vector field.pdf

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 Fields

13 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 in R392 6.3 Vector Potential of a Vector Field in R396 6.4 Helmholtzs 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 GREENSIDENTITIES

107 A1.1 Partial Integration111 A1.2 Greens 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 in Rn

124 References126 Index

ForewordThese 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 Vektorikentt, 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 analysisclassical as well as modernhas 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. This generally 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 concepts and 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 classics TALLQVIST and VISL, too.

iii

manifolds and form fields, and Cliffords 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 & HUBBARDhave 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 & STERNBERG might be mentioned,too.

Keijo Ruohonen

Chapter 1

POINT. VECTOR. VECTOR FIELD

One need only know that geometric objects in spacetimeare entities that exist independently of coordinate systems

or reference frames.

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

1.1 Geometric PointsA point in space is a physico-geometric primitive, and is not given any particular definitionhere. Let us just say that dealing with points, lines, planes and solids is mathematically partof 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 byd(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 thatB(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 ballB(R,P ) (R > 0) contains both a point of A and a point of the complement of A. Inparticular the boundary of the empty set is empty. A set is thus open if and only if it doesnot 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 thusclosed. Since the boundaries of a set and its complement clearly are the same, a set isclosed 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 theseoperations are defined. But in reality they are then the corresponding operations for vectors, aswill 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 VectorsThe directed line segment connecting the two points P (the initial point) and Q (the terminalpoint) is denoted byPQ. Two such directed line segmentsPQ andRS are said to be equivalentif they can be obtained from each other by parallel transform, i.e., there is a parallel transformwhich takes P to R and Q to S, or vice versa.

Directed line segments are thus partitioned into equivalence classes. In each equivalenceclass directed line segments are mutually equivalent, while those in different equivalence classesare 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 vectorwith 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, directionand length do not depend on the choice of the representative.

In the sequel geometric vectors will be denoted by small italic letters equipped with anoverarrow: ~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 vectorwith 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 tothe direction of the geometric vectorunless ~r is the zero vectorand the magnitude ~r is givenin 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 hasno attached physical unit, or has an empty unit. Physical units can be multiplied, divided andraised 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 aproper 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 theyare 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. SpringerVerlag (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 beparenthesized in any (correct) way, or written totally without parenteheses, without theresult 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 a positive scalar, then ~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 |~r |. Then ~r = PR.

In addition it is agreed that ~0 = ~0 and 0~r = ~0.This operation is multiplication of vector by scalar. Defining further ()~r = (~r ) weget multiplication by a negative scalar. Evidently

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

If the physical scalar and physical vector ~r have their physical units, the unit of ~r istheir product. With a bit of work the following laws of calculation can be (geometrically)verified:

1(2~r) = (12)~r ,

(1 + 2)~r = 1~r + 2~r and(~r1 + ~r2) = ~r1 + ~r2,

where , 1, 2 are scalars and ~r, ~r1, ~r2 are vectors.Division of a vector ~r by a scalar 6= 0 is multiplication of the vector by the inverse 1/,denoted by ~r/.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 choiceof representatives) is the angle between the directed line segments PQ and PR given inradians in the interval [0, ] rad. It is obviously assumed here that ~r, ~s 6= ~0. Moreover anangle 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) isd(~r, ~s ) = d(Q,R) = |~r ~s |.

In particular d(~r,~0 ) = |~r |. This distance also satisfies the triangle inequalityd(~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 (~s+ ~t ) = (~r ~s ) + (~r ~t ),

where and are scalars. Geometrically commutativity is obvious. Bilinearity on theother hand requires a somewhat complicated geometric proof. Using coordinates makesbilinearity straightforward.The unit of the dor 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 alwayszero.)

The cross product (or vector product) of the vectors ~r and ~s is the vector ~r ~s given bythe following. First, if ~r or ~s is ~0, or (~r, ~s ) is 0 or , then ~r ~s = ~0. Otherwise ~r ~s isthe unique vector ~t satisfying

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

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 (~s+ ~t ) = (~r ~s ) + (~r ~t ),where and are scalars. Geometrically anticommutatitivy is obvious, handidnesschanges. Bilinearity again takes a complicated geometric proof, but is fairly easily seenusing coordinates.


Cross product is an information-dense operation, involving lengths of vectors, angles andhandidness. 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 theproduct 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 theparallelepiped 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. (Asspecial 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 Lagranges 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 ageometric intuition is then not as easily obtained.

1.3 Coordinate Points and VectorsIn basic courses on mathematics points are thought of as coordinate points, i.e. triples (a, b, c) ofreal numbers. In the background there is then an orthonormal right-handed coordinate systemwith 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 is0 = (0, 0, 0).

A coordinate system is determined by the corresponding coordinate function which mapsgeometric points to the triples of R3. We denote coordinate functions by small boldface Greekletters, and their components by the corresponding indexed letters. If the coordinate function is, then the coordinates of the point P are

(P ) =(1(P ), 2(P ), 3(P )

).

A coordinate function is bijective giving a one-to-one correspondence between the geometricspace and the Euclidean space R3.

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

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

thend(P,Q) =

(P ) (Q) =(x1 x2)2 + (y1 y2)2 + (z1 z2)2.A coordinate function also gives a coordinate representation for vectors. The coordinate

version of the vector ~r = PQ is

(~r ) =((Q) (P ))T =

1(Q) 1(P )2(Q) 2(P )3(Q) 3(P )

.

Note that this is a column array. As is easily seen, this presentation does not depend on choicesof the representative directed line segments. In particular the zero vector has the representation(~0 ) = (0, 0, 0)T = 0T. Also the distance of the vectors ~r = PQ and ~s = PR (note thechoice of representatives) is obtained from the Euclidean norm of R3:

d(~r, ~s ) =(~r ) (~s ) = (Q) (R) = d(Q,R).

And then |~r | = d(~r,~0) = (~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. Inthe background there of course is a fixed coordinate system and a coordinate function. The rowarray versions of these are also used as coordinate points.

Familiar operations of coordinate vectors now correspond exactly to the geometric vectoroperations in the previous section. Let us just recall that if

(~r ) = r =

a1b1c1

and (~s ) = s =

a2b2c2

,

then~r ~s = r s = a1a2 + b1b2 + c1c2


and

(~r ~s ) = r s = b1c2 b2c1c1a2 c2a1

a1b2 a2b1

.

The latter is often given as the more easily remembered formal determinanti a1 a2j b1 b2k c1 c2

,to be expanded along the first column.

A coordinate transform changes the coordinate function. If and are two availablecoordinate functions, then they are connected via a coordinate transform, that is, there is a 33orthogonal matrix3 Q and a coordinate vector b such that

(P ) = (P )Q+ b

and(P ) = (P )QT bQT.

The coordinate representation of a vector ~r = PQ is transformed similarly:

(~r ) =

((Q) (P ))T = ((Q)Q+ b (P )Q b)T

= QT((Q) (P ))T = QT(~r )

and(~r ) = Q(~r ).

Note that b is the representation of the origin of the old coordinate system in the newsystem, and that the columns of QT are the representations of the basis vectors i, j,k of the oldcoordinate system in the new system. Similarly bQT is the representation of the origin ofthe new coordinate system in the old system, and the columns of Q are the representationsof the new basis vectors i, j,k in the old system.

1.4 Tangent Vectors. Vector Fields. Scalar FieldsGeometrically a tangent vector4 is simply a directed line segment PQ, the point P is its pointof action.

It is however easier to think a tangent vector (especially a physical one) as a pair [P,~r ] whereP is the point of action and ~r is a vector. It is then easy to apply vector operations to tangentvectors: just operate on the vector component ~r. If the result is a vector, it may be thought of asa tangent vector, with the original (joint) point of action, or as just a vector without any point ofaction. Moreover, in the pair formulation, a coordinate representation is simply obtained usinga coordinate function :

([P,~r ]

)=[(P ),(~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 Q1 = 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 onlythe vector component of the pair is used, usually in a coordinate representation. This is whatwe 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 pointof action). Mostly we denote this just by ~F . A vector field may not be defined for all points ofthe geometric space, i.e., its domain of definition may be smaller.

In the coordinate representation given by the coordinate function we denote

r = (P ) and F(r) = (~F (P )

),

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

r = rQ+ b (i.e. (P ) = (P )Q+ b)the vector field F = (~F ) (that is, its representation) is transformed to the the field F = (~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 thentaken to other coordinate systems using the transform formula. On the other hand, definitionof a physico-geometric vector field cannot possibly depend on any coordinate system, the fieldexists without any coordinates, and will automatically satisfy the transform formula.

A coordinate vector field is the vector-valued function of three arguments familiar frombasic 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 scalarfields are denoted by italic letters, usually small. In the coordinate representation we denoter = (P ) and just f(r) (instead of the correct f(1(r))). In the coordinate transform

r = rQ+ b (i.e. (P ) = (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 coordinatesystems using the transform formula. But a physico-geometric scalar field exists without anycoordinate systems, and will automatically satisfy the transform formula.

Again a coordinate scalar field is the real-valued function of three arguments familiar frombasic 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 avector field.


1.5 Differential Operations of FieldsNaturally, partial derivatives cannot be defined for physico-geometric fields, since they are in-trinsically connected with a coordinate system. In a coordinate representation partial derivativescan be defined, as was done in the basic courses.

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

f =(fx

,f

y,f

z

)and gradient

grad(f) = f = f T =

f

xf

yf

z

,

and the derivative or Jacobian (matrix) of a vector field F =F1F2F3

F =

F 1F 2F 3

=

F1x

F1y

F1z

F2x

F2y

F2z

F3x

F3y

F3z

.

Using the chain rule5 we get the transforms of the derivatives in a coordinate transformr = rQ + b:

f (r) =(f((r b)QT)) = f ((r b)QT)Q

andF(r) =

(QTF

((r b)QT)) = QTF((r b)QT)Q.

Despite partial derivatives depending on the coordinate system used, differentiability itselfis coordinate-free: If a field has partial derivates in one coordinate system, it has them in anyother coordinate system, too. This is true for second order derivatives as well. And it is true forcontinuity: Continuity in one coordinate system implies continuity in any other system. Andfinally it is true for continuous differentiability: If a field has continuous partial derivatives (firstor second order) in one coordinate system, it has them in any other system, too. All this followsfrom 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 derivativeof 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 , andits Laplacian

f = (f) = 2f = 2f

x2+2f

y2+f 2

z2,

and for a vector field F =

F1F2F3

its divergence

div(F) = F = F1x

+F2y

+F3z

and curl

curl(F) = F =

F3y

F2z

F1z

F3x

F2x

F1y

=

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 thedirection of the gradient is the direction of fastest growth for a scalar field, and its length is thisspeed of growth (given as a directional derivative). For divergence and curl the situation is notat all as obvious.

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

F1F2F3

.

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 1 and 2 are constant scalars, then e.g.

(1f + 2g) = 1f + 2g and

(1F+ 2G) = 1 F+ 2 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 scalarfield f or a vector field F = (F1, F2, F3)T as follows:

(G )f = G (f) = G1fx

+G2f

y+G3

f

z

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

(G )F =(G )F1(G )F2(G )F3

= FG.

These are both coordinate-free and hence fields. Coordinate-freeness of (G)F follows fromthe nabla rules below (or from the coordinate-freeness of FG).

Let us tabulate the familiar nabla-calculus rules:

(i) (fg) = gf + fg

(ii) 1f= 1

f 2f

(iii) (fG) = f G+ f G

(iv) (fG) = f G+ fG

(v) (FG) = F G F G

(vi) (FG) = (G )F ( F)G+ ( G)F (F )G

(vii) (F G) = (G )F ( F)G (G) F+ (F )GIn matrix notation(F G) = FTG+GTF.

(viii) ( F)G = (F FT)G

(ix) ( F) = 0

(x) f = 0

(xi) ( F) = ( F)F (so-called double-curl expansion)

(xii) (fg) = fg + gf + 2f g

In formulas (ix), (x), (xi) we assume F and f are twice continuously differentiable. Theseformulas are all symbolical identities, and can be verified by direct calculation, or e.g. using theMaple 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 operatorsthen means the following:

1. (f((r b)QT)) = QTf(r) (gradient)Subtracting b and multiplying by QT we move from the new coordinates r to the oldones, get the value of f , and then the gradient in the new coordinates. The result must bethe same when the gradient is obtained in the old coordinates and then transformed to thenew ones by multiplying by QT.

2. (QTF((r b)QT)) = F(r) (divergence)Subtracting b and multiplying by QT we move from the new coordinates r to the oldones, get F, transform the result to the new coordinates by multiplying by QT, and getthe divergence using the new coordinates. The result must remain the same when thedivergence is obtained in the old coordinates.

3. (QTF((r b)QT)) = QT( F(r)) (curl)Subtracting b and multiplying by QT we move from the new coordinates r to the oldones, get F, transform the result to the new coordinates by multiplying by QT, and get thecurl using the new coordinates. The result must be the same when the curl is obtained inthe 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))TQ and F(r) = QTF(r)Q.

This immediately gives formula 1. since

f (r) = f (r)T = QTf(r).

To show formula 2. we use the trace of the Jacobian. Let us recall that the trace of a squarematrix A, denoted trace(A), is the sum of the diagonal elements. A nice property of trace6 isthat if the product matrix AB is squarewhence BA is square, toothen

trace(AB) = trace(BA).

Sincetrace(F) =

F1x

+F2y

+F3z

= F,

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

trace(AB) =n

k=1

ckk =n

k=1

ml=1

aklblk =ml=1

nk=1

blkakl =ml=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 firstcomponent of the curlF(r). Using the transform formula for the Jacobian, nabla formula(viii) and rules for the scalar triple product we get

( F(r))1=

F 3y

F2

z

= qT3F(r)q2 qT2F(r)q3 (because F = QTFQ)

= qT3F(r)q2 qT3F(r)Tq2 (qT2Fq3 is a scalar)

= qT3(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 othercomponents are dealt with similarly.

Coordinate-freeness of the Laplacian follows directly from that of gradient and divergencefor 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 = 12

( (FG) + ( 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 thematrix form (G )F = FG, but the formula above is of other interest, too!)

1.6 Nonstationary Scalar and Vector FieldsPhysical fields naturally are often time-dependent or dynamical, that is, in the definition of thefield 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 pointof 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 callednonstationary, 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 forthe 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) = Ft

G+ F Gt

(2) t

(FG) = Ft

G+ F Gt

(3) t

(fF) =f

tF+ f

F

t

(4) t

(F GH) = Ft

GH+ F Gt

H+ F G Ht

(5) t

(F (GH)) = F

t (GH) + F

(Gt

H)+ F

(G H

t

)

Another kind of time dependence in a coordinate representation is obtained by allowing amoving coordinate system (e.g. a rotating one, as in a carousel). If the original representation ina fixed coordinate system is f(r) (a scalar field) or F(r) (a vector field), then at time t we havea 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 representationand 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 getthe 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 themoving coordinate system.

Chapter 2

MANIFOLD

The intuitive picture of a smooth surface becomes analyticwith the concept of a manifold. On the small scale a manifold

looks like Euclidean space, so that infinitesimal operationslike differentiation may be defined on it.

(WALTER THIRRING: A Course in Mathematical Physics)

2.1 Graphs of FunctionsThe graph of a function f : A Rm, where A Rk is an open set, is{(

r, f(r)) r A},

a subset of Rk+m. The graph is often denotedusing a slight abuse of notationas follows:

s = f(r) (r A).

Here r contains the so-called active variables and s the so-called passive variables. Aboveactive variables precede the passive ones in the order of components. In a graph we also allow asituation where the variables are scattered. A graph is smooth,1 if f is continuously differentiablein its domain of definitionA. In the sequel we only deal with smooth graphs. Note that a graphis specifically defined using a coordinate system and definitely is not coordinate-free.

A familiar graph is the graph of a real-valuedunivariate 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 allcurves or surfaces are graphs, however, e.g. circlesand 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 positionand time, that is 1, 2 and 3, whence k +m is 2, 3 or 4. On the other hand, degrees of freedomin mechanical systems etc. may lead to some very high dimensions

As a limiting case we also allow m = 0. In Rm = R0 there is then only one element (theso-called empty vector ()). Moreover, then Rk+m = Rk, thus all variables are active and thegraph of the function f : A Rm is A. Open subsets of the space are thus always graphs, andagreed 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 graphconsists 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

g1(C) = {r g(r) C}.Note that this has little to do with inverse functions, indeed the function g need not have aninverse at all, and C need not be included in B.

For a continuous function defined in an open setA the inverse image of an open set is alwaysopen.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) orm = 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 ManifoldsA subsetM of Rn is a k-dimensional manifold, if it is locally a smooth graph of some functionof k variables.5

Locally means that for every point p of M there is an open set Bp of Rn containing thepoint p such that M Bp is the graph of some function fp of some k variables. For differentpoints p the set Bp may be quite different, the active variables chosen in a different way, alwaysnumbering 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 smallatlas 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, becauseg(r) g(r0)2 = r r02 + 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 itsn-dimensional manifold, and each single point is a 0-dimensional manifold. If a space curve isa 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 theabsolute 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 ofTheorem 2.1 we getTheorem 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 LociOne way to define a manifold is to use loci. A locus is simply a set of points satisfying somegiven conditions. For instance, the P -centered circle of radius R is the locus of points havingdistance R from P . As was noted, it is a manifold.

In general a locus is determined via a coordinate representation, and the conditions are givenas 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 functionof n variables. The locus of points satisfying the condition is then the set of points (r, s) inRn 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 thepassive ones in the component order, active variables may be scattered, roles of variables maydiffer locally, actives changing to passives, etc.


Example. A circle and a sphere are loci of this kind when we set the conditionsF (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| = 0is 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 theorigin (but rather of two functions: y = x). Actually the condition

(y x)2 = 0does 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, andsomehow uniqueness of solution should be ensured, too, at least locally. In classical analysisthere is a result really taylor-made for this, the so-called Implicit Function Theorem, long knownand useful in many contexts. Here it is used to make the transition from local loci6 to graphs. Itshould be mentioned that in the literature there are many versions of the theorem,7 we choosejust one of them.Implicit Function Theorem. Assume that the function F : S Rnk, 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 (whence

also 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 are

linearly 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

rF

s

)in block form.)

Then there is an open subset B of Rkcontaining the point r0, and a uniquely determinedfunction f : B Rnk such that

6This is not pleonasm, although it might seem to be since local could be constructed to mean relating to alocus etc.

7See e.g. KRANTZ, S.G. & PARKS, H.R.: The Implicit Function Theorem. History, Theory, and Applications.Birkhuser (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) = Fs

(r, f(r)

)1F

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

Fr

(r, f(r)

)+ F

s

(r, f(r)

)f (r) = O.

Using the Implicit Function Theorem (and Theorem 2.1) we immediately get definition ofmanifolds 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 Rnk such that the conditions 1. 4. in the Implicit Function Theorem aresatisfied, and the locus condition F(p) = 0 defines the set MS, 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 Rnk such that the conditions 1. 4. of the ImplicitFunction Theorem are satisfied and the locus condition F(p) = 0 defines the set MS.Proof. Let us just see the case k > 0. (The case where k = 0 is similarreally a special case.)If M is a k-dimensional manifold of Rn, then locally in some open set containing the point p0it 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 mayassume that the active variables precede the passive ones.

Choose now the set S to be the Cartesian product A Rnk, i.e.

S = {(r, s) r A and s Rnk},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 thenF =

(f Ink )is of full rank (Ink is the (n k) (n k) identity matrix).

Excluding the n-dimensional manifolds, manifolds of Rn are thus exactly all sets which arelocal 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 therepresentation using local graphs of functionsas in the original definition is called an explicitrepresentation.9

Example. 2-dimensional manifolds in R3 are smooth surfaces. Locally such a surface is definedas the set determined by a condition

F (x, y, z) = 0,

where in the points of interest

F =(Fx

,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 locusof 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) = 08If B is the r0-centered ball of radius R in A and s0 Rnk, then the (r0, s0)-centered open ball of radius R

in Rn is included in S, because in this ballr r02 r r02 + s s02 =

(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 W1 (redlower branch), see the figure below (Maple).Since here

y = xey and ln2 > 1/e,

this means that

22

2

2.

exists, which may seem odd because2 > 1.

Actually it has the value

W0( ln2 ) ln2 = 2.

2.4 Mapping Manifolds. Coordinate-FreenessManifolds 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 thatthe resulting set really is a manifold. For parametrized manifolds this is frequently easier, seethe 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 full

rank (i.e. linearly independent rows),then the inverse image g1(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 surfaceF2(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. ThenM is an open subset of Rn and its inverse image g1(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 g1(M), i.e. a point suchthat g(r0) M. Locally near the point p0 = g(r0) the manifold M can be determined asa locus, by Theorem 2.4. More specifically, there is an open subset S of Rn and a functionF : S Rnk, 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 g1(S) is open. The locus condition

F(g(r)

)= 0

determines locally some part of the set g1(M). In the open set g1(S) the composite functionF 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, g1(M) is a manifold of Rm and its dimension ism (n k) (the dimension of r minus the dimension of F).

So far we have not considered coordinate-freeness of manifolds. A manifold is alwaysexpressly defined in some coordinate system, and we can move frome one system to anotherusing coordinate transforms. But is a manifold in one coordinate system also a manifold in anyother, and does the dimension then remain the same?

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

r = rQ+ b,

and choose m = n andg(r) = (r b)QT

in the theorem. Then a manifold M in coordinates r is the inverse image of the manifold incoordinates r, and thus truly a manifold. Dimension is preserved as well. Being a manifold inone coordinate system guarantees being a manifold in any other coordinates. Manifoldness isa coordinate-free property.

2.5 Parametrized ManifoldsWhenever a manifold can be parametrized it will be in many ways easier to handle.12 Forinstance, defining and dealing with integrals over manifolds then becomes considerably simpler.

A parametrization13 of a k-dimensional manifoldM of Rn consists of an open subset U ofRk (the so-called parameter domain), and a continuously differentiable bijective function

: U M

having a derivative matrix of full rank. Since the derivative is an nk matrix and k n, itis 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 ONEILL.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 (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 .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 (u) in the circle moves along the circumference ineiher direction. It cannot stop or turn back because is a bijection and has full rank. Thusalso the limiting point

p = limua+

(u)

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

open intervals in U , andas abovewe see that a sufficiently short open interval (v, v+) ismapped to an open arc of Y containing the point p. This would mean that 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 :

(x, y) = () = (R cos ,R sin) (0 < < 2).Now

() = (R cos ,R sin)

is a continuously differentiable bijection, and

() =

(R sinR cos

)

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

= 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

2 + arctany

xfor x > 0 and y < 0

+ arctany

xfor x < 0

2for x = 0 and y > 0

3

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 continuouslydifferentiableexcluding the negative x-axis,see the figure on the right (by Maple)since(verify!)

atan(x, y)

x= y

x2 + y2

andatan(x, y)

y=

x

x2 + y2.

0

2

4

6

8

10

2

1

1

2

y

2

1

1

2x

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

x2 + z2 = R2 , y = 0 , x 0is removed then a manifold is obtained which can be parametrized by the familiar sphericalcoordinates as

(x, y, z) = (, ) = (R sin cos ,R sin sin,R cos )

and the parameter domain is the open rectangle

U : 0 < < , 0 < < 2.The derivative matrix

(, ) =

R cos cos R sin sin R cos sin R sin cos

R sin 0

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

z

R

= atan(x, y).

Example. Parametrizarion of a general smooth space curve is of the formr = (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 is continuously differentiable and 6= 0, which guarantees that the curve has a tangent everywhere.

Example. Parametrization of a general smooth surface is of the formr = (u) (u U),

where U is an open subset of R2. Here is continuously differentiable and 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 (u)).


Parametrization is at the same time more restrictive and more extensive than our earlierdefinitions of manifolds: Not all manifolds can be parametrized and not all parametrizationsdefine manifolds. On the other hand, as noted, parametrization makes it easier to deal withmanifolds. In integration restrictions of parametrizations can be mostly neglected since they donot affect the values of the integrals, as we will see later. Let us note, however, that if a set isparametrized 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 continuouslydifferentiable bijective function : U M with a derivative of full rank, then there is anopen subset V of U such that u0 V and (V) is a k-dimensional manifold of Rn.Proof. Consider a point

p0 = (u0)

of M. Then the columns of (u0) are linearly independent, and thus some k rows are linearlyindependent, too. Reordering, if necessary, we may assume that these rows are the first k rowsof (u0).

Let us first consider the case k < n. For a general point p = (r, s) of M, r contains the kfirst components. We denote further by 1 the function consisting of the first k components of, and r0 = 1(u0). Similarly, taking the last n k components of we get the function 2.The function

F(u, r) = r 1(u)defined in the open set S = U Rk (cf. the proof of Theorem 2.4) then satisfies the conditions1. 4. of the Implicit Function Theorem. Thus there is a continuously differentiable functiong : B Rk, defined in an open set B, whose graph u = g(r) is included in S, such that

r = 1(g(r)

).

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

f(r) = 2(g(r)

).

Finally we choose V = 11 (B), an open set. (And where, if anywhere, do we need bijectivityof ?)

The case k = n is similar. The function

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

p = (g(p)

).

Thus B M. Again we choose V = 1(B), an open set.Parametrization of a manifoldM by

: U Mmay be exchanged, a so-called reparametrization, as follows. Take a new parameter domainV Rk, and a continuously differentiable bijective function

: V U ,


such that the derivative is nonsingular. The new parametrization is then by the compositefunction , that is, as

r = ((v)

) (v V).This really is a parametrization since, by the chain rule, is continuously differentiable andits derivative

((v)

)(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) = (, , ) = ( sin cos, sin sin , cos ),

where the parameter domain is the open rectangular prism

V : 0 < < R , 0 < < , 0 < < .z

y

x

= 0

= R

= parameter domain

R

pi

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 = 1(u) (u U) and r = 2(v) (v V).An obvious candidate for the reparametrization is the one using = 11 2, as

u = 11(2(v)

).

This function is bijective, we just must show that it is continuously differentiable. For this letus first define

G(u,v) = 1(u) 2(v).


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

Consider then a pointr0 = 1(u0) = 2(v0)

ofM. Since the k columns of 1(u0) are linearly independent, then some k rows of 1(u0) arealso linearly independent. Picking from G the corresponding k components we get the functionF. In the open set S = U V the function F satisfies the conditions 1. 4. of the ImplicitFunction Theorem, and then the obtained function f clearly is .

Since the point v0 was an arbitrary point of V , is continuously differentiable. On the otherhand, is also nonsingular because 2 = 1 , and by the chain rule

2(v) =

1

((v)

)(v).

If now would be singular in some point of V , then 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 againa 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 aboveusing a parameter domain U and a function : U N which is continuously differentiableand whose derivative is of full rank, but do not require that is bijective. If now for eachpoint p of N there is an open subset Up of U such that

p = (u) for some point u of Up, and is bijective when restricted into Up,

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

(x, y) = () =(r() cos, r() sin

) (0 < < 2),where

r() = ecos 2 cos 4+ sin5 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) = ()T (0 < < 2).It shows that is of full rank (the curve does not pass through the origin). It also indicatesthe parameter value = 0 (or = 2) 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 < < 4containg the parameter value = 2. 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 < < 24, outside of which it starts to repeat itself:


2.6 Tangent SpacesLocally, near a point p0 of Rn, a k-dimensional manifold M is a graph of some k-variatefunction 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 touchingM in p0. The dimensions k = n and k = 0 are dealt with separately. In the former the tangentspace consists of all vectors, in the latter of only the zero vector. In the sequel we assume that0 < 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 andits 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 inthe point p0. Geometrically this graph is part of a k-dimensional hyperplane, or a plane or aline 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 theterminal 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 zerovector 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 correspondingmatrix.

Note that replacing the graph s = f(r) by another graph t = h(u) (as needs to be done whenmoving from one chart to another) simply corresponds to a local reparametrization u = (r)and change of basis of the tangent space using the matrix (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 M of 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 graphz = 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

(h1h2

)((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 conditionF(r, s) = 0

(assuming a proper order of variables)? According to Corollary 2.3, then M is given locallynear the point p0 = (r0, s0) also as a graph s = f(r) and (cf. the Implicit Function Theorem)

f (r0) = Fs(r0, f(r0)

)1F

r

(r0, f(r0)

)where

F =(F

rF

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

Fs

(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 alocal locus (Theorem 2.4). More specifically, if we take a coordinate transform p = pQ + band denote

F(p) = F((p b)QT) and m = QTm,

then (cf. Section 1.5)F(p0)m

= F((p0 b

)QT)QQTm = F(p0)m.

Moreover, 0-dimensional and n-dimensional manifolds of Rn (points and open sets) ofcourse are coordinate-free.

Example. The tangent space of a circleF (x, y) = x2 + y2 R2 = 0

in the point (x0, y0) is then the null space of the 1 2 matrixF (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 = 0in the point (x0, y0, z0) is the null space of the 1 3 matrix

F (x0, y0, z0) = (2x0, 2y0, 2z0).

It consists of vectorshkl

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 equation

F (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)x

h+F (p0)

yk +

F (p0)

zl = 0.

As a further consequence of Theorem 2.8 we see that if a manifold is parametrized, then itstangent space can be parametrized, too.

Corollary 2.10. If the k-dimensional manifold M of Rn has the parametrization : U Mand p0 = (u0), then the tangent space Tp0(M) consists of exactly all vectors of the form

(u0)v (v Rk),

that is, Tp0(M) is the column space of (u0).Proof. Locally near the point p0 the manifold M can be given as a locus determined by somesuitable condition F(p) = 0. Thus the equation

F((u)

)= 0

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

F((u))(u) = O,

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

F(p0)(u0) = O,

showing that columns of (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 (u0)

are linearly independent, the columns of (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 = (u) (u U),then its tangent space Tr0(C) in the point r0 = (u0) consists of exactly all vectors

h (u0) (h R).Example. If the parametrization of a smooth surface of R3 (a 2-dimensional manifold) is

r = (u) (u U),then its tangent space in the point r0 = (u0) consists of exactly all vectors

(u0)h =

1(u0)

u1

1(u0)

u2

2(u0)

u1

2(u0)

u2

3(u0)

u1

3(u0)

u2

(h1h2

)=

((u0)

u1

(u0)

u2

)h (h R2).


2.7 Normal SpacesGeometrically the normal space of a k-dimensional manifoldM of Rn in the point p0, denotedNp0(M), consists of exactly all (tangent) vectors perpendicular to to all vectors of the tangentspace. 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 thenormal space consists of the the zero vector only, and in the latter of all vectors. Vectors in anormal space are called normals or normal vectors.

Basic properties of a normal space Np0(M) follow fairly directly from those of the tangentspace. We just list them here. From basic courses of mathematics we remember that the nullspace of a matrix is the orthogonal complement of the column space of its transpose, and thatthe 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

Ink

)k (k Rnk).

If the k-dimensional manifold M of Rn is near the point p0 locally given as a locusdetermined by the condition F(p) = 0 (with the assumptions of Corollary 2.3), thenits normal space Np0(M) is the column space of the matrix F(p0)T, i.e., it consists ofexactly all vectors

F(p0)Tk (k Rnk).

If the k-dimensional manifoldM of Rn is parametrized by : U M and p0 = (u0),then the normal space Np0(M) is the null space of (u0)T, i.e., it consists of exactly allvectors n satisfying

(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 manifoldM 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 0

0 1

(k1

k2

)(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 graphz = 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)y

k

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) isr = (u) (u U),

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

n =

hkl

satisfying

(u0)

Tn =

1(u0)

u1

2(u0)

u1

3(u0)

u1

1(u0)

u2

2(u0)

u2

3(u0)

u2

hkl

= (0

0

).

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

t((u0)

u1 (u0)

u2

)(t R).

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

For instance, the normal space of a sphere parametrized by spherical coordinates as(x, y, z) = (, ) = (R sin cos,R sin sin ,R cos )

(0 < < , 0 < < 2)in the point (0, 0) consists of the vectors

t

R cos 0 cos 0R cos 0 sin0

R sin 0

R sin 0 sin0R sin 0 cos 0

0

= t

R2 sin2 0 cos0R2 sin2 0 sin0

R2 sin 0 cos 0

(t R),

i.e. vectorst(0, 0)

T (t R).


2.8 Manifolds and Vector FieldsTo deal with the spaces we choose bases for the tangent space Tp0(M) and the normal spaceNp0(M) of a k-dimensional manifold M of Rn in the point p0:

t1, . . . , tk and n1, . . . ,nnk,

respectively. (These bases need not be normalized nor orthogonal.) Note that it was alwayseasy to get a basis for one of the spaces above.

Since the two spaces are orthogonal complements of each other, combining the bases givesa 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 fieldthrough the manifold M in the point p0. (Cf. flow of a liquid through a surface.) It naturallysuffices to have one of these, the other is obtained by subtraction.

From the bases we get the matrices

T = (t1, . . . , tk) and N = (n1, . . . ,nnk)

and further the nonsingular matrices (so-called Gramians)G = TTT = (Gij) , where Gij = ti tj , andH = NTN = (Hij) , where Hij = ni nj .

Let us further denote

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

a1.

.

.

ak

,

whereai = F(p0) ti = Ftan(p0) ti,

and

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

b1.

.

.

bnk

,

wherebi = 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 = TG1a and

Fnorm(p0) = N(NTN)1NTF = NH1b.

14The least squares formulas, probably familiar from many basic courses of mathematics. Note that TTT isnonsingular 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 = n2 ,b = F(p0) n

and (as is familiar from basic courses)

Fnorm(p0) =(F(p0) nn

) nn = n

1

n2(F(p0) n

).

x

y

p0

z

tangent plane

Fnorm

F

Ftan

normal vector

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 SetsGeometrical 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. Forthis we need first two related concepts. An outer cover P of A 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 coverP 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 = infPPout

|P | and |A|in = supPPin

|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

Jordans 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., amanifold R parametrized by

(u) = b+

ki=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 Rk in an obvious way in Rn, and thus define the k-dimensionalvolume of a bounded subset of R. (The n-dimensional volume of these subsets in Rn is = 0, asis 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 aJordan measure.

The inner and outer covers used in defining Jordans 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 vectorsa1, 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 JordanPeano measure or Jordans 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 Rn, 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 R3, with its sides given by 3-dimensionalvectors.

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

|A|out = infPPout

|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 helpfulgeneral result is

Theorem 3.2. If M is a k-dimensional manifold of Rn and k < n, then bounded subsets of Mare 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 in M. But near the center of the cube M 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 setsas well as the k-null-sets to be defined beloware 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 interior A 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. If M is a k-dimensional manifold of Rn and k < m n, then bounded subsetsof M 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 ManifoldsIn general, the k-dimensional volume of a k-dimensional manifold M of Rn is a difficult con-cept. For a parametrized manifold it is considerably easier. If the parametrization of M is

r = (u) (u U),

then|M|k =

U

det

( (u)T (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 = (u) of the manifold k-dimensional volumes of parallelepipeds whose edges are given bythe vectors

dui(u)

ui(i = 1, . . . , k).

Moreover, (as a directed line segment) the vector

dui(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 manifoldsdefinition, computing, etc.othersremain. 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 maysometimes be given a parametrization such that the above integral does not exist.

Things thus depend on the parametrization. In the sequel we tacitly assume s

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