^{m} d y^{n}) .$ Internal integrals use the Riemannian volume density $d \hat{μ} = \sqrt{\det \hat{g}} ∣ d y^{1} \dots d y^{D - d} ∣$ (and its induced boundary density), so no orientation is assumed. The lower-dimensional Newton coefficient is finite, with $G_{N, d}^{- 1}$ proportional to $\int_{X} \sqrt{\hat{g}} Ω^{D - 2}$ . Consider the adopted two-derivative action $S_{D} = \frac{1}{2 κ_{D}^{2}} \int d^{D} x \sqrt{- g} R - \int d^{D} x \sqrt{- g} [\frac{1}{2} h_{a b} (ϕ) \partial_{M} ϕ^{a} \partial^{M} ϕ^{b} + V (ϕ) + \sum_{n = 2}^{D - 1} \frac{Z_{n} (ϕ)}{2 n!} F_{(n)}^{2}],$ where $κ_{D}^{2}$ has units of length $^{D - 2}$ , the matter density has units of length $^{- D}$ , $h_{a b}$ is positive definite, $Z_{n} > 0$ , $V \leq 0$ on the solution's scalar range, and $F_{(n)}$ is a real massless $n$ -form with $F_{(n)}^{2} = F_{M_{1} \dots M_{n}} F^{M_{1} \dots M_{n}}$ . On the smooth branch, scalars and forms are smooth on $X$ ; on the singular branch, they are smooth up to the boundary of each regular superlevel domain defined below. Assume each antisymmetric tensor $Q$ in the adopted field equations has the invariant decomposition $Q = Q_{e} + {vol}_{g_{d}} \land Q_{l}$ , with $Q_{e}$ and $Q_{l}$ internal forms of ranks $rank (Q)$ and $rank (Q) - d$ ; set $Q_{l} = 0$ when $rank (Q) < d$ . Thus no component has only some external indices. There are no localized stress terms or boundary terms in the field equations on the regular locus. The trace-reversed Einstein equation is $R_{M N} = κ_{D}^{2} (T_{M N} - \frac{g_{M N}}{D - 2} T^{P}_{P}),$ with $T_{M N}$ obtained by varying the displayed action.

Define the full external and total traces using $g$ and set $\tilde{T} = κ_{D}^{2} (- T^{μ}_{μ} + \frac{d}{D - 2} T^{P}_{P}) .$ Then $\tilde{T} \geq 0$ and the correctly normalized external trace equation is $\frac{{\hat{\nabla}}^{2} Ω^{D - 2}}{(D - 2) Ω^{D - 2}} = \frac{R_{d} + Ω^{2} \tilde{T}}{d} .$ On the smooth compact branch, $Ω$ is constant on each connected component; de Sitter is excluded, and in Minkowski only scalar-gradient terms and an external rank- $(D - 1)$ form can contribute in the zero-trace sector, while internal forms of rank $n > 1$ must vanish. On the singular branch, assume $Ω$ is bounded above. Call $ϵ > 0$ a regular value when $d Ω_{y} \neq 0$ at every $y \in X_{r e g}$ with $Ω (y) = ϵ$ . For arbitrarily small such $ϵ$ , assume the closed superlevel set $K_{ϵ} : = {Ω \geq ϵ} \subset X_{r e g}$ is a compact smooth manifold with boundary $\partial K_{ϵ} = {Ω = ϵ}$ ; write $R_{ϵ} : = {Ω > ϵ}$ for its interior. Assume its inward unit normal $n_{i n}$ satisfies $- \int_{\partial K_{ϵ}} Ω^{D - 2} n_{i n} \cdot \hat{\nabla} (Ω^{D - 2}) d \hat{Σ} \leq 0.$ The singular-branch integration excludes invariant de Sitter. In Minkowski space it forces $Ω$ to be constant on each connected component of $X_{r e g}$ ; disconnected regular components may have different constants. This theorem does not cover positive potentials, higher-derivative/string corrections, other source or boundary terms, or time-dependent external metrics.

Statement references

Maldacena and Nuñez, §6.1–6.3

Van Riet and Zoccarato, §4.2

Facts & Assumptions

Given: The adopted action, compactness and regularity hypotheses, the trace-reversed field equation in the Statement, and exactly one of the stated warp-boundary branches.

[A1]

The scalar metric $h_{a b}$ and every $Z_{n}$ are positive, and $V \leq 0$ along the solution. The forms are real and preserve all external isometries.

[F1]

Conformal warped-product external Ricci trace: for $g = Ω^{2} (g_{d} + \hat{g})$ with $ω = \log Ω$ depending only on $X$ , the external Ricci block is $R_{μ ν} (g) = R_{μ ν} (g_{d}) - (g_{d})_{μ ν} [\hat{Δ} ω + (D - 2) ∣ d ω ∣^{2}]$ , and $\hat{Δ} Ω^{D - 2} / ((D - 2) Ω^{D - 2}) = \hat{Δ} ω + (D - 2) ∣ d ω ∣^{2}$ .

[F2]

Finite chart localization gives choice-free integration and compact Stokes: the integral of an exact compactly supported top form on a boundaryless oriented manifold is zero; this supplier proves only the boundaryless case used on the smooth branch.

[F3]

The general Stokes theorem: assuming countable choice, for an oriented smooth $n$ -manifold with boundary, $n \geq 1$ , and a compactly supported smooth $(n - 1)$ -form $η$ , $\int_{M} d η = \int_{\partial M} j^{*} η$ with the outward-normal-first boundary orientation.

[A2]

The Axiom of Choice: every family of nonempty sets has a choice function. Its restriction to sequences gives the countable-choice hypothesis in [F3]; the only use of choice in this proof is Step 5.1.

Proof

1.1givenA1algebra

Vary the matter terms with respect to $g^{M N}$ , using $δ \sqrt{- g} = - \frac{1}{2} \sqrt{- g} g_{M N} δ g^{M N}$ . For the scalar kinetic term, $δ (h_{a b} \partial_{P} ϕ^{a} \partial^{P} ϕ^{b}) = h_{a b} \partial_{M} ϕ^{a} \partial_{N} ϕ^{b} δ g^{M N}$ ; for an $n$ -form, antisymmetry makes the $n$ inverse-metric variations in $δ F_{(n)}^{2}$ equal, giving $δ F_{(n)}^{2} = n F_{M L_{2} \dots L_{n}} F_{N}^{L_{2} \dots L_{n}} δ g^{M N}$ . Hence $T_{M N} = h_{a b} (\partial_{M} ϕ^{a} \partial_{N} ϕ^{b} - \frac{1}{2} g_{M N} \partial_{P} ϕ^{a} \partial^{P} ϕ^{b}) - V g_{M N} + \sum_{n} Z_{n} [\frac{1}{(n - 1)!} F_{M L_{2} \dots L_{n}} F_{N}^{L_{2} \dots L_{n}} - \frac{g_{M N}}{2 n!} F_{(n)}^{2}]$ . Varying the Einstein–Hilbert term gives $G_{M N} = κ_{D}^{2} T_{M N}$ ; tracing gives $R = - 2 κ_{D}^{2} T^{P}_{P} / (D - 2)$ , and substitution yields the trace-reversed equation stated.

2.1givenF1step 1.1algebra

Apply [F1] to the external block and trace the trace-reversed equation with $(g_{d})^{μ ν}$ . Since $g_{μ ν} = Ω^{2} (g_{d})_{μ ν}$ , the matter side becomes $Ω^{2} [T^{μ}_{μ} - d T^{P}_{P} / (D - 2)]$ . Rearranging and using [F1] gives $\hat{Δ} Ω^{D - 2} / ((D - 2) Ω^{D - 2}) = (R_{d} + Ω^{2} \tilde{T}) / d$ .

2.2A1step 1.1algebra

For scalar gradients, external invariance gives $T^{μ}_{μ} = - \frac{d}{2} K - d V$ and $T^{P}_{P} = (1 - \frac{D}{2}) K - D V$ , where $K = Ω^{- 2} h_{a b} {\hat{g}}^{m n} \partial_{m} ϕ^{a} \partial_{n} ϕ^{b} \geq 0$ . Substitution cancels the $K$ terms and gives ${\tilde{T}}_{s c a l a r} = - 2 κ_{D}^{2} d V / (D - 2) \geq 0$ .

2.3step 1.1A1algebra

For a wholly internal $n$ -form, $F_{(n)}^{2} \geq 0$ , $T^{μ}_{μ} = - d Z_{n} F_{(n)}^{2} / (2 n!)$ , and $T^{P}_{P} = Z_{n} (n - D / 2) F_{(n)}^{2} / n!$ . Thus ${\tilde{T}}_{i n t} = κ_{D}^{2} Z_{n} d (n - 1) F_{(n)}^{2} / [n! (D - 2)] \geq 0$ , with zero contribution for $n = 1$ and positive contribution for any nonzero internal form of rank $n > 1$ .

2.4step 1.1A1algebra

For a component with all $d$ external indices, it exists only if $n \geq d$ , has $F_{(n)}^{2} \leq 0$ , and antisymmetry gives $g^{μ ν} F_{μ L_{2} \dots L_{n}} F_{ν}^{L_{2} \dots L_{n}} = (d / n) F_{(n)}^{2}$ . Substitution in the stress trace gives ${\tilde{T}}_{e x t} = - κ_{D}^{2} Z_{n} d (D - 1 - n) F_{(n)}^{2} / [n! (D - 2)] \geq 0$ ; it vanishes at rank $n = D - 1$ . Summing these separate invariant components proves $\tilde{T} \geq 0$ .

3.1F1F2step 2.1step 2.2step 2.3step 2.4

First justify integration without orientability. For a smooth vector field $V$ define $\overset{}{div} V = (\det \hat{g})^{- 1 / 2} \partial_{m} ((\det \hat{g})^{1 / 2} V^{m})$ . The finite chart-cutoff construction of [F2] uses bumps and compactness independently of orientation. Choose such cutoffs $χ_{i}$ summing to one on $X$ . In each chart, integration of the compactly supported coordinate derivative gives $\int_{X} \overset{}{div} (χ_{i} V) d \hat{μ} = 0$ . Summing and using $\sum_{i} d χ_{i} = 0$ proves $\int_{X} \overset{}{div} V d \hat{μ} = 0$ . Applying this to $V = \hat{\nabla} f$ and $V = f \hat{\nabla} f$ yields the two identities used below. On the smooth branch put $f = Ω^{D - 2} > 0$ . Multiply step 2.1 by $(D - 2) f$ and integrate on $X$ ; [F2] gives $\int_{X} \hat{Δ} f d \hat{μ} = 0$ , hence $\int_{X} f R_{d} d \hat{μ} + \int_{X} f Ω^{2} \tilde{T} d \hat{μ} = 0$ . If $R_{d} > 0$ , the first integral is strictly positive, a contradiction. If $R_{d} = 0$ , the second integrand is continuous and nonnegative, so zero integral forces $\tilde{T} = 0$ pointwise. Step 2.1 then gives $\hat{Δ} f = 0$ ; applying the just-proved divergence identity to $f \hat{\nabla} f$ gives $\int_{X} ∣ d f ∣^{2} d \hat{μ} = 0$ , so $f$ and therefore $Ω$ are constant on each connected component.

4.1A1step 2.2step 2.3step 2.4step 3.1

Since every contribution computed in steps 2.2–2.4 is nonnegative, $\tilde{T} = 0$ forces $V = 0$ and excludes every nonzero form contribution except an external rank- $(D - 1)$ form. Scalar gradients are not forced to vanish by this trace argument.

5.1F3A2givenstep 2.1step 2.2step 2.3step 2.4step 4.1algebra∎

On the singular branch fix an arbitrarily small regular $ϵ > 0$ with $R_{ϵ} \neq \emptyset$ and put $f = Ω^{D - 2}$ on $K_{ϵ}$ . Use finite chart cutoffs near $K_{ϵ}$ , with charts straightening its smooth boundary. Give each chart its coordinate orientation and apply [F3] to the compactly supported form $ι_{χ_{i} f \hat{\nabla} f} {\hat{v o l}}_{i}$ there, where ${\hat{v o l}}_{i}$ is the metric volume form for that chart orientation. The coordinate orientation signs cancel against those in the positive volume density; the outward-normal-first convention gives the outward normal flux. Summing cancels the cutoff derivatives as in step 3.1 and proves the density divergence theorem on $K_{ϵ}$ ; its outward unit normal is $- n_{i n}$ , so $\int_{K_{ϵ}} f \hat{Δ} f d \hat{μ} = - \int_{\partial K_{ϵ}} f n_{i n} \cdot \hat{\nabla} f d \hat{Σ} - \int_{K_{ϵ}} ∣ \hat{\nabla} f ∣^{2} d \hat{μ}$ . Multiplying the trace equation in step 2.1 by $(D - 2) f^{2}$ and integrating over $K_{ϵ}$ also gives $\int_{K_{ϵ}} f \hat{Δ} f d \hat{μ} = \frac{D - 2}{d} \int_{K_{ϵ}} f^{2} (R_{d} + Ω^{2} \tilde{T}) d \hat{μ} \geq 0$ . The boundary assumption makes the boundary term in the Stokes expression nonpositive, so both equalities force $\hat{\nabla} f = 0$ on $R_{ϵ}$ and the right-hand integral to vanish. Since $R_{ϵ}$ is nonempty, $R_{d} > 0$ would make that integral strictly positive, excluding de Sitter. For Minkowski, $f$ is constant on each connected component $C$ of $R_{ϵ}$ ; if $\partial C \neq \emptyset$ , continuity and $Ω = ϵ$ on $\partial C$ contradict $Ω > ϵ$ in $C$ . Each remaining boundaryless $C$ is open in $X$ , compact, and therefore closed in $X$ and a connected component of $X$ . For any point of $X_{r e g}$ , choose such an $ϵ < Ω$ at that point; its connected component of $X$ then lies in $X_{r e g}$ and has constant $Ω$ . Since arbitrarily small admissible regular values are available, these domains cover $X_{r e g}$ , so the zero-trace conclusions of step 4.1 hold throughout it.

Maldacena–Nuñez constraints on Minkowski warping and de Sitter compactifications

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources