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Stationary transport of vanishing from a section

Statement

Use the IK AF package, with exterior $E$ and section $Σ_{0}$ , and let $Φ : R \times E \to E$ be the complete stationary flow. Let $S$ be a smooth tensor field on $E$ of any fixed type $(r, s)$ , where $r, s$ are nonnegative integers, with real or complex values. Use the diffeomorphism pullback specified below, extended to complex values by applying it to the real and imaginary parts. If $Φ_{t}^{*} S = S$ for every real $t$ and $S = 0$ on $Σ_{0} \cap E$ , then $S = 0$ on $E$ .

For application to the Mars–Simon tensor, additionally establish or import $E \subseteq N = {1 - σ \neq 0}$ , where $σ$ is the supplied Ernst potential, and establish or import flow invariance of that tensor. Its definition supplies a complex covariant four-tensor only on $N$ . Neither domain inclusion nor preservation follows merely from vanishing on the section.

Statement references

Exact asymptotic and orbit package for the IK criterion

Ernst sublevels and the Mars–Simon tensor interface

Time-t flow maps are diffeomorphisms between open domains

A smooth tensor field

A type (r,s) tensor on a finite-dimensional vector space

The pullback of a covariant tensor field

A diffeomorphism pulls back tensor fields and forms isomorphically

The differential of a diffeomorphism is an isomorphism

Facts & Assumptions

[F1]

Every stationary orbit through $E$ is complete and meets the section. The flow in the Statement is on $E$ , so its section intersection lies in $Σ_{0} \cap E$ . Exact asymptotic and orbit package for the IK criterion

[F2]

Each time- $t$ map of a smooth maximal flow is a diffeomorphism between its open domains, with inverse the time- $(- t)$ map. Applied to the complete flow on $E$ , these domains are both $E$ . Time-t flow maps are diffeomorphisms between open domains

[F3]

A smooth type- $(r, s)$ tensor field is a smooth section of its tensor bundle; each real fiber tensor is a multilinear map on $r$ cotangent and $s$ tangent arguments. Complex values mean a pair of real tensors, with all constructions applied componentwise. A smooth tensor field A type (r,s) tensor on a finite-dimensional vector space

[F4]

For covariant fields, $(f^{*} S)_{q} (v_{1}, \dots, v_{s}) = S_{f (q)} (d f_{q} v_{1}, \dots, d f_{q} v_{s})$ . If $f$ is a diffeomorphism, this pullback has inverse $(f^{- 1})^{*}$ . The pullback of a covariant tensor field A diffeomorphism pulls back tensor fields and forms isomorphically

[F5]

For a diffeomorphism $f$ the differential $A = d f_{q}$ is a linear isomorphism. The mixed-tensor pullback convention used here is $(f^{*} S)_{q} (α_{1}, \dots, α_{r}, v_{1}, \dots, v_{s}) = S_{f (q)} (α_{1} \circ A^{- 1}, \dots, α_{r} \circ A^{- 1}, A v_{1}, \dots, A v_{s})$ , where $α_{i} \in T_{q}^{*} E$ and $v_{j} \in T_{q} E$ . This defines a multilinear tensor and agrees with the covariant construction when $r = 0$ . The differential of a diffeomorphism is an isomorphism A type (r,s) tensor on a finite-dimensional vector space

[F6]

The Mars–Simon tensor formula defines a smooth complex covariant four-tensor on the open set $N = {1 - σ \neq 0}$ ; its restriction to all of $E$ therefore requires $E \subseteq N$ . Ernst sublevels and the Mars–Simon tensor interface

Proof

Given: The flow-invariant real or complex type- $(r, s)$ tensor on $E$ and its vanishing on $Σ_{0} \cap E$ .

Goal: Show $S_{p} = 0$ for every $p \in E$ .

1.1givenF2F5

Each $Φ_{t} : E \to E$ is a diffeomorphism with inverse $Φ_{- t}$ . In particular $d Φ_{t} ∣_{q} : T_{q} E \to T_{Φ_{t} (q)} E$ is a linear isomorphism at every $q \in E$ .

1.2givenF1choose

Fix $p \in E$ . By the orbit-section assumption, choose $q \in Σ_{0} \cap E$ and $t \in R$ such that $p = Φ_{t} (q)$ . Invariance and section vanishing give $(Φ_{t}^{*} S)_{q} = S_{q} = 0$ .

2.1F3F4F5step 1.1step 1.2algebra∎

Set $A = d Φ_{t} ∣_{q}$ . For arbitrary $β_{i} \in T_{p}^{*} E$ and $w_{j} \in T_{p} E$ , insert $α_{i} = β_{i} \circ A$ and $v_{j} = A^{- 1} w_{j}$ into the mixed-pullback formula. It gives $S_{p} (β_{1}, \dots, β_{r}, w_{1}, \dots, w_{s}) = (Φ_{t}^{*} S)_{q} (β_{1} \circ A, \dots, β_{r} \circ A, A^{- 1} w_{1}, \dots, A^{- 1} w_{s}) = 0$ . Thus every evaluation is zero, so $S_{p} = 0$ . This also covers $r = 0$ with the covariant pullback and its inverse, and $r = s = 0$ with empty argument lists. Applying the same argument to the real and imaginary parts proves the complex case. Since $p$ was arbitrary, $S = 0$ on $E$ .

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