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Leading stationary modes and a spatial normalization of the same maximal end

Statement

Work in ZFC on the smooth stationary vacuum Lorentzian four-dimensional end supplied by the same-slice polyhomogeneous gauge lemma in [F1]. Retain its asymptotically horizontal maximal stationary slicing, with $T = \partial_{τ}$ future timelike near infinity, $N \to 1$ , $X \to 0$ , and the same supplied CW maximal hypersurface at $τ = 0$ . Write $h$ , $k$ , $N$ and $X$ for its induced metric, second form, lapse and shift, with $k = (2 N)^{- 1} L_{X} h$ . Use the supplied asymptotically Cartesian quotient-harmonic spatial coordinates $y$ , with $r = ∣ y ∣$ and $ω = y / ∣ y ∣$ , full differentiated polyhomogeneous expansions and $h - δ, N - 1, X = O_{\infty} (r^{- 1})$ . The normalized maximal vacuum KID identities are those supplied in [F2].

In the source spatial harmonic gauge, write the first coefficients as $N = 1 + r^{- 1} n_{1} (ω) + \dots$ , $X^{i} = r^{- 1} A^{i} (ω) + \dots$ , and $h_{i j} = δ_{i j} + r^{- 1} H_{i j} (ω) + \dots$ . Then $n_{1} = - m$ is constant, A vanishes, and $H_{i j} = m (3 δ_{i j} - ω_{i} ω_{j})$ . The spatial radial change $z = y + (m / 2) y / ∣ y ∣$ preserves the same time coordinate and the same maximal slice and preserves full polyhomogeneity. In the resulting chart, $h_{i j} = δ_{i j} + 2 m δ_{i j} / r + O_{\infty} (r^{- 2} \log^{J} r), N = 1 - m / r + O_{\infty} (r^{- 2} \log^{J} r), X = O_{\infty} (r^{- 2} \log^{J} r),$ with finite logarithmic powers at each retained order. The final spatial chart is not asserted still to be quotient-harmonic.

Statement references

The source maximal-gauge expansion can be used on the same CW slice

Independent stationary maximal-gauge asymptotic expansion and KID input

Facts & Assumptions

[F1]

The same maximal end has full differentiated expansions and order-minus-one initial lapse, shift and metric corrections in quotient-harmonic spatial coordinates. The source maximal-gauge expansion can be used on the same CW slice

[F2]

The normalized maximal vacuum KID system supplies div $X = 0$ , the lapse and vector equations and the stationary Ricci equation. Independent stationary maximal-gauge asymptotic expansion and KID input

[F3]

The sphere Laplacian has constant kernel, degree-one eigenvalue minus two and the stated self-adjoint spectral decomposition. Spherical harmonic modes are orthogonal, complete for continuous data, and have polynomial evaluation bounds

[F4]

Work in ZFC, retaining the analytic choice assumptions inherited through [F1] and [F2]; the coefficient calculation and explicit radial map make no additional use of choice. The Axiom of Choice

Proof

Given: The supplied stationary vacuum maximal end, normalization, quotient-harmonic chart and differentiated expansions in the Statement and [F1]–[F2], working in ZFC as in [F4].

Goal: Prove the stated same-slice end consequence.

1.1F1F2F3givenalgebra

For a spherical function a, $Δ_{0} (r^{σ} a) = r^{σ - 2} (Δ_{S^{2}} + σ (σ + 1)) a$ . There are no log terms at order r to the power minus one, since the source O( $r^{- 1}$ ) bound forbids them. The maximal identity gives $k = (2 N)^{- 1} L_{X} h = O (r^{- 2})$ . In $Δ_{h} N = N ∣ k ∣^{2}$ , the order $r^{- 3}$ coefficient is therefore $Δ_{S^{2}} n_{1} = 0$ , so $n_{1} = - m$ . In the vector KID equation the order $r^{- 3}$ coefficient likewise gives $Δ_{S^{2}} A^{i} = 0$ componentwise, so A is a constant vector. The order $r^{- 2}$ coefficient in div $X = 0$ is $- A \cdot ω = 0$ , which forces $A = 0$ .

2.1F1F2F3step 1.1algebra

The stationary Ricci equation now has leading part $Ric (h)_{i j} = \partial_{i} \partial_{j} (- m / r) + O (r^{- 4} \log^{J} r)$ . The quotient metric differs from h only by a shift-square term, so its spatial harmonic condition gives the linearized h harmonic condition at order $r^{- 2}$ . The Ricci coordinate formula consequently gives $Δ_{S^{2}} H_{i j} = 2 m (3 ω_{i} ω_{j} - δ_{i j})$ . Since $Δ_{S^{2}} (ω_{i} ω_{j}) = 2 δ_{i j} - 6 ω_{i} ω_{j}$ , all solutions have $H_{i j} = - m ω_{i} ω_{j} + C_{i j}$ for a constant symmetric matrix C. The linearized harmonic condition is $- C ω + \frac{1}{2} tr (C) ω - \frac{3}{2} m ω = 0$ , which gives $C = 3 m δ$ .

3.1F1step 2.1algebra∎

The map z is a spatial diffeomorphism sufficiently far out. Its inverse has $y = z - (m / 2) z / ∣ z ∣ + O (r^{- 1})$ , whose leading derivative correction changes $m (3 δ - ω \otimes ω) / r$ to $2 m δ / r$ . It leaves the leading lapse coefficient unchanged and cannot restore an absent $r^{- 1}$ shift coefficient. The radial relation is exactly $∣ z ∣ = ∣ y ∣ + m / 2$ , so substitution and the chain rule preserve integer-power/logarithmic expansions and all differentiated remainders. This is a spatial normalization after the source harmonic-gauge analysis; time and the slice have not changed.

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