ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Extremal cylindrical puncture end

Example

Let $m = ∣ q ∣ > 0$ in the one-hole Reissner–Nordström (RN) conformal-data family; this equality defines its extremal parameter case. Use the punctured spatial manifold $Σ = (0, \infty) \times S^{2}$ . Then $ψ^{2} = 1 + m / r$ , $h : = ψ^{4} (d r^{2} + r^{2} γ_{S^{2}})$ , and the centered sphere is $S_{r} : = {r} \times S^{2}$ , with areal radius $ρ (r) : = r ψ^{2} = r + m$ . On each $S_{r}$ use the outward unit normal $ν : = ψ^{- 2} \partial_{r}$ and define the mean-curvature scalar $H : = {div}_{h} ν$ ; a sphere is minimal exactly when $H = 0$ . With this convention, no centered coordinate sphere is minimal. The proper radial coordinate

$s (r) : = r + m \log (r / m)$

has range $R$ and gives

$h = d s^{2} + (r (s) + m)^{2} γ_{S^{2}} .$

As $r ↓ 0$ , $s \to - \infty$ and $r (s) + m \to m$ , so this end is asymptotic to the product cylinder $R \times S^{2}$ with round-sphere radius $m$ . The puncture is omitted; no point-source interpretation or global event-horizon claim is part of this example.

Example references

RN conformal data on a punctured exterior

Facts & Assumptions

Given: $m = ∣ q ∣ > 0$ , the isotropic conformal data on $(0, \infty) \times S^{2}$ , and the formulas in the Statement.

[F1]

RN conformal data on a punctured exterior: $ψ^{2} = (1 + m / (2 r))^{2} - q^{2} / (4 r^{2})$ , $K = 0$ , and $R_{a r e a l} = r ψ^{2}$ ; the puncture $r = 0$ is excluded from the smooth data manifold.

Verification

1.1givenF1algebra

Substituting $q^{2} = m^{2}$ into the conformal factor gives $ψ^{2} = (1 + m / (2 r))^{2} - m^{2} / (4 r^{2}) = 1 + m / r$ , which is positive for every $r > 0$ . Therefore $ρ = r ψ^{2} = r + m$ and $ρ^{'} (r) = 1$ .

2.1givenstep 1.1algebra

The radial line element of $h = ψ^{4} (d r^{2} + r^{2} γ_{S^{2}})$ is $ψ^{2} d r = (1 + m / r) d r$ . Since $s^{'} (r) = 1 + m / r = ψ^{2} > 0$ , $s$ is a proper radial coordinate; moreover $s \to - \infty$ as $r ↓ 0$ and $s \to + \infty$ as $r \to \infty$ , so its range is $R$ .

3.1step 1.1step 2.1algebra

Replacing $ψ^{4} d r^{2}$ by $d s^{2}$ and using $ρ = r + m$ gives $h = d s^{2} + ρ (s)^{2} γ_{S^{2}}$ . As $s \to - \infty$ , $r (s) \to 0$ , so $ρ (s)^{2} \to m^{2}$ and the spherical coefficient converges to that of the product cylinder of radius $m$ .

4.1step 1.1step 2.1algebra∎

The unit normal to $S_{r}$ is $ν = ψ^{- 2} \partial_{r}$ . Its mean curvature is $H = {div}_{h} ν = 2 ρ^{'} / (ψ^{2} ρ) = 2 r / (r + m)^{2} > 0$ for every $r > 0$ . Thus no centered coordinate sphere is minimal, while the infinite negative range of $s$ proves that the puncture is an infinite-distance cylindrical end.

Depends on

RN conformal data on a punctured exterior