CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck pass

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.

A transverse-traceless planar profile travelling slower than light fails the linearized vacuum equations

Statement refuted

Statement refuted: "For every real $v$ with $- 1 < v < 1$ and every smooth $f : R \to R$ , the symmetric perturbation $h$ with components

$h_{11} (x) = f (x^{3} - v x^{0}), h_{22} (x) = - f (x^{3} - v x^{0}), h_{12} = h_{21} = 0, all other components 0,$

solves the linearized vacuum equations of the flat linearization, $G_{a b}^{(1)} (h) = 0$ ."

Here the phase parameter is defined by the level sets $x^{3} - v x^{0} = c o n s t a n t$ ; their coordinate ratio is $∣ d x^{3} / d x^{0} ∣ = ∣ v ∣ < 1$ , while the null phase has coordinate ratio $1$ in the same flat chart.

Facts & Assumptions

Given: The flat model with inertial chart, the number $v = \frac{1}{2}$ , the bounded smooth function $f (s) = \cos s$ , the function $U (x) : = x^{3} - \frac{1}{2} x^{0}$ , and the perturbation $h$ with $h_{11} = \cos U$ , $h_{22} = - \cos U$ , all other components zero. The bound $∣ \cos U ∣ \leq 1$ makes $g (λ) = η + λ h$ Lorentzian on $R^{4}$ for $∣ λ ∣ < \frac{1}{2}$ ; taking $q = 0$ and $ε_{1} = \frac{1}{2}$ makes $h$ an admissible first-order perturbation of a flat linearized family.

[L1]

In harmonic gauge the first-order Einstein tensor is $G_{a b}^{(1)} (h) = - \frac{1}{2} □ {\overset{ˉ}{h}}_{a b}$ , and the trace reversal is ${\overset{ˉ}{h}}_{a b} = h_{a b} - \frac{1}{2} η_{a b} η^{c d} h_{c d}$ ; the flat d'Alembertian is $□ = - \partial_{0}^{2} + \partial_{1}^{2} + \partial_{2}^{2} + \partial_{3}^{2}$ (Flat linearized perturbations: families, trace reversal, gauge transformation and harmonic gauge clauses (b)-(c), First-order curvature of a flat linearized family, gauge transformations and the quadratic remainder clause (d)).

[L2]

For $U_{v} (x) : = x^{3} - v x^{0}$ , the chain rule gives $\partial_{0} (f \circ U_{v}) = - v f^{'} (U_{v})$ , $\partial_{3} (f \circ U_{v}) = f^{'} (U_{v})$ , $\partial_{0}^{2} (f \circ U_{v}) = v^{2} f^{''} (U_{v})$ and $\partial_{3}^{2} (f \circ U_{v}) = f^{''} (U_{v})$ ; $\partial_{1}, \partial_{2}$ derivatives vanish and derivatives of products obey the usual rules (The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

[L3]

For every real $s$ , $∣ \cos s ∣ \leq 1$ and $\frac{d^{2}}{d s^{2}} \cos s = - \cos s$ (Parity and the Pythagorean identity for sine and cosine, The derivatives of sine and cosine are cosine and minus sine).

Counterexample

Take $v = \frac{1}{2}$ and $f (s) = \cos s$ . Then

$h_{11} (x) = \cos (x^{3} - \frac{1}{2} x^{0}), h_{22} (x) = - \cos (x^{3} - \frac{1}{2} x^{0}), h_{13} = h_{23} = h_{33} = 0,$

and the perturbation is traceless and in harmonic gauge, yet at each point where $U (x) = 0$

$G_{11}^{(1)} (h) = \frac{3}{8} \neq 0$

so it does not solve the linearized vacuum equations on $R^{4}$ .

1.1L2L3givenalgebra

(Trace, harmonic gauge and admissibility.) The trace is $η^{a b} h_{a b} = h_{11} + h_{22} = \cos U - \cos U = 0$ , so $\overset{ˉ}{h} = h$ . The divergence components are $\partial^{a} h_{a 0} = 0$ ; for $b = 1$ , $\partial^{a} h_{a 1} = η^{00} \partial_{0} h_{01} + η^{11} \partial_{1} h_{11} + η^{33} \partial_{3} h_{31} = 0 + 0 + 0 = 0$ , since $h$ does not depend on $x^{1}$ and $h_{31} = 0$ ; similarly for $b = 2, 3$ . Hence $\partial^{a} {\overset{ˉ}{h}}_{a b} = 0$ : the perturbation is in harmonic gauge. Also, [L3] gives $g_{11} = 1 + λ \cos U > 0$ and $g_{22} = 1 - λ \cos U > 0$ for $∣ λ ∣ < \frac{1}{2}$ , with $g_{00} = - 1$ and $g_{33} = 1$ , so $g (λ)$ is Lorentzian and $h$ is an admissible first-order perturbation.

1.2L2L3algebra

(The flat d'Alembertian of the profile.) By [L2]-[L3], $□ f (U) = (- v^{2} + 1) f^{''} (U) = \frac{3}{4} (- \cos U) = - \frac{3}{4} \cos U$ . Hence $□ h_{11} = - \frac{3}{4} \cos U$ , $□ h_{22} = \frac{3}{4} \cos U$ and $□ h_{a b} = 0$ for all other pairs.

2.1L1step 1.1step 1.2algebra∎

(The linearized Einstein tensor.) By [L1] and steps 1.1-1.2, $G_{11}^{(1)} (h) = - \frac{1}{2} □ {\overset{ˉ}{h}}_{11} = \frac{3}{8} \cos U$ , which equals $\frac{3}{8}$ wherever $U = 0$ . Therefore $G_{a b}^{(1)} (h)$ is not identically zero and the perturbation does not solve the linearized vacuum equations on $R^{4}$ , refuting the statement.

Remarks

Why the phase rate matters. For a profile $f (x^{3} - v x^{0})$ one computes $□ f = (1 - v^{2}) f^{''}$ , which vanishes for every $f$ exactly when $v^{2} = 1$ . The distinguished null profile of the vacuum plane waves is $u = x^{0} - x^{3}$ , for which the profiles with the same spatial structure do solve the equations (TT plane waves with arbitrary bounded profiles solve the linearized vacuum equations, are in harmonic gauge, and give Lorentzian small-amplitude metrics with a quadratic Einstein-tensor residual clause (b)); the counterexample uses the sub-null coordinate ratio $v = \frac{1}{2}$ and exhibits the failure quantitatively.
The perturbation is not a solution of the exact field law either. Since $G_{11}^{(1)} \neq 0$ , the perturbation is not even a first-order vacuum solution, so no exact vacuum metric is obtained by "correcting" it at higher order in the displayed family; the counterexample concerns the stated universal claim only.

A transverse-traceless planar profile travelling slower than light fails the linearized vacuum equations

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources