vector fields - tutmath.tut.fi/~ruohonen/vf.pdf · these lecture notes form the base text for the...

VECTOR FIELDS

Keijo Ruohonen

2013

Contents

1 I POINT. VECTOR. VECTOR FIELD1 1.1 Geometric Points2 1.2 Geometric Vectors5 1.3 Coordinate Points and Vectors7 1.4 Tangent Vectors. Vector Fields. Scalar Fields9 1.5 Differential Operations of Fields13 1.6 Nonstationary Scalar and Vector Fields

16 II MANIFOLD16 2.1 Graphs of Functions17 2.2 Manifolds18 2.3 Manifolds as Loci22 2.4 Mapping Manifolds. Coordinate-Freeness23 2.5 Parametrized Manifolds30 2.6 Tangent Spaces34 2.7 Normal Spaces36 2.8 Manifolds and Vector Fields

38 III VOLUME38 3.1 Volumes of Sets41 3.2 Volumes of Parametrized Manifolds43 3.3 Relaxed Parametrizations

46 IV FORMS46 4.1 k-Forms52 4.2 Form Fields54 4.3 Forms and Orientation of Manifolds59 4.4 Basic Form Fields of Physical Fields

62 V GENERALIZED STOKES’ THEOREM63 5.1 Regions with Boundaries and Their Orientation69 5.2 Exterior Derivatives73 5.3 Exterior Derivatives of Physical Form Fields77 5.4 Generalized Stokes’ Theorem

83 VI POTENTIAL83 6.1 Exact Form Fields and Potentials87 6.2 Scalar Potential of a Vector Field inR3

92 6.3 Vector Potential of a Vector Field inR3

96 6.4 Helmholtz’s Decomposition97 6.5 Four-Potential98 6.6 Dipole Approximations and Dipole Potentials

i

ii

102 VII PARTIAL DIFFERENTIAL EQUATIONS102 7.1 Standard Forms102 7.2 Examples

107 Appendix 1: PARTIAL INTEGRATION AND GREEN’SIDENTITIES

107 A1.1 Partial Integration111 A1.2 Green’s Identities

111 Appendix 2: PULLBACKS AND CURVILINEAR COORDINATES111 A2.1 Local Coordinates113 A2.2 Pullbacks115 A2.3 Transforming Derivatives of Fields117 A2.4 Derivatives in Cylindrical and Spherical Coordinates

118 Appendix 3: ANGLE118 A3.1 Angle Form Fields and Angle Potentials119 A3.2 Planar Angles120 A3.3 Solid Angles122 A3.4 Angles inRn

124 References126 Index

Foreword

These lecture notes form the base text for the course ”MAT-60506 Vector Fields”. They aretranslated from the older Finnish lecture notes for the course ”MAT-33351 Vektorikentät”, withsome changes and additions.

These notes deal with basic concepts of modern vector field theory, manifolds, (differential)forms, form fields, Generalized Stokes’ Theorem, and various potentials. A special goal is aunified coordinate-free physico-geometric representation of the subject matter. As a sufficientbackground, basic univariate calculus, matrix calculus and elements of classical vector analysisare assumed.

Classical vector analysis is one of the oldest areas of mathematical analysis.1 Modellingstructural strength, fluid flow, thermal conduction, electromagnetics, vibration etc. in the three-space needs generalization of the familiar concepts and results of univariate calculus. Thereseem to be a lot of these generalizations. Indeed, vector analysis—classical as well as modern—has been largely shaped and created by the many needs of physics and various engineeringapplications. For the latter, it is central to be able to formulate the problem as one where fastand accurate numerical methods can be readily applied. Thisgenerally means specifying thelocal behavior of the phenomenon using partial differential equations (PDEs) of a standard type,which then can be solved globally using program libraries. Here PDEs are not extensively dealtwith, mainly via examples. On the other hand, basic conceptsand results having to do withtheir derivation are emphasized, and treated much more extensively.

Modern vector analysis introduces concepts which greatly unify and generalize the manyscattered things of classical vector analysis. Basically there are two machineries to do this:

1There is a touch of history in the venerable Finnish classicsTALLQVIST and VÄISÄLÄ , too.

iii

manifolds and form fields, and Clifford’s algebras. These notes deal with the former (the latteris introduced in the course ”Geometric Analysis”).

The style, level and order of presentation of the famous textbook HUBBARD & H UBBARD

have turned out to be well-chosen, and have been followed here, too, to an extent. Many tediousand technical derivations and proofs are meticulously worked out in this book, and are omittedhere. As another model the more advanced book LOOMIS & STERNBERGmight be mentioned,too.

Keijo Ruohonen

Chapter 1

POINT. VECTOR. VECTOR FIELD

”Time is Nature’s way of keeping everything from

happening all at once. Space is Nature’s way of keeping

everything from happening all at the same place.”

(VARIOUS ATTRIBUTIONS AND PARAPHRASINGS)

1.1 Geometric Points

A point in space is a physico-geometric primitive, and is not given any particular definition

here. Let us just say that dealing with points, lines, planes and solids is mathematically part

of so-called solid geometry. In what follows points will be denoted by capital italic letters:

P,Q,R, . . . and P1, P2, . . . , etc. The distance between the points P and Q is denoted by

d(P,Q). Obviously d(P, P ) = 0, d(P,Q) = d(Q,P ) and

d(P,R) ≤ d(P,Q) + d(Q,R) (triangle inequality).

An open P -centered ball of radius R is the set of all points Q with d(P,Q) < R, and is

denoted by B(R,P ). Further:

• The point set A is open, if for each point P in it there is a number RP > 0 such that

B(RP , P ) ⊆ A. In particular the empty set ∅ is open.

• The boundary ∂A of the point set A is the set of all points P such that every open ball

B(R,P ) (R > 0) contains both a point of A and a point of the complement of A. In

particular the boundary of the empty set is empty. A set is thus open if and only if it does

not contain any of the points of its boundary.

• The point set A is closed, if it contains its boundary. In particular the empty set is thus

closed. Since the boundaries of a set and its complement clearly are the same, a set is

closed if and only if its complement is open.

• The closure1 of the point set A is the set A = A∪ ∂A and the interior is A = A− ∂A.

The interior of an open set is the set itself, as is the closure of a closed set.

Geometric points naturally cannot be added, subtracted, or be multiplied by a scalar (a real-

valued constant). It will be remembered from basic calculus that for coordinate points these

operations are defined. But in reality they are then the corresponding operations for vectors, as

will be seen soon. Points and vectors are not the same thing.

Note. Here and in the sequel only points, vectors and vector fields in space are explicitly dealtwith. The concepts may be defined for points, vectors and vector fields in plane or real axis.For instance, an open ball in plane is an open circle, in real axis an open ball is an open finiteinterval, and so on. To an extent they can be defined in higher dimensions, too.

1Do not confuse this with the complement which is often also denoted by an overbar!

1

CHAPTER 1. POINT. VECTOR. VECTOR FIELD 2

1.2 Geometric Vectors

The directed line segment connecting the two points P (the initial point) and Q (the terminal

point) is denoted by−→PQ. Two such directed line segments

−→PQ and

−→RS are said to be equivalent

if they can be obtained from each other by parallel transform, i.e., there is a parallel transform

which takes P to R and Q to S, or vice versa.

Directed line segments are thus partitioned into equivalence classes. In each equivalence

class directed line segments are mutually equivalent, while those in different equivalence classes

are not. The equivalence class containing the directed line segment−→PQ is denoted by 〈−→PQ〉.

The directed line segment−→PQ is then a representative of the class. Each class has a representa-

tive for any given initial (resp. terminal) point.

Geometric vectors can be identified with these equivalenc classes. The geometric vector

with a representative−→PQ has a direction (from P to Q) and length (the distance d(P,Q)).

Since representatives of an equivalence class are equivalent via parallel transforms, direction

and length do not depend on the choice of the representative.

In the sequel geometric vectors will be denoted by small italic letters equipped with an

overarrow: ~r, ~s, ~x . . . and ~r0, ~r1, . . . , etc. For convenience the zero vector ~0 will be included,

too. It has no direction and zero length. The length of the vector ~r is denoted by |~r|. A vector

with length = 1 is called a unit vector.A physical vector often has a specific physical unit L (sometimes also called dimension),

e.g. kg/m2/s. In this case the geometric vector ~r couples the direction of the physical action to

the direction of the geometric vector—unless ~r is the zero vector—and the magnitude ~r is given

in physical units L. Note that if the unit L is a unit of length, say metre, then a geometric vector

~r may be considered as a physical vector, too. A physical vector may be unitless, so that it has

no attached physical unit, or has an empty unit. Physical units can be multiplied, divided and

raised to powers, the empty unit has the (purely numerical) value 1 in these operations.2

Often, however, a physical vector is completely identified with a geometric vector (with a

proper conversion of units). In the sequel we just generally speak about vectors.

Vectors have the operations familiar from basic courses of mathematics. We give the geo-

metric definitions of these in what follows. Geometrically it should be quite obvious that they

are well-defined, i.e., independent of choices of representatives.

• The opposite vector of the vector ~r = 〈−→PQ〉 is the vector

−~r = 〈−→QP 〉.

In particular −~0 = ~0. The unit of a physical vector remains the same in this operation.

• The sum of the vectors

~r = 〈−→PQ〉 and ~s = 〈−→QR〉(note the choice of the representatives) is the vector

~r + ~s = 〈−→PR〉.

In particular we define

~r +~0 = ~0 + ~r = ~r

2See e.g. GIBBINGS, J.C.: Dimensional Analysis. Springer–Verlag (2011) for many more details of dimen-

sional analysis.


and

~r + (−~r ) = (−~r ) + ~r = ~0.

Only vectors sharing a unit can be physically added, and the unit of the sum is this unit.

Addition of vectors is commutative and associative, i.e.

~r + ~s = ~s + ~r and ~r + (~s+ ~t ) = (~r + ~s ) + ~t.

These are geometrically fairly obvious. Associativity implies that long sums may be

parenthesized in any (correct) way, or written totally without parenteheses, without the

result changing.

The difference of the vectors ~r and ~s is the vector

~r − ~s = ~r + (−~s ).

For physical vectors the units again should be the same in this operation.

• If ~r = 〈−→PQ〉 is a vector and λ a positive scalar, then λ~r is the vector obtained as follows:

– Take the ray starting from P through the point Q.

– In this ray find the point R whose distance from P is λ|~r |.– Then λ~r = 〈−→PR〉.

In addition it is agreed that λ~0 = ~0 and 0~r = ~0.

This operation is multiplication of vector by scalar. Defining further (−λ)~r = −(λ~r ) we

get multiplication by a negative scalar. Evidently

1~r = ~r , (−1)~r = −~r , 2~r = ~r + ~r , etc.

If the physical scalar λ and physical vector ~r have their physical units, the unit of λ~r is

their product. With a bit of work the following laws of calculation can be (geometrically)

verified:λ1(λ2~r) = (λ1λ2)~r ,

(λ1 + λ2)~r = λ1~r + λ2~r and

λ(~r1 + ~r2) = λ~r1 + λ~r2,

where λ, λ1, λ2 are scalars and ~r, ~r1, ~r2 are vectors.

Division of a vector ~r by a scalar λ 6= 0 is multiplication of the vector by the inverse 1/λ,

denoted by ~r/λ.

A frequently occurring multiplication by scalar is normalization of a vector where a vector

~r 6= ~0 is divided by its length. The result ~r/|~r | is a unit vector (and unitless).

• The angle ∠(~r, ~s ) spanned by the vectors ~r = 〈−→PQ〉 and ~s = 〈−→PR〉 (note the choice

of representatives) is the angle between the directed line segments−→PQ and

−→PR given in

radians in the interval [0, π] rad. It is obviously assumed here that ~r, ~s 6= ~0. Moreover an

angle is always unitless, the radian is not a physical unit.


• The distance of the vectors

~r = 〈−→PQ〉 and ~s = 〈−→PR〉

(note the choice of representatives) is

d(~r, ~s ) = d(Q,R) = |~r − ~s |.

In particular d(~r,~0 ) = |~r |. This distance also satisfies the triangle inequality

d(~r, ~s ) ≤ d(~r,~t ) + d(~t, ~s ).

• The dot product (or scalar product) of the vectors ~r 6= ~0 and ~s 6= ~0 is

~r • ~s = |~r ||~s | cos∠(~r, ~s ).

In particular

~r •~0 = ~0 • ~r = 0 and ~r • ~r = |~r |2.

Dot product is commutative, i.e.

~r • ~s = ~s • ~r,and bilinear, i.e.

~r • (λ~s+ η~t ) = λ(~r • ~s ) + η(~r • ~t ),where λ and η are scalars. Geometrically commutativity is obvious. Bilinearity on the

other hand requires a somewhat complicated geometric proof. Using coordinates makes

bilinearity straightforward.

The unit of the dot product of physical vectors is the product of their units. Geometrically,

if ~s is a (unitless) unit vector, then

~r • ~s = |~r | cos∠(~r, ~s )

is the (scalar) projection of ~r on ~s. (The projection of a zero vector is of course always

zero.)

• The cross product (or vector product) of the vectors ~r and ~s is the vector ~r × ~s given by

the following. First, if ~r or ~s is ~0, or ∠(~r, ~s ) is 0 or π, then ~r× ~s = ~0. Otherwise ~r× ~s is

the unique vector ~t satisfying

– |~t | = |~r ||~s | sin∠(~r, ~s ),– ∠(~r,~t ) = ∠(~s,~t ) =

π

2, and

– ~r, ~s,~t is a right-handed system of vectors.

Cross product is anticommutative, i.e.

~r × ~s = −(~s× ~r ),

and bilinear, i.e.

~r × (λ~s+ η~t ) = λ(~r × ~s ) + η(~r × ~t ),

where λ and η are scalars. Geometrically anticommutatitivy is obvious, handidness

changes. Bilinearity again takes a complicated geometric proof, but is fairly easily seen

using coordinates.


Cross product is an information-dense operation, involving lengths of vectors, angles and

handidness. It is easily handled using coordinates, too. Geometrically

|~r × ~s | = |~r ||~s | sin∠(~r, ~s )

is the area of the parallelogram with lengths of sides |~r | and |~s | and spanning angle

∠(~r, ~s ). If ~r and ~s are physical vectors, then the unit of the cross product ~r × ~s is the

product of their units.

• Combining these products we get the scalar triple product

~r • (~s× ~t )

and the vector triple products

(~r × ~s )× ~t and ~r × (~s× ~t ).

There being no danger of confusion, the scalar product is usually written without paren-

theses as ~r • ~s× ~t.

Scalar triple product is cyclically symmetric, i.e.

~r • ~s× ~t = ~s • ~t× ~r = ~t • ~r × ~s.

By this and commutativity of scalar product, the operations • and × can be interchanged,

i.e.

~r • ~s× ~t = ~r × ~s • ~t.

Geometrically it is easily noted that the scalar triple product ~r •~s×~t is the volume of the

parallelepiped with edges incident on the vertex P being the vectors

~r = 〈−→PR〉 , ~s = 〈−→PS〉 and ~t = 〈−→PT 〉,

with a positive sign if ~r, ~s,~t is a right-handed system, and a negative sign otherwise. (As

special cases situations where the scalar triple product is = 0 should be included, too.)

Cyclic symmetry follows geometrically immediately from this observation.

The triple vector product expansions (also known as Lagrange’s formulas) are

(~r × ~s )× ~t = (~r • ~t )~s− (~s • ~t )~r and

~r × (~s× ~t ) = (~r • ~t )~s− (~r • ~s )~t.

These are somewhat difficult to prove geometrically, proofs using coordinates are easier.

Exactly as for points we can define an ~r-centered open ball ~B(R,~r ) of radius R for vectors,

open and closed sets of vectors, and boundaries, closures and interiors of sets of vectors, but a

geometric intuition is then not as easily obtained.

1.3 Coordinate Points and Vectors

In basic courses on mathematics points are thought of as coordinate points, i.e. triples (a, b, c) of

real numbers. In the background there is then an orthonormal right-handed coordinate system

with its axes and origin. Coordinate points will be denoted by small boldface letters: r, s,


x, . . . and r0, r1, . . . , etc. The coordinate point corresponding to the origin of the system is

0 = (0, 0, 0).A coordinate system is determined by the corresponding coordinate function which maps

geometric points to the triples of R3. We denote coordinate functions by small boldface Greek

letters, and their components by the corresponding indexed letters. If the coordinate function is

κ, then the coordinates of the point P are

κ(P ) =(

κ1(P ), κ2(P ), κ3(P ))

.

A coordinate function is bijective giving a one-to-one correspondence between the geometric

space and the Euclidean space R3.

Distances are given by the familiar Euclidean norm of R3. If

κ(P ) = (x1, y1, z1) and κ(Q) = (x2, y2, z2),

then

d(P,Q) =∥

∥κ(P )− κ(Q)∥

∥ =√

(x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2.

A coordinate function κ also gives a coordinate representation for vectors. The coordinate

version of the vector ~r = 〈−→PQ〉 is

κ(~r ) =(

κ(Q)− κ(P ))T

=

κ1(Q)− κ1(P )κ2(Q)− κ2(P )κ3(Q)− κ3(P )

.

Note that this is a column array. As is easily seen, this presentation does not depend on choices

of the representative directed line segments. In particular the zero vector has the representation

κ(~0 ) = (0, 0, 0)T = 0T. Also the distance of the vectors ~r = 〈−→PQ〉 and ~s = 〈−→PR〉 (note the

choice of representatives) is obtained from the Euclidean norm of R3:

d(~r, ~s ) =∥

∥κ(~r )− κ(~s )∥

∥ =∥

∥κ(Q)− κ(R)∥

∥ = d(Q,R).

And then |~r | = d(~r,~0) =∥

∥κ(~r )∥

∥.

In the sequel also coordinate representations, or coordinate vectors, will be denoted by

small boldface letters, but it should be remembered that a coordinate vector is a column vector.

Certain coordinate vectors have their traditional notations and roles:

i =

100

, j =

010

, k =

001

and r =

xyz

.

The vectors i, j,k are the basis vectors and the vector r is used as a generic variable vector. In

the background there of course is a fixed coordinate system and a coordinate function. The row

array versions of these are also used as coordinate points.

Familiar operations of coordinate vectors now correspond exactly to the geometric vector

operations in the previous section. Let us just recall that if

κ(~r ) = r =

a1b1c1

and κ(~s ) = s =

a2b2c2

,

then

~r • ~s = r • s = a1a2 + b1b2 + c1c2


and

κ(~r × ~s ) = r× s =

b1c2 − b2c1c1a2 − c2a1a1b2 − a2b1

.

The latter is often given as the more easily remembered formal determinant

∣

∣

∣

∣

∣

∣

i a1 a2j b1 b2k c1 c2

∣

∣

∣

∣

∣

∣

,

to be expanded along the first column.

A coordinate transform changes the coordinate function. If κ and κ∗ are two available

coordinate functions, then they are connected via a coordinate transform, that is, there is a 3×3orthogonal matrix3 Q and a coordinate vector b such that

κ∗(P ) = κ(P )Q+ b

and

κ(P ) = κ∗(P )QT − bQT.

The coordinate representation of a vector ~r = 〈−→PQ〉 is transformed similarly:

κ∗(~r ) =

(

κ∗(Q)− κ

∗(P ))

T

=(

κ(Q)Q+ b− κ(P )Q− b)

T

= QT(

κ(Q)− κ(P ))

T

= QTκ(~r )

and

κ(~r ) = Qκ∗(~r ).

Note that b is the representation of the origin of the ”old” coordinate system in the ”new”

system, and that the columns of QT are the representations of the basis vectors i, j,k of the ”old”

coordinate system in the ”new” system. Similarly −bQT is the representation of the origin of

the ”new” coordinate system in the ”old” system, and the columns of Q are the representations

of the ”new” basis vectors i∗, j∗,k∗ in the ”old” system.

1.4 Tangent Vectors. Vector Fields. Scalar Fields

Geometrically a tangent vector4 is simply a directed line segment−→PQ, the point P is its point

of action.It is however easier to think a tangent vector (especially a physical one) as a pair [P,~r ] where

P is the point of action and ~r is a vector. It is then easy to apply vector operations to tangent

vectors: just operate on the vector component ~r. If the result is a vector, it may be thought of as

a tangent vector, with the original (joint) point of action, or as just a vector without any point of

action. Moreover, in the pair formulation, a coordinate representation is simply obtained using

a coordinate function κ:

κ(

[P,~r ])

=[

κ(P ),κ(~r )]

.

Often in the pair [P,~r ] we consider ~r as a physical vector operating in the point P .

3Here and in the sequel matrices are denoted by boldface capital letters. Since handidness needs to be preserved,

we must have here det(Q) = 1. Recall that orthogonality means that Q−1 = QT which implies that det(Q)2 = 1.4Thus called because it often literally is a tangent.


If the point of action is clear from the context, or is irrelevant, it is often omitted and only

the vector component of the pair is used, usually in a coordinate representation. This is what

we will mostly do in the sequel.

A vector field is a function mapping a point P to a tangent vector[

P, ~F (P )]

(note the point

of action). Mostly we denote this just by ~F . A vector field may not be defined for all points of

the geometric space, i.e., its domain of definition may be smaller.

In the coordinate representation given by the coordinate function κ we denote

r = κ(P ) and F(r) = κ(

~F (P ))

,

thus coordinate vector fields are denoted by capital boldface letters. Note that in the coordinate

transform

r∗ = rQ+ b (i.e. κ∗(P ) = κ(P )Q+ b)

the vector field F = κ(~F ) (that is, its representation) is transformed to the the field F∗ = κ∗(~F )

given by the formula

F∗(r∗) = QTF(

(r∗ − b)QT)

.

A vector field may of course be defined in fixed coordinates in one way or another, and then

taken to other coordinate systems using the transform formula. On the other hand, definition

of a physico-geometric vector field cannot possibly depend on any coordinate system, the field

exists without any coordinates, and will automatically satisfy the transform formula.

A coordinate vector field is the vector-valued function of three arguments familiar from

basic courses of mathematics

F(r) =

F1(r)F2(r)F3(r)

,

with its components. Thus all operations and concepts defined for these apply, limits, continuity,

differentiability, integrals, etc.

A scalar field is a function f mapping a point P to a scalar (real number) f(P ), thus scalar

fields are denoted by italic letters, usually small. In the coordinate representation we denote

r = κ(P ) and just f(r) (instead of the correct f(

κ−1(r)

)

). In the coordinate transform

r∗ = rQ+ b (i.e. κ∗(P ) = κ(P )Q+ b)

a scalar field f is transformed to the scalar field f ∗ given by the formula

f ∗(r∗) = f(

(r∗ − b)QT)

.

A scalar field, too, can be defined in fixed coordinates, and then transformed to other coordinate

systems using the transform formula. But a physico-geometric scalar field exists without any

coordinate systems, and will automatically satisfy the transform formula.

Again a coordinate scalar field is the real-valued function of three arguments familiar from

basic courses of mathematics. Thus all operations and concepts defined for these apply, limits,

continuity, differentiability, integrals, etc.

An important observation is that all scalar and vector products in the previous section ap-

plied to vector and scalar fields will again be fields, e.g. a scalar field times a vector field is a

vector field.


1.5 Differential Operations of Fields

Naturally, partial derivatives cannot be defined for physico-geometric fields, since they are in-

trinsically connected with a coordinate system. In a coordinate representation partial derivatives

can be defined, as was done in the basic courses.

In this way we get partial derivatives of a scalar field f , its derivative

f ′ =(∂f

∂x,∂f

∂y,∂f

∂z

)

and gradient

grad(f) = ∇f = f ′T =

∂f

∂x∂f

∂y∂f

∂z

,

and the derivative or Jacobian (matrix) of a vector field F =

F1

F2

F3

F′ =

F ′1

F ′2

F ′3

=

∂F1

∂x

∂F1

∂y

∂F1

∂z∂F2

∂x

∂F2

∂y

∂F2

∂z∂F3

∂x

∂F3

∂y

∂F3

∂z

.

Using the chain rule5 we get the transforms of the derivatives in a coordinate transform

r∗ = rQ + b:

f ∗′(r∗) =(

f(

(r∗ − b)QT))′

= f ′(

(r∗ − b)QT)

Q

and

F∗′(r∗) =(

QTF(

(r∗ − b)QT))′

= QTF′(

(r∗ − b)QT)

Q.

Despite partial derivatives depending on the coordinate system used, differentiability itself

is coordinate-free: If a field has partial derivates in one coordinate system, it has them in any

other coordinate system, too. This is true for second order derivatives as well. And it is true for

continuity: Continuity in one coordinate system implies continuity in any other system. And

finally it is true for continuous differentiability: If a field has continuous partial derivatives (first

or second order) in one coordinate system, it has them in any other system, too. All this follows

from the transform formulas.

5Assuming f and g differentiable, the familiar univariate chain rule gives the derivative of the composite func-

tion:(

f(

g(x)))

′

= f ′(

g(x))

g′(x).

More generally, assuming F and G continuously differentiable (and given as column arrays), we get the derivative

of the composite function as(

F(

G(r)T))

′

= F′(

G(r)T)

G′(r).

The arguments are here thought of as row arrays. The rule is valid in higher dimensions, too.


The common differential operations for fields are the gradient (nabla) of a scalar field f , and

its Laplacian

∆f = ∇ • (∇f) = ∇2f =∂2f

∂x2+

∂2f

∂y2+

∂f 2

∂z2,

and for a vector field F =

F1

F2

F3

its divergence

div(F) = ∇ • F =∂F1

∂x+

∂F2

∂y+

∂F3

∂z

and curl

curl(F) = ∇× F =

∂F3

∂y− ∂F2

∂z∂F1

∂z− ∂F3

∂x∂F2

∂x− ∂F1

∂y

=

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

i∂

∂xF1

j∂

∂yF2

k∂

∂zF3

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

∣

.

(As the cross product, the curl can be given as a formal determinant.)

As will be verified shortly, gradient, divergence and curl are coordinate-free. Thus ∇ • F

can be interpreted as a scalar field, and, as already indicated by the column array notation,

∇f and ∇× F as vector fields.

For the gradient coordinate-freeness is physically immediate. It will be remembered that the

direction of the gradient is the direction of fastest growth for a scalar field, and its length is this

speed of growth (given as a directional derivative). For divergence and curl the situation is not

at all as obvious.

It follows from the coordinate-freenes of the gradient that the directional derivative of a

scalar field f in the direction n (a unit vector)

∂f

∂n= n • ∇f

is also coordinate-free and thus a scalar field.

The Laplacian may be applied to a vector field as well as follows:

∆F =

∆F1

∆F2

∆F3

.

This ∆F is coordinate-free, too, and can be interpreted as a vector field.

A central property, not to be forgotten, is that all these operations are linear, in other words,

if λ1 and λ2 are constant scalars, then e.g.

∇(λ1f + λ2g) = λ1∇f + λ2∇g and

∇ • (λ1F+ λ2G) = λ1∇ • F+ λ2∇ •G , etc.

The following notational expression appears often:

G • ∇ = G1

∂

∂x+G2

∂

∂y+G3

∂

∂z,


where G = (G1, G2, G3)T is a vector field. This is interpreted as an operator applied to a scalar

field f or a vector field F = (F1, F2, F3)T as follows:

(G • ∇)f = G • (∇f) = G1

∂f

∂x+G2

∂f

∂y+G3

∂f

∂z

and (taking F1, F2, F3 to be scalar fields)

(G • ∇)F =

(G • ∇)F1

(G • ∇)F2

(G • ∇)F3

= F′G.

These are both coordinate-free and hence fields. Coordinate-freeness of (G•∇)F follows from

the nabla rules below (or from the coordinate-freeness of F′G).

Let us tabulate the familiar nabla-calculus rules:

(i) ∇(fg) = g∇f + f∇g

(ii) ∇ 1

f= − 1

f 2∇f

(iii) ∇ • (fG) = ∇f •G+ f∇ •G

(iv) ∇× (fG) = ∇f ×G+ f∇×G

(v) ∇ • (F×G) = ∇× F •G− F • ∇ ×G

(vi) ∇× (F×G) = (G • ∇)F− (∇ • F)G+ (∇ •G)F− (F • ∇)G

(vii) ∇(F •G) = (G • ∇)F− (∇× F)×G− (∇×G)× F+ (F • ∇)G

In matrix notation ∇(F •G) = F′TG+G′TF.

(viii) (∇× F)×G = (F′ − F′T)G

(ix) ∇ • (∇× F) = 0

(x) ∇×∇f = 0

(xi) ∇× (∇× F) = ∇(∇ • F)−∆F (so-called double-curl expansion)

(xii) ∆(fg) = f∆g + g∆f + 2∇f • ∇g

In formulas (ix), (x), (xi) we assume F and f are twice continuously differentiable. These

formulas are all symbolical identities, and can be verified by direct calculation, or e.g. using the

Maple symbolic computation program.


Let us, as promised, verify coordinate-freeness of the operators. In a coordinate transform

r∗ = rQ+ b

we denote the nabla in the new coordinates by ∇∗. Coordinate-freeness for the basic operators

then means the following:

1. ∇∗(

f(

(r∗ − b)QT))

= QT∇f(r) (gradient)

Subtracting b and multiplying by QT we move from the new coordinates r∗ to the old

ones, get the value of f , and then the gradient in the new coordinates. The result must be

the same when the gradient is obtained in the old coordinates and then transformed to the

new ones by multiplying by QT.

2. ∇∗ •(

QTF(

(r∗ − b)QT))

= ∇ • F(r) (divergence)


ones, get F, transform the result to the new coordinates by multiplying by QT, and get

the divergence using the new coordinates. The result must remain the same when the

divergence is obtained in the old coordinates.

3. ∇∗ ×(

QTF(

(r∗ − b)QT))

= QT(

∇× F(r))

(curl)


ones, get F, transform the result to the new coordinates by multiplying by QT, and get the

curl using the new coordinates. The result must be the same when the curl is obtained in

the old coordinates and then transformed to the new ones by multiplying by QT.

Theorem 1.1. Gradient, divergence, curl and Laplacian are coordinate-free. Furthermore, ifF and G are vector fields (and thus coordinate-free), then so is (G • ∇)F.

Proof. By the above

f ∗′(r∗) =(

∇f(r))

T

Q and F∗′(r∗) = QTF′(r)Q.

This immediately gives formula 1. since

∇∗f ∗(r∗) = f ∗′(r∗)T = QT∇f(r).

To show formula 2. we use the trace of the Jacobian. Let us recall that the trace of a square

matrix A, denoted trace(A), is the sum of the diagonal elements. A nice property of trace6 is

that if the product matrix AB is square—whence BA is square, too—then

trace(AB) = trace(BA).

Since

trace(F′) =∂F1

∂x+

∂F2

∂y+

∂F3

∂z= ∇ • F,

6Denoting A = (aij) (an n×m matrix), B = (bij) (an m× n matrix), AB = (cij) and BA = (dij) we have

trace(AB) =n∑

k=1

ckk =n∑

k=1

m∑

l=1

aklblk =m∑

l=1

n∑

k=1

blkakl =m∑

l=1

dll = trace(BA).


formula 2. follows:

∇∗ • F∗(r∗) = trace(

F∗′(r∗))

= trace(

QTF′(r)Q)

= trace(

QQTF′(r))

(take B = Q)

= trace(

F′(r))

= ∇ • F(r).

To prove formula 3. we denote the columns of Q by q1,q2,q3. Let us consider the first

component of the curl ∇∗×F∗(r∗). Using the transform formula for the Jacobian, nabla formula

(viii) and rules for the scalar triple product we get

(

∇∗ × F∗(r∗))

1=

∂F ∗3

∂y∗− ∂F ∗

2

∂z∗

= qT

3F′(r)q2 − qT

2F′(r)q3 (because F∗′ = QTF′Q)

= qT

3F′(r)q2 − qT

3F′(r)Tq2 (qT

2F′q3 is a scalar)

= qT

3

(

F′(r)− F′(r)T)

q2 (extract the factors qT3 and q2)

= q3 •(

∇× F(r))

× q2 (formula (viii))

= q2 • q3 ×(

∇× F(r))

(cyclic symmetry)

= q2 × q3 •(

∇× F(r))

(interchange • and ×)

= q1 •(

∇× F(r))

(here q2 × q3 = q1)

=(

QT(

∇× F(r)))

1.

Note that q2 × q3 = q1 since the new coordinate system must be right-handed, too. The other

components are dealt with similarly.

Coordinate-freeness of the Laplacian follows directly from that of gradient and divergence

for scalar fields, and for vector fields from formula (xi).

Adding formulas (vi) and (vii) on both sides we get an expression for (G • ∇)F:

(G • ∇)F =1

2

(

∇× (F×G) + (∇ • F)G− (∇ •G)F+∇(F •G)

+ (∇× F)×G+ (∇×G)× F)

.

All six terms on the right hand side are coordinate-free and thus vector fields. The left hand side

(G • ∇)F then also is coordinate-free and a vector field. (This is also easily deduced from the

matrix form (G • ∇)F = F′G, but the formula above is of other interest, too!)

1.6 Nonstationary Scalar and Vector Fields

Physical fields naturally are often time-dependent or dynamical, that is, in the definition of the

field a time variable t must appear.

A scalar field is then of the form f(P, t) and a vector field of the form ~F (P, t). (The point

of action is omitted here, even though that, too, may be time-dependent.) In a coordinate rep-

resentation these forms are respectively f(r, t) and F(r, t). Time-dependent fields are called

nonstationary, and time-independent fields are called stationary.


From the coordinate representation, interpreted as a function of the four variables x, y, z, t,we again get the concepts continuity, differentiability, etc., familiar from basic courses, also for

the time variable t. In a coordinate transform r∗ = rQ+b the time variable is not transformed,

i.e.

f ∗(r∗, t) = f(

(r∗ − b)QT, t)

and

F∗(r∗, t) = QTF(

(r∗ − b)QT, t)

.

Thus for the time derivatives we get the corresponding transform formulas

∂i

∂tif ∗(r∗, t) =

∂i

∂tif(

(r∗ − b)QT, t)

and

∂i

∂tiF∗(r∗, t) = QT

∂i

∂tiF(

(r∗ − b)QT, t)

,

which shows that they are fields.

In addition to the familiar partial derivative rules we get for the time derivatives e.g. the

following rules, which can be verified by direct calculation:

(1)∂

∂t(F •G) =

∂F

∂t•G+ F • ∂G

∂t

(2)∂

∂t(F×G) =

∂F

∂t×G+ F× ∂G

∂t

(3)∂

∂t(fF) =

∂f

∂tF+ f

∂F

∂t

(4)∂

∂t(F •G×H) =

∂F

∂t•G×H+ F • ∂G

∂t×H+ F •G× ∂H

∂t

(5)∂

∂t

(

F× (G×H))

=∂F

∂t× (G×H) + F×

(∂G

∂t×H

)

+ F×(

G× ∂H

∂t

)

Another kind of time dependence in a coordinate representation is obtained by allowing a

moving coordinate system (e.g. a rotating one, as in a carousel). If the original representation in

a fixed coordinate system is f(r) (a scalar field) or F(r) (a vector field), then at time t we have

a coordinate transform

r∗(t) = rQ(t) + b(t)

and the representations of the fields are

f ∗(r∗, t) = f((

r∗(t)− b(t))

Q(t)T)

and

F∗(r∗, t) = Q(t)TF((

r∗(t)− b(t))

Q(t)T)

.

Note that now the fields are stationary, time-dependence is only in the coordinate representation

and is a consequence of the moving coordinate system.


Similarly for nonstationary fields f(r, t) and F(r, t) in a moving coordinate system we get

the representations

f ∗(r∗, t) = f((

r∗(t)− b(t))

Q(t)T, t)

and

F∗(r∗, t) = Q(t)TF((

r∗(t)− b(t))

Q(t)T, t)

.

Now part of the time-dependence comes from the time-dependence of the fields, part from the

moving coordinate system.

Chapter 2

MANIFOLD

”The intuitive picture of a smooth surface becomes analytic

with the concept of a manifold. On the small scale a manifold

looks like Euclidean space, so that infinitesimal operations

like differentiation may be defined on it.’

(WALTER THIRRING: A Course in Mathematical Physics)

2.1 Graphs of Functions

The graph of a function f : A → Rm, where A ⊆ R

k is an open set, is

(

r, f(r)) ∣

∣ r ∈ A

,

a subset of Rk+m. The graph is often denoted—using a slight abuse of notation—as follows:

s = f(r) (r ∈ A).

Here r contains the so-called active variables and s the so-called passive variables. Above

active variables precede the passive ones in the order of components. In a graph we also allow a

situation where the variables are scattered. A graph is smooth,1 if f is continuously differentiable

in its domain of definition A. In the sequel we only deal with smooth graphs. Note that a graph

is specifically defined using a coordinate system and definitely is not coordinate-free.A familiar graph is the graph of a real-valued

univariate function f in an open interval (a, b), i.e.,

the subset of R2 consisting of the pairs

(

x, f(x))

(a < x < b),

the curve y = f(x). Another is the graph of a real-

valued bivariate function f , i.e., the surface

z = f(x, y) ((x, y) ∈ A)

in the space R3 (see the figure on the right). Not all

curves or surfaces are graphs, however, e.g. circles

and spheres are not (and why not?).

x

y

z = f(x,y)

z

(x,y)

A

The most common dimensions k and m are of course the ones given by physical position

and time, that is 1, 2 and 3, whence k +m is 2, 3 or 4. On the other hand, degrees of freedom

in mechanical systems etc. may lead to some very high dimensions

As a limiting case we also allow m = 0. In Rm = R

0 there is then only one element (the

so-called empty vector ()). Moreover, then Rk+m = R

k, thus all variables are active and the

graph of the function f : A → Rm is A. Open subsets of the space are thus always graphs, and

agreed to be smooth, too.

1In some textbooks smoothness requires existence of continuous derivatives of all orders.

16

CHAPTER 2. MANIFOLD 17

Similarly we allow k = 0. Then f has no arguments, and so it is constant, and the graph

consists of one point.2 Again it is agreed that such graphs are also smooth.

In what follows we will need inverse images of sets. For a function g : A → B the inverseimage of the set C is the set

g−1(C) =

r∣

∣g(r) ∈ C

.

Note that this has little to do with inverse functions, indeed the function g need not have an

inverse at all, and C need not be included in B.

For a continuous function defined in an open set A the inverse image of an open set is always

open.3 This implies an important property of graphs:

Theorem 2.1. If a smooth graph of a function is intersected by an open set, then the result iseither empty or a smooth graph.

Proof. This is clear if the intersection is empty, and also if k = 0 (the intersection is a point) or

m = 0 (intersection of two open sets is open). Otherwise the intersection of the graph

s = f(r) (r ∈ A)

and the open set C in Rk+m is the graph

s = f(r) (r ∈ D),

where D is the inverse image of C for the continuous4 function

g(r) =(

r, f(r))

.

2.2 Manifolds

A subset M of Rn is a k-dimensional manifold, if it is locally a smooth graph of some function

of k variables.5

”Locally” means that for every point p of M there is an open set Bp of Rn containing the

point p such that M ∩ Bp is the graph of some function fp of some k variables. For different

points p the set Bp may be quite different, the active variables chosen in a different way, always

numbering k, however, and the function fp may be very different.

The functions fp are called charts, and the set of all charts is called an atlas. Often a small

atlas is preferable.

Example. A circle of radius R centered in the origin is a 1-dimensional manifold of R2, since(see the figure below) its each point is in an open arc delineated by either black dots or whitedots, and these are smooth graphs of the functions

y = ±√R2 − x2 and x = ±

√

R2 − y2

(atlas) in certain open intervals.

2Here we may take A to be the whole space R0, an open set.

3This in fact is a handy definition of continuity. Continuity of g in the point r0 means that taking C to be an

arbitrarily small g(r0)-centered open ball, there is in A a (small) r0-centered open ball which g maps to C.4It is not exactly difficult to see that if f is continuous in the point r0, then so is g, because

∥

∥g(r) − g(r0)∥

∥

2

= ‖r− r0‖2 +∥

∥f(r) − f(r0)∥

∥

2

.

5There are several definitions of manifolds of different types in the literature. Ours is also used e.g. in HUB-

BARD & HUBBARD and NIKOLSKY & VOLOSOV. They are also often more specifically called ”smooth mani-

folds” or ”differentiable manifolds”. With the same underlying idea so-called abstract manifolds can be defined.


In a similar fashion the sphere x2 + y2 + z2 = R2 is seento be a 2-dimensional manifold of R3. Locally it is a smoothgraph of one of the six functions

x = ±√

R2 − y2 − z2 , y = ±√R2 − x2 − z2

andz = ±

√

R2 − x2 − y2

(atlas) in properly chosen open sets.

x

y

Of course, each smooth graph itself is a manifold, in particular each open subset of Rn is its

n-dimensional manifold, and each single point is a 0-dimensional manifold. If a space curve is

a smooth graph, say of the form

(y, z) =(

f1(x), f2(x))

(a < x < b),

where f1 and f2 are continuously differentiable, then it will be a 1-dimensional manifold of R3.

Also the surface

z = f(x, y) ((x, y) ∈ A)

is a manifold of R3 if f is continuously differentiable. On the other hand, e.g. the graph of the

absolute value function y = |x| is not smooth and therefore not a manifold of R2.

A manifold can always be restricted to be more local. As an immediate consequence of

Theorem 2.1 we get

Theorem 2.2. If a k-dimensional manifold of Rn is intersected by an open set, then the resultis either empty or a k-dimensional manifold.

Note. Why do we need manifolds? The reason is that there is an unbelievable variety of looselydefined curves and surfaces, and there does not seem to be any easy general global method todeal with them. There are continuous curves which fill a square or a cube, or which intersectthemselves in all their points, continuous surfaces having normals in none of their points, etc.The only fairly easy way to grasp this phenomenon is to localize and restrict the conceptssufficiently far, at the same time preserving applicability as much as possible.

Finding and proving global results is then a highly demanding and challenging area ofalgebro-topological mathematics in which many Fields Medals have been awarded.

2.3 Manifolds as Loci

One way to define a manifold is to use loci. A locus is simply a set of points satisfying some

given conditions. For instance, the P -centered circle of radius R is the locus of points having

distance R from P . As was noted, it is a manifold.

In general a locus is determined via a coordinate representation, and the conditions are given

as mathematical equations. A condition then is of the form

F(r, s) = 0,

where r has dimension k, s has dimension n − k, and F is a n − k-dimensional function

of n variables. The locus of points satisfying the condition is then the set of points (r, s) in

Rn determined as solutions of the equation, solving for s. As indicated by the notation used,

r is purported to be active and s passive. Even though here active variables appear before the

passive ones in the component order, active variables may be scattered, roles of variables may

differ locally, actives changing to passives, etc.


Example. A circle and a sphere are loci of this kind when we set the conditions

F (x, y) = R2 − x2 − y2 = 0

andF (x, y, z) = R2 − x2 − y2 − z2 = 0

(centered in the origin and having radius R). In the circle one of the variables is always active,in the sphere two of the three variables.

Not all loci are manifolds. For instance, the locus of points of R2 determined by the condi-

tion

y − |x| = 0

is not, and neither is the locus of points satisfying the condition

y2 − x2 = 0.

The former is not smooth in the origin, and the latter is not a graph of any single function in the

origin (but rather of two functions: y = ±x). Actually the condition

(y − x)2 = 0

does not determine a proper manifold either. The locus is the line y = x, but counted twice!

So, in the equation F(r, s) = 0 surely F then should be continuously differentiable, and

somehow uniqueness of solution should be ensured, too, at least locally. In classical analysis

there is a result really taylor-made for this, the so-called Implicit Function Theorem, long known

and useful in many contexts. Here it is used to make the transition from local loci6 to graphs. It

should be mentioned that in the literature there are many versions of the theorem,7 we choose

just one of them.

Implicit Function Theorem. Assume that the function F : S → Rn−k, where 0 ≤ k < n,

satisfies the following conditions:

1. The domain of definition S is an open subset of Rn.

2. F is continuously differentiable.

3. F′ is of full rank in the point p0, i.e., the rows of F′(p0) are linearly independent (whencealso some n− k columns are also linearly independent).

4. F(p0) = 0 and the n − k columns of F′(p0) corresponding to the variables in s arelinearly independent.

Denote by r variables other than the ones in s. By changing the order of variables, if necessary,we may assume that p = (r, s), and especially p0 = (r0, s0). Denote further the derivative ofF with respect to the variables in r by F′

r, and with respect to the variables in s by F′

s. (Thus

F′ =(

F′r

F′s

)

in block form.)Then there is an open subset B of Rkcontaining the point r0, and a uniquely determined

function f : B → Rn−k such that

6This is not pleonasm, although it might seem to be since ’local’ could be constructed to mean ’relating to a

locus’ etc.7See e.g. KRANTZ, S.G. & PARKS, H.R.: The Implicit Function Theorem. History, Theory, and Applications.

Birkhäuser (2012).


(i) the graph of f is included in S,

(ii) p0 =(

r0, f(r0))

,

(iii) F(

r, f(r))

= 0 in B,

(iv) f is continuously differentiable, the matrix F′s

(

r, f(r))

is nonsingular in B, and

f ′(r) = −F′s

(

r, f(r))−1

F′r

(

r, f(r))

(this only for k > 0).

Proof. The proof is long and tedious, and is omitted here, see e.g. APOSTOL or HUBBARD &

HUBBARD or NIKOLSKY & VOLOSOV. The case k = 0 is obvious, however. Then r0 is the

empty vector and f is the constant function p0. The derivative of f in item (iv) is obtained by

implicit derivation, i.e., applying the chain rule to the left hand side of the identity

F(

r, f(r))

= 0

in B, and then solving for f ′(r) the obtained equation

F′r

(

r, f(r))

+ F′s

(

r, f(r))

f ′(r) = O.

Using the Implicit Function Theorem (and Theorem 2.1) we immediately get definition of

manifolds using local loci:

Corollary 2.3. If for any point p0 in the subset M of Rn there is a subset S of of Rn and afunction F : S → R

n−k such that the conditions 1.– 4. in the Implicit Function Theorem aresatisfied, and the locus condition F(p) = 0 defines the set M∩S, then M is a k-dimensionalmanifold of Rn.

The converse holds true, too, i.e., all manifolds are local loci:

Theorem 2.4. If M is a k-dimensional manifold of Rn and k < n, then for each point p of Mthere is a set S and a function F : S → R

n−k such that the conditions 1.— 4. of the ImplicitFunction Theorem are satisfied and the locus condition F(p) = 0 defines the set M∩S.

Proof. Let us just see the case k > 0. (The case where k = 0 is similar—really a special case.)

If M is a k-dimensional manifold of Rn, then locally in some open set containing the point p0

it is the graph of some continuously differentiable function f

s = f(r) (r ∈ A)

for some choice of the k variables of r (the active variables). Reordering, if necessary, we may

assume that the active variables precede the passive ones.

Choose now the set S to be the Cartesian product A× Rn−k, i.e.

S =

(r, s)∣

∣ r ∈ A and s ∈ Rn−k

,

and F to be the function

F(r, s) = s− f(r).


Then S is an open subset8 of Rn and F is continuously differentiable in S. Moreover then

F′ =(

−f ′ In−k

)

is of full rank (In−k is the (n− k)× (n− k) identity matrix).

Excluding the n-dimensional manifolds, manifolds of Rn are thus exactly all sets which are

local loci. In particular conditions of the form

G(p) = c or G(p)− c = 0,

where c is a constant, define a manifold (with the given assumptions), the so-called level mani-fold of G.

Representation of a manifold using loci is often called an implicit representation and the

representation using local graphs of functions—as in the original definition– is called an explicitrepresentation.9

Example. 2-dimensional manifolds in R3 are smooth surfaces. Locally such a surface is defined

as the set determined by a condition

F (x, y, z) = 0,

where in the points of interest

F ′ =(∂F

∂x,∂F

∂y,∂F

∂z

)

6= 0.

In particular surfaces determined by conditions of the form G(x, y, z) = c, where c is constant,are level surfaces of G.

For such surfaces it is often quite easy to check whether or not a point p0 = (x0, y0, z0) is inthe surface. Just calculate (locally) F (x0, y0, z0) and check whether or not it is = 0, of courseeven this could turn out to be difficult.

Example. 1-dimensional manifolds in R3 are smooth space curves. Locally a curve is the locus

of points satisfying a condition

F(x, y, z) = 0 or

F1(x, y, z) = 0

F2(x, y, z) = 0,

where in the curve the derivative matrix

F′ =

(

F ′1

F ′2

)

is of full rank, i.e., its two rows are linearly independent. Locally we have then the curve as theintersection of the two smooth surfaces

F1(x, y, z) = 0 and F2(x, y, z) = 0

8If B is the r0-centered ball of radius R in A and s0 ∈ Rn−k, then the (r0, s0)-centered open ball of radius R

in Rn is included in S, because in this ball

‖r− r0‖2 ≤ ‖r− r0‖2 + ‖s− s0‖2 =∥

∥(r, s)− (r0, s0)∥

∥

2

< R,

so that r ∈ B.9There is a third representation, so-called parametric representation, see Section 2.5.


(locus conditions, cf. the previous example). It may be noted that the curve of intersection oftwo smooth surfaces need not be a smooth manifold (curve), for this the full rank property isneeded.10

Example. The conditionF (x, y) = yey − x = 0

defines a 1-dimensional manifold (a smooth curve, actually a graph) of R2. On the other hand,

F ′(x, y) =(

− 1, (1 + y)ey)

,

so, except for the point (−1/e,−1), the corresponding local graph can be taken as(

f(y), y)

where f is one of the so-called Lambert W functions11 W0 (green upper branch) or W−1 (redlower branch), see the figure below (Maple).Since here

y = xe−y and − ln√2 > −1/e,

this means that

√2√2

√

2

√

2·.·

exists, which may seem odd because√2 > 1.

Actually it has the value

W0

(

− ln√2)

− ln√2

= 2.

2.4 Mapping Manifolds. Coordinate-Freeness

Manifolds are often ”manipulated” by mapping them by some functions in one way or another.

For manifolds defined as in the previous sections it then is not usually at all easy to show that

the resulting set really is a manifold. For parametrized manifolds this is frequently easier, see

the next section.

On the other hand, inverse images come often just as handy:

Theorem 2.5. If

• M is a k-dimensional manifold of Rn included in the open set B,

• A is an open subset of Rm, where m ≥ n, and

• g : A → B is a continuously differentiable function having a derivative matrix g′ of fullrank (i.e. linearly independent rows),

then the inverse image g−1(M) is an m− n + k-dimensional manifold of Rm.10For instance, the intersection of the surface F1(x, y, z) = z − xy = 0 (a saddle surface) and the surface

F2(x, y, z) = z = 0 (the xy-plane) is not a manifold (and why not?).11Very useful in many cases.


Proof. The case k = n is clear. Then M is an open subset of Rn and its inverse image g−1(M)is an open subset of Rm, i.e. an m-dimensional manifold of Rm.

Take then the case k < n. Consider an arbitrary point r0 of g−1(M), i.e. a point such

that g(r0) ∈ M. Locally near the point p0 = g(r0) the manifold M can be determined as

a locus, by Theorem 2.4. More specifically, there is an open subset S of Rn and a function

F : S → Rn−k, such that the conditions 1.– 4. of the Implicit Function Theorem are satisfied.

For a continuous function defined in an open set the inverse image of an open set is open,

so g−1(S) is open. The locus condition

F(

g(r))

= 0

determines locally some part of the set g−1(M). In the open set g−1(S) the composite function

F g now satisfies the conditions 1.– 4. of the Implicit Function Theorem since (via chain rule)

its derivative is

(F g)′(r0) = F′(

g(r0))

g′(r0) = F′(p0)g′(r0),

and it is of full rank. Thus, by Corollary 2.3, g−1(M) is a manifold of Rm and its dimension is

m− (n− k) (the dimension of r minus the dimension of F).

So far we have not considered coordinate-freeness of manifolds. A manifold is always

expressly defined in some coordinate system, and we can move frome one system to another

using coordinate transforms. But is a manifold in one coordinate system also a manifold in any

other, and does the dimension then remain the same?

As a consequence of Theorem 2.5 the answer is positive. To see this, take a coordinate

transform

r∗ = rQ+ b,

and choose m = n and

g(r∗) = (r∗ − b)QT

in the theorem. Then a manifold M in coordinates r∗ is the inverse image of the manifold in

coordinates r, and thus truly a manifold. Dimension is preserved as well. Being a manifold in

one coordinate system guarantees being a manifold in any other coordinates. ”Manifoldness” is

a coordinate-free property.

2.5 Parametrized Manifolds

Whenever a manifold can be parametrized it will be in many ways easier to handle.12 For

instance, defining and dealing with integrals over manifolds then becomes considerably simpler.

A parametrization13 of a k-dimensional manifold M of Rn consists of an open subset U of

Rk (the so-called parameter domain), and a continuously differentiable bijective function

γ : U → M

having a derivative matrix γ′ of full rank. Since the derivative γ ′ is an n×k matrix and k ≤ n, it

is the columns that are linearly independent. These columns are usually interpreted as vectors.

(It is naturally assumed here that k > 0.)

12In many textbooks manifolds are indeed defined using local parametrization, see e.g. SPIVAK or O’NEILL.

This usually requires the so-called transition rules to make sure that the chart functions are coherent. Our definition,

too, is a kind of local parametrization, but not a general one, and not requiring any transition rules.13This is often called a smooth parametrization.


Evidently, if a manifold is the graph of some function, i.e.

s = f(r) (r ∈ A),

it is parametrized, we just take

U = A and γ(r) =(

r, f(r))

.

Also an n-dimensional manifold of Rn, i.e. an open subset A, is parametrized in a natural way,

we take A itself as the parameter domain and the identity function as the function γ.

Example. A circle Y : x2 + y2 = R2 is a 1-dimensional manifold of R2 which cannot beparametrized. To show this, we assume the contrary, i.e., that Y actually can be parametrized,and derive a contradiction. The parameter domain U is then an open subset of the real line,that is, it consists of disjoint open intervals. Consider one of these intervals, say (a, b) (wherewe may have a = −∞ and/or b = ∞). Now, when a point u in the interval (a, b) moves tothe left towards a, the corresponding point γ(u) in the circle moves along the circumference ineiher direction. It cannot stop or turn back because γ is a bijection and γ

′ has full rank. Thusalso the limiting point

p = limu→a+

γ(u)

is in the circle Y .But we cannot have in U a point v such that p = γ(v). Such a point would be in one of the

open intervals in U , and—as above—we see that a sufficiently short open interval (v−ǫ, v+ǫ) ismapped to an open arc of Y containing the point p. This would mean that γ cannot be bijective,a contradiction.

On the other hand, if we remove one of the points in Y , say (R, 0), it still remains a manifold(why?) and can be parametrized using the familiar polar angle φ:

(x, y) = γ(φ) = (R cos φ,R sinφ) (0 < φ < 2π).

Nowγ(φ) = (R cos φ,R sinφ)

is a continuously differentiable bijection, and

γ′(φ) =

(

−R sinφ

R cosφ

)

is always 6= 0.Also the inverse parametrization is easily obtained:

φ = atan(x, y),

where atan is the bivariate arctangent, i.e., an arc tangent giving correctly the quadrant andalso values in the coordinate axes, that is

atan(x, y) =

arctany

xfor x > 0 and y ≥ 0

2π + arctany

xfor x > 0 and y < 0

π + arctany

xfor x < 0

π

2for x = 0 and y > 0

3π

2for x = 0 and y < 0.


It can be found, in one form or in another, injust about all mathematical programs. atanis actually continuous and also continuouslydifferentiable—excluding the negative x-axis,see the figure on the right (by Maple)—since(verify!)

∂atan(x, y)

∂x= − y

x2 + y2

and∂atan(x, y)

∂y=

x

x2 + y2.

0

2

4

6

8

10

–2

–1

1

2

y

–2

–1

1

2

xExample. A sphere x2+y2+ z2 = R2 cannotbe parametrized either. However, if, say, thehalf great circle

x2 + z2 = R2 , y = 0 , x ≥ 0

is removed then a manifold is obtained which can be parametrized by the familiar sphericalcoordinates as

(x, y, z) = γ(θ, φ) = (R sin θ cos φ,R sin θ sinφ,R cos θ)

and the parameter domain is the open rectangle

U : 0 < θ < π , 0 < φ < 2π.

The derivative matrix

γ′(θ, φ) =

R cos θ cosφ −R sin θ sin φR cos θ sin φ R sin θ cosφ−R sin θ 0

is of full rank, and the inverse parametrization is again easy to find:

θ = arccosz

R

φ = atan(x, y).

Example. Parametrizarion of a general smooth space curve is of the form

r = γ(u) (u ∈ U),

where U is an open interval. (The parameter domain might actually consist of several openintervals but then the curve can be divided similarly.) Here γ is continuously differentiable andγ′ 6= 0, which guarantees that the curve has a tangent everywhere.

Example. Parametrization of a general smooth surface is of the form

r = γ(u) (u ∈ U),

where U is an open subset of R2. Here γ is continuously differentiable and γ′ has full rank,

i.e., its two columns are linearly independent. This guarantees that the surface has everywherea normal (the cross product of the two columns of γ ′(u)).


Parametrization is at the same time more restrictive and more extensive than our earlier

definitions of manifolds: Not all manifolds can be parametrized and not all parametrizations

define manifolds. On the other hand, as noted, parametrization makes it easier to deal with

manifolds. In integration restrictions of parametrizations can be mostly neglected since they do

not affect the values of the integrals, as we will see later. Let us note, however, that if a set is

parametrized then at least it is a manifold in a certain localized fashion:

Theorem 2.6. If M ⊆ Rn, U is an open subset of Rk, u0 ∈ U , and there is a continuously

differentiable bijective function γ : U → M with a derivative γ′ of full rank, then there is an

open subset V of U such that u0 ∈ V and γ(V) is a k-dimensional manifold of Rn.

Proof. Consider a point

p0 = γ(u0)

of M. Then the columns of γ ′(u0) are linearly independent, and thus some k rows are linearly

independent, too. Reordering, if necessary, we may assume that these rows are the first k rows

of γ ′(u0).Let us first consider the case k < n. For a general point p = (r, s) of M, r contains the k

first components. We denote further by γ1 the function consisting of the first k components of

γ, and r0 = γ1(u0). Similarly, taking the last n − k components of γ we get the function γ2.

The function

F(u, r) = r− γ1(u)

defined in the open set S = U × Rk (cf. the proof of Theorem 2.4) then satisfies the conditions

1.– 4. of the Implicit Function Theorem. Thus there is a continuously differentiable function

g : B → Rk, defined in an open set B, whose graph u = g(r) is included in S, such that

r = γ1

(

g(r))

.

The chart function f in the point p0 is then obtained by taking

f(r) = γ2

(

g(r))

.

Finally we choose V = γ−1

1 (B), an open set. (And where, if anywhere, do we need bijectivity

of γ?)

The case k = n is similar. The function

F(u,p) = p− γ(u)

defined in the open set S = U ×Rn then satisfies conditions 1.– 4. of the Implicit Function The-

orem. Hence there is an open set B, containing the point p0, and a continuously differentiable

function g : B → Rn, whose graph u = g(p) is included in S, such that

p = γ(

g(p))

.

Thus B ⊆ M. Again we choose V = γ−1(B), an open set.

Parametrization of a manifold M by

γ : U → M

may be exchanged, a so-called reparametrization, as follows. Take a new parameter domain

V ⊆ Rk, and a continuously differentiable bijective function

η : V → U ,


such that the derivative η′ is nonsingular. The new parametrization is then by the composite

function γ η, that is, as

r = γ(

η(v))

(v ∈ V).

This really is a parametrization since, by the chain rule, γ η is continuously differentiable and

its derivative

γ′(

η(v))

η′(v)

is of full rank. n-dimensional manifolds of Rn, i.e. open subsets, often benefit from reparam-

etrization.

Example. 3-dimensional manifolds of R3, i.e. open subsets or ’solids’, are often given usingparametrizations other than the trivial one by the identity function.

Familiar parametrizations of this type are those using cylindrical or spherical coordinates.For instance, the slice of a ball below (an open set) can be parametrized by spherical coordi-nates as

r = (x, y, z) = γ(ρ, θ, φ) = (ρ sin θ cosφ, ρ sin θ sin φ, ρ cos θ),

where the parameter domain is the open rectangular prism

V : 0 < ρ < R , 0 < θ < π , 0 < φ < α.

z

y

x

φ = 0

ρ = R

φ = α parameter domain

α

Rρ

θ

φ

π

U

Different parametrizations of a manifold may come separately, without any explicit repara-

metrizations. Even then in principle, reparametrizations are there.

Theorem 2.7. Different parametrizations of a manifold can always be obtained from each otherby reparametrizations.

Proof. Consider a situation where the k-dimensional manifold M of Rn has the parametriza-

tions

r = γ1(u) (u ∈ U) and r = γ2(v) (v ∈ V).

An obvious candidate for the reparametrization is the one using η = γ−1

1 γ2, as

u = γ−1

1

(

γ2(v))

.

This function η is bijective, we just must show that it is continuously differentiable. For this let

us first define

G(u,v) = γ1(u)− γ2(v).


Then the columns of the derivative G′ corresponding to the variable u, i.e. γ ′1, are linearly

independent.

Consider then a point

r0 = γ1(u0) = γ2(v0)

of M. Since the k columns of γ ′1(u0) are linearly independent, then some k rows of γ ′

1(u0) are

also linearly independent. Picking from G the corresponding k components we get the function

F. In the open set S = U × V the function F satisfies the conditions 1.– 4. of the Implicit

Function Theorem, and then the obtained function f clearly is η.

Since the point v0 was an arbitrary point of V , η is continuously differentiable. On the other

hand, η′ is also nonsingular because γ2 = γ1 η, and by the chain rule

γ′2(v) = γ

′1

(

η(v))

η′(v).

If now η′ would be singular in some point of V , then γ

′2 would not have full rank there.

A parametrized manifold may be localized also in the parameter domain: Take an open

subset U ′ of U and interprete it as a new parameter domain. The thus parametrized set is again

a manifold, and it has a parameter representation (cf. Theorem 2.6).

This in fact also leads to a generalization of manifolds. Just parametrize a set N as above

using a parameter domain U and a function γ : U → N which is continuously differentiable

and whose derivative γ′ is of full rank, but do not require that γ is bijective. If now for each

point p of N there is an open subset Up of U such that

• p = γ(u) for some point u of Up, and

• γ is bijective when restricted into Up,

then as in Theorem 2.6, the parametrization defines a manifold when restricted into Up. The set

N itself then need not be a manifold. Generalized manifolds of this kind are called trajectorymanifolds. A trajectory manifold can reparametreized exactly as the usual manifold.

Example. The subset of R2 parametrized by the polar angle φ given by

(x, y) = γ(φ) =(

r(φ) cosφ, r(φ) sinφ)

(0 < φ < 2π),

where

r(φ) = ecos φ − 2 cos 4φ+ sin5φ

12,

is a complicated plane curve, but not a manifold since it passes through the origin six times, seethe left figure below (Maple). It is however a 1-dimensional trajectory manifold. The figure onthe right is the hodograph

(x, y) = γ′(φ)T (0 < φ < 2π).

It shows that γ ′ is of full rank (the curve does not pass through the origin). It also indicatesthe parameter value φ = 0 (or φ = 2π) could have been included locally. This would notdestroy smoothness of the curve. This is common in polar parametrizations. It should alsobe remembered that even though atan is discontinuous, sin

(

atan(x, y))

and cos(

atan(x, y))

are continuously differentiable. So the parameter interval could have been, say, 0 < φ < 4πcontaing the parameter value φ = 2π. Note also that the polar parametrization allows evennegative values of the radius!


–3

–2

–1

0

1

2

3

–1 1 2 3

–8

–6

–4

–2

0

2

4

6

–6 –4 –2 2 4 6

Many more self-intersections appear when the curve is drawn for the ”full” parameterinterval 0 < φ < 24π, outside of which it starts to repeat itself:


2.6 Tangent Spaces

Locally, near a point p0 of Rn, a k-dimensional manifold M is a graph of some k-variate

function f , i.e.

s = f(r) (r ∈ A)

in Rn, and in particular

p0 =(

r0, f(r0))

.

Geometrically the tangent space of M in the point p0 consists of all tangent vectors touching

M in p0. The dimensions k = n and k = 0 are dealt with separately. In the former the tangent

space consists of all vectors, in the latter of only the zero vector. In the sequel we assume that

0 < k < n.

Locally, near the point r0, f comes close to its affine approximation, i.e.

f(r) ∼= f(r0) + (r− r0)f′(r0)

T.

The affine approximation of a function in a point gives correctly the values of the function and

its derivative in this point. Let us denote

g(r) = f(r0) + (r− r0)f′(r0)

T

(whence g′(r0) = f ′(r0)). Then s = g(r) is a graph which locally touches the manifold M in

the point p0. Geometrically this graph is part of a k-dimensional hyperplane, or a plane or a

line in lower dimensions.

The tangent space of M in the point p0, denoted by Tp0(M), consists of all (tangent)

vectors such that the initial point of their representative directed line segments is p0 and the

terminal point is in the graph s = g(r), i.e., it consists of exactly all vectors

((

r, g(r))

−(

r0, f(r0)))T

=(

r− r0, (r− r0)f′(r0)

T)T

=

(

(r− r0)T

f ′(r0)(r− r0)T

)

=

(

Ik

f ′(r0)

)

(r− r0)T,

where r ∈ Rk and Ik is the k × k identity matrix. In particular r = r0 is included, so the zero

vector always is in a tangent space.

In a sense the above tangent space is thus the graph of the vector-valued function

T(h) = f ′(r0)h

of the vector variable h. Apparently T is a linear function and f ′(r0) is the corresponding

matrix.

Note that replacing the graph s = f(r) by another graph t = h(u) (as needs to be done when

moving from one chart to another) simply corresponds to a local reparametrization u = η(r)and change of basis of the tangent space using the matrix η

′(r0) (cf. Theorem 2.7 and its proof).

The space itself remains the same, of course.

Example. A smooth space curve or a 1-dimensional manifold MÊof R3 is locally a graph

(y, z) = f(x) =(

f1(x), f2(x))

(or one of the other two alternatives). The tangent space of M in the point p0 = (x0, y0, z0),where (y0, z0) = f(x0), consists of exactly all vectors


hf ′1(x0)h

f ′2(x0)h

(h ∈ R).

Geometrically the vectors are directedalong the line r = p0 + tv (t ∈ R),

where

v =(

1, f ′1(x0), f

′2(x0)

)

.x

y

p0 = (x0 , f1(x0) , f2(x0))

space curve + tangent vector

z

Example. A smooth surface in R3 is a 2-dimensional manifold M. Locally M is the graph

z = f(x, y) (or then one of the other two alternatives). The tangent space of M in the pointp0 = (x0, y0, z0), where z0 = f(x0, y0), consists of exactly all vectors

1 00 1

∂f(x0, y0)

∂x

∂f(x0, y0)

∂y

(

h1

h2

)

((h1, h2) ∈ R2).

Geometrically the vectors are thus in the plane

r = p0 + t1v1 + t2v2 (t1, t2 ∈ R),

where

v1 =(

1, 0,∂f(x0, y0)

∂x

)

and

v2 =(

0, 1,∂f(x0, y0)

∂y

)

.

x

y

p0 = (x0 , y0 , f(x0,y0))

z v1tangent plane

v2

What about when a manifold M is given by local loci, say locally by the condition

F(r, s) = 0

(assuming a proper order of variables)? According to Corollary 2.3, then M is given locally

near the point p0 = (r0, s0) also as a graph s = f(r) and (cf. the Implicit Function Theorem)

f ′(r0) = −F′s

(

r0, f(r0))−1

F′r

(

r0, f(r0))

where

F′ =(

F′r

F′s

)

.

The tangent space Tp0(M) consists of the vectors

(

Ik

f ′(r0)

)

h.

But these are exactly all vectors

m =

(

h

k

)


satisfying the condition

F′s

(

r0, f(r0))

k+ F′r

(

r0, f(r0))

h = 0,

i.e.

F′(p0)m = 0.

So we get

Theorem 2.8. If a manifold M is locally near the point p0 given as the locus defined by thecondition F(p) = 0 (with the assumptions of Corollary 2.3), then the tangent space Tp0

(M) isthe null space of the matrix F′(p0).

In practice it of course suffices to find a basis for the tangent space (or null space). Coordi-

nate-freeness of tangent spaces has not been verified yet, but as a consequence of the theorem,

Corollary 2.9. Tangent spaces of manifolds are coordinate-free.

Proof. This follows because a null space is coordinate-free, and a manifold can be given as a

local locus (Theorem 2.4). More specifically, if we take a coordinate transform p∗ = pQ + b

and denote

F∗(p∗) = F(

(p∗ − b)QT)

and m∗ = QTm,

then (cf. Section 1.5)

F∗′(p∗0)m

∗ = F′(

(p∗0 − b

)

QT)QQTm = F′(p0)m.

Moreover, 0-dimensional and n-dimensional manifolds of Rn (points and open sets) of

course are coordinate-free.

Example. The tangent space of a circle

F (x, y) = x2 + y2 −R2 = 0

in the point (x0, y0) is then the null space of the 1× 2 matrix

F ′(x0, y0) = (2x0, 2y0).

It consists of vectors

(

hk

)

satisfying

2x0h+ 2y0k = 0

(cf. the equation of a line).Similarly the tangent space of a sphere

F (x, y, z) = x2 + y2 + z2 − R2 = 0

in the point (x0, y0, z0) is the null space of the 1× 3 matrix

F ′(x0, y0, z0) = (2x0, 2y0, 2z0).

It consists of vectors

hkl

satisfying

2x0h+ 2y0k + 2z0l = 0

(cf. the equation of a plane).Tangent spaces of second degree curves and surfaces are sometimes called polar spaces.


Example. In general, the tangent space of a smooth surface (manifold) M in R3 defined im-

plicitly by the equationF (x, y, z) = 0

in the point p0 = (x0, y0, z0) is the null space of the 1× 3 matrix F ′(p0), i.e., the set of vectorsm = (h, k, l)T satisfying

F ′(p0) •m =∂F (p0)

∂xh+

∂F (p0)

∂yk +

∂F (p0)

∂zl = 0.

As a further consequence of Theorem 2.8 we see that if a manifold is parametrized, then its

tangent space can be parametrized, too.

Corollary 2.10. If the k-dimensional manifold M of Rn has the parametrization γ : U → Mand p0 = γ(u0), then the tangent space Tp0

(M) consists of exactly all vectors of the form

γ′(u0)v (v ∈ R

k),

that is, Tp0(M) is the column space of γ ′(u0).

Proof. Locally near the point p0 the manifold M can be given as a locus determined by some

suitable condition F(p) = 0. Thus the equation

F(

γ(u))

= 0

is an identity valid in a neighborhood of the parameter value u0. Applying the chain rule we get

another identity

F′(γ(

u))

γ′(u) = O,

where O is a zero matrix of appropriate dimensions. Substituting u = u0 we get the equation

F′(p0)γ′(u0) = O,

showing that columns of γ ′(u0) are in the null space of F′(p0).On the other hand, since the dimension of the null space is k and the k columns of γ ′(u0)

are linearly independent, the columns of γ ′(u0) span the tangent space Tp0(M).

Example. If the parametrization of a smooth space curve C (a 1-dimensional manifold of R3)is

r = γ(u) (u ∈ U),

then its tangent space Tr0(C) in the point r0 = γ(u0) consists of exactly all vectors

hγ ′(u0) (h ∈ R).

Example. If the parametrization of a smooth surface of R3 (a 2-dimensional manifold) is

r = γ(u) (u ∈ U),

then its tangent space in the point r0 = γ(u0) consists of exactly all vectors

γ′(u0)h =

∂γ1(u0)

∂u1

∂γ1(u0)

∂u2

∂γ2(u0)

∂u1

∂γ2(u0)

∂u2

∂γ3(u0)

∂u1

∂γ3(u0)

∂u2

(

h1

h2

)

=

(

∂γ(u0)

∂u1

∂γ(u0)

∂u2

)

h (h ∈ R2).


2.7 Normal Spaces

Geometrically the normal space of a k-dimensional manifold M of Rn in the point p0, denoted

Np0(M), consists of exactly all (tangent) vectors perpendicular to to all vectors of the tangent

space. In other words, the normal space is the orthogonal complement of the tangent space.

Again the cases k = n and k = 0 are special and are omitted in the sequel. In the former the

normal space consists of the the zero vector only, and in the latter of all vectors. Vectors in a

normal space are called normals or normal vectors.Basic properties of a normal space Np0

(M) follow fairly directly from those of the tangent

space. We just list them here. From basic courses of mathematics we remember that the null

space of a matrix is the orthogonal complement of the column space of its transpose, and that

the column space is the orthogonal complement of the null space of its transpose.

• If the k-dimensional manifold M of Rn is near the point p0 locally a graph s = f(r),then its normal space Np0

(M) consists of exactly all vectors

( −f ′(r0)T

In−k

)

k (k ∈ Rn−k).

• If the k-dimensional manifold M of Rn is near the point p0 locally given as a locus

determined by the condition F(p) = 0 (with the assumptions of Corollary 2.3), then

its normal space Np0(M) is the column space of the matrix F′(p0)

T, i.e., it consists of

exactly all vectors

F′(p0)Tk (k ∈ R

n−k).

• If the k-dimensional manifold M of Rn is parametrized by γ : U → M and p0 = γ(u0),then the normal space Np0

(M) is the null space of γ ′(u0)T, i.e., it consists of exactly all

vectors n satisfying

γ′(u0)

Tn = 0.

• A normal space is coordinate-free.

• The dimension of the a normal space of a k-dimensional manifold of Rn is always n− k.

Example. A smooth space curve, i.e. a 1-dimensional manifold M of R3, is locallya graph

(y, z) = f(x) =(

f1(x), f2(x))

(or one of the other two alternatives). Thenormal space of M in the point p0 =(x0, y0, z0), where (y0, z0) = f(x0), thenconsists of exactly all vectors

x

y

p0 = (x0 , f1(x0) , f2(x0))

space curve + normal plane

z

−f ′1(x0) −f ′

2(x0)

1 00 1

(

k1k2

)

(k1, k2 ∈ R).

Geometrically these vectors are in the plane

(x− x0) + f ′1(x0)(y − y0) + f ′

2(x0)(z − z0) = 0.


Some normals are more interesting than others, dealing with curvature, torsion, and the planemost accurately containing the curve near p0.

Example. A smooth surface in R3 is a 2-dimensional manifold M. Locally M is a graph

z = f(x, y) (or one of the other two alternatives). The normal space of M in the pointp0 = (x0, y0, z0), where then z0 = f(x0, y0), consists of exactly all vectors

−∂f(x0, y0)

∂xk

−∂f(x0, y0)

∂yk

k

(k ∈ R).

Geometrically these vectors are in the line r = p0 + tv (t ∈ R), where

v =(

− ∂f(x0, y0)

∂x,−∂f(x0, y0)

∂y, 1)

.

Example. If the parametrization of a smooth surface (a 2-dimensional manifold of R3) is

r = γ(u) (u ∈ U),

then its normal space in the point r0 = γ(u0) consists of exactly all vectors

n =

hkl

satisfying

γ′(u0)

Tn =

∂γ1(u0)

∂u1

∂γ2(u0)

∂u1

∂γ3(u0)

∂u1

∂γ1(u0)

∂u2

∂γ2(u0)

∂u2

∂γ3(u0)

∂u2

hkl

=

(

00

)

.

The basis vector of the null space is now obtained in the usual way using cross product, thusthe vectors are

t(∂γ(u0)

∂u1

× ∂γ(u0)

∂u2

)

(t ∈ R).

(The cross product is not the zero vector because the columns of γ ′(u0) are linearly indepen-dent.)

For instance, the normal space of a sphere parametrized by spherical coordinates as

(x, y, z) = γ(θ, φ) = (R sin θ cosφ,R sin θ sin φ,R cos θ)

(0 < θ < π , 0 < φ < 2π)

in the point γ(θ0, φ0) consists of the vectors

t

R cos θ0 cos φ0

R cos θ0 sinφ0

−R sin θ0

×

−R sin θ0 sinφ0

R sin θ0 cos φ0

0

= t

R2 sin2 θ0 cosφ0

R2 sin2 θ0 sinφ0

R2 sin θ0 cos θ0

(t ∈ R),

i.e. vectorstγ(θ0, φ0)

T (t ∈ R).


2.8 Manifolds and Vector Fields

To deal with the spaces we choose bases for the tangent space Tp0(M) and the normal space

Np0(M) of a k-dimensional manifold M of Rn in the point p0:

t1, . . . , tk and n1, . . . ,nn−k,

respectively. (These bases need not be normalized nor orthogonal.) Note that it was always

easy to get a basis for one of the spaces above.

Since the two spaces are orthogonal complements of each other, combining the bases gives

a basis of Rn. A vector field F may be projected to these spaces in the point p0:

F(p0) = Ftan(p0) + Fnorm(p0).

Here Ftan(p0) is the flux of the field in the manifold M and Fnorm(p0) is the flux of the field

through the manifold M in the point p0. (Cf. flow of a liquid through a surface.) It naturally

suffices to have one of these, the other is obtained by subtraction.

From the bases we get the matrices

T = (t1, . . . , tk) and N = (n1, . . . ,nn−k)

and further the nonsingular matrices (so-called Gramians)

G = TTT = (Gij) , where Gij = ti • tj , and

H = NTN = (Hij) , where Hij = ni • nj .

Let us further denote

a = TTF(p0) = TTFtan(p0) =

a1...

ak

,

where

ai = F(p0) • ti = Ftan(p0) • ti,and

b = NTF(p0) = NTFnorm(p0) =

b1...

bn−k

,

where

bi = F(p0) • ni = Fnorm(p0) • ni.

Since dot product is coordinate-free, elements of these matrices and vectors are so, too.

The components being in their corresponding spaces we can write

Ftan(p0) = Tc and Fnorm(p0) = Nd

for vectors c and d. Solving these from the equations

a = TTTc and b = NTNd

we see that the components of the field are given by the formulas14

Ftan(p0) = T(TTT)−1TTF = TG−1a and

Fnorm(p0) = N(NTN)−1NTF = NH−1b.

14The least squares formulas, probably familiar from many basic courses of mathematics. Note that TTT is

nonsingular because otherwise there would be a nonzero vector c such that cTTTTc = 0, i.e. Tc = 0.


Example. The flux of a vector field in asmooth surface (a 2-dimensional manifold ofR

3) and through it is obtained by project-ing the field to a (nonzero) normal vector n.Here

N = n , H = n • n = ‖n‖2 ,

b = F(p0) • n

and (as is familiar from basic courses)

Fnorm(p0) =(

F(p0) •n

‖n‖) n

‖n‖ = n1

‖n‖2(

F(p0) • n)

.

x

y

p0

z

tangent plane

Fnorm

F

Ftan

normal vector

Note. A k-dimensional manifold in Rn together with its k-dimensional tangent spaces may

be thought of, at least locally, as a 2k-dimensional manifold (in Rn+k), the so-called tan-

gent bundle, and the tangent spaces are its fibres. Similarly the manifold together with itsn− k-dimensional normal spaces can be thought of as an n-dimensional manifold (in R

2n−k),the normal bundle.

The flux of a vector field in the manifold may then be seen locally as a cross-section of thetangent bundle parametrized by the manifold, and similarly the flux through the manifold asa cross-section of the normal bundle. This kind of geometric view of vector fields (and moregenerally tensor fields) is popular in modern physics, but is pursued no further here.

A more general concept is the fiber bundle. See e.g. ABRAHAM & MARSDEN & RATIU.

Chapter 3

VOLUME

”Calculating surface area is a foolhardy enterprise;fortunately one seldom needs to know the area of a

surface. Moreover, there is a simple expression for dAwhich suffices for theoretical considerations.”

(MICHAEL SPIVAK: Calculus on Manifolds)

3.1 Volumes of Sets

Geometrical things such as the area of a square or the volume of a cube can be defined usinglengths of their sides and edges. If the objects are open sets, these are already examples ofvolumes of manifolds, as is length of an open interval. A square may be situated in three-dimensional space, squares having the same length of sides are congruent and they have thesame area. In the n-dimensional space the n-dimensional volume of an n-dimensional cube (orn-cube) is similar: If the edge length is h, then the volume is hn. Such a cube may be situatedin an even higher-dimensional space and still have the same (n-dimensional) volume. Suchvolumes may be thought of as geometric primitives.

A way to grasp the volume of a bounded set A ⊂ Rn is the so-called Jordan measure. For

this we need first two related concepts. An outer cover P ofA consists of finitely many similarn-cubes in an n-dimensional grid, the union of which contains A. Adjacent n-cubes share aface but no more or no less. The family of all outer covers is denoted by Pout. An inner cover

P consists of finitely many similar n-cubes in a grid, the union of which is contained in A. Thefamily of all inner covers is denoted by Pin. (This Pin may well be empty.) Note that lengths ofedges or orientations of the grid are in no way fixed, and neither is any coordinate system.

The volume of a cover P , denoted by |P |, is the sum of the volumes its n-cubes. (And thisis a geometric primitive.) The volume of the empty cover is = 0. Quite obviously, the volumeof any inner cover is at most the volume of every outer cover, outer covers covering all innercovers.

outer cover inner cover

The Jordan outer and inner measures of the set A are

|A|out = infP∈Pout

|P | and |A|in = supP∈Pin

|P |,

respectively. A set A is Jordan measurable if |A|out = |A|in, and the common value |A| is its

38

CHAPTER 3. VOLUME 39

Jordan’s measure1. This measure is now defined to be the volume of A. Such a volume clearlyis coordinate-free.

The precise same construct can be used in a k-dimensional space embedded in an n-dimen-sional space (where n > k). The k-dimensional space is then an affine subspace of Rn, i.e., amanifoldR parametrized by

γ(u) = b+

k∑

i=1

uivi (u ∈ Rk),

where v1, . . . ,vk are linearly independent. Taking b as the origin and, if needed, orthogonaliz-ing v1, . . . ,vk we may embed R

k in an obvious way in Rn, and thus define the k-dimensional

volume of a bounded subset ofR. (The n-dimensional volume of these subsets in Rn is = 0, as

is easily seen.) Such affine subspaces are e.g. planes and lines of R3.

Note. Not all bounded subsets of Rn have a volume! There are bounded sets not having a

Jordan measure.

The inner and outer covers used in defining Jordan’s inner and outer measures remind usof the n-dimensional Riemann integral, familiar from basic courses of mathematics, with itspartitions and lower and uppersums. It is indeed fairly easy tosee that whenever the volume ex-ists, it can be obtained by inte-grating the constant 1, with im-proper integrals, if needed:

|A| =∫

A

1 dr.

An important special case is thevolume of a parallelepiped of Rn

(of R3 on the right).

a1

a2

a3

Theorem 3.1. The volume of the parallelepiped P in Rn with edges given by the vectors

a1, a2, . . . , an (interpreted as line segments) is

|P| =∣

∣ det(A)∣

∣,

where A = (a1, a2, . . . , an) is the matrix with columns a1, a2, . . . , an.

Proof. The case is clear if a1, a2, . . . , an are linearly dependent, the volume is then = 0. Letus then consider the more interesting case where a1, a2, . . . , an are linearly independent. Thevolume is given by the integral above, with a change of variables by r = uAT + b. A well-known formula then gives

|P| =∫

P

1 dr =

∫

C

∣

∣ det(A)∣

∣du =∣

∣ det(A)∣

∣|C|,

where C is the unit cube in Rn given by 0 ≤ ui ≤ 1 (i = 1, . . . , n). The volume of this cube is

= 1.

1Also called Jordan–Peano measure or Jordan’s content.


We already know volumes are coordinate-free. For the parallelepiped this is also clearbecause the volume can be written as

|P| =√

det(G),

where G is the Gramian (cf. Section 2.8)

G = ATA = (Gij) and Gij = ai • aj .

Being dot products, elements of a Gramian are coordinate-free. The same formula is validwhen we are dealing with the k-dimensional volume of a k-dimensional parallelepiped P aspart of Rk embedded as an affine subspace in R

n, and P is given by the n-dimensional vectorsa1, a2, . . . , ak. The Gramian is here, too, a k × k matrix. Note that we need no coordinatetransforms or orthogonalizations, the volume is simply given by the n-dimensional vectors.An example would be a parallelogram embedded in R

3, with its sides given by 3-dimensionalvectors.

A bounded subset A of Rn is called a null set if its volume (or Jordan’s measure) is = 0.For a null set A then

|A|out = infP∈Pout

|P | = 0

(the empty inner cover is always available). An unbounded subset A is a null set if its all(bounded) subsets

r∣

∣ r ∈ A and ‖r‖ ≤ N

(N = 1, 2, . . . ) are null sets.Using the above definition it is often quite difficult to show a set is a null set. A helpful

general result is

Theorem 3.2. IfM is a k-dimensional manifold of Rn and k < n, then bounded subsets ofMare null sets of Rn.

Proof. The proof is based on a tedious and somewhat complicated estimation, see e.g. HUB-BARD & HUBBARD. Just about the only easy thing here is that if the volume of a boundedsubset A exists, then it is = 0. Otherwise |A|in > 0 and some inner cover of A would have atleast one n-dimensional cube contained inM. But near the center of the cubeM is locally agraph of a function, which is not possible.

Example. Bounded subsets of smooth surfaces and curves of R3 (1- and 2-dimensional mani-

folds) are null sets.

Null sets—as well as the k-null-sets to be defined below—are very important for integrationsince they can be included and excluded freely in the region of integration without changingvalues of integrals.

Rather than the ”full” n-dimensional one, it is possible to define a lower-dimensional vol-ume for subsets of Rn, but this is fairly complicated in the general case.2 On the other hand, itis easy to define the k-dimensional volume of an n-cube: It is hk if the edge length of the cubeis h. This gives us the k-dimensional volume of an outer cover P of a subsetA of Rn, and then,via the infimum, the k-dimensional Jordan outer measure. The corresponding inner measure isof course always zero if A does not contain an open set, i.e. its interiorA is empty.

Thus, at least it is easy to define the k-dimensional zero volume for a subset A of Rn, thatis, the k-null-set of Rn: It is a set having a zero k-dimensional Jordan outer measure. Againthis definition is not that easy to use. Theorem 3.2 can be generalized, however, the proof thenbecoming even more complicated (see HUBBARD & HUBBARD):

2And has to do e.g. with fractals.


Theorem 3.3. IfM is a k-dimensional manifold of Rn and k < m ≤ n, then bounded subsets

ofM are m-null-sets of Rn.

For instance, bounded subsets of smooth curves of R3 (1-dimensional manifolds) are 2-null-setsof R3.

3.2 Volumes of Parametrized Manifolds

In general, the k-dimensional volume of a k-dimensional manifoldM of Rn is a difficult con-cept. For a parametrized manifold it is considerably easier. If the parametrization ofM is

r = γ(u) (u ∈ U),

then

|M|k =∫

U

√

det(

γ ′(u)Tγ ′(u))

du,

or briefly denoted

|M|k =∫

M

dr.

Such a volume may be infinite. Note that inside the square root there is a Gramian determinantwhich is coordinate-free. The whole integral then is coordinate-free.

Comparing with the k-dimensional volume of a parallelepiped

|P|k =√

det(ATA)

in the previous section we notice that in a sense we obtain |M|k by ”summing” over the pointsr = γ(u) of the manifold k-dimensional volumes of parallelepipeds whose edges are given bythe vectors

dui∂γ(u)

∂ui(i = 1, . . . , k).

Moreover, (as a directed line segment) the vector

dui∂γ(u)

∂ui

approximatively defines the movement of a point in the manifold when the parameter ui changesa bit the other parameters remaining unchanged.

Note. In a certain sense definition of the volume of a parametrized manifold then is natural,

but we must remember that it is just that, a definition. Though various problems concern-

ing volumes are largely solved for parametrized manifolds—definition, computing, etc.—others

remain. There are parametrizations giving a volume for a manifold even when it does not oth-

erwise exist (e.g. as Jordan measure). On the other hand, a manifold having a volume may

sometimes be given a parametrization such that the above integral does not exist.

Things thus depend on the parametrization. In the sequel we tacitly assume such ”patho-

logical” cases are avoided.


It is important to check that the definition above gives the open parallelepiped P the samevolume as before. The parallelepiped is a manifold and can be parametrized naturally as

r = γ(u) = b+

k∑

i=1

uiaT

i (U : 0 < u1, . . . , uk < 1),

whence

γ ′(u) = A and |P|k =∫

U

√

det(ATA) du =√

det(ATA).

Another important fact to verify is freeness of parametrization.

Theorem 3.4. The volume of a parametrized manifold does not depend on the parametrization.3

Proof. By Theorem 2.7 different parametrizations can be obtained from each other by repara-metrization. When reparametrizing a k-dimensional manifold M originally parametrized bysome γ : U → M a new parameter domain V ⊆ R

k is taken and a continuously differentiablebijective function η : V → U having a nonsingular derivative matrix η′. The new parametriza-tion then is given by the composite function δ = γ η as

r = γ(

η(v))

= δ(v) (v ∈ V).

Via the chain ruleδ′(v) = γ ′

(

η(v))

η′(v).

In the integral giving |M|k this corresponds to the change of variables u = η(v) and thecorresponding transform of the region of integration from U to V . Thus we only need to check4

the form of the new integrand:

|M|k =∫

U

√

det(

γ ′(u)Tγ ′(u))

du

=

∫

V

√

det(

γ ′(

η(v))T

γ ′(

η(v))) ∣

∣ det(

η′(v))∣

∣dv

=

∫

V

√

det(

γ ′(

η(v))T

γ ′(

η(v)))(

det(

η′(v)))2

dv

=

∫

V

√

det(

η′(v)T)

det(

γ ′(

η(v))T

γ ′(

η(v)))

det(

η′(v))

dv

=

∫

V

√

det(

δ′(v)Tδ′(v))

dv.

Example. The 1-dimensional volume of a smooth parametrized space curve

C : r = γ(u) (u ∈ U)

is its length since

|C|1 =∫

U

√

det(

γ ′(u)Tγ ′(u))

du =

∫

U

√

γ ′(u) • γ ′(u) du =

∫

U

∥

∥γ ′(u)∥

∥du.

3Remember that we tacitly assume all parametrizations do give a volume.4Recall that for square matrices det(AB) = det(A) det(B) and det(AT) = det(A).


Example. Similarly the 2-dimensional volume of a parametrized surface

S : r = γ(u) (u ∈ U)

is

|S|2 =∫

U

√

det(

γ ′(u)Tγ ′(u))

du.

This is the same as the familiar area

∫

U

∥

∥

∥

∂γ(u)

∂u1× ∂γ(u)

∂u2

∥

∥

∥du

since the area (2-dimensional volume) of the parallelogram in the integrand (a 2-dimensional

parallelepiped) can be given as the length of the cross product.

Note. The k-dimensional volume of a k-dimensional trajectory manifold of Rn is defined analo-

gously, and it, too, does not depend on the parametrization. Of course, the difficulties mentioned

in the previous note remain true for trajectory manifolds, too.

3.3 Relaxed Parametrizations

Since the k-dimensional volume of k-null-sets of Rn is = 0, adding such sets in a k-dimensionalmanifold (via union) does not change the k-dimensional volume of the manifold.

When defining a parametrized manifold as in Section 2.5 often parts of the manifold mustbe removed first. For instance, parametrizing a sphere using spherical coordinates does notwork as such, but only after parts of the surface are removed (e.g. a half great circle). Properlyextending parametrization these excluded parts may be included again, at least as far as volumesand other integrations are concerned.

A relaxed parametrization5 of a k-dimensional manifoldM of Rn is obtained by spesifyingan extended parameter domain U ⊆ R

k, an exception set X ⊂ U , and a continuous functionγ : U → R

n, such that

1. M⊆ γ(U) and γ(U − X ) ⊆M (frequentlyM = γ(U)).

2. The boundary ∂U of U is a null set of Rk.6 Often ∂U is at least partly included in theexception set X .

3. X is a null set of Rk.

4. γ(X ) is a k-null-set of Rn.

5. The setM′ = γ(U − X ) is a k-dimensional manifold of Rn parametrized by

r = γ(u) (u ∈ U − X )

(in the sense of Section 2.5).

5This is in no way a standard concept in the literature, where several kinds of relaxed parametrizations ap-pear. Here the concept is so general that according to HUBBARD & HUBBARD every manifold has an relaxedparametrization.

6This strange-looking condition is needed to exclude certain ”pathological” situations. Open subsets of Rk mayactually have a boundary that is not a null set.


Note that item 5. implies γ is continuously differentiable and bijective in U − X . In U it is justcontinuous and not necessarily bijective. Furthermore, U − X is an open set but U might notbe. Thus, should U contain points of its boundary ∂U , they must be in X , too.

Example. Parametrization of a sphere by spherical coordinates

in a form where the parametrization is extended to the whole

sphere is an example of a relaxed parametrization:

r = γ(θ, φ) = (R sin θ cosφ,R sin θ sin φ,R cos θ)

(U : 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2π).

The exception set X is here the boundary ∂U of U . The corre-

sponding ball (a 3-dimensional manifold of R3) in turn is obtained

by the relaxed parametrizationθ

φ2π

π

UX = U

r = γ(ρ, θ, φ) = (ρ sin θ cos φ, ρ sin θ sin φ, ρ cos θ)

(U : 0 ≤ ρ ≤ R , 0 ≤ θ ≤ π , 0 ≤ φ ≤ 2π).

U is a rectangular prism whose boundary is the exception set.

Example. After removing the four vertices, the perimeter of

a square is a 1-dimensional manifoldM of R2, consisting

of four separate parts, with the relaxed parametrization

γ(u) =

(u+ 3,−1) for −4 ≤ u ≤ −2(1, u+ 1) for −2 ≤ u ≤ 0

(1− u, 1) for 0 ≤ u ≤ 2

(−1, 3− u) for 2 ≤ u ≤ 4,

M

x

y

11

1

1

where the parameter domain is U : −4 ≤ u ≤ 4. The exception set is X = −4,−2, 0, 2, 4.Similarly, relaxed parametrizations could be obtained for surfaces of cubes, or more gener-

ally, surfaces of polyhedra.

In volume computations the exception setX has no contribution since it is mapped to γ(X ),a k-null-set of Rn:

|M|k = |M′|k =∫

U−X

√

det(

γ ′(u)Tγ ′(u))

du.

If the integrand√

det(

γ ′(u)Tγ ′(u))

has a continuous extension onto the whole U , as is some-times the case, we can write further

|M|k =∫

U

√

det(

γ ′(u)Tγ ′(u))

du,

since X is a null set of Rk. Improper integrals are here allowed, too, giving even more ”re-laxation”. Thus relaxed parametrizations can be used in volume computations and integrationsmore or less as the ”usual” parametrizations of Section 2.5.


Even relaxed reparametrization7 is possible and preserves volumes. A new extended pa-rameter domain V ⊆ R

k, an exception set Y ⊂ V , and a continuous function η : V → U arethen sought, such that

1. restricted to Y , η is a bijective function Y → X .

2. restricted to V −Y , η is a continuously differentiable bijective function V −Y → U −X ,and its derivative η′ is nonsingular (i.e. η then gives the ”usual” reparametrization).

In such a relaxed reparametrization the exception set is mapped to an exception set, whichguarantees preservation of the volume in reparametrization (as a consequence of Theorem 3.4).To make reparametrization in this sense possible, exception sets often need to be modified.

Example. The circle x2 + y2 = 1 is a 1-dimensional manifold of R2 having e.g. the following

two relaxed parametrizations:

r = γ1(φ) = (cosφ, sinφ) (0 ≤ φ < 2π)

(the usual polar coordinate parametrization with the exception set 0), and

r = γ2(u) =

(

− u− 2,√

1− (u+ 2)2)

for −3 ≤ u ≤ −1(

u,−√1− u2

)

for −1 ≤ u < 1(U : −3 ≤ u < 1)

(the exception set is −3,−1). These two relaxed parametrizations cannot be directly repara-

metrized to each other. If, however, in the exception set of the first parametrization the ”unnec-

essary” number π is added, then a reparametrization is possible by the function

u = η(φ) =

−2 − cosφ for 0 ≤ φ ≤ π

cosφ for π ≤ φ < 2π.

Note. Relaxed parametrizations are possible for trajectory manifolds, too. And volumes will

then also be preserved in relaxed reparametrizations.

7Again in literature this is defined in many different ways.

Chapter 4

FORMS

”Gradient a 1-form? How so? Hasn’t one always knownthe gradient as a vector? Yes, indeed, but only becauseone was not familiar with the more appropriate 1-form

concept. The more familiar gradient is the vectorcorresponding to the 1-form gradient.”

(C.W. MISNER & K.S. THORNE & J.A. WHEELER: Gravitation)

4.1 k-Forms

Whenever vector fields are integrated—the result being a scalar—the fields first need to be”prepared” in one way or another, so that the integrand is scalar-valued. A pertinent propertyof a vector field is its direction, which then must be included somehow in the preparation.Integration regions are here parametrized manifolds, possibly in a relaxed sense. Directionsrelated to the manifold can be included via tangent spaces or normal spaces. And in manycases orientation of the manifold should be fixed, too. Examples of integrals of this type are thefamiliar line and surface integrals

∫

C

F(r) • ds and∫

S

F(r) • dS.

A general apparatus for the ”preparation” is given by so-called forms. In integration theyappear formally as a kind of differentials, and are therefore often called differential forms. Invarious notations, too, differentials appear for this reason. Other than this, they do not have anyreal connection with differentials.

An n-dimensional k-form is a function φ mapping n-dimensional vectors1 r1, . . . , rk (thusk vectors, and the order is relevant!) to a real number φ(r1, . . . , rk), satisfying the conditions

1. φ is antisymmetric, i.e., interchanging any two vectors ri and rj changes the sign of thevalue φ(r1, . . . , rk). Thus if for instance k = 4, then

φ(r3, r2, r1, r4) = −φ(r1, r2, r3, r4).

2. φ is multilinear, i.e., it is linear with respect to any argument position (vector ri):

φ(r1, . . . , ri−1, c1ri1 + c2ri2, ri+1, . . . , rk) = c1φ(r1, . . . , ri−1, ri1, ri+1, . . . , rk)

+ c2φ(r1, . . . , ri−1, ri2, ri+1, . . . , rk)

(similarly the cases i = 1 and i = n).

As a limiting case, 0-forms can be included, too: they are constant functions which do not haveany variable vectors at all.

1Often tangent vectors where only the vector part is used. Note that in 1-, 2- and 3-dimensional spaces geomet-ric vectors could be used.

46

CHAPTER 4. FORMS 47

Thinking about the intended use of k-forms—as devices for coupling vector fields andk-dimensional volumes for integration—this definition looks fairly minimal. Antisymmetrymakes it possible the change direction by ”reflecting”, that is, interchanging two vector vari-ables. Multilinearity on the one hand guarantees the volume of a combination of two disjointsets is duly obtained by adding the volumes of the two parts, and on the other hand implies thatscaling works correctly, i.e., if the vector variables are multiplied (scaled) by the scalar λ, thenthe value is scaled by λk.

Example. A familiar example of an n-dimensional n-form is the determinant det. Take n vec-

tors r1, . . . , rn, consider them as columns of a matrix: (r1, . . . , rn), and compute its determinant

det(r1, . . . , rn). It follows directly from basic properties of determinants that this is an n-form.

Example. An equally familiar example of an n-dimensional 1-form is dot product with a fixed

vector. If a is a fixed constant vector, then the function φ(r) = a • r is a 1-form. Note that

antisymmetry is not relevant here because there is only on vector variable.

The dot product r1 • r2 is however not a 2-form (and why not?).

Example. A further familiar example of a 3-dimensional 2-form is the first component (or any

other component) of the cross product r1 × r2. Properties of the cross product immediately

imply that

(r2 × r1)1 = (−r1 × r2)1 = −(r1 × r2)1

(antisymmetry) and that

(

r1 × (c1r21 + c2r22))

1= (c1r1 × r21 + c2r1 × r22)1 = c1(r1 × r21)1 + c2(r1 × r22)1

(multilinearity).

Let us list some basic properties of forms:

Theorem 4.1. (i) If φ(r1, . . . , rk) is an n-dimensional k-form and c is a constant, then

cφ(r1, . . . , rk) is also a k-form.

(ii) If φ1(r1, . . . , rk) and φ2(r1, . . . , rk) are n-dimensional k-forms, then so is their sum

φ1(r1, . . . , rk) + φ2(r1, . . . , rk).

(iii) If φ(r1, . . . , rk) is an n-dimensional k-form and c is a constant vector, then

φ1(r1, . . . , rk−1) = φ(r1, . . . , rk−1, c)

is an n-dimensional k−1-form. (And similarly when c is in any other argumant position.)

(iv) If the vectors a1, . . . , ak are linearly dependent, then the value of a k-form φ(a1, . . . , ak)is = 0. In particular, if some ai is the zero vector, then φ(a1, . . . , ak) = 0.

Proof. Items (i), (ii) and (iii) are immediate, so let us proceed to item (iv). We notice first thatif two of the vectors a1, . . . , ak are the same, then the value is = 0. Indeed, interchanging thesevectors the value changes its sign, and also remains the same. If then a1, . . . , ak are linearlydependent, then one of them can be expressed as a linear combination of the others, say,

ak =k−1∑

i=1

ciai.

CHAPTER 4. FORMS 48

But then by the multilinearity

φ(a1, . . . , ak) =k−1∑

i=1

ciφ(a1, . . . , ak−1, ai),

which is = 0. In particular, if say a1 = 0, then

φ(0, a2, . . . , ak) = φ(0 · 0, a2, . . . , ak) = 0φ(0, a2, . . . , ak) = 0.

As a consequence of item (iv), an n-dimensional k-form where k > n is rather uninteresting: italways has the value 0.

Forms in fact are a kind of generalizations of determinants. To see this, let us see how ann-dimensional k-form φ(r1, . . . , rk) is represented when the vectors r1, . . . , rk are representedin an orthonormalized basis e1, . . . , en as

ri =

n∑

j=1

xj,iej (i = 1, . . . , k).

Multilinearity of the last argument position rk gives

φ(r1, . . . , rk) =

n∑

j=1

xj,kφ(r1, . . . , rk−1, ej).

Continuing using the representation of rk−1 in the basis gives further

φ(r1, . . . , rk) =

n∑

j=1

n∑

l=1

xl,k−1xj,kφ(r1, . . . , rk−2, el, ej).

Note that all terms having i = j can be omitted in the sum. In this way we finally get arepresentation of φ(r1, . . . , rk) as a sum of terms of the form

xj1,1xj2,2 · · ·xjk,kφ(ej1, ej2, . . . , ejk)

where j1, j2, . . . , jk are disjoint indices.Let us then collect together terms where the indices j1, j2, . . . , jk are the same, just possibly

in a different order. As an example take the case where the indices are the numbers 1, 2, . . . , kin various permuted orders. The other cases are of course similar. These then are the k! terms

xj1,1xj2,2 · · ·xjk,kφ(ej1 , ej2, . . . , ejk),

where j1, j2, . . . , jk are the numbers 1, 2, . . . , k in some order. Interchanging the vectors inφ(ej1, ej2, . . . , ejk) repeatedly to make the order ”correct” this term can be written as

±axj1,1xj2,2 · · ·xjk,k,

where a = φ(e1, e2, . . . , ek) and the sign ± is determined by the parity of the number of usesof antisymmetry (’even’ = +, ’odd’ = −). On the other hand, this is exactly the way thedeterminant

∣

∣

∣

∣

∣

∣

∣

∣

∣

x1,1 x1,2 · · · x1,kx2,1 x2,2 · · · x2,k

......

. . ....

xk,1 xk,2 · · · xk,k

∣

∣

∣

∣

∣

∣

∣

∣

∣

CHAPTER 4. FORMS 49

is expanded as a sum of products, the only difference is that then a is = 1.The above indicates that in a coordinate representation, certain k-forms of a special type

seem to be central. These forms, the so-called elementary k-forms, are defined as follows.Of the indices 1, 2, . . . , n take k indices in increasing order: j1 < j2 < · · · < jk. Then theelementary k-form2

dxj1 ∧ dxj2 ∧ · · · ∧ dxjk

is defined by

(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

xj1,1 xj1,2 · · · xj1,kxj2,1 xj2,2 · · · xj2,k

......

. . ....

xjk,1 xjk,2 · · · xjk,k

∣

∣

∣

∣

∣

∣

∣

∣

∣

.

Thus we get the value of the form by taking elements having the indices j1, j2, . . . , jk fromr1, . . . , rk, forming the corresponding k× k determinant, and computing the value of the deter-minant. As a special case an elementary 0-form is included, it always has the value = 1.

Obviously there are(

n

k

)

=n!

k!(n− k)!elementary n-dimensional k-forms. In particular there is one elementary 0-form and one ele-mentary n-form. This makes it possible to ”embed” single scalars in forms in two ways. On theother hand there are n elementary 1-forms, and also n elementary n− 1-forms, which makes itpossible to ”embed” vectors in forms via the coefficients, again in two ways (even when n = 2,in a sense!).

Note. The definition of the above k-form

dxj1 ∧ dxj2 ∧ · · · ∧ dxjk

using a determinant actually works even when the indices j1, j2, . . . , jk are not in order of

magnitude. This possibility is a convenient one to allow. For instance, interchanging dxjland dxjh interchanges the ith and the hth rows in the determinant and thus changes its sign.

Furthermore, if two (or more) of the indices j1, j2, . . . , jk are the same, then we naturally agree

that the k-form dxj1 ∧ dxj2 ∧ · · · ∧ dxjk has the value 0.

Example. n-dimensional elementary 1-forms are simply the component functions. If

r =

x1x2...

xn

,

then (dxj)(r) = xj (j = 1, . . . , n). As determinants these are 1× 1 determinants.

2This particular notation is traditional. It does not have much to do with differentials. The symbol ’∧’ is read”wedge”. The corresponding binary operation, the so-called wedge product (or sometimes the exterior product), isgenerally available for forms, as will be seen.

CHAPTER 4. FORMS 50

Example. The first component of the cross product r1 × r2 is a 3-dimensional elementary

2-form. If

r1 =

x1y1z1

and r2 =

x2y2z2

,

then the first component is

(r1 × r2)1 = y1z2 − z1y2 =∣

∣

∣

∣

y1 y2z1 z2

∣

∣

∣

∣

= (dy ∧ dz)(r1, r2).

Other components are 2-forms, too:

r1 × r2 =

(dy ∧ dz)(r1, r2)(dz ∧ dx)(r1, r2)(dx ∧ dy)(r1, r2)

.

Here in the 2-form dz ∧ dx the variables are not in the correct order, cf. the above note.

Example. n-dimensional elementary n-forms are simply n× n determinants:

(dx1 ∧ dx2 ∧ · · · ∧ dxn)(r1, r2, . . . , rn) = det(r1, r2, . . . , rn).

If the variables are not in the correct order and/or are repeated, then

(dxj1 ∧ dxj2 ∧ · · · ∧ dxjn)(r1, r2, . . . , rn) = det(rj1, rj2, . . . , rjn),

since it does not matter whether it is the rows or the columns that are interchanged.

The construct above gives a representation of forms by elementary forms:

Theorem 4.2. If φ is an on n-dimensional k-form, then

φ(r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk),

where

aj1,j2,...,jk = φ(ej1 , ej2, . . . , ejk).

The representation is unique, i.e., the coefficients aj1,j2,...,jk can only be chosen in one way (and

why is that?).

Example. The n-dimensional 1-forms are exactly all forms

φ(r) =n∑

j=1

ajxj = a • r,

i.e., exactly all forms obtained by taking a dot product with a constant vector a = (a1, . . . , an)T.

It might be a bit confusing to consider this using elementary 1-forms:

a1 dx1 + a2 dx2 + · · ·+ an dxn.

CHAPTER 4. FORMS 51

Example. The 3-dimensional elementary 2-forms are exactly all forms

φ(r1, r2) = a • r1 × r2,

where a is a constant vector. This mostly explains the central role of the scalar triple product

(and the cross product) in vector analysis. This can also be written using determinants as

φ(r1, r2) = det(a, r1, r2).

In general, thinking about expanding a determinant along the first column, the n-dimensional

n− 1-forms are exactly all forms

φ(r1, . . . , rn−1) = det(a, r1, . . . , rn−1),

where a is a constant vector.

Example. The n-dimensional n-forms are exactly all forms

φ(r1, . . . , rn) = a det(r1, . . . , rn),

where a is a scalar constant, or, given by elementary n-forms,

φ = a dx1 ∧ dx2 ∧ · · · ∧ dxn.

By Theorem 4.2, the general wedge product can now be defined. If the forms are givenusing elementary forms as

φ(r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk)

(a k-form) and

ψ(s1, . . . , sl) =∑

1≤h1<h2<···<hl≤n

bh1,h2,...,hl(dxh1

∧ dxh2∧ · · · ∧ dxhl

)(s1, . . . , sl)

(an l-form), then their wedge product is the k + l-form

(φ ∧ ψ)(r1, . . . , rk, s1, . . . , sl)

whose representation by elementary forms is

φ ∧ ψ =∑

1≤j1<j2<···<jk≤n1≤h1<h2<···<hl≤n

aj1,j2,...,jkbh1,h2,...,hldxj1 ∧ dxj2 ∧ · · · ∧ dxjk ∧ dxh1

∧ dxh2∧ · · · ∧ dxhl

,

with the proviso of the above note concerning order and/or repetetition of indices. If k = 0, thatis φ is a 0-form (constant), this simply is multiplication by a constant. Note especially that thewedge product is associative and distributive, i.e.,

φ ∧ (ψ ∧ ξ) = (φ ∧ ψ) ∧ ξ ,φ ∧ (ψ + ξ) = φ ∧ ψ + φ ∧ ξ and

(φ+ ψ) ∧ ξ = φ ∧ ξ + φ ∧ ξ.

Note. The concept of a form was defined without using any coordinate systems. On the other

hand, the wedge product above was defined via elementary forms, and thus using a coordinate

system. It can, however, also be defined without elementary forms, starting from the original

definition of forms, and so is coordinate-free. See e.g. HUBBARD & HUBBARD.

CHAPTER 4. FORMS 52

4.2 Form Fields

A k-form field3 of Rn is a function which for each point p gives a k-form:

Φ(p; r1, . . . , rk).

Here r1, . . . , rk are the vector variables for the k-form given by the form field for the point p.Often they are considered as tangent vectors with the point of action p. By Theorem 4.2, todefine a form field Φ it suffices to give the coefficients in the presentation by elementary forms,and they will depend on p:

Φ(p; r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(p)(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk),

where the needed

(

n

k

)

coefficients are

aj1,j2,...,jk(p) = Φ(p; ej1 , ej2, . . . , ejk).

Often a form field is not defined in the whole space Rn but only in a subset, e.g. in a manifold.

Example. A constant k-form field is one depending only on the vector variables and not on the

point p, i.e. a form φ(r1, . . . , rk).

Example. The n-dimensional 1-form fields are the fields

Φ(p; r) = F(p) • r,

where F is a vector-valued function (vector field) defined in a suitable subset of Rn (say, a

manifold).

Example. A general 2-form field of R3 has the form

Φ(p; r1, r2) = F(p) • r1 × r2,

where F is a vector field defined in a suitable subset of Rn (say, a manifold).

Example. The n-dimensional n− 1-form fields are the fields

Φ(p; r1, . . . , rn−1) = det(

F(p), r1, . . . , rn−1)

,

where F is a vector-valued function (vector field) defined in a suitable subset of Rn (say, a

manifold).

Example. A general n-form field of Rn has the form

Φ(p; r1, . . . , rn) = a(p) det(r1, . . . , rn),

where a is a real-valued valued function (scalar field) defined in a suitable subset of Rn (say, a

manifold).

3For some reason form fields are also called differential forms in some literature, though they have little to dowith differentials.

CHAPTER 4. FORMS 53

These examples already indicate how manifolds, vector and scalar fields and forms areconnected: Using forms vector and scalar fields are transformed into form fields which thenare integrated over a region, often a manifold or a parameter domain, possibly in a relaxedparametrization. In the general case such integrals are hard to deal with. For parametrizationsit is easier, in principle anyway.

If a k-dimensional manifoldM of Rn has a (relaxed) parametrization

r = γ(u) (u ∈ U)

and Φ(p; r1, . . . , rk) is an n-dimensional k-form field defined inM, then its integral overM is∫

M

Φ =

∫

U

Φ(

γ(u);γ ′(u))

du.

Here the k columns of γ ′(u) are interpreted as the vector variables of the form field. Manyintegrals familiar from basic courses on mathematics are of this type, and they now have auniform formulation.

Example. The line integral of a vector field F over the smooth curve

C : r = γ(u) (u ∈ U),

i.e.,∫

C

F(r) • ds =∫

U

F(

γ(u))

• γ ′(u) du

is the integral of the 1-form field

Φ(p; r) = F(p) • rover C. The curve C could here also be only piecewise smooth via a relaxed parametrization. It

certainly should be obvious why this integral often is given (in R3) as

∫

C

F1(r) dx+ F2(r) dy + F3(r) dz.

Example. Similarly the surface integral of the vector field F over the smooth surface

S : r = γ(u) (u ∈ U),

i.e.,∫

S

F(r) • dS =

∫

U

F(

γ(u))

• ∂γ(u)∂u1

× ∂γ(u)

∂u2du

is the integral of the 2-form field

Φ(p; r1, r2) = F(p) • r1 × r2

over S. Here, too, the surface S could be only piecewise smooth via a relaxed parametrization.

Thinking about the connection of the cross product and elementary 2-forms, it should not be

surprising that this integral is sometimes given as∫

S

F1(r) dy ∧ dz + F2(r) dz ∧ dx+ F3(r) dx ∧ dy,

or even∫

S

F1(r) dy dz + F2(r) dz dx+ F3(r) dx dy.

CHAPTER 4. FORMS 54

Example. The integral of a real-valued function f over an n-dimensional parametrized mani-

fold of Rn

M : r = γ(u) (u ∈ U)

(possibly in a relaxed parametrization), i.e.

∫

M

f(r) dr =

∫

U

f(

γ(u))∣

∣ det(

γ ′(u))∣

∣du

(recall change of variables in an integral), is not the integral of a form field overM, because

of the absolute value of the determinant. But the integral

∫

U

f(

γ(u))

det(

γ ′(u))

du

is, i.e., it is the integral of the n-form field

Φ(p; r1, . . . , rn) = f(p) det(r1, . . . , rn)

overM.

And what about the case of a 0-form field Φ of Rn? It simply is a real-valued function Φ(p)of n variables. On the other hand, a 0-dimensional manifold M of R

n consists of separatepoints (possibly infinitely many), so

∫

M

Φ =∑

p∈M

Φ(p).

4.3 Forms and Orientation of Manifolds

Geometrically a surface of R3 can be oriented by giving in each point a normal vector whichvaries continuously, and never equals the zero vector, and thus cannot jump or move to the otherside of the surface. Similarly, a curve can be oriented by giving in each point a tangent vectorwhich varies continuously and never equals the zero vector, and thus cannot turn around.

In general, a k-dimensional manifold M of Rn can be oriented with a similar idea, al-

though it might be difficult to accomplish. The orientation is obtained by taking, if possible, ann-dimensional k-form field

Φ(p; r1, . . . , rk)

such that

1. in each point p0 of the manifold the k-form

Φ(p0; r1, . . . , rk)

is defined in the tangent space Tp0(M). I.e., Φ(p0; r1, . . . , rk) is defined for all vectors

r1, . . . , rk in the tangent space Tp0(M), cf. Section 2.6. Of course, the k-form may well

be defined for other vectors as well but they are not needed here.

CHAPTER 4. FORMS 55

2. Φ is continuous with respect to the variable p inM and does not change sign.

More specifically, for every point p ofM there exist vectors

t1(p) , . . . , tk(p)

of the tangent space Tp(M) such that

Φ(

p; t1(p), . . . , tk(p))

is continuous with respect to p in M, and is either positive in M or negative in M.The sign can then used for orientation. Note that the vectors t1(p), . . . , tk(p) then mustconstitute a basis for the tangent space Tp(M), otherwise the form would have the value= 0. The question thus is whether or not the tangent space can always be chosen a basiswith the same orientation (same ”handedness”).

If a manifoldM is parametrized by

p = γ(u) (u ∈ U),

then it is possible to use this to orientM. Item 1. remains the same but item 2. is replaced by

2.’ Φ(

γ(u);γ ′(u))

is continuous with respect to the parameter u, and does not change sign.

This is not a different type of orientation than the general one above, parametrization is just usedto choose the bases for the tangent spaces. As was seen in the proof of Theorem 2.6, locallythe parameter u can be given as a continuous—and even continuously differentiable—functionu = g(r) of some components r of p. Thus γ−1 is continuous inM, and according to item 2.,the orientation is given by the value Φ

(

p;γ ′(

γ−1(p)))

.Not all manifolds can be oriented. Classical examples are various Möbius’ bands. An

example is given by the relaxed parametrization

γ1(u) =(

1 + u1 cosu22

)

cosu2

γ2(u) =(

1 + u1 cosu22

)

sin u2

γ3(u) = u1 sinu22

(U : −1 < u1 < 1, 0 ≤ u2 < 2π).

If the parameter value u2 = 0 is omitted, the bandis ”severed” and is a parametrized 2-dimensionalmanifold of R3. (Proving this takes a bit of tediouscalculation.) The severed band can be oriented us-ing the parametrization. On the other hand, if inthe relaxed parametrization we take another inter-val for the parameter u2, say π ≤ u2 < 3π, thebreak would move elsewhere. Since ”manifold-ness” is a local property, this basically means thatthe ”unsevered” Möbius band is a manifold, too—but not a parametrized one. Looking at the fig-ure on the right (Maple) it is at least geometricallyfairly obvious that the band cannot be oriented (theexact proof again being tedious).

Manifolds that can be oriented are called orientable. If the orientation of an orientablemanifoldM is fixed (using some form field), this is indicated by denoting

−→M. The oppositeorientation is then often denoted by −−→M (sometimes by

←−M).

CHAPTER 4. FORMS 56

Orientations of surfaces and curves are actually examples of this method of orientation.

Example. A smooth parametrized surface

S : r = γ(u) (u ∈ U)

can be oriented remembering that its tangent

space in the point γ(u) is the column space

of γ ′(u) (Corollary 2.10), i.e., the two

columns of γ ′(u) constitute a basis of the

space. Agreeing that the first column is the

first basis vector t1, and the second column

the second basis vector t2 already orients

the surface. Since the surface is smooth (a

x

y

z t1tangent plane

t2

2-dimensional parametrized manifold of R3), these vectors are always linearly independent,

and thus cannot be ”interchanged” (see the figure below). Geometrically, the normal vector

∂γ(u)

∂u1× ∂γ(u)

∂u2

always points to the same side of the surface.

But what is the 2-form field needed for the orientation? The field is known to be of the form

Φ(p; r1, r2) = F(p) • r1 × r2,

where F is a vector field defined in the surface. We simply choose

F(p) = F(

γ(u))

=∂γ(u)

∂u1× ∂γ(u)

∂u2.

Then the value of the 2-form is∥

∥

∥

∂γ(u)

∂u1× ∂γ(u)

∂u2

∥

∥

∥

2

and it is always positive. Other choices are of course possible, too, but maybe not quite as

natural.

Let us try the orientation on the torus. A relaxed parametrization of a torus is

γ1(u) = (2 + cosu2) cosu1

γ2(u) = (2 + cosu2) sin u1

γ3(u) = sin u2,

where the parameter domain is U : 0 ≤ u1, u2 < 2π. As

for the Möbius band, it may be noticed that this torus is a

manifold, but not a parametrized one. For the orientation we

omit the value u1 = 0 (although the whole torus is of course

orientable, too). See the figure on the right (Maple). Basis–4

–3

–2

–1

0

1

2

3

4

–4–3

–2–1

12

34

–4–3

–2–1

12

34

vectors of the tangent space are (in this order)

∂γ(u)

∂u1=

−(2 + cosu2) sin u1(2 + cosu2) cosu1

0

and∂γ(u)

∂u2=

− sin u2 cosu1− sin u2 sin u1

cos u2

.

CHAPTER 4. FORMS 57

Which side of the torus is the one where the basis vectors are positively ordered (as in the

xy-coordinate system)? Take the basis vectors for the parameter values u1 = u2 = π, i.e. in the

point (−1, 0, 0) of the x-axis:

∂γ(π, π)

∂u1=

0−10

and∂γ(π, π)

∂u2=

00−1

.

These basis vectors are ordered as the negative y- and z-axes, thus this side is the ”outside”.

(In the ”inside” view the axes are interchanged).

Example. A parametrized smooth space curve

C : p = γ(u) (u ∈ U)

can be oriented similarly. In the point γ(u) the basis vector γ ′(u) of the tangent space is chosen,

and the curve is oriented. The 1-form field needed is of the form

Φ(p; r) = F(p) • r

and we choose

F(

γ(u))

= γ ′(u),

the value of the 1-form is then∥

∥γ ′(u)∥

∥

2> 0. We could have chosen, say,

F(

γ(u))

= 2γ ′(u)

(which gives the same orientation) or

F(

γ(u))

= −γ ′(u)

(which gives the opposite orientation).

If a manifold consists of several disjoint parts, orientation in these parts can be chosenarbitrarily, the continuity condition does not connect the parts in any way. If for instance in a3-dimensional manifold of R3 consisting of two disjoint parts, coordinate system in one of theparts is right-handed, nothing prevents us from taking a left-handed system for the other part(unless it is the general preference for right-handed systems).

For a connected manifold orientation is coherent all over the manifold. A manifoldM ofR

n is said to be connected, if for each two points p0 and p1 of the manifold there is a continuousfunction f : [0, 1]→M such that

f(0) = p0 and f(1) = p1.

In other words, any two points of the manifold can always be connected by a continuous curvein the manifold.

Theorem 4.3. If the oriented manifoldM is connected then an orientation in one point gives

the orientation everywhere in the manifold.

Proof. SupposeM is oriented as above. The function of the point p of the manifold in item 2.

Φ(

p; t1(p), . . . , tk(p))

CHAPTER 4. FORMS 58

is then in the chosen point p0 either positive or negative. Any other point p1 of the manifoldcan be connected to p0 using some continuous function f : [0, 1] →M, where p0 = f(0) andp1 = f(1). But then the function

Φ(

f(s); t1(

f(s))

, . . . , tk(

f(s)))

is a continuous function of s in the interval [0, 1] and does not attain the value 0. Thus

Φ(

p0; t1(p0), . . . , tk(p0))

and Φ(

p1; t1(p1), . . . , tk(p1))

have the same sign.

In reparametrization the orientation of a parametrized manifold may be changed for theopposite. This can be prevented by setting a further condition:

Theorem 4.4. If the oriented parametrized k-dimensional manifold

M : r = γ(u) (u ∈ U)

of Rn is reparametrized by

u = η(v) (v ∈ V),

where

det(

η′(v))

> 0,

then the orientation remains the same. If, on the other hand, det(

η′(v))

< 0, then the orien-

tation will change. (It is naturally assumed here that the orientation is defined using the same

k-form field.)

Proof. As above, the manifoldM is oriented using some k-form field Φ(p; r1, . . . , rk). Con-sider then orientation in some point p0 = γ(u0) ofM, given by the value Φ

(

p0;γ′(u0)

)

. ByTheorem 4.2 the k-form Φ(p0; r1, . . . , rk) can be represented as a linear combination of ele-mentary k-forms. Let us show that these forms will be multiplied by the same positive numberin reparametrization.

The elementary k-forms above are of the form

(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk) =

∣

∣

∣

∣

∣

∣

∣

∣

∣

xj1,1 xj1,2 · · · xj1,kxj2,1 xj2,2 · · · xj2,k

......

. . ....

xjk,1 xjk,2 · · · xjk,k

∣

∣

∣

∣

∣

∣

∣

∣

∣

.

On the other hand, in the new parametrization a basis for the tangent space is given by thecolumns of the matrix

γ ′(u0)η′(v0)

(via the chain rule), where u0 = η(v0), and the value of the corresponding new elementaryk-form is

(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(

γ ′(u0)η′(v0)

)

= (dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(

γ ′(u0))

det(

η′(v0))

.

(Remember matrix multiplication and that for square matrices det(AB) = det(A) det(B).)All elementary k-forms appearing here are thus multiplied by the same positive number

det(

η′(v0))

. So

Φ(

p0;γ′(

η(v0))

η′(v0))

= Φ(

p0;γ′(u0)

)

det(

η′(v0))

,

and the orientation does not change.Showing that the orientation does change in case det

(

η′(v))

< 0 is analogous.

CHAPTER 4. FORMS 59

Except for sketching the geometric picture, orientation is important for integration in repara-metrization.

Theorem 4.5. Orientation preserving reparametrization of a parametrized oriented manifold

does not change integrals of form fields over the manifold. If, on the other hand, reparametriza-

tion changes the orientation, integrals change sign.

Proof. If the parametrized oriented manifold

−→M : r = γ(u) (u ∈ U)

is reparametrized preserving orientation, then in the integration variable in the integral

u = η(v) (v ∈ V)

is changed and the integral∫

−→M

Φ =

∫

U

Φ(

γ(u);γ ′(u))

du =

∫

V

Φ(

γ(

η(v))

;γ ′(

η(v)))∣

∣ det(

η′(v))∣

∣dv

is obtained. Since the reparametrization was orientation preserving, det(

η′(v))

must be posi-tive by the previous theorem. Thus the integral can be written as∫

−→M

Φ =

∫

V

Φ(

γ(

η(v))

;γ ′(

η(v)))

det(

η′(v))

dv =

∫

V

Φ(

γ(

η(v))

;γ ′(

η(v))

η′(v))

dv

(cf. the previous proof), and this is the integral of the form field Φ overM given in the newparametrization.

On the other hand, if the reparametrization does change the orientation, then det(

η′(v))

isnegative, and the sign is changed.

Change of sign of an integral by change of orientation is denoted as∫

−−→M

Φ =

∫

←−M

Φ = −∫

−→M

Φ.

4.4 Basic Form Fields of Physical Fields

The basic form fields of a vector field F(p) of Rn are the work form field

ΦF–work(p; r) = F(p) • r

(a 1-form field), and the flux form field

ΦF–flux(p; r1, . . . , rn−1) = det(

F(p), r1, . . . , rn−1)

(an n− 1-form field).The basic form fields of a scalar field f(p) are the field itself (as a 0-form field), and the

density form field

Φf–density(p; r1, . . . , rn) = f(p) det(r1, . . . , rn)

(an n-form field, also called the mass form field).

CHAPTER 4. FORMS 60

Physically, in a work form field F is typically a force field—e.g. gravitation—and in a fluxform field it is a flux—e.g. the flux of a fluid flow. The scalar field of a density form field is adensity quantity—e.g. mass density. Integrals of these over manifolds give the correspondingnet quantities—work needed to move a particle in gravitational field, net flow of fluid througha surface, mass of a solid body, etc.

Using wedge product the familiar operations of fields in R3 can be transferred to form fields:

1. For scalar fields f and g and vector field F we of course have f ∧ g = fg, and also

f ∧ ΦF–work = ΦfF–work ,

f ∧ ΦF–flux = ΦfF–flux and

f ∧ Φg–density = Φfg–density.

2. For vector fields F and G

ΦF–work ∧ ΦG–work = ΦF×G–flux and

ΦF–work ∧ ΦG–flux = ΦF•G–density.

As an example, let us verify the formula

ΦF–work ∧ ΦG–flux = ΦF•G–density,

leaving the rest for the reader as an exercise:

(F1 dx+ F2 dy + F3 dz) ∧ (G1 dy ∧ dz +G2 dz ∧ dx+G3 dx ∧ dy)

= F1G1 dx ∧ dy ∧ dz + F2G2 dy ∧ dz ∧ dx+ F3G3 dz ∧ dx ∧ dy

= (F •G) dx ∧ dy ∧ dz.

Further convenience is given by the Hodge star operator ∗ which transforms a k-form fieldΦ in an n-dimensional space to an n− k-form field ∗Φ (the Hodge dual) as follows. If

Φ =∑

j1<···<jk

aj1,...,jk(p) dxj1 ∧ · · · ∧ dxjk ,

then ∗Φ =∑

j1<···<jk

±aj1,...,jk(p) dxjk+1∧ · · · ∧ dxjn ,

where in each summand the indices j1, . . . , jn are exactly all numbers 1, . . . , n and the sign ±is determined by the condition

dxj1 ∧ · · · ∧ dxjn = ±dx1 ∧ · · · ∧ dxn.

For a 0-form field (scalar field) f and an n-form field Φf–density we define ∗f = Φf–density and∗Φf–density = f . In R

3 we then see (check!) that ∗ΦF–work = ΦF–flux and ∗ΦF–flux = ΦF–work. Allthis makes it easy to change types of fields. Note also that for a k-form field Φ of Rn the dualof the dual is ∗∗Φ = (−1)k(n−k)Φ, and in particular in R

3 then ∗∗Φ = Φ.

CHAPTER 4. FORMS 61

There are physical fields which are neither scalar nor vector fields. For instance, the electricfield E is not a vector field (and of course not a scalar field either). Lorentz’s law says that theforce acting on a charged particle with charge q in the point r is

q(

E(r) + v ×B(r))

,

where v is the speed of the particle and B is the magnetic flux density. If there is no ambientelectric field, the particle does not move, and the coordinate frame is stationary, there is no force(i.e. the force is a zero vector). In a coordinate system moving with a constant speed (i.e. aninertial frame) the force remains a zero vector (the particle does not no accelerate), but now v

is not a zero vector, the moving charged particle creates a magnetic field and a compensatingelectric field. Seen from a moving frame then E is not a zero field. No physical vector fieldcan behave in this way, indeed, electric and magnetic fields should be dealt with together inspace-time (in R

4, the xyzt-space).Even though electro-magnetic fields then cannot be modelled as vector fields, they can be

modelled as 2-form fields of R4, the so-called Faraday form field and Maxwell form field 4

E1 dx ∧ dt+ E2 dy ∧ dt+ E3 dz ∧ dt +B1 dy ∧ dz +B2 dz ∧ dx+B3 dx ∧ dy

and

−c2B1 dx ∧ dt− c2B2 dy ∧ dt− c2B3 dz ∧ dt+ E1 dy ∧ dz + E2 dz ∧ dx+ E3 dx ∧ dy,

where c is the speed of light in vacuum, often written simply—with only a slight abuse ofnotation—as

ΦFaraday = ΦE–work ∧ dt+ ΦB–flux

andΦMaxwell = ΦE–flux − c2ΦB–work ∧ dt.

4Michael Faraday and James Clerk Maxwell certainly did not consider form fields like these. The namesprobably originated in the celebrated book MISNER & THORNE & WHEELER. The exact definitions of theseforms seem to vary a bit in the literature, the gist remaining the same, of course.

Chapter 5

GENERALIZED STOKES’ THEOREM

”Stokes theorem was stated by Sir George Stokes asa Cambridge examination question, having been raised

by Lord Kelvin in a letter to Stokes in 1850.”

(A Dictionary of Science, Oxford University Press, 1999)

Many integration theorems in basic courses on mathematics share a common overall form:

b∫

a

df(x)

dxdx = f(b)− f(a)

(Fundamental Theorem of Integral Calculus)r2∫

r1

∇f • ds = f(r2)− f(r1) (Gradient Theorem)

∫

A

(∂F2(r)

∂x− ∂F1(r)

∂y

)

dr =

∮

−→∂A

F(r) • ds (Green’s Theorem)

∫

A

∇ • F(r) dr =∮

−→∂A

F(r) • dS (Gauß’ Theorem)

∫

−→S

∇× F(r) • dS =

∮

−→∂S

F(r) • ds (Stokes’ Theorem)

The common features are the following:

• In the left hand side the integration region is a manifold, possibly in a relaxed parametriza-tion, and oriented.

• In the left hand side the integrand is a derivative of some sort, coupled with the manifoldvia tangent space and a suitable form (as a form field).

• In the right hand side the integration region (or summation region) is the oriented ”bound-ary” of the left hand side manifold, possibly in a relaxed parametrization. Note that inthe two first theorems an oriented boundary also appears, and one of the points has a plussign and the other one a minus sign (orientation).

• In the right hand side the integrand (or summand) is a field appearing in the left hand sidedifferentiated in some way.

To get a coherent form for all these theorems, and others, we must define a general deriva-tive for form fields, and the concept of an oriented boundary over which form fields can beintegrated. The final result then is the Generalized Stokes Theorem.

62

CHAPTER 5. GENERALIZED STOKES’ THEOREM 63

5.1 Regions with Boundaries and Their Orientation

In general the boundary1 is of a k-dimensional manifoldM of Rn is easily defined. It consistsof all points p satisfying the following two conditions:

(i) p is not inM.

(ii) Any open subset containing p also contains a point of M. In particular this is true forp-centered open balls.

The boundary of a manifold may be empty. E.g. the boundary of a sphere in R3 is empty. On

the other hand, the boundary of a manifold need not be a (lower-dimensional) manifold. E.g.the boundary of the open square 0 < x, y < 1 of R2 is the perimeter of the square, and thus nota manifold. In fact, the boundary of a bounded manifold can have a very complicated structure,it may e.g. not have a Jordan measure. A more serious problem often is existence of tangentspaces, needed for the orientation.

A better way to start is to define the boundary of a subset within a manifold. This is so,because inside the manifold its smoothness and tangent spaces can be used for the orientation.

The boundary of a subset A ⊂ Mwithin the manifold, denoted by ∂MA,consists of exactly all points of M suchthat in every open set containing the pointthere is a point ofA and a point ofM−A.(Cf. the condition (ii) above.) Note that apoint of the boundary ∂MA may itself bein A.

Consider then a point p of the bound-ary ∂MA. Since the point p0 is in themanifoldM, for some open subsetB con-taining the point the intersectionM∩ Bis a local locus, cf. Theorem 2.4. In other

M

A

a nonsmooth point

a smooth point

MA

of the boundary

of the boundary

words, there is a continuously differentiable function Fp0: B → R

n−k such that F′p0has full

rank and p is inM∩ B if and only if Fp0(p) = 0. There may be several such functions Fp0

to choose from. The point p0 is called a smooth point of the boundary ∂MA if, in addition tothe function Fp0

, another continuously differentiable function gp0: B → R can be chosen such

that

1. g′p0cannot be given as a linear combination of rows of F′p0

.

2. the point p is in the intersection A∩ B exactly when

Fp0(p) = 0 and gp0

(p) ≥ 0.

The subset of the boundary ∂MA consisting of exactly all smooth points is called the smooth

boundary of A in the manifoldM, and denoted by ∂ sMA.

The definition is easily interpreted even when k = n. Item 1. is then not needed and theonly condition is gp0

(p) ≥ 0.

1Note that this (usually) is not the boundary as a boundary of a set in Rn.


Example. The locus condition

F (x, y, z) = z − xy = 0

determines in R3 a smooth surface (a 2-di-

mensional manifold) S, one of the so-called

saddle surfaces. The condition ‖p‖ ≤ 1specifies a subset of A. The boundary ∂SAis then the locus given by the conditions –1–0.500.51

y

–1

0

1

x

–0.4

–0.2

0

0.2

0.4

z − xy = 0

1− x2 − y2 − z2 = 0,

in fact a 1-dimensional manifold. See the figure above (Maple). All points of the boundary ∂SAare smooth points, since g can be chosen as

g(x, y, z) = 1− x2 − y2 − z2,

and

F ′(x, y, z) = (−y,−x, 1) and g′(x, y, z) = (−2x,−2y,−2z)are locally linearly independent (check!). Thus ∂ s

SA = ∂SA.

Example. The square in R3 given by

N : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z = 1

is a subset of the plane z = 1 (a 2-dimensional manifold). Its boundary is the perimeter, i.e.,

the union of the four line segments

0 ≤ x ≤ 1 , y = 0 , z = 1 ,

0 ≤ x ≤ 1 , y = 1 , z = 1 ,

x = 0 , 0 ≤ y ≤ 1 , z = 1 and

x = 1 , 0 ≤ y ≤ 1 , z = 1.

With the exception of the vertices

(0, 0, 1) , (0, 1, 1) , (1, 0, 1) and (1, 1, 1),

points of the boundary are smooth, the function g can be taken as

y , 1− y , x and 1− x,

respectively, and F as

F (x, y, z) = z − 1.

It is to be anticipated in these examples that

Theorem 5.1. The smooth boundary ∂ sMA of a subsetA of a k-dimensional manifoldM of Rn

is a k − 1-dimensional manifold (or empty).


Proof. We use the (local) notation above, and consider an arbitrary point p0 of the smoothboundary. By Corollary 2.3, the condition

Fp0(p) = 0

gp0(p) = 0

then determines a k − 1-dimensional manifold K in some open subset B1 ⊆ B containing p0.We need to show now that the intersection K∩B1 equals the intersection ∂ s

MA∩B1. Sincep0 was an arbitrary point of the smooth boundary, this shows that ∂ s

MA is a local locus and thusa manifold.

Take first a point p1 in the intersection ∂ sMA∩B1, and show that it is also in the intersection

K ∩ B1. Let us assume the contrary: p1 is not in K ∩ B1. Then

gp0(p1) 6= 0.

(Since p1 is inM, we have Fp0(p1) = 0.) By continuity then gp0

(p) 6= 0 also in some (small)open subset B2 containing p1. Thus in the set B2, gp0

has only positive values or only negativevalues. But then B2 cannot contain both points ofM and points ofM−A, and p1 cannot be apoint of the boundary of A inM. This contradiction shows that p1 ∈ K ∩ B1.

Proving inclusion in the other direction we take a point p2 in the intersection K ∩ B1, andshow that it is also in the intersection ∂ s

MA∩B1. Assume again the contrary: p2 /∈ ∂ sMA∩B1.

Now gp0(p2) = 0, and in some (small) open subset B3 containing p2, the function gp0

has eitheronly positive values or only negative values. Otherwise p2 would be a point of the smoothboundary (take Fp2

= Fp0and gp2

= gp0). This means that p2 is a local solution of the

constrained optimization problem

gp0(p) = extremum!

Fp0(p) = 0

(either a point of local maximum or local minimum, the local extremum value being = 0).Using the method of Lagrange’s multipliers, we introduce the Lagrange function

L(p,λ) = gp0(p) + λFp0

(p)T,

where λ is the (row) vector of Lagrange multipliers. In the point (p2,λ2) of local extremum itsderivative is zero:

∂L(p2,λ2)

∂p= g′p0

(p2) + λ2F′p0(p2) = 0

∂L(p2,λ2)

∂λ= Fp0

(p2) = 0.

This cannot possibly be true since the first equation tells that g′p0(p2) is a linear combination of

the rows of F′p0(p2). This contradiction proves that p2 ∈ ∂ s

MA ∩ B1.

Finally we can now define what would be a proper region of integration. A subset A of ak-dimensional manifoldM of Rn is its region with boundary, if

(1) A is a closed subset of Rn.

All points of the boundary ∂MA are there in A.


(2) The set ∂MA − ∂ sMA (i.e. nonsmooth points of the

boundary) is a k − 1-null-set of Rn, see Section 3.1.Nonsmooth points do not affect integrals over theboundary of A.

This condition is needed, because it is even possiblethat all points of the boundary ∂MA are nonsmooth.E.g., in the figure on the right is the so-called Koch

snowflake, its boundary is everywhere nonsmooth andhas infinite length.

(3) For each positive number ǫ (however small) thereis an open subset Bǫ such that Bǫ contains all non-smooth points of the boundary ∂MA and that the k−1-dimensional volume of ∂ s

MA∩Bǫis < ǫ. Note that ∂ s

MA ∩ Bǫ is a k − 1-dimensional manifold by Theorem 2.2. Thisstrange-looking condition is needed because it is possible that the k − 1-dimensionalvolume of the smooth boundary is = ∞, even when restricted to arbitrarily small opensubsets containing a smooth point. In some sense the condition prevents integrals beingimproper.

For instance, the obviously nonsmooth bound-ary points of the subset (of R2) in the figure onthe right (Maple) are the two vertices, and theorigin where the two spiralling curves

r =1

φand r =

1

1 + φ(1 ≤ φ <∞)

(hyperbolic spirals, in polar coordinate param-etrization) converge to. Now conditions (1) and(2) are satisfied, but not the above condition,since the spirals have infinite length in any openset containing the origin.

An important rôle in integration is played by orientation of the boundary ∂ sMA and the way

it is related to the orientation of the manifoldM. Of course, the boundary can be oriented onlyin its smooth points.

In the smooth point p0 a vector t of the tangent space Tp0(M) of the manifoldM is called

an exterior vector of A, ifg′p0

(p0) • t < 0,

and an interior vector of A, ifg′p0

(p0) • t > 0.

These names are apt. The function g increases when we move locally near the point p0 overthe boundary of the region A from outside to inside. Outside gp0

has a negative value, inside apositive one, and in the boundary it is = 0. Thus, if

g′p0(p0) • t < 0,

then the direction of the vector t is opposite to the direction of increasing gp0, that is, away from

the region A. Similarly, ifg′p0

(p0) • t > 0,


then the direction of the vector t is the direction of increasing gp0, into the regionA.

Using exterior vectors we can couple an orientation for the smooth boundary ∂ sMA and an

orientation ofM. Cf. Section 4.3. Consider then the case of a manifoldM oriented by a k-formfield Φ. Thus in its points p vectors

t1(p) , . . . , tk(p),

of its tangent space Tp(M) can be chosen such that

Φ(

p; t1(p), . . . , tk(p))

is continuous with respect to p and 6= 0 inM.Orienting the boundary ∂ s

MA in its smooth points p then means choosing some k − 1-formfield Φboundary, the smooth boundary here being a k − 1-dimensional manifold, and vectors

s1(p) , . . . , sk−1(p)

of the tangent space Tp(∂ sMA) (depending on p) such that

Φboundary

(

p; s1(p), . . . , sk−1(p))

is continuous with respect to p in the smooth boundary and 6= 0. We choose

Φboundary(p; s1, . . . , sk−1) = Φ(

p; texterior(p), s1, . . . , sk−1)

,

where texterior(p) is a suitable exterior vector depending on p. This choice couples orientationsof the boundary ∂ s

MA and the manifoldM. This oriented boundary of a region with boundary

of an oriented manifold−→M is denoted by

−→∂ s−→MA.

Note that the vectorss1(p), . . . , sk−1(p)

are in the tangent space of the manifoldM in the point p, so that the form field Φ is available.Indeed, the tangent space of of the smooth boundary ∂ s

MA in the point p is the null space of thematrix

(

F′p0(p)

g′p0(p)

)

,

cf. Theorem 2.8 and the proof of Theorem 5.1. Thus it is included in the null space of the matrixF′p0

(p), which again is the tangent space of the manifoldM. What then is needed is a basis

texterior(p), s1(p), . . . , sk−1(p)

for the tangent space ofM in the points p of its smooth boundary having the same ”handidness”as the basis t1(p), . . . , tk(p).

Example. A parametrized smooth space curve

C : p = γ(u) (u ∈ U)

is oriented using a 1-form field

Φ(p; r) = F(p) • rchoosing

F(

γ(u))

= γ ′(u).


The resulting 1-form has the value∥

∥γ ′(u)∥

∥

2> 0.

The smooth boundary of a region with orienta-

tionA of C (part of the curve) consists of disjoint

points. A ”positive” orientation in the smooth

point p = γ(u) of the boundary takes place by

the form field condition

Φboundary(p) = γ ′(u) • texterior(p) > 0.

In particular, if the smooth boundary of A con-x

yp

z

texterior(p)A

sists of two points, the initial point and the terminal point, the initial point is ”negative” and

the terminal point is ”positive”. See the figure above.

Example. A smooth parametrized surface

S : r = γ(u) (u ∈ U)

is oriented using a 2-form field

Φ(p; r1, r2) = n(p) • r1 × r2,

where

n(p) = n(

γ(u))

=∂γ(u)

∂u1× ∂γ(u)

∂u2

x

y

zn(p)

s(p)

texterior(p)

Ap

is the normal vector in the point p. The value of the 2-form is then

∥

∥

∥

∂γ(u)

∂u1× ∂γ(u)

∂u2

∥

∥

∥

2

and it is always positive. The smooth boundary of a region with orientation A of S consists of

smooth space curves, and it is oriented in a smooth point p = γ(u) by the form field condition

Φboundary

(

p; s(p))

= n(p) • texterior(p)× s(p) > 0,

where s(p) is the tangent vector of the boundary (curve) in the point p. The vectors

n(p) , texterior(p) and s(p)

then form a right-handed system, and the orientation is given by the familiar ”right hand rule”,

see the figure below.

Example. The tangent space of a 3-dimensional manifold (or open subset) M of R3 always

consists of the whole vector space R3. A region with boundary (or a solid body) A ofM can

be oriented, say, by the 3-form field condition

Φ(p; r1, r2, r3) = det(r1, r2, r3) > 0.

The boundary ∂ sMA consists of smooth surfaces. The exterior vector in the point p could be

the exterior normal n(p) of the surface, and the tangent vectors s1(p) and s2(p) could be the

spanning vectors of its tangent plane. If the orienting condition is

det(

n(p), s1(p), s2(p))

= n(p) • s1(p)× s2(p) > 0,


then it is all about orienting the boundary (surface) in such a way that the normal vector n(p)is the exterior normal, and that

n(p) , s1(p) and s2(p)

form a right-handed system.

Example. Similarly, orienting an n-dimensional manifoldM of Rn and the smooth boundary

of its region with boundaryA can be done by the condition

det(

n(p), s1(p), . . . , sn−1(p))

> 0,

where n(p) is a vector (exterior normal) of the normal space and s1(p), . . . , sn−1(p) form a

basis of the tangent space of the smooth boundary ∂ sMA in the point p.

5.2 Exterior Derivatives

We still need a sufficiently general derivative, the so-called exterior derivative. The usualderivative of a univariate function f is defined as the limit of the difference quotient:

f ′(x) = limh→0

f(x+ h)− f(x)h

.

In basic courses of calculus also the corresponding 1-form field, the so-called differential

(df)(x; r) = f ′(x)r,

is introduced, and it, too, can be defined as a difference quotient:

f ′(x)r = limh→0

f(x+ hr)− f(x)h

.

The exterior derivative is an extension of this for k-form fields. Note in particular that thefunction f is a 0-form field and that the differential df is a 1-form field. Taking an exteriorderivative increases the degree of a form by one.

The interval[x, x+ h] (or [x+ h, x])

appearing in the difference quotient is replaced by a k + 1-dimensional (closed) parallelepiped

P(p; r1, . . . , rk+1) =

p+k+1∑

i=1

uirT

i

∣

∣

∣

∣

0 ≤ u1, . . . , uk+1 ≤ 1

.

For the real line this is a closed interval, in the plane R2 a parallelogram and in the space R

3

a ”geometric” parallelepiped. Here k + 1 ≤ n, so we could be dealing with a parallelepipedembedded in a higher-dimensional space.

Embedded in the affine subspace

A(p; r1, . . . , rk+1) =

p+

k+1∑

i=1

uirT

i

∣

∣

∣

∣

u1, . . . , uk+1 ∈ R


of Rn (a k + 1-dimensional manifold of Rn) the parallelepiped P(p; r1, . . . , rk+1) is a regionwith boundary where the boundary consists of faces. There are 2(k + 1) faces and they appearin pairs, the ith pair is

T −i =

p+

k+1∑

i=1

uirT

i

∣

∣

∣

∣

0 ≤ u1, . . . , uk+1 ≤ 1 and ui = 0

and

T +i =

p+

k+1∑

i=1

uirT

i

∣

∣

∣

∣

0 ≤ u1, . . . , uk+1 ≤ 1 and ui = 1

.

In the former face ri is an interior vector and in the latter it is an exterior vector. These facesthus have opposite orientations, and it is agreed thatA(p; r1, . . . , rk+1) is oriented according tothe orientations of the faces T +

i by the exterior vectors ri.The smooth boundary consists of faces minus their edges. By Theorem 3.3 the edges are

k-null-sets, and by Theorem 5.1 the smooth boundary is a k-dimensional manifold. The orientedsum in the difference quotient, where f(x+ hr) has a plus-sign and f(x) a minus-sign, is nowreplaced by an integral over the oriented (smooth) boundary of the parallelepiped. Alternationof the signs in included, too, since the faces T −i and T +

i always have opposite orientations.We thus get the exterior derivative of a k-form field Φ as

(dΦ)(p; r1, . . . , rk+1) = limh→0

1

hk+1

∫

−→∂ −→

APh

Φ,

where, for brevity, we denoted

A = A(p; r1, . . . , rk+1) and Ph = P(p; hr1, . . . , hrk+1).

Note that the integration region can be given by a relaxed parametrization, so that the integralis of the form in Section 4.2. (As an oriented sum if k = 0.) In addition it is agreed that ifk + 1 > n, then the exterior derivative is = 0.

Exterior derivative calculation is tedious starting from the very definition, as often is calcu-lation of the usual univariate derivative, too. To make things simpler, differentiation rules canbe given. Together they form a sufficiently extensive toolbox to make derivative calculationsrelatively easy. Proving these rules then can be quite tedious.

Let us give a basic collection of differentiation rules. The vector variables r1, . . . , rk aremostly omitted for brevity.

(I) If Φ and Ψ are k-form fields, and c1 and c2 are constants, then

d(c1Φ+ c2Ψ) = c1(dΦ) + c2(dΨ).

Taking an exterior derivative thus is a linear operation. This follows directly from thedefinition by the integral.

(II) If a k-form field Φ is constant (i.e. a constant form), then dΦ is a k + 1-form field havingonly the value 0 (i.e. a zero form field).

As is usual, derivative of a constant is zero. This, too, follows directly from the definitionby the integral, since as a consequence of the opposite orientations of the pairs of facesthe integral then is always = 0. (The integrand having the same value in the matchingpoints of the faces.)


(III) The exterior derivative of a function f or a 0-form field is

(df)(p; r) = f ′(p)r =n∑

i=1

∂f(p)

∂xidxi(r).

According to the definition this exterior derivative is the limit

limh→0

f(p+ hrT)− f(p)h

=( d

dtf(p+ trT)

)

t=0= f ′(p)r

(by the chain rule). In particular, if ‖r‖ = 1, this is the usual directional derivative in thedirection of r.

(IV) If f is a function, then

d(f(p) dxj1 ∧ · · · ∧ dxjk) = (df)(p; ·) ∧ dxj1 ∧ · · · ∧ dxjk

=n∑

i=1

∂f(p)

∂xidxi ∧ dxj1 ∧ · · · ∧ dxjk .

This is the key rule since, combined with the previous rules, it makes it possible to cal-culate the exterior derivative starting from the expansion of the form field as a linearcombination of elementary form fields, and in fact also shows that the exterior derivativeis a k + 1-form field. The proof of this rule (even via Theorem 6.1) is a bit tedious, seee.g. HUBBARD & HUBBARD.

(V) The k+ 2-form d2Φ = d(dΦ) field obtained by taking the exterior derivative of a k-formfield Φ twice is a constant form field having the value 0 (a zero form field).

Form fields are thus always ”of the first degree” in a sense. Differentiating twice alwaysgives zero. This is a consequence of the previous rules, since if f is a function, then

(d2f)(p; ·, ·) =n∑

i=1

d(∂f(p)

∂xidxi

)

=

n∑

j=1

n∑

i=1

∂2f(p)

∂xj∂xidxj ∧ dxi = 0.

(Recall that dxi ∧ dxj = −dxj ∧ dxi and dxi ∧ dxi = 0.)

Using these rules calculation of the exterior derivative of a k-form field

Φ(p; r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(p)(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk)

is basically straightforward and is brought back to calculation of partial derivatives.

Note. As a matter of fact, the whole exterior derivative could be defined via these rules. It is,

however, nice to see that it is a generalization of the ordinary univariate derivative. And the

definition by the integral does give a ”geometric” way of considering the exterior derivative,

not just a symbolical machinery.

Example. Let us calculate the exterior derivative of the 2-form field

Φ = x1x2 dx2 ∧ dx4 − x22 dx3 ∧ dx4


of R4. (For brevity p = (x1, x2, x3, x4) as well as r1 and r2 are omitted.) Applying the rules we

get

dΦ =(I) d(x1x2 dx2 ∧ dx4)− d(x22 dx3 ∧ dx4)

=(IV)

(∂(x1x2)

∂x1dx1 +

∂(x1x2)

∂x2dx2 +

∂(x1x2)

∂x3dx3 +

∂(x1x2)

∂x4dx4

)

∧ dx2 ∧ dx4

−(∂(x22)

∂x1dx1 +

∂(x22)

∂x2dx2 +

∂(x22)

∂x3dx3 +

∂(x22)

∂x4dx4

)

∧ dx3 ∧ dx4

= (x2 dx1 + x1 dx2) ∧ dx2 ∧ dx4 − (2x2 dx2 ∧ dx3 ∧ dx4)

= x2 dx1 ∧ dx2 ∧ dx4 + x1 dx2 ∧ dx2 ∧ dx4 − 2x2 dx2 ∧ dx3 ∧ dx4

= x2 dx1 ∧ dx2 ∧ dx4 − 2x2 dx2 ∧ dx3 ∧ dx4.

For exterior derivatives of wedge products of form fields we have the famous

Cartan’s Magic Formula.2 The exterior derivative of the wedge product of a k-form field Φand an l-form field Ψ is given by

d(Φ ∧Ψ) = (dΦ) ∧Ψ+ (−1)kΦ ∧ (dΨ).

Proof. By the definition of wedge product it suffices to show the rule for elementary form fields

Φ = a(p) dxj1 ∧ · · · ∧ dxjk and Ψ = b(p) dxh1∧ · · · ∧ dxhl

and their wedge product

Φ ∧Ψ = a(p)b(p) dxj1 ∧ · · · ∧ dxjk ∧ dxh1∧ · · · ∧ dxhl

.

By rule (IV)

d(Φ ∧Ψ) =

n∑

i=1

∂(

a(p)b(p))

∂xidxi ∧ dxj1 ∧ · · · ∧ dxjk ∧ dxh1

∧ · · · ∧ dxhl

=

n∑

i=1

(∂a(p)

∂xib(p) +

∂b(p)

∂xia(p)

)

dxi ∧ dxj1 ∧ · · · ∧ dxjk ∧ dxh1∧ · · · ∧ dxhl

=(

n∑

i=1

∂a(p)

∂xidxi ∧ dxj1 ∧ · · · ∧ dxjk

)

∧(

b(p) dxh1∧ · · · ∧ dxhl

)

+(

a(p) dxj1 ∧ · · · ∧ dxjk)

∧ (−1)k(

n∑

i=1

∂b(p)

∂xidxi ∧ dxh1

∧ · · · ∧ dxhl

)

= (dΦ) ∧Ψ+ (−1)kΦ ∧ (dΨ).

2La ”formule magique de Cartan”. Élie Cartan was the father of form theory.


5.3 Exterior Derivatives of Physical Form Fields

As noted in Section 4.4, the basic form fields of a vector field F(p) in R3 are the work form

fieldΦF–work(p; r) = F(p) • r (a 1-form field)

and the flux form field

ΦF–flux(p; r1, r2) = det(

F(p), r1, r2)

= F(p) • r1 × r2

(a 2-form field), and for a scalar field f(p) the basic form field (in addition to itself as a 0-formfield) is the density form field

Φf–density(p; r1, r3, r3) = f(p) det(r1, r2, r3)

(a 3-form field). The connection between these basic form fields and exterior derivatives is thefollowing:

(A) df = Φ∇f–work (df is the work form field of the gradient ∇f .)

(B) dΦF–work = Φ∇×F–flux (dΦF–work is the flux form field of the curl ∇× F.)

(C) dΦF–flux = Φ∇•F–density (dΦF–flux is the density form of the divergence∇ • F.)

All these can be proved by straightforward calculation. (A) follows immediately from rule (III).(B) is shown as follows:

dΦF–work = d(F1 dx+ F2 dy + F3 dz) = (dF1) ∧ dx+ (dF2) ∧ dy + (dF3) ∧ dz

=(∂F1

∂xdx+

∂F1

∂ydy +

∂F1

∂zdz)

∧ dx+(∂F2

∂xdx+

∂F2

∂ydy +

∂F2

∂zdz)

∧ dy

+(∂F3

∂xdx+

∂F3

∂ydy +

∂F3

∂zdz)

∧ dz

=(∂F3

∂y− ∂F2

∂z

)

dy ∧ dz +(∂F1

∂z− ∂F3

∂x

)

dz ∧ dx+(∂F2

∂x− ∂F1

∂y

)

dx ∧ dy

= Φ∇×F–flux.

Item (C) is similar (left as an exercise for the reader).The whole thing can be given as a condensed scheme:

scalar field 0-form field

ygradient

yd

vector fieldwork form field−−−−−−−−−→ 1-form field

ycurl

yd

vector fieldflux form field−−−−−−−−→ 2-form field

ydivergence

yd

scalar fielddensity form field−−−−−−−−−−→ 3-form field


According to the scheme, curl of a gradient corresponds to exterior derivation twice, as doesdivergence of a curl. By the rule (V) these are zero form fields, and indeed

∇×∇f = 0 and ∇ • (∇× F) = 0.

On the other hand, the other second order derivatives ∇ • (∇f) = ∇2f = ∆f (Laplacian),∇(∇•F) and∇× (∇×F) (double curl) do not correspond to second exterior derivatives, andin general do not vanish.

Cartan’s Magic Formula gives fairly directly nabla formulas (i), (iii), (iv) and (v) in Section1.5 since, as noted in Section 4.4, for scalar fields f and g, and vector fields F and G we have

f ∧ g = fg , ΦF–work ∧ ΦG–work = ΦF×G–flux ,

f ∧ ΦF–work = ΦfF–work , ΦF–work ∧ ΦG–flux = ΦF•G–density ,

f ∧ ΦF–flux = ΦfF–flux.

For instance, the formula

(v) ∇ • F×G = ∇× F •G− F • ∇ ×G

is obtained as follows:

Φ∇•F×G–density = dΦF×G–flux = dΦF–work ∧ ΦG–work + (−1)1ΦF–work ∧ dΦG–work

= Φ∇×F–flux ∧ ΦG–work − ΦF–work ∧ Φ∇×G–flux

= Φ∇×F•G–density − ΦF•∇×G–density

= Φ(∇×F•G−F•∇×G)–density.

Note. Formulas (vi) and (vii) in Section 1.5, on the other hand, are of a quite different form,

and do not follow directly from the Magic Formula. They are more closely related to another

rule of derivation given in Theorem 6.1.

Let us then give some physical intuition for the differential operators. Thinking about thedefinition of the exterior derivative as the limit of the integral in the previous section, it maybe noticed that in a very small parallelogram Ph(p; r1, r2) (where h is small) embedded in R

3

hr1 x hr2

hr1

hr2

p Ph

hr1

hr2

hr3

Ph

p

exterior normal

the scaled projection of the curl on the normal of the parallelogram (see the figure on the leftabove) is

(

∇× F(p))

• (hr1)× (hr1) ∼=∫

−→∂ −→

APh

ΦF–work =

∫

U

F(

γ(u))

• γ ′(u) du =

∮

−→∂ −→

APh

F(r) • ds,


where the (relaxed) parametrization of−→∂ −→APh is r = γ(u) (u ∈ U). The projection thus is

approximatively the line integral of the vector field around the perimeter of the parallelogramin the direction given by the right hand rule. For this reason the curl is sometimes also calledvorticity or vortex density.

Similarly in a very small (meaning h is small) parallelepiped Ph(p; r1, r2, r3) of R3 the

scaled divergence is

(

∇ • F(p))

det(hr1, hr2, hr3) ∼=∫

−→∂ −→

APh

ΦF–flux =

∮

−→∂ −→

APh

F(r) • dS.

Locally the divergence thus is the flux through the boundary of a very small parallelepiped fromthe inside to the outside, see the figure on the right above. This could be a change of mass ofa fluid in the parallelepiped. For this reason the divergence is also called source density. It is atypical density since det(hr1, hr2, hr3) is the volume of the parallelepiped.

Combining the Hodge star (see Section 4.4) and exterior derivation gives a handy tool, anda way to get higher derivatives. We see (check!) that in R

3

d∗ΦF–work = Φ∇•F–density and ∗d∗ΦF–work =∇ • F ,

d∗d∗ΦF–work = Φ∇(∇•F)–work and ∗d∗d∗ΦF–work = Φ∇(∇•F)–flux ,

d∗ΦF–flux = Φ∇×F–flux and ∗d∗ΦF–flux = Φ∇×F–work ,

d∗d∗ΦF–flux = Φ∇×(∇×F)–flux and ∗d∗d∗ΦF–flux = Φ∇×(∇×F)–work ,

d∗Φf–density = Φ∇f–work and ∗d∗Φf–density = Φ∇f–flux ,

d∗d∗Φf–density = Φ∆f–density and ∗d∗d∗Φf–density = ∆f.

Electro-magnetic fields are governed by the famous Maxwell equations:

(M1) ∇ •D = ρ (Gauß’ law)

(M2) ∇ •B = 0 (Gauß’ law for magnetism)

(M3) ∇× E = −∂B∂t

(Faraday’s law)

(M4) ∇×H = J+∂D

∂t(Ampère’s law)

The various fields are connected by the so-called matter laws3

D = εE , B = µH , J = σE (Ohm’s law),

and

∇ • J = −∂ρ∂t

(continuity equation).

As noticed in Section 4.4 these really are not physical vector fields, and certain 2-form fieldswould seem to better fit the bill. Let us compute the exterior derivatives of these form fields.

3We are dealing with the isotropic case only. Here E is the electric field [V/m], D is the electric flux density[As/m2], H is the magnetic field [A/m], B is the magnetic flux density [Vs/m2] (tesla), ε is the permittivity[As/V/m], µ is the permeability [Vs/A/m], ρ is the charge density [As/m3], J is the current density [A/m2], and σ

is the conductivity [A/V/m].


First the Faraday 2-form field:

dΦFaraday = d(ΦE–work ∧ dt) + dΦB–flux

= d(E1dx ∧ dt) + d(E2dy ∧ dt) + d(E3dz ∧ dt)

+ d(B1dy ∧ dz) + d(B2dz ∧ dx) + d(B3dx ∧ dy)

= Φ∇×E–flux ∧ dt + Φ∇•B–density

+∂B1

∂tdt ∧ dy ∧ dz +

∂B2

∂tdt ∧ dz ∧ dx+

∂B3

∂tdt ∧ dx ∧ dy

=(M2) Φ∇×E–flux ∧ dt + Φ∂B∂t

–flux ∧ dt =(M3) 0.

The exterior derivative of the Maxwell 2-form field is obtained similarly:

dΦMaxwell = d(−c2ΦB–work ∧ dt) + dΦE–flux

= −c2Φ∇×B–flux ∧ dt + Φ∇•E–density + Φ∂E∂t

–flux ∧ dt.

Assuming constant ε and µ (homogeneity), recalling that then

c =1√εµ,

and applying the matter laws and (M1) and (M4), the exterior derivative is simplified (check!)to

dΦMaxwell =1

ε(Φρ–density − ΦJ–flux ∧ dt).

This exterior derivative, the so-called current density form field, is often denoted as a Hodgedual ∗J . (What is then J?). Second exterior derivatives being zeros, we have then also

d∗J = 0 (charge conservation law).

All in all, we then note that (in the homogeneous case) Maxwell’s equations can be writtenusing the Faraday form field and the Maxwell form field in an utterly simple form:

dΦFaraday = 0

dΦMaxwell = ∗J.

(We did need all Maxwell’s equations in deriving these two equations!)


5.4 Generalized Stokes’ Theorem

We have now collected all pieces for

Generalized Stokes’ Theorem. If−→M is a k + 1-dimensional oriented manifold of Rn, A its

bounded region with boundary, the smooth boundary−→∂ s−→MA correspondingly oriented, and Φ a

k-form field, continuously differentiable in some open subset containingA, then4

∫

A

dΦ =

∫

−→∂ s

−→MA

Φ.

In case the smooth boundary−→∂ s−→MA is empty, the right-hand side integral is = 0.

Proof. A rigorous proof consists of a long and complicated estimation5, see e.g. HUBBARD &HUBBARD. The rough idea however is the following. We approximateA by a set of very smallk + 1-dimensional parallelepipeds6:

Pj = Pj(pj ; rj,1, . . . , rj,k+1) (j = 1, . . . , N).

Parametrized these are

Pj : γj(u) = pj +k+1∑

i=1

uirT

j,i (0 ≤ u1, . . . , uk+1 ≤ 1),

whenceγ ′j(u) =

(

rj,1, · · · , rj,k+1

)

.

The larger the number of parallelepipeds is, the better the approximation:

4This is often written in the short form∫

−→A

dΦ =

∫

−→∂ A

Φ or even∫

A

dΦ =

∫

∂A

Φ

assuming all conventions about orientations, boundaries, etc.5But cf. the example in Section A2.2.6Why not just triangulation or similar? A well-known counterexample in R

2 is Schwarz’s lantern.


(figure by Maple). Thus, by the definition of the parametrized integral as a Riemann integral,

∫

A

dΦ ∼=N∑

j=1

(dΦ)(pj ; rj,1, . . . , rj,k+1).

On the other hand, by the definition of the exterior derivative (taking h = 1, since the paral-lelepipeds are very small),

(dΦ)(pj ; rj,1, . . . , rj,k+1) ∼=∫

−→∂ s

−→MPj

Φ.

Exterior normals of faces shared by adjacent parallelepipeds are opposite, i.e. the faces haveopposite orientations. Thus integrals over such faces are cancelled, and only integrals overfaces approximating the smooth boundary of A remain. So

N∑

j=1

∫

−→∂ s

−→MPj

Φ ∼=∫

−→∂ s

−→MA

Φ.

Proving convergence of the various approximations is quite tedious.

Let us take the standard integral theorems of vector analysis as examples,

Example. (Gauß’ Theorem or Divergence Theorem) As the manifold take an open subsetMof R3 (oriented by the right hand rule), and as the region a correspondingly oriented subset (a

solid)−→K whose smooth boundary

−→∂ s−→MK is oriented by exterior normals. Then

dΦF–flux = Φ∇•F–density,

and so, according to the Generalized Stokes Theorem,

∫

−→K

∇ • F(r) dr =∫

−→K

Φ∇•F–density =

∮

−→∂ s

−→MK

ΦF–flux =

∮

−→∂ K

F(r) • dS

(note the brief notation). It is important here that the exterior normal corresponds to the right-

handedness of the coordinate system.

Example. (General Divergence Theorem) The Divergence Theorem is valid in any dimension

in the form∫

−→K


∮

−→∂ s

−→MK

ΦF–flux

(cf. also the example in Section A2.2). The 1-dimensional version is the usual Fundamental

Theorem of Integral Calculus.

In particular the Divergence Theorem is valid in two dimensions. Then

ΦF–flux = −F2(p)dx+ F1(p)dy

and so

d(

− F2(p)dx+ F1(p)dy)

= ∇ • F(p) dx ∧ dy.


Thus, according to the Generalized Stokes Theorem,

∫

−→A


∫

−→A

∇ • F(r) dr =∮

−→∂ A

(

−F2(r)F1(r)

)

• ds.

Here the manifold is an open subset of R2, and the regionA is its subset whose smooth boundary

is oriented by exterior normals. Note (check!) that then integration around the boundary (curve)

ofA is in the counterclockwise direction. The right hand side is sometimes written in one of the

forms∮

−→∂ A

F(r) • dn =

∮

−→∂ A

−F2(r) dx+ F1(r) dy =

∮

−→∂ A

∣

∣

∣

∣

F1(r) dxF2(r) dy

∣

∣

∣

∣

.

Example. (Green’s Theorem) Applying the 2-dimensional Divergence Theorem above to the

vector field

G(r) =

(

F2(r)−F1(r)

)

(i.e., F(r) rotated by −90) we get Green’s Theorem:

∫

−→A

(∂F2

∂x− ∂F1

∂y

)

dr =

∫

−→A

∇ •G(r) dr =

∮

−→∂ A

(

−G2(r)G1(r)

)

• ds =∮

−→∂ A

F(r) • ds.

Again integration around the boundary (curve) of A is in the counterclockwise direction (cor-

responding to the exterior normals).

Example. (Stokes’ Theorem) As the manifold take an oriented surface−→S of R3, and as the

region its correspondingly oriented part−→A whose smooth boundary

−→∂ s−→SA is oriented by the

right-hand rule. Since

dΦF–work = Φ∇×F–flux,

then according to the Generalized Stokes Theorem

∫

−→A

∇× F(r) • dS =

∫

−→A

Φ∇×F–flux =

∮

−→∂ s

−→SA

ΦF–work =

∮

−→∂ A

F(r) • ds.

In case A = S is a closed surface having no smooth boundary, the right hand side is = 0.

These formulations of the classical integral theorems are fairly general, relaxed parametriza-tions and sufficiently continuous integrands is more or less what is needed. Manifolds and theirregions with boundaries may consist of several separate parts, as may their smooth boundaries.Note however, that in Stokes’ Theorem the surface S must be orientable, so e.g. the Möbiusband does not qualify.

There are also nonorientable closed surfaces, e.g. the so-called Klein bottles, a version ofwhich is depicted below (Maple). This surface intersects itself, in R

4 there however are versionsof Klein’s bottles which do not. Gauß’ Theorem is not applicable to Klein’s bottles.


These classical integral theorems are notthe only ones, other similar results can beobtained by applying the Generalized StokesTheorem to form fields of various kinds.

Example. (Vectoral Gauß’ Theorem) If in

the 3-form field

f(p)r1 • r2 × r3

the vector variable r1 is fixed as a constant

vector a, we get the 2-form field

Φ(p; r2, r3) =(

f(p)a)

• r2 × r3

(flux form field). Applying the Generalized Stokes Theorem to this, as we did for the Gauß’

Theorem, we get

∫

−→K

Φ∇•(fa)–density =

∫

−→K

∇ •(

f(r)a)

dr =

∮

−→∂ K

(

f(r)a)

• dS.

Since

∇ •(

f(r)a)

= ∇f(r) • aand a was an arbitrary constant vector (choosing a = i, j,k we get the coordinates), this result

is often written in the vectoral form

∫

−→K

∇f(r) dr =∮

−→∂ K

f(r)dS.

The classical integral theorems, and others, find their use in the manipulation of physicalformulas.

Example. By the Maxwell equation (M4), in the stationary case

∇×H = J.

If the closed curve−→C (loop) is sufficiently smooth and oriented, there is an oriented surface

−→S ,

having−→C as its oriented boundary (via the right hand rule) and satisfying the assumptions of

Stokes’s Theorem, and

∮

−→C

H(r) • ds =∫

−→S

∇×H(r) • dS =

∫

−→S

J(r) • dS.

The left hand side is the integral of the magnetic field around the loop, and the right hand side

tells it is equal to the net current through the loop. This is the so-called Ampère Piercing law.

Example. Similarly, by the Maxwell equation (M3)

∇×E = −∂B∂t.


So, as above, around a loop satisfying sufficient assumptions

∮

−→C

E(r, t) • ds =∫

−→S

∇× E(r, t) • dS = −∫

−→S

∂B(r, t)

∂t• dS = − d

dt

∫

−→S

B(r, t) • dS.

Thus the electromotive force around a loop equals the time derivative of the net magnetic flux

through the loop.

Example. As a first approximation—and often as a last one, too—in many fluid dynamics prob-

lems the fluid is assumed to be ideal, inviscid and incompressible. Let us consider a stationary

flow. The so-called Thomson (Kelvin) circulation law tells us that, with certain assumptions, if

the curl of the velocity is a zero vector initially it will remain so afterwards. Thus then all the

time

∇× v = 0.

If C is a closed curve (loop) and a boundary of a surface satisfying the assumptions of the Stokes

Theorem, and v is defined and continuously differentiable, then the integral around the loop is

zero:∮

−→C

v(r) • ds = 0.

Especially in fluid flow mod-

elling it is often not possible to

assume existence of such a surface,

even a virtual one. The reason is

mostly an obstacle in the flow, a

solid object, e.g. a wing, inside

which there is no fluid and of course

no flow either. (In electric fields a

similar situation occurs when there

is insulating material or a hole

inside conducting material.) Wings

z

x

y

C

x

y

C

z

and other obstacles in the flow can often be assumed to be infinitely long, say in the z-axis

direction, which means a loop encircling the obstacle cannot be the boundary of a surface

where the flow velocity is defined.

If, however, it is known the flow velocity satisfies ∇× v = 0, it can be shown that for any

loop encircling the obstacle, the line integral

Γ =

∮

v(r) • ds

always has the same value, and this can then be calculated using the best loop for the purpose.

This is important e.g. when the Zhukovsky7 lift law is applied. The law tells the lift per unit of

length of an airfoil is

F = ρΓv∞

for air of density ρ flowing from infinity with speed v∞. The force is perpendiculat to the velocity,

as is familiar from airplane wings.

Independence of the value of the line integral of the loop is seen roughly as follows. If−→C1 and−→C2 are two loops encircling the same (infinitely long) obstacle, they can be cut and

7Often Joukowsky or Joukowski or Schukowski etc.!


recombined using a double connecting line L to a single closed oriented curve−→C as in the

figure below. (The connecting line is drawn here as two lines for clarity.) The closed loop

C then is a boundary curve of a surface where the velocity v is defined, and since the curl

vanishes,∮

−→C

v(r) • ds = 0.

On the other hand, in the line integral

over−→C the connecting line L is traversed

twice in opposite directions. These two

integrals over L cancel each other, and

so (think about the orientations)∮

−→C

v(r) • ds =∮

−→C2

v(r) • ds−∮

−→C1

v(r) • ds = 0.

x

y

C1

z

C2

x

y

C

z

L

Example. In an even-dimensional space R2m the 2-form field

Φ =

m∑

i=1

dxi ∧ dxm+i

is called the (standard) symplectic form. We see immediately that Φ = dΨ where

Ψ =

m∑

i=1

xi dxm+i.

Ψ is called the (standard) symplectic potential8 (see the next chapter). Consider then a

2-dimensional oriented manifold−→M of R2m and its bounded region A, where the boundary

of A has the (relaxed) parametrization

C : r =(

γ1(u),γ2(u))

(u ∈ U)

and γ1 contains the first m coordinates. By the Generalized Stokes Theorem then

∫

−→A

Φ =

∫

−→A

dΨ =

∮

−→C

Ψ =

∫

U

γ1(u)γ′2(u) du.

In R2 this also follows from Green’s Theorem (and then the integral equals the area of A).

Symplectic forms are important in mechanics where the phase spaces may have a quite high

(even) dimension.

Note. The foremost problem in results like these is that in our formalism the regions of inte-

gration must in principle be embedded in larger manifolds, and that the fields must exist (as

continuous or continuously differentiable) in regions larger than the region of integration. In

many practical situations such extensions are not readily available, not even theoretical or ar-

tificial ones (cf. the famous Nash Embedding Theorem etc.).

8And also Liouville’s 1-form, Poincaré’s 1-form, the canonical 1-form or the tautological 1-form, etc. As theysay: ”A dear child has many names.”

Chapter 6

POTENTIAL

”Truth in the world resides only inmathematical proofs and physics labs.”

(ANTONIO GARCÍA MARTÍNEZ: Chaos Monkeys)

"Mathematics is not an absolute measure ofthe natural world, but a relativistic parallel.”

(BRIAN CLEGG: The Reality Frame)

6.1 Exact Form Fields and Potentials

For definite integrals the Generalized Stokes Theorem gives a characterization quite similar tothe familiar Fundamental Theorem of Integral Calculus. But what about indefinite integrals?From basic calculus we know that

d

dx

∫f(x) dx =

∫df(x)

dxdx = f(x)

which says indefinite integral is the inverse derivative, that is, the antiderivative. For form fieldsthis then would correspond to inverse exterior derivative. Such an operation would come handyin many cases, since exterior differentiation increases the degree of the form field by one, andthus makes it more complex.

A form field Φ is said to be exact if it is the exterior derivative of another form field Ψ, i.e.Φ = dΨ. The form field Ψ is then called the potential of Φ. Since always dΓ = 0, wheneverΓ is a constant form field (rule (II) in Section 5.2), then Ψ + Γ is another potential of Φ. Apotential thus is not unique, but then neither is an indefinite integral. And often there are evenmore potentials of Φ.

Evidently, if Φ is exact, then dΦ = 0, because

dΦ = d2Ψ = 0.

Thus not all form fields are exact. On the other hand, as will be seen shortly, if dΦ = 0, then atleast locally Φ is exact.1

In this chapter, to not clutter the notation too much, we do not consistently use the columnarray notation for vectors. Note, however, that mostly the rules of thumb ”point minus point isa vector” and ”point plus vector is a point” hold.

Example. The Newton form field

Φ(p; r) = −C (p− p1) • r‖p− p1‖3

(a 1-form field of R3), where C is a constant and p1 a fixed point, is exact in any region not

containing p1. (The difference p−p1 is here interpreted as a vector.) It is the exterior derivative

of the scalar field (0-form field)

f(p) =C

‖p− p1‖(Newton’s potential),

1A form field whose exterior derivative is = 0, is often called closed.

83

CHAPTER 6. POTENTIAL 84

i.e. the work form field ∇f(p) • r. On the other hand, we have dΦ = 0, since the exterior

derivative is the flux form field of the curl

∇× −C(p− p1)

‖p− p1‖3= 0 (check!).

Here of course

−C (p− p1)

‖p− p1‖3= ∇f(p)

is the familiar Newton vector field. It will be remembered that according to Newton’s law of

gravitation Earth attracts a point mass m in p by the force

−GmM p− p1

‖p− p1‖3,

where p1 is the center of Earth, M is Earth’s mass, and G is the gravitational constant.

Example. Newton’s vector field can also be connected to the 2-form field

Φ(p; r1, r2) = −C (p− p1) • r1 × r2

‖p− p1‖3.

This form field is exact, too, since it is a flux form field and its exterior derivative is the density

form field of the divergence

∇ • −C(p− p1)

‖p− p1‖3= 0 (check!).

Here, however, existence of a potential may actually depend on the region, as will be seen.

To deal with general potentials one more exterior differentiation rule is needed, in additionto those in Section 5.2. This could also be used as a (coordinate-free) definition of the exteriorderivative.

Theorem 6.1. The exterior derivative of a k-form field Φ can be written as

(dΦ)(p; r1, . . . , rk+1) =

k+1∑

i=1

(−1)i−1Φ′(p; r1, . . . , ri−1, ri, ri+1, . . . , rk+1)ri,

where Φ′ is the derivative of Φ with respect to the point variable p and the circumflex above a

vector variable means that it is omitted from the list.

Proof. By Theorem 4.2, Φ can be expanded as a combination of elementary form fields as

Φ(p; r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(p)(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk),

and it suffices to prove the formula for each term

aj1,j2,...,jk(p) dxj1 ∧ dxj2 ∧ · · · ∧ dxjk .

By rule (IV) of Section 5.2,

d(aj1,j2,...,jk(p) dxj1 ∧ · · · ∧ dxjk

)=

n∑

l=1

∂aj1,j2,...,jk(p)

∂xldxl ∧ dxj1 ∧ · · · ∧ dxjk .


On the other hand, by the definition of an elementary form,

(dxl ∧ dxj1 ∧ · · · ∧ dxjk)(r1, . . . , rk+1) =

∣∣∣∣∣∣∣∣∣

xl,1 xl,2 · · · xl,k+1

xj1,1 xj1,2 · · · xj1,k+1...

.... . .

...xjk,1 xjk,2 · · · xjk,k+1

∣∣∣∣∣∣∣∣∣.

When this determinant is expanded along the first row, we get the sum

k+1∑

i=1

(−1)i−1xl,i(dxj1 ∧ · · · ∧ dxjk)(r1, . . . , ri−1, ri, ri+1, . . . , rk+1).

Exchanging the summings∑n

l=1 and∑k+1

i=1 we then get the differentiation formula.

The general result will be stronger than just a local one: It will deal with the so-called star-shaped sets. A subset A of Rn is said to be star-shaped with respect to the point p0, if whenevera point p is in A, then the whole line segment connecting it to p0, i.e. the set

p0 + t(p− p0)

∣∣ 0 ≤ t ≤ 1,

is in A. Some familiar star-shaped regions (with respect to which points?) are

and some familiar non-star-shaped regions are

The operation corresponding to indefinite integration of a k+1-form field Φ in a star-shapedregion with respect to the point p0 is

(IΦ)(p; r1, . . . , rk) =

1∫

0

tkΦ(p0 + t(p− p0);p− p0, r1, . . . , rk

)dt.

Poincaré’s Lemma. If dΦ = 0 in a star-shaped region A with respect to the point p0, then Φis exact in the region and Φ = d(IΦ).

Proof. The proof is a bit tricky. For brevity of notation, let us just consider the case p0 = 0, thegeneral case is quite similar. Take the coordinate representation

p =

n∑

l=1

plel,

whence by multilinearity

Φ(tp;p, r1, . . . , rk) =n∑

l=1

plΦ(tp; el, r1, . . . , rk)


and by differentiating

(Φ(tp;p, r1, . . . , rk)

)′= q+

n∑

l=1

tplΦ′(tp; el, r1, . . . , rk) = q+ tΦ′(tp;p, r1, . . . , rk),

whereq =

(Φ(tp; e1, r1, . . . , rk), . . . ,Φ(tp; en, r1, . . . , rk)

).

Let us first calculate d(IΦ) as in Theorem 6.1 using the above derivative and exchangingexterior derivation and integration2:

(d(IΦ))(p; r1, . . . , rk+1)

=k+1∑

i=1

(−1)i−1(IΦ)′(p; r1, . . . , ri−1, ri, ri+1, . . . , rk+1)ri

=

k+1∑

i=1

(−1)i−1

1∫

0

tkΦ(tp; ri, r1, . . . , ri−1, ri, ri+1, . . . , rk+1) dt

+

k+1∑

i=1

(−1)i−1

1∫

0

tk+1Φ′(tp;p, r1, . . . , ri−1, ri, ri+1, . . . , rk+1)ri dt.

Note thatqri = Φ(tp; ri, r1, . . . , ri−1, ri, ri+1, . . . , rk+1).

The first sum is simplified a lot by antisymmetry when the ri’s are moved to their ”correct”positions:

k+1∑

i=1

(−1)i−1

1∫

0

tkΦ(tp; ri, r1, . . . , ri−1, ri, ri+1, . . . , rk+1) dt = (k+1)

1∫

0

tkΦ(tp; r1, . . . , rk+1) dt.

Similarly using Theorem 6.1 we can calculate I(dΦ):

(I(dΦ)

)(p; r1, . . . , rk+1) =

1∫

0

tk+1(dΦ)(tp;p, r1, . . . , rk+1) dt

=

1∫

0

tk+1Φ′(tp; r1, . . . , rk+1)p dt

+k+1∑

i=1

(−1)i1∫

0

tk+1Φ′(tp;p, r1, . . . , ri−1, ri, ri+1, . . . , rk+1)ri dt.

(Note the powers tk+1 and (−1)i. Why these?) Combining these two results we see that(d(IΦ)

)(p; r1, . . . , rk+1) +

(I(dΦ)

)(p; r1, . . . , rk+1)

=

1∫

0

(k + 1)tkΦ(tp; r1, . . . , rk+1) dt+

1∫

0

tk+1Φ′(tp; r1, . . . , rk+1)p dt

=

1∫

0

d

dt

(tk+1Φ(tp; r1, . . . , rk+1)

)dt = Φ(p; r1, . . . , rk+1).

Since we assumed that dΦ = 0 in A, the theorem follows.2This is allowed in a fairly general situation.


Example. If the n-form field Φ is the density form field of the sclar field f , i.e. of the form

Φ(p; r1, . . . , rn) = Φf–density(p; r1, . . . , rn) = f(p) det(r1, . . . , rn),

then dΦ = 0 and Φ is exact. By Poincaré’s Lemma in a star-shaped region with respect to some

point p0 then Φ has the potential

(IΦ)(p; r1, . . . , rn−1) =

1∫

0

tn−1f(p0 + t(p− p0)

)dt det(p− p0, r1, . . . , rn−1),

which is the flux-form field ΦF–flux(p; r1, . . . , rn−1) of the vector field

F(p) =

1∫

0

tn−1f(p0 + t(p− p0)

)dt (p− p0).

In a sense this is the one and only analogue of the integral function of a univariate function.

For instance, in R3 Gauß’ Theorem implies

∫

−→K

f(r) dr =

∫

−→K

∇ • F(r) dr =∮

−→∂ K

F(r) • dS

for any (sufficiently regular) region K.

As a concrete example, let us calculate the potential in R3 for

f(x, y, z) = x+ y + z and p0 = 0.

The result is

F(x, y, z) =

1∫

0

t2(tx+ ty + tz) dt

xyz

=

x+ y + z

4

xyz

.

6.2 Scalar Potential of a Vector Field in R3

By Poincaré’s Lemma the 1-form field

Φ(p; r) = F(p) • r

(the work form field of the vector field F) has a potential in star-shaped regions with respect tosome point p0 whenever

dΦ = Φ∇×F–flux = 0,

i.e. whenever ∇× F = 0, or the field is irrotational, and one such potential is

f(p) =

1∫

0

F(p0 + t(p− p0)

)• (p− p0) dt.

The potential (a 0-form field, or a scalar field) f is then called a scalar potential of the vectorfield F, and F = ∇f .


In engineering applications the vector field F is in general unknown, and the task is to findit—analytically or numerically—based on given initial data. This would mean finding threescalar-valued functions (the components of the field). In the irrotational case the unknownvector field can be given as the gradient of an unknown scalar field: F = ∇f . To find the vectorfield it thus suffices to find only one scalar-valued function, the potential f . Irrotationalityusually follows from the statement of the problem (by Thomson’s circulation law, Maxwell’sequations, etc.).

As noticed, if f is a scalar potential and c is a constant, then f + c is another scalar potentialfor the same vector field, because

∇(f + c) = ∇f.On the other hand, if f and g both are scalar potentials of a vector field, then ∇(f − g) = 0 andf − g is thus constant. A scalar potential is thus unique modulo a scalar additive constant.

There is a connection between existence of a scalar potential and line integrals:

Theorem 6.2. The following three conditions are equivalent for a continuous vector field F

defined in an open subset (manifold) K of R3.

(i) F has in K a scalar potential f .

(ii) For any closed curve (1-dimensional manifold in relaxed parametrization) C in K∮

C

F(r) • ds = 0.

(iii) If the points r0 and r1 are in K and the curve (1-dimensional oriented manifold in relaxed

parametrization)−→C in K connects them, i.e., in the orientation r0 is a fixed initial point

and r1 a variable terminal point, then the value of integral

∫

−→C

F(r) • ds = h(r1)

depends on the choice of the curve−→C only through the terminal point r1, and thus defines

a function h(r1) of r1.

The function h in item (iii) is a potential of F in K.

Proof. Let us first establish the implication (i) ⇒ (iii). So, let us assume item (i), and take arelaxed parametrization r = γ(u) (a ≤ u ≤ b) of

−→C . Using the assumed potential f we get

∫

−→C

F(r) • ds =∫

−→C

∇f(r) • ds =b∫

a

f ′(γ(u)

)γ ′(u) du

=

b∫

a

d

duf(γ(u)

)du =

b/

a

f(γ(u)

)= f(r1)− f(r0),

which implies item (iii) when we take

h(r1) = f(r1)− f(r0).

(Note that this h now is a potential.)


Next we establish the implication (iii) ⇒ (i). Weonly need to show that the function h given by item(iii) really is a potential. For this, let us denote r1 =(x1, y1, z1) and show, as an example, that

∂h(r1)

∂x1= F1(r1),

the other partial derivatives being treated similarly.For this, assume r1 is in the manifold K and take

−→C tobe an oriented curve where the last ”piece” is a shortline segment

y

x

r0

r1r2

C1

C2

z

−→C2 : r = r2 ± u(1, 0, 0) (0 < u ≤ ±(x1 − x2))

parallel to the x-axis, where r2 = (x2, y2, z2) and the sign depends on the direction (i.e., plus ifx1 > x2, minus otherwise), see the figure below. Clearly ±i is a tangent vector of

−→C2. The valueof the integral does not depend on the path, so this choice is possible. Since K is an open subset,there is the space available, too. Let us denote the initial part of

−→C by−→C1, so

−→C =−→C1 +

−→C2.If the direction is from left to right (meaning that the sign is plus, the case of a minus sign issimilar), then

h(r1) =

∫

−→C

F(r) • ds =∫

−→C1

F(r) • ds+∫

−→C2

F(r) • ds

=

∫

−→C1

F(r) • ds +x1−x2∫

0

F(r2 + u(1, 0, 0)

)• i du

=

∫

−→C1

F(r) • ds +x1−x2∫

0

F1(r2 + u(1, 0, 0)) du.

Thus, taking the derivative with respect to x1,

∂h(r1)

∂x1= F1(r1).

The implication (ii) ⇒ (iii) is fairly obvious. As-sume item (ii) holds and take in item (iii) two curves−→C and

−→C ′, of the kind given. Reverse the orientationof

−→C ′ and connect the resulting curve −−→C ′ to−→C , see

the figure on the right. For the resulting closed curve−→C −−→C ′,r0

r1

C

C

r0

r1

C C

0 =

∮

−→C −

−→C ′

F(r) • ds =∫

−→C

F(r) • ds−∫

−→C ′

F(r) • ds.

by item (ii).Implication (iii) ⇒ (ii) is also fairly obvious. Assume item (iii) and take a closed curve

−→C ,and two points r0 and r1 of C. Thus C is divided into two parts, the oriented curve

−→C ′ connectingr0 and r1, and the oriented curve

−→C ′′ connecting r1 and r0. Reversing the orientation of the lattercurve we get two curves

−→C ′ and −−→C ′′, as in item (iii), see the figure below. It then follows from


item (iii) that∮

−→C

F(r) •ds =∫

−→C ′

F(r) •ds−∫

−−→C ′′

F(r) •ds = 0.

A vector field satisfying item (iii) of the theoremis called conservative. For a conservative vector fieldF we often write just

r0

r1

CC

r0

r1

C

∫

−→C

F(r) • ds =r1∫

r0

F(r) • ds,

the value of the integral depending only on the endpoints.Existence of a scalar potential does not depend only on the vector field F but also on the

manifold (or solid body) K. By Poincaré’s Lemma the potential does exist in star-shaped man-ifolds, and that goes fairly far already. But what about more general manifolds? The resultdepends on the topology of the manifold, i.e., what can be done to to curves and surfaces in themanifold using continuous transforms.

We say that a manifold K is simply connected, if any closed curve C (a 1-dimensional man-ifold in a relaxed parametrization) in K can be shrunk to a point using continuous transformswithout leaving K. More exactly, there is in K a surface with a relaxed parametrization

S : r = γ(t, u) (U : a(u) ≤ t ≤ b(u), 0 < u < 1)

such that

• a(u) and b(u) are both continuous in the interval [0, 1),

• the curve Cu : r = γ(t, u) (a(u) ≤ t ≤ b(u)) is closed for all 0 < u < 1,

• C is one of the curves Cu (0 < u < 1), and

• the ”curve” C0 : r = γ(t, 0) (a(0) ≤ t ≤ b(0)) is a point (a degenerate curve).

A prototype of such a surface is the segment of a spherewith radius R parametrized in spherical coordinates as

S : r = γ(t, u) = (R sin u cos t, R sin u sin t, R cos u)

(0 ≤ t ≤ 2π, 0 < u < 1).

Its boundary circle can be shrunk into the centre poleby continuous transforms, see the figure on the right(Maple).

The pertinent result is now

Theorem 6.3. If F is an irrotational continuously differentiable vector field in a simply con-

nected manifold (solid body) of R3, then it has there a scalar potential.

Proof. The proof is tedious and technical, see e.g. NIKOLSKY & VOLOSOV. The basic ideaof course is to apply the (Generalized) Stokes Theorem to the above line C, a surface S, and asubsurface having C as its boundary. The vector field being irrotational, its curl is a zero vector,so the line integral around C is vanishes, and Theorem 6.2 becomes available.

The problems come with the orientation. For instance, even though the Möbius band cannotbe oriented, its boundary is a single closed curve (see the figure below: Maple). But of coursethere are other orientable surfaces having this boundary curve.


It may also be noticed that the surface S can be globally extremely complicated, just thinkof a case where C is a knot (figure: Maple):

Example. As an example of a manifold which is not simply connected take the inside of a torus,

say the one parametrized as

x = (R1 +R2 cosu2) cosu1

y = (R1 +R2 cos u2) sin u1

z = R2 sin u2

(0 ≤ u1, u2 ≤ 2π).

See the figure on the right (Maple).

–4

–3

–2

–1

0

1

2

3

4

–4–3

–2–1

12

34

–4–3

–2–1

12

34

Then e.g. the vector field

F(r) =1

x2 + y2

−yx0

does not have a potential even though it is easily verified that it is irrotational (check!). This

follows from Theorem 6.2 since for e.g. the centre circle

−→C : γ(u1) = (R1 cosu1, R1 sin u1, 0) (0 ≤ u1 ≤ 2π)

we have

∮

−→C

F(r) • ds =2π∫

0

1

R21

−R1 sin u1R1 cosu1

0

•

−R1 sin u1R1 cos u1

0

du1 =

2π∫

0

du1 = 2π 6= 0.


On the other hand, after removing the part of the torus in the right xy-plane the potential

appears, it is atan (see the example in Section 2.5).

If the manifold is not simply connected, an irrotational vector field may still have a localscalar potential, but it need not be globally unique modulo an additive constant. Since nu-merical methods can only find unique solutions, the manifold must be made artificially simplyconnected by ”cutting it open” using a suitable surface, as was done in the above example.

Note. The condition in Theorem 6.3 is sufficient but not necessary. In some cases an irrotational

vector field has a unique scalar potential (modulo the additive constant) even when the manifold

is not simply connected. This is the case e.g. for some problems dealing with the electric field

E when the magnetic field is stationary and there are no electromotive forces. For reasons of

energy conservation, the integral around a closed curve C is then∮

C

E(r) • ds = 0.

Thus the integralr1∫

r0

E(r) • ds

is path-independent and defines a (unique) electric potential.

A scalar potential may also be obtained in an approximate sense, cf. Section 6.6.

6.3 Vector Potential of a Vector Field in R3

By Poincaré’s Lemma the 2-form field

Φ(p; r1, r2) = F(p) • r1 × r2

(the flux form field of F) has a potential in star-shaped regions with respect to the point p0, if

dΦ = Φ∇•F–density = 0,

i.e. ∇ • F = 0, or the vector field is solenoidal, and one such potential is

U(p) • r =1∫

0

tF(p0 + t(p− p0))× (p− p0) • r dt.

(the work form field of the vector field U). A vector field U such that ΦU–work is a potential ofΦ is called a vector potential of the vector field F, and then F = ∇×U.

Example. Let us find a vector potential for the vector field

F(x, y, z) =

xy

−2z

in R3. It is easily verified that this is solenoidal. Choosing p0 = 0 and integrating we get

(omitting the vector variable r)

U(x, y, z) =

1∫

0

t

txty

−2tz

×

xyz

dt =

1∫

0

t2

3yz−3xz

0

dt =

yz−xz

0

.


Example. As another example let us take a more complicated but sometimes very useful case

(cf. Appendix 3). Let us find a vector potential for the Newton vector field

F(p) = −C p− p1

‖p− p1‖3,

in the manifold which we get by removing from R3 the ray (so-called Dirac’s string)

p1 + u(p1 − p0) (u ≥ 0).

The resulting region is star-shaped with respect to the point p0 assuming—as is done—that

p0 6= p1. Note that the point p1 is then removed but the point p0 is not. For brevity we denote

a = ‖p0 − p1‖2 = (p0 − p1) • (p0 − p1) ,

b = 2(p0 − p1) • (p− p0) = 2(p0 − p1) • (p− p1)− 2a ,

c = ‖p− p0‖2 = ‖p− p1‖2 − 2(p0 − p1) • (p− p1) + a and

d = 4ac− b2 = 4‖p0 − p1‖2‖p− p0‖2 − 4((p0 − p1) • (p− p0)

)2.

Some simple vector manipulation then shows that

4a+ 2b = 4(p0 − p1) • (p− p1) ,

a + b+ c = ‖p− p1‖2 and

d = 4‖p0 − p1‖2‖p− p1‖2 − 4((p0 − p1) • (p− p1)

)2.

Calculate then the vector potential (omitting the vector variable r and the constant −C):

U(p) =

1∫

0

tp0 + t(p− p0)− p1∥∥p0 + t(p− p0)− p1

∥∥3 × (p− p0) dt

= (p0 − p1)× (p− p1)

1∫

0

t dt

(a+ bt + ct2)3/2

= (p0 − p1)× (p− p1)

1/

0

−2(2a+ bt)

d√a+ bt + ct2

= (p0 − p1)× (p− p1)4√a√a+ b+ c− (4a+ 2b)

d√a + b+ c

= (p0 − p1)× (p− p1)‖p0 − p1‖‖p− p1‖ − (p0 − p1) • (p− p1)

‖p− p1‖(‖p0 − p1‖2‖p− p1‖2 −

((p0 − p1) • (p− p1)

)2)

=(p0 − p1)× (p− p1)

‖p0 − p1‖‖p− p1‖2 +((p0 − p1) • (p− p1)

)‖p− p1‖

.

This result is in many ways the best possible, it is scarcely possible to remove anything less than

the ray from R3 to get the vector potential. (See the example in Section 3 of Appendix 3.)


An unknown solenoidal vector field can then be given by as the curl of another (also so-farunknown) vector field, the vector potential: F = ∇ × U. Solenoidality usually follows fromthe statement of the problem. For form fields, often a work form field is somehow ”simpler”than a flux form field.

If U is a vector potential of the vector field F and ∇f is a gradient field, then U + ∇f isalso a vector potential of F, because

∇× (U+∇f) = ∇×U+∇×∇f = ∇×U.

On the other hand, if both U and V are vector potentials of the same vector field, then∇ × (U − V) = 0 and U − V is irrotational. A vector potential is thus unique modulo anadditive irrotational vector field. In case the manifold allows existence of a scalar potential—e.g. being simply connected—vector potential is unique modulo an additive gradient field. Butif scalar potentials do not exist, then this might not be true. For instance, in the torus of theexample in the previous section the zero vector field 0 surely has a vector potential, one such isof course 0, but

F(r) =1

x2 + y2

−yx0

is also a vector potential of 0, and it is not a gradient field.There is a connection between existence of vector potentials and surface integrals given by

the Generalized Stokes Theorem, once we remember that a surface integral is the integral ofthe flux form field of the curl of the vector potential, and that the flux form field is the exteriorderivative of the work form field of the vector potential:

Theorem 6.4. If a continuously differentiable vector field F has a vector potentialU in the open

subset K of R3, and S is a closed oriented surface in−→K (a 2-dimensional oriented manifold in

extended parametrization having an empty smooth boundary), then

∮

−→S

F(r) • dS =

∮

−→S

∇×U(r) • dS = 0.

As was the case for the scalar potential, existence of a vector potential does not depend onlyon the field but also on the manifold K. In star-shaped manifolds solenoidal vector fields have avector potential by Poincaré’s Lemma. In other cases the situation is more complicated. In anycase, the necessary condition of the theorem above must hold true.

Example. A typical example of a case where a solenoidal vector field does not have a vector

potential is the Newton vector field

F(r) =r− r0

‖r− r0‖3

in a ”punctured” manifold K, where the point r0 is removed but which otherwise completely

”surrounds” the point r0. As was noted before, this vector field is solenoidal. Integrating over

a small r0-centered sphere

S : r = γ(θ, φ) = r0 + δ(sin θ cosφ, sin θ sinφ, cos θ) (0 ≤ θ ≤ π, 0 ≤ φ < 2π)

of radius δ (oriented by the exterior normal) we get


∮

−→S

F(r) • dS =

2π∫

0

π∫

0

1

δ2

sin θ cosφsin θ sinφcos θ

•

sin θ cos φsin θ sinφ

cos θ

δ2 sin θ dθ dφ

=

2π∫

0

π∫

0

sin θ dθ dφ = 4π 6= 0.

The conclusion of Theorem 6.4 is thus not valid, so no vector potential exists.

Gauß’ Theorem, too, has its consequences for vector potentials. If the closed oriented sur-face

−→S in Theorem 6.4 encloses a solid L which is included in the manifold K, then, in linewith the theorem, according to Gauß’ Theorem

∮

−→S

F(r) • dS =

∫

L

∇ • F(r) dr = 0,

because in order to have a vector potential the vector field F must be solenoidal.But what if not all of L is contained in the manifold K, but K has ”cavities” which are

included in L? Taking a sufficiently small surface S we may assume that L contains only onesuch cavity. Let us denote by N the part of the manifold K included in L, and the boundaryof the cavity by

−→T . Then−→T is a closed surface, oriented by the exterior normal directed to the

inside of the cavity. When we now apply Gauß’ Theorem to N and its boundary−→S +

−→T (where−→S is oriented using an exterior normal) and also Theorem 6.4 we get

0 =

∫

N

∇ • F(r) dr =∮

−→S+

−→T

F(r) • dS =

∮

−→S

F(r) • dS+

∮

−→T

F(r) • dS =

∮

−→T

F(r) • dS.

Thus even through the boundary of each cavity inside K the flux of the vector field must be = 0,in order for the field to have a vector potential in K.

Ideally this necessary condition for existence of a vector potential should also be sufficientfor solenoidal vector fields. The situation is however complicated. In the literature there aresome sufficient conditions of this classical type—in addition to the case of star-shaped mani-folds—e.g. the following:

• The vector field F is continuously differentiable and solenoidal. Each manifold L en-closed by a closed surface inside K ∪ ∂K (manifold + boundary) is included in K (nocavities) (APOSTOL).

• A magnetic field B in vacuum where there are moving charges (FEYNMAN, R.P. et al.:The Feynman Lectures on Physics. Volume II. Addison–Wesley (1998)).

• The vector field F is differentiable and its flux through any closed piecewise smoothsurface inside the manifold K vanishes. K can be deformed by continuous transformsto a star-shaped manifold possibly with finitely many ball-shaped cavities (TON, T.-C.:On the Potential of a Solenoidal Vector Field. Journal of Mathematical Analysis and

Applications 151 (1990), 557–580).

As a more physical treatment, the classic reference MISNER, C.W. & WHEELER, J.A.: Clas-sical Physics as Geometry: Gravitation, Electromagnetism, Unquantified Charge, and Mass as


Properties of Curved Empty Space. Annals of Physics 2 (1957), 525–603, gives a relativity-theoretic abstract characterization for existence of a vector potential in some fairly generalcases.

Advances have been made using manifold theory, too. A classic reference is WEYL, H.: TheMethod of Orthogonal Projection in Potential Theory. Duke Mathematical Journal 7 (1940),411–444. More modern ideas are based on the fundamental work of Georges de Rham. SeeDE RHAM, G.H.: Differentiable Manifolds. Springer–Verlag (1984). It could be said that inthis way we are already very close to the above mentioned ideal characterization for existenceof vector potential.

6.4 Helmholtz’s Decomposition

As was noted, a vector potential of F(r), if it exists, is only unique modulo an additive irrota-tional vector field. This gives a possibility for the vector potential to satisfy extra conditions.

Consider first the case where continuously differentiable irrotational vector fields have ascalar potential in the manifold K. Now, what if we want for F a solenoidal vector potentialU1, when we already know some vector potential U. Denoting ∇ •U = g, the required vectorpotential is of the form U1 = U+∇φ and

0 = ∇ •U1(r) = ∇ •U(r) + ∆φ(r) = g(r) + ∆φ(r).

(The case g = 0 is of course clear anyway, and can be omitted.) Thus for the needed scalar fieldφ we get the equation

∆φ(r) = −g(r).This is a partial differential equation, a so-called Poisson equation. The solution of the equationis not unique, in fact any function φ+h, where ∆h = 0, is also a solution. The partial differentialequation

∆h(r) = 0

in turn is called Laplace’s equation, and its solutions are called harmonic functions 3, see alsoChapter 7 and Appendix 1. So any φ (if it exists) is unique modulo an additive harmonicfunction. To fix a φ, some extra conditions—so-called boundary conditions—for the Poissonequation need to be set, an important topic not dealt with here any further, however.

All in all, since Poisson’s equations have solutions in very general situations, if there is avector potential, there also is a solenoidal vector potential.4

A similar idea can be used to get an irrotational vector field U2 with a given divergence f .Then U2 = ∇ψ and

∇ •U2(r) = ∆ψ(r) = f(r).

So again we have a Poisson equation whose solution ψ is unique up to an additive harmonicfunction. And finally, if we want a vector field U with a given curl ∇ × U = F and a givendivergence ∇ • U = f , then it is simply U = U1 + U2. To fix U uniquely, extra (boundary)conditions are needed.

From these we get the celebrated Helmholtz’s decomposition. In a fairly general situation avector field U can be expressed as a sum of a solenoidal vector field and an irrotational vectorfield. On way to get decomposition would be the following:

3Harmonic functions have a central rôle in the investigation and modelling of scalar fields.4And as a consequence, if F has a vector potential, then it also has a double vector potential V such that

F = ∇× (∇×V), and a triple vector potential W such that F = ∇×(∇× (∇×W)

), and so on.


1. Take F = ∇×U and find the solenoidal vector potential U1 above.

2. Write the vector field U asU = U1 + (U−U1).

Then U1 is solenoidal and (U−U1) is irrotational.

Another way would be:

1. Take f = ∇ •U and find the irrotational vector field U2 above.

2. Write the vector field U asU = (U−U2) +U2

Then U−U2 is solenoidal and U2 is irrotational.

The field U can thus be written as the sum of gradient and a curl.It should be emphasized, of course, that all this works with the assumption that in the man-

ifold K scalar and vector potentials exist, and that the Poisson equations have solutions.

Note. The double-curl expansion rule in Chapter 1 (in a bit different form)

∆F = ∇(∇ • F) +∇×(∇× (−F)

)

gives one such Helmholtz’s decomposition.

For form fields there is a corresponding decomposition, the so-called Hodge decomposition,see e.g. ABRAHAM & MARSDEN & RATIU.

6.5 Four-Potential

A four-potential is the potential of a 4-dimensional 2-form field Φ. By Poincaré’s Lemma itexists in a star-shaped region if dΦ = 0.

It was noted in Section 5.3 that dΦFaraday = 0. Thus in a star-shaped region the Faradayform field has a potential A, the so-called electromagnetic four-potential. This potential is a1-form field, (traditionally) written as

A = A1dx+ A2dy + A3dz − φ dt.

Let us compute the exterior derivative and find the coefficients A1, A2, A3 and φ. By rule(IV) in Section 5.2,

dA =(∂A1

∂xdx+

∂A1

∂ydy +

∂A1

∂zdz +

∂A1

∂tdt)∧ dx

+(∂A2

∂xdx+

∂A2

∂ydy +

∂A2

∂zdz +

∂A2

∂tdt)∧ dy

+(∂A3

∂xdx+

∂A3

∂ydy +

∂A3

∂zdz +

∂A3

∂tdt)∧ dz

−(∂φ∂x

dx+∂φ

∂ydy +

∂φ

∂zdz +

∂φ

∂tdt)∧ dt,

and this must be

E1dx ∧ dt + E2dy ∧ dt+ E3dz ∧ dt+B1dy ∧ dz +B2dz ∧ dx+B3dx ∧ dy.


Expanding and comparing we see that

E = −∂A∂t

−∇φ and B = ∇×A , where A =

A1

A2

A3

.

Thus we getA by first finding a vector potentialA for the magnetic flux densityB (by Maxwell’sequation (M2) B is solenoidal), and then a scalar potential φ for the vector field

−∂A∂t

− E.

This is possible since by Maxwell’s equations (and some weak extra conditions)

∇×(− ∂A

∂t− E

)= − ∂

∂t(∇×A)−∇×E = −∂B

∂t+∂B

∂t= 0.

The four-potential A is not unique, we can add a gradient field ∇ψ in A, and replace φ byφ − ∂ψ/∂t, without the exterior derivative changing. Often A and φ are chosen to satisfy theso-called Lorenz gauge condition

∇ •A+1

c2∂φ

∂t= 0.

In a sense this separates A and φ. If originally

∇ •A+1

c2∂φ

∂t= g 6= 0,

then Lorenz’s gauge condition will be satisfied if ψ is a solution of the partial differential equa-tion

∆ψ(r, t)− 1

c2∂2ψ(r, t)

∂t2= −g(r, t)

(check!). This equation is a so-called wave equation and it has a solution in some very generalsituations. Lorenz’s gauge condition can then be assumed quite generally.

6.6 Dipole Approximations and Dipole Potentials

A dipole field is a vector field generated by two potentialsof opposite signs. A typical situation is a pair of electriccharges of equal magnitude but opposite sign, separated bya small distance, and their electrostatic field.

Let us first consider this electric dipole. The electriccharges +q and −q are close to each other in the pointsr′ + h and r′, see the figure on the right. The electrostaticfield is observed in the point r. For the dipole approxima-tion to work the point r must be far from the charges, that is‖r− r′‖ ≫ ‖h‖.

Electrophysics tells us that the Coulomb potential of

r

r

r + hq

+q

these charges in the point r is

φ(r) =q

4πε

( 1

‖r− r′ − h‖ − 1

‖r− r′‖).


On the other hand, it was noticed in Section 5.3 that the exterior derivative of the function (or0-form field)

f(r′) =q

4πε

1

‖r− r′‖is the work form field of the gradient of f , i.e.

(df)(r′;h) = ∇′f(r′) • h ∼= f(r′ + h)− f(r′) = φ(r),

where the derivation is with respect to r′ (the primed nabla). The approximation follows imme-diately from the definition of the exterior derivative. Thus in the point of observation r

φ(r) ∼= q

4πε

(∇′ 1

‖r− r′‖)• h =

q

4πε

r− r′

‖r− r′‖3 • h.

Hereqh = p′

is called the electric dipole moment. With this notation we get the usual dipole approximation

of the potential as

φ(r) ∼= 1

4πε

p′ • (r− r′)

‖r− r′‖3 ,

where the connection between r′ and p′ is indicated by the primes.But what about the electric field in r? In electrophysics negative potentials are used, so

E(r) = −∇φ(r), and

E(r) ∼= − 1

4πε∇p′ • (r− r′)

‖r− r′‖3 .

The gradient is is taken with respect to the variable r. The vector p′ does not depend on r.Applying the derivation rule (i) in Section 1.5 we get

∇p′ • (r− r′)

‖r− r′‖3 =(p′ • (r− r′)

)∇ 1

‖r− r′‖3 +1

‖r− r′‖3∇(p′ • (r− r′)

)

= −3p′ • (r− r′)

‖r− r′‖5 (r− r′) +p′

‖r− r′‖3 .

Thus the dipole approximation of the electric field is

E(r) ∼= 1

4πε‖r− r′‖5(− ‖r− r′‖2p′ + 3

(p′ • (r− r′)

)(r− r′)

)

Except for the dipoles themselves, such an approximation can be useful for other fields closelyresembling dipoles. The dipole moment is then obtained by physical considerations.

A magnetic dipole is very similar to the electric dipole above. We approach it using a currentloop remembering the definition of curl via exterior derivation in Section 5.3.

So, let us consider a small parallelogram-shaped current loop−→∂P with current I . A normal

vector for the plane of the parallelogram is given by r1 × r2 = An, where A is the area of theparallelogram and n is the corresponding unit normal vector, correctly oriented with respect tothe direction of the current. The point of observation r is far away compared with the size ofthe loop. We use the vertex r0 to characterize the position of the loop, see the figure below.Points of the loop are denoted by the variable r′. The point of observation r being ”far away”then means that ‖r− r′‖ ≫ ‖r′ − r0‖.


According to the Biot–Savart law in electromagnetics the mag-netic field in the point r is given by

H(r) =I

4π

∮

−→∂P

r′ − r

‖r− r′‖3 × ds′.

Thus, for a constant vector a,

a •H(r) =I

4π

∮

−→∂P

a× (r′ − r)

‖r− r′‖3 • ds′.

r1 x r2

r1

r2

r0 P

I

This is the integral of the vector field

I

4π

a× (r′ − r)

‖r− r′‖3

around the loop and—as noted in Section 5.3—it is approximately the flux form field of the curlin the point r0, i.e.

a •H(r) ∼=(∇′ ×

( I4π

a× (r′ − r)

‖r− r′‖3))

r′=r0

• r1 × r2 =AI

4πn •(∇′ × a× (r′ − r)

‖r− r′‖3)r′=r0

,

where ∇′ is with respect to r′. Using the derivation rule (vi) in Section 1.55 and rememberingthat the Newton vector field is solenoidal, the curl will be seen to be

− a

‖r− r′‖3 + 3a • (r− r′)

‖r− r′‖5 (r− r′)

(check!). Thus

a •H(r) ∼= IA

4π

(− a • n

‖r− r0‖3+ 3

(r− r0) • n‖r− r0‖5

(a • (r− r0))).

The constant vector a was arbitrary, so

H(r) ∼= IA

4π

(− n

‖r− r0‖3+ 3

(r− r0) • n‖r− r0‖5

(r− r0)).

Adopting the corresponding magnetic dipole moment

m′ = IAn

of the loop, we get the dipole approximation of the magnetic field in a far away point r as

H(r) ∼= 1

4π‖r− r0‖5(− ‖r− r0‖2m′ + 3

((r− r0) •m′

)(r− r0)

).

This expression is of exactly the same form as it was for the electric dipole field! So, reverseargumentation as above shows that the magnetic field has here an approximative scalar potentialφ, i.e.

H(r) ∼= −∇φ(r),5Or using Theorem 6.1.


where

φ(r) =1

4π

m′ • (r− r0)

‖r− r0‖3.

Using the derivation rule (iii) in Section 1.5 we may check that the dipole approximationfield is solenoidal. Thus it also has a vector potential (in appropriate regions). We could findit more or less as we found a vector potential for the Newton vector field in the example inSection 6.3. Looking at the curl expression above it should become clear, however, that thevector potential is

A(r) =1

4π

m′ × (r− r0)

‖r− r0‖3.

Note. Dipole approximation is just a part of the more general multipole expansion of a vector

field. Its first term is the so-called monopole approximation, a Newton vector field. The second

term is the dipole approximation of a combination of two Newton vector fields, as above. The

third term is the quadrupole approximation of the combination of four Newton vector fields,

with pairwise opposite signs. The fourth term is the octupole approximation, and so on.

Chapter 7

PARTIAL DIFFERENTIAL EQUATIONS

”The miracle of the appropriateness of thelanguage of mathematics for the formulation

of the laws of physics is a wonderful gift whichwe neither understand nor deserve.”

(EUGENE (JENO) WIGNER: The UnreasonableEffectiveness of Mathematics in the Natural Sciences.

Communications in Pure and Applied

Mathematics 13 No. 1 (1960))

7.1 Standard Forms

Numerical solvers for partial differential equations (PDEs) are especially suitable for equationsof the form

∇ •(k(r)∇u(r)

)= F (r) (elliptic PDE)

in the stationary case, or of the form

∇ •(k(r)∇u(r, t)

)= f(r)

∂u

∂t+ g(r)

∂2u

∂t2+ F (r, t),

where k(r) > 0, in the nonstationary case, with the proper boundary and initial conditions(which are not dealt with here). The coefficient functions k, f and g may also depend on u, andthe function g possibly also on the gradient ∇u. The function F is a so-called forcing function

representing exterior forces etc. In the nonstationary case the different basic types are

• f > 0 and g = 0: parabolic PDE, heat equation, diffusion equation,

• f < 0 and g = 0: reverse heat equation, Black–Scholes equation,

• f = 0 and g > 0: hyperbolic PDE, wave equation,

• f > 0 and g > 0: telegraph equation, lossy wave equation, hyperbolic heat equation.

The order of a PDE is the order of the highest partial derivative appearing in the equation.The above PDEs are all of second order. There are, of course, many important kinds of PDEsof first order, e.g. advection equations, and also of order higher than two, e.g. the biharmonicequation (elasticity), Korteweg–de Vries equation (shallow water waves), and Dym’s equation(solitons). Theories and numerical solution methods for these other orders are very differentfrom the ones for the above PDEs.

7.2 Examples

Let us take examples of field modelling problems leading to second order PDEs of the abovetypes. For electric and magnetic fields (M1)–(M4) refer to the Maxwell equations in Section5.3.

102

CHAPTER 7. PARTIAL DIFFERENTIAL EQUATIONS 103

Example. (Electrostatic field) Since in the stationary case the electric field E is irrotational

(M3), we have ∇× E = 0 and the field has a scalar potential: E = −∇V . (In electrophysics

the negative potential Φ = −V is used.)

On the other hand (M1),

∇ •D = ∇ • (εE) = ρ (charge density).

Here ε is the permittivity. If the charge density—and of course the permittivity—is known,

∇ •(ε(r)∇V (r)

)= −ρ(r).

This is a PDE of a standard form.

Example. (Stationary electric current) By Ohm’s law, the current density is J = σE, where

σ is the conductance. According to the Kirchhoff law electricity is not accumulated anywhere,

i.e., the net current through a closed surface is zero. For the boundary (a closed surface)−→S of

the solid K, ∮

−→S

J(r) • dS = 0 =

∫

K

∇ • J(r) dr

(by Gauß’ Theorem). In these problems the field usually is assumed continuously differentiable,

thus since K is arbitrary we have further ∇ • J = 0.

The field E being irrotational (M3) it has a scalar potential. i.e. E = −∇V , and so

∇ • J(r) = ∇ •(σ(r)∇V (r)

)= 0,

and again we have a standard form PDE.

Example. (Magnetic field with a scalar potential) In a region with no conductive material

charge density is zero, and (M4) ∇ × H = J = 0. There is thus a scalar potential, i.e.

H = ∇Φ.

On the other hand (M2)

∇ •B = ∇ • (µH) = 0,

where µ is the permeability. We have again a PDE of the same type since now

∇ •(µ(r)∇Φ(r)

)= 0.

Example. (Stationary incompressible irrotational fluid flow) Here the flow is known to be

irrotational by nature, that is ∇× v = 0. The velocity thus has a scalar potential: v = ∇φ. In

an incompressible flow fluid is not accumulated in any body K bounded by the closed surface−→S . What comes in also goes out, and so fluid leaves the body at zero net rate;

∮

−→S

v(r) • dS =

∫

K

∇ • v(r) dr = 0.

Since the body K is arbitrary, then as above ∇ • v = 0, and we have the Laplace equation

∇ •∇φ(r) = ∆φ(r) = 0.


Example. (Stationary heat conduction) The heat flow is v = −k∇T , where T is the temper-

ature and k is the thermal conductivity. Here the model is an empirical one, valid for isotropic

material (one where local heat flow is the same in all directions).

If the temperature is stationary, heat is not accumulated anywhere, and as above with the

fluid flow or with the Kirchhoff law, ∇ • v = 0. Thus we have

∇ •(k(r)∇T (r)

)= 0.

Example. (Nonstationary heat conduction) As above, the heat flow is v = −k∇T . The net

heat flow out of a body K with boundary surface−→S is

∮

−→S

v(r, t) • dS,

i.e. heat is accumulated there at the rate (power)

dE

dt= −

∮

−→S

v(r, t) • dS =

∫

K

∇ • (k(r)∇T (r, t)) dr

(again by Gauß’ Theorem). Energy conservation forces this to be the same as the power needed

to raise the temperature of the body.

The power per volume unit needed to raise the temperature at the rate∂T

∂tis

C(r)ρ(r)∂T (r, t)

∂t,

where C(r) is the thermal capacity and ρ(r) the mass density in the point r. For the whole body

the power is thusdE

dt=

∫

K

C(r)ρ(r)∂T (r, t)

∂tdr.

Comparing the powers (and again remembering that K was arbitrary) we deduce that the inte-

grands must be the same, and we get the heat equation

∇ •(k(r)∇T (r, t)

)= C(r)ρ(r)

∂T (r, t)

∂t.

It is interesting how mathematical modelling in different areas leads to the precise sametypes of PDEs. Some PDEs contain second time derivatives originating from Newton’s secondlaw. This would be the case e.g. for nonstationary fluid flow. Our example, however, is anacoustic wave equation.

Example. (Small amplitude acoustic plane wave) A plane wave is a planar wave front. We

set the coordinates in such a way that the wave front ”proceeds” to the direction of the positive

x-axis, and is thus parallel to the yz-plane. We then need only the x-coordinate and time t,within the front there is no change. This would correspond to a sound source far away in the

direction of the negative x-axis.

Let us denote the position of an air molecule at time t, initially in the point x, by x+u(x, t).Thus u(x, t) is the deviation from the initial position, and u(x, 0) = 0. Initially the left edge of

a layer of thickness dx is in the point x, and at time t it is in the point x+u(x, t). The thickness

of the layer is then

dx+∂u(x, t)

∂xdx,

see the figure below.


The mass of the layer per area unit remains the same,

i.e.

ρ0 dx = ρ(x, t)(1 +

∂u(x, t)

∂x

)dx,

where ρ0 is air density at time t = 0 and ρ(x, t) at

time t. We see thus that

ρ(x, t)

ρ0=(1 +

∂u(x, t)

∂x

)−1

.

In an adiabatic process, where there is no heat loss

nor heat gain, is is known that

p(x, t)

p0=(ρ(x, t)

ρ0

)γ,

x

y

z

u(x,t)

dx dx + ux

x

dx

ρ0 ρ

where p0 is the air pressure at time t = 0 and p(x, t) at time t, and γ is the so-called adiabatic

constant, for air γ = 1.40. Differentiating we get

∂p(x, t)

∂x= p0

∂

∂x

(1 +

∂u(x, t)

∂x

)−γ

= −γp0(1 +

∂u(x, t)

∂x

)−γ−1 ∂2u(x, t)

∂x2

= −γp0ρ0

ρ(x, t)(1 +

∂u(x, t)

∂x

)−γ ∂2u(x, t)

∂x2.

For small amplitudesγp0ρ0

= c2

is constant, speed of sound squared. This follows from the ideal gas law, since for small ampli-

tudes temperature is constant and density is inversely proportional to volume. For air, at 1 atand at sea level, this gives c = 340.3 ms−1.

A layer is moved by force dictated by the pressure difference (per area unit) between the

edges, and this must be the same as the force given by Newton’s second law:

∂p(x, t)

∂xdx = −c2ρ(x, t)

(1 +

∂u(x, t)

∂x

)−γ ∂2u(x, t)

∂x2dx

= −ρ(x, t)(1 +

∂u(x, t)

∂x

)∂2u(x, t)∂t2

dx.

Note the sign, molecules are moving against the pressure difference. Thus finally we get the

PDE∂2u(x, t)

∂x2=

1

c2

(1 +

∂u(x, t)

∂x

)γ+1∂2u(x, t)

∂t2,

and it is of the type indicated. The coefficient function g depends here also on ∂u/∂x.

For small amplitudes and acoustic frequencies ∂u(x, t)/∂x is also small, so at least approx-

imatively∂2u(x, t)

∂x2=

1

c2∂2u(x, t)

∂t2.

This last PDE is a so-called wave equation. Its general solution is of the form

u(x, t) = f(x− ct) + g(x+ ct)

where f and g are arbitrary twice continuously differentiable functions.


Including the so-far omitted y and z coordinates, we have of course

∂u(r, t)

∂y=∂u(r, t)

∂z= 0.

Thus∂2u(r, t)

∂y2=∂2u(r, t)

∂z2= 0 and

∂2u(r, t)

∂x2= ∆u(r, t),

and the wave equation can be written as

∆u(r, t) =1

c2∂2u(r, t)

∂t2.

The Laplacian is coordinate-free, so this equation holds true for any direction of the planar

wave front.

A general wave equation in R3 has solutions other than the plane waves, e.g. the so-called

(r0-centered) spherical waves

u(r, t) =1

‖r− r0‖f(‖r− r0‖ − ct

)+

1

‖r− r0‖g(‖r− r0‖+ ct

),

where f and g are arbitrary twice continuously differentiable functions (check, and see also

Section A2.4). These may be interpreted as expanding/contracting spherical wave fronts with a

point source in r0. Strangely such ”spreading sharp signals” are only possible in odd dimen-

sions1—though in another form in higher dimensions—and thus in particular not in R2.

Often the units are chosen so that c = 1, and the notation

u = ∆u− ∂2u

∂t2(so-called d’Alembertian)

is adopted.2 The wave equation is then simply

u = 0 or u = F (r, t),

the latter in case a forcing function is required.

It is again interesting that the same wave equation is obtained for small amplitude vibra-tions in general, for vibrating strings and membranes, for longitudinal, transverse and torsionalvibrations of rods, for electromagnetic waves (as a consequence of Maxwell’s equations), foroscillating electric circuits and mechanical systems, for vibrating columns of air and all kindsof acoustic pressure waves, etc.

1As a consequence of the Hyugens Principle.2Sometimes with the signs reversed or denoted by

2.

Appendix 1

PARTIAL INTEGRATION AND

GREEN’S IDENTITIES

hi i need some help here

it’s due next monday

how to use green’s first identity to show if a function

is harmonic on D(a surface), then the line intergral of

”normal derivative of” is equal to zero.

i totally have no idea,

what does harmonic mean?

and what is normal derivative?

thanks!

(A desperate plea in Math Help Forum)

A1.1 Partial Integration

There are two common integration techniques well drilled in basic courses of calculus, thechange of variable method (essentially just our reparametrization) and partial integration.3 Thefamiliar univariate partial integration

b∫

a

u′(x)v(x) dx =

b/

a

u(x)v(x)−b∫

a

u(x)v′(x) dx

can be generalized4 using the Generalized Stokes Theorem and Cartan’s magic formula as∫

A

dΦ ∧Ψ =

∫

−→∂ s

−→M

A

Φ ∧Ψ+ (−1)k+1

∫

A

Φ ∧ dΨ,

where Φ is a k-form field.As noted in Section 5.3, for physical scalar fields f and g and vector fields F and G the

basic wedge products are

∧ g Φg–density ΦG–work ΦG–flux

f fg Φfg–density ΦfG–work ΦfG–flux

ΦF–work — — ΦF×G–flux ΦF•G–density

and the exterior derivatives are

df = Φ∇f–work ,

dΦF–work = Φ∇×F–flux and

dΦF–flux = Φ∇•F–density.

These can be combined into four partial integration formulas (the remaining two are uninterest-ing: 0 = 0± 0):

3There is actually a third one, probably not as well drilled as the other two: inverse integration∫f−1(x) dx = xf−1(x)− F

(f−1(x)

)where F (y) =

∫f(y) dy.

4There is a generalization for indefinite integrals (potentials) as well but it does not seem to be that useful.

107

APPENDIX 1. PARTIAL INTEGRATION AND GREEN’S IDENTITIES 108

1.∫

A

Φg∇f–work =

∫

−→∂ s

−→M

A

fg −∫

A

Φf∇g–work

I.e., for an oriented curve−→C with end vertices r1 (initial) and r2 (terminal),

∫

−→C

g(r)∇f(r) • ds = f(r2)g(r2)− f(r1)g(r1)−∫

−→C

f(r)∇g(r) • ds.

2.∫

A

Φ∇f×G–flux =

∫

−→∂ s

−→M

A

ΦfG–work −∫

A

Φf∇×G–flux

I.e., for an oriented surface−→S ,

∫

−→S

∇f(r)×G(r) • dS =

∮

−→∂ S

f(r)G(r) • ds−∫

−→S

f(r)∇×G(r) • dS.

3.∫

A

ΦG•∇f–density =

∫

−→∂ s

−→M

A

ΦfG–flux −∫

A

Φf∇•G–density

I.e., for an oriented solid body−→K ,

∫

−→K

G(r) • ∇f(r) dr =∮

−→∂ K

f(r)G(r) • dS−∫

−→K

f(r)∇ •G(r) dr.

4.∫

A

ΦG•∇×F–density =

∫

−→∂ s

−→M

A

ΦF×G–flux +

∫

A

ΦF•∇×G–density

I.e., for an oriented solid body−→K ,

∫

−→K

G(r) • ∇ × F(r) dr =

∮

−→∂ K

F(r)×G(r) • dS+

∫

−→K

F(r) • ∇ ×G(r) dr.

Of these formula 3. is the most commonly used.When vector fields computed numerically, say by the finite element method (FEM), integrals

of the type ∫

K

v(r)∇ •(k(r)∇u(r)

)dr

are ubiquitous. The usual assumptions then are that the functions k and v are continuouslydifferentiable and that u is twice contiunuously differentiable. Partial integration formula 3. isnow applicable when we choose f = v and G = k∇u (and denote by

−→S the boundary surfaceof K oriented by exterior normals):

∫

K

v(r)∇ •(k(r)∇u(r)

)dr = −

∫

K

k(r)∇u(r) • ∇v(r) dr+∮

−→S

v(r)k(r)∇u(r) • dS

= −∫

K

k(r)∇u(r) • ∇v(r) dr+∫

S

v(r)k(r)∂u(r)

∂ndS,


where∂u(r)

∂n= ∇u(r) • n

is the normal derivative of the function u, i.e., its directional derivative in the direction of theexterior unit normal n.

Partial integration is significant in two ways when PDEs are solved using FEM. First, knownboundary conditions render the right hand side surface integral known. Hence boundary condi-tions are explicitly included in the solution. Second, the integral

∫

K

k(r)∇u(r) • ∇v(r) dr

only requires continuity of the function k and the partial derivatives of the functions u and v, ifnot even that!5 Using partial integration makes it possible to have much less stringent continuityrequirements.

A1.2 Green’s Identities

Let us assume that the functions u and v in the previous section are twice continuously differ-entiable. Taking k(r) = 1 we get Green’s first identity

∫

K

v(r)∆u(r) dr = −∫

K

∇u(r) • ∇v(r) dr+∫

S

v(r)∂u(r)

∂ndS.

Exchanging the functions u and v we similarly get∫

K

u(r)∆v(r) dr = −∫

K

∇v(r) • ∇u(r) dr+∫

S

u(r)∂v(r)

∂ndS,

and subtracting these equalities on both sides we get Green’s second identity

∫

K

(v(r)∆u(r)− u(r)∆v(r)) dr =

∫

S

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS.

To get the third identity we need to first derive a property of harmonic functions. Considera scalar field u which is twice continuously differentiable in K and harmonic, i.e. ∆u = 0 (cf.Section 6.4). Take a point r0 in the interior of the solid body K, and take v to be the Newtonpotential

v(r) =1

‖r− r0‖.

Since v has a singularity (the point r0) in K, we specify another solid body K1 by removingfrom K a small r0-centered ball K2 of radius δ. The corresponding sphere S2 is oriented usingits exterior normal, that is the normal pointing into the body K1.

5In FEM the functionsu and v would be so-called element functions, and the approximate solution is formulatedas a linear combination of the element functions ui.


The body K1 has two boundary surfaces, the inner one and the outer one. In the bodyK1 both functions u and v are harmonic (it will be remembered that Newton’s vector field issolenoidal). We now apply Green’s second identity to the body K1:

0 =

∫

S

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS −

∫

S2

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS.

The sign of the latter term is because of the orientation of S2.We continue by fixing K to be ‖r − r0‖ < R, i.e., an open r0-centered ball of radius R.

The outer boundary surface S then is an r0-centered sphere of radius R. In this outer boundarysurface the function v is the constant 1/R. The normal derivative in this surface in turn is

∂v(r)

∂n= ∇ 1

‖r− r0‖• n =

r0 − r

‖r− r0‖3• r− r0

‖r− r0‖= − 1

‖r − r0‖2= − 1

R2.

For any harmonic function∫

S

∂u(r)

∂ndS =

∮

−→S

∇u(r) • dS =

∫

K

∆u(r) dr = 0,

(by Gauß’ Theorem), so that∫

S

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS =

∫

S

1

R2u(r) dS

= 4π× (mean value of u in the sphere S).

The inner boundary surface is a ball of radius δ, and the above is valid for it, too (rememberingthe orientation). Thus

∫

S2

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS =

∫

S2

1

δ2u(r) dS

= 4π× (mean value of u in the inner sphere).

The function u is continuous, so the limit of the mean value is u(r0), when δ → 0+.Returning now to the earlier equality

0 =

∮

S

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS −

∮

S2

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS

the following remarkable property of harmonic functions follows:

Mean Value Theorem for Harmonic Functions. The value of a harmonic function in the

centre of a sphere is the mean value of the function in the sphere.

And finally, returning to our original body K and its boundary surface S,∫

S

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS =

∫

S2

(v(r)

∂u(r)

∂n− u(r)

∂v(r)

∂n

)dS = 4πu(r0),

we get Green’s third identity

1

4π

∫

S

( 1

‖r− r0‖∂u(r)

∂n− u(r)

∂

∂n

1

‖r− r0‖)dS = u(r0).

Appendix 2

PULLBACKS AND CURVILINEAR

COORDINATES

”There was a little girl,Who had a little curl

Right in the middle of her forehead.When she was good,

She was very good indeed,But when she was bad she was horrid.”

(HENRY WADSWORTH LONGFELLOW: There Was a Little Girl )

A2.1 Local Coordinates

Parametrization of an n-dimensional manifold (open subset) of Rn as

M : r = γ(u) = (γ1(u), . . . , γn(u)

)(u ∈ U)

constitutes a so-called curvilinear coordinate system. To get the axes (curves) in the point γ(u1)just fix values of the parameters to u = u1 except for one.

The tangent space Tr(M) of M in the point r = γ(u) is the whole Rn. On the other hand itis the column space of the derivative matrix γ ′(u), see Section 2.6. Thus the columns of γ ′(u)give a basis for Rn which depends on the point r, and will change (in general) with the point.This basis gives a so-called local coordinate system.

In what follows we only deal with the case where the local coordinate system is orthogonal,i.e., the columns of γ ′(u) are mutually orthogonal and positively oriented. Let us denote byQ(u) the matrix which we get by normalizing the columns of γ ′(u), i.e. dividing each columnby its length. Then Q(u) is an orthogonal matrix, i.e. Q(u)−1 = Q(u)T, and (for the positiveorientation) det

(Q(u)

)= 1. We denote

Λ(u) =

∥∥∥∂γ(u)∂u1

∥∥∥ 0 0

0∥∥∥∂γ(u)∂u2

∥∥∥ 0

0 0∥∥∥∂γ(u)∂u3

∥∥∥

.

Thus the local coordinate transform matrix is

Q(u) = γ ′(u)Λ(u)−1.

Example. Parametrization of a 3-dimensional manifold of R3 using cylindrical coordinates has

the form

r = γ(r, φ, z) = (r cosφ, r sinφ, z) ((r, φ, z) ∈ U),

111

APPENDIX 2. PULLBACKS AND CURVILINEAR COORDINATES 112

where U is a parameter domain. Thus

γ ′(r, φ, z) =

cosφ −r sinφ 0sin φ r cosφ 00 0 1

, Λ(r, φ, z) =

1 0 00 r 00 0 1

and

Q(r, φ, z) =

cosφ − sinφ 0sinφ cosφ 00 0 1

.

Apparently Q(r, φ, z) is orthogonal and det(Q(r, φ, z)

)= 1. The local coordinate vectors are

then

er =

cos φsinφ0

, eφ =

− sin φcosφ0

and ez =

001

.

Example. Parametrization of a 3-dimensional manifold of R3 using spherical coordinates has

the form

r = γ(ρ, θ, φ) = (ρ sin θ cosφ, ρ sin θ sin φ, ρ cos θ) ((ρ, θ, φ) ∈ U),

where U is the parameter domain. Thus

γ ′(ρ, θ, φ) =

sin θ cosφ ρ cos θ cosφ −ρ sin θ sin φsin θ sin φ ρ cos θ sinφ ρ sin θ cosφ

cos θ −ρ sin θ 0

, Λ(r, φ, z) =

1 0 00 ρ 00 0 ρ sin θ

and

Q(ρ, θ, φ) =

sin θ cosφ cos θ cosφ − sinφsin θ sin φ cos θ sin φ cos φ

cos θ − sin θ 0

.

Here, too, Q(ρ, θ, φ) is orthogonal and

det(Q(ρ, θ, φ)

)= 1 (check!). The local

coordinate vectors (see the figure on the

right) are

eρ =

sin θ cos φsin θ sinφ

cos θ

,

eθ =

cos θ cosφcos θ sinφ− sin θ

and

eφ =

− sin φcosφ0

.

z

x

y

eρ

eθ

eφ

In addition to representing a vector field F as a function of the curvilinear coordinates, i.e. inthe form F

(γ(u)

), often the field should be given in the local coordinates in the point γ(u) (or


in the tangent space in the point). Columns of the matrix Q(u) are the new coordinate vectorsin the old coordinates. The required representation is thus (cf. Section 1.3)

G(u) = Q(u)TF(γ(u)

).

For a scalar field f the representation is simply

g(u) = f(γ(u)

).

The gradient and the Laplacian of a scalar field f in curvilinear local coordinates are thus

Q(u)T∇f(γ(u)

)and ∆f

(γ(u)

),

and the curl and the divergence of a vector field F are

Q(u)T∇× F(γ(u)

)and ∇ • F

(γ(u)

).

A2.2 Pullbacks

The correct tool for dealing with curvilinear coordinates is the so-called pullback. If Φ is ak-form field of Rn and δ : Rm → R

n is a continuously differentiable function, then the k-formfield

(δ∗Φ)(u; r1, . . . , rk) = Φ(δ(u); δ′(u)r1, . . . , δ

′(u)rk)

of Rm is the so-called pullback form field of Φ with respect to the function δ. The notation δ∗Φis conventional. Note in particular that the δ′(u)ri’s are n-dimensional vectors, as they shouldbe. Pullbacks make it easy to define new form fields, more or less as Theorem 2.5 was usedto define manifolds. However, to make the pullback of a continuously differentiable form fieldcontinuously differentiable, the function δ should be twice continuously differentiable.

Pullbacks are clearly left-distributive over addition, i.e.

δ∗(Φ + Ψ) = δ∗Φ + δ∗Ψ,

but they have other nice properties:

Theorem A2.1. Pullbacks are left-distributive over wedge product and they commute with ex-

terior differentiation, i.e.

δ∗(Φ ∧Ψ) = δ∗Φ ∧ δ∗Ψ and d(δ∗Φ) = δ∗(dΦ).

Proof. By Theorem 4.2, a form field Φ can be expanded as a combination of elementary formfields as

Φ(p; r1, . . . , rk) =∑

1≤j1<j2<···<jk≤n

aj1,j2,...,jk(p)(dxj1 ∧ dxj2 ∧ · · · ∧ dxjk)(r1, . . . , rk).

It is seen then that it suffices to prove the properties for form fields of the form

a(p) dx1 ∧ dx2 ∧ · · · ∧ dxk.

Left distributivity over the wedge product of such form fields follows fairly directly from thedeterminant definition of elementary forms (we leave the details to the reader but cf. also theproof of Theorem 4.4).


We use induction on the degree k to prove commutativity of pullbacks and exterior differ-entiation. If k = 0 then (δ∗a)(u) is the composite function a(δ

(u)). Using chain rule and the

definition of pullbacks we get

d(δ∗a)(u; r) = a′(δ(u)

)δ′(u)r =

(δ∗(da)

)(u; r),

so the commutativity holds true for k = 0.Assume then that the theorem holds for degrees k ≤ m (induction hypothesis) and consider

the case of degree k = m+1. We may write the m+1-form field as Φ∧ dxm+1, where Φ is anm-form field. Using Cartan’s Magic Formula we see first that

d(Φ ∧ dxm+1) = dΦ ∧ dxm+1 + (−1)mΦ ∧ ddxm+1 = dΦ ∧ dxm+1 + 0 = dΦ ∧ dxm+1.

By the induction hypothesis, d(δ∗Φ) = δ∗(dΦ). Again applying the Magic Formula we get

d(δ∗(Φ ∧ dxm+1)

)= d(δ∗Φ ∧ δ∗dxm+1) = d(δ∗Φ) ∧ δ∗dxm+1 + (−1)mδ∗Φ ∧ d(δ∗dxm+1)

= δ∗(dΦ) ∧ δ∗dxm+1 + (−1)mδ∗Φ ∧ dd(δ∗xm+1)

= δ∗(dΦ) ∧ δ∗dxm+1 + 0 = δ∗(dΦ ∧ dxm+1) = δ∗(d(Φ ∧ dxm+1)

).

As an example of the many uses of pullbacks we take a reduction of the Generalized StokesTheorem.

Example. For the notation we refer to the Generalized Stokes Theorem in Section 5.4. Assume

that the manifold M has the (relaxed) parametrization p = γ(u) (u ∈ U), and further that the

bounded region with boundary A has the (relaxed) parametrization p = γ(u) (u ∈ V).Let us first consider orientations. The manifold M is oriented by a k-form field Ψ via

the sign of Ψ(p; t1, . . . , tk) where t1, . . . , tk are in the tangent space Tp(M), as explained in

Section 4.3. Writing ti = γ ′(u)vi (i = 1, . . . , k) we get the corresponding orientation for U(and V) by the pullback k-form field

Ψ(γ(u);γ ′(u)v1, . . . ,γ

′(u)vk

)= (γ∗Ψ)(u;v1, . . . ,vk).

The smooth boundary ∂ sMA is defined locally, near the point p0 = γ(u0), by a continuously

differentiable function gp0via the sign of gp0

(p): + (p inside A), − (p outside A), 0 (p in

the boundary), see Section 5.1. For the orientation of the smooth boundary, an exterior vector

texterior(p), satisfying g′p0(p)texterior(p) < 0, is chosen as t1, and the k − 1-form field

Ψ(p; texterior(p), t2, . . . , tk

).

is used. Returning to the parameter domain, for orienting ∂ sUV the function gp0

(p) is replaced

by hu0(u) = gp0

(γ(u)

). Writing texterior(p) = γ ′(u)vexterior(u) we see first that

h′u0(u)vexterior(u) = g′p0

(γ(u)

)γ ′(u)vexterior(u) = g′p(p0)texterior(p) < 0,

i.e. vexterior(u) will indeed be an exterior vector, and second that ∂ sUV is oriented by the pullback

k − 1-form field

Ψ(γ(u);γ ′(u)vexterior(u),γ

′(u)v2, . . . ,γ′(u)vk

)= (γ∗Ψ)

(u;vexterior(u),v2, . . . ,vk

).

Thus all four orientations, that of M, of ∂ sMA, of U , and of ∂ s

UV , agree.


Second we consider the integrals in the theorem. For any k-form field Ξ defined on M we

have ∫

−→A

Ξ =

∫

V

Ξ(γ(u);γ ′(u)

)du =

∫

V

(γ∗Ξ)(u; Ik) du =

∫

−→V

γ∗Ξ

(we used here the identity parametrization for U for which the derivative is the k × k identity

matrix Ik). Assuming γ is twice continuously differentiable, applying this to dΦ we thus get, by

Theorem A2.1, ∫

−→A

dΦ =

∫

−→V

γ∗(dΦ) =

∫

−→V

d(γ∗Φ).

Let then the smooth boundary ∂ sUV have the (relaxed) parametrization u = ε(s) (s ∈ S),

whence ∂ sMA has the (relaxed) parametrization p = γ

(ε(s)

)(s ∈ S). So

∫

−→∂ A

Φ =

∫

S

Φ(γ(ε(s)

);γ′(ε(s)

)ε′(s)

)ds =

∫

S

(γ∗Φ)(ε(s); ε′(s)

)ds =

∫

−→∂ V

γ∗Φ.

This means we can prove the Generalized Stokes Theorem by first taking the pullback to the

parameter domain and then proving the theorem there which is very much easier (the theorem

becomes essentially the Generalized Divergence Theorem). This however works only with the

annoying assumption of γ being twice continuously differentiable.

A2.3 Transforming Derivatives of Fields

Let us then return to the curvilinear coordinates and consider the basic physical form fields andtheir derivatives, assuming, as required, that γ is twice continuously differentiable. We firstconsider the gradient. By Theorem A2.2

d(γ∗f)(u; r) =(γ∗(df)

)(u); r) = (γ∗Φ∇f–work)(u; r)

= ∇f(γ(u)

)•(γ ′(u)r

)=(γ ′(u)T∇f

(γ(u)

))• r.

On the other hand,

(γ∗f)(u) = f(γ(u)

)and d(γ∗f)(u; r) = ∇uf

(γ(u)

)• r,

where ∇u operates on u, and so

γ ′(u)T∇f(γ(u)

)= ∇uf

(γ(u)

).

Note that here γ only needs to be once continuously differentiable. Normalizing γ ′(u) we thenget the gradient in the local coordinates:

Q(u)T∇f(γ(u)

)= Λ(u)−1∇ug(u).

Let us take next the curl. By Theorem A2.2

d(γ∗ΦF–work)(u; r1, r2) =(γ∗(dΦF–work)

)(u; r1, r2) = (γ∗Φ∇×F–flux)(u; r1, r2)

= det(∇× F

(γ(u)

),γ ′(u)r1,γ

′(u)r2)

= det(γ ′(u)

)det(γ ′(u)−1∇× F

(γ(u)

), r1, r2

)

= det(γ ′(u)

)(γ ′(u)−1∇× F

(γ(u)

))•(γ ′(u)r1

)×(γ ′(u)r2

).


On the other hand,

(γ∗ΦF–work)(u; r) = F(γ(u)

)•(γ ′(u)r

)=(γ ′(u)TF

(γ(u)

))• r

= Φγ ′(u)TF(γ(u))–work(u; r),

and

d(γ∗ΦF–work)(u; r1, r2) = Φ∇u×(γ′(u)TF(γ(u)))–flux(u; r1, r2)

=(∇u ×

(γ ′(u)TF

(γ(u)

)))• r1 × r2,

and comparing with the above

γ ′(u)−1∇× F(γ(u)

)=

1

det(γ ′(u)

)∇u ×(γ ′(u)TF

(γ(u)

)).

Normalizing γ ′(u) we get the desired curl given in the local coordinates:

Q(u)T∇× F(γ(u)

)=

Λ(u)

det(γ ′(u)

)∇u ×(Λ(u)G(u)

).

The divergence is treated similarly. By Theorem A2.2

d(γ∗ΦF–flux)(u; r1, r2, r3) =(γ∗(dΦF–flux)

)(u; r1, r2, r3) = (γ∗Φ∇•F–density)(u; r1, r2, r3)

= ∇ • F(γ(u)

)det(γ ′(u)r1,γ

′(u)r2,γ′(u)r3

)

= det(γ ′(u)

)∇ • F

(γ(u)

)det(r1, r2, r3).

On the other hand,

(γ∗ΦF–flux)(u; r1, r2) = det(F(γ(u)

),γ′(u)r1,γ

′(u)r2)

= det(γ ′(u)

)det(γ ′(u)−1F

(γ(u)

), r1, r2

)

= det(γ ′(u)

)(γ ′(u)−1F

(γ(u)

))• r1 × r2

= Φdet(γ ′(u))(γ ′(u)−1F(γ(u)))–flux(u; r1, r2)

and

d(γ∗ΦF–flux)(u; r1, r2, r3) = Φ∇u•(det(γ′(u))(γ ′(u)−1F(γ(u))))–density(u; r1, r2, r3)

= ∇u •(det(γ ′(u)

)(γ ′(u)−1F

(γ(u)

)))det(r1, r2, r3),

and thus

∇ • F(γ(u)

)=

1

det(γ ′(u)

)∇u •(det(γ ′(u)

)(γ ′(u)−1F

(γ(u)

))).

The divergence in the local coordinates is then obtained by normalizing γ ′(u):

∇ • F(γ(u)

)=

1

det(γ ′(u)

)∇u •(det(γ ′(u)

)Λ(u)−1G(u)

).


Combining the representations of the gradient and the divergence we also get the Laplacianin the local coordinates:

∆f(γ(u)

)=

1

det(γ ′(u)

)∇u •(det(γ ′(u)

)Λ(u)−2∇ug(u)

).

Note that this expression contains both second order and first order partial derivatives.

A2.4 Derivatives in Cylindrical and Spherical Coordinates

Application of the above formulas is basically easy but somewhat tedious. A symbolic compu-tation program, e.g. Maple, will come handy.

We collect here the results for cylindrical and spherical coordinates (clearly the correspond-ing parametrizations are twice continuously differentiable). First the cylindrical coordinateswhere the local basis vectors are er, eφ and ez. Writing

G = Grer +Gφeφ +Gzez

we have

∇f =∂g

∂rer +

1

r

∂g

∂φeφ +

∂g

∂zez

∇ • F =1

rGr +

∂Gr

∂r+

1

r

∂Gφ

∂φ+∂Gz

∂z

∇× F =(1r

∂Gz

∂φ− ∂Gφ

∂z

)er +

(∂Gr

∂z− ∂Gz

∂r

)eφ +

1

r

(Gφ + r

∂Gφ

∂r− ∂Gr

∂φ

)ez

∆f =1

r

∂g

∂r+∂2g

∂r2+

1

r2∂2g

∂φ2+∂2g

∂z2

These representations of the derivative operations are very handy for axially symmetric fields.The spherical coordinates in turn are very handy for radially symmetric fields. The local

basis vectors are eρ, eθ and eφ. Writing

G = Gρeρ +Gθeθ +Gφeφ

we have

∇f =∂g

∂ρeρ +

1

ρ

∂g

∂θeθ +

1

ρ sin θ

∂g

∂φeφ

∇ • F =2

ρGρ +

∂Gρ

∂ρ+

1

ρ tan θGθ +

1

ρ

∂Gθ

∂θ+

1

ρ sin θ

∂Gφ

∂φ

∇× F =1

ρ sin θ

(cos θ Gφ + sin θ

∂Gφ

∂θ− ∂Gθ

∂φ

)eρ +

1

ρ

( 1

sin θ

∂Gρ

∂φ−Gφ − ρ

∂Gφ

∂ρ

)eθ

+1

ρ

(Gθ + ρ

∂Gθ

∂ρ− ∂Gρ

∂θ

)eφ

∆f =2

ρ

∂g

∂ρ+∂2g

∂ρ2+

1

ρ2 tan θ

∂g

∂θ+

1

ρ2∂2g

∂θ2+

1

ρ2 sin2 θ

∂2g

∂φ2

Appendix 3

ANGLE

”For Angling may be said to be so like the Mathematics that it cannever be fully learned...”

(IZAAK WALTON: The Compleat Angler)

A3.1 Angle Form Fields and Angle Potentials

The flux form field (n− 1-form field)

ΦF–flux(p; r1, . . . , rn−1) = ‖p‖−n det(pT, r1, . . . , rn−1)

of the vector field F(p) = ‖p‖−npT in Rn is called the angle form field, denoted by Φn−angle.

Its integral over an n− 1-dimensional oriented manifold (or region with boundary)−→M of Rn is

called the angle spanned by−→M from the origin. (And it is assumed that the origin itself is not

in−→M.)Since Φn−angle is the flux form field of the vector field

F(p) = ‖p‖−npT,

its exterior derivative is the density form field of the divergence ∇ • F, and this divergence iszero (check!), Poincaré’s Lemma says that the angle form field Φn−angle has a potential Ψn−angle

(an n− 2-form field) in star-shaped regions, a so-called angle potential.

By the (Generalized) Stokes Theorem, then the angle spanned by the n − 1-dimensionaloriented manifold

−→M is∫

−→M

Φn−angle =

∫

−→M

dΨn−angle =

∫

−→∂ M

Ψn−angle.

The angle is thus determined by the boundary of the manifold, and can be computed by integra-tion over it.

The case n = 1 is included here, although it is very simple and subject to interpretation.Then

Φ1−angle(p) =p

|p| = signum(p),

and an oriented 0-dimensional oriented manifold is a finite set−→M = p1, . . . , pm of oriented

points. The orientation of a point pi is denoted by ω(pi). When chosen as the sign of pi it

contributes positively to the angle, negatively otherwise. Then the (net) angle spanned by−→M is

∫

−→M

Φ1−angle =m∑

i=1

ω(pi)signum(pi).

Note that if the orientation of each point pi is its sign, then the angle is = m.

118

APPENDIX 3. ANGLE 119

A3.2 Planar Angles

Origins of the names in the previous section will be apparent when we take a look at the familiarcase of n = 2, the planar angle. Then

Φ2−angle(p; r1) = ‖p‖−2 det(pT, r1) =xy1 − x1y

x2 + y2,

when we denote p = (x, y) and r1 =

(x1y1

), i.e.

Φ2−angle =−y

x2 + y2dx+

x

x2 + y2dy.

This is the exterior derivative of the function (0-form field) atan in the star-shaped region whichwe get by removing from R

2 the positive x-axis and the origin, cf. the example in Section 2.5.Even though atan itself is not continuous in the positive x-axis—and not even defined in theorigin—its derivative atan′ is continuous everywhere except in the origin.

Let us integrate Φ2−angle over the oriented curve

−→C : r = γ(u) (u ∈ U)

(a relaxed parametrization). We assume in addition that the curve does not contain the origin.Then ∫

−→C

Φ2−angle =

∫

U

atan′(γ(u)

)γ ′(u) du =

∫

U

d

duatan

(γ(u)

)du.

This is the net angle that a point moving along the curvescans seen from the origin, counted as positive in the posi-tive rotation and negative in the negative rotation. Note thatdiscontinuity of atan does not matter because when mov-ing over the positive x-axis the value of atan changes by2π, but its derivative is continuous. The star-shaped regionindicated above could be replaced by any region we get byremoving from R

2 a ray starting from the origin, and mak-ing the corresponding changes in the definition of atan.

x

yC

The angle potential in R2 is a scalar potential, essentially atan or any function obtained by

adding a constant. The boundary is formed of separatepoints, and the net angle spanned is always in principle ob-tained by addition/subtraction.

Example. Let us compute the angle spanned by the the

curve

−→C : r = γ(u) =(u cos(cosu), u sin(cosu)

)(1 ≤ u ≤ 5),

see the figure on the right (Maple). Then

–2

–1

0

1

y

1 2 3 4x

atan′(γ(u)

)γ ′(u) = − u sin(cosu)

u2(cos(cosu) + u sin(cosu) sinu

)

+u cos(cosu)

u2(sin(cosu)− u cos(sin u) sinu

)

= − sin u


and the net angle is5∫

1

(− sin u) du = cos 5− cos 1 ∼= −0.26 rad,

as it should.

Note. For those familiar with complex analysis this brings to mind the complex logarithm and

its derivative.

A3.3 Solid Angles

An angle potential in R3 is the work form field of a vector potential of the Newton vector field

F(p) =pT

‖p‖3 ,

and Φ3−angle is the flux form field of F. The angle here is the so-called solid angle.

Geometrically the solid angle spannedby the surface

−→S seen from the origin isthe area of the part of the origin-centeredunit sphere P that exactly covers

−→S (see thefigure on the right). The general situationis however more complex. We must allowthe possibility that some part of the sphereis seen ”twice” or even more times. Wealso count as positive those points p of thesurface where the normal n points awayfrom the origin, i.e. p • n > 0, and asnegative points where the normal pointstowards the origin, i.e. p•n < 0. The wholespace obviously spans the solid angle 4π (the

S

y

x

z

P

area of a unit sphere). The unit of solid angle is the steradian (sr).Mathematically the solid angle seen from the origin spanned by the oriented 2-dimensional

manifold −→S : r = γ(u) (u ∈ U)

(an oriented surface in a relaxed parametrization) of R3 is

Ω(−→S ) =

∫

U

δ(u) • ∂δ(u)∂u1

× ∂δ(u)

∂u2du,

where

δ(u) =γ(u)∥∥γ(u)

∥∥ .

Here we again assume that S does not contain the origin. Note that the values of δ(u) are in theunit sphere x2 + y2 + z2 = 1.

On the other hand, −→P : r = δ(u) (u ∈ U)


is not (necessarily) a parametrized manifold, even in a relaxed sense. A part of the unit spheremay appear several times, possibly with opposite orientations, depending on how many times aray starting from the origin intersects the surface

−→S and in which direction. Generally however−→P can be considered as an oriented trajectory manifold in a relaxed parametrization.The unit normal of the origin-centered unit sphere in the point δ(u) is either δ(u)T (the

exterior normal) or −δ(u)T (the interior normal). Thus

∂δ(u)

∂u1× ∂δ(u)

∂u2= ±

∥∥∥∥∂δ(u)

∂u1× ∂δ(u)

∂u2

∥∥∥∥ δ(u),

where the sign is dictated by the particular orientation, and further

δ(u) • ∂δ(u)∂u1

× ∂δ(u)

∂u2= ±

∥∥∥∥∂δ(u)

∂u1× ∂δ(u)

∂u2

∥∥∥∥ .

The thus defined Ω(−→S ) is then just the solid angle corresponding to the geometrical idea.

That indeed

Ω(−→S ) =

∫

−→S

Φ3−angle

follows as special case of the general result in the next section (but can be verified quite easilyseparately, too).

Example. Using the result in the example in Section 6.3 we get a vector potential U for the

vector field

F(p) =p

‖p‖3in the star-shaped region that we get by removing from the space one of the rays −up0 (u ≥ 0)where p0 6= 0, as

U(p) =p0 × p

‖p0‖‖p‖2 + (p0 • p)‖p‖(we chose p1 = 0).6 The corresponding angle potential is the 1-form field

Ψ3−angle(p; r) = U(p) • r = p0 × p • r‖p0‖‖p‖2 + (p0 • p)‖p‖

.

The solid angle spanned by the oriented surface−→S is thus obtained by Stokes’ Theorem by

integrating over the correspondingly oriented boundary curve−→C of

−→S :

Ω(−→S ) =

∮

−→C

U(p) • ds =∮

−→C

p0 × p

‖p0‖‖p‖2 + (p0 • p)‖p‖• ds.

We can choose the ray (the Dirac string) removed from the space to be one pointing to the

opposite of the direction of sight, i.e. p0 is in the direction of sight.

As a concrete example, let us compute the solid angle spanned by the ellipse

−→C : p = γ(u) = (cosu, 2 sinu, 2) (0 ≤ u ≤ 2π)

6For simplicity we again use the same notation for points and vectors.


—or any surface having it as a boundary curve—from the ori-

gin. We choose p0 = (0, 0, 1). See the figure on the right

(Maple). Then

p0 × γ(u) =

−2 sin ucosu0

and γ ′(u) =

− sin u2 cosu

0

and the solid angle is

Ω =

2π∫

0

2 du

cos2 u+ 4 sin2 u+ 4 + 2√cos2 u+ 4 sin2 u+ 4

=

2π∫

0

2 du

3 sin2 u+ 5 + 2√3 sin2 u+ 5

∼= 1.1 sr.

1

2

3

z

–2–1.5

–1–0.5

0.511.5

2

x

–1–0.50.51 y

(The integral here is a so-called elliptic integral and it cannot be given by elementary functions.

Numerical calculation is of course possible and easy.)

The part of the space not spanned by the ellipse then determines the solid angle

4π − Ω ∼= 11.5 sr.

Note how the ray removed from the space—here the negative z-axis—deftly chooses which one

of these two conjugate solid angles, Ω or 4π − Ω, is computed!

A3.4 Angles in Rn

Mathematically the angle ”seen” from the origin spanned by the oriented n − 1-dimensionalmanifold −→M : r = γ(u) (u ∈ U)

(an oriented hypersurface in a relaxed parametrization) of Rn is

Ω(−→M) =

∫

U

±√

det(δ′(u)Tδ′(u)

)du,

where

δ(u) =γ(u)

‖γ(u)‖and the sign is chosen by the particular local orientation. Here we again assume that S does notcontain the origin. Thus Ω(

−→M) is the volume of the part P of a unit hypersphere ‖p‖ = 1 ofR

n exactly covering M as ”seen” from the origin. Here, too,

−→P : r = δ(u) (u ∈ U)

is not (necessarily) a parametrized manifold, even in a relaxed sense, but it may be consideredas an oriented trajectory manifold in a relaxed parametrization.

Writing first

δ(u) =γ(u)

(γ(u) • γ(u)

)1/2


it is straightforward to verify (check!) that

δ′(u) =1∥∥γ(u)∥∥ γ ′(u)− 1∥∥γ(u)

∥∥3 γ(u)Tγ(u)γ ′(u)

and further (check this, too!) that

δ′(u)Tδ′(u) =1∥∥γ(u)∥∥2 γ

′(u)Tγ ′(u)− 1∥∥γ(u)∥∥4 γ

′(u)Tγ(u)Tγ(u)γ ′(u).

Thus we have the formula

Ω(−→M) =

∫

U

± 1∥∥γ(u)∥∥n−1

√det(γ ′(u)Tγ ′(u)− 1∥∥γ(u)

∥∥2 γ′(u)Tγ(u)Tγ(u)γ ′(u)

)du.

The Gauß elimination formula for block matrices is (check!)(A B

C D

)=

(A O

O I

)(I O

C I

)(I A−1B

O D−CA−1B

)

for zero matrixesO and identity matrices I of appropriate sizes, provided that A is a nonsingularsquare matrix. Taking determinants we immediately have the so-called Schur identity7

det

(A B

C D

)= det(A) det(D−CA−1B),

which we will apply to the matrix

(γ(u)T

∣∣γ ′(u))T(

γ(u)T∣∣γ ′(u)

)=

(γ(u)

γ ′(u)T

)(γ(u)T

∣∣γ ′(u))=

( ∥∥γ(u)∥∥2 γ(u)γ ′(u)

γ ′(u)Tγ(u)T γ ′(u)Tγ ′(u)

),

taking A to be the 1× 1 matrix∥∥γ(u)

∥∥2. The determinant of this matrix is then

det(γ(u),γ ′(u)

)2=∥∥γ(u)

∥∥2 det(γ ′(u)Tγ ′(u)− 1∥∥γ(u)

∥∥2 γ′(u)Tγ(u)Tγ(u)γ ′(u)

).

Thus indeed

Ω(−→M) =

∫

U

1∥∥γ(u)∥∥n det

(γ(u),γ ′(u)

)du =

∫

−→M

Φn−angle.

Note that we did not indicate here the sign in the integrand. The sign choice is now automaticsince the sign of the determinant tells whether or not the normal points away from the origin(+) or towards it (−). We leave it as an exercise for the reader to verify the not so difficult factthat det

(γ(u),γ ′(u)

)is a dot product of γ(u) and a nonzero normal vector. The determinant

also takes care of situations where the normal and γ(u) are perpendicular and the contributionto the angle, and of the determinant, is zero. Indeed, large parts of the hypersurface M maywell be parts of conical hypersurfaces having the apex in the origin.

The full n-dimensional angle is the angle spanned by an origin-centered hypersphere, i.e.,the volume of a unit hypersphere. A somewhat difficult induction proof shows that it is

2πn/2

Γ(n/2),

where Γ is the gammafunction. Since Γ(1/2) =√π, Γ(1) = 1 and Γ(3/2) =

√π/2, the

formula gives correct values for n = 2 and n = 3, but also the value 2 of the full 1-dimensionalangle.

7Here D − CA−1

B is the so-called Schur complement of D. Schur’s complements are very useful in manyother contexts, too, see e.g. ZHANG, F. (Ed.): The Schur Complement and Its Applications. Springer (2005).

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Index 126

Index

active variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16addition of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . .2advection equation . . . . . . . . . . . . . . . . . . . . . . . . . 102affine approximation . . . . . . . . . . . . . . . . . . . . . . . . 30affine subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39Ampère’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75,81angle form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118angle potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3,118atan . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25,92,119anticommutativity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4antisymmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46atlas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1biharmonic equation. . . . . . . . . . . . . . . . . . . . . . . .102bilinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Biot–Savart law . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101Black–Scholes equation . . . . . . . . . . . . . . . . . . . . 102boundary . . . . . . . . . . . . . . . . . . . . . . . . . . 1,43,63,115Cartan’s Magic Formula . . . . . . . . . . . . . . . . . 72,107chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9charge conservation law . . . . . . . . . . . . . . . . . . . . . 76chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18closed form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83closed set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1closure of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1connected manifold . . . . . . . . . . . . . . . . . . . . . . . . . 58conservative vector field . . . . . . . . . . . . . . . . . . . . . 90continuity law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75coordinate function . . . . . . . . . . . . . . . . . . . . . . . . . . . 6coordinate point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5coordinate transform . . . . . . . . . . . . . . . . . . . . . . . . 7,9coordinate vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6coordinate-freeness . . . . . . . . . . . . . . . . . . . . . . . 12,22cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4,7curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10,74,115current density form field . . . . . . . . . . . . . . . . . . . . 76curvilinear coordinate system . . . . . . . . . . . . . . . 111cyclical symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5cylindrical coordinates . . . . . . . . . . . . . . . . . . 111,117d’Alembertian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106density form field . . . . . . . . . . . . . . . . . . . . . . . . 59,73derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9,70difference of vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 3differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70differential form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46diffusion equation . . . . . . . . . . . . . . . . . . . . . . . . . . 103dipole approximation . . . . . . . . . . . . . . . . . . . . . . . . 98Dirac’s string . . . . . . . . . . . . . . . . . . . . . . . . . . . 93,121

direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2directional derivative . . . . . . . . . . . . . . . . . . . . . . . . 10distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,4Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . 78divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 10,74,116division by scalar . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3dot product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4,6double-curl expansion . . . . . . . . . . . . . . . . . . . . 11,97Dym’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102dynamical . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13electric dipole moment . . . . . . . . . . . . . . . . . . . . . . 99elementary form . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49elliptic PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102exact form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83exception set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43explicit representation . . . . . . . . . . . . . . . . . . . . . . . 21extended parameter domain . . . . . . . . . . . . . . . . . . 43exterior derivative . . . . . . . . . . . . . . . . . . . . . 69,70,83exterior vector . . . . . . . . . . . . . . . . . . . . . . . . . . 66,114face . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70Faraday’s form field . . . . . . . . . . . . . . . . . . . . . . 61,76Faraday’s law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75FEM. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109fiber . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37fiber bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37finite element method . . . . . . . . . . . . . . . . . . . . . . 110flux form field . . . . . . . . . . . . . . . . . . . . . . . . . . . 59,73flux . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36forcing function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46four-potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97Fundamental Theorem of Integral Calculus 62,83Gauß’ law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75Gauß’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 62,78General Divergence Theorem. . . . . . . . . . . . . 78,115Generalized Stokes’ Theorem . . . . . . . . . 62,77,114geometric vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9,73,115Gradient Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 62Gramian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37,40,41graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16Green’s first identity . . . . . . . . . . . . . . . . . . . . . . . . 109Green’s second identity . . . . . . . . . . . . . . . . . . . . . 109Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 62,79Green’s third identity . . . . . . . . . . . . . . . . . . . . . . . 110harmonic function . . . . . . . . . . . . . . . . . . . . . . . 97,109heat equation . . . . . . . . . . . . . . . . . . . . . . . . . . 103,104Helmholtz’s decomposition . . . . . . . . . . . . . . . . . . 96Hodge’s decomposition . . . . . . . . . . . . . . . . . . . . . . 97

Index 127

Hodge’s dual . . . . . . . . . . . . . . . . . . . . . . . . . . . . .60,75Hodge’s star . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60,75hyperbolic heat equation . . . . . . . . . . . . . . . . . . . . 102hyperbolic PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102Implicit Function Theorem . . . . . . . . . . . . . . . . . . . 19implicit representation . . . . . . . . . . . . . . . . . . . . . . . 21inner cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53interior of a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1interior vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66inverse image . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17irrotational . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88,103Jacobian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Jordan measurable . . . . . . . . . . . . . . . . . . . . . . . . . . 38Jordan’s inner measure . . . . . . . . . . . . . . . . . . . . . . 38Jordan’s measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Jordan’s outer measure . . . . . . . . . . . . . . . . . . . . . . 38k-form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46k-form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52Klein’s bottle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79k-null set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40Koch’s snowflake . . . . . . . . . . . . . . . . . . . . . . . . . . . 66Korteweg–de Vries equation . . . . . . . . . . . . . . . . 102Lagrange’s formulas . . . . . . . . . . . . . . . . . . . . . . . . . . 5Lambert’s W function . . . . . . . . . . . . . . . . . . . . . . . 22Laplace’s equation . . . . . . . . . . . . . . . . . . . . . . 96,103Laplacian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10,117length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2level manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21local coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . 111locus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18Lorenz’s gauge condition . . . . . . . . . . . . . . . . . . . . 98lossy wave equation . . . . . . . . . . . . . . . . . . . . . . . . 102magnetic dipole moment . . . . . . . . . . . . . . . . . . . . 100manifold . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17,24,29mass form field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59Maxwell’s equations . . . . . . . . . . . . . . . . . . . . .75,102Maxwell’s form field . . . . . . . . . . . . . . . . . . . . . 61,76Mean Value Theorem for Harmonic Functions 110Möbius’ band . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55,90monopole approximation . . . . . . . . . . . . . . . . . . . 101multilinearity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46multiplication by scalar . . . . . . . . . . . . . . . . . . . . . . . 3multipole expansion . . . . . . . . . . . . . . . . . . . . . . . . 101nabla rules. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11Nash’s Embedding Theorem . . . . . . . . . . . . . . . . . 82n-cube . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38Newton’s form field . . . . . . . . . . . . . . . . . . . . . . . . . 83Newton’s potential . . . . . . . . . . . . . . . . . . . . . . . . . . 83Newton’s vector field . . . . . . . . . . . . . . . . . . . . . . . . 93nonstationary . . . . . . . . . . . . . . . . . . . . . . . 13,103,105normal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

normal bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37normal derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 109normal space . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34,37normal vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34normalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3null set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40octupole approximation . . . . . . . . . . . . . . . . . . . . . 101Ohm’s law. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75open ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1open set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1opposite vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .102orientable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55orientation . . . . . . . . . . . . . . . . . . . . . . . . . . . 54,67,114outer cover . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38parabolic PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102parallelepiped . . . . . . . . . . . . . . . . . . . . . . . . . 39,41,42parameter domain . . . . . . . . . . . . . . . . . . . . . . . . . . . 24parametrization . . . . . . . . . . . . . . . . . . . . . . . 23,41,43parametrized manifold . . . . . . . . . . . . . . . . . 23,42,43partial differential equation . . . . . . . . . . . . . . . . . 103partial integration . . . . . . . . . . . . . . . . . . . . . . . . . . 107passive variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16PDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103planar angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119plane wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104Poincaré’s Lemma . . . . . . . . . . . . . . . . . . . . . 85,87,92point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1,5point of action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Poisson’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . 96polar space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83projection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4pullback form field . . . . . . . . . . . . . . . . . . . . . . . . . 113quadrupole approximation . . . . . . . . . . . . . . . . . . 101region with boundary . . . . . . . . . . . . . . . . . . . . . . . . 65relaxed parametrization . . . . . . . . . . . . . . . . . . . . . . 43reparametrization . . . . . . . . . . . . . . . . . . . 26,42,45,59reverse heat equation . . . . . . . . . . . . . . . . . . . . . . . 102Riemann’s integral . . . . . . . . . . . . . . . . . . . . . . . . . . 39scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8scalar potential . . . . . . . . . . . . . . . . . . . . . . . . . . 87,104scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4scalar triple product . . . . . . . . . . . . . . . . . . . . . . . . . . 5Schur’s identity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123simply connected . . . . . . . . . . . . . . . . . . . . . . . . . . . 90smooth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16smooth boundary . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63smooth point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63solenoidal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92solid angle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120source density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Index 128

spherical coordinates . . . . . . . . . . . . . . . . . . . 112,117spherical wave . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106standard forms of PDEs. . . . . . . . . . . . . . . . . . . . .102star-shaped . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85stationary. . . . . . . . . . . . . . . . . . . . . . . . . . .13,103,104steradian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 62,79symplectic form. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82symplectic potential . . . . . . . . . . . . . . . . . . . . . . . . . 82tangent bundle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37tangent space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30,37tangent vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7telegraph equation . . . . . . . . . . . . . . . . . . . . . . . . . 103Thomson’s circulation law . . . . . . . . . . . . . . . . . . . 81torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56,91trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12trajectory manifold . . . . . . . . . . . . . . . . . . . . . . . 29,43triangle inequality . . . . . . . . . . . . . . . . . . . . . . . . . . 1,4triple vector product . . . . . . . . . . . . . . . . . . . . . . . . . . 5unit vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2,6vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8,36vector potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92vector product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Vectoral Gauß’ Theorem . . . . . . . . . . . . . . . . . . . . . 80volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38,39,41vortex density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .75vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75wave equation . . . . . . . . . . . . . . . . . . . . . . . . . 102,106wedge product . . . . . . . . . . . . . . . . . . . . . . . . . . . 49,51work form field . . . . . . . . . . . . . . . . . . . . . . . . . . 59,73zero vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2Zhukovsky’s lift law . . . . . . . . . . . . . . . . . . . . . . . . . 82

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