2 Vector Fields & Differential Forms
In preparation for our study of surfaces, we further develop the notion of
a tangent vector. To permit easy differentiation, throughout this section all
functions are assumed to be smooth (infinitely differentiable) and U ⊆ R
n
will denote a connected open set: (informally) a region consisting of a single
piece without edge points. As previously, n will always be 1, 2 or 3: when
n = 1, U = (a, b) is an open interval; the picture illustrates n = 2.
U
2.1 Directional Derivatives, Tangent Vectors & Vector Fields
First recall some basic objects and facts from elementary multivariable calculus.
Definition 2.1. The gradient of f : U ⊆ R
n
→ R is the function ∇f : U → R
n
defined by
∇f (x
1
, . . . , x
n
) =
∂ f
∂x
1
, . . . ,
∂ f
∂x
n
Given a point p ∈ U, a vector v = (v
1
, . . . , v
n
) ∈ R
n
, and a function f : U → R, the directional
derivative of f at p in the direction v is the scalar
D
v
f (p) :=
n
∑
k=1
v
k
∂ f
∂x
k
p
= v ·
∇f (p)
Example 2.2. Suppose f (x, y, z) = x
2
−z cos y, p = (1, π, 0), and v = (3, 5, 1). Then
∇f =
2x
z sin y
−cos z
=⇒ D
v
f (p) =
3
5
1
·
2
0
1
= 7
The directional derivative describes the rate of change of the value of f in a given direction.
Lemma 2.3. 1. By the chain rule, if x(t) is a curve such that x(0) = p and x
′
(0) = v, then
d
dt
t=0
f
x(t)
=
n
∑
k=1
∂ f
∂x
k
p
x
′
k
(0) = D
v
f (p)
is the rate of change of f at p as one travels along the curve.
2. If t is small, then f (p + tv) ≈ f (p) + D
v
f (p) t.
3. If v is a unit vector making angle θ with ∇f (p), then
D
v
f (p) = v ·∇f (p) =
||
∇f (p)
||
cos θ
R
t
p
v
U
0 f (p)
x
f
is maximal when v points in the same direction as ∇f (p). Otherwise said, ∇f (p) points in the
direction of greatest increase of f at p; its magnitude measures the rate of change.
32
By placing the function f at the end of the directional derivative, we are tempted to create an operator
D
v
|
p
=
n
∑
k=1
v
k
∂
∂x
k
p
which takes a function f : U → R and returns the scalar D
v
f (p). This operator is a map (function)
from the set of smooth functions f : U → R to the real numbers. It is even more tempting to drop
the point p and allow the components of v to be smooth functions. This yields a new definition of an
old concept.
Definition 2.4. The set of directional derivative operators D
v
|
p
is the tangent space T
p
R
n
at p ∈ R
n
.
A vector field v on U ⊆ R
n
is a smooth choice for each p ∈ U of an element of T
p
R
n
: that is
v =
n
∑
k=1
v
k
∂
∂x
k
where each v
k
: U → R is smooth
Each operator
∂
∂x
k
is termed a co-ordinate vector field.
If f : U → R is smooth, we write v[ f ] =
∑
v
k
∂ f
∂x
k
for the result of applying the vector field v to f ; this
is itself a smooth function v[ f ] : U → R.
Each tangent space T
p
R
n
is a vector space, with natural basis
∂
∂x
1
p
, . . . ,
∂
∂x
n
p
. In this brave new
world, a tangent vector v
p
=
∑
v
k
∂
∂x
k
p
corresponds to our previous notion v
p
= (v
1
, . . . , v
n
). While
this might seem artificially complicated, the rational is simple: the purpose of tangent vectors is to
measure how functions change in given directions (Lemma 2.3!).
Examples 2.5. 1. The vector field v = 3x
∂
∂x
+ 2xz
∂
∂y
− x
∂
∂z
on R
3
corresponds to the vector-valued
function v(x, y, z) = (3x, 2xz, −x). Given f (x, y, z) = xy
2
+ z, we have
v[ f ] = 3x
∂ f
∂x
+ 2xz
∂ f
∂y
− x
∂ f
∂z
= 3xy
2
+ 4x
2
yz − x
which, as expected, is a smooth function v[ f ] : R
3
→ R.
2. Suppose, in R
2
, that we are given a vector field v = y
2
∂
∂x
− x
∂
∂y
, a function f (x, y) = x
2
y, and a
point p = (2, −1). These may be combined in various ways, for instance:
Vector field on R
2
f v = x
2
y
y
2
∂
∂x
− x
∂
∂y
= x
2
y
3
∂
∂x
− x
3
y
∂
∂y
Tangent vector ( f v)(p) = f (p) v
p
= −4
∂
∂x
p
+ 8
∂
∂y
p
∈ T
p
R
2
Function R
2
→ R v[ f ] = y
2
∂
∂x
(x
2
y) − x
∂
∂y
(x
2
y) = 2xy
3
− x
3
Number
v[ f ]
(p) = −4 −8 = −12
Note the use of different brackets! Note also that f v denotes the vector field obtained by multi-
plying v by the value of f at each point. It does not mean apply the function f to the vector field v,
which makes no sense!
33
Here are the basic rules of computation for vector fields. These are all essentially trivial if you take
v =
∑
v
k
∂
∂x
k
, etc., as in Definition 2.4. Just be careful with notation!
Lemma 2.6. Let v, w be vector fields on U, let f , g : U → R be smooth, and a, b ∈ R constant. Then,
1. f v + gw is a vector field: at each p ∈ U, ( f v + gw)(p) := f (p)v
p
+ g(p)w
p
2. Vector fields act linearly on smooth functions: v[a f + bg] = av[ f ] + bv[g]
3. (Leibniz rule) Vector fields obey a product rule: v[ f g] = f v[g] + gv[ f ]
Examples 2.7. 1. We verify the Leibniz rule for the vector field v =
∂
∂x
−xy
∂
∂y
and functions f (x, y) =
x and g(x, y) = ye
x
.
v[ f g] =
∂
∂x
− xy
∂
∂y
[xye
x
] = ye
x
+ xye
x
− x
2
ye
x
f v[g] + gv[ f ] = x
∂
∂x
− xy
∂
∂y
[ye
x
] + ye
x
∂
∂x
− xy
∂
∂y
[x] = x(ye
x
− xye
x
) + ye
x
2. (Polar co-ordinates) Let U be the plane without the non-positive x-axis. On U, the standard
rectangular co-ordinates (x, y) are related to the polar co-ordinates ( r, θ) via
(
x = r cos θ
y = r sin θ
↭
(
r =
p
x
2
+ y
2
θ = tan
−1
y
x
(or ±
π
2
if x = 0)
The chain rule tells us that the co-ordinate vector fields
∂
∂x
,
∂
∂y
,
∂
∂r
,
∂
∂θ
are related via
∂
∂r
=
∂x
∂r
∂
∂x
+
∂y
∂r
∂
∂y
= cos θ
∂
∂x
+ sin θ
∂
∂y
=
1
p
x
2
+ y
2
x
∂
∂x
+ y
∂
∂y
∂
∂θ
=
∂x
∂θ
∂
∂x
+
∂y
∂θ
∂
∂y
= −r sin θ
∂
∂x
+ r cos θ
∂
∂y
= −y
∂
∂x
+ x
∂
∂y
r
θ
x
y
∂
∂x
p
∂
∂y
p
∂
∂r
p
∂
∂θ
p
p
At p, these point in the direction of maximal increase for the corresponding co-ordinate.
We could similarly compute
∂
∂x
and
∂
∂y
by differentiating. For variety, we instead use linear
algebra:
∂
∂r
∂
∂θ
=
cos θ sin θ
−r sin θ r cos θ
∂
∂x
∂
∂y
!
=⇒
∂
∂x
∂
∂y
!
=
1
r
r cos θ −sin θ
r sin θ cos θ
∂
∂r
∂
∂θ
=⇒
∂
∂x
= cos θ
∂
∂r
−
sin θ
r
∂
∂θ
∂
∂y
= sin θ
∂
∂r
+
cos θ
r
∂
∂θ
34
The first matrix is the familiar Jacobian
∂(x,y)
∂(r,θ)
from multivariable calculus. Strictly, we are view-
ing U as subsets of two different versions of R
2
:
• In rectangular co-ordinates, U = R
2
\{(x, 0) : x ≤ 0} is a cut plane.
• In polar co-ordinates, U = (0, ∞) × (−π, π) is an infinite open rectangle.
In practice, particularly since we are so familiar with polar co-ordinates, it is easier to stick to
the first interpretation and draw all four co-ordinate tangent vectors on the same picture.
Exercises 2.1. 1. You are given the following vector fields and functions
u = 7
∂
∂x
−3
∂
∂y
v = x
∂
∂x
+ 2y
∂
∂y
w = sin x
∂
∂x
−2 cos x
∂
∂y
f (x, y) = xy
2
g(x, y) = −y
Compute the functions:
(a) u[ f ] (b) v[ f ] (c) w[ f ]
(d) v[ f g] (e) f u[g] (f) v
w[g]
2. Revisit Example 2.7.2 on polar co-ordinates.
(a) Use the chain rule to compute
∂
∂x
and
∂
∂y
directly in terms of r, θ,
∂
∂r
and
∂
∂θ
and verify that
you obtain the same expressions as the linear algebra approach.
(b) Suppose T
p
R
2
is equipped with the standard dot product so that
∂
∂x
p
and
∂
∂y
p
are con-
sidered orthonormal.
i. Show that
∂
∂r
p
and
∂
∂θ
p
are perpendicular.
ii. What are the lengths of
∂
∂r
p
and
∂
∂θ
p
?
3. Consider the spherical polar co-ordinate system
x = r cos θ cos ϕ
y = r sin θ cos ϕ
z = r sin ϕ
where r > 0, 0 < θ < 2π and −
π
2
< ϕ <
π
2
Show that
∂
∂r
=
1
r
x
∂
∂x
+ y
∂
∂y
+ z
∂
∂z
4. Prove the Leibniz rule (Lemma 2.6 part 3).
5. If f , g, h are smooth functions and v is a vector field, expand v[ f gh] using the Leibniz rule.
6. Let s = x
2
−y
2
and t = 2xy. Compute
∂
∂s
,
∂
∂t
in terms of
∂
∂x
and
∂
∂y
.
(Hint: use the chain rule to find
∂
∂x
and
∂
∂y
, then invert the Jacobian)
35
2.2 Differential 1-forms
Make sure you are comfortable with vector fields before you tackle this section and the next! There is
a lot of new notation to get used to here, but with a little practice it is very easy to use.
Definition 2.8. Let (x
1
, . . . , x
n
) be co-ordinates on U ⊆ R
n
and p ∈ U. The (co-ordinate) 1-form dx
k
at
p is the linear map
13
dx
k
: T
p
R
n
→ R defined by
dx
k
∂
∂x
j
p
!
= δ
jk
=
(
1 j = k
0 j = k
A 1-form α =
n
∑
k=1
a
k
dx
k
on U is a smooth assignment (a
k
: U → R smooth) of 1-forms.
If v is a vector field on U, we write α(v) for the function U → R obtained by mapping p 7→ α(v
p
).
Examples 2.9. 1. Consider the vector field v = xy
∂
∂x
−2
∂
∂y
on R
2
. At each p ∈ R
2
, the components
xy and −2 are scalars and thus ignored by the linear map dx : T
p
R
2
→ R. We therefore obtain
a function dx(v) : R
2
→ R
dx(v) = dx
xy
∂
∂x
−2
∂
∂y
= xy dx
∂
∂x
−2 dx
∂
∂y
= xy
2. Again on R
2
, let α = 2x dx + dy and v = x
2
y
∂
∂x
−e
xy
∂
∂y
. Then
α(v) = ( 2x dx + dy)
x
2
y
∂
∂x
−e
xy
∂
∂y
= 2x
3
y −e
xy
Remember that a 1-form α is linear when restricted to each tangent space T
p
R
n
: if v
p
∈ T
p
R
n
and
f : U → R
n
, we obtain a real number
α
p
f (p)v
p
= f (p)α
p
v
p
∈ R
by pointwise multiplication by the value of f . Taken over all points p, this means that scalar functions
come straight through a 1-form: if v is a vector field on U, then
α( f v) = f α(v)
Definition 2.10. Let f : U → R be smooth. The exterior derivative of f is the 1-form
d f =
n
∑
k=1
∂ f
∂x
k
dx
k
=
∂ f
∂x
1
dx
1
+ ···+
∂ f
∂x
n
dx
n
If a 1-form is the exterior derivative of a function, we say that it is exact.
13
For those who’ve met dual vector spaces in linear algebra, the set of 1-forms at p is the cotangent space T
∗
p
R
n
, or the
space of covectors. At each p, {dx
1
, . . . , dx
n
} is the dual basis to
∂
∂x
1
p
, . . . ,
∂
∂x
n
p
.
36
Our approach essentially splits a derivative into two pieces: for each k, we have d f
∂
∂x
k
=
∂ f
∂x
k
.
Moreover, since a linear map (d f
p
: T
p
R
n
→ R) is determined by what it does to a basis, the exterior
derivative d f is the unique 1-form with the property that d f (v) = v[ f ] for all vector fields v on U.
This says that the definition is co-ordinate independent (does not depend on x
1
, . . . , x
n
).
Examples 2.11. 1. Let f (x, y) = x
2
y, then d f = α = 2xy dx + x
2
dy. As a sanity check, consider a
general vector field v = a
∂
∂x
+ b
∂
∂y
(remember that a, b are smooth functions!) and compute
d f (v) = 2axy + bx
2
= v[x
2
y]
2. If α = 4xy
2
dx + (4x
2
y + 1) dy = f
x
dx + f
y
dy is exact, then ‘partial integration’ forces
f (x, y) =
Z
4xy
2
dx = 2x
2
y
2
+ g(y) =
Z
4x
2
y + 1 dy = 2x
2
y
2
+ y + h(x)
for some functions g, h. Plainly g, h must be constant and α = d(2x
2
y
2
+ y).
3. We could a similar game to see that α = 3x
2
y dx + 2 dy is not exact on R
2
. Alternatively, note
that if α = d f = f
x
dx + f
y
dy, we obtain a contradiction by observing that the mixed partial
derivative is simultaneously
3x
2
=
∂ f
x
∂y
= f
xy
= f
yx
=
∂ f
y
∂x
= 0
See Exercise 6 for the general result.
Lemma 2.12. If f , g are smooth functions, then
1. d( f + g) = d f + dg
2. d( f g) = f dg + g d f
3. d f = 0 ⇐⇒ f is a constant function
Proof. These follow straight from the definition of d f . For instance
d f = 0 ⇐⇒
∂ f
∂x
j
= d f
∂
∂x
j
= 0 for all j = 1, . . . , n ⇐⇒ f is constant
Example (2.7.2 cont). The exterior derivative and part 2 of the Lemma make it easy to compute the
relationship between the 1-forms dx, dy, dr, dθ:
(
x = r cos θ
y = r sin θ
=⇒
(
dx = cos θ dr −r sin θ dθ
dy = sin θ dr + r cos θ dθ
=⇒
(
dr =
1
r
(x dx + y dy)
dθ =
1
r
2
( −y dx + x dy)
We may also verify directly that the dual basis relations hold; for instance,
dr
∂
∂r
=
1
r
(x dx + y dy)
cos θ
∂
∂x
+ sin θ
∂
∂y
=
1
r
(x cos θ + y sin θ)
= cos
2
θ + sin
2
θ = 1
37
Elementary Calculus & Line Integrals
It is worth reviewing some staples from basic calculus in our new language.
If f : R → R is differentiable, then its exterior derivative d f = f
′
(x) dx feels familiar.
14
To make
sense of this as a relation between 1-forms we need vector fields: the derivative of f isn’t the ratio of
two 1-forms, rather it is the application of the 1-form d f to the vector field
d
dx
:
d f
dx
=
d
dx
[ f ] = d f
d
dx
Vector fields in R are written with a straight d rather than partial ∂ since there is only one direction
in which to differentiate!
You’ve seen 1-forms before when integrating: we integrate 1-forms over oriented curves.
Definition 2.13. Let α be a 1-form on U ⊆ R
n
and suppose x : [a, b] → U parametrizes a smooth
curve C. Our usual identification (Definition 2.4) produces the tangent vector field
x
′
( t) = x
′
1
( t)
∂
∂x
1
+ ···+ x
′
n
( t)
∂
∂x
n
along the curve. Now define the integral of α along C by
Z
C
α :=
Z
b
a
α
x
′
( t)
dt =
Z
b
a
α
x
′
1
( t)
∂
∂x
1
+ ···+ x
′
n
( t)
∂
∂x
n
dt
Examples 2.14. 1. We integrate α = x dy over the unit-circle x(t) = (cos t, sin t) counter-clockwise.
Differentiate to obtain the tangent vector field x
′
( t) = −sin t
∂
∂x
+ cos t
∂
∂y
, then
Z
C
α =
Z
2π
0
α
x
′
( t)
dt
Z
2π
0
cos
2
t dt =
1
2
Z
2π
0
1 + cos 2t dt = π
2. Integrate α = y
2
dx − x
2
dy over the curve x(t) = (t, t
2
) between (0, 0) and (1, 1):
Z
C
α =
Z
1
0
α
x
′
( t)
dt =
Z
1
0
α
∂
∂x
+ 2t
∂
∂y
dt =
Z
1
0
y(t)
2
−2t
x( t)
2
dt
=
Z
1
0
t
4
−2t
3
dt =
1
5
−
1
2
= −
3
10
Lemma 2.15. The integral of a 1-form along a curve is independent of the choice of (orientation-
preserving) parametrization.
Otherwise said, if x(t) = y
s(t)
parametrizes the same curve where s
′
( t) > 0, then
Z
b
a
α
x
′
( t)
dt =
Z
s(b)
s(a)
α
y
′
( s)
ds
The proof is an easy exercise in interpreting old material (the chain rule/substitution).
14
Consider the equivalence of notations
d f
dx
= f
′
(x), linear approximations (differentials) & integration by substitution.
38
Our final result from elementary calculus shows that integrals of exact forms are independent of
path. This is essentially the fundamental theorem of calculus for curves.
Theorem 2.16 (Fundamental Theorem of Line Integrals). If f is a function on U ⊆ R
2
and C is a
curve in U, then the integral of d f depends only on the values of f at the endpoints of C:
Z
C
d f = f
end of C
− f
start of C
The converse also holds: if
R
C
α is independent of path, then α is exact.
Proof. Suppose x : [a, b] → U parametrizes C, then
Z
C
d f =
Z
b
a
d f (x
′
) dt =
Z
b
a
x
′
[ f ] dt =
Z
b
a
x
′
1
( t)
∂ f
∂x
1
+ ···+ x
′
n
( t)
∂ f
∂x
n
dt
=
Z
b
a
d
dt
f (x(t))
dt = f
x(b)
− f
x(a)
The converse is sketched in an exercise.
In elementary multivariable calculus this result was written
R
C
∇f ·dx = f
x(b)
− f
x(b)
which
comports with our new notation when we view dx as a vector of 1-forms:
∇f ·dx =
∂ f
∂x
1
.
.
.
∂ f
∂x
n
·
dx
1
.
.
.
dx
n
=
∂ f
∂x
1
dx
1
+ ···+
∂ f
∂x
n
dx
n
= d f
The exterior derivative d f is just the gradient in disguise!
Example 2.17. If α = cos(xy)(y dx + x dy), find the integral of α over any curve C joining the points
π,
1
3
and
1
2
, π
. Since α = d sin(xy) is exact on R
2
, we see that
Z
C
α = sin(xy)
(
1
2
,π
)
(
π,
1
3
)
= sin
π
2
−sin
π
3
= 1 −
√
3
2
Summary
• Tangent vectors & vector fields encode directional derivatives, measuring how functions change
in given directions.
• Vector fields and 1-forms break standard derivatives into two pieces: the result is a more flexible
and extensible language for describing familiar results from multi-variable calculus.
The real pay-off comes once our new language is applied to surfaces and higher-dimensional objects.
Here is a pr
´
ecis. A parametrized surface is a function x : U ⊆ R
2
→ E
3
; its exterior derivative dx is a
vector-valued 1-form which, at each point p ∈ U, describes a linear map between tangent spaces
dx
p
: T
p
R
2
→ T
x(p)
E
3
which maps the co-ordinate fields
∂
∂x
,
∂
∂y
on U to corresponding vector fields tangent to the surface.
39
Exercises 2.2. 1. In R
2
, let α = 2y dx −3 dy and v = 3x
2
∂
∂x
+
∂
∂y
. Compute α(v), and v
α(v)
.
2. On R
3
, suppose f (x, y, z) = x
2
cos(yz) and v = e
x
∂
∂x
+ 2y
∂
∂z
. Verify that d f (v) = v[ f ].
3. Find dr directly by taking the exterior derivative of the equation r
2
= x
2
+ y
2
.
4. Prove parts 1 and 2 of Lemma 2.12.
5. Continuing Example 2.7.2, verify that dθ
∂
∂θ
= 1, and dr
∂
∂θ
= 0 = dθ
∂
∂r
.
6. Suppose that α =
∑
a
k
dx
k
is exact. Prove that
∂a
k
∂x
j
=
∂a
j
∂x
k
for all j, k.
7. Decide whether the 1-forms α are exact on R
2
. If yes, find a function f such that α = d f .
(a) α = 2x dx + dy (b) α = dx + 2x dy
(c) α = cos(x
2
y)(2y dx + x dy) (d) α = x cos(x
2
y)(2y dx + x dy)
8. We consider a partial converse to Exercise 6.
(a) Suppose α = a dx + b dy is a 1-form on a rectangle [p, q] × [r, s], where
∂a
∂y
=
∂b
∂x
. Define
f (x, y) :=
Z
x
p
a(s, y) ds +
Z
y
r
b(p, t) dt
Prove that df = α is exact.
(b) Let α =
−y dx+x dy
x
2
+y
2
= a dx + b dy be defined on the punctured plane R
2
\{(0, 0)}.
Show that
∂a
∂y
=
∂b
∂x
but that α is not exact: the full converse to Exercise 6 is therefore false.
(Hint: α = dθ except on the non-positive real axis; why is this a problem?)
9. Evaluate the integral
R
C
α given C and α.
(a) α = dx − x
−1
dy, where C is parametrized by x( t) = (t
2
, t
3
), 0 ≤ t ≤ 1.
(b) α = 2x tan
−1
y dx +
x
2
1+y
2
dy, where C is parametrized by x(t) =
1
t+1
, 1
, 0 ≤ t ≤ 2.
(c) α = cos x dx + dy, with C the graph of y = cos x over one period of the curve.
10. Which of the integrals in the previous question are path-independent?
11. Prove Lemma 2.15. Moreover, show that if we reverse the orientation of the curve (s
′
( t) < 0)
then the order of the limits is reversed and
R
α becomes −
R
α.
12. Let p ∈ U ⊆ R
2
and let α = a dx + b dy be a 1-form on U. For each q define f (q) :=
R
C
α where
we additionally assume this value is independent of the path C joining p to q.
Let h be small and C
h
the straight line from q to q + hi. Integrate over C
h
to show that
∂ f
∂x
q
= lim
h→0
f
(
q + hi
)
− f (q)
h
= a(q)
Make a similar argument to conclude that α = d f is exact.
13. (If you’ve done complex analysis) Let f (x, y) = u(x, y) + iv(x, y) be a complex-valued func-
tion f : R
2
→ C where u, v are real-valued. Viewing z = x + iy and z = x −iy as co-ordinates
on R
2
, prove that df
∂
∂
¯
z
= 0 if and only if u, v satisfy the Cauchy-Riemann equations:
u
x
= v
y
, v
x
= −u
y
40
2.3 Higher-degree Forms
We introduce a new operation on forms which generalizes the cross product of vectors.
Definition 2.18. Given 1-forms α, β on U, their wedge product α ∧ β is the function which takes two
vector fields and returns the smooth function
α ∧ β
u, v
= det
α(u) α(v)
β(u) β(v)
: U → R
We call α ∧ β a 2-form.
Example 2.19. Let x, y be the usual co-ordinates on R
2
. The standard area form is the object dx ∧ dy
which takes two vector fields u
1
∂
∂x
+ u
2
∂
∂y
and v = v
1
∂
∂x
+ v
2
∂
∂y
and returns the determinant
dx ∧dy
u, v
=
dx(u) dx(v)
dy(u) dy(v)
=
u
1
v
1
u
2
v
2
This gets its name since, at each point p, it returns the (signed) area of the
parallelogram spanned by the tangent vectors u
p
, v
p
.
For instance, if u = 3x
∂
∂x
+ 2y
∂
∂y
and v = y
∂
∂x
+ 4x
∂
∂y
, then
dx ∧dy
u, v
=
3x y
2y 4x
= 12x
2
−2y
2
u
p
v
p
p
0 x x+y x+3x
0
y
y+4x
y+2y
Recall that determinants change sign if you switch its rows or columns, and that they are linear
functions of both their rows and columns. This has two consequences for α ∧ β.
Lemma 2.20. 1. (Columns) At each p ∈ U, a wedge product of 1-forms is an alternating, bilinear
function α ∧ β : T
p
R
n
× T
p
R
n
→ R: given vector fields u, v, w and functions f , g : U → R,
α ∧ β
v, u
= −α ∧ β
u, v
(alternating)
α ∧ β
f u + gv, w
= f α ∧ β
u, w
+ g α ∧ β
v, w
(linear in 1
st
slot)
2. (Rows) Wedge products are alternating and addition distributes over ∧
β ∧α = −α ∧ β and α ∧ α = 0 (alternating)
( α + γ) ∧ β = α ∧ β + γ ∧ β (distributivity in 1
st
slot)
Linearity/distributivity in the second slot is similar in both cases.
The linearity and alternating properties tell us that every wedge product of 1-forms on R
2
may be
written
α ∧ β =
a
1
dx + a
2
dy
∧
b
1
dx + b
2
dy
= (a
1
b
2
− a
2
b
1
) dx ∧dy
Notice the determinant again!
41
For higher order forms, we extend the same approach.
Definition 2.21. The wedge product of 1-forms α
1
, . . . , α
k
on U ⊆ R
n
takes k vector fields and returns a
smooth function:
α
1
∧···∧α
k
v
1
, . . . , v
k
=
α
1
(v
1
) ··· α
1
(v
k
)
.
.
.
.
.
.
.
.
.
α
k
(v
1
) ··· α
k
(v
k
)
: U → R
Let x
1
, . . . , x
n
be co-ordinates on U. A k-form on U (alternating form of degree k) is an expression
α =
∑
a
I
dx
i
1
∧···∧dx
i
k
, a
I
: U → R smooth
where we sum over all increasing multi-indices I = {i
1
< i
2
< ··· < i
k
} ⊆ {1, 2, . . . , n} of length k.
The wedge product of a k-form α and an l-form β is the (k + l)-form
α ∧ β =
∑
I,J
a
I
b
J
dx
i
1
∧···∧dx
i
k
∧dx
j
1
∧···∧dx
j
l
where the 1-forms dx may be rearranged/cancelled using the alternating property (Lemma 2.20.2).
By convention, a 0-form is a smooth function f : U → R, whose wedge product with anything is
pointwise multiplication. At each point p ∈ U, the k-forms comprise the vector space of alternating
multilinear maps with basis {dx
i
1
∧···∧dx
i
k
: i
1
< ··· < i
k
} and dimension
(
n
k
)
=
n!
k!(n−k)!
.
In this course we’ll never have reason to work in more then three dimensions!
The table describes all k-forms in 2 and 3 di-
mensions written in standard co-ordinates.
Analogous to Example 2.19, dx ∧dy ∧dz is
the standard volume form on R
3
.
k R
2
R
3
0 function f f
1 f dx + g dy f dx + g dy + h dz
2 f dx ∧dy f dx ∧dy + g dx ∧dz + h dy ∧dz
3 None f dx ∧dy ∧dz
4+ None None
Examples 2.22. 1. Given 1-forms α = 2 dx −3x dy and β = y
2
dx + y dy on R
2
,
α ∧ β = (2 dx −3x dy) ∧
y
2
dx + y dy
= 2y
2
dx ∧dx + 2y dx ∧dy −3xy
2
dy ∧dx −3xy dy ∧dy
=
2y −3xy
2
dx ∧dy
2. Given the 1-forms α = dx + 2 dy + x dz and 2-form β = 3z dx ∧dy −dy ∧dz on R
3
, the wedge
product α ∧ β is the 3-form
α ∧ β = dx ∧(−dy ∧dz) + 3xz dz ∧dx ∧dy
= (3xz − 1) dx ∧dy ∧dz
Note how dz ∧ dx ∧ dy = −dx ∧ dz ∧ dy = dx ∧ dy ∧ dz requires two swaps, so the sign is
ultimately unchanged!
42
Lemma 2.23. For any forms α, β,
β ∧α = (−1)
deg α deg β
α ∧ β
where deg α = k means that α is a k-form.
This is true by definition when α, β are 1-forms, and trivially true when α is a 0-form. Check the
previous examples to make sure they agree.
Example 2.24 (Polar co-ordinates). Changing to polar co-ordinates, the standard area form on R
2
becomes
dx ∧dy = (cos θ dr −r sin θ dθ) ∧ (sin θ dr + r cos θ dθ) = r dr ∧dθ
This should remind you of change of variables in integration: if f (x, y) = g(r, θ), then
Z
f (x, y)dxdy =
Z
g(r, θ)r drdθ
The example illustrates one of the advantages of forms: change of variables (Jacobians) are built in!
The Exterior Derivative Just as with functions, we can apply ‘d’ to forms.
Definition 2.25. The exterior derivative of a k-form α =
∑
a
I
dx
i
1
∧···∧dx
i
k
is the (k + 1)-form
dα =
∑
da
I
∧dx
i
1
∧···∧dx
i
k
where da
I
=
∑
j
∂a
∂x
j
dx
j
is the usual exterior derivative of a function (Definition 2.10).
Example 2.26. In R
3
, let α = xy
2
z dx − xz dz. Then
dα = d(xy
2
z) ∧ dx −d(xy) ∧dz
= (y
2
z dx + 2xyz dy + xy
2
dz) ∧dx −(z dx + x dz) ∧dz
= −2xyz dx ∧dy −(xy
2
+ z) dx ∧dz
Since dx ∧dx = 0 = dz ∧dz, there was no need to write the blue terms.
Theorem 2.27. Let α, β be forms:
1. d(α + β) = dα + dβ (α, β must have the same degree)
2. d(α ∧ β) = dα ∧ β + (−1)
deg α
α ∧dβ
3. d(dα) = 0. This is often written
15
d
2
α = 0, or just d
2
= 0.
15
A k-form α is closed if dα = 0, and exact if ∃β such that α = dβ. The result says that every exact form is closed. Poincare’s
Lemma gives a partial converse: every closed form on an open ball/hypercube is exact. Exercise 2.2.8 is a simple version.
43
Example (2.26 cont). We verify that d
2
α = 0:
d( dα) = d( −2xyz) ∧dx ∧dy − d(xy
2
+ z) ∧ dx ∧dz
= −2xy dz ∧dx ∧dy −2xy dy ∧ dx ∧dz = 0
Proof. This is very easy to prove explicitly for the only forms we’ll ever see (up to 3-forms in R
3
).
Here are general arguments that work in any dimension.
For simplicity of notation, write dx
I
= dx
i
1
∧···∧dx
i
k
, whenever I = {i
1
< ··· < i
k
}. Then
d(α + β) =
∑
I
da
I
∧dx
I
+ db
I
∧dx
I
=
∑
I
(da
I
+ db
I
) ∧dx
I
= dα + dβ
Part 2 is an exercise. For part 3, we extend Exercise 2.2.6 which in fact shows that d
2
f = 0 for any
function (0-form)
d( dα) = d
∑
I
da
I
∧dx
I
= d
∑
j∈I
∂a
I
∂x
j
dx
j
∧dx
I
=
∑
i,j∈I
∂
2
a
I
∂x
i
x
j
dx
i
∧dx
j
∧dx
I
=
∑
i<j∈I
∂
2
a
I
∂x
i
x
j
−
∂
2
a
I
∂x
j
x
i
dx
i
∧dx
j
∧dx
I
= 0
since mixed partial derivatives commute.
A New Take on Vector Calculus
The standard vector calculus operations of div, grad and curl in E
3
are closely related to the exterior
derivative. For instance, compare the curl of a vector field v = a
1
i + a
2
j + a
3
k with the exterior
derivative of the 1-form α = a
1
dx + a
2
dy + a
3
dz:
∇×v =
∂
∂x
∂
∂y
∂
∂z
×
a
1
a
2
a
3
=
∂a
3
∂y
−
∂a
2
∂z
i +
∂a
1
∂z
−
∂a
3
∂x
j +
∂a
2
∂x
−
∂a
1
∂y
k
dα =
∂a
3
∂y
−
∂a
2
∂z
dy ∧dz +
∂a
1
∂z
−
∂a
3
∂x
dz ∧dx +
∂a
2
∂x
−
∂a
1
∂y
dx ∧dy
Comparing coefficients gives part of the dictionary for comparing forms and traditional vector fields.
function f
_
d
←→ function f
_
∇
grad
a
1
dx + a
2
dy + a
3
dz
_
d
←→ a
1
i + a
2
j + a
3
k
_
∇×
curl
b
1
dy ∧dz + b
2
dz ∧dx + b
3
dx ∧dy
_
d
←→ b
1
i + b
2
j + b
3
k
_
∇·
div
c dx ∧dy ∧ dz ←→ function c
44
The exterior derivative d is div, grad and curl all in one tidy package! Moreover:
• The identity d
2
= 0 translates to two familiar results from vector calculus:
∇×(∇f ) = 0 and ∇ · (∇ × v) = 0
• Under the above identification, the wedge product of 1-forms corresponds to the cross product,
and the wedge product of a 1-form and a 2-form to the dot product. Various identities may be
obtained this way: for instance, if α is a 1-form, then
d( f α) = d f ∧α + f dα ←→ ∇× f v = ∇f ×v + f ∇ × v
• Changes of co-ordinates are built into forms (e.g. Example 2.24).
• The exterior derivative and wedge product apply in any dimension, thus extending standard
vector calculus and the cross product to arbitrary dimensions.
None of what we’ve done in this chapter is strictly necessary for the analysis of surfaces in E
3
. How-
ever, forms are the language of modern differential geometry (and other things besides) and it is
easier to meet them first in a familiar setting. And if you want to do higher-dimensional geometry
(e.g., general relativity), this new language becomes almost essential.
Exercises 2.3. 1. Compute α(u, v), given α = dx ∧dy + z dy ∧dz, u =
∂
∂x
−
∂
∂z
and v = y
∂
∂y
+
∂
∂z
.
2. Let α = y
2
dx ∧dz −dy ∧ dz and u = x
∂
∂x
+ xy
∂
∂y
−
∂
∂z
and v = −y
∂
∂x
+ y
3
∂
∂y
.
(a) Compute α(u, v).
(b) Find the 3-form dα.
3. Given s = x
2
−y
2
and t = 2xy, compute ds ∧ dt in terms of dx ∧dy
4. Revisit Lemma 2.20. State what it means for a wedge product of 1-forms α ∧ β to be linear in
the second slot.
5. Let f , g be functions and consider the 1-form α = g d f . Show that α ∧dα = 0. Can the 1-form
dx + y dz be written in the form g d f ?
6. (a) Check the claim that the wedge product of 1-forms on R
3
corresponds to the cross product.
(b) Suppose α is a 2-form on R
3
. To what vector calculus identity does d( f α) = d f ∧ α + f dα
correspond?
(c) State an expression using forms, d and ∧which corresponds to the vector calculus identity
∇·( u ×v) = (∇ × u) ·v −u ·(∇ × v)
7. Let r, θ, ϕ be the spherical polar co-ordinate system in Exercise 2.1.3. Show that
dx ∧dy ∧dz = r
2
cos ϕ dr ∧dθ ∧ dϕ
45
8. A 2-form is decomposable if it can be written as a wedge product α ∧ β for some 1-forms α, β.
(a) Show that every 2-form on R
3
is decomposable.
(b) If w, x, y, z are co-ordinates on R
4
, show that the 2-form dw ∧ dx + dy ∧dz is not decom-
posable.
(Hint: if a 2-form γ is decomposable, what is γ ∧γ?)
9. (Hard) Suppose α, β are forms, sketch an argument for why
α ∧ β = (−1)
deg α deg β
β ∧α
Now prove that
d(α ∧ β) = dα ∧ β + (−1)
deg α
α ∧dβ
10. (Hard) Given vector fields u, v, their Lie bracket [u, v] is the vector field such that
[u, v][ f ] := u
v[ f ]
−v
u[ f ]
for all functions f .
(a) Compute [u, v][ f ] where u = 3x
∂
∂x
+
∂
∂y
and v =
∂
∂x
− x
∂
∂y
and f (x, y) = x
2
y.
(b) If u =
∑
u
j
∂
∂x
j
and v =
∑
v
k
∂
∂x
k
, show that [u, v] really is a vector field by explicitly comput-
ing [u, v][ f ] in the form
∑
c
j
∂ f
∂x
j
: how do the coefficients c
j
of the vector field [u, v] depend
on those of u, v? Find the field [u, v] when u, v are as in part (a).
(c) If α is a 1-form and u, v are vector fields, prove that
dα
u, v
= u
α(v)
−v
α(u)
−α
[u, v]
This provides a co-ordinate-free definition of dα; similar expressions exist for k-forms
(Hint: Write everything out as sums over j, k so that all differentiations of scalars are with respect
to the single variable x
k
; now compare!)
46
