6.7 Differential forms and the one big theorem

A differential form is the thing under an integral sign, a recipe that eats tiny oriented pieces of a curve, surface or solid; the wedge product and the exterior derivative d turn gradient, curl and divergence into one operation, d squared is zero says curl grad and div curl vanish, and a single theorem, the integral of d omega over a region equals the integral of omega over its boundary, contains the fundamental theorem of calculus, Green, Stokes and Gauss.

In 6.1 you wrote $\int_CP\,dx + Q\,dy$ and in 6.2 you called $P\,dx + Q\,dy$ an exact differential. We never said what $dx$ is. This chapter does. The answer unifies the whole unit. An object called a differential form sits under every integral sign of Unit 6; a single operation $d$ turns gradient, curl and divergence into one rule; the identities curl grad $= 0$ and div curl $= 0$ become one identity, $d^2 = 0$ ; and the four theorems of the unit, the fundamental theorem of calculus, Green, Stokes and Gauss, become one theorem. None of this is needed for the MATH 376 exam. It is here so that the pattern is visible, because it is what a course on manifolds starts from.

Before you start

6.6: $\operatorname{curl}\nabla f = 0$ and $\operatorname{div}\operatorname{curl}F = 0$ . 6.5: flux through a parametrized surface is $\iint F\cdot(r_u\times r_v)\,du\,dv$ . 6.4: the Jacobian determinant is the area stretching factor. 6.1: $\int_CP\,dx + Q\,dy = \int(Px' + Qy')\,dt$ .

Quick checkFor a path

r(t) = (x(t), y(t))

, the symbol

\int_CP\,dx + Q\,dy

means which integral?

Quick checkWhat is

\operatorname{curl}(\nabla f)

for a

C^2

function

f

\mathbb R^3

Quick checkA map sends the parameter square with area

A

in the

uv

-plane to a region in the

xy

-plane. Its Jacobian determinant is

3

everywhere. The area of the region is

The picture: what a form eats

Think about what an integral sign needs. In $\int_CF\cdot dr$ it needs, at each point of the curve, a small displacement vector, and it returns a number: the part of $F$ along the displacement. In $\iint_SF\cdot n\,dS$ it needs a tiny piece of surface, a tiny parallelogram with two edge vectors $a$ and $b$ in a definite order, and it returns a number: the amount of $F$ crossing the piece, $F\cdot(a\times b)$ . In $\iiint f\,dV$ it needs a tiny parallelepiped with three edges, and returns $f$ times the signed volume $\det[a, b, c]$ . A differential form is exactly that: a rule that eats tiny pieces and returns numbers, linear in each edge, and changing sign when two edges are swapped (because swapping edges reverses orientation).

Here is the whole structure in a table, for three dimensions. The columns are the degree of the form, what it eats, and what vector calculus calls it.

Degree	Typical form	Eats	What it returns	Same as
0	$f$	a point	$f$ at the point	a scalar field
1	$P\,dx + Q\,dy + R\,dz$	a vector $v$	$P v_1 + Q v_2 + R v_3 = F\cdot v$	a vector field $F=(P,Q,R)$ , via the line integral
2	$A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$	two vectors $a, b$	$(A, B, C)\cdot(a\times b)$	a vector field $G = (A,B,C)$ , via the flux
3	$f\,dx\wedge dy\wedge dz$	three vectors $a, b, c$	$f\,\det[a, b, c]$	a scalar field, via the volume

Read down the ‘same as’ column and notice the pattern: scalars and vector fields appear twice. That is why gradient, curl and divergence look unrelated and are not: they are one operation, $d$ , acting on degrees $0\to1$ , $1\to2$ , $2\to3$ .

Where this came from· Where forms came from

Hermann Grassmann in 1844 built an algebra of geometric objects in which the product of two vectors depends on their order, with $ab = -ba$ ; it is the algebra of the wedge product below. The step to calculus was Élie Cartan’s, in a long 1899 paper in the Annales scientifiques de l’École Normale Supérieure, ‘Sur certaines expressions différentielles et le problème de Pfaff’. The Pfaff problem asks: given an expression $a_1\,dx_1 + \dots + a_n\,dx_n$ , can a change of variables put it in the simplest possible form? Cartan introduced the operation we now call the exterior derivative to answer it, and the paper is where it first appears. The unification of Green, Stokes and Gauss into one statement came later, as the language of forms spread through geometry in the twentieth century.

Forms, and the wedge product

Definition B

One-form

At each point $p$ , $dx$ is the linear function that sends a vector $v = (v_1, v_2, v_3)$ to $v_1$ , $dy$ sends it to $v_2$ , and $dz$ to $v_3$ . A 1-form is $\omega = P\,dx + Q\,dy + R\,dz$ , where $P, Q, R$ are functions of position: at $p$ it sends $v$ to $P(p)v_1 + Q(p)v_2 + R(p)v_3 = F(p)\cdot v$ , with $F = (P, Q, R)$ . So 1-forms and vector fields carry the same information. The line integral of the vector field $F$ along a path is the integral of the 1-form $\omega_F = F\cdot dr$ .

In one lineIn $\mathbb R^3$ a 1-form is an expression $\omega = P\,dx + Q\,dy + R\,dz$ with functions $P, Q, R$ ; at each point it takes a vector $v$ to the number $Pv_1 + Qv_2 + Rv_3$ .

Why it existsIt is the thing under a line integral. $dx$ is the 1-form that reads off the $x$ component of a displacement, so $P\,dx + Q\,dy$ is 'the part of the field along the step'.

Definition B

Two-form

The symbol $dx\wedge dy$ (’ $dx$ wedge $dy$ ’) is the 2-form that sends a pair of vectors $(a, b)$ to $a_1b_2 - a_2b_1$ , the signed area of the shadow of the parallelogram on the $xy$ -plane. Likewise $dy\wedge dz$ sends $(a, b)$ to $a_2b_3 - a_3b_2$ and $dz\wedge dx$ to $a_3b_1 - a_1b_3$ . These are exactly the three components of $a\times b$ . A 2-form is a combination $\omega = A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$ , and it sends $(a, b)$ to $(A, B, C)\cdot(a\times b)$ . The flux of $G = (A, B, C)$ through a surface is the integral of the 2-form $A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$ . This is the notation of Apostol 12.9, which the course listed, and it is the one noted in 6.5.

In one lineIn $\mathbb R^3$ a 2-form is $A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$ ; at each point it takes two vectors $a, b$ to $(A, B, C)\cdot(a\times b)$ , the amount of the field $(A, B, C)$ crossing the parallelogram on $a$ and $b$ .

Why it existsIt is the thing under a surface integral: it eats a tiny oriented parallelogram. The sign changes when you swap $a$ and $b$ , which is what orientation means.

Definition B

Differential form

A differential form of degree $k$ (a $k$ -form) on a region of $\mathbb R^n$ is a sum $\omega = \sum f_{i_1\dots i_k}\,dx_{i_1}\wedge\dots\wedge dx_{i_k}$ over increasing index lists, with functions $f_{i_1\dots i_k}$ . At each point it eats $k$ vectors and returns a number, linear in each vector and changing sign when two vectors are swapped. In $\mathbb R^3$ the degrees are $0, 1, 2, 3$ ; in $\mathbb R^2$ they are $0, 1, 2$ with the only 2-form $f\,dx\wedge dy$ ; there are no $k$ -forms for $k$ larger than $n$ .

In one lineA $k$ -form is a rule that assigns to each point a function of $k$ vectors that is linear in each and changes sign when two of them are swapped. In coordinates it is a sum of functions times $dx_{i_1}\wedge\dots\wedge dx_{i_k}$ . A 0-form is a function.

Why it existsIt is the single object that can be integrated over a $k$ -dimensional piece, for every $k$ : points, curves, surfaces, solids.

Definition B

Wedge product

The wedge product $\omega\wedge\eta$ of a $j$ -form and a $k$ -form is a $(j + k)$ -form. The rules: it is bilinear (it distributes over sums and constants and functions come out), it is associative, and the basic rule is that swapping two 1-forms flips the sign: $dx\wedge dy = -dy\wedge dx$ , and hence $dx\wedge dx = 0$ (since $dx\wedge dx = -dx\wedge dx$ ). Two 1-forms always anticommute; a function (0-form) commutes with everything. If you evaluate on vectors, $(dx\wedge dy)(a, b) = dx(a)\,dy(b) - dx(b)\,dy(a) = a_1b_2 - a_2b_1$ .

In one lineThe product of forms, bilinear and associative, with $\omega\wedge\eta = (-1)^{jk}\eta\wedge\omega$ for degrees $j, k$ ; in particular $dx_i\wedge dx_j = -dx_j\wedge dx_i$ and $dx_i\wedge dx_i = 0$ .

Why it existsIt is how a 1-form eats two vectors at once (it is the signed area of a shadow) and how forms of degrees $j$ and $k$ combine into degree $j + k$ .

Everything about the wedge product follows from $dx_i\wedge dx_j = -dx_j\wedge dx_i$ . Expand, kill the terms with a repeated $dx_i$ , and reorder with a sign.

Worked example

Compute

(2\,dx - 3\,dy)\wedge(dx + 4\,dy)

in the plane, and

(dx + 2\,dy + 3\,dz)\wedge(dy + dz)

in space. Compare the second with a cross product.

Solution.

Plane, expand. $2\,dx\wedge dx + 8\,dx\wedge dy - 3\,dy\wedge dx - 12\,dy\wedge dy$ . Distribute: four products.
Plane, simplify. $dx\wedge dx = 0 = dy\wedge dy$ and $-3\,dy\wedge dx = +3\,dx\wedge dy$ , so the result is $(8 + 3)\,dx\wedge dy = 11\,dx\wedge dy$ . This is the determinant $\det\begin{bmatrix}2&-3\\1&4\end{bmatrix} = 11$ : the wedge of two 1-forms in the plane is a $2\times2$ determinant.
Space, expand. $dx\wedge dy + dx\wedge dz + 2\,dy\wedge dy + 2\,dy\wedge dz + 3\,dz\wedge dy + 3\,dz\wedge dz$ .
Space, simplify. Drop $dy\wedge dy$ and $dz\wedge dz$ . Rewrite $dx\wedge dz = -dz\wedge dx$ and $3\,dz\wedge dy = -3\,dy\wedge dz$ . The result is $(2 - 3)\,dy\wedge dz - dz\wedge dx + dx\wedge dy = -dy\wedge dz - dz\wedge dx + dx\wedge dy$ .
Compare. In the dictionary this is the vector $(-1, -1, 1)$ , and $(1, 2, 3)\times(0, 1, 1) = (2 - 3,\ 0 - 1,\ 1 - 0) = (-1, -1, 1)$ .

Answer. $11\,dx\wedge dy$ and $-dy\wedge dz - dz\wedge dx + dx\wedge dy$ . The wedge product of two 1-forms is the cross product: that is why the cross product is anti-symmetric and why it exists only in three dimensions as a vector.

Quick checkWhat is

dy\wedge dx + dx\wedge dy

Quick checkWhat is

(x\,dx)\wedge(x\,dx)

The exterior derivative

Definition B

Exterior derivative

The exterior derivative $d$ takes a $k$ -form to a $(k+1)$ -form. For a function $f$ : $df = f_x\,dx + f_y\,dy + f_z\,dz$ (the total differential of 5.3). For a form $\omega = f\,dx_{i_1}\wedge\dots\wedge dx_{i_k}$ with one term: $d\omega = df\wedge dx_{i_1}\wedge\dots\wedge dx_{i_k}$ . For a sum, add. Because $dx_i\wedge dx_i = 0$ , only the derivatives of $f$ in directions not already present survive.

In one lineThe operation $d$ that raises the degree of a form by one: on a function $df = f_x\,dx + f_y\,dy + f_z\,dz$ , and on $f\,dx_I$ it is $df\wedge dx_I$ , extended by linearity.

Why it existsIt is the one derivative that is built to be integrated: $\int_Md\omega$ over a region equals $\int_{\partial M}\omega$ over its boundary. It contains gradient, curl and divergence.

Theorem B Rudin 10.20 and the dictionary of vector calculus; Spivak ch. 4

d is gradient, curl and divergence

Under the dictionary of the table above, $d$ on 0-forms is the gradient, on 1-forms the curl, and on 2-forms the divergence.

In one lineIn $\mathbb R^3$ : $df = f_x\,dx + f_y\,dy + f_z\,dz$ (gradient); $d(P\,dx + Q\,dy + R\,dz) = (R_y - Q_z)\,dy\wedge dz + (P_z - R_x)\,dz\wedge dx + (Q_x - P_y)\,dx\wedge dy$ (curl); $d(A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy) = (A_x + B_y + C_z)\,dx\wedge dy\wedge dz$ (divergence).

Why it is trueIt shows that the three operators of 6.6 are one operator, and it makes their relations, such as the product rules, automatic.

ProofRudin 10.20

The idea: Apply the rule d(f dx_I) = df wedge dx_I term by term, then use dx wedge dx = 0 and the anticommuting rule to put the survivors in the standard order.

df = f_x\,dx + f_y\,dy + f_z\,dz

, the gradient

\nabla f = (f_x, f_y, f_z)

in the dictionary.

The 0-form case is the definition.

d\omega = dP\wedge dx + dQ\wedge dy + dR\wedge dz

Apply the rule to each of the three terms of

\omega = P\,dx + Q\,dy + R\,dz

= (P_y\,dy + P_z\,dz)\wedge dx + (Q_x\,dx + Q_z\,dz)\wedge dy + (R_x\,dx + R_y\,dy)\wedge dz

Expand each differential:

dP = P_x\,dx + P_y\,dy + P_z\,dz

, and so on.

= (R_y - Q_z)\,dy\wedge dz + (P_z - R_x)\,dz\wedge dx + (Q_x - P_y)\,dx\wedge dy

Terms with a repeated

dx_i

vanished already (

P_x\,dx\wedge dx = 0

and so on). Reorder with

dy\wedge dx = -dx\wedge dy

dz\wedge dy = -dy\wedge dz

and

dx\wedge dz = -dz\wedge dx

.The three survivors in the standard order

dy\wedge dz

dz\wedge dx

dx\wedge dy

are the cyclic order that matches the cross product.

Add the three:

d\omega = (A_x + B_y + C_z)\,dx\wedge dy\wedge dz

\blacksquare

For a 2-form, only the derivative in the missing direction survives:

d(A\,dy\wedge dz) = A_x\,dx\wedge dy\wedge dz

since

A_y\,dy\wedge dy\wedge dz = 0

and

A_z\,dz\wedge dy\wedge dz = 0

. Similarly

d(B\,dz\wedge dx) = B_y\,dy\wedge dz\wedge dx = B_y\,dx\wedge dy\wedge dz

(a cyclic shift is an even permutation) and

d(C\,dx\wedge dy) = C_z\,dz\wedge dx\wedge dy = C_z\,dx\wedge dy\wedge dz

In the plane the same computation gives $d(P\,dx + Q\,dy) = (Q_x - P_y)\,dx\wedge dy$ : the integrand of Green’s theorem. In one variable $d(f) = f'\,dx$ .

Worked example new (compare MATH 376 worksheets 13, 14 and 28)

Compute

d

of: (a)

f = x^2yz

; (b)

\omega = -y\,dx + x\,dy

; (c)

\omega = yz\,dx + xz\,dy + xy\,dz

; (d)

\omega = x^2\,dy\wedge dz + y^2\,dz\wedge dx + z^2\,dx\wedge dy

Solution.

(a) $df = 2xyz\,dx + x^2z\,dy + x^2y\,dz$ . The three partial derivatives, as in 5.2.
(b) $d\omega = (\partial_x(x) - \partial_y(-y))\,dx\wedge dy = 2\,dx\wedge dy$ . $Q_x - P_y = 1 - (-1)$ . Green’s theorem will turn this into twice the area.
(c) $P = yz$ , $Q = xz$ , $R = xy$ : $R_y - Q_z = x - x = 0$ , $P_z - R_x = y - y = 0$ , $Q_x - P_y = z - z = 0$ . So $d\omega = 0$ . It is the field of worksheet 28: curl zero, and indeed $\omega = d(xyz)$ .
(d) $A = x^2$ , $B = y^2$ , $C = z^2$ , so $d\omega = (2x + 2y + 2z)\,dx\wedge dy\wedge dz$ . The divergence of $(x^2, y^2, z^2)$ .

Answer. As in steps 1 to 4. Note (c): a form with $d\omega = 0$ that is also $d$ of something is the situation of the next theorem.

Theorem B Rudin 10.20(b); Spivak ch. 4

d squared is zero

If $\omega$ has $C^2$ coefficients then $d(d\omega) = 0$ .

In one lineFor every $C^2$ form $\omega$ , $d(d\omega) = 0$ . For $\omega = f$ a function in $\mathbb R^3$ this is $\operatorname{curl}\nabla f = 0$ , and for a 1-form it is $\operatorname{div}\operatorname{curl}F = 0$ .

Why it is trueOne identity, one proof, replaces both identities of 6.6. It is the reason closed and exact forms are different questions, and the starting point of de Rham cohomology.

ProofRudin 10.20

The idea: Differentiating twice gives second partial derivatives f_ij paired with dx_j wedge dx_i. Each pair appears twice with opposite signs: once as f_ij dx_j wedge dx_i and once as f_ji dx_i wedge dx_j, and f_ij = f_ji while the wedge is anti-symmetric.

For a function

f

df = \sum_if_i\,dx_i

, where

f_i = \partial f/\partial x_i

Definition:

d

of a function is the sum of its partials times the basic 1-forms.

d(df) = \sum_i\sum_jf_{ij}\,dx_j\wedge dx_i

, where

f_{ij} = \partial_j\partial_if

Apply

d

to each term

f_i\,dx_i

d(f_i\,dx_i) = df_i\wedge dx_i

= \sum_{j<i}f_{ij}\,dx_j\wedge dx_i + \sum_{j<i}f_{ji}\,dx_i\wedge dx_j

Split the double sum into the terms with

j<i

and

j>i

(the terms with

j = i

vanish because

dx_i\wedge dx_i = 0

), and rename in the second sum.

= \sum_{j<i}f_{ij}\big(dx_j\wedge dx_i - dx_j\wedge dx_i\big) = 0

Equality of mixed partials for

C^2

functions (5.5):

f_{ij} = f_{ji}

. And

dx_i\wedge dx_j = -dx_j\wedge dx_i

.These are the same two facts as in the proofs of 6.6, now stated once.

d(d\omega) = 0

for every

C^2

form.

\blacksquare

For a general form

f\,dx_I

d(f\,dx_I) = df\wedge dx_I

, and applying

d

again gives

d(df)\wedge dx_I = 0

by the previous steps, because

d(dx_I) = 0

(

dx_I

has constant coefficients).

Proof workshopRebuild the proof that d(df) = 0 for a function of three variablesRudin 10.20

Tap the lines in the order a proof would write them. Each line can only use what is already above it.

Why· One identity for two

Under the dictionary, $d\circ d = 0$ on a function is $\operatorname{curl}\circ\nabla = 0$ , and on a 1-form it is $\operatorname{div}\circ\operatorname{curl} = 0$ . In the plane, $d(df) = (f_{yx} - f_{xy})\,dx\wedge dy = 0$ is the test of 6.2: $P_y = Q_x$ for $P = f_x$ , $Q = f_y$ . So the ‘necessary condition for a gradient field’ of 6.2 is now one line: if $\omega = df$ then $d\omega = d(df) = 0$ .

Closed and exact

Definition B

Closed form

A form $\omega$ is closed if $d\omega = 0$ . For $\omega = P\,dx + Q\,dy$ this is $Q_x - P_y = 0$ . For $\omega = P\,dx + Q\,dy + R\,dz$ it is $\operatorname{curl}F = 0$ . For a 2-form it is $\operatorname{div}G = 0$ .

In one lineA form $\omega$ with $d\omega = 0$ .

Why it existsThis is the test that can be computed. For a 1-form in the plane it is $P_y = Q_x$ , the necessary condition of 6.2.

Definition B

Exact form

A form $\omega$ is exact if $\omega = d\eta$ for some form $\eta$ defined on the same region. A 1-form $\omega_F$ is exact exactly when $F$ is a gradient field, $F = \nabla f$ (6.2). The line integral of an exact 1-form between two points depends only on the points. A first order equation $P\,dx + Q\,dy = 0$ is an ‘exact differential equation’ (6.2) when the form $P\,dx + Q\,dy$ is exact.

In one lineA form $\omega$ that is $d\eta$ for some form $\eta$ (for a 1-form: $\omega = df$ , a gradient).

Why it existsThis is what we want: a potential. In 6.2 it meant a gradient field. By $d^2 = 0$ , every exact form is closed, and the question is when the converse holds.

By $d^2 = 0$ every exact form is closed. The converse is the question of 6.2 in new clothes, and the answer depends on the shape of the region.

Theorem B Rudin ch. 10; Spivak ch. 4; stated for 1-forms in 6.2

Poincare lemma

Let $U\subset\mathbb R^n$ be star-shaped (there is a point $p$ such that the segment from $p$ to any point of $U$ lies in $U$ ) and $\omega$ a closed $C^1$ form of degree $k\ge1$ on $U$ . Then $\omega = d\eta$ for a form $\eta$ of degree $k-1$ on $U$ .

In one lineOn a star-shaped region (such as a ball or all of $\mathbb R^n$ ) every closed form of degree $\ge1$ is exact.

Why it is trueIt is the converse of $d^2 = 0$ on nice regions, and it explains why a gradient field exists wherever curl is zero on $\mathbb R^3$ and why a vector potential exists wherever divergence is zero on $\mathbb R^3$ .

For 1-forms the proof is a formula, the integral along rays from the centre; the stretch problems carry it out. On a region with a hole the lemma fails, and this is exactly where closed and exact part ways.

Worked example new (compare MATH 376 homework 5 problem 4.1 and 6.2)

The angle form. Show that

\omega = \dfrac{-y\,dx + x\,dy}{x^2 + y^2}

is closed on the punctured plane, and that it is not exact there.

Solution.

Components. $P = \dfrac{-y}{x^2 + y^2}$ , $Q = \dfrac x{x^2 + y^2}$ .
Partial derivatives. $P_y = \dfrac{-(x^2 + y^2) + 2y^2}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}$ and $Q_x = \dfrac{(x^2 + y^2) - 2x^2}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}$ .
Closed. $d\omega = (Q_x - P_y)\,dx\wedge dy = 0$ on $\mathbb R^2\setminus\{0\}$ .
Not exact. On the unit circle $r(t) = (\cos t, \sin t)$ we have $\int\omega = \int_0^{2\pi}(\sin^2t + \cos^2t)\,dt = 2\pi\ne0$ . If $\omega = df$ , the line integral around every closed path would be $0$ (6.2).
What it is. In polar coordinates $\omega = d\theta$ , the change in angle. A function whose differential is $d\theta$ would be $\theta$ itself, which increases by $2\pi$ round the origin and cannot be a single-valued function.

Answer. Closed, not exact on the punctured plane (though exact on any region not circling the origin, such as the right half plane, where $\omega = d\arctan(y/x)$ ). The hole is what makes ‘closed’ differ from ‘exact’. Counting holes this way is called de Rham cohomology.

Worked example new (MATH 376 Midterm 1 problem 1 style, 6.2)

Decide whether

\omega = (2xy + y^3)\,dx + (x^2 + 3xy^2)\,dy

is closed and exact on

\mathbb R^2

, and find a potential. Then do the same for

y\,dx - x\,dy

Solution.

Closed test. $P_y = 2x + 3y^2$ and $Q_x = 2x + 3y^2$ . They are equal, so $d\omega = 0$ .
Exact. The plane is star-shaped, so by the Poincare lemma $\omega$ is exact.
Find the potential. $f_x = 2xy + y^3$ gives $f = x^2y + xy^3 + h(y)$ . Then $f_y = x^2 + 3xy^2 + h'(y)$ must equal $Q = x^2 + 3xy^2$ , so $h' = 0$ . The method of 6.2: integrate one equation, differentiate, compare.
Check. $df = (2xy + y^3)\,dx + (x^2 + 3xy^2)\,dy$ .
Second form. $y\,dx - x\,dy$ has $P_y = 1$ and $Q_x = -1$ , so $d\omega = (Q_x - P_y)\,dx\wedge dy = -2\,dx\wedge dy\ne0$ . It is not closed, hence not exact: $\int$ around a counterclockwise circle of radius $r$ is $-2\pi r^2$ .

Answer. $f = x^2y + xy^3$ (plus a constant) for the first; the second is not closed.

Your answerLet

\omega = (3x^2y + z)\,dx + x^3\,dy + x\,dz

. Find

\int_C\omega

along any path from

(0, 0, 0)

(1, 2, 3)

Pullback: how a form is integrated

To integrate a form over a curve or surface given by a parametrization, we substitute the parametrization into the form. That substitution has a name.

Definition B

Pullback

For $\varphi(u, v) = (x(u, v), y(u, v), z(u, v))$ , the pullback $\varphi^*\omega$ is the form in $du, dv$ obtained by replacing $x, y, z$ in the coefficients by $x(u, v), y(u, v), z(u, v)$ , and each $dx$ by $x_u\,du + x_v\,dv$ (similarly $dy$ , $dz$ ), and expanding with $du\wedge du = 0$ and $dv\wedge du = -du\wedge dv$ . For a curve $r(t)$ the same with $dx = x'(t)\,dt$ .

In one lineIf $\varphi$ is a map from parameters $(u, v)$ to $(x, y, z)$ , the pullback $\varphi^*\omega$ of a form $\omega$ is obtained by substituting $x = x(u, v)$ and so on into the coefficients and $dx = x_u\,du + x_v\,dv$ into the basic forms, then expanding with the wedge rules.

Why it existsIt is the exact meaning of 'substitute the parametrization', and it gives integrals over any parametrized piece in one rule.

The integral of a form over a parametrized piece is the integral of the pullback over the parameter region, with its orientation: for a curve $\int_C\omega = \int_a^b r^*\omega$ , and for a surface $\iint_S\omega = \iint_Tr^*\omega$ , where $f\,du\wedge dv$ is integrated as $\iint f\,du\,dv$ .

Proposition B Rudin ch. 10 and Spivak ch. 4; computation below

Forms contain the integrals of 6.1 and 6.5

Let $F = (P, Q, R)$ and $G = (A, B, C)$ .

In one lineFor a curve, $r^*(P\,dx + Q\,dy + R\,dz) = F(r)\cdot r'\,dt$ , so $\int_C\omega$ is the line integral. For a surface, $r^*(A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy) = G(r)\cdot(r_u\times r_v)\,du\wedge dv$ , so $\iint_S\omega$ is the flux.

Why it is trueIt shows that the two integrals of 6.1 and 6.5 are the same operation on different forms, and that the components of the fundamental vector product are the three wedge coefficients.

Proofdirect computation

The idea: Substitute and expand. The cross product appears because the coefficient of du wedge dv in dy wedge dz is the determinant y_u z_v minus y_v z_u.

r^*(P\,dx + Q\,dy + R\,dz) = (P(r)x' + Q(r)y' + R(r)z')\,dt = F(r)\cdot r'\,dt

Pullback of a 1-form along a curve:

dx = x'\,dt

dy = y'\,dt

dz = z'\,dt

r^*(dy\wedge dz) = (y_u\,du + y_v\,dv)\wedge(z_u\,du + z_v\,dv)

Pullback of

dy\wedge dz

along a surface:

dy = y_u\,du + y_v\,dv

and

dz = z_u\,du + z_v\,dv

= (y_uz_v - y_vz_u)\,du\wedge dv

Expand;

du\wedge du = dv\wedge dv = 0

and

dv\wedge du = -du\wedge dv

r^*(dz\wedge dx) = (z_ux_v - z_vx_u)\,du\wedge dv

and

r^*(dx\wedge dy) = (x_uy_v - x_vy_u)\,du\wedge dv

This is the first component of

r_u\times r_v = (y_uz_v - z_uy_v,\ z_ux_v - x_uz_v,\ x_uy_v - y_ux_v)

. The other two pullbacks are the cyclic copies.

r^*\omega = G(r)\cdot(r_u\times r_v)\,du\wedge dv

\blacksquare

Add with coefficients

A, B, C

evaluated at

r(u, v)

So the flux integral of 6.5 is $\iint G(r)\cdot(r_u\times r_v)\,du\,dv$ , and the sign: swapping the roles of $u$ and $v$ changes $du\wedge dv$ into $dv\wedge du = -du\wedge dv$ , which is why swapping the parameters reverses the orientation (6.5). The fundamental vector product is the pullback made visible.

Worked example new (compare 6.4, polar and spherical Jacobians)

Find the pullback of

dx\wedge dy

under polar coordinates

x = r\cos\theta

y = r\sin\theta

, and of

dx\wedge dy\wedge dz

under spherical coordinates. Use it to explain the Jacobian factors.

Solution.

Differentials. $dx = \cos\theta\,dr - r\sin\theta\,d\theta$ and $dy = \sin\theta\,dr + r\cos\theta\,d\theta$ .
Wedge. $dx\wedge dy = \cos\theta\cdot r\cos\theta\,dr\wedge d\theta + (-r\sin\theta)\sin\theta\,d\theta\wedge dr = r\cos^2\theta\,dr\wedge d\theta + r\sin^2\theta\,dr\wedge d\theta$ . The terms $dr\wedge dr$ and $d\theta\wedge d\theta$ vanish, and $d\theta\wedge dr = -dr\wedge d\theta$ .
Simplify. $dx\wedge dy = r\,dr\wedge d\theta$ .
Spherical. With $x = \rho\sin\varphi\cos\theta$ , $y = \rho\sin\varphi\sin\theta$ , $z = \rho\cos\varphi$ the same computation gives $dx\wedge dy\wedge dz = \det\dfrac{\partial(x, y, z)}{\partial(\rho, \varphi, \theta)}\,d\rho\wedge d\varphi\wedge d\theta = \rho^2\sin\varphi\,d\rho\wedge d\varphi\wedge d\theta$ .

Answer. $dx\wedge dy = r\,dr\wedge d\theta$ and $dx\wedge dy\wedge dz = \rho^2\sin\varphi\,d\rho\wedge d\varphi\wedge d\theta$ . In general the pullback of $dx_1\wedge\dots\wedge dx_n$ under a map is the Jacobian determinant times $du_1\wedge\dots\wedge du_n$ . That is the change of variables formula of 6.4: the $|J|$ factor is the coefficient of the pullback, and the absolute value appears when you turn an oriented integral into an area integral for a map that preserves orientation.

Your answerPull

dx\wedge dy

back under

x = u^2 - v

y = u + v^2

. The result is

J\,du\wedge dv

. What is

J

(u, v) = (1, 2)

Proposition B Rudin ch. 10 and Spivak ch. 4

Pullback commutes with d

Let $\varphi$ be a $C^2$ map and $\omega$ a $C^1$ form. Then $\varphi^*(d\omega) = d(\varphi^*\omega)$ .

In one lineFor a $C^2$ map $\varphi$ and a form $\omega$ , $\varphi^*(d\omega) = d(\varphi^*\omega)$ .

Why it is trueIt is the chain rule for forms, and it is the step that lets a theorem proved on a rectangle of parameters hold on every parametrized surface.

The case of a function $f$ is the chain rule: $\varphi^*(df) = \varphi^*(f_x\,dx + f_y\,dy + \dots) = (f_x\,x_u + f_y\,y_u + \dots)\,du + (\dots)\,dv = d(f\circ\varphi)$ . Higher degrees follow because $\varphi^*$ respects the wedge product and $d(d(\cdot)) = 0$ . As a check, the form $\omega = x\,dy$ and the map $x = uv$ , $y = u + v$ have $\varphi^*\omega = uv\,(du + dv)$ , so $d(\varphi^*\omega) = (v\,du + u\,dv)\wedge(du + dv) = (v - u)\,du\wedge dv$ after expanding ( $v\,du\wedge dv + u\,dv\wedge du$ ), while $d\omega = dx\wedge dy$ and $\varphi^*(dx\wedge dy) = (v\,du + u\,dv)\wedge(du + dv) = (v - u)\,du\wedge dv$ . They agree.

The one theorem

Theorem B Spivak ch. 4 and Rudin 10.33; the special cases are 6.2, 6.4 and 6.6

Generalized Stokes theorem

Let $M$ be a compact oriented $k$ -dimensional region (an interval, a curve, a plane region, a surface, a solid) with piecewise smooth boundary $\partial M$ oriented by the rule ‘outward normal first’: a basis of the boundary is positive if, with the outward normal in front, it is positive for $M$ . Let $\omega$ be a $C^1$ form of degree $k - 1$ on a neighbourhood of $M$ . Then $\int_{\partial M}\omega = \int_Md\omega$ .

In one lineIf $M$ is a compact oriented $k$ -dimensional region with boundary $\partial M$ carrying the induced orientation and $\omega$ is a $C^1$ $(k-1)$ -form on a neighbourhood of $M$ , then $\displaystyle\int_{\partial M}\omega = \int_Md\omega$ .

Why it is trueOne statement contains the fundamental theorem of calculus, the fundamental theorem for line integrals, Green's theorem, Stokes' theorem and the divergence theorem. It says the integral of a derivative over a region is the integral of the function over the boundary.

The special cases are exactly the theorems of the unit. Each line is a theorem you have used.

$M$	$\omega$	$d\omega$	$\int_{\partial M}\omega = \int_Md\omega$ is
an interval $[a, b]$ , $\partial M = \{b\} - \{a\}$	a function $f$	$f'\,dx$	$f(b) - f(a) = \int_a^bf'(x)\,dx$ , the fundamental theorem of calculus
a curve from $A$ to $B$	a function $\varphi$	$d\varphi = \nabla\varphi\cdot dr$	$\varphi(B) - \varphi(A) = \int_C\nabla\varphi\cdot dr$ (6.2)
a region $D$ in the plane	$P\,dx + Q\,dy$	$(Q_x - P_y)\,dx\wedge dy$	$\oint_{\partial D}P\,dx + Q\,dy = \iint_D(Q_x - P_y)\,dA$ , Green (6.4)
a surface $S$ in space	$P\,dx + Q\,dy + R\,dz$	the 2-form of $\operatorname{curl}F$	$\oint_{\partial S}F\cdot dr = \iint_S\operatorname{curl}F\cdot n\,dS$ , Stokes (6.6)
a solid $E$	$A\,dy\wedge dz + B\,dz\wedge dx + C\,dx\wedge dy$	$(\operatorname{div}G)\,dx\wedge dy\wedge dz$	$\iint_{\partial E}G\cdot n\,dS = \iiint_E\operatorname{div}G\,dV$ , Gauss (6.6)

The orientation rule ‘outward normal first’ reproduces every rule of 6.4 and 6.6: counterclockwise for a plane region, the right-hand rule for a surface, outward for a solid.

Why it is true

ProofSpivak ch. 4

The idea: Reduce to a cube. On the unit square, the boundary integral is four one-dimensional integrals and the area integral is two, and the fundamental theorem of calculus matches them. Then pull a general piece back to a cube with a parametrization, using that pullback commutes with d.

\displaystyle\int_{\partial I^2}\omega = \int_0^1P(x, 0)\,dx + \int_0^1Q(1, y)\,dy - \int_0^1P(x, 1)\,dx - \int_0^1Q(0, y)\,dy

The case of the unit square

I^2 = [0, 1]^2

and

\omega = P\,dx + Q\,dy

\partial I^2

is the bottom (

y = 0

, left to right), the right side (

x = 1

, up), the top (right to left) and the left side (down).

\displaystyle\int_{I^2}d\omega = \int_0^1\!\!\big[Q(1, y) - Q(0, y)\big]dy - \int_0^1\!\!\big[P(x, 1) - P(x, 0)\big]dx

On the other side

d\omega = (Q_x - P_y)\,dx\wedge dy

, and Fubini with the fundamental theorem of calculus in each variable.

The two sides of the theorem agree on the square. The same argument, with one pair of opposite faces per coordinate, proves it on a

k

-cube.

Compare the two lines: the four terms are the same.In dimension

k

the same happens: for each coordinate there are two opposite faces with opposite orientations, and the fundamental theorem of calculus in that coordinate pairs them. The sign pattern is the orientation rule.

For a piece

M = \varphi(I^k)

\int_{\partial M}\omega = \int_{\partial I^k}\varphi^*\omega

Definition of the integral over a parametrized piece

\varphi: I^k\to M

and its boundary:

\int_{\partial\varphi}\omega = \int_{\partial I^k}\varphi^*\omega

= \int_{I^k}d(\varphi^*\omega)

The cube case applied to the form

\varphi^*\omega

= \int_{I^k}\varphi^*(d\omega)

Pullback commutes with

d

= \int_Md\omega

\blacksquare

Definition of the integral of

d\omega

over

M

.A general region is cut into finitely many such pieces. The integrals over the common faces cancel, because neighbouring pieces traverse a shared face in opposite directions, as in the 'why the inner edges cancel' box of 6.6.

Proof workshopGive the reason for each line: from the cube to a parametrized pieceSpivak ch. 4

Each line of the proof is shown in order. Choose the reason that justifies it, then press Check.

$\displaystyle\int_{\partial M}\omega = \int_{\partial I^k}\varphi^*\omega$ .
$\displaystyle= \int_{I^k}d(\varphi^*\omega)$ .
$\displaystyle= \int_{I^k}\varphi^*(d\omega)$ .
$\displaystyle= \int_Md\omega$ .

Worked example new (the proof of Stokes via forms)

Recover Stokes' theorem of 6.6 from Green's theorem and pullback.

Solution.

Setup. Let $r: T\to S$ be a smooth parametrization on a rectangle $T$ , with $r_u\times r_v$ pointing the way the normal $n$ does (so that $\partial T$ run counterclockwise maps to $\partial S$ with the induced orientation), and $\omega = P\,dx + Q\,dy + R\,dz$ the 1-form of $F$ .
Boundary side. $\oint_{\partial S}F\cdot dr = \int_{\partial T}r^*\omega$ , since a line integral is the integral of the pullback.
Green on the rectangle. $\int_{\partial T}r^*\omega = \iint_Td(r^*\omega)$ . Green’s theorem for the form $r^*\omega = p\,du + q\,dv$ , which is a planar 1-form on $T$ .
Pullback commutes with $d$ . $d(r^*\omega) = r^*(d\omega)$ .
The 2-form $d\omega$ is the curl. $d\omega = (R_y - Q_z)\,dy\wedge dz + \dots$ , and its pullback is $(\operatorname{curl}F)(r)\cdot(r_u\times r_v)\,du\wedge dv$ by the proposition above.

Answer. $\oint_{\partial S}F\cdot dr = \iint_T(\operatorname{curl}F)(r)\cdot(r_u\times r_v)\,du\,dv = \iint_S(\operatorname{curl}F)\cdot n\,dS$ : Stokes’ theorem. The proof in 6.6 for a graph was this argument with the chain rule written out.

Worked example new (compare 6.6 and 6.4)

Verify the generalized Stokes theorem for

\omega = x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy

on the ball of radius

R

, and for

\omega = \frac12(x\,dy - y\,dx)

on the ellipse

\frac{x^2}{a^2} + \frac{y^2}{b^2}\le1

Solution.

Ball, the sphere side. $r^*\omega = F(r)\cdot(r_\varphi\times r_\theta)\,d\varphi\wedge d\theta$ with $F = (x, y, z) = R\hat\rho$ and $r_\varphi\times r_\theta = R^2\sin\varphi\,\hat\rho$ outward, so $F\cdot N = R^3\sin\varphi$ and $\iint = 2\pi\cdot2\cdot R^3 = 4\pi R^3$ .
Ball, the volume side. $d\omega = (1 + 1 + 1)\,dx\wedge dy\wedge dz = 3\,dx\wedge dy\wedge dz$ , and $\iiint3\,dV = 3\cdot\frac43\pi R^3 = 4\pi R^3$ . They agree.
Ellipse, the volume (area) side. $d\omega = \frac12(dx\wedge dy - dy\wedge dx) = dx\wedge dy$ , and $\iint dx\,dy = \pi ab$ .
Ellipse, the boundary side. With $x = a\cos t$ , $y = b\sin t$ : $r^*\omega = \frac12(a\cos t\cdot b\cos t + b\sin t\cdot a\sin t)\,dt = \frac12ab\,dt$ and $\int_0^{2\pi} = \pi ab$ . They agree.

Answer. $4\pi R^3$ and $\pi ab$ . The second is the area-by-line-integral formula of 6.4; the first is the volume of the ball as a third of the flux of the position vector.

Your answerUse the generalized Stokes theorem to compute

\iint_Sz\,dx\wedge dy

over the unit sphere

S

with the outward orientation.

Quick checkIn the generalized Stokes theorem for a solid

E

\mathbb R^3

, what degree is

\omega

Quick checkFor

M = [a, b]

and

\omega = f

a function, what is

\int_{\partial M}\omega

Quick checkWhy can you not apply the generalized Stokes theorem to the angle form

d\theta

on the unit disc?

Python labStokes numerically on a wavy patch, and the sign flip when the parameters are swapped

import numpy as np

def F(x, y, z):     return (y * z, x**2, np.sin(x) + y * z)            # P, Q, R
def curlF(x, y, z): return (z, y - np.cos(x), 2 * x - z)               # (R_y - Q_z, P_z - R_x, Q_x - P_y), by hand
def surf(u, v):     return (u, v, 0.3 * np.sin(2 * u) * np.cos(3 * v) + 0.2 * u * v)
def d(f, u, v, i, h=1e-6):                                              # partial derivative of the surface, finite differences
    return (np.array(f(u + h * (i == 0), v + h * (i == 1))) - np.array(f(u - h * (i == 0), v - h * (i == 1)))) / (2 * h)

def both_sides(r, u0, u1, v0, v1, n=400):
    u = u0 + (np.arange(n) + .5) * (u1 - u0) / n; v = v0 + (np.arange(n) + .5) * (v1 - v0) / n
    U, V = np.meshgrid(u, v, indexing="ij")
    N = np.cross(d(r, U, V, 0), d(r, U, V, 1), axis=0)                  # r_u x r_v
    c = curlF(*r(U, V)); surface = sum(ci * Ni for ci, Ni in zip(c, N)).sum() * (u1 - u0) * (v1 - v0) / n**2
    def edge(a, t0, t1, along):                  # pull back F . dr to one edge: along = 0, u moves; along = 1, v moves
        t = t0 + (np.arange(n) + .5) * (t1 - t0) / n
        uu, vv = (t, a + 0 * t) if along == 0 else (a + 0 * t, t)
        rr = d(r, uu, vv, along); f = F(*r(uu, vv)); return sum(fi * ri for fi, ri in zip(f, rr)).sum() * (t1 - t0) / n
    line = edge(v0, u0, u1, 0) + edge(u1, v0, v1, 1) - edge(v1, u0, u1, 0) - edge(u0, v0, v1, 1)
    return round(line, 5), round(surface, 5)

print("r(u, v):  (circulation, flux of curl) =", both_sides(surf, 0, 1, 0, 2))
print("r(v, u):  (circulation, flux of curl) =", both_sides(lambda u, v: surf(v, u), 0, 2, 0, 1))

Try this. Run it. The surface is the graph z = 0.3 sin(2u) cos(3v) + 0.2uv over the rectangle [0,1] by [0,2], and the field is F = (yz, x^2, sin x + yz). The first number is the line integral of F round the edge of the patch, computed by pulling F . dr back to the four edges of the parameter rectangle. The second is the integral of curl F . (r_u x r_v) over the rectangle, from the cross product of finite-difference tangent vectors. Both are 1.2596: Stokes' theorem on a surface with no symmetry. The second line swaps the parameters (u and v exchange roles, the rectangle becomes [0,2] by [0,1]): the same surface, but both numbers become -1.2596, because the orientation reversed. Change the surface, the field (update the hand-computed curl, P_z, R_x and so on) or the rectangle, and check that the two numbers agree every time.

Where it is used

Used inside maths· Inside maths: counting holes, and complex analysis

The angle form $d\theta$ is closed and not exact on the punctured plane. The set of closed 1-forms modulo exact ones is a vector space, the first de Rham cohomology group, and its dimension counts the independent holes of the region: $1$ for the punctured plane, $0$ for the plane. In complex analysis, $dz/z = dr/r + i\,d\theta$ and $\oint_{\lvert z\rvert=1}dz/z = i\int_0^{2\pi}d\theta = 2\pi i$ because the integral of the exact part $dr/r = d(\ln r)$ around a closed path is $0$ and the angle form contributes $2\pi$ . That single computation is the heart of the residue theorem. In the other direction, the divergence-free field $G = (x, y, z)/\rho^3$ of 6.6 is closed and not exact as a 2-form on $\mathbb R^3\setminus\{0\}$ : its integral over the unit sphere is $4\pi$ , whereas the integral of an exact 2-form over a closed surface is $0$ .

Used in the real world· Used in the real world: heat is not exact, entropy is

For one mole of an ideal gas, the heat absorbed in a small change of temperature $T$ and volume $V$ is the 1-form $\delta Q = C_V\,dT + \dfrac{RT}V\,dV$ , with $R = 8.314$ J/(mol K) and $C_V = \frac32R$ for a monatomic gas. Is $\delta Q$ closed? $d(\delta Q) = \dfrac RV\,dT\wedge dV\neq0$ (the coefficient of $dV$ has a $T$ -derivative $R/V$ , and the coefficient of $dT$ has none). So heat is not exact: there is no function $Q(T, V)$ whose change is the heat absorbed, and the heat depends on the path, as the line integral of a non-closed form does. Divide by $T$ : $\dfrac{\delta Q}T = C_V\dfrac{dT}T + R\dfrac{dV}V$ has $d\big(\dfrac{\delta Q}T\big) = 0$ because the coefficient of $dV$ does not depend on $T$ and that of $dT$ does not depend on $V$ . It is exact: $\dfrac{\delta Q}T = d(C_V\ln T + R\ln V)$ . That function is the entropy $S$ , and $1/T$ is an integrating factor in the sense of the exact differential equations of 6.2. From $300$ K and $1$ litre to $600$ K and $2$ litres, with $C_V = \frac32R$ , the change is $\Delta S = (\frac32R + R)\ln2 = 2.5\cdot8.314\cdot0.693\approx14.4$ J/K, along any path. A heat engine designer uses this: the area enclosed in a temperature-entropy diagram is the net heat of a cycle.

Your answerOne mole of a monatomic ideal gas (

C_V = \frac32R

R = 8.314

J/(mol K)) goes from

(T, V) = (300\text{ K}, 1\text{ L})

(600\text{ K}, 2\text{ L})

. The form

\delta Q/T

is exact. What is the entropy change in J/K?

A preview· Preview: Maxwell's equations in two lines

In 6.8 the four Maxwell equations are written with $\nabla\cdot$ and $\nabla\times$ . In the language of this chapter the electric and magnetic fields together make up one 2-form $F$ in four-dimensional spacetime, and two of the equations (no magnetic monopoles, and Faraday’s law) say $dF = 0$ , while the other two (Gauss’s law and Ampere’s law with Maxwell’s correction) say that $d$ of the Hodge star of $F$ equals the current 3-form. That $dF = 0$ implies that $F = dA$ locally, with $A$ the four-potential, by the Poincare lemma. The integral forms of the equations in 6.8 are the generalized Stokes theorem for those forms.

Traps

Watch out· The wedge is not a product of numbers

$dx\wedge dy = -dy\wedge dx$ and $dx\wedge dx = 0$ . A form $dx\wedge dy$ is not the same as the area element $dx\,dy$ : the first has an orientation and the second does not. Integrating $dy\wedge dx$ over a region gives minus its area, unless you swap the orientation of the region at the same time. Reordering without the sign is the most common error.

Watch out· Substitute dx too, not only x

To pull back, replace the coefficients and also each differential: $dx = x_u\,du + x_v\,dv$ . Leaving $dx$ unchanged, or replacing it by $x'(t)$ without the $dt$ , is the form version of the slip of dotting $F(x, y)$ with $r'(t)$ before substituting the path (see the traps of 6.1). The method of 6.1 (parametrize, differentiate, substitute, dot, integrate) is the pullback.

Watch out· Closed does not mean exact, and d squared needs C two

$d\omega = 0$ is a local test; $\omega = df$ is global. The angle form is closed everywhere on the punctured plane and exact nowhere around the hole. Also $d^2 = 0$ needs continuous second partials: it is the equality of mixed partials in disguise.

Watch out· Notation of components

On the 376 midterm, a student wrote the components of $F$ as $F_x, F_y, F_z$ , which invites confusion with partial derivatives. In the language of forms the coefficients are $P, Q, R$ (or $F_1, F_2, F_3$ ), and a subscript $x$ always means a derivative. Write the component names first and the derivatives after.

Problems

check

Compute $d\omega$ for $\omega = x^2y\,dx + xy^2\,dy$ .

Solution

$P = x^2y$ , $Q = xy^2$ . $d\omega = (Q_x - P_y)\,dx\wedge dy = (y^2 - x^2)\,dx\wedge dy$ .

check

Compute $(dx + dy)\wedge(dx - dy)$ and $(x\,dy + y\,dx)\wedge dx$ .

Solution

$(dx + dy)\wedge(dx - dy) = dx\wedge dx - dx\wedge dy + dy\wedge dx - dy\wedge dy = -dx\wedge dy - dx\wedge dy = -2\,dx\wedge dy$ . And $(x\,dy + y\,dx)\wedge dx = x\,dy\wedge dx + y\,dx\wedge dx = -x\,dx\wedge dy$ .

checknew (the pullback of 6.1)

Compute the pullback of $\omega = y\,dx + x\,dy$ under $r(t) = (t^2, t^3)$ and integrate from $t = 0$ to $1$ . Compare with the potential.

Solution

$dx = 2t\,dt$ , $dy = 3t^2\,dt$ , so $r^*\omega = t^3\cdot2t\,dt + t^2\cdot3t^2\,dt = 5t^4\,dt$ and the integral is $1$ . The potential is $f = xy$ with $f(1, 1) - f(0, 0) = 1$ .

check

Which of these 1-forms on $\mathbb R^2$ is closed: $y\,dx + x\,dy$ , $y\,dx - x\,dy$ , $x\,dx + y\,dy$ ?

Solution

$d$ of $P\,dx + Q\,dy$ is $(Q_x - P_y)\,dx\wedge dy$ : first $1 - 1 = 0$ (closed, $= d(xy)$ ); second $-1 - 1 = -2$ (not closed); third $0 - 0 = 0$ (closed, $= d\frac{x^2 + y^2}2$ ).

usenew (Final 3 of MATH 376 in the language of forms)

Show that $\omega = yz\cos(xyz)\,dx + xz\cos(xyz)\,dy + xy\cos(xyz)\,dz$ is exact, and find $\int_C\omega$ for $C$ the unit circle in the plane $z = 0$ .

Solution

$\omega = d\big(\sin(xyz)\big)$ by the chain rule: $\partial_x\sin(xyz) = yz\cos(xyz)$ , and the same for $y$ , $z$ . So $\omega$ is exact and the integral around a closed curve is $0$ (the function returns to its value). Also $d\omega = d(d\sin(xyz)) = 0$ by $d^2 = 0$ : this is the curl computed term by term in 6.6, in one line.

usenew (generalized Stokes, a surface integral by boundary)

Compute $\iint_Sx\,dy\wedge dz$ where $S$ is the unit sphere with the outward orientation.

Solution

$\omega = x\,dy\wedge dz$ has $d\omega = dx\wedge dy\wedge dz$ . By the generalized Stokes theorem with $M$ the unit ball, $\iint_S\omega = \iiint_Bd\omega = \operatorname{Vol}(B) = \frac{4\pi}3$ . (Equivalently the flux of $(x, 0, 0)$ , with divergence $1$ .)

usenew (pullback with a Jacobian)

Pull $dx\wedge dy$ back under $x = 2u + v$ , $y = u - 3v$ and compute the area of the image of the square $[0, 1]^2$ .

Solution

$dx = 2\,du + dv$ , $dy = du - 3\,dv$ . $dx\wedge dy = 2\cdot(-3)\,du\wedge dv + 1\cdot1\,dv\wedge du = (-6 - 1)\,du\wedge dv = -7\,du\wedge dv$ . This is $\det\begin{bmatrix}2&1\\1&-3\end{bmatrix} = -7$ . The map reverses orientation, and the area is $\lvert-7\rvert\cdot1 = 7$ .

usenew (thermodynamics, 6.7)

Show that $\omega = p\,dV + V\,dp$ is exact and $\omega' = p\,dV$ is not closed on the $(V, p)$ -plane. What does the second say about the work $\int p\,dV$ done by a gas?

Solution

Write $\omega = p\,dV + V\,dp$ : in coordinates $(V, p)$ with the form $P\,dV + Q\,dp$ , $P = p$ , $Q = V$ , $d\omega = (Q_V - P_p)\,dV\wedge dp = (1 - 1)\,dV\wedge dp = 0$ , and $\omega = d(pV)$ . For $\omega' = p\,dV$ : $P = p$ , $Q = 0$ , $d\omega' = (0 - 1)\,dV\wedge dp = -dV\wedge dp\neq0$ . So the work $\int p\,dV$ depends on the path in the $(V, p)$ -plane: for a cycle it equals the area enclosed, which is why a heat engine can do net work.

provenew (rule for d of a product)

Prove that for a function $f$ and a 1-form $\omega$ , $d(f\omega) = df\wedge\omega + f\,d\omega$ .

Solution

Write $\omega = \sum_iP_i\,dx_i$ . Then $f\omega = \sum_ifP_i\,dx_i$ , and by the definition $d(fP_i\,dx_i) = d(fP_i)\wedge dx_i = (f\,dP_i + P_i\,df)\wedge dx_i$ by the product rule for functions. Sum over $i$ : $d(f\omega) = f\sum_idP_i\wedge dx_i + df\wedge\sum_iP_i\,dx_i = f\,d\omega + df\wedge\omega$ .

provenew (exact implies closed, with the line integral)

Prove that if $\omega = P\,dx + Q\,dy$ is exact with $C^2$ potential $f$ , then $\oint_C\omega = 0$ for every closed piecewise smooth path, and $d\omega = 0$ .

Solution

$\omega = df$ means $P = f_x$ and $Q = f_y$ . Then $d\omega = (Q_x - P_y)\,dx\wedge dy = (f_{yx} - f_{xy})\,dx\wedge dy = 0$ by equality of mixed partials. For a closed path $r$ on $[a, b]$ , $\oint\omega = \int_a^b(f_xx' + f_yy')\,dt = \int_a^b\frac{d}{dt}f(r(t))\,dt = f(r(b)) - f(r(a)) = 0$ , by the chain rule and the fundamental theorem of calculus.

provenew (determinant)

Prove that for a linear map $A$ of the plane, $A^*(dx\wedge dy) = \det A\,dx\wedge dy$ . Deduce that a linear map multiplies areas by $\lvert\det A\rvert$ .

Solution

Let $A = \begin{bmatrix}a&b\\c&d\end{bmatrix}$ , so the new coordinates are $x' = au + bv$ , $y' = cu + dv$ in terms of old ones. $dx' = a\,du + b\,dv$ and $dy' = c\,du + d\,dv$ . Then $dx'\wedge dy' = ad\,du\wedge dv + bc\,dv\wedge du = (ad - bc)\,du\wedge dv$ . The area of the image of a region $D$ is $\iint_D\lvert\det A\rvert\,du\,dv = \lvert\det A\rvert\operatorname{area}(D)$ , since the area counts regardless of orientation.

stretchnew (the Poincare lemma for 1-forms)

Let $\omega = \sum_iP_i\,dx_i$ be a closed $C^1$ 1-form on a ball centred at $0$ in $\mathbb R^n$ , meaning $\partial_jP_i = \partial_iP_j$ . Prove that $f(x) = \int_0^1\sum_iP_i(tx)x_i\,dt$ satisfies $df = \omega$ .

Solution

Differentiate under the integral sign: $\partial_jf = \int_0^1\Big[P_j(tx) + \sum_itx_i(\partial_jP_i)(tx)\Big]dt$ . By closedness, $\partial_jP_i = \partial_iP_j$ , so the sum is $\sum_itx_i(\partial_iP_j)(tx) = t\,\frac d{dt}\big[P_j(tx)\big]$ (chain rule: $\frac d{dt}P_j(tx) = \sum_ix_i(\partial_iP_j)(tx)$ ). So $\partial_jf = \int_0^1\Big[P_j(tx) + t\frac d{dt}P_j(tx)\Big]dt = \int_0^1\frac d{dt}\big[tP_j(tx)\big]dt = P_j(x)$ . Hence $df = \omega$ . The integral is the line integral of $\omega$ along the ray from $0$ to $x$ , the method of 6.2.

stretchnew (the cube proof in three variables)

Verify the generalized Stokes theorem on the unit cube $[0,1]^3$ for $\omega = f\,dy\wedge dz$ , and say what the other two kinds of term do.

Solution

$d\omega = f_x\,dx\wedge dy\wedge dz$ , so $\int_{[0,1]^3}d\omega = \iint\big[f(1, y, z) - f(0, y, z)\big]dy\,dz$ by Fubini and the fundamental theorem of calculus in $x$ . On the boundary, $dy\wedge dz$ restricted to the faces $y =$ const or $z =$ const is $0$ (one of $dy$ , $dz$ is zero there), so only the faces $x = 1$ (outward normal $+x$ , orientation $+$ ) and $x = 0$ (outward $-x$ , orientation $-$ ) contribute, giving $\iint f(1, y, z)\,dy\,dz - \iint f(0, y, z)\,dy\,dz$ , the same. The terms $g\,dz\wedge dx$ and $h\,dx\wedge dy$ are the same with the roles of the coordinates cycled, and adding the three gives the divergence theorem on the cube.

Summary

A differential form is the thing under an integral sign: it eats tiny oriented pieces (vectors, parallelograms, boxes) and returns numbers. 1-forms are line-integral integrands $P\,dx + Q\,dy + R\,dz$ (vector fields), 2-forms are flux integrands (vector fields again), 3-forms are volume integrands.
The wedge product is bilinear with $dx_i\wedge dx_j = -dx_j\wedge dx_i$ and $dx_i\wedge dx_i = 0$ . On 1-forms in space it is the cross product; in the plane it is the determinant.
The exterior derivative $d$ raises the degree by one: gradient on functions, curl on 1-forms, divergence on 2-forms. In the plane $d(P\,dx + Q\,dy) = (Q_x - P_y)\,dx\wedge dy$ .
$d^2 = 0$ , because mixed partials are equal and the wedge is anti-symmetric. This is both $\operatorname{curl}\nabla f = 0$ and $\operatorname{div}\operatorname{curl}F = 0$ .
A form is closed if $d\omega = 0$ and exact if $\omega = d\eta$ . Exact implies closed; the converse holds on star-shaped regions (Poincare lemma) and fails around holes: the angle form $d\theta$ .
The pullback substitutes a parametrization, with $dx = x_u\,du + x_v\,dv$ . It turns forms into the integrals of 6.1 and 6.5, the cross product $r_u\times r_v$ appears as the coefficient of $du\wedge dv$ , and the Jacobian determinant is the pullback of $dx\wedge dy$ .
Generalized Stokes: $\int_{\partial M}\omega = \int_Md\omega$ . It is proved on a cube by the fundamental theorem of calculus and extended by pullback. The cases are the fundamental theorem of calculus, the fundamental theorem for line integrals, Green, Stokes and Gauss.

Watch and read

The same ideas explained by someone else. Use these when an explanation here does not land.

6.7 Differential forms and the one big theorem

The picture: what a form eats

Forms, and the wedge product

One-form

Two-form

Differential form

Wedge product

The exterior derivative

Exterior derivative

d is gradient, curl and divergence

d squared is zero

Closed and exact

Closed form

Exact form

Poincare lemma

Pullback: how a form is integrated

Pullback

Forms contain the integrals of 6.1 and 6.5

Pullback commutes with d

The one theorem

Generalized Stokes theorem

Why it is true

Where it is used

Traps

Problems

Summary

Watch and read

Where this sits

Comes from

Leads to

Same idea elsewhere