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Discrete de Rham complexes and edge-element foundations — Examples

1 · Prerequisites

2 · Summary

The examples compute one oriented Nedelec edge circulation and its face curl flux, exhibit a zero-curl nodal vector mode outside the compatible nodal-gradient image, and give an azimuthal curl-free field with nonzero circulation on a solid torus. The nodal example diagnoses a mismatch in that chosen finite element pair; the handle example shows why a cohomology or harmonic component must be retained on noncontractible domains.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

An oriented Nedelec edge circulation on one tetrahedron

Example

On the reference tetrahedron with vertices x0=0, x1=e1, x2=e2, x3=e3, take the Whitney edge field ϕ01=λ0∇λ1−λ1∇λ0 and orient the face [0,1,2] by the vertex order shown. Compute the circulation on [0,1] and the curl flux through [0,1,2].

Facts & Assumptions

Given: The local edge basis and oriented edge and face moments of Lowest-order Whitney and Nedelec de Rham spaces, their unisolvence from Lowest-order Nedelec edge circulation unisolvence, and Stokes commutation from Whitney edge, face, and cell moments commute with grad, curl, and div.

[F1]

The barycentric edge field is ϕ01=λ0∇λ1−λ1∇λ0, and its edge circulation is the canonical degree of freedom. (Lowest-order Whitney and Nedelec de Rham spaces)

[F2]

The six local edge circulations form the dual basis for the Whitney edge fields. (Lowest-order Nedelec edge circulation unisolvence)

[F3]

On the oriented face [0,1,2], the curl flux equals the signed sum of its three boundary edge circulations. (Whitney edge, face, and cell moments commute with grad, curl, and div)

Verification

1.1F1constructalgebra

In the reference coordinates, λ1=x1, λ0=1−x1−x2−x3, so ϕ01=(1−x2−x3,x1,x1). Along [0,1], parameterized by x(t)=(t,0,0), this field has tangent component 1; therefore its oriented circulation is ∫011 dt=1.

2.1F1F2F3step 1.1algebra∎

Differentiating the displayed polynomial gives curl⁡ϕ01=(0,−2,2). The oriented face [0,1,2] lies in x3=0 with normal e3 and area 1/2, so its curl flux is 2⋅(1/2)=1. Its oriented boundary is [1,2]−[0,2]+[0,1]; by [F2], the first two circulations vanish and the last is one, agreeing with the flux by [F3].

ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-6-luna)Open item page →

A nodal curl-free mode outside the compatible gradient image

Example

On one tetrahedron, compare the naive vector nodal space [P1(K)]3 with the compatible scalar-to-edge gradient map ∇Wh0 of Lowest-order Whitney and Nedelec de Rham spaces. The vector field v(x)=x1e1 lies in the nodal vector space, has zero curl, and is not the gradient of a continuous piecewise-affine scalar. Thus the naive nodal curl-curl matrix has a zero-energy mode outside that compatible gradient image. This identifies a discrete kernel mismatch; it does not claim that gradients are absent from the continuum curl kernel.

Facts & Assumptions

Given: Countable Choice for the real L2 gauge convention; a reference tetrahedron, the scalar nodal space and compatible edge space of Lowest-order Whitney and Nedelec de Rham spaces, and the gauge formulation of Kernel quotient and gauge for discrete curl-curl.

[A1]

Countable Choice is assumed by the real L2 product and the stable gauge supplier; no full Axiom of Choice is used. (The Axiom of Countable Choice (ACω))

[F1]

The scalar nodal space consists of affine functions on the tetrahedron, so every member has constant gradient. (Lowest-order Whitney and Nedelec de Rham spaces)

[F2]

The pure curl-curl energy vanishes on its curl kernel, and the mixed gauge formulation fixes a representative by orthogonality to that kernel. (Kernel quotient and gauge for discrete curl-curl, Stable mixed curl-curl formulation with kernel gauge)

Verification

1.1A1givenF2algebra

The field v(x)=(x1,0,0) is vector-valued affine, so it belongs to [P1(K)]3. Its componentwise curl is zero because its only nonzero component depends only on x1. Hence (curl⁡v,curl⁡v)L2(K)=0.

2.1A1F1F2step 1.1algebra∎

By [F1], every qh∈Wh0=P1(K) has constant gradient, while v is nonconstant on the interior of K. Thus v∉∇Wh0. The kernel of the naive nodal curl-curl energy is therefore strictly larger than the compatible discrete gradient image, and its representative is not unique without a gauge as in [F2].

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A handle-domain harmonic field with nonzero circulation

Example

Let Ω={(x,y,z):(x2+y2−R)2+z2<r2} be a solid torus with R>r>0. Define u(x,y,z)=(−yx2+y2,xx2+y2,0). This is a smooth curl-free and divergence-free field on Ω‾, tangent to the boundary, whose circulation around the core circle is 2π. Here “harmonic” means curl and divergence both vanish with zero normal boundary component. It is a concrete nonzero-period field caused by the handle; the nonzero period proves it is not the gradient of a single-valued C1 potential. It is the continuous model for the nontrivial cohomology components described in Discrete de Rham cohomology and contractible-mesh exactness.

Facts & Assumptions

Given: Countable Choice for the graph-space and real L2 conventions; R>r>0, the H(curl) convention of The real H(curl) graph space on a polyhedron, the topology-dependent cohomology convention of Discrete de Rham cohomology and contractible-mesh exactness, and the core loop γ(θ)=(Rcos⁡θ,Rsin⁡θ,0), 0≤θ≤2π.

[A1]

Countable Choice is assumed by the cited H(curl) and cohomology interfaces; the explicit field and circulation computation use no stronger choice axiom. (The Axiom of Countable Choice (ACω))

[F1]

H(curl) consists of square-integrable vector fields with square-integrable distributional curl. (The real H(curl) graph space on a polyhedron)

[F2]

The classical coordinate divergence and curl are given by the usual partial-derivative formulas for a C1 vector field. (Divergence and curl of a C1 vector field)

[F3]

On a noncontractible realization, curl-free edge fields modulo gradients have the first cohomology dimension and must be retained as a harmonic component. (Discrete de Rham cohomology and contractible-mesh exactness)

Verification

1.1A1F1F2givenalgebra

On Ω‾, x2+y2≥R−r>0, so every component of u and its first derivatives are bounded; hence u∈L2(Ω;R3) and its classical curl is also in L2. Direct differentiation by [F2] gives div⁡u=2xy/(x2+y2)2−2xy/(x2+y2)2=0 and curl⁡u=0. Thus [F1] places u in H(curl) with zero curl.

2.1A1F2F3F4givenalgebra∎

Write eθ=(−sin⁡θ,cos⁡θ,0). On the torus boundary the normal is a linear combination of the meridional radial vector (cos⁡θ,sin⁡θ,0) and e3, whereas u=eθ/x2+y2; hence u⋅n=0. Along the core curve, u(γ(θ))=eθ/R and γ′(θ)=Reθ, so ∫γu⋅dℓ=∫02π1 dθ=2π. A single-valued C1 potential would have zero integral around every closed curve by [F4], so this field is not such a gradient and represents a nonzero period. This is the type of harmonic component retained by [F3].

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