03.04.E2 · modern-geometry / differential-forms

Differential forms and Stokes exercise pack (Shifrin / Arnold supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

This pack drills the core machinery of the exterior calculus: the algebra of alternating forms and the wedge product; pullback $f^{*}$ and its commutation with $d$ ; the exterior derivative; and the generalized Stokes theorem $\int_{M} d ω = \int_{\partial M} ω$ together with its three classical specializations — Green's theorem, the Kelvin-Stokes curl theorem, and the divergence theorem. The latter half exercises de Rham cohomology computations (closed-versus-exact, the angle form, punctured spaces) and integration on oriented manifolds.

The problems — three easy, four medium, three hard — follow Shifrin's Chapter 8 in spirit, with one cohomology problem aligned to Bott-Tu and one polynomial-form computation aligned to the Arnold-Falk-Winther de Rham complex (the bridge to finite element exterior calculus). Each problem has a hint and a full solution. The pack is read alongside its prerequisite units.

Conventions: forms live on an oriented manifold; $d$ is the unique antiderivation with $d^{2} = 0$ that restricts to the differential on functions; pullback satisfies $f^{*} (d ω) = d (f^{*} ω)$ and $f^{*} (α \land β) = f^{*} α \land f^{*} β$ ; and $\partial M$ carries the induced (outward-normal-first / Stokes) orientation, so that the generalized Stokes theorem holds with no sign correction.

Key theorem with full solution Intermediate

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

Lead exercise. Derive the classical divergence theorem from the generalized Stokes theorem.

Solution. Let $Ω \subset R^{3}$ be a compact region with smooth boundary surface $S = \partial Ω$ , and let $F = (P, Q, R)$ be a smooth vector field. Associate to $F$ the 2-form

ω_{F} = P d y \land d z + Q d z \land d x + R d x \land d y .

Its exterior derivative is

d ω_{F} = (\frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}) d x \land d y \land d z = (\nabla \cdot F) d V,

since, for instance, $d (P d y \land d z) = \frac{\partial P}{\partial x} d x \land d y \land d z$ (the $d y$ - and $d z$ -derivatives wedge to zero against the repeated factors).

The generalized Stokes theorem gives $\int_{Ω} d ω_{F} = \int_{\partial Ω} ω_{F}$ . The left side is $\int_{Ω} (\nabla \cdot F) d V$ . On the right, restricting $ω_{F}$ to $S$ with its outward orientation reproduces the flux integral: $ω_{F} ∣_{S} = F \cdot n d A$ , where $n$ is the outward unit normal and $d A$ the surface area form. Hence

\int_{Ω} (\nabla \cdot F) d V = \int_{S} F \cdot n d A,

the divergence theorem. $□$

The same pattern produces the other classical theorems: Green's theorem is generalized Stokes for a 1-form on a planar region, and the Kelvin-Stokes curl theorem is generalized Stokes for a 1-form on a surface (Exercise 6). The single statement $\int_{M} d ω = \int_{\partial M} ω$ unifies the three vector-calculus integral theorems, each recovered by choosing the dimension and the form-degree.

Exercises Intermediate

Exercise 3 (easy, symbolic). $d^{2} = 0$ on a 1-form.

Verify directly that $d (d α) = 0$ for the 1-form $α = f d x + g d y$ on $R^{2}$ , given $f, g \in C^{2}$ .

Hint

Compute $d α$ , then $d$ of that. Equality of mixed partials forces the cancellation.

Answer

First $d α = df \land d x + d g \land d y = (\frac{\partial f}{\partial y} d y) \land d x + (\frac{\partial g}{\partial x} d x) \land d y = (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) d x \land d y$ (the $d x \land d x$ and $d y \land d y$ terms vanish).

Then $d (d α) = d (\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}) \land d x \land d y$ . Any 1-form wedged with $d x \land d y$ on $R^{2}$ is a 3-form, which is zero in two variables. So $d (d α) = 0$ automatically here; more structurally, the coefficient is a function and $d h \land d x \land d y = 0$ in $R^{2}$ .

To see $d^{2} = 0$ where it is not automatic, note the engine is equality of mixed partials: $\frac{\partial ^{2} f}{\partial x \partial y} = \frac{\partial ^{2} f}{\partial y \partial x}$ , which is exactly what makes $d^{2} = 0$ hold in every dimension for $C^{2}$ forms.

Exercise 4 (medium, proof). Pullback commutes with $d$ .

Let $f : M \to N$ be smooth and $ω$ a $k$ -form on $N$ . Prove $f^{*} (d ω) = d (f^{*} ω)$ .

Hint

Reduce to coordinates. It suffices to check on $ω = h d y_{i_{1}} \land \dots \land d y_{i_{k}}$ and on functions, where it is the chain rule.

Answer

Both sides are linear and local, so check on a monomial $ω = h d y_{I}$ with $d y_{I} = d y_{i_{1}} \land \dots \land d y_{i_{k}}$ in local coordinates $y$ on $N$ .

On functions ( $k = 0$ ): $f^{*} (d h) = f^{*} (\sum_{j} \frac{\partial h}{\partial y _{j}} d y_{j}) = \sum_{j} (\frac{\partial h}{\partial y _{j}} \circ f) d (y_{j} \circ f) = d (h \circ f) = d (f^{*} h)$ , which is precisely the chain rule.

General $k$ : Using $f^{*} (α \land β) = f^{*} α \land f^{*} β$ and $f^{*} (d y_{I}) = d (y_{i_{1}} \circ f) \land \dots$ , note $f^{*} (d y_{I})$ is itself exact, hence closed: $d (f^{*} d y_{I}) = 0$ since $f^{*} d y_{i_{j}} = d (y_{i_{j}} \circ f)$ and $d^{2} = 0$ . Then $$ d(f^\omega)=d\big((h\circ f),f^dy_I\big)=d(h\circ f)\wedge f^dy_I=f^(dh)\wedge f^(dy_I)=f^(dh\wedge dy_I)=f^(d\omega), $$ using the function case in the middle step. So $f^ $an d$ d $co mm u t e, w hi c hi s w ha t mak es$ f^ $a c hainma p o n t h e d e R ham co m pl e x an d in d u ces$ f^\colon H^_{\mathrm{dR}}(N)\to H^_{\mathrm{dR}}(M)$.

Exercise 5 (medium, numeric). Closed but not exact: the angle form.

On $R^{2} ∖ {0}$ let $ω = \frac{- y d x + x d y}{x ^{2} + y ^{2}}$ . Show $d ω = 0$ , compute $\oint_{S^{1}} ω$ , and conclude $ω$ is not exact.

Hint

In polar coordinates $ω = d θ$ locally. An exact form integrates to zero around any loop.

Answer

Write $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ . A direct computation gives $\frac{\partial Q}{\partial x} = \frac{y ^{2} - x ^{2}}{( x ^{2} + y ^{2} ) ^{2}} = \frac{\partial P}{\partial y}$ , so $d ω = (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x \land d y = 0$ . Thus $ω$ is closed.

In polar coordinates $x = r cos θ$ , $y = r sin θ$ , one checks $ω = d θ$ (locally). Integrating around the unit circle, parametrise $θ \in [0, 2 π)$ :

\oint_{S^{1}} ω = \int_{0}^{2 π} d θ = 2 π \neq = 0.

If $ω = df$ were exact, Stokes (or the fundamental theorem on the closed loop) would force $\oint_{S^{1}} ω = 0$ . Since the integral is $2 π$ , $ω$ is closed but not exact: $[ω]$ is a nonzero class generating $H_{dR}^{1} (R^{2} ∖ {0}) ≅ R$ . The obstruction is the puncture; $θ$ is not a single-valued function on the punctured plane.

Exercise 6 (medium, proof). Kelvin-Stokes from generalized Stokes.

Let $S \subset R^{3}$ be a compact oriented surface with boundary curve $\partial S$ , and $F = (P, Q, R)$ a vector field. Derive the curl theorem $\oint_{\partial S} F \cdot d r = \iint_{S} (\nabla \times F) \cdot n d A$ from $\int_{S} d ω = \int_{\partial S} ω$ .

Hint

Take the work 1-form $ω = P d x + Q d y + R d z$ . Show $d ω$ is the flux 2-form of $\nabla \times F$ .

Answer

Let $ω = P d x + Q d y + R d z$ , the 1-form dual to $F$ , so $\int_{\partial S} ω = \oint_{\partial S} F \cdot d r$ .

Compute $d ω$ . Each term contributes, e.g. $d (P d x) = \frac{\partial P}{\partial y} d y \land d x + \frac{\partial P}{\partial z} d z \land d x$ . Collecting all terms and using antisymmetry,

d ω = (\frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}) d y \land d z + (\frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}) d z \land d x + (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}) d x \land d y .

The three coefficients are exactly the components of $\nabla \times F$ , so $d ω = ω_{\nabla \times F}$ , the flux 2-form of the curl. Restricting to $S$ with its orientation gives $\int_{S} d ω = \iint_{S} (\nabla \times F) \cdot n d A$ .

Generalized Stokes $\int_{S} d ω = \int_{\partial S} ω$ then reads

\iint_{S} (\nabla \times F) \cdot n d A = \oint_{\partial S} F \cdot d r,

the Kelvin-Stokes curl theorem. Green's theorem is the planar case $R = 0$ , $S \subset {z = 0}$ , where the curl reduces to $\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}$ .

Exercise 7 (medium, numeric). De Rham cohomology of the 2-torus.

Compute $H_{dR}^{*} (T^{2})$ and exhibit generators as explicit forms.

Hint

$T^{2} = S^{1} \times S^{1}$ . Use the Künneth formula, or write down $d θ_{1}$ , $d θ_{2}$ , $d θ_{1} \land d θ_{2}$ and check they are closed and non-exact.

Answer

Use angular coordinates $θ_{1}, θ_{2}$ on $T^{2} = S^{1} \times S^{1}$ (each $θ_{i}$ defined mod $2 π$ ; the 1-forms $d θ_{i}$ are globally defined and closed, though $θ_{i}$ is not single-valued).

By the Künneth formula $H_{dR}^{*} (T^{2}) = H^{*} (S^{1}) \otimes H^{*} (S^{1})$ :

H^{0} = R ⟨ 1 ⟩, H^{1} = R^{2} ⟨[d θ_{1}], [d θ_{2}]⟩, H^{2} = R ⟨[d θ_{1} \land d θ_{2}]⟩ .

Each generator is closed (every $d θ_{i}$ is closed and their wedge is too). None is exact: $\oint_{γ_{i}} d θ_{j} = 2 π δ_{ij}$ over the two generating loops shows $[d θ_{1}], [d θ_{2}]$ are independent and nonzero, and $\int_{T^{2}} d θ_{1} \land d θ_{2} = (2 π)^{2} \neq = 0$ shows $[d θ_{1} \land d θ_{2}] \neq = 0$ . The Betti numbers $(1, 2, 1)$ match the singular homology of the torus and have alternating sum $χ (T^{2}) = 0$ .

Exercise 8 (hard, proof). Poincaré lemma via the homotopy operator.

Prove the Poincaré lemma: every closed $k$ -form ( $k \geq 1$ ) on a star-shaped open set $U \subset R^{n}$ is exact.

Hint

Build a homotopy operator $K$ with $d K + K d = id$ on positive-degree forms, by integrating along the radial contraction $H (t, x) = t x$ .

Answer

Assume $U$ is star-shaped about the origin, so $H : [0, 1] \times U \to U$ , $H (t, x) = t x$ , is a smooth homotopy from the constant map (at $t = 0$ ) to $id_{U}$ (at $t = 1$ ). For a $k$ -form $ω$ , define the homotopy operator $K ω$ by pulling back $ω$ under $H$ and integrating out the $d t$ direction:

(K ω) = \int_{0}^{1} ι_{\partial_{t}} (H^{*} ω) d t,

a $(k - 1)$ -form on $U$ , where $ι_{\partial_{t}}$ contracts the $d t$ factor.

The standard computation (Cartan's formula $L_{\partial_{t}} = d ι_{\partial_{t}} + ι_{\partial_{t}} d$ applied to $H^{*} ω$ , integrated over $t \in [0, 1]$ ) yields the homotopy identity

d K ω + K d ω = H_{1}^{*} ω - H_{0}^{*} ω = ω - 0 = ω (k \geq 1),

since $H_{1} = id$ and $H_{0}$ is constant (so $H_{0}^{*} ω = 0$ for $k \geq 1$ ).

Now if $ω$ is closed, $d ω = 0$ , so the identity collapses to $ω = d K ω$ . Thus $ω$ is exact, with explicit primitive $K ω$ . Hence $H_{dR}^{k} (U) = 0$ for $k \geq 1$ on any star-shaped (in particular convex) $U$ . This is the local vanishing that makes de Rham cohomology a global invariant: on any manifold, a good cover is one all of whose intersections are contractible, hence have vanishing positive-degree cohomology by this lemma.

Exercise 9 (hard, numeric). Solid angle and the degree of the Gauss map.

Let $σ = \frac{x d y \land d z + y d z \land d x + z d x \land d y}{( x ^{2} + y ^{2} + z ^{2} ) ^{3/2}}$ on $R^{3} ∖ {0}$ . Show $d σ = 0$ and compute $\int_{S^{2}} σ$ over the unit sphere.

Hint

$σ$ restricts to the area form on the unit sphere. Closedness is a direct (if tedious) computation; the integral is the total solid angle.

Answer

On the unit sphere $r = 1$ the denominator is $1$ and $σ ∣_{S^{2}} = x d y \land d z + y d z \land d x + z d x \land d y$ , which is exactly the standard area 2-form of $S^{2}$ (the contraction $ι_{r} (d x \land d y \land d z)$ restricted to the sphere). Hence

\int_{S^{2}} σ = Area (S^{2}) = 4 π .

Closedness: $σ$ is the pullback of the area form under the Gauss map $x \mapsto x /∣ x ∣$ , and away from the origin $d σ = 0$ by a direct computation (each term's $r^{- 3}$ scaling cancels the polynomial derivatives; equivalently $σ$ is the flux form of the inverse-square field $x /∣ x ∣^{3}$ , which is divergence-free off the origin).

Because $\int_{S^{2}} σ = 4 π \neq = 0$ , $σ$ is closed but not exact on $R^{3} ∖ {0}$ , generating $H_{dR}^{2} (R^{3} ∖ {0}) ≅ R$ . Normalising by $4 π$ gives the generator measuring the degree of a map into $S^{2}$ via $\frac{1}{4 π} \int f^{*} σ$ — the three-dimensional analogue of the winding number from Exercise 5.

Exercise 10 (hard, symbolic). Polynomial de Rham complex on the line: the Koszul bridge to FEEC.

On $R^{1}$ the polynomial spaces are $P_{r} Λ^{0} = P_{r}$ (polynomials of degree $\leq r$ ) and $P_{r} Λ^{1} = P_{r} d x$ . Show $d : P_{r} Λ^{0} \to P_{r - 1} Λ^{1}$ together with the Koszul map $κ : P_{r - 1} Λ^{1} \to P_{r} Λ^{0}$ , $κ (p d x) = \int$ -style contraction, realise the Poincaré lemma at polynomial level, and verify the dimension count.

Hint

On $R^{1}$ the Koszul operator is $κ (p (x) d x) = x p (x)$ (contraction with the Euler/radial vector field). Use $d κ + κ d = (r + k) \cdot id$ on homogeneous pieces.

Answer

In one variable $d (p (x)) = p^{'} (x) d x$ , and the Koszul (contraction-with- $x \partial_{x}$ ) operator is $κ (q (x) d x) = x q (x)$ , a $0$ -form. On a homogeneous polynomial $x^{m}$ (degree $m$ , form-degree $0$ ): $κ d (x^{m}) = κ (m x^{m - 1} d x) = m x^{m}$ , and $d κ (x^{m}) = 0$ since $κ$ of a $0$ -form is $0$ . So $(d κ + κ d) (x^{m}) = m x^{m} = (m + k) x^{m}$ with $k = 0$ . On a 1-form $x^{m} d x$ : $κ (x^{m} d x) = x^{m + 1}$ , $d κ (x^{m} d x) = (m + 1) x^{m} d x$ , and $κ d (x^{m} d x) = 0$ , giving $(d κ + κ d) = (m + 1) id = (m + k) id$ with $k = 1$ . This is the Koszul homotopy identity $d κ + κ d = (de g + form-degree) id$ , the polynomial Poincaré lemma: on homogeneous pieces of positive total degree it is invertible, so the polynomial de Rham complex is exact there.

Dimension check via the AFW formula $dim P_{r} Λ^{k} (R^{n}) = (r + k r + n) (k r + k)$ with $n = 1$ : $dim P_{r} Λ^{0} = (r r + 1) (0 r) = r + 1$ (the polynomials $1, x, \dots, x^{r}$ ) and $dim P_{r} Λ^{1} = (r + 1 r + 1) (1 r + 1) = r + 1$ ( $d x, x d x, \dots, x^{r} d x$ ). The map $d : P_{r} Λ^{0} \to P_{r - 1} Λ^{1}$ has rank $r$ (kernel the constants), and the reduced space $P_{r}^{-} Λ^{1} = P_{r - 1} Λ^{1}$ has dimension $r$ , matching $dim P_{r}^{-} Λ^{k} = (r + k - 1 r + n - 1) (k r + k - 1)$ at $n = 1, k = 1$ . This homogeneous Koszul splitting is the algebraic engine behind the finite element exterior calculus spaces.

Exercise pack. Shifrin Ch. 8 / Arnold-Falk-Winther supplement: exterior algebra and the wedge product, pullback and its commutation with $d$ , the generalized Stokes theorem and its classical specializations (Green / Kelvin-Stokes / divergence), de Rham cohomology of the punctured plane and the torus, the Poincaré lemma, and the polynomial (Koszul) de Rham complex bridging to FEEC.

Prerequisites

03.04.02 pending
03.04.03 pending
03.04.04 pending
03.04.05 pending
03.04.06 pending
03.04.20 pending
03.04.21 pending

Tier anchors

beginner
intermediate: Shifrin, Multivariable Mathematics, Ch. 8 exercises (exterior algebra, pullback, the generalized Stokes theorem and its classical specializations, integration on manifolds); Arnold-Falk-Winther, FEEC (Bull. AMS 2010) §3-§4 — the polynomial de Rham complex as a worked exterior-calculus setting
master

References

Shifrin — Multivariable Mathematics: Linear Algebra, Multivariable Calculus, and Manifolds · Ch. 8 (exterior algebra, $k$-forms and the wedge product, pullback $f^*$, the exterior derivative $d$, the generalized Stokes theorem $\int_M d\omega=\int_{\partial M}\omega$ and its reduction to Green / divergence / classical Stokes, closed and exact forms)
Arnold-Falk-Winther — Finite Element Exterior Calculus (Bull. AMS 47, 2010) · §3-§4 — the de Rham complex, the Koszul operator $\kappa$, the polynomial spaces $\mathcal{P}_r\Lambda^k$, integration by parts on a domain
Spivak — Calculus on Manifolds · Ch. 4-5 — the wedge product, pullback, and Stokes' theorem on chains
Bott-Tu — Differential Forms in Algebraic Topology · §1-§4 — the de Rham complex, the Poincaré lemma, and Mayer-Vietoris for cohomology computations

Estimated time

intermediate: 5h