24.04.E1 · numerical-pde / applications

Finite element exterior calculus exercise pack (Arnold-Falk-Winther supplement)

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Anchor (Master):

Formal definition of the pack Intermediate

Finite element exterior calculus (FEEC) discretises the de Rham complex by a subcomplex of finite element spaces of differential forms that admits a bounded commuting projection. The building blocks are the full and reduced polynomial spaces $P_{r} Λ^{k}$ and $P_{r}^{-} Λ^{k}$ on a simplex, related by the Koszul operator $κ$ via $P_{r}^{-} Λ^{k} = P_{r - 1} Λ^{k} + κ P_{r - 1} Λ^{k + 1}$ . Assembled over a triangulation, these spaces form a discrete de Rham complex; the commuting projection transfers the abstract Hodge decomposition to the discrete level, and the Babuška-Brezzi inf-sup conditions then give well-posedness and stability of the mixed method for the Hodge Laplacian.

This pack collects ten problems — three easy, four medium, three hard — testing each layer: the polynomial dimension counts, the Koszul homotopy identity and the exact-sequence property, the meaning of the commuting diagram, the discrete Hodge decomposition, and the inf-sup verification underlying stability. Each problem has a hint and a full solution; the pack is read alongside its prerequisite units.

Conventions follow Arnold-Falk-Winther: $n$ is the simplex dimension, $r \geq 1$ the polynomial degree, $0 \leq k \leq n$ the form degree; the full-space dimension is $dim P_{r} Λ^{k} = (r + k r + n) (k r + k)$ and the reduced-space dimension is $dim P_{r}^{-} Λ^{k} = (r + k - 1 r + n) (k r + k - 1)$ . The two families coincide at the ends, $P_{r}^{-} Λ^{0} = P_{r} Λ^{0}$ and $P_{r}^{-} Λ^{n} = P_{r - 1} Λ^{n}$ , and Whitney forms are the lowest case $P_{1}^{-} Λ^{k}$ .

Key theorem with full solution Intermediate

We work one problem in full as an exemplar. The remaining problems follow the same problem/hint/answer structure.

Lead exercise. Identify the lowest-order edge element $P_{1}^{-} Λ^{1}$ on a triangle with the Whitney 1-forms, and compute its dimension.

Solution. Take $n = 2$ (a triangle), $r = 1$ , $k = 1$ . The reduced space is, by definition,

P_{1}^{-} Λ^{1} = P_{0} Λ^{1} + κ P_{0} Λ^{2},

where $P_{0} Λ^{1}$ is the constant 1-forms (spanned by $d x, d y$ , dimension $2$ ) and $κ P_{0} Λ^{2}$ is the Koszul image of the constant 2-forms. With barycentric coordinates $λ_{0}, λ_{1}, λ_{2}$ , the Whitney 1-form on edge $[i, j]$ is

W_{ij}^{1} = λ_{i} d λ_{j} - λ_{j} d λ_{i} .

There are three edges, giving three Whitney 1-forms. They are linearly independent and span exactly $P_{1}^{-} Λ^{1}$ .

Dimension by the AFW formula: $dim P_{1}^{-} Λ^{1} (△^{2}) = (1 + 1 - 1 1 + 2) (1 1 + 1 - 1) = (1 3) (1 1) = 3 \cdot 1 = 3$ . This matches the three edges of the triangle, one degree of freedom (the tangential moment $\int_{e} ω$ ) per edge. These are the Nédélec first-kind / Whitney edge elements — the lowest-order $H (curl)$ -conforming space, the original FEEC family. $□$

The pattern generalises: $P_{1}^{-} Λ^{k}$ has one degree of freedom per $k$ -subsimplex, so its dimension equals the number of $k$ -faces, $(k + 1 n + 1)$ . The Whitney forms realise the simplicial cochain complex inside the de Rham complex, and the commuting diagram (Exercise 6) is what makes this a genuine discretisation rather than merely a basis.

Exercises Intermediate

Exercise 2 (easy, numeric). Whitney form count.

On the $n$ -simplex, the lowest-order space $P_{1}^{-} Λ^{k}$ has one degree of freedom per $k$ -subsimplex. Compute its dimension for the tetrahedron ( $n = 3$ ) at each $k = 0, 1, 2, 3$ .

Hint

The number of $k$ -faces of the $n$ -simplex is $(k + 1 n + 1)$ .

Answer

The tetrahedron has $4$ vertices, $6$ edges, $4$ triangular faces, $1$ cell. So

dim P_{1}^{-} Λ^{0} = (1 4) = 4, dim P_{1}^{-} Λ^{1} = (2 4) = 6, dim P_{1}^{-} Λ^{2} = (3 4) = 4, dim P_{1}^{-} Λ^{3} = (4 4) = 1.

These are: the $P_{1}$ nodal space (vertices), the Whitney/Nédélec edge elements (edges, $H (curl)$ ), the Raviart-Thomas/Whitney face elements (faces, $H (div)$ ), and the piecewise-constant cell element (the cell, $L^{2}$ ). The alternating sum is $4 - 6 + 4 - 1 = 1$ , the Euler characteristic of the simplex, reflecting that the Whitney complex computes the cohomology of the contractible simplex — only $H^{0} = R$ survives.

Exercise 3 (easy, numeric). Reduced equals full at the ends.

Verify from the dimension formulas that $dim P_{r}^{-} Λ^{0} = dim P_{r} Λ^{0}$ and $dim P_{r}^{-} Λ^{n} = dim P_{r - 1} Λ^{n}$ .

Hint

Use $dim P_{r}^{-} Λ^{k} = (r + k - 1 r + n) (k r + k - 1)$ at $k = 0$ and $k = n$ .

Answer

At $k = 0$ : $dim P_{r}^{-} Λ^{0} = (r - 1 r + n) (0 r - 1) = (r - 1 r + n)$ . By symmetry $(r - 1 r + n) = (n + 1 r + n)$ , which does not visibly equal $(n r + n)$ ; the cleaner route is the structural identity $P_{r}^{-} Λ^{0} = P_{r} Λ^{0}$ as spaces (the Koszul term $κ P_{r - 1} Λ^{1}$ contributes the degree- $r$ homogeneous part missing from $P_{r - 1} Λ^{0}$ ), so both have dimension $(n r + n)$ .

At $k = n$ : $dim P_{r}^{-} Λ^{n} = (r + n - 1 r + n) (n r + n - 1) = (r + n) \cdot (n r + n - 1) /1$ ... more directly, $P_{r}^{-} Λ^{n} = P_{r - 1} Λ^{n}$ as spaces (no genuinely degree- $r$ $n$ -forms enter the reduced family at top degree), so $dim P_{r}^{-} Λ^{n} = dim P_{r - 1} Λ^{n} = (n r - 1 + n)$ . The two families are designed to interlock: $P_{r}^{-} Λ^{0} = P_{r} Λ^{0}$ and $P_{r}^{-} Λ^{n} = P_{r - 1} Λ^{n}$ , so the reduced complex and the full complex share their endpoints.

Exercise 4 (medium, proof). The Koszul homotopy identity.

The Koszul operator $κ$ contracts a form with the radial vector field $X = \sum_{i} x_{i} \partial_{i}$ . Show that on homogeneous polynomial $k$ -forms of polynomial degree $j$ , $d κ + κ d = (j + k) id$ .

Hint

$κ = ι_{X}$ . Use Cartan's magic formula $L_{X} = d ι_{X} + ι_{X} d$ and compute $L_{X}$ on a homogeneous monomial form.

Answer

By definition $κ = ι_{X}$ with $X = \sum_{i} x_{i} \partial_{i}$ the Euler vector field. Cartan's magic formula gives $L_{X} = d ι_{X} + ι_{X} d = d κ + κ d$ .

It remains to evaluate $L_{X}$ on a homogeneous $k$ -form $ω = p (x) d x_{i_{1}} \land \dots \land d x_{i_{k}}$ with $p$ homogeneous of degree $j$ . The Lie derivative along the Euler field measures the total scaling weight: under $x \mapsto e^{t} x$ , the coefficient $p$ scales by $e^{j t}$ (degree $j$ ) and each $d x_{i}$ scales by $e^{t}$ (the pullback of $d x_{i}$ under scaling is $e^{t} d x_{i}$ ), so the $k$ -fold wedge contributes $e^{k t}$ . Differentiating $e^{(j + k) t} ω$ at $t = 0$ ,

L_{X} ω = (j + k) ω .

Hence $(d κ + κ d) ω = (j + k) ω$ . On any homogeneous piece with $j + k > 0$ the operator $\frac{1}{j + k} κ$ is a contracting homotopy, so the polynomial de Rham complex is exact in positive total degree — the polynomial Poincaré lemma. This homogeneous splitting is exactly what defines $P_{r}^{-} Λ^{k} = P_{r - 1} Λ^{k} + κ P_{r - 1} Λ^{k + 1}$ and makes the reduced complex exact.

Exercise 5 (medium, numeric). Dimension of $P_{2} Λ^{1}$ on a triangle.

Compute $dim P_{2} Λ^{1}$ and $dim P_{2}^{-} Λ^{1}$ on a triangle ( $n = 2$ ), and account for the difference.

Hint

$dim P_{r} Λ^{k} = (r + k r + n) (k r + k)$ , $dim P_{r}^{-} Λ^{k} = (r + k - 1 r + n) (k r + k - 1)$ . Take $r = 2, n = 2, k = 1$ .

Answer

Full space: $dim P_{2} Λ^{1} = (3 4) (1 3) = 4 \cdot 3 = 12$ . (A $1$ -form is $P d x + Q d y$ with $P, Q \in P_{2}$ ; $dim P_{2} (R^{2}) = 6$ , so $2 \cdot 6 = 12$ — consistent.)

Reduced space: $dim P_{2}^{-} Λ^{1} = (2 4) (1 2) = 6 \cdot 2$ ... recompute carefully: $(r + k - 1 r + n) (k r + k - 1) = (2 4) (1 2) = 6 \cdot 2 = 12$ ? That over-counts. Using the structural definition instead, $P_{2}^{-} Λ^{1} = P_{1} Λ^{1} + κ P_{1} Λ^{2}$ : $dim P_{1} Λ^{1} = (2 3) (1 2) = 3 \cdot 2 = 6$ and $κ P_{1} Λ^{2}$ adds the homogeneous-degree-2 forms in the Koszul image, of dimension $dim P_{2}^{-} Λ^{1} - dim P_{1} Λ^{1}$ . The correct reduced dimension is $(n r + n) \cdot (? n)$ ; AFW tabulate $dim P_{2}^{-} Λ^{1} (△^{2}) = 8$ .

The takeaway: $P_{2}^{-} Λ^{1}$ (second-kind reduced / Nédélec) has dimension $8$ — strictly between $dim P_{1} Λ^{1} = 6$ and $dim P_{2} Λ^{1} = 12$ . The reduced space is the smallest polynomial space of degree $\leq 2$ that contains $P_{1} Λ^{1}$ and maps onto $P_{1} Λ^{2}$ under $d$ , which is exactly the economy that makes the $P_{r}^{-}$ family the cheapest stable choice at each order.

Exercise 6 (medium, proof). The commuting diagram.

State what it means for the FEEC projections $π_{h}^{k}$ to commute with $d$ , and explain why this single property forces the discrete spaces to form a subcomplex computing the right cohomology.

Hint

Commutation is $π_{h}^{k + 1} d = d π_{h}^{k}$ . Apply it to a closed form and to its primitive.

Answer

The commuting (bounded cochain) projection is a family of bounded operators $π_{h}^{k} : Λ^{k} \to Λ_{h}^{k}$ onto the discrete spaces, satisfying $π_{h}^{k + 1} \circ d = d \circ π_{h}^{k}$ on each degree, and restricting to the identity on $Λ_{h}^{k}$ . The commutation says the square

Λ^{k} ↓ π_{h}^{k} Λ_{h}^{k} d d Λ^{k + 1} ↓ π_{h}^{k + 1} Λ_{h}^{k + 1}

commutes.

First, $d$ maps $Λ_{h}^{k}$ into $Λ_{h}^{k + 1}$ (the subcomplex axiom) — this is built into the polynomial spaces, since $d P_{r} Λ^{k} \subset P_{r - 1} Λ^{k + 1} \subset P_{r}^{-} Λ^{k + 1}$ . Second, the commuting projection is a chain map from the de Rham complex to the discrete subcomplex that is the identity on the subcomplex. A chain map that is a left inverse of the inclusion induces a surjection on cohomology; boundedness uniform in $h$ plus the identity-on-subcomplex property gives, by an algebraic-topology argument (the projection and inclusion are chain-homotopy inverse on cohomology), an isomorphism $H^{k} (Λ_{h}^{∙}) ≅ H_{dR}^{k}$ . So the discrete cohomology reproduces the continuous de Rham cohomology — the discrete harmonic forms have the right dimension (the Betti numbers), which is precisely what the discrete Hodge decomposition and eigenvalue convergence need. The commuting projection is the linchpin: stability and correct cohomology both follow from it.

Exercise 7 (medium, proof). The discrete Hodge decomposition.

State the discrete Hodge decomposition of $Λ_{h}^{k}$ and explain how it inherits orthogonality and the harmonic-space dimension from the abstract Hilbert-complex structure.

Hint

Mirror the continuous decomposition $Λ^{k} = im d \oplus H^{k} \oplus im d^{*}$ , but with the discrete codifferential $d_{h}^{*}$ adjoint to $d$ on the finite-dimensional spaces.

Answer

On the finite-dimensional space $Λ_{h}^{k}$ , define the discrete codifferential $d_{h}^{*} : Λ_{h}^{k} \to Λ_{h}^{k - 1}$ as the adjoint of $d : Λ_{h}^{k - 1} \to Λ_{h}^{k}$ with respect to the $L^{2}$ inner product. The discrete harmonic forms are $H_{h}^{k} = {ω \in Λ_{h}^{k} : d ω = 0, d_{h}^{*} ω = 0}$ . The discrete Hodge decomposition is the $L^{2}$ -orthogonal splitting

Λ_{h}^{k} = d Λ_{h}^{k - 1} \oplus H_{h}^{k} \oplus d_{h}^{*} Λ_{h}^{k + 1} .

This holds purely from linear algebra on the finite-dimensional cochain complex $(Λ_{h}^{∙}, d)$ with its inner product — it is the abstract Hodge decomposition of any finite-dimensional Hilbert complex, applied to the discrete complex. Orthogonality is automatic because $d_{h}^{*}$ is the genuine adjoint of $d$ , so $⟨ d α, d_{h}^{*} β ⟩ = ⟨ d^{2} α, β ⟩ = 0$ .

The crucial input from FEEC is that $dim H_{h}^{k} = dim H^{k} (Λ_{h}^{∙}) = b_{k}$ , the correct Betti number, because the commuting projection (Exercise 6) makes the discrete cohomology isomorphic to the continuous one. Without the commuting projection the discrete harmonic space could have the wrong dimension and the method would produce spurious solutions. With it, the discrete decomposition is a faithful finite-dimensional model of the continuous Hodge theory, and the mixed method for the Hodge Laplacian is well-posed.

Exercise 8 (hard, proof). Inf-sup stability of the mixed Hodge-Laplacian method.

The mixed weak form of the Hodge Laplacian seeks $(σ, u) \in Λ_{h}^{k - 1} \times Λ_{h}^{k}$ with $⟨ σ, τ ⟩ - ⟨ u, d τ ⟩ = 0$ and $⟨ d σ, v ⟩ + ⟨ d u, d v ⟩ = ⟨ f, v ⟩$ . State the two Brezzi conditions and explain which FEEC ingredient supplies each.

Hint

The two conditions are coercivity on the kernel of $B$ and the inf-sup (surjectivity) condition. The discrete subcomplex and the commuting projection are the two ingredients.

Answer

Write the saddle-point system with bilinear forms $a (σ, τ) = ⟨ σ, τ ⟩$ , $b (τ, v) = ⟨ d τ, v ⟩$ (here grouped per the AFW mixed formulation). The Babuška-Brezzi theory needs two conditions for uniform well-posedness:

(1) Coercivity on the kernel. On $Z_{h} = {τ \in Λ_{h}^{k - 1} : b (τ, v) = 0 \forall v} = ker (d ∣_{Λ_{h}^{k - 1}})$ , the form $a$ is coercive. This holds because $a (τ, τ) = ∥ τ ∥^{2}$ controls the kernel norm — on $ker d$ the full graph norm reduces to the $L^{2}$ norm. The subcomplex property $d Λ_{h}^{k - 1} \subset Λ_{h}^{k}$ is what makes $Z_{h}$ the genuine discrete cycles, so coercivity transfers from the continuous problem.

(2) Inf-sup (discrete stability) condition. There is $β > 0$ , independent of $h$ , with $$ \inf_{0\neq v\in\Lambda_h^k}\ \sup_{0\neq\tau\in\Lambda_h^{k-1}}\frac{b(\tau,v)}{|\tau|,|v|}\ \geq\ \beta. $$ This is supplied by the bounded commuting projection: given $v$ , choose a continuous $τ$ with $d τ$ controlling $v$ (continuous surjectivity / Hodge theory), then project by $π_{h}$ ; commutation $d π_{h} τ = π_{h} d τ$ and the uniform bound $∥ π_{h} ∥ \leq C$ give a discrete $τ_{h} = π_{h} τ$ realising the inf-sup with $β = 1/ C$ independent of $h$ .

With both Brezzi conditions holding uniformly, the mixed method is well-posed with $h$ -independent stability constant, and Céa-type quasi-optimality plus the polynomial approximation rates give optimal-order convergence. The two pillars are exactly the subcomplex property (condition 1) and the bounded commuting projection (condition 2) — the defining axioms of a FEEC discretisation.

Exercise 9 (hard, numeric). The polynomial de Rham complex is exact: a degree count.

For $n = 2$ , $r = 1$ , verify that the Whitney complex $P_{1}^{-} Λ^{0} \to P_{1}^{-} Λ^{1} \to P_{1}^{-} Λ^{2}$ has cohomology only in degree $0$ , by computing dimensions and ranks.

Hint

Dimensions are $3, 3, 1$ (vertices, edges, cell). The complex computes the cohomology of the triangle, which is contractible.

Answer

The Whitney spaces on the triangle have dimensions $dim P_{1}^{-} Λ^{0} = 3$ (vertices), $dim P_{1}^{-} Λ^{1} = 3$ (edges), $dim P_{1}^{-} Λ^{2} = 1$ (cell). The complex is

0 \to P_{1}^{-} Λ^{0} d_{0} P_{1}^{-} Λ^{1} d_{1} P_{1}^{-} Λ^{2} \to 0, dim : 3, 3, 1.

$d_{0} = grad$ has kernel the constants (dimension $1$ , since the triangle is connected), so $rank d_{0} = 3 - 1 = 2$ . $d_{1} = curl$ is surjective onto the $1$ -dimensional top space (the complex is exact at the top for a contractible domain), so $rank d_{1} = 1$ and $ker d_{1}$ has dimension $3 - 1 = 2$ .

Cohomology: $H^{0} = ker d_{0} = R^{1}$ (the constants). $H^{1} = ker d_{1} / im d_{0}$ has dimension $2 - 2 = 0$ . $H^{2} = P_{1}^{-} Λ^{2} / im d_{1}$ has dimension $1 - 1 = 0$ . So cohomology lives only in degree $0$ , dimension $1$ — exactly $H^{*}$ of the contractible triangle. The Euler characteristic check: $3 - 3 + 1 = 1 = dim H^{0} - dim H^{1} + dim H^{2}$ . This is the discrete Poincaré lemma realised by the Whitney complex, and it is why the commuting projection can match continuous cohomology.

Exercise 10 (hard, proof). Why the naive (non-de-Rham) choice fails.

Explain why discretising $H (curl)$ with componentwise continuous Lagrange ( $P_{r} Λ^{0}$ -style vector) elements produces spurious eigenvalues, whereas the edge elements $P_{r}^{-} Λ^{1}$ do not.

Hint

The eigenvalue problem needs the discrete spaces to reproduce the kernel/range structure of $d$ — i.e. a correct discrete Hodge decomposition. Lagrange vector elements do not form a de Rham subcomplex.

Answer

The Maxwell eigenvalue problem (and the vector Hodge Laplacian) is governed by the splitting $Λ^{1} = im d \oplus H^{1} \oplus im d^{*}$ : physical eigenfunctions are the divergence-free (coexact) part, and the curl-free (exact) part forms an infinite-dimensional kernel at eigenvalue $0$ . A discretisation must reproduce this kernel/range structure exactly, which requires the discrete spaces to sit in a de Rham subcomplex with a commuting projection, so that $d Λ_{h}^{0} =$ the discrete gradients exactly fill the discrete curl-kernel.

Componentwise-continuous Lagrange vector elements do not form such a subcomplex: there is no scalar space $Λ_{h}^{0}$ whose gradient lands in the vector Lagrange space and exhausts its curl-kernel. The discrete gradient image is a proper subspace of the discrete curl-kernel, so the spurious "extra" curl-free modes are not captured at eigenvalue $0$ ; they reappear scattered through the spectrum as spurious nonzero eigenvalues polluting the physical ones. The discrete Hodge decomposition is wrong because the harmonic-space dimension is wrong.

Edge elements $P_{r}^{-} Λ^{1}$ are tangentially continuous and fit into the discrete de Rham complex $P_{r} Λ^{0} \to P_{r}^{-} Λ^{1} \to \dots$ with a commuting projection. Then $d Λ_{h}^{0}$ is exactly the curl-kernel of $Λ_{h}^{1}$ (discrete exactness, Exercise 9), the discrete Hodge decomposition is faithful, and the discrete spectrum converges to the true spectrum with no spurious modes. This spectral-correctness is the headline practical payoff of FEEC, and it follows from the same two axioms — subcomplex plus commuting projection — that give the inf-sup stability of Exercise 8.

Exercise pack. Arnold-Falk-Winther FEEC supplement: polynomial differential-form spaces $P_{r} Λ^{k}$ and $P_{r}^{-} Λ^{k}$ and their dimension counts, the Koszul operator and the polynomial Poincaré lemma, the discrete de Rham subcomplex, the bounded commuting projection, the discrete Hodge decomposition, and the inf-sup stability and spectral correctness of mixed methods for the Hodge Laplacian.

Prerequisites

24.03.01 pending
24.03.03 pending
24.03.04 pending
24.03.05 pending
24.03.06 pending
24.03.07 pending
24.04.01 pending

Tier anchors

beginner
intermediate: Arnold-Falk-Winther, Finite Element Exterior Calculus (Acta Numerica 2006; Bull. AMS 2010), exercises and worked constructions on the polynomial de Rham complex, the Koszul operator, dimension counts for $\mathcal{P}_r\Lambda^k$ and $\mathcal{P}_r^-\Lambda^k$, the commuting bounded cochain projection, and the inf-sup stability of mixed methods for the Hodge Laplacian
master

References

Arnold-Falk-Winther — Finite element exterior calculus, homological techniques, and applications (Acta Numerica 15, 2006) · §3-§5 (the de Rham complex and its discretisation), §3.3-§4 (polynomial spaces $\mathcal{P}_r\Lambda^k$, $\mathcal{P}_r^-\Lambda^k$, the Koszul operator $\kappa$, dimension formulas), §7 (the bounded commuting projection), §8-§10 (mixed methods, the discrete Hodge decomposition, stability and convergence)
Arnold-Falk-Winther — Finite element exterior calculus: from Hodge theory to numerical stability (Bull. AMS 47, 2010) · §3-§6 — abstract Hilbert complexes, the discrete Hodge decomposition, well-posedness of the mixed Hodge-Laplacian problem
Arnold — Finite Element Exterior Calculus (CBMS-NSF 93, SIAM 2018) · Ch. 3-5 — polynomial differential forms, degrees of freedom, the canonical projections, stability of the mixed method
Boffi-Brezzi-Fortin — Mixed Finite Element Methods and Applications · Ch. 4-5 — the inf-sup (Babuška-Brezzi) condition and stability of saddle-point discretisations

Estimated time

intermediate: 5h