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The PEC full-kernel gauge is infinite-dimensional

Statement

Let Ω⊂R3 be any nonempty open set, let V=H0(curl⁡;Ω), let K={u∈V:curl⁡u=0}, and put X=V∩K⊥ using the L2 inner product. Then X is infinite-dimensional.

Facts & Assumptions

Given: a nonempty open Ω⊂R3, the distributional curl convention, and a ball Br(x0)⊂Ω.

[F1]

A nonempty open subset of Euclidean space contains a nondegenerate ball.

[F2]

For every pair 0<ρ<R there is a smooth bump equal to one on the smaller closed ball and supported in the larger ball. (A smooth bump between concentric Euclidean balls)

[F3]

For k∈H(curl⁡), ∫Ωk⋅curl⁡A dx=∫Ωcurl⁡k⋅A dx for every A∈Cc∞(Ω;R3), by the distributional definition. (The real H(curl) graph space on a polyhedron)

Proof

Given: the nonempty open set, a ball Br(x0)⊂Ω, and the full-kernel gauge X.

1.1F1F2givenchooseconstruct

For j≥1, set cj=x0+r(1−2−j)e1 and rj=r2−j−4. The balls Bj=Brj(cj) lie in Br(x0): ∣cj−x0∣+rj=r(1−15⋅2−j−4)<r. If l>j, ∣cl−cj∣=r(2−j−2−l)≥r2−j−1, while rj+rl≤3r2−j−5<r2−j−1, so these balls are pairwise disjoint. By [F2], choose ρ∈Cc∞(B1(0)) equal to one near the origin. Set ψ(ξ)=ξ1ρ(ξ), so ∂1ψ(0)=1, and put ψj(x)=ψ((x−cj)/rj) and Aj=(0,0,ψj). Then vj:=curl⁡Aj=(∂2ψj,−∂1ψj,0) is a nonzero smooth vector field supported in Bj.

2.1F3step 1.1algebra∎

Each vj has compact support in Ω, hence vj∈V. For every k∈K, [F3] gives (vj,k)L2=∫Ωk⋅curl⁡Aj dx=∫Ωcurl⁡k⋅Aj dx=0. Thus vj∈X. Pairwise disjoint supports and vj≠0 imply linear independence: restricting any finite vanishing linear combination to Bj forces its jth coefficient to be zero. The gauge space therefore contains infinitely many linearly independent vectors and is infinite-dimensional.

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