Eötvös Loránd University

Faculty of Science

András Csirik

Differential Forms

and Applications

Bachelor's Thesis in Mathematics

Supervisor:

Dr. Sándor Kovács

Department of Numerical Analysis

Budapest, 2019

Acknowledgement

I would like to express my gratitude to my supervisor, Dr. Sándor Kovács for his

contributions to my thesis. His guidance and expert advice have been invaluable

throughout all stages of the work.

1

Contents

1 Introduction 3

2 Dierential forms 5

2.1 The elements of multilinear algebra . . . . . . . . . . . . . . . . . . 5

2.2 Dierential forms in Rn . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2.1 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2.2.2 The induced form . . . . . . . . . . . . . . . . . . . . . . . . 24

2.2.3 Closed and exact forms . . . . . . . . . . . . . . . . . . . . . 28

2.2.4 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.3 The Poincaré-Stokes theorem . . . . . . . . . . . . . . . . . . . . . 34

3 Applications 41

3.1 Maxwell's equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.1.1 The classical form of the Maxwell's equations . . . . . . . . 41

3.1.2 Minkowski spacetime . . . . . . . . . . . . . . . . . . . . . . 42

3.1.3 The Hodge-star operator . . . . . . . . . . . . . . . . . . . . 44

3.1.4 Maxwell's equations in terms of dierential forms . . . . . . 49

3.2 Brouwer's Fixed Point Theorem . . . . . . . . . . . . . . . . . . . . 54

2

Chapter 1

Introduction

The theory of dierential forms is a relatively new eld in mathematics. It was

rst introduced at the beginning of the 20th century with the pioneer work of

H. Poincare, E. Goursat and E. Cartan. Dierential forms contributed to the

evolution of many elds in mathematics, such as geometry or topology. Moreover,

they provide an essential tool in modern physics, for example in the areas of

classical mechanics, electrodynamics and general relativity.

One of the main advantages of using dierential forms is that it does not require

coordinates. Originally, the laws of classical physics were described by using vector

calculus. In this case one rst had to choose a coordinate system for calculations.

But nature does not come "equipped" with a coordinate system. This is merely a

human construction in order to make the computations less dicult. However, the

laws of nature are general truths, independent from any chosen coordinate system.

Therefore we can say that describing the laws of physics with the language of

dierential forms captures the real and essential properties of nature.

Another advantage of this formalism in comparison with tensors is that tensor

elds "do not behave themselves" under mappings. This means that if Φ is a map

from the X space to the Y space and there is a given tensor eld on the X space

then there is no naturally induced eld on the Y space. However, with dierential

forms it can be dened quite naturally. So if we have a dierential form in a space,

we can automatically dene it in other spaces using various maps.

3

From a theoretical point of view, dierential forms can be considered as an elegant

generalization of vector calculus. This general, more abstract theory involves a

lot about what an undergraduate student comes across in their studies. Consider

the main theorems of multi-variable calculus:, the Green-, Stokes- and the Gauss-

theorems. All these theorems can be derived as a special case from the more

general Poincaré-Stokes theorem.

The purpose of this thesis is to construct the theoretical background of dierential

forms and to present some elegant applications of this formalism. As a physical

application we will rewrite Maxwell's equations in terms of dierential forms and

as a mathematical we will present a proof for Brouwer's xed point theorem.

4

Chapter 2

Dierential forms

2.1 The elements of multilinear algebra

Denition 2.1.1. Let k ∈ N and let X1, ..., Xk, Y be vector spaces over the eld

K. We say that the

f : X1 × ...×Xk → Y

map is k-linear (in notation: f ∈ Lk(X1, ...Xk;Y )), if for all i ∈ 1, ..., k andfor all aj ∈ Xj (j ∈ 1, ..., k, j 6= i) the map

φi : Xi → Y, φi(x) := f(a1, ...ai−1, x, ai+1, ..., ak)

is linear. In the special case

X := X1 = ... = Xk, Y := K.

f is called a k-form and we use the notation f ∈ Lk(Xk, K). By convention

L0(X0, K) := K

Examples.

1. If k ∈ N, and X1, ..., Xk, Y are vector spaces then

f0 : X1 × ...×Xk → Y, f0(a1, ..., ak) := 0 ∈ Y

5

is obviously a k-linear map.

2. If a ∈ Rn, then

fa : Rn → R, fa(x) := 〈a, x〉 :=n∑i=1

aixi

is a 1-form. If e1, ..., en is the canonical basis of Rn, then for all i ∈ 1, ..., n

fei(x) = 〈ei, x〉 = xi (x ∈ Rn)

is the 1-form which returns the ith coordinate of a vector. So for an arbitrary

a = (a1, ..., an) ∈ Rn vector

fa(x) = 〈a, x〉 =n∑i=1

aixi =n∑i=1

aifei(x) (x ∈ Rn)

which means

fa =n∑i=1

aifei .

So the 1-forms fe1 , ..., fen compose the basis of the dual space Rn)′. The

following notations are often used as well for the dual basis:

fei =: ei =: dxi.

In this thesis we will mainly use dxi.

3. If k, n ∈ N, N := 1, ..., n, i = (i1, ..., ik) ∈ Nk, then

∆n,ki : (Rn)k → R, ∆n,k

i (x1, ...xk) := det

x1i1 . . . xki1...

...

x1ik . . . xkik

(x1, ..., xk ∈ Rn)

is a k-form.

These ∆n,ki forms can be constructed in more general spaces. In the euclidean

space Rn there is a scalar product. For an x ∈ Rn element the ith coordinate

6

of x can be expressed as 〈x, ei〉, so the above formula can be written as:

∆n,ki (x1, ...xk) := det

〈x1, ei1〉 . . . 〈xk, ei1〉

......

〈x1, eik〉 . . . 〈xk, eik〉

(x1, ..., xk ∈ Rn).

And it can be generalized even more. In a so-called pseudo-euclidean space

V (which we will cover in chapter 3) there is no scalar product only a g

non-degenerate symmetric bilinear map. In these spaces the denition is:

∆n,ki (x1, ...xk) := det

g(x1, ei1) . . . g(xk, ei1)

......

g(x1, eik) . . . g(xk, eik)

(x1, ..., xk ∈ V ).

Due to the properties of the determinant if i is not an injective multiindex:

∃r, s ∈ N, r 6= s : ir = is,

then

∆n,ki = f0 (∈ Lk(Rn)k),R).

If i is a permutation of i then

∆n,ki = ±∆n,k

i

Thus, we only need to focus on the strictly increasing multiindice:

Nk∗ =

N := 1, ..., n (k = 1),

i ∈ Nk : i1 < ... < ik (k > 1).

Theorem 2.1.1. Let e1, ..., en be a basis in X. Then any f ∈ Lk(Xk,R) form

is uniquely determined by its f(ei1 , ...eik) (i ∈ 1, ..., nk) values on the k-tuples of

basis vectors.

Proof. Let x1, ..., xk be arbitrary elements in X. Since e1, ..., en is a basis of X

7

these elements can be expressed as follows:

xi =n∑

j1=1

aijieji (aijl ∈ R).

Using the k-linearity of f we get:

f(x1, ..., xk) = f

(n∑

j1=1

a1j1ej1 , ...,

n∑jk=1

a1jkejk

)=

n∑j1=1

a1j1 , ...,

n∑jk=1

a1jkf(ej1 , ..., ejk).

On the other hand if we are given the numbers f(ei1 , ...eik), the above expression

determines a k-form.

Denition 2.1.2. Let 2 ≤ n ∈ N, X be a vector space over the eld K and

f ∈ Lk(Xk, K). f is called alternating if it changes sing whenever two of its

arguments are interchanged:

f(a1, ..., ai, , ..., aj, ..., ak) = −f(a1, ..., aj, ..., ai, ..., ak)

for all i, j ∈ N, i 6= j, ai ∈ X.

The set of all alternating k-forms

Ak(X) :=

K (k = 0),

X ′ (k = 1),

f ∈ Lk(Xk, K) : f is alternating (k > 1).

builds a subspace in Lk(Xk, K). Examples.

1. The previously dened ∆n,ki forms are alternating: ∆n,k

i ∈ Ak(Rn).

2. If φ, ψ ∈ A1(X) then their wedge product is φ ∧ ψ ∈ A2(X):

φ ∧ ψ(x1, x2) := φ(x1)ψ(x2)− φ(x2)ψ(x1) (x1, x2 ∈ X).

8

3. If x1, ..., xn ∈ Rn, then

f(x1, ..., xn) := det(x1, ...xn)

is an alternating n-form: f ∈ An(Rn).

Denition 2.1.3. The wedge product of k 1-form is dened as the following.

Let φ1, ..., φk ∈ A1(X), x1, ..., xk ∈ X. Then

φ1 ∧ ... ∧ φk ∈ Ak(X), φ1 ∧ ... ∧ φk(x1, ..., xk) := det(φi(xj))ni,j=1.

This denition coincides with the one given before for two 1-forms. Moreover the

following equation holds:

∆n,ki = dx1 ∧ ... ∧ dxk.

Theorem 2.1.2. Let e1, ..., en be a basis in X. Then the set ∆n,ki : i ∈ Nk

∗ isa basis of Ak(X).

Proof.

Step 1

First we show that these elements are linearly independent. Let's suppose

that some linear combination of these forms is zero and show that all the

coecients must be zero. ∑i∈Nk

∗

ai∆n,ki = f0.

For an arbitrary l ∈ Nk∗ multiindex the following equation holds:

0 = f0(el1 , ..., elk) =∑i∈Nk

∗

ai∆n,ki (el1 , ..., elk) = al.

The last equality is true, because

∆n,ki (el1 , ..., elk) =

1 (i = l),

0 (i 6= l).

9

(If i = l then we get the determinant of the identity matrix and if i 6= l there

will be a full zero column in the determinant.)

Step 2

Now we prove that the ∆n,ki forms span the space, meaning that any arbitrary

f ∈ Ak(Rn) element can be expressed as their linear combination:

f =∑i∈Nk

∗

ai∆n,ki

with some ai coecients. For this set

g =∑i∈Nk

∗

f(ei1 , ..., eik)∆n,ki .

This way g ∈ Ak(Rn) and

g(ei1 , ..., eik) = f(ei1 , ..., eik)

for all i ∈ Nk∗ . Which means f = g, and if we choose the coecients

ai := f(ei1 , ..., eik) we get the desired format for f .

We will almost always use the basis of strictly increasing multiindice except in the

case n = 3, k = 2. Then we will use the basis (∆3,2(2,3),∆

3,2(3,1),∆

3,2(1,2)).

Denition 2.1.4. Let f ∈ Ak(X) and g ∈ Al(X). The wedge product or

exterior product of f and g is the f ∧ g ∈ Ak+l(X) alternating form dened as

follows:

f ∧ g(x1, ...xk+l) :=1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l)).

This denition is made in a way that it coincides with the previous ones. Moreover

the following equation holds:

∆n,kj ∧∆n,l

i = ∆n,k+lj,i (j ∈ Nk

∗ , i ∈ N l∗).

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The next theorem summarizes the properties of the exterior product.

Theorem 2.1.3. Suppose that f1, f2, f ∈ Ak(X), g1, g2, g ∈ Al(X), h ∈ Am(X)

and a ∈ K. Then

1. (f1 + f2) ∧ g = f1 ∧ g + f2 ∧ g,

2. f ∧ (g1 + g2) = f ∧ g1 + f ∧ g2,

3. (af) ∧ g = f ∧ (ag) = a(f ∧ g),

4. (f ∧ g) ∧ h = f ∧ (g ∧ h),

5. f ∧ g = (−1)klg ∧ f .

Proof.

1. Let x1, ..., xk+l ∈ X arbitrary elements. Then (f1 + f2) ∧ g(x1, ..., xk+l) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)(f1 + f2)(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l)) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)f1(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l))+

+1

k!l!

∑σ∈Sk+l

sgn(σ)f2(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l)) =

=f1 ∧ g(x1, ..., xk+l) + f2 ∧ g(x1, ..., xk+l).

2. For x1, ..., xk+l ∈ X we have f ∧ (g1 + g2)(x1, ..., xk+l) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))(g1 + g2)(xσ(k+1), ..., xσ(k+l)) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))g1(xσ(k+1), ..., xσ(k+l))+

+1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))g2(xσ(k+1), ..., xσ(k+l)) =

=f ∧ g1(x1, ..., xk+l) + f ∧ g2(x1, ..., xk+l).

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3. If x1, ..., xk+l ∈ X, then (af) ∧ g(x1, ...xk+l) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)(af)(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l)) =

= a1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))g(xσ(k+1), ..., xσ(k+l)) =

= a(f ∧ g)(x1, ..., xk+l) =

=1

k!l!

∑σ∈Sk+l

sgn(σ)f(xσ(1), ...xσ(k))(ag)(xσ(k+1), ..., xσ(k+l)) =

= f ∧ (ag)(x1, ..., xk+l).

4. Set again any elements. x1, ..., xk+l+m ∈ X. From the denition we get

((f ∧ g) ∧ h)(x1, ..., xk+l+m) =

=1

(k + l)!m!

∑σ∈Sk+l+m

sgn(σ)(f ∧ g)(xσ(1), ...xσ(k+l))h(xσ(k+l+1), ..., xσ(k+l+m)).

Now we decompose the permutation group Sk+l+m into so-called residual

classes with respect to the subgroup Sk+l. This means grouping the permu-

tations of Sk+l+m based on how they act on the last m element. This way

each class contains (k + l)! elements. Let C denote the set of these classes.

Let R be an arbitrary class and σR a particular element in R. Each element

σ ∈ R can be decomposed as σ = σR π where π ∈ Sk+l. Then the above

sum can be written as

1

(k + l)!m!

∑R∈C

sgn(σR)

(∑σ∈R

sgn(π)(f ∧ g)(xσ(1), ..., xσ(k+l))

)·

·h(xσR(k+l+1), ..., xσR(k+l+m)).

Now note that all terms in the big parenthesis are equal, because they are

all the permutation π from a xed ordering given by σR. Since all classes in

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C have (k + l)! elements we can write the equation

1

m!

∑R∈C

sgn(σR)(f ∧ g)(xσR(1), ..., xσR(k+l))h(xσR(k+l+1), ..., xσR(k+l+m)).

Now calculating the wedge product of f and g we obtain

1

k!l!m!

∑R∈C

sgn(σR)∑τ∈Sk+l

sgn(τ)f(xτ(σR(1)), ..., xτ(σR(k)))g(xτ(σR(k+1)), ..., xτ(σR(k+l)))·

·h(xσR(k+l+1), ..., xσR(k+l+m))

But again, all permutations σ ∈ Sk+l+m can be decomposed as σ = τ σRand, since τ acts as the identity on the last m indices, for the indices used

in the argument of h σ = σR is true. Thus we nally obtain:

((f ∧ g) ∧ h)(x1, ..., xk+l+m) =1

k!l!m!

∑σ∈Sk+l+m

sgn(σ)f(xσ(1), ..., xσ(k))·

·g(xσ(k+1), ..., xσ(k+l))h(xσ(k+l+1), ..., xσ(k+l+m)).

Because this result does not depend on the order we associate the operations

the statement is proven.

5. Let's write f and g as the following:

f =:∑i∈Nk

∗

aidxi1 ∧ ... ∧ dxik (ai ∈ R),

g =:∑j∈N l

∗

bjdxj1 ∧ ... ∧ dxjl (bj ∈ R).

Then

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f ∧ g =

∑i∈Nk

∗

aidxi1 ∧ ... ∧ dxik

∧∑j∈N l

∗

bjdxj1 ∧ ... ∧ dxjl

=

=∑i∈Nk

∗

∑j∈N l

∗

aibjdxi1 ∧ ... ∧ dxik ∧ dxj1 ∧ ... ∧ dxjl

Now we want to interchange the dxi forms with the dxj ones and use the

fact that

dxi ∧ dxj = −dxj ∧ dxi.

All k dxi-s have to "move through" l dxj-s, thus the expression changes sign

kl times:

f ∧ g =∑i∈Nk

∗

∑j∈N l

∗

aibj(−1)kdxj1 ∧ (dxi1 ∧ ... ∧ dxik) ∧ (dxj2 ∧ ... ∧ dxjl) =

=∑i∈Nk

∗

∑j∈N l

∗

aibj(−1)kl(dxj1 ∧ ... ∧ dxjl) ∧ (dxi1 ∧ ... ∧ dxik) =

= (−1)kl∑i∈Nk

∗

∑j∈N l

∗

bjai(dxj1 ∧ ... ∧ dxjl) ∧ (dxi1 ∧ ... ∧ dxik) =

= (−1)kl

∑j∈N l

∗

bjdxj1 ∧ ... ∧ dxjl

∧∑i∈Nk

∗

aidxi1 ∧ ... ∧ dxik

=

= (−1)klg ∧ f.

14

Remarks.

1. The dimension of Ak(X) is:

dim(Ak(X)) =

(nk

)(k ≤ n),

0 (k > n).

2. The A(X) symbol denotes the direct sum of the Ak(X) vector spaces:

A(X) := A0(X)⊕ ...⊕An(X)⊕ ... := ⊕m∈NAm(X).

It is clear that if dim(X) = n, then:

dim(A(X)) = 1 +

(n

1

)+ ...+

(n

n

)+ 0 + ... = 2n.

If r, s ∈ N0 and φ ∈ Ar(X), ψ ∈ As(X), then let

f :=∞∑r=0

φr ∈ A(X), g :=∞∑s=0

ψs ∈ A(X)

and

f ∧ g :=∞∑l=1

∑r+s=l

φr ∧ ψs ∈ A(X).

Then the A(X) vector space equipped with the exterior product constitutes

the (A(X),∧) exterior algebra.

2.2 Dierential forms in Rn

Denition 2.2.1. Let n ∈ N, k ∈ N0, ∅ 6= V ⊂ Rn be an open set. Then the map:

ω : V → Ak(Rn)

is called a dierential form of degree k (over Rn).

Since ∆n,ki (i ∈ Nk

∗ ) is a basis in Ak(Rn) it is clear, that for every x ∈ V exist the

15

unique numbers

ωi(x) ∈ R (i ∈ Nk∗ )

such that

ω(x) =∑i∈Nk

∗

ωi(x)∆n,ki .

The ωi : V → R maps are called the coordinate functions of the dierential

form ω. The above discussed notations can be written shortly as:

ω =∑i∈Nk

∗

ωi∆n,ki .

This is the so-called canonical form of ω.

Denition 2.2.2. Let r ∈ N. The space Λrk(V ) denotes the dierential forms of

degree k, whose coordinate functions are r-times continuously dierentiable:

Λrk(V ) := ω : V → Ak(Rn) : ω ∈ Cr

The property ω ∈ Cr is equivalent to ωi ∈ Cr ∀i ∈ Nk∗ .

The following operations are dened between dierential forms:

Denition 2.2.3. Let f : V → R, ω, w ∈ Λrk(V ), σ ∈ Λr

l (V ). Then:

1. (fω)(x) := f(x)ω(x),

2. (ω + w)(x) := ω(x) + w(x),

3. (ω ∧ σ)(x) := ω(x) ∧ σ(x).

Remarks.

1. In the special case k = 0 A0(Rn) = R, hence a dierential 0-form is an

ω : V → R n-variable function (scalar eld).

2. If k > n then Ak(Rn) = f0.

16

3. It is clear that if α : V → R, ω, w ∈ Λrk(V ) and σ ∈ Λr

l (V ), then

αω =∑i∈Nk

∗

(aωi)∆n,ki ∈ Λr

k(V ), ω + w =∑i∈Nk

∗

(ωi + wi)∆n,ki ∈ Λr

k(V )

and

ω ∧ σ =∑i∈Nk

∗

∑j∈N l

∗

(ωiσi)∆n,ki ∧∆n,l

j ∈ Λrk+l(V ).

4. Let us introduce the following somewhat simplifying notation:

ω(x;x1, ..., xk) := ω(x)(x1, ..., xk) =∑i∈Nk

∗

ωi(x)∆n,ki (x1, ..., xk)

(x ∈ V, x1, ..., xk ∈ Rn).

5. If e1, ..., en are the canonical basis vectors of Rn, then for all i ∈ Nk∗ the

coordinate functions of a dierential k-form can be computed as follows:

ω(x; ei1 , ..., eik) =∑j∈Nk

∗

ωj(x)∆n,kj (ei1 , ..., eik) = ωi(x) (x ∈ V ).

2.2.1 Derivation

Denition 2.2.4. The exterior derivative of a dierential form:

d : Λrk(V )→ Λr−1

k+1(V ), dω := d(ω) :=

∑

j∈N ∂jω∆n,1j (k = 0),

∑j∈N

∑i∈Nk

∗∂jωi∆

n,k+1(j,i) (k > 0)

where (j, i) := (j, i1, ..., ik) ∈ N ×Nk∗ .

Remarks.

1. If k = 0 and ω ∈ C1(V,R), then for all x ∈ V , and ξ ∈ Rn we have

dω(x)(ξ) =n∑j=1

∂jω(x)∆n,1j (ξ) = 〈gradω(x), ξ〉.

17

With the commonly used notation

∂jω =∂ω

∂xjand ∆n,1

j = dxj

we get

dω =n∑j=1

∂ω

∂xjdxj,

which is the so-called dierential of the ω scalar eld at the point x.

2. If k ∈ N , then using the equality ∆n,k+1(j,i) = ∆n,1

j ∧∆n,ki we have

dω =∑j∈N

∑i∈Nk

∗

∂jωi∆n,1j ∧∆n,k

i =∑i∈Nk

∗

∑j∈N

∂jωi∆n,1j ∧∆n,k

i =∑i∈Nk

∗

dωi ∧∆n,ki .

Theorem 2.2.1. The exterior derivative has the following properties:

1. The d operator is linear, meaning if ω,w,∈ Λrk(V ) and α ∈ R, then

d(ω + w) = dω + dw and d(αω) = αdω.

2. If ω ∈ Λrk(V ) and σ ∈ Λr

l (V ), then

d(ω ∧ σ) = dω ∧ σ + (−1)kω ∧ (dσ).

3. If ω ∈ Λ2k(V ), then

d(dω) = 0 (∈ Λ0k+2(V )).

Proof.

1. (a) First we prove the additivity.

18

• If k = 0 then

d(ω + w) =∑j∈N

∂j(ω + w)∆n,1j =

∑j∈N

(∂jω + ∂jw)∆n,1j =

=∑j∈N

∂jω∆n,1j +

∑j∈N

∂jw∆n,1j = dω + dw.

• If k > 0 then

d(ω + w) =∑j∈N

∑i∈Nk

∗

∂j(ωi + wi)∆n,k+1(j,i) =

=∑j∈N

∑i∈Nk

∗

(∂jωi + ∂jwi)∆n,k+1(j,i) =

=∑j∈N

∑i∈Nk

∗

∂jωi∆n,k+1(j,i) +

∑j∈N

∑i∈Nk

∗

∂jwi∆n,k+1(j,i) =

= dω + dw.

(b) Now we prove the homogeneity, which goes analogously.

• If k = 0 then

d(αω) =∑j∈N

∂j(αω)∆n,1j = α

∑j∈N

∂jω∆n,1j = αdω.

• If k > 0 then

d(αω) =∑j∈N

∑i∈Nk

∗

∂j(αωi)∆n,k+1(j,i) = α

∑j∈N

∑i∈Nk

∗

∂jωi∆n,k+1(j,i) = αdω.

19

2. Using the linearity of d:

d(ω ∧ σ) = d

∑i∈Nk

∗

∑j∈N l

∗

(ωiσj)∆n,ki ∧∆n,l

j

=∑i∈Nk

∗

∑j∈N l

∗

d(ωiσj) ∧∆n,ki ∧∆n,l

j =

=∑i∈Nk

∗

∑j∈N l

∗

(σjdωi + ωidσj) ∧∆n,ki ∧∆n,l

j =

=∑i∈Nk

∗

∑j∈N l

∗

(σjdωi) ∧∆n,ki ∧∆n,l

j +∑i∈Nk

∗

∑j∈N l

∗

(ωidσj) ∧∆n,ki ∧∆n,l

j =

=

∑i∈Nk

∗

dωi ∧∆n,ki

∧∑j∈N l

∗

σj∆n,lj

+

+∑i∈Nk

∗

∑j∈N l

∗

ωi(dσj ∧∆n,ki ) ∧∆n,l

j =

= (dω ∧ σ) +∑i∈Nk

∗

∑j∈N l

∗

ωi((−1)k∆n,ki ∧ dσj) ∧∆n,l

j =

= (dω ∧ σ) + (−1)k

∑i∈Nk

∗

ωi∆n,ki

∧∑j∈N l

∗

dσj ∧∆n,lj

=

= dω ∧ σ + (−1)kω ∧ dσ.

3. • If k = 0 then by using the linearity of d and Young's theorem

20

d(dω) = d

(∑j∈N

∂jω∆n,1j

)=∑j∈N

d(∂jω) ∧∆n,1j =

=∑j∈N

(∑i∈N

∂i(∂jω)∆n,1i

)∧∆n,1

j =∑i,j∈N

∂ijω∆n,1i ∧∆n,1

j =

=∑i<j

(∂ijω − ∂ijω)∆n,1i ∧∆n,1

j = 0 ∈ Λ02(V ).

• If k > 0 then using the linearity, the rule for dierentiating an exterior

product and the k = 0 case we have

d(dω) = d

∑i∈Nk

∗

dωi ∧∆n,ki

=∑i∈Nk

∗

d(dωi ∧∆n,ki ) =

=∑i∈Nk

∗

[d(dωi) ∧∆n,ki − (d∆n,k

i ) ∧ (dωi)] = 0 ∈ Λ0k+2(V ).

The last equation holds, since for all i ∈ Nk∗ : ωi ∈ Λ2

0(V ), so because of

the k = 0 case we have

d(dωi) = 0 ∈ Λ02(V )

and

d∆n,ki = d(1∆n,k

i ) =∑j∈N

(∂j1)∆n,k+1(j,i) =

∑j∈N

0∆n,k+1(j,i) = 0 ∈ Λ1

k+1 (i ∈ Nk∗ ).

21

Denition 2.2.5. Let fi : V → R (i ∈ Nk∗ ) and f := (fi, i ∈ Nk

∗ ) : V → R(nk),

then the dierential form generated by f is:

ωf :=∑i∈Nk

∗

fi∆n,ki

meaning that:

ωf (x;x1, ..., xk) :=∑i∈Nk

∗

fi(x)∆n,ki (x1, ..., xk) (x ∈ V, x1, ..., xk ∈ Rn).

Special cases.

1. If k = 0, then f : V → R, so ωf = f , thus:

dωf =∑j∈N

∂jf∆n,1j = ωgrad(f).

2. If n = 2, k = 1, then f : V → R2, so ωf = f1∆2,11 + f2∆2,1

2 , thus:

dωf =2∑j=1

2∑i=1

∂jfi∆2,2(j,i) =

2∑j=1

(∂jf1∆2,2

(j,1) + ∂jf2∆2,2(j,2)

)=

= ∂1f1∆2,2(1,1) + ∂1f2∆2,2

(1,2) + ∂2f1∆2,2(2,1) + ∂2f2∆2,2

(2,2) =

= 0 + ∂1f2∆2,2(1,2) − ∂2f1∆2,2

(1,2) + 0 = ω∂1f2−∂2f1 .

3. If n = 3, k = 1, then f : V → R3, so ωf = f1∆3,11 + f2∆3,1

2 + f3∆3,13 , thus

22

dωf =3∑j=1

3∑i=1

∂jfi∆3,2(j,i) =

3∑j=1

(∂jf1∆3,2

(j,i) + ∂jf2∆3,2(j,2) + ∂jf3∆3,2

(j,3)

)=

=∂1f1∆3,2(1,1) + ∂1f2∆3,2

(1,2) + ∂1f3∆3,2(1,3)+

+ ∂2f1∆3,2(2,1) + ∂2f2∆3,2

(2,2) + ∂2f3∆3,2(2,3)+

+ ∂3f1∆3,2(3,1) + ∂3f2∆3,2

(3,2) + ∂3f3∆3,2(3,3) =

=(∂2f3 − ∂3f2)∆3,2(2,3) + (∂3f1 − ∂1f3)∆3,2

(3,1) + (∂1f2 − ∂2f1)∆3,2(1,2) =

=ωrot(f).

4. If n = 3, k = 2, then f : V → R3, so

ωf = f1∆3,2(2,3) + f2∆3,2

(3,1) + f3∆3,2(1,2),

thus

dωf =3∑j=1

3∑i=1

∂jfi∆3,3(j,i) =

3∑j=1

(∂jf1∆3,3

(j,2,3) + ∂jf2∆3,3(j,3,1) + ∂jf3∆3,3

(j,1,2)

)=

= (∂1f1∆3,3(1,2,3) + 0 + 0) + (0 + ∂2f2∆3,3

(2,3,1) + 0) + (0 + 0 + ∂3f3∆3,3(3,1,2)) =

= (∂1f1 + ∂2f2 + ∂3f3)∆3,3(1,2,3) = ωdiv(f).

23

2.2.2 The induced form

One of the most important features of dierential forms is the way they behave

under dierentiable maps.

Denition 2.2.6. Let r ∈ N0, n,m ∈ N, k ∈ 0, 1, ..., n, U ⊂ Rm, V ⊂ Rn be

open sets, ω ∈ Λrk(V ) and Φ ∈ Cr+1(U, V ). Let us dene the induced form or the

pullback of ω as follows:

ω ∗ Φ ∈ Λrk(U).

If k = 0, then

ω ∗ Φ := ω Φ.

If k > 0, then

ω ∗ Φ(x;x1, ...xk) := ω(Φ(x); Φ′(x)x1, ...,Φ′(x)xk) (x ∈ U, x1, ...xk ∈ Rm).

This denition makes sense since Φ(x) ∈ V , Φ′(x) ∈ Rn×m, thus Φ′(x)xi ∈ Rn.

Theorem 2.2.2. Let U ⊂ Rm, V ⊂ Rn be open subsets, Φ ∈ Cr+1(Rm,Rn)

k ∈ 0, 1, ..., n, ω, σ ∈ Λrk(Rn) and g ∈ Λr

0(Rn). Then:

1. (ω + σ) ∗ Φ = ω ∗ Φ + σ ∗ Φ,

2. (gω) ∗ Φ = g Φ · ω ∗ Φ,

3. if ω1, ..., ωk ∈ Λr1(Rn), then (ω1 ∧ ... ∧ ωk) ∗ Φ = ω1 ∗ Φ ∧ ... ∧ ωk ∗ Φ.

Proof.

1. If x ∈ U and x1, ..., xk ∈ Rm then

((ω + σ) ∗ Φ)(x;x1, ..., xk) = (ω + σ)(Φ(x); Φ′(x)x1, ...,Φ′(x)xk) =

= ω(Φ(x); Φ′(x)x1, ...,Φ′(x)xk) + σ(Φ(x); Φ′(x)x1, ...,Φ

′(x)xk) =

= ω ∗ Φ(x;x1, ..., xk) + σ ∗ Φ(x;x1, ..., xk).

24

2. For x ∈ U and x1, ..., xk ∈ Rm we have

((gω) ∗ Φ)(x;x1, ..., xk) = (gω)(Φ(x); Φ′(x)x1, ...Φ′(x)xk) =

= g(Φ(x))ω(Φ(x); Φ′(x)x1, ...Φ′(x)xk) = g Φ(x) · ω ∗ Φ(x1, ..., xk).

3. Let x ∈ U and x1, ..., xk ∈ Rm. Then

((ω1 ∧ ... ∧ ωk) ∗ Φ)(x;x1, ..., xk) = (ω1 ∧ ... ∧ ωk)(Φ(x); Φ′(x)x1, ...,Φ′(x)xk) =

= det(ωi(Φ(x; Φ′(x)xj))) = det(ωi ∗ Φ(x;xj)) =

= (ω1 ∗ Φ ∧ ... ∧ ωk ∗ Φ)(x;x1, ..., xk).

Now we can interpret the meaning of the induced form. Let (x1, ..., xn) be co-

ordinates in Rn and (y1, ..., ym) be coordinates in Rm. Let Φ : Rn → Rm be a

dierentiable function which "substitutes" the coordinates:

y1 = Φ1(x1, ..., xn), ..., ym = Φm(x1, ..., xn).

Let ω =∑

i∈Nk∗ωidyi1∧...∧dyik be a k-form in Rm. Then using the above properties

of the induced form we obtain

ω ∗ Φ =

∑i∈Nk

∗

ωidyi1 ∧ ... ∧ dyik

∗ Φ =∑i∈Nk

∗

(ωidyi1 ∧ ... ∧ dyik) ∗ Φ =

=∑i∈Nk

∗

(ωi Φ) · (dyi1 ∧ ... ∧ dyik) ∗ Φ =∑i∈Nk

∗

(ωi Φ) · dyi1 ∗ Φ ∧ ... ∧ dyik ∗ Φ.

25

Since for all x ∈ U and for all v ∈ Rm

(dyj ∗ Φ)(x; v) = dyj(Φ(x); Φ′(x)v) = dyj(Φ′(x)v) = (Φ′(x)v)j =

= ∂1Φj(x)v1 + ...+ ∂nΦj(x)vn = 〈Φ′j(x), v〉 = dΦj(x)(v),

we have

ω ∗ Φ =∑i∈Nk

∗

(ωi Φ)dΦi1 ∧ ... ∧ dΦik .

Thus the pullback of ω is the same as substituting the xi variables and their dxi

dierentials by the functions of xk and dxk.

Theorem 2.2.3. Let Φ ∈ Cr+1(Rm,Rn), ω, σ ∈ Λrk(Rn) and Ψ ∈ Cr+1(Rp,Rm).

Then

1. (ω ∧ σ) ∗ Φ = ω ∗ Φ ∧ σ ∗ Φ,

2. ω ∗ (Φ Ψ) = (ω ∗ Φ) ∗Ψ,

3. dω ∗ Φ = d(ω ∗ Φ).

Proof.

Let (y1, ..., yn) = (Φ1(x1, ..., xm), ...,Φn(x1, ..., xm)) ∈ Rn, (x1, ..., xm) ∈ Rm, and

ω =∑i∈Nk

∗

ωidyi1 ∧ ... ∧ dyik , σ =∑j∈Nk

∗

σidyj1 ∧ ... ∧ dyjk .

1. Then (ω ∧ σ) ∗ Φ =

=

∑i∈Nk

∗

∑j∈Nk

∗

ωiσj(dyi1 ∧ ... ∧ dyik) ∧ (dyj1 ∧ ... ∧ dyjk)

∗ Φ =

=∑i∈Nk

∗

∑j∈Nk

∗

ωi Φ · σj Φ · (dΦi1 ∧ ... ∧ dΦik) ∧ (dΦj1 ∧ ... ∧ dΦjk) =

26

=

∑i∈Nk

∗

ωi Φ · dΦi1 ∧ ... ∧ dΦik

∧∑j∈Nk

∗

σj Φ · dΦj1 ∧ ... ∧ dΦjk

=

= ω ∗ Φ ∧ σ ∗ Φ.

2. Let x, x1, ..., xk ∈ Rp. Then

(ω ∗ (Φ Ψ))(x;x1, ..., xk) = ω(Φ Ψ(x); (Φ Ψ)′(x)x1, ..., (Φ Ψ)′(x)xk) =

= ω(Φ(Ψ(x)); Φ′(Ψ(x))Ψ′(x)x1, ...,Φ′(Ψ(x))Ψ′(x)xk) =

= ω ∗ Φ(Ψ(x); Ψ′(x)x1, ...,Ψ′(x)xk) =

= (ω ∗ Φ) ∗Ψ(x;x1, ..., xk).

3. Step 1

We will rst prove the statement for 0-forms. This case ω : Rn → R is

a dierentiable function.

dω ∗ Φ =

(n∑i=1

∂ωi∂yi

dyi

)∗ Φ =

n∑i=1

m∑j=1

∂ωi∂yi Φ · ∂Φi

∂xjdxj =

m∑j=1

∂(ω Φ)

∂xjdxj =

= d(ω Φ) = d(ω ∗ Φ).

Step 2 Now we consider the k > 0 case.

d(ω ∗ Φ) = d

∑i∈Nk

∗

ωi Φ · (dyi1 ∗ Φ) ∧ ... ∧ (dyik ∗ Φ)

=

27

=∑i∈Nk

∗

d(ωi Φ · (dΦi1 ∧ ... ∧ dΦik)) =

=∑i∈Nk

∗

d(ωi Φ) ∧ (dΦi1 ∧ ... ∧ dΦik) =

=∑i∈Nk

∗

(dωi ∗ Φ) ∧ (dyi1 ∗ Φ) ∧ ... ∧ (dyik ∗ Φ) =

=

∑i∈Nk

∗

dωi ∧ dyi1 ∧ ... ∧ dyik

∗ Φ = dω ∗ Φ.

2.2.3 Closed and exact forms

From the theory of the Riemann-integral it is a well-known fact, that every contin-

uous function on R has a primitive function. In the language of dierential forms

it means that if ω ∈ Λ01(I) and has the form: ω(x) = f(x)dx with some continuous

function f, there exists some function F such that dF = ω.

Now we want to ask the analogous question of whether any dierential form has

a "primitive form". We would like to give a necessary and sucient condition for

the existence of this form.

Denition 2.2.7. The ω ∈ Λrk(V ) dierential form is said to be

1. closed if dω = 0,

2. exact if there is an η ∈ Λr+1k−1 such that dη = ω.

Since for all ω ∈ Λ2k(V ) : ddω = 0 it is clear that if a form is exact, then it is

closed. The following "lemma" presents a necessary condition for closed forms to

be exact.

28

Theorem 2.2.4 (Poincaré-lemma). Let k ∈ N, 2 ≤ r ∈ N ∪ ∞ and V be a

star-shaped1 domain. In this case ω ∈ Λrk(V ) is closed if and only if it is exact.

Proof. It only remains to show that if ω is closed, then it is exact. Let us suppose

that the star-point of V is 02, and the closed form ω is the folowing:

ω =∑i∈Nk

∗

ωi∆n,ki .

We will dene an operation P : Λrk(V ) → Λr+1

k−1(V ) for which dP (ω) = ω, thus

proves, that ω is exact. Let x ∈ V and P be the following:

P (ω)(x) :=∑i∈Nk

∗

k∑l=1

(−1)l−1

(∫ 1

0

tk−1ωi(tx)dt

)xildxi1∧...∧dxil−1

∧dxil+1∧...∧dxik .

Now if we use the rule for the derivative of a wedge product for the coordinate

functions we get: dP (ω)(x) =

=∑i∈Nk

∗

k∑l=1

(∫ 1

0

tk−1ωi(tx)dt

)∆n,ki +

+n∑j=1

∑i∈Nk

∗

k∑l=1

(−1)l−1

(∫ 1

0

ttk−1∂jωi(tx)dt

)xildxi1 ∧ ... ∧ dxil−1

∧ dxil+1∧ ... ∧ dxik =

=k∑i∈Nk

∗

(∫ 1

0

tk−1ωi(tx)dt

)∆n,ki +

+n∑j=1

∑i∈Nk

∗

k∑l=1

(−1)l−1

(∫ 1

0

ttk−1∂jωi(tx)dt

)xildxi1 ∧ ... ∧ dxil−1

∧ dxil+1∧ ... ∧ dxik .

1It means that there exists an a ∈ V (so-colled star-point) such that for all x ∈ V

[a, x] := a + t(x− a) ∈ Rn : t ∈ [0, 1] ⊂ V.

2If 0 6= a is a star-point of V (and 0 is not) we can apply the transformation y := x− a.

29

Now let's calculate P (dω)(x). Since

dω =n∑j=1

∑i∈Nk

∗

∂jωi∆n,k(j,i)

we have P (dω)(x) =

=n∑j=1

∑i∈Nk

∗

(∫ 1

0

tk−1ωi(tx)dt

)xj∆

n,ki −

−n∑j=1

∑i∈Nk

∗

k∑l=1

(−1)l−1

(∫ 1

0

ttk−1∂jωi(tx)dt

)xildxi1 ∧ ... ∧ dxil−1

∧ dxil+1∧ ... ∧ dxik .

If we add the two formulas there are a lot of therms which cancel each other out,

thus we get:

(P (dω) + dP (ω))(x) = k∑i∈Nk

∗

(∫ 1

0

tk−1ωi(tx)dt

)∆n,ki +

+n∑j=1

∑i∈Nk

∗

(∫ 1

0

tk−1ωi(tx)dt

)xj∆

n,ki =

=∑i∈Nk

∗

(∫ 1

0

d

dt[tkωi(tx)]dt

)∆n,ki =

=∑i∈Nk

∗

[1 · ωi(x)− 0 · ωi(0)] ∆n,ki = ω.

So we have

P (dω) + dP (ω) = ω.

Using our assumption that ω is closed this simplies to dP (ω) = ω, which means

that P (ω) is the form we were looking for.

30

2.2.4 Integration

Let us introduce the following notation:

Ik := [0, 1]k (k ∈ N)

the closed unit cube in Rk. By convention

I := I1 and I0 := 0.

Denition 2.2.8. Let k ∈ N0, 2 ≤ n ∈ N, r ∈ N ∪ ∞ and ∅ 6= V ⊂ Rn be an

open subset. Consider a map

Φ : Ik → V.

Φ is called a k-cube, if it is continuous. If Φ is also r-times continuously dier-

entiable, then it is called an (r-times) smooth k-cube.

Denition 2.2.9. Let k, n ∈ N, ∅ 6= V ∈ Rn be an open subset and Φ be a k-cube.

If j ∈ 1, ..., k and s ∈ 0, 1, then Φjs denotes the following k − 1-cube:

Φjs : Ik−1 → V, Φjs(x1, ..., xk−1) :=

Φ(s) (k = 1),

Φ(x1, ...xj−1, s, xj+1, ..., xk−1) (k > 1).

The set

∂Φ := Φjs : j ∈ 1, ..., k, s ∈ 0, 1

is called the boundary of the k-cube Φ.

Since dim(Ak(Rk)) =(kk

)= 1, Ak(Rk) has only one basis element: ∆k,k

(1,...,k). So if

k ∈ 1, ..., n and Φ ∈ Cr+1(Ik, V ), then ω ? Φ ∈ Λrk(Ik), which means the induced

form has the following form:

ω ? Φ = Ω∆k,k(1,...,k)

with some function Ω ∈ Cr+1(Ik,R).

Denition 2.2.10. Let k ∈ 0, ..., n, ω ∈ Λrk(V ) and Φ ∈ Cr+1(Ik, V ). We dene

the integral of the k-form ω over the k-cube Φ as follows:

31

If k = 0, then ∫Φ

ω := ω(Φ(0)),

If k > 0, then ∫Φ

ω :=

∫Ik

Ω.

Special cases.

• The case k = 1.

In this case ω and Φ are the following:

ω = ωf =n∑i=1

fi∆n,1i , Φ ∈ Cr+1([0, 1], V ),

where f = (f1, ..., fn) ∈ Cr(V,Rn). Let's compute the ωf ∗ Φ ∈ Λr1([0, 1])

induced form. Let x ∈ [0, 1] and y ∈ R.

(ωf ∗ Φ)(x)(y) =n∑i=1

fi(Φ(x))∆n,1i (Φ′(x)y) =

n∑i=1

fi(Φ(x))(Φ′(x)y)i =

=n∑i=1

fi(Φ(x))Φ′i(x)y =n∑i=1

fi(Φ(x))Φ′i(x)∆1,11 (y) =

= 〈f Φ,Φ′〉 (x)∆1,11 (y) =

(〈f Φ,Φ′〉∆1,1

1

)(x)(y)

So in this case Ω = 〈f Φ,Φ′〉, thus∫Φ

ωf =

∫[0,1]

〈f Φ,Φ′〉 =

∫Φ

f (line integral).

• The case k = n.

In this case ω and Φ are the following:

ω = ωf = f∆n,n(1,...,n), Φ ∈ Cr+1(In, V ),

where f ∈ Cr(V,R). Let x ∈ In, x1, ..., xn ∈ Rn and suppose that Φ is

32

regular.3

(ωf ∗ Φ)(x)(x1, ..., xn) = f(Φ(x))∆n,n(1,...,n)(Φ

′(x)x1, ...,Φ′(x)xn) =

= f(Φ(x)) det(Φ′(x)x1, ...,Φ′(x)xn) = f(Φ(x)) det(Φ′(x)) det(x1, ..., xn) =

= f(Φ(x)) det(Φ′(x))∆n,n(1,...,n)(x1, ..., xn) = ((f Φ) det(Φ′))(x)∆n,n

(1,...,n)(x1, ..., xn) =

= (((f Φ) det(Φ′))∆n,n(1,...,n))(x)(x1, ..., xn).

So in this case Ω = (f Φ) det(Φ′). Thus∫Φ

ωf =

∫In

(fΦ) det(Φ′) =

∫In

(fΦ)| det(Φ′)| =∫RΦ

f (multiple integral)

• The case n = 3, k = 2.

In this case ω and Φ are the following:

ω = ωf = f1∆3,2(2,3) + f2∆3,2

(3,1) + f3∆3,2(1,2), Φ ∈ Cr+1(I2, V ),

where f = (f1, f2, f3) ∈ Cr(V,R3). Here we take a dierent approach to

compute Ω. We use the formula for computing the coordinate functions of a

dierential form. Since ωf ∗ Φ = Ω∆2,2(1,2), we have:

Ω(x) =(ωf ∗ Φ)(x; e1, e2) = ωf (Φ(x); Φ′(x)e1,Φ′(x)e2) = ωf (Φ(x); ∂1Φ(x), ∂2Φ(x))

=f1(Φ(x))∆3,2(2,3)(∂1Φ(x), ∂2Φ(x)) + f2(Φ(x))∆3,2

(3,1)(∂1Φ(x), ∂2Φ(x))+

+ f3(Φ(x))∆3,2(1,2)(∂1Φ(x), ∂2Φ(x)) = f1(Φ(x))(∂1Φ2∂2Φ3 − ∂2Φ2∂1Φ3)+

+ f2(Φ(x))(∂1Φ1∂2Φ3 − ∂2Φ1∂1Φ3) + f3(Φ(x))(∂1Φ1∂2Φ2 − ∂2Φ1∂1Φ2) =

=(f Φ)1(x)(∂1Φ× ∂2Φ)1(x) + (f Φ)2(x)(∂1Φ× ∂2Φ)2(x)+

+ (f Φ)3(x)(∂1Φ× ∂2Φ)3(x) = 〈f Φ, ∂1Φ× ∂2Φ〉(x) (x ∈ I2).

3It means that det(Φ′(x)) > 0 for all x ∈ V .

33

Thus ∫Φ

ωf =

∫I2〈f Φ, ∂1Φ× ∂2Φ〉 =

∫Φ

f (surface integral).

Denition 2.2.11. Let k ∈ 0, ..., n, ω ∈ Λrk(V ) and Φ ∈ Cr+1(Ik+1, V ). We

dene the integral of the k-form ω over the boundary of the k+1-cube Φ

as follows: ∫∂Φ

ω :=k+1∑j=1

1∑s=0

(−1)j+s∫

Φjs

ω.

2.3 The Poincaré-Stokes theorem

Theorem 2.3.1. Let ∅ 6= V ⊂ Rn be an open set and ω ∈ Λrk(V ) (k = 0, ..., n−1).

Let Φ ∈ Cr+1(Ik+1, V ) be a k + 1-cell. Then∫Φ

dω =

∫∂Φ

ω.

Proof.

Step 1 First we prove the k = 0 case.

In this case ω : V → R and Φ : [0, 1] → V . Using the Newton-Leibniz-rule

for line integrals we have:∫∂Φ

ω =

∫Φ11

ω −∫

Φ10

ω = ω(Φ(1))− ω(Φ(0)) =

∫Φ

gradω =

∫Φ

dω.

Step 2 Now we prove the k > 0 case where Φ is the identity map.

V = Ik+1, Φ : Ik+1 → Ik+1 Φ(x) = x (x ∈ Ik+1).

Let us denote the the multiindex i∗ := (1, ..., i− 1, i+ 1, ..., k+ 1). This way

34

ω ∈ Λrk(Ik+1) can be expressed as:

ω =k+1∑i=1

ωi∆k+1,ki∗ .

The derivative of ω is then

dω =k+1∑i=1

∑j∈(K+1)k∗

∂jωi∆k+1,k+1(j,i∗) .

The only non-zero terms are when j = i in which case the number i needs

to be interchanged with i−1 other to reach the identity permutation. So we

have

dω =

(k+1∑i=1

(−1)i−1∂iωi

)∆k+1,k+1

(1,...,k+1).

Hence, by the denition of the integral

∫Φ

dω =k+1∑i=1

(−1)i−1

∫Ik+1

∂iωi.

Let's compute the other side of the equation. The boundary of Φ by denition

is

∂Φ = Φjs : j ∈ 1, ..., k + 1; s ∈ 0, 1,

where

Φjs : Ik → Ik+1 Φjs(x1, ..., xk) = Φ(x1, ..., xj−1, s, xj+1, ..., xk)

(x1, ..., xk ∈ [0, 1]).

Since

ω ∗ Φjs ∈ Λrk(Ik)

it can be written as

ω ∗ Φjs = Ωjs∆k,k(1,...,k) (Ωjs : Ik → R)

35

Let x = (x1, ..., xk) ∈ Ik and let e1, ..., ek be the canonical basis of Rk.

Using the formula for the coordinate functions we have

Ωjs(x) = (ω ∗ Φjs)(x; e1, ..., ek) = ω(Φjs(x); Φjs′(x)e1, ...,Φjs

′(x)ek) =

=k+1∑i=1

ωi(Φjs(x))∆k+1,ki∗ (∂1Φjs(x), ..., ∂kΦjs(x))

Here the partial derivatives are

∂iΦjs(x1, ..., xk) = ∂iΦ(x1, ..., xj−1, s, xj+1, ..., xk) = (0, ..., 1, ..., 0) (i ∈ 1, ..., k),

where the 1 is in the ith position if i < j and in the (i + 1)th position if

i ≥ j. It means that the argument of ∆k+1,ki∗ is a matrix with k + 1 row and

k column, which is almost the "identity" matrix, but the jth row is lled

with 0-s. It follows that the only non-zero term in the summation is where

i = j for which

∆k+1,kj∗ (∂1Φjs(x), ..., ∂kΦjs(x)) = 1,

the determinant of the identity matrix. It follows that

Ωjs(x) = ωj(Φjs(x)) = ωj(Φjs(x1, ..., xk)) = ωj(Φ(x1, ..., xj−1, s, xj+1, ..., xk)).

And since Φ is the identity map we get

Ωjs(x) = ωj(x1, ..., xj−1, s, xj+1, ..., xk),

so

(ω ∗ Φjs)(x) = ωj(x1, ..., xj−1, s, xj+1, ..., xk)∆k,k(1,...,k).

36

Now we can compute the integral over the boundary:

∫∂Φ

ω =k+1∑j=1

1∑s=0

(−1)j+s∫

Φjs

ω =k+1∑j=1

(−1)j+1

(∫Φj1

ω −∫

Φj0

ω

)=

=k+1∑j=1

(−1)j−1

(∫Ikωj(x1, ..., xj−1, 1, xj+1, ..., xk)−

−∫Ikωj(x1, ..., xj−1, 0, xj+1, ..., xk)

)=

=k+1∑j=1

(−1)j−1

∫Ik

(ωj(x1, ..., xj−1, 1, xj+1, ..., xk)−

− ωj(x1, ..., xj−1, 0, xj+1, ..., xk)) =

=k+1∑j=1

(−1)j−1

∫Ik

(∫ 1

0

∂jωj(x1, ..., xj−1, t, xj+1, ..., xk)dt

)=

=k+1∑j=1

(−1)j−1

∫Ik+1

∂jωj

which is the same result we got earlier.

Step 3 Now we prove the theorem for arbitrary Φ : Ik+1 → V k+1-cubes. We use

the fact that the exterior derivative commutes with the pullback of forms.∫Φ

dω =

∫Ik+1

dω ∗ Φ =

∫Ik+1

d(ω ∗ Φ)

37

Since d(ω ∗ Φ) ∈ Λr−1k+1(Ik+1) we can use the result from step 2:

∫Ik+1

d(ω ∗ Φ) =

∫∂Ik+1

ω ∗ Φ =k+1∑j=1

1∑s=0

∫Ik+1js

ω ∗ Φ =k+1∑j=1

1∑s=0

∫ΦIk+1

js

ω =

∫∂Φ

ω.

Special cases.

• The case k = 0.

In this case

ω ∈ Λr0 = Cr(V,R) Φ ∈ Cr+1([0, 1], V ).

We have already shown that if

f ∈ Cr(V,R) ω := ωf

then

dω = ωgrad(f).

Thus∫Φ

ωgrad(f) =

∫Φ

dω =

∫∂Φ

ω =1∑s=0

(−1)1+s

∫Φ1s

ω =1∑s=0

(−1)1+sω(Φ1s(0)) =

=1∑s=0

(−1)1+sω(Φ(s)) = ω(Φ(1))− ω(Φ(0)).

This is the same as∫Φ

grad(f) = f(Φ(1))−f(Φ(0)) (Newton-Leibniz formula for line integrals).

• The case n = 2, k = 2.

In this case

ω ∈ Λr1(V ), Φ ∈ Cr+1(I2, V ).

38

We have already shown that if

f = (f1, f2) ∈ Cr(V,R2), ω := ωf = f1∆2,11 + f2∆2,1

2

then

dω = ω∂1f2−∂2f1 .

Thus ∫RΦ

∂1f2 − ∂2f1 =

∫Φ

ω∂1f2−∂2f1 =

∫Φ

dω =

∫∂Φ

ω =

∫∂Φ

f

This is the Green-theorem.

• The case n = 3, k = 2.

In this case

ω ∈ Λr1(V ), Φ ∈ Cr+1(I2, V ).

We have already shown that if

f = (f1, f2, f3) ∈ Cr(V,R3), ω := ωf = f1∆3,11 + f2∆3,1

2 + f3∆3,13

then

dω = ωrot(f).

Thus if Φ is regular∫Φ

rot(f) =

∫Φ

ωrot(f) =

∫Φ

dω =

∫∂Φ

ω =

∫∂Φ

f.

This is the Stokes-theorem.

• The case n = 3, k = 3.

In this case

ω ∈ Λr2(V ), Φ ∈ Cr+1(I3, V ).

We have already shown that if

f = (f1, f2, f3) ∈ Cr(V,R3), ω := ωf = f1∆3,2(2,3) + f2∆3,2

(3,1) + f3∆3,2(1,2)

39

then

dω = ωdiv(f)

thus if Φ is regular∫RΦ

div(f) =

∫Φ

ωdiv(f) =

∫Φ

dω =

∫∂Φ

ω =

∫∂Φ

f.

This is the Gauss-theorem.

40

Chapter 3

Applications

3.1 Maxwell's equations

3.1.1 The classical form of the Maxwell's equations

In his work published in 1865 Maxwell worked out a unied theory which con-

nected the seemingly dierent phenomena of electricity and magnetism. His four

equations describe the impact of an electromagnetic eld on a distribution of elec-

trical charges in space as well as the interaction between the electric eld and the

magnetic eld.

Physical quantities can be modeled by dierent mathematical objects. In this sec-

tion rst we introduce the model most commonly used in classical electrodynamics.

Then we present another approach: a model using dierential forms.

Denition 3.1.1. Let Ω ⊂ R3 be a domain. The electric eld and the magnetic

eld are described by the time dependent dierentiable vector elds E and B dened

on the domain Ω:

E : Ω× R→ R3, B : Ω× R→ R3.

The electric charge density ρ is described by a time dependent scalar eld and

41

the current density j by a time dependent vector eld on the domain Ω:

ρ : Ω× R→ R, j : Ω× R→ R3.

The electric and magnetic eld has an impact on the electric charge and the current

density. Also electric charges and currents generate the electric and magnetic eld.

Maxwell's equations describe how the interaction between these quantities work.

Having chosen units in which µ0 = ε0 = c = 1 the equations take the following

form:

divE = ρ, rotE = −∂B∂t,

divB = 0, rotB = j +∂E

∂t.

3.1.2 Minkowski spacetime

One of the most important principles in physics is that every law describing nature

needs to be expressed with equations which are independent from the location of

the observer. Physical phenomena do not depend on the coordinate system in

which we describe it. Thus the equations must be invariant under changes of co-

ordinate systems.

This principle led to the birth of the theory of special relativity and the concept

of spacetime, since Maxwell's equations are not invariant. However if we don't

distinguish time from the spacial coordinates we can overcome this trouble. To

read more about the topic see [8].

Now we construct the mathematical background for the Minkowski spacetime.

Denition 3.1.2. Let V be a nite dimensional vector space. V is called a

pseudo-euclidean space if on V there is symmetric bilinear non-degenerate map

g : V × V → R

42

such that for all u, v, w ∈ V and λ ∈ R the following properties are satised:

1. g(λu+ v, w) = λg(u,w) + g(v, w),

2. g(u, λv + w) = λg(u,w) + g(u,w) (bilinear),

3. g(u, v) = g(v, u) (symmetric),

4. g(u, v) = 0 ∀u ∈ V ⇐⇒ v = 0 (non-degenerate).

For a given basis e1, ...en in V let's dene the matrix:

M(g) := (g(ei, ej))ni,j=1 := (gij)

ni,j=1.

We will denote this pseudo-euclidean space (V,g).

A pseudo-euclidean space is "less" than an euclidean space, since the g map is not

positive-denite. If it were, it would be a scalar product. However, in physics'

literature g is often called (mistakenly) a scalar product, since their algebraic

properties are quite similar.

Denition 3.1.3. Let (V, g) be a pseudo-euclidean space and e1, ...en a basis in

V . We say that e1, ...en is an orthonormal basis if

g(ei, ej) =

±1 (i = j),

0 (i 6= j)..

Theorem 3.1.1 (Sylvester). Let (V, g) be a pseudo-euclidean space. Then there

exists a basis e1, ...en in V such that the matrix M(g) is diagonal and only

contains +1's and −1's, i.e.

M(g) =

1. . . 0

1

−1

0. . .

−1

.

43

The numbers of +1's (p) and −1's (q) are independent of the particular basis. The

pair (p,q) is called the signature of g.

We will not prove this theorem here, for a proof see for example [9].

Since the matrix M(g) is diagonal, and has no 0 elements it follows that it is

invertable. We will denote the entries of the inverse matrix gij.

Denition 3.1.4. The pseudo-euclidean space (R4, g) is calledMinkowski-spacetime

if g has a signature (3,1).

A point in Minkowski-spacetime can be given by four coordinates:

x = (x1, x2, x3, ct) ∈ R4,

where the rst three coordinates are the usual "spatial" coordinates, and the fourth

one is the time coordinate, c being the speed of light in vacuum. In theoretical

works one frequently chooses a unit system where c = 1 (we will also do it).

Another commonly used notation is x0 := t, so in computations the 0 index always

refers to the time coordinate.

3.1.3 The Hodge-star operator

In this section we introduce the Hodge-star operator, which is an isomorphism

between the Ak(V ) and An−k(V ) spaces. It will be an essential tool for modeling

the electric and magnetic elds with dierential forms. It will turn out that these

quantities are strongly depend on each other and it is better to model them as one

object.

First we extend the g map on a pseudo-euclidean space (V, g) to theAk(V ). Adding

this new structure will make them pseudo-euclidean spaces as well. This extension

will be done relative to an orthonormal basis.

Denition 3.1.5. Let (V, g) be a pseudo-euclidean space and e1, ...en an or-

44

thonormal basis in it. Let f1, f2 ∈ Ak(V ). Then

gk : Ak(V )×Ak(V )→ R,

gk(f1, f2) :=∑i∈Nk

∗

gi1i2 ...gikikf1(ei1 , ..., eik)f2(ei1 , ..., eik).

Theorem 3.1.2. The above dened gk map is non-degenerate, symmetric and

bilinear. Moreover if e1, ..., en is an orthonormal basis in V (with respect to g),

then ∆n,ki (i ∈ Nk

∗ ) is an orthonormal basis in Ak(V ).

Proof. The bilinearity and symmetry of g is clear, so we focus on the non-degenerativity.

Let h ∈ Ak(V ) and suppose that

gk(f, h) =∑i∈Nk

∗

gi1i2 ...gikikf(ei1 , ..., eik)h(ei1 , ..., eik) = 0 ∈ R ∀f ∈ Ak(V ).

We need to show that

h = 0 ∈ Ak(V ).

Since a k-form is uniquely determined by its value on the k-tuples of vectors with

strictly increasing multiindex let's choose f as the following:

f(ei1 , ..., eik) := gi1i2 ...gikikh(ei1 , ..., eik).

This way

gk(f, h) =∑i∈Nk

∗

(gi1i2 ...gikik

)2h(ei1 , ..., eik)2 = 0.

This is only possible if all the terms in the summation are 0. Since the numbers

gijij are not 0 it follows that all h(ei1 , ...eik) must be 0. Which means that h =

0 ∈ Ak(V ), thus g is non-degenerate.

To prove the orthonormality set two multiindex i, j ∈ Nk∗ .

gk(∆n,ki ,∆n,k

j ) =∑l∈Nk

∗

gl1l1 ...gklkl∆n,ki (el1 , ..., elk)∆n,k

j (ej1 , ..., ejk)

This expression does not equal 0 if and only if i = j = l, in which case it is ±1,

45

thus the basis ∆n,ki (i ∈ Nk

∗ ) is in fact orthonormal.

Apart from the g map we x an orientation on V . For this consider the set B(V )

of all ordered bases B = (v1, ..., vn). For two ordered bases B = (v1, ..., vn) and

B′ = (v′1, ..., v′n) there exists a linear transformation A(B,B′) = (aij)

ni,j=1 which

transforms one into the other:

vi =n∑j=1

aijv′j.

We can dene an equivalence relation ∼ on the set B(V ) by requiring that

B ∼ B′ ⇐⇒ det(A(B,B′)) > 0.

This way we have two equivalence classes.

Denition 3.1.6. The orientation of V is a choice of one of the two equivalence

classes in the set B(V ).

Denition 3.1.7. Let (V, g) be an oriented pseudo-euclidean space. Let e1, ...enbe a basis in V such that the matrix M(g) has the diagonal format of Sylvester's

theorem, and (e1, ..., en) is positively oriented. Now we dene the volume-form

dV ∈ An(V ) as follows:

dV (v1, ..., vn) := det(g(vi, ej))ni,j=1 = ∆n,n

(1,...,n)(v1, ..., vn) (v1, ..., vn ∈ V ).

Theorem 3.1.3. Let (V, g) be a pseudo-euclidean space and f : V → R be a linear

function. Then for all α ∈ V there exists a unique β ∈ V such that f(α) = g(α, β).

Proof.

Step 1 (uniqueness)

Suppose that f(α) = g(α, β) = g(α, γ). We have

g(α, β)− g(α, γ) = 0.

46

Using the bilinearity of g we get

g(α, β − γ) = 0.

Since g is non-degenerate it follows that

β − γ = 0 =⇒ β = γ.

Step 2 (existence)

Let e1, ...en be an orthonormal basis in V , expand α in terms of the basis

elements:

α =n∑i=1

αiei

and let β be the following:

β :=n∑j=1

g(ej, ej)f(ej)ej.

Using the bilinearity g we get:

g(α, β) = g

(n∑i=1

αiei,n∑j=1

g(ej, ej)f(ej)ej

)=

n∑i=1

n∑j=1

αig(ej, ej)f(ej)g(ei, ej).

Because the basis is orthonormal this simplies to:

g(α, β) =n∑i=1

αig(ei, ei)2f(ei).

Since g(ei, ei)2 = 1, and again using the bilinearity

g(α, β) =n∑i=1

αif(ei) = f

(n∑i=1

αiei

)= f(α).

Now let's use this result for the vector space An−k(V ). Let λ ∈ Ak(V ) a xed

47

k-form. Then for any θ ∈ An−k(V ) we have λ ∧ θ ∈ An(V ). Since An(V ) is one

dimensional, and the basis element is the volume-form there exists a unique a ∈ Rsuch that λ∧ θ = a · dV . Using this equality we can dene the following function:

fλ : An−k(V )→ R, fλ(θ) := a.

With this denition fλ is a linear function, thus we can apply the previous theorem,

which tells us that there exists a unique element φ ∈ An−k(V ) such that

fλ(θ) = g(θ, φ) (θ ∈ An−k(V )).

We can nally dene the Hodge-dual ?λ of λ to be

?λ := (−1)qφ ∈ An−k(V ).

Denition 3.1.8. The Hodge-star operator is a map between k-forms and n−kforms

? : Ak(V )→ An−k(V )

dened by the following equation:

λ ∧ θ = (−1)qg(θ, ?λ)dV (θ ∈ An−k(V )).

This quite abstract denition is not sucient to actually compute the Hodge-dual

of some form. In order to do this we use a dierent approach. Since the Hodge-

star operator is linear it suces to compute the duals of the basis elements, then

expend the results in a linear and alternating way.

So we wish to compute ?∆n,ki (i ∈ Nk

∗ ). Let j ∈ Nn−k∗ . Then from the denition

of ?:

∆n,ki ∧∆n,n−k

j = (−1)qg(∆n,n−kj , ?∆n,k

i )∆n,n(1,...,n).

Note that the left side of the equation diers from 0 only if j is the complementary

index of i. Since the ∆n,n−kj (j ∈ Nn−k

∗ ) basis elements are orthonormal the right

48

side tells us that ?∆n,ki has the following form:

?∆n,ki = c∆n,n−k

j

where j is the complementary index and c ∈ R. Putting this back into the originalequation yields:

sgn(τ)∆n,n(1,...,n) = ∆n,k

i ∧∆n,n−kj = (−1)qg(∆n,n−k

j , c∆n,n−kj )∆n,n

(1,...,n) =

= (−1)qcgj1j1 ...gjn−k,jn−k∆n,n(1,...,n).

Where τ is the number of inversions in the permutation which takes the sequence

(i1, ..., ik, j1, ..., jn−k) to (1, ..., n). Using the fact that gjrjr = 1gjrjr

we obtain

c = (−1)q sgn(τ)gj1j1 ...gjn−kjn−k.

Hence

?∆n,ki = (−1)q sgn(τ)gj1j1 ...gjn−kjn−k

∆n,n−kj .

3.1.4 Maxwell's equations in terms of dierential forms

First we consider the homogeneous Maxwell equations:

divB = 0, rotE +∂B

∂t= 0.

We have shown that divergence of a vector eld can be thought of as the exterior

derivative of a dierential 2-form, and the rotation of a vector eld as the exterior

derivative of a 1-form. Therefore it seems logical to represent the magnetic eld

as a 2-form and the electric eld as a 1-form:

B := B1dx2 ∧ dx3 +B2dx3 ∧ dx1 +B3dx1 ∧ dx2,

E := E1dx1 + E2dx2 + E3dx3.

Now, these forms are the inhabitants of spacetime, and we can combine them into

one dierential form.

49

Denition 3.1.9. The electromagnetic eld F is a dierential 2-form on space-

time dened by the following equation:

F ∈ Λ12(R4) F := B + E ∧ dx0

where dx0 denotes the time coordinate of spacetime.

The full form of F is the following:

F =E1dx1 ∧ dx0 + E2dx2 ∧ dx0 + E3dx3 ∧ dx0+

+B1dx2 ∧ dx3 +B2dx3 ∧ dx1 +B3dx1 ∧ dx2.

Now let's take the exterior derivative of F:

dF =∂2E1dx2 ∧ dx1 ∧ dx0 + ∂3E1dx3 ∧ dx1 ∧ dx0 + ∂1E2dx1 ∧ dx2 ∧ dx0+

+ ∂3E2dx3 ∧ dx2 ∧ dx0 + ∂1E3dx1 ∧ dx3 ∧ dx0 + ∂2E3dx2 ∧ dx3 ∧ dx0+

+ ∂1B1dx1 ∧ dx2 ∧ dx3 + ∂0B1dx0 ∧ dx2 ∧ dx3 + ∂2B2dx2 ∧ dx3 ∧ dx1+

+ ∂0B2dx0 ∧ dx3 ∧ dx1 + ∂3B3dx3 ∧ dx1 ∧ dx2 + ∂0B3dx0 ∧ dx1 ∧ dx2.

Now collecting the terms and using the anti-symmetry of the wedge product we

get:

dF =(∂1B1 + ∂2B2 + ∂3B3)dx1 ∧ dx2 ∧ dx3+

+ (∂2E3 − ∂3E2 + ∂0B1)dx0 ∧ dx2 ∧ dx3+

+ (∂3E1 − ∂1E3 + ∂0B2)dx0 ∧ dx3 ∧ dx1+

+ (∂1E2 − ∂2E1 + ∂0B3)dx0 ∧ dx1 ∧ dx2.

50

Note that dF = 0 is the same as

∂1B1 + ∂2B2 + ∂3B3 = 0

and

∂1B1 + ∂2B2 + ∂3B3 = 0,

∂3E1 − ∂1E3 + ∂0B2 = 0,

∂1E2 − ∂2E1 + ∂0B3 = 0.

The rst equality corresponds with divB = 0 and next three with

rotE +∂B

∂t= 0.

Thus the homogeneous equations can be written in the compact form of:

dF = 0

which means that the electromagnetic eld 2-form is closed.

For the inhomogeneous equations let's compute Hodge-dual of F :

?(dx1 ∧ dx0) = − ? (dx0 ∧ dx1) = − ((−1)q sgn(τ)g22g33dx2 ∧ dx3)

= −((−1)1 · 1 · 1 · 1

)dx2 ∧ dx3 = dx2 ∧ dx3,

?(dx2 ∧ dx0) = − ? (dx0 ∧ dx2) = − ((−1)q sgn(τ)g11g33dx1 ∧ dx3)

= −((−1)1 · (−1) · 1 · 1

)dx2 ∧ dx3 = dx3 ∧ dx2,

?(dx3 ∧ dx0) = − ? (dx0 ∧ dx3) = − ((−1)q sgn(τ)g11g22dx1 ∧ dx2)

= −((−1)1 · 1 · 1 · 1

)dx2 ∧ dx3 = dx1 ∧ dx2,

?(dx2 ∧ dx3) = (−1)q sgn(τ)g00g11dx0 ∧ dx1 =

= (−1)1 · 1 · (−1) · 1 · dx0 ∧ dx1 = −dx1 ∧ dx0,

51

?(dx3 ∧ dx1) = − ? (dx1 ∧ dx3) = − ((−1)q sgn(τ)g00g22dx0 ∧ dx2) =

= −((−1)1 · (−1) · (−1) · 1

)dx0 ∧ dx2 = −dx2 ∧ dx0,

?(dx1 ∧ dx2) = (−1)q sgn(τ)g00g33dx0 ∧ dx3

= (−1)1 · 1 · (−1) · 1 · dx0 ∧ dx3 = −dx3 ∧ dx0.

So the Hodge-dual of F is:

dF =−B1dx1 ∧ dx0 −B2dx2 ∧ dx0 −B3dx3 ∧ dx0+

+ E1dx2 ∧ dx3 + E2dx3 ∧ dx1 + E3dx1 ∧ dx2.

We can also unify the electric charge and current density into a 1-form:

J := j1dx1 + j2dx2 + j3dx3 − ρdx0.

Let's compute the exterior derivative of ?F :

d ? F =− ∂2B1dx2 ∧ dx1 ∧ dx0 − ∂3B1dx3 ∧ dx1 ∧ dx0 − ∂1B2dx1 ∧ dx2 ∧ dx0−

− ∂3B2dx3 ∧ dx2 ∧ dx0 − ∂2B3dx2 ∧ dx3 ∧ dx0 − ∂1B3dx1 ∧ dx3 ∧ dx0+

+ ∂0E1dx0 ∧ dx2 ∧ dx3 + ∂1E1dx1 ∧ dx2 ∧ dx3 + ∂0E2dx0 ∧ dx3 ∧ dx1+

+ ∂2E2dx2 ∧ dx3 ∧ dx1 + ∂0E3dx0 ∧ dx1 ∧ dx2 + ∂3E3dx3 ∧ dx1 ∧ dx2.

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Collecting the terms we obtain:

d ? F =(∂1E1 + ∂2E2 + ∂3E3)dx1 ∧ dx2 ∧ dx3+

+ (∂3B2 − ∂2B3 + ∂0E1)dx0 ∧ dx2 ∧ dx3+

+ (∂3B1 − ∂1B3 − ∂0E2)dx0 ∧ dx1 ∧ dx3+

+ (∂2B1 − ∂1B2 + ∂0E3)dx0 ∧ dx1 ∧ dx2.

Now let's take the dual of d ? F

?d ? F =− (∂1E1 + ∂2E2 + ∂3E3)dx0 + (∂3B2 − ∂2B3 + ∂0E1)dx1+

+ (∂3B1 − ∂1B3 − ∂0E2)dx2 + (∂2B1 − ∂1B2 + ∂0E3)dx3.

Now ?d ? F = J corresponds to

∂1E1 + ∂2E2 + ∂3E3 = ρ

and

∂3B2 − ∂2B3 + ∂0E1 = j1,

∂3B1 − ∂1B3 − ∂0E2 = j2,

∂2B1 − ∂1B2 + ∂0E3 = j3.

The rst equation is the same as divE = ρ and the next three is the same as

rotB − ∂E∂t

= j. Thus the inhomogeneous equations take the form:

?d ? F = J.

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3.2 Brouwer's Fixed Point Theorem

Theorem 3.2.1. In the Euclidean space (Rn, 〈·, ·〉) every continous map

f : B1(0)→ B1(0)

from the closed unit ball to itself has a xed point.

Proof. Step 1 First we prove the theorem for f ∈ C1 functions. Indirectly let's

suppose that

f : B1(0)→ B1(0) (f ∈ C1)

is a function with no xed points:

f(x) 6= x (x ∈ B1(0)).

Now we construct the function which assigns to each x ∈ B1(0) point the

point of intersection from f(x) through x with the sphere ∂B1(0). The line

from x through f(x) can be parameterized as follows:

L = x+ t(x− f(x)) : t ∈ R.

This line intersects with ∂B1(0) if for some λ ∈ R the following condition is

satised:

‖x+ λ(x− f(x))‖2 = 1.

Let's compute the norm squared of this expression:

‖x+ λ(x− f(x))‖2 = 〈x+ λ(x− f(x)), x+ λ(x− f(x))〉 =

= 〈x, x〉+ 2λ〈x, x− f(x)〉+ λ2〈x− f(x), x− f(x)〉 =

= ‖x‖2 + 2λ〈x, x− f(x)〉+ λ2‖x− f(x)‖2.

So we need to solve the following quadratic equation for λ:

‖x− f(x)‖2λ2 + 2〈x, x− f(x)〉λ+ ‖x‖2 − 1 = 0.

54

The solution is

λ±(x) =〈x, f(x)− x〉 ±

√〈x, f(x)− x〉2 + (1− ‖x‖2)‖f(x)− x‖2

‖f(x)− x‖2.

Now we can construct our function:

F = (F1, ..., Fn) : B1(0)→ ∂B1(0)

F (x) := x+〈x, f(x)− x〉 ±

√〈x, f(x)− x〉2 + (1− ‖x‖2)‖f(x)− x‖2

‖f(x)− x‖2(x−f(x)).

From the formula we can see that F is a C1 function. Moreover F acts on

the boundary of the unit ball as the identity map:

F (x) = x (x ∈ ∂B1(0)).

Since the image of any x is on the boundary of the unit ball the following

relation holds for all x ∈ B1(0):

‖F (x)‖2 =n∑i=1

Fi(x)2 = 1.

Dierentiating the above formula yields:

2n∑i=1

Fi(x)F ′i (x) = 2n∑

i,j=1

Fi(x)∂jFi(x)dxj = 0

and it follows that for each index j

n∑i=1

Fi(x)∂jFi(x) = 0.

This equality shows that the system of equations

n∑i=1

αi∂jFi(x) = 0.

55

has a non-trivial solution. (α1, ...αn) = (F1(x), ..., Fn(x)) 6= (0, ..., 0). Hence

the determinant of the following matrix vanishes.

det (∂jFi(x)) = 0

Now we dene an ω ∈ Λrn−1 dierential form and using the observation made

above conclude that its derivative vanishes. If

ω := F1 · dF2 ∧ ... ∧ dFn

then

dω = dF1 ∧ dF2 ∧ ... ∧ dFn = det (∂jFi(x)) dx1 ∧ ... ∧ dxn = 0.

Now we use Stokes' theorem and the fact that F acts on the boundary of

the unit ball as the identity map and arrive at a contradiction.

0 =

∫B1(0)

dω =

∫∂B1(0)

ω =

∫∂B1(0)

x1dx2 ∧ ... ∧ dxn =

∫B1(0)

dx1 ∧ ... ∧ dxn =

= µ(B1(0)).

Which would mean that the volume of the unit ball is zero, thus this is a

contradiction.

Step 2 Now we consider the general case, where f is continuous. This case will be

reduced to the previous one by using the Stone-Weierstrass approximation

theorem. This theorem states (see [3]) that if Ω ⊂ Rn is compact, f : Ω→ Ris continuous, then for all ε > 0 there exists a p : Ω → R polynomial such

that |f(x) − p(x)| < ε for all x ∈ Ω. In our case B1(0) is compact, so

by applying this theorem to the f1, ..., fn coordinate functions of f we can

conclude that for each ε > 0 there exists a

p : B1(0)→ Rn (p ∈ C1)

56

polynomial such that

‖f(x)− p(x)‖ < ε.

Now consider the normalized function:

p(x) :=p(x)

1 + ε.

Because of

‖p(x)‖ − ‖f(x)‖ ≤ ‖p(x)− f(x)‖ ≤ ε and ‖f(x)‖ ≤ 1

we have ‖p(x)‖ ≤ 1 + ε, thus ‖p(x)‖ ≤ 1. So p is a map from the unit ball

to itself:

p : B1(0)→ B1(0).

Moreover p can be estimated against f :

‖f(x)− p(x)‖ ≤ ‖f(x)− p(x)‖+ ‖p(x)− p(x)‖ ≤ ε+ ‖p(x)‖∣∣∣∣1− 1

1 + ε

∣∣∣∣ ≤≤ ε+ (1 + ε)

ε

1 + ε≤ 2ε.

To sum up we have proved so far that for each ε > 0 there exists a map

p : B1(0)→ B1(0) (p ∈ C1) such that for all x ∈ B1(0) the following estimate

holds:

‖f(x)− p(x)‖ ≤ 2ε.

Now let's suppose that the continuous map f : B1(0) → B1(0) has no xed

points. This means that

0 < µ := infx∈B1(0)

‖f(x)− x‖

Now for ε = µ/2 we choose a smooth map p with the above discussed proper-

ties. Since p ∈ C1 we can apply the rst part of this proof to p and conclude

that it has a xed point x0 ∈ B1(0). But this means

‖f(x0)− p(x0)‖ = ‖f(x0)− x0‖ < µ

57

which contradicts the denition of µ. So our initial assumption (that f has

no xed points) is false.

58

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