02.13.E1 · analysis / pde

Integration and currents exercise pack (Whitney Ch. I, IX, XI supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

Whitney's geometric integration theory rebuilds the calculus of chains and forms so that Stokes' theorem survives for non-smooth objects. The starting algebra is the exterior power: $r$ -vectors in $Λ_{r} R^{n}$ and $r$ -covectors in $Λ^{r} R^{n}$ paired by the wedge product and the duality $⟨ α, ξ ⟩$ . A current is a continuous linear functional $T$ on the space of compactly supported smooth $r$ -forms; its boundary $\partial T$ is defined by $\partial T (ω) = T (d ω)$ , dual to the exterior derivative, so that Stokes' theorem $T (d ω) = \partial T (ω)$ holds by definition. The size of a current is measured by its mass $M$ and, for convergence and approximation, by the flat norm $F$ (Whitney's $♭$ ) and the sharp norm $♯$ ; flat chains are the completion of polyhedral chains in the flat norm. The Whitney extension theorem supplies the smooth functions whose level sets and graphs feed the construction.

This pack collects nine exercises — two easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units. The exercises group loosely by topic: exterior-algebra and wedge-product computations (easy/medium); mass, flat and sharp norms and the boundary operator (medium); and Stokes for currents, the constancy theorem, and the Whitney extension theorem (hard).

Conventions follow Whitney and Federer: $D^{r}$ denotes compactly supported smooth $r$ -forms and $D_{r}$ its dual, the $r$ -currents; mass is the comass-dual norm $M (T) = sup {T (ω) : ∥ ω ∥_{comass} \leq 1}$ ; the boundary lowers dimension by one, $\partial : D_{r} \to D_{r - 1}$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Define the current $[[M]]$ associated with a compact oriented smooth $m$ -submanifold $M \subset R^{n}$ with boundary, and prove that $\partial [[M]] = [[\partial M]]$ .

Solution. A compact oriented $m$ -dimensional submanifold $M$ defines an $m$ -current $[[M]] \in D_{m} (R^{n})$ by integration of forms: $$ \llbracket M \rrbracket(\omega) = \int_M \omega, \qquad \omega \in \mathcal{D}^m(\mathbb{R}^n). $$ This is linear in $ω$ and continuous (bounded by $M ([[M]]) = H^{m} (M) < \infty$ times the comass of $ω$ ), so it is a genuine current. Its mass is the $m$ -area of $M$ .

The boundary of a current is defined by duality with the exterior derivative: $\partial T (η) = T (d η)$ for $η \in D^{m - 1}$ . Applying this to $T = [[M]]$ : $$ \partial\llbracket M \rrbracket(\eta) = \llbracket M \rrbracket(d\eta) = \int_M d\eta. $$ By the classical Stokes theorem for the smooth oriented manifold-with-boundary $M$ , $$ \int_M d\eta = \int_{\partial M} \eta = \llbracket \partial M \rrbracket(\eta), $$ where $\partial M$ carries the boundary orientation. Since this holds for every test form $η \in D^{m - 1}$ , the currents agree: $\partial [[M]] = [[\partial M]]$ . $□$

This is the design principle of currents: the boundary operator is built as the adjoint of $d$ precisely so that Stokes' theorem becomes the definition rather than a theorem to be reproved. Two immediate consequences: $\partial^{2} = 0$ (dual to $d^{2} = 0$ ), and the mass of $\partial [[M]]$ equals $H^{m - 1} (\partial M)$ , the area of the boundary.

Exercises Intermediate

Exercise 2 (easy, symbolic). Decomposability via the wedge.

Show that a $2$ -covector $ω \in Λ^{2} R^{4}$ is decomposable (a single wedge $α \land β$ ) iff $ω \land ω = 0$ .

Hint

Write $ω = \sum_{i < j} c_{ij} d x_{i} \land d x_{j}$ and compute $ω \land ω$ in $Λ^{4} R^{4} ≅ R$ ; it is a Pfaffian.

Answer

In $R^{4}$ , $ω = c_{12} d x_{1} d x_{2} + c_{13} d x_{1} d x_{3} + c_{14} d x_{1} d x_{4} + c_{23} d x_{2} d x_{3} + c_{24} d x_{2} d x_{4} + c_{34} d x_{3} d x_{4}$ (wedges suppressed). Then $$ \omega\wedge\omega = 2(c_{12}c_{34} - c_{13}c_{24} + c_{14}c_{23}),dx_1 dx_2 dx_3 dx_4. $$ The scalar $Pf (ω) = c_{12} c_{34} - c_{13} c_{24} + c_{14} c_{23}$ is the Pfaffian of the skew coefficient matrix.

If $ω = α \land β$ is decomposable, then $ω \land ω = α \land β \land α \land β = 0$ since $α$ repeats. Conversely, if $ω \land ω = 0$ then $Pf (ω) = 0$ , which is exactly the condition for the rank- $2$ skew matrix $(c_{ij})$ to have rank $2$ (not $4$ ); a rank- $2$ skew form is $α \land β$ for a suitable pair, so $ω$ is decomposable. Hence decomposability $⟺ ω \land ω = 0$ in $Λ^{2} R^{4}$ .

Exercise 3 (medium, symbolic). Comass of a simple covector.

Compute the comass $∥ ω ∥^{*}$ of the $2$ -covector $ω = d x_{1} \land d x_{2}$ on $R^{4}$ , defined as the supremum of $⟨ ω, ξ ⟩$ over unit simple $2$ -vectors $ξ$ .

Hint

A unit simple $2$ -vector is $ξ = u \land v$ with $u, v$ orthonormal. $⟨ d x_{1} \land d x_{2}, u \land v ⟩$ is a $2 \times 2$ minor; bound it by the orthonormality.

Answer

For orthonormal $u, v \in R^{4}$ , the pairing is the determinant of the projection onto the $(x_{1}, x_{2})$ -plane: $$ \langle dx_1\wedge dx_2,\ u\wedge v\rangle = \det\begin{pmatrix} u_1 & v_1 \ u_2 & v_2 \end{pmatrix} = u_1 v_2 - u_2 v_1. $$ This is the signed area of the parallelogram spanned by the projections of $u$ and $v$ onto the $(x_{1}, x_{2})$ -plane. Since projection does not increase area and $u \land v$ has unit area, $∣ u_{1} v_{2} - u_{2} v_{1} ∣ \leq 1$ , with equality when $u, v$ lie in the $(x_{1}, x_{2})$ -plane (e.g. $u = e_{1}$ , $v = e_{2}$ ). Hence $$ |\omega|^* = \sup_{\xi}\langle\omega,\xi\rangle = 1. $$ The comass of a simple unit covector is $1$ ; this is the dual norm that makes the mass $M (T) = sup {T (ω) : ∥ ω ∥^{*} \leq 1}$ of the current $[[M]]$ equal the area of $M$ .

Exercise 4 (medium, proof). Boundary squares to zero for currents.

Using the duality definition $\partial T (η) = T (d η)$ , prove $\partial (\partial T) = 0$ for every current $T$ .

Hint

Apply the definition twice and use $d \circ d = 0$ on forms.

Answer

Let $T \in D_{r}$ and let $η \in D^{r - 2}$ be any test form. Apply the boundary definition twice: $$ \partial(\partial T)(\eta) = (\partial T)(d\eta) = T\bigl(d(d\eta)\bigr) = T(d^2\eta). $$ On smooth forms $d^{2} = 0$ (the exterior derivative satisfies $d \circ d = 0$ ), so $d^{2} η = 0$ and therefore $T (d^{2} η) = T (0) = 0$ . Since $η \in D^{r - 2}$ was arbitrary, $\partial (\partial T) = 0$ as an element of $D_{r - 2}$ .

The boundary operator on currents is nilpotent because it is the adjoint of the nilpotent exterior derivative. This makes $(D_{*}, \partial)$ a chain complex dual to the de Rham cochain complex $(D^{*}, d)$ , and it is the abstract reason a boundary has no boundary, generalizing $\partial\partial M = \emptyset$ for manifolds.

Exercise 5 (medium, numeric). Flat norm of a small loop versus the disk it bounds.

Let $T = [[\partial D_{ε}]]$ be the boundary $1$ -current of the disk $D_{ε}$ of radius $ε$ in $R^{2}$ . Estimate the flat norm $F (T)$ and explain why it tends to $0$ as $ε \to 0$ , even though the mass $M (T)$ does not.

Hint

The flat norm is $F (T) = in f {M (T - \partial S) + M (S)}$ over $S$ . Take $S = [[D_{ε}]]$ , so $\partial S = T$ and the first term vanishes.

Answer

The flat norm of an $r$ -current is $$ \mathbf{F}(T) = \inf\bigl{ \mathbf{M}(T - \partial S) + \mathbf{M}(S) : S \in \mathcal{D}{r+1}\bigr}. $$ Choose $S = \llbracket D\varepsilon\rrbracket $, t h e d i s k f i l l in g t h e l oo p . T h e n$ \partial S = T $, so$ T - \partial S = 0 $an d$ \mathbf{M}(T - \partial S) = 0 $. T h er e mainin g t er mi s$ \mathbf{M}(S) = \mathcal{H}^2(D_\varepsilon) = \pi\varepsilon^2$. Hence $$ \mathbf{F}(T) \le \pi\varepsilon^2 \xrightarrow[\varepsilon\to 0]{} 0. $$ By contrast the mass is the length of the loop, $M (T) = 2 π ε$ , which also tends to $0$ but only linearly — and a unit-radius loop has $F \leq π$ while $M = 2 π$ . The point of the flat norm is that small boundaries are flat-small because they bound small-area regions: it sees that a tiny loop is close to $0$ in the homological sense. This is why the flat norm, not the mass, is the right topology for the compactness and approximation theorems (a flat-Cauchy sequence of polyhedral chains has a flat-chain limit).

Exercise 6 (medium, proof). Sharp norm bounds the flat norm.

Recall Whitney's sharp norm $♯$ controls how a chain pairs with forms of bounded $C^{1}$ -norm. Show $F (T) \leq M (T)$ and explain the inclusion of norm topologies $M \geq F \geq ♯$ (mass strongest, sharp weakest).

Hint

For $F \leq M$ , take $S = 0$ in the flat-norm infimum.

Answer

$F (T) \leq M (T)$ . In the infimum $F (T) = in f_{S} {M (T - \partial S) + M (S)}$ , take the admissible choice $S = 0$ : this gives $M (T - 0) + M (0) = M (T)$ . The infimum is at most this value, so $F (T) \leq M (T)$ .

Ordering of topologies. Mass measures a chain against all comass- $\leq 1$ forms; the flat norm relaxes this by allowing a piece of the chain to be discarded as the boundary of a small-mass filling, so flat-small chains include all small-area boundaries that mass would still see as large. The sharp norm relaxes further: it tests against forms whose values and whose variation (Lipschitz/ $C^{1}$ constant) are bounded, so it tolerates small geometric perturbations (a chain slightly translated is sharp-close to the original). Thus $$ \mathbf{M}(T) \ge \mathbf{F}(T) \ge c,\sharp(T) $$ on bounded sets, with mass the finest and sharp the coarsest. Convergence in mass implies convergence in flat norm implies convergence in sharp norm, but none of the reverse implications holds — which is exactly why each norm exists: mass for size, flat for homological/compactness arguments, sharp for the most flexible continuity.

Exercise 7 (hard, proof). Stokes for a flat chain.

Let $T$ be a flat $r$ -chain (a limit of polyhedral chains in the flat norm) and $ω$ a smooth $r$ -form with $d ω$ also smooth. Prove the Stokes identity $T (d ω) = (\partial T) (ω)$ persists from polyhedral chains to $T$ .

Hint

For polyhedral chains, $T (d ω) = \partial T (ω)$ is classical Stokes. Both sides are continuous in the flat norm; pass to the limit.

Answer

For a polyhedral $r$ -chain $P$ , classical Stokes gives $P (d ω) = (\partial P) (ω)$ — the sum over oriented simplices of $\int_{σ} d ω = \int_{\partial σ} ω$ , with interior faces cancelling.

Let $P_{k} \to T$ in the flat norm, $F (T - P_{k}) \to 0$ . Two facts make the identity pass to the limit:

Pairing is flat-continuous. For a fixed smooth $ω$ with $∥ ω ∥^{*} \leq c_{0}$ and $∥ d ω ∥^{*} \leq c_{1}$ , the flat-norm definition gives $∣ S (ω) ∣ \leq max (c_{0}, c_{1}) F (S)$ for any flat chain $S$ . Hence $P_{k} (d ω) \to T (d ω)$ and $P_{k} (ω) \to T (ω)$ .
Boundary is flat-continuous. The boundary operator is $1$ -Lipschitz in the flat norm: $F (\partial S) \leq F (S)$ follows from the definition (any filling for $S$ provides one for $\partial S$ ). So $F (\partial T - \partial P_{k}) = F (\partial (T - P_{k})) \leq F (T - P_{k}) \to 0$ , giving $(\partial P_{k}) (ω) \to (\partial T) (ω)$ .

Therefore, passing to the limit in $P_{k} (d ω) = (\partial P_{k}) (ω)$ , $$ T(d\omega) = \lim_k P_k(d\omega) = \lim_k (\partial P_k)(\omega) = (\partial T)(\omega). $$ Stokes' theorem holds for every flat chain. This is the payoff of Whitney's construction: the flat-norm completion is designed exactly so the two operations in Stokes — pairing with $d ω$ and taking the boundary — are continuous, so the smooth identity survives the passage to non-smooth chains.

Exercise 8 (hard, proof). Constancy theorem in dimension one.

Let $T$ be a $1$ -current in an interval $(a, b) \subset R$ with $\partial T = 0$ and locally finite mass. Show $T = c [[(a, b)]]$ for a constant $c$ (the one-dimensional constancy theorem).

Hint

$\partial T = 0$ means $T (f^{'}) = 0$ for every compactly supported smooth $f$ . Interpret $T$ as a distribution and use that a distribution with zero derivative is a constant.

Answer

A $1$ -current on $(a, b)$ acts on $1$ -forms $ω = g d x$ with $g \in C_{c}^{\infty} (a, b)$ ; write $T (g d x) = ⟨ μ, g ⟩$ for a distribution $μ$ (a measure, by the finite-mass hypothesis). A $0$ -form (test function) is $f \in C_{c}^{\infty} (a, b)$ , with $df = f^{'} d x$ .

The hypothesis $\partial T = 0$ means $0 = (\partial T) (f) = T (df) = T (f^{'} d x) = ⟨ μ, f^{'} ⟩$ for all $f \in C_{c}^{\infty} (a, b)$ . By definition of the distributional derivative, $⟨ μ, f^{'} ⟩ = - ⟨ μ^{'}, f ⟩$ , so $⟨ μ^{'}, f ⟩ = 0$ for all $f$ , i.e. $μ^{'} = 0$ as a distribution.

A distribution on an interval with zero derivative is a constant: $μ = c$ for some $c \in R$ . (Proof: pick $f_{0}$ with $\int f_{0} = 1$ ; any $g$ writes as $g = (\int g) f_{0} + h^{'}$ with $h \in C_{c}^{\infty}$ , and $⟨ μ, g ⟩ = (\int g) ⟨ μ, f_{0} ⟩$ since $⟨ μ, h^{'} ⟩ = - ⟨ μ^{'}, h ⟩ = 0$ ; so $⟨ μ, g ⟩ = c \int g$ with $c = ⟨ μ, f_{0} ⟩$ .) Therefore $T (g d x) = c \int_{a}^{b} g d x = c [[(a, b)]] (g d x)$ , i.e. $T = c [[(a, b)]]$ .

This is the simplest case of Federer's constancy theorem: a current of top dimension in a connected open set with zero boundary is a constant multiple of the ambient orientation current.

Exercise 9 (hard, proof). Whitney extension on a closed set.

State the Whitney extension theorem for $C^{1}$ data and use it to show that any closed set $K \subset R^{n}$ is the zero set of some $C^{\infty}$ function $f : R^{n} \to [0, \infty)$ .

Hint

For the zero-set claim, the relevant Whitney result is the smooth Urysohn/zero-set construction: cover the open complement by countably many balls and sum bump functions. State the $C^{1}$ extension theorem, then give the zero-set corollary.

Answer

Whitney extension theorem ( $C^{1}$ form). Given a closed set $K \subset R^{n}$ , a function $f_{0} : K \to R$ , and a candidate "derivative" field $L : K \to (R^{n})^{*}$ satisfying the first-order Whitney compatibility condition $$ f_0(y) - f_0(x) - L(x)(y - x) = o(|y - x|) \quad \text{uniformly as } |y-x|\to 0,\ x,y\in K, $$ there exists $F \in C^{1} (R^{n})$ with $F ∣_{K} = f_{0}$ and $D F ∣_{K} = L$ . The extension is built by interpolating $f_{0}$ across the complement using a Whitney partition of unity subordinate to a dyadic cube decomposition of $R^{n} ∖ K$ .

Zero-set corollary. Let $K$ be closed and $U = R^{n} ∖ K$ open. Apply the Whitney cube decomposition of $U$ : countably many dyadic cubes ${Q_{i}}$ with $diam Q_{i} \approx dist (Q_{i}, K)$ and a subordinate $C^{\infty}$ partition of unity ${φ_{i}}$ , $supp φ_{i} \subset 2 Q_{i} \subset U$ . Choose weights $c_{i} > 0$ small enough (depending on $sup$ -norms of derivatives of $φ_{i}$ on $Q_{i}$ ) that $$ f = \sum_i c_i,\varphi_i $$ converges in $C^{\infty}$ on $U$ ; the controlled decay near $\partial U$ makes $f$ and all its derivatives extend by $0$ across $K$ , giving $f \in C^{\infty} (R^{n})$ with $f > 0$ on $U$ and $f \equiv 0$ on $K$ (and one may take $f \geq 0$ throughout). Hence $K = f^{- 1} (0)$ : every closed set is the zero set of a nonnegative smooth function. This is the existence-of-smooth-cutoffs facet of Whitney's theory, the same machinery that supplies the smooth functions parametrizing the chains and currents of the rest of the pack.

Exercise pack — Whitney Geometric Integration Theory Ch. I, IX-XI supplement: exterior forms and the wedge product, mass / flat / sharp norms, Stokes' theorem for currents, and the Whitney extension theorem. Nine exercises, two easy / four medium / three hard.

Prerequisites

02.13.05 pending
02.13.07 pending
02.13.11 pending
02.07.09 pending
03.04.02 pending
03.04.04 pending
03.04.05 pending

Tier anchors

beginner
intermediate: Whitney *Geometric Integration Theory* Ch. I (exterior algebra), Ch. IX-X (integration over flat chains), Ch. XI (flat and sharp norms); Federer §4.1 exercises (currents and Stokes)
master

References

Whitney — Geometric Integration Theory (Princeton 1957) · Ch. I (exterior algebra, $r$-vectors and $r$-covectors), Ch. VI (mass and flat norm on polyhedral chains), Ch. XI (the flat norm $\flat$ and sharp norm $\sharp$), Ch. IX-X (integration of forms over flat chains, Stokes)
Federer — Geometric Measure Theory (Springer 1969) · §1.5 (exterior algebra), §4.1.1-4.1.7 (currents, boundary, mass), §4.1.12 (slicing), §4.1.20 (constancy theorem)
Morgan — Geometric Measure Theory: A Beginner's Guide (5th ed., Academic Press 2016) · Ch. 4 (currents, the boundary operator, mass and flat norm), Ch. 6 (rectifiable and integral currents)

Estimated time

intermediate: 6h