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LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05
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Characteristic hypersurfaces are independent of the defining function

Statement

Let Σ∩U={ϕ=0}={ψ=0} be a C1 hypersurface with dϕ≠0 and dψ≠0 on Σ∩U. If pm is the principal symbol of an order-m scalar operator, then

pm(x,dϕ(x))=0⟺pm(x,dψ(x))=0

for every x∈Σ∩U.

Facts & Assumptions

Given: Two defining functions ϕ,ψ for the same hypersurface Σ, and the principal symbol pm.

[L1]

Characteristic covectors and characteristic hypersurfaces are defined by vanishing of the principal symbol on the conormal (Characteristic covectors, hypersurfaces, and noncharacteristic data).

[L2]

The principal symbol is homogeneous of degree m in the covector variable (Principal part and principal symbol of a scalar PDE).

Proof

technique · direct
1.1given

Fix x∈Σ∩U. Any tangent vector v∈TxΣ is the velocity of a C1 curve in Σ, so dϕ(x)(v)=dψ(x)(v)=0 because both defining functions vanish on Σ; thus dϕ(x) and dψ(x) have the same kernel, namely the tangent hyperplane TxΣ. Since both covectors are nonzero, the annihilator of that hyperplane is one-dimensional, so there is a unique scalar h(x)≠0 with dψ(x)=h(x) dϕ(x).

2.1L1L2step 1.1∎

By [L2], homogeneity gives pm(x,dψ(x))=pm(x,h(x)dϕ(x))=h(x)mpm(x,dϕ(x)) for every x∈Σ∩U; since h(x)≠0, these values vanish together, so [L1] shows that the characteristic property is independent of the chosen defining function.

Depends on

Used by

Cited to discharge well-definedness by Characteristic covectors, hypersurfaces, and noncharacteristic data.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources