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A compact temporal covector persists under metric perturbation

Statement

Let $g$ be a $C^{2}$ Lorentz metric on an open subset of a smooth four-manifold, $N$ a continuous timelike orientation field, $K$ compact, and $α$ a continuous covector field near $K$ with $g^{- 1} (α, α) < 0$ and $α (N) > 0$ on $K$ . There is $δ > 0$ such that every Lorentz metric $\tilde{g}$ whose coefficients differ from those of $g$ by less than $δ$ on a fixed finite compact chart cover of $K$ satisfies $\tilde{g} (N, N) < 0$ , ${\tilde{g}}^{- 1} (α, α) < 0$ and $α (N) > 0$ on $K$ . Orient $\tilde{g}$ by $N$ on $K$ . Thus if $α = d τ$ for a smooth real function $τ$ , the $\tilde{g}$ -gradient of $τ$ is past timelike on $K$ . This asserts persistence on the compact set; it asserts no causal property away from that set.

Statement references

Causal exterior, chronological boundaries and Cauchy data

The inverse metric, musical isomorphisms and index raising

Compact sets admit relatively compact open neighborhoods and finite chart pieces

A temporal covector has uniform compact negativity and orientation margins

Inversion is uniformly stable on compact chart families

Smooth manifolds and their smooth charts

Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces

In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones

Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace

A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact

Facts & Assumptions

[F1]

Use a finite compact chart cover and the compact margins and component bounds for $α, N$ . Compact sets admit relatively compact open neighborhoods and finite chart pieces A temporal covector has uniform compact negativity and orientation margins

[F2]

Uniformly small matrix perturbations make inverse coefficients uniformly close. Inversion is uniformly stable on compact chart families

[F3]

The inverse metric raises covectors, and the past-timelike convention is fixed by pairing positively with $N$ . The inverse metric, musical isomorphisms and index raising Causal exterior, chronological boundaries and Cauchy data

[F4]

A smooth manifold has an underlying topological manifold, which is Hausdorff. Every compact subset of a Hausdorff space is closed. Closed sets in a subspace are traces of ambient closed sets, and a closed subspace of a compact space is compact. Smooth manifolds and their smooth charts Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact

Proof

Given: The hypotheses in the Statement.

Goal: Establish the conclusions in the Statement.

1.1F1F3F4

If $K$ is empty the conclusion is vacuous. Otherwise write $C_{j}$ for the pieces of the fixed finite compact chart cover and set $K_{j} = C_{j} \cap K$ . The ambient smooth manifold is Hausdorff by [F4]. Each compact $C_{j}$ is therefore closed in it, so $K_{j} = C_{j} \cap K$ is closed in the compact subspace $K$ and hence compact by [F4]. These compact pieces still cover $K$ and lie where $α, N$ are defined. Choose $m, a, b > 0$ from [F1], and a common bound $R \geq 1$ for all components of $α$ and $N$ on the restricted pieces $K_{j}$ . For a coefficient perturbation less than $δ$ , the change in $g (N, N)$ has absolute value at most $16 δ R^{2}$ . Choose $δ < b / (32 R^{2})$ to retain $\tilde{g} (N, N) < 0$ .

2.1F1F2F3step 1.1∎

Apply [F2] on the restricted pieces $K_{j}$ with $η = m / (32 R^{2})$ and reduce $δ$ to its resulting threshold. The change in $g^{- 1} (α, α)$ then has absolute value at most $16 η R^{2} = m / 2$ , so it remains negative. The pairing $α (N) \geq a$ does not change with the metric. For $α = d τ$ , [F3] identifies these inequalities with a past-timelike $\tilde{g}$ -gradient.

Depends on

Used by · 2

Dependency tree · 2 levels

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