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Extremal horizon conservation and conditional derivative growth — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)Open item page →

Nonzero horizon charge alone does not force this growth

Example

Let $m > 0$ , let $c \in R ∖ {0}$ , and on the fixed extremal RN background take the constant neutral spherical test scalar $ψ (v, r) = c$ . Then $□_{g} ψ = 0$ , $ϕ (v, r) = c r$ , and its horizon charge is $H = c \neq 0$ , while $ϕ (v, m) = c m$ does not tend to zero and $\partial_{r}^{2} ϕ (v, m) = 0$ for all $v$ . Hence nonzero horizon charge alone does not satisfy the extra decay hypothesis of the conditional horizon-growth proposition and does not force its linear-growth conclusion.

Statement references

Spherical scalar waves in regular extremal RN coordinates

Exact spherical horizon charge

Second derivative growth under an explicit horizon decay hypothesis

Facts & Assumptions

Given: $m > 0$ , $c \in R ∖ {0}$ , the fixed extremal RN metric, and ${A C}_{ω}$ for the inherited metric-connection background.

[F1]

Spherical scalar waves in regular extremal RN coordinates: an admissible scalar has a real reduced $C^{3}$ profile $ψ (v, r)$ with full lift $\tilde{ψ} (v, r, ω) = ψ (v, r)$ solving the extended trace equation $□_{g} \tilde{ψ} = 0$ . The metric and Maxwell field $F_{i n} = - q_{e} r^{- 2} d v \land d r + q_{m} ω_{S^{2}}$ are fixed, with no scalar coupling or backreaction; the horizon is the null level $r = m$ generated by $ξ = \partial_{v}$ with zero signed surface gravity. Set $ϕ = r ψ$ and take radial derivatives in the regular ingoing chart.

[F2]

Exact spherical horizon charge: for such a solution, $H = \partial_{r} ϕ (v, m)$ is constant in $v$ .

[F3]

Second derivative growth under an explicit horizon decay hypothesis: the derivative-growth conclusion additionally requires $H \neq 0$ and $ϕ (v, m) \to 0$ .

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): the metric-connection suppliers inherit only ${A C}_{ω}$ ; the explicit constant solution uses no choice.

Verification

1.1givenF1A1algebra

The constant profile $ψ (v, r) = c$ has full lift $\tilde{ψ} (v, r, ω) = c$ , is real and $C^{3}$ , and has all first and second coordinate derivatives zero. Its scalar Hessian and metric trace therefore vanish, so $□_{g} \tilde{ψ} = 0$ and it is admissible by [F1]. The definition $ϕ = r ψ$ gives $ϕ (v, r) = c r$ .

2.1step 1.1F2algebra

Differentiating $ϕ = c r$ gives $\partial_{r} ϕ = c$ and $\partial_{r}^{2} ϕ = 0$ . Thus [F2] identifies its constant horizon charge as $H = c \neq 0$ , and its horizon trace is $ϕ (v, m) = c m$ .

3.1givenF3step 2.1algebra∎

Since $c \neq 0$ and $m > 0$ , $c m \neq 0$ , so the horizon trace is constant nonzero and does not tend to zero. The extra hypothesis of [F3] therefore fails, while step 2.1 shows the second transverse derivative vanishes identically. The conditional proposition remains consistent with this example.

Remarks

The constant solution checks why nonzero charge alone is insufficient for the conditional growth conclusion. It makes no assertion about decay for solutions generated by global initial data.