Introdution to differential forms

Introduction to differential forms
Donu Arapura
May 6, 2016
The calculus of differential forms give an alternative to vector calculus which
is ultimately simpler and more flexible. Unfortunately it is rarely encountered
at the undergraduate level. However, the last few times I taught undergraduate
advanced calculus I decided I would do it this way. So I wrote up this brief
supplement which explains how to work with them, and what they are good for,
but the approach is kept informal. In particular, multlinear algebra is kept to
a minimum, and I don’t define manifolds or anything like that. By the time I
got to this topic, I had covered a certain amount of standard material, which is
briefly summarized at the end of these notes.
My thanks to João Carvalho, John Crow, Matúˇs Goljer and Josh Hill for
catching some typos.
Contents
1 1-forms 2
1.1 1-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2 Exactness in R2
. . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Parametric curves . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Line integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.5 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.6 Green’s theorem for a rectangle . . . . . . . . . . . . . . . . . . . 8
1.7 Exercise Set 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2 2-forms 10
2.1 Wedge product . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.2 2-forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
2.3 Exactness in R3
and conservation of energy . . . . . . . . . . . . 13
2.4 Derivative of a 2-form and divergence . . . . . . . . . . . . . . . 14
2.5 Poincaré’s lemma for 2-forms . . . . . . . . . . . . . . . . . . . . 15
2.6 Exercise Set 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
3 Surface integrals 18
3.1 Parameterized Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 18
3.2 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
1

3.3 Surface Integrals (continued) . . . . . . . . . . . . . . . . . . . . 22
3.4 Length and Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3.5 Exercise Set 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4 Stokes’ Theorem 29
4.1 Green and Stokes . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
4.2 Proof of Stokes’ theorem . . . . . . . . . . . . . . . . . . . . . . . 30
4.3 Cauchy’s theorem* . . . . . . . . . . . . . . . . . . . . . . . . . . 32
4.4 Exercise Set 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
5 Gauss’ theorem 37
5.1 Triple integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.2 Gauss’ theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
5.3 Proof for the cube . . . . . . . . . . . . . . . . . . . . . . . . . . 38
5.4 Gravitational Flux . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5.5 Laplace’s equation* . . . . . . . . . . . . . . . . . . . . . . . . . . 41
5.6 Exercise Set 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6 Beyond 3 dimensions* 44
6.1 Beyond 3D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6.2 Maxwell’s equations in R4
. . . . . . . . . . . . . . . . . . . . . . 47
7 Further reading 49
A Review of multivariable calculus 50
A.1 Differential Calculus . . . . . . . . . . . . . . . . . . . . . . . . . 50
A.2 Integral Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1 1-forms
1.1 1-forms
A differential 1-form (or simply a differential or a 1-form) on an open subset of
R2
is an expression F(x, y)dx + G(x, y)dy where F, G are R-valued functions on
the open set. A very important example of a differential is given as follows: If
f(x, y) is C1
R-valued function on an open set U, then its total differential (or
exterior derivative) is
df =
∂f
∂x
dx +
∂f
∂y
dy
It is a differential on U.
In a similar fashion, a differential 1-form on an open subset of R3
is an
expression F(x, y, z)dx+G(x, y, z)dy +H(x, y, z)dz where F, G, H are R-valued
functions on the open set. If f(x, y, z) is a C1
function on this set, then its total
differential is
df =
∂f
∂x
dx +
∂f
∂y
dy +
∂f
∂z
dz
2

At this stage, it is worth pointing out that a differential form is very similar
to a vector field. In fact, we can set up a correspondence:
Fi + Gj + Hk ↔ Fdx + Gdy + Hdz
where i, j, k are the standard unit vectors along the x, y, z axes. Under this set
up, the gradient f corresponds to df. Thus it might seem that all we are doing
is writing the previous concepts in a funny notation. However, the notation is
very suggestive and ultimately quite powerful. Suppose that that x, y, z depend
on some parameter t, and f depends on x, y, z, then the chain rule says
df
dt
=
∂f
∂x
dx
dt
+
∂f
∂y
dy
dt
+
∂f
∂z
dz
dt
Thus the formula for df can be obtained by canceling dt.
1.2 Exactness in R2
Suppose that Fdx + Gdy is a differential on R2
with C1
coefficients. We will
say that it is exact if one can find a C2
function f(x, y) with df = Fdx + Gdy
Most differential forms are not exact. To see why, note that the above equation
is equivalent to
F =
∂f
∂x
, G =
∂f
∂y
.
Therefore if f exists then
∂F
∂y
=
∂2
f
∂y∂x
=
∂2
f
∂x∂y
=
∂G
∂x
But this equation would fail for most examples such as ydx. We will call a
differential closed if ∂F
∂y and ∂G
∂x are equal. So we have just shown that if a
differential is to be exact, then it had better be closed.
Exactness is a very important concept. You’ve probably already encountered
it in the context of differential equations. Given an equation
dy
dx
=
F(x, y)
G(x, y)
we can rewrite it as
Fdx − Gdy = 0
If Fdx − Gdy is exact and equal to say, df, then the curves f(x, y) = c give
solutions to this equation.
These concepts arise in physics. For example given a vector field F =
F1i + F2j representing a force, one would like find a function P(x, y) called
the potential energy, such that F = − P. The force is called conservative (see
section 2.3) if it has a potential energy function. In terms of differential forms,
F is conservative precisely when F1dx + F2dy is exact.
3

1.3 Parametric curves
Before discussing line integrals, we have to say a few words about parametric
curves. A parametric curve in the plane is vector valued function C : [a, b] → R2
.
In other words, we let x and y depend on some parameter t running from a to
b. It is not just a set of points, but the trajectory of particle travelling along the
curve. To begin with, we will assume that C is C1
. Then we can define the the
velocity or tangent vector v = (dx
dt , dy
dt ). We want to assume that the particle
travels without stopping, v = 0. Then v gives a direction to C, which we also
refer to as its orientation. If C is given by
x = f(t), y = g(t), a ≤ t ≤ b
then
x = f(−u), y = g(−u), −b ≤ u ≤ −a
will be called −C. This represents the same set of points, but traveled in the
opposite direction.
Suppose that C is given depending on some parameter t,
x = f(t), y = g(t)
and that t depends in turn on a new parameter t = h(u) such that dt
du = 0.
Then we can get a new parametric curve C
x = f(h(u)), y = g(h(u))
It the derivative dt
du is everywhere positive, we want to view the oriented curves
C and C as the equivalent. If this derivative is everywhere negative, then −C
and C are equivalent. For example, the curves
C : x = cos θ, y = sin θ, 0 ≤ θ ≤ 2π
C : x = sin t, y = cos t, 0 ≤ t ≤ 2π
represent going once around the unit circle counterclockwise and clockwise re-
spectively. So C should be equivalent to −C. We can see this rigorously by
making a change of variable θ = π/2 − t.
It will be convient to allow piecewise C1
curves. We can treat these as unions
of C1
curves, where one starts where the previous one ends. We can talk about
parametrized curves in R3
in pretty much the same way.
1.4 Line integrals
Now comes the real question. Given a differential Fdx + Gdy, when is it exact?
Or equivalently, how can we tell whether a force is conservative or not? Checking
that it’s closed is easy, and as we’ve seen, if a differential is not closed, then
it can’t be exact. The amazing thing is that the converse statement is often
(although not always) true:
4

THEOREM 1.4.1 If F(x, y)dx + G(x, y)dy is a closed form on all of R2
with
C1
coefficients, then it is exact.
To prove this, we would need solve the equation df = Fdx + Gdy. In other
words, we need to undo the effect of d and this should clearly involve some kind
of integration process. To define this, we first have to choose a parametric C1
curve C. Then we define:
DEFINITION 1.4.2
C
Fdx + Gdy =
b
a
F(x(t), y(t))
dx
dt
+ G(x(t), y(t))
dy
dt
dt
If C is piecewise C1
, then we simply add up the integrals over the C1
pieces.
Although we’ve done everything at once, it is often easier, in practice, to do
this in steps. First change the variables from x and y to expresions in t, then
replace dx by dx
dt dt etc. Then integrate with respect to t. For example, if we
parameterize the unit circle c by x = cos θ, y = sin θ, 0 ≤ θ ≤ 2π, we see
−
y
x2 + y2
dx +
x
x2 + y2
dy = − sin θ(cos θ) dθ + cos θ(sin θ) dθ = dθ
and therefore
C
−
y
x2 + y2
dx +
x
x2 + y2
dy =
2π
0
dθ = 2π
From the chain rule, we get
LEMMA 1.4.3
−C
Fdx + Gdy = −
C
Fdx + Gdy
If C and C are equivalent, then
C
Fdx + Gdy =
C
Fdx + Gdy
While we’re at it, we can also define a line integral in R3
. Suppose that
Fdx + Gdy + Hdz is a differential form with C1
coeffients. Let C : [a, b] → R3
be a piecewise C1
parametric curve, then
DEFINITION 1.4.4
C
Fdx + Gdy + Hdz =
b
a
F(x(t), y(t), z(t))
dx
dt
+ G(x(t), y(t), z(t))
dy
dt
+ H(x(t), y(t), z(t))
dz
dt
dt
5

The notion of exactness extends to R3
automatically: a form is exact if it
equals df for a C2
function. One of the most important properties of exactness
is its path independence:
PROPOSITION 1.4.5 If ω is exact and C1 and C2 are two parametrized
curves with the same endpoints (or more acurately the same starting point and
ending point), then
C1
ω =
C2
ω
It’s quite easy to see why this works. If ω = df and C1 : [a, b] → R3
then
C1
df =
b
a
df
dt
dt
by the chain rule. Now the fundamental theorem of calculus shows that the
last integral equals f(C1(b)) − f(C1(a)), which is to say the value of f at the
endpoint minus its value at the starting point. A similar calculation shows that
the integral over C2 gives same answer. If the C is closed, which means that
the starting point is the endpoint, then this argument gives
COROLLARY 1.4.6 If ω is exact and C is closed, then C
ω = 0.
Now we can prove theorem 1.4.1. If Fdx + Gdy is a closed form on R2
, set
f(x, y) =
C
Fdx + Gdy
where the curve is indicated below:
(0,0) (x,0)
(x,y)
We parameterize both line segments seperately by x = t, y = 0 and x =
x(constant), y = t, and sum to get
f(x, y) =
x
0
F(t, 0)dt +
y
0
G(x, t)dt
6

Then we claim that df = Fdx + Gdy. To see this, we differentiate using the
fundamental theorem of calculus. The easy calculation is
∂f
∂y
=
∂
∂y
y
0
G(x, t)dt
= G(x, y)
Slightly trickier is
∂f
∂x
=
∂
∂x
x
0
F(x, 0)dt +
∂
∂x
y
0
G(x, t)dt
= F(x, 0) +
y
0
∂G(x, t)
∂x
dt
= F(x, 0) +
y
0
∂F(x, t)
∂t
dt
= F(x, 0) + F(x, y) − F(x, 0)
= F(x, y)
The same proof works if if we replace R2
by an open rectangle. However, it
will fail for more general open sets. For example,
−
y
x2 + y2
dx +
x
x2 + y2
dy
is C1
1-form on the open set {(x, y) | (x, y) = (0, 0)} which is closed. But it is
not exact (see exercise 6). In more advanced treatments, this failure of closed
forms to be exact can be measured by something called the de Rham cohomology
of the set.
1.5 Work
Line integrals have many important uses. One very direct application in physics
comes from the idea of work. If you pick up a rock off the ground, or perhaps
roll it up a ramp, it takes energy. The energy expended is called work. If you’re
moving the rock in straight line for a short distance, then the displacement can
be represented by a vector d = (∆x, ∆y, ∆z) and the force of gravity by a vector
F = (F1, F2, F3). Then the work done is simply
−F · d = −(F1∆x + F2∆y + F3∆z).
On the other hand, if you decide to shoot a rocket up into space, then you would
have to take into account that the trajectory c may not be straight nor can the
force F be assumed to be constant (it’s a vector field). However as the notation
suggests, for the work we would now need to calculate the integral
−
c
F1dx + F2dy + F3dz
7

One often writes this as
−
c
F · ds
(think of ds as the “vector” (dx, dy, dz).)
1.6 Green’s theorem for a rectangle
Let R be the rectangle in the xy-plane with vertices (0, 0), (a, 0), (a, b), (0, b).
Let C be the boundary curve of the rectangle oriented counter clockwise. Given
C1
functions P(x, y), Q(x, y) on R, the fundamental theorem of calculus yields
D
∂Q
∂x
dxdy =
b
0
[Q(a, y)) − Q(0, y)]dy =
C
Q(x, y)dy
D
∂P
∂y
dydx =
a
0
[P(x, b) − P(x, 0)]dx = −
C
P(x, y)dx
Subtracting yields Green’s theorem for R
THEOREM 1.6.1
C
Pdx + Qdy =
R
∂Q
∂x
−
∂P
∂y
dxdy
Our goal is to understand, and generalize to 3 dimensions, the operation
which takes the one form Pdx+Qdy to the integrand on the right. In traditional
vector calculus this is handled using the curl ( ×) which a vector ﬁeld deﬁned
so that
× (Pi + Qj + Rk) · k =
∂Q
∂x
−
∂P
∂y
is the integrand of the right in Green’s theorem. In general, one can discover
the formula for the other components of × (Pi + Qj + Rk) by expressing the
integrals of Pi + Qj + Rk around the boundaries of rectangles in the xz and yz
planes and rewriting them as double integrals. To make a long story short,
× (Pi + Qj + Rk) = (Ry − Qz)i + (Qx − Py)k + (Pz − Rx)j
(In practice, this is often written as a determinant
× (Pi + Qj + Rk) =
i j k
∂
∂x
∂
∂y
∂
∂z
P Q R
.
But this should really be treated as a memory aid and nothing more.)
8

1.7 Exercise Set 1
1. Determine which of the following 1-forms on R2
are exact. Express the
exact 1-forms in the form df.
(a) 3ydx + xdy
(b) ydx + xdy
(c) ex
ydx + ex
dy
(d) −ydx + xdy
2. It is sometimes possible make a 1-form exact by multiplying it by a nonzero
function of x and y called an integrating factor. In each of the nonexact
1-forms of problem 1, ﬁnd an integrating factor of the form xn
, for some
integer n.
3. An immediate consequence of Green’s theorem is that the area of a rect-
angle enclosed by C is C
xdy. Check this by direct calculation.
4. Let C be a circle of radius r centered at 0 oriented counterclockwise. Check
that C
xdy gives the area of the circle.
5. Let C1 be the helix x = cos θ, y = sin θ, z = θ, 0 ≤ θ ≤ 2π, and C2 be the
line segment connecting (0, 0, 0) to (0, 0, 2π). Calculate C1
zdx+xdz and
C2
zdx + xdz.
6. Let C be the unit circle centered at 0 oriented counterclockwise. Calculate
the integral C
− y
x2+y2 dx+ x
x2+y2 dy and check that it is nonzero. Conclude
that this 1-form is not exact.
7. Let r, θ be polar coordinates, so x = r cos θ, y = r sin θ. Convert dx and dy
to polar coordinates. Use this to calculate C
dx+dy where C is r = sin θ,
0 ≤ θ ≤ π.
8. Use the previous problem to solve for dr and dθ. Observe that dθ is the
same 1-form as in problem 6. So dθ isn’t exact, in spite of the way it is
written! How do you explain this?
9

2 2-forms
2.1 Wedge product
The cross product of vectors u × v is a very useful operation in 3 dimensional
geometry. It determines the area of the parallelogram containing these vectors
and the plane containing it. While there is no direct analogue of the cross
product in higher dimensions, there is an operation which determines the last
two things. Given (row) vectors u, v ∈ Rn
, define the matrix
u ∧ v =
1
2
(uT
v − vT
u) (1)
(∧ is pronounced “wedge”). In R3
(a, b, c) ∧ (d, e, f) =
1
2


0 ae − bd af − cd
−ae + bd 0 bf − ce
−af + cd −bf + ce 0


We can see that the nonzero entries are basically the same as for the cross
product,
(a, b, c) × (d, e, f) = (bf − ce)i + (−af + cd)j + (ae − bd)k
So these two operations are in some sense equivalent. The big difference is,
of course, that the wedge product produces a matrix, but not just any matrix.
Recall that a matrix A = (aij) is a skew symmetric if AT
= −A, i.e. aji = −aij.
Let ∧2
Rn
denote the space of all skew symmetric n × n real matrices.
THEOREM 2.1.1 The wedge product of two vectors lies in ∧2
Rn
.
This should be clear when n = 3 from the above formula. In general, we can
use standard facts from linear algebra to see that
1
2
(uT
v − vT
u)T
=
1
2
(vT
uT T
− uT
vT T
)
=
1
2
(vT
u − uT
v) = −
1
2
(uT
v − vT
u)
The following properties are easy to check from the definition.
v ∧ u = −u ∧ v (2)
u ∧ u = 0 (3)
and
c(u ∧ v) = (cu) ∧ v = u ∧ (cv) (4)
u ∧ v + u ∧ w = u ∧ (v + w) (5)
In practice, we will use these rules rather than the definition (1) for calculations.
To really be convinced that the wedge captures the essential geometric features
of the cross product, we note the following non-obvious fact.
10

THEOREM 2.1.2 The product u∧v determines the area of the parallelogram
spanned by u and v and the plane containing these vectors, when there is a
unique such plane.
It will be useful to work in a bit more generality. Recall that a vector space
V is an abstraction of Rn
, where elements of V , to be thought of as vectors,
can be added and multiplied by numbers and these operations satisfy standard
rules of algebra. Given a vector space V , we can construct a new vector space
of 2-vectors
∧2
V = {
i
ui ∧ vi | ui, vi ∈ V }
where the rules (2), (3), (4) and (5) are imposed, and only those rules. To see
how this compares to the earlier description. We can expand a vector in Rn
as
(a1, . . . , an) = a1e1 + . . . + anen
uniquely, where
e1 = (1, 0, . . . , 0), e2 = (0, 1, . . . , 0), . . .
In other words, the vectors ei form a basis of Rn
. Any 2-vector can be expanded
uniquely as an expression
n
i=1
i−1
j=1
aijei ∧ ej =
1
2
n
i=1
n
j=1
aijei ∧ ej
where the coefficients satisfy aij = −aji. The upshot is that we may identify
∧2
Rn
with the space of skew symmetric n × n matrices as we did above.
2.2 2-forms
A 2-form is an expression built using wedge products of pairs of 1-forms. On
R3
, this would be an expression:
F(x, y, z)dx ∧ dy + G(x, y, z)dy ∧ dz + H(x, y, z)dz ∧ dx
where F, G and H are functions defined on an open subset of R3
. Any wedge
product of two 1-forms can be put in this format. For example, using the above
rules, we can see that
(3dx + dy) ∧ (ex
dx + 2dy) = 3ex
dx ∧ dx + 6dx ∧ dy + ex
dy ∧ dx + 2dy ∧ dy
= (6 − ex
)dx ∧ dy
To be absolutely clear, we now allow c to be a function in (4).
The real significance of 2-forms will come later when we do surface integrals.
A 2-form will be an expression that can be integrated over a surface in the same
way that a 1-form can be integrated over a curve.
11

Earlier, we learned how to convert a vector field to a 1-form:
F1i + F2j + F3k ↔ F1dx + F2dy + F3dz
We can also convert vector fields to 2-forms and back
F1i + F2j + F3k ↔ F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy
Here we are matching i with dy ∧ dz, j with dz ∧ dx, and k with dx ∧ dy. A rule
that is easy to remember is as follows. To convert a 2-form to a vector, replace
dx, dy, dz by i, j, k and take cross products.
i~dx
j~dy
k~dz
To complete the picture, we can interchange 1-forms and 2-forms using the so
called Hodge star operator.
∗(F1dx + F2dy + F3dz) = F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy
∗(F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy) = F1dx + F2dy + F3dz
Given a 1-form F(x, y, z)dx + G(x, y, z)dy + H(x, y, z)dz. We want to define
its derivative dω which will be a 2-form. The rules we use to evaluate it are:
d(α + β) = dα + dβ
d(fα) = (df) ∧ α + fdα
d(dx) = d(dy) = d(dz) = 0
where α and β are 1-forms and f is a function. Recall that
df = fxdx + fydy + fzdz
where fx = ∂f
∂x and so on. Putting these together yields a formula
d(Fdx+Gdy +Hdz) = (Gx −Fy)dx∧dy +(Hy −Gz)dy ∧dz +(Fz −Hx)dz ∧dx
If we start with a vector field V = Fi + Gj + Hk, replace it by a 1-form
Fdx + Gdy + Hdz, apply d, then convert it back to a vector field, we end up
with the curl of V
× V = (Hy − Gz)i + (Gx − Fy)k + (Fz − Hx)j
12

2.3 Exactness in R3
and conservation of energy
A C1
1-form ω = Fdx + Gdy + Hdz is called exact if there is a C2
function
(called a potential) such that ω = df. A 1-form ω is called closed if dω = 0, or
equivalently if
Fy = Gx, Fz = Hx, Gz = Hy
These equations must hold when
F = fx, G = fy, H = fz
Therefore:
THEOREM 2.3.1 Exact 1-forms are closed.
We have a converse statement which is sometimes called “Poincaré’s lemma”.
THEOREM 2.3.2 If ω = Fdx + Gdy + Hdz is a closed form on R3
with C1
coefficients, then ω is exact. In fact if f(x0, y0, z0) = C
ω, where C is any
piecewise C1
curve connecting (0, 0, 0) to (x0, y0, z0), then df = ω.
This can be rephrased in the language of vector fields. If F = Fi + Gj + Hk
is C1
vector field representing a force, then it is called conservative if there is a
C2
real valued function P, called potential energy, such that F = − P. The
theorem implies that a force F, which is C1
on all of R3
, is conservative if and
only if × F = 0. P(x, y, z) is just the work done by moving a particle of unit
mass along a path connecting (0, 0, 0) to (x, y, z).
To appreciate the importance of this concept, recall from physics that the
kinetic energy of a particle of constant mass m and velocity
v =
dx
dt
,
dy
dt
,
dz
dt
is
K =
1
2
m||v||2
=
1
2
mv · v.
Also one of Newton’s laws says
m
dv
dt
= F
If F is conservative, then we can replace it by − P above, move it to the other
side, and then dot both sides by v to obtain
mv ·
dv
dt
+ v · P = 0 (6)
which can be simplified (exercise 6) to
d
dt
(K + P) = 0. (7)
This implies that the total energy K + P is constant.
13

2.4 Derivative of a 2-form and divergence
Earlier we defined wedge products of pairs of vectors. Now we extend it to
triples. Given three vectors u, v, w ∈ R3
, we may think of the 3-vector u∧v∧w
as the oriented volume of parallelopiped with u, v, w as the first, second and
third sides. Oriented volume is the usual volume if u, v, w is right-handed,
otherwise it is minus the usual volume. With these rules, we see that
u ∧ v ∧ w = −v ∧ u ∧ w = v ∧ w ∧ u = . . .
and that
u ∧ v ∧ w = 0
if any two of the vectors are equal. In addition, we have the distributive law
(u1 + u2) ∧ v ∧ w = u1 ∧ v ∧ w + u2 ∧ v ∧ w = 0
A 3-form is simply an expression
f(x, y, z)dx ∧ dy ∧ dz = −f(x, y, z)dy ∧ dx ∧ dz = f(x, y, z)dy ∧ dz ∧ dx = . . .
These are things that will eventually get integrated over solid regions. The
important thing for the present is an operation which takes 2-forms to 3-forms
once again denoted by “d”.
d(Fdy ∧ dz + Gdz ∧ dx + Hdx ∧ dy) = (Fxdx + Fydy + Fzdz) ∧ dy ∧ dz
+ (Gxdx + Gydy + Gzdz) ∧ dz ∧ dx
+ (Hxdx + Hydy + Hzdz) ∧ dx ∧ dy
This simplifies to
d(Fdy ∧ dz + Gdz ∧ dx + Hdx ∧ dy) = (Fx + Gy + Hz)dx ∧ dy ∧ dz
It’s probably easier to understand the pattern after converting the above 2-
form to the vector field V = Fi + Gj + Hk. Then the coefficient of dx ∧ dy ∧ dz
is the divergence
· V = Fx + Gy + Hz
So far we’ve applied d to functions (also called 0-forms) to obtain 1-forms,
and then to 1-forms to get 2-forms, and finally to 2-forms.
{0-forms}
d //

{1-forms}
d //

{2-forms}
d //

{3-forms}

{functions} // {vector fields}
×
// {vector fields}
· // {functions}
The real power of this notation is contained in the following simple-looking
formula
14

PROPOSITION 2.4.1 d2
= 0
What this means is that given a C2
real valued function defined on an open
subset of R3
, then d(df) = 0, and given a 1-form ω = Fdx + Gdy + Hdz with
C2
coefficents defined on an open subset of R3
, d(dω) = 0. Both of these are
quite easy to check:
d(df) = (fyx − fxy)dx ∧ dy + (fzy − fyz)dy ∧ dz + (fxz − fzx)dz ∧ dx = 0
d(dω) = [Gxz − Fyz + Hyx − Gzx + Fzy − Hxy]dx ∧ dy ∧ dz = 0
In terms of standard vector notation this is equivalent to
× ( f) = 0
· ( × V) = 0
2.5 Poincaré’s lemma for 2-forms
The analogue of theorem 2.3.2 holds:
THEOREM 2.5.1 If ω is a 2-form on R3
such that dω = 0, then there exists
a 1-form ξ such that dξ = ω.
In terms of vector calculus, this says that a vector field on R3
is a gradient
if it satisfies × F = 0. We will outline the proof, which gives a method for
finding ξ. Let
ω(x, y, z) = f(x, y, z)dx ∧ dy + g(x, y, z)dz ∧ dx + h(x, y, z)dz ∧ dy
The equation dω = 0 implies
fz = hx − gy (8)
The first step is to integrate out z. We define an operation called a homotopy
by
Hz(ω) =
z
0
g(x, y, z)dz dx +
z
0
h(x, y, z)dz dy
Differentiating, using the fundamental theorem of calculus and (8) yields
dHzω =
z
0
(hx − gy)dz dx ∧ dy
+ [g(x, y, z) − g(x, y, 0)]dz ∧ dy + [h(x, y, z) − h(x, y, 0)]dz ∧ dx
=
z
0
fzdz dx ∧ dy + [g(x, y, z) − g(x, y, 0)]dz ∧ dy + [h(x, y, z) − h(x, y, 0)]dz ∧ dx
= [f((x, y, z) − f(x, y, 0)]dx ∧ dy + [g(x, y, z) − g(x, y, 0)]dz ∧ dy + [h(x, y, z) − h(x, y, 0)]dz ∧ dx
We can write this as
dHzω = ω − ω(x, y, 0)
15

This doesn’t solve the problem but it simplifies it, because ω(x, y, 0) doesn’t
depend on z. Differentiating both sides of the last equation, shows that ω(x, y, 0)
is also closed. So if we can find ξ1 so that
ω(x, y, 0) = dξ1
then
ξ = ξ1 + Hzω
solves the original problem. To find ξ1, we reduce the problem further by inte-
grating out y and then x as above, to get
dHyω(x, y, 0) = ω(x, y, 0) − ω(x, 0, 0)
dHzω(x, 0, 0) = ω(x, 0, 0) − ω(0, 0, 0)
So now we have reduced the problem to finding ξ2 such that ω(0, 0, 0) = dξ2;
this is easy (exercise 7). Then using the above equations, we see that
ξ = ξ2 + Hzω + Hyω(x, y, 0) + Hxω(x, 0, 0)
gives a solution.
2.6 Exercise Set 2
Let α = xdx + ydy + zdz, β = zdx + xdy + ydz and γ = xydz in the following
problems.
1. Calculate
(a) α ∧ β
(b) α ∧ γ
(c) β ∧ γ
(d) (α + γ) ∧ (α + γ)
2. Calculate
(a) dα
(b) dβ
(c) d(α + γ)
(d) d(xα)
3. Given ω = fdx + gdy + hdz such that ω ∧ dz = 0, what can we conclude
about f, g and h?
4. (a) Let ω = Fdx + Gdy + Hdz and let C be the straight line connecting
(0, 0, 0) to (x0, y0, z0) show that
C
ω =
1
0
(F(x0t, y0, z0) + G(x0, y0t, z0) + H(x0, y0, z0t))dt
16

(b) Use this to prove theorem 2.3.2
5. Check that the following 1-forms are exact, and express them as df.
(a) dx + 2ydy + 3z2
dz
(b) zy cos(xy)dx + zx cos(xy)dy + sin(xy)dz.
6. Check that equations (6) and (7) in the text are equivalent.
7. Prove the last step of the proof of theorem 2.5.1 that a 2-form with con-
stant coeﬃcients is exact. Hint: observe that d(xdy) = dx ∧ dy etc.
8. Find a solution of dξ = ω, when ω = zdx ∧ dz + dy ∧ dz.
17

3 Surface integrals
3.1 Parameterized Surfaces
Recall that a parameterized curve is a C1
function from an interval [a, b] ⊂ R1
to
R3
. If we replace the interval by subset of the plane R2
, we get a parameterized
surface. Let’s look at a few of examples
1) The upper half sphere of radius 1 centered at the origin can be parame-
terized using cartesian coordinates



x = u
y = v
z =
√
1 − u2 − v2
u2
+ v2
≤ 1
2) The upper half sphere can be parameterized using spherical coordinates



x = sin(φ) cos(θ)
y = sin(φ) sin(θ)
z = cos(φ)
0 ≤ φ ≤ π/2, 0 ≤ θ 2π
(Since there is more than one convention, we should probably be clear about
this. For us, θ is the “polar” angle, and φ is the angle measured from the positve
z-axis.)
3) The upper half sphere can be parameterized using cylindrical coordinates



x = r cos(θ)
y = r sin(θ)
z =
√
1 − r2
0 ≤ r ≤ 1, 0 ≤ θ 2π
An orientation on a curve is a choice of a direction for the curve. For a
surface an orientation is a choice of “up” or “down”. The easist way to make
this precise is to view an orientation as a choice of (an upward, downward,
outward or inward pointing) unit normal vector field n on S. A parameterized
surface S 


x = f(u, v)
y = g(u, v)
z = h(u, v)
(u, v) ∈ D
is called smooth provided that f, g, h are C1
, the function that they define from
D → R3
is one to one, and the tangent vector fields
Tu =
∂x
∂u
,
∂y
∂u
,
∂z
∂u
Tv =
∂x
∂v
,
∂y
∂v
,
∂z
∂v
18

are linearly independent. In this case, once we pick an ordering of the variables
(say u ﬁrst, v second) an orientation is determined by the normal
n =
Tu × Tv
||Tu × Tv||
T
u
T v
n
FIGURE 1
S
v=constant
u=constant
If we look at the examples given earlier. (1) is smooth. However there is
a slight problem with our examples (2) and (3). Here Tθ = 0, when φ = 0 in
example (2) and when r = 0 in example (3). To deal with scenario, we will
consider a surface smooth if there is at least one smooth parameterization for
it.
Let C be a closed C1
curve in R2
and D be the interior of C. D is an example
of a surface with a boundary C. In this case the surface lies ﬂat in the plane,
but more general examples can be constructed by letting S be a parameterized
surface 


x = f(u, v)
y = g(u, v)
z = h(u, v)
(u, v) ∈ D ⊂ R2
then the image of C in R3
will be the boundary of S. For example, the boundary
19

of the upper half sphere S



x = sin(φ) cos(θ)
y = sin(φ) sin(θ)
z = cos(φ)
0 ≤ φ ≤ π/2, 0 ≤ θ 2π
is the circle C given by
x = cos(θ), y = sin(θ), z = 0, 0 ≤ θ ≤ 2π
In what follows, we will need to match up the orientation of S and its boundary
curve. This will be done by the right hand rule: if the ﬁngers of the right hand
point in the direction of C, then the direction of the thumb should be “up”.
n
S
C
FIGURE 2
3.2 Surface Integrals
Let S be a smooth parameterized surface



x = f(u, v)
y = g(u, v)
z = h(u, v)
(u, v) ∈ D
with orientation corresponding to the ordering u, v. The symbols dx etc. can
be converted to the new coordinates as follows
dx =
∂x
∂u
du +
∂x
∂v
dv
20

dy =
∂y
∂u
du +
∂y
∂v
dv
dx ∧ dy =
∂x
∂u
du +
∂x
∂v
dv ∧
∂y
∂u
du +
∂y
∂v
dv
=
∂x
∂u
∂y
∂v
−
∂x
∂v
∂y
∂u
du ∧ dv
=
∂(x, y)
∂(u, v)
du ∧ dv
In this way, it is possible to convert any 2-form ω to uv-coordinates.
DEFINITION 3.2.1 The integral of a 2-form on S is given by
S
Fdx∧dy+Gdy∧dz+Hdz∧dx =
D
F
∂(x, y)
∂(u, v)
+ G
∂(y, z)
∂(u, v)
+ H
∂(z, x)
∂(u, v)
dudv
In practice, the integral of a 2-form can be calculated by ﬁrst converting it
to the form f(u, v)du ∧ dv, and then evaluating D
f(u, v) dudv.
Let S be the upper half sphere of radius 1 oriented with the upward normal
parameterized using spherical coordinates, we get
dx ∧ dy =
∂(x, y)
∂(φ, θ)
dφ ∧ dθ = cos(φ) sin(φ)dφ ∧ dθ
So
S
dx ∧ dy =
2π
0
π/2
0
cos(φ) sin(φ)dφdθ = π
On the other hand if use the same surface parameterized using cylindrical
coordinates
dx ∧ dy =
∂(x, y)
∂(r, θ)
dr ∧ dθ = rdr ∧ dθ
Then
S
dx ∧ dy =
2π
0
1
0
rdrdθ = π
which leads to the same answer as one would hope. The general result is:
THEOREM 3.2.2 Suppose that an oriented surface S has two diﬀerent smooth
C1
parameterizations, then for any 2-form ω, the expression for the integrals of
ω calculated with respect to both parameterizations agree.
(This theorem needs to be applied to the half sphere with the point (0, 0, 1)
removed in the above examples.) Let us give the proof. Suppose that u, v
is a parameterization as above, and u = p(s, t), v = q(s, t) for one to one C1
-
functions p, q on a domain E in pq-plane. We also want to assume that ∂(u,v)
∂(s,t) 0
on E. This will ensure that s, t gives a new parameterization of S with the
21

correct orientation. Let us assume for simplicity that we have ω = Fdx ∧ dy.
Then computing S
ω in two ways gives either
D
F
∂(x, y)
∂(u, v)
dudv
or
E
F
∂(x, y)
∂(s, t)
dsdt
We have to prove that these integrals are equal. The chain rule can be expressed
as a matrix equation
∂x
∂s
∂x
∂t
∂y
∂s
∂y
∂t
=
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
∂u
∂s
∂u
∂t
∂v
∂s
∂v
∂t
Take the determinant of both sides to obtain
∂(x, y)
∂(s, t)
=
∂(x, y)
∂(u, v)
∂(u, v)
∂(s, t)
(9)
The change of variables formula for double integrals tells us that
D
F
∂(x, y)
∂(u, v)
dudv =
E
F
∂(x, y)
∂(u, v)
∂(u, v)
∂(s, t)
dsdt
We can drop the absolute value sign on the second Jacobian because it is posi-
tive. Now substitute (9) to obtain
D
F
∂(x, y)
∂(u, v)
dudv =
E
F
∂(x, y)
∂(s, t)
dsdt
3.3 Surface Integrals (continued)
Complicated surfaces may be divided up into nonoverlapping patches which can
be parameterized separately. The simplest scheme for doing this is to triangulate
the surface, which means that we divide it up into triangular patches as depicted
below. Each triangle on the surface can be parameterized by a triangle on the
plane.
22

boundary
We will insist that if any two triangles touch, they either meet only at a
vertex, or they share an entire edge. In addition, any edge lies on at most two
triangles. So the picture below is not a surface from our point of view. We deﬁne
the boundary of a surface to be the union of all edges which are not shared. The
surface is called closed if the boundary is empty.
Given a surface which has been divided up into patches, we can integrate a
2-form on it by summing up the integrals over each patch. However, we require
that the orientations match up, which is possible if the surface has “two sides”.
Below is a picture of a one sided, or nonorientable, surface called the Mobius
strip.
23

Once we have picked an orientation of S, we get one for the boundary using
the right hand rule.
In many situations arising in physics, one needs to integrate a vector field
F = F1i + F2j + F3k over a surface. The resulting quantity is often called a
flux, which is usually written as S
F·dS or S
F·n dS. As a typical example,
consider a fluid such as air or water. Associated to this, there is a scalar field
ρ(x, y, z) which measures the density, and a vector field v which measures the
velocity of the flow (e.g. the wind velocity). Then the rate at which the fluid
passes through a surface S is given by the flux integral S
ρv · dS.
So how does one actually define or compute the flux? For our purposes, we
can simply define it as
DEFINITION 3.3.1
S
F · dS =
S
F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy
It is probably easier to view this as a two step process, first convert F to a
2-form as follows:
F1i + F2j + F3k ↔ F1dy ∧ dz + F2dz ∧ dx + F3dx ∧ dy,
then integrate. We will say more about the traditional way of defining the flux
in the next section.
3.4 Length and Area
It is important to realize some line and surface integrals are not expressible as
integrals of differential forms in general. Two notable examples are the arclength
and area integrals.
24

DEFINITION 3.4.1 The arclength of C : [a, b] → R3
is given by
C
ds =
b
a
dx
dt
2
+
dy
dt
2
+
dz
dt
2
dt
The symbol ds is not a 1-form in spite of the notation. For example
−C
ds =
C
ds
whereas for 1-forms the integral would change sign. Nevertheless, ds (or more
accurately its square) is a sort of generalization of a diﬀerential form called a
tensor. To get a sense what this means, let us calculate the arclength of a curve
lying on a surface. Suppose that S is a parameterized surface given by



x = f(u, v)
y = g(u, v)
z = h(u, v)
(u, v) ∈ D
and suppose that C lies on S. This means that there are functions k, : [a, b] →
R such that x = f(k(t), (t)), . . . determines C. We can calculate the arclength
of C by applying the chain to the above integral all at once. Instead, we want
to break this down into a series of steps.
dx =
∂x
∂u
du +
∂x
∂v
dv
dy =
∂y
∂u
du +
∂y
∂v
dv
dz =
∂z
∂u
du +
∂z
∂v
dv
For the next step, we introduce a new product (indicated by juxtaposition)
which is distributive and unlike the wedge product is commutative. We square
the previous formulas and add them up. (The objects that we are getting are
tensors.)
dx2
=
∂x
∂u
2
du2
+ 2
∂x
∂u
∂x
∂v
dudv +
∂x
∂v
2
dv2
dy2
=
∂y
∂u
2
du2
+ 2
∂y
∂u
∂y
∂v
dudv +
∂y
∂v
2
dv2
dz2
=
∂z
∂u
2
du2
+ 2
∂z
∂u
∂z
∂v
dudv +
∂z
∂v
2
dv2
dx2
+ dy2
+ dz2
= Edu2
+ 2Fdudv + Gdv2
(10)
where
E = ||Tu||2
, F = Tu · Tv, G = ||Tv||2
25

in the notation of section 3.1. The expression in (10) is called the metric tensor
of the surface. We can easily deduce a formula for arclength in terms of it:
C
ds =
b
a
E
du
dt
2
+ 2F
du
dt
dv
dt
+ G
dv
dt
2
dt
The area of S can also be expressed in terms of the metric tensor. First
recall that
DEFINITION 3.4.2 The area of S is given by
S
dS =
D
||Tu × Tv||dudv
THEOREM 3.4.3 The area is given by
S
dS =
D
EG − F2dudv
The proof is as follows
||Tu × Tv||2
= ||Tu||2
||Tv||2
sin2
θ
= ||Tu||2
||Tv||2
(1 − cos2
θ)
= ||Tu||2
||Tv||2
− (Tu · Tv)2
= EG − F2
If S is sphere of radius 1 parameterized by spherical coordinates, a straight
forward calculation gives the metric tensor as
sin2
φ dθ2
+ dφ2
which yields
area(S) =
2π
0
π
0
sin φ dφdθ = 4π
as expected.
More generally, if f(x, y, z) is a scalar valued function, we can define
DEFINITION 3.4.4
S
f(x, y, z)dS =
D
f(x, y, z)||Tu × Tv||dudv
We can now explain the usual method of introducing the flux. Given a vector
field F, the dot product F·n with the unit normal field gives a scalar field. This
can now be integrated over S as above to obtain S
F · ndS. This coincides
with flux.
THEOREM 3.4.5 S
F · dS = S
F · ndS.
26

If F = F1i + F2j + F3k, then can compute
S
F · dS =
D
F1
∂(x, y)
∂(u, v)
+ F2
∂(y, z)
∂(u, v)
+ F3
∂(z, x)
∂(u, v)
dudv
=
D
F ·
∂(x, y)
∂(u, v)
i +
∂(y, z)
∂(u, v)
j +
∂(z, x)
∂(u, v)
k dudv
=
D
F · (Tu × Tv)dudv =
D
F ·
Tu × Tv
||Tu × Tv||
||Tu × Tv||dudv
=
S
F · ndS
In practice, however, this is usually not a very eﬃcient method for computing
the ﬂux.
3.5 Exercise Set 3
Let α = xdx + ydy + zdz, β = zdx + xdy + ydz and γ = xydz in the following
problems.
1. Let D be the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 1 oriented with the upward
normal. Calculate
(a) D
α ∧ β
(b) D
α ∧ γ
(c) D
β ∧ γ
2. Let S be the surface of the unit sphere x2
+ y2
+ z2
= 1 oriented with the
outward normal. Calculate
(a) S
zdx ∧ dy
(b) S
(i + j) · ndS
(c) S
dγ
3. Let S be the hyperboloid z2
− x2
− y2
= 1, −1 ≤ z ≤ 1 oriented with
outward normal. Calculate
S
xdx ∧ dz
x2 + y2
+
ydy ∧ dz
x2 + y2
(Try using cylindrical coordinates.)
4. Calculate the area of the intersection of the sphere x2
+ y2
+ z2
= 4 and
z ≥ 1.
5. Fix numbers a b 0. Draw a circle of radius b in the xz-plane centered
at (a, 0, 0), and rotate it about the z-axis to get a torus or doughnut shape
T.
27

Calculate the surface area of T. It may help to parameterize it by the angle
0 ≤ θ ≤ 2π which has the same meaning as in cylindrical coordinates, and
the angle 0 ≤ ψ ≤ 2π indicated in the picture. A little trigonometry gives
x = (a + b cos ψ) cos θ
y = (a + b cos ψ) sin θ
z = b sin ψ
6. With T as in the last problem, compute T
zdx ∧ dy.
28

4 Stokes’ Theorem
4.1 Green and Stokes
Stokes’ theorem is really the fundamental theorem of calculus for surface inte-
grals. We assume that the surfaces can be triangulated.
THEOREM 4.1.1 (Stokes’ theorem) Let S be an oriented smooth surface
with smooth boundary curve C. If C is oriented using the right hand rule, then
for any C1
1-form ω on an open set of R3
containing S,
S
dω =
C
ω
If the surface lies in the plane, it is possible make this very explicit:
THEOREM 4.1.2 (Green’s theorem) Let C be a closed C1
curve in R2
oriented counterclockwise and D be the interior of C. If P(x, y) and Q(x, y) are
both C1
functions then
C
Pdx + Qdy =
D
∂Q
∂x
−
∂P
∂y
dxdy
As an application, Green’s theorem shows that the area of D can be com-
puted as a line integral on the boundary
D
dxdy =
C
ydx
If S is a closed oriented surface in R3
, such as the surface of a sphere, Stoke’s
theorem shows that any exact 2-form integrates to 0, where a 2-form is exact if
it equals dω for some 1-form ω. To see this write S as the union of two surfaces
S1 and S2 with common boundary curve C. Orient C using the right hand
rule with respect to S1, then orientation coming from S2 goes in the opposite
direction. Therefore
S
dω =
S1
dω +
S2
dω =
C
ω −
C
ω = 0
In vector notation, Stokes’ theorem is written as
S
× F · n dS =
C
F · ds
where F is a C1
-vector field.
In physics, there a two fundamental vector fields, the electric field E and the
magnetic field B. They’re governed by Maxwell’s equations, one of which is
× E = −
∂B
∂t
29

where t is time. If we integrate both sides over S, apply Stokes’ theorem and
simplify, we obtain Faraday’s law of induction:
C
E · ds = −
∂
∂t S
B · n dS
To get a sense of what this says, imagine that C is wire loop and that we are
dragging a magnet through it. This action will induce an electric current; the
left hand integral is precisely the induced voltage and the right side is related to
the strength of the magnet and the rate at which it is being dragged through.
Stokes’ theorem works even if the boundary has several components. How-
ever, the inner an outer components would have opposite directions.
S
C
1
C
2
THEOREM 4.1.3 (Stokes’ theorem II) Let S be an oriented smooth sur-
face with smooth boundary curves C1, C2 . . .. If Ci is oriented using the right
hand rule, then for any C1
1-form ω on an open set of R3
containing S,
S
dω =
C1
ω +
C2
ω + . . .
4.2 Proof of Stokes’ theorem
Suppose that
ω = fdx + gdy + hdz
is a 1-form in R3
, and S is a surface parameterized by u, v. Let us denote by
π∗
ω = (f
∂x
∂u
+ g
∂y
∂u
+ h
∂z
∂u
)du + (f
∂x
∂v
+ g
∂y
∂v
+ h
∂z
∂v
)dv
the conversion of ω to uv-coordinates. Up to now, we have not used any special
notation for this process, but it is easy to get very confused at this point if we
don’t.
THEOREM 4.2.1
d(π∗
ω) = π∗
dω
ie. converting ω to uv-coordinates and diﬀerentiating is the same as diﬀerenti-
ating and then converting to uv-coordinates.
30

Both sides of the putative equation decompose over sums. So it is enough
to treat the case where ω consists of a single term, say fdx. Even then, it’s a
bit messy, but we only have to do this once. We have
dω = −fydx ∧ dy − fzdx ∧ dz
So
π∗
dω = −fy(xudu + xvdv) ∧ (yudu + yvdv) − fz(xudu + xvdv) ∧ (zudu + zvdv)
= (fy(xvyu − xuyv) + fz(xvzu − xuzv))du ∧ dv
On the other hand,
d(π∗
ω) = d(fxudu + fxvdv)
= (−fvxu −¨¨¨fxuv + fuxv +¨¨¨fxvu)du ∧ dv
= (−(fxxv + fyyv + fzzv)xu + (fxxu + fyyu + fzzu)xv)du ∧ dv
= ($$$$$
−fxxvxu − fyyvxu − fzzvxu +$$$$fxxuxv + fyyuxv + fzzuxv)du ∧ dv
= (fy(xvyu − xuyv) + fz(xvzu − xuzv))du ∧ dv
which is the same.
We are now ready to prove Stokes’ theorem. We ﬁrst assume that S consists
of a single triangle.
PROPOSITION 4.2.2 Stokes’ theorem holds when S has a single triangular
patch.
We parameterize S by a triangle D in the uv-plane. After a translation, we
can assume that one of the vertices is the origin. Then after a linear change of
variables
u = au + bv
v = cu + dv
we can assume that D is the triangle with vertices (0, 0), (1, 0), (0, 1).
Let us start with a 1-form
ω = fdx + gdy + hdz
and write
π∗
ω = P(u, v)du + Q(u, v)dv
Let C denote the boundary of D, which consists of line segments u = 0, 0 ≤ v ≤
1; v = 0, 0 ≤ u ≤ 1 and u + v = 1, 0 ≤ u, v ≤ 1. Therefore
C
ω =
C
Pdu + Qdv
=
1
0
P(u, 0)du −
1
0
P(u, 1 − u)du −
1
0
Q(0, v)dv +
1
0
Q(1 − v, v)dv
31

On the other hand,
S
dω =
D
(Qu − Pv)dudv =
1
0
1−v
0
Qududv −
1
0
1−u
0
Pvdvdu
=
1
0
Q(1 − v, v) − Q(0, v)dv −
1
0
P(u, 1 − u) − P(u, 0)du
So these are equal.
We are now ready to prove the full version of Stokes’ theorem. We triangulate
the surface S. Thus S is covered by triangular patches Si. We have
S
dω =
i Si
dω
By the last proposition, we can write this as
i Ci1
ω +
Ci2
ω +
Ci3
ω
where Ci1, Ci2, Ci3 are the edges of Si. We separate these into exterior curves,
which are contained in the boundary C, and interior curves. A curve Ci1 is an
interior curve if it also occurs as an edge, say Cj1 of an adjacent triangle Sj. If
we take into account the orientations, then we see that Cj1 = −Ci1.
j1Ci1 C
Therefore the sum of integrals along the interior curves cancel. The conclu-
sion is that
S
dω =
exterior Cik
Cik
ω =
C
ω
4.3 Cauchy’s theorem*
Recall that a complex number is an expression z = a + bi where a, b ∈ R and
i =
√
−1, so that i2
= −1. The components a and b are called the real and
imaginary parts of z. We can identify the set of complex numbers C with the
plane R2
= {(a, b) | a, b ∈ R}. Addition and subtraction of complex numbers
correspond to the usual vector operations:
(a + bi) ± (c + di) = (a ± c) + (b ± d)i
32

However, we can do more, such as multiplication and division:
(a + bi)(c + di) = (ac − bd) + (ad + bc)i
a + bi
c + di
=
(a + bi)(c − di)
(c + di)(c − di)
=
ac + bd
c2 + d2
+
bc − ad
c2 + d2
i
The next step is calculus. The power of complex numbers is evident in the
beautiful formula of Euler
eiθ
= cos(θ) + i sin(θ)
which uniﬁes the basic functions of calculus. Given a function f : C → C, we
can write it as
f(z) = f(x + yi) = u(x, y) + iv(x, y)
where x, y are real and imaginary parts of z ∈ C, and u, v are the real and
imaginary parts of f. f is continuous at z = a + bi if u and v are continuous
at (a, b) in the usual sense. So far there are no surprises. However, things get
more interesting when we deﬁne the complex derivative
f (z) = lim
h→0
f(z + h) − f(z)
h
Notice that h is a complex number. For the limit to exist, we should get the
same value no matter how it approaches 0. If h = ∆x approaches along the
x-axis, we get
f (z) = lim
∆y→0
[u(x + ∆x, y) − u(x, y)] + i[v(x + ∆x, y) − v(x, y)]
∆x
=
∂u
∂x
+ i
∂v
∂x
If h = ∆y i approaches along the y-axis, then
f (z) = lim
∆y→0
[u(x, y + ∆y) − u(x, y)] + i[v(x, y + ∆y) − v(x, y)]
i∆y
= −i
∂u
∂y
+
∂v
∂y
Setting these equal leads to the Cauchy-Riemann equations
∂u
∂x
=
∂v
∂y
,
∂v
∂x
= −
∂u
∂y
These equations have to hold when the complex derivative f (z) exists, and in
fact f (z) exists when they do. f is called analytic at z = a + bi when these
hold at that point. For example, z2
= x2
+ y2
+ 2xyi and
ez
= ex
cos(y) + iex
sin(x)
33

are analytic everywhere. But f(z) = ¯z = x − iy is not analytic anywhere.
A complex differential form is an expression α+iβ where α, β are differential
forms in the usual sense. Complex 1-forms can be integrated by the rule
C
α + iβ =
C
α + i
C
β
Suppose that f is analytic. Then expanding
f(z)dz = (u + iv)(dx + idy) = [udx − vdy] + i[vdx + udy]
Differentiating and applying the Cauchy-Riemann equations shows
d(f(z)dz) = −
∂u
∂y
+
∂v
∂x
dx ∧ dy + i
∂u
∂x
−
∂v
∂y
dx ∧ dy = 0
Therefore Stokes’ theorem implies what may be thought of as the fundamental
theorem of complex analysis:
THEOREM 4.3.1 (Cauchy’s theorem) If f(z) is analytic on a region with
boundary C then
C
f(z)dz = 0
Suppose we replace f(z) by g(z) = f(z)
z . This is analytic away from 0.
Therefore the theorem applies to the boundary of any region not containing
0. If C is a closed positively oriented curve whose interior U contains 0, then
applying Cauchy’s theorem to a U − Dr ,where Dr is a disk of small radius r in
U, shows that
C
g(z)dz =
Cr
g(z)dz (11)
Here Cr is a circle of radius r around 0. We can parameterize this with the help
of Euler’s formula by
z = r cos(θ) + ri sin(θ) = reiθ
, 0 ≤ θ ≤ 2π
Then dz = rieiθ
dθ, so that
Cr
g(z)dz = ri
2π
0
f(reiθ
)
reiθ
eiθ
dθ = i
2π
0
f(reiθ
)dθ
As r → 0, f(reiθ
) → f(0), therefore the above integral approaches 2πr. Since
(11) holds for all small r, it follows that this equality holds on the nose. There-
fore:
THEOREM 4.3.2 (Cauchy’s Integral Formula) If f(z) is analytic in the
interior of positively oriented closed curve C, then
f(0) =
1
2πi C
f(z)
z
dz
34

Using a change of variable z → z − a, we get a more general formula:
COROLLARY 4.3.3 (Cauchy’s Integral Formula II) With the same as-
sumptions, for any point a in the interior of C
f(a) =
1
2πi C
f(z)
z − a
dz
This formula has many uses. Among other things, it can be used to eval-
uate complicated definite integrals. For example, even if we didn’t know that
dx
x2+1 = tan−1
x, we could still integrate this from −∞ to ∞ as follows. Let C
be the union of the line segment [−R, R] with the semicircle C2 of radius R in
the upper half of the complex plane. Here we choose R large, and in particular
R 1. We can factor
1
z2 + 1
=
1
z + i
1
z − i
Set f(z) to the first factor on the right, and apply Cauchy’s formula with a = i
to obtain
C
1
z2 + 1
= 2πif(i) =
2πi
i + i
= π
The integral can be rewritten as
R
−R
dx
x2 + 1
+
C2
1
z2 + 1
Now let R → ∞, the first integral goes to what we want to evaluate, and the
second goes to 0 because its absolute value is bounded by a constant times 1/R.
Therefore ∞
−∞
dx
x2 + 1
= π
4.4 Exercise Set 4
1. Given a region D ⊂ R2
with a smooth boundary C.
(a) Show that the area of D is 1
2 C
xdy − ydx.
(b) Show that the area is 1
2 C
r2
dθ in polar coordinates.
2. Use the formula of 1(b) to compute the area contain in r = sin θ, 0 ≤ θ ≤
π.
3. Calculate S
× F · ndS, where S is the paraboloid z = x2
+ y2
, z ≤ 1,
and F = xi + yj + xk.
4. Let S be the part of the sphere of radius 2 centered at the origin below
the plane z = 1. We orient this using the outward normal. Calculate
S
zdy∧dz. (You can either do this directly, or try to use Stokes’ theorem
which might be easier.)
35

5. T be the half torus 0 ≤ θ ≤ π, 0 ≤ ψ ≤ 2π (see problem 5 of exercise set
3 for the notation). Calculate T
zdy ∧ dz (same hint as above).
6. Calculate
∞
−∞
dx
x2+x+1 as above.
36

5 Gauss’ theorem
5.1 Triple integrals
Recall that a 3-form is an expression ω = f(x, y, z)dx ∧ dy ∧ dz. Given a solid
region V ⊂ R3
, we define
V
f(x, y, z)dx ∧ dy ∧ dz =
V
f(x, y, z)dxdydz
Orientations are also implicit here. To be consistent with earlier rules, we have
to take
V
f(x, y, z)dy ∧ dx ∧ dz = −
V
f(x, y, z)dxdydz
for example. The rationale may be clearer if we distinguish left and right handed
coordinate systems. The system (x, y, z) is right handed because the thumb of
the right hand points in the direction of the third coordinate axis, when the
fingers are curled from first to second. If we reorder, say to (y, x, z) then we get
a left handed system. For more general coordinate systems, such as spherical,
it may be harder to visualize what right handed means. So let us give a more
precise mathematical definition. Suppose that u, v, w is a new ordered system
of coordinates, we will say that it is right handed if
∂(x, y, z)
∂(u, v, w)
0
holds everywhere; left handed means that this strictly negative. Now given ω,
we can convert it to an expression g(u, v, w)du ∧ dv ∧ dw. Let W be the region
corresponding to V in uvw-space. Then
V
ω = ±
W
g(u, v, w)dudvdw
where we choose a plus sign if these are right handed coordinates, and a minus
sign if these are left handed.
5.2 Gauss’ theorem
THEOREM 5.2.1 (Gauss’ theorem) Let V be the interior of a smooth closed
surface S oriented with the outward pointing normal. If ω is a C1
2-form on an
open subset of R3
containing V , then
V
dω =
S
ω
This is also called the divergence theorem. The reason for this name becomes
clear if we express this in standard vector notation, where it reads
V
· F dV =
S
F · n dS
37

where F is a C1
vector field.
As an application, consider a fluid with density ρ and velocity v. If S is
the boundary of a solid region V with outward pointing normal n, then the flux
S
ρv·n dS is the rate at which matter flows out of V . In other words, it is minus
the rate at which matter flows in, and this equals −∂/∂t V
ρdV . On the
other hand, by Gauss’ theorem, the above flux integral equals S
· (ρv) dV .
Therefore
V
· (ρv)dV = −
∂
∂t V
ρdV
which yields
V
· (ρv) +
∂ρ
∂t
dV = 0.
The only way this can hold for all possible regions V is that the integrand
· (ρv) +
∂ρ
∂t
= 0 (12)
This is one of the basic laws of fluid mechanics.
We can extend Gauss’ theorem to solids with disconnected boundary. Sup-
pose that S2 is a smooth closed oriented surface contained inside another such
surface S1. We use the outward pointing normal S1 and the inner pointing
normal on S2. Let V be region in between S1 and S2. Then,
THEOREM 5.2.2 (Gauss’ theorem II) If ω is a C1
2-form on an open
subset of R3
containing V , then
V
dω =
S1
ω +
S2
ω
5.3 Proof for the cube
We will just give the proof for the easiest case, where we can choose right handed
coordinates u, v, w, so that V corresponds to the the unit cube W = {(u, v, w) |
0 ≤ u, v, w ≤ 1}. The boundary consist of 6 pieces, the top
T = {(1, v, w) | 0 ≤ v, w ≤ 1}
the bottom
B = {(0, v, w) | 0 ≤ v, w ≤ 1}
and 4 sides that we won’t label.
The proof is similar to the proof of Stokes’ theorem. Let denote by π∗
ω the
conversion of ω to uvw-coordinates.
THEOREM 5.3.1 π∗
dω = dπ∗
ω
38

We skip the proof which would involve a messy computation. Now let us
write
π∗
ω = Pdv ∧ dw + Qdw ∧ du + Rdu ∧ dv
Then
S
ω =
T
Pdv ∧ dw −
B
Pdv ∧ dw + . . .
=
1
0
1
0
P(1, u, v)dudv −
1
0
1
0
P(0, u, v)dudv + . . . (13)
By the previous theorem
π∗
dω = (Pu + Qv + Rw)du ∧ dv ∧ dw
The integral
V
dω =
W
(Pu + Qv + Rw)dudvdw
By the fundamental theorem of calculus,
W
Pududvdw =
1
0
1
0
P(1, u, v)dudv −
1
0
1
0
P(0, u, v)dudv
and similar expression for the integrals of Qu and Rw. Adding these up gives
us (13).
5.4 Gravitational Flux
Place a “point particle” of mass m at the origin of R3
, then this generates a
force on any particle of unit mass at r = (x, y, z) given by
F =
−mr
r3
where r = ||r|| = x2 + y2 + z2. This has singularity at 0, so it is a vector ﬁeld
on R3
− {0}. The corresponding 2-form is given by
ω = −m
xdy ∧ dz + ydz ∧ dx + zdx ∧ dy
r3
(14)
Let BR be the ball of radius R around 0, and let SR be its boundary. Since the
outward unit normal to SR is just n = r/r. One should expect that the ﬂux
SR
F · ndS = −
m
R2
SR
dS
= −
m
R2
area(SR) = −
m
R2
(4πR2
)
39

This can be justified by theorem 3.4.5. This can also be checked directly. We
work in spherical coordinates. SR is given by



x = ρ sin(φ) cos(θ)
y = ρ sin(φ) sin(θ)
z = ρ cos(φ)
ρ = R
0 ≤ φ ≤ π, 0 ≤ θ 2π
Rewriting ω in these coordinates and simplifying:
ω = −m sin(φ)dφ ∧ dθ
Therefore,
SR
F · ndS =
SR
ω = −
2π
0
π
0
sin(φ)dφdθ = 4πm
as hoped. We claim that if S is any closed surface not containing 0
S
ω =
−4πm if 0 lies in the interior of S
0 otherwise
We can see that dω = 0 (exercise 5). Therefore, Gauss’ theorem yields
S
ω =
V
dω = 0
if the interior V does not contain 0. On the other hand, if V contains 0, let BR
be a small ball contained in V , and let V − BR denote part of V lying outside
of BR. We use the second form of Gauss’ theorem
S
ω −
SR
ω =
V −BR
dω = 0
We are subtracting the second surface integral, since we are supposed to use the
inner normal for SR. Thus
S
ω = −4πm
From here, we can easily extract an expression for the flux for several par-
ticles or even a continuous distribution of matter. If F is the force of gravity
associated to some mass distribution, for any closed surface S oriented by the
outer normal, then the flux
S
F · ndS
is −4π times the mass inside S. For a continuous distribution with density µ,
this is given by µdV . Applying Gauss’ theorem again, in this case, yields
V
( · F + 4πµ)dV = 0
for all regions V . Therefore
· F = −4πµ (15)
40

5.5 Laplace’s equation*
The Laplacian is a partial differential operator defined by
∆f =
∂2
f
∂x2
+
∂2
f
∂y2
+
∂2
f
∂z2
This can expressed using previous operators as ∆f = · ( f). As an example
of where this arises, suppose that F is a gravitational force, this is known to
be conservative so that F = − P. Substituting into (15) yields the Poisson
equation
∆P = 4πµ
In a vacuum, this reduces to Laplace’s equation
∆P = 0
A solution to Laplace’s equation is called a harmonic function. These are of
fundamental importance both in pure and applied mathematics. If we write
r = x2 + y2 + z2, then
P = −
m
r
+ Const.
is the potential energy associated to particle of mass m at 0. This is harmonic
away from the singularity 0.
We express ∆ in terms of forms as
∆fdx ∧ dy ∧ dz = d ∗ df
or simply
∆f = ∗d ∗ df
once we define ∗(gdx ∧ dy ∧ dz) = g. This last formula also works in the plane
provided we define
∗(fdx + gdy) = fdy − gdx
∗(fdx ∧ dy) = f
(The ∗-operator in n dimensions always takes p-forms to (n − p)-forms.)
As an exercise, let us work out the Laplace equation in polar coordinates,
and use this to determine the radially symmetric harmonic functions on the
plane. The key is the determination of the ∗-operator:
dr =
∂r
∂x
dx +
∂r
∂y
dy =
x
r
dx +
y
r
dy
Similarly
dθ = −
y
r2
dx +
x
r2
dy
So that
∗dr =
x
r
dy −
y
r
dx = rdθ
41

∗dθ = −
y
r2
dy −
x
r2
dx = −
1
r
dr
∗(dr ∧ dθ) = ∗(
1
r
dx ∧ dy) =
1
r
Thus
∆f = ∗d ∗
∂f
∂r
dr +
∂f
∂θ
dθ
= ∗d r
∂f
∂r
dθ −
1
r
∂f
∂θ
dr
=
1
r
∂f
∂r
+
∂2
f
∂r2
+
1
r2
∂2
f
∂θ2
If f is radially symmetric, then it depends only on r so we obtain
1
r
df
dr
+
d2
f
dr2
=
1
r
d
dr
r
df
dr
= 0
This diﬀerential equation can be solved using standard techniques to get
f(r) = C + D log r
for constants C, D. By a similar, but more involved, calculation we ﬁnd that
f(r) = C +
D
r
are the only radially symmetric harmonic functions in R3
, where as above we
write r instead of ρ for the distance from the origin. These are precisely the
physical solutions written at the beginning of this section.
5.6 Exercise Set 5
1. Calculate E
dx ∧ dy + 2xdy ∧ dz on the elliposoid x2
+ y2
+ 1
4 z2
= 1.
2. Give two proofs that if S is a closed surface then S
× F · ndS = 0.
3. Show that if S is a closed surface then S
zdx ∧ dy is the volume of the
region enclosed by S. Use this to calculate the area of the tetrahedron
bounded by planes x = 0, y = 0, z = 0 and x + y + z = h, where h 0.
4. Show that if S is a closed surface then S
zrdr ∧ dθ, using cylindrical
coordinates, is the volume of the region enclosed by S. Calculate the
volume of the solid x2
+ y2
≤ 1, 0 ≤ z ≤ 4 − x2
− y2
.
5. Prove that the volume of a region enclosed by S is
(a) 1
3 S
xdy ∧ dz + ydz ∧ dx + zdx ∧ dy
(b) 1
3 S
ρ3
sin(φ)dφ ∧ dθ in spherical coordinates.
42

6. Show that for any ﬁxed value of m = 0, ω given in (14) is closed i.e.
dω = 0 but not exact. Therefore theorem 2.5.1 fails for R3
− {0}.
7. Prove Green’s identities
(a) S
f ∗ dg = V
(df ∧ ∗dg + fd ∗ dg)
(b) df ∧ ∗dg = (fxgx + fygy + fzgz)dx ∧ dy ∧ dz = dg ∧ ∗df
(c) S
(f ∗ dg − g ∗ df) = V
(fd ∗ dg − gd ∗ df)
Conclude that S
(f ∗ dg − g ∗ df) = 0 if both f and g are harmonic.
43

6 Beyond 3 dimensions*
6.1 Beyond 3D
It is possible to do calculus in Rn
with n 3. Here the language of differential
forms comes into its own. While it would be impossible to talk about the curl
of a vector field in, say, R4
, the derivative of a 1-form or 2-form presents no
problems; we simply apply the rules we’ve already learned. For example, if
x, y, z, t are the coordinates of R4
, then a 1-form is a linear combination of the
4 basic 1-forms
dx, dy, dz, dt
a forms is a linear combination of the 6 basic 2-forms
dx ∧ dy = −dy ∧ dx
dx ∧ dz = −dz ∧ dx
. . .
dz ∧ dt = −dt ∧ dz
and a 3 form is a linear combination of
dx ∧ dy ∧ dz = −dy ∧ dx ∧ dz = −dx ∧ dz ∧ dy = dy ∧ dz ∧ dx = . . .
. . .
dy ∧ dz ∧ dt = −dz ∧ dy ∧ dt = . . .
The higher dimensional analogue of a surface is a k-manifold. A parameter-
ized k-manifold M in Rn
is given by a collection of C1
functions



x1 = f1(u1, . . . uk)
x2 = f2(u1, . . . uk)
. . .
xn = fn(u1, . . . uk)
(u1, . . . uk) ∈ D ⊆ Rk
open
such that the map D → Rn
is one to one and the tangent vectors (∂x1
∂u1
, . . . ∂xn
∂u1
), . . .
( ∂x1
∂uk
, . . . ∂xn
∂uk
) are linearly independent for all values of the coordinates (u1, . . . uk).
Given a k-form α on Rn
, we can express it as a linear combination of k-fold
wedges of dx1 . . . dxn, and then rewrite it as g(u1, . . . uk)du1 ∧ . . . ∧ duk . The
“surface” integral is defined as
M
α = . . .
D
g(u1 . . . uk)du1 . . . duk (16)
Notice that number of integrations is usually suppressed in the notation on the
left, since it gets too cumbersome after a while. In practice, for computing inte-
grals, it’s convenient to relax the conditions a bit by allowing D to be nonopen
and allowing some degenerate points where the map D → Rn
isn’t one to one.
44

More generally, a k-manifold M is obtained by gluing several parameterized
manifolds as we did for surfaces. To be more precise, a closed set M ⊂ Rn
is a k-manifold, if each point of M lies in the image of a parameterized k-
manifold called a chart. As with curves and surfaces, it is important to specify
orientations. Things are a little trickier since we can no longer rely on our
geometric intuition to tell us which way is “up” or “down”. Instead we can
think that an orientation is a rule for specifying whether a coordinate system
on a chart is right or left handed. We’ll spell this out in an example below. The
integral M
α can be deﬁned by essentially summing up (16) over various non-
overlapping right handed charts (we can use left handed charts provided we use
the opposite sign). A k-manifold with boundary M is a closed set which can be
decomposed as a union of a (k −1)-manifold ∂M, called the boundary, and a k-
manifold M −∂M. We orient this by the rule that a coordinate system u2, . . . uk
of ∂M is right handed if it can be completed to right handed coordinate system
u1, u2, . . . uk of M such that the tangent vector (∂x1
∂u1
, . . . ∂xn
∂u1
) “points out”.
Then the ultimate form of Stokes’ theorem is:
THEOREM 6.1.1 (Generalized Stokes’ theorem) If M is an oriented k
manifold with boundary ∂M and if α is a (k − 1)-form deﬁned on (an open set
containing) M, then
M
dα =
∂M
α
In order to get a feeling for how this works, let’s calculate the “volume” V
of the 4-dimensional ball B = {(x, y, z, t) | x2
+ y2
+ z2
+ t2
≤ R} of radius
R in R4
in two ways. This is a 4-manifold with boundary S = {(x, y, z, t) |
x2
+ y2
+ z2
+ t2
= R}.
V can be expressed as the integral
V =
B
dxdydzdt =
B
dx ∧ dy ∧ dz ∧ dt
We use extended spherical coordinates σ, ψ, φ, θ, where σ measures the distance
of (x, y, z, t) to the origin in R4
, and ψ the angle to the t-axis. So that
t = σ cos ψ
and
ρ = σ sin ψ
is the distance from the projection (x, y, z) to the origin. Then letting φ, θ be
the remaining spherical coordinates gives
x = ρ sin φ cos θ = σ sin ψ sin φ cos θ
y = ρ sin φ sin θ = σ sin ψ sin φ sin θ
z = ρ cos φ = σ sin ψ cos φ
45

ρ
t
ψ
x
y
σ
z
In these coordinate B is described as



0 ≤ ψ ≤ π
0 ≤ φ ≤ π
0 ≤ θ ≤ 2π
0 ≤ σ ≤ R
To simply computations, we note that form will get multiplied by the Jacobian
when we change coordinates:
dx ∧ dy ∧ dz ∧ dt =
∂(x, y, z, t)
∂(σ, ψ, θ, φ)
dσ ∧ dψ ∧ dθ ∧ dφ
= σ3
sin2
ψ sin φ dσ ∧ dψ ∧ dθ ∧ dφ
Note that the Jacobian is positive, and this what it means to say the coordinate
system σ, ψ, θ, φ is right handed or positively oriented. The volume is now easily
computed
R
0
σ3
dσ
π
0
sin2
ψdψ
2π
0
dθ
π
0
sin φdφ =
1
2
π2
R4
Alternatively, we can use Stokes’ theorem, to see that
V =
B
dx ∧ dy ∧ dz ∧ dt = −
S
tdx ∧ dy ∧ dz
The parameter x, y, z gives a left hand coordinate system on the upper hemi-
sphere U = S ∩ {t 0}. It is left handed because t, x, y, z is left handed on R4
.
46

For similar reasons, x, y, z gives a right handed system on the lower hemisphere
L where t 0. Therefore
V = −
U
tdx ∧ dy ∧ dz −
L
tdx ∧ dy ∧ dz
= 2
x2+y2+z2≤R
R2 − x2 − y2 − z2dxdydz
= 2
R
0
ρ2
R2 − ρ2dρ
π
0
sin φ dφ
2π
0
dθ
= 8π
π/2
0
R4
sin2
α cos2
αdα
=
1
2
π2
R4
6.2 Maxwell’s equations in R4
As exotic as higher dimensional calculus sounds, there are many applications of
these ideas outside of mathematics. For example, in relativity theory one needs
to treat the electric E = E1i+E2j+E3k and magnetic fields B = B1i+B2j+B3k
as part of a single “field” on space-time. In mathematical terms, we can take
space-time to be R4
- with the fourth coordinate as time t. The electromagnetic
field can be represented by a 2-form
F = B3dx ∧ dy + B1dy ∧ dz + B2dz ∧ dx + E1dx ∧ dt + E2dy ∧ dt + E3dz ∧ dt
If we compute dF using the analogues of the rules we’ve learned:
dF =
∂B3
∂x
dx +
∂B3
∂y
dy +
∂B3
∂z
dz +
∂B3
∂t
dt ∧ dx ∧ dy + . . .
=
∂B1
∂x
+
∂B2
∂y
+
∂B3
∂z
dx∧dy ∧dz +
∂E2
∂x
−
∂E1
∂y
+
∂B3
∂t
dx∧dy ∧dt+. . .
Two of Maxwell’s equations for electromagnetism
· B = 0, × E = −
∂B
∂t
can be expressed very succintly in this language as dF = 0. The analogue of
theorem 2.5.1 holds for Rn
, and shows that
F = d(A1dx + A2dy + A3dz + A4dt)
for some 1-form called the potential. Thus we’ve reduced the 6 quantites to just
4. In terms of vector analysis this amounts to the more complicated looking
equations
B = × (A1i + A2j + A3k), E = A4 −
∂A1
∂t
i +
∂A2
∂t
j +
∂A3
∂t
k
47

There are two remaining Maxwell equations
· E = 4πρ, × B =
∂E
∂t
+ 4πJ
where ρ is the electric charge density, and J is the electric current. The first law
is really an analog of (15) for the electric field. After applying the divergence
theorem, it implies that the electric flux through a closed surface equals (−4π)
times the electric charge inside it. These last two Maxwell equations can also
be replaced by the single equation d∗F = 4πJ of 3-forms. Here
∗F = E3dx ∧ dy + E1dy ∧ dz + E2dz ∧ dx − B1dx ∧ dt − B2dy ∧ dt − B3dz ∧ dt
and
J = ρdx ∧ dy ∧ dz − J3dx ∧ dy ∧ dt − J1dy ∧ dz ∧ dt − J2dz ∧ dx ∧ dt
(We have been relying on explicit formulas to avoid technicalities about the
definition of the ∗-operator. In principle however, it involves a metric, and in
this case we use the so called Lorenz metric.)
Let’s see how the calculus of differential forms can be used to extract a
physically meaningful consequence of these laws. Proposition 2.4.1 (in extended
form) implies that dJ = 1
4π d2
∗F = 0. Expanding this out yields
∂ρ
∂t
dt ∧ dx ∧ dy ∧ dz −
∂J3
∂z
dz ∧ dx ∧ dy ∧ dt + . . . =
−
∂ρ
∂t
+ · J dx ∧ dy ∧ dz ∧ dt = 0
Thus the expression in brackets is zero. This really an analog of the equa-
tion (12). To appreciate the meaning integrate ∂ρ
∂t over a solid region V with
boundary S. Then this equals
−
V
· JdV = −
S
J · ndS
In other words, the rate of change of the electric charge in V equals minus the
flux of the current accross the surface. This is the law of conservation of electric
charge.
48

7 Further reading
For more information about differential forms, see the books [Fl, S, W]. Most
the physics background – which is only used here for illustration, in any case
– can be found in [Fe]. A standard reference for complex analysis is [A]. The
material of the appendix can be found in any book on advanced calculus. For
a rigorous treatment, see [R, S].
References
[A] L. Ahlfors, Complex Analysis
[Fe] R. Feynman et. al., Lectures on Physics
[Fl] H. Flanders, Differential forms and applications to the physical sciences
[R] W. Rudin, Principles of mathematical analysis
[S] M. Spivak, Calculus on manifolds
[W] S. Weintraub, Differential forms: theory and practice
49

A Review of multivariable calculus
A.1 Differential Calculus
To simplify the review, we’ll stick to two variables, but the corresponding state-
ments hold more generally. Let f(x, y) be a real valued function defined on open
subset of R2
. Recall that the limit
lim
(x,y)→(a,b)
f(x, y) = L
means that f(x, y) is approximately L whenever (x, y) is close to (a, b). The
precise meaning is as follows. If we specified 0 (say = 0.0005), then
we could pick a tolerance δ 0 which would guarantee that |f(x, y) − L|
(i.e. f(x, y) agrees with L up to the first 3 digits for = 0.0005) whenever the
distance between (x, y) and (a, b) is less than δ. A function f(x, y) is continuous
at (a, b) if
lim
(x,y)→(a,b)
f(x, y)
exists and equals f(a, b). It is continuous if it is so at each point of its domain.
We say that f is differentiable, if near any point p = (x0, y0, f(x0, y0)) the
graph z = f(x, y) can approximated by a plane passing through p. In other
words, there exists quantities A, B such that we may write
f(x, y) = f(x0, y0) + A(x − x0) + B(y − y0) + remainder
with
lim
(x,y)→(x0,y0)
|remainder|
|(x, y) − (x0, y0)|
= 0
The last condition says that as the distance |(x, y) − (x0, y0)| goes to zero, the
remainder goes to zero at an even faster rate. We can see that the coefficients
are nothing but the partial derivatives
A =
∂f
∂x
(x0, y0) = lim
h→0
f(x0 + h, y0) − f(x0, y0)
h
B =
∂f
∂y
(x0, y0) = lim
h→0
f(x0, y0 + h) − f(x0, y0)
h
There is a stronger condition which is generally easier to check. f is called
continuously differentiable or C1
if it and its partial derivatives exist and are
continuous. Consider the following example
f(x, y) =
x3
x2+y2 if (x, y) = (0, 0)
0 if (x, y) = (0, 0)
This is continuous, however
∂f
∂x
=
3x2
x2 + y2
−
2x4
(x2 + y2)
2
50

has no limit as (x, y) → (0, 0). To see this, note that along the x-axis y = 0, we
have ∂f
∂x = 1. So the limit would have to be 1 if it existed. On the other hand,
along the y-axis x = 0, ∂f
∂x = 0, which shows that there is no limit. So f(x, y)
is not C1
.
Partial derivatives can be used to determine maxima and minima.
THEOREM A.1.1 If (a, b) is local maximum or minimum of a C1
function
f(x, y) , then (a, b) is a critical point, i.e.
∂f
∂x
(a, b) =
∂f
∂y
(a, b) = 0
THEOREM A.1.2 (Chain Rule) If f, g, h : R2
→ R are C1
functions, then
f(g(u, v), h(u, v)) is also C1
and if z = f(x, y), x = g(u, v), y = h(u, v) then
∂z
∂u
=
∂z
∂x
∂x
∂u
+
∂z
∂y
∂y
∂u
∂z
∂v
=
∂z
∂x
∂x
∂v
+
∂z
∂y
∂y
∂v
A function f(x, y) is twice continously differentiable or C2
if is C1
and if its
partial derivatives are also C1
. We have the following basic fact:
THEOREM A.1.3 If f(x, y) is C2
then the mixed partials
∂2
f
∂y∂x
=
∂
∂y
∂f
∂x
∂2
f
∂x∂y
=
∂
∂x
∂f
∂y
are equal.
If f is C2
, then we have a Taylor approximation
f(x, y) ≈ f(a, b) +
∂f
∂x
(a, b) (x − a) +
∂f
∂y
(a, b) (y − b) +
1
2
∂2
f
∂x2
(a, b) (x − a)2
+
∂2
f
∂y∂x
(a, b) (x − a)(y − b) +
1
2
∂2
f
∂y2
(a, b) (y − b)2
More precisely, the remainder, which is the difference of left and right, should
go to zero faster than |(x, y) − (a, b)|2
goes to zero. Since it is relatively easy to
determine when quadratic polynomials have maxima or minima, this leads to
the second derivative test.
THEOREM A.1.4 A critical point (a, b) of a C2
function f(x, y) is a local
minimum (respectively maximum) precisely when the matrix
∂2
f
∂x2 (a, b) ∂2
f
∂y∂x (a, b)
∂2
f
∂y∂x (a, b) ∂2
f
∂y2 (a, b)
is positive (respectively negative) definite.
51

The above conditions are often formulated in a more elementary but ad hoc
way in calculus books. Positive definiteness is equivalent to requiring
∂2
f
∂x2
(a, b) 0
∂2
f
∂x2
(a, b)
∂2
f
∂y2
(a, b) −
∂2
f
∂y∂x
(a, b)
2
0
A.2 Integral Calculus
Integrals can be defined using Riemann’s method. This has some limitations
but it’s the easiest to explain. Given a rectangle R = [a, b] × [c, d] ⊂ R2
, choose
integers m, n 0 and let ∆x = b−a
m ∆y = d−c
n . Choose a set of sample points
P = {(x1, y1), . . . (xm, yn)} ⊂ R with
(xi, yj) ∈ Rij = [a + (i − 1)∆x, a + i∆x] × [b + (j − 1)∆y, b + j∆y]
The Riemann sum
S(m, n, P) =
i,j
f(xi, yj)∆x∆y
Then the double integral is
R
f(x, y)dxdy = lim
m,n→∞
S(m, n, P)
This definition is not really that precise because we need to choose P for each
pair m, n. For the integral to exist, we really have to require that the limit
exists for any choice of P, and that any two choices lead to the same answer.
The usual way to resolve the above issues is to make the two extreme choices.
Define upper and lower sums
U(m, n) =
i,j
Mij∆x∆y
L(m, n) =
i,j
mij∆x∆y
where
Mij = max{f(x, y) | (x, y) ∈ Rij}
mij = min{f(x, y) | (x, y) ∈ Rij}
In the event that the maxima or minima don’t exist, we should use the greatest
lower bound and least upper bound instead. As m, n → ∞ the numbers L(m, n)
tend to increase. So their limit can be understood as the least upper bound, i.e.
the smallest number L ≥ L(m, n). Likewise we define the limit U as the largest
number U ≤ U(m, n). If these limits coincide, the common value is taken to be
R
f(x, y)dxdy = L = U
otherwise the (Riemann) integral is considered to not exist.
52

THEOREM A.2.1 R
f(x, y)dxdy exists if f is continuous.
The integral of
f(x, y) =
1 if (x, y) has rational coordinates
0 otherwise
would be undefined from the present point of view, because L = 0 and U = 1.
Although, in fact the integral can be defined using the more powerful Lebesgue
theory [R]; in this example the Lebesgue integral is 0.
For more complicated regions D ⊂ R, set
D
f(x, y)dxdy =
R
f(x, y)χD(x, y)dxdy
where χD = 1 inside D and 0 elsewhere. The key result is
THEOREM A.2.2 (Fubini) If D = {(x, y) | a ≤ x ≤ b, g(x) ≤ y ≤ h(x)}
with f, g, h continuous. Then the double integral exists and
D
f(x, y)dxdy =
b
a
h(x)
g(x)
f(x, y)dy dx
A similar statement holds with the roles of x and y interchanged.
This allows one to compute these integrals in practice.
The final question to answer is how double integrals behave under change of
variables. Let T : R2
→ R2
be a transformation given by C1
functions
x = f(u, v), y = g(u, v)
We think of the first R2
as the uv-plane and the second as the xy-plane. Given
a region D in the uv-plane, we can map it to the xy-plane by
T(D) = {(f(u, v), g(u, v)) | (u, v) ∈ D}
The Jacobian
∂(x, y)
∂(u, v)
=
∂x
∂u
∂y
∂v
−
∂x
∂v
∂y
∂u
= det
∂x
∂u
∂x
∂v
∂y
∂u
∂y
∂v
.
THEOREM A.2.3 If T is a one to one function, h is continuous and D a
region of the type occurring in Fubini’s theorem, then
T (D)
h(x, y)dxdy =
D
h(f(u, v), g(u, v))
∂(x, y)
∂(u, v)
dudv
53

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