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The Lumenary · Advanced undergraduate

Differential forms and Stokes' theorem

Name: ______________________

Date: ______________________

You will learn to

Compute with the wedge product, the exterior derivative and pullbacks, and verify $d \circ d = 0$
State and prove Stokes' theorem for cubes and chains, and recover the fundamental theorem of calculus and the theorems of Green, Stokes and Gauss as special cases
Show that $d\theta$ on the punctured plane is closed but not exact, and prove the Poincaré lemma for 1-forms on star-shaped domains
Define de Rham cohomology and explain why $H^1$ of the punctured plane is non-zero

1.
Evaluate the 2-form $dx \wedge dy$ on the pair of vectors $u = (1, 2)$ , $v = (3, 4)$ , in that order.
2.
Which of these forms on $\mathbb{R}^3$ are equal to $0$ ? (Choose all that apply.)
- A. $dx \wedge dx$
- B. $dx \wedge dy + dy \wedge dx$
- C. $dx \wedge dy - dy \wedge dx$
- D. $(dx + dy) \wedge (dx + dy)$
- E. $dx \wedge dy \wedge dz$
- F. $dz \wedge (dx \wedge dy) - (dx \wedge dy) \wedge dz$
3.
Compute $d(x^2 y\, dz)$ on $\mathbb{R}^3$ .
- A. $2xy\, dx \wedge dz + x^2\, dy \wedge dz$
- B. $2xy\, dx \wedge dz + x^2\, dy \wedge dz + x^2 y\, dz \wedge dz$
- C. $2xy\, dz \wedge dx + x^2\, dz \wedge dy$
- D. $2xy + x^2$
4.
Let $R$ be the rectangle $[0, 2] \times [0, 3]$ with its boundary oriented anticlockwise. Use Stokes' theorem to compute $\oint_{\partial R} (x\,dy - y\,dx)$ .
5.
Green's theorem $\iint_D (Q_x - P_y)\,dx\,dy = \oint_{\partial D} P\,dx + Q\,dy$ is the general Stokes theorem $\int_M d\omega = \int_{\partial M}\omega$ with which choice of $M$ and $\omega$ ?
- A. $M = D$ , a region in $\mathbb{R}^2$ ( $k = 2$ ), and $\omega = P\,dx + Q\,dy$ , so $d\omega = (Q_x - P_y)\,dx \wedge dy$
- B. $M = D$ and $\omega = Q - P$ , a function
- C. $M = \partial D$ , a curve, and $\omega = P\,dx + Q\,dy$
- D. $M$ a solid in $\mathbb{R}^3$ and $\omega$ a 2-form
6.
Away from the $y$ -axis, $d\theta = d\big(\arctan(y/x)\big)$ . Why does this not show that $d\theta$ is exact on $\mathbb{R}^2 \setminus \{0\}$ ?
- A.Because $\arctan(y/x)$ , and indeed any polar angle function, cannot be defined continuously on all of $\mathbb{R}^2 \setminus \{0\}$ ; exactness requires a single function on the whole domain.
- B.Because $d\theta$ is not closed.
- C.Because the derivative of $\arctan(y/x)$ is not $d\theta$ .
- D.Because exact forms must be defined at the origin.
7.
Which Lean statement correctly says that $d\theta$ is not exact on the punctured plane? (dTheta p is the linear functional $d\theta$ at the point p.)
- A.¬ ∃ f : ℝ × ℝ → ℝ, ∀ p : ℝ × ℝ, p ≠ 0 → HasFDerivAt f (dTheta p) p
- B.¬ ∃ f : ℝ × ℝ → ℝ, ∀ p : ℝ × ℝ, HasFDerivAt f (dTheta p) p
- C.∃ f : ℝ × ℝ → ℝ, ∀ p : ℝ × ℝ, p ≠ 0 → HasFDerivAt f (dTheta p) p
- D.∀ p : ℝ × ℝ, p ≠ 0 → ¬ ∃ f : ℝ × ℝ → ℝ, HasFDerivAt f (dTheta p) p
8.
A 1-form on $\mathbb{R}^3$ is a triple $(a_1, a_2, a_3)$ meaning $a_1\,dx + a_2\,dy + a_3\,dz$ , and a 2-form is a triple $(c_1, c_2, c_3)$ meaning $c_1\,dy\wedge dz + c_2\,dz \wedge dx + c_3\,dx \wedge dy$ . Write wedge(a, b) returning the 2-form $a \wedge b$ as such a triple. What familiar operation do you get?
Computational — Practice
9.
On which of these open sets is every closed 1-form exact? (Choose all that apply.)
- A. $\mathbb{R}^2$
- B.The slit plane $\mathbb{R}^2$ minus the closed negative $x$ -axis
- C. $\mathbb{R}^2 \setminus \{0\}$
- D.The open annulus $1 \lt |x| \lt 2$
- E. $\mathbb{R}^3 \setminus \{0\}$

Answer key — Differential forms and Stokes' theorem

1.
$(dx \wedge dy)(u, v) = 1 \cdot 4 - 3 \cdot 2 = -2$ , the determinant of the matrix with columns $u, v$ . Its absolute value $2$ is the area of the parallelogram spanned by $u$ and $v$ ; the negative sign says that going from $u$ to $v$ is a clockwise turn (Lean: dx_wedge_dy).
2.
Zero: $dx \wedge dx$ , $dx \wedge dy + dy \wedge dx$ , $(dx + dy) \wedge (dx + dy)$ and $dz \wedge dx \wedge dy - dx \wedge dy \wedge dz$ (a 1-form and a 2-form commute). Not zero: $dx \wedge dy - dy \wedge dx = 2\,dx \wedge dy$ and the volume form $dx \wedge dy \wedge dz$ .
3.
$d(x^2 y\,dz) = d(x^2 y) \wedge dz = (2xy\,dx + x^2\,dy) \wedge dz = 2xy\,dx \wedge dz + x^2\,dy \wedge dz$ . In vector-calculus terms, this is the curl of $(0, 0, x^2 y)$ , namely $(x^2, -2xy, 0)$ , written as a 2-form $x^2\,dy\wedge dz - 2xy\,dz\wedge dx$ — the same thing, since $dz \wedge dx = -dx \wedge dz$ .
4.
$d(x\,dy - y\,dx) = 2\,dx \wedge dy$ , so $\oint_{\partial R} (x\,dy - y\,dx) = \iint_R 2\,dx\,dy = 2 \cdot 6 = 12$ . Halving gives the area: this is the principle of the planimeter. (The second experiment checks the same identity numerically on a square and on circles.)
5.
Take $M = D$ and $\omega = P\,dx + Q\,dy$ . Then $d\omega = dP \wedge dx + dQ \wedge dy = P_y\,dy \wedge dx + Q_x\,dx \wedge dy = (Q_x - P_y)\,dx \wedge dy$ , and Stokes' theorem is exactly Green's theorem. The other classical theorems arise the same way with $k = 1$ and $k = 3$ (see the table in the lesson).
6.
$\arctan(y/x)$ is undefined on the $y$ -axis. Patching it with constants gives a continuous angle on any slit plane, but going once around the origin always forces a jump of $2\pi$ . If $d\theta = df$ for a function $f$ on the whole punctured plane, the integral around the unit circle would be $f(\text{end}) - f(\text{start}) = 0$ ; in fact it is $2\pi$ . So $d\theta$ is locally but not globally exact (Lean: dTheta_not_exact).
7.
The correct statement is ¬ ∃ f, ∀ p ≠ 0, HasFDerivAt f (dTheta p) p: no single function has derivative $d\theta$ throughout the punctured plane. Moving ∃ f inside ∀ p gives a false statement (local primitives exist), and requiring the derivative also at 0 gives a weaker one. The Lean proof composes f with the circle, gets derivative 1, and contradicts g(2π) = g(0).
8.
```
def wedge(a, b):
    return (a[1] * b[2] - a[2] * b[1],
            a[2] * b[0] - a[0] * b[2],
            a[0] * b[1] - a[1] * b[0])
```
This is the cross product. In $\mathbb{R}^3$ (with its metric and orientation) 2-forms can be identified with vectors, and under that identification $\wedge$ of 1-forms becomes $\times$ . The wedge product needs no metric and works in every dimension, while the cross product only exists in $\mathbb{R}^3$ .
9.
Exact: $\mathbb{R}^2$ and the slit plane (star-shaped), and $\mathbb{R}^3 \setminus \{0\}$ (simply connected: every loop can be shrunk, so by Stokes every closed 1-form has zero integral around loops). Not exact: the punctured plane and the annulus, where $d\theta$ has integral $2\pi$ around the hole. Star-shaped is sufficient, not necessary; for 1-forms what matters is $H^1_{\mathrm{dR}}$ , which vanishes on simply connected domains.